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Chemistry-for-CAPE-Examinations

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Helen Jacobs
Novelette Sadler-McKnight Stewart McLean
Patrice Piggot-Cumberbatch
Graeme Corbin Mike Taylor
CHEMISTRY
®
FOR CAPE EXAMINATIONS
Chemistry
For CAPE® Examinations
The Higgs boson
In the early 19th century atoms were thought to be the ultimate, indivisible, components of
matter. Then came the discovery of the atomic nucleus, made up of protons and neutrons,
and the electrons that surround the nucleus. We now know that electrons are indeed
indivisible, but that protons and neutrons are themselves made up of ‘quarks’: it takes three
quarks to make either a proton or a neutron.
We have also learned that matter and energy are one and the same thing, according to
Einstein’s famous equation E = mc2. But this is a concept outside our everyday experience,
which we can accept and use but not picture happening.
Another such concept is the idea of ‘action at a distance’. Two magnets attract or repel
one another, but why and how do they do this? We get round it by saying that there is
a magnetic ‘field’ between them, with ‘lines of force’ that change shape and produce the
attractive or repulsive effect. But is that really an explanation?
Today’s explanation (called the standard model) says that there is a region of space round
the magnets (the ‘field’) inside which the magnets throw other particles to and fro between each
other. It is like children who throw and catch a ball: as long as they do so, they stay together.
If the ball is very heavy the thrower staggers backwards after throwing and so does the
catcher. It is the exchange of energy packets which is the ‘force’.
There are only four forces in nature. These are gravity, electromagnetism, the ‘strong force’
(which holds nuclear particles together) and the ‘weak force’ (responsible for effects such as
magnetism). Each one comprises a field and an associated exchange particle.
But there is a problem with the standard model. Nothing in it makes possible the property
that we call mass. It was this problem which led Professor Higgs and his co-workers to
suggest that the whole universe is filled with a ‘mass field’ and that particles interact with
that field to acquire ‘mass’. If that is the case, the mass field, like any other, must have an
exchange particle associated with it. Some very subtle physics showed that the particle must
have a particular property; it must behave as though it had no spin. Such a particle-cumenergy-packet is called a boson. The exchange particle which goes with the mass field is
called the Higgs boson.
Other calculations showed that such a particle could only exist for, at most, nanoseconds
and that it would carry (or be) a large amount of energy. The only way to find it would be
to smash two other particles together at speeds near to that of the speed of light. For this
you need the apparatus called the Large Hadron Collider (LHC). It was in the LHC on July
4th 2012 that collisions were seen which could only have come from the decay of a Higgs
boson. The tracks that you see in the cover picture of this book (a computer visualization of
the response of the huge array of detectors in the LHC) are some that can only have come
from the decay of the Higgs boson. Any other particle could not produce exactly these traces
as it decayed.
Chapter 1 Atomic structure
Chemistry
®
For CAPE Examinations
Helen Jacobs, Novelette McKnight, Stewart
McLean, Patrice Piggot-Cumberbatch,
Graeme Corbin, Mike Taylor
CAPE® is a registered trade mark of the Caribbean
Examinations Council (CXC).
Chemistry For CAPE® Examinations is an independent
publication and has not been authorized, sponsored,
or otherwise approved by CXC.
III
Macmillan Education
4 Crinan Street, London N1 9XW
A division of Macmillan Publishers Limited
Companies and representatives throughout the world
ISBN 978-0-230-48315-6 AER
Text © Helen Jacobs, Novelette McKnight, Stewart McLean,
Patrice Piggott-Cumberbatch and Mike Taylor
SBA chapter contributed by Graeme Corbin and © Macmillan
Publishers Limited.
Design and illustration © Macmillan Publishers Limited 2014
First published 2014
All rights reserved; no part of this publication may be reproduced,
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without the prior written permission of the publishers.
Designed and typeset by Mike Brain Graphic Design Limited
Illustrated by Peter Harper
Cover design by Clare Webber and Macmillan Education
Cover photograph by Pasieka, Science Photo Library
The publishers would like to thank Tim Jackson for his help in
developing and editing the manuscript for publication.
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Contents
Preface
Chapter 4
UNIT 1
Chemical principles and applications I
Module 1
Fundamentals in chemistry
Chapter 1
Chapter 2
Chapter 3
Chemical bonding
33
Introduction
33
Formation of bonds
33
Types of chemical bond
33
Bond formation and energy changes
34
Ionic bonding
35
Covalent bonding
35
Atomic structure
2
The hydrogen bond
37
The atom: introduction
2
The metallic bond
38
Early ideas about the atom: classical models
2
van der Waals forces
39
Atomic structure
4
The periodic table and bond type
40
Electrons in atoms
7
Properties associated with different bond types 40
Summary
10
Mixed bonds
41
Review questions
10
Summary
42
Answers to ITQs
10
Review questions
42
Answers to Review questions
11
Answers to ITQs
42
The quantum atom and the periodic table
12
Shapes of covalent molecules
43
The quantum atom
12
Lewis structures
43
Developing the periodic table
16
Molecular geometry
44
The modern periodic table
17
Hybrid orbitals
47
Periodicity
18
Resonance
48
Periodic properties in atomic ‘size’
18
Molecular polarity
49
Summary of general periodic trends
24
Two common misconceptions
50
Summary
25
Summary
51
Review questions
25
Review questions
51
Answers to ITQs
26
Answers to ITQs
52
Answers to Review questions
27
Answers to Review questions
52
Radioactivity
28
An introduction to the mole
53
Introduction: the alchemists’ dream
28
Relative atomic mass of elements, Ar
53
Nuclear transitions
29
Radioactive decay
30
Relative formula mass and relative molecular
mass of compounds
53
Properties of α, β and γ rays
30
The mole
54
Problems caused by radiation
30
Molar mass
54
Uses of radioisotopes
31
Writing chemical equations
55
Summary
32
Calculations involving the mole
56
Review questions
32
The concept of the limiting reagent
57
Answers to ITQs
32
Empirical and molecular formulae
59
The mole concept applied to solutions
61
Titrimetric (volumetric) analysis
63
The mole concept applied to gases
64
Summary
66
Review questions
66
Answers to ITQs
67
Answers to Review questions
67
Chapter 5
Chapter 6
VI
Contents
Chapter 7
Chapter 8
Gases
68
Behaviour of gases
68
What are acids and bases?
104
Gas laws
69
Acid/base reactions
104
Behaviour of real gases
71
Strong and weak acids and bases
105
Kinetic-molecular theory
72
pKa
105
Summary
72
pH
105
Review questions
73
Changes in pH in acid/base titrations
106
Answers to Review questions
73
Buffer solutions
107
Thermochemistry
74
Introduction to thermodynamics
74
Heat and heat capacity
75
Latent heat
76
Heat and the kinetic-molecular theory
77
Heat and work
77
The first law of thermodynamics
78
Introduction
112
Calorimetry
79
Electrode potential
113
Enthalpy
80
Summary
86
Galvanic cells: using redox reactions to
generate electricity
113
Review questions
87
The standard hydrogen electrode (S.H.E.)
114
Answers to ITQs
87
Measuring standard electrode potentials
115
87
Measuring the E
non-metals
Answers to Review questions
Module 2
Kinetics and equilibria
Chapter 9
Chapter 11 Acid/base equilibria
104
Solubility product
108
Summary
110
Review questions
110
Answers to ITQs
110
Answers to Review questions
111
Chapter 12 Redox equilibria
112
of half-cells involving
116
Uses of standard electrode potentials
116
The effect of concentration on electrode
potential
119
Chemical kinetics
88
Energy storage devices
119
Collision theory
88
Summary
123
Reaction rate
89
Review questions
123
Determining the order of reaction
93
Answers to ITQs
125
Summary
95
Answers to Review questions
125
Review questions
95
Answers to ITQs
97
Answers to Review questions
97
Module 3
Chemistry of the elements
98
Chapter 13 Elements and periodicity: period 3
Chapter 10 Chemical equilibrium
126
Reversible reactions
98
Introduction
126
Gas reactions
99
Atomic properties
126
129
Le Chatelier’s principle
100
Bulk properties
Changes in the value of K
101
Chemical properties
131
Summary
102
Summary
133
Review questions
102
Review questions
134
103
Answers to ITQs
135
Answers to ITQs
Contents
Chapter 14 Elements and periodicity: Group II
136
Introducing Group II
136
Physical properties
136
Chemical reactions
137
Uses of magnesium and calcium compounds
139
Summary
142
Review questions
142
Answers to ITQs
143
Answers to Review questions
143
UNIT 2
Chemical principles and applications II
Module 1
The chemistry of carbon compounds
Chapter 19 Alkanes
182
Introduction to carbon compounds
182
Alkanes
185
144
Physical properties, sources and uses
of alkanes
188
144
Reactions of alkanes: an introduction
189
Variation in physical properties
144
Summary
191
The Group IV tetrachlorides
145
Review questions
192
The Group IV oxides
146
Answers to ITQs
192
Bonding and the ‘inert pair’ effect
147
Stability of the +2 and +4 oxidation states
147
Introduction
193
Silicon
148
Alkenes
193
Summary
149
Alkynes
195
Review questions
150
Answers to ITQs
151
Physical properties, sources and uses
of alkenes and alkynes
196
152
An introduction to the reactions of alkenes
and alkynes
196
Introducing the Group VII elements
152
Summary
198
Variation in physical properties
152
Review questions
198
Bonding types
153
Answers to ITQs
199
Chemical properties and reactivity
153
Summary
157
Chapter 15 Elements and periodicity: Group IV
Introducing Group IV
Chapter 16 Elements and periodicity: Group VII
Chapter 20 Alkenes and alkynes
Chapter 21 Alcohols and amines
193
200
Review questions
158
Introduction
Answers to ITQs
159
Haloalkanes
200
Alcohols
200
Chapter 17 The first row transition elements
160
200
160
Amines – RNH2
205
Introduction to the transition elements
Summary
208
Electronic configurations
160
Review questions
208
Answers to ITQs
209
Trends across the period of transition elements 161
Characteristic properties
162
The oxidation states of vanadium
169
Chapter 22 Stereochemistry
210
Summary
170
Introduction
Review questions
171
Structural isomers
210
Answers to ITQs
172
Geometric isomers
210
Answers to Review questions
172
Summary
214
Review questions
214
Answers to ITQs
214
Chapter 18 Qualitative inorganic analysis
173
Introducing inorganic analysis
173
Identification of cations
173
Flame tests
174
Identification of anions
175
Testing for gases
177
Review questions
179
Answers to ITQs
180
Answers to Review questions
180
210
VII
VIII
Contents
Chapter 23 Aldehydes and ketones
215
Introduction
215
Introduction
258
Nomenclature of aldehydes and ketones
216
Homolytic and heterolytic cleavage
258
Bonding in the carbonyl group
216
Homolysis and radical reactions
259
General properties of aldehydes and ketones
217
Heterolysis and ionic reactions
261
Preparation of aldehydes and ketones
218
Nucleophilic substitution reactions
263
Reactions of aldehydes and ketones
218
Summary
265
Summary
222
Review questions
265
Review questions
222
Answers to ITQs
266
Answers to ITQs
222
Chapter 24 Carboxylic acids and derivatives
224
Chapter 27 Reaction mechanisms
258
Module 2
Analytical methods and separation techniques
Introduction
224
Nomenclature
224
General properties
224
Introduction
268
Preparation of carboxylic acids
225
Defining some terms
268
Acidity of carboxylic acids
225
Uncertainty in single determinations
269
Amino acids
226
Uncertainty in addition and subtraction
270
Reactions of carboxylic acids and their
derivatives
Significant figures
271
228
Glassware used for measuring volume
271
Summary
232
Measuring mass
272
Review questions
232
Summary
273
Answers to ITQs
233
Review questions
273
Answers to ITQs
273
Chapter 25 Aromatic compounds
234
Chapter 28 Measurement in chemical analysis
Chapter 29 Gravimetric analysis
268
274
Introduction
234
Characteristics of aromatic compounds
235
Introduction
274
The stability of benzene
235
The precipitation method
274
The electron structure of benzene
236
Aromaticity
236
Apparatus and glassware for gravimetric
analysis
275
Nomenclature of benzene derivatives
237
Volatilization methods
276
Properties and uses of aromatic compounds
237
Applications of gravimetric analysis
278
Reactions of benzene
238
Summary
279
Properties and reactions of aniline
240
Review questions
279
Properties and reactions of phenol
241
Answers to ITQs
280
Summary
243
Review questions
243
Introduction
281
Answers to ITQs
244
Acid/base titrations
281
245
Back titrations in acid/base titrimetric analysis 284
Introduction
245
Polymerization
245
Titrations monitored by measurement of pH
(potentiometric titrations)
286
Addition polymerization
245
Thermometric and conductimetric titrations
287
Condensation polymerization
248
Primary standards
287
Carbohydrates
251
Redox titrations
287
Plastics in the environment
253
Summary
294
Waste management
254
Review questions
295
Summary
255
Answers to ITQs
295
Review questions
256
Answers to ITQs
257
Chapter 26 Macromolecules
Chapter 30 Titrimetric analysis
281
Contents
Chapter 31 Introduction to spectroscopy
296
Chapter 35 Phase separations
322
Introduction to spectroscopy: resonance
296
Introduction
322
Electromagnetic radiation
297
Simple distillation
322
Regions of the electromagnetic spectrum
298
Fractional distillation
323
The interaction of electromagnetic radiation
with atoms and molecules
Vacuum distillation
327
298
Steam distillation
327
Effects of irradiation
299
Solvent extraction
328
Summary
300
Summary
330
Review questions
300
Review questions
331
Answers to ITQs
300
Answers to ITQs
333
Chapter 32 Ultraviolet–visible spectroscopy
Molecular orbitals in covalent molecules
301
Chapter 36 Chromatography
301
Absorption of energy by electrons in molecular
orbitals
302
Studying absorption of UV–visible radiation
303
The Beer–Lambert law
304
334
Introduction
334
Chromatography
334
Elution
334
Locating individual substances
335
Ion-exchange chromatography
336
Identifying peaks in a chromatogram
336
Uses of chromatography
337
Gas-liquid chromatography
337
Summary
338
308
Review questions
338
Introduction
308
Answers to ITQs
338
How organic molecules absorb infrared
radiation
308
Interpreting infrared spectra
309
Obtaining infrared spectra
311
Applications of ultraviolet–visible spectroscopy 304
Summary
306
Review questions
307
Answers to ITQs
307
Chapter 33 Infrared spectroscopy
Infrared absorption in climate and the
environment
312
Summary
312
Review questions
313
Answers to ITQs
313
Chapter 34 Mass spectrometry
314
Introduction
314
Mass spectra of atoms
314
Mass spectra of molecules
315
How mass spectra are obtained
316
Applications of mass spectrometry
316
Summary
320
Review questions
320
Answers to ITQs
321
Module 3
Industry and the environment
Chapter 37 Environmental effects
339
Locating industrial plants
339
Water
341
The atmosphere
346
Solid waste
356
Answers to ITQs
362
Chapter 38 Chemical industry
364
Aluminium
364
Crude oil
366
Ammonia
372
Ethanol
374
Chlorine
378
Sulfuric acid
381
Answers to ITQs
384
CAPE SBA
385
Index
399
IX
X
Preface
This series of textbooks for CAPE Sciences follows directly from
Macmillan’s CSEC Science series. The books in the series will be
especially valuable for students who have completed CSEC Science
examinations successfully, and wish to continue their studies at a higher
level, to gain employment in a scientific field or to extend their education
at degree level.
CAPE subject studies are each divided into two units, and students have
the option to study either or both of them. These books are designed
to be used with any of these three options. In each unit the material is
based on the knowledge and skills that the student will have gained in
CSEC studies.
The move to higher-level studies is not without its pitfalls. To minimize
these problems the books have several new and innovative features.
In the sciences a good diagram is worth a thousand words. Diagrams
in these books are carefully presented to convey the maximum
understanding with the minimum of extraneous detail. Their captions
are comprehensive, to help the reader to integrate the visual material as
fully and easily as possible with the text.
As in the familiar CSEC series, use is made of ‘In-text Questions’ (ITQ),
but at this level they provoke analytical thought rather than confirm
comprehension.
All science teachers are aware of fundamental misconceptions that are
commonly held. Throughout these books, notes are provided to highlight
and dispel these misapprehensions.
Dr Mike Taylor
Adviser
Chapter 1 Atomic structure
Unit 1
Chemical principles and applications I
1
2
Module 1
Fundamentals in chemistry
Chapter 1
Atomic structure
Learning objectives
■ Compare the properties of electrons, protons and neutrons in terms of their relative charges, masses
■
■
■
■
■
■
and behaviour in electric and magnetic fields.
Distinguish between atomic number and mass number.
Discuss the concept of isotopes and give examples.
Summarize Dalton’s, Thomson’s and Rutherford’s models of the atom.
Outline the Bohr theory and model of the atom and explain how it accounted for the absorption and
emission spectra of hydrogen.
Perform calculations using energy, wavelength and frequency of electromagnetic radiation, using E = hv.
State and explain the origins of the Lyman, Balmer and Paschen series in the hydrogen spectrum.
The atom: introduction
The study of the atom is at the core of chemistry. By
studying the atom we gain a greater understanding of the
physical, chemical and structural properties of compounds.
The concept of the atom has undergone many changes,
from the early ideas of it being indivisible, as proposed by
the Greeks, to the modern quantum mechanical model in
which it behaves both as a particle and as a wave. The atom,
which was once thought to be hollow on the inside, has now,
through experiments, been shown to contain electrons,
protons and neutrons, with these particles being formed
of other sub-atomic particles. With the use of sophisticated
instruments, for example a scanning tunnelling microscope,
images of individual atoms have been produced.
In this chapter we will review the classical models proposed
to explain the structure of the atom. Chapter 2 goes on to
look at the quantum mechanical model of the atom. The
development of these models is an excellent example of
how scientists (chemists, physicists, mathematicians) work
together as a team to design and conduct experiments, to
test hypotheses and develop or refute models and theories.
As experimental evidence becomes available, existing
models are confirmed, modified or abandoned.
Early ideas about the atom: classical
models
The Ancient Greek and Roman philosophers debated about
the composition of matter. Some, for example Plato and
Aristotle, argued that matter was continuous. Others,
for example Democritus, believed that matter could be
divided in such a way that an ultimate particle could be
attained beyond which any further sub-division would be
impossible. Democritus (460–370 BCE) named this ultimate
particle atomos, the Greek word meaning ‘indivisible’. This
concept of an ultimate particle was maintained for about
2000 years without any scientific experiments being
conducted to prove or disprove it. This approach to science
changed in the eighteenth century, as chemists began to
make measurements of the changes in mass and volume
that could be observed.
Early quantitative ideas
The works of Joseph Priestley, an English theologian and
chemist (1733–1804), and Antoine Lavoiser, a French
nobleman and chemist (1743–1794), centred on the
process of combustion. Their work led to the development
of two key fundamental principles of chemistry: the
Chapter 1 Atomic structure
law of conservation of mass and the law of definite
proportions.
The law of mass conservation states that mass is neither
created nor destroyed during a chemical reaction.
This means that the combined masses of the products of a
reaction are always equal to the combined masses of the
starting reactants.
The law of definite proportions states that different samples
of a pure substance always contain the same proportions of
the same elements by mass.
This statement codifies the idea of a ‘pure’ substance. For
example, every sample of water, regardless of its origin,
will contain eight parts of oxygen to one part of hydrogen
by mass.
Dalton’s theory
In the early 1800s, John Dalton, an English schoolteacher
and chemist (1766–1844), proposed a new atomic theory.
Unlike the ideas proposed by the Greeks, Dalton’s work
consisted of statements about the atom that could be
tested through experiments. He carried out experiments
in which he combined elements in different ratios to form
compounds. He noticed that certain patterns resulted. For
example, when he combined hydrogen with oxygen to
form water, the ratio of the mass of oxygen to the mass of
hydrogen was 8:1, whereas in hydrogen peroxide, the ratio
of the mass of oxygen to the mass of hydrogen was 16:1.
On the basis of these and other experiments, he proposed
the following:
■ all matter is composed of tiny indestructible particles
called atoms that cannot be created, destroyed or
subdivided;
■ atoms of one element cannot be converted into atoms
of another element;
■ all atoms of a given element are identical, in weight
and other properties, and are different from atoms of
any other elements (nowadays we would say ‘mass’,
not ‘weight’);
■ atoms can combine with each other in simple whole
number ratios.
Dalton’s theory incorporated the idea of an indestructible
atom, suggested by Democritus, as well as the law of
conservation of mass and the law of definite proportions,
proposed earlier by Lavoisier and Priestley.
According to Dalton’s theory, the small size of an individual
atom makes it impossible for us to determine its mass
directly so its mass must be measured relative to the mass of
other atoms. Since hydrogen was the lightest (least dense)
element known at the time, he assigned it an ‘atomic
weight’ of 1. The theory was tested by other investigators
in their attempts to determine the relative atomic masses of
elements from their mass ratios in compounds.
Chemists thought that water consisted of roughly 15%
hydrogen and 85% oxygen by mass. Dalton assumed that
atoms combined in the simplest ratios. As a result, he gave
water the formula HO, and calculated that oxygen had an
atomic weight of 5.7.
These results were challenged by Joseph Gay-Lussac, a
French chemist (1778–1850). He, like Dalton, conducted
experiments with gases. Instead of measuring the mass
of the gasses, Gay-Lussac used their volumes. When he
combined hydrogen gas and oxygen gas to form water
vapour, he observed that:
■ 2 dm3 of hydrogen gas combined with 1 dm3 of oxygen
gas;
■ 2 dm3 of water vapour was formed for every 2 dm3 of
hydrogen gas that reacted;
■ 2 dm3 hydrogen gas + 1 dm3 oxygen gas →
2 dm3 water vapour.
At that time all elements were thought to be composed of
single atoms. Many thought that equal volumes of gases
contained equal numbers of particles. Dalton used symbols
for atoms and would have suggested this equation:
+
H
=
+
=
O
1 vol
HO
1 vol
should give1 vol
However, Gay-Lussac’s results suggested that water is
H2O, and not OH as was suggested from Dalton’s theory.
Problems like these caused confusion amongst chemists for
some years.
In 1811 Amadeo Avogadro, an Italian physicist (1776–
1856), suggested that individual elements could exist as
‘molecules’, such as O2, rather than as single atoms. For
example, a hydrogen gas ‘particle’ consisted of two atoms
that could split apart to form two separate atoms. The same
is true for each oxygen ‘particle’. The separate atoms of
hydrogen and oxygen then recombine to form molecules
of water vapour.
+
2H2
2 vol
+
=
O2
1 vol
=
2H2O
2 vol
3
4
Unit 1 Module 1 Fundamentals in chemistry
Avogadro’s suggestion fitted exactly with Gay-Lussac’s
result. This idea was eventually accepted and we now have
no problem with the idea of elemental molecules. Later,
Avogadro’s ideas were used to confirm the relative atomic
mass of oxygen as 16 and to unify a system of ‘atomic
weights’ on which all chemists could agree.
Chemists had begun to realize that atoms might have a
more complex structure than that of a solid ball that Dalton
had suggested.
Limitations of Dalton’s theory
Dalton’s theory could not explain the following:
■ why elements combined in the specific ratios observed;
■ the electrical nature of particles, which were being
observed in other experiments.
As an example, in 1807 Humphry Davy, an English chemist
(1778–1829), had connected a sample of ‘potash’ (what we
would now call soluble potassium compounds) between
the plates of a battery and observed a ‘vivid action’ taking
place. Dalton’s theory didn’t explain why electricity would
have such an effect.
electric current was passed through them. Based on this, he
proposed the existence of ‘an electromotive force’ holding
elements together in chemical compounds.
In 1832 Davy’s student, Michael Faraday (1791–1867),
established that the amount of substance produced by a
chemical reaction during electrolysis is proportional to the
quantity of electricity that passed through the electrolysis
cell.
In the late 1860s, Sir William Crookes, an English physicist
(1832–1919), designed experiments in which he passed
an electric current through sealed evacuated tubes. These
tubes contained metals as both the positive and negative
electrodes, connected to an external source of electricity.
He observed ‘rays’ that travelled in a straight line from
the negative electrode (cathode) to the positive electrode
(anode), regardless of the metal used as the cathode. He
named these ‘cathode rays’. His conclusion was ‘that
cathode rays are negatively charged’.
screen
cathode
Table 1.1 Dalton’s theory and its limitations
Dalton’s theory
What is now known
Matter consists of tiny indivisible particles
(atoms) that cannot be created or
destroyed.
Atoms of one element cannot be
converted into atoms of another element.
During a chemical reaction, reacting
molecules separate into atoms that
recombine to form different molecules.
Atoms of a given element are identical
in mass and other properties and are
different from atoms of any another
element.
Atoms of different elements combine
with each other in simple specific whole
number ratios to give combined atoms
(molecules). This is the law of multiple
proportions.
Atoms are not indivisible and are
composed of sub-atomic particles.
+
–
battery
During a chemical reaction this idea
is true. However, in nuclear reactions
atoms of one element can change
into atoms another element.
Atoms of an element can have
slight differences in their mass and
properties. Isotopes will be discussed
later in this chapter.
True for most compounds, but there
are a few compounds in which there
are slight variations in their atomic
ratios.
Atomic structure
Discovery of the electron
Although Dalton’s ideas were only partly correct, they
could be tested experimentally. They paved the way for
the development of more ideas and theories. This process
of one theory suggesting new ideas and new theories
suggesting further experiments is how science develops.
As mentioned above, in the early 1800s Humphry Davy
observed that some substances were decomposed when an
Figure 1.1 Crookes’ tube, showing the shadow image caused by
the cathode rays.
Other scientists found that the rays cast sharp-edged
shadows, as shown in Figure 1.1. In 1869, Johann Hittorf, a
German physicist (1824–1914), recognized that this could
be explained if the ‘rays’ were in fact a stream of particles.
In 1897, John Joseph Thomson, an English physicist
(1856–1940), investigated these rays further and found
that they could be deflected by electric and magnetic fields.
He showed that they were made up of very light particles.
The results of his experiments enabled him to calculate
their ratio of charge to mass (e/m). He found this to be
1.8 × 1011 C kg−1. The accurate value that we now use is
e/m = 1.759 × 1011 C kg−1. Thomson further estimated that
the mass of the electron was approximately 1/1840 that of
the mass of one atom of the lightest element, hydrogen.
The observation of cathode rays coming from atoms
contradicted Dalton’s theory that atoms are indivisible and
suggested that all atoms contained these negatively charged
particles of extremely tiny mass. The particles came to be
called ‘electrons’.
Chapter 1 Atomic structure
Thomson concluded that since matter as a whole is
electrically neutral, the atom could be broken down into
two parts; one negatively charged and the other positively
charged. The negatively charged parts (electrons) were
observed to be easily removed from the atom but the
positively charged parts were not.
further used radioactivity as a tool to study the structure of
matter. In one experiment he and his research students,
Hans Geiger and Ernest Marsden, designed and carried out
experiments in which a very thin gold foil was bombarded
with alpha particles. (Alpha particles are positively charged
helium nuclei – see Chapter 3, page 29.)
In 1904, Thomson proposed that the atom was ‘jelly-like’,
consisting mainly of positive charge distributed throughout
the jelly with electrons embedded in this jelly rather
like plums in a pudding. This model is called the ‘plum
pudding’ model (Figure 1.2). The negative charges balance
the positive charges so the atom is electrically neutral.
Thomson therefore provided the experimental evidence
that the atom did posses an internal structure, directly
against Dalton’s model of the indivisible atom. The radius
of an atom based on this model was 10−10 m.
They observed that:
■ most of the particles passed through the gold foil with
little or no deflection;
■ a few particles were deflected off course slightly;
■ a very few particles were deflected backwards, almost
in the direction from which they came (Figure 1.3).
electrons
nucleus
alpha
particles
atom
Figure 1.3 A representation of Rutherford’s gold foil experiment,
showing the deflection observed when a beam of alpha particles
hit a metal target.
Figure 1.2 Thompson’s plum pudding model. The electrons are
the ‘plums’ mixed throughout the positively charged ‘pudding’.
Thomson’s conclusions were that:
■ all matter is made up of negatively charged particles
‘enclosed in a sphere of uniform positive electrification’;
■ the negative particles contribute only a tiny fraction of
the mass of the atom.
Rutherford and the nuclear atom
Ernest Rutherford, a New Zealand-born British chemist
(1871–1937) was working with radioactive materials. He
was building on the earlier discoveries of X-rays by Wilhelm
Roentgen in 1895 and radioactivity by Henri Becquerel in
1896 (work then developed by Marie and Pierre Curie).
In 1899, Rutherford proposed that there are three major
types of spontaneous radioactivity: alpha particles (α),
beta particles (β) and gamma rays (γ). These particles/
rays display distinctly different properties when subjected
to electric and magnetic fields. Alpha particles and beta
particles bend in opposite directions in an electric field
while gamma radiation is unaffected by an electric field.
Alpha particles are positively charged helium nuclei, while
beta particles are negatively charged high speed electrons
and gamma rays are uncharged (see Chapter 3). Rutherford
These results were quite stunning and totally unexpected.
Rutherford’s recollection of the results is worth quoting:
‘It was quite the most incredible event that has ever happened to
me in my life. It was almost as incredible as if you fired a 15-inch
shell at a piece of tissue paper and it came back and hit you.’
If Thomson’s plum pudding model of the atom was true,
then all the positively charged alpha particles should pass
through the jelly-like diffuse positive charges with only
slight, occasional deflections. Rutherford and his team
concluded that since most of the alpha particles passed
through the gold foil without being deflected, then an atom
is comprised of mainly empty space in which the electrons
are located. The few positively charged alpha particles that
were deflected backwards in the direction from which they
came, resulted from the repulsion by a positive charge
concentrated in a tiny volume, much smaller than the
atom itself.
Rutherford’s recollection continues:
‘On consideration, I realized that this scattering backward must be
the result of a single collision, and when I made calculations I saw
that it was impossible to get anything of that order of magnitude
unless you took a system in which the greater part of the mass of
the atom was concentrated in a minute nucleus. It was then that I
had the idea of an atom with a minute massive centre, carrying a
charge.’
5
6
Unit 1 Module 1 Fundamentals in chemistry
Rutherford’s experiment suggested that the positive charge
of the atom is essentially concentrated in a ‘nucleus’ (from
the Latin nucleum meaning ‘kernel’) that was extremely
small (the diameter of the nucleus was estimated to be
≈10−15 m) and therefore very dense. Surrounding the
nucleus was mainly empty space containing the electrons.
In this model of the atom the positively charged nucleus
was at the centre, with the electrons revolving around the
nucleus in a manner similar to the Sun and the planets,
with most of the mass of the atom being in the nucleus.
The positive charges in the nucleus are exactly balanced
by the negative charges on the surrounding electrons,
resulting in the atom being electrically neutral.
The number of positive charges in the nucleus of an atom
is called the atomic number (Z). This value is different
for each element and is characteristic of it. The positive
nucleus of the hydrogen atom was called a proton (from
the Greek word proton meaning ‘first’).
Rutherford speculated about structure of the nucleus,
suggesting in 1920 that another particle was involved.
When alpha particles (helium nuclei) were used to
bombard elements such as beryllium, an intensely
penetrating radiation was emitted. At first it was thought
that this was gamma radiation (see Chapter 3, page 30),
but then it was realized that it consisted of neutral particles
with a mass only slightly greater than that of the proton.
Finally, in 1932, James Chadwick, an English physicist
(1891–1974), discovered the third sub-atomic particle.
Because the particles were electrically neutral they were
called ‘neutrons’.
Sub-atomic particles
Atoms were therefore shown to be made up of smaller
particles, with properties summarized in Table 1.2.
Table 1.2 Sub-atomic particles
Name and
symbol
electron, e−
proton, p+
neutron, n0
Charge
Mass
1
relative: 1837
relative: −1
absolute: −1.602 × 10−19 C
absolute: 9.109 × 10−28 g
relative: +1
relative: 1
absolute: +1.602 × 10−19 C absolute: 1.673 × 10−24 g
relative: 0
relative: 1
absolute: 0
absolute: 1.675 × 10−24 g
Location
has an atomic number that is unique to it. Each atom of a
given element has the same atomic number. For example,
all nitrogen atoms have an atomic number of 7 since they
each have seven protons. If they didn’t have seven protons
then they wouldn’t be nitrogen atoms.
The mass number of an atom is the total number of its
protons and neutrons. The mass of the electron is not
included since its mass is only 1/1837 of a mass unit
whereas each proton has a relative mass of 1 unit and each
neutron also has a relative mass of 1 unit. A nitrogen atom,
for example, that has seven protons and seven neutrons
will have a mass number of (7 + 7) =14.
In the shorthand atomic symbol of an element, the mass
number is written on the left as a superscript and the atomic
number on the left as a subscript. For example, the isotope
of uranium used in nuclear reactors can be symbolized as
238
92 U.
The number of neutrons can be found by subtracting the
atom’s atomic number from the mass number. The example
above represents an atom with 92 protons (uranium),
while the protons and neutrons total 238. Therefore the
atom contains 146 neutrons.
number of neutrons = mass number − atomic number
=A−Z
Isotopes of elements
One of John Dalton’s ideas was that ‘all the atoms of a
particular element are identical in weight’. This is only
partly true. The factor that fixes the identity of an element
is the number of protons in its nucleus (Z). However, the
nucleus also contains neutrons; the neutrons contribute
mass but have almost no effect on chemical properties.
Therefore, it is possible to have atoms of an element,
all with the same number of protons, but with different
numbers of neutrons and, hence, different masses. These
variants are called isotopes of the element.
outside
nucleus
Isotopes of an element have same number of protons
(atomic number, Z) but different number of neutrons and
therefore different mass numbers (A).
inside
nucleus
inside
nucleus
Relative atomic mass and isotopic mass
The atomic number of an element (Z) is equal to the
number of protons in the nucleus of its atoms. Each element
ITQ 1 Carbon has three naturally occurring isotopes, 12C, 13C and
14C. Work out the number of protons, electrons and neutrons in
each isotope of carbon.
Chemists in the nineteenth century referred to the ‘atomic
weight’ of an element. What they were really talking about
was the ‘relative atomic weight’, taking the weight of a
ITQ 2 The relative atomic masses of most elements are not whole
numbers. Why is this so?
Chapter 1 Atomic structure
hydrogen atom as 1. Today we refer to the relative atomic
mass (RAM) of an element. We use as our benchmark
atom the isotope 12C, which is taken as having a value of
12. This is a relative value (one mass divided by another),
so RAM has no units.
The relative atomic mass (RAM) of an element is the mass
1
of one atom of the element relative to 12 the mass of one
atom of the carbon-12 isotope (12C).
Relative isotopic mass is the mass of an atom of a specific
isotope of the element relative to the mass of the standard
carbon-12 isotope.
Every element occurs naturally as a mixture of isotopes. The
relative atomic mass is therefore a weighted average of all
the stable relative isotopic masses, taking their abundance
into account. For example, chlorine occurs as a mixture of
35Cl (roughly 75%) and 37Cl (roughly 25%). The weighted
average of chlorine’s RAM is:
75
25
3550
= 35.5
× 35 +
× 37 =
100
100
100
Actual atomic mass
Actual atomic mass, as opposed to relative atomic mass, is
1
measured in atomic mass units (amu). One amu is 12 the
mass of a neutral atom of 12C. The atomic mass unit (amu)
has now been changed to the Dalton (Da), so one atom of
carbon-12 has a mass of 12 Da. However, it is to be hoped
that you realize the error in saying that any individual
atom of chlorine has an atomic mass 35.5 Da!
Electrons in atoms
Rutherford correctly placed the nucleus of an atom at
its centre, but the placement of the electrons posed him
problems. Electrons could not be placed at a distance from
the nucleus and also be stationary because the electrostatic
attractive force of the positively charged nucleus would
pull them towards the nucleus. On the other hand, if
the electrons were to move, then based on the laws of
classical physics, they would continuously radiate energy
and eventually collapse into the nucleus. The electron can
apparently, therefore, be neither stationary nor in motion.
The Bohr model
The arrangement of electrons around the nucleus of the
atom remained a great challenge for scientists for many
years. This dilemma was resolved in 1913 by the Danish
physicist Niels Bohr (1885–1962). Bohr was aware that
if hydrogen gas is enclosed in a tube at low pressure and
Figure 1.4 The Bohr planetary model of the atom.
subjected to a high voltage, it gives off a pink light. When
you look at that light with a spectroscope, you see that it
is a mixture of coloured lines, not a continuous spectrum
like a rainbow.
Bohr suggested that it was possible for electrons to remain
stable in orbits around a nucleus provided that the electrons
had specific energies (Figure 1.4). He also suggested that an
atom jumping from one energy state to another would give
off light of a specific colour. He suggested that the electrons
were held to the nuclei both by gravitational and electric
(coulombic) forces. Spectroscopic evidence, the existence
of characteristic lines, showed that the energy levels were
the same in all hydrogen atoms. Bohr called these energy
levels ‘stationary states’.
Bohr’s model for the hydrogen atom was based on the
following assumptions:
■ Electrons revolve around the nucleus in circular
stationary states called orbits.
■ The energy of an electron in these stationary states
(orbits) is dependent on the distance (r) of the orbit
from the nucleus.
■ Only specific electron orbits of certain radii and certain
energies are allowed. No orbits exist between these
allowed orbits.
■ The absorption of light energy by an electron can result
in a transition from an orbit of lower energy to one
of higher energy if the frequency of the light energy
supplied corresponds exactly to the energy difference
between the orbits (ΔE).
ΔE = (Ehigher − Elower) = h ν (the Bohr frequency
condition)
ITQ 3 Why should an electron in a Bohr orbit NOT be stable,
according to classical physics?
7
8
Unit 1 Module 1 Fundamentals in chemistry
■ Similarly, a transition from an orbit of a higher energy
to one of lower energy will result in the emission
of light where the frequency of the emitted light
is such that the energy emitted exactly equals the
energy difference between the energy levels of the
orbits (Figure 1.5). Einstein had showed that light
rays (regarded until then purely as wave motion) can
behave as particles (photons) carrying fixed amounts
of energy, and that the energy of a photon is related
to its frequency by the expression E = hv, where h is a
constant and v is the frequency of the light.
n=3
n=2
n=1
ΔE = hv
+ Ze
Figure 1.5 The Bohr atom emitting a photon.
Notice that the last of these assumptions is meant to explain
why the observed emission spectra contain sharp lines.
Conversely, the sharp lines in emission spectra provide
evidence that Bohr’s theory is at least partly correct.
Bohr also suggested that chemical properties of atoms were
based on the electrons in the outermost orbit. Bohr’s model
received strong criticism from the scientific community,
especially since he offered no explanation as to why these
stationary states existed. Bohr’s model gained greater
acceptance only after it stood up to several specially designed
experiments (especially that of Franck and Hertz) and after
it turned out to form the basis of the revolutionary new
theory of quantum mechanics (see Chapter 2, page 12).
the next energy level. Each Bohr orbit was referred to as
a shell. (There is no interaction between electrons in each
shell.)
Successes of the Bohr model
The Bohr model allowed for the derivation of the Rydberg
constant (see footnote), as well as the accurate calculation
of the wavelengths of experimentally observed lines in
the electromagnetic absorption and emission spectra of
one-electron species such as hydrogen and hydrogen-like
species.
The Bohr model predicted the radius of the n = 1 orbit of
the hydrogen atom as 5.3 × 10−11 m and the ionization
energy (the energy required to completely remove an
electron from an atom of hydrogen) as 2.18 × 10−18 J.
Both of these results were in excellent agreement with the
experimentally determined values.
The idea that chemical properties are based on the
outermost electrons fits with the arrangement of elements
in the periodic table. The periodic table is discussed in
Chapter 2 (page 17).
The most stable energy level of an atom is the one with
the lowest energy and is referred to as the ground state
of the atom. In the case of the hydrogen atom this state is
the n = 1 energy level. The other energy states for which
n > 1 are referred to as excited states. The electron in the
n = 1 energy state of the hydrogen atom can jump from the
ground state to an excited state provided that the correct
amount of energy is supplied. Similarly, when the electron
is in the n = 2 energy state it can jump to another excited
state. These energy states and the electron ‘jumps’ are
shown in Figure 1.6.
Each series in Figure 1.6 is named after its discoverer.
Pfund
series
The Bohr model and many electron atoms
The hydrogen atom contains only one electron, which
greatly simplifies the situation. Bohr’s model of the
hydrogen atom was applied to atoms containing many
electrons. According to Bohr, each orbit can accommodate
a certain maximum number of electrons. When that orbit
is filled then the additional electrons occupy an orbit in
■ The Rydberg constant occurs in an expression which Bohr used to
predict the frequency of spectral lines emitted by excited hydrogen gas
but which had been known empirically since the late nineteenth century.
Brackett
series
n=4
Paschen
series
n=3
Balmer
series
n=2
Lyman
series
n=1
n=5
n=7
n=6
n=5
n=4
n=3
n=2
n=1
Figure 1.6 The energy level changes in the Bohr hydrogen atom.
Chapter 1 Atomic structure
■ The Lyman series (discovered in 1906) has
Beyond Bohr …
wavelengths in the ultraviolet (UV) spectrum, resulting
from electrons dropping from higher energy levels into
the n = 1 orbit.
■ The Balmer series (discovered in 1885) has
wavelengths in the visible light spectrum, resulting
from electrons falling from higher energy levels into
the n = 2 orbit.
■ The Paschen series (discovered in 1908) has
wavelengths in the infrared spectrum, resulting from
electrons falling from higher energy levels into the
n = 3 orbit.
The energy difference between the lines in the series
decreases as the energy increases (or the wavelength
decreases). Eventually the lines become so close that they
form a continuous band called a continuum. This can
be seen towards the top of Figure 1.6. At this point and
beyond it is impossible to distinguish between the lines
and a convergence limit is said to be reached (Figure
1.7). At the convergence limit, the electron will no longer
experience the effect of the nuclear attraction. In other
words, it is ‘free’ from the influence of the nucleus and the
atom that has lost the electron has become ionized.
A(g) → A+(g) + e–
15
ionized continuum above 13.6 eV
n=3
Energy (eV)
discrete bound states
10
n=2
5
0
ground state
n=1
Figure 1.7 The convergence limit.
Limitations of the Bohr model
■ No explanation was provided by Bohr as to why the
electron absorbs or emits radiation when it moves
from one energy level to another.
■ The Bohr model did not predict accurately all the lines
in the spectra of multi-electron atoms.
■ It did not predict the intensity of the lines.
■ It assumed that an electron occupies a specific orbit at
a specific distance (radius) from the nucleus. This was
shown by Heisenberg to be incorrect.
■ It could not explain why the frequency of spectral lines
is changed by an external magnetic field.
Bohr’s theory, like Dalton’s a century before, built a
foundation on which new knowledge and understanding
could be based. The single-electron atom (hydrogen) could
be fitted to the Bohr model with some precision. Multielectron atoms were not so easy to deal with. The many
interactions between the electrons in their ‘orbits’ were too
complex. Bohr was reduced to suggesting that (by analogy
with the Sun and planets), orbits could be circular, or
elliptical to various degrees, that the orbits rotated about
the nucleus, and that an orbit could contain more than one
electron. These ideas are developed further in Chapter 2.
9
10
Unit 1 Module 1 Fundamentals in chemistry
Review questions
Summary
1
Naturally occurring boron consists of two isotopes,
10B (19.9%) with an atomic mass of 10.0129 Da and
11B (80.1%) with an atomic mass of 11.00931 Da.
Calculate the atomic mass of naturally occurring
boron.
2
Describe the contributions made by (a) Dalton,
(b) Thomson and (c) Rutherford to our understanding
of the structure of the atom.
3
Calculate the energy in joules of 1 mole of photons
with (a) a frequency of 2.6 × 10–14 Hz and (b) a
wavelength of 546 nm.
4
Sodium vapour lamps have a characteristic yellow
colour. Given that the wavelength is 589 nm, calculate
the frequency of this light.
5
Complete the following table:
✓ Atoms have quantized energy levels.
✓ An electron can move from one energy level to
another only if it absorbs or emits a photon of
energy equal to the difference in the energy of
the two energy levels.
✓ Each element has a unique line spectrum which
results from the emission of photons of specific
energy (and hence of specific frequency) as its
electrons move from higher to lower energy
levels.
✓ Bohr’s model is successful in explaining the
line spectrum of hydrogen and many other
one-electron species but failed when applied
to the spectra of atoms with more than one
electron.
Symbol
✓ An atom consists of a nucleus containing
Atomic Mass Number of Number of Number of
Charge
number number protons
neutrons electrons
15N
positively charged proton(s) and neutral
neutron(s).
19
20
1
✓ The nucleus of the atom is surrounded by the
11
protons (or electrons).
10
6
A microwave oven was used to warm a meal. If the
frequency of the radiation is 2.0 × 109 s–1, determine
the energy of one photon of this microwave radiation.
7
Explain why the spacing between a series of spectral
lines decreases as the wavelength becomes shorter.
✓ The mass number of the atom is the sum of the
protons and the neutrons.
12
Sr2+
of protons, resulting in the atom being neutral.
✓ The atomic number of the atom is the number of
0
90
negatively charged electron(s).
✓ The number of electrons is equal to the number
+1
2
Answers to ITQs
✓ All atoms of a given element contain the same
1
number of protons/electrons but can contain
different numbers of neutrons.
✓ The atomic mass of an element is the average
1
mass of one atom of the element relative to 12 th
the mass of one atom of the carbon-12 isotope
(12C).
✓ An element occurs as a mixture of isotopes.
Isotopes are atoms with the same number of
protons but different numbers of neutrons. Each
isotope has a different, characteristic relative
atomic mass (RAM).
12
C: 6 protons, 6 neutrons, 6 electrons
C: 6 protons, 7 neutrons, 6 electrons
14
C: 6 protons, 8 neutrons, 6 electrons
13
2
The atomic mass of an element is a weighted average
of the mass of its isotopes. This is unlikely to be a
whole number.
3
As a moving particle in an electric field, the electron
‘should’ radiate energy and so move in an orbit of
ever-decreasing radius.
Chapter 1 Atomic structure
Answers to Review questions
1
10.811 Da
3
(a) 1 × 105 J; (b) 2.19 × 105 J
4
ν = 5.09 x 1014 Hz
5
Symbol
Atomic Mass Number of Number of Number of
Charge
number number protons
neutrons electrons
15N
7
15
7
8
7
0
39 +
19
39
19
20
18
+1
3H+
K
1
3
1
2
0
+1
+
Na
11
23
11
12
10
+1
90Sr2+
38
90
38
52
36
+2
23
6
1.33 × 10–24 J
7
The difference in energy for the ninitial to the nfinal gets
progressively less with increasing distance from the
nucleus.
11
12
Chapter 2
The quantum atom and the periodic table
Learning objectives
■ State the Pauli exclusion principle and the Aufbau principle.
■ Write the electron configuration of a given atom or ion given its atomic number.
■ Illustrate electron configuration using an electron diagram.
■ Sketch the periodic table, illustrating the blocks and the elements in each group.
■ Discuss periodic trends in atomic and ionic radii.
■ Define ionization energy.
■ Define electron affinity.
■ Explain the general periodic trends in ionization energy and electron affinity among the main group
elements.
■ Define electronegativity and state the periodic trends in electronegativity.
■ Explain, using the elements of period 3 as an example, how ionization data can provide evidence for
sub-shells.
■ Predict the electronic configuration of an element from data on successive ionization energy.
The quantum atom
Bohr’s theory (discussed in Chapter 1), and the equations
he developed in 1913, were based purely on the idea of the
electron as a particle. Bohr saw the electron as something
that had a definite position in space and moved in a
discernible fixed path. He offered no explanation of why
an electron moving in an ‘orbit’ should thwart the laws of
classical physics by remaining stable.
Matter and waves
In 1905 Albert Einstein, the famous German-born
theoretical physicist (1879–1955), had suggested a new idea
about light. Although light is regarded as a wave motion
because it can be refracted and can suffer interference and
diffraction, sometimes it shows the properties of a particle.
For example, some chemical reactions are light-initiated,
but light below a certain frequency does not help the
reaction, however intense that light may be. Einstein called
these ‘light-particles’ photons.
In 1924 Louis de Broglie, a French physicist (1892–1987),
extended this idea by suggesting the reverse: that material
particles can have wave properties. If de Broglie’s suggestion,
that all particles travel in waves, is correct then electrons
should also exhibit wave-like properties such as diffraction.
According to de Broglie, every particle is associated with a
wave. Its wavelength (λ) is given by the relationship:
h
λ= p
where h is a constant (known as Planck’s constant,
h = 6.63 × 10−34 m2 kg s−1) and p is the momentum of the
particle (mass × velocity). His thesis was quickly supported
by the observation that electrons, so far regarded as
particles, could be diffracted. Diffraction is a fundamental
property of a wave.
This phenomenon is called wave–particle duality. The
particular properties, either wave or particle, which are
displayed depend on the nature of the experiment being
used. The idea that anything can have two totally different
sets of properties at one and the same time is outside our
everyday experience. If I throw a black stone into a pond I
do not expect it to re-appear as a scarlet ibis and fly away.
But in everyday life, wave–particle duality is not of much
importance. The wavelength of a cricket ball coming from
a fast-bowler’s hand is roughly 1 × 10−34 m, some 1028
times smaller even than an atom. So we do not expect the
ball to be diffracted around the edge of the bat.
In contrast, the wavelength of an electron is of the order of
tens of picometres (10−9 m), which is comparable with the
Chapter 2 The quantum atom and the periodic table
size of an atom. An electron is stable when its wavelength,
or a multiple of it, is an exact fit around a nucleus at a
particular distance, but not otherwise. Figure 2.1 shows an
electron wave with three wavelengths around a nucleus at
distance r3 and four wavelengths at distance r4. An electron
with this wavelength could not fit in an orbit with a value
of r between these.
r4
r3
The higher the frequency of the wave, the more energy
each photon carries. Because frequency is inversely
proportional to wavelength this means that photons with
longer wavelengths carry less energy.
This equation also tells us that the energy carried by a
stream of photons cannot have just any value. The energy
comes in ‘packets’ of fixed size (hv) because each packet
comes from an identical electron transition between energy
levels. Such a packet is called a quantum of energy; the
plural of quantum is quanta. Einstein’s idea that energy
is quantized rather than continuous has given us the
quantum theory of matter.
The uncertainty principle
Figure 2.1 Electron waves fitting exactly around a nucleus at two
different distances. The wavelength is the same in both cases.
The same wavelength cannot fit in an orbit with a value of r
between these two radii.
Energy and wavelength
The wavelength of an electromagnetic wave and its
frequency (i.e. the number of waves which pass a given
point in one second) are related by the expression
c=v×λ
where c is the velocity of light, v is the frequency and λ is
the wavelength (Figure 2.2).
Not content with having to understand the idea of wave–
particle duality, we also have to cope with a built-in lack
of precision in measurements at the atomic level. In
1927 Werner Heisenberg, a German theoretical physicist
(1901–1976), realized that because the devices we use to
measure particle properties (for example, photons) are
very similar in size to the particles themselves, the very
interaction between the particle and the measuring device
modifies the original properties of the particle. Heisenberg’s
uncertainty principle says that the more precisely
the position of a particle is known (or experimentally
determined), the less precisely can its momentum be found
in the same experiment. The uncertainty principle is stated
mathematically as
Δp.Δx ≥ =
Figure 2.2 Frequency and wavelength. The diagram shows two
waves, one with half the wavelength of the other. If both are
travelling at the same speed then in a given time, twice as many
of the short wavelength waves will pass a point in the direction of
travel than will the long wavelength waves.
Einstein’s idea of the photon included the statement that the
energy carried by a single photon is given by the expression
E=h×v
where E is the photon energy, h is Planck’s constant and v
is the frequency of the wave associated with the photon.
ITQ 1 A cricket ball has a mass of 160 g and a fast bowler can bowl
at 140 km/hour (38.9 m s−1). Use the de Broglie equation to confirm
the wavelength of the cricket ball. h = 6.63 × 10−34 m2 kg s−1.
h
4π
where the symbol Δ means ‘the uncertainty in’. Notice that h
(Planck’s constant) is involved yet again. Since momentum
(p) is a measure of particle energy, this equation tells us
that on the sub-atomic scale we cannot accurately know
both the position and the energy of a particle. In turn this
means that Bohr’s idea of a ‘particle’ travelling in a fixed
‘orbit’ has to be modified.
Orbits are dead, long live orbitals!
Erwin Schrödinger, an Austrian physicist (1887–1961),
created a set of equations that help us to describe the
behaviour of electrons around atomic nuclei. Instead of
saying exactly where an electron exists at any moment in
time, the solutions of Schrödinger’s equations define the
probability of finding an electron. They can be used to
ITQ 2 How is the rise of the quantum theory like the rise of
Dalton’s atomic theory?
13
Unit 1 Module 1 Fundamentals in chemistry
describe volumes of space in which there is a high probability
of finding the electron. The most useful graphical solutions
are called radial distribution plots.
The radial distribution plot shows the probability of finding
an electron at a given radial distance r from the nucleus,
plotted against distance from the nucleus. For the one
electron of the hydrogen atom, the radial distribution plot
is shown in Figure 2.3.
Figure 2.4 A computer
generated simulation of the
probability of finding an electron
in its lowest energy state in a
hydrogen atom. The maximum
probability is at 53 pm from the
nucleus.
that an orbital can only contain one electron. To define
the energy of an electron exactly we need four numbers.
The first three define quite large differences in energy (the
numbers 1, 2, 3, … as used in Chapter 1), but the fourth is
different. It is called the spin of the electron. It has a value
of 12 and it makes only a minute difference to the energy.
So the orbital containing an electron with spin + 12 and the
orbital containing the electron with spin − 12 are so similar
to each other that the difference is lost in the fuzziness.
An orbital can contain two electrons, but they must have
opposite spins.
Radial distribution function
14
Radius, r
Figure 2.3 A plot showing the probability of finding the electron
of a hydrogen atom anywhere in a shell of radius r plotted
against r, where r is the distance from the nucleus of the atom.
The hydrogen atom is spherically symmetrical; this means
that there exists a sphere around the nucleus in which
there is a high probability of finding the electron. Because
of the uncertainty principle, this sphere is not sharp-edged,
but ‘fuzzy’. This ‘fuzziness’ is shown in Figure 2.4. For the
hydrogen atom, the maximum probability of finding the
electron is at a distance of 53 pm from the nucleus, which
coincides with the Bohr radius for this atom.
The volume of space that contains a good probability
of finding an electron is called an orbital. We shall be
concerned with the numbers and shapes of the orbitals
as well as the electrons they contain. Usually we think of
an orbital as the volume of space which contains a 95%
probability of containing the electron. Notice that nowhere
in this discussion do we try to describe the electron itself,
only its position.
The orbital map
Because we cannot know exactly what space an electron
occupies, the uncertainty principle tells us that we can
know its energy with great precision. Although the actual
energies of electrons around a nucleus vary with the
nucleus, the pattern of their magnitudes is always the same.
We will compare the pattern with the one we developed in
Chapter 1. The energies can be shown on a ‘ladder’ with
energy levels as rungs. The rungs are numbered 1 to 8,
but we will only need to look at the first five. These are
shown in outline in Figure 2.5. This number is called the
principal quantum number and is given the symbol n.
n=5
n=4
n=3
n=2
Pauli’s exclusion principle
Wolfgang Pauli, an Austrian theoretical physicist (1900–
1958), pointed out that in any one atom, no two electrons
can have exactly the same energy. This statement is known
as the Pauli exclusion principle.
An orbital like the one in Figure 2.4 is defined using the
energy of the electron as one factor, but that does not mean
n=1
Figure 2.5 An energy ladder showing the first five main divisions
of energy levels. The lowest energy level is shown at the bottom.
The levels get closer as they get nearer to the top.
Chapter 2 The quantum atom and the periodic table
Any electron that occupies any one of these levels is found
in a spherical orbital like that in Figure 2.4. These orbitals are
called s orbitals. (Memory hint: think of ‘s’ for spherical.)
3d
3p
For level 1 there are no other energy levels.
3s
For level 2 and all subsequent levels, there is another
energy level a little way above the s level. It is not spherical.
It has three components, all at right angles to each other.
We say that they are along the x-, y- and z-axes. Each of the
three components is at exactly the same energy level. Each
associated orbital looks a bit like an egg-timer (Figure 2.6).
We usually draw these orbitals as shown in Figure 2.6, but
notice that the two lobes together make up one orbital.
Like any other orbital it can accommodate two electrons,
so the energy level as a whole can hold six electrons. In
the old terminology of ‘shells’ (page 8, Chapter 1), these
would be ‘sub-shells’. These orbitals are called p orbitals.
(Memory hint: they are shaped like propellers.)
Figure 2.7 shows these p orbitals added to the first three
levels of the energy ladder.
These energy levels are sufficient to define the electron
configurations for the first 18 elements (from hydrogen to
argon) in the periodic table.
Above the level of the third group, another set of levels
appears. In it there are five levels which are separate but
identical in energy. Look at the pattern of numbers of levels:
level 1 has 1, level 2 has 1,3 and level 3 has 1,3,5, …). The
associated orbitals are called d orbitals. (Memory hint: one
of them is shaped like a donut.) Figure 2.8 shows the energy
ladder with the 3d levels added.
2p
2s
1s
Figure 2.8 The first three levels of the energy ladder with d levels
added. The levels are identified as 1s, 2s, 2p, 3s, 3p and 3d.
The fourth main level has another set of levels above its d
levels. The associated orbitals are called f orbitals. (Memory
hint: they have funny shapes.) We shall not need to
consider these orbitals any further.
When we draw the energy ladder we take account of the
fact that the 4s level is lower than the 3d level (Figure 2.9).
This has some interesting consequences, which we will see
in Chapter 17 about the transition elements.
6p
5d
6s
5p
5s
4p
4s
Energy
3s
2p
4d
3d
3p
3p
z
4f
3s
2p
2s
2s
y
1s
x
Figure 2.9 Relative energy levels for n = 1 to 6.
1s
Figure 2.6 Shape of
p orbitals, both in outline
and in probability.
Figure 2.7 The first three
levels of the energy ladder
with p levels added. The
levels are identified as 1s, 2s,
2p, 3s and 3p.
The aufbauprinzip principle
Bohr and Pauli formulated the aufbauprinzip (which means
‘building up’ principle). This principle governs the order in
which energy levels are filled when building up electrons
around an atom. Each orbital can hold two electrons, with
opposite spin. One spin gives slightly less energy than the
other so, at any level, electrons half-fill each orbital first.
15
Unit 1 Module 1 Fundamentals in chemistry
To build up the electron structure for an atom, fill a diagram
like that in Figure 2.9 one electron at a time. Start at the
bottom of the ‘ladder’ (1s) and the left-hand side of each
rung. Move on to the next rung horizontally and then
the next vertically until all the electrons are added. Using
sodium (Z = 11) as an example, the order of filling is
1s1 → 1s2 → 2s1 → 2s2 → 2px1 → 2py1 → 2pz1 → 2px2 →
2py2 → 2pz2 → 3s1
Figure 2.10 shows the diagram for a sodium atom.
6p
5d
4f
6s
5p
4d
5s
4p
structure of the preceding noble gas. The sodium structure
would then be written as Ne 3s1.
The electronic structure of atoms is linked to their position
in the periodic table. This topic is looked at next in this
chapter.
Developing the periodic table
During the middle of the nineteenth century more
elements were being discovered. With this discovery came
the challenge of memorizing their names and properties.
Attempts were made to develop a systematic manner of
grouping and classifying the elements according to their
properties. What we now are familiar with as the periodic
table developed over many years.
One of the earliest attempts at ordering elements was
made by Johann Wolfgang Dobereiner, a German chemist
(1780–1849). In 1829 he published work that grouped
elements with similar properties in blocks of threes (triads),
as shown in Figure 2.13. He had noticed that the average
atomic mass of lithium and potassium was similar to the
atomic mass of sodium, and that these three elements had
similar properties. He found that this observation was true
for other groups of three elements.
3d
4s
3p
Energy
16
3s
2p
2s
1s
Figure 2.10 The electron ladder for a sodium atom.
Li
Ca
S
Cl
Mn
Writing electron structures
Na
Sr
Se
Br
Cr
K
Ba
Te
I
Fe
The series of electrons shown above can very neatly be
abbreviated as 1s2 2s2 2p6 3s1. This shows the energy level,
the orbital type and the number of electrons in those
orbitals, using the symbols shown in Figure 2.11.
number of
electrons
energy level
2p6
type of
orbital
Figure 2.11 Symbols showing the occupancy of an atomic orbital.
Figure 2.12 illustrates a readable way of showing an
electron structure. This approach, which has the advantage
of also showing electron spin, is to draw the orbitals as a
horizontal series of boxes, and fill them one at a time. Work
from left to right and always put in the first arrow pointing
downwards to indicate the low-spin state. You will also see
this notation used on page 161.
1s
2s
2p
3s
Figure 2.12 Electrons in boxes for Na, Z = 11.
When larger atoms are involved, not much of this needs
to be spelt out. You only need to show detail above the
Figure 2.13 Dobereiner’s triads – a very early version of the
periodic table.
Quite a few years later (in 1865), John Newlands, an English
chemist (1837–1898), gave a lecture outlining his idea that
when the elements were listed in order of increasing atomic
mass, some similar elements were separated by intervals of
eight (Law of octaves). His work wasn’t accepted because
of some odd groupings. For example, he put oxygen, sulfur
and iron in the same group.
The most successful arrangement of the elements, however,
was the one developed in 1869 by two outstanding scientists
and teachers Dmitri Ivanovich Mendeleev (Russian,
1834–1907) and Lothar Meyer (German, 1830–1895).
Working independently (and without knowing about
Newlands’ work), they both discovered that when the
known elements were arranged in order of increasing
atomic weights, certain similarities in properties were
repeated in cycles of eight. Mendeleev published his version
of the periodic table before Meyer and so he is credited for
its discovery.
Chapter 2 The quantum atom and the periodic table
I
VIII
1
Atomic number
Hydrogen
Element name
Alkali metals
Other metals
Alkali-earth metals
Semimetals
III
Transition metals
Non-metals
5
6
7
8
9
4.0
10
Boron
Carbon
Nitrogen
Oxygen
Fluorine
Neon
Rare earths
Noble gases
Radioactive rare earths
Hydrogen
H
1.0
3
II
Symbol
4
Lithium
Beryllium
Atomic weight
(mean relative mass)
Li
Be
6.9
11
9.0
12
Sodium
Magnesium
Na
Mg
23.0
19
24.3
20
21
22
23
24
25
Potassium
Calcium
Scandium
Titanium
Vanadium
Chromium
K
Ca
Sc
Ti
V
Cr
39.1
37
40.1
38
45.0
39
47.9
40
50.9
41
52.0
42
Rubidium
Strontium
Yttrium
Zirconium
Niobium
Rb
Sr
Y
Zr
Nb
Mo
Tc
85.5
55
87.6
56
88.9
57
91.3
72
93.0
73
95.9
74
Caesium
Barium
Lanthanum
Hafnium
Tantalum
Cs
Ba
La
Hf
Ta
132.9
87
137.3
88
138.9
89
178.5
104
Francium
Radium
Actinium
Fr
Ra
Ac
[223.0]
[226.0]
[227.0]
ACTINIDES
IV
V
VI
VII
He
B
C
N
O
F
Ne
10.8
13
12.0
14
14.0
15
16.0
16
19.0
17
20.2
18
Aluminium
Silicon
Phosphorous
Sulfur
Chlorine
Argon
Al
Si
P
S
Cl
Ar
30
27.0
31
28.1
32
31.0
33
32.1
34
35.4
35
39.9
36
Krypton
26
27
28
29
Manganese
Iron
Cobalt
Nickel
Copper
Zinc
Gallium
Germanium
Arsenic
Selenium
Bromine
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
54.9
43
55.9
44
58.9
45
58.7
46
63.5
47
65.4
48
69.7
49
72.6
50
74.9
51
79.0
52
79.9
53
83.8
36
Ruthenium
Rhodium
Palladium
Silver
Cadmium
Indium
Tin
Antimony
Tellurium
Iodine
Xenon
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
[97.9]
75
101.1
76
102.9
77
106.4
78
107.9
79
112.4
80
114.8
81
119.0
82
121.8
83
127.6
84
126.9
85
131.3
86
Tungsten
Rhenium
Osmium
Iridium
Platinum
Gold
Mercury
Thallium
Lead
Bismuth
Polonium
Astatine
Radon
W
Re
Os
Ir
Pt
Au
Hg
Ti
Pb
Bi
Po
At
Rn
181.0
105
183.8
106
186.2
107
190.2
108
192.2
109
195.1
110
197.0
111
200.6
112
204.4
113
207.2
114
209.0
115
[209.0]
116
[210.0]
117
[222.00]
118
Rutherfordium
Dubnium
Seaborgium
Bohrium
Hassium
Ununtrium
Flerovium
Rf
Db
Sg
Bh
Hs
Mt
Ds
Rg
Cn
Uut
Fl
Uup
Lv
Uus
Uuo
[265.1]
[268.1]
[271.1]
[270.0]
[277.2]
[276.2]
[281.2]
[280.2]
[285.2]
[284.2]
[289.2]
[288.2]
[293.0]
[294.0]
[294.0]
59
60
61
58
LANTHANIDES
2
Helium
Cerium
Molybdenum Technetium
Praseodymium Neodymium Promethium
Meitnerium Darmsladtium Roentgenium Copernicium
Ununpentium Livermorium Ununseptium Ununoctium
62
63
64
65
66
67
68
69
70
71
Samarium
Europium
Gadolinium
Terbium
Dysprosium
Holmium
Erbium
Thulium
Ytterbium
Lutetium
Ce
Pr
Nd
Pm
Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
Lu
140.12
90
140.91
91
144.24
92
[144.91]
93
150.36
94
151.96
95
157.25
96
158.93
97
162.50
98
164.93
99
167.26
100
168.93
101
173.04
102
174.97
103
Thorium
Protactinium
Uranium
Neptunium
Plutonium
Americium
Curium
Brekelium
Californium
Einsteinium
Fermium
Mendelevium
Nobelium
Lawrencium
Th
Pa
U
Np
Pu
Am
Cm
Bk
Cf
Es
Fm
Md
No
Lr
232.04
231.04
238.03
[237.05]
[244.06]
[243.06]
[247.07]
[247.07]
[251.08]
[252.08]
[257.10]
[258.10]
[259.10]
[262.11]
Figure 2.14 A modern version of periodic table.
Unlike Newlands, Mendeleev’s work was accepted because
he deliberately left spaces in his table and then made
predictions about the existence and properties of elements
that would be discovered to fill the spaces. Several years
later, elements were discovered with properties that bore
remarkable similarity to his predictions.
As nuclear reactions were studied, tiny amounts of a
manganese-like element fitting into the space were found,
and it was named technetium. It is radioactive, and the
half-life of its most stable isotope is 4.2 million years. Since
the age of the Earth is estimated as 4.5 × 109 years, its
present-day absence is unsurprising.
In a modern periodic table, the elements are arranged
in order of increasing atomic number instead of atomic
weight. However, at the time of Mendeleev’s work, atomic
numbers were not known because protons and electrons
had not yet been discovered. The concept of isotopes was
also unknown, so many of the atomic weights used were
incorrect.
Since then at least 30 other elements have been discovered
that are results of the radioactive decay of ‘natural’ elements,
or created through nuclear reactions. All these elements are
themselves radioactive. Some are familiar, such as element
86, radon (which was until 2002 the heaviest of the noble
gases). However, most are transient, with half-lives often
measured in seconds. Some are unbelievably rare: only 4
atoms of element 118 (Uuo) have ever been produced. Its
neighbour in the periodic table, Uus (ununseptium, Latin
for 117) has only very recently been announced (2013).
It is the last member of the halogen series and it would
presumably be a metal, if enough atoms could ever be made.
The modern periodic table
A modern form of the periodic table is shown in Figure 2.14.
Synthetic elements
Until 1936, nearly 30 years after Mendeleev’s death, one of
the gaps he had left remained unfilled. Element 43, which
would be placed between molybdenum and ruthenium
and directly below manganese, was still undiscovered. The
element, which Mendeleev called ‘ekamanganese’, could
not be found on Earth.
Structure of the periodic table
Elements in the periodic table are arranged in order of
increasing atomic number. Horizontal rows of elements are
called periods and vertical columns of elements are called
groups. The elements in a group have similar electronic
configurations and similar properties.
17
18
Unit 1 Module 1 Fundamentals in chemistry
Groups
Older systems assigned a Roman numeral (I–VIII) to each
group and added the suffix A for the main groups and B for
the transition elements (in the centre of the periodic table).
In 1985, the International Union of Pure and Applied
Chemistry (IUPAC) adopted a simpler system in which the
columns are labelled from 1 to 18 from left to right. This
system is used by the majority of the international scientific
community. Also in use is a numbering system where the
Roman numerals are converted to Arabic numbers, so that
Group VII is referred to as Group 7. The periodic table in
the CAPE Chemistry Syllabus uses the I–VII system; we
will use this system in this book, but on occasions give the
IUPAC group number in brackets.
Blocks
For convenience, the elements are referred to as being in
blocks.
■ Elements in Groups I and II (1 and 2) are in the
s-block since these elements have configurations ns1
and ns2 respectively.
Periodicity
When chemical and physical properties of atoms are
plotted against atomic numbers, repeat patterns (known as
periodicity) are often observed. These patterns are due to
the fact that the electronic configurations of the atoms are
being repeated. Elements in a group have similar chemical
properties since all members of the group have similar
valence shell electronic configurations. For example, all the
elements in Group I have one electron in the outer shell.
It is the outermost electrons (electrons in highest energy
levels) of the atom that have the greatest influence on its
properties since these electrons are the easiest to remove
or become attracted by nuclei of other atoms or molecules.
Also it is easier to add electrons to the outermost energy
levels.
Changes to properties resulting from addition of electrons to
inner orbitals, for example d orbitals, are usually much less
marked than those associated with the addition of electrons
to outer s and p orbitals. Across a period there are significant
differences in the chemical properties of the elements.
■ The p-block refers to the elements in Groups III–VIII
(13–18). These have electronic configurations ranging
from ns2 np1 for Group III to ns2 np6 for Group VIII.
■ The d-block consists of elements in Groups IIIB–IIB
(3–12) with the configuration (n−1)d1 ns2 to
(n−1)d10 ns2. There are some irregularities in this
block.
■ The f-block elements generally (though there are
some irregularities) have electronic configurations
(n−2)f1 (n−1)d10 ns2 to (n−2)f14 (n−1)d10 ns2.
Elements in the s- and p-blocks are referred to as main
group (or representative) elements. Among the main
group elements are the alkali metals (Group I or Group
1), the alkaline earth metals (Group II or Group 2), the
halogens (Group VII or Group 17) and the noble gases
(Group VIII or Group 18). Elements in the d-block are
the transition elements and the f-blocks contain the
lanthanides and actinides.
Periods
Each period starts with an element with the ns1
configuration and ends with one with the ns2 or ns2 np6
noble gas configuration.
The period number refers to the principal quantum level n,
which contains the outermost electrons.
The law of chemical periodicity
The properties of elements are a periodic function of their
atomic numbers.
Periodic patterns are observed for several physical
properties of the individual atoms of the elements. Here
are some examples:
■ the sizes of the atoms and ions (atomic and ionic radii);
■ the strength with which the outermost electrons are
held by the isolated atom (ionization energy);
■ the ease with which electrons can be added to these
isolated atoms (electron affinity).
The magnitude of these properties are related to the net
attractions between the outer shell electrons.
Periodic properties in atomic ‘size’
Atomic radius
An atom does not exist in isolation neither does it have a
sharply defined boundary (see page 17 above). The size of
an atom can be estimated by assuming that it is spherical
and that, when identical atoms are bonded together, the
radius of one atom can be approximated to be half the
distance between neighbouring atoms. These radii can be
determined from X-ray scattering measurements. Since
atoms bond with each other in a variety of ways (see
Chapter 4), different types of atomic radii can be defined.
Chapter 2 The quantum atom and the periodic table
K
200
Ar
2r
2r
Figure 2.15 Covalent radius.
Figure 2.16 Metallic radius.
Atomic radius / pm
Ne
■ Metallic radius is half the distance between the
nuclei of adjacent metal atoms in a crystal of a metal
(Figure 2.16).
Ca
Na
He
Mg
Li
Si
Al
100
S
P
Be
C
B
Cl
O
N
F
50
H
■ Covalent radius is half the distance between nuclei of
adjacent identical atoms bonded together in a covalent
molecule (Figure 2.15).
150
0
0
2
4
6
8
10
12
14
16
18
20
Element
Figure 2.18 Trends in atomic radius, showing the changes
across periods 2 and 3. Note that the noble gases don’t form
compounds, so the values shown are van der Waals radii.
atomic size decreases
■ Ionic radius is the radius of ions in crystalline ionic
compounds, related to the distance between the nuclei
of neighbouring cations and anions in the crystal
(Figure 2.17).
I–
Li
+
iodide ion
diameter
K
+
II
III
IV
Be
B
C
V
VI
VII
H
Li
atomic size increases
I–
I
He
Ne
N
Na
VIII
Mg
Al
Si
O
F
S
Cl
P
Ar
K
Ca
Ga
Ge
As
Se
Br
Kr
Rb
Sr
In
Sn
Sb
Te
I
Xe
Cs
Ba
Tl
Pb
Bi
Po
At
Rn
+
sum of K
and I– radii
Figure 2.17 Ionic radii are measured indirectly be comparing
internuclear distances in salts whose positive ions vary in size.
Here we can measure the diameter of the iodide ion directly and
so calculate the ionic radius of the potassium ion.
■ van der Waals radius is the radius of adjacent atoms
which are not chemically bonded in a solid but are in
‘contact’ with each other. In the noble gases (Group
VIII), for example, atoms are not chemically bonded to
each other and their van der Waals radii are estimated
as half the distance between the nuclei of adjacent
atoms of solidified samples of the gases. These radii
are much larger than covalent radii and are sometimes
not included in discussions of periodic trends of the
elements.
The above definitions of atomic and ionic radii are only
approximations. Atoms approach each other more closely
in the bonded than the non-bonded state. The atomic
radius of an element may vary depending on the compound
in which it is present. For example, the atomic radius of
carbon in diamond is 77 pm, in ethane (C–C) it is also
77 pm, in ethene (C=C) it is 69 pm and in ethyne (C≡C) it
is 60 pm.
Figure 2.19 Periodic trends in atomic radius for the main group
elements. The trends both down groups and across periods can
be seen. Note that the noble gases don’t form compounds, so
the values shown are van der Waals radii.
An element such as the metal sodium will have a metallic
radius, an ionic radius when it is bonded in sodium chloride
and a covalent radius when present in its vapour phase.
The discussions of periodic trends in atomic radius of the
elements that follow refer to the covalent radii of the
elements.
There are some distinct trends in the variation of atomic
radius (Figures 2.18 and 2.19):
■ atomic radius generally increases down a group;
■ atomic radius generally decreases across a period from
left to right.
We can explain these trends by consider two opposing factors
that contribute to the size of atoms. These are the number
of occupied electron shells, i.e. the principal quantum
number (n) of the orbitals in which the outermost electrons
are present, and the effective nuclear charge.
19
20
Unit 1 Module 1 Fundamentals in chemistry
A higher principal quantum number (n) correspond to
larger orbitals that lie further from the nucleus. With
increasing value of n, electrons are therefore at greater
distances from the nucleus.
When an electron is added to the neutral atom the total
number of electrons increases but the number of protons
remains the same. The result is an excess of negative charge
and a negatively charged species or an anion is formed.
The nuclear charge is equal to the number of the protons
in the nucleus of the atom. The effective nuclear charge
is the residual net charge felt by the outer valence electrons.
For atoms containing more than one electron, the effective
nuclear charge (Zef) is less than the full nuclear charge (Znc).
Electrons in inner shells ‘shield’ the outer electrons from
the nucleus as a result of electron–electron repulsions.
■ A positive ion (cation) is smaller than the neutral atom
The higher the effective nuclear charge the greater is
the attractive force between the outer electrons and the
positively charged nucleus, hence the smaller the atomic
radius will be (for a fixed value of n).
Down a group, electrons are present in orbitals with
progressively higher principal quantum numbers, which
are progressively further from the nucleus. Though the
nuclear charge is increasing down the group, the outer
electrons will not experience the full attractive force of the
nucleus as a result of shielding from electrons in shells of
lower n. The main contributing factor to the increase in
atomic radius down the group, therefore, is the increased
value of the principal quantum number. Among the Group
I elements, for example, Li has its outermost electron in the
2s orbital whereas for Na this electron is in the 3s orbital
(n = 3). Similarly, for K, its outermost electron is in the
4s orbital. The order of increasing atomic radius for these
atoms is K > Na > Li.
Atomic radius decreases from left to right across a period.
Across a period, from left to right, there is no change
in the principal quantum number, n, of the outermost
orbitals. However, the nuclear charge of each element
increases across a period. Electrons are being added to s
or p orbitals in same principal quantum level, and these
orbitals are at approximately the same distance from the
nucleus. Shielding from other electrons in the same shell
is insignificant. The effective nuclear charge therefore
increases across the period, resulting in electrons being
pulled closer to the nucleus.
Ionic radii
A neutral atom contains an equal number of protons and
electrons. When an electron is removed from a neutral
atom, the number of remaining electrons decreases but
the number of protons (nuclear charge) remains the same.
This excess of protons results in the atom being positively
charged and this positively charged ion is called a cation.
from which it was derived.
■ A negative ion (anion) is larger than the neutral atom
from which it was derived.
Removal of an electron from the neutral atom results in a
reduction in the number of electrons and hence a reduction
of electron–electron repulsions. Since the number of
protons in the positively charged nucleus of the atom has
not been changed, the electron–nuclear attractions in the
cation will be strengthened. The cation has a smaller size
than the neutral parent atom. For example, the Na+ ion
(95 pm) is smaller than the neutral Na atom (186 pm). For
the same value of n, the greater the positive charge on the
cation, the smaller is its radius. For example, the Na+ ion
(95 pm) is larger than the Mg2+ ion (65 pm), which in turn
is larger than the Al3+ ion (50 pm).
Table 2.1 Atomic radii of Li and Na compared with cation radii
Atom/ion
Electronic
configuration
Nuclear
charge
Number of Atomic
electrons radius / pm
Li
1s2 2s1
+3
3
152
Li+
1s2
+3
2
60
Na
1s2
+11
11
186
Na+
1s2 2s2 2p6
+11
10
95
2s2
2p6
3s1
For a negative ion, addition of one or more electrons to the
neutral atom increases the electron–electron repulsions.
The electrons are more spread out and the anion is larger
than the corresponding neutral atom. The greater the
negative charge on the anion, the larger is its radius. For
example, the N3− ion (171 pm) is larger than the O2− ion
(140 pm), which in turn is larger than the F− ion (136 pm).
Table 2.2 Atomic and anionic radii for nitrogen, oxygen and
flourine
Atom/ion
Electronic
Nuclear
configuration charge
Number of
electrons
Atomic/ionic
radius / pm
N
1s2 2s2 2p3
+7
7
70
N3−
1s2 2s2 2p6
+7
10
171
O
1s2
2p4
+8
8
66
O2−
1s2 2s2 2p6
+8
10
140
F
1s2
2p5
+9
9
64
F−
1s2 2s2 2p6
+9
10
136
2s2
2s2
Atoms and ions with the same number of electrons are
isoelectronic. O2−, F−, Ne, Na+ and Mg2+ all have 10
electrons arising from their configuration 1s2 2s2 2p6. Since
these ions have the same number of electrons, they all have
Chapter 2 The quantum atom and the periodic table
Li
+
Be
2+
N
3–
O
2–
F
■ The Group VIII elements have the highest ionization
–
energies in their period.
60 152
+
Na
95 186
K
+
Mg
+
2+
65 160
Ca
133 231
Rb
31 111
2+
99 197
Sr
2+
171
Al
3+
50 143
Ga
3+
62 122
In
3+
70
140
S
66
2–
136
Cl
184 104
Se
2–
198
Te
117
2–
–
181
Br
99
–
185
I
64
114
–
■ There is a general increase in ionization energy from
left to right across a period from Group I to Group VIII.
The value then falls again after Group VIII to the start of
the next period. The pattern resembles that of sawtooth
(Figure 2.21), similar to that observed for atomic radii.
■ There is a gradual decrease in ionization energy down
a group.
113 215
81 162
221 137
216 133
Figure 2.20 Ionic radii. The ions are coloured red (cations) and
blue (anions). The parent atoms are coloured brown. The radii are
given in picometres.
the same amount of electron–electron repulsions. However,
they have different nuclear charges and sizes. Mg2+, with
the greatest nuclear charge in the series, will experience
the greatest electron–nuclear attraction and is therefore
the smallest. O2−, with the smallest nuclear charge, will
experience the lowest electron–nuclear attraction and is
therefore the largest.
For an isoelectronic group of ions, the greater the nuclear
charge on the ion, the smaller is the ion.
Ionization energy
The formation of bonds between atoms depends on the size
of the energy changes as an individual atom gains, loses or
shares electrons.
There are some exceptions to these trends. Group III
elements have lower first ionization energies than the
preceding Group II elements. For example, the ionization
energy of boron, B, is lower than that of beryllium, Be,
despite it having a greater effective nuclear charge. The
ionization energy of the Group VI elements are lower than
those of Group V. For example, oxygen, even though it
has a greater effective nuclear charge than nitrogen, has a
lower first ionization energy.
Ne
2000
F
1600
Ar
N
Cl
O
C
H
1200
P
Be
Mg
800
Si
S
Ca
B
Li
400
Ionization energy (IE) is the minimum energy required to
remove an electron from an atom in its gaseous phase.
A(g) → A+(g) + e−
He
2400
Ionization energy / kJ mol –1
148 244
Al
Na
K
0
0
2
4
6
8
10
12
14
16
18
20
Nuclear charge, Z
Figure 2.21 Ionization energies of the first 20 elements.
IE = E(A+) − E(A), where E stands for energy
Removing an electron from the outermost orbital of an
atom requires an input of energy in order to overcome
the electron–nucleus attraction. The first ionization energy
(IE1) is the energy required to removed the least tightly
bound (outermost) electron from the neutral atom. The
second ionization energy (IE2) is the energy required to
ionize the cation resulting from IE1. An atom will have as
many ionization energies as it has electrons.
The general trends in ionization energy among the main
group elements are summarized as follows:
■ The elements of Group I have the lowest ionization
Generally, across a period there is an increase in the effective
nuclear charge while n remains constant and the atom gets
smaller. Removal of an electron from the smaller atom will
become progressively harder so ionization energy increases
across the period.
Down a group, outermost electrons are present in orbitals
that are progressively further from the nucleus. These
outer electrons are also shielded from the nucleus by inner
electrons. As a result, they are more easily removed so
ionization energy decreases down a group.
In general, small atoms have large ionization energies and
large atoms have small ionization energies.
energies in their period.
ITQ 3 Arrange the following elements
in order of increasing atomic radius:
Na, Cl, K, Br
ITQ 4 Consider the following ions: Na+, Cl−, O2−.
(a) Which ion will be the smallest?
(b) Which ion will be the largest?
ITQ 5 Suggest a reason why the gas
phase is used for measurements of
ionization energy.
21
22
Unit 1 Module 1 Fundamentals in chemistry
Worked example 2.1
Q
Which of the following atoms would have the lowest first
ionization energy?
A [He] 2s2 2p4
B [Ne] 3s2 3p5
C [Xe] 6s2
D [Ne] 3s1
A
The correct answer is C, since its outermost electron is in the
n = 6 shell. This shell is furthest from the nucleus, and so
experiences greater shielding from inner shells and a lower
effective nuclear charge. It would require less energy to remove
an electron from this, the largest atom of those in the example.
furthest from the nucleus. The cation Na+ resulting from this
process has electrons in the n = 2 shell, which is closer to the
nucleus and has the electronic configuration of the noble
gas neon. It is not surprising therefore that the removal
of the second electron (IE2) from this stable inner core
requires a significantly higher energy than the first. Similar
patterns will be observed for all elements of Group I since
they all have the general [X] ns1 electron configuration and
the same single outer valence electron.
Aluminium is a Group III (13) element, with three valence
electrons. A big change in ionization energy occurs after
the third ionization process since this requires removal
of an electron from the Al3+ cation which has a noble gas
electronic configuration.
Variation in successive ionization energies
A(g) → A+(g) + e−
IE1
Electron affinity
A+(g) → A2+(g) + e−
IE2
A2+(g) → A3+(g) + e−
IE3
Ionization energies (IE) are the energy changes involved
in the formation of positive ions. Ionization energies are
always positive, meaning that energy is taken into the
system:
Successive ionization energies of an element increase in
the order IE1 < IE2 < IE3 < IE4. Removal of the first electron
from the neutral atom gives a smaller positively charged
ion. Removal of an electron from this smaller positively
charged ion will require more energy so IE2 > IE1. Successive
ionization energies will increase since the electron is being
removed from a more positive ion.
The greatest difference between successive ionization
energies for a given atom occurs after the removal of all
the outer valence electrons. Here is the data for the first
four ionization energies of sodium.
Na(g) (1s2 2s2 2p6 3s1) → Na+(g) (1s2 2s2 2p6) + e−
IE1 = 495.8 kJ mol−1
Na+(g)
(1s2
2s2
2p6)
→
Na2+(g)
(1s2
2s2
2p5)
e−
+
IE2 = 4562 kJ mol−1
Na2+(g) (1s2 2s2 2p5) → Na3+(g) (1s2 2s2 2p4) + e−
IE3 = 6910 kJ mol−1
Na3+(g) (1s2 2s2 2p4) → Na4+(g) (1s2 2s2 2p3) + e−
IE4 = 9543 kJ mol−1
As is expected, the values of the successive ionization
energies increase as the species produced from the ionization
process becomes progressively more positively charged and
smaller. Removal of the first electron from the 3s orbital of
the sodium atom is the easiest process, since this orbital is
ITQ 6 Suggest an explanation for the decrease in first ionization
energy between P and S. You need to think about which electron
is being removed. Why might it be easier to remove the electron
from S?
M → M+ + e− + ΔH
ΔH is always positive.
However, many elements, especially those towards the
right-hand side of the periodic table, are able to form
anions. When one mole of a monatomic species gains one
electron to form an anion, the energy change is called the
first electron affinity (EA) of the element.
The first electron affinity is an exothermic process, meaning
that in the reaction
X + e– → X– + ΔE
the value of ΔE is negative.
For example, the first electron affinity (EA1) of chlorine is
–349 kJ mol–1.
The first electron affinity refers to the energy released on
the addition of one electron to a neutral atom, the second
electron affinity refers to the energy change when a second
electron is added to the negatively charged ion, and so on.
These subsequent values of electron affinity (EA2, EA3,
…) are always positive, meaning that heat is taken in as
an anion becomes multiply charged. The change from
negative to positive is because for later EA changes, the
electrons are being forced in against the repulsion of a
negatively charged ion.
Chapter 2 The quantum atom and the periodic table
Periodic trends in electron affinity
Worked example 2.2
Variations in electron affinity follow a similar trend as
ionization energy and parallels variation in atomic radius. In
general the smaller atoms (except for the period 2 elements)
with the greater effective nuclear charges will attract an
added electron and so will have greater electron affinities
than the larger atoms with smaller effective nuclear charge
and greater electron–electron repulsions. In general:
Q
Suggest a reason for the large decrease in electron affinity
between lithium and beryllium.
A
Li has the electronic configuration 1s2 2s1 and Be has the
electronic configuration 1s2 2s2. Li has a smaller nuclear
charge than Be. One would expect the Be atom with the
greater nuclear charge to attract an additional electron more
readily than Li. This is not what is observed, however, and
Be has a lower electron affinity than Li. If we consider the
electronic configuration of each atom, we will see that the
additional electron to be added to Li will enter the half-filled 2s
orbital and fill it. For Be the additional electron will enter a 2p
orbital which is of higher energy and further from the nucleus
than the 2s orbital. The 2p orbital will also experience greater
electron–electron repulsions from the two electrons already
present in the 2s orbital. The result of this is the lower electron
affinity of Be.
Be(g) (1s2 2s2) + e− → Be−(g) (1s2 2s2 2p1)
Li(g) (1s2 2s1) + e− → Li−(g) (1s2 2s2)
■ electron affinity increases across a period (Figure 2.22);
Electron affinity / kJ mol –1
400
F
300
200
C
O
100
H
Li
He
Be
B
N
Ne
0
0
1
2
3
4
5
6
7
8
9
10
Atomic number
Electronegativity
Figure 2.22 Electron affinities for period 2.
Electronegativity is the tendency of an atom to pull the
electrons in a bond towards itself.
400
Cl
Br
Electron affinity / kJ mol –1
F
I
300
At
Se
S
200
Te
Po
O
100
0
1
2
3
4
5
6
7
Period (row) on periodic table
Linus Pauling, an American chemist and biochemist
(1901–1994), developed a scale of electronegativity in
which elements are assigned different values (Figure 2.24).
Fluorine has the highest value and Group I (Group 1)
elements have the lowest. In general, the electronegativity
of an element increases across a period from left to
right and decreases down a group. Elements with high
electronegativities will readily attract electrons in a bond.
This concept will be further discussed in Chapter 4 when
we look at chemical bonding.
Figure 2.23 Electron affinities for Groups VI and VII.
■ elements with the electronic configuration
np3
ns2
np6
(Group II/2),
(Group V/15) or
(Group VIII/18)
have the lowest electron affinities, e.g. Be, N, Ne;
■ elements with the electronic configuration np4
(Group VI/16) and np5 (Group VII/17) have the
highest electron affinities, e.g. O, F.
ITQ 7 The electronic configurations for three neutral atoms are
given below.
I
1s2
II
1s2 2s2 2p6 3s2 3p3
2
2s2
2
2p6
6
3s2
2
I
increasing electronegativity
H
2.1
II
III
IV
V
VI
VII
Li
Be
B
C
N
O
F
1.0
1.5
2.0
2.5
3.0
3.5
4.0
Na
Mg
Al
Si
P
S
Cl
0.9
1.2
1.5
1.8
2.1
2.5
3.0
K
Ca
Br
0.8
1.0
2.8
I
increasing electronegativity
■ electron affinity decreases down a group (Figure 2.23);
2.5
Figure 2.24 The Pauling scale of electronegativity.
6
III 1s 2s 2p 3s 3p
Which of these atoms would have the largest 3rd ionization
energy?
ITQ 8 The electron affinity of nitrogen is less than that of carbon.
Suggest a reason for this observation.
23
24
Unit 1 Module 1 Fundamentals in chemistry
Summary of general periodic trends
Among the main group elements of the periodic table:
■ atomic radius decreases across a period and increases
down a group (with some exceptions);
■ the first ionization energy increases across a period and
decreases down a group (with some exceptions);
■ successive ionization energies show the greatest increase
when the inner core electrons are being removed;
■ elements of Group I (Group 1) and Group II (Group 2)
have the lowest ionization energies and the smallest
electron affinities. They will therefore lose electron(s)
readily to form +1 and +2 ions;
■ elements of Group VI (Group 16) and Group VII
(Group 17) have high ionization energies and large
electron affinities and so they will easily gain electrons
to form negatively charged species;
■ the noble gases (Group VIII / Group 18) have very
high ionization energies and very low electron
affinities (endothermic process) indicating that they
neither lose nor gain electrons.
■ Periodic properties such as atomic radius, ionization
of the electron configuration of the element combined
with the competing electron–nuclear attraction and
electron–electron repulsion. The stronger the electron–
nucleus attraction the smaller the atom is and the
greater the ionization energy and the electron affinity.
On the other hand, if the electron–electron repulsions
are greater than the electron–nucleus attraction,
the larger the atom is and the smaller the ionization
energy and the electron affinity.
■ As a result, we can predict the properties of elements
based on their position in the periodic table and so we
do not have to memorize information on each and
every element (see Table 2.3).
Table 2.3 Periodic trends in summary
Property
Across period
Down group
Zeff
increases
increases slightly
n (principal quantum number)
remains constant
increases
atomic radius
decreases
increases
ionization energy
increases
decreases
electron affinity
increases
decreases
electronegativity
increases
decreases
energy and electron affinity can be explained in terms
Quantum effects in chemical bonds
In the next chapter we shall study several types of chemical bond, all based on the idea of electrostatic attraction
between positively charged and negatively charged regions of atoms. For example, protons in atomic nuclei ‘attract’
electrons shared with other atoms. However, modern calculations show that other factors must sometimes be at
work. Here is one example.
Intriguingly, the covalent component of the interatomic force between two fluorine atoms in the fluorine molecule
is repulsive, not cohesive. Some other effect must be at work between the two atoms, overcoming this repulsion
and making an overall attraction.
When two or more versions of the same molecule exist, which differ only in the possible arrangement of their
electrons, the molecule is more stable than predicted. We say that the different electronic versions of the molecule
resonate with each other. The molecule exists, at any one moment in time, as a fusion of the various possible
extremes. You will see this in action when you read about the structure of benzene (page 235).
Quantum calculations show that in ‘covalent’ bonds such as F–F and hydrazine (H2N–NH2) the electron density
in the bond region is very little different from the rest of the molecule. In these molecules the bond is reinforced
by a strong resonance between covalent ‘electron share’ and ionic ‘electron transfer’ forms. The bond is not a
consequence of the low energy of the molecule but is a consequence of the resonance energy.
This resonance stabilization is found increasingly as the bonded atoms move from left to right and from bottom to
top in the periodic table. The effect has been named the charge-shift bond.
ITQ 9 Account for the fact that the first electron affinity, EA1, can
be either positive or zero whereas the second electron affinity,
EA2, is always negative (requires an input of energy).
Chapter 2 The quantum atom and the periodic table
Summary
✓ Electromagnetic radiation has the properties of
both waves and particles.
✓ Light exists as photons (quanta) which have
energy proportional to their frequency.
✓ According to quantum theory, an atom can have
only certain amounts of energy (E = nhν) which
can only be changed if the atom absorbs or emits
light.
✓ Allowed energy levels are related to allowed
wavelengths of the electron’s motion.
✓ Electrons exhibit diffraction patterns (as do
waves of energy), and photons exhibit transfer of
momentum (as do particles of mass).
✓ The wave–particle duality of matter and energy
is observable only on the atomic scale.
✓ According to the uncertainty principle, we
cannot know simultaneously the exact position
and velocity of an electron.
✓ The electron’s wave function (atomic orbital) is a
mathematical description of the electron’s wavelike motion in an atom.
✓ Each wave function is associated with one of the
atom’s allowed energy states.
✓ An electron density plot and a radial probability
distribution plot show the electron occupies the
space near the nucleus for a particular energy
level.
✓ Three features of the atomic orbital are described
by quantum numbers: size (n), shape (l)
orientation (ml).
✓ Orbitals are part of sub-shells defined by n and l,
which are part of an energy level defined by n.
✓ An orbital can accommodate a maximum of two
electrons. More generally, this can be expressed
as no two electrons can have the same four
quantum numbers simultaneously. That is, in an
orbital containing two electrons, the electrons
must have opposite spins.
✓ The quantum mechanical atom provides the
theoretical foundation for the experimentally
based periodic table.
✓ All physical and chemical behaviour of each
element in the periodic table is based on the
electron configuration of its atoms.
✓ The elements in a group have similar outer
electron configurations and similar chemical
properties.
✓ Atomic radius increases down a group and
decreases across a period.
✓ Across the transition series the atomic radius
remains approximately constant.
✓ The first ionization energy decreases down a
group and increases across a period.
✓ Successive ionization energies show a largest
increase when inner core electrons are being
removed.
Review questions
1 Arrange the following atoms in order of increasing
atomic radius:
Al, P, C, K, Na
2 The first electron affinity of the oxygen atom releases
141 kJ mol−1 of energy but the second electron
affinity requires an energy input of 844 kJ mol−1.
Suggest an explanation for this observation.
3 The diagram represents a section of the periodic table,
with elements W, X, Y and Z marked.
W
X
Y
Z
(a) Which element has the highest electron affinity?
(b) Which element has the largest radius?
(c) Which element has the largest first ionization
energy?
(d) Which element has the smallest first ionization
energy?
25
26
Unit 1 Module 1 Fundamentals in chemistry
4 Provide explanations for each of the following
observations.
(a) Peaks in first ionization energies occur with atoms
with atomic numbers 4, 7, 10, 18, 36.
(b) Elements with atomic numbers 9, 17, 35 have
large electron affinities whereas those with atomic
numbers 10, 18, 36 have very small or zero
electron affinities.
5 Successive ionization energies for elements P, Q and R
are given in Table 2.4.
Table 2.4
Element 1st IE
2nd IE
3rd IE
4th IE
5th IE
X
786
1577
3229
4356
16080
Y
738
1450
7732
10550
13620
Z
577
1816
2744
11580
15030
(a) Where would each element would be found in the
periodic table? Explain your answer.
(b) Given that all three elements are in the same
period, which of the three elements would have
largest atomic radius?
6 The sizes of the species Ne, F−, N3− and Na+ are
represented by the spheres shown below. Match each
species with an appropriate sphere. Explain your
answer.
Answers to ITQs
1 mass = 160 g = 0.16 kg; velocity = 38.9 m s−1; so
p = 6.2 kg m s−1; h = 6.63 × 10−34 m2 kg s−1
λ=
2 Both theories were preceded by ‘continuous’ theories:
one that matter is infinitely divisible and the other
that energy is infinitely divisible also.
3 Cl < Na < Br < K
Na and Cl are in the third period, n = 3, with Cl
having the greater nuclear charge so the Cl atom is
smaller than Na atom (Cl < Na). K and Br are both in
the fourth period, n = 4, with Br having the greater
nuclear charge and the smaller radius (Br < K).
Both K and Br are larger than Na and Cl since their
outermost electrons are in a shell further from the
nucleus.
4 (a) Na+
(b) Cl−
Consider the electronic configuration of the ions:
Nuclear
charge
Number of
electrons
Na+ 1s2 2s2 2p6
+11
10
Cl−
+17
18
+8
10
Ion
O2−
A
B
C
Electronic
configuration
1s2 2s2 2p6 3s2 3p6
2
2
1s 2s 2p
6
D
7 Plot and interpret a graph of ionization energy
(kJ mol−1) on the y-axis against the number of
electrons removed (x-axis) for the elements Li, C, O
and S.
8 Which of the following ions would lose an electron
most easily?
S2−, Cl−, Ar, K+, Ca2+
9 Which of these atoms would have the largest 1st
ionization energy?
[Ne] 3s2 3p1, [Ne] 3s2 3p2, [Ne] 3s2 3p3, [Ne] 3s2 3p4,
[Ar] 3d10 4s2 4p3
10 Place these elements in order of increasing ionization
energies?
Al, B, C, K, Na
h
6.63 × 10−34
=
≈ 1 × 10−34 m
p
6.2
Na+ and O2− have the same electronic configuration
(are isoelectronic); however, Na+ has the greater
nuclear charge and so its electrons will experience
a greater nuclear attraction than O2−. Na+ will be
smaller than O2−. The outer electrons in the Cl− ion
are in the n = 3 shell and so are further from the
nucleus than those of the Na+ and O2− ions, whose
outer electrons are in the n = 2 shell. The Cl− ion is
therefore the largest of the three ions given.
5 In the gas phase there is little interaction between
atoms and hence it is the best representation of an
isolated atom. If the atom was bonded to another (as
is the case for liquid or solid phases), then some of the
energy applied to the atom in the ionization process
would be absorbed in breaking bonds.
Chapter 2 The quantum atom and the periodic table
6 P = [Ne] 3s2 3p3; P: ҘғҘҘҘ
S = [Ne] 3s2 3p4; S: ҘғҘғҘҘ
The P atom has three 3p orbitals each containing an
unpaired electron arranged with parallel spins. This
arrangement of electrons minimizes electron–electron
repulsions among the electrons. In the S atom one of
the p orbitals contains a pair of electrons with opposite
spins. These electrons sharing the same region of
space will experience added repulsion from each other
in addition to that experienced from electrons in the
other orbitals. Removing an electron from the S atom
on ionization results in a reduction of the electron–
electron repulsions hence the process is favourable
and hence the lower ionization energy of S.
7 Atom I, since after the removal of the first two
electrons the species formed has a noble gas
configuration.
I (1s2 2s2 2p6 3s2) → I+ (1s2 2s2 2p6 3s1) + e−;
electron lost is from 3s
IE1
9 The equations for the first and second electron affinity
of an atom X are:
EA1
X + e− → X−
X− + e− → X2−
EA2
The second electron affinity involves adding an
electron to a negatively charged ion. There will be
significant electron–electron repulsions to overcome
so energy would have be input in order to force the
electron on the negatively charged ion.
Answers to Review questions
1 C, P, Al, Na, K
2 Second electron added to a negatively charged O− ion
so there is increased electron–electron repulsions
making it harder to add an electron.
3 (a)
(b)
(c)
(d)
I+ (1s2 2s2 2p6 3s1) → I2+ (1s2 2s2 2p6) + e−;
electron lost is from 3s
IE2
X
Y
X
Y
5 (a) X is Group IV (14); Y is Group II (2),
Z is Group III (13)
I2+ (1s2 2s2 2p6) → I3+ (1s2 2s2 2p5) + e−;
electron lost is from 2p
IE3
(b) X
8 Since N is smaller and has a greater nuclear charge
than C, we would expect N to have a greater electron
affinity than C. The opposite is observed, which
suggests that we need to consider other factors. One
such factor is the electron–electron repulsions in the
resulting ions.
The electron configurations of the neutral and charged
ions of each atom are as follows:
C ([He] 2s2 2p2) + e− → C− ([He] 2s2 2p3)
Ҙғ Ҙ Ҙ
Ҙғ Ҙ Ҙ Ҙ
EA1
N ([He] 2s2 2p3) + e− → N− ([He] 2s2 2p4)
Ҙғ Ҙ Ҙ Ҙ
Ҙғ Ҙғ Ҙ Ҙ
EA1
Carbon has an empty p orbital which can
accommodate an additional electron with an opposite
spin (reducing repulsive effects). Nitrogen has an
unpaired electron in each of its p orbitals so an
additional electron would have to be accommodated
in an orbital which is already occupied. This electron
would experience increased repulsions from the
electrons already present which offsets the increased
nuclear attraction due to its greater nuclear charge.
The net result of this is the nitrogen atom has a lower
electron affinity than the carbon atom.
6 A = N3−; B = F−; C = Ne; D = Na+
8 S2−
9 [Ne] 3s2 3p3
10 C, B, Al, Na, K
27
28
Chapter 3
Radioactivity
Learning objectives
■ Appreciate that some atomic nuclei are unstable.
■ Understand the words ‘isotope’ and ‘nuclide’.
■ Distinguish between α, β and γ radiations.
■ Explain the origins of α, β and γ radiations.
■ Write symbols for sub-atomic particles and nuclides.
■ Write simple nuclear equations.
■ Describe selected uses of radioactive nuclides.
Introduction: the alchemists’ dream
Dalton’s theory (see page 3, Chapter 1) seemed to explain
why they failed: atoms, according to Dalton’s theory,
were immutable and indivisible. Both Bohr’s theory and
Thomson’s ‘plum pudding’ model of the atom also offered
no hint that atoms could change. However, Rutherford’s
nuclear model put matters in a different light, as well as
raising some more questions.
How can several positively charged particles possibly exist
so close to each other in the nucleus of an atom? The
electrostatic force pushing them apart, which doubles in
strength as the distance between the particles is halved,
is enormous. Some hitherto unknown force must oppose
that repulsion and make the nucleus stable. This force is
called simply the strong force.
The strong force is one of the four fundamental forces of
nature: the others are the electromagnetic force, gravity
and the ‘weak force’. The strong force operates between
quarks, the particles which themselves make up protons
and neutrons. All these forces are non-contact forces.
As well as protons, the nucleus also contains neutrons. The
strong force can be thought of as operating between the
protons and the neutrons, transferring energy back and
forth, rather like children playing together, throwing and
catching a ball. As long as the numbers of good throwers and
good catchers roughly balance, the group stays together.
Similarly, in nuclei, as long as the neutron : proton ratio
does not depart too far from 1, the nucleus is likely to be
stable – though there are other factors to take into account.
The neutron : proton ratio is close to 1 : 1 up to about
Z = 20, then increases slowly up to about 1.6 : 1 (Figure 3.1).
120
110
band of stability
100
90
Number of neutrons, N
In Europe, in the Middle Ages, chemists (known then
as alchemists) tried to invent a magical potion that they
called the ‘philosopher’s stone’. This potion would change
base metals, such as iron, into gold and also give the gift of
eternal life. They amassed a huge store of what we would
now call ‘industrial chemistry’ and a fair knowledge of
medicine, but they failed to find the philosopher’s stone.
80
N=Z
70
60
50
40
30
20
10
0
0
10
20
30
40
50
60
70
80
Number of protons, Z
Figure 3.1 Ratio of protons to neutrons in stable nuclei.
90
Chapter 3 Radioactivity
Nuclear transitions
Isotopes are atoms of any element X which have the same
proton number but different numbers of neutrons (see
page 2, Chapter 1). You should note that the word ‘isotope’
can only be used if you say which element you are talking
about. If you don’t give the name of the element then the
word to use is nuclide. So we may speak of ‘the isotopes of
chlorine’ but ‘nuclides with different numbers of neutrons’.
For nuclear stability, even numbers of neutrons are
preferred to odd, and the same applies to the number of
protons, and even numbers of both are best of all (Table
3.1). However, when the n : p ratio is within allowable
limits and no other rules are broken, the nucleus is stable.
Table 3.1 Numbers of stable nuclides with odd and even
numbers of protons and neutrons
Z
N
Number of stable nuclides
odd
odd
4
odd
even
50
even
odd
57
even
even
168
Stable nuclides are shown within the green area in Figure
3.1. Outside these limits, the nucleus is unstable (it contains
too much inherent energy) and in the manner of all hyperenergetic systems, changes towards a more stable state by
losing some energy. This change is called a decay.
A proton has a mass of approximately 1 amu (atomic mass
mass
X,
unit) and a charge of +1. Using the convention charge
1
we write a proton a 1 p. Similarly, we write a neutron
as 10 n and an electron as −10 e. We can think of a neutron as
a proton plus an electron. This can be written as a nuclear
equation:
1
0
n = 11p + –10e
β emission
If a nucleus has too many neutrons to be stable, then one of
them can change to a proton plus an electron. The neutron
number goes down by one and the proton number goes
up by one. The n : p ratio has changed towards the stable
value. However, electrons cannot exist inside the nucleus,
so the electron, the second product of the nuclear change,
is ejected. The ejected electrons are called beta rays (β
rays), but you should use the term beta particle if you
are talking about just one.
ITQ 1 Which element would you need to start with in order to
transform it into gold by β emission?
The alchemists’ dream of transforming one element into
another has been realized. The atomic number of the
product atom is one greater than the original atom. It is an
atom of a different element. The element has moved along
one place in the periodic table.
α emission
What about an atom that has too few neutrons to be
stable? The nearest electrons are too far away from the
nucleus to be captured, so the nucleus cannot change by
forming a neutron from a proton and an electron. Rather
than ejecting a proton, the nucleus can eject a helium
nucleus, 42 He, which is a very stable particle. This type of
decay is confined to heavier nuclei. Lighter nuclei decay by
emitting antimatter. (Antimatter is a topic that is outside
the scope of your course but if you want to know more,
look up ‘positron’ on the web!)
Emitting a helium nucleus reduces the atomic number
by two units so the new element is two places earlier in
the periodic table. The helium nuclei that are ejected are
called alpha particles (α particles). Even though both
the proton number and the neutron number have gone
down by two, the n : p ratio will have increased.
Nuclear equations
The changes to atoms and molecules in a chemical reaction
can be recorded in a chemical equation. In the same way,
the nuclei and sub-atomic particles in a nuclear reaction
can be recorded in a nuclear equation. The same rules
apply: the total mass and the total charge must be the same
on both sides.
Worked example 3.1
Q
A platinum nucleus with relative atomic mass 195 loses a
β particle to form a nucleus of gold. Write an equation to
show this change.
A
(i) Look up the values of Z for platinum (Pt) and gold (Au). They
are 78 and 79, respectively.
(ii) The new nucleus of gold must also have A = 195 because
a beta particle has no mass. You can now write symbols for
this nuclear reaction:
195
78
Pt
195
79
Au + –10e
(iii) Check that mass (195 = 195 + 0) and charge
(78 = 79 + −1) both balance. This is fine.
ITQ 2 Why is the helium nucleus stable?
29
30
Unit 1 Module 1 Fundamentals in chemistry
Radioactive decay
Properties of α, β and γ rays
Radioactive decay is a first-order reaction (see page 90,
Chapter 9). The rate of radioactive decay is unaffected by
changes in temperature, pressure or chemical environment.
One feature of first-order reactions is that they have a
fixed half-life (t 12 ). This means that the time taken for
the amount of a material to halve (that is, change from 1
1
1
to 2 to 4 and so on) is constant. Half-lives of radioactive
nuclides vary from fractions of a second (e.g. 315Po, t 12
= 3 × 10−5 s) to billions of years (e.g. 238U, t 12 = 1 × 1011
years). Each nuclide has its own characteristic half-life.
Some nuclides with very long half-lives are found in the
environment around us. They give rise to the low levels
of background radiation that surrounds us constantly –
and has done since the beginning of time.
These three types of radiation are all used to our advantage,
although they can also be harmful. Table 3.2 shows some
of their important properties.
Energy changes in radioactive decay
If sufficient energy is emitted in a small length of time, the
outcome can be an explosion. The most famous examples
are the two bombs dropped on Japan in August 1945 to
bring an end to the Second World War in the Pacific.
We sometimes assume that protons and neutrons have a
mass of exactly 1 amu and that electrons have no mass at
all. This is only an approximation to the truth, although
it does work well much of the time. In fact, neutrons are
a little heavier than protons and electrons do have some
mass.
If we use Einstein’s famous equation, E = mc2, we see that
mass (given by m) and energy (given by E) are really the
same thing. This means that when we say that mass is
conserved in a nuclear change, we should really say that
the total of (mass + energy) is conserved. For example, in
the change from a neutron to a proton plus an electron,
a tiny amount of mass, less than one thousandth of an
atomic mass unit, is lost: this ‘lost’ mass appears as energy
– the electron is ejected at very high speed.
In many other changes, the final nucleus (the daughter
nucleus) is left in an excited (high-energy) state. The
excess energy is then emitted as electromagnetic radiation
with wavelengths roughly 100 times those of X-rays. These
radiations are called gamma rays (γ rays). We do not
include them in equations of nuclear change because they
carry neither mass nor charge.
ITQ 3 A nucleus has Z = 86 and N = 132.
(a) What is its n : p ratio?
(b) What is the new n : p ratio after an alpha particle is emitted?
(c) What were the two nuclides?
ITQ 4 Tritium is an isotope of hydrogen and it has two neutrons
in its nucleus. Write the nuclear equation describing the decay of
tritium (3H) to helium-3 (3He).
Table 3.2 Properties of α, β and γ rays
Alpha
Beta
Gamma
Character
mostly as a
particle
particle or ray
mostly as rays
Relative mass
/ amu
4
approx. 1/2000
almost zero
Penetrating
power
penetrates skin, thin penetrates lead
stopped by skin or
sheets of aluminium sheet, human body
tissue paper
or concrete
or Perspex
Ionizing power
high
medium
almost zero
Problems caused by radiation
More recent examples of the danger of radiation can be seen
following the explosion at the Chernobyl power station
(April 1986) and the nuclear meltdown in the Japanese
nuclear reactor at Fukushima (March 2011). In both of
these the danger arose from highly active radionuclides
scattered in the explosions rather than the disruptive
power of the explosions themselves.
The harmful effects of these radiations come mainly from
their ability to ionize atoms. If ionization happens within
a living cell there is the risk that the damage will not be
repaired, and therefore the cell ceases to function. If
enough cells are damaged, illness or even death can be the
result. This is known as radiation sickness.
If the damage is in the gametes (i.e. in a sperm or an egg),
then damaged DNA can be passed down through the
generations. Inherited genetic mutations are not often seen
in humans because the individual damaged sperm or ovum
must be involved in reproduction, and also because most
mutations produced in this way are in recessive genes, not
dominant genes.
In the plant world, however, radiation-induced changes
are more common. For example, new varieties of food
crops such as rice and the ornamental Chrysanthemum have
been produced by gamma irradiation. The process is called
‘mutation breeding’ and can be beneficial.
ITQ 5 Write the nuclear equation for the transition mentioned in
ITQ 3.
Chapter 3 Radioactivity
Uses of radioisotopes
Radiotherapy
The ionizing effect of radiation can be used to kill cancer
cells in the body. This is known as ‘radiotherapy’. Either
a beam of radiation is targeted onto the cancer so that
surrounding tissue is not so much affected, or tiny ‘seeds’
of nuclides producing short-range radiations are implanted
directly into the tumour. X-rays and γ-rays are used, as are
protons and β particles.
Fission reactions
In the nuclear reactions called fission reactions (so-called
because large nuclei split into two or more smaller
fragments), large quantities of energy are released. Fission
reactions are triggered by the absorption of neutrons
into susceptible nuclei such as 238U. If this absorption
of neutrons is uncontrolled and the mass of fissionable
material is sufficient, then an explosion is the result. But
neutrons are easily controlled, and so the release of energy
can be slowed down to make a useful, safe, power source. In
many countries, nuclear power stations provide significant
proportions of the total energy supply. At present there are
no nuclear power stations in the Caribbean region, though
they are common in the USA.
In most cases the heat energy released by the fission
reaction is used to boil water, and then the steam is used
to drive electricity generators. The downside of the process
is that it is hard to dispose safely of the radioactive waste
from the power station.
α particles
The decay of nuclides releasing α particles is used in
batteries that must remain unattended for long periods.
Such batteries are used in spacecraft or heart pacemakers.
238
Pu (plutonium) decays in this way. Some of the energy
released in the decay appears as heat, which is used to warm
one side of a bank of thermocouples. The thermocouples
then produce an electrical output. The output from 1 g of
plutonium is about 0.5 W.
Why is such a system useful? The α particles are easily
absorbed by thin shielding and therefore present no health
hazard and the rate of decay of 238Pu is slow (t 12 = 88 years)
so the batteries have a long life.
A more familiar use is in smoke detectors. Here a tiny
amount of 241Am (americium) ionizes the air between two
electrodes, allowing a tiny current to pass. When smoke
enters the detector the ionization is interrupted, the current
is reduced, and an alarm sounds.
β sources
Beta sources are valuable as tracers, especially in medicine.
Some nuclides become concentrated at particular sites in
the body. An example is 131I (iodine), which gathers in the
thyroid gland. The distribution of the nuclide in the gland
can be detected because it undergoes β decay, and this
helps in the diagnosis of problems.
By incorporating an active nuclide into a molecule and
then tracing its pathway through subsequent reactions, we
can get information about the mechanism of the reactions.
Because β rays have greater penetrating power than α
particles, they can be used to monitor and control the
thickness of sheet material produced in a rolling mill. A
source such as 37Cs (caesium) is held on one side of the
sheet and a detector is positioned on the other side. If the
sheet produced is too thick the signal is reduced and more
pressure is applied to the rollers producing the sheet. If
the sheet is too thin, the signal will be increased and the
pressure needs to be reduced.
Beta radiation can be used to date ancient organic material.
Atmospheric CO2 contains trace amounts of the β emitter
14
C. 14C is photosynthesized in the same way as 12C,
becoming part of living material. When the organism dies,
the 14C is no longer replaced. The amount in the organism
decays, with a half-life of 5370 years. By measuring the
beta activity of an ancient material and comparing it to the
activity expected in a living sample, the age of the material
can be worked out. The method is used for carboncontaining materials with ages up to about 60 000 years
and is commonly used in archaeological investigations.
Gamma rays
Gamma radiation is highly penetrating and affects
photographic film in the same way as X-rays. It is used
in industry to check for flaws in heavy components such
as remote pipelines. The source of the rays is often 60Co
(cobalt): such a source can be made portable and can be
used in locations where no power is available for X-ray
machines.
In medicine, gamma rays can be used to destroy cancer
cells, gamma radiation can be used to sterilize pre-packed
surgical instruments, and has been used to kill the pupae
of the cotton boll-weevil, which is a serious pest in the
American cotton-growing industry.
31
32
Unit 1 Module 1 Fundamentals in chemistry
Summary
✓ When an atomic nucleus changes into a different
nucleus, the process is called radioactivity.
✓ In a radioactive change the nucleus may emit
alpha, beta and or gamma rays.
✓ Each radioactive decay has a characteristic halflife.
✓ Alpha, beta and gamma rays differ significantly
in penetrating power.
✓ Atomic radiation can be dangerous but it can
also be highly beneficial.
Review questions
Answers to ITQs
1
1
Gold has Z = 79. The element with Z = 78 is platinum.
2
The helium nucleus contains even numbers of protons
(2) and neutrons (2) and its n : p ratio is 1.
3
(a) Original n : p ratio is 1.535 : 1.
(b) The n : p ratio is 1.548 : 1. Not a great change but
a move in the required direction.
(c) The original nuclide was 218
86 Rn, which become
214
84 Po.
4
3
1H
5
218
86 Rn
2
(a) Explain what is meant by the statement ‘Chlorine
has two isotopes, chlorine-35 and chlorine-37’.
(b) Another isotope of chlorine, chlorine-38, can be
made but it has a half-life of 37 minutes.
(i) What is meant by the term ‘half-life’?
(ii) Why does this isotope of chlorine not exist in
nature?
(a) Atoms of the nuclide strontium-88 are stable.
Atoms of its isotope strontium-89 are radioactive.
What difference between the nuclei of the two
makes the heavier atom unstable?
(b) The heavier isotope has a half-life of 50 days. It
emits ˚ particles.
(i) What is a ˚ particle?
(ii) Atomic nuclei contain only protons and
neutrons. Explain the origin of the ˚ particle.
(iii) Why does emitting the ˚ particles make the
strontium-89 atoms more stable?
(iv) How long would it take 15/16 of a sample of
strontium-89 to decay?
→ 32 He +
0
−1 e
4
→ 214
84 Po + 2 He
33
Chapter 4
Chemical bonding
Learning objectives
■ Explain the origin of the forces which act as chemical bonds.
■ Describe ionic, covalent, hydrogen, van der Waals and metallic bonds.
■ Explain the ways in which some physical properties of materials are
related to the bonds that they contain.
Introduction
In Chapter 2 we considered the concepts of ionization of
the elements, looking at ionization energy and electron
affinity, and we have also already discussed atomic size. In
this chapter, you will see the role that these concepts play
in bond formation.
One driving force behind the formation of stable
molecules is the attainment of minimum potential
energy.
Types of chemical bond
We need to consider five types of chemical bonds:
Formation of bonds
Chemical bonds are forces that hold atoms or ions together
in a compound. Without the formation of chemical bonds
we would have only naturally occurring ‘free’ elements
rather than the millions of different molecules that exist. In
nature ‘free’ elements are rarely found. Instead, we have
compounds containing elements bonded to each other or
to other elements.
■ ionic bonding;
■ covalent bonding;
■ hydrogen bonding;
■ metallic bonding;
■ van der Waals forces.
Bonding mechanisms need to explain a range of situations,
bearing in mind that bond formation is directly related to
energy changes.
It is therefore clear that in the natural world a combined
state is preferred. Why is this?
■ Some substances, for example sodium chloride, NaCl,
We say that the combined state is ‘more stable’ and this
turns out to mean ‘a state of lower potential energy’.
■ In a substance such as water, H2O, the bonds are
Think about the process of bringing atoms together:
■ the positively charged nucleus of each atom attracts
its own negatively charged electrons, as well as the
electrons of nearby atoms
■ the positively charged nuclei will repel each other
have bonds between oppositely charged ions.
between two apparently neutral atoms.
■ Water boils at a temperature well above that which its
molecular mass would suggest.
■ Metals conduct heat and electricity.
■ Noble gas atoms, normally existing in isolation, do
stick together at low enough temperatures.
■ the electrons of the atoms will repel each other.
In other words, when atoms approach each other, there is a
set of attractive and repulsive interactions. If the attractive
interactions exceed the repulsive ones, then the atoms
combine to form a stable particle of lower potential energy.
The atoms are said to have formed a chemical bond.
ITQ 1 Suggest why the cold water in a kettle does not
spontaneously come to its boiling point.
34
Unit 1 Module 1 Fundamentals in chemistry
Bond formation and energy changes
The likelihood of an ionic compound being formed is
related to the stability of the compound that is formed.
We can get an indication of the stability of the compound
from its standard enthalpy of formation (ΔH f ). If
ΔH f is negative, the formation reaction is exothermic
(heat is given out), the compound formed is stable and
likely to be formed.
As an example, let’s look at the formation of potassium
bromide, KBr.
Combining solid potassium with liquid bromine (Br2)
results in a violent reaction. Solid KBr is produced, as well
as a great amount of heat. The equation for this reaction is:
K(s) +
1
2 Br2(l)
→ KBr(s)
ΔH
f
= −389.9 kJ mol−1
You are probably asking ‘Why is the formation exothermic’?
‘Why is the formation of KBr so favourable?’ ‘Does it arise
from the transfer of electrons from K to Br?’
If we look at the enthalpy change for the electron transfer
process only, we will see that this is not an exothermic
process:
K(g) + Br(g) → K+(g) + Br−(g)
ΔHf = +77 kJ mol−1
Note that separated atoms are considered to be similar to
gaseous phase atoms, hence the (g) representation.
Although ΔHf cannot be measured directly, it can be
calculated using other known enthalpy values – as you will
see later.
Clearly, the above endothermic reaction does not give the
complete story. However, we can consider all the steps that
might be involved in the reaction. If the enthalpy changes
for these steps are known, then by applying Hess’s law (see
page 81, Chapter 8), the sum of the enthalpies of these
individual steps should be equal to the formation enthalpy
of KBr. The application of Hess’s law to thermochemical
analysis cycles is called a Born–Haber cycle (see page 84,
Chapter 8).
the process of ionization can occur through electron
transfer and the charged species once formed will attract
each other to form the crystal. These steps are summarized
below, and values for their enthalpy changes are given.
1
1
2 Br2(l) → 2 Br2(g)
1
2 Br2(g) → Br(g)
Step 1
Step 2
ΔH = +15.0 kJ
ΔH = +96.6 kJ
Step 3 K(s) → K(g)
ΔH = +89.9 kJ
Step 4 Br(g) + e → Br (g)
ΔH = −341.4 kJ
Step 5 K(g) → K+(g) + e−
ΔH = +418.4 kJ
Total so far:
ΔH = +278.5 kJ
−
−
As we saw above:
1
K(s) + 2 Br2(l) → KBr(s)
ΔH = −389.9 kJ mol−1
Therefore, the difference must be due to the formation of
the solid lattice:
Step 6 K+(g) + Br−(g) → KBr(s)
ΔH = −668.4 kJ mol−1
The enthalpy change for Step 4 is the electron affinity and
for Step 5 is the ionization energy. These enthalpy changes
are discussed in more detail in Chapter 8.
Most of the steps, except 4 and 6 above, are endothermic
processes. This means that they require an input of energy.
If you sum all the steps up to the formation of the gaseous
ion (Steps 1–5), you obtain a value of ΔH = +278.5 kJ.
The enthalpy of formation is the sum of all the enthalpy
changes for the six steps listed.
+89.9
step 5
step 3
+96.6
+418.4
step 4
–341.4
step 2
+15.0
initial state
step 1
step 6
Let us look at some of the possible steps.
Before any atom of potassium can combine with any atom
of bromine, they must first be separated from other atoms
to which they are bonded. That is, in the solid metallic K
there are several K atoms bonded to each other and in the
liquid Br2 there are Br2 molecules bonded to each other
and Br atoms bonded together within them. One way of
separating the atomic interactions in each reactant is by
taking them to a gas phase. Once the atoms are separated,
–668.4
cumulative energy changes for the reaction:
K(s) +
1
2
Br 2 (l)
KBr(s)
final state
–389.9
Chapter 4 Chemical bonding
Since the formation of gaseous (free) ions from solid K and
liquid Br2 is an overall endothermic process, it is the last
step (step 6), where the ions attract each other to form the
lattice (crystalline network), that makes the overall process
exothermic.
Thus the major contributing factor to the stability of KBr is
the strong force of attraction between the ions that give rise
to the formation of the crystal. The energy change associated
with the formation of the solid crystal from these gaseous
ions is referred to as the lattice energy. If this is large and
positive then the compound will be stable. If it is small it
may not be sufficient to supply the energy needed to reach
the intermediate state and so the compound is unstable.
Similar arguments can be applied to the formation of
any type of bond. If the overall energy change between
reactants and products is negative (that is to say, the
reaction is exothermic), then the products are favoured
over the reactants. This is frequently the most important
factor, but it is not the only one.
When sodium solid burns in chlorine gas the white
crystalline solid sodium chloride is produced. This
compound contains positively charged sodium ions and
negatively charged chloride ions. Sodium has lost its 3s
electron to the chlorine atom.
1s2
Na
2s2 2p6 3s1
Cl
1s 2s 2p6 3s2 3p5
2
→
1s2
+ e− →
Na+
2s2 2p6
+ e−
Cl−
1s 2s 2p6 3s2 3p6
2
2
2
Notice that for both ions the electron configuration
acquired after the electron transfer is that of a noble gas,
that of neon in the case of sodium and that of argon in the
case of chlorine.
A charged ion can interact with another ion approaching
from any direction, and so the attraction between the
oppositely charged Na+ and Cl− ions gives rise to a threedimensional structure or lattice (Figure 4.2).
a
b
■ Ammonium nitrate dissolves readily in water but the
reaction is endothermic.
■ Citric acid reacts with sodium bicarbonate, but the
reaction is endothermic.
■ Solid barium hydroxide reacts with solid ammonium
nitrate to form an intensely cold slurry.
Something must be over-riding the energy change to make
these reactions proceed. The ‘something’ is a change in
entropy, but this topic is outside the scope of this book.
Ionic bonding
Metal atoms have between one and three outer electrons.
Non-metals usually have one or two unfilled gaps in theirs.
When a metal atom transfers an electron to a non-metal
atom, the metal acquires a net positive charge and the
non-metal a net negative charge. These oppositely charged
ions will attract each other. It is the electrostatic force of
attraction between the ions that constitutes a bond. This
type of bond is the ionic bond.
As an example, we will look at sodium chloride, NaCl
(Figure 4.1). Note that the positive ion is smaller than its
parent atom. The negative ion is larger than its parent ion.
Na
Cl
+
Na
Figure 4.1 The electrostatic nature of the ionic bond.
Cl
–
Figure 4.2 (a) The 3D structure of solid sodium chloride. (b) This
structure shows the positions of the ion nuclei.
A crystal of sodium chloride does not consist of one single
sodium ion bonded to one chloride ion. On the contrary,
it consists of a lattice of several positively charged sodium
ions surrounded by and attracted simultaneously to
several negatively charged chloride ions. The formula
NaCl gives the ratio of Na+ to Cl− that exists within the
lattice. A crystal of NaCl can therefore be regarded as a
giant molecule of the compound.
Covalent bonding
If two elements are very similar in their outer electron
configurations, or if achieving a noble gas configuration
would produce a high ionic charge (e.g. C4+), atoms can
produce stable compounds without ionization. Water
provides an example.
ITQ 2 Why would you expect a stronger ionic bond between
lithium and fluorine than between sodium and chlorine?
35
Unit 1 Module 1 Fundamentals in chemistry
Oxygen atoms have two vacant places in their 2p shell: the
electron structure is 1s2 2s2 2p4. Because they are mutually
repulsive, the p electrons are arranged as px2 py1 pz1.
A hydrogen atom has the electron structure 1s1. If a
hydrogen atom is moved progressively towards an oxygen
atom in the direction of either the py or the pz electron,
it will at first be repelled (electron/electron interaction).
However, as the distance between the nuclei decreases
a point is reached where this repulsion is balanced by
the two nucleus-to-electron attractive forces. Once this
equilibrium is achieved, changing the interatomic distance
either way needs an input of energy. The balance point is
the position of least potential energy, which means that a
chemical bond has been formed (Figure 4.3). This type of
bond is a covalent bond.
electron repulsion
Energy
36
nuclear repulsion
bond position
Particle separation
Figure 4.3 Energy changes in covalent bond formation.
Py
Px
Remember: bond formation releases energy: bond
rupture needs energy input.
Although no ionization is involved, the covalent bond, like
the ionic bond, relies on electrostatic forces. The electron
of the O atom in the O–H bond is now attracted to both the
oxygen nucleus and the nucleus of the H atom. Similarly
the electron from the H atom is attracted to the nucleus
of the O atom. The pair of electrons acts like the jam in a
sandwich, holding the two nuclei together.
The covalent bond always comprises a pair of electrons.
We can show a covalent bond by using × as an electron.
Then the reaction H + H → H2 is shown as
H
H H
++
+
+
+
H
To show which atom originally contained each electron
we can use × and •:
For example, the structure
of the ammonia (NH3)
molecule becomes
This is called a ‘dot-and-cross’
diagram.
H
N
H
H
Sometimes a low energy state can be obtained if orbitals (i.e.
energy levels) co-operate with each other. For example,
the electron structure of carbon is 1s2 2s2 2p2. Figure 4.4
shows how the electrons are distributed.
These electrons all repel each other. To reach a state of
low energy they must be as far apart from each other as
possible. This happens if one electron is at each point of
a tetrahedral pyramid surrounding the atom. For this to
happen the original three p orbitals and the s orbital combine
Pz
Figure 4.4 Hybrid orbitals in the carbon atom. The s orbital
contains two electrons, the py and the px orbitals one each. The
pz orbital is empty.
and produce the same total number (four) of equivalent
orbitals. Since there are four electrons, one electron will
exist in each of these new orbitals. They are called hybrid
orbitals and since these were made up from one s orbital
and three p orbitals, they are called sp3 hybrids.
When a carbon atom combines with hydrogen it therefore
needs four hydrogen atoms, so that each orbital contains
a pair of electrons. When
H
we draw the ‘dot-and-cross’
diagram of the bonds we
H C H
do not try to draw in 3D.
H
Instead we imagine the
structure flattened onto the Figure 4.5 The ‘dot-and-cross’
page, as shown in Figure 4.5. structure of CH4, methane.
Although a covalent bond is usually formed using one
electron from each combining atom, it is possible for an
atom with an unused pair of electrons (a lone pair) to
bond with an atom that has an empty electron shell. The
simplest example is the combination of ammonia with
hydrogen ions:
NH3 + H+ → NH4+
In this reaction the nitrogen atom has a lone pair of
electrons (2p2) whereas the hydrogen ion has an empty
shell (1s0). The effect is exactly the same, but the bond is
sometimes called the dative covalent bond since both
electrons are ‘donated’ by the one atom.
Chapter 4 Chemical bonding
In the formation of an
ionic bond, electrons
are transferred from one
atom to another. For
covalent bond formation
N
H
H
electrons are controlled
by the nuclei of both
atoms
simultaneously
H
– often referred to as
‘sharing’ of electrons. The
Figure 4.6 Electron structure of
ionic and covalent models
the ammonium ion NH4+. The ‘×’
of bonding represent the
electrons come from the nitrogen
atom and the ‘•’ electrons come
extremes of electron
from the hydrogen atoms.
transfer and electron
sharing. Most actual bonds fall somewhere between these
two extremes. A covalent bond between two particles of
different electronegativity (for example carbon and oxygen,
as in the carboxylate ion –COO−) inevitably means that in
the C–O bond the bonding pair lies nearer to the nucleus of
the oxygen atom, giving it a small negative charge (δ−) and
leaving a small positive charge (δ+) on the carbon atom.
H
dative
bond
The hydrogen bond
Liquid water has two features which appear anomalous.
■ Its melting point and boiling point are much higher
than its molecular mass would suggest (Figure 4.7).
■ Its density does not change regularly as its temperature
falls, especially around the freezing point (Figure 4.8).
Boiling point / ˚C
100
0
–100
Each molecule of water is held together by two covalent
O–H bonds. Oxygen, however, has a greater capacity for
attracting electrons to its nucleus than does hydrogen. On
the Pauling scale of electronegativity (see Chapter 2, page
23), oxygen has the value 3.5 but hydrogen is only 2.1.
The effect of this is that the pair of electrons making up
each O–H bond are not central between the nuclei, but
are displaced toward the oxygen. This in turn means
that each molecule has an electric dipole – a small but
significant separation of electric charge. You can think of
it as a covalent bond with a small degree of ionic character
(Figure 4.9).
–
+
O
H
+
H
+
–
O
+
H
H
Figure 4.9 Hydrogen bonds in water, showing the polarity.
Although the molecules in liquid water are in constant
random motion, any two molecules will be, for a short
period, close to each other. If they are oriented as shown in
Figure 4.10, an electrostatic attractive force exists between
them. It is only about 10% the strength of either an ionic
bond or a covalent bond. However, it is strong enough that,
for the short time that the molecules are close together, the
effective molecular mass of the molecule is greater than
when the molecules are solitary.
The effect need not be restricted to two molecules. It is
restricted to an interaction between the positively charged
hydrogen and a lone pair on the oxygen, so one water
molecule can potentially form four hydrogen bonds at the
same time. The strength of the hydrogen bond is greater
than might be thought because the small size of the
polarized H atom allows close approach of the polarized O
atom (Figure 4.10). The effect is electrostatic and therefore
obeys the inverse square law.
–200
CH4
NH3
H2O
HF
Figure 4.7 Boiling points of water and other molecules of similar
molecular mass. Water is very much the ‘odd one out’.
198 pm
Density / g cm –3
1.000
H
96 pm
0.99
O
H
0.98
Figure 4.10 The lengths of the covalent O–H bond (96 pm) and
the O...H hydrogen bond (198 pm).
0.97
0.96
0
4
20
40
60
80
Temperature / ˚C
Figure 4.8 The density of water does not change regularly as its
temperature falls.
ITQ 3 Draw a diagram showing the electron structure of the
hydrated proton H3O+.
37
Unit 1 Module 1 Fundamentals in chemistry
Hydrogen bonds are formed between molecules which
contain polarizable hydrogen atoms and those which have
an atom with higher electronegativity (such as oxygen,
nitrogen or fluorine) that have at least one lone pair
(Figure 4.11). Hydrogen fluoride fits these criteria but
each molecule can only form two H-bonds. Therefore, the
increase in boiling point is not so pronounced as in water.
100
R
R
H2O
80
Boiling point / ˚C
38
H bond
60
R
40
20
HF
H2Te
0
–20
–40
H2Se
NH3
H2S
AsH3
–60
–80
–100
Period 2
HCl
PH 3
Period 3
SbH3
R
Hl
R
HBr
Period 4
Period 5
Figure 4.11 Hydrogen bonding in NH3, H2O and HF means that
the boiling points for these molecules are higher than expected
as compared with similar molecules.
The formation of hydrogen bonds also accounts for the
anomalous density changes in water as it is cooled. As
the temperature falls, thermal agitation becomes less and
more H…O interactions are effective in holding molecules
together. The molecules become closer together, as we
would expect. But at 4 °C the equilibrium between their
formation and break up tips in favour of formation, and
the H-bonded state begins to predominate. Networks
of water molecules mimicking fragments of ice begin to
have transient existence, and the structure is one which
occupies more volume than the normal liquid state. Hence
as the temperature falls towards 0 °C, the structure of the
‘liquid’ becomes more and more open, meaning that its
density falls. 4 °C is the temperature at which water has its
maximum density, 999.8395 kg m−3.
Hydrogen bonds can form between different molecular
species, such as water and an alcohol or water and a
sugar, or within a complex molecule such as a protein or
DNA. The release of energy as H-bonds form is the factor
controlling the solubility of alcohols and sugars in water.
The H-bonding within a protein or DNA molecule may
lock the conformation and hence the shape of the molecule
(Figure 4.12).
The metallic bond
True metals have properties that cannot be explained on
the basis of covalent, ionic or hydrogen bonding. They
conduct electricity very well in the solid state, obeying
Ohm’s law. They conduct heat better than most other
Figure 4.12 Hydrogen bonding is important in giving proteins
their 3D structure.
materials, although this is not an absolute rule. For
example, silver is an excellent electrical conductor but
its thermal conductivity is less than that of diamond, an
electrical insulator.
Sodium can be taken as an example of a substance that has
metallic bonding. Sodium has the electron structure 1s2 2s2
2p6 3s1. The 1s, 2s and 2p electrons are strongly bound to
the nucleus and take no part in the metallic bond. When
two sodium atoms come together, their 3s atomic orbitals
fuse into one molecular orbital. (See page 47 for more
about this concept.) The total number of orbitals cannot
change so a second orbital is created at a higher energy
level.
single orbitals
a
3s
3s
molecular orbitals
b
Figure 4.13 3s orbitals for sodium. (a) Two orbitals form separate
atoms; (b) two atoms forming a molecular orbital.
As more sodium atoms are added they add they form a
second, almost identical molecular orbital. No two electrons
can have exactly the same energy, so this orbital is at a
ITQ 4 The structure of glucose can be represented as
OHC–(H–C–OH)4–CH2OH. Why is it so soluble in water?
Chapter 4 Chemical bonding
very slightly different level. When many atoms are added,
the energy levels of all the molecular orbitals fuse together
into a ‘band’ extending across the whole of the metal. The
higher-level orbitals do the same.
The upper level and lower level bands are called the
‘conduction band’ and the ‘valence band’ (Figure 4.14).
Electron energy
There are three types of attractive force between molecules.
They are generally taken together and called van der Waals
forces. You may also find the second and third types called
‘London’ forces after Fritz London, who first suggested
them. The three types are:
■ dipole–dipole interactions;
conduction
band
■ dipole–induced dipole interactions;
■ induced dipole–induced dipole interactions.
overlap
band gap
valance
band
metal
forces but the intramolecular covalent bonds within the
naphthalene do not break even at its boiling point of 218 °C.
semiconductor
insulator
Figure 4.14 Valence and conduction bands in metals,
semiconductors and insulators.
In a metal the two bands overlap and so electrons from the
valence band can enter the conduction band. This extends
across the whole of the metal. Electrons in it are free from
their original nuclei and are free to move anywhere. The
metal is often described as having ‘a sea of electrons’ drifting
amongst the metal cations. An applied voltage causes the
electrons to flow, forming an electrical current. Because no
energy is used separating electrons from nuclei, the current
obeys Ohm’s law: the current is proportional to the applied
voltage.
If the conduction band and the valence band do not overlap
then a band gap exists. If this is small, the material is a
semiconductor. If the gap is large, so that no electrons can
jump up across it, then the material is an insulator.
van der Waals forces
Johannes Diderik van der Waals (1837–1923) was the
first Professor of Physics at the Municipal University of
Amsterdam. At a time when some scientists doubted the
very existence of atoms, he published work in which he
assumed not only the existence of molecules of finite size,
but also that they attracted one another – though he had
no idea why. You have only to warm a solid substance like
naphthalene and watch it melt without any decomposition
to realize that the forces holding the molecules together in
the solid state must be much weaker than those within the
molecules. Warming is enough to break the intermolecular
Dipole–dipole interactions
We have already seen one example of dipole–dipole
interactions in the hydrogen bond. But very few molecules
are electrically neutral everywhere, unless neighbouring
atoms have closely similar electronegativities.
For example, a hydrocarbon chain such as butane, C4H10,
is scarcely polar. However, 1-fluorobutane contains a C–F
bond and is significantly polar, with the fluorine atom
carrying a small negative charge (written δ− to distinguish
it from a whole ‘one-electron’ charge). The corresponding
positive charge is distributed over the nearby atoms. Table
4.1 gives some Pauling electronegativites. Note how similar
the values are for carbon (2.4) and hydrogen (2.1), but also
note how electronegative fluorine is (4.0).
If two butane molecules come together there is little or no
interaction between them. With almost no cohesive forces
between molecules, the boiling point of butane is low
(−1 °C). 1-Fluorobutane, by contrast, boils at 32 °C.
Table 4.1 Some Pauling electronegativity values
Element
Pauling electronegativity
hydrogen, H
2.1
carbon, C
2.4
sulfur, S
2.5
chlorine, Cl
3.0
nitrogen, N
3.0
oxygen, O
3.5
fluorine, F
4.0
Dipole–induced dipole interactions
If a polarized molecule approaches a neutral molecule at
first sight it would seem that there would be no interaction.
But the electron distribution in a molecule is fluid: even
the electrons in a covalent bond are not locked tightly in
place. As a charged molecule (however small the charge)
approaches either a neutral molecule or the neutral portion
of another identical molecule, the charge interacts with
39
40
Unit 1 Module 1 Fundamentals in chemistry
nearby electrons and either attracts them (if positive) or
repels them (if positive). The effect is only momentary since
the two particles are in constant motion: but it is enough
to form a weak, transient bond. You might expect to find
this type of bond both in a mixture of ammonia (which is
polar) and liquid butane (which is not).
Induced dipole–induced dipole interaction
These forces, although the weakest of all, are the easiest
to illustrate. A ‘neutral’ molecule is only neutral when
averaged over a period of time. At any instant the random
motion of the electrons within it makes the particle, for
that instant, slightly polar. If at that instant the molecule is
near to another, it will induce, for that instant, an opposing
dipole and the two particles will have an attractive force
between them.
This is the force responsible for the formation of solid
helium, albeit at a temperature of 0.95 K (−272 °C). The
larger the particle, the more electrons there are to create
the force and the stronger it is. Table 4.2 gives data for the
melting points of all the noble gases.
Table 4.2 The increase in melting point for the noble gases
correlates with the number of electrons in each atom
Element Atomic number
He
2
Melting point / K
0.95
Ne
10
25
Ar
18
84
Kr
36
116
Xe
54
161
Rn
86
202
H
He
Li
Be
Na
Mg
Al
Si
P
S
Cl
Ar
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
Rb
Sr
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
Hf
Ta
W
Re
Os
Ir
Pt
Au
Hg
Tl
Pb
Bi
Po
At
Rn
B
Cs
Ba
La
Fr
Ra
Ac
C
N
O
F
Ne
Figure 4.15 The electronegativity of atoms varies with their
position in the periodic table.
In summary:
■ metals form ionic compounds with non-metals;
■ elements close to each other in the periodic table tend
to form covalent bonds – examples are carbon and
hydrogen forming alkanes and silicon and oxygen
forming silicones.
Properties associated with different bond
types
Ionic compounds
■ Composed of separate cations (+) and anions (−) in a
regular lattice. No separate molecules exist.
■ Force is non-directional: ions attract whatever their
orientation with respect to each other. The number of
neighbours of an ion is determined by the geometry
of the structure rather than the atom’s electron
structures. An ion can attract any number of ions of
opposite charge.
■ High melting point: only the strong ionic bonds are
present and extend in three dimensions.
The periodic table and bond type
We can use the periodic table to predict the principle bond
type between a given pair of atoms.
Ionic bonds involve the transfer of electrons from one
atom to the other. This is linked to a large difference in
electronegativity between the two atoms (Figure 4.13).
Ionic bonds are more likely between elements in Groups
I and II and elements towards the top right of the periodic
table – an example is sodium chloride. Less clearly, ionic
bonds also form between elements at the top right and
elements in the bottom right of the periodic table – lead
oxide is an example (see Chapter 15, page 146).
ITQ 5 Hexane, CH3(CH2)4CH3, has a melting point of −100 °C.
Polythene, CH3(CH2)nCH3, softens just below +100 °C. The value
of n in polythene may be several thousand. Suggest a reason for
the difference in melting points.
■ Substance does not conduct electricity when solid
because ions are immobile, but does conduct when
molten or in aqueous solution. Ions are then mobile
and can move under an applied voltage.
■ Soluble in water rather than organic solvents. Ions
become solvated – form transient bonds to the polar
water molecules – which liberates energy sufficient to
separate more ions from the lattice.
Covalent compounds
■ Composed of separate molecules within which strong
covalent (intramolecular) forces hold atoms together.
■ Covalent forces are directional, existing between
specific atomic nuclei. Any atom can form a specific
number of bonds.
■ Molecules are held to each another by much weaker
(intermolecular) forces.
Chapter 4 Chemical bonding
■ Substances have a low melting point because the
Mixed bonds
intermolecular forces are weak.
■ Substances do not decompose at their melting
point because the energy supplied is not enough to
overcome the covalent intramolecular forces.
■ Do not conduct electricity when solid, molten or in
solution because the structure contains no charged
particles.
■ There may be single, double or triple bonds.
We have already seen that ionic and covalent bonds are
the two extremes of a spectrum. We have also seen that
solid covalent materials contain both covalent and van der
Waals forces – iodine, I2, is a good example. The individual
molecules are covalently bonded to form the I2 molecules
but the molecules are held together by van der Waals forces
(Figure 4.16).
a
b
Hydrogen bonds
■ A transient electrostatic attractive force between a
polar molecule and the oppositely charged polar region
of a water-like substance.
shorter
distance,
covalent
bond
■ The positive region is normally a polar hydrogen atom
within a molecule. The negative region is frequently a
polar oxygen atom, but can also involve nitrogen and
fluorine atoms.
■ Strength is roughly 10% that of the average ionic or
covalent bond.
■ Are sufficient to cause anomalously high melting and
boiling points for the molecules concerned, which are
momentarily connected in groups.
Metallic bond
■ Produces high electrical and thermal conductivity.
■ Normally a strong bond.
■ Metals normally have high density.
■ Bond can be visualized as a sea of unattached electrons
washing between positive metal ions. Electrons are free
of their original atoms but act as a ‘glue’ between them.
Figure 4.16 The 3D structure of iodine, showing the covalent
bonds between iodine atoms in an iodine molecule and the van
der Waals forces between iodine molecules.
Some lattices contain two different bonds within
themselves. Graphite (an allotrope of carbon) has physical
properties quite different from those of diamond (another
allotrope of carbon). Although graphite has a similarly high
melting point, it is soft (not hard), it is a good electrical
conductor (not an insulator) and it is a good lubricant (not
an abrasive). These differences can be explained by looking
at the structure of graphite (Figure 4.17) and diamond.
Graphite is made up of layers of covalently bonded carbon
atoms, with the layers held together by weaker forces
produced by wandering electrons (which also give the
electrical conductivity).
■ The free electrons are responsible for both electrical
and thermal conductivity.
■ The spherical ions pack together in the most space-
saving way.
van der Waals forces
■ Weak forces between either transient or permanent
dipoles within, or between, molecules.
■ Produce either low or very low melting points and
boiling points (which can be near absolute zero).
covalent
bonds
ITQ 6 Diamond is composed only of carbon atoms (Z = 6). Boron
nitride, BN, shares many of its properties. Suggest why this may
be so. (B, Z = 5; N, Z = 7)
larger
distance,
van der
Waals force
Figure 4.17 The structure of graphite.
41
42
Unit 1 Module 1 Fundamentals in chemistry
Summary
✓ A bond is formed when the attractive forces
✓ Five types of bond are:
■
ionic
covalent
■ hydrogen
■ metallic
■ van der Waals
between particles outweigh the repulsive forces.
■
✓ The forces involved in bonding are all
electrostatic.
✓ Bond formation normally results in the release of
energy.
✓ Each bond type produces characteristic
properties.
Review questions
1
2
3
A student claimed that ‘ionic, covalent and hydrogen
bonds are all the same, because they all depend on
electrostatic attraction’. Explain:
(a) why it is true that all three bonds depend on
electrostatic attraction;
(b) how the properties of substances containing these
bonds show that the bonds are different from each
other.
In single atoms of a metal, such as sodium, the
electrons are firmly bonded to the atomic nucleus.
Sodium metal in bulk is a good conductor of
electricity, which means that an electric current can
flow through it.
(a) What do we mean by ‘an electric current’?
(b) Give a description of a metal structure which
makes clear why such a current can flow, and use
it to explain why the metal atoms do not fly apart.
(c) Sodium metal dissolves in liquid ammonia to give
a blue, electrically conducting solution containing
free electrons and solvated sodium ions Na(NH3)x+.
How is this ion similar to the sodium ion found in
an aqueous solution of sodium chloride, and why?
(a) State two properties of gasoline that make it likely
that it is:
(i) a mixture
(ii) a mixture of covalent molecules.
(b) Contrast the properties of gasoline with the
properties of molten sodium chloride. Explain
how the bonding in sodium chloride gives rise to
the properties of sodium chloride.
4
Poly(ethene) (‘polythene’) is a solid. It is made up of
saturated hydrocarbon molecules.
(a) Which force is responsible for holding the
molecules one to another?
(b) Explain the origin of this force.
(c) Explain why water boils at a temperature much
higher than would be expected from its low
relative molecular mass (RMM).
Answers to ITQs
1
To do so would need the kettle to move spontaneously
from a low energy state to a higher energy state. We
accept as an axiom that this does not happen.
2
In lithium fluoride the notional charge is the same
as in sodium chloride but the ions are smaller. The
electrostatic force between them (which obeys the
inverse square law) is therefore stronger.
+
3
H
O
H
H
4
The molecule contains a number (5) of –OH groups
which can form hydrogen bonds with water
molecules, releasing energy to break down the van
der Waals forces between the glucose molecules.
5
The very much longer chain in polythene will be able
to form many more van der Waals forces thus needing
much more energy to separate the molecules, as
compared to the small chain in hexane.
6
There are the same number of bonding electrons
between C–C as for B–N and the atoms are of similar
sizes.
43
Chapter 5
Shapes of covalent molecules
Learning objectives
■ Write the Lewis structures for covalent molecules
and ions.
■ Predict the geometrical shape and bond angles
of simple molecules and ions using the VSEPR theory.
■ Outline the basic principles of hybridization.
■ Based on its molecular structure, predict
whether a molecule is polar or non-polar.
Lewis structures
Drawing Lewis structures
Gilbert N. Lewis (1875–1946) was an American physical
chemist who developed the earliest successful model of
the chemical bond. According to Lewis’s model, each pair
of electrons involved in bonding lies in a region of space
between two neighbouring atoms. He represented the pair
of electrons as a line or a pair of dots. Electron pairs shared
between atoms are referred to as bond pairs. Electrons
present on only one atom are referred to as non-bonding
pairs or lone pairs. Covalent bonds can be denoted as
A–B (or A:B), A=B (or A::B) and A≡B (or A:::B), consisting
of one, two and three bonded electron pairs respectively.
Non-bonding electron pairs do not contribute directly to
the bonding but they influence the shape and chemical
properties of molecules.
Follow the steps in this list in order to draw the Lewis
structure of a molecule.
Lewis structures account for the formulae of covalent
molecules; they do not indicate the three-dimensional
shapes of molecules.
■ Determine the number of electrons to be included in
the structure by counting all the valence electrons for
each atom. Valence electrons are the electrons in the
outermost orbitals. Add an electron for a negatively
charged ion and subtract an electron for a positively
charged ion.
■ Select a central atom. Note that the central atom is
usually the least electronegative element. Then decide
which atoms are bonded to which. Place atoms around
the central atom using two dots to represent a single
bond. Hydrogen is never a central atom.
■ Complete the octet of electrons on the atoms
surrounding the central atom, using non-bonding
electron pairs (an octet is eight). In the case of
hydrogen, there is a maximum of two electrons.
■ After completing the octets, place any remaining
electrons on the central atom as non-bonding electron
pairs.
■ If the central atom has less than eight electrons then
form double bonds or triple bonds by removing one or
two non-bonding pairs of electron from outer atom(s).
Oxygen can form single or double bonds; carbon and
nitrogen can form single, double or triple bonds.
44
Unit 1 Module 1 Fundamentals in chemistry
Worked example 5.2
Worked example 5.1
−
Q
Draw the Lewis structure for BF4 .
Q
Work out the Lewis structure of carbon dioxide, CO2.
A
Step 1: First, determine the total number of valence electrons.
■ Each boron (B) atom has the electronic configuration
1s2 2s2 2p1 and has three valence electrons. The electrons
in the inner 1s shell are not included.
■ Each fluorine (F) atom (1s2 2s2 2p5) has seven valence
electrons, giving a total of (4 × 7) or 28 electrons.
■ Since the ion has a single negative charge, add an
additional electron.
■ The total number of valence electrons is therefore equal to
(3 + 28 + 1) = 32
A
Step 1: Count the valence electrons
one carbon atom
4
two oxygen atoms (2 × 6)
12
total
16
Step 2: The skeletal structure is
O—C—O
Step 3: Distribution of the 16 electrons, using two per bond and
completing the octet of the outer atoms, gives the following:
Step 2: The formula of the ion tells us
that there are four fluorine atoms bonded
to one boron atom. B is the central atom
(the less electronegative) and the fluorine
atoms are placed around it:
F
F
B
F
F
F
B
F
F
O
–
F
B
F
This is the Lewis structure for BF4−. Note that the negative
charge is ascribed to the ion as a whole and not to an individual
atom.
–
F
F
B
F
F
Showing a molecule like this can be very useful because you
can see at a glance how many non-bonding pairs there are,
and where they are in the molecule.
ITQ 1 How many lone pairs of electrons are present on (a) Sn in
SnCl2 and (b) Br in BrF2−?
ITQ 2 What would be the Lewis structure of the SO42− ion?
C
O
You will notice that these structures are very like the
‘dot-and-cross’ structures used in Chapter 4, except that in
a Lewis structure all the electrons are written as dots.
Molecular geometry
F
The electrons forming the
bonds can be shown as lines.
O
Now each atom has a full octet of eight electrons.
F
F
C
Step 4: At the moment the central atom C has only four
electrons. If a pair of non-bonding electrons is removed from
each oxygen atom to form two double bonds between carbon
and each oxygen then the carbon atom will have eight electrons.
Step 3: In the next step two dots are
used to represent each bond between
B and F.
We have used up eight
electrons of the 32 electrons so
far. Complete the octet of each
of the F atom:
O
The shape of a molecule affects its physical and chemical
properties. For example, the differences in boiling points
of several molecules with similar formulae result from
differences in their structures. In biological reactions, only
molecules of a certain shape and size take part in certain
reactions. Small changes in the sizes and structures of
molecules used as drugs may alter their activity, usefulness
and toxicity.
Molecular geometry refers to the three-dimensional
arrangement of the atoms or groups of atoms in a compound.
Predicting molecular shapes
This section covers the basis of the valence shell electron
pair repulsion theory (VSEPR).
The valence shell is the outermost electron-occupied
region of an atom; it contains the electrons that participate
in bonding with other atoms. In a covalent bond, two
atoms share a pair of electrons, referred to as a bond pair
of electrons. In polyatomic atoms where two or more
atoms are bonded to a central atom, electrostatic repulsions
among the bonding electron pairs cause them to remain as
Chapter 5 Shapes of covalent molecules
far apart as possible. The shape that the molecule assumes,
as defined by positions of all the atoms, is such that these
repulsions are at a minimum.
In the VSEPR model, double and triple bonds are treated
as single bonds. This is an approximation for qualitative
purposes. For accurate predictions of bond angles there
is a difference between single bonds and multiple bonds.
Also, if two or more resonance forms (see page 217) are
possible for a molecule, we apply the VSEPR model to
any one of them. For simple molecules, the VSEPR model
gives excellent agreement between predicted shapes and
the shapes determined from experimental data such as
spectroscopic information, bond angle determinations and
X-ray crystallographic data. However, the predictions the
model makes about electron density in the molecule are
less reliable.
The basic molecular shapes
This first section looks at molecules where the central atom
doesn’t have any lone pairs (non-bonding pairs). All of the
shapes can be modelled by using balloons. Take a balloon,
inflate it with air and tie the end. Repeat this with a second
balloon and then tie the two balloons together. Notice the
shape they take up. As you continue to add balloons, you
will be able to observe the corresponding changes in the
geometry.
Two electron pairs
According to the VSEPR model, a central atom with two
electron pairs only have those two electron pairs in a
straight line. The angle between them will be 180°. An
example is the BeF2 molecule with two electron pairs
around the central Be atom. The linear arrangement puts
the bonding electron pairs as far apart as possible. Other
examples include CO2, HCN and XeF2.
180˚
F
Be
Four electron pairs
Four electron pairs round a central atom will lie at the
corners of a regular tetrahedron with all the bond angles
being 109.5°. The central atom is positioned at the centre of
the tetrahedron. An example is CH4 and another example
is NH4+.
H
H
109.5˚
H
C
H
C
H
H
H
Five electron pairs
Five electron pairs round a central atom will lie at the
corner of a trigonal bipyramid. You can imagine the shape
to be made from three atoms at 120° in one plane (as in
the three electron pairs) with an atom above and the final
atom below. An example is PCl5.
Cl
Cl
P
Cl Cl
Cl
Cl
120˚
90˚
Cl
P
Cl Cl
Cl
For this geometry, there are two types of positions. You
have the axial positions, above and below the trigonal
plane, and the equatorial positions, within the plane. The
bond angle between an axial atom and an equatorial atom
is 90° and between two equatorial atoms is 120°.
Six electron pairs
Six electron pairs around a central atom will lie at the
corners of a octahedron. Every bond angle has a value of
90°. An example is SF6.
F
axial
Three electron pairs
Three electron pairs round a central atom will lie in the
same plane at the corners of an equilateral triangle in order
to maximize separation and minimize repulsions. The bond
angles will all be equal to 120°. An example of a molecule
containing three electron pairs on the central atom is BF3.
F
120˚
B
F
H
F
F
F
S
axial
equatorial
F
F
F
90˚
F
45
46
Unit 1 Module 1 Fundamentals in chemistry
■ In the case of NH3 one position in the tetrahedron will
Summary
be occupied by a lone pair of electrons. This pair repels
the bonding pairs more strongly than they repel each
other. Thus, the molecule has a trigonal pyramidal
geometry.
Table 5.1 summarizes these basic molecular geometries.
Table 5.1 Basic molecular geometries
Number of
electron pairs
Molecular geometry
Bond angles
Examples
2
Linear
180°
BeCl2
3
Trigonal planar
120°
BF3
4
Tetrahedral
109.5°
CH4, NH4+
5
Trigonal bipyramidal
120°, 90°
PCl5
6
Octahedral
90°
SF6
■ In the case of OH2, with only three atoms, the
molecule is planar and V shaped.
Molecules with lone pairs
In the examples given so far, the electron pairs around
the central atom are all bonded. However, the shape of
a molecule is governed by the arrangement of both the
bonding pairs and the lone pairs of electrons. The attractive
forces on the nuclei of two bonded atoms localize the
bonding electron pairs between them. Lone electron pairs
on the other hand, are on one atom, are less restricted
and can occupy a larger region of space. This means that
lone electron pairs will experience greater repulsion from
neighbouring lone pairs than from bond pairs. These
strengths of these repulsions fall in the order:
lone pair–lone pair repulsions > lone pair–bond pair
repulsions > bond pair–bond pair repulsions
How this affects the shapes of molecules can be shown by
comparing the molecules CH4, NH3 and OH2. The Lewis
structures for these molecules (showing the lone pairs) are
as follows:
H
In CH4, where there are no lone pairs, the bond angles
are all 109.5°, the normal value observed in regular
tetrahedral shapes. For NH3, however, the H–N–H bond
angles are 107.3° due to the presence of the lone pair with
larger volume requirements forcing the bond pairs closer
together. In the OH2 molecule, where there are two lone
pairs, the bond angles are even smaller (104.5°).
H
109.5˚
H
C
H
H
tetrahedral
N
H
107.3˚
H
O
H
H
trigonal pyramidal
104.5˚
H
V shape
Hydrogen ions in water form the oxonium ion, H3O+. The
hydrogen ion adds onto one of the lone pairs of the water
molecule without adding any other electrons. The electron
structure is then exactly the same as in the NH3 molecule.
The two molecules are said to be ‘isoelectronic’. The
oxonium ion is trigonal pyramidal.
In general:
■ molecules with a general formula of AB4, in which
the central atom (A) is surrounded by four bonded
electron pairs (B), have a tetrahedral geometry
■ molecules with the general formula AB3E, where A
H
C
H
H
N
H
O
H
H
H
H
In all three cases, there are four electron pairs on the
central atom. These electron pairs, according to the
VSEPR model, should result in the molecules adopting a
tetrahedral geometry. However, experiments confirm this
to be true only for CH4. In the case of NH3 three of the four
electron pairs on the central atom are bonded and is one
non-bonded. In the case of OH2, two are bonded and two
non-bonded. The presence of the non-bonded electron
pairs in NH3 and OH2 results in some adjustments to the
tetrahedral geometry.
ITQ 3 Predict the shape of the xenon fluoride molecule XeF2.
(Hint: xenon can hold 10 electrons in its valence shell.)
is the central atom, B is a bonding electron pair and
E a non-bonded pair of electrons, have a trigonal
pyramidal shape
■ molecules with the general formula AB2E2 adopt a V
shape or angular shape.
Guidelines for applying the VSEPR model
1 Write the Lewis structure of the molecule.
2 Count the total number of electron pairs (bond pairs
and lone pairs) on the central atom.
3 Treat double and triple bonds as though they are single
bonds, i.e. one pair of electrons.
4 Predict the overall arrangement of electrons and
geometry of the molecule based on the number of
electron pairs on the central atom.
Chapter 5 Shapes of covalent molecules
5 In predicting bond angles remember that:
Hybrid orbitals
lone pair–lone pair repulsion > lone pair–bond pair
repulsion > bond pair–bond pair repulsion
For organic molecules, when VSEPR is used to describe
the shape, it is applied to each individual carbon, nitrogen
and oxygen atom. Ethane, ethanal, ethene and propanone
molecules are shown here as examples.
H
H
C
H
H
C
tetrahedral
trigonal planar
H
H
tetrahedral
trigonal planar
C
H
H
When the atom forms bonds with hydrogen, the electron
pairs are furthest apart if they exist at the corners of a
regular tetrahedron. The 2s and the 2p orbitals lose their
identity and in their place, four equivalent identical orbitals
are formed. These are called hybrid orbitals and since they
are formed from one s and three p orbitals, they are called
sp3 hybrids.
C
H
O
H
H
H
H
O
H
C
C
C
C
H
C
H
H
H
H
H
Similarly, for molecules with more than one central atom,
VSEPR is applied to each central atom.
F
F
H
B
N
F
H
H
When two atoms combine, pairs of electrons come under
the control of both nuclei. The orbitals in which they exist
overlap with each other. Linus Pauling proposed that the
bonding orbitals of atoms do not just overlap, but combine
to form hybrid atomic orbitals. In the compound, these
are called molecular orbitals. An s orbital, for example,
can overlap end-on with a p orbital to form an sp hybrid
orbital. If the overlap is end-on-end then it is referred to
as a sigma bond (σ). A single bond is made up of one
sigma bond. There is free rotation about a single bond.
When p orbitals overlap in a side-to-side manner they form
pi bonds (π).
Worked example 5.3
Q
Use the VSEPR theory to predict the shapes of the PF5 molecule.
A
P has five valence electrons and five F atoms have a total of 35
valence electrons, giving a total of 40 valence electrons.
The Lewis structure is:
F
F
F
B
F
F
The central atom, P, has five bonded
electron pairs so it will have a
trigonal bipyramidal geometry:
We saw in Chapter 4 that atomic orbitals can ’co-operate’
with each other to form an equivalent number of hybrid
orbitals. The reason for the change lies in the lower energy
states that are produced. A good example is the isolated
carbon atom with the electron structure 1s2 2s2 2p2. There
are four vacant spaces available if the atom is to have the
noble gas structure of neon (1s2 2s2 2p6).
Ethane
F
F
P
F
A double bond consists of one sigma bond and one pi bond.
A triple bond consists of one sigma bond and two pi bonds.
F
F
ITQ 4 On the basis of the VSEPR theory, what would be the
shape of the following molecules?
(a) ClF3
(b) NHCl3+
(c) XeF4
In ethane, C2H6, each C atom forms four sp3 hybrid orbitals.
Two of these overlap end-on to form a sigma orbital. This
contains two electrons, one from each atom, and constitutes
a single bond. This is the carbon–carbon single bond. The
ITQ 5 Which of the species in ITQ 4 would have the smallest
bond angles?
47
48
Unit 1 Module 1 Fundamentals in chemistry
remaining three sp3 orbitals form sigma bonds with the
s orbital of hydrogen atoms, forming carbon–hydrogen
single bonds in the two –CH3 groups.
H
H
H
H
C
C
H
H
Ethene
In ethene, C2H4, the 2s and 2p orbitals of each carbon
combine to produce three sp2 hybrid orbitals that lie in a
plane 120° from each other. This leaves one unhybridized
2p orbital. The three sp2 hybrid orbitals form two sigma
(σ) bonds with the two hydrogen atoms and one with a
carbon atom. A pi bond is also formed between the carbon
atoms from the overlap of their unhybridized p orbitals.
Each carbon atom is bonded to two hydrogen atoms and a
carbon atom, arranged with trigonal planar geometry.
H
H
Table 5.2 Molecular geometry and hybridization of orbitals
Molecular shape
H
H
C
C
H
Table 5.2 summarizes some geometries that result from the
hybridization of other orbitals.
Hybridization
A common structure which relies on resonance for its
stability is benzene. For many years after its formula of
C6H6 was determined, chemists could not write a structural
formula for the compound using single, double and triple
carbon–carbon bonds. The story is that Friedrich Kekulé, a
German organic chemist (1829–1896), was puzzling on the
problem and dreamed one night about a snake gripping its
own tail. He realized that by writing the structure as a ring
with alternating double and single bonds, each carbon atom
could then show its normal valency of four (Figure 5.1).
H
C
C
Other examples of hybridization
General formula Electron pairs
Transforming one structure into the other involves only the
transfer of one electron from one oxygen atom to the other.
In practice, we can imagine that the moving electron takes
up an intermediate position. The two extremes shown in
the diagram are called canonical forms of the structure.
The structure exhibits resonance. Experiment shows that
a resonant structure is more stable than would be expected
from either of its canonical forms.
C
C
C
H
C
C
H
H
H
H
C
H
C
C
H
H
H
C
C
C
C
H
C
C
H
H
Figure 5.1 The two possible Kekulé structures of benzene.
■ Although this satisfies the demands of valency, it does
not represent the properties of benzene very well.
AB
1
linear
AB2
2
sp
linear
trigonal planar
■ The double bonds should give reactivity: in fact
AB3
3
sp2
AB2E
3
sp2
bent, angular, V shaped
AB4
4
sp3
tetrahedral
benzene is relatively inert.
■ Double C=C bonds are shorter than single C–C bonds
yet studies show no such thing in the molecule.
■ Benzene is roughly 150 kJ mol−1 more stable than can
Resonance
be predicted from calculation.
In some molecules, electrons can occupy two or more
equivalent positions. One example is the carboxylate ion,
–COO−. Drawn as structural formulae we can have either
of these two structures:
O
O–
C
C
O
O–
ITQ 6 What is the type of hybridization on the N atom in NO2−
and in NO3−?
From Figure 5.1, you can see that benzene has two
canonical forms. This would explain the extra stability.
But we need to look at the electron distribution to explain
the other problems. Any of the C atoms forms two single
bonds and a double bond. This is the same as in ethene (see
above). The three single bonds lie in one plane at 120° to
each other and the unused p orbital stands at right angles
to the plane so that its lobes overlap on top and beneath it.
The overlapping orbitals merge into hybrids forming one
ring of charge above the plane of the carbon atoms and one
below it (Figure 5.2).
ITQ 7 What is the type of hybridization on the C atom in HCN?
Chapter 5 Shapes of covalent molecules
as CCl4, for example, the geometry is such that individual
bond dipoles cancel each other out. In other cases, the
dipoles do not cancel, resulting in a polar molecule. The
arrangement of non-bonding electron pairs on the central
atom also contributes to the polarity of a molecule since
these contribute to the shape of the molecule.
Figure 5.2 The hybridization of p orbitals in benzene and the
resulting molecular shape.
From this we see that there are no separate double bonds.
This explains the lack of ethene-like reactivity. The carbon
atoms in the molecule form a symmetrical six-membered
ring.
Molecular polarity
In Chapter 4 we saw how bonds between atoms can be
polar or non-polar depending on the electronegativities
of the atoms involved. The greater the electronegativity
difference between the atoms, the more polar the bond.
A polar bond acts as an electric dipole with a negative (δ−)
region or ‘pole’ and a positive (δ+) pole separated by a
distance d. The more electronegative atom in the bond has
the partial negative charge and the bond dipole is directed
towards this end, shown on a diagram by the arrowhead.
The crossed tail of the arrow is the partial positive pole.
+
–
A
B
d
The dipole moment is a measure of the polarity of a
molecule.
dipole moment = charge × distance between charges
μ=q×d
Dipole moments are determined from experiments. A polar
molecule has a dipole moment greater than zero and a
non-polar molecule has a dipole moment equal to zero. The
greater the dipole moment, the more polar is the molecule.
A bond dipole is a vector quantity, which means that it has
a magnitude and a direction. Since a molecule consists of
one or more bonds, the overall polarity of a molecule will
be the resultant of the vector combination of the dipole(s)
of its individual bond(s).
■ A molecule that consists of only non-polar bonds and
has no non-bonding electrons will be non-polar.
■ Molecules containing polar bonds can be either polar
or non-polar.
Molecules containing polar bonds can be non-polar
because of the shape of the molecule. For molecules such
In order to predict the polarity of a molecule, we must know
how its polar bonds and non-bonding electrons on the
central atom are arranged. In other words, the molecular
geometry of the molecule must first be determined.
Diatomic molecules
Diatomic molecules can be either homonuclear molecules
(A2) or heteronuclear molecules (AB).
Diatomic molecules consisting of atoms of only one element
(homonuclear), for example H2, Cl2, O2, N2, are linear
and non-polar. In these molecules the covalent bond is
non-polar since the atoms have identical electronegativity.
Diatomic molecules containing two different atoms, for
example CO, HF, NO, are polar. For carbon monoxide, CO,
the measured dipole moment μ = 0.1 D. (The unit used
here, the debye, symbol D, is a measure of the electric
dipole moment.)
Triatomic molecules
Let us consider a linear molecule of general formula AB2, in
which A is the central atom and B is more electronegative
than A. Each A–B bond is polar, with A having a partial
positive (δ+) charge and B having a partial negative charge
(δ−), The bond dipole is written with the arrowhead
pointing towards the negative end, B. Each bond dipole has
a magnitude and a direction. In this linear arrangement, the
dipoles are exactly equal in magnitude but are opposite in
direction. They therefore cancel, resulting in the molecule
being non-polar. The dipole moment is zero. Examples of
molecules of this type are BeH2 and CO2.
–
+
–
B
A
B
In the CO2 molecule, the C=O bond is polar. The CO2
molecule, however, is non-polar since the dipoles are
exactly equal in magnitude but exactly opposite in direction
(one dipole points to the left and the other to the right).
–
+
–
O
C
O
■ The electronegativity values are O = 3.5 and C = 2.5
■ The bond polarity is 3.5 − 2.5 = 1.0.
■ The dipole moment is 1.0 + (−1.0) = 0.0
49
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Unit 1 Module 1 Fundamentals in chemistry
Trigonal planar geometry
1.57 D
An example of trigonal planar geometry is BF3.
F
–
F
F
120˚
–
+
B
1.87 D
1.01 D
F
B
F
CH3Cl
B
F
CHCl 3
Guidelines
F
–
CH2Cl2
F
Here are some guidelines on how to predict the polarity of
simple covalent molecules.
F
Each B–F bond is polar; the electronegativity difference
between B and F is 2.0. These three bond dipoles are
symmetrical and cancel each other out, resulting in the BF3
molecule being non polar (μ = 0).
Tetrahedral geometry
NH3 is not trigonal planar because there are three bonded
atoms and one lone pair of electrons on the central atom.
NH3 is trigonal pyramidal when you consider the N–H
bonds. However, when you consider the lone pair too, its
overall structure is tetrahedral.
The nitrogen atom is more electronegative that each
hydrogen atom, so the dipoles point toward the nitrogen
and partially reinforce each other. The NH3 molecule is not
symmetrical: the nitrogen centre is more electron rich and
the dipoles directions do not cancel.
1 Based on the electronegativity difference between the
atoms in the bonds, determine whether polar or nonpolar bonds are present.
2 Use the VSEPR theory to predict the shape of the
molecule.
3 Based the molecular shape of the molecule, determine
whether or not the bond dipoles cancel.
4 If there are lone pairs on the central atom, determine
whether these are arranged so that they cancel.
5 If there are no polar bonds or lone pairs present then
the molecule is non-polar.
These guidelines are shown in Figure 5.3 as a flow diagram.
Are there polar bonds present?
Are the polar
bonds arranged so
that they cancel?
N
Are there lone pairs
on the central atom?
N
H
H
H
1.47 D
H
H
H
Molecule
is polar
Tetrahedral molecules with no lone pairs are non-polar
since their bond dipoles cancel. Examples include CCl4,
CH4 and SiCl4.
C
Cl
109.5˚
+
–
Cl
Cl
Cl
C
Cl
Molecule is
non-polar
Figure 5.3 Flow chart for determining whether a molecule is polar
or non-polar.
Two common misconceptions
–
Cl
Are these lone pairs
arranged so that
they cancel?
Cl
–
Cl
–
Some tetrahedral molecules have more than one type of
atom bonded to the central atom. Molecules of this type
are polar since their bond dipoles are not equal and so
do not cancel. Good examples of this type of molecular
arrangement are CH3Cl, CH2Cl2 and CHCl3.
We will conclude with two common misconceptions about
molecular geometry and polarity.
■ Molecules with similar chemical formulae must have
the same geometry. This is not necessarily true. You
cannot predict the shape of a molecule just by looking
at its formula. For example, CO2 is linear but SO2 is
angular (V shaped).
■ Molecules that have polar bonds are must be polar.
This is also not necessarily true. You need to consider
the geometry of the molecule.
Chapter 5 Shapes of covalent molecules
Summary
Review questions
1
✓ Lewis structures indicate the arrangement of
Show that the Lewis structures of N2, O2 and NH4+ are
as follows:
electrons in a molecule.
✓ The VSEPR (valence shell electron pair
N
repulsion) model can be used to predict the
approximate shape of molecules.
O
H
N
H
For each of the following molecules and ions, write
the Lewis diagram and then use valence shell electron
pair repulsion theory (VSEPR) to predict its shape.
(a) XeO3
(b) IO4−
(c) TeCl4
(d) BrF3
(e) CH2Cl2
(f) SnCl62−
3
NO2+, NO2 and NO2− have O–N–O bond angles
of 180°, 134° and 115°, respectively. Provide an
explanation for this variation in bond angles.
4
Arrange the following molecules and ions in order of
increasing bond angles:
CH3+ NF3 NH4+ XeF4
5
Explain why the PF3 molecule has a dipole moment of
1.03 D but the BF3 molecule has a dipole moment of
0 D.
6
(a) Draw Lewis electron dot representations for the
following molecules and ions:
AlBr4− PCl6− BrF5 NH2Cl2+
(b) Predict the shapes of each of the above molecules
and ions using the VSEPR theory. Draw diagrams,
give names of the shapes and give estimates of
bond angles.
(c) Predict, based on bond dipoles and shapes,
whether each of the above molecules and ions
would have a net dipole moment.
7
Which of the following molecules would you expect
to have a net molecular dipole (polar)? Show your
reasoning.
SiCl4 SiHCl3 SiH2Cl2 SiH3Cl SiH4
8
Discuss in terms of bond polarities, why NaOH is basic
and ClOH is acidic.
different shapes to form lower energy systems.
✓ Two structures resonate if they differ only in the
arrangement of their electrons.
✓ A predicted shape may be modified if resonance
is possible within the molecule.
polarity of bonds within it.
O
2
✓ Atomic orbitals can reconfigure (hybridize) into
✓ The shape of a molecule is also affected by the
N
H
✓ VSEPR uses repulsions between electron pairs to
predict the molecular shape.
+
H
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Unit 1 Module 1 Fundamentals in chemistry
Answers to ITQs
Answers to Review questions
1
2
(a)
(b)
(c)
(d)
(e)
(f)
3
NO2+ has 16 valence electrons, two bond pairs and no
lone pairs so the ion is linear with O–N–O bond angle
of 180°.
(a) There are two lone pairs on the Sn atom.
(b) There are three lone pairs on the Br atom.
2
2–
O
O
S
O
O
3
A xenon atom has 8 valence electrons: 2 of these
are used forming bond pairs with the fluorine atoms
so 3 lone pairs are left. These repel each other more
strongly than they repel the bonding electrons (i.e.
they occupy more space) and so the bonding electrons
are forced away from them. The final structure is
linear.
F
Xe
NO2 has 17 valence electrons. Central N atom has a
lone electron and two bond pairs. Shape based on
the trigonal planar geometry. The single non-bonded
electron (orbital half filled) takes up less space so the
O–N–O bond opens outward beyond the ideal 120°.
NO2− has 18 valence electrons. Central N atom has a
lone pair of electrons and two bond pairs. Shape based
on the trigonal planar geometry. The non-bonded
electron pair takes up more space so the O–N–O bond
is pushed inward reducing the bond angle from its
ideal value of 120° to 115°.
F
4
(a) ClF3 is T shaped
(b) NHCl3+ is tetrahedral
(c) XeF4 is square planar
5
ClF3
6
sp2 for both
7
sp
XeO3 is trigonal pyramidal
IO4− is tetrahedral
TeCl4 is distorted tetrahedral
BrF3 is T shaped
CHCl2 is tetrahedral
SnCl62− is octahedral
4
XeF4 NF3 NH4+ CH3+
5
PF3 is trigonal pyramidal and hence is polar. BF3 is
trigonal planar and hence is non-polar.
6
AlBr4−, tetrahedral, non-polar
PCl6−, octahedral, non-polar
BrF5, square pyramidal, polar
NH2Cl2+, tetrahedral, polar
7
All the silicon compounds given are tetrahedral.
SiHCl3, SiH2Cl2 and SiH3Cl would be expected to be
polar and SiH4 and SiCl4 would be expected to be
non-polar.
8
Hint: the Na–O bond in NaOH is more polar than the
O–H bond. It is the more polar bond that is likely to
break when the ions form. This means that the OH−
ion is likely to form.
In ClOH the O–H bond is more polar than the Cl–O
bond. This means that the O–H bond is likely to break
when the ions form and so the H+ ion is likely to form.
53
Chapter 6
An introduction to the mole
Learning objectives
■ Define the mole and the term ‘molar mass’.
■ Use relative atomic mass to calculate the relative molecular mass.
■ Write balanced molecular and ionic equations.
■ Interconvert mass, moles and number of particles.
■ Understand the concept of the limiting reagent.
■ Calculate empirical and molecular formulae.
■ Calculate mass and molar concentrations of solutions from theoretical and experimental examples.
Relative atomic mass of elements, Ar
Matter may be described as being made up of entities
which may be atoms, combinations of atoms (molecules)
or electrically charged species (ions). The masses of these
entities are infinitesimally small, too small to be measured
by even the most precise balance. As a result of these small
masses, scientists have devised a system of measure called
the relative atomic mass scale, in which the mass of an
atom is expressed relative to a standard. The standard used
is the carbon-12 isotope, 126 C, which is assigned a mass of
exactly 12.00 atomic mass units (amu).
The relative atomic mass of an element, abbreviated to Ar,
is defined as the ratio of the mass of an atom compared to
1
12 of the mass of an atom of carbon-12.
Relative atomic mass is given by this formula:
Ar =
mass of 1 atom of the element
1
12 the
mass of 1 atom of carbon-12
This can be re-arranged to give this formula:
Ar =
mass of 1 atom of the element
mass of 1 atom of carbon-12
×12
As Ar is a ratio of two masses with the same units the
units cancel out. This means that Ar has no units (it is
dimensionless).
Worked example 6.1
Q
A
Calculate the Ar of hydrogen, oxygen and magnesium, given
that their respective actual atomic masses are as follows:
(a) hydrogen = 1.67 × 10−27 kg
(b) oxygen = 2.66 × 10−26 kg
(c) magnesium = 4.00 × 10−26 kg
To put an atom’s mass in context, 1 atom of carbon-12 weighs
2.00 × 10−26 kg.
1.67 × 10−27
× 12 = 1.002
2.00 × 10−26
2.66 × 10−26
(b) oxygen: A r =
× 12 = 15.96
2.00 × 10−26
(a) hydrogen: A r =
4.00 × 10−26
× 12 = 24.00
2.00 × 10−26
The values for Ar are very close to the mass numbers for the
atoms.
(c) magnesium: A r =
Relative formula mass and relative
molecular mass of compounds
Compounds can be ionic or covalent (see page 35). Ionic
compounds are made up of formula units while covalent
compounds are made up of molecules.
The masses of formula units in ionic compounds and of
molecules in covalent compounds are also compared using
the carbon-12 isotope as the standard.
54
Unit 1 Module 1 Fundamentals in chemistry
The relative formula mass is the mass of one formula
1
unit of the compound compared to 12 of the mass of one
atom of carbon-12.
The relative molecular mass is the mass of one molecule
1
of the compound compared to 12 of the mass of one atom
of carbon-12.
You will find that chemists in real life often refer to formula
units as molecules – this is not precise, but does avoid having
to worry about what type of compound you are talking about
when working out its formula mass or molecular mass. As a
result, the term ‘relative molecular mass’ (abbreviation Mr)
is commonly used to describe the mass of one formula unit
as well as the mass of one molecule.
To determine the Mr of a compound, you need to follow
these four steps.
1 Write the chemical formula of the compound.
2 Identify the atom or atoms present.
3 Multiply the Ar of each atom by the total number of
each atom present.
of these entities. It doesn’t matter if you are talking about
atoms, molecules, ions or even electrons. As with Ar and
Mr, the mole is also related to the carbon-12 isotope as a
standard. A mole is defined as the amount of substance
that contains as many entities as there are atoms in 12 g of
carbon-12. So, let us now calculate how many atoms there
are in 12 g of carbon-12.
1 atom of carbon-12 weighs 2.00 × 10−26 kg (2.00 × 10−23 g)
1
therefore 12 g of carbon-12 =
×12 atoms,
2.00×10−23
which is approximately 6.0 × 1023 atoms of carbon-12.
(
)
This value of 6.0 × 1023 is the number of entities in one mole
of substance and is referred to as the Avogadro constant.
In summary, 1 mole of carbon-12 contains 6.0 × 1023 atoms
and has a mass of 12 g.
Avogadro’s constant is the number of particles in 1 mole
of that particle. The value of the Avogadro constant is
6.022 × 1023.
4 Add up the numbers obtained in step 3.
Worked example 6.3
Values of Ar are often rounded to the nearest whole number.
An exception is chlorine, which is taken as 35.5. For more
accurate work, values given to 4 significant figures can be
used, as given in the data booklet for the CAPE course.
Examples are Cl = 35.45, Fe = 55.85 and Cu = 63.55.
Q
Use the data from Worked example 6.1 (page 53) to calculate
the mass of 6.0 × 1023 atoms (in g) for
(a) hydrogen;
(b) oxygen;
(c) magnesium.
A
(a) hydrogen: 1.67 × 10−24 × 6.0 × 1023 = 1.002 g
(b) oxygen: 2.66 × 10−23 × 6.0 × 1023 = 15.96 g
(c) magnesium: 4.00 × 10−23 × 6.0 × 1023 = 24.00 g
Worked example 6.2
Q
Calculate the Mr of C6H12O6.
A
C: 6 × 12 = 72
H: 12 × 1 = 12
O: 6 × 16 = 96
Total = 180
The relative molecular mass of C6H12O6 is 180. This is often
abbreviated as: Mr [C6H12O6] = 180.
The mole
We use special words to mean particular numbers. For
example:
pair = 2
score = 20
mole = 6.0 × 1023
decade = 10 (years)
century = 100
dozen = 12
gross = 144
Chemists have defined the mole (symbol ‘mol’) to refer
to a specific number of entities, regardless of the nature
Molar mass
The mass in grams of 1 mole of a substance is called its
molar mass (M). Molar mass has units of g mol−1. So,
for example, 1 mole of oxygen atoms (O) contains
6.0 × 1023 oxygen atoms and has a mass of 16 g. Thus, the
Ar of oxygen is 16 and the molar mass is 16 g mol−1. Note
that relative atomic mass has no units but that molar mass
has units of g mol−1. The mass of 6.0 × 1023 atoms (1 mole)
of an element expressed in grams is numerically equal to
the Ar of the element. Although the relative atomic mass of
an element and its molar mass are numerically equal, the
terms should not be used interchangeably.
Table 6.1 In summary …
Carbon
Oxygen
Magnesium
A r = 12
A r = 16
A r = 24
M = 12 g mol−1
M = 16 g mol−1
M = 24 g mol−1
contains 6.0 ×1023 atoms contains 6.0 ×1023 atoms contains 6.0 ×1023 atoms
Chapter 6 An introduction to the mole
Here are some examples of everyday reactions:
The physical state of each reactant or product is often given
(in parentheses) next to the formula; these are called state
symbols. If we add state symbols to equation 6.2, it becomes:
■ brushing your teeth – sodium fluoride in the
HCl(aq) + Na2CO3(aq) → NaCl(aq) + H2O(l) + CO2(g)(6.3)
Writing chemical equations
toothpaste reacts in your mouth to rebuild enamel and
control bacteria;
■ cooking on a gas stove – burning chemicals to release
■ solid = (s)
energy;
■ liquid = (l)
■ taking antacid tablets – react with the acid in your
■ gas = (g)
stomach to reduce the acidity;
■ photosynthesis – plants use sunlight energy to produce
■ aqueous solution = (aq); this refers to a substance
dissolved in water
sugar.
Chemical reactions take place all around us and even
inside us! Some reactions are simple, whilst others may
be quite complex. During your early studies of chemistry,
you would have used a ‘word equation’ to summarize the
chemical reaction between substances.
Chemical reactions can be summarized in words. For
example, the acid/base reaction in which hydrochloric acid
reacts with sodium carbonate to produce sodium chloride
solution, water and carbon dioxide gas can be summarized
in a word equation as:
hydrochloric acid + sodium carbonate →
sodium chloride + water + carbon dioxide (6.1)
Note the following points:
■ the arrow means ‘produces’ or ‘yields’;
■ information written above or below the arrow
indicates the reaction conditions;
■ the substances on the left of the arrow are the reactants;
■ the substances on the right of the arrow are the
products.
Although word equations are useful, it is more convenient
to use the chemical symbols of the substances. When
formulae are used, the reaction is represented by a
chemical equation, which is a symbolic representation of
what actually happens in a chemical reaction.
Chemical equations fall into two categories: molecular and
ionic. We will first look at molecular equations.
Molecular equations
Replacing chemical names in equation 6.1 with formulae
gives:
HCl + Na2CO3 → NaCl + H2O + CO2
(6.2)
(b) Cl2
(c) CO2
(d) Al2O3
Equation 6.3 is written as a molecular equation in which
the complete neutral formulae for every compound are
shown in the reaction.
But look more closely at equation 6.3. For instance, how
many hydrogen atoms are on each side of the equation?
There is one hydrogen atom on the left-hand side and two
on the right-hand side. Where did this additional hydrogen
atom come from? Notice also that there are two sodium
atoms on the left-hand side and one on the right-hand side.
Do you think that something needs to be sorted out here?
During a chemical reaction, all that changes is the way in
which the atoms are joined together. The total number
of atoms in a chemical reaction remains the same. There
must be the same number of the same types of atoms on
both sides of the equation. To ensure that this happens
we create a balanced equation, where coefficients are
added as needed in front of the formulae of the reactants
and products. This results in a change in the number of
molecules in the equation.
To balance the equation for this acid/base reaction, we put
the coefficient ‘2’ before HCl and NaCl as follows:
2HCl(aq) + Na2CO3(aq) →
2NaCl(aq) + H2O(l) + CO2(g) (6.4)
Check to ensure that the equation is balanced by adding up
the total number of each type of atom on both sides of the
equation. If there is no coefficient, a 1 is implied.
2HCl(aq) + Na2CO3(aq) → 2NaCl(aq) + H2O(l) + CO2(g)
2 H atoms
2 Cl atoms
2 Na atoms
1 C atom
3 O atoms
2 H atoms
2 Cl atoms
2 C atoms
1 C atom
3 O atoms
The equation is balanced since the numbers of each type
of atom on both sides of the equation are equal.
ITQ 1 Calculate Mr for the following:
(a) NaCl
The abbreviations that can be used to indicate the states
are:
(e) H2SO4
(f) (NH4)2SO4
55
56
Unit 1 Module 1 Fundamentals in chemistry
What does the balanced molecular equation 6.4 tell us?
The equation ‘2HCl(aq) + Na2CO3(aq) → 2NaCl(aq) +
H2O(l) + CO2(g)’ therefore tells us the following:
■ 2 mol HCl react with 1 mol Na2CO3 to produce 2 mol
NaCl, 1 mol H2O and 1 mol CO2;
■ 73 g HCl react with 106 g Na2CO3 to produce 117 g
NaCl, 18 g H2O and 44 g CO2.
Chemical equations may be of three main types, namely
molecular, complete ionic and net ionic:
■ molecular equations show the complete, neutral
formulae for every compound in the reaction;
■ complete ionic equations show all of the species as
they are actually present in solution;
■ net ionic equations show only the species that actually
Ionic equations
participate in the reaction.
Reactions occurring in aqueous solution can be written to
show that ionic compounds normally dissociate and form
ions which are free in aqueous solution. For example, the
balanced molecular equation shown above as equation 6.4
can be represented by the following expanded equation:
2H+(aq) + 2Cl−(aq) + 2Na+(aq) + CO32−(aq) →
2Na+(aq) + 2Cl−(aq) + H2O(l) + CO2(g) (6.5)
In equation 6.5, the reactants and products are shown as they
are actually present in solution; this is called a complete
ionic equation. We should notice that in this complete
ionic equation, some of the ions occur unchanged on both
sides of the equation. These ions are seen highlighted in
red in equation 6.5 and are called spectator ions because
they do not participate in the reaction. They can be omitted
from the complete ionic equation, leaving us with:
2H+(aq) + CO32−(aq) → H2O(l) + CO2(g)
Summary
(6.6)
Equation 6.6 shows only the species that actually participate
in the reaction. An equation such as this is called a net
ionic equation.
It should be noted that when writing complete ionic
equations, certain substances are written in molecular
form. Such substances produce few or no ions in solution
and include:
■ metals;
■ solids and precipitates formed upon mixing aqueous
solutions of ionic compounds. e.g. AgCl(s), CaCO3(s);
■ non-electrolytes and weak electrolytes such as glucose,
water and ammonia solution;
■ gases such as H2(g), N2(g) and CO2(g).
Calculations involving the mole
When carrying out chemical reactions, what we are able
to measure is mass and volume of reactants and products.
In order to understand these reactions, and so able to
represent them in chemical equations, we need to be able
to convert these measurements to moles and back again.
So far, we have used three ways to express the amount of
substance:
■ mass
■ moles
■ number of atoms or particles.
Any one of these can be converted to the others, as shown
in the following worked examples.
Do you need a memory aid for the relationship between
moles, mass and molar mass?
In this triangular arrangement,
the horizontal line in the
triangle implies that you need
to divide whilst the vertical
line implies that you need
to multiply.
mass
moles
Three expressions result from this triangle:
(a) Mg(s) + O2(g) → MgO(s)
■ mass = moles × molar mass
(c) Al(s) + HCl(aq) → AlCl3(aq) + H2(g)
(d) Na(s) + H2O(l) → NaOH(aq) + H2(g)
(e) MnO2(s) + Al(s) → Mn(s) + Al2O3(s)
(f) Ca(OH)2(aq) + HCl(aq) → CaCl2(aq) + H2O(l)
(g) CaCO3(s) + H3PO4(aq) → Ca3(PO4)2(aq) + CO2(g) + H2O(l)
molar
mass
To find the expression
for a given term, you
‘cover up’ that term
mulitply
(i.e. pretend it’s not
there), and evaluate the remaining terms. For example,
to determine the expression for mass, you cover up ‘mass’
and what you are left with is ‘moles × molar mass’.
ITQ 2 Balance the following equations:
(b) Al(s) + N2(g) → AlN(s)
divide
■ moles =
mass
molar mass
■ molar mass =
mass
moles
Chapter 6 An introduction to the mole
Worked example 6.4: converting mass to moles
Q
Calculate the number of moles of sodium chloride formula units
present in 1.17 g of sodium chloride.
A
1 What is the conversion required? The quantity given is mass
and the quantity asked for is moles.
2 Write a statement of known facts connecting both
quantities.
The molar mass of NaCl is 58.5 g mol−1,
i.e. M [NaCl] = 58.5 g mol−1.
So, 58.5 g of NaCl contains 1 mole of NaCl formula units.
3 Perform the conversion using simple proportions.
mass of element or compound
number of moles =
molar mass of element or compound
1.17 g NaCl will contain
1.17
moles
58.5
= 0.02 moles of NaCl formula units
The concept of the limiting reagent
The mole concept may be likened to the concept of
following a cooking recipe. When making pancakes, for
example, there is a recipe for how the ingredients come
together to produce the perfect pancake. In much the same
way, in a balanced equation, there is a ‘recipe’ for how the
reactants come together to form products. Let us consider
the following pancake recipe:
3 cups flour + 2 eggs + 1 tbsp baking powder → 12 pancakes
+
+
We can see various relationships between the ingredients
and the number of pancakes:
■ 3 cups flour ≡ 12 pancakes
■ 2 eggs ≡ 12 pancakes
Worked example 6.5: converting moles to mass
■ 1 tbsp baking powder ≡ 12 pancakes
Q
What is the mass of 0.25 moles of H2SO4?
These relationships are written as equivalences, where the
‘≡’ sign means ‘is equivalent to’.
A
You are asked to convert moles to mass. You know that
M [H2SO4] = 98 g mol−1. 1 mole of H2SO4 weighs 98 g.
mass = number of moles × molar mass
So, 0.25 moles of H2SO4 will weigh (98 × 0.25) g = 24.5 g
In this recipe, there are numerical relationships between
the pancake ingredients and the number of pancakes.
For instance, if you have 3 cups of flour and enough of
everything else, you can make 12 perfect pancakes.
3 cups flour ≡ 12 pancakes
Worked example 6.6: converting moles to number
of particles
Q
What is the number of sodium atoms in 0.5 moles of the
element?
A
The conversion is from moles to particles: 1 mole of Na
contains 6.0 × 1023 Na atoms.
number of particles = Avogadro’s constant × number of moles
So, 0.5 moles of Na will contain 6.0 × 1023 × 0.5
= 3.0 × 1023 Na atoms
This equivalence relationship implies that the ‘flour to
pancake’ ratio must be 1 : 4 to produce perfect pancakes.
So, how many pancakes can you make if you have 6 cups
of flour and enough of everything else?
If 3 cups flour ≡ 12 pancakes then 6 cups flour ≡ 24 pancakes
The pancake recipe also provides us with relationships
among the ingredients themselves, for example:
3 cups flour ≡ 2 eggs
3 cups flour ≡ 1 tbsp baking powder
Worked example 6.7: converting number of
particles to moles
Q
Calculate the number of moles of magnesium that contains
1.2 × 1024 Mg atoms.
A
The conversion is from particles to moles. There are
6.0 × 1023 Mg atoms in 1 mole of Mg.
number of moles =
number of particles
Avogadro’s constant
So, 1.2 × 1024 atoms contains moles
= 2 moles of Mg
1
× 1.2 × 1024
6.0 × 1023
2 eggs ≡ 1 tbsp baking powder
ITQ 3 Balance the following molecular equations (if necessary)
and then write their net ionic equations.
(a) HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
(b) HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(l)
(c) AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
(d) Ba(NO3)2(aq) + H2SO4(aq) → BaSO4(s) + HNO3(aq)
(e) HBr(aq) + Ca(OH)2(aq) → CaBr2(aq) + H2O(l)
(f) Pb(NO3)2(aq) + LiCl(aq) → PbCl2(s) + LiNO3(aq)
(g) AgNO3(aq) + K2CrO4(aq) → Ag2CrO4(s) + KNO3(aq)
57
58
Unit 1 Module 1 Fundamentals in chemistry
How many eggs are required to go with 6 cups of flour?
■ Calculate the number of moles of each reactant.
If 3 cups flour ≡ 2 eggs then 6 cups flour ≡ 4 eggs
■ Calculate the number of moles of product that should
The pancake recipe is much like a balanced equation. For
perfect pancakes, all the ingredients must be mixed in strict
accordance with the recipe. Similarly, when a chemical
reaction goes to completion, all the reactants are consumed
or used up, provided that they are mixed in the mole ratio
as shown by the balanced equation. This ratio is referred to
as the stoichiometric ratio.
Let us consider the following situation where we set out
to make some pancakes by combining all the pancake
ingredients that are available in our kitchen. We have 6
cups of flour, 8 eggs and 5 tbsp baking powder. How many
perfect pancakes can we make?
be obtained from each reactant using the mole ratio
shown by the balanced equation.
■ The reactant that yields the lowest number of moles of
product is the limiting reagent.
Let us now cook our pancakes! Remember that we should
get 24 pancakes, but we accidentally burn two and then
one fell on the floor! So, we are now left with 21 perfect
pancakes, which is our actual yield. Our percentage yield
would be given by the formula:
percentage yield =
=
Using the equivalence relationships from our perfect
pancake recipe:
■ 3 cups flour ≡ 12 pancakes then 6 cups flour ≡ 24
pancakes, so we have enough flour for 24 pancakes
■ 2 eggs ≡ 12 pancakes then 8 eggs ≡ 48 pancakes, so we
have enough eggs for 48 pancakes
Worked example 6.8
Q
Consider the following equation:
Cu2O(s) + C(s) → Cu(s) + CO(g)
When 114.5 g of Cu2O are allowed to react with 11.5 g of C,
87.4 g of Cu are obtained.
(a) Balance the above equation.
Now determine:
(b) the limiting reagent
(c) the theoretical yield of Cu
(d) the percentage yield of Cu
A
(a) balanced equation
Cu2O(s) + C(s) → 2Cu(s) + CO(g)
mass / g
114.5
11.5
87.4
−1
molar mass / g mol
143.0
12.0
63.5
number of moles
0.80
0.96
1.38
theoretical mole ratio 1
1
2
1
(b) 0.80 mol Cu2O should produce 1.60 mol Cu
0.96 mol C should produce 1.92 mol Cu
Cu2O is the limiting reagent since it produces the lower
number of moles of product (1.60 compared to 1.92).
(c) mass = number of moles × molar mass of Cu
= 1.60 mol × 63.5 g mol−1
the theoretical yield of Cu = 101.6 g
powder ≡ 60 pancakes, so we have enough baking
powder for 60 pancakes
Unless we get more ingredients, we can make only 24
pancakes – this is our theoretical yield. The flour limits
the number of pancakes we can make. Let us now see
which ingredients will be leftover or in excess.
Since the flour limits the number of pancakes we make, the
flour is called the limiting reagent or the limiting reactant.
Observe that the amount of product obtained is controlled
by the quantity of the limiting reagent available.
In most chemical reactions, one reactant may be in excess
whilst the others are used up. The reactant that is used up
in a chemical reaction is called the limiting reagent or
the limiting reactant. The limiting reagent is the reactant
that limits the amount of product in a chemical reaction, or
in other words, the reactant which produces the smallest
yield of products.
What is the easiest way to determine the limiting reagent
in a reaction?
21 pancakes
×100
24 pancakes
= 87.5%
■ 1 tsp baking powder ≡ 12 pancakes. Then 5 tbsp baking
Flour
Eggs Baking powder
the perfect recipe: 3 cups
2
1 tbsp
6 cups
8
5 tbsp
we mixed:
based on the flour limitation, the perfect ratio would have
been:
6 cups
4
2 tbsp
excess ingredients: 0 cups
4
3 tbsp
actual yield
×100
theoretical yield
(d) percentage yield =
=
actual yield
×100
theoretical yield
87.4 g
×100
101.6 g
= 86.0%
the percentage yield of Cu = 86.0%
Chapter 6 An introduction to the mole
ITQ 4
(a) Given that Ar [Na] = 23.0, calculate the following:
(i) the mass of 5 moles of Na;
(ii) the number of moles of Na atoms in 57.5 g of the element;
(iii) the mass of 1.2 × 1025 atoms of Na;
(iv) the number of particles in 34.5 g of the element.
(b) How many molecules of CO2 are present in 880 g of the
compound?
(c) What is the mass of 3 × 1021 molecules of HNO3?
ITQ 5
(a) A technician allowed 14.4 g of calcium oxide and 13.8 g of
carbon dioxide to react. This reaction produced 19.4 g of
CaCO3. The equation is as follows:
CaO(s) + CO2(g) → CaCO3(s)
(i) Which is the limiting reagent?
(ii) What is the theoretical yield of CaCO3?
(iii) What is the percentage yield for the reaction?
(b) Ammonia can be manufactured according to the following
general equation:
NO(g) + H2(g) → NH3(g) + H2O(g)
(i) Balance the equation.
45.8 g of NO and 12.4 g of H2 are used to produce ammonia.
(ii) Which is the limiting reagent?
(iii) What is the maximum amount of ammonia that can be
produced?
(c) Ethylene glycol [C2H4(OH)2] can be prepared by the reaction of
ethylene oxide and water, as shown in the following equation:
C2H4O + H2O → C2H4(OH)2
If 200 g of ethylene oxide reacts with 90 g of water and 260
g of ethylene glycol are produced, determine the percentage
yield of ethylene glycol.
(d) Urea (CN2H4O) is prepared by the reaction of ammonia and
carbon dioxide, as shown in the following equation:
NH3 + CO2 → CN2H4O + H2O
(i) Balance the equation.
125 g of ammonia reacts with 135 g of carbon dioxide.
(ii) Work out the number of moles of each reactant.
(iii) Which is the limiting reagent?
(iv) What is the yield of urea?
(e) A student spilled 50 cm3 of 1.0 M hydrochloric acid on the floor
in the lab. He attempted to neutralize the acid by adding 50 g
of solid Na2CO3 to the spill according to the following reaction:
Na2CO3(s) + HCl(aq) → NaCl(aq) + H2O(l) + CO2(g)
(i) Was the HCl completely neutralized? If not, what mass of
hydrochloric acid was neutralized?
(ii) Did the student add too much or too little Na2CO3? Justify
your answer.
Empirical and molecular formulae
The mole concept may also be used to determine the
formulae of compounds. When describing compounds,
several types of formulae can be used. Two such formulae
are the molecular formula and the empirical formula.
The molecular formula gives the actual number of atoms or
ions that are present in one molecule or one formula unit
of a compound.
The empirical formula gives the simplest whole number
ratio of atoms or ions in a compound.
The molecular and empirical formulae of a given compound
may be the same or they may be different.
■ Instances where they are the same: CO2, for example,
has the same molecular and empirical formulae, as do
NaCl and H2O.
■ Instances where they are different: the molecular
formula of glucose is C6H12O6, so its empirical formula
is CH2O. In such instances, the molecular formula
is always a whole-number multiple of the empirical
formula.
molecular formula = empirical formula × n,
where n = 1, 2, 3, …
Using the example of glucose again, C6H12O6 = CH2O × 6.
The subscript of each atom in the empirical formula CH2O
is multiplied by 6 to obtain the molecular formula C6H12O6.
Note that different compounds with different molecular
formulae may have the same empirical formula. For
example, the molecules C2H4 and C4H8 have the same
empirical formula CH2.
The empirical formula of a compound can be calculated
from either:
■ combustion data, which gives the experimentally
determined masses of the various elements which
make up a known mass of the compound, or
■ the percentage composition by mass which gives the
percentage by mass of each element in 1 mole of the
compound.
Once the empirical formula is calculated, the molecular
formula can then be found provided that the molar mass
of the compound is known. Worked example 6.9 outlines
the procedure for determining the empirical formula and
the molecular formula of a compound.
59
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Unit 1 Module 1 Fundamentals in chemistry
Worked example 6.9
Q
A 0.450 g sample of an organic acid containing the elements carbon, hydrogen and oxygen was subjected
to combustion analysis and the following data were obtained:
■ mass of C = 0.279 g
■ mass of H = 0.0467 g
■ mass of O = 0.124 g
The molar mass of the acid is 116 g mol−1. What are the empirical and molecular formulae of the acid?
A
Step 1: write down the mass of each element present in the sample of the compound.
If percentage compositions by mass are given, assume a 100 g sample and compute the masses of each
element from the given percentages.
C
H
O
0.279 g
0.0467 g
0.124 g
Step 2: convert each of the masses in Step 1 to moles.
number of moles =
C
mass
Ar
H
O
0.279
0.0467
0.124
= 0.02325 mol
= 0.0467 mol
= 0.00775 mol
12
1
16
Step 3: divide throughout by smallest number of moles.
C
H
O
0.02325 mol
=3
0.00775 mol
0.0467 mol
=6
0.00775 mol
0.00775 mol
=1
0.00775 mol
Therefore, the ratio of C : H : O in the acid is 3 : 6 : 1 and so the empirical formula is C3H6O.
The empirical formula molar mass = (3 × 12) + (6 × 1) + (1 × 16) = 58 g mol−1
We are told that the molar mass of the acid is 116 g mol−1.
The number of empirical formula units in 116 g is given by the formula:
molar mass of compound
116 g mol−1
=
=2
empirical formula molar mass
58 g mol−1
There are 2 empirical formula units per mole of acid. The molecular formula is obtained by multiplying the
subscripts of each atom in the empirical formula by this value.
C3H6O × 2 → C6H12O2
The molecular formula of the acid is C6H12O2.
ITQ 6
ITQ 7
(a) The rotten smell of a decaying animal carcass may be attributed to
putresine, a compound containing carbon, hydrogen and nitrogen. The
results of an elemental analysis of putresine indicate that it contains
54.50% C, 13.73% H and 31.77% N. Calculate the empirical formula of
putresine.
(a) Work out the concentration in mol dm−3 of the
following solutions:
(b) Nicotine, the main factor responsible for the dependence-forming
properties of tobacco smoking, has the following percentage composition
by mass: 74.03% C, 8.70% H and 17.27% N. The molar mass of nicotine
is 162.23 g mol−1. Calculate the molecular formula of nicotine.
(c) Estradiol is the predominant sex hormone present in females and has a
critical impact on reproductive and sexual functioning. Estradiol has the
elemental composition: 79.37% C, 8.88% H and 11.75% N. Determine
the empirical formula of estradiol. Given that the molar mass of estradiol
is 272.37 g mol−1, calculate the molecular formula of estradiol.
(i) 25 cm3 of nitric acid which contains
2.5 × 10−3 mol HNO3
(ii) 500 mL of sodium hydroxide which contains
1.0 mol NaOH
(iii) 6.5 dm3 of sodium carbonate which contains
9.5 mol Na2CO3
(b) Calculate the number of moles of solute in the
following volumes of solution:
(i) 25 mL of 0.1 mol dm−3 HCl
(ii) 500 cm3 of 0.2 mol dm−3 H2SO4
(iii) 3 dm3 of 0.05 mol dm−3 NaOH
Chapter 6 An introduction to the mole
The mole concept applied to solutions
The mole concept can be used to measure the strength of
solutions. In everyday life, we describe the strength of a
solution as dilute, concentrated or saturated. Additionally,
the word ‘concentration’ is often used generally to refer
to the amount of a particular substance (called the solute)
dissolved in another substance (called the solvent).
However, there are a couple ways of precisely expressing
the concentration of a solution.
Worked example 6.10
Q
Find the concentration in mol dm−3 of:
(a) a solution containing 3.5 mol H2SO4 in 5.0 dm3 of solution;
(b) a solution containing 0.1 mol H2SO4 in 450 cm3 of solution;
(c) a solution containing 0.020 mol H2SO4 in 25 mL of solution.
A
(a) 5.0 dm3 of solution ≡ 3.5 mol H2SO4
3.5
1.0 dm3 of solution ≡ 5.0 mol H2SO4 = 0.7 mol H2SO4
The concentration of this H2SO4 solution is
0.7 mol dm−3.
(b) 450 cm3 of solution ≡ 0.1 mol H2SO4
0.1
1 cm3 of solution ≡ 450 mol H2SO4
0.1
1000 cm3 of solution ≡ 450 ×1000 mol H2SO4
≡ 0.22 mol H2SO4
The H2SO4 solution has a concentration of
0.22 mol dm−3.
(c) 25 mL of solution ≡ 0.020 mol H2SO4
0.020
1 mL of solution ≡ 25 mol H2SO4
0.020
1000 mL of solution ≡ 25 ×1000 H2SO4
≡ 0.8 mol H2SO4
The concentration of the solution is 0.8 mol dm−3.
■ Molar concentration or molarity, which gives the
number of moles of solute in 1 dm3 (1000 cm3) of
solution. The unit of molar concentration can be
written as mol dm−3, mol/dm3, mol/L or M. This book
will use mol dm−3 but M is still quite commonly used.
■ Mass concentration, which gives the mass (in g) of
solute dissolved in 1 dm3 of solution. The unit is
written as g dm−3 or g/dm3.
For molar concentration, we always refer to the volume of
the final solution, not to the volume of solvent that may be
added.
When dealing with the subject of volume as it relates to
concentration, we often encounter a confusion of different
units. Let us try and sort this out …
1 dm = 10 cm
If we cube both sides:
Mass concentration
We have defined mass concentration as the mass of solute
dissolved in 1 dm3 of solution:
mass of solute in g
mass concentration
=
in g dm−3
volume of solution in dm3
13 dm3 = 103 cm3
1 dm3 = 1000 cm3
= 1000 mL
=1L
We can rearrange this formula to find the mass of solute
present in a measured volume of solution once the mass
concentration of the solution is known:
You may come across four different units of volume that
are all equivalent:
mass of solute in g =
mass concentration in g dm−3 × volume of solution in dm3
1 dm3 = 1000 cm3 = 1000 mL = 1 L
Molar concentration
We have already defined molar concentration as the
number of moles of solute dissolved in 1 dm3 of solution:
molar conentration number of moles of solute
=
in mol dm−3
volume of solution in dm3
We can rearrange this formula to find the number of moles
of solute present in a measured volume of solution once
the concentration of the solution is known:
number of moles of solute =
molar concentration in mol dm−3 × volume of solution in dm3
Worked example 6.11
Q
Find the mass concentration in g dm−3 of:
(a) a solution containing 1.0 g CuSO4 in 50 cm3 of solution;
(b) a solution containing 120 g HNO3 in 3 dm3 of solution.
A
(a) 50 cm3 of solution ≡ 1.0 g CuSO4
1.0
1 cm3 of solution ≡ 50 g CuSO4
1.0
1000 cm3 of solution ≡ 50 × 1000 g CuSO4 ≡ 20 g CuSO4
The concentration of this CuSO4 solution is 20 g dm−3.
(b) 3 dm3 of solution ≡ 120 g HNO3
120
1 dm3 of solution ≡ 3 g HNO3 ≡ 40 g HNO3
The HNO3 solution has a concentration of 40 g dm−3.
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Unit 1 Module 1 Fundamentals in chemistry
The relationship between molar concentration
and mass concentration
Here is a useful memory aid for the relationship between
molar concentration, mass concentration and molar mass:
Worked example 6.14
Q
Determine the molar concentration of a potassium nitrate
solution containing 20.2 g dm−3 of KNO3.
A
Mr [KNO3] = 39 + 14 + (3 × 16) = 101
molar concentration =
mass
concentration
molar
concentration
1 mass concentration =
molar concentration × molar mass
3 molar mass =
mass concentration
molar mass
Worked example 6.15
Q
350 mL of a sodium chloride solution contains 16.38 g NaCl.
Calculate the molar concentration of this solution.
A
Firstly, determine the mass concentration of the NaCl solution:
350 mL of solution ≡ 16.38 g NaCl
16.38
1 mL of solution ≡ 350 g NaCl
16.38
1000 mL of solution ≡ 350 × 1000 g NaCl = 46.8 g NaCl
The mass concentration of this NaCl solution is 46.8 g dm−3.
Then we determine the molar concentration of the NaCl
solution:
mass concentration
molar concentration =
molar mass
mass concentration
molar concentration
We already know that number of moles of substance =
mass of substance (in g)
molar mass of substance
If we apply this relationship to solutions, specifically to
1 dm3 of solution, we can infer that:
=
number of moles of substance in 1 dm3 =
mass of substance (in g) in 1 dm3
molar mass of substance
Worked example 6.12
Q
Calculate the concentration in g dm−3 of a sodium chloride
solution whose molar concentration is 0.2 mol dm−3.
(M [NaCl] = 58.5 g mol−1)
A
mass concentration = molar concentration × molar mass
= 0.2 mol dm−3 × 58.5 g mol−1
= 11.7 g dm−3
Q
What is the mass concentration in g dm−3 of a solution of
0.04 mol dm−3 HNO3?
A
We first need to calculate the relative molecular mass of HNO3.
Mr [HNO3] = 1 + 14 + (3 × 16) = 63
mass concentration = molar concentration × molar mass
= 0.04 mol dm−3 × 63 g mol−1
= 2.52 g dm−3
46.8 g dm−3
58.5 g mol−1
= 0.8 mol dm−3
Worked example 6.16
Q
A
Given that a solution of potassium manganate(VII) has a mass
concentration of 1.58 g dm−3 and a molar concentration of
0.01 mol dm−3, calculate the mass of 1 mol of potassium
manganate(VII).
molar mass =
=
Worked example 6.13
20.2 g dm−3
101 g mol−1
= 0.2 mol dm−3
molar
mass
Three expressions result from this triangle:
2 molar concentration =
=
mass concentration
molar mass
mass concentration
molar concentration
1.58 g dm−3
0.01 mol dm−3
= 158 g mol−1
ITQ 8 Work out the concentration in g dm−3 of the following
solutions:
(a) 25 cm3 of sodium chloride solution which contains 20 g NaCl;
(b) 50 mL of sodium hydroxide solution which contains 0.04 mol
NaOH;
(c) 5 dm3 of sodium carbonate which contains 0.5 kg Na2CO3.
Chapter 6 An introduction to the mole
Worked example 6.17
Q
A
Deduce the value of x in the formula FeSO4·xH2O, given that
a solution of FeSO4·xH2O has a mass concentration of
30.58 g dm−3 and a molar concentration of 0.11 mol dm−3.
M [FeSO4·xH2O] =
mass concentration
30.58 g dm−3
=
molar concentration
0.11 mol dm−3
= 278 g mol−1
Using the relative molecular mass of 278, we can now deduce
the value of x in the formula:
Mr [FeSO4·xH2O]
= Ar [Fe] + Ar [S] + (4 × Ar [O]) + (x × Mr [H2O])
= 278
⇒ 56 + 32 + (4 × 16) + 18x = 278
⇒ 152 + 18x = 278
⇒ 18x = 278 − 152 = 126
126
=7
⇒x=
18
The formula is FeSO4·7H2O.
Titrimetric (volumetric) analysis
So far we have been dealing with theoretical means of
determining the concentrations of solutions. We have been
carrying out calculations based on the mole concept. The
concentrations of solutions may also be experimentally
determined in a common laboratory method called a
titration. Titration is the process of adding a reactant
a little at a time until the desired result is reached. The
process is sometimes described as a titrimetric analysis and,
because volume measurements play a key role in titration,
it is also known as volumetric analysis. The main aspect
of volumetric analysis involves measuring the volume of
a solution of accurately known concentration (called the
standard solution) which then reacts quantitatively with
another solution whose concentration is unknown (called
the analyte).
In a titration, the standard solution is added to a titration
flask (usually a conical flask) from a pipette. The solution
of the analyte is then added to the standard solution
from a burette until the reaction is complete. The point at
which the titration is complete is known as the end-point
or equivalence point. This point occurs when the two
solutions just react and neither is in excess. An indicator is
often added to the contents of the titration flask to detect
the end-point of the titration. The colour of the indicator
depends on the acidity of the solution. The volume of
standard solution added from the burette is called the titre.
Based on the results of the titration, the concentration of
the unknown solution can be calculated. Worked example
6.18 shows how.
During acid/base titrations, you can determine the
concentration of an acid by titrating a known volume of
the acid with a suitable base of known concentration (for
example sodium hydroxide). Similarly, the concentration of
a base may be calculated by titration with a suitable acid of
known concentration (such as hydrochloric acid or sulfuric
acid). An indicator such as methyl orange or phenolphthalein
(Table 6.2) can be used to signal the end-point of this titration
– the point in the titration where the number of moles of H+
equals the number of moles of OH−.
Table 6.2 The colour changes of some common indicators
Colour with acid
Methyl orange
Phenolphthalein
Universal
red
colourless
red
Colour with water
orange
colourless
green
Colour with base
yellow
pink
purple
Methyl orange can be ‘screened’ by adding a dye, xylene cyanol, to it. The indicator
is then red in acid, grey about pH7, and green in alkali.
In redox titrations, oxidizing agents react with reducing
agents. The oxidizing agent of choice is often potassium
manganate(VII) (KMnO4), owing to its stability, ease of
storage and the fact that the intense purple colour of its
solution provides its own indication of the end-point of the
titration.
In a redox titration to determine the concentration of a
reducing agent (e.g. Fe2+(aq) or I−(aq)) a solution of
purple acidified potassium manganate(VII) of known
concentration is added from a burette to the reducing
agent in a titration flask until the first faint appearance
of a persisting pink colour. This pink colour shows that
there is a small excess of KMnO4 and hence indicates the
end-point of the titration. Some other oxidizing agents
include potassium dichromate(VI) and iodide thiosulfate (a
ITQ 9
(a) A solution of QCl2 has a mass concentration of
0.475 g dm−3 and a molar concentration of 0.005 mol dm−3.
Use this information to determine the relative atomic mass of
Q. Then, by using the periodic table, identify element Q.
(b) A solution of FeSO4·xH2O was made up by dissolving
55.6 g of solute in 2 dm3 of solution. The molar concentration
of the solution is 0.1 mol dm−3. Determine the value of x in the
formula FeSO4·xH2O.
(c) A solution of hydrated barium chloride, BaCl2·xH2O, has a
molar concentration of 0.2 mol dm−3. 40.0 cm3 of this solution
contains 1.952 g of the salt. Calculate:
(i) the mass concentration of the solution;
(ii) the value of x in the formula BaCl2·xH2O.
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Unit 1 Module 1 Fundamentals in chemistry
combination of potassium iodide and sodium thiosulfate).
Hydrogen peroxide can act either as an oxidizing agent or
a reducing agent, depending on the conditions.
Worked example 6.18
Q
A standardized solution of sodium carbonate is made by
dissolving a 1.49 g sample of sodium carbonate in distilled
water and making up to 250 cm3 of solution. Three
25.0 cm3 aliquots of this solution are pipetted and titrated
against a solution of sulfuric acid of unknown concentration
using screened methyl orange as the indicator. The average
volume of sulfuric acid used for the titration is found to be
24.65 cm3.
(a) Calculate the number of moles of sodium carbonate used
for the titration, if the concentration of the stock solution is
5.62 × 10−2 mol dm−3.
(b) Calculate the accurate concentration of the standardized
solution of sulfuric acid in mol dm−3.
A
(a) The concentration of the standardized solution of sodium
carbonate = 5.62 × 10−2 mol dm−3
Standard solutions
Standard solutions can be prepared in two ways:
■ by dissolving an accurately measured mass of solute in
a known volume of solvent;
■ by diluting a more concentrated solution.
Standard solutions should have the following properties:
■ high purity – of at least 99.8%;
■ stability towards air – the composition must not
change when exposed to air and so should not be
oxidized by air, absorb water vapour, or react with
carbon dioxide;
■ absence of water of crystallization – it should be stable
to drying, i.e. the composition should not alter when
heated;
■ relatively large formula weight – this minimizes
relative errors associated with weighing;
■ it should be soluble in common titration media.
The mole concept applied to gases
The amount of a gas is usually measured by its volume.
Amadeo Avogadro, an Italian physicist, suggested in 1811
that ‘equal volumes of ideal gases at the same temperature
and pressure contain the same number of particles.’ This is
now called Avogadro’s law.
The volume of a gas which contains one mole of gas under
standard conditions of temperature and pressure is called
its molar volume. The molar volume of gas contains the
Avogadro number of particles (see page 54). Avogadro’s
number is 6.022 × 1023.
1000 cm3 of solution ≡ 5.62 × 10−2 mol Na2CO3
5.62 × 10−2
mol Na2CO3
1000
5.62 × 10−2
≡
× 25 mol Na2CO3
1000
1 cm3 of solution ≡
25 cm3 of solution
≡ 1.41 × 10−3 mol Na2CO3
(b) This is an acid/base titration, so we need a balanced
molecular equation for the reaction:
H2SO4(aq) + Na2CO3(aq) → Na2SO4(aq) + H2O(l) + CO2(g)
Based on the equation, 1 mol of H2SO4 reacts with 1 mol of
Na2CO3. From part (a), we calculated that 1.41 × 10−3 mol
Na2CO3 were used for the titration.
∴1.41 × 10−3 mol Na2CO3 ≡ 1.41 × 10−3 mol H2SO4
1.41 × 10−3 mol H2SO4 were contained in the average titre
of 24.65 cm3.
Thus, 24.65 cm3 of solution ≡ 1.41 × 10−3 mol H2SO4.
1.41 × 10−3
mol H2SO4
24.65
1.41 × 10−3
solution ≡ 24.65 × 1000
1 cm3 of solution ≡
1000 cm3 of
mol H2SO4
≡ 0.0572 mol H2SO4
The concentration of the sulfuric acid is 0.0572 mol dm−3.
The volume of a mass of gas is influenced by both
temperature and pressure and so the molar volume of a
gas is usually quoted under one of two sets of conditions.
Through this chapter we have been accumulating ways
of measuring chemicals. So, in summary, we can say the
following about 1 mole of nitrogen gas, N2:
■ Standard temperature and pressure (s.t.p.) is a
■ contains 2 mol of nitrogen atoms;
temperature of 273.15 K (0 °C) and a pressure of
■ has a mass of (2 × 14) = 28 g;
101 kPa (1 atm). One mole of a gas at s.t.p. occupies
22.4 dm3 (22 400 cm3).
■ contains 6.0 × 1023 N2 molecules;
■ Room temperature and pressure (r.t.p.) is taken to be a
■ occupies a volume of 22.4 dm3 at s.t.p.;
■ occupies a volume of 24 dm3 at r.t.p.
temperature of 298 K (25 °C) and a pressure of
101 kPa (1 atm). One mole of a gas at r.t.p. occupies
24 dm3 (24 000 cm3).
ITQ 10 20.0 cm3 of a sulfuric acid solution required 22.87 cm3 of
0.158 mol dm−3 KOH for complete neutralization. Calculate the
molar concentration of the H2SO4 in mol dm−3.
Chapter 6 An introduction to the mole
Worked example 6.19
Worked example 6.23
Q
What volume is occupied by 8 g of oxygen (O2) at s.t.p.?
Q
A
1 mol of oxygen has a mass of (2 × 16) = 32 g
32 g of oxygen occupy 22.4 dm3 at s.t.p.
22.4
∴ 8 g of oxygen occupy 32 × 8 dm3 at s.t.p. = 5.6 dm3
Calculate the volume of gaseous product formed when
1.35 moles of butane undergoes complete combustion at room
temperature and pressure.
A
Avogadro’s law applies only to gases.
The balanced equation for the complete combustion of butane:
2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l)
2 mol C4H10 reacts with 13 mol O2 to produce 8 mol CO2.
2 mol C4H10 ≡ 8 mol CO2
8
1.35 mol C4H10 ≡ 2 × 1.35 ≡ 5.4 mol CO2
1 mol CO2 ≡ 24.0 dm3 at r.t.p.
5.4 mol ≡ 24.0 × 5.4 = 129.6 dm3 CO2
Worked example 6.20
Q
How many molecules are present in 4.2 dm3 of carbon dioxide
at s.t.p.?
A
1 mol of carbon dioxide occupies 22.4 dm3 at s.t.p. and
contains 6.0 × 1023 molecules.
6.0 × 1023
therefore 4.2 dm3 ≡ 22.4 × 4.2 = 1.125 × 1023 molecules
Worked example 6.21
Q
Calculate the molar mass of gas X provided that 0.367 g of the
gas occupies 200 cm3 at r.t.p.
A
1 mol of gas X occupies 24 000 cm3 at r.t.p. and 200 cm3 of
gas X weighs 0.367 g.
24 000 cm3 weighs 0.367 ×
24 000
200
cm3 ≡ 44 g
The molar mass of gas X is 44 g mol−1.
Worked example 6.22
Q
Calculate the volume of oxygen required for the complete
combustion of 100 cm3 of propane. Also find the volume of
carbon dioxide produced.
A
The balanced equation for the complete combustion of propane:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l)
From the balanced equation we can deduce that 1 mol C3H8
reacts with 5 mol O2 to produce 3 mol CO2. So, by Avogadro’s
law, 1 volume of C3H8 reacts with 5 volumes of O2 to produce
3 volumes of CO2. Hence, 100 cm3 of C3H8 will react with
500 cm3 of O2 to produce 300 cm3 of CO2 at the same
temperature and pressure.
Worked example 6.24
Q
60 cm3 of oxygen is required for the complete combustion
of 10 cm3 of a hydrocarbon; 40 cm3 of carbon dioxide is
produced. Determine the formula of the hydrocarbon and hence
deduce the balanced equation for the reaction. All volumes are
measured at the same temperature and pressure.
A
Let CxHy represent the unknown hydrocarbon.
10 cm3 of CxHy requires 60 cm3 of O2 to produce 40 cm3 of
CO2. Using Avogadro’s law, 10 molecules of CxHy reacts with
60 molecules of O2 to produce 40 molecules of CO2. Hence,
1 molecule of CxHy reacts with 6 molecules of O2 to produce
4 molecules of CO2.
CxHy(g) + 6O2(g) → 4CO2(g) + ?H2O(l)
The number of H atoms can be determined indirectly from the
number of O atoms. There are a total of 12 O atoms on the
left-hand side of the equation and so there should also be 12
on the right-hand side. The 4CO2 account for 8 such O atoms,
and so that leaves 4 O atoms to be in the H2O. Therefore, there
must be 4H2O.
CxHy(g) + 6O2(g) → 4CO2(g) + 4H2O(l)
For the equation to be balanced, x = 4 and y = 8. The
hydrocarbon is C4H8. The balanced equation is:
C4H8(g) + 6O2(g) → 4CO2(g) + 4H2O(l)
65
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Unit 1 Module 1 Fundamentals in chemistry
Summary
✓ Masses of atoms and molecules are measured
relative to the mass of an atom of 12C, the mass
of which is taken as 12.
✓ A mole (mol) of a substance contains 6.022 × 1023
particles of that substance. This number is called
the Avogadro constant (L).
✓ To be valid, chemical equations must contain
the same number of atoms of each individual
element on both sides.
✓ Ionic equations must show the same total
electrical charge on both sides.
✓ The empirical formula of a compound shows the
elements present and the simplest ratio between
the atoms of each.
✓ The molecular formula is a simple multiple of
the empirical formula (the multiplier can be 1).
✓ The concentration of a solution is quoted in
–3)
moles per litre (mol dm
(g dm–3).
or in grams per litre
✓ Indicators are substances which change colour
according to the acidity (pH) of their solution.
✓ A standard solution is one with an accurately
known concentration that is unlikely to change
with time.
Review questions
1 What mass of oxygen contains the same number of
atoms as 48 g of carbon?
2 What is the percentage by mass of magnesium in
magnesium sulfate?
3 How many moles of substance do each of the
following represent?
(a) 23 g of sodium metal
(b) 35.5 g of chlorine gas
(c) 132 g of ammonium sulfate
(d) 160 g of iron(III) oxide
4 What volume of a 0.15 mol dm−3 HCl solution is
required to obtain 0.006 moles of the solute?
5 200 g of sodium hydroxide is reacted with 300 g of
sulfuric acid.
(a) Write a balanced equation for the reaction
occurring.
(b) Calculate the number of moles of each reactant to
1 d.p.
(c) Calculate the limiting reagent.
(d) Calculate the mass of salt formed.
6 The following reaction is used to obtain iron from iron
ore:
Fe2O3(s) + CO(g) → Fe(s) + CO2(g)
(a) Balance the equation.
(b) The reaction of 185 g of Fe2O3 with 93.5 g of CO
produces 87.4 g of Fe. Find the percentage yield of
iron.
7 The titration of a 25.0 cm3 of a H2SO4 solution of
unknown concentration requires 40.0 cm3 of a
0.035 mol dm−3 KOH solution to reach the end-point.
Calculate the molar concentration of the H2SO4 in
mol dm−3.
8 During the complete combustion of an unknown
hydrocarbon, 3.52 g of carbon dioxide and 1.62 g of
water are collected.
(a) Calculate the mass of:
(i) carbon in 3.52 g of carbon dioxide;
(ii) hydrogen in 1.62 g of water.
(b) Use your answers obtained in part (a) to calculate
the empirical formula of the hydrocarbon.
9 A compound X whose molar mass is 46 g mol−1
contains the elements carbon, hydrogen and oxygen.
When 0.544 g of the compound was burnt in oxygen,
0.637 g of water and 1.039 g of carbon dioxide
were produced. Determine both the empirical and
molecular formula of compound X.
10 Mass spectrometric analysis of a compound Q
reveals a relative molecular mass of 108. Elemental
analysis shows the presence of the elements carbon,
hydrogen and one other element. When 1.08 g of
this compound is completely burnt in oxygen, the
products contain 1340 cm3 of carbon dioxide and
448 cm3 of nitrogen dioxide, measured at s.t.p. Find
the molecular formula of compound Q.
Chapter 6 An introduction to the mole
Answers to ITQs
1 (a)
(b)
(c)
(d)
(e)
(f)
Mr [NaCl] = 58.5
Mr [Cl2] = 71
Mr [CO2] = 44
Mr [Al2O3] = 102
Mr [H2SO4] = 98
Mr [(NH4)2SO4] = 132
2 (a)
(b)
(c)
(d)
(e)
(f)
(g)
2Mg(s) + O2(g) → 2MgO(s)
2Al(s) + N2(g) → 2AlN(s)
2Al(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2(g)
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
3MnO2(s) + 4Al(s) → 3Mn(s) + 2Al2O3(s)
Ca(OH)2(aq) + 2HCl(aq) → CaCl2(aq) + 2H2O(l)
3CaCO3(s) + 2H3PO4(aq) →
Ca3(PO4)2(aq) + 3CO2(g) + 3H2O(l)
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
H+(aq) + OH−(aq) → H2O(l)
HNO3(aq) + NaOH(aq) → NaNO3(aq) + H2O(l)
H+(aq) + OH−(aq) → H2O(l)
Ba(NO3)2(aq) + H2SO4(aq) →
BaSO4(s) + 2HNO3(aq)
Ba2+(aq) + SO42−(aq) → BaSO4(s)
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
Ag+(aq) + Cl−(aq) → AgCl(s)
2HBr(aq) + Ca(OH)2(aq) → CaBr2(aq) + 2H2O(l)
H+(aq) + OH−(aq) → H2O(l)
Pb(NO3)2(aq) + 2LiCl(aq) → PbCl2(s) + 2LiNO3(aq)
Pb2+(aq) + 2Cl−(aq) → PbCl2(s)
2AgNO3(aq) + K2CrO4(aq) →
Ag2CrO4(s) + 2KNO3(aq)
2−
+
2Ag (aq) + CrO4 (aq) → Ag2CrO4(s)
3 (a)
(b)
(c)
(d)
(e)
(f)
(g)
4 (a) (i) 115 g
(ii) 2.5 moles
(iii) 460 g
(iv) 9 × 1023 particles
(b) 1.2 × 1025 molecules
(c) 0.315 g
5 (a) (i) CaO
(ii) 26 g
(iii) 74.6%
(b) (i) 2NO(g) + 5H2(g) → 2NH3(g) + 2H2O(g)
(ii) NO
(iii) 26.01 g
(c) 92.2%
(d) (i) 2NH3 + CO2 → CN2H4O + H2O
(ii) 7.35 mol NH3, 3.07 mol CO2
(iii) CO2
(iv) 184.2 g
(e) The HCl was completely neutralized
The student spilled 50/1000 mol of HCl. He used
50/106 mol of Na2Co3, which is a large excess.
6 (a) C2H6N
(b) empirical formula, C3H6O;
molecular formula, C3H6O
(c) C10H14N2
7 (a) (i) 0.1 mol dm−3
(ii) 2.0 mol dm−3
(iii) 1.5 mol dm−3
(b) (i) 0.0025 mol
(ii) 0.1 mol
(iii) 0.15 mol
8 (a) 800 g dm−3
(b) 32 g dm−3
(c) 100 g dm−3
9 (a) magnesium, Mg
(b) x = 7
(c) (i) 48.8 g dm−3
(ii) 2
10 9.03 × 10−2 mol dm−3 H2SO4
Answers to Review questions
1 64 g
2 20%
3 (a)
(b)
(c)
(d)
1 mole of sodium metal (Na)
0.5 mol of chlorine gas (Cl2)
1 mol of (NH4)2SO4(s)
1 mol of Fe2O3
4 40 mL
5 (a) 5.0 mol of NaOH, 3.1 mol H2SO4
(b) NaOH is the limiting reagent
(c) 355 g
6 70%
7 0.028 mol dm−3
8 C8H18
9 C2H6O
10 C6H8N2
67
68
Chapter 7
Gases
Learning objectives
■ Interconvert units of pressure.
■ Outline how the pressure of a gas is determined using a manometer.
■ Use the ideal gas law to calculate pressure, volume, moles of gas or temperature,
given the other three variables.
■ Perform stoichiometric calculations relating the mass of a reactant to the mass,
moles, volume or pressure of a gaseous product.
■ Use the ideal gas law to calculate the molar mass of a gas.
■ Use the kinetic molecular theory of gases to explain each of the gas laws.
■ Explain the difference between real and ideal gases.
Behaviour of gases
Matter exists in three different states:
about fixed positions. Liquids have moderate disorder and
molecules are relatively free to move. Gases have extreme
disorder and molecules have almost complete freedom of
motion and are randomly arranged.
■ solids
■ liquids
■ gases.
In solids – atoms, molecules and ions are held rigidly
together resulting in a specific shape and volume. Solids
keep their own volume and shape.
In liquids – atoms, molecules and ions are held together
less strongly than solids. This gives liquids a specific volume
but indefinite shape. They have their own volume but take
the shape of the container.
In gases – atoms and molecules have little attraction for
each other and are free to move about in any volume or
space. As a result, gases have no specific shape or volume.
A gas takes both the volume and the shape of its container.
These states of matter are shown in Figure 7.1. Solids have
an ordered arrangement where the molecules can vibrate
Gases are homogenous: they mix thoroughly because
their constituent particles are free to move about. Gases
are also compressible: the constituent particles (atoms and
molecules) are far apart. Generally the particles occupy less
than 0.1% of the total volume of the container, so when
pressure is applied they are able to move closer together.
Pressure
Gases exert a measurable pressure on the walls of the
container they occupy. When a gas particle bounces off a
container wall its direction of travel has changed. Therefore
a force has acted on it and so the particle has, in its turn,
exerted a force on the container.
force = mass × acceleration
F = ma
The SI unit of force is the newton (N): 1 N = 1 kg m s−2
Pressure, P, is defined as the force, F, exerted per unit
area, A.
P=
solid
liquid
Figure 7.1 A comparison of solids, liquids and gases.
gas
F
m×a
=
A
A
The SI unit of pressure is the pascal: 1 Pa = 1 N m−2.
Chapter 7 Gases
Atmospheric pressure
Boyle’s law: volume and pressure relationship
The mass of the atmosphere, under the influence of
gravity, exerts a pressure (a force per unit area) on the
Earth’s surface. This is called atmospheric pressure
and can be quoted in the unit of standard atmospheres.
Atmospheric pressure was originally measured using a
mercury barometer, as shown in Figure 7.2. A pressure
of one atmosphere (1 atm) supports a mercury column
with a height of 760 mm.
Boyle’s law states that the volume of a fixed mass of gas
at constant temperature is inversely proportional to the
pressure.
What this means is that the product PV is constant
when n and T are kept constant. This can be expressed
mathematically:
1
P
■ PV = k at constant n and T (k is a constant)
■ V∝
■ P1V1 = P2V2
If the volume of a gas is halved, the pressure is doubled
(Figure 7.3).
P = 2.0 atm
P = 1.0 atm
760 mm
atmospheric
pressure
mercury
filled
dish
increase
V = 1.0 dm
3
pressure
V = 0.5 dm3
Figure 7.2 A simple mercury barometer.
One standard atmosphere (1 atm) = 760 mmHg =
101 325 Pa. Note that ‘standard pressure’ is often quoted
as 101 kPa. This is not a convenient figure for simple
calculations and so the ‘bar’ is often used.
Figure 7.3 Boyle’s law in practice.
a
b
P
1 bar = 100 000 Pa ≈ 1 atm (strictly, 100 000 Pa = 0.986 atm)
The bar and its associated unit the millibar is the unit of
pressure that is often used in weather forecasting.
0
P
0
V
The mmHg is also called a torr: 1 mmHg = 1 torr.
c
1
V
d
Gas laws
The gas laws define the relationships between the four
variables that determine the physical properties of any
ideal gas:
■ pressure (P)
■ temperature (T)
■ volume (V)
■ number of moles (n)
PV
0
PV
V
0
P
Figure 7.4 The pressure/volume relationship for an ideal gas.
(a) The pressure/volume plot shows that P is inversely
proportional to V. (b) A plot of P versus 1/V is linear.
(c, d) A plot of the product of pressure and volume (PV) versus V
or P is constant.
69
70
Unit 1 Module 1 Fundamentals in chemistry
Charles’ law
P = 1.0 atm
P = 1.0 atm
Charles’ law states that the volume of a fixed mass of
gas at constant pressure is directly proportional to its
absolute temperature.
This means that V divided by T is constant when n and P
are kept constant. This can be expressed mathematically:
add
gas
■ V∝T
■
V = 44.8 dm3
V = 22.4 dm3
V
= k at constant n and P (k is a constant)
T
V1 V2
■
=
T1
T2
n = 1 mol
n = 2 mol
Figure 7.6 Avogadro’s law in practice.
If the temperature of a gas is doubled, the volume is
doubled; if the temperature of the gas is halved, the
volume is halved. Remember that Charles’ law mentions
the absolute temperature, which is measured in kelvin
(K). To get a numerical value for the absolute temperature,
you add 273 (or, more accurately, 273.15) to the Celsius
temperature.
Volume
1 mole of any gas occupies 22.4 dm3 (more accurately
22.414 dm3) at 0 °C (273.15 K) and 1.00 atm pressure.
This volume is called the standard molar volume. There
is more about Avogadro’s law in Chapter 6 (page 64).
Ideal gas law
The ideal gas law is a combination of all three of the gas
laws discussed so far. It describes how the volume of a gas is
affected by changes in pressure, temperature and number
of moles. The ideal gas law assumes that the particles of a
gas have no volume and exert no force on one another –
hence the word ‘ideal’.
The ideal gas law states that PV = nRT, where R is the
gas constant.
absolute zero
Temperature / K
Figure 7.5 The volume/temperature relationship for an ideal gas
at constant pressure. The volume/temperature plot shows that V
is directly proportional to T.
Avogadro’s law
Avogadro’s law states that the volume of a gas at a fixed
pressure and temperature is directly proportional to the
number of moles of gas present.
This means that V divided by n is constant when T and P
are kept constant. This can be expressed mathematically:
■ V∝n
■
V
= k at constant T and P (k is a constant)
n
■
V1 V2
=
n1 n2
If the number of moles of a gas is doubled, the volume
doubles; if the number of moles of the gas is halved, the
volume is halved (Figure 7.6).
The ideal gas law can be rearranged to show the three gas
laws:
■ Boyle’s law: PV = nRT = constant
when n and T are constant
V
nR
■ Charles’ law:
=
= constant
T
P
when n and P are constant
V
RT
■ Avogadro’s law:
=
= constant
n
P
when T and P are constant
A value for the gas constant, R, can be calculated because
1 mole of an ideal gas occupies 22.4 dm3 at 0 °C (273 K)
and 101 kPa (1 atm) pressure.
R=
PV
101 000 Pa × 22.4 × 10−3 m3
=
= 8.31 J K−1 mol−1
nT
1 mol × 273
Note that SI units are used: pressure in Pa, volume in m3
and temperature in K.
Using the ideal gas law
Consider the production of hydrogen from the reaction of
magnesium with water:
Mg(s) + 2H2O(l) → Mg(OH)2(aq) + H2(g)
Chapter 7 Gases
A real gas deviates from an ‘ideal gas’ in two important ways.
Worked example 7.1
Q
What volume of H2 is produced at s.t.p. from the reaction of
4.69 g of Mg?
A
number of moles of Mg =
4.69 g
= 0.193 mol Mg
24.31 g mol−1
From the equation, 1 mol of Mg produces 1 mol of H2,
therefore, 0.193 mol of Mg produces 0.193 mol of H2.
The volume of the gas produced can now be found from the
ideal gas law. It can help in this type of calculation to list all the
information you are given and convert the values into SI units:
■ P = 1 atm = 101 000 Pa
■ V = to be found
■ n = 0.193 mol
−1
−1
■ R = 8.31 J K mol
■ T = 0 °C = 273 K
PV = nRT
nRT
0.193 mol × 8.31 J K−1 mol−1 × 273 K
=
P
101 000 Pa
= 4.33 × 10−3 m3
= 4.33 dm3
V=
volume of H2 = 4.33 dm3 at s.t.p.
For an ideal gas the volume of a gas is mostly empty space,
and the volume of the molecules themselves is negligible.
In contrast, the molecules in a ‘real’ gas take up part of
the total space, which is called the ‘excluded volume’. The
assumption about an ideal gas is valid at low pressures:
at 1 atm (0 °C), volume of molecules is 0.05%. However,
this assumption is not valid at high pressures: at 500 atm
(0 °C), the volume of molecules is 20%.
For an ideal gas there is no force of attraction between
gas particles or between the particles and the walls of
the container. This is valid at low pressures because the
particles are so far apart. At high pressures particles are
pushed closer to each other and attractive forces become
significant; as a result, the assumption becomes invalid.
Attractions reduce the frequency of collisions with the
walls of the container and therefore a real gas will exert a
lower pressure.
We can show this graphically by finding the values of PV
for a real gas at various temperatures and pressures. The
gas law PV = nRT implies that the product PV will always be
the same. Taking methane as an example, we get the result
shown in Figure 7.7.
Worked example 7.2
Calculate the molar mass of an unknown hydrocarbon gas with
density 1.96 kg m−3 at s.t.p.
A
The number of moles of the gas can be calculated using the
ideal gas law.
■ P = 1 atm = 101 000 Pa
3
■ V=1m
■ n = to be found
−1
−1
■ R = 8.31 J K mol
■ T = 0 °C = 273 K
ideal
observed
6
5
PV / kJ
Q
7
200 ˚C
4
50 ˚C
3
–70 ˚C
2
1
PV = nRT
n=
PV
101 000 Pa × 1 m3
=
= 44.5 mol
RT
8.31 J K−1 mol−1 × 273 K
This number of moles corresponds to 1.96 kg of the unknown
gas.
molar mass =
1.96 × 1000 g
= 44.0 g mol−1
44.5 mol
The unknown gas is propane, C3H8.
Behaviour of real gases
All real gases deviate slightly from the behaviour predicted
by the ideal gas law because their molecules do interact
with each other and they do have a finite volume.
0
0
200
600
400
800
1000
P / atm
Figure 7.7 The value of PV at various temperatures and
pressures for methane, a real gas.
At high pressures, PV is always above the ideal value. This
is because the molecules are crammed together, and the
total molecular volume is a significant fraction of the total
volume. PV exhibits a positive deviation from the ideal.
As the temperature falls the molecules approaching the
walls and about to exert pressure on them are held back
by the attraction of molecules in the bulk. The lower the
temperature the more noticeable this effect becomes. PV
exhibits a negative deviation from the ideal.
71
72
Unit 1 Module 1 Fundamentals in chemistry
Kinetic-molecular theory
The kinetic-molecular theory is a century-old model
proposed to help in our understanding of the behaviour of
gases. The theory is based on the following assumptions:
■ a gas consists of tiny particles (atoms/molecules) in
■
■
■
■
constant random motion;
the volume of particles is negligible when compared to
the total volume the gas occupies; most of the volume
of a gas is empty space;
gas particles act independently of each other; there are
no attractive or repulsive forces between particles;
collisions of gas particles with themselves or the walls
of the container are elastic, i.e. the total kinetic energy
of the particles is constant at constant T;
the average kinetic energy of the gas particles is
proportional to the absolute temperature of the sample.
Kinetic-molecular theory can be used to explain the
individual gas laws.
Boyle’s law (V
∝ 1/P)
The gas pressure measures the number and forcefulness
of the collisions between gas particles and the walls of the
container. Therefore the smaller the volume (at constant T
and n), the closer together the particles are and the greater
the number of collisions.
P
↑
V
↓
Charles’ law (V
∝ T)
Temperature measures the average kinetic energy of gas
particles. The higher the temperature (at constant n and P),
the faster the movement of the particles and the more space
required to avoid an increase in the number of collisions
with the walls of the container.
T
↑
V
↑
Avogadro’s law (V
∝ n)
The more particles in a gas (at constant T and P), the more
volume the particles need to avoid increasing the number
of collisions with the walls of the container.
n
↑
V
↑
Dalton’s law (Ptotal = P1 + P2 + P3 + …)
The chemical identity of the particles is irrelevant. The
pressure of a fixed volume of gas is dependent on T and n.
The pressure exerted by a specific kind of particle depends
on the mole fraction of that kind of particle in the gas
mixture. In a mixture the gases do not act on each other.
Summary
✓ The state or condition of a gas is described by the
variables: pressure, volume, temperature and the
quantity of the gas.
✓ At constant pressure, the volume of an ideal gas
varies directly with its absolute temperature
(Charles’ law).
✓ At constant temperature, the pressure of an ideal
gas varies inversely with its absolute volume
(Boyle’s law).
✓ The kinetic-molecular theory accounts for the
properties of an ideal gas by making the following
assumptions about the nature of the gas:
■ the molecules are in constant, random motion
■ the volume of the gas molecules is negligible in
relation to the volume of its container
■ there are no attractive forces among the gas
molecules
■ the average kinetic energy of the gas molecules
is proportional to the absolute temperature.
✓ The ideal gas equation PV = nRT, where T is in
kelvin, is the equation of state for an ideal gas.
Most simple gases obey the ideal gas equation
at pressures of about 1 atm and temperatures of
300 K and above.
✓ Departure from ideal gas behaviour increases as
the pressure increases and as the temperature
decreases.
✓ The ideal gas equation is useful for calculating P,
V, T and n (number of moles).
✓ Real gases depart from ideal behaviour because
their molecules have a finite volume and
experience attractive forces for one another on
collision.
Chapter 7 Gases
Review questions
1 A student collected natural gas from a laboratory gas
jet at 25 °C in a 500 cm3 flask until the pressure of the
gas was 0.722 atm. The gas sample weighed 0.236 g at
25 °C. Calculate the molar mass of the gas.
2 A deep breath of air has a volume of 1.05 dm3 at a
pressure of 740 mmHg. Considering that the body’s
temperature is 37 °C, calculate the number of
molecules in the breath. (1 atm = 760 mmHg.)
3 A cylinder is filled with 70.0 g propane gas (C3H8) at
a pressure of 840 Torr. What would be the pressure of
the cylinder in (a) Torr and (b) Pa if 10.0 g of propane
gas is removed at constant temperature?
(1 Torr = 133.3 Pa; 760 Torr = 1 atm)
4 The pressure of a sample of gas with an initial pressure
of 3.50 atm and a temperature of 100 K is to be
increased to twice its original value at a constant
volume. To what temperature must the gas be
brought?
5 An engineer designed a new piston-cylinder device
and pumps air in at 25 °C. The volume at this
temperature is 70.2 cm3. At what temperature (in
kelvin) will the volume be 120.0 cm3?
6 Sodium, a Group I metal, reacts vigorously with
chlorine gas to produce sodium chloride. Calculate the
mass of sodium chloride that will be produced when
5.25 dm3 of chlorine gas at 0.95 atm reacts with 17 g
of sodium at 293 K. (Hint: consider which reactant is
limiting.)
7 A tank is filled with 600 dm3 of nitrogen gas at
20 °C. After sometime it is discovered that the tank
has developed a leak. The tank is resealed. If the
pressure of the tank after sealing it is 2.40 atm,
determine the mass of nitrogen that remains in the
tank.
8 The Blue Mountain peak in Jamaica is 7402 ft above
sea level. Would you expect the boiling point of water
at the peak to be greater or less than 100 °C? Explain
your answer.
9 Use the kinetic-molecular theory to predict what
would happen if a balloon is filled with air, tied and
(a) placed in a refrigerator and then (b) removed from
the refrigerator.
10 Use the kinetic-molecular theory to explain how a
pressure cooker works.
Answers to Review questions
1 16.0 g mol−1, gas is methane
2 2.42 × 1022 gas molecules
3 (a) 720 Torr; (b) 95992 Pa
4 200 K
5 509 K
6 24.4 g
7 1.68 × 103 g N2
8 Less than 100 °C since atmospheric pressure will be
less at the higher altitude.
9 (a) In the refrigerator, air is cooled so volume
decreases.
(b) Removing from the refrigerator, air is heated so it
will expand.
10 The increase in pressure in the pressure cooker allows
food to be cooked at temperatures greater than 100 °C
without the water boiling away, thus reducing the
cooking time.
73
74
Chapter 8
Thermochemistry
Learning objectives
■ Explain the concept of temperature.
■ Distinguish heat and heat capacity.
■ Explain what is meant by heat and work.
■ Distinguish between heat and work.
■ State the first law of thermodynamics; conservation of energy.
■ Define state functions.
■ Describe how to measure heat with a calorimeter.
■ Contrast enthalpy and internal energy.
■ Define enthalpies of reaction.
■ Define standard state enthalpies of reaction.
■ State and use Hess’s law.
■ Illustrate, using energy profile diagrams, the concepts of exothermic
and endothermic reactions.
Introduction to thermodynamics
Thermodynamics is the branch of science that describes
the relationship between heat, work and other forms of
energy. It allows us to predict whether or not a physical
or chemical change is possible, but gives no information
about the rate at which changes can take place. Rates of
reaction are studied in Chapter 9.
Energy is defined as the capacity to do work.
The prefix thermo originates from the Greek word that
means heat and dynamics comes from the word for force.
By combining these two meanings, we could say that
thermodynamics focuses on the force of heat or other forms
of energy. Thermochemistry is a branch of thermodynamics
that focuses on the study of energy changes that take place
during physical and chemical processes. Our focus in this
chapter is on thermochemistry.
The unit of work is the joule. This unit is named in
recognition of the work done by James Joule (1818–1889),
a British physicist and brewer, who carried out much work
in the field of thermodynamics. 1 joule is the work done
when a force of 1 newton (kg m s−1) is used to move an
object by 1 metre in the direction of the force.
Thermochemistry describes the absorption or release of
energy associated with a chemical reaction.
Changes in energy content accompany all physical and
chemical processes. Unlike matter, which can be seen and
touched, one cannot see or touch energy. Like matter,
energy can neither be created nor destroyed but energy
can be changed from one form to another. Note that matter
and energy can be interconverted in nuclear reactions (see
Chapter 3).
In chemistry we commonly speak of ‘heat’. Heat is the
energy that flows to a region of low temperature from a
region of higher temperature.
Units of heat, work and energy
1J=1Nm
Joule discovered that work can be converted to heat and
vice versa, so the joule is used to measure energy in the
form of both heat and work. Energy is measured in units
of joules (J) or kilojoules (kJ). Older units, still used for the
energy content of food, include calories (cal) or kilocalories
(kcal).
1 cal = 4.184 J
1 kcal = 4.184 kJ = 1 Cal; this is the ‘calorie’ used in food
data
Chapter 8 Thermochemistry
Energy change in a reaction
Matter always exists with energy that enables it to do
work. For example, some of the energy in oil is liberated
(as heat) during combustion. This heat energy can be used
to make machines do work. The net energy change of a
reaction is the energy difference between the reactants and
the products.
ΔE = Eproducts − Ereactants
where Δ means ‘the change in’
Stored energy
Different forms of stored energy fall into two main
categories:
■ kinetic energy, which is due to the motion of the
particles;
■ potential energy, which is stored in bonds and is also
related to the position of the object.
In chemistry, potential energy results from attractions
and repulsions between charges; for example protons and
electrons in an atom. The magnitude of this energy differs
with the distance between the charges. Potential energy
changes therefore take place during the transfer of electrons
or the sharing of electrons when bonds are formed between
atoms, or the smaller interactions making up van der Waals
bonds. It is sometimes called chemical energy. Some of this
energy is released when substances react to form something
more stable. The study of thermochemistry tells us how
energy changes are observed, measured and predicted for
chemical and physical processes. Definitions of the terms
heat, temperature, energy and work are central to the
understanding of this topic.
Temperature
Atoms and molecules are in constant motion. Even
bodies that appear stationary to the naked eye contain
atoms moving with random motion. Not all particles in
an object will move at the same speed; some have lower
kinetic (movement) energy and so move more slowly
than others. The average kinetic energy of all the particles
in an object is proportional to the absolute temperature
(in kelvin) of the object. The temperature of an object is
therefore determined by the kinetic energy of its particles
ITQ 1 What kind of energy change do you associate with each of
the following?
(a) car battery;
(b) steam engine;
(c) furnace.
and serves as a measure of it. At low temperatures, the
average kinetic energy of the particles is lower than that
at higher temperatures; molecular motion is less at the
lower temperatures. An increase in temperature indicates
an increase in kinetic energy. Atoms and molecules at
low temperatures (slower moving particles) are perceived
as being ‘cold’ while those at high temperatures (faster
moving particles) are perceived as being ‘hot’.
Temperature is an intensive property, that is, it is not
dependent on the size of the sample. For example, if we
have a container with 50 cm3 of water at 30 °C and we
combine it with 30 cm3 of water from another container
at the same temperature, the temperature of the 80 cm3 of
water is still 30 °C even though the volume has increased.
Quantities that are dependent on size for example, mass
and volume, are extensive properties.
When two bodies at different temperatures come in
contact with each other heat will flow from the hotter to
the cooler until they reach the same temperature called the
thermal equilibrium temperature. The zeroth law of
thermodynamics states that:
If two bodies are in thermal equilibrium with a third body
then they are in thermal equilibrium with each other.
Heat and heat capacity
Don’t confuse the concept of heat with that of temperature.
If you have swum for too long and you feel cold, would
you rather lie in a bath that is full of warm water or have
someone hold a lighted match under you? The water in
the warm bath has lots of heat but does not have a high
temperature. The lighted match doesn’t have much heat
but does have a high temperature.
Heat causes the temperature of an object to change. When
an object absorbs heat, its temperature may increase, or it
may melt or boil. When it loses heat to its surroundings its
temperature may decrease or it may condense or freeze.
The amount of heat needed to raise the temperature of
100 cm3 of water by 1 °C is not the same as the heat needed
to do the same to 100 cm3 of other liquids.
The heat capacity, C, of an object is the quantity of heat
required to raise its temperature by 1 kelvin.
heat exchanged
heat capacity =
temperature change
ΔH
C=
where ΔT = Tfinal − Tinitial
ΔT
The quantity of heat exchanged (q) is given by:
q = C × ΔT
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Unit 1 Module 1 Fundamentals in chemistry
ΔT can have a negative or positive value. It will have the
same numerical value whether it is in degrees Celsius or
kelvin. For example, if the final temperature is 50 °C and
the initial temperature is 40 °C, then
Table 8.1 Specific heat capacities of some common substances
Substance
Specific heat capacity / J g−1 K−1
H2O(l)
4.1840
NH3(l)
4.70
Fe(s)
0.449
Al(s)
0.901
Cu(s)
0.38
ΔT = 323.15 K − 313.15 K = 10 K
stainless steel
0.51
Regardless of the units, the temperature changes by 10
units.
ethanol
2.42
N2(g)
1.040
NaCl(s)
0.8641
ΔT = 50 °C − 40 °C = 10 °C
or, in kelvin,
The value of the heat capacity varies with the size or
amount of substance (it is an extensive property). Larger
amounts of a substance will have a greater heat capacity
than smaller amounts of the same substance. Heat capacity
is more usefully described in terms of an intensive property
– specific heat capacity or molar heat capacity.
The specific heat capacity (specific heat) c, is the amount of
heat necessary to raise the temperature of 1 g of a substance
by 1 °C (or 1 K). The SI units of c are J g−1 K−1.
c=
C
mass
The equation for heat change then becomes:
q = c × m × ΔT
The greater the mass of a substance the more heat is
required to raise its temperature.
By convention, if heat is absorbed into a system, the heat
change (Δq) is positive. If heat is lost from a system to the
surroundings, Δq is negative.
We may express heat change in terms of the molar heat
capacity, which is the heat required to raise the temperature
of one mole of a pure substance by 1 K (1 °C):
cm =
C
number of moles
q = cm × number of moles × ΔT
The equations given above involving specific heat
are used only when there is a change in temperature
without any phase change or chemical reaction. For
example, when ice at 0 °C absorbs heat it melts to form
liquid water at 0 °C. In this case ΔT is equal to zero but q
for the process is not zero.
Table 8.1 lists the specific heat capacities of some common
substances. Note that water has a very large value, second
only to liquid ammonia.
ITQ 2 Suggest an important consequence of the large value of
the specific heat capacity of water.
Worked example 8.1
Q
The specific heat capacity of iron is 0.449 J g−1 K−1. Calculate
the heat capacity in J °C−1 of 3.00 g of iron.
A
heat capacity = specific heat × mass
= 0.449 J g−1 K−1 × 3.00 g
= 1.347 J K−1 = 1.347 J °C−1
Latent heat
Heating a substance does not always make it get hotter.
While at first sight that seems like an odd statement,
consider what happens when we heat pure water at 1 atm
pressure. The temperature increases until the water boils
at 100 °C. The water then remains at this temperature,
even with added heat, until it all completely evaporates
(changes to the gas phase).
Heat is entering the system without being detectable as
a change in temperature. The heat entering the system
provides the energy required to overcome the attractive
forces between the particles in the liquid to enable the
particles to become a gas.
This kind of heat is called latent heat (‘latent’ in this
context means ‘hidden’). Latent heat is involved when
there is a change in the phase (state) of the substance.
The changes of state are when a substance changes from
solid to liquid (melts), liquid to solid, (freezes), boils (liquid
to vapour) or condenses (vapour to liquid). These are all
shown on Figure 8.1.
ITQ 3 The specific heat capacities of aluminium and water are
0.902 J g−1 °C−1 and 4.184 J g−1 °C−1 respectively. Equal masses
of water and aluminium were heated with 1 kJ of electrical energy
at room temperature. Which of the two substances (water or
aluminium) would you expect to get hotter? Explain why.
Chapter 8 Thermochemistry
6 E = E final – E initial
latent heat
(vaporization)
liquid / gas
gas
surroundings
Energy out of system
to surroundings: – sign
condensation
vaporization
T / ˚C
liquid
freezing
system
Energy into system from
surroundings: + sign
melting
solid / liquid
solid
latent heat
(fusion)
Heat added
Figure 8.1 Latent heat of fusion and vaporization.
Heat and the kinetic-molecular theory
Thermodynamics divides the universe into a ‘system’ and
its ‘surroundings’ (Figure 8.2). The system is the small part
that we are interested in investigating. For example, it could
be the reaction mixture in a beaker. The surroundings
are everything else outside the system, i.e. the rest of
the universe. A boundary separates the system and its
surroundings. This boundary can be real or imaginary, rigid
or elastic. For example, it could be the glass of the beaker
or a cylinder wall. The boundary can be an ideal conductor
of heat or an insulator.
Figure 8.3 The movement of energy between a system and its
surroundings.
Heat and work
When a force moves an object, work is done. The quantity
of work done is defined as the force multiplied by the
distance moved in the direction of the force:
work = force × distance = F × d
The unit of work is the newton-metre, usually called the
joule (J).
Chemical reactions can do work electrical work by forcing
an electric current through a wire, e.g. an electric current
through the filament in a light bulb. In other cases, chemical
reactions can do work of expansion because of the changes
in volume of the system during the reaction.
In order to better understand how work of expansion is
done by a chemical reaction, consider the combustion of
ethane gas with oxygen occurring in a cylinder with a
piston (a closed system), as shown in Figure 8.4.
surroundings
system
P= F
A
system boundary
P= F
A
Figure 8.2 The surroundings and the system – that’s all there is,
according to thermodynamics.
d
A system can be described further as open, closed or
isolated.
reaction
w =Fxd
=–PxAxd
= – P6V
■ In an open system both matter and energy are
exchanged with the surroundings. Some examples
include an open reaction flask, the human body and
the engine of an car.
■ A sealed flask would be a closed system since it has a
fixed amount of matter but is able to exchange energy
with the surroundings.
■ An isolated system is sealed and insulated from the
surroundings and can exchange neither matter nor
energy with the surroundings. A sealed vacuum flask
is an example of an isolated system.
before reaction
(initial state)
after reaction
(final state)
Figure 8.4 Calculating work done during expansion.
2C2H6(g) + 7O2(g) → 6H2O(g) + 4CO2(g)
Before the start of the reaction, the pressure of the gases
pushing up on the piston is balanced by the weight of the
ITQ 4 In cold climates, blossom on fruit trees is sometimes spoilt
by sub-zero temperatures before dawn. The farmer may spray
his trees with water, which freezes on the branches. Suggest how
this helps to preserve the blossom.
77
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Unit 1 Module 1 Fundamentals in chemistry
piston and so the total volume of the gases is constant.
9 moles of reactants (2 mol ethane plus 7 mol of oxygen)
react to produce 10 moles of products. The volume of gas
after the reaction is therefore greater than the volume
before the reaction. If the reaction takes place in a sealed
container at constant pressure and temperature, with a
movable piston, then the increase in the volume of the gas
would push the piston outward against the force of gravity.
Work is therefore done by the system (and in opposition to
atmospheric pressure, thus −Pext) and is given by
w=F×d
where d is the distance moved by the piston.
(8.1)
The pressure exerted by the gas on the piston (Pext) is equal
to the force (F) with which it pushes against the piston
divided by the surface area (A) of the piston:
F
Pext =
or F = Pext × A
A
If this expression for F is substituted in equation 8.1, the
work done is
w = (P × A) × d = P × (A × d)
The product A × d is equal to the change in volume ΔV
when the gas expands. Therefore:
w = Pext ΔV
In an expansion (an increase in volume) ΔV is positive,
work is done by the system, so the system loses energy and
w = −Pext ΔV is negative.
In a contraction (a decrease in volume) ΔV is negative,
work is done by the surroundings on the system so the
system gains energy and w = −Pext ΔV is positive.
If there is no change in volume, then no work is done.
The first law of thermodynamics
Energy conservation
The first law of thermodynamics is the result of the
observations of the British physicist William Thomson (later
Lord Kelvin) and the German physicist Rudolf Clausius.
Working independently, they observed that whilst neither
heat nor work is conserved in nature, energy is conserved.
The first law of thermodynamics states that energy is
conserved in a closed system. Energy is neither created
nor is it destroyed. It is converted from one form to
another.
The first law of thermodynamics can be restated as: ‘the
total energy of an isolated system is constant’. A system
can lose or gain energy but any changes in the energy of
the system must be accompanied by an equivalent change
in the energy of the surroundings so as to ensure that the
total energy of the universe is constant.
ΔEsystem + ΔEsurroundings = 0
In thermodynamics, the energy of a system is referred to
as internal energy, U. Internal energy is proportional
to the sum of the system’s kinetic energy (energy due to
motion of atoms, molecules, etc.) and the potential energy
(energy due to the position or ‘stored’ energy in the case of
chemical reactions involving bonds).
The internal energy of an ideal gas at temperature T has
zero potential energy so its internal energy is the kinetic
energy at T which is given by the following equation:
U=
3RT
2
where R is the universal gas constant (see Chapter 7) and
T is in kelvin.
For non-ideal and more complex systems, the internal
energy cannot be determined directly. However, changes
in the internal energy (ΔU) can be determined from
temperature changes of the system and are defined as
the difference between the initial and final values of the
internal energy.
ΔU = Ufinal − Uinitial
Since the internal energy is proportional to its temperature,
an increase in temperature results in an increase in internal
energy and ΔU will be positive.
Inter-conversion of heat and work
According to the first law of thermodynamics, for a
closed system, energy transfer between the system and
its surroundings can only be in the form of heat or work.
The change in the internal energy is therefore a balance
between the heat (q) and the work (w) that cross the
boundary between the system and its surroundings.
ΔU = q + w
This equation also shows that the amount of heat generated
by a system can be limitless as long as enough work is done
on it to compensate for its loss of internal energy. Similarly,
the amount of work done by a system can be boundless if
enough heat is pumped into it to compensate for the load
on its internal energy. If either heat or work is removed
from the system without any compensation, then the
internal energy of the system will decrease until the system
can longer generate heat or do work.
Chapter 8 Thermochemistry
State functions
Worked example 8.2
Q
A
A gas absorbs 600 J of heat energy and is compressed from
50 dm3 to 40 dm3 by an opposing pressure of 4.0 atm.
Calculate the change in the internal energy (ΔU ) for this
process.
ΔU = q + w
q = +600 J (heat absorbed by the system)
4 atm = 4 × 101 000 Pa
40
50
40 dm3 =
m3 = 0.04 m3; 50 dm3 =
m3 = 0.05 m3
1000
1000
w = −PΔV = Popposing × (Vfinal − Vinitial )
= −4.0 × 101 000 Pa × (0.04 m3 − 0.05 m3)
= 4040 J (work done on the system is positive)
ΔU = q + w = 600 J + 4040 J = 4640 J
The internal energy of the system increases by 4640 J due to
absorption of 300 J of heat and 4040 J of work energy.
We can now explain the concept of latent heat using the
first law of thermodynamics. When heat is applied to a
system it can increase the temperature (ΔU is positive) or
it can do work. Whereas the increase in temperature can
be detected, work on the system (ΔU = 0) cannot and it is
therefore ‘latent’.
There is a convention that we follow:
■ ΔU is negative when the system loses energy to its
surroundings (Ufinal < Uinitial) or when the system does
work on its surroundings, i.e. when q is negative and
w is negative;
■ ΔU is positive when heat enters the system from its
surroundings (Ufinal > Uinitial) or when work is done on
the system by the surroundings, i.e. when q is positive
and w is positive.
Work (w)
Heat (q)
A state function is property of a system whose value
depends only on the state of the system and not on the
path used to arrive to that state.
Temperature, for example, is a state function because the
net change in temperature of a system, ΔT, depends only
on the initial and final values regardless of how many
times the system may have been heated or cooled. ΔT =
Tfinal − Tinitial. Internal energy is also a state function since it
is proportional to temperature.
State functions are reversible; that is, they can return to
their initial condition. The overall change in a state function
is zero when the system returns to its original condition.
Conventions
In equations, state functions are represented using upper
case symbols, such as T, P, V and U. Properties that are not
state functions are represented using lower case symbols,
such as w, q, m and n.
Calorimetry
The amount of heat absorbed or given off during a chemical
reaction can be measured using a calorimeter. A calorimeter
is an insulated container with a thermometer.
From the first law of thermodynamics, the change in
internal energy of a system is affected by both heat and
work, according to the equation
ΔU = q + w
For a system in which no work is being transferred between
the system and its surroundings (i.e. w = 0), the internal
energy change will be equal to the heat absorbed or lost by
the system to its surroundings.
ΔU = q (if w = 0)
work done on system
6U = +
6U = +
system
work done by system
6U = –
6U = –
surroundings
Figure 8.5 Sign convention. The work done on the system (w)
and the heat absorbed by the system (q) are positive as both
increase the internal energy of the system. The work done on
the system (–w) and the heat lost by the system (–q) are both
negative as both decrease the internal energy of the system.
ΔU can therefore be found by measuring the temperature
change accompanying the heat loss or gain by the system
and the surroundings when w = 0.
How can we ensure that w = 0? In most chemical reactions,
the work we are interested in is that of expansion (PΔV).
Work of expansion is given by
w = −PΔV
Substituting this in the equation for ΔU gives
ΔU = q + w = q − PΔV
ITQ 5 Which of the following are state functions?
(a) heat
(b) work
(c) pressure
(d) volume
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Unit 1 Module 1 Fundamentals in chemistry
No work of expansion will be done by the system or on the
system therefore if there is no change in the volume of the
system (ΔV = 0). Under these conditions of constant V, the
change in internal energy is equal to the heat lost or gained
by the system.
ignition wires
thermometer
insulation
ΔU = qV (V is constant so w = 0)
A bomb calorimeter is used to make measurements of
heat given off or absorbed during a chemical reaction at
constant volume (Figure 8.6). The chemical reactants
(the system) are placed in the calorimeter, which is then
sealed and placed in another container with a known
volume of water (the surroundings). An electric current
is passed through an ignition wire which is used to initiate
the chemical reaction. The reaction will release heat to the
surroundings, resulting in an increase in the temperature
of the entire apparatus. The amount of heat transferred
from the system to the surroundings is determined from
the mass of the water, the specific heat capacity of the
water and the temperature change.
q = m × c × ΔT
This amount of heat will be equal to the change in internal
energy of the system since no work was done by or on the
system at constant volume:
ΔE = qV = m × c × ΔT
Worked example 8.3
stirrer
bomb
sample holder
water in inner container
Figure 8.6 A bomb calorimeter.
Enthalpy
Most chemical reactions are done in open flasks and not
in sealed containers at constant volume (Figure 8.7).
Under these conditions, where gases can enter or leave the
system, the total pressure remains constant and equal to
the atmospheric pressure.
thermometer
Q
A calorimeter containing 300 g water is heated with 250 kJ of
heat. The temperature of the water increases from 30 °C to
80 °C. Calculate the heat capacity of the calorimeter in J °C−1.
The specific heat capacity of the water is 4.184 J g−1 °C−1.
A
We will first calculate the heat gained by the water in the
calorimeter. The remaining heat will be used to heat the
calorimeter so we can use it to determine the heat capacity of
the calorimeter.
temperature change = 80 °C − 30 °C = 50 °C
heat gained by the water in the calorimeter = mass of the water
× specific heat capacity of the water × temperature change
heat gained by the water = 300 g × 4.184 J g−1 °C−1 × 50 °C
= 62760 J = 62.76 kJ
We are told that the total heat added was 250 kJ. Therefore:
heat absorbed by the calorimeter = 250 kJ − 62.76 kJ
= 187.24 kJ
heat absorbed by the calorimeter =
heat capacity of the calorimeter × temperature change
stirrer
insulated vessel
heat absorbed by the calorimeter
heat capacity of the
=
calorimeter
temperature change
=
187240 J
= 3744.8 J °C−1
50 °C
reaction mixture
Figure 8.7 Reactions in real life – in an open flask and not at
constant volume.
The heat change taking place under constant pressure is
called the change in enthalpy of the system. Enthalpy
is given the symbol H and the change in heat content of
a system at constant pressure – the change in enthalpy – is
written ΔH.
In practice, a change in enthalpy is simply a change in heat
content and the words are used interchangeably. However,
in thermodynamic calculations, where conditions must be
specified, it is correct only to say ‘enthalpy’. A quantity
of heat energy can properly be called ‘heat’ and referred
to as ‘energy’ but a change in heat content of a system at
constant pressure is a change in its enthalpy.
Chapter 8 Thermochemistry
■ When ΔH is positive,
the process is said to be
endothermic. Heat is
absorbed by the system and
it gains energy.
■ When ΔH is negative,
the process is said to be
exothermic. Heat is released
from the system and it loses
energy.
b
Ea
Ea
products
Energy
Chemical reactions are classified
as exothermic if they give off
heat or endothermic if they
absorb heat.
a
Energy
Enthalpy of reaction
reactants
6H
6H
reactants
products
Reaction coordinate
Reaction coordinate
Figure 8.8 Reaction profiles for (a) an endothermic reaction and (b) an exothermic reaction.
The reaction between ammonium nitrate (NH4NO3) and
water is an example of an endothermic reaction. Some
cooling packs, used to treat sporting injuries, contain
ammonium nitrate and water – separated from each other
by a thin membrane. When the membrane is broken, they
combine and the ammonium nitrate dissolves in the water.
This reaction absorbs heat from the surroundings, the pack
gets cold and can be used in the place of ice to cool items or
treat sport injuries.
NH4NO3(s) + H2O(l) → NH4+(aq) + NO3−(aq)
ΔH = 25.7 kJ mol−1
Thermodynamic standard state enthalpy of
reaction
For a given reaction the enthalpy change for the chemical
reaction depends on the temperature, the pressure and other
conditions under which the reaction was carried out. In order
to allow for the comparison of the results of experiments
done under different conditions, a standard set of conditions
is defined for all thermodynamic measurements.
The standard state refers to a pure substance in a specified
state, usually its most stable form, at 1 bar pressure,
at a specified temperature, usually 25 °C (298 K) and
1 mol dm−3 concentration for all substances in solution.
The thermodynamic standard state defines the specific
physical state of reactants and set of conditions for all
chemical reactions.
The enthalpy change that is reported for a reaction is the
amount of heat released when reactants are converted
to products in the molar amounts represented in the
stoichiometric equation.
For example, the combustion of methane gas produces
890 kJ of heat. This reaction (thermochemical equation) is
represented as:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
ΔH = −890.4 kJ
This means that the heat released when 1 mole of methane
reacts with 2 moles of oxygen to give 1 mole of carbon
dioxide and 2 moles of water is 890.4 kJ. ΔH is negative
because enthalpy is lost from the system to the surroundings.
Notice that the equation specifies water in its standard
state (liquid) rather than as steam. The combination of
the balanced equation and the molar enthalpy change of
the reaction gives the thermochemical equation for the
chemical reaction.
An enthalpy change determined under standard conditions
for a reaction is called the standard enthalpy of reaction,
ΔH (note the superscript ).
2H2(g) + O2(g) → 2H2O(g)
ΔH
= −484 kJ
For a given reaction, the enthalpy change of the reverse
reaction has the same magnitude but the opposite sign. If
the forward reaction is endothermic, the reverse reaction
is exothermic.
Hess’s law
The difference between the initial and final values of
enthalpy does not depend on the path taken by the
reaction. This conclusion was reached in 1840 by Germain
Hess (a Swiss-born Russian chemist, 1802–1850), based on
his experiments. A general law, now known as Hess’s law
was proposed.
Hess’s law: the enthalpy change of a reaction (ΔH ) is the
same regardless of whether the reaction occurs in one step
or several steps. The overall enthalpy change for a reaction
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Unit 1 Module 1 Fundamentals in chemistry
is therefore equal to the sum of the enthalpy changes for
the individual steps in the reaction.
M–X(g) → M(g) + X(g)
If enthalpy was not a state function it would be possible
for a system to change from state 1 to another state 2 via
a path requiring a certain amount of energy. If returning
from state 2 to state 1 via a different path released more
energy, this would mean that each time we go from state
1 to state 2 then back to state 1 we would have energy
to spare. We would have produced a device that creates
energy, known as a perpetual motion machine, a situation
that is contrary to the law of conservation of energy.
These values are positive. The products contain more
energy than the original compound.
The industrial production of ammonia from hydrogen and
nitrogen (the Haber process) can be proposed to proceed
via the following hypothetical steps:
Step 1 2H2(g) + N2(g)
Step 2 N2H4(g) + H2(g)
→N2H4(g)
→2NH3(g)
ΔH 1 = ?
ΔH 2 = −187.6 kJ
N2(g) + 3H2(g) + N2H4(g)→ N2H4(g) + 2NH3(g)
Overall reaction
3H2(g) + N2(g)
→2NH3(g) ΔH reaction = −92.2 kJ
H2(g) → H(g) + H(g)
The enthalpy change of atomization or enthalpy of
atomization, ΔHatm, is the amount of energy required to
break all the bonds in a gaseous molecule into its neutral
gaseous atoms. ΔHatm is therefore the sum of all the bond
energies in the molecule. For example, ΔHatm for methane,
CH4, is the energy required for the process:
CH4(g) → C(g) + 4H(g)
The values are positive.
For a simple diatomic molecule, e.g. I2 or H2, ΔHatm is the
same as ΔHdiss since there is only one bond between the
atoms.
The enthalpy change of combustion or enthalpy of
combustion, ΔHcomb, is the energy released by 1 mole of a
given substance when it is completely burned in oxygen.
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
Both steps of the reaction add to give the overall reaction
and, according to Hess’s law, the sum of the enthalpy
changes for each step gives the overall enthalpy change.
Values here are negative since energy is lost from the
system.
The enthalpy changes for step 2 and the overall reaction
are known. However, it is difficult to measure the enthalpy
change for step 1. The enthalpy change for step 1 can be
calculated using Hess’s law as follows:
The enthalpy change of neutralization or enthalpy of
neutralization, ΔHneut, is the energy released when 1 mole
of an acid or base is completely neutralized. For all strong
acids or bases, its value is approximately −57.5 kJ mol−1.
ΔH reaction → ΔH 1 + ΔH 2
The enthalpy change of fusion (melting) or enthalpy
of fusion, ΔHfus, is the energy required to melt 1 mole of a
substance at its melting point to give the liquid at the same
temperature.
ΔH 1 = ΔH reaction − ΔH 2 = +95.4 kJ
Rules for applying Hess’s law
■ If a reaction is reversed, the sign of ΔH must be reversed.
■ If the reaction is multiplied or divided by a factor, ΔH
must also be multiplied or divided by the same factor.
Examples of enthalpy changes
There are special terms assigned to enthalpy changes
for particular types of chemical reactions and physical
processes. We will use the convention that if heat energy
is lost from the system, the value of the change is negative
and if heat energy is gained by the system, its value is
positive. Tables of values supplied in other books and on
the internet do not always use this convention, so be sure
that you get the signs (+ or –) right
The enthalpy change of bond dissociation or bond
dissociation enthalpy, ΔHdiss, is the amount of energy
required to break a chemical bond in 1 mole of an isolated
molecule in the gaseous state.
M(s) → M(l)
Values are positive.
The enthalpy change of vaporization or enthalpy
of vaporization, ΔHvap, is the energy required to convert
1 mole of a substance from the liquid to the gas phase at
the same temperature.
M(l) → M(g)
Values are positive.
The enthalpy change of sublimation, ΔHsub, is the
energy required for the sublimation of 1 mole of a substance
from solid to gas at the same temperature.
ΔHsub = ΔHfus + ΔHvap
Values are positive.
Chapter 8 Thermochemistry
The enthalpy change of hydration or heat of solution,
ΔHhyd, is the energy change during the hydration of 1 mole
of a gaseous ion.
Na+(g) + water → Na+(aq)
The lattice enthalpy, ΔHlat, is the heat energy given out
when 1 mole of a substance is formed from its gaseous ions.
Na+(g) + Cl−(g) →NaCl(s)
This can be positive (if heat is released) or negative (if heat
is absorbed).
The enthalpy change of solution or heat of solution,
ΔHsoln, is the energy change when 1 mole of a substance is
dissolved in a solvent at infinite dilution, e.g. water.
NaCl(s) + water → NaCl(aq) or
NaCl(s) + water → Na+(aq) + Cl−(aq)
The value can be positive if the solution absorbs heat
energy or negative if heat energy is released.
In order for a substance to dissolve in a solvent, energy must
be supplied to the ions in the substance to separate them
from each other (ΔHlat). The ‘free’ ions are then solvated/
hydrated (surrounded by solvent) due to electrostatic
attraction between the solvent and the ions. If the solvent
is water, these interactions could be hydrogen bonding,
ion–dipole interactions or dipole–dipole interactions. This
is discussed further below.
The enthalpy change of solution is therefore the sum of
the lattice enthalpy of the compound and the hydration
enthalpies of the ions. For example:
+
ΔHsoln(NaCl)
NaCl(s) + water → Na
ΔHhyd(Na+)
Na+(g) + water → Na+(aq)
−(g) + water → Cl−(aq)
ΔHhyd(Cl−)
Cl
ΔHlat(NaCl)
NaCl(s) → Na+(g) + Cl−(g)
+
−
ΔHsoln = ΔHhyd(Na ) + ΔHhyd(Cl ) + ΔHlat(NaCl)
+(aq)
Lattice enthalpy
Cl−(aq)
The enthalpy change of first ionization, ΔHIE1, is the
energy required during the removal of an electron from
1 mole of a gaseous atom.
Na(g) → Na+(g) + e−
Values are positive.
The enthalpy change of electron gain (or electron
affinity), ΔHea, is the energy change during the addition of
an electron to 1 mole of a gaseous atom.
Cl(g) + e− → Cl−(g)
If heat energy is released during this process then the value
is negative. If heat energy is absorbed then the value is
positive. First electron affinities are negative, but many
data tables quote them as positive values.
Lattice enthalpy depends on:
■ the charges on the ions;
■ the distance between the ions.
The lattice enthalpy is proportional to the product of the
charges on the ions. The greater the charge on the ions, the
greater is the attraction between the ions and the greater is
the lattice enthalpy. For a given set of ions of similar size,
the lattice enthalpy is greatest for the most highly charged
ions. The lattice enthalpy of a compound with charges
M2+X2− is grater than one with M+X−. As an example, the
lattice enthalpy of LiCl is less than the lattice enthalpy of
MgCl2.
The lattice enthalpy is inversely proportional to the distance
between the ions. So for compounds with the same charges
on their ions, the smaller the distance between the ions, the
greater is the attraction between the ions and the greater
is the lattice enthalpy. For example the lattice enthalpy
increases in the order
NaF > NaCl > NaBr > NaI
LiF > LiCl > LiBr
LiF > NaF > KF
Calculating enthalpy changes
The standard enthalpy of formation is the enthalpy
change ΔH f for the formation of 1 mole of a substance
in its standard state from its constituent elements in their
standard (or reference) states (pressure = 1 bar and 25 °C/
298 K).
The reaction to form a substance from its elements is
hypothetical. This definition gives a reference level from
which all changes are measured. Each substance involved
in the reaction must be in its standard state: usually
the most stable form at 1 bar pressure and the specified
temperature (usually 298 K).
The standard enthalpy change for any chemical reaction is
the difference between the sum of the heats of formation
of all reactants and the sum of the heats of formation of all
products:
ΔH reaction → ΔH f (products) − ΔH f (reactants)
83
84
Unit 1 Module 1 Fundamentals in chemistry
Consider the general reaction:
The Born–Haber cycle
aA + bB + … → cC + dD + …
The Born–Haber cycle is a method of assessing the
stability of ionic compounds. The application of Hess’s law
can be extended to calculating the overall energy change
for the formation of ionic crystals.
ΔH reaction = [cΔH f (C) + dΔH f (D) + …]
− [aΔH f (A) + bΔH f (B) + …]
Now consider the reaction:
CO(g) + ½O2(g) → CO2(g)
ΔH reaction = ? kJ
The hypothetical steps for this reaction are as follows:
Step 1
C(s) + ½O2(g) → CO(g)
ΔH f (CO) = −110 kJ mol−1
Step 2
C(s) + O2(g) → CO2(g)
ΔH f (CO2) = −393.5 kJ mol−1
Step 3
O2(g) → O2(g)
We can get an indication of the stability of a compound
from its standard enthalpy of formation (ΔH f ). If ΔH f
is negative, the formation reaction is exothermic and the
compound formed is stable and likely to be formed.
Combining solid Na with gaseous Cl2 results in a violent
reaction and NaCl solid is produced, as well as a great
amount of heat. The equation for this reaction is:
Na(s) + ½Cl2(g) → NaCl(s)
ΔH f (O2) = 0 kJ mol−1
In accordance with the general formula given above, we get:
ΔH reaction = ΔH f (CO2) − [ΔH f (CO) + ΔH f (O2)]
ΔH reaction = −393.5 kJ mol−1 −
[(−110 kJ mol−1) + (0 kJ mol−1)]
= −283.5 kJ mol−1
Using bond dissociation energies
For reactions where ΔHf values are not available, it is
possible to get approximate values for enthalpy changes by
using average bond dissociation energies (ΔH diss ).
Bond dissociation energies are always positive, as energy
must be supplied to break a chemical bond. For example,
the standard enthalpy change for the reaction Cl2(g) →
2Cl(g) is +243 kJ mol−1:
Cl–Cl → 2Cl
ΔH diss = +243 kJ mol−1
ΔH diss can be abbreviated to D.
By application of Hess’s law, it is possible to calculate
an approximate enthalpy change for any reaction by
subtracting the total energy of bonds formed in the products
from the total energy of bonds broken in the reactants.
= −411 kJ mol−1
If we look at the enthalpy changes for the electron transfer
process only, we will see that this is not an exothermic
process. To get a better picture of the reaction between
Na and Cl2, we can consider all the steps that might be
involved. If the enthalpy changes for each of these steps are
known then we can apply Hess’s law. The sum of enthalpies
of these individual steps should be equal to the formation
enthalpy of NaCl.
The reaction is between solid Na and Cl2 gas so, before
an atom of Na can combine with an atom of Cl, it must
first be separated from other atoms to which it is bonded.
That is, in the solid metallic Na there are several Na atoms
bonded to each other and in the gas Cl2 there are Cl atoms
bonded to each other as molecules. One way of separating
the atomic interactions in each reactant is by taking them
to the gas phase. Once the atoms are separated the process
of ionization can occur through electron transfer and the
charged species once formed will attract each other to form
the crystal. These steps are summarized below and values
for their enthalpy changes are given: The overall energy
change, ΔH reaction , is calculated by considering the series of
steps:
1: enthalpy of sublimation
Na(s) → Na(g)
2: bond dissociation enthalpy ½Cl2(g) → Cl(g)
H2(g) + Cl2(g) → 2HCl(g)
3: enthalpy of ionization
Bonds broken: H–H(g) and Cl–Cl(g)
4: enthalpy of electron affinity Cl(g) + e− → Cl−(g)
Bonds formed: 2H–Cl(g)
f
The question we need to ask is: ‘Why is the formation of
NaCl so favourable?’
ΔH reaction = ∑D(bonds broken) − ∑D(bonds formed)
Let’s look at an actual reaction:
ΔH
Na(g) → Na+(g) + e−
ΔH IE1
= +495.8 kJ mol−1
ΔH
ea
= −348.6 kJ mol−1
→ NaCl(s) ΔH
lat
= −787 kJ mol−1
5: lattice energy
Na+(g)
6: ΔH reaction
Na(s) + ½Cl2(g) → NaCl(s)
+
Cl−(g)
−1
ΔH sub = +107.3 kJ mol
−1
ΔH diss = +122 kJ mol
= −411 kJ mol−1
ΔH reaction = [D(H2) + D(Cl2)] − (2D(HCl)]
= (243 kJ + 436 kJ) − (2 × 432 kJ)
= −185 kJ
ITQ 6 The enthalpy change of combustion of acetylene gas,
C2H2, at 25 °C is 310.5 kJ mol−1. Work out the enthalpy of
formation of acetylene gas.
Chapter 8 Thermochemistry
The five steps that are involved
in the formation of sodium
chloride, and the overall energy
change, can be represented in
a thermochemical cycle called
a Born–Haber cycle. The Born–
Haber cycles for the formation of
sodium chloride and magnesium
chloride are shown in Figures 8.9
and 8.10.
The enthalpy changes for steps 3
and 4 are the ionization energy
and electron affinity respectively.
Note that all the steps except 4
and 5 are endothermic processes.
In fact, if the enthalpy changes
for the steps up to the formation
of the gaseous ion (steps 1–4)
are combined, a value of ΔH
= +376.2 kJ mol−1 is obtained,
indicating that that the formation
of gaseous (free) ions from solid
Na and gaseous Cl2 is an overall
endothermic process. The addition
of the last step (step 5), where the
ions attract each other to form the
lattice (crystalline network), make
the overall process exothermic.
One can conclude therefore that
the major contributing factor to
the stability of NaCl is the strong
force of attraction between the
ions that give rise to the formation
of the crystal (ΔH lat ).
The enthalpy of formation is
therefore the sum of all the
enthalpy changes for the steps
listed.
ΔH
f
Na+(g) + e–
1st ionization: Na(g)
–348.6 kJ mol –1
electron affinity: Cl(g) + e–
Cl –
495.8 kJ mol –1
bond dissociation:
1
2 Cl2(g)
Cl(g)
sublimation: Na(s)
Na(s) +
122 kJ mol –1
Na(g)
107.3 kJ mol –1
lattice energy
1
2 Cl2(g)
–787 kJ mol
–1
net reaction
–411kJ mol –1
NaCl(s)
Figure 8.9 The Born–Haber cycle for NaCl.
2nd ionization: Mg+(g)
Mg 2+(g) + e –
–697.2 kJ mol –1
electron affinity: 2Cl(g) + 2e –
2Cl –
1450.7 kJ mol –1
Mg+(g) + e–
1st ionization: Mg(g)
737.7 kJ mol –1
lattice energy
–2524 kJ mol–1
bond dissociation: Cl 2(g)
sublimation: Mg(s)
Mg(g)
2Cl(g)
243 kJ mol –1
147.7 kJ mol –1
Mg(s) + Cl2(g)
net reaction
Mg(s) + Cl2(g)
MgCl2(s)
–642 kJ mol –1
MgCl2(s)
Figure 8.10 The Born–Haber cycle for MgCl2.
= ΔH vap + ΔH diss + ΔH sub + ΔH IE1 + ΔH ea + ΔH latt
The enthalpy of formation of an ionic compound can be
determined experimentally and the value compared to
that obtained theoretically using Hess’s law. Similarly,
lattice energy values can be obtained from Hess’s law
calculations as well as from a theoretical equation. If there
is good agreement between the value calculated from the
Born–Haber cycle and that obtained experimentally, then
one can say that the ionic model of bonding proposed
is a good one for the compound under consideration. If
the agreement is poor that one can conclude that there
might be some other type of bonding occurring, perhaps
covalent. Tables 8.2–8.4 show some values obtained for
theoretical and experimental values of lattice enthalpy for
some compounds. NaCl shows excellent agreement and
is therefore a good model of almost pure ionic bonding.
For AgCl, which is larger, the discrepancy is greater due to
greater polarizability of the ions resulting in more covalent
character in this compound.
85
86
Unit 1 Module 1 Fundamentals in chemistry
Table 8.2 Lattice enthalpies for Group I halides
Compound
ΔH lat / kJ mol−1
Compound
ΔH lat / kJ mol−1
Compound
ΔH lat / kJ mol−1
Compound
ΔH lat / kJ mol−1
LiF
1037
LiCl
852
LiBr
815
LiI
761
NaF
936
NaCl
787
NaBr
747
NaI
705
KF
821
KCl
717
KBr
689
KI
649
Table 8.3 Lattice enthalpies for some other ionic compounds
Compound
ΔH lat / kJ mol−1
Compound
ΔH lat / kJ mol−1
Compound
ΔH lat / kJ mol−1
Compound
ΔH lat / kJ mol−1
MgO
3850
CaO
3461
SrO
3283
BaO
3114
MgS
3406
CaS
3119
SrS
2974
BaS
2832
Al2O3
15900
Table 8.4 Theoretical and experimental values of ΔH lat in kJ mol−1 for some compounds
Compound
ΔH lat / kJ mol−1, theoretical
ΔH lat / kJ mol−1, experimental
NaCl
−769
−787
NaBr
−732
−747
AgCl
−830
−904
Summary
✓ Thermochemistry is the branch of
thermodynamics that focuses on the energy
changes that occur during chemical and physical
processes.
✓ A system can transfer energy to its surroundings
in the form of heat that can be measured in
terms of temperature changes.
✓ The total energy of the system and its
surroundings is, however, constant. When a
system expands, the sign of the work done is
negative and when the system contracts, the
work done has a positive sign.
✓ The heat given off or absorbed by a closed system
at constant volume can be measured using a
bomb calorimeter.
✓ Reactions that give off heat are exothermic and
those that absorb heat are endothermic.
✓ A state function is a property whose value
depends only of the present state of the system
and not on the path taken to arrive at that state.
Internal energy, temperature and pressure are
state functions.
✓ Enthalpy change is the heat change in a system
at constant pressure.
✓ Hess’s law states that the overall enthalpy
change for a reaction is equal to the sum of the
enthalpy changes for the individual steps in
the reaction. Application of Hess’s law allows
for the determination of enthalpy of formation
of reactions from enthalpy changes for other
related reactions.
✓ Bond dissociation enthalpies can be used to
calculate enthalpy of formation for a compound.
Chapter 8 Thermochemistry
Review questions
1
2
3
50 cm3 of 1.0 mol dm−3 HCl at 30 °C at was mixed
with 50 cm3 of 1.0 mol dm−3 NaOH at 30 °C in a
Styrofoam calorimeter. The temperature of the
calorimeter rose by 4.5 °C. Calculate the heat of
reaction per mol of H2O(l) formed. (The heat capacity
of the calorimeter is 50 J °C−1.)
15.0 g of methanol (CH3OH) was burned in a
bomb calorimeter containing 2.40 kg of water. The
temperature of the water and calorimeter rose from
35.00 to 38.10 °C. The specific heat capacity of water
is 4.18 J g−1 K−1. Assuming negligible heat loss to the
calorimeter, determine:
(a) the amount of heat evolved in the reaction;
(b) the molar heat of combustion for methanol.
8
From ΔHcomb values for methane, ethane and propane
(which are −885.4, −1547.3 and −2210.5 kJ mol−1,
respectively) estimate the increase in ΔHcomb per
added CH2 group in a hydrocarbon. Predict ΔHcomb
for octane on this basis, and compare this value to the
usually accepted value of −5468 kJ mol−1.
9
The heat released on neutralization of NaOH with all
strong acids is 56.2 kJ mol−1. The heat released on
neutralization of NaOH with HF (a weak acid) is
68.9 kJ mol−1. Calculate ΔH for the ionization of HF
in water.
Answers to ITQs
1
(a) chemical energy to electric energy and vice versa
(b) heat energy to kinetic energy
(c) chemical energy to heat energy
2
Effect of oceans moderating the weather, cooling of
our bodies when we perspire.
Use this data:
4C(s) + 4H2(g) + O2(g) → C3H7COOH(l) ΔH = −522 kJ
3
Aluminium since it requires less heat to change its
temperature by 1 °C.
C(s) + O2(g) → CO2(g)
4
The water freezes on the cold branches, releasing
latent heat. This raises the temperature of the blossom
sufficiently to preserve it from the intense cold.
5
Work is not a state function since it depends on the
distance travelled, which depends on the path taken
from the initial to final state. Heat also depends on
the path taken so it not a state function. Pressure and
volume are state functions.
6
54.2 kJ mol−1
Determine the heat of reaction for the following
combustion:
C3H7COOH(l) + 5O2(g) → 4CO2(g) + 4H2O(l)
ΔH = −392.9 kJ
2H2(g) + O2(g) →2H2O(g) ΔH = −241.6 kJ
H2O(l) → H2O(g)
4
ΔH = 43.9 kJ
Calculate the enthalpy change of formation for MgO,
ΔHf(MgO(s)) given the following data:
ΔH = 117 kJ
MgCO3(s) → MgO(s) + CO2(g)
ΔHf(MgCO3(s)) = −1113 kJ mol−1
ΔHf(CO2(g)) = −394 kJ mol−1
5
Given the following information:
ΔH f (SO2(g)) = −297 kJ mol−1
ΔH f (CaO(s)) = −635.5 kJ
mol−1
ΔH f (CaSO4(s)) = −1433 kJ mol−1
Calculate the enthalpy change for the reaction:
CaSO4(s) → CaO(s) + SO2(g) + ½O2(g)
6
7
Name the enthalpy changes for the following processes:
(a) H2O(s) → H2O(l)
(b) I2(s) → I2(g)
(c) 2K(s) + Cl2(g) → 2KCl(s)
(d) C3H8(g) + O2(g) → 3CO2(g) + 4H2O(l)
Calculate the heat produced when 1.00 gal of octane,
C8H18, reacts with oxygen to form carbon monoxide
and water vapour at 25 °C.
Data: density of octane = 0.7025 g cm−3;
1 gal = 3.785 dm3, ΔHcombustion(C8H18) = −1302 kcal
Answers to Review questions
1
42.2 kJ mol–1
2
(a) –31.09 kJ
(b) –66.3 kJ mol–1
3
–1708.4 kJ
4
–602 kJ mol–1
5
500.5 kJ
6
(a)
(b)
(c)
(d)
7
15 520 kcal
8
–5522.5 kJ
9
–12.7 kJ
fusion
sublimation
formation
combustion
87
88
Module 2
Kinetics and equilibria
Chapter 9
Chemical kinetics
Learning objectives
■ Explain what is meant by the rate of a reaction.
■ Apply the collision theory to explain factors affecting reaction rate.
■ Use the concept of a Boltzmann distribution in reaction rates.
■ Understand the information contained within a reaction profile.
■ Explain what is meant by the order of a reaction.
■ Distinguish between first-order and pseudo first-order reactions.
■ Explain what is meant by a reaction half-life.
■ Describe how to find the order of a reaction from experimental data.
Collision theory
Is there a way to explain why some reactions, such as
explosions, are fast and some, like the rusting of an old car,
are slow?
You need to remember that matter is made of particles. These
particles may be atoms, ions, molecules, micelles, colloidal
dispersions or anything else. What ever the particle, there
are only two ways in which a reaction can happen:
■ a particle can absorb energy (as heat, light or UV
energy) and split apart;
■ two particles can collide, making fragments which
recombine either in the same form or as something
different.
The second of these possibilities is the more common.
(movement from place to place). For a massive particle,
such as an atom or molecule, the movement energy
(kinetic energy) is the largest component. The magnitude
of kinetic energy is given by
1 2
mv
2
where KE is the kinetic energy, m the mass and v the
velocity.
KE =
We can show that for an ideal monatomic gas
3
kT
2
where T is the absolute temperature (in kelvin) and k is a
constant. Ideal gases are introduced in Chapter 7.
KE =
Combining the two equations gives
Now we can ask what factors are at work to regulate it.
–
1 2
3
mv = kT , so v ∝ √T
2
2
As the temperature rises, the particles rush around more
quickly. This applies in general to all particles, not just the
atoms of an ideal gas.
Internal energy
Bond energy
At any temperature above absolute zero, particles (whatever
they are) contain movement energy. This movement
energy can be in the form of vibration, spin or translation
We can widen the argument to consider molecules. When
the atoms or ions making up the molecule came together,
they stayed together because a bond formed between
We can set up a general chemical reaction:
AB + CD → AC + BD
Chapter 9 Chemical kinetics
them. Another way of saying this is that in the ‘together’
state, the molecule contained less energy than when its
constituents were separate. When bonds are formed, the
energy content of the system falls.
To split the particle up again, an equal amount of energy
must be supplied.
Collision effects
Number of molecules
Not all collisions will have the same energy. In a fluid (gas
or liquid) at a fixed temperature, some particles are moving
slowly, most rather more quickly and some very quickly.
This distribution of energies, known as the Boltzmann
distribution, is shown in Figure 9.1.
If the energy available in the collision is below ΔE there can
be no reaction.
If the energy is above ΔE then the reacting particles are
separated into their components and at that moment there
is an intermediate state containing reactive fragments:
AB + CD → A + B + C + D
At this point either the fragments will recombine, giving
the original reactants, or they will change partners and
combine giving the new materials AC and BD.
Product stability
We only say that a reaction has taken place if noticeable
amounts of AC and BD are produced. This happens if (AC
+ BD) is more stable than (AB + CD); this is the case if the
energy released in forming AC and BD is greater than the
energy released in recombining the original AB and CD.
Effect of molecular shape
Energy
6E
Figure 9.1 Particle energy (x-axis) against number of particles
having that energy (y-axis).
Most particles have an energy near to the red line. The
number of particles having an energy greater than a value
such as the dotted line is proportional to the area under
the curve to the right of that line, as shown by the shaded
portion.
If the proportion of collisions with enough energy to cause
a reaction goes up, the number of fruitful collisions per
second goes up and the reaction proceeds more quickly.
In a gas, or in a solution, the particles are moving largely
independently of each other. They move in straight lines
except when they bounce off the walls of the container
or into each other. When they collide, the sum of their
available energies can either be sufficient to split up the
molecules into fragments, or insufficient to do so. We can
show this energy as the value ΔE.
If the reacting particles are spherical it does not matter
which way they collide. If the particles have a reactive site
(for example, the –OH group in an alcohol) then the other
reagent must strike that site. The proportion of fruitful
collisions goes down.
Reaction rate
We have a verbal scale for reaction speed. We can say ‘doesn’t
react’ or ‘slow’ or ‘fast’ or ‘explosive’. However, we can also
measure the change in a reaction by measuring the change
in quantity of a reactant or product over time. In this case
we will record our results in mol s−1. We can also measure
some physical property of the reaction mixture, such as the
volume of a gas evolved, or the depth of colour or the pH.
Because the majority of reactions only take place when
particles collide, anything that changes the effective collision
frequency between particles will affect the reaction rate.
Factors to consider are:
■ concentration of a solution or the pressure of a gas;
■ temperature;
ITQ 1 What might happen if the collision energy was enough to
split up AB but not enough to split up CD?
■ surface area of a solid;
■ use of a catalyst.
Concentration
ITQ 2 Why are inorganic reactions usually faster than organic
reactions?
ITQ 3 What property of a gas do we use to express its
concentration?
If the concentration of a reactant is increased, the rate of
reaction may also increase. If we use the simple collision
theory outlined above, we could say that the reacting particles
are crowded together more, so they collide more often.
89
90
Unit 1 Module 2 Kinetics and equilibria
Experiments show that for the reaction
A+B→X+Y
one factor controlling the reaction rate is the concentration
of the reactants:
This reaction is first order with respect to both 2-bromo-2methylpropane and to water and so is second order overall
(see below). But if the water is present in a large excess
(as the solvent) then its concentration does not change
appreciably and the rate equation becomes
rate = k [A] [B]
rate = k [(CH3)3CBr]
However, the reactants may not react in a simple way. In
practice, the general rate equation is:
The reaction now appears to be first order – hence the term
‘pseudo’ first-order.
rate = k [A]a [B]b
where a and b are small numbers, usually 0, 1 or 2. The
constant k is called the rate constant for that reaction.
Order of reaction
The integers a and b in the rate equation define the order
of the reaction. In our example, the reaction is of order
a with respect to A, and of order b with respect to B. The
overall order is (a + b).
First-order reactions
Consider the reaction
A → products
The rate equation is found to be
rate = k [A]
Suppose that after a certain time t the rate has fallen to ½
the original value. After another interval t the rate will be
½ × ½ = ¼ and so-on. The time for the rate to reduce by a
factor of 2 is called the half-life of the reaction. Radioactive
decay proceeds in this way (see Chapter 3).
Second-order reactions
There are two types of second-order reactions:
■ those in which two different species react together, the
reaction being first order with respect to each;
■ those where there is only one reactant in the rate
equation, but the reaction is second order with respect
to that species.
An example of the first type is the alkaline hydrolysis of an
ester such as ethyl ethanoate:
CH3COOC2H5 + OH− → CH3COO− + C2H5OH
For this reaction, the rate equation is
rate = k [CH3COOC2H5] [OH−]
Notice that because only one mole of each reactant is
involved in the change, each term in the equation is only
to the first order.
An example of the second type, where there is only one
reactant, is the decomposition of nitrogen dioxide:
2NO2 → 2NO + O2
rate = k [NO2]2
Worked example 9.1
Q
The half-life of tritium (3H) is 12.3 years. A sample is decaying
at the rate of 1000 atoms per second. How long will it be
before the decay rate is only 125 atoms per second?
A
125
1
1
1
1
= = × × so three half-lives are needed.
1000 8
2
2
2
Half-life is 12.3 years, so decay will take 3 × 12.3 = 36.9 years
Pseudo first-order reactions
When a reagent is present in considerable excess, its
concentration does not change appreciably during the
reaction, and so the order with respect to that reagent is
effectively zero.
Consider the reaction
(CH3)3CBr + H2O → products
Here, because the reactant appears to the second order in
the chemical equation for the reaction, a first suggestion
would be that the corresponding term in the rate equation
is squared.
You could write the reaction as
NO2 + NO2 → 2NO + O2
This link between the chemical equation for the reaction
and the rate equation is true of any single-step reaction.
However, you must be aware that many reactions take
place in a series of steps; if this is the case then the order of
the reaction can only be found by experiment – which is
discussed later in this chapter (page 93).
ITQ 4 If the reaction between (CH3)3CBr and H2O is performed
with methanol as the solvent instead of water, suggest the rate
equation for the reaction.
Chapter 9 Chemical kinetics
Zero-order reaction
2NH3 → N2 + 3H2
on a molybdenum catalyst, the surface of the catalyst
becomes covered with ammonia molecules whatever the
gas pressure. Therefore the reaction rate is independent of
the gas pressure.
An example of the second type is the reaction between
iodine and acetone (propanone).
CH3COCH3(aq) + I2(aq) → CH2ICOCH3(aq) + HI(aq)
The rate of this reaction does not change if the concentration
of the iodine is changed. The reaction is said to be zero
order with respect to iodine. This tells us nothing about the
order of reaction overall.
Never assess the order of a reaction from the equation
alone. Dinitrogen oxide decomposes to give nitrogen and
oxygen:
2N2O → 2N2 + O2
but the reaction is first order. This is because many reactions
take place in stages, through intermediates, and one stage
may be slower than others and so dominate the kinetics.
Temperature
Increasing the temperature of a reaction has two effects:
Potential energy
The first type is a gas reaction catalysed by a metal surface.
In the reaction
Ea
reactants
Reaction coordinate
Figure 9.2 A reaction profile for an exothermic reaction showing
the activation energy (Ea). The y-axis shows the potential energy
of a pair of reactant particles. The x-axis shows the course of the
reaction, which is called the reaction coordinate.
For a reaction to take place, a collision must provide energy
equal to, or greater than, the energy at the peak of the
profile. This energy (Ea) is called the activation energy of
the reaction.
If the temperature is increased we can see from the diagram
of the Boltzmann distribution that a higher percentage of
particles will have more than a particular amount of energy,
and therefore the percentage of collisions with more than
the activation energy will also be higher (Figure 9.3).
■ the energy of the particles is increased, so the
proportion of the collisions that offer equal or more
than the activation energy of the reaction increases
(activation energy is discussed in detail below);
■ the reacting particles travel faster and therefore collide
more frequently.
We can draw an energy profile for a reaction (Figure 9.2).
This type of graph shows the potential energy of a colliding
pair of reactant particles. Initially much of the energy is
kinetic energy, but as the reactant particles approach
each other that energy is converted to potential energy
to overcome the repulsion of the electron clouds. This
potential energy reaches a maximum at the point where
the old bonds are breaking and the new bonds are forming.
This is called the transition state.
ITQ 5 In Figure 9.3, what is the relationship between the total
area under the green line and the total area under the red line?
6H
Number of molecules
There are two types of reaction that appear to be
independent of the pressure or concentration of a reactant.
transition state
Energy
E
Figure 9.3 The Boltzmann distribution, showing the effect of
raising the temperature of a fixed number of particles. The
red line represents the Boltzmann distribution at a higher
temperature. The area under the red line to the right of the
activation energy (Ea) is greater than the green shaded area. The
hotter substance has a higher proportion of particles with more
than the activation energy.
Catalysis
A catalyst is a substance that speeds up a reaction without
itself being used up in the process. A catalyst may act by
forming an intermediate product or by providing a surface
on which reacting molecules can come together. In either
case, the catalyst does not lower the activation energy of
the reaction: it gives an alternative reaction path with a
lower activation energy (Figure 9.4).
91
Unit 1 Module 2 Kinetics and equilibria
astonishing efficiency at or near room temperature and
at the pH of the living organism of which they form
part. Enzymes are highly specific. For example, human
saliva contains amylase, which catalyses the conversion
of starch into sugars but has no other effect.
transition state
Ea
Potential energy
92
reactants
6H
Reaction coordinate
Figure 9.4 The reaction profile for a reaction. The green line shows
the uncatalysed reaction and the red line the catalysed reaction.
The final energy of the system is unchanged but, since Ea is lower
in the presence of the catalyst, more collisions will be fruitful.
Examples of catalysis
■ Dilute acid catalysing the hydrolysis of an ester
Enzyme molecules are coiled up into specific shapes.
One theory for their action, and for their specificity, is
that the enzyme and reagent molecules come closely
together, and the shape of the enzyme molecule fits
neatly onto the active part of the reactant molecule. This
is called the lock-and-key mechanism (Figure 9.5).
Enzymes can be very effective: one enzyme working in
the replication of a DNA molecule not only catalyses a
polymerization, but checks that the resultant polymer
molecule is correct.
substrate
(Chapter 25, page 205):
CH3CH2CH2COOC2H5 + H2O → CH3CH2CH2COOH + C2H5OH
ester
acid
alcohol
The acid provides a high concentration of protons
(H+), which are the catalyst. This reaction takes place
in solution. There is only one liquid phase, and so the
effect is called homogeneous catalysis.
■ Finely divided nickel metal (‘Raney nickel’). The metal
is actually an alloy of nickel and aluminium, from
which much of the aluminium has been dissolved
out. The nickel remains as a spongy mass with a high
surface area. A spoonful of Raney nickel has a surface
area of about 100 m2.
Raney nickel is used to catalyse the hydrogenation of
vegetable oils, converting C=C bonds to –C–C– bonds
and so turning the liquid oil into solid materials such as
margarine.
The metal surface combines with both the doublebonded molecule and the hydrogen to give an
intermediate; the molecules then react together.
In this example, more than one phase (liquid, solid or
gas) is present and the process is called heterogeneous
catalysis.
The conversion of nitrogen and hydrogen into ammonia
in the Haber process is similar, using an iron catalyst
(see Chapter 37, page 372).
■ Platinum metal is used in motor car exhausts to make
possible the conversion of nitrogen oxides (NOx) and
carbon monoxide into nitrogen and carbon dioxide.
■ Enzymes are proteins – long, linear chains of amino
acids that fold to produce a three-dimensional
product. They act as biological catalysts and work with
enzyme
ES complex
Figure 9.5 An enzyme fits directly onto a reactant – known as the
substrate – to form an intermediate complex in a reaction.
Catalyst poisoning
Many catalysts, particularly heterogeneous catalysts,
become much less active in the presence of oxygen or
sulfur compounds. It is likely that the oxygen or sulfur
compounds bond to the catalyst surface and so prevent it
from working. This is called catalyst poisoning.
In the Haber process for the production of ammonia,
methane (CH4) is used as the source of hydrogen. The
methane is first cleaned of hydrogen sulfide. Then the
hydrogen needs to be cleaned of the carbon monoxide
produced in the first stage of the process:
CH4 + H2O → CO + 3H2
Both hydrogen sulfide and carbon monoxide are powerful
catalyst poisons.
Unusual catalytic effects
Strangely, some catalysts reduce the rate of reactions,
which contradicts the definition of a catalyst! How can this
happen? When two catalysts are added to a reaction mixture
they are usually more effective than one alone. But if one
catalyst happens to form an intermediate compound which
reacts more slowly, then the other catalyst accelerates the
formation of this unwanted intermediate and the reaction
slows down.
Chapter 9 Chemical kinetics
Light-induced reactions
At the beginning of the chapter, we said that most reactions
proceed through collision. Some reactions, however, take
place differently in the dark and in sunlight. A glass flask
filled with a mixture of equal parts of hydrogen gas and
chlorine gas is stable at room temperature in a dark room.
Throw the flask out through the window into the sunlight
and it will explode before it hits the ground.
The photons (energy packets) of UV light carry enough
energy to split chlorine molecules into atoms:
Rate measurements
The rate of a reaction is found from a graph of concentration
(or pressure of a gas) against time. The slope of this graph
at any time is the rate of change at that time. From such
a graph we can derive a plot of rate against concentration
of a reactant. Also, the shape of the graph can tell us the
order of the reaction with respect to that substance. If the
substance is a reactant, its concentration falls with time
(Figure 9.6). If it is a product, its concentration rises.
6A
A / mol dm–3
Some reactions produce their own catalyst as they proceed.
For example permanganate ions (MnO4−) are slowly
reduced by oxalic acid to near-colourless Mn2+ ions. The
reaction is catalysed by Mn2+ ions, so as they are formed
in a permanganate/oxalic acid mixture, the reaction rate
increases progressively.
Cl–Cl → Cl + Cl
t
Time / s
The two electrons of the Cl–Cl bond are shared equally
between the two atoms. We show this by using dots for the
electrons, so the reaction is better written as:
Cl–Cl → Cl• + Cl•
The fragments carrying the odd number of electrons are
called free radicals. Free radicals are very reactive and a
chain reaction is set up:
H–H + Cl• → HCl + H•
H• + Cl2 → HCl + Cl• and so on
Determining the order of reaction
Think about our general reaction:
A + B → products
It seems that this is likely to be a second-order reaction
with a rate equation
rate =k [A]a [B]b, where a and b can be 0, 1, 2 …
■ However, as we noted above (page 90), the order of
Figure 9.6 Calculating the rate of a reaction.
Follow the steps to find the order of a reaction.
■ Draw a graph of concentration of reagent against time.
■ Draw a tangent to the curve at a selected time. You can
use t = 0 to find the initial rate.
■ Use the tangent as the hypotenuse of a triangle that
cuts both axes.
■ Divide the value of [A] at the point where your
tangent intercepts the axis by the corresponding value
of t.
Δ[A]
■ This gives you
, the rate of reaction.
t
■ You can repeat this for various values of t, to draw
another graph showing the variation of rate with time
■ Alternatively, you can repeat the experiment several
times, starting with different concentrations of
reagent, and use the initial reaction rates to draw
a graph showing the variation of rate with reagent
concentration.
reaction does need to be determined by experiment:
■ if changing the initial concentration of A between two
experiments does not change the initial rate, then
a = 0 and the reaction is zero order with respect to A;
■ if doubling the initial concentration of A doubles the
initial reaction rate, then a = 1 and the reaction is the
first order with respect to A;
■ if doubling the initial concentration of A quadruples
the initial reaction rate, then a = 2.
ITQ 6
What are the units of [A] and
Δ[A]
t
?
93
Unit 1 Module 2 Kinetics and equilibria
Worked example 9.2
This data was derived from an experiment in which a reagent A
was progressively used up. Calculate the order of the reaction
with respect to A.
A
Table 9.1 gives data on how the concentration of A alters with
time. Plotting this data gives Figure 9.7.
Tangents are drawn on Figure 9.7 when the concentration
of A is 1, 0.6 and 0.2 mol dm−3.
Table 9.1 Step 1, the raw data
Concentration of A / mol dm−3
1.0
0.84
0.71
0.59
0.5
0.42
0.35
0.25
0.21
Time /s
0
5
10
15
20
25
30
40
45
1.2
1.0
Concentration of A / mol dm–3
Q
0.8
0.6
0.4
0.2
0
0
20
40
60
Time / s
Figure 9.7 Step 2, plotting the data and drawing the tangents.
Data from the tangents is collected in Table 9.2.
Table 9.2 Step 3, data derived from Figure 9.7 and plotted
on Figure 9.8.
Concentration
1
0.6
0.2
Rate
0.035
0.022
0.007
0.040
0.035
Step 1 shows the raw data.
■ Step 2 shows the concentration versus time graph with
some tangents drawn.
■ Step 3 shows the values of the rates and concentrations.
■ Step 4 shows the rate versus concentration graph.
The graph in Figure 9.8 is very nearly a straight line. The
deviation might be because:
■ the original data is not quite accurate;
■ data is only given to two significant figures;
■ the tangents were not easy to draw with complete accuracy.
The graph in Figure 9.8 is a straight line within our limits of
accuracy. It is certainly not a curve. Therefore we can conclude
that the rate is proportional to the concentration of the reagent.
The reaction is first order with respect to that reagent.
■
0.030
0.025
Rate
94
0.020
0.015
0.010
0.005
0
0
0.5
1.0
Concentration / mol dm–3
Figure 9.8 Step 4, plotting the rate data against time. The graph
is very nearly a straight line.
Chapter 9 Chemical kinetics
Finding half-life
Half-life is constant for a first-order reaction.
The half-life is the time it takes for half the reagent to be
removed.
Simple inspection of the data in Worked example 9.2
shows that the half-life is 20 seconds (look back at Table
9.1 and look at the data for t = 20 s and t = 40 s). If the data
is less accommodating, then look at the graph (Figure 9.7).
Here you need to pick any two convenient concentrations
that differ by a factor of two and read off the time interval
between them.
The mathematical rule for a first-order reaction is that
Summary
✓ The rate of a reaction can be measured in units
of mol s−1 of a reagent.
✓ Most reactions occur when particles collide.
✓ Collision frequency is altered by concentration
and temperature.
✓ Collisions must provide an activation energy
sufficient to break existing bonds.
✓ Activation energy may be lowered by a catalyst.
✓ Some reactions are triggered by energy input
rate = k [A]
from visible or UV light.
and the half-life is given by the expression
✓ Enzymes are highly efficient biological catalysts.
0.693
t 12 =
k
At this stage you need not know how the expression is
derived.
✓ The general rate equation is
rate = k [A]a [B]b … [N]n.
✓ The order of a reaction is given by the sum
(a + b + … n).
Review questions
1
The following data were obtained for the reaction
A+B→P
ITQ 7 The reaction between carbon monoxide and nitrogen
dioxide is a first-order reaction with respect to carbon monoxide.
CO + NO2 → CO2 + NO
In an experiment using this reaction, NO2 was continually added
to the reaction mixture to keep its concentration in the reaction
vessel constant. It was found that the concentration of CO fell
from its initial value to half that value in 3.5 seconds.
(a) What is the value of k in the rate equation:
−d[CO]
= k[CO]
dt
(b) What are the units of k?
Experiment
Initial [A]
/ mol dm–3
Initial [B]
/ mol dm–3
1
2
3
4
0.10
0.20
0.20
0.40
0.10
0.10
0.20
0.20
Initial rate of
formation of
P / mol dm–3 s–1
4.0 × 10–6
16.0 × 10–6
32.0 × 10–6
?
Use this data to determine:
(a) The order of the reaction with respect to reactant A.
(b) The order of the reaction with respect to reactant B.
(c) The overall order of the reaction.
(d) The rate equation for the reaction.
(e) The rate constant for the reaction.
(f) The initial rate of formation of P for experiment 4.
95
Unit 1 Module 2 Kinetics and equilibria
2
Consider the following reaction and its associated rate
law:
A+B→C
6
rate = k[A]2
D
Discuss the effect that you would expect on the rate of
the reaction of each of the following changes:
(a) Increasing the concentration of A.
(b) Increasing the concentration of B.
(c) Increasing the temperature of the reaction.
3
A
C
Reaction progress
(a) Indicate which letter represents each of the
following:
(i) the activation energy of the forward reaction
(ii) the heat of reaction
(iii) the activation energy of the reverse reaction?
(b) Is the reaction exothermic or endothermic?
Explain your answer.
D → E + C (slow)
Write:
(a) the rate equation for this reaction;
(b) the overall equation for the reaction.
Consider the reaction:
2H2(g) + 2NO(g) → 2H2O(g) + N2(g)
7
A large piece of calcium carbonate is reacted with
dilute hydrochloric acid at a constant temperature.
Explain how grinding the mass of calcium carbonate
into smaller particles would affect the rate of the
reaction.
8
Give an example of an enzyme that is very important
in the human body and explain its role in the body.
9
Grain that is stored in a grain elevator is unreactive.
However, if a spark is placed near fine dust in the
silos, an explosive reaction will occur. Explain this
observation.
The rate law is rate = k [H2] [NO]2. At 700 °C, the
rate of the reaction is 3 × 10–3 mol dm–3 s–1 when the
concentration of hydrogen is 1 × 10–3 mol dm–3 and
that of NO is 6 × 10–3 mol dm–3.
Determine:
(a) the rate constant for the reaction;
(b) the order of the reaction.
5
B
The proposed mechanism for a reaction is
A + B + C → D (fast)
4
Use the following diagram to answer the questions
below.
Potential energy
96
At room temperature, hydrogen peroxide decomposes
to produce oxygen and water.
(a) Write a balanced equation to represent this
reaction.
(b) Suggest an experimental method that one could
use to determine the rate of this reaction.
(c) What effect would adding manganese(IV) oxide
have on the rate of this reaction?
(d) Catalase, an enzyme found in liver, can cause
a change in the decomposition of hydrogen
peroxide. What effect would you expect if a piece
of liver was placed in hydrogen peroxide solution?
10 Use the collision theory and suitable diagrams to
explain how the following affect reaction rates:
(a) surface area
(b) temperature
(c) concentration
(d) catalyst
Chapter 9 Chemical kinetics
11 The following graph represents the concentration
of H2O2(aq) over time for the decomposition of
hydrogen peroxide into water.
[H2O2 ] / mol dm–3
2H2O2(l) → 2H2O(l) + O2(g)
A
B
Answers to ITQs
1
There would be no reaction.
2
Inorganic reaction are often between ions which are
spherical or nearly so. Organic reactions are often
between molecules with complex shapes.
3
Pressure.
4
rate = k [(CH3)3CBr] [H2O]
5
They are the same.
6
mol L−1 or mol dm−3; mol dm−3 s−1
7
(a) Half-life is 3.5 s
Time, s
(a) Explain how the rate of reaction at A can be
determined.
(b) How would you expect the rate of the reaction at
A to compare with that at B?
(c) On the same graph, sketch the general shape of a
curve, showing the concentration of H2O versus
time.
12 In an experiment, a large excess of zinc was added to
100 cm3 of 0.2 M hydrochloric acid to produce
240 cm3 of hydrogen.
(a) Determine the volume of hydrogen that would be
produced by each of the following changes to the
reaction:
(i) adding 100 cm3 of water to the reaction vessel;
(ii) adding an additional 10 g of zinc the reaction
vessel;
(iii) adding an additional 50 cm3 of the 0.2 M HCl
to the reaction vessel.
(b) Describe the effect that each of the changes in part
(a) would have on the reaction rate.
k=
0.693
= 0.190
t 12
(b) Units are
constant
= s−1
time
Answers to Review questions
2
(a) rate should increase
(b) no change since rate independent of [B]
(c) increase in rate
3
(a) rate = k[D]
(b) A + B → E
4
(a) 8.3 × 104 mol–2 dm6 s–1
(b) third order
5
(a) 2H2O2(l) → 2H2O(l) + O2(g)
(b) Determine the volume of oxygen gas produced.
Use a gas syringe, inverted measuring cylinder full
of water, balance (to record mass lost).
(c) Increase the rate of decomposition.
(d) Increase the rate of decomposition.
6
(a) (i) A
(ii) C
(iii) B
(b) endothermic
11 (a) Draw a tangent to the curve at A and find the
slope of the tangent.
(b) Rate at B is less than at A.
(c) Increases with time.
97
98
Chapter 10
Chemical equilibrium
Learning objectives
■ Explain what is meant by ‘chemical equilibrium’.
■ State the characteristics of a chemical reaction
when it is at equilibrium.
■ Explain the meaning of the ‘equilibrium constant’.
■ State and use Le Chatelier’s principle.
Reactions take place when particles meet. As a reaction
continues, the number of reactant particles falls and so they
meet less often. As a result, the rate of the reaction falls.
There are no products at the start of the reaction, so only
the forward reaction can take place. As product particles
are formed, the reverse reaction becomes possible. The rate
constant for the reverse reaction will be different from that
of the forward reaction. The reverse reaction, however slow
or fast, will accelerate as time goes on as the concentration
of the products increases (Figure 10.1). See Chapter 9 for
more about rates of reaction.
Concentration
products
Rate
Reversible reactions
forward
reverse
Time
Figure 10.2 Progress of a reaction: rates of forward and reverse
reactions.
This is dynamic equilibrium because both reactions are
continuing. Neither reaction has stopped. By contrast,
your book is in static equilibrium on your desk: neither
the book nor the desk is moving.
Let’s consider a general reaction:
A+BҡC+D
reactants
■ For the forward reaction:
rate = k1 [A] [B]
Time
Figure 10.1 Progress of a reaction: reagent concentration.
Sooner or later, the forward reaction and the reverse
reaction will proceed at the same rate (Figure 10.2). This
means that products are formed at the same rate as they
are lost. At this point the reaction is said to be in dynamic
equilibrium. The characteristic of dynamic equilibrium is
that the reaction appears to stop.
■ For the reverse reaction:
rate = k2 [C] [D]
At equilibrium the rates are equal:
rate = k1 [A] [B] = rate = k2 [C] [D]
Re-arranging this means we get:
k1 [C] [D]
=
k2 [A] [B]
Chapter 10 Chemical equilibrium
Because k1 and k2 are both constants, this becomes
K=
[C] [D]
[A] [B]
where K is the equilibrium constant for the reaction.
■ If K is defined in terms of concentrations of substances
we refer to Kc.
■ If K is defined in terms of the pressures of gases in a
The expression contains p(NH3)2 and not p(NH3) because
there are two molecules of ammonia in the equation.
The equation could have been written as
N2(g) + H2(g) + H2(g) + H2(g) ҡ NH3(g) + NH3(g)
Similarly the partial pressure of the hydrogen appears
as p(H2)3.
mixture we refer to Kp.
It is customary to put the concentrations of the products as
the numerator of the ratio (the terms on top; [C] and [D]
in our example) and the concentrations of the reactants as
the denominator (the terms underneath; [A] and [B] in
our example).
Worked example 10.1
Q
In a reaction between ethyl alcohol (EtOH) and ethanoic acid
(HAc), the products are ethyl ethanoate (EtAc) and water:
EtOH + HAc ҡ EtAc + H2O
Under certain conditions a mixture of 2.0 mol EtOH and 1.0 mol
HAc reacted to give, at equilibrium, 0.80 mol of EtAc.
Calculate the equilibrium constant for this reaction under these
conditions.
A
If 0.80 mol EtAc were produced then 0.80 mol EtOH and 0.80
mol HAc were used.
The remaining quantities of each are therefore:
(2.0 − 0.8) = 1.2 mol EtOH
(1.0 − 0.8) = 0.2 mol HAc
The equilibrium constant is given by:
Here are three important points about any equilibrium
constant:
■ it is only constant at a given temperature – its value
will be different at any other temperature;
■ it is unaffected by the presence of a catalyst – the
catalyst will shorten the time needed to reach
equilibrium, but does not affect its position;
■ it is determined as the ratio of two rates – if the
forward reaction is fast compared to the reverse
reaction, the equilibrium constant will be small, and
vice versa.
Gas reactions
When considering gas reactions it is possible to express
concentrations exactly, as if you were dealing with a
solution. The units would be mol dm−3. However, it is usual
to express gas concentrations in terms of the pressure of
each gas in a mixture. Each gas contributes its own partial
pressure to the total pressure. As an example, let’s look at
the reaction between hydrogen and nitrogen to produce
ammonia:
N2(g) + 3H2(g) ҡ 2NH3(g)
We can write:
Kp =
Kc =
[EtAc] [H2O]
[EtOH] [HAc]
If the volume of the mixture is V cm3 then we may write:
(
(
EtAc × 1000
V
Kc =
EtOH × 1000
V
All the factors
Kc =
)(
)(
H2O × 1000
V
HAc × 1000
V
)
)
1000
cancel out so the expression reduces to:
V
0.8 × 0.8
= 2.7 to 2 sig. figs
1.2 × 0.2
Note that in this example the factors converting quantity to
concentration all cancel out – this may not always be the case.
Similarly the value of Kc is a pure number – again, this may not
always be the case.
p(NH3)2
p(N2) p(H2)3
ITQ 1 If a reaction reaches equilibrium when only a small fraction
of one reactant has been used, is the equilibrium constant low or
high?
ITQ 2 Consider this reaction:
A+BҡC
What are the units of the equilibrium constant?
99
100
Unit 1 Module 2 Kinetics and equilibria
A system in dynamic equilibrium will change to diminish
any alteration applied to it.
Worked example 10.2
Q
A
At a particular temperature, 1.00 mol of hydrogen and
1.00 mol of iodine were allowed to react to equilibrium.
The total pressure in the reaction vessel at equilibrium was
2.00 atm. The quantity of HI present was then 1.70 mol.
Calculate the equilibrium constant for the reaction under these
conditions.
H2 + I2 ҡ 2HI
Kp =
p(HI)2
p(H2) p(I2)
0.15
atm
2
1.70
atm
2
Substituting into the equation for Kp gives:
Kp =
Applying Le Chatelier’s principle
Let’s use the reaction between nitrogen and hydrogen,
producing ammonia, as an example:
N2(g) + 3H2(g) ҡ 2NH3(g)
From the equation, we find that 1.70/2 = 0.85 mol of hydrogen
and 0.85 mol of iodine were used up, so 1.00 − 0.85 = 0.15 mol
of each was left. The total pressure was 2.00 atm.
The partial pressures were therefore:
H2
I2
HI
0.15
atm
2
If a small change is made to a system in equilibrium, the
system will alter to minimize the change.
( )
( )( )
1.70
2
2
0.15
0.15
×
2
2
= 128
These examples all refer to chemical changes. But the
concept of dynamic equilibrium can be applied more widely.
■ A liquid in a closed container will evaporate until the
space above it is saturated with vapour. When the
vapour is saturated the rate at which molecules leave
the surface is equal to the rate at which they re-enter.
■ If the rate at which water runs into a bath is equal to
the rate at which it runs out, the level of water in the
bath will not change.
■ If at a road junction, vehicles arrive at the same rate as
they leave, there will be no traffic jam. If more vehicles
arrive per second than leave, traffic will build up at the
junction.
The characteristic of a dynamic equilibrium is that the
system appears to stop changing: the composition of the
mixture does not change with time.
Amongst other things, the equilibrium point is governed by
the rate at which hydrogen molecules collide with nitrogen
molecules. Now consider what happens if we now inject
extra hydrogen into the system, keeping the volume the
same.
■ The total pressure and the partial pressure of hydrogen
go up.
■ The collision rate between hydrogen and nitrogen will
go up.
■ The rate of production of ammonia molecules will go up.
After a while, there will be more ammonia molecules in
the mixture than before.
■ The reverse reaction, producing hydrogen and nitrogen
accelerates.
■ A new equilibrium is established.
Nothing in that argument tells us whether the concentration
of ammonia will go up or down. However, Le Chatelier’s
principle demands that the original change (the increase
in pressure caused by the addition of hydrogen) must be
minimized. To do this, the total number of molecules in the
reaction must be reduced.
The total number of molecules in the reaction will fall
when a molecule of nitrogen reacts completely with three
molecules of hydrogen: four molecules in total are lost and
only two molecules of ammonia are produced. So, the new
equilibrium must be one containing more ammonia than
the original.
ITQ 3
(a) State Le Chatelier’s principle.
Le Chatelier’s principle
Henry Louis Le Chatelier, a French chemist (1850–1936),
investigated the effect of small changes of concentration,
pressure and temperature on systems in equilibrium. His
result (sometimes ennobled as a ‘law’) can be stated in
many ways. Because it can be difficult to understand, here
are two.
(b) State two factors that can disturb the equilibrium of a system.
(c) The equation for the calcination (decomposition) of limestone
can be represented as:
CaCO3(s) ҡ CaO(s) + CO2(g)
Applying Le Chatelier’s principle, state the effect on the
equilibrium when:
i)
a small quantity of CaCO3(s) is added.
ii) some CO2(g) is removed.
Chapter 10 Chemical equilibrium
The common-ion effect
The converse is true for an endothermic reaction.
A second example of how to apply Le Chatelier’s principle
is in the ‘common-ion effect’. The common-ion effect is
when the concentration of an ion, already existing as part
of an equilibrium mixture, is increased.
Equilibrium constants for gas reactions (Kp) do not change
with pressure if the total number of molecules is the same
on both sides of the equation.
A solution of copper sulfate in pure water is often slightly
cloudy because of the reaction
H2(g) + Cl2(g) ҡ 2HCl(g)
Cu2+(aq) + 2H2O(l) ҡ Cu(OH)2(s) + 2H+(aq)
Kc =
If an amount of a strong acid is added, the solution clears.
The concentration of the H+(aq) ion has been increased.
Le Chatelier’s principle tells us that the increase must be
minimized. This can only happen if the reaction proceeds
in reverse, using up some of the added hydrogen ions
and so reducing their concentration. The insoluble copper
hydroxide vanishes from the mixture.
However, in the Haber process reaction, shown earlier:
Looked at from the point of view of the equilibrium
constant, the equation leads to the expression
Kc =
[H+]2
[Cu2+]
We can ignore the concentration of the water, because it
hardly changes, and the concentration of Cu(OH)2 as it is
constant.
If [H+] is increased, the ratio Kc goes up. Le Chatelier’s
principle says that it must be reduced again. This can only
happen if either [H+] goes down or [Cu2+] goes up. When
the Cu(OH)2 dissolves, both these things happen. Some H+
is removed and some copper ions pass back into solution.
There is nothing in Le Chatelier’s principle to say how
quickly any changes will take place. It is possible that
they will happen very quickly, or that they will happen so
slowly that they are un-noticeable. This is called kinetic
hindrance.
Changes in the value of K
The position of equilibrium of a reaction, and hence the
value of K, is unchanged by the addition of a catalyst. A
catalyst speeds up the rate at which equilibrium is reached
but has no effect on its position.
Equilibrium constants (Kc) vary with temperature. If a
reaction is exothermic the reaction mixture heats up. If
a mixture is at equilibrium and is heated, Le Chatelier’s
principle demands that the temperature rise be minimized
and so the reaction tends to reverse: the concentration
of products goes down and the value of K at the higher
temperature falls below its value at the lower temperature.
For example:
[HCl]2
[H2] [Cl2]
N2(g) + 3H2(g) ҡ 2NH3(g)
If the pressure at equilibrium is changed then the number
of molecules (and hence their concentrations) will change
to restore the original value of Kc.
101
102
Unit 1 Module 2 Kinetics and equilibria
Summary
Review questions
1
✓ A system comes to equilibrium when the rates of
the forward and reverse reactions become equal.
✓ At equilibrium, the overall reaction appears to
stop.
(a) State what is meant by Le Chatelier’s principle.
(b) Describe how the following equilibria would be
affected by the proposed changes.
(i) An increase in pressure on the reaction
ΔH = −ve
2SO2(g) + O2(g) ҡ 2SO3(g)
(ii) A decrease in temperature on the equilibrium
N2(g) + 3H2(g) ҡ 2NH3(g)
ΔH = −ve
✓ The position of equilibrium can be defined using
an ‘equilibrium constant’.
(iii) An increase in pressure on the system
H2(g) + I2(g) ҡ 2HI(g)
✓ The equilibrium constant can be in terms of
concentration (Kc) or, for gases, in terms of
partial pressures (Kp).
(c) Write expressions for the equilibrium constant, Kc,
for each of the reactions in part (b), stating units
where appropriate.
(d) State how the value of Kc for each of the reactions
in part (b) is affected by:
(i) change in temperature
(ii) change in pressure
(iii) the addition of a catalyst.
(e) A mixture of SO2 and O2 of concentrations
a mol dm−3 and b mol dm−3 respectively reach
equilibrium when x moles of SO2 had reacted.
Derive an expression for Kc.
✓ The position of equilibrium is affected by
variables such as pressure and temperature.
✓ Le Chatelier’s principle states that ‘if a small
change is applied to a system in equilibrium,
the system will change so as to minimize the
change’.
✓ Studies of equilibria give no information about
reaction rates.
2
(a) BiCl3(l) + H2O(l) ҡ BiOCl(s) + 2HCl(aq)
Given the equilibrium above, describe how
the effect of changes in concentration can be
demonstrated.
(b) When chlorine and nitrogen(II) oxide are mixed,
an endothermic reaction occurs. The reaction can
be represented by the equation:
Cl2(g) + 2NO(g) ҡ 2NOCl(g)
(i) Deduce an expression for Kp of this reaction.
(ii) Explain the effect of an increase in
temperature on the value of Kp.
(c) Explain the effect on the position of equilibrium
caused by:
(i) an increase in pressure
(ii) the use of a catalyst.
Chapter 10 Chemical equilibrium
3
The second step in the Contact process used for the
manufacture of sulfuric acid involves the dynamic
equilibrium represented below:
ΔH = –98 kJ mol−1
2SO2(g) + O2(g) ҡ 2SO3(g);
The reaction is carried out at 420 °C under
approximately 2 atmospheres pressure and the gases
are passed over beds of vanadium(V) oxide catalyst.
Excess oxygen is used.
(a) List four characteristics of a system that is in
dynamic equilibrium.
(b) Explain what effect (if any):
(i) the vanadium(V) oxide has on the position of
equilibrium
(ii) using a temperature of 600 °C would have
on the equilibrium concentration of sulfur
trioxide
(iii) a change in pressure on the equilibrium
system
(iv) a catalyst on the equilibrium system.
Answers to ITQs
1
2
3
Low: the concentrations of products are small and the
concentrations of reactants have hardly changed.
1
(mol dm−3)
=
= dm3 mol−1
(mol dm−3) (mol dm−3) (mol dm−3)
(a) Le Chatelier’s principle states that if a change in
conditions is made to a system in equilibrium,
the system moves in the direction which will
oppose the change. The system always proceeds to
re-establish equilibrium.
(b) Changes in pressure, concentration.
(c) (i) the equilibrium shifts to the right.
(ii) the equilibrium shifts to the right.
103
104
Chapter 11
Acid/base equilibria
Learning objectives
■ Explain what is meant by an acid and a base.
■ Explain the meaning of ‘strong’ and ‘weak’ when applied to acids and bases.
■ Calculate the pH of an acidic solution.
■ Describe the function of a pH indicator.
■ Explain what is meant by a ‘buffer solution’.
■ Calculate the pH of a specific buffer.
■ Have an understanding of the common-ion effect.
■ Use the concept of a solubility product.
What are acids and bases?
Acid/base reactions
By 1889, an acid was defined as a substance that releases
hydrogen ions in aqueous solution. However, a hydrogen
ion – which is just a proton – is so small that the electric
field around it is very strong. The proton interacts with
neighbouring water molecules to give what is called the
oxonium ion, H3O+:
Dry hydrogen chloride gas has no acidic properties. For
example, it has no effect on dry indicator paper. In water,
however, hydrogen chloride reacts as an acid:
H+(aq) + H2O(l) ҡ H3O+(aq)
In 1923, Brönsted and Lowry proposed the definition
that an acid is a substance that donates a proton to another
substance. Working from this proposition, a base is defined
as a substance that accepts a proton from another substance.
In an acid/base reaction, a proton is exchanged between
the two substances.
+
acid ҡ base + H
base + H+ ҡ acid
This means that no substance can act as an acid unless
there is a base to accept the proton. The equations we have
just used do not represent real reactions because the bare
proton cannot exist in solution. In an equation where we
would have written H+, we can now write H3O+.
The acid and the base that are related in this way are called
a conjugate pair.
HCl(g) + H2O(l) ҡ H3O+(aq) + Cl−(aq)
Here, the water is acting as a base because it accepts a
proton from the HCl. In aqueous solution it is the H3O+
that provides the proton for reactions.
Similarly, dry ammonia gas reacts with water:
H2O(l) + NH3(g) ҡ NH4+(aq) + OH−(aq)
Here, the water is acting as an acid. It donates a proton to
the NH3, leaving the OH− ion to react.
In most aqueous reactions the water is present in excess
and only a little takes part in the reactions.
If we add together the products from the two equations we
have looked at so far, as if we were reacting HCl and NH3,
we end up with:
HCl(g) + NH3(g) + 2H2O(l) →
H3O+(aq) + Cl−(aq) + NH4+(aq) + OH−(aq)
The products of this equation simplify to
Cl−(aq) + NH4+(aq) + 2H2O(l)
and we see that the only reaction in this neutralization is
between H3O+ and OH− ions.
Chapter 11 Acid/base equilibria
A substance which, like water, can act as both an acid
and as a base is called amphoteric. The hydrogensulfate
ion, HSO4−(aq), is also amphoteric; it can react with a
base to form the sulfate ion:
HSO4−(aq) + H2O(l) ҡ H3O+(aq) + SO42−(aq)
or with an acid to form sulfuric acid:
HSO4−(aq) + H3O+(aq) ҡ H2SO4(aq) + H2O(l)
Strong and weak acids and bases
In Chapter 10 we saw that the position of equilibrium can
be specified by an equilibrium constant. This idea can be
used for acids and bases.
For ethanoic acid, Ka = 1.8 × 10−5, showing that dissociation
is limited. We refer to acids such as ethanoic acid as weak
acids.
Acids such as hydrochloric acid are 100% dissociated in
water and we refer to them as strong acids.
■ A strong acid has a high concentration of H3O+.
■ A weak acid has a low concentration of H3O+.
Similarly:
■ A strong base has a high concentration of OH−.
■ A weak base has a low concentration of OH−.
pKa
Water itself is only slightly ionized:
In many situations, we don’t use Ka itself, but a value
known as pKa.
2H2O(l) ҡ H3O+(aq) + OH–(aq)
pKa = −log10 Ka
Kw = [H3O+] [OH−]
The term [H2O]2 does not appear in the expression for
the equilibrium constant. Because water is only slightly
ionized, its concentration is nearly constant.
In pure water, the concentrations of the oxonium ion
[H3O+] and the hydroxyl ion [OH−] are equal. This value is
found to be 1 × 10−7 mol dm−3.
Worked example 11.1
Q
For ethanoic acid, Ka = 1.8 × 10−5. What is the value of pKa for
ethanoic acid?
A
pKa = −log10 Ka
−log10 1.8 × 10−5 = 4.74
Therefore:
[H3O+] [OH−] = 1 × 10−14 mol2 dm−6
Worked example 11.2
This is called the ionic product for water. It is true for
any aqueous solution, not just for pure water.
Q
Oxalic acid has Ka = 6 × 10−2. What is the value of pKa for
oxalic acid?
Acid/base equations contain the equilibrium sign (ҡ)
rather than a reaction arrow (→) between reactants and
products. The position of the equilibrium can be anywhere
from the far left to the far right. When an substance such as
ethanoic acid is in aqueous solution, it dissociates:
A
pKa = −log10 Ka
−log10 6.0 × 10−2 = 1.22
CH3COOH(l) + H2O(l) ҡ H3O+(aq) + CH3COO−(aq)
If this equilibrium lies far to the left it means that the
solution will not contain a high concentration of H3O+. If
the equilibrium lies far to the right, the concentration of
H3O+ will be high. This is expressed by using the equilibrium
constant for the dissociation, again leaving out [H2O] as
this remains effectively constant during the dissociation.
The constant is given the symbol Ka.
Ka =
[H3O+] [CH3COO−]
[CH3COOH]
ITQ 1 Why is the heat of reaction between hydrochloric acid and
sodium hydroxide roughly the same as that between nitric acid
and potassium hydroxide?
pH
The concentration of H3O+ ions in an aqueous solution
can be talked about by a using a quantity called the pH of
the solution. You will have used pH on many occasions in
your study of chemistry, but perhaps only in a rough and
ready way. You may have used the colour changes shown
by universal indicator paper to gain an idea of the pH of a
solution. However, pH does have a precise meaning.
The pH of a solution is the negative logarithm to base 10 of
the concentration of the H3O+ ions in the solution.
pH = −log10 [H3O+], when [H3O+] is given in mol dm−3
■ Although the bare proton exists as the oxonium ion in water, the symbol
H+ is frequently used in its place
105
106
Unit 1 Module 2 Kinetics and equilibria
You need to note the following about pH:
■ Logarithms are pure numbers, so pH has no units.
■ A change of +1 unit of pH means that the ion
concentration has gone down by a factor of 10. A
change of −1 means that the concentration has gone
up by a factor of 10.
Worked example 11.4
Q
What is the pH of a solution of hydrochloric acid of
concentration 0.05 mol dm−3?
A
Concentration of H3O+ is 0.05 mol dm−3.
pH = −lg 0.05 mol dm−3 = 1.3
pH of the solution is 1.3.
■ The range of pH for aqueous solutions is taken as 0 to
14.
■ The pH of a neutral solution, in which [H3O+] = [OH−],
is 7.
■ [H3O+] [OH−] = 1 × 10−14 mol2 dm−6, so pH + pOH = 14
■ ‘log10’ is often written as lg.
Table 11.1 shows the [H3O+], [OH−] and pH values for
aqueous solutions, ranging from strongly acidic to strongly
alkaline.
Worked example 11.5
Q
What is the pH of a solution of sodium hydroxide of
concentration 0.01 mol dm−3?
A
The approach here is to use the ionic product for water to link
[OH−] and [H3O+].
[H3O+] [OH−] = 1 × 10−14 mol2 dm−6
Table 11.1 [H3O+], [OH−] and pH values for aqueous solutions
[H3O+] / mol dm−3 [OH−] / mol dm−3
[H3O+] =
pH
1 × 10−14 1 × 10−14
=
= 1 × 10−12 mol dm−3
[OH−]
0.01
lg 1 × 10−12 = 12
pH =12
1
10−1
10−2
10−3
Changes in pH in acid/base titrations
10−4
10−5
An indicator is a substance which changes colour according
to its pH. Indicators can be used alone or as a mixture such
as universal indicator. Single indicators usually have only
one colour change, such as red to blue, but this can happen
either over a very small pH range or over a greater range.
10−6
10−7
10−8
10−9
10−10
The natural substance flavin, responsible for the red colour
of fruit such as plums, is red in strong acid and green in
alkali, but between the two appears purple. Notice that:
10−11
10−12
10−13
■ indicators change colour over a range of pH;
10−14
■ indicators do not necessarily change colour at pH 7.
Table 11.2 gives information about a range of indicators.
Worked example 11.3
Q
What is the pH of a solution of hydrochloric acid of
concentration 0.10 mol dm−3 ?
A
Hydrochloric aid is a strong acid, and is completely ionized in
solution. Concentration of H3O+ is 0.10 mol dm−3.
pH = −lg 0.10 mol dm−3 = 1
pH of the solution is 1.
Table 11.2 Common acid/base indicators
Indicator
Colour in acid Colour in alkali pH range
methyl orange
red
yellow-orange
3.1–4.4
screened methyl orange red
green
3.1–4.4
bromophenol blue
yellow
blue
3.0–4.6
litmus
red
blue
4.5–8.3
phenolphthalein
colourless
magenta
8.3–10
The end-point of an acid/base titration is not necessarily at
pH 7. This is because the product of the reaction (the salt)
may itself be hydrolysed. For example, sodium ethanoate
(NaEt) ionizes in water:
NaEt(aq) → Na+(aq) + Et−(aq)
ITQ 2 Show that the pH of pure water is 7.0.
Chapter 11 Acid/base equilibria
but the ethanoate ion (Et−) reacts with water:
Et−(aq) + H2O(l) ҡ HEt(aq) (ethanoic acid) + OH−(aq)
Added OH− ions are largely removed because they combine
with the acid to form its salt:
OH−(aq) + CH3COOH(aq) ҡ CH3COO−(aq) + H2O(l)
and the solution is therefore alkaline.
When we titrate ethanoic acid with sodium hydroxide we
therefore expect the equivalence point (the ‘end-point’) to
be at a pH above 7.
Added H+ ions are removed because they combine with
existing ethanoate ions:
H+(aq) + CH3COO−(aq) ҡ CH3COOH(aq)
12
Buffer solutions in nature
11
Buffer solutions are very common in nature.
10
phenolphthalein
9
■ The activity within a body cell is dependent on
8
maintenance of a constant pH in the cell fluid.
The fluid contains a buffer system based on the
dihydrogenphosphate ion, H2PO4−. This ion can ionize
further:
pH
7
6
ethanoic acid
5
4
methyl orange
H2PO4−(aq) ҡ H+(aq) + HPO42−(aq)
3
2
HCl
1
0
0
5
10
15
20
25
30
Volume acid added / cm 3
Figure 11.1 Titration curves for the reaction of 25 cm3 sodium
hydroxide solution with ethanoic acid (a weak acid) and
hydrochloric acid (a strong acid). All solutions are 1.0 mol dm−3.
Using a strong acid there is a large and sudden pH change
at the equivalence point: almost any of the usual indicators
will change colour within one drop of the equivalence
point.
With ethanoic acid, a weak acid, methyl orange would
change colour long before the equivalence point is reached.
Instead, an indicator which changes colour at above pH 7
is needed. Phenolphthalein would be a good choice in this
case.
Any extra H+ ions entering the cell are removed as
H2PO4− ions. Any OH− ions entering the cell react with
H2PO4− ions to form HPO42− ions.
■ There is a similar system in blood plasma, this time using
carbonic acid, H2CO3, and its conjugate base, HCO3−:
H2CO3(aq) ҡ H+(aq) + HCO3−(aq)
Added OH− ions are removed by reaction with the free
acid (H2CO3). H+ ions are removed by reconverting
the HCO3− ion to H2CO3. The buffer keeps the pH at
around pH 7.4 (see Worked example 11.6 below).
■ Amino acids contain both the weakly basic –NH2 group
and the weakly acidic –COOH group. They can ionize
as either acids or bases, depending on the surrounding
pH. Either group can function as a buffer.
These are all good applications of le Chatelier’s principle.
pH of buffer solutions
Buffer solutions
A solution that resists changes in pH when small volumes
of an acid or a base is added to it is called a buffer solution.
A solution containing a weak acid and its conjugate base
(e.g. as the sodium salt) will act in just this way.
Look again at Figure 11.1. Half way along the ethanoic
acid titration curve, addition of further OH− ions produces
only a small pH change, and adding H+ ions reverses the
pH equally gently.
■ screened methyl orange contains an added dye (xylene cyanol) – the
end-point is grey and easier to see than that of methyl orange itself;
■ for bromophenol blue it is easy to see the mid-range colour (green);
■ litmus is a poor indicator because its colour change is over a big range
of pH;
The pH of a buffer solution containing a weak acid (HA)
and its sodium salt (NaA) can be calculated from the
equation for Ka:
Ka =
[H3O+] [A−]
[HA]
A weak acid is only slightly ionized, so we can say that all
the anions come from the salt, which is completely ionized.
We can also say that the concentration of the acid itself is
largely unchanged, since it is mostly un-ionized.
ITQ 3 Use Figure 11.1 to estimate the percentage difference in
the end-points of a titration of 1.0 mol dm−3 sodium hydroxide
solution with 1.0 mol dm−3 hydrochloric acid using (a) methyl
orange and (b) phenolphthalein as the indicator.
107
108
Unit 1 Module 2 Kinetics and equilibria
The equation can be arranged to become:
[H3O+] = Ka
[HA]
[A−]
which is the same as
[H3O+] = Ka
[acid]
[salt]
If we now take logarithms:
log10 [H+] = log10 [Ka] + log10
[acid]
[salt]
−log10 [H+] = −log10 [Ka] – log10
pH = pKa + log10
[acid]
[salt]
[salt]
[acid]
This is characteristic of an equilibrium situation; ions are
leaving the solid lattice at the same rate as they are being
re-bonded to it. For silver chloride, the solubility is about
6 × 10−6 mol dm−3 at 10 °C.
For such sparingly soluble substances a more useful
quantity is the solubility product (symbol Ksp). This is
the product of the concentrations of the component ions in
that solvent at that temperature. For silver chloride:
AgCl(s) ҡ Ag+(aq) + Cl−(aq)
At 10 °C the solubility is 6 × 10−6 mol dm−3. This is therefore
the concentration of each ion at that temperature.
Ksp = [Ag+] [Cl−]
= (6 × 10−6 mol dm−3) × (6 × 10−6 mol dm−3)
= 3.6 × 10−11 mol2 dm−6
Worked example 11.6
Q
A
Calculate the pH of the biological phosphate buffer when the
concentrations of the two phosphate species are the same.
pKa = 7.21.
pH = pKa + log10
[salt]
[acid]
If [H2PO4−] = [HPO42−] then their ratio is 1 and log10 1 = 0
pH = pKa which is 7.21.
In mammals, cellular fluids need to be between pH 6.9 and
pH 7.4. The buffer is therefore suited to its task!
Worked example 11.7
Q
A
In blood plasma, the concentration of hydrogencarbonate ions
is about 10 times that of the free acid. What is the pH of blood?
pKa of carbonic acid is 6.35.
pH = pKa + log10
[salt]
[acid]
[salt]
= 10
[acid]
pH = 6.35 + log10 10
pH = 6.35 + 1.00 = 7.35
Solubility product
The solubility of a material in water is commonly expressed
in mol dm−3. Sodium chloride, for example, has a solubility
in water at 30 °C of about 19 mol dm−3.
When a partially soluble material such as silver chloride is
placed in water, it slowly dissolves and the concentration
of dissolved silver ions and chloride ions rises. Ultimately,
if sufficient solid is present, the dissolving appears to stop.
For most compounds, the value of the solubility product
is dependent on the temperature: at 100 °C the solubility
product for AgCl is 2 × 10−9 mol2 dm−6.
The units of solubility product depend on the number of
ions involved. For example, zinc hydroxide dissolves to a
tiny extent in water at 20 °C.
Zn(OH)2(s) ҡ Zn2+(aq) + 2OH−(aq)
Ksp = [Zn2+] [OH−]2 = 1.8 × 10−14 mol3 dm−9
Although a saturated solution is in equilibrium, the solid
substance does not appear in the equilibrium expression.
This is because the ions are dissolved whilst the substance
itself is solid, i.e. in a different state. In these cases only the
substances on the right-hand side of the equilibrium (and
hence the numerator of the equilibrium equation) are used.
Worked example 11.8
Q
What is the solubility, in mol dm−3, of barium sulfate at 25 °C?
A
The solubility product of barium sulfate is
1.10 × 10−10 mol2 dm−6 at 25 °C.
BaSO4(s) ҡ Ba2+(aq) + SO42−(aq)
[Ba2+(aq)] = [SO42−(aq)]
Ksp = [Ba2+] [SO42−] = 1.10 × 10−10 mol2 dm−6
[Ba2+(aq)] = [SO42−(aq)] = 1.10 × 10−10 = 1.05 × 10−5 mol dm−3
ITQ 4 Calculate the pH of a buffer solution containing 4.0 g
benzoic acid C6H5COOH and 5.0 g sodium benzoate
C6H5COONa in 1.0 dm−3 of solution.
Ka benzoic acid = 6.3 × 10−5 mol dm−3
Hint: sodium benzoate is fully ionized in solution.
Chapter 11 Acid/base equilibria
Worked example 11.9
Q
A
Plants such as banana and hibiscus do not thrive in soils where
the pH is much above 7. This is in part because they demand a
good supply of the ion Fe2+(aq).
The ionic product for water = 1.0 × 10–14 mol2 dm–6
If the solubility product for Fe(OH)2 is 2 × 10–14 mol3 dm–9,
what is the maximum concentration of Fe2+(aq) ions in the soil
at (a) pH 7 and (b) pH 8?
(a) at pH 7: [H+] = [OH–] = 1 × 10–7
solubility product = [Fe2+] [OH–]2 = 2 × 10–14 mol3 dm–9
[Fe2+] =
2 × 10−14
= 2 mol dm–3
(1 × 10−7)2
This is plenty for the needs of the plant.
(b) at pH 8: [H+] = 1 × 10–8 mol dm–3
so [OH–] = 1 × 10–6 mol dm–3
[Fe2+] =
2 × 10−14
= 2 × 10–2 mol dm–3
(1 × 10−6)2
This is insufficient for the needs of this type of plant.
Testing for metal cations
In tests for metal cations, we often use a solution of
ammonium hydroxide. Laboratory reagents are commonly
made to a concentration of 0.1 mol dm−3, but ammonium
hydroxide is a weak base and so is only partially ionized.
It is diluted further because we only add a few drops
to the test, so the actual concentration of OH− ions is
much reduced. Its value concentration may be around
1.0 × 10−3 mol dm−3.
The metal cation solutions we are testing are often made
up at about 1.0 mol dm−3.
■ Calcium ions don’t form a precipitate with ammonium
hydroxide.
■ Magnesium ions do form a precipitate with
ammonium hydroxide.
■ The solubility product for magnesium hydroxide is
−11
1.1 × 10
3
−9
mol dm .
■ The solubility product for calcium hydroxide is
5.5 × 10−6 mol3 dm−9.
■ The concentration of OH− in the test solution is
1.0 × 10−3 mol dm−3.
Explain why magnesium ions give a precipitate with
ammonium hydroxide but calcium ions don’t.
For calcium:
Ksp = [Ca2+] [OH−]2 = 5.5 × 10−6 mol3 dm−9
[Ca2+] =
5.5 × 10−6
= 5.5 mol dm−3
(1.0 × 10−3)2
A precipitate of calcium hydroxide will only form if
the concentration of the calcium solution is above 5.5
mol dm−3, which is unlikely to be the case with normal
laboratory solutions.
For magnesium:
Ksp = [Mg2+] [OH−]2 = 1.1 × 10−11 mol3 dm−9
[Mg2+] =
1.1 × 10−11
= 1.1 × 10−5 mol dm−3
(1.0 × 10−3)2
A precipitate of magnesium hydroxide will form if
the concentration of the magnesium solution is above
1.1 × 10−5 mol dm−3.
Common-ion effect
In any solution the value of the solubility product for any
one solute cannot be exceeded. For example, if solid silver
chloride is put into a dilute solution of hydrochloric acid
of concentration 0.01 mol dm−3 the solubility of the silver
chloride is changed.
The maximum value for the product [Ag+] [Cl−] cannot
exceed 3.6 × 10−11 mol2 dm−6.
Ksp = [Ag+] [Cl−] = 3.6 × 10−11 mol2 dm−6
[Cl−] = 1.0 × 10−2 (from the acid)
[Ag+] =
3.6 × 10−11
= 3.6 × 10−9 mol dm−3
1.0 × 10−2
Above this concentration the silver ions will form a
precipitate.
In the absence of the acid, the concentration of Ag+ at
equilibrium is 6.0 × 10−6 mol dm−3.
The hydrochloric acid has reduced the solubility of the
silver chloride through the common-ion effect.
109
110
Unit 1 Module 2 Kinetics and equilibria
Summary
Review questions
1
(a) Explain what is meant by an acid and a base.
(b) Define the pH of a solution.
(c) In pure water, the concentration of H3O+ ions is
1 × 10–7 mol dm–3. Calculate the pH of pure water.
(d) The dissociation constant for ethanoic acid,
CH3COOH, is 1.8 × 10−5 whilst that for
1-chloroethanoic acid is 1.4 × 10–3. Explain which
is the stronger acid and calculate the pH of a
solution of that acid with a concentration of 1.0
mol dm–3.
2
Note that you need to understand the idea of a
logarithm to attempt this question.
(a) Define a ‘buffer solution’.
(b) Show that for a buffer solution containing a weak
acid and its sodium salt, the pH of the solution is
given by
[salt]
pH = pKa + log
.
[acid]
[H+][A−]
.
Start from the statement that Ka =
[HA]
✓ An acid is a substance that donates a proton to
another substance.
✓ In aqueous solution, hydrogen ions (protons)
form the oxonium ion H3O+.
✓ Strong acids and strong bases are substantially
ionized in solution.
✓ Weak acids and weak bases are only slightly
ionized in solution.
✓ The pH of a solution is the logarithm to base 10
of the H+ concentration, with the sign changed
(usually from minus to plus).
✓ pH indicators are substances which change
colour according to the pH of their solution.
✓ The scale of pH for aqueous solutions runs from
0 to 14; the neutral point is pH 7.
✓ In an aqueous solution, the product [H+] [OH−] is
always 1 × 10−14.
(c) Calculate the pH of a buffer solution containing
3.0 mol dm–3 ethanoic acid and 2.0 mol dm–3
sodium ethanoate. (Ka for ethanoic acid is
1.8 × 10–5)
✓ A buffer solution is one that resists changes in
pH.
✓ The solubility product of a substance is the
product of the concentrations of its cations and
anions in a saturated solution.
3
✓ The solubility product can be used to predict
whether a substance will be precipitated from a
solution by the addition of a common ion.
(a) What is meant by the ‘solubility product’ of a
compound?
The usual test for the iodide ion is to add a
solution of lead nitrate to the test solution. A
yellow precipitate indicates the presence of I– ions.
(b) Ksp for lead iodide, PbI2, is given in tables as
9.8 × 10–9. What are the units of this number?
(c) What is the concentration of iodide ions in a
saturated solution of lead iodide?
Answers to ITQs
1
Both are reactions between H+ and OH− ions. Other
ions take no part in the neutralization.
2
If [H3O+] [OH−] = 1 × 10−14 mol2 dm−6 and [H3O+] =
[OH−] for pure water, then [H+] = 1 × 10−7.
pH = −log10 [H+] = 7.0 (to 2 sig. figs)
3
The true end-point is when 25.0 cm3 of acid is added.
This is the result gained with phenolphthalein. Using
methyl orange, the end-point appears to be when
24.0 cm3 of acid is added.
The difference is therefore
25 − 24
× 100 = 4%
25
Chapter 11 Acid/base equilibria
4
Molecular mass for sodium benzoate = 144 g mol−1
Molecular mass for benzoic acid = 122 g mol−1
Concentration of sodium benzoate =
Concentration of benzoic acid is
5
mol dm−3 = 0.035 mol dm−3
144
4
mol dm−3 = 0.033 mol dm−3
122
[H+] = 0.033 mol dm−3
Sodium benzoate is fully ionized, so [A−] = 0.035
[H3O+] = Ka
[acid]
[salt]
[H3O+] = 6.3 × 10−5
[0.033]
= 5.94 × 10−5 mol dm−3
[0.035]
pH = −log10 [H3O]+ = 4.22
Answers to Review questions
1
(c) pH = –log10 [H+]
if [H+] = 1 × 10–7 then –log10 [H+] = –7; pH = 7.0.
However, the data had only 1 significant figure, so
the final answer is: pH = 7.
(d) 1-chloroethanoic acid is the stronger acid.
Ka =
[H+][CH2ClCOO−]
= 1.4 × 10−3
[CH3COOH]
The dissociation is tiny so [CH3COOH] is virtually
1.0 in this solution; also [H+] = [CH2ClCOO–]
Substituting into the expression for Ka gives
1.4 × 10−3 =
[H+]2
; [H+] = 3.7 × 10−2
1
pH = –log10 [H+] = –log10 (3.7 × 10–2)
pH = 1.43
2
(c) pH = pKa + log10
[salt]
[acid]
= –log10 (1.8 × 10–5 ) + log10 (3/2)
= 4.74 + 0.18
pH = 4.92
3
(c) Pb2+(aq) + 2I–(aq) → PbI2(s)
Ksp for lead iodide is 9.8 × 10–9 = [Pb2+][I–]2 mol3 dm–9
1
In the solution [Pb2+(aq)] = 2 [I–(aq)]
1
2
× [I–] × [I–]2 = 9.8 × 10–9
[I–]3 = 2 × 9.8 × 10–9, [I–] = 2.70 × 10–3 mol dm–3
111
112
Chapter 12
Redox equilibria
Learning objectives
■ Understand the reactions that take place in an electrochemical cell.
■ Describe the standard hydrogen electrode.
■ Describe methods used to measure standard electrode potentials.
■ Calculate standard cell potentials from standard electrode potentials.
■ Use standard electrode potentials to determine electron flow direction and feasibility of reaction.
■ Predict how the value of an electrode potential varies with concentration.
■ Apply the principles of redox processes to energy storage devices.
Introduction
The term ‘redox’ is an abbreviation for a process in which
reduction and oxidation occur simultaneously. Redox
reactions are common in nature, in industry and everyday
processes. For example, photosynthesis, respiration, the
bleaching of hair and the rusting of iron all involve redox
reactions.
The most useful definition of oxidation and reduction is
given in terms of electron transfer:
■ oxidation – a reaction in which there is loss of
electron(s);
■ reduction – a reaction in which there is gain of
electron(s).
You can remember this by using the memory aid ‘OIL RIG’:
■ OIL – Oxidation Is Loss
■ RIG – Reduction Is Gain
The electrons lost from one reagent in an oxidation must
be gained by another reagent in a reduction. Oxidation and
reduction happen together.
■ The substance that is oxidized is called the reducing
agent.
■ The substance that is reduced is called the oxidizing
agent.
Redox reactions play an important role in chemistry. Their
importance lies in the fact that the transfer of electrons
between species forms the basis for the functioning of
several devices we use in everyday life. However, in the
laboratory, devices known as electrochemical cells
can be used as vessels to carry out redox reactions. An
electrochemical cell is an apparatus which is used either for
generating electrical energy from redox reactions, or uses
electrical energy to propel redox reactions.
An electrochemical cell that spontaneously produces a
current is called a galvanic cell or a voltaic cell. Such a
cell consists of an electrical conductor (e.g. a metal strip),
which is called an electrode, dipping into an electrolyte,
which can be either a molten ionic compound, an aqueous
solution of an ionic compound or a polar covalent
compound which dissolves in water to produce ions.
Electrolytes conduct electricity due to the presence of
mobile or ‘free’ ions. When electrolytes conduct an electric
current, the positive and negative ions, which are free to
move independently, move in opposite directions. The
loss or gain of electrons in the formation of ions or atoms
is a process described as the discharge of ions, and such
reactions take place at the electrodes. Equations can be
written to represent these reactions taking place at each
electrode.
Chapter 12 Redox equilibria
Electrode potential
potential difference
M
When a metal is in contact with a solution of its ions, metal
atoms tend to lose electrons (they are oxidized) and pass
into solution as aqueous ions:
M(s) → Mn+(aq) + ne−
Conversely, the aqueous ions in solution may gain these
electrons (they are reduced) and re-form the metal:
Mn+(aq) + ne− → M(s)
In the reaction, a redox equilibrium is established when
the rate at which electrons are leaving the surface of the
metal is exactly equal to the rate at which they are being
discharged on it again. The equation for the redox reaction
at equilibrium is written with a reversible arrow:
n+(aq)
M
+
ne−
ҡ M(s)
Reactive metals form ions readily.
Less reactive metals do not form ions readily.
The ease with which a metal loses electrons is referred to as
its ‘reactivity’. For example, if a strip of the fairly reactive
zinc metal is placed in a solution of its ions, the zinc metal
loses electrons readily (is oxidized) to form Zn2+ ions:
Zn(s) → Zn2+(aq) + 2e−
The electrons stay on the surface of the metal, which
acquires a negative charge.
Also, some of the Zn2+ ions in solution accept these
electrons from the surface of the metal and are discharged
as Zn atoms:
Zn2+(aq) + 2e− → Zn(s)
We can combine these two half-reactions:
Zn2+(aq) + 2e− ҡ Zn(s)
Electrons are, by convention, written on the left-hand side
of the redox equilibrium.
Zinc is a fairly reactive metal. It loses electrons readily and
the equilibrium lies well to the left. Many electrons are
released and they stay on the metal; the metal acquires
a considerable negative charge. The solution in contact
with the zinc becomes positively charged since extra Zn2+
ions have been released into it. This charge difference that
develops between the negatively charged zinc strip and
the positively charged zinc solution is called a potential
difference; the zinc strip is said to have a negative
potential (Figure 12.1). A potential difference develops
when a metal is placed in contact with a solution of its ions.
The sign and size of this electrode potential depends on:
e
M+
positive metal ions
–
–
e–
M
negative ions
e–
M+
–
+
–
M
+
M+
e–
M+
M+
–
M+
–
Figure 12.1 A metal in contact with a solution of its ions and the
generation of a potential difference.
■ the relative reactivity of the metal;
■ the concentration of the ions in solution at equilibrium.
A less reactive metal, such as copper, does not form ions as
readily and so the equilibrium
Cu2+(aq) + 2e− ҡ Cu(s)
lies much further to the right.
Thus, there is a greater tendency for the Cu2+ ions in
solution to accept electrons from the Cu metal to re-form
Cu atoms. The metal develops a positive charge, giving it a
positive potential.
The absolute potential of a single electrode cannot be
measured in isolation. How we do measure electrode
potential will be discussed later.
Galvanic cells: using redox reactions to
generate electricity
We have established that redox reactions involve the
transfer of electrons from one species to another. This
transfer of electrons may be viewed as ‘a flow of electric
charge’, which is in fact an electric current. Therefore,
it stands to reason that redox reactions can be used to
generate electric currents, and that such electric currents
could be used to do electrical work.
Look at an example of a galvanic cell that can be set up in the
laboratory (Figure 12.2). In this electrochemical cell, a solid
Zn strip is placed into a zinc solution to form a half-cell.
Similarly, a solid Cu strip is placed into a copper solution
to form a second half-cell. The reactions that occur in each
half-cell are called half-cell reactions. The equations are
shown in Figure 12.2. These two half-cells are connected
by attaching a wire from the Zn strip through a voltmeter
(which measures voltage) to the Cu strip. Once this circuit is
complete, there is a flow of electrons between the half-cells.
113
114
Unit 1 Module 2 Kinetics and equilibria
■ the K+ or Na+ ions flow to the reduction half-cell to
1.10 V
e–
NO3
–
offset the accumulation of negative charge at the
cathode.
e–
K+
KNO3, KCl and NaCl are suitable salts for use in a salt
bridge because:
Zn
electrode
Cu
electrode
■ they are soluble in water;
■ they do not react with other ions commonly used in
Zn(s)
Zn2+(aq) + 2e–
oxidation
Cu 2+(aq) + 2e–
Cu(s)
reduction
Figure 12.2 A galvanic cell.
We know that like charges repel. Therefore, it stands to
reason that since electrons are negatively charged, electrons
will tend to flow away from the negative electrode and
towards the positive electrode in an electrochemical cell. As
a result, the following occurs:
■ electrons flow out of the left-hand half-cell through
the Zn strip (labelled with a negative sign); this strip
is called the anode and since electrons are lost here,
oxidation occurs at the anode
■ electrons flow into the right-hand half-cell through the
Cu metal strip (labelled with a positive sign); this strip
is called the cathode and since electrons are gained
here, reduction occurs at the cathode.
As electrons flow out of the oxidation half-cell through
the Zn strip, Zn2+ ions form in this cell. These electrons
then pass through the wire into the reduction half-cell
via the Cu strip. The Cu2+ ions in the reduction half-cell
are deposited as neutral Cu atoms. As electrons leave one
half-cell and flow to the other, a difference in charge is
established. Normally, this charge difference would prevent
further flow of electrons. However, a device known as a
salt bridge addresses this problem. This device is often an
inverted U-shaped tube and contains a strong electrolyte
such as KNO3, KCl or NaCl. The electrolyte is often jellified
with agar to help prevent intermixing of fluids.
Note that ions pass through the salt bridge but electrons
flow through the wire.
The salt bridge serves the following functions:
■ it joins the two half-cells and allows the flow of ions to
maintain a balance in charge between these cells;
■ it keeps the contents of each cell separate;
■ the NO3− or Cl− ions within the salt bridge flow to
the oxidation half-cell to offset the accumulation
of positive charge at the anode and hence maintain
electrical neutrality;
electrochemical cells.
With the charge difference balanced, electrons can flow
once again, and the redox reactions can proceed.
The anode, which is the more negative electrode,
is conventionally placed on the left when drawing
diagrams of these cells.
■ The anode is negatively charged since the
spontaneous oxidation at the anode is the source of
the cell’s electrons or negative charge.
■ The cathode is positively charged.
Electrical voltage is a measure of the tendency of electrons
to flow:
V = IR (voltage = current × resistance)
A flow of electrons in the external circuit indicates a
difference in potential between the two electrodes; this
difference in potential is called an electromotive force,
e.m.f. or E (if measured under standard conditions)
and is measured in volts. The e.m.f. is determined using
a high-resistance voltmeter which uses negligible current
in the external circuit, and therefore the cell registers its
maximum potential difference. The e.m.f. of the zinc/
copper cell is 1.10 V.
The standard hydrogen electrode (S.H.E.)
We pointed out earlier that the absolute potential of a single
electrode cannot be measured in isolation. E.m.f.s can only
be measured for a complete circuit with two electrodes. In
other words, only differences in potentials are measurable.
However, it would be useful if we could in fact assign a
characteristic electrode potential value to half-cells. To
achieve this, electrode potentials are measured relative
to the standard hydrogen electrode, S.H.E., which is
assigned a potential of 0.00 V (Figure 12.3). The half-cell
under test is connected to the S.H.E. and the potential
difference of the cell is measured; this difference is called
the standard electrode potential, E of the cell.
ITQ 1 Why do electrons flow through the voltmeter from left to
right in Figure 12.2?
Chapter 12 Redox equilibria
high-resistance voltmeter
V
hydrogen in
at 1 bar
pressure
H2(g) at 298 K
and 1 atm
salt bridge
acid solution
containing
1.0 mol dm–3
H+(aq)
platinum
electrode
glass tube with holes
in to allow bubbles of
H2(g) to escape
S.H.E.
half-cell under test
Figure 12.3 The standard hydrogen electrode (S.H.E.).
Figure 12.4 Measuring the standard electrode potential of a
half-cell.
The S.H.E. consists of hydrogen gas bubbling around a
platinum electrode immersed in a solution of H+ ions under
standard conditions. Hydrogen is adsorbed on the platinum
and an equilibrium is established between the adsorbed
layer of H2 gas and H+(aq) ions in the solution:
The potential of the half-cell under test is equal to the
e.m.f. of the cell.
2H+(aq) + 2e− ҡ H2(g)
E
= 0.00 V
The platinum electrode is ‘platinized’, which means that
it is coated with a layer of finely divided platinum. This
serves to increase its surface area so that the equilibrium
between the H2(g) and H+(aq) can be established as quickly
as possible.
The platinum electrode has two important properties:
If, however, the S.H.E. forms the positive electrode in the
cell, then:
E
= E(positive electrode) − E(negative electrode)
= 0.00 − E(negative electrode)
= −E(positive electrode)
In this case, the potential of the half-cell under test is
numerically equal to the e.m.f. of the cell, but has a
negative value.
■ temperature of 25 °C (298 K);
For example, the standard electrode potential of the
Cu2+(aq)/Cu(s) half-cell is measured by connecting it to
the S.H.E., as shown in Figure 12.4. The standard electrode
potential of the Cu2+(aq)/Cu(s) half-cell is thus the potential
difference between the electrodes of a cell consisting of the
S.H.E. and the standard Cu2+(aq)/Cu(s) half-cell.
■ gases at 1 atm pressure (101.3 kPa ,1 bar)
In this cell:
■ it is inert and does not form platinum ions;
■ H2 gas is readily adsorbed onto its surface.
Standard conditions are as follows:
■ 1.00 mol dm−3 solutions.
Measuring standard electrode potentials
The standard electrode potential, E , of a standard half-cell
is the potential of that half-cell relative to the S.H.E. under
standard conditions. The e.m.f. of an electrochemical cell
is given by:
E
= E(positive electrode) − E(negative electrode)
■ the reactions occurring at the electrodes are:
H2(g) + 2e− ҡ 2H+(aq)
Cu2+(aq) + 2e− ҡ Cu(s)
■ electrons flow from the S.H.E. to the Cu2+(aq)/Cu(s)
half-cell;
■ the Cu electrode is positive;
■ the S.H.E. is negative;
■ the e.m.f. is 0.34 V.
The ‘minus’ sign is because we regard the two half-cells as
working against one another.
Remember: electrons flow away from the negative
electrode and towards the positive electrode.
If the S.H.E. forms the negative electrode in the cell, then:
= E(positive electrode) − 0.00
Since the Cu electrode is positive, the standard electrode
potential for the Cu2+(aq)/Cu(s) half-cell is +0.34 V. This is
written as:
= E(positive electrode)
Cu2+(aq) + 2e− ҡ Cu(s)
E
= E(positive electrode) − E(negative electrode)
E
= +0.34 V
115
116
Unit 1 Module 2 Kinetics and equilibria
Now consider what happens when a standard zinc half-cell
is connected to the S.H.E.:
■ the reactions occurring at the electrodes are:
solution containing
equal concentrations
of Fe 2+(aq) and Fe3+(aq)
2H+(aq) + 2e− ҡ H2(g)
Zn(s) + 2e− ҡ Zn2+(aq)
■ electrons flow from the Zn2+(aq)/Zn(s) half-cell to the
platinum
electrode
S.H.E.;
■ the Zn electrode is negative;
Figure 12.5 The Fe3+(aq)/Fe2+(aq) half-cell.
■ the S.H.E. is positive;
In this system:
■ the e.m.f. is 0.76 V.
■ there is no metal to allow electron transfer;
Since the Zn electrode is negative, the standard electrode
potential for the Zn2+(aq)/Zn(s) half-cell is −0.76 V. This is
written as:
2+(aq)
Zn
+
2e−
ҡ Zn(s)
E
= −0.76 V
■ iron exists in two different oxidation states;
■ equimolar solutions are used;
■ platinum serves as an inert electron carrier.
Other examples:
Conventionally, the S.H.E. is placed on the left-hand side
of the electrochemical cell. In this way, the sign of the
standard electrode potential of a half-cell indicates whether
that electrode is positive or negative.
■ the Br2(aq)/Br−(aq) half-cell is set up with a platinum
The hydrogen electrode is not an easy device to manipulate
and in practice the calomel cell, which is based on
mercury chloride, Hg2Cl2, is used as a secondary standard.
The calomel cell has E = +0.24 V.
half-cell consists of chlorine gas bubbling over a
platinum electrode immersed in a solution containing
1.0 mol dm−3 of Cl− ions.
electrode immersed in a solution containing
1.0 mol dm−3 of Br− ions;
■ the Cl2(g)/Cl−(aq) half-cell involves a gas and so the
Uses of standard electrode potentials
Measuring the E
non-metals
of half-cells involving
Thus far, we have dealt only with redox reactions that
involve metals. However, some redox reactions do not
involve metals. Instead, non-metal ions of the same
elements exist in different oxidation states. If a half-cell
doesn’t involve a metal, then remember that a platinum
electrode can be used as an electron carrier. The platinum
electrode is immersed in a solution containing 1.0 mol
dm−3 of metal ions. The standard electrode potential of
the half-cell under test is measured by connecting it to the
S.H.E. (or a secondary standard) under standard conditions.
Figure 12.5 shows a half-cell involving a non-metal.
The Fe3+(aq)/Fe2+(aq) half-cell is set up with a platinum
electrode immersed in a solution containing 1.0 mol dm−3
of Fe3+ ions and 1.0 mol dm−3 of Fe2+ ions.
Relative oxidizing and reducing powers
Standard electrode potentials can be arranged in order of
their values from the most negative through to the most
positive to produce a list known as the electrochemical
series (Table 12.1).
In Table 12.1, the electrode reactions are written as
reduction processes, so that electrons are added on the
left-hand side.
■ Li+(aq), found at the top of Table 12.1, is the strongest
reducing agent but is the weakest oxidizing agent. As
you go down the list, the oxidizing strength increases.
■ F2(g), found at the bottom of Table 12.1, is the strongest
oxidizing agent but is the weakest reducing agent. As
you go up the list, the reducing strength increases.
■ Hydrogen is found mid-way down Table 12.1, with a
E
ITQ 2
(a) Draw a labelled diagram to show how the E
Zn2+(aq)/Zn(s) electrode can be found.
(b) Which is the positive electrode?
(c) Indicate the direction in which electrons flow.
value for the
value of 0.00 V.
ITQ 3 Will iodide ions react with chlorine gas? Explain your
answer.
ITQ 4 Which one of the following metals is capable of reducing
Sn4+ to Sn2+? Cu, Zn or Ag.
Chapter 12 Redox equilibria
Table 12.1 Standard electrode potentials arranged in an
electrochemical series
Electrode process
E
Li+(aq) + e− ҡ Li(s)
−3.03
Rb+(aq) + e− ҡ Rb(s)
−2.93
K+(aq) + e− ҡ K(s)
−2.92
Sr2+
−
−2.89
Ca2+
−
−2.87
(aq) + 2e ҡ Sr(s)
(aq) + 2e ҡ Ca(s)
Na+
−
(aq) + e ҡ Na(s)
/V
−2.71
Mg2+
−
−2.37
Be2+
−
−1.85
(aq) + 2e ҡ Mg(s)
(aq) + 2e ҡ Be(s)
Al3+(aq) + 3e− ҡ Al(s)
−1.66
Mn2+
−1.19
−
(aq) + 2e ҡ Mn(s)
Zn2+(aq)
2e−
ҡ Zn(s)
−0.76
Cr2+(aq) + 3e− ҡ Cr(s)
−0.74
2CO2(g) + 2H+(aq) + 2e− ҡ H2C2O4(aq)
−0.49
Fe2+(aq)
−0.44
+
Cr3+(aq)
+
2e−
+
e−
Ti3+
ҡ Fe(s)
ҡ
Cr2+(aq)
−
−0.41
2+
(aq) + e ҡ Ti (aq)
−0.37
Co2+(aq) + 2e− ҡ Co(s)
−0.28
Ni2+(aq)
−0.25
2e−
+
Sn2+
ҡ Ni(s)
−
(aq) + 2e ҡ Sn(s)
Pb2+(aq)
H
+(aq)
+
+
2e−
e−
−0.14
ҡ Pb(s)
−0.13
1
2 H2(g)
0.00
ҡ
Sn4+(aq) + 2e− ҡ Sn2+(aq)
+0.15
Cu2+(aq) + e− ҡ Cu+(aq)
+0.15
Cu2+(aq)
1
2 O2(g)
+
2e−
ҡ Cu(s)
+ H2O(l) +
+(aq)
2e−
+
e−
ҡ Cu(s)
+
e−
ҡ
I−(aq)
MnO4 (aq) +
e−
Cu
1
2 I2(aq)
−
+0.34
ҡ
2OH−(aq)
+0.40
+0.52
+0.54
2−
ҡ MnO4 (aq)
+0.56
Fe3+(aq) + e− ҡ Fe2+(aq)
+0.77
Ag+(aq)
+0.80
+
e−
ҡ Ag(s)
NO3−(aq) + 4H+(aq) + 3e− ҡ NO(g) + 2H2O(l)
+0.96
1
2 Br2(g)
+1.09
+ e− ҡ Br−(aq)
1
IO3−(aq) + 6H+(aq) + 5e− ҡ 2 I2(aq) + 3H2O(l)
MnO2(s) +
4H+(aq)
1
2−
2 Cr2O7 (aq)
1
2 Cl2(aq)
+
2e−
ҡ
Mn2+(aq)
+1.19
If we compare the positions of Zn and Cu in the
electrochemical series, as well as their E values, we can
see that Zn is higher than Cu and has the greater negative
E value:
Zn2+(aq) + 2e− → Zn(s)
Cu2+(aq) + 2e− → Cu(s)
+1.36
■ Zn is a stronger reducing agent than Cu;
+1.51
Pb4+(aq)
+1.69
Co3+(aq) + e− ҡ Co2+(aq)
+1.81
S2O82−(aq) + 2e− ҡ 2SO42− (aq)
+2.01
1
2 F2(aq)
+2.87
E
E
= −0.76 V
= +0.34 V
From these numbers, we can deduce the following:
■ Zn is more reactive than Cu;
MnO4−(aq) + 8H+(aq) + 5e− ҡ Mn2+(aq) + 4H2O(l)
+ e− ҡ F−(aq)
Note that this does not mean that for a reaction to occur,
the E value for one half of the reaction must be positive
and the must be other negative. All that is needed is that
one value must be more negative than the other to act as
the reducing agent, while the more positive of the two acts
as the oxidizing agent.
+1.33
+1.51
ҡ
The data can act as a measure of the oxidizing and
reducing powers of species. Negative E values show that
the species loses electrons (i.e. it is oxidized) more readily
than hydrogen and hence acts as a reducing agent. On the
contrary, positive E values show that the species gains
electrons (i.e. it is reduced) more readily than hydrogen
and hence acts as an oxidizing agent.
7
Mn3+(aq) + e− ҡ Mn2+(aq)
+
You need to bear in mind that individual reactions
may be kinetically hindered. Just as thermodynamic
data can tell us whether a reaction is feasible but give
no hint about the reaction rate, the difference between
the electrode potentials of two half-reactions can give
information about the position of equilibrium of their
combined reaction, but tells us nothing about how
quickly that equilibrium is reached.
+1.23
+ e− ҡ Cl−(aq)
Pb2+(aq)
The data can act as a guide to the reactivity of species. The
most reactive metals are at the top of the series, whilst
the most reactive non-metallic species are to the bottom.
This makes sense. Metals always react by losing electrons,
and species which lose electrons most readily will have
large negative electrode potentials. On the other hand,
non-metals generally form ions by gaining electrons.
+ 2H2O(l)
+ 7H+(aq) ҡ Cr3+(aq) + 2 H2O(l)
2e−
The data in Table 12.1 provides invaluable information on
redox systems.
■ Cu2+ is a stronger oxidizing agent than Zn2+.
Calculating standard cell potentials
Follow this sequence of steps in order to calculate the
standard cell potential, E cell .
1 Write the two half-cell reactions and their respective
E values.
117
118
Unit 1 Module 2 Kinetics and equilibria
2 The half-cell carrying the more negative (or less
positive) E value forms the anode. Re-write this
anodic reaction as an oxidation reaction, i.e. loss of
electrons. Change the sign of the voltage.
3 Balance the loss/gain of electrons (if necessary). Note
that the E value is independent of the number of
electrons transferred.
4 Add the anode half-cell to the cathode half-cell.
The convention of writing the anode process on the
left-hand side in the cell diagram (Figure 12.6) leads to a
positive cell voltage. This represents the flow of electrons
from left to right in the cell.
If the cell diagram were written the other way around,
we would get a negative cell voltage, which indicates
that the cell reaction as written cannot take place.
5 When two half-cells are connected under standard
conditions, the resulting electrochemical cell registers
its maximum potential difference or e.m.f., which is
called the standard cell potential, E cell .
The zinc/copper half-cell arrangement we have been
studying closely resembles one of the first practical cells
to be used – known as the Daniell cell. The Daniell cell is
discussed in more detail later (page 120).
This standard cell potential can be calculated from the
standard electrode potentials of the half-cells.
Feasibility of reactions
Worked example 12.1
Q
Calculate the standard cell potential of the zinc/copper cell.
A
1 Write the half-cell equations and find their standard
electrode potentials (see Table 12.1).
Zn2+(aq) + 2e− → Zn(s)
E = −0.76 V
Cu2+(aq) + 2e− → Cu(s)
E = +0.34 V
2 The zinc half-cell carries the more negative E value,
therefore zinc is more reactive than copper and becomes
the anode. We need to re-write the Zn half-cell as an
oxidation-type reaction:
Zn(s) − 2e− → Zn2+(aq)
E = +0.76 V
Since the reaction is now written as an oxidation instead of
a reduction, the sign of the voltage becomes positive.
3 The loss and gain of electrons are already equal, i.e. two
electrons are lost and two are gained.
4 The anode half-cell is added to the cathode half-cell:
Zn(s)
− 2e− → Zn2+(aq)
+0.76 V anode half-cell
2+
Cu (aq) + 2e− → Cu(s)
+0.34 V cathode half-cell
2+
2+
Zn(s) + Cu (aq) → Zn (aq) + Cu(s) +1.10 V E cell
The cell voltage is 1.10 V.
The combination of half-cells and the resulting cell e.m.f.
can be summarized in the following cell diagram (Figure
12.6).
Zn 2+(aq)
anode process
salt
bridge
Figure 12.6 A cell diagram.
Cu 2+(aq) Cu(s)
E cell = 1.10 V
cathode process
e.m.f.
’
Zn(s)
A reaction is feasible only if the standard cell potential is
positive. By extension, a negative e.m.f. implies that the
reaction is not feasible. The standard cell potential for the
zinc/copper cell, as calculated above, is +1.10 V, suggesting
that the reaction is energetically feasible.
Worked example 12.2
Q
Will a cell containing a copper electrode and a silver electrode
be feasible? If it is, what is its standard potential?
Note: E values are independent of the number of electrons
transferred.
A
Cu2+(aq) + 2e− → Cu(s)
E = +0.34 V
+
−
Ag (aq) + e → Ag(s)
E = +0.80 V
The copper half-cell carries the less positive E value, therefore
copper is more reactive than silver and becomes the anode.
Re-write the Cu half-cell as an oxidation type reaction:
Cu(s) − 2e− → Cu2+(aq)
E = −0.34 V
Since the reaction is now written as an oxidation instead of a
reduction, the sign of the voltage becomes negative.
Since two electrons are lost in the copper half-cell, two
electrons should be gained in the silver half-cell. Therefore, the
silver half-cell reaction is multiplied by two to give:
2Ag+(aq) + 2e− → 2Ag(s)
E = +0.80 V
The anode half-cell is added to the cathode half-cell:
Cu(s)
− 2e− → Cu2+(aq)
E = −0.34 V
+
−
2Ag (aq) + 2e → 2Ag(s)
E = +0.80 V
+
2+
Cu(s) + 2Ag (aq) → Cu (aq) + 2Ag(s)
E cell = +0.46 V
The positive e.m.f. value of +0.46 V indicates that the reaction
is feasible.
ITQ 5 The standard copper half-cell is connected to a standard silver half-cell.
(a) What is the term used to describe the potential difference obtained?
(b) Which cell acts as the negative electrode?
(c) Calculate the potential difference obtained when the two half-cells are connected.
ITQ 6 Will aluminium metal displace
copper(II) ions from an aqueous solution?
Write balanced equations for this reaction
and describe any changes observed.
Chapter 12 Redox equilibria
Kinetic feasibility
There are times when the calculation gives a standard cell
potential that is positive. However, in practice, the reaction
may be too slow to notice. E values relate only to the
relative stabilities of reactants and products, and therefore
only indicate the feasibility of a reaction from an energetic
standpoint. E values give no information about the rate
of a reaction or its kinetic feasibility.
Take, for example, the E values that predict Cu2+(aq)
should oxidize H2(g) to H+(aq). Let us calculate the standard
cell potential for this redox system:
Cu2+(aq) + 2e− → Cu(s)
= +0.34 V
E
2H+(aq) + 2e− → H2(g)
E
= 0.00 V
The hydrogen half-cell carries the less positive E value.
We need to re-write the hydrogen half-cell as an oxidation
type reaction:
H2(g) − 2e− → 2H+(aq)
E
= 0.00 V
The loss and gain of electrons are already equal.
The anode half-cell is added to the cathode half-cell:
H2(g)
Cu2+(aq)
−
2e− → 2H+(aq)
+
2e−
H2(g) + Cu2+(aq)
→ Cu(s)
E
= 0.00 V
E
= +0.34 V
→ 2H+(aq) + Cu(s) E cell = +0.34 V
The positive e.m.f. value of +0.34 volts indicates that
the reaction is theoretically feasible. However, nothing
happens when hydrogen is bubbled into copper(II) sulfate.
In this instance, the reaction is so slow that the reaction
rate is in effect zero.
Let us now compare the reaction below and determine if it
is energetically feasible.
Cu2+(aq) + 2Ag(s) → Cu(s) + 2Ag+(aq)
The e.m.f. is calculated as follows:
Cu2+(aq)
2Ag(s)
Cu2+
+
2e− → Cu(s)
−
2e−
(aq) + 2Ag(s)
→ 2Ag (aq)
+
E
= +0.34 V
E
= −0.80 V
→ Cu(s) + 2Ag (aq) E cell = −0.46 V
+
This reaction will not occur because the overall e.m.f. is
negative.
It was earlier in this chapter that standard electrode
potentials were measured under standard conditions.
Therefore, if these conditions are changed, i.e. if there are
changes in temperature, pressure or concentration, the
electrode potential values will also change. It is possible
to maintain the standard conditions of temperature and
pressure. However, maintaining the standard conditions
of concentration (1.0 mol dm−3) is impossible in practice
because as soon as a reaction begins, the concentrations of
both the reactants and the products change.
Let us look at the effect of concentration on the value of
the standard electrode potential.
Cu2+(aq) + 2e− ҡ Cu(s)
E
= +0.34 V
If the concentration of Cu2+(aq) decreases, then, according
to Le Chatelier’s principle, the equilibrium will shift to
the left in order to restore the concentration of Cu2+(aq)
and release more electrons. Electrons are negatively
charged and therefore when the equilibrium moves to the
left the E value becomes more negative (less positive).
Consequently, the electrode potential of copper in contact
with 0.10 mol dm−3 Cu2+(aq), for example, rather than the
standard 1.00 mol dm−3, is reduced. The value is found to
be 0.31 V.
Zn2+(aq) + 2e− ҡ Zn(s)
E
= −0.76 V
If the concentration of Zn2+(aq) decreases, then, according
to Le Chatelier’s principle, the equilibrium will shift to
the left, as with the copper example above. The E value
becomes more negative. As a result, the electrode potential
of zinc in contact with 0.10 mol dm−3 Zn2+(aq) is more
negative. The value is found to be −0.79 V.
Energy storage devices
Electrochemical cells use redox reactions to generate
electrical energy; examples are batteries and cells. A
battery consists to two or more cells connected in series
or parallel; however; the term is sometimes used for single
cells. Batteries and cells convert chemical energy into
electrical energy. Batteries can be of two types, primary
and secondary.
■ Primary cells produce an e.m.f. from irreversible
The effect of concentration on electrode
potential
In Chapter 10 we were introduced to Le Chatelier’s principle.
This states that, if a change in conditions is made to a
system in equilibrium, the system moves in the direction
that will oppose the change. The system always proceeds to
re-establish equilibrium.
chemical reactions. Once the chemicals are used up,
the cell cannot be restored or recharged and must be
discarded.
■ Secondary cells produce e.m.f. from reversible
chemical reactions. The chemicals are restored by
passing an electric current through the battery in the
opposite direction of normal cell operation.
119
120
Unit 1 Module 2 Kinetics and equilibria
cells
batteries
primary
secondary
non-rechargeable
one time use
rechargeable
can be reused
Daniell cell
Leclanché cell
Dry-cell batteries
Dry-cell batteries are so-called because they do not contain
large amounts of solution. They are widely used in small
electrical appliances such as flashlights, radios, bicycle
lamps and electric bells. There are several common types of
dry-cell batteries. They include the Leclanché dry cell and
the alkaline battery.
lead-acid
accumulator
Leclanché dry-cell
One of the most common, inexpensive and convenient
types of dry cell is the Leclanché dry cell (Figure 12.9).
This was invented in 1867 by Georges Leclanché, a French
electrical engineer (1839–1882).
alkaline battery
fuel cell
Figure 12.7 Types of cells.
Figure 12.7 summarizes the types of electrochemical cells
that will be discussed in this section of the chapter.
The Daniell cell
The Daniell cell was one of the first practical and reliable
cells to be used as a source of electricity. It was invented
in 1836 by John Frederic Daniell, an English chemist
(1790–1845), and closely resembles the Zn2+(aq)/Zn(s) and
Cu2+(aq)/Cu(s) half-cell set-up. The only difference lies in
the way in which the ions are allowed to flow between
the two half-cells. The Daniell cell (Figure 12.8) uses a
porous pot to allow the flow of ions whilst stopping the
two solutions from mixing. The laboratory Zn2+(aq)/Zn(s)
and Cu2+(aq)/Cu(s) half-cell set-up uses a salt bridge.
In the Daniell cell, a central zinc anode dips into a porous
pot containing zinc sulfate solution. The porous pot is
immersed in a solution of copper sulfate contained in a
copper can, which acts as the cell’s cathode. The Daniell
cell is a primary cell and, as we saw previously, it produces
1.10 V.
In the Leclanché dry cell the cathode (positive terminal) is a
central carbon rod surrounded by a mixture of manganese
dioxide and carbon powder (usually graphite powder)
(Figure 12.9). The manganese dioxide prevents the buildup of hydrogen gas bubbles on the terminal, which would
reduce its efficiency. It does this by oxidizing the hydrogen
produced at the electrode to water. The carbon powder
increases the surface area of the positive terminal to
increase the electrical conductivity.
metal cap (+)
carbon rod
(positive electrode)
zinc case
(negative electrode)
moist paste of
ammonium chloride
(electrolyte)
manganese(IV) oxide
metal bottom (–)
zinc rod
Figure 12.9 The Leclanché dry cell.
copper can
porous pot
copper sulfate
solution
The next layer is the electrolyte, which consists of a paste
of ammonium chloride (as a source of H+ ions) and zinc
chloride dissolved in water. And all of this is contained
within the anode (negative terminal) which is made of zinc
and also serves as the outside shell of the battery.
The Leclanché dry cell makes use of these two reactions:
zinc sulfate
solution
Figure 12.8 The Daniell cell.
at the anode: Zn(s) → Zn2+(aq) + 2e−
at the cathode: 2MnO2(s) + 2H+(aq) + 2e− →
Mn2O3(s) + H2O(l)
ITQ 7 Suggest changes which could be made to the zinc/copper
cell to cause the e.m.f. to be greater than 1.10 V?
Chapter 12 Redox equilibria
The H+ ions are in turn provided by ammonium ions, NH4+,
through the reaction:
NH4+(aq)
+ water →
H+(aq)
+ NH3(aq)
The Leclanché dry cell can be summarized as:
Zn(s) | Zn2+(aq) || 2NH4+(aq) | [2NH3(g) + H2(g)] | C(graphite)
E cell = +1.5 V
When the dry cell is in use, the zinc casing becomes thinner
as the zinc is oxidized to zinc ions and the electrolyte oozes
out of the battery. Even when the cell is not in use, the zinc
casing is eaten away since the ammonium chloride inside
the battery is acidic and reacts with the zinc.
A number of variants subsequently followed, and by
1889 (22 years after its discovery), there were at least six
well-known dry batteries in use.
The lithium-ion cell
Nowadays, lithium-ion cells are more and more used,
because they can deliver more energy per gram and can be
recharged time after time with no loss of function.
They are based on the half-reaction:
Li+ + e− → Li
E
= −3.03 V
The lithium-ion cell has one of the most negative electrode
potentials known.
One electrode is the compound LiCoO2 and the other is
carbon. In the charge/discharge cycle, lithium ions move
from the lithium/cobalt compound through a micromesh
electrode separator to the carbon cathode and back again.
These batteries can be recharged without the ‘memory
effect’ which limits others such as the Ni-H cell.
Alkaline battery
Fuel cells
Alkaline batteries are comparable to the Leclanché dry cell.
As opposed to the acidic ammonium chloride/zinc chloride
electrolyte found in the Leclanché dry cell, alkaline batteries
use an alkaline electrolyte of potassium hydroxide; hence
they acquired the name ‘alkaline battery’.
Fuel cells use fuels such as hydrogen, hydrocarbons and
alcohols as a source of chemical energy and convert them
to electrical energy. The reaction between a fuel supply
and an oxidizing agent generates electricity. Fuel cells are
primary cells but differ from conventional electrochemical
cells in that the fuel producing the electricity can be
constantly replenished. Therefore, fuel cells can operate
continuously provided the necessary reactants are provided
continuously.
In the alkaline battery:
■ the cathode is made of manganese dioxide;
■ the anode is made of zinc powder, which gives more
surface area for increased current.
The half reactions are:
at the anode: Zn(s) + 2OH−(aq) → ZnO(s) + H2O(l) + 2e−
at the cathode: 2MnO2(s) + H2O(l) + 2e− →
Mn2O3(s) + 2OH−(aq)
The different electrolyte system in the alkaline battery
accounts for the higher electrochemical efficiency as
compared to the Leclanché counterpart. This increased
electrochemical efficiency gives rise to its increased
capacity, longer storage life and better performance at both
high and low temperatures. The nominal voltage of a fresh
alkaline cell is also 1.5 V. Although this voltage falls every
time it is used, the rate at which it declines is not as steep
as in the Leclanché cell.
There are many types of fuel cells, but the mode of
operation and characteristic features are all the same. The
design of all fuel cells features three compartments which
are sandwiched together: the anode, the electrolyte and
the cathode. The electrodes are coated with a catalyst. One
of the most important is the hydrogen/oxygen fuel cell
(Figure 12.10).
e–
e–
H2O(g)
porous
carbon anode
containing Ni
porous carbon
cathode containing
Ni and NiO
H2(g)
O2(g)
reduction:
O2(g) + 2H2O(l) + 4e–
4OH–(aq)
oxidation:
2H2(g) + 4OH–(aq)
4H2O(l) + 4e –
ITQ 8 Why is the Leclanché dry cell more convenient and
portable than the Daniell cell?
warm KOH
solution
Figure 12.10 A hydrogen/oxygen fuel cell.
ITQ 9 Why is the NH4Cl/ZnCl2 electrolyte used as a paste as
opposed to a dry solid?
121
122
Unit 1 Module 2 Kinetics and equilibria
In this cell:
Lead–acid accumulator (car battery)
■ hydrogen is the fuel;
An ‘accumulator’ is a storage device. To supply electrical
power in motor vehicles a battery is used that is made up
of either three or six secondary cells based on the reaction
between lead and sulfuric acid. Each cell generates a
voltage of about 2 V: the battery therefore produces either
6 V or 12 V. Each lead–acid cell (Figure 12.11) consists of
a lead anode and a lead(IV) oxide cathode immersed in
dilute sulfuric acid (roughly 5 mol dm–3) as the electrolyte.
Although these batteries are heavy and have a low charge/
volume ratio, they can provide large currents for short
times and as such are ideal for powering starter motors.
■ oxygen (usually from the air) serves as the oxidizing
agent;
■ the anode and cathode are porous graphite electrodes
impregnated with either nickel or platinum catalysts;
■ the electrolyte is warm potassium hydroxide solution.
H2(g) enters the negative compartment and diffuses through
the porous anode. The nickel catalyst on the anode breaks
down the H2(g) to H+(aq) ions and electrons. The H+(aq)
ions enter into the KOH solution where they combine with
the OH−(aq) ions in solution to produce H2O(l) as a waste
chemical.
The electrons flow through an external circuit to the
cathode, creating an electrical current.
positive terminal
gas vents
lead
negative terminal
The overall reaction at the anode is:
lead oxide
2H2(g) + 4OH−(aq) → 4H2O(l) + 4e−
O2(g) enters the positive compartment and diffuses through
the porous cathode. At the cathode, the oxygen combines
with water and electrons to form hydroxide ions:
sulfuric acid
insulating case
O2(g) + 2H2O(l) + 4e− → 4OH−(aq)
Figure 12.11 A lead–acid battery.
The overall reaction (obtained by adding the anode and
cathode reactions) is:
During cell operation, the lead anode dissolves to form
lead(II) ions:
2H2(g) + O2(g) → 2H2O(l)
Pb(s) → Pb2+(aq) + 2e−
As long as there is a supply of hydrogen and oxygen to
this cell, it will continue to operate and produce electrical
energy. The potassium hydroxide is kept warm so that the
water produced by the cell reaction evaporates just as fast
as it is formed. Otherwise, the water will gradually dilute
the potassium hydroxide, rendering the cell inoperative.
At the positive terminal, the lead(IV) oxide cathode reacts
with the H+ ions in the sulfuric acid, also forming lead(II)
ions and water:
Fuel cells have a significant advantage over all other
devices that convert chemical energy to electrical energy:
their efficiency. Compared to an internal combustion
engine (25% efficient) and a steam engine (35% efficient),
the H2/O2 cell can operate at an efficiency of 45%. Owing
to this high efficiency, coupled with the fact that they
are pollution-free, many possible developments and uses
for fuel cells have been proposed as alternative energy
resources as well as solutions to current energy problems.
Such propositions include their use in homes, industries
and vehicles as a means of generating electricity and power.
Apollo Moon vehicles and the American Gemini Space
probes used fuel cells as their primary source of electrical
energy. They are also used in spacecraft to provide drinking
water, heat and electricity for astronauts.
PbO2(s) + 4H+(aq) + 2e− → Pb2+(aq) + 2H2O(l)
The Pb2+ ions formed at the electrodes react with SO42−
ions in the acid to form insoluble lead(II) sulfate:
Pb2+(aq) + SO42−(aq) → PbSO4(s)
As the PbSO4 precipitate forms, it coats both the electrodes,
which reduces the efficiency of the battery. The equation
for the overall reaction is:
Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l)
When recharging the battery: the reverse reaction occurs.
Since a PbSO4 build-up is responsible for the reduction
in the cell’s efficiency, the reaction that produces this
precipitate is reversed, thereby restoring the battery to
its original condition. This restoration process essentially
involves converting the PbSO4 back into Pb on the anode
ITQ 10 In the operation of a lead–acid battery, lead exists in three
oxidation states. List these states.
Chapter 12 Redox equilibria
and back into PbO2 on the cathode. This is achieved with
the help of the alternator in the vehicle, which passes
an electric current through the battery in the opposite
direction of the cell reactions.
As the battery discharges for a long time, the concentration
of sulfuric acid decreases. Over a long period of time, the fine
PbSO4 precipitate forms a coarse, inert and non-reversible
layer. In this case, recharging the battery cannot restore it
to its original condition. A modern lead–acid battery lasts
for about four years or more.
Summary
✓ Oxidation is the removal of electrons from a
✓ A system with a more negative potential than
another will act as a reducing agent towards it.
substance.
✓ Reduction is the addition of electrons to a
✓ A system with a more positive potential than
another will act as an oxidizing agent towards it.
substance.
✓ Oxidation/reduction (‘redox’) reactions involve
the transfer of electrons between substances.
✓ The tendency of a substance to gain (or lose)
electrons in an aqueous system is measured by
its standard electrode potential.
✓ The more negative the standard electrode
✓ Electrode potential measurements give no
information about rates of reactions.
✓ The flow of electrons in a redox system is an
electric current and can do electrical work.
✓ A redox system in which electron flow is
prevented can act as a storage cell.
potential, the better reducing agent the
substance is.
✓ Standard electrode potentials are measured
against that of the standard hydrogen electrode,
which is given the value 0.00 V.
Review questions
1
John tries to determine the electrode potential for
MnO4−(aq)/Mn2+(aq) using the set-up shown in
Figure 12.12.
Pt wire
Fe
1.0 mol dm–3 FeSO4(aq)
1.0 mol dm–3 KMnO4(aq)
Figure 12.12 Apparatus for determining standard electrode
potential.
The experiment fails to produce any results. However,
if a few modifications are made, the experiment
would be successful.
(a) Define the term ‘standard electrode potential’.
(b) What are THREE modifications that need to be
made to the apparatus in Figure 12.12 in order
to measure the standard cell potential of a cell
comprising an Fe2+(aq)/Fe(s) half-cell and an
MnO4−(aq)/Mn2+(aq) half-cell?
(c) Write equations for the reactions occurring in each
half-cell of the cell indicated in part (b) and hence
write an equation for the overall reaction that
occurs.
123
124
Unit 1 Module 2 Kinetics and equilibria
(d) Calculate the standard cell potential, E cell , for the
cell.
(e) Determine one metal that produces an E cell
greater than that produced by Fe.
(f) Determine one metal that produces an E cell less
than that produced by Fe.
(g) What effect (if any) would replacing the
1.00 mol dm−3 FeSO4 with a 2.00 mol dm−3 FeSO4
solution have on the E cell value calculated in
part (d). Explain your answer.
(h) Draw a fully labelled diagram of the apparatus
needed to measure the standard electrode
potential for MnO4−(aq)/Mn2+(aq) and indicate on
your diagram the direction of electron flow in the
external circuit.
2
(a) (i) Define the term ‘standard electrode potential’.
(ii) State one use of standard electrode potentials.
(iii) Draw a labelled diagram to describe how the
standard electrode potential of the Fe3+(aq)/
Fe2+(aq) half-cell can be determined.
(iv) Write the equation for the reaction occurring
in each half-cell.
(b) (i) Write a balanced equation for the reaction
between aqueous Fe2+(aq) and acidified
manganate(VII) ions.
(ii) Calculate the e.m.f. of the cell represented by
this reaction.
(c) Consider the cell represented below:
Zn(s) | Zn2+(aq) || Ag+(aq) | Ag(s)
(i) Write the equation for the reaction occurring
at the silver half-cell.
(ii) Explain the effect of an increase in
concentration of Zn2+(aq) on the e.m.f. of the
cell.
(d) Discuss the feasibility of the reaction below:
Ni2+(aq) + 2Cl−(aq) → Ni(s) + Cl2(g)
5
A Zn2+(aq)/Zn(s) half-cell is connected to a Ag+(aq)/
Ag(s) half-cell as shown below.
Zn(s) | Zn2+(aq) || Ag+(aq) | Ag(s)
(a) Write the equations for the reactions occurring at
each half-cell.
(b) Calculate the standard e.m.f. of the cell.
(c) David sets up the Ag+(aq)/Ag(s) half-cell using
a solution of 0.1 mol dm−3 Ag+ instead of
1.0 mol dm−3. Suggest how this would affect the
e.m.f. of the cell. Give a reason for your answer.
6
The lead storage battery that is used in motor vehicles
is one of the most common and useful batteries.
Anode – made of lead
Cathode – made of lead(IV) oxide (PbO2)
Electrolyte – sulfuric acid
(a) (i) Write the equations for the reactions occurring
at each electrode during discharge.
(ii) Calculate the standard cell potential.
(b) A primary source of electrical supply on the
Apollo moon flights was the fuel cell (H2/O2).
Such a cell uses porous electrodes into which
streams of hydrogen (at the anode) and oxygen
(at the cathode) are introduced. Deduce the
useful by-product of the reaction. Include relevant
equations.
Which of the two cells described below produces the
larger potential difference? Show how you arrived at
your answer.
CELL I Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
CELL II Co(s) | Co2+(aq) || Ag+(aq) | Ag(s)
Predict which one of the metals above, if any, is
capable of reducing Sn4+ to Sn2+.
E = +0.15 V
Sn4+ + 2e− ҡ Sn2+
3
4
(a) Use the following electrochemical data to
construct the labelled cell diagram for the
combined half-cells.
E = +0.34 V
Cu2+/Cu
+
E = +0.80 V
Ag /Ag
(b) Write the relevant half-equations for the changes
taking place at the anode and cathode.
(c) Write the equation for the overall cell reaction.
(d) At which electrode in the electrochemical cell
does reduction take place? Give a reason for your
answer.
(e) Calculate the cell potential.
(f) (i) Describe three changes you would observe
if you replaced the Ag half-cell with a Zn
half-cell in the cell diagram in part (a).
(ii) Suggest one reason for the changes observed
in (f)(i).
(iii) Identify an electrolyte that could be used in
the zinc half-cell.
Chapter 12 Redox equilibria
Answers to ITQs
Answers to Review questions
1 Zinc is the anode/negative electrode; placed on
the left of the cell. Copper is the cathode/positive
electrode; placed on the right. Electrons are negatively
charged and will tend to flow away from the negative
electrode and towards the positive electrode in the
cell. Hence electrons flow from left to right.
1
(c) Anode reaction: Fe(s) − 2e− → Fe2+(aq)
Cathode reaction: MnO4−(aq) + 8H+(aq) + 5e− →
Mn2+(aq) + 4H2O(l)
Overall cell reaction:
5Fe(s) + 2MnO4−(aq) + 16H+(aq) →
Fe2+(aq) + 2Mn2+(aq) + 8H2O(l)
(d) E cell = +1.95 V
3
(b) Anode reaction: Cu(s) − 2e− → Cu2+(aq)
Cathode reaction: 2Ag+(aq) + 2e− → 2Ag(s)
(c) Overall cell reaction: Cu(s) + 2Ag+(aq) →
Cu2+(aq) + 2Ag(s)
(e) E cell = +0.46 V
4
(a) (iv) Anode reaction: H2(g) + 2e− → 2H+(aq)
Cathode reaction: Cu2+(aq) + 2e− → Cu(s)
(b) (i) 5Fe2+(aq) + MnO4−(aq) + 8H+(aq) →
5Fe3+(aq) + Mn2+(aq) + 4H2O(l)
(ii) E cell = +0.74 V
(c) (i) Ag+(aq) + e− → Ag(s)
5
(a) Anode reaction: Zn(s) − 2e− → Zn2+(aq)
Cathode reaction: Ag+(aq) + e− → Ag(s)
(b) E cell = +1.56 V
6
(a) (i) Anode reaction: Pb(s) → Pb2+(aq) + 2e−
Cathode reaction: PbO2(s) + 4H+(aq) + 2e− →
Pb2+(aq) + 2H2O(l)
(ii) E cell = 1.82 V
2 (a)
high-resistance voltmeter
V
H2(g) at 298 K
and 1 atm
salt bridge
acid solution
containing
1.0 mol dm–3
H+(aq)
platinum
electrode
zinc strip
solution of
Zn2+(aq)
(1.0 mol dm–3 )
(b) The standard hydrogen electrode is the positive
electrode.
(c) Electrons flow from right to left.
3 YES because E for I−/I2 (+0.54 V) is less positive
that that for Cl−/Cl2 (+1.36 V) so chlorine is a better
oxidizing agent than iodine.
4 Zinc.
5 (a) Electromotive force, e.m.f.
(b) Cu acts as the negative electrode.
(c) E = (−0.34) + (+0.80) = +0.46 V
6 E = (+1.66) + (−0.34) = +1.32 V. The positive e.m.f.
value of +1.32 V indicates that the reaction is feasible.
However, the rate of this reaction is very slow and
does not take place unless a small amount of NaCl is
added to the solution of Cu2+ ions.
7 Increase the concentration of the Cu2+(aq) ions, which
will shift the equilibrium to the right
Cu2+(aq) + 2e− ҡ Cu(s)
8 A dry cell does not contain any sloshing liquid that
might leak or drip when inverted or handled roughly.
9 Dry ammonium chloride will not conduct electricity.
The electrolyte is used as a paste so that enough
moisture is provided to allow the current to flow.
10 0 (in Pb), +2 (in PbSO4) and +4 (in PbO2).
125
126
Module 3
Chemistry of the elements
Chapter 13
Elements and periodicity: period 3
Learning objectives
■ Describe the variations in physical properties of the period 3 elements in terms of structure and
■
■
■
■
bonding.
Describe the reactions of the elements with oxygen, water and chlorine.
Predict the types of chemical bonding present in the oxides and chlorides.
Describe the reactions of the oxides and chlorides with water.
Explain the trend in acid/base behaviour of the oxides and chlorides.
Introduction
The arrangement of elements in the periodic table reveals
that a considerable number of physical and chemical
properties of these elements vary periodically with atomic
number. This concept of ‘periodicity’ embodied in the
periodic table has facilitated rapid progress in understanding
the properties of all the elements. On account of this,
the periodic table has been extremely instrumental in
the classification and arrangement of our accumulated
chemical knowledge.
The physical properties of the elements vary throughout
the periodic table and these properties may be divided into
two categories: atomic properties and bulk properties.
Atomic properties depend only upon the structure of the
atoms, and any variation in the properties may be explained
purely in terms of single isolated atoms. Bulk properties,
however, not only depend on the characteristics of the
separate atoms, but also on how they are linked or packed
together into the structure of the element.
This chapter serves to highlight some of the properties of
the period 3 elements, as shown in Figure 13.1.
elemental properties
Atomic properties
atomic
bulk
depend on
structure only
depend on structure
and bonding
electronic configuration
melting point
atomic radii
electrical conductivity
ionic radii
density
ionization energy
electronegativity
Figure 13.1 Atomic and bulk properties of the elements.
Electronic configuration
The atomic number gives the number of electrons in an
atom. From this we can work out the arrangement of
the electrons in an atom, which is called the electronic
configuration. Table 13.1 shows the shortened version
of the electronic configurations of the eight elements in
period 3 of the periodic table. We can see that:
■ the 3s and 3p sub-levels are being filled with electrons;
■ the elements have the same number of electrons in
their inner shells – they all have the [Ne] structure;
■ the number of valence electrons increases.
Chapter 13 Elements and periodicity: period 3
Table 13.1 Electronic configurations of the period 3 elements
internuclear distance between similar atoms either
joined by a covalent bond or in a metallic crystal lattice
(Figure 13.2b). Covalent radius may be measured for
most elements, since even metals in the vapour phase
often exist as diatomic molecules. Metallic radius is
restricted to those elements which form metallic lattices.
3s1
[Ne]
Mg
[Ne] 3s2
Al
[Ne] 3s2 3px1
Si
[Ne] 3s2 3px1 3py1
P
[Ne] 3s2 3px1 3py1 3pz1
S
[Ne] 3s2 3px2 3py1 3pz1
Cl
[Ne] 3s2 3px2 3py2 3pz1
Ar
[Ne] 3s2 3px2 3py2 3pz2
It is these valence electrons that determine the structure,
bonding and properties of the elements.
Atomic radii
You will remember from your basic geometry that the radius
of a circle is defined as the distance between the centre of
the circle and its circumference. Assuming that an atom is
circular in shape, the atomic radius may be defined as the
distance between the centre of the nucleus and the outermost
electron (valence) shell. However, according to quantum
theory (see Chapter 2), there is no well-defined ‘outeredge’ valence shell since valence electrons do not reside in
a specific orbit. Rather, an atom’s electrons are described as
occupying regions of space (i.e. atomic orbitals), which are
given by a probability distribution. As a consequence, the
size of an atom is difficult to define and measure accurately.
There are different ways to define the size of an atom;
the definition depends on both the method used and the
conditions under which they are measured. Consider an
automobile tyre: the radius of the tyre is different when
measured to the top of the tyre than when measured to the
bottom of the tyre resting on the ground. The former value
will be greater than the latter. In much the same way, the
radius of an atom will vary depending on whether it is free,
or bonded to other atoms – different values for the sizes of
atoms are obtained.
■ van der Waals radius – half the internuclear distance
between non-bonded similar atoms at their closest
approach (Figure 13.2a). This is most easily determined
for non-metals, and is particularly useful for the noble
gases as they do not form chemical bonds.
a
Regardless of the measure of atomic radius that is chosen,
the general trend across period 3 remains the same. The
two primary factors that affect the atomic radius are:
■ the size of the charge on the nucleus; and
■ the number of shells between the nucleus and the
outermost electrons (inner shells).
As you go along period 3 from left to right, electrons are
going into the same main outer shell and therefore the
number of inner shells remains the same. However, due to
the increase in the number of protons in the nucleus, the
nuclear charge increases. The resultant effect can be seen
in Figure 13.3.
0.20
0.18
0.16
0.14
0.12
0.10
0.08
Na
Mg
Al
Si
Element
P
S
Cl
Figure 13.3 Atomic radii of the elements from period 3. The pull
of the nucleus on the outermost electrons increases, resulting in
a decrease in atomic radius.
Ionic radii
b
2r
Clearly, the type of bonding will influence the size of the
atom. Hence, the various atomic radii will have different
values for the same element. Generally, the covalent radius
is the smallest and the van der Waals radius is the largest.
For instance, the covalent radius for Na is 0.157 nm and
the metallic radius is 0.186 nm. When chemists speak of
atomic radii, they usually refer to covalent radii.
Atomic radius / nm
Element Electronic configuration
Na
■ Covalent or metallic radius – half the shortest
2r
Figure 13.2 (a) van der Waals radius; (b) covalent or metallic
radius.
Like the atomic radius, the ionic radius is determined
by measuring the internuclear distance between adjacent
nuclei. This distance gives the sum of the radii of the cation
and anion. By comparing the internuclear distances for
a range of compounds, the radii of individual ions are
established.
127
Unit 1 Module 3 Chemistry of the elements
Elements to the left of period 3 (Na, Mg, Al) are metals
and hence lose electrons to form cations. Cations are
always smaller than their parent atoms. A cation has fewer
electrons than the atom and so the nucleus attracts the
remaining electrons more strongly, causing a reduction in
size.
Elements to the right of period 3 (P, S, Cl) are non-metals
and tend to gain electrons to form anions. Anions are
always larger than their parent atoms. An anion has more
electrons than the atom and so the attraction of the nucleus
for each electron is reduced, resulting in expansion.
The trend in ionic radii across period 3 shows:
■ a decrease in ionic radii for the cations of Na, Mg and Al;
■ a decrease in ionic radii for the anions of P, S and Cl;
■ a large jump in ionic radii between Al3+ and P3−.
When thinking about Na+, Mg2+ and Al3+, each resulting
cation has the same electronic configuration as neon;
they are isoelectronic. However, the number of protons
is increasing. This increases the nuclear attraction on the
remaining electrons. Therefore, the size of the metallic ion
decreases across the period (Table 13.2).
Figure 13.4 summarizes this data.
0.25
Ionic radius / nm
Before we describe the trend in ionic radius across period 3,
we must first appreciate a couple of points.
0.15
0.10
0
Na+
Mg 2+
Al 3+
P3–
S 2–
Cl –
Period 3 ions
Figure 13.4 Ionic radii for period 3. Note the decreases at the left
and the right of the period and the large jump in the centre.
First ionization energy
In general, the ease with which an atom loses electrons
to form a cation describes the ionization energy of the
atom. First ionization energy is defined as the energy
required to remove the most loosely held electron from a
neutral atom in the gaseous state to form a cation.
Na(g) − e− → Na+(g)
This equation is more commonly written as
Na(g) → Na+(g) + e−
Table 13.2 Cationic radii for period 3
1600
Cation
Na+
Mg2+
Al3+
Number of protons
11
12
13
Electronic configuration of ion 2,8
2,8
2,8
Ionic radius / nm
0.072
0.054
0.102
0.20
0.05
When thinking about P3−, S2− and Cl−, the anions are
isoelectronic with argon. There is now an extra inner shell
of electrons. However, the progressively larger nuclear
charge exerts a greater force of attraction on the valence
shell electrons. As a result, the size of the anion decreases
(Table 13.3).
First ionization energy / kJ mol –1
128
1400
1200
1000
800
600
400
200
0
Na
Mg
Al
Si
P
Element
S
Cl
Ar
Figure 13.5 The change in first ionization energy across period 3.
Table 13.3 Anionic radii
Anion
P3−
S2−
Cl−
Number of protons
15
16
17
Electronic configuration of ion 2,8,8
2,8,8
2,8,8
Ionic radius / nm
0.184
0.181
0.212
You should also notice the big jump in ionic radius between
Al3+ (0.054 nm) and P3− (0.212 nm). This huge disparity is
attributed to the addition of an inner shell of electrons.
ITQ 1 Suggest a second reason for the
reduction in radius going from Na to Na+
other than Na+ having fewer electrons.
There is a general increase in the first ionization energy
across a period, as can be seen from Figure 13.5. On passing
across a period, the nuclear charge steadily increases and the
atomic radius falls. Both of these factors result in an increase
in nuclear attraction, which means that the electrons are
held more tightly. This results in an increase in ionization
energy across a period. If the electron is held more tightly,
it is harder to pull away and thus more energy is required.
ITQ 2 Describe the relationship between
atomic radius and first ionization energy.
ITQ 3 Suggest two other pairs of elements
which should exhibit differences in their first
ionization energies as Mg/Al and P/S.
Chapter 13 Elements and periodicity: period 3
The ionization energy does not increase smoothly on passing
across period 3. There are two irregularities, which occur at
aluminium and sulfur, and these can be explained on the
basis of sub-levels. Let us look at each irregularity in turn.
The first ionization energy of aluminium is smaller than
magnesium (Table 13.4). The outermost electron of
magnesium is located in a filled 3s orbital, which is relatively
close to the nucleus. However, aluminium’s outermost
electron is unpaired, resides further away from the nucleus
in a 3p orbital and is partially screened by the 3s electrons.
Therefore, it requires less energy to remove the less tightly
held, unpaired 3px1 electron in aluminium than it is to
remove an electron from the filled 3s orbital in magnesium.
Table 13.4 Comparing the first ionization energies for Mg and Al
Electronic configuration
ΔHi1 / kJ mol
3s2
738
[Ne] 3s2 3px1
578
magnesium [Ne]
aluminium
−1
The first ionization energy of sulfur is smaller than
phosphorus (Table 13.5). For both sulfur and phosphorus,
the electron is being removed from the same orbital and
the screening is identical. In the case of phosphorus, the
three electrons in the 3p orbitals all have parallel spins.
However, in sulfur, the electron being removed is one of
the 3px2 pair. The repulsion between the two electrons in
the same orbital makes this electron easier to remove.
Table 13.5 Comparing the first ionization energies for P and S
Electronic configuration
sulfur
[Ne]
3px1
3py1
[Ne]
3s2
3px2
3py1
Bulk properties
Melting point
The melting point of a substance may be defined as the
temperature at which the pure solid breaks down and is
in equilibrium with pure liquid at standard atmospheric
pressure. Melting points depend on both the structure and
bonding in a substance. More specifically, there are four
factors which influence the melting point:
■ the type of forces holding the solid state together,
whether they are ionic, covalent, metallic or van der
Waals;
■ the strength of attraction between the particles in the
structure;
■ the stability of the lattice;
■ the size of the molecule.
ΔHi1 / kJ mol−1
3pz
1
1012
3pz
1
1000
2000
Electronegativity
3.5
Pauling electronegativity value
Figure 13.6 shows a graph of electronegativity values
plotted against atomic number for the elements Na to Cl.
As the atomic radius decreases across the period (see Figure
13.3), the attractive force of the nucleus is felt more strongly
by the electrons in the bond. As a result, electronegativity
increases across a period.
Melting point / K
phosphorus
3s2
of an atom is expressed relative to a standard. The standard
used is fluorine – the most electronegative element – which
is assigned an electronegativity of 4.0.
3.0
1500
1000
500
2.5
0
2.0
Na
1.5
Mg
Al
Si
P
Element
S
Cl
Ar
Figure 13.7 A graph of melting point for period 3.
1.0
0.5
0
Na
Mg
Al
Si
Element
P
S
Cl
Figure 13.6 Change in electronegativity across period 3.
The electronegativity of an atom provides a numerical
measure of the power of an atom to attract electron pairs
to itself in a covalent bond. Electronegativity cannot be
measured directly. As a result, Pauling devised a system called
the electronegativity scale, in which the electronegativity
As period 3 is traversed from sodium to argon, the melting
point generally rises sharply from sodium through to silicon
(Figure 13.7). It then undergoes a rapid fall between silicon
and phosphorus. Between phosphorus and argon there is a
general decrease in the melting point. These trends may be
explained in more detail.
■ Sodium, magnesium and aluminium are metals and
hence exhibit metallic bonding in which electrostatic
attractive forces exist between the positively charged
metals ions and a ‘sea’ of delocalized electrons. On
129
Unit 1 Module 3 Chemistry of the elements
moving from sodium to aluminium, the charge on the
metal ion increases from +1 to +3 and therefore the
number of delocalized electrons increases. With the
nuclei getting more positively charged and the ‘sea’
getting more negatively charged, the strength of the
attraction increases which causes an increase in the
strength of the metallic bond and hence the melting
point increases.
■ Silicon is a metalloid, showing both metallic and
non-metallic properties. Silicon forms a giant threedimensional lattice structure similar to that of
diamond. Each silicon atom is covalently bonded to
four other silicon atoms in a tetrahedral arrangement.
Considerable energy is required to break these bonds,
so the melting point of silicon is very high.
■ Phosphorus, sulfur and chlorine are non-metals
The trend in electrical conductivity across period 3 elements
shows a dramatic increase between sodium and aluminium,
followed by a drastic fall at silicon; the remaining elements
do not conduct electricity (Figure 13.9). These trends can
be explained on the same basis that was used to explain
melting point.
1.0
Relative electrical conductivity
130
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
and their structures contain simple, small covalent
molecules held together only by weak van der Waals
forces of attraction. As a result of such weak forces, the
melting points of these elements are low since little
energy is required to overcome them.
■ Argon is a noble gas and therefore exists as individual
argon atoms. The extent of van der Waals attractions is
very limited and so the melting point is extremely low.
When thinking about phosphorus, sulfur and chlorine, the
magnitude of their melting points are governed entirely
by the size of the molecules (Figure 13.8). The larger the
molecule, the more van der Waals forces present and hence
the greater the melting point.
■ Sulfur has the highest melting point; it exists as S8
molecules.
■ Phosphorus has the next highest; it exists as P4
molecules.
■ Chlorine has the lowest melting point; it exists as Cl2
molecules.
Na
Mg
Al
Si
P
Element
S
Cl
Ar
Figure 13.9 Relative electrical conductivity of the period 3
elements (Al = 1.00).
■ Sodium, magnesium and aluminium are metals and
hence exhibit metallic bonding in which positive
metal ions are attracted to delocalized electrons. These
delocalized electrons serve as mobile charge carriers.
In going from sodium to aluminium, the number
of mobile delocalized electrons increases. Therefore
there are more charge carriers and so the electrical
conductivity increases.
■ Silicon is a metalloid and is a semiconductor; the full
explanation for this semiconductivity is beyond the
scope of the CAPE syllabus. We should know that
within the tetrahedral structure, the four electrons in
each silicon atom are held strongly in covalent bonds.
However, there are a few delocalized electrons which
accounts for the poor conductivity of silicon. Unlike a
true metal, the conductivity increases with an increase
in temperature.
■ Phosphorus, sulfur, chlorine and argon are non-metals
S8
P4
Cl2
Figure 13.8 S8, P4 and Cl2 molecules.
Electrical conductivity
Electrical conductivity is a measure of a material’s
ability to conduct an electric current. An electric current
may be described as a flow of electric charge and, as such,
an element can conduct electricity provided that it contains
electrons (charge carriers) that are free to move. Generally,
metals are good conductors of electricity whilst non-metals
are poor conductors.
and do not conduct electricity because there are no
free delocalized electrons within their structures to
convey an electric current. In phosphorus, sulfur and
chlorine the outer electrons are not free to move and
carry charge because they are held strongly in covalent
bonds. In argon, the outer electrons are not free to
move and carry charge because they are held strongly
in a stable third energy level.
ITQ 4 Carbon exists as graphite and as diamond (allotropes).
Graphite is an excellent conductor of heat and electricity whereas
diamond is an insulator. What does this suggest about the
electron arrangement in the two allotropes?
Chapter 13 Elements and periodicity: period 3
Table 13.6 Period 3 elements and their reactions with oxygen, water and chlorine
Element Reaction with oxygen
Na
Reaction with water
burns with an orange flame, producing a mixture of an exothermic reaction with cold water, producing
Na2O and Na2O2
NaOH and H2
burns with a brilliant white flame, producing MgO
reacts slowly with cold water, giving Mg(OH)2 and H2,
but exothermically with steam, giving MgO and H2
will burn only when finely divided, forming Al2O3
aluminium powder heated in steam produces Al2O3
and H2
will burn if heated strongly enough, producing SiO2 no reaction
Mg
Al
Si
P
no reaction
burns vigorously with a brilliant pinkish-white
flame, forming P4O6 and P4O10. White P burns
spontaneously in air; red P needs heating
burns in air or oxygen on gentle heating with a
brilliant blue flame, giving SO2 and SO3
S
Cl
burns with an intense white flame, giving MgCl2
burns, forming AlCl3
reacts, producing SiCl4
no reaction
reacts, forming S2Cl2. Unstable SCl2 and SCl4 can
also be formed but they readily decompose to
form S2Cl2
a disproportionation reaction to produce HCl and HOCl no reaction
does not react directly with oxygen; however, there
are several chlorine oxides, e.g. Cl2O7, Cl2O
■ For Si, P and S, which are solids with more open
structures, it is the structure which controls the density.
The density of a material is defined as:
■ For the small covalent molecules Cl and Ar, which are
mass
density =
volume
For an element, this is dependent on the density of the
individual atoms, and how many of them are in packed
into any particular volume. This, in turn, depends on the
structure of the element and the strength of the forces
between the atoms.
An atom is mostly empty space, with a mass (the nucleus)
at its centre. We can work out the relative densities of
individual atoms by dividing the atomic mass of the
element by its volume, which is in turn proportional to its
atomic radius. When we do this we see that one atom in
period 3 differs regularly from the next (see the red line
in Figure 13.10). The actual density of the element at STP
varies as shown by the blue line.
■ For Na, Mg and Al (the first three elements), all of
which have close-packed metal structures, the atom
density is the controlling factor.
3.0
25
2.0
20
15
1.0
10
relative
atomic density
Relative atomic density
bulk
density
5
0
0
Mg
Al
Si
P
S
Cl
gases at STP, it is the absence of strong forces between
molecules which gives rise to the very low density.
Chemical properties
Trends in chemical properties
In much the same way that the elements of period 3 show
trends in physical properties, they also show trends in
chemical properties. Some of these trends are shown in
Table 13.6.
Variation in oxidation states of the oxides and
chlorides
There is a trend within the formulae of the oxides and
chlorides of period 3 elements. Table 13.7 allows us to recap
the formulae of the oxides. Some of the period 3 elements
form other oxides. Here we are interested in the oxidation
state of the ‘highest’ oxide.
Table 13.7 Formulae of the period 3 oxides and oxidation state of
the highest oxides
30
Na
burns with a bright orange flame, producing NaCl
burns, producing a mixture of PCl3 and PCl5
Density
Bulk density / g cm –3
Reaction with chlorine
Ar
Figure 13.10 Bulk density and relative atomic density for period
3 elements.
Highest oxide
Na
Mg
Al
Na2O
MgO
Al2O3 SiO2
P4O10 SO3
Cl2O7
+3
+5
+6
+7
P4O6
SO2
Cl2O
Oxidation state of highest
+1
+2
oxide
Other oxides
Na2O2
Si
+4
P
S
Cl
The oxygen in the peroxide ion (O22−) is assumed to have an oxidation state of −1.
Therefore, the oxidation state of sodium in Na2O2 is +1.
In the highest oxides, the period 3 element is in its highest
oxidation state. We can see that the highest oxidation state
of the element is equal to the number of valence electrons on
the atom, which is in effect the group number. This implies
that in these oxides, all the valence electrons in the element
131
132
Unit 1 Module 3 Chemistry of the elements
are used in bonding. This ranges from one valence electron
on sodium to all seven valence electrons on chlorine.
Physical and chemical properties of the
chlorides of period 3
In the highest chlorides, the maximum oxidation state
ranges from +1 to +5 (Table 13.8).
The physical and chemical properties of the period 3
chlorides are determined by their structure, bonding and
reaction with water. As with the oxides, the physical
properties of the chlorides also reveal a bonding pattern that
ranges from ionic bonding on the left-hand side to covalent
bonding on the right-hand side. Aluminium chloride and
phosphorus(V) chloride are complicated; they change their
structure from ionic to covalent when the solid turns to a
liquid or vapour.
Table 13.8 Oxidation states of the highest period 3 chlorides
Na
Mg
Si
P
Highest chloride
NaCl
MgCl2 AlCl3
Al
SiCl4
PCl5
Oxidation state
+1
+2
+4
+5
+3
Sulfur has been omitted from Table 13.8 because the
highest oxidation state of sulfur is in SCl4. If you look back
to Table 13.6, SCl4 is unstable and readily decomposes to
form S2Cl2.
Aluminium chloride shows additional characteristics:
■ it sublimes at around 180 °C and ordinary atmospheric
Physical and chemical properties of the
oxides of period 3
The physical and chemical properties of the period 3 oxides
are determined by their structure, bonding and reaction
with water. These oxides show variations that demonstrate
periodic patterns.
The physical properties of the oxides reveal a bonding
pattern that ranges from ionic bonding on the left-hand
side to covalent bonding on the right-hand side. This
bonding pattern in turn determines the trend in acid/base
behaviour from strongly basic oxides on the left-hand side
to strongly acidic ones on the right via an amphoteric oxide.
The trend in the pH of the solutions formed from the
oxides goes from alkaline to acidic. Table 13.9 highlights
some properties of the oxides of the period 3 elements.
As we see from Table 13.9, the bonding pattern of the
oxides of period 3 change from a giant ionic lattice to small
covalent molecules. This change in bond type is attributed
to the decreasing electronegativity difference between that
of oxygen and the element as period 3 is traversed from
left to right. Generally, a large electronegativity difference
between two elements in a bond implies that the bond is
primarily ionic whilst a small electronegativity difference
implies that the bond is primarily covalent.
As period 3 is traversed, the trend is from strongly basic
oxides on the left to an amphoteric oxide in the middle
to strongly acidic oxides on the right. This pattern occurs
because the chemical character changes from metallic
oxides (which are basic) to non-metallic oxides (which are
generally acidic). Hence the acid/base behaviour changes
from basic to amphoteric to acidic.
ITQ 5 Account for the difference in pH of the aqueous solutions
of the oxides of Na and Mg.
pressure;
■ if the pressure is raised to just over 2 atm, it melts at
192 °C;
■ it exists in some instances as a dimer (two molecules
joined together), Al2Cl6.
The reactions of the period 3 elements with water show
that the ionic chlorides (sodium chloride and magnesium
chloride) dissolve in water without any reaction; the other
chlorides react in a variety of ways in a reaction known
as hydrolysis. The trend shows that ionic metal chloride
salts give nearly neutral solutions whilst covalent metal
and non-metal chlorides hydrolyse to give acidic solutions.
Table 13.10 highlights some properties of the chlorides of
the period 3 elements.
Chapter 13 Elements and periodicity: period 3
Table 13.9 Some properties of the oxides of the period 3 elements
Oxides
Bonding
Solubility in water
Acid/base behaviour
Reaction with water
pH of solution formed
Na2O
ionic
soluble
strong base
Na2O(s) + H2O(l) → 2NaOH(aq)
13
MgO
ionic
almost insoluble
weak base
MgO(s) + H2O(l) → Mg(OH)2(aq)
8
Al2O3
ionic with covalent character
insoluble
amphoteric
no reaction
–
SiO2
giant covalent
insoluble
weak acid
no reaction
–
P4O6
simple covalent
soluble
strong acid
P4O6(s) + 6H2O(l) → 4H3PO3(aq)
2
P4O10
simple covalent
soluble
moderately strong acid
P4O10(s) + 6H2O(l) → 4H3PO4(aq)
1
SO2
simple covalent
soluble
moderately strong acid
SO2(aq) + H2O(l) → H2SO3(aq)
1
SO3
simple covalent
soluble
moderately strong acid
SO3(g) + H2O(l) → H2SO4(aq)
0–1
Cl2O7
simple covalent
soluble
strong acid
Cl2O7(l) + H2O(l) → 2HClO4(aq)
1
Table 13.10 Some properties of period 3 chlorides
Chlorides Bonding
NaCl
ionic
Solubility in
water
soluble
Acid/base
behaviour
neutral
Reaction with water
pH of solution formed
Dissolves in water
NaCl(aq) → Na+(aq) + Cl−(aq)
Dissolves in water
MgCl2(aq) → Mg2+(aq) + 2Cl−(aq)
With limited water, gives acid fumes
AlCl3(s) + 3H2O(l) → Al(OH)3(s) + 3HCl(g)
In excess water, gives weakly acidic solution due to the acidity of [Al(H2O)6]3+
AlCl3(s) + 6H2O(l) → [Al(H2O)6]3+(aq) + 3Cl−(g)
7
MgCl2
ionic
soluble
faintly acidic
AlCl3
ionic with covalent
character
soluble
strong acid
SiCl4
giant covalent
soluble
strong acid
SiCl4(l) + 2H2O(l) → SiO2(s) + 4HCl(aq)
2
PCl3
simple covalent
soluble
strong acid
PCl3(l) + 3H2O(l) → H3PO3(aq) + 3HCl(aq)
2
PCl5
simple covalent
soluble
strong acid
PCl5(s) + 4H2O(l) → H3PO4(aq) + 5HCl(aq)
2
S2Cl2
simple covalent
soluble
strong acid
reacts slowly to produce a complex mixture of S, HCl, H2S, H2SO3, H2SO4
2
6.5
3
Summary
✓ The position of an element in the periodic table
is related directly to its electron structure.
✓ The magnitudes of physical and chemical
properties of the elements follow clear trends
across periods and down groups in the table.
✓ The magnitudes of these properties are modified
by any substantial change in the electron
structures concerned.
✓ Similar trends are found in the bulk physical
properties of elements as are found in their
atomic properties.
✓ These effects are shown in the elements of
period 3 (Na–Ar) by such properties as ionization
energy, melting point, density, oxidation state
and the reactivities of the oxides and chlorides of
the elements.
133
134
Unit 1 Module 3 Chemistry of the elements
Review questions
1
4
(a) Describe the observations made when Na and S
are heated separately with chlorine under suitable
conditions.
(b) Two other elements are found to react similarly
when heated in dry chlorine. Discuss the variation
in electrical conductivity of these elements, in
terms of structure and bonding.
5
(a) The oxides and chlorides of the period 3 elements
show variations that demonstrate periodic patterns
or trends. With reference to named examples,
explain the periodic variation in the oxidation
numbers of the oxides of the elements.
(b) Write a balanced equation for the reaction of each
of the following with water:
(i) a metal chloride
(ii) a non-metallic oxide
(c) Aluminium chloride is a metallic chloride, yet its
solution is acidic.
(i) How is this observation different from that of
chlorides of Group I and Group II elements?
(ii) Suggest an explanation for the acidic nature of
the aluminium chloride solution.
(d) (i) Explain the trend in acid/base behaviour of the
oxides of the elements of period 3 in terms of
structure and bonding.
(ii) Write a balanced equation to represent each of
the following:
a) the acidic nature of a selected oxide of the
elements of period 3;
b) the basic nature of a selected oxide of the
elements of period 3.
6
The physical and chemical properties of the period
3 oxides and chlorides are determined by their
structure, bonding and reaction with water.
(a) Copy and complete the following table by writing
in the type of bonding for each compound.
Table 13.11 gives the atomic radii and melting points
of the elements in period 3.
Table 13.11 Some properties of elements in period 3
Na
Mg
Al
Si
P
S
Cl
Atomic radius / nm 0.157 0.136 0.125 0.117 0.110 0.104 0.099
Melting point / °C
98
651
660
1410 44
114
−101
(a) State and account for the trend in the values of
the atomic radii across the period from Na to Cl.
(b) The trend in the melting points of the elements
in Table 13.11 is related to their structure and
bonding. Describe the trend in the structure of
the elements, and the trend in the bonding of the
elements in Table 13.11.
2
(a) Describe the reaction of sulfur with:
(i) oxygen;
(ii) water.
(b) Write the chemical equations for the reactions in
part (a).
(c) Sulfur dioxide is an acidic gas that dissolves in
water to form sulfurous acid. Moist sulfur dioxide
(H2SO3) can act as a strong reducing agent and as
a bleaching agent.
(i) Write an ionic equation illustrating the action
of the sulfite (SO32−) ion as a reducing agent.
(ii) Write an equation for the reaction between
sulfur dioxide and sodium oxide.
(iii) Suggest the chemistry involved in the action of
the sulfur dioxide as a bleaching agent.
3
The pH values of the oxides of the period 3 elements
are given in Table 13.12.
Table 13.12
pH of aqueous solution of the oxide
Na
Mg
Al
Si
P
S
13
8
7
7
2
3
(a) Account for the difference in pH of the aqueous
solutions of the oxides of Na and Mg.
(b) Describe the trend in acid/base nature of the
oxides of period 3.
(c) (i) Explain in terms of bonding why aluminium
oxide is described as an amphoteric oxide and
not as a neutral oxide.
(ii) Write one chemical equation to illustrate
either the acidic or basic nature of aluminium
oxide.
Compound
Type of bonding
Na2O
MgO
Al2O3
AlCl3
SiCl4
PCl5
(b) State the acid/base character of:
(i) MgO
(ii) Al2O3
Chapter 13 Elements and periodicity: period 3
(c) Write balanced equations to show the reaction
between water and each of the following period 3
compounds:
(i) Na2O
(ii) SiCl4
(iii) PCl5
(d) Give the approximate pH of the solution formed
from the reactions in part (c)(i) and part (c)(ii).
(e) SiCl4 is a liquid at room temperature and pressure
whilst SiO2 is a solid with a high melting point.
Explain these observations in terms of structure
and bonding of the silicon compounds.
7
8
(a) Define the term ‘electronegativity’.
(b) Consider the chlorides of the elements in period
3 of the periodic table, and answer the following
questions.
(i) Describe the structure of the chlorides.
(ii) Describe the difference in pH of the solutions
formed when the chlorides react with water.
(iii) Write the equation for the reaction of
silicon(IV) chloride and water.
(a) Describe the variation in melting points and
electrical conductivities of the elements sodium
to chlorine, which are found in period 3 of the
periodic table. In each case, explain the variation
in terms of the bonding and structure of the
elements.
(b) Compounds A and B are the chlorides of the
elements in period 3. Some physical properties of
A and B are given in Table 13.13.
Table 13.13
Compound A
Compound B
melts at 801 °C
sublimes at 178 °C
insoluble in organic solvents
dissolves in most organic solvents
soluble in water and its solution
has a pH of 7
dissolves in water and its solution
has a pH of 3
(i) Explain the differences in the observed
properties identified in Table 13.13, of these
two compounds A and B.
(ii) a) Suggest the identities of A and B.
b) Write the equation to explain the
formation of the solution with a pH of 3.
Answers to ITQs
1
In Na+ there are no electrons in the original outer
shell.
2
Atomic radius is inversely related to first ionization
energy. As the atomic radius decreases, the outer
electrons are held more tightly, thereby resulting in
an increase in ionization energy. If the electron is held
more tightly, it is harder to pull away.
Conversely, as the atomic radius increases, the outer
electrons are held less tightly, thereby resulting in a
decrease in ionization energy. If the electron is held
less tightly, it is easier to pull away.
3
Beryllium / boron and nitrogen / oxygen.
4
The electrons are free to move in graphite but are
localized in diamond.
5
Sodium oxide is a strongly basic oxide. It contains the
oxide ion, O2−, which is a very strong base with a high
tendency to combine with hydrogen ions. The solid is
held together by attractions between 1+ and 2− ions.
Magnesium oxide is also a basic oxide and contains
oxide ions. However, it is not as strongly basic as
sodium oxide because the oxide ions are not very
free. In the case of MgO, the solid is held together by
attractions between 2+ and 2− ions which are stronger
and require more energy to overcome.
135
136
Chapter 14
Elements and periodicity: Group II
Learning objectives
■ Explain the variations in properties of the elements in terms of structure and bonding.
■ Describe the reactions of the elements with oxygen, water and dilute acids.
■ Explain the variation in solubility of the sulfates.
■ Explain the variation in the thermal decomposition of the carbonates and nitrates.
■ Discuss the uses of some of the compounds of magnesium and calcium.
Introducing Group II
In this chapter we will explore the trends in physical
properties as well as some chemical reactions of the Group
II elements – beryllium, magnesium, calcium, strontium,
barium and radium. Radium is radioactive and will not
be considered during discussions. This group is generally
called the ‘alkaline earth metals’. These elements are also
sometimes referred to as part of the s-block elements, since
in all these metals, the only electrons in the outermost
shell occupy an s sub-level.
Physical properties
Atomic radius
There are two primary factors that affect the atomic radius:
■ the size of the charge in the nucleus;
■ the number of shells between the nucleus and the
outermost electrons (inner shells).
As Group II is descended, both the size of the nuclear
charge and the number of inner shells increases. However,
the increasing number of inner shells (which shield the
outermost electrons from the attraction of the nucleus)
outweighs the increase in nuclear charge. Consequently,
the attraction of the nucleus for the outermost electrons
becomes weaker as the group is descended and the outcome
is an increase in atomic radius (Figure 14.1).
Atomic radius / nm
0.2
0.1
0
Be
Mg
Ca
Element
Sr
Ba
Figure 14.1 Atomic radius for Group II, showing the increase in
radius from Be to Ba.
900
First ionization energy / kJ mol –1
Do you remember from Chapter 2 that there different
numbering schemes for the groups in the periodic table? In
this chapter we are following the CAPE Chemistry Syllabus
and calling this group of elements Group II. You may also
see the group referred to as Group IIA and Group 2.
0.3
800
700
600
500
400
300
200
100
0
Be
Mg
Ca
Element
Sr
Ba
Figure 14.2 First ionization energy for Group II, showing the
reduction in first ionization energy from Be to Ba.
First ionization energy
First ionization energy is defined as the energy required
to remove an electron from a neutral atom in the gaseous
state. On passing down Group II, the increasing number
of inner shells causes the outermost electrons to be farther
Chapter 14 Elements and periodicity: Group II
away from the nucleus. These electrons are therefore held
less tightly and the ionization energy decreases down the
group (Figure 14.2).
producing a small amount of magnesium hydroxide.
However, the reaction soon stops because the Mg(OH)2
formed is almost insoluble in water and forms a coating on
the magnesium which prevents further reaction.
Chemical reactions
Mg(s) + 2H2O(l) → Mg(OH)2(s) + H2(g)
The chemical properties of Group II elements are dominated
by the high reactivity and strong reducing power of the
elements. Standard electrode potential values for Group II
elements (Table 14.1) reveal that the elements are high in
the electrochemical series and become increasingly reactive
on descending the group.
Calcium, strontium and barium react with cold water to
give the metal hydroxide and hydrogen according to the
general equation:
Table 14.1 Standard electrode potential values for the Group II
elements
E
Reaction
2+(aq)
Be
+
2e−
ҡ Be(s)
/V
−1.85
Mg2+(aq) + 2e− ҡ Mg(s)
−2.37
Ca2+(aq) + 2e− ҡ Ca(s)
−2.87
Sr2+(aq) + 2e− ҡ Sr(s)
−2.89
Ba2+(aq)+
−2.91
2e−
ҡ Ba(s)
The greater the negative value of the standard electrode
potential, E , the greater is the reactivity and the reducing
power.
Reactions with oxygen
On the whole, the Group II elements burn readily in oxygen
to form a metallic oxide. These are highly exothermic
reactions and can be represented by a general equation:
M(s) + 2H2O(l) → M(OH)2(s or aq) + H2(g)
The metal hydroxides are not very soluble. However,
solubility increases down the group, and therefore less
precipitate is formed as more of the hydroxide dissolves in
water.
Reactions with acids
All the metals, with the exception of beryllium, reduce
acids to form salts and hydrogen gas. As expected, the
reactivity increases down the group.
Solubility of the Group II sulfates
When an ionic solid dissolves in water, an enthalpy change
occurs and this can be described by an enthalpy cycle
(Figure 14.3).
+
–
M X (s)
6H s
Group II owes its name ‘alkaline earth metals’ to their
oxides. These oxides are basic (alkaline) (although BeO is
amphoteric). Furthermore, ‘earth’ is an old term applied by
early chemists to substances that are insoluble in water and
resistant to heating – properties shared by these oxides.
These oxides melt at such high temperatures that they
remain solids (‘earths’) in fires.
Reactions with water
Beryllium has no reaction with water.
Magnesium reacts only with steam, to form magnesium
oxide and hydrogen gas:
Mg(s) + H2O(g) → MgO(s) + H2(g)
Magnesium has a very slight reaction with cold water,
–
6H h
–6H l
+
2M(s) + O2(g) → 2MO(s)
All the Group II metals except beryllium and magnesium
tarnish rapidly in air as a layer of oxide is formed on the
surface of the metal. Barium is so reactive it is stored under
oil.
+
M (aq) + X (aq)
–
M (g) + X (g)
Figure 14.3 An enthalpy cycle for the dissolution of M+X− in
water.
■ ΔHs, the enthalpy change of solution, is the enthalpy
change when 1 mol of ionic solid dissolves in so much
water that more dilution produces no further enthalpy
change:
M+X−(s) → M+(aq) + X−(aq)
■ −ΔHl, the reverse of the lattice energy, is the enthalpy
needed to convert 1 mol of ionic lattice into gaseous
ions:
M+X−(s) → M+(g) + X−(g)
■ ΔHh, the enthalpy change of hydration of each ion,
M+(g) and X−(g), is the enthalpy change when 1 mol
of gaseous ions are dissolved in so much water that
further dilution produces a negligible enthalpy change:
M+(g) + X−(g) → M+(aq) + X−(aq)
ITQ 1 Using E values, explain why the reaction with water would
be expected to occur more vigorously with barium than beryllium.
137
138
Unit 1 Module 3 Chemistry of the elements
These three steps form an enthalpy cycle for the dissolution
of the ionic solid M+ X− in water, as shown in Figure 14.3.
Now, Hess’s law (see Chapter 8) states that the enthalpy
change during a reaction is independent of the route
followed. Applying this to Figure 14.3 we get:
ΔHs [M+X−(s)] = −ΔHl [M+X−(s)] + ΔHh [M+(g)] + ΔHh [X−(g)]
From this equation, we see that we have two possibilities:
■ the enthalpy of solution will be exothermic if the sum
of the hydration enthalpies is numerically less than the
lattice enthalpy;
■ the enthalpy of solution will be endothermic if the
sum of the hydration enthalpies is numerically greater
than the lattice enthalpy.
Table 14.2 Ionic radii and hydration energies
Ions
Ionic radii / nm Hydration energy / kJ mol−1
2+
0.065
−1981
Ca2+
0.099
−1562
Sr2+
0.113
−1414
Ba2+
0.135
Mg
SO42−
−1273
−1115
Table 14.3 Solubility of Group II sulfates
Compound
Lattice energy
/ kJ mol−1
Sum of the hydration
energies / kJ mol−1
Solubility / moles per
100 g water
MgSO4
−2959
−3096
0.02
CaSO4
−2653
−2677
1.1 × 10−3
SrSO4
−2603
−2529
6.2 × 10−5
BaSO4
−2423
−2388
9.0 × 10−7
The practical out-working of these possibilities is as follows:
■ if ΔHs is endothermic, then the ionic solid will be
sparingly soluble;
■ if ΔHs is exothermic, then the ionic solid will be soluble.
From these statements, we can gather that the more
endothermic (or less exothermic) the enthalpy of solution,
the less soluble the compound. When the lattice of ions in the
Group II sulfates is broken up, energy has to be supplied, and
when these ions form bonds with water, energy is released.
Lattice energy is governed by the inter-ionic distance
between the cation and anion. For the Group II sulfates,
the anion SO42− is constant but as the group is descended,
the cations get bigger. Thus, as we go down the group, this
distance increases, which means weaker forces holding
them together and hence less energy is needed to break
the lattice. As such, the lattice energy decreases down the
Group II sulfates.
Hydration energy is governed by the strength of the total
attractive force between the ion and the water molecules.
As the cationic size increases down the group, the attraction
to water becomes weaker (Table 14.2). Weaker attractions
mean decreasing amounts of energy released as the ions
bond to water molecules. Hence, the hydration energy
decreases down Group II.
We see that both the lattice and hydration energies decrease
as the group is descended. This gives no indication as to
whether the energy of solution will be endothermic or
exothermic. The deciding factor is how fast they fall relative
to each other and this is determined by the size of the anion,
which in turn affects the lattice energy. The sulfate anion
ITQ 2 Would you expect greater lattice energies amongst Group I
sulfates or Group II sulfates?
is large and this greatly controls the inter-ionic distance;
increases in the cationic size have small effects on the
distance. Consequently, if the increase in inter-ionic distance
is small, the decrease in the lattice energy will also be small.
The outcome is that the hydration energy falls faster than
the lattice energy. Therefore, if the sum of the hydration
energies is greater than the lattice energy as Group II is
descended, the enthalpy of solution will become more
endothermic as the group is descended. This translates to
mean that the Group II sulfates become less soluble as the
group is descended (Table 14.3).
Thermal stability of the Group II carbonates
and nitrates
The thermal stability of a compound relates to the effect
of heat on the compound. When a compound is heated, it
splits up and is said to undergo thermal decomposition.
The compounds of Group II elements have different
thermal stabilities, inferring that they decompose at
different temperatures when heated. Thermal stability of
the compound M+X−(s) is dependent on two factors: the
charge and size of its ions. Let us now look at the effect of
heat on the carbonates and nitrates of Group II elements.
Carbonates
The Group II carbonates undergo thermal decomposition
to give the metal oxide and carbon dioxide gas:
MCO3(s) → MO(s) + CO2(g)
The carbonates, as well as the metal oxides formed, are
white solids. As the group is descended, the carbonates
ITQ 3 A barium sulfate meal is often fed to patients in preparation
for X-ray analysis of the digestive tract. Explain why the use of
barium sulfate is acceptable even though Ba2+ ions are toxic.
Chapter 14 Elements and periodicity: Group II
The delocalized electrons
are pulled towards the
positive ion
O
+
2
O
C
This end of the ion
is on its way to
breaking away and
becoming carbon
dioxide
The nitrates as well as the metal oxides formed are white
solids. As you go down the group, the nitrates have to be
heated more strongly before they decompose. This means
that the thermal stability of the nitrates also increases down
the group.
Figure 14.4 The polarizing effect of the cation (M2+).
The explanation for this trend is the same as it is for the
carbonates. The small M2+ ions at the top of the group
polarize/distort the nitrate ions more than the larger M2+
ions at the bottom. Therefore, as the group is descended,
the nitrates become more thermally stable.
become more stable and therefore higher temperatures are
required to decompose them. This means that the thermal
stability of the carbonates increases down the group.
Uses of magnesium and calcium
compounds
O
This oxygen atom is
well on the way to
becoming an oxide ion
The explanation for this trend involves the charge/size ratio
or charge density of the metal cations; the carbonate anion
is constant. As you go down Group II, the charge of the
cation remains as 2+ but the size increases. At the top of
the group, the small cation has a high charge packed into a
small volume of space. The outcome is a high charge density
cation which will powerfully attract electrons in a nearby
carbonate ion. This attraction will weaken covalent bonds
in the anion. This causes the anion to become distorted,
promoted by what is called the polarizing effect of the
cation; the anion is said to be polarized.
At the bottom of Group II, the larger M2+ ion has a high
charge packed into a large volume of space. The outcome is
a low charge density cation which will cause less distortion
to nearby anions. Figure 14.4 shows what happens when a
M2+ ion is placed next to the carbonate ion.
This section focuses on the uses of some of the compounds
of magnesium and calcium, specifically magnesium oxide,
calcium oxide, calcium hydroxide and calcium carbonate.
We will now discuss the uses of each compound in turn.
Magnesium oxide
Magnesium oxide, otherwise known as magnesia, is found
naturally in white powder form in metamorphic rocks. The
useful properties of this material make it an ideal tool for a
wide range of activities and, as such, it is found in an array
of household and industrial items.
Medical
Magnesium oxide in its hydrated form is magnesium
hydroxide, which is a base.
■ It is commonly used as an antacid which neutralizes
As we see from Figure 14.4, if the carbonate is heated, the
carbon dioxide breaks free to leave the metal oxide. The
quantity of heat required to effect this dissociation depends
on the extent to which the anion was polarized. If it was
highly polarized, less heat is required than if it was only
slightly polarized. As you go down the group, the cationic
size increases and the charge density decreases. This results
in a decreasing polarizing effect of the cation and hence a
decreased distorting effect on the carbonate anion. With
decreasing distortion to the anion, more heat has to be
supplied to the compound in order to persuade the carbon
dioxide to break free and leave the metal oxide. Thus,
as the group is descended, the carbonates become more
thermally stable.
Magnesium oxide is also used as a dietary supplement for
animals.
Nitrates
Drying agent
The nitrates of Group II undergo thermal decomposition to
produce the metal oxide and brown nitrogen dioxide gas
together with oxygen:
Magnesium oxide in its powder form is hygroscopic in
nature, thereby allowing it to absorb water molecules from
surrounding objects, and keep them dry.
M(NO3)2(s) → 2MO(s) + 4NO2(g) + O2(g)
excess acid in the stomach, and treats indigestion.
■ It has short-term laxative effects and is used for
temporary relief from constipation. Magnesium
hydroxide is the active ingredient in common overthe-counter drugs such as milk of magnesia, Mylanta
and Maalox.
■ It is also used as medication to relieve heartburn and
sour stomach.
■ It serves as a dietary supplement in the human body
as it is important to maintain the systems between the
muscles and the nerves.
139
140
Unit 1 Module 3 Chemistry of the elements
■ Libraries and paper storage facilities often use
magnesium oxide to help preserve paper as it reacts
with ambient moisture to dry the book storage areas.
■ Rock climbers also use the compound as a way to
reduce moisture build-up from perspiration on their
hands and handheld equipment.
Refractory and insulation
A refractory material is one that is physically and chemically
stable at high temperatures.
■ Magnesium oxide is used as a refractory material
in the making of crucibles – containers made with
the intention of being placed in extremely high
temperatures in order to heat the contents.
■ Owing to its heat-resistant properties, magnesium
oxide makes an excellent insulator in industrial cables.
It is used for the protection of critical electrical circuits,
for example in fire protection devices such as alarms
and smoke control systems.
■ Due to its average thermal conductivity and high
dielectric strength, it is used extensively in heating as a
component of ‘CalRod’-styled heating elements.
Cement and construction
■ Magnesium oxide is one of the raw materials for
making cement in dry process plants; specifically
Portland cement.
■ As a construction material, magnesium oxide
wallboards have attractive characteristics such as
strength and resistance to fire, moisture, mould and
mildew.
■ It is used as a principal fireproofing ingredient in
construction materials.
Colorimetry
As a result of its remarkable diffusing and reflectivity
properties, magnesium oxide is used as a reference white
colour in colorimetry.
Calcium carbonate (limestone)
Calcium carbonate is found naturally in minerals and
rocks, and is the major constituent of the shells of marine
organisms, snails and pearls. When rocks and minerals
dissolve, calcium carbonate is added to natural water
sources, resulting in hard water; calcium carbonate is usually
the primary cause of hard water. Calcium carbonate, also
known as limestone, has many uses in industry, agriculture
and medicine.
Industry
The primary use of calcium carbonate is in the construction
industry. Some of the most popular construction materials,
such as marble and limestone, are originally formed from
calcium carbonate. The main constituent of limestone is
calcite, which is one of the most abundant minerals. Calcite
can form marble when exposed to suitable conditions of
heat and pressure.
Here is a list showing the wide range of uses of calcium
carbonate:
■ making mortar, which is used in bonding bricks,
concrete blocks, stones and tiles;
■ road building, or as filler in cement and paints;
■ in adhesives, stained glass windows and putty;
■ in the purification of iron from iron ore in a blast
furnace;
■ in the oil industry, added to drilling fluids as a
formation-bridging and filter-cake sealing agent;
■ sometimes used as blackboard chalk, although this is
often made of calcium sulfate;
■ as the active ingredient in agricultural lime,
neutralizing soil which is too acidic to grow crops;
■ in swimming pools, to offset the acidic properties of
the disinfectant agent.
Health and dietary uses
Since calcium is essential for healthy bones and teeth,
calcium carbonate is used as a dietary calcium supplement.
■ It is effective in treating certain ailments related to
calcium deficiency, for example osteoporosis.
■ It is used as an inert filler for tablets and other
pharmaceuticals, in the production of toothpaste, in
grocery products such as baking powder, dry-mix
dessert mixes.
■ It is also used as a source of dietary calcium in some
soy milk products.
Environmental uses
Calcium carbonate is used in water treatment to reduce
acidity and as a flocculent.
It is also used to desulfurize waste gases and to neutralize
acidic effluents.
Calcium oxide (quicklime)
Calcium oxide, commonly known as quicklime, has many
properties that make it quite valuable. One of its oldest
uses is its ability to react with carbon dioxide to regenerate
calcium carbonate.
Chapter 14 Elements and periodicity: Group II
Calcium oxide reacts exothermically with water to form
calcium hydroxide and the reaction produces sufficient
heat to ignite combustible materials in some instances.
When calcium oxide is mixed with water and sand, the
result is lime mortar, which was used in construction to
secure bricks, blocks and stones together. Nowadays,
cement is more often used, except for repairs to older
buildings.
Perhaps one of the most important modern uses of calcium
oxide relies on its ability to form solutions with silicates.
Silicates are used in the production of iron and steel from
their ores, which are rocks that contain iron oxides. When
calcium oxide is mixed with the ore and the mixture melted,
these silicates combine with the calcium oxide forming a
solution called slag. Slag is immiscible with molten iron,
which allows the silicates to be removed from the iron by
draining off the slag.
Calcium oxide is also used in the production of other
metals. For instance, it is used to remove silicates from
alumina prior to the alumina being reduced to aluminium
metal.
Calcium hydroxide (slaked lime)
Calcium hydroxide, informally referred to as slaked lime,
is a compound formed from the reaction between calcium
oxide and water. The resultant substance is a colourless
crystal or white powder which is strongly alkaline. Owing
to such basic properties, the compound has many and
varied uses in food manufacturing, hair care products,
dental work and leather production. Calcium hydroxide
may be applied to some of the same uses as calcium oxide,
including steel manufacture, cement and mortar.
■ Slaked lime has been used for centuries to modify soils
and make them more productive. ‘Lime’ is effective in
raising the pH of soil, helps to break up heavy clays,
and maintains the stability of pH in soil throughout the
growing season.
■ Food industry: in addition to its basic properties,
calcium hydroxide also has a low toxicity and is
therefore widely used in the food industry. For
example, it is used for processing water for alcoholic
beverages and soft drinks, to fortify fruit drinks and
baby formula, home food preservation in the making
of pickles and as an alternative to baking soda.
ITQ 4 Suggest a reason for adding slaked lime to agricultural
land.
■ Hair care products: calcium hydroxide is the active
alkaline ingredient in some hair relaxer products
which are designed to straighten curly hair; many of
these products contain lye, a caustic soda.
■ Dental uses: calcium hydroxide is commonly used in
dental work. For example, it can aid in disinfecting
teeth and can be used as a temporary treatment for
pain relief and swelling in preparation for ‘root canal’
surgery.
■ Leather production: calcium hydroxide makes an
effective solution for separating hair from animal hides
in preparation for the production of leather.
141
142
Unit 1 Module 3 Chemistry of the elements
Summary
✓ Periodic trends down Group II (Be to Ba) are
similar to those shown across period 3.
✓ Properties such as reaction with water, solubility
and thermal stability of compounds also show
trends down Group II.
✓ Some elements (e.g. Mg, Ca) are widely used in
industry, the environment and the home.
Review questions
1
2
3
The Group II elements, specifically beryllium to
barium, and their compounds show distinct trends/
patterns in properties and behaviour.
(a) Explain the trend in the first ionization energy
with atomic radii for the Group II elements.
(b) For a named element in Group II, write an
equation for:
(i) the first ionization energy;
(ii) the reaction with water.
(a) Describe the reaction of calcium with cold water
and write an equation for the reaction that takes
place.
(b) Radium (Ra) is a member of the Group II elements
and is located at the bottom of the group. Predict:
(i) its reaction with cold water;
(ii) the ease of reaction between radium and
oxygen;
(iii) the thermal solubility of RaCO3 relative to the
other Group II carbonates;
(iv) the effect of heat on radium nitrate.
(c) Write an equation for the reaction in part (b)(iv).
(d) (i) Write the formula of radium hydroxide.
(ii) Comment on the solubility of radium
hydroxide in water and the pH of any solution
formed.
Two calcium salts, A and B, were heated and the
following observations were made:
Compound A decomposed to produce a gas which
formed a white precipitate on being bubbled into an
aqueous solution of calcium hydroxide.
Compound B decomposed to produce two gases. One
gas rekindled a glowing splint and the other gas was
brown in colour.
(a) Identify the gas evolved on heating Compound A.
(b) Identify the two gases evolved in heating
Compound B.
(c) Deduce the molecular formulae of the two calcium
salts A and B.
4
The following observations are made about elements
in a specific group in the periodic table.
■ Observation 1: A crystalline metal nitrate melts
on gentle heating and decomposes at ~200 °C to
produce the metal oxide as a solid residue and a
gaseous mixture that is reddish brown in colour.
■ Observation 2: A second metal nitrate, which is
produced from a metal within the same group, is
anhydrous and requires temperatures over 800 °C
to produce the metal oxide residue and the same
gaseous mixture.
[All the nitrates of the elements in this group are
decomposed to produce the oxide as the solid
residue.]
(a) Suggest an explanation for the two observations.
(b) Write a balanced equation for the reaction stated
in either Observation 1 or Observation 2.
5
Calcium compounds are often used as structural
materials by organisms, and in the construction
industry.
(a) (i) The shells of shellfish contain a calcium
compound. Name this compound.
(ii) State a property of the above named
compound which makes it suitable for its
mentioned role.
(b) State and explain the trend:
(i) of the variation in the thermal decomposition
of the carbonates of Group II elements;
(ii) in the solubility of the sulfates of the elements
in Group II of the periodic table.
Chapter 14 Elements and periodicity: Group II
(c) Based on the explanation provided in part (b)(ii),
design an experiment that can be executed by a
chemist to identify four bottles of Group II metal
sulfates. The chemist does not have access to flame
test rods.
(d) In the 19th century, wooden boats were used
to transport quicklime. The crew often had to
extinguish fires that occurred during the journey.
Suggest an explanation for the occurrence of these
fires and write a balanced equation to illustrate
this explanation.
Answers to ITQs
1
Be2+(aq) + 2e− ҡ Be(s)
2+
−
Ba (aq) + 2e ҡ Ba(s)
E
= −1.85 V
E
= −2.91 V
The greater the negative value of E , the greater
is the reactivity. Barium has the more negative E
value. Hence, barium reacts more vigorously with
water than beryllium.
2
Group II compounds have double the charge of Group
I compounds and are slightly smaller. Both factors
produce higher forces between anion and cation and,
therefore, higher lattice energies.
3
The low solubility of barium sulfate protects the
patient from absorbing harmful amounts of the metal.
4
Some plants grow better in soil having a higher pH.
Slaked lime is a base and will raise soil pH, but it is
not so reactive as quicklime and so less likely to harm
plants.
Answers to Review questions
2
(a) Ca(s) + 2H2O(l) → Ca(OH)2(s) + H2(g)
(c) 2Ra(s) + O2(g) → 2RaO(s)
(d) (i) Ra(OH)2
3
(a) carbon dioxide
(b) oxygen and nitrogen dioxide
(c) Compound A: CaCO3
Compound B: Ca(NO3)2
4
(b) M(NO3)2(s) → 2MO(s) + 4NO2(g) + O2(g)
5
(a) (i) Calcium carbonate
(d) CaO(s) + H2O(l) → Ca(OH)2(aq) + heat
143
144
Chapter 15
Elements and periodicity: Group IV
Learning objectives
■ Explain the variations in properties of the Group IV elements in terms of structure and bonding.
■ Describe the bonding of the Group IV tetrachlorides.
■ Explain the reactions of the Group IV tetrachlorides with water.
■ Discuss the trends in bonding, acid/base character and thermal stability of the Group IV oxides of
oxidation states +2 and +4.
■ Discuss the relative stabilities of the oxides and aqueous cations of the Group IV elements in their
higher and lower oxidation states.
■ Discuss the uses of ceramics based on silicon(IV) oxide.
Introducing Group IV
In this chapter we are following the CAPE Chemistry
Syllabus and calling this group of elements Group IV. You
may also see the group referred to as Group IVA and Group
4 (and even Group 14).
The similarities between elements in the same group that
was so obvious in Group II is much less apparent here in
Group IV. The Group IV elements are carbon (C), silicon
(Si), germanium (Ge), tin (Sn) and lead (Pb). Although all
have four electrons in their valence shell, these elements
show, perhaps, the widest variation in properties of any
group in the periodic table.
There is considerable change in the character of the
elements as the group is descended:
■ carbon is a non-metal;
■ silicon and germanium are metalloids;
■ tin and lead are typical metals.
However, this does fit with the general rule that metallicity
increases down a group or towards the left-hand side of the
periodic table.
Variation in physical properties
As with all groups, atomic radius increases on passing
down Group IV. This is expected since the screening effect
associated with an increasing number of inner shells filled
with electrons greatly outweighs the increase in nuclear
charge. As a result, the attraction of the nucleus for the
outermost electrons becomes weaker and the outcome is
an increase in atomic radius down the group (Table 15.1).
The increasing atomic radius down Group IV means that
the outer electrons are further away from the nucleus and
therefore less energy is required to remove an outer electron.
As a consequence, the ionization energy decreases down
the group. The ionization energy decreases considerably
from carbon to silicon. After silicon, the difference is
relatively small, as shown in Table 15.1. The reason for
this is as follows: after silicon, the ‘d’ and ‘f’ sub-levels are
being filled and these sub-levels do not screen the nucleus
as efficiently as the ‘s’ or ‘p’ sub-levels and thus there is a
larger increase in the effective nuclear charge. However,
this increase in effective nuclear charge is counterbalanced
by the increase in atomic radius down the group, which
results in little difference in ionization energy after silicon.
The physical properties of the Group IV elements vary
more from one element to the next than with Group
II. This variation is related to the change in structure of
the elements from giant molecular in carbon, silicon and
germanium to giant metallic in tin and lead.
Carbon exists in at least three different physical forms; a
phenomenon described as allotropy. The three allotropic
forms are diamond, graphite and the oddly named
buckminsterfullerene, all of which are macromolecular
(Figure 15.1).
Chapter 15 Elements and periodicity: Group IV
Table 15.1 Some physical properties of the Group IV elements
Property
C
Atomic number
6
Electronic configuration
[He]
Si
Ge
14
2s2
2p
2
Sn
32
2
2
50
2
[Ne] 3s 3p
Pb
2
[Ar] 4s 4p
82
2
2
[Xe] 6s2 6p2
[Kr] 5s 5p
Atomic radius / nm
0.077
0.117
0.122
0.141
0.154
Electronegativity
2.5
1.8
1.8
1.8
1.8
First ionization energy / kJ mol−1
1090
786
762
707
716
Structure
giant molecular
giant molecular
giant molecular
giant metallic
giant metallic
Melting point / °C
diamond: 3730 (sublimes)
1410
937
232
327
Boiling point / °C
diamond: 4830
2680
2830
2270
1730
Density / g cm−3
graphite: 2.26
diamond: 3.51
graphite: fairly good
diamond: non-conductor
2.33
5.32
7.3
11.44
semi-conductor
semi-conductor
conductor
conductor
1 × 10−6
2 × 10−6
8 × 10−6
5 × 10−6
Conductivity
Electrical conductivity / ohm−1 m−1
a
−
b
c
Figure 15.1 The structures
of the allotropes of carbon:
(a) diamond, (b) graphite and
(c) buckminsterfullerene.
Carbon can also be formed
into long cylinders called
carbon nanotubes.
■ In diamond, each carbon atom is covalently bonded to
four other carbon atoms in a tetrahedral arrangement.
■ Graphite has a layered, planar structure which consists
of macromolecular sheets in which each carbon atom
is covalently bonded to three others in a hexagonal
arrangement; the fourth electron of each carbon atom
forms a delocalized electron cloud.
■ Buckminsterfullerene was discovered in 1985
and consists of C60 molecules (sometimes called
‘buckyballs’ because their structure is like that of a
soccer ball).
Silicon and germanium crystallize in the same structure as
diamond whilst tin and lead have distorted close-packed
structures.
As the group is descended and atomic radius increases, the
atoms get larger and the inter-atomic bonding becomes
weaker. Consequently, the attraction of the nucleus for the
electrons in the covalent bond gets weaker, which results in
electrons becoming delocalized. The delocalized electrons are
attracted to the positively charged nucleus and hence bonding
changes from covalent to metallic. This change in bonding
brings about general decreases in melting and boiling points
as well as general increases in density and conductivity as the
group is descended (see data in Table 15.1).
The Group IV tetrachlorides
A Group IV atom may achieve a noble gas configuration by
sharing four electrons, i.e. by forming four covalent bonds.
A Group IV atom contains only two unpaired electrons,
but an electron can be ‘promoted’ from the 2s sub-level
into the empty 2p sub-level. As a result of this promotion,
the atom is said to have moved from its ground state to
its excited state. This promotion is illustrated in Figure
15.2 using the carbon atom. The 2s and all three of the 2p
orbitals are ‘mixed’ to produce four equivalent sp3 hybrid
atomic orbitals.
C
ground
state
He
2s
2p
2s
2p
He
sp 3 hybrid orbitals
Figure 15.2 Promotion of the 2s electron.
C
excited
state
145
146
Unit 1 Module 3 Chemistry of the elements
All the elements of Group IV react with chlorine to form
tetrachlorides with the formula MCl4 (where M represents
the Group IV atom). They are all simple covalent molecules
with a tetrahedral shape, as shown in Figure 15.3. All the
tetrachlorides have low melting and boiling points and are
non-polar, volatile liquids at room temperature. As the
group is descended, the M–Cl bond becomes longer and
weaker, and the tetrachlorides get less stable.
Cl
Cl
Cl
C
Cl
Group IV dioxides
Physical properties
Table 15.2 highlights some of the physical properties of the
dioxides of the Group IV elements.
Table 15.2 Properties of the dioxides of Group IV elements
CO2
Structure
simple
molecular
Boiling point −78
/ °C
SiO2
GeO2
giant
molecular
2590
intermediate between giant molecular
and ionic
1200
1900
decomposes
on heating
SnO2
PbO2
Thermal stability
PbO2 decomposes to PbO on warming:
Figure 15.3 Tetrahedral shape of carbon tetrachloride.
PbO2(s) → PbO(s) + ½O2(g)
All of the tetrachlorides (except CCl4) are readily hydrolysed
to form the hydrated metal(IV) dioxide and fumes of HCl:
All the other dioxides are stable, even at high temperatures.
MCl4(l) + 4H2O(l) → M(OH)4(s) + 4HCl(g)
Acid/base nature
M(OH)4(l) → MO2.2H2O(s)
For a water molecule to react with a molecule of MCl4,
the oxygen atom on the water molecule must first attach
itself to the Group IV atom via the oxygen’s lone pair. This
lone pair is donated to the d orbitals on the Group IV atom
to form a dative covalent bond. As the M–O bond forms,
the M–Cl bond weakens and breaks. Bonds are formed
and broken one at a time until all four chlorine atoms are
displaced.
Carbon tetrachloride is immiscible in water and does not
undergo hydrolysis. The bonding electrons used by carbon
are from the 2s and 2p sub-levels. The carbon atom has no
available orbitals to accommodate lone pairs of electrons
from oxygen atoms on water molecules; the 3s orbitals
are energetically too far away and 2d orbitals do not exist.
Hence, carbon tetrachloride does not react with water.
All other Group IV elements have available d orbitals which
are energetically close to the occupied p orbitals. These
d orbitals accept lone pairs of electrons from the oxygen
atoms in water molecules to form dative covalent bonds.
Thus, hydrolysis of the tetrachloride occurs.
The resistance of CCl4 to hydrolysis results from the inability
of the carbon atom to act as an electron pair acceptor
The Group IV oxides
All the oxides of Group IV are solid, except for those of
carbon which are gaseous at room temperature and
pressure. This difference in physical property reflects a
difference in structure and bonding between the oxides of
carbon and those of the rest of the group.
CO2 and SiO2 are acidic. They react with alkalis to form
salts:
CO2 + 2NaOH → Na2CO3 + H2O
sodium carbonate
SiO2 + 2NaOH(hot, conc.) → Na2SiO3 + H2O
sodium silicate
GeO2, SnO2 and PbO2 are amphoteric, which means that
they show both acidic and basic properties.
Their acidic properties are shown by the reactions with
alkalis:
GeO2 + 2OH− + 2H2O → [Ge(OH)6]2−
germanate
SnO2 + 2OH−(conc.) + 2H2O → [Sn(OH)6]2−
stannate
PbO2 + 2NaOH(molten) → H2O + Na2PbO3
plumbate
Their basic properties are shown in the reaction with
concentrated HCl to form +4 salts. The use of concentrated
acid suppresses the hydrolysis of the chloride produced.
MO2 + 4HCl → MCl4 + 2H2O, where M = Ge, Sn, Pb
■ In the case of SnO2, the SnCl4 dissolves in excess HCl
to form the complex [SnCl6]2−;
■ in the case of PbO2, the reaction has to be done at
temperatures below 0 °C because the PbCl4 formed is
unstable and decomposes to give PbCl2 and Cl2 gas.
ITQ 1 Explain why the tetrachlorides of Group IV elements are
non-polar.
Chapter 15 Elements and periodicity: Group IV
Group IV monoxides
Structure
CO and SiO are simple molecular.
nuclear attraction towards them is greater since the d and f
orbitals do not screen the nucleus as effectively as the s and
p orbitals. Consequently, the ionization energy required for
their removal is quite large.
GeO, SnO and PbO are predominantly ionic.
Acid/base nature
Stability of the +2 and +4 oxidation states
CO and SiO are neutral oxides; they react with neither
acids nor alkalis.
The oxidation states shown by elements in Group IV are
+2 and +4.
GeO, SnO and PbO are amphoteric.
The typical oxidation state is +4. A piece of evidence in
support of this relates to the fact that if the elements are
heated in oxygen, they all (with the exception of lead),
form oxides with oxidation state +4. However, as the group
is descended, and the bonding changes from covalent to
ionic, the +2 oxidation state becomes predominant.
Their acidic character is illustrated by their reaction with
alkalis to form salts:
MO + 2OH− → MO22− + H2O, where M = Ge, Sn, Pb
Their basic character is shown by the reaction with
concentrated HCl to form +2 salts:
MO + 2HCl → MCl2 + H2O, where M = Ge, Sn, Pb
In the case of PbO, insoluble PbCl2 is formed. This dissolves
in excess HCl to form the soluble complex [PbCl4]2−.
Thermal stability
PbO is stable. The others are readily oxidized to the dioxide.
Bonding and the ‘inert pair’ effect
As we already know and saw in Table 15.1, all the elements
in Group IV have four electrons in their valence shell,
conforming to an electronic configuration of ns2 np2.
Therefore, it should not be surprising that they show a
well-defined oxidation state of +4. However, none of
the elements form the M4+ cation in its solid compounds
due to the high ionization energies required to remove
four successive electrons from the atom. Instead, it is
more energetically feasible to share electrons via covalent
bonding.
Ionic bonding is present amongst the Group IV elements.
Going down the group, germanium, tin and lead tend
to form ionic compounds, e.g. GeO, SnO, PbO, PbF2 and
PbCl2, in which the Group IV element has an oxidation
state of +2. In such ionic compounds, the Ge2+, Sn2+ and
Pb2+ ions are formed by the loss of the two electrons in the
p sub-shell. The two electrons in the s sub-shell remain
relatively inert. This phenomenon is referred to as the
‘inert pair’ effect which states that on passing down
Group IV, there is an increasing tendency for the pair of
electrons in the valence s sub-level to remain inert, i.e. not
to take part in bonding, resulting in greater stability of the
divalent compounds. The inertness and relative stability
of the ns2 electrons arises out of the fact that the effective
Let us now look at the relative stabilities of the +2 and +4
oxidation states for each element in turn.
Carbon
The +4 state is more stable for carbon than the +2 state.
The only common example of carbon in a +2 state occurs
in carbon monoxide. CO is a strong reducing agent because
it is easily oxidized to CO2 where the oxidation state is the
more thermodynamically stable +4.
Silicon
With silicon, again the +4 state is more stable than the +2
state. SiO, which is silicon in a +2 oxidation state, does not
exist under normal conditions.
Germanium
Germanium tends to form ionic compounds and forms
oxides in both +2 and +4 states. However, GeO2 (with Ge
in the +4 state) is rather more stable than GeO (with Ge in
the +2 state); GeO is readily converted to GeO2.
Tin
In tin compounds, the +4 state is only slightly more stable
than the +2 state. However, the +2 state is increasingly
common, with a variety of both Sn2+ and Sn4+ compounds.
The closeness in stability of tin(II) and tin(IV) means that
it will be fairly easy to convert tin(II) compounds into
tin(IV) compounds. Evidence of this is provided by the fact
that aqueous Sn2+ ions function as reducing agents. For
example, Sn2+ will reduce:
■ iodine to iodide ions;
■ iron(III) ions to iron(II) ions;
■ mercury(II) ions to mercury.
147
148
Unit 1 Module 3 Chemistry of the elements
Sn2+ ions are also easily oxidized by powerful oxidizing
agents such as acidified potassium manganate(VII)
(MnO4−).
Lead
The +2 state in lead is undoubtedly more stable than the
+4 state. PbO is relatively stable whilst PbO2 is a strong
oxidizing agent. PbO2 can oxidize
Silicon
Silicon is second only to oxygen as the most abundant
element on the Earth. It exists as silicon(IV) oxide, formula
SiO2, also known as silica. It is most commonly found in
nature as sand, sandstone and quartz. It is estimated that
about 90% of the Earth’s crust is made up of silica.
O–
■ hydrochloric acid to chlorine;
■ hydrogen sulfide to sulfur.
Compounds of lead(IV) are easily reduced to lead(II).
Evidence of the increased stability of the +2 oxidation
state relative to the +4 state is provided by the following
examples:
■ lead(IV) chloride decomposes at room temperature to
give lead(II) chloride and chlorine gas;
■ lead(IV) oxide decomposes on heating to give lead(II)
oxide and oxygen.
Oxidation state summary
There is a steady increase in stability of the +2 oxidation
state on descending Group IV. This is due to the ‘inert
pair’ effect. The following conclusions can also be made
regarding the oxidation states in Group IV:
O
Si
–
O–
O–
Figure 15.4 Representing the silicon tetrahedron.
Silicates are compounds that contain a silicon-bearing
anion but we will focus on the basic chemical unit of
silicates, SiO44−, the silicate ion. In the silicate ion, the Si
atom shows tetrahedral coordination, with four oxygen
atoms surrounding a central Si atom (Figure 15.4). On
average, all four oxygen atoms of the SiO4 tetrahedra are
shared with others, thereby forming either chains, sheets or
ring structures (Figure 15.5). The strong covalent bonding
between atoms goes on and on in three dimensions to give
rise to a giant three-dimensional structure.
■ +4 is the most stable state for C, Si, Ge and Sn, but is
the least stable for Pb;
■ +2 is most unstable for C and Si, but it is most stable
for Pb.
The elements in their higher +4 oxidation state can only
form covalent bonds whereas in the lower +2 oxidation
state they are expected to form ionic bonds. Hence, the
bonding changes from covalent to ionic as the group is
descended.
The marked increased stability of the +2 state relative to
the +4 state as Group IV is descended is well illustrated by
the standard electrode potentials of the M4+(aq)/M2+(aq)
systems for germanium, tin and lead (Table 15.3). As the
standard electrode potentials get more positive from Ge4+
to Pb4+, the oxidized form is more readily reduced to the
+2 oxidation state.
Table 15.3 E
values for some Group IV elements
Reaction
E
Ge4+(aq) + 2e− ҡ Ge2+(s)
−1.60
/V
Sn4+(aq) + 2e− ҡ Sn2+(s)
+0.15
Pb4+(aq) + 2e− ҡ Pb2+(s)
+1.80
Figure 15.5 The SiO2 tetrahedron is the basis for a wide variety
of silicate structures. These are formed by sharing oxygen atoms.
The structure is made electrically neutral by the inclusion of a
wide variety of metal cations.
Chapter 15 Elements and periodicity: Group IV
Silica is used primarily in the production of glass and is a
widely used ceramic material, both as a precursor to the
fabrication of other ceramic products and as a material
on its own. Ceramics are heat-resistant, non-metallic,
inorganic solids that are made up of compounds formed
from metallic and non-metallic elements. Ceramics tend to
also be corrosion-resistant and hard, but brittle, and serve
as good insulators as they can withstand high temperatures.
It is these properties that have led to their use in virtually
every aspect of life.
Ceramics fall into two main categories, namely traditional
and advanced. Traditional ceramics include materials which
are made from clay and cement and have been hardened
by heating at high temperatures. Traditional ceramics are
used in wall tiles, flowerpots, dishes, crockery and roof
tiles. Advanced ceramics are geared towards crystalline
structures with superior properties to the traditional
ceramics that were produced years ago. Such ceramics
include:
■ carbides, such as silicon carbide, SiC;
■ nitrides, such as silicon nitride, Si3N4;
■ oxides, such as aluminium oxide, Al2O3;
■ mixed oxide ceramics that can act as superconductors.
Advanced ceramics are designed to have properties such
as hardness, strength and the ability to withstand high
temperatures for applications such as heat shields in
spacecraft and armour in military vehicles. Designing such
properties requires modern processing techniques and the
development of these techniques has led to major advances
in engineering and medicine.
Summary
✓ Periodic trends in Group IV are similar to those
shown across period 3.
✓ In this group, effects determined by variations in
electron structures such as hybridization become
important.
✓ The elements in the group show a clear variation
from non-metal through metalloid to true metal
as the group is descended.
✓ This is evidenced through the nature of the
chlorides, the acid/base nature of the oxides, and
the increasing importance of the divalent ion as
the group is descended.
✓ Silicon is of particular importance as the element
forms the backbone of many terrestrial silicate
structures.
149
150
Unit 1 Module 3 Chemistry of the elements
Review questions
1
Complete Table 15.4 by identifying the products of
the reactions of the following Group IV metal oxides
and hence describe the nature of the oxides. Write
equations for any reaction occurring.
(iv) Write an equation for the reaction of the oxide
of A with alkali.
(v) Comment on the difference in the thermal
stability of the oxides of A and B.
4
Table 15.4 Reactions of some Group IV metal oxides
Compound
Reaction with
HNO3(aq)
Reaction with
NaOH(aq)
Nature of oxide
CO
SiO2
no reaction
PbO
2
Table 15.5 shows the melting points for the Group IV
elements carbon to lead.
Table 15.7 Group IV elements in the +2 oxidation state
CO
Table 15.5
Element
C
Si
Ge
Sn
Pb
Melting point / °C
3730
1410
937
232
327
(a) With reference to structure and bonding, account
for the variation in the melting points of the
elements.
(b) The Group IV tetrachlorides are non-polar, volatile
liquids.
(i) a) Draw the molecular shape of SiCl4.
b) Use your diagram to explain why the
tetrachlorides are non-polar.
(ii) State the trend in volatility of the
tetrachlorides and give one reason for the
variation.
(c) Elements A and B are members of Group IV. A
has a density of 2.33 g cm−3 whilst B has a density
of 11.44 g cm−3. There is a general gradation in
the properties of the Group IV elements from
non-metallic to metallic.
(i) What does the difference in density indicate
about the nature of A and B? Give a reason
for your answer.
The oxides of A and B of oxidation state +4
exhibit the properties given in Table 15.6.
Table 15.6
Boiling point / °C Nature
Oxide of A 2590
Oxide of B
Group IV elements exhibit +2 and +4 oxidation states
in many compounds.
(a) Explain the relative stability of the oxides of the
Group IV elements of oxidation states +2 and +4.
(b) Tables 15.7 and 15.8 provide information on some
properties of the oxides of Group IV elements.
Use this information to answer the questions that
follow.
decomposes on
warming
acidic
Thermal stability
stable even at high
temperatures
amphoteric decomposes to BO
(ii) Suggest identities for A and B.
(iii) How would you expect the +4 oxides of A and
B to respond to treatment with:
a) acid;
b) alkali.
SiO
Acid/base
neutral
neutral
nature
simple
Structure
unknown
molecular
GeO
SnO
amphoteric
amphoteric
intermediate between giant molecular
and ionic
Table 15.8 Group IV elements in the +4 oxidation state
CO2
SiO2
GeO2
Acid/base
acidic
acidic
amphoteric
nature
simple
giant
Structure
predominantly ionic
molecular molecular
SnO2
amphoteric
(i) State the trend observed in the nature of the
oxides formed from the elements in the +4
oxidation state.
(ii) Suggest the nature of PbO2 and give a reason
for your answer.
(iii) Suggest a possible structure for SiO.
(c) In an attempt to illustrate the acidic nature of
silicon dioxide, write a balanced equation for the
reaction between SiO2 and aqueous alkali.
5
(a) The properties of the Group IV elements vary from
non-metallic to metallic as the group is descended,
but the tetrachlorides are all covalent compounds.
(i) Describe the bonding in a named Group
IV tetrachloride and explain why all of the
tetrachlorides are covalent compounds.
(ii) Silicon tetrachloride is stable at room
temperature with respect to dissociation into
the constituent elements. However, lead
tetrachloride decomposes into lead(II) chloride
and chlorine. Account for this difference in
stability.
(b) SiCl4 can be converted to an intermediate
compound, SiCl3OH, on reaction with water.
(i) Write an equation for the formation of this
intermediate compound SiCl3OH.
Chapter 15 Elements and periodicity: Group IV
(ii) The electronegativity value of silicon is 1.8
whilst that of chlorine is 3.0. Explain how this
difference promotes the reaction of SiCl4 with
the water molecule.
(iii) The reaction between SiCl4 and water can
yield the final product as hydrated silicon(IV)
oxide. Name the process that occurs to
produce hydrated silicon(IV) oxide.
(c) Owing to their chemical durability, ceramics
are used in many applications. Some properties
of ceramics include hardness, heat resistance,
corrosion resistance and super conductivity at high
temperatures.
(i) Explain how the structure and bonding of
silicon(IV) oxide makes it suitable as a base for
ceramics with the above properties.
(ii) Comment on the heat resistance and
corrosion resistance of a ceramic based on
germanium(IV) oxide, relative to one based on
silicon(IV) oxide.
6
Table 15.9 shows the variation in some properties of
the Group IV elements.
Table 15.9
Element
C
Melting point / °C
3730 1410
Electrical conductivity /
ohm−1 m−1
Melting point of dioxide
/ °C
Si
Ge
Sn
Pb
937
232
327
—
1×
10−6
2 × 10−6
8 × 10−6
−56
1610
1115
5 × 10−6
1630
290
(a) Describe the trend in electrical conductivity from
silicon to tin and suggest a reason for this trend.
(b) Account for the variation in the melting points
from C to Sn in terms of structure and bonding.
(c) With reference to the melting point of the +4
oxides, suggest the type of structure and bonding
exhibited by these oxides.
(d) Suggest an explanation for the relatively low
melting point of PbO2 compared to the oxides of Si
to Sn.
7
(a) Describe the trend in electrical conductivities
of the Group IV elements and relate it to their
physical structure.
(b) State the type of bonding found in the following
Group IV dioxides: CO2, SiO2, GeO2, PbO2.
(c) State the acid/base nature of the Group IV
dioxides and explain how it relates to the type of
bonding.
(d) Use the following standard electrode potentials to
comment on the relative stability of the +4 and +2
oxidation states of Ge, Sn and Pb.
E = −1.6 V
Ge4+ + 2e− ҡ Ge2+
4+
−
2+
E = +0.15 V
Sn + 2e ҡ Sn
E = +1.8 V
Pb4+ + 2e− ҡ Pb2+
(e) Sn2+ ions will reduce orange Cr2O72− to green Cr3+,
but Pb2+ ions will not. Using standard electrode
potentials, discuss this statement, using suitable
half-equations to illustrate your answer.
(f) SiCl4 fumes and forms a white precipitate with
water whilst CCl4 is immiscible with water.
Explain these observations and write balanced
equations for the reactions occurring.
Answers to ITQs
1
Chlorides such as CCl4 form tetrahedral molecules.
The overall effect of each polarized C–Cl bond is
cancelled out by the combined effect of the other
three C–Cl bonds.
151
152
Chapter 16
Elements and periodicity: Group VII
Learning objectives
■ Explain the variations in properties of the Group VII elements in terms of structure and bonding.
■ Describe the reactions of the Group VII elements with hydrogen, water and dilute acids.
■ Explain the relative reactivities of the Group VII elements as oxidizing agents.
■ Explain the relative stabilities of the hydrides of the Group VII elements.
■ Describe the reactions of chlorine and the halide ions.
Introducing the Group VII elements
Variation in physical properties
In this chapter we are following the CAPE Chemistry
Syllabus and calling this group of elements Group VII. You
may also see the group referred to as Group VIIA, Group 7
and Group 17.
As Group VII is descended, the increasing number of inner
shells filled with electrons greatly outweighs the increase
in nuclear charge. Consequently, the attraction of the
nucleus for the outermost electrons becomes weaker and
the outcome is an increase in atomic radius down the
group (Table 16.2).
This chapter focuses on the trends in physical properties
as well as some chemical reactions of the elements of
Group VII, known collectively as the halogens – fluorine,
chlorine, bromine, iodine and astatine. All the isotopes of
astatine are highly unstable and intensely radioactive with
short half-lives, and therefore will not be considered during
discussions. It has been estimated that there is less than
30 g of astatine on Earth at any one time!
The elements of Group VII are all non-metals. They all
have an outer electron shell containing seven electrons,
which correspond to an ‘ns2 np5’ electronic configuration
(Table 16.1). This similar electron structure is responsible
for the similarities in their reactions. However, as a group,
halogens exhibit highly variable physical and chemical
properties, as well as showing distinct trends in behaviour
down the group.
Table 16.1 Electronic configuration of the Group VII elements
Element
Symbol
Electronic configuration
fluorine
F
[He] 2s2 2p5
chlorine
Cl
[Ne]
3s2
bromine
Br
[Ar] 4s2 4p5
iodine
I
[Kr] 5s2 5p5
astatine
At
[Xe] 6s2 6p5
3p5
The elements of Group VII all exist as diatomic molecules,
X2; the two atoms are linked by a covalent bond. The
intermolecular attractions between one molecule and its
neighbours are called van der Waals dispersion forces,
and it is these forces which allow us to explain certain
trends down Group VII. As the molecules get bigger,
there are more electrons which can move around and set
up the temporary dipoles which create these attractions.
Therefore, as the relative molecular mass increases down
the group, the attractive forces increase, which in turn
affects their physical properties.
■ Physical state: Group VII is the only group within the
periodic table that contains elements in all three states
of matter. At room temperature, the halogens range
from gaseous (F2 and Cl2) to liquid (Br2) to solid (I2).
■ Melting and boiling points: the stronger attractive
forces down the group mean that more energy is
required to break the bonds between the molecules.
Therefore the melting and boiling points increase as
the group is descended.
■ Volatility: owing to the increasing intermolecular
forces of attraction, the halogens become less volatile
going down the group.
Chapter 16 Elements and periodicity: Group VII
Table 16.2 Properties of the halogens
Fluorine
Chlorine
Bromine
Iodine
Atomic radius
/ nm
Molecular
formula
Model
0.072
0.099
0.114
0.133
F2
Cl2
Br2
I2
State at 20 °C
gas
gas
liquid
solid
Colour
pale yellow pale green
red-brown
dark purple/black
Melting point / °C −220
−101
−7
114
Boiling point / °C −188
−35
59
184
cm−3
1.11
1.56
3.12
4.93
Electronegativity
4.00
2.85
2.75
2.20
Density / g
may achieve a stable noble gas configuration in two ways,
as discussed below.
Ionic bonding
All the halogen atoms form ionic halides with electropositive
metals. The halogen atom, X, achieves a noble gas
configuration by gaining one electron (from the metal) to
form a halide ion, X−.
X + e− → X−
ns2 np5
ns2 np6
Polyhalide ions are also known; these contain one or
more halogen molecules attached to a halide ion. For
example, I3−, is formed on dissolving iodine in aqueous
iodide solution:
I2(aq) + I−(aq) ҡ I3−(aq)
■ Density: as the attractive forces between the atoms
increase, the atoms pack closer together, thereby
occupying a smaller volume. Thus, the mass per unit
volume, i.e. the density, increases.
The halogens also have some other characteristic features.
■ Colour: the depth of colour of the halogen molecules
at room temperature increases as the group is
descended. Fluorine is a pale yellow gas whilst chlorine
is a pale green gas. Bromine is a red-brown liquid.
Iodine is a shiny black solid which sublimes upon
heating to give a purple vapour.
■ Electronegativity: fluorine is the most electronegative
element in the periodic table; the other halogens
have some of the highest electronegativities of all the
elements. Electronegativity values decrease down the
group.
Table 16.2 summarizes the physical properties of the Group
VII elements.
Bonding types
All the atoms of Group VII have the outer electronic
configuration of ‘ns2 np5’, i.e. seven electrons in their
outermost shell. They have one electron less than the
noble gas which follows them in the periodic table and so
ITQ 1
(a) List the physical state and colour of the elements chlorine,
bromine and iodine at room temperature.
(b) With reference to their respective structures and bonding,
explain your answer in part (a).
(c) Explain the trend in volatility of the halogens as the group is
descended in terms of atomic size and intermolecular bonding.
Covalent bonding
All the elements in Group VII may achieve a noble gas
configuration by forming a single covalent bond. The
formation of hydrogen halides provides one such example
of the halogens participating in covalent bonding.
H
Cl
H
Cl
Figure 16.1 The formation of hydrogen chloride involving a
covalent bond.
Chemical properties and reactivity
All the halogens are one electron short of a stable octet
and, as such, their chemistry is dominated by a tendency
to gain a completely filled outermost electron shell,
thereby making them highly reactive species. This high
reactivity is also attributed to the high electronegativity
of the atoms on account of their high effective nuclear
charge. The reactivity of the halogen decreases markedly
down the group. Fluorine is, in fact, the most reactive of
all non-metals. Let us now look at some of the chemical
reactions involving the halogens.
The halogens as oxidizing agents
An oxidizing agent can be defined as a species that is
capable of accepting or gaining electrons in a chemical
reaction. All the halogens act as oxidizing agents when
they combine with metals or non-metals.
When the halogens combine with metals to form ionic
compounds, they gain electrons from the metal to form
negative halide ions. The elements they react with have
positive oxidation numbers in the resultant compounds.
153
154
Unit 1 Module 3 Chemistry of the elements
For example:
Oxidation reactions
overall reaction: 2Na(s) + Cl2(g) → 2Na+Cl−(s)
oxidation:
2Na
→ 2Na+ + 2e−
reduction:
Cl2 + 2e−
→ 2Cl−
Fluorine, chlorine and bromine will oxidize iron(II) ions to
iron(III) ions:
In this reaction, Cl2 accepts two electrons (from the sodium
metal) and acts as an oxidizing agent. The oxidation state
of Na in the resultant compound is +1.
X2(aq) + 2e− → 2X−(aq)
Since fluorine is the most reactive halogen, it is by extension
the most powerful oxidizing agent. The oxidizing power
decreases as we descend the group and therefore the order
of decreasing power as oxidizing agents is F2 > Cl2 > Br2 > I2.
Both electronegativity values (see Table 16.2) and standard
electrode potentials (Table 16.3) become less positive from
fluorine to iodine and this reflects the decreasing oxidizing
power.
Table 16.3 Standard electrode potentials of the halogens
E
Reaction
/V
F2(aq) + 2e− ҡ F−(aq)
+2.87
2e−
Cl−(aq)
+1.36
Br2(aq) + 2e− ҡ Br−(aq)
+1.09
Cl2(aq) +
−
ҡ
−
I2(aq) + 2e ҡ I (aq)
+0.54
The decreasing oxidizing power on passing down the
halogen group is further emphasized by other reactions.
Displacement reactions
Any halogen will oxidize a halide ion below it in Group VII.
This, however, does not apply to fluorine, which is such a
powerful oxidizing agent that it oxidizes water to oxygen
directly, making it impossible to perform aqueous reactions
with fluorine.
2F2(g) + 2H2O(l) → 4HF(aq) + O2(g)
2Fe2+(aq) → 2Fe3+(aq) + 2e−
However, iodine is too weak an oxidizing agent to carry out
this oxidation.
All the halogens except for iodine can oxidize thiosulfate
ions (S2O32−(aq)) to sulfate ions (SO42−(aq)); again, iodine
is too weak an oxidizing agent.
4X2(aq) + S2O32−(aq) + 5H2O(l) →
8X−(aq) + 2SO42−(aq) +10H+(aq)
Iodine, however, oxidizes thiosulfate ions to tetrathionate
(S4O62−(aq)) ions:
I2(aq) + S2O32−(aq) → 2I−(aq) + S4O62−(aq)
There are other reactions in which the halogens demonstrate
their oxidizing power.
■ All halogens can oxidize sulfite ions (SO32−(aq)) to
sulfate ions (SO42−(aq)).
■ All halogens oxidize hydrogen sulfide (H2S) to sulfur
(S).
■ Fluorine and chlorine can oxidize many coloured
dyes to colourless substances. Thus, indicators such as
litmus and Universal Indicator are decolorized when
exposed to fluorine and chlorine.
■ Chlorine acts as an oxidizing agent when it is used for
bleaching.
Table 16.4 summarizes these reactions.
Table 16.4 Oxidizing reactions of halogens
Chlorine can oxidize bromide ions to bromine as well as
iodide ions to iodine.
Oxidizing agent
Reaction
All halogens
When chlorine water (Cl2(aq)) is added to aqueous KBr,
yellow-orange bromine is formed:
F2, Cl2, Br2
SO32− → SO42−
H2S → S
Fe2+ → Fe3+
S2O32− → SO42−
Cl2(aq) + 2Br−(aq) → 2Cl−(aq) + Br2(aq)
I2
When Cl2(aq) is added to aqueous KI, the iodine produced
dissolves in the KI to give a red-brown solution:
Cl2(aq) + 2I−(aq) → 2Cl−(aq) + I2(aq)
Bromine can only oxidize iodide ions to iodine:
Br2(aq) + 2I−(aq) → 2Br−(aq) + I2(aq)
Iodine is the last halogen in the group, if we disregard
astatine, and is the weakest oxidizing agent. Iodine does
not oxidize any of the halide ions above it.
S2O32− → S4O62−
The halogens and aqueous sodium hydroxide
Chlorine, bromine and iodine undergo similar reactions
with aqueous sodium hydroxide. However, the products
obtained depend on the temperature at which the reaction
ITQ 2 Using Table 16.3, describe and explain the reactions
between:
(a) bromine and iodide ions;
(b) bromine and chloride ions.
Include balanced equations in your answers.
Chapter 16 Elements and periodicity: Group VII
is carried out. Chlorine will be used as the representative
element of the halogen group when writing balanced
equations during these discussions.
Table 16.5 Compounds of chlorine and their oxidation number
Oxidation number Compounds
+7
Cl2O7, NaClO4
Chlorine reacts with cold, dilute sodium hydroxide at about
15 °C to produce a mixture of sodium chloride and sodium
chlorate(I) (sodium hypochlorite), NaClO.
+6
ClO3
+5
NaClO3
+4
ClO2
Cl2(g) + 2NaOH(aq) → NaCl(aq) + NaClO(aq) + H2O(l)
+3
KClO2
+1
Cl2O, NaClO
The sodium chlorate(I) which is produced in the first
reaction then slowly decomposes to form sodium chloride
and sodium chlorate(V), NaClO3.
0
Cl2
−1
NaCl
This second reaction is, however, quite rapid at 70 °C.
Therefore, NaClO can be obtained by passing chlorine
into sodium hydroxide at 15 °C, whilst NaClO3 is obtained
by carrying out the same reaction at 70 °C. The overall
reaction occurring at 70 °C is as follows:
Fluorine never exhibits a positive oxidation number. Owing
to the fact that it is the most electronegative element,
fluorine can never form a compound in which it is the less
electronegative element. In this regard, fluorine reacts with
alkalis differently to the other halogens – it forms a mixture
of fluoride and oxygen difluoride. In both compounds, the
oxidation state of fluorine is −1.
3Cl2(g) + 6NaOH(aq) → 5NaCl(aq) + NaClO3(aq) + 3H2O(l)
2F2(g) + 2OH−(aq) → OF2(g) + 2F−(aq) + H2O(l)
With bromine, both reactions are rapid at 15 °C, but
decomposition of NaBrO is slow at 0 °C.
The halogens and hydrogen
3NaClO(aq) → 2NaCl(aq) + NaClO3(aq)
With iodine, decomposition of NaIO occurs rapidly even at
0 °C, so it is difficult to prepare NaIO free from NaIO3.
In these two reactions of halogens with sodium hydroxide,
the halogen molecule is simultaneously oxidized and
reduced; such reactions are called disproportionation
reactions. Let us look at the two reactions of chlorine that
we have discussed; the oxidation number of the respective
chlorine atom is shown in the equations.
The halogens, with the exception of fluorine, form
compounds in which they have positive oxidation numbers
up to +7. Chlorine, for instance, exhibits variable oxidation
states in its many compounds. Table 16.5 demonstrates the
range of oxidation states of chlorine. As expected, the most
stable oxidation state for halogens is −1.
All the halogens react directly with hydrogen to produce
hydrogen halides according to the general equation:
H2(g) + X2(g) → 2HX(g)
The reaction conditions as well as the speed at which the
reaction occurs vary for each halogen as follows:
■ with fluorine, the reaction is explosive even at low
temperatures;
■ with chlorine, the reaction is slow in the dark and
explosive in sunlight;
■ bromine combines with H2(g) at high temperatures in
the presence of a catalyst;
■ the reaction with iodine is slow and reversible, giving a
low yield.
Stability of the hydrides
reduction
Cl2(g) + 2NaOH(aq)
NaCl(aq) + NaClO(aq) + H2O(l)
0
–1
+1
oxidation
All the hydrogen halides are colourless gases at room
temperature and pressure with the exception of hydrogen
fluoride which boils just below room temperature. The
boiling points of the halides should generally increase with
increasing relative molecular mass, indicating an increase
in intermolecular attraction. However, hydrogen fluoride
reduction
3NaClO(aq)
+1
2NaCl(aq) + NaClO3(aq)
–1
+5
ITQ 3 Chlorine reacts with cold, dilute NaOH and with hot, conc.
NaOH. Explain these reactions, including in your answer:
(a) the type of reactions undergone;
oxidation
Figure 16.2 Disproportionation reactions involving chlorine.
(b) the temperature of the reactions;
(c) the ionic equations representing the reactions.
155
156
Unit 1 Module 3 Chemistry of the elements
has by far the highest boiling point due to the existence of
extremely strong intermolecular hydrogen bonding in the
liquid state.
The length of the H–X bond increases as we descend Group
VII. Longer bonds are weaker bonds and hence the strength
of the H–X bond decreases down the group (Table 16.6).
Table 16.6 How the H–X bond links with boiling point and Ka
Hydrogen
halide
Bond length
/ nm
Boiling point
/ °C
Bond enthalpy, Ka in aqueous
H–X / kJ mol−1 solution
10−4
HF
0.092
20
+567
7×
HCl
0.128
−85
+431
107
HBr
0.141
−69
+366
>107
HI
0.160
−35
+298
>107
The decreasing bond strength has an impact on:
■ the thermal stability of the hydrides;
■ the strength of the hydrohalic acids.
Table 16.7 Some reactions of aqueous halide ions
Reagent
Pb(NO3)2(aq)
AgNO3(aq)
F−(aq)
Cl−(aq)
Br−(aq)
I−(aq)
white ppt of
PbF2
no ppt; AgF is
soluble in water
white ppt of
PbCl2
white ppt of
AgCl
white ppt of
PbBr2
cream ppt of
AgBr
white ppt of
PbI2
yellow ppt
of AgI
soluble
insoluble
insoluble
soluble
soluble
insoluble
white AgCl
turns purplegrey
cream AgBr
turns greenyellow
no effect
Solubility of
silver halide in –
dil. NH3
Solubility of
silver halide in –
conc. NH3
Effect of
sunlight on the no effect
silver halide
ppt = precipitate
Solid halides react with concentrated sulfuric acid to first
form fumes of the hydrogen halide according to the general
equation:
X−(s) + H2SO4(l) → HX(g) + HSO4−(s), where X = Cl, Br, I
The thermal stability of the hydrides decreases as the group
is descended. This is especially noticeable with hydrogen
iodide, which readily dissociates into its elements on
heating (about 30% dissociation at 1000 °C):
Concentrated H2SO4 is also an oxidizing agent and is
sufficiently powerful to oxidize HBr to Br2 and HI to I2.
However, it is not powerful enough to oxidize HF and HCl.
HI(g) ҡ H2(g) + I2(g)
2HBr(g) + H2SO4(l) → Br2(g) + 2H2O(l) + SO2(g)
The hydrogen halides are highly soluble in water and form
strong acid solutions according to the equation:
2HI(g) + H2SO4(l) → I2(g) + 2H2O(l) + SO2(g)
HX(g) + H2O(l) →
H3O+(aq)
+
X−(aq)
The strength of the acid increases down the group, with
HI being the strongest acid; the acid dissociation constants,
Ka, given in Table 16.6 reveal this trend. This trend occurs
because the H–X bond strength decreases down the group
and so they dissociate more easily.
The polar covalent hydrogen halides are appreciably soluble
without dissociation in organic solvents such as benzene.
Reactions of halide ions
In view of the fact that the halides are so common, it is
important that we are able to identify the presence of each
ion. Most metal halides are soluble, except all lead halides,
AgCl, AgBr and AgI. Therefore, solutions of Pb2+(aq) and
Ag+(aq) ions can be used to test for the presence of halide
ions in solution since the halides are precipitated. For
example:
Pb (aq) + 2Cl (aq) → PbCl2(s)
2+
−
HCl can be oxidized to Cl2 when conc. H2SO4 is used in
conjunction with a stronger oxidizing agent such as MnO2.
HF is still not oxidized to F2.
4HCl(l) + MnO2(s) → Cl2(g) + MnCl2(aq) + 2H2O(l)
Since conc. H2SO4 oxidizes both HBr and HI, it is not
possible to use this reagent to selectively prepare these
hydrogen halides. Instead, concentrated phosphoric(V)
acid, H3PO4, is used since it is a relatively poor oxidizing
agent.
X−(s) + H3PO4(l) → HX(g) + H2PO4−(s), where X = Br, I
Table 16.8 highlights the products formed during the
reactions of the solid halides.
ITQ 4 Explain the trend in stability of the hydrides of the elements
of Group VII.
Ag+(aq) + Cl−(aq) → AgCl(s)
Some reactions of the aqueous halide ions are summarized
in Table 16.7. During the test with AgNO3(aq), dilute nitric
acid is added to prevent precipitation of other silver salts
(e.g carbonate).
ITQ 5
(a) Describe what you would see when an aqueous solution of
silver nitrate is added to a solution containing iodide ions
followed by aqueous ammonia.
(b) Write balanced equations for the reactions occurring in part (a).
Chapter 16 Elements and periodicity: Group VII
Table 16.8 Products formed during the reactions of solid halides
Reagent
F−(s)
Cl−(s)
Br−(s)
I−(s)
Summary
conc. H2SO4
HF(g)
HCl(g)
HBr(g) and
some Br2(g)
HI(g) and some I2(g)
✓ Periodic trends down Group VII (F to I) are
conc. H2SO4 + MnO2
HF(g)
Cl2(g)
Br2(g)
I2(g)
conc. H3PO4
HF(g)
HCl(g)
HBr(g)
HI(g)
Halogen chemistry is all around us.
■ Fluoride is in toothpaste because it is believed that it
helps people avoid dental cavities.
■ Fluorine compounds are found in non-stick coatings
on pans and in some aerosols propellants.
■ Chlorine is added to drinking water and swimming
pools as it useful in killing harmful bacteria, viruses
and fungi.
■ The most important chlorine compound, sodium
chloride, commonly known as ‘table salt’, was and still
is used to preserve food.
■ Bromide is used photographic film.
■ Iodine and its compounds are used in medicines,
photographic film and dyes.
ITQ 6
(a) List and explain the observations occurring when conc.
H2SO4 is added to solid potassium iodide.
(b) Describe, with the aid of equations, the reactions of conc.
H2SO4 with sodium bromide.
similar to those shown down Groups II and IV.
✓ In Group VII the clearest periodic trends are
those in physical properties and oxidizing
potential.
✓ All the elements react with sodium hydroxide by
disproportionation to form the (XO)– ion.
✓ The stability of the hydrides (HF to HI) gives a
good illustration of the effect of the hydrogen
bond.
157
158
Unit 1 Module 3 Chemistry of the elements
Review questions
1
3
All the halogens exist as diatomic molecules, which
persist in the gaseous, liquid and solid states. Table
16.9 lists some properties of the halogens. Use this
table to answer the questions that follow.
Br2(l) + 2OH−(aq) ҡ Br−(aq) + BrO−(aq) + H2O(l)
When excess Ag+ ions are added to solution G, a
suspension is formed which is filtered. On heating
the colourless filtrate, a cream-coloured precipitate
is formed. (NB silver(I) salts of bromate ions are
soluble in water.)
(i) Use the equation to explain the term
‘disproportionation’.
(ii) Explain what happened to solution G after excess
silver(I) ions are added to the filtrate from the
suspension and heated.
(b) When a sample, H, consisting of two compounds
was treated with conc. H2SO4 and warmed, a
purple gas was liberated. Aqueous AgNO3 was
added to an aqueous solution of H and a creamcoloured precipitate was formed. When dilute
ammonia was added to this precipitate, part of it
dissolved leaving a yellow precipitate. Deduce the
ions present in H, giving reasons for your answer.
Table 16.9
Atomic number
Fluorine
Chlorine
Bromine
Iodine
9
17
35
53
State at 20 °C
gas
gas
liquid
solid
Colour
pale yellow
pale green
red-brown
black
colourless
colourless
red-brown
violet
colourless
colourless
brown
brown
Colour in non-polar
solvents
Colour in polar
solvents
Melting point / °C
−220
−101
−7
113
Boiling point / °C
−188
−35
59
183
242
193
151
Bond energy / kJ mol−1 158
(a) State and explain the trend in the colour of the
halogens in each of the following types of solvents:
(i) polar solvents;
(ii) non-polar solvents.
(b) (i) Describe the reactions of the halogens with
hydrogen, writing balanced equations for the
reactions.
(ii) Discuss the relative stabilities of the hydrides
formed in part (b)(i).
(c) Discuss the trend in volatility of the halogens
down the group.
(d) The halogen beneath iodine in the period table
is astatine. State the colour and physical state of
astatine at room temperature.
(e) Which of the halogens is the strongest oxidizing
agent. Give a reason for your answer.
2
Referring to the E values provided in Table 16.10,
answer the questions which follow.
Table 16.10
Reaction
E
Cl2 + 2e− ҡ 2Cl−
+1.36
Br2 +
2e−
ҡ
2Br−
+1.09
I2 + 2e− ҡ 2I−
2−
/V
+0.54
−
2−
−
2SO3 + 3H2O + 4e ҡ S2O3 + 6OH
+0.58
S4O62− + 2e− ҡ 2S2O32−
+0.09
(a) Explain the similarities and differences in the
behaviour of the halogens with the thiosulfate
(S2O32−) ion.
(b) Based on your answer in part (a), which of the
halogens (chlorine, bromine or iodine) would be
most suitable for use in quantitative estimation of
thiosulfate solutions.
(a) When brown bromine is added to cold aqueous
sodium hydroxide, a colourless solution, G, is
formed. The following equation represents the
disproportionation reaction.
4
(a) State what is observed during the following
reactions involving halides and write balanced
ionic equations.
(i) Aqueous silver nitrate is added to aqueous
sodium chloride followed by ammonia
solution.
(ii) Warm concentrated sulfuric acid is added to
solid potassium iodide. What precaution must
be taken when performing this reaction in the
laboratory?
(iii) Aqueous chlorine is added to aqueous
potassium bromide and the mixture shaken.
(b) NaX is a sodium halide which gives the following
results on testing.
I Bubbling Cl2 into an aqueous solution of NaX
gives a red-brown solution. When starch is added,
a blue-black colour forms.
II When AgNO3 is added to NaX, a yellow
precipitate is formed which is insoluble in aqueous
ammonia.
Identify element X and explain the reactions
taking place in each of the tests, providing
balanced equations.
Chapter 16 Elements and periodicity: Group VII
Answers to ITQs
1
(a) gas, pale green: liquid, orange: solid, black.
(b) & (c) Chlorine exists as Cl2 molecules. They are
relatively small and have low mass so the van
der Waals forces between them are small. Hence
they exist in the gas state at STP. Bromine exists
as Br2 molecules which are larger and heavier
than chlorine molecules. The forces between them
are larger and they need more energy to separate
the molecules and exist as a volatile liquid at STP.
Iodine atoms are heavier and the forces between
them larger. They therefore exist as a crystalline
solid at STP.
2
(a) Bromine is a more powerful oxidizing agent than
iodine and therefore the reaction
Br2 + 2I− → 2Br− + I2
is feasible and proceeds from left to right.
(b) The reaction
Br2 + 2Cl− → 2Br− + Cl2
is not feasible because chlorine has the more
negative E value – i.e. is a more powerful
oxidizing agent than bromine. The reaction can
only go from right to left.
159
160
Chapter 17
The first row transition elements
Learning objectives
■ Explain what is meant by a transition element.
■ Describe the electron configurations of typical transition elements.
■ Describe the characteristic chemical properties of the transition elements.
■ Describe the colour of the compounds and the variety of oxidation states.
■ Explain what is meant by a coordination compound and describe selected properties.
Introduction to the transition elements
Transition metals have played an important part in our
history and still continue to play a role in present everyday
life. For example, coins were originally minted from gold
and silver, but in more recent times, these precious metals
have become important in modern-day jewellery.
it is from this latter categorization that a formal definition
arises. In ‘d’ block elements, the final electron enters the
d sub-level. A transition metal generally refers to an
element which has an atom, or forms at least one ion, with
a partially filled d sub-level. It is interesting to note that
zinc, although a member of the d block, is not regarded as
a transition element. Since both the physical and chemical
properties of transition metals are significantly dependent
on their electronic configurations, these are discussed first.
Transition metals are a category of elements that form a
particular section of the periodic table of the elements. The
periodic table can be broadly divided into main-group
elements and transition elements, as can be seen in
Figure 17.1.
Electronic configurations
The transition elements used to be described as ‘B’ group
elements, located between Groups IIA (also known as Group
II or Group 2, i.e. Be, Mg, Ca, etc.) and IIIA (also known as
Group III, Group 3 or Group 13, i.e. B, Al, Ga, etc.). In the
most up-to-date versions of the periodic table, the transition
elements are numbered as Groups 3 to 12. However, they
are more commonly described as d-block elements, and
There are three rows of transition elements. However, for
the purposes of the CAPE syllabus, you are only required
to know the electronic configurations of the first row
transition elements from scandium (Sc) to zinc (Zn). The
electronic configurations and the respective ‘electrons-inboxes’ representations are shown in Table 17.1.
Main group
H
He
Transition
Li
Be
Na
Mg
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Rb
Sr
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Cs
Ba
La
Hf
Ta
W
Re
Os
Ir
Fr
Ra
Ac
Rf
Db
Sg
Bh
Hs
Ce
Pr
Nd
Th
Pa
U
B
C
N
O
F
Ne
Al
Si
P
S
Cl
Ar
Zn
Ga
Ge
As
Se
Br
Kr
Ag
Cd
In
Sn
Sb
Te
I
Xe
Pt
Au
Hg
Tl
Pb
Bi
Po
At
Rn
Mt
Ds
Rg
Cn
Uut
Fl
Uup
Lv
Uus
Uuo
Pm
Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
Lu
Np
Pu
Am
Cm
Bk
Cf
Es
Fm
Md
No
Lr
Inner transition
Figure 17.1 The transition
elements form the central part
of the periodic table, and are
coloured here in blue.
Chapter 17 The first row transition elements
Table 17.1 The electronic configurations of the first row transition elements
Transition element Symbol
Electronic configuration
Electrons-in-boxes representation
scandium
Sc
[Ar] 3d1 4s2
[Ar]
titanium
Ti
[Ar] 3d2 4s2
[Ar]
vanadium
V
[Ar] 3d3 4s2
[Ar]
chromium
Cr
[Ar] 3d5 4s1
[Ar]
manganese
Mn
[Ar] 3d5 4s2
[Ar]
iron
Fe
[Ar] 3d6 4s2
[Ar]
cobalt
Co
[Ar] 3d7 4s2
[Ar]
nickel
Ni
[Ar] 3d8 4s2
[Ar]
copper
Cu
[Ar] 3d10 4s1
[Ar]
zinc
Zn
[Ar] 3d10 4s2
[Ar]
3d
Notice the unexpected electronic configurations for
chromium and copper. Table 17.2 gives the actual and
expected electron arrangements for these two elements.
Table 17.2 Electron arrangements for chromium and copper
Actual
3d5
Expected
4s1
Chromium
[Ar]
Copper
[Ar] 3d10 4s1
[Ar] 3d4 4s2
[Ar] 3d9 4s2
The explanation of these anomalies is based on the fact that
filled and half-filled d sub-levels have lower energies than
the expected configurations (Table 17.2) and hence are
more stable. The extra stability of a half-filled d sub-level
(where each orbital contains one electron) or of a filled
d sub-level (where each orbital contains two electrons)
results in a symmetrical distribution of charge around the
atom.
Trends across the period of transition
elements
There are relatively small changes in atomic radii, ionic
radii and ionization energies of the elements across the
period of the transition elements.
4s
Atomic radius
The atomic radius may be defined as the distance between
the centre of the nucleus and the outermost electron
(valence) shell. The two primary factors that affect the
atomic radius are:
■ the size of the charge on the nucleus;
■ the number of shells between the nucleus and the
outermost electron (inner shells).
You may also recall that the nucleus is positively charged,
and it attracts electrons which are negatively charged
(opposites attract). The bigger the charge on the nucleus,
the greater is the force of attraction for the outermost
electrons. However, we also have to consider that the
greater the number of shells between the nucleus and the
outermost electrons, the less these outermost electrons feel
the attractive force of the nucleus.
To be able to appreciate the relatively small change in
atomic radii across the period of transition elements, we
need to compare this trend to that observed with another
period of elements, say period 3 from sodium to argon
(see Chapter 13). As period 3 is traversed, electrons are
161
162
Unit 1 Module 3 Chemistry of the elements
going into the same main outer shell and therefore the
number of inner shells remains the same. However, due to
the increase in the number of protons, the nuclear charge
increases. Consequently, the pull of the nucleus on the
outermost electrons increases, thereby resulting in a sharp
decrease in atomic radius.
As the period of first row transition elements is traversed,
the increasing nuclear charge is more or less offset by the
electrons that are being added to an inner 3d sub-level.
These inner 3d electrons ‘screen’ the outer 4s electrons
from the increasing nuclear charge and consequently, the
atomic radii decrease much less rapidly. The outcome is
that there is a slight contraction at the beginning of the
transition metal series, but generally the atoms are all
much the same size (see Table 17.3).
Ionic radius
Ionic radii for transition metals are quite complicated owing
to the fact that transition metals show variable oxidation
states. Also, a proper comparison cannot be made since the
ions do not have the same electronic structure, i.e. they
are not isoelectronic. However, there is a small decrease
in ionic radii as the period of the transition elements is
traversed. This, and the small changes in ionization energy
across the group, arise in the same way as the small changes
in atomic radius.
Ionization energy
Metals tend to lose their electrons easily to form cations.
In general, the ease with which an atom loses electrons
to form a cation describes the ionization energy of the
atom. Ionization energy is defined as the energy required
to remove an electron from a neutral atom to form a cation,
for example:
Na − e− → Na+
which is more commonly written as
Na → Na+ + e−
Ionization energy increases across the period of transition
elements from left to right (Table 17.3), but only marginally
compared with the increase observed across period 3 from
sodium to argon. The trend can be explained in terms of
the changes in nuclear charge and atomic radius.
ITQ 1 Transition metals have higher melting points and higher
densities than s-block metals. Explain this statement with
reference to atomic size and structure.
The trend for ionization energy is related to that observed
for the atomic radius. We have learnt that the atomic
radius decreases marginally across the period of transition
elements. This means that the electrons are held slightly
more tightly, thereby resulting in an increase in ionization
energy across the period. If the electron is held more tightly,
it is harder to pull away.
Table 17.3 Atomic radii and first ionization energies for the
elements Sc to Zn
Sc
Atomic radius
/ nm
0.16
First ionization energy
/ kJ mol−1
+630
Ti
0.15
+660
V
0.14
+650
Cr
0.13
+650
Mn
0.14
+720
Fe
0.13
+760
Co
0.13
+760
Ni
0.13
+740
Cu
0.13
+750
Zn
0.13
+910
Characteristic properties
Like other Group I and Group II metals (the s-block metals),
transition metals are strong, hard, shiny, malleable, ductile
and are also good conductors of heat and electricity. They
have higher densities as well as higher melting and boiling
points than a typical s-block metal. However, transition
metals exhibit unique characteristics which distinguish
them from s-block metals. Having discussed the electronic
structures of transition metals, let us now highlight some
such characteristics.
Variable oxidation states
Firstly, recall that the oxidation number or oxidation
state is a positive or negative number which indicates the
real or theoretical number of electrons lost or gained by
an element in a given compound. Secondly, we have seen
that transition metals have electrons in both the 3d and
4s sub-levels, and the similar energies of these sub-levels
implies that the electrons also have similar energies.
Consequently, it is quite easy for any transition element to
form ions of roughly the same stability by losing different
numbers of electrons. Thus, many transition elements
exhibit more than one positive oxidation state. Figure 17.2
shows the oxidation states for the elements Sc to Zn, and
also highlights the more common oxidation states.
Chapter 17 The first row transition elements
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
+3
+4
+5
+6
+7
+6
+5
+4
+3
+2
+3
+4
+5
+6
+5
+4
+3
+2
+2
+3
+4
+5
+4
+3
+2
+1
+1
+2
+3
+4
+3
+2
+1
+1
+2
+3
+2
+1
+1
+2
+1
Energy
main energy
level
n=4
n=3
sub-level
notation
4f
4d
4p
3d
4s
3p
3s
+1
2p
n=2
Figure 17.2 Oxidation states of the elements Sc to Zn (common
oxidation states are highlighted).
In some instances, the metal and its oxidation state (Mn+)
can be written by stating the name of the metal, followed
by the charge of the cation in Roman numerals, either in
parentheses or as a superscript. For example, the Mn2+ ion
is also expressed as Mn(II) or MnII.
Some generalizations can be made from Figure 17.2.
■ All the transition metals from titanium to copper
exhibit oxidation numbers of +1, +2 and +3 in their
compounds.
■ The common oxidation states for each element include
+2 and/or +3. +3 states are more common at the
beginning of the series whilst +2 states are common
towards the end.
■ The highest oxidation states from Sc to Mn correspond
to the respective numbers of all the electrons outside
the [Ar] core (3 for Sc, 4 for Ti, 5 for V, 6 for Cr and
7 for Mn). After Mn, the increasing nuclear charge
causes a greater pull on the d electrons, and as such,
the weakly held 4s electrons are relatively easier to
remove, giving rise to the common oxidation state of
+2 in the elements from Fe to Zn.
■ When transition metals form ions, they lose electrons
from the 4s sub-level before the 3d sub-level. This may
seem strange because we should recall that the 4s sublevel is lower in energy than the 3d sub-level and is
therefore more stable (Figure 17.3).
Why are electrons lost from the energetically more stable
4s sub-level? This happens because the energy of the 4s
sub-level rises as electrons are added to the 3d sub-level. In
effect, electrons in the 3d sub-level penetrate closer to the
nucleus, thereby ‘screening’ the 4s electrons and making
these 4s electrons relatively easier to remove.
■ V, which is [Ar] 3d3 4s2, forms the V3+ ion, [Ar] 3d2.
The 4s electrons are lost first, followed by one of the
3d electrons.
■ Cr, which is [Ar] 3d5 4s1, forms the Cr3+ ion, [Ar] 3d3.
To form the 3+ ion, the 4s electron is lost first, followed
by two of the 3d electrons.
2s
n=1
1s
Figure 17.3 The positions of energy sub-levels in an atom.
■ Mn, which is [Ar] 3d5 4s2, forms the Mn2+ ion,
[Ar] 3d5. The 2+ ion is formed by the loss of the two 4s
electrons.
■ Fe, which is [Ar] 3d6 4s2, forms the Fe3+ ion, [Ar] 3d5.
The 4s electrons are lost first, followed by one of the
3d electrons.
■ Cu, which is [Ar] 3d10 4s1, forms the Cu2+ ion,
[Ar] 3d9. An electron is first lost from the 4s sub-level
and then one electron from the 3d sub-level.
Formation of complex ions
Introduction
An ion that consists of a central metal ion bonded to anions
or neutral groups is called a complex ion; when the metal
is a transition element, the ion is called a transition metal
complex ion. The anions and neutral groups are usually
considered to be ‘electron-rich’, meaning that they either
possess a lone pair(s) or a negative charge; such species
are called ligands. Some examples of ligands include H2O,
NH3, Cl− and CN−. Note that these are considered to be
ligands only when they are bonded to transition metal ions.
In transition metal complexes, the ligand donates its lone
pairs of electrons to the vacant d orbitals on the transition
metal ion to form dative covalent bonds to the metal ion.
ITQ 2
(a) List the electron configurations and draw the ‘electrons-inboxes’ diagrams of V, V4+ and Ni2+.
(b) List the electron configurations of Fe, Fe2+ and Fe3+.
(c) (i) Comment on the stability of the Fe2+ and Fe3+ ions with
reference to your answers in part (b).
(ii) Explain why Fe2+ ions are readily oxidized to Fe3+ ions.
(d) (i) Write the electron configurations of Mn, Mn2+ and Mn3+.
(ii) Explain why Mn2+ ions are not readily oxidized to Mn3+ ions.
163
164
Unit 1 Module 3 Chemistry of the elements
The number of dative covalent bonds on the metal ion gives
the coordination number (CN) of the central ion. In the
case of transition metals, the most common coordination
number is six, but four and two are not uncommon.
Types of ligands
CN –
CN –
NC
–
Fe3+
NC
One of the ways in which a ligand may be classified is
according to the number of bonds it forms with the central
metal ion. This classification is described as the denticity
of the ligand. The word ‘denticity’ is derived from dentis,
the Latin word for ‘tooth’. Thus, the ligand is thought of as
‘biting’ the metal at linkage point(s). The ligands that we
have encountered so far can only form one bond with the
central metal ion and are described as monodentate. Table
17.4 lists some types of ligands.
–
CN –
CN –
Figure 17.4 Coordination number 6 gives an octahedral complex.
■ Most complexes with coordination number 4 tend
to be tetrahedral (Figure 17.5), with bond angles of
109.5°. A few show square planar shapes (Figure 17.6).
Bond angles are 90°.
NH3
Table 17.4 Some typical ligands
Number of bonds formed
Ligand type
Example
1
monodentate
H2O, NH3, Cl−, CN−
2
bidentate
NH2–CH2–CH2–NH2
(1,2-diaminoethane)
The bidentate ligand 1,2-diaminoethane (often given the
abbreviation ‘en’) has the following structure:
Zn2+
NH3
H3N
NH3
Figure 17.5 Coordination number 4 giving a tetrahedral complex.
NH3
H3N
Cu2+
NH3
H3N
Figure 17.6 Coordination number 4 giving a square planar
complex.
You can imagine the shape of this molecule by thinking of
a pair of headphones. The ear pieces are the N atoms of the
two NH2 groups whilst the piece that goes over your head
represents the –CH2–CH2– group. You can then imagine
the two N atoms forming dative covalent bonds with your
head!
■ Complexes with coordination number 2 give a linear
shape (Figure 17.7). The bond angles are 180°.
H3N
Ag+
NH3
Shapes of complex ions
Figure 17.7 Coordination number 2 gives a linear complex.
The shape of a transition metal complex ion depends on
the number of ligands bonded to the central metal ion.
Octahedral complexes can form with bidentate ligands;
these form ring complexes called chelates. The name
is derived from the Greek word chelè meaning ‘claw’. In
these complexes, the ligands imitate the claw-like grip (of
a lobster) on the central metal ion. Since the central ion is
held firmly by the ligands, these complexes show enhanced
stability. The complex ions [Ni(en)3]2+ and [Cr(en)3]3+ are
examples of chelates; they contain three bidentate ligands
coordinated around the respective central metal ion in an
octahedral manner. A typical chelate is shown in Figure
17.8.
■ Complexes with coordination number 6 tend to be
octahedral in shape; four of the ligands are in one plane,
with the fifth and sixth ligands above and below the
plane respectively (Figure 17.4). Bond angles are 90°.
ITQ 3
In an aqueous solution of chromium(III) chloride (CrCl3(aq)),
chromium forms the complex ion [Cr(H2O)4Cl2]+(aq). Deduce the
likely shape and bond angles in this complex ion.
Chapter 17 The first row transition elements
Table 17.7 Naming the metal part of a complex ion
N
N
N
M
N
N
N
Figure 17.8 A chelate formed between a central M ion and three
‘en’ ligands.
Naming complex ions
The name of a complex ion has four parts.
■ The first part addresses the number of ligands and the
normal prefixes apply (Table 17.5).
Table 17.5 Prefixes indicating the number of ligands
Metal
Name changed to
cobalt
cobaltate
copper
cuprate
nickel
nickelate
iron
ferrate
aluminium
aluminate
chromium
chromate
vanadium
vanadate
■ The fourth part gives the oxidation state of the central
metal ion. This is written in Roman numerals enclosed
in brackets, e.g. (III).
Some examples will help …
■ [Cu(H2O)6]2+ is hexaaquacopper(II). [Cu(H2O)6]2+ is
made up of 6 (hexa) waters (aqua) around copper in
an overall positive ion (copper). The copper has an
oxidation state of +2 (II).
Number of ligands
Prefix
1
mono
2
di
■ [NiCl4]2− is tetrachloronickelate(II). [NiCl4]2− is made
3
tri
4
tetra
5
penta
up of 4 (tetra) chlorines (chloro) around nickel in an
overall negative ion (nickelate). The nickel has an
oxidation state of +2 (II).
6
hexa
■ The second part names the ligand (Table 17.6). Note
that H2O, NH3 and CO are neutral ligands and have
zero charge. If a complex ion contains more than one
type of ligand, the names of the ligands are written in
alphabetical order. When working out the alphabetical
order, you ignore any prefixes.
Table 17.6 Names of ligands to be used in naming complex ions
Ligand
Name
H2O
aqua
NH3
ammine
■ [Cu(NH3)4(H2O)2]2+ is tetraamminediaquacopper(II).
[Cu(NH3)4(H2O)2]2+ is made up of 4 (tetra) ammonias
(ammine) as well as 2 (di) waters (aqua) around
copper in an overall positive ion (copper). The copper
has an oxidation state of +2 (II).
■ [Al(H2O)2(OH)4]− is diaquatetrahydroxoaluminate(III).
[Al(H2O)2(OH)4]− is made up of 2 (di) waters (aqua)
as well as 4 (tetra) hydroxyls (hydroxo) around
aluminium in an overall negative ion (aluminate).
The aluminium has an oxidation state of +3 (III).
(Aluminium always forms the +3 oxidation state and
therefore the oxidation state is frequently left out.)
CO
carbonyl
Cl−
chloro
Acidity of hexaaqua ions
F−
fluoro
CN−
cyano
OH−
hydroxo
Complex ions of the type [M(H2O)6]n+ are acidic. In the
hexaaquairon(III) ion, [Fe(H2O)6]3+, each of the six water
molecules are attached to the central Fe3+ ion via a dative
covalent bond using one of the lone pairs on the oxygen
part of the water molecule. If we consider one of the dative
covalent bonds, the high charge density of the Fe3+ ion
powerfully attracts the electrons in the Fe–O bond which
in turn has the effect that the electrons in the O–H bond
are pulled towards the oxygen more than usual, thereby
weakening the covalent bonds in the aqua ligand (Figure
17.9). The overall effect is that the hydrogen atoms attached
to the aqua ligands are sufficiently positive that they can be
pulled off by a water molecule in solution, i.e. when the
complex ion is dissolved in water. Therefore the aqueous
■ The third part names the metal (Table 17.7). The
name of the metal depends on whether the complex
is cationic (positively charged) or anionic (negatively
charged). If the complex is cationic, the English name
of the metal is used. If the complex is anionic, the
Latin name of the metal with suffix –ate is used. The
suffix –ate shows that the metal now forms part of a
negative ion.
165
166
Unit 1 Module 3 Chemistry of the elements
solution of the transition metal complex ion acts as an acid
and donates a proton.
These electron pairs are all being pulled
away from the oxygens towards the 3+ ion.
H2O
OH2
H2O
+
Fe3+
H
O
H2O
H2O
H
That causes the electron pairs
in the O–H bond to be pulled
even closer to the oxygen
than normal.
+
That makes the hydrogen atoms even more positive than
they normally are when they are attached to oxygen.
Figure 17.9 The polarizing effect of the central ion in a hexaaqua
complex ion.
[Fe(H2O)6]3+(aq) ҡ [Fe(H2O)5(OH)]2+(aq) + H+(aq)
Successive loss of hydrogen ions can occur from the
remaining five aqua ligands.
Ligand exchange reactions
Ligand exchange is a reaction in which one ligand in a
complex ion is replaced by another ligand. This replacement
occurs since some ligands form stronger bonds with a
particular metal ion than other ligands do. Thus, stronger
ligands may displace weaker ligands from a complex.
For example, if ammonia solution is added to a solution
containing hexaaquacopper(II) ions, [Cu(H2O)6]2+, four of
the aqua ligands eventually become replaced by ammonia
molecules to give [Cu(NH3)4(H2O)2]2+. Even though the
four aqua ligands get replaced one at a time, it is more
convenient to write an equilibrium expression for the
overall ligand displacement reaction, as follows:
[Cu(H2O)6]2+(aq) + 4NH3(aq) ҡ
[Cu(NH3)4(H2O)2]2+(aq) + 4H2O(l)
Like any other equilibrium reaction, this one has an
associated equilibrium constant called the stability
constant, Kstab.
[Cu(NH3)4(H2O)22+]
Kstab =
[Cu(H2O)62+][NH3]4
Recall that the sign ҡ represents that the reaction is
reversible, i.e. it can proceed in both directions and the
reaction is in ‘balance’ or in equilibrium.
ITQ 4
Name the following complex ions:
(a) [Al(H2O)6]3+
(b) [CuCl4]2−
(c) [CoCl2(NH3)4]+
(d) [Co(H2O)4(OH)2]
The stability constant in this instance measures the stability
of the formed [Cu(NH3)4(H2O)2]2+ species relative to
[Cu(H2O)6]2+, and in effect is indicative of the ease with
which the ammonia molecule can replace the aqua ligand.
The higher the value of the stability constant, the more
stable is the complex formed. Since Kstab values occur over a
very wide range, they are often expressed as the logarithm,
log10 Kstab so that numbers are simplified and patterns can
be more easily recognized.
Ligand exchange reactions are effectively demonstrated
by experiments in chemistry since colour changes usually
occur when the incoming ligands replace the existing
ligands in a complex.
Anhydrous copper(II) sulfate is a white solid that forms
blue hexaaquacopper(II) ions [Cu(H2O)6]2+ in solution. On
adding a small amount of ammonia, the ammonia functions
as a base and accepts H+ ions from the aqua ligands to form
a pale blue precipitate of copper(II) hydroxide. Copper(II)
hydroxide is normally written as Cu(OH)2 with the aqua
ligands omitted, but for completeness it is written here as
the complex ion [Cu(H2O)4(OH)2].
[Cu(H2O)6]2+(aq) + 2NH3(aq) ҡ
[Cu(H2O)4(OH)2](aq) + 2NH4+(aq)
In excess ammonia, the ammonia now acts as a ligand as four
of the aqua ligands in [Cu(H2O)6]2+ are replaced by ammonia
molecules to form deep blue tetraamminediaquacopper(II)
ions. This deep blue colour is so very intense that this
reaction is used as a sensitive test for copper(II) ions.
[Cu(H2O)6]2+(aq) + 4NH3(aq) ҡ
[Cu(NH3)4(H2O)2]2+(aq) + 4H2O(l)
The four ammine ligands all lie in one plane whilst the two
aqua ligands lie above and below the plane respectively.
Water molecules and ammonia molecules are very similar
in size and therefore the coordination number remains as 6.
The position of equilibrium for the reaction of [Cu(H2O)6]2+
ions in excess ammonia lies to the right.
[Cu(H2O)6]2+(aq) + 4NH3(aq)
[Cu(NH3)4(H2O)2]2+(aq) + 4H2O(l)
Therefore, as [Cu(H2O)6]2+ ions are used up in the reaction,
the [Cu(H2O)4(OH)2] precipitate dissolves to restore the
[Cu(H2O)6]2+ ions.
[Cu(H2O)6]2+(aq) + 2NH3(aq)
[Cu(H2O)4(OH)2](aq) + 2NH4+(aq)
On adding a small amount of concentrated hydrochloric
acid to blue [Cu(H2O)6]2+ ions in solution, the solution turns
green and then yellow. The chloride ions gradually replace
Chapter 17 The first row transition elements
the aqua ligands to form a yellow solution of [CuCl4]2−
ions. The green colour is as a result of the mixture of the
blue [Cu(H2O)6]2+ ions and the yellow [CuCl4]2− ions.
[Cu(H2O)6]2+(aq)
+
4Cl−(aq)
ҡ
[CuCl4]2−(aq)
+ 6H2O(l)
The coordination number of copper changes from 6 to 4.
Chloride ions are bigger than water molecules and there is
not enough room around the central copper ion to fit six
chloride ions.
−
[Cu(H2O)6] (aq) + 4Cl (aq)
2−
[CuCl4] (aq) + 6H2O(l)
CH3
O
H3C
N
N
Fe l l
H2C
N
N
OH
Adding water to the system shifts the equilibrium to the
left.
2+
OH
H2C
CH3
CH3
O
Figure 17.10 Structure of heme.
Water molecules replace the chloro ligands once again, and
the solution returns to blue.
O
O
Haemoglobin
N
Haemoglobin is the iron-containing protein found in red
blood cells that transports oxygen from the lungs to the
rest of the body.
In a simplified description, haemoglobin is an assembly of
four globular protein units. At the centre of each protein
unit is an iron(II) complex known as heme. A heme group
consists of an Fe2+ ion held in a heterocyclic ring known
as porphyrin (Figure 17.10). The central FeII ion has a
coordination number of 6. Within this six-coordination
sphere (Figure 17.11), iron is bound to:
■ four nitrogen atoms of the porphyrin – these lie in one
plane;
N
heme
Fe
N
N
N
protein
(globin)
NH
Figure 17.11 Coordination sphere of iron in haemoglobin.
oxygen-binding
site (heme group)
■ a nitrogen from a surrounding globular protein via the
ring of a histidine residue – this bond occurs below the
plane of the porphyrin ring;
■ either molecular oxygen or water.
Haemoglobin also carries carbon dioxide, but the carbon
dioxide binds to the protein chains of the structure.
II
In the lungs, oxygen binds to the Fe ion, forming
oxyhaemoglobin. In other parts of the body, the
bound oxygen is replaced by an aqua ligand forming
deoxyhaemoglobin. This provides the means by which
haemoglobin is able to transport oxygen from the lungs to
the rest of the body; haemoglobin is capable of transporting
four molecules of oxygen at a time owing to the four
globular protein assembly.
Unfavourable ligand exchange reactions can occur with
haemoglobin inside the body. Stronger ligands such as
carbon monoxide and cyanide can compete with and
replace the oxygen at the oxygen-bonding site. This
reaction is irreversible, thereby preventing haemoglobin
Figure 17.12 Haemoglobin, showing the three-dimensional
structure of the rest of the protein.
from transporting oxygen. For instance, haemoglobin
is 200 times more likely to bond with carbon monoxide
forming a very bright red form of haemoglobin called
carboxyhaemoglobin and this accounts for the poisonous
nature of these substances.
Coloured compounds
Many transition metal compounds are coloured due to the
fact that the d sub-levels in their ions are incompletely filled.
Recall that d sub-levels in the same major energy level are
degenerate, i.e. they have equal energies. Before bonding,
the transition metal ion has degenerate d sub-levels;
167
168
Unit 1 Module 3 Chemistry of the elements
however, bonding with ligands causes the d sub-levels to
split into two groups having different energies, i.e. they
are no longer degenerate. In an octahedral complex, these
groups are t2g and eg (Figure 17.13) whilst in a tetrahedral
complex the groups are e and t2.
This difference in energy (ΔE) between the split d sub-levels
in the metal ions in their complexes are of the same
order of magnitude as the energies of the components of
white light. The energy absorbed (ΔE) is used to promote
electrons from the lower energy state (t2g) to the higher
energy state (eg) once a vacancy (or vacancies) exist in
the higher energy state. These are called d–d transitions
and the atom or ion absorbing the energy changes from its
ground state to an excited state.
Energy
higher
energy state
6E
Transition metals and their compounds are important
catalysts, both in biological systems and in industry. Many
transition metal ions of chromium, manganese, iron, cobalt,
nickel and copper are essential for the successful catalytic
activity of various enzymes in humans, plants and animals.
These metal ions are required in very small amounts and,
as such, are referred to as trace elements. For instance,
cytochrome oxidase, one of the most important coppercontaining enzymes, is involved when energy is obtained
from the oxidation of food. If copper is absent, this enzyme
is completely inhibited and the organism is unable to
metabolize food effectively.
Some industrially important reactions in which transition
metal ions function as catalysts include the Contact process,
the Haber process and the hydrogenation of alkenes.
The Contact process
In the manufacture of sulfuric acid, sulfur dioxide is
oxidized to sulfur trioxide by air at 450 °C in the presence
of a vanadium(V) oxide, V2O5, catalyst:
2SO2(g) + O2(g) → 2SO3(g)
unsplit d sub-levels
lower
energy state
Figure 17.13 Splitting of the d sub-levels in an octahedral
complex.
The absorbed energy coincides with the wavelength for
a specific colour component of white light and so the
complex ion absorbs that particular colour. The observed
colour of the complex is the complementary colour of
the one absorbed.
For example, if white light passes through a CuII solution,
the ions in solution absorb in the orange and red parts of
the spectrum. The light passing through is a mixture of the
remaining colours and appears as the familiar blue-green.
Catalytic properties
A substance that alters (usually accelerates) the rate of a
chemical reaction is described as a catalyst (see Chapter
9 for more information about catalysis). Transition metals
and their compounds catalyse reactions because they
introduce a totally new reaction pathway which has a lower
activation energy than that of the uncatalysed reaction.
Since the activation energy of the catalysed reaction is
lower, the reaction rate is faster.
Chemists believe that it is the ease with which transition
metal ions can accept and donate electrons whilst changing
their oxidation state that often makes them very useful
catalysts.
The Haber process
Nitrogen and hydrogen combine under pressure at ~450 °C
in the presence of a finely divided nickel catalyst:
N2(g) + 3H2(g) ҡ 2NH3(g)
Hydrogenation of alkenes
Liquid oils can be solidified into margarine by treatment
with hydrogen at 200 °C using a nickel catalyst. Hydrogen
is added across some or all of the double bonds.
RCH=CH2 + H2 → RCH2CH3
Magnetic properties
Magnetic properties relate to the effects of exposure
of a substance to an external magnetic field. Magnetic
properties which concern the attraction or repulsion of a
substance by a magnetic field arise principally from:
■ the charge of the electrons;
■ the spin angular momentum of the electrons,
associated with the spin of the electron about its axis;
■ the orbital angular momentum of the electrons,
associated with the rotation about the nucleus.
The above factors are responsible for the existence of
magnetic fields in an atom. Each of these internal magnetic
fields interacts with one another and with external ones
as well. A material can exhibit one of five distinct types
of magnetism depending on the way these magnetic
Chapter 17 The first row transition elements
Table 17.8 Differences between diamagnetism and paramagnetism
Diamagnetism
Paramagnetism
Universal property of all forms of matter since all substances contain some, if not
all, electrons in closed shells, i.e. paired electrons. In closed shells, there is no net
angular momentum since the spin and orbital angular momenta cancel each other.
Arises from the spin and orbital angular momenta of unpaired electrons in a
substance, which gives rise to permanent magnetic moments that align themselves
with an applied field. Note: observed only in the presence of an external magnetic
field.
Material attracted by a magnetic field since the field’s capacity to force alignment
dominates the thermal tendency toward randomness.
Material repelled by a magnetic field because a magnetic moment is induced in
opposition to the direction of the applied field.
applied field
Example is CuI, which has a 3d10 configuration; all the electrons in the 3d sub-level
are paired.
fields interact with each other. However, for the purposes
of this syllabus, we will only focus on two such types:
paramagnetic and diamagnetic.
Substances which are attracted to a magnetic field
are described as paramagnetic substances, whereas
those which are repelled by a magnetic field are called
diamagnetic substances. Paramagnetism is a property
that occurs as a result of the presence of unpaired
electrons. We have seen from Table 17.1 that the transition
metals generally contain one or more unpaired electrons
in the d sub-level. Thus, most of the transition metals are
paramagnetic. The paramagnetic character increases as
the number of unpaired electrons increases. On the other
hand, the transition metals that contain paired electrons
behave as diamagnetic substances. Table 17.8 highlights the
differences between diamagnetism and paramagnetism.
The oxidation states of vanadium
Now that we have an understanding of the characteristic
properties of transition metals and their compounds, we can
appreciate one of the most effective demonstrations of the
range of oxidation states and colours of a transition metal:
the brilliant colours characteristic of vanadium complexes.
Firstly, let us recall that vanadium’s ground state electronic
configuration is [Ar] 3d3 4s2. Vanadium has five electrons
outside of the [Ar] core that can be lost; vanadium is able
to exhibit the four common oxidation states +5, +4, +3
and +2, each of which can be distinguished by its colour.
These colour changes can be shown by shaking a solution
containing vanadium(V) with zinc and dilute acid.
applied field
Example is MnII, which has a 3d5 configuration, i.e. five unpaired electrons. The d5
configuration has the maximum number of unpaired electrons.
The solution of vanadium(V) is made by dissolving ~3 g
of ammonium vanadate(V), NH4VO3, in 40 cm3 of 2 mol
dm−3 NaOH (since it is not very soluble in water) and
then acidifying with 80 cm3 of 1 mol dm−3 H2SO4. This
ammonium vanadate(V) solution is yellow in colour and
contains dioxovanadium(V) ions, VO2+, in acid solution.
VO3−(aq) + 2H+(aq) → VO2+(aq) + H2O(l)
yellow solution; oxidation state +5
The solution can be successively reduced with granulated
zinc or zinc amalgam to obtain the different colours of
vanadium in the four oxidation states. The original yellow
colour changes gradually through green (mixture of VO2+
and VO2+) to blue oxovanadium(IV) ions, VO2+(aq).
VO2+(aq) + 2H+(aq) + e− → VO2+(aq) + H2O(l)
blue solution; oxidation state +4
The VO2+(aq) then changes to green vanadium(III) ions,
V3+(aq).
VO2+(aq) + 2H+(aq) + e− → V3+(aq) + H2O(l)
green solution; oxidation state +3
The solution finally changes to violet vanadium(II) ions,
V2+(aq).
V3+(aq) + e− → V2+(aq)
violet solution; oxidation state +2
The feasibility of vanadium reactions
We have learnt from our study of redox equilibria (see
Chapter 12) that a reaction is deemed feasible only if the
standard cell potential is positive. By extension, a negative
e.m.f. implies that the reaction is not feasible. Let’s use
169
170
Unit 1 Module 3 Chemistry of the elements
E values to show that the vanadium reactions showing
the different colours of vanadium are all feasible.
Summary
■ When zinc is added to a solution of VO2+ ions, the
✓ Transition elements have electrons in d orbitals.
anode and cathode reactions respectively are:
Zn(s) − 2e− → Zn2+(aq)
E
= +0.76 V
2VO2+(aq) + 4H+(aq) + 2e− → 2VO2+(aq) + 2H2O(l)
E = +1.00 V
The overall reaction is:
2VO2+(aq) + 4H+(aq) + Zn(s) →
2VO2+(aq) + 2H2O(l) + Zn2+(aq)
E cell = +1.76 V
■ The VO2+ ions are further reduced to V3+ ions; the
anode and cathode reactions respectively are:
Zn(s) − 2e− → Zn2+(aq)
2+(aq)
2VO
+
4H+(aq)
+
E
2e−
→
2V3+(aq)
= +0.76 V
+ 2H2O(l)
E = +0.34 V
The overall reaction is:
2VO2+(aq) + 4H+(aq) + Zn(s) →
2V3+(aq) + 2H2O(l) + Zn2+(aq)
E cell = +1.10 V
■ The V3+ ions so formed are reduced further to V2+ ions;
the anode and cathode reactions respectively are:
Zn(s) − 2e− → Zn2+(aq)
E
= +0.76 V
2V3+(aq) + 2e− → 2V2+(aq)
E
= −0.26 V
The overall reaction is:
2V3+(aq) + Zn(s) → 2V2+(aq) + Zn2+(aq) E cell = +0.50 V
The standard cell potential for all these reactions is positive,
suggesting that the reactions are energetically feasible. Zinc
is able to gradually reduce vanadium from the +5 oxidation
state to the +2 oxidation state.
✓ In all the transition elements, except for zinc, d
orbitals are not completely full.
✓ These elements are all metals.
✓ The elements in this series show only a small
diminution in radius, one to the next, and a
smaller change that might be expected in their
first ionization energies.
✓ The transition elements exhibit a wide range of
oxidation states.
✓ The transition elements, except for zinc, have
coloured compounds.
✓ Transition elements have unpaired electrons and
are diamagnetic.
✓ Transition elements form complex ions by
accepting lone pairs of electrons from anions or
neutral species.
✓ The shape of such complex ions can be inferred
from the number of ligands involved.
✓ Transition elements and their compounds often
have catalytic effects.
Chapter 17 The first row transition elements
Review questions
1
(a) Write the electronic configuration of Zn and Zn2+.
(b) State the reason why zinc is not considered to be a
transition metal.
2
(a) Complete the electronic configuration of:
(i) a chromium atom, Cr, 1s2 2s2 2p6 3s2 3p6
(ii) a chromium ion, Cr3+, 1s2 2s2 2p6 3s2 3p6
(b) It has been observed that a solution of aqueous
chromium(III) ions, [Cr(H2O)6]3+(aq), is weakly
acidic. Suggest an explanation for this observation.
3
(a) List four characteristic properties of transition
elements.
The reaction shown in Figure 17.15 illustrates the
reaction occurring between oxyhaemoglobin and
carbon monoxide to form carboxyhaemoglobin.
O2
N
5
Write the formula of the coloured species present in
each of the following:
(a) ammonia solution is added to aqueous copper(II)
sulfate and a light blue precipitate is formed;
(b) excess ammonia solution is added to (a) and a
deep-blue solution is formed.
6
(a) Complete the figure by writing the colour of the
species labelled A, B, C, D and E in Figure 17.16.
(b) Write the formula of species E.
CO
N
N
+ CO
Fe
N
(ii) The gradual addition of a concentrated
solution of sodium chloride to aqueous
copper(II) sulfate leads to the formation of a
green solution. A colour change from green to
yellow is observed on further addition of the
sodium chloride solution.
(b) The complex ion Z is obtained on adding a
concentrated solution of NaCN to aqueous NiCl2.
The ion Z has the percentage composition of
36.1% Ni, 29.5% C and 34.4% N.
(i) Determine the formula of the complex ion Z.
(ii) Draw the shape of the complex ion Z.
N
Histidine
N
Fe
N
N
Histidine
Cu(OH)2
Figure 17.15
(b) Explain what is meant by the term ligand and
identify two ligands in the haemoglobin structures
in Figure 17.15.
(c) The presence of carbon monoxide in the blood can
prevent oxygen from reaching the tissues.
(i) State and explain the principle on which the
reaction in Figure 17.15 is based.
(ii) Using the information in Figure 17.15, account
for the toxic effect of carbon monoxide at high
concentrations.
(iii) Suggest a treatment for a patient suffering
from exposure to carbon monoxide and give a
reason for your suggestion.
4
(a) Provide explanations for each of the following in
terms of the characteristic properties of transition
metals and their complexes:
(i) Anhydrous copper(II) sulfate is a white solid
that gradually turns blue on the drop-wise
addition of water. Further addition results in
the solid dissolving, with the formation of a
blue solution.
OH
CuSO4 (anhy)
H2O
A
Cu(H2O)4
heat
E
C
–
+
H
2+
B
Cl
–
H2O
CuCl4
2–
D
Figure 17.16
7
(a) (i) Write the colour of the species labelled A, B, C
and D in Figure 17.17.
Co(NH3 )6
C
NH3
CoCl2 (s)
H2O
A
Co(H2O)6
2+
2+
B
Cl
–
I
CoCl4
2–
D
Figure 17.17
(ii) State the reagent used for the conversion in
reaction 1 (D → B).
(b) Iron forms a complex ion with cyanide ions
(CN−). The formula of the complex is [Fe(CN)6]4−.
Explain how an aqueous solution of iron(II) sulfate
functions as an antidote for cyanide poisoning.
171
172
Unit 1 Module 3 Chemistry of the elements
Answers to ITQs
Answers to Review questions
(a) V is [Ar] 3d3 4s2
V4+ is [Ar] 3d1
Ni2+ is [Ar] 3d8 4s2
(b) Fe is [Ar] 3d6 4s2
Fe2+ is [Ar] 3d6
Fe3+ is [Ar] 3d5
(d) (i) Mn is [Ar] 3d5 4s2
Mn2+ is [Ar] 3d5
Mn3+ is [Ar] 3d4
1
(a) Zn is [Ar] 3d10 4s2
Zn2+ is [Ar] 3d10
2
(a) (i) Cr is 1s2 2s2 2p6 3s2 3p6 3d5 4s1
(ii) Cr3+ is 1s2 2s2 2p6 3s2 3p6 3d3
4
(b) (i) Z is Ni(CN)4
5
(a) [Cu(H2O)4(OH)2]
(b) [Cu(NH3)4(H2O)2]2+
6
(a) A is white, B is blue, C is pale blue, D is yellow, E
is black
(b) CuO
7
(a) (i) A is sky blue, B is pink, C is brown, D is blue
(ii) water
2
3
Shape is octahedral; bond angles are 90°.
4
(a)
(b)
(c)
(d)
hexaaquaaluminium(III) ion
tetrachlorocuprate(II) ion
dichlorotetraamminecobalt(III) ion
tetraaquadihydroxocobalt(II)
173
Chapter 18
Qualitative inorganic analysis
Learning objectives
■ Describe tests to identify specified anions, cations and gases.
■ Explain the basis of each test.
■ Interpret the results of reported tests.
Introducing inorganic analysis
Identification of cations
Qualitative inorganic analysis seeks to identify the
elements found within inorganic compounds through the
use of various reagents. The primary focus in this chapter
is identifying ions in aqueous solutions. The general
procedure is that the unknown solution, when treated
with a certain reagent, is converted into a new compound
that has a characteristic and diagnostic colour, solubility or
other visible change. Reagents should be added gradually
until no further change is observed. Some reactions
liberate gases and confirmatory tests for these gases should
be performed.
Cations are usually identified using two main reagents:
aqueous sodium hydroxide and aqueous ammonia (Table
18.2).
You are required to know how to test for and identify the
ions and gases listed in Table 18.1.
Table 18.1 Cations, anions
and gases to be identified
In some instances, aqueous sodium carbonate is used
as a reagent to precipitate insoluble carbonates. A white
precipitate is produced in all instances of cations listed in
Table 18.2, except for Cr3+ (green), Mn2+ (pink), Fe2+ (green),
Fe3+ (brown) and Cu2+ (blue). NH3(g) is evolved with NH4+.
The precipitates are formed during the general reaction:
Mn+(aq) + nOH−(aq) → M(OH)n(s)
where ‘M’ represents the metal atom and ‘n’ represents the
size of the charge on the cation.
Table 18.2 Summary of reactions of cations with NaOH(aq) and NH3(aq)
Reaction with NaOH(aq)
Reaction with NH3(aq)
Cations
Anions
Gases
Mg2+
white ppt; insoluble in excess NaOH(aq)
white ppt; insoluble in excess NH3(aq)
Mg2+
CO32−
CO2
Ca2+
white ppt; insoluble in excess NaOH(aq)
no reaction
Al3+
NO3−
H2
Ba2+ white ppt from concentrated solutions only
Ca2+
SO42−
HCl
Al3+
Cr3+
SO32−
H2S
Pb2+ white ppt; soluble in excess NaOH(aq) to form the complex Pb(OH)42−(aq) white ppt; insoluble in excess NH3(aq)
Mn2+
Cl−
NH3
Fe2+
Br−
NO2
Fe3+
I−
O2
Cu2+
CrO42−
SO2
Zn
2+
Cl2
Ba2+
Br2 / HBr
Pb2+
I2
NH4+
Cr3+
Mn2+
Fe2+
no reaction
−
white ppt; soluble in excess NaOH(aq) to form the complex Al(OH)4 (aq)
grey-green ppt; soluble in excess NaOH(aq) to give a dark green solution
grey-green ppt; insoluble in excess NH3(aq)
containing Cr(OH)4−(aq)
white ppt which rapidly turns light brown;
white ppt which rapidly turns light brown; insoluble in excess NaOH(aq)
insoluble in excess NH3(aq)
dirty green ppt which changes to brown in air; insoluble in excess
dirty green ppt which changes to brown in
NaOH(aq)
air; insoluble in excess NH3(aq)
Fe3+
red-brown ppt; insoluble in excess NaOH(aq)
Cu2+
pale blue ppt; insoluble in excess NaOH(aq); ppt turns black on heating
Cu(OH)2(s) → CuO(s) + H2O(l)
Zn2+
white ppt; soluble in excess NaOH(aq) to form the complex ion
[Zn(OH)4]2−(aq)
solution remains colourless; on heating, pungent colourless ammonia
gas evolved which turns moist red litmus blue
NH4+(aq) + OH− → NH3(g) + H2O(l)
NH4+
white ppt; sparingly soluble in excess NH3(aq)
red-brown ppt; insoluble in excess NH3(aq)
pale blue ppt soluble in excess NH3(aq) to
give a deep-blue solution containing the
complex ion Cu(NH3)42+(aq)
white ppt; soluble in excess NH3(aq) to form
the complex ion [Zn(NH)3]42+(aq)
–
’ppt’ is used as the abbreviation for precipitate; Al3+ can be distinguish from Pb2+ by the insolubility of the lead(II) halides or the
flame test.
174
Unit 1 Module 3 Chemistry of the elements
The information in Table 18.2
is also provided in the form
of flow charts in Figures 18.1
and 18.2. You can follow these
flow charts when detecting
cations.
add NaOH(aq)
drop by drop
Cation in
aqueous
solution
The procedure for performing
a flame test is as follows:
■ clean a platinum wire by
dipping it repeatedly into
conc. hydrochloric acid
and heating it until the
flame remains colourless
(alternatives to platinum
wire include wooden
splints and the tip of a lead
pencil);
■ dip the end of the wire
into the acid and then into
the sample of the element
or compound to be tested;
■ heat the wire in the
colourless flame and
observe any flame colour
which is produced.
white
precipitate
does not
dissolve Ca2+
white precipitate
possibly
Zn2+, Pb 2+, Ca2+, Al 3+
yellow
precipitate
possibly Pb 2+
precipitate
dissolves
possibly
Zn2+, Pb2+ or Al 3+
blue precipitate
Cu2+
Flame tests
When some metal ions
are heated in a colourless
flame, the ions become
excited causing them to
emit visible light. This light
emission is responsible for
the characteristic colour that
the respective ion turns the
flame. Therefore, burning a
substance in a flame test is
a technique used to visually
determine the presence of
certain metal ions.
add excess
NaOH(aq)
red-brown
precipitate
Fe 3+
no
precipitate
Zn2+
no precipitate
possibly
Zn 2+ , Al 3+
add Kl(aq) to a
fresh sample
of solution
green precipitate
darkening in colour
on standing Fe2+
white
precipitate
Al 3+
add excess NH3(aq)
to a fresh sample
of solution
no smell of
ammonia
+
+
Na or K
no precipitate
possibly
+
+
+
NH4 , Na , K
add excess
NaOH(aq) then heat
smell of ammonia,
red litmus turns
+
blue NH4
Figure 18.1 Tests for cations using aqueous sodium hydroxide.
add NH3(aq)
drop by drop
Cation in
aqueous
solution
add excess
NH3(aq) and stir
white precipitate
possibly
Zn2+, Pb 2+, Al 3+
white
precipitate does
not dissolve
Pb 2+ or Al 3+
white precipitate
soluble in
excess
NH3(aq) Zn2+
blue precipitate
possibly Cu 2+
add excess
NH3(aq) and stir
red-brown
precipitate
possibly Fe3+
yellow
precipitate Pb 2+
add excess Kl(aq)
to a fresh sample
of solution
precipitate
dissolves to give
a deep blue
solution Cu 2+
precipitate does
not dissolve
Fe 3+
green precipitate
possibly Fe 2+
precipitate does
not dissolve but
darkens on
exposure to air
Fe 2+
no precipitate
Ca 2+, NH4+, Na+ , K +
add excess
NH3(aq) and stir
Figure 18.2 Tests for cations using aqueous ammonia.
ITQ 1
ITQ 2
(a) State what is observed when aqueous ammonia is added
drop-wise to a solution of copper(II) nitrate until the ammonia
is present in excess.
(a) Using NaOH(aq) and NH3(aq) only, describe how these
reagents can be used to distinguish between Ca2+(aq),
Zn2+(aq) and Al3+(aq) ions.
(b) Write a balanced equation to represent the overall reaction.
(b) Write balanced equations to represent the reactions between:
(i) Ca2+(aq) and NaOH(aq)
(ii) Al3+(aq) with excess ammonia solution
Chapter 18 Qualitative inorganic analysis
Sodium is a contaminant in many compounds and its
spectrum tends to dominate over others. To alleviate this
problem, flame colours are often viewed through cobalt
blue glass to filter out the yellow flame colour of sodium
and hence allow for easier viewing of other metal ions.
Some characteristic flame colorations are given in Table
18.3.
Table 18.3 Flame tests
Colour of flame
Illustration of colour
Inference
lilac (purple through blue
glass)
K+
bright yellow (invisible
through blue glass)
Na+
Nitrate(V), NO3−
Four tests are available.
■ Add conc. H2SO4: on warming, HNO3(g) and red-
brown NO2(g) are given off. The HNO3(g) formed
undergoes thermal decomposition to produce NO2(g)
and O2(g).
NO3−(s) + H2SO4(l) → HNO3(g) + HSO4−(aq)
4HNO3(g) → 2H2O(l) + 4NO2(g) + O2(g)
■ Add conc. H2SO4 in the presence of Cu: NO2(g) and
HNO3(g) are liberated. The HNO3(g) reacts with Cu to
produce NO2(g) and the resulting solution is greenblue in colour due to the presence of Cu2+(aq) ions.
4HNO3(g) + Cu(s) → Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)
■ Add powdered Al or Zn (reducing agents) and
brick-red
Ca2+
NaOH(aq): on heating, NH3(g) is given off. The
NO3−(aq) ions are reduced by Al/Zn.
NO3−(s) + 3Al(s) + 3OH−(aq) + 6H2O(l) →
3[Al(OH)4]−(aq) + NH3(g)
yellow-green
Ba2+
NO3−(s) + 4Zn(s) + 7OH−(aq) + 6H2O(l) →
4[Zn(OH)4]2−(aq) + NH3(g)
■ Add iron(II) sulfate and conc. H2SO4: crystals of FeSO4
blue-green
Cu2+
Identification of anions
We shall now describe the reactions of some anions with
different reagents.
Carbonates, CO32−
Two tests are available.
■ Add dilute HCl or conc. H2SO4: CO2(g) is liberated with
effervescence. Carbonates of the cations Pb2+, Ca2+ and
Ba2+ do not react with H2SO4 as these cations form
insoluble sulfates.
CO32−(s) + 2H+(aq) → H2O(l) + CO2(g)
■ Add Ba2+(aq) or Ca2+(aq) followed by dilute acid: a
white precipitate of BaCO3(s) or CaCO3(s) is formed.
The precipitate is soluble in dilute acid with the
liberation of CO2(g).
are mixed with the nitrate solution and then conc.
H2SO4 is gently added to the mixture such that the
H2SO4 forms a layer above the aqueous solution. A
brown ring forms at the junction of the two liquids
owing to the presence of Fe(NO)SO4; this test is
known as the brown ring test. The NO3− ion is reduced
by iron(II) which is oxidized to iron(III) and forms a
nitrosyl complex.
NO3−(aq) + 3Fe2+(s) + 4H+(aq) →
NO(g) + 3Fe3+(aq) + 2H2O(l)
NO(g) + FeSO4(s) → Fe(NO)SO4(s)
Sulfate(VI), SO42−
Two tests are available.
■ Add Ba2+(aq)/H+(aq): a white precipitate of BaSO4(s) is
formed. It is soluble in warm conc. HCl.
Ba2+(aq) + SO42−(aq) → BaSO4(s)
This test is carried out in an acidified medium to
inhibit the precipitation of carbonate and sulfate(IV)
(sulfite).
■ Add Pb2+(aq): a white precipitate of PbSO4(s) is formed
M2+(aq) + CO32−(aq) → MCO3(s)
which is soluble in hot conc. H2SO4.
Ba2+ ions can be obtained from BaCl2(aq) and
Ba(NO3)2(aq); Ca2+ ions can be obtained from CaCl2(aq).
Pb2+(aq) + SO42−(aq) → PbSO4(s)
Pb2+ ions can be obtained from lead(II) ethanoate.
175
176
Unit 1 Module 3 Chemistry of the elements
Sulfate(IV) (sulfite), SO32−
Iodides, I−
Three tests are available
Three tests are available.
■ Add dilute HCl or conc. H2SO4: SO2(g) is evolved on
■ Add AgNO3(aq) followed by NH3(aq): a yellow or
warming.
SO32−(aq)
+ 2H (aq) → SO2(g) + H2O(l)
+
■ Add Ba2+(aq): a white precipitate of BaSO3(s) is
formed which is readily soluble in dilute HCl with the
liberation of SO2(g).
Ba2+(aq) + SO32−(aq) → BaSO3(s)
■ Add AgNO3(aq): a white precipitate of Ag2SO3(s) is
formed which turns from grey to black on warming as
a result of the decomposition to silver.
2Ag+(aq) + SO32−(aq) → Ag2SO3(s)
Chlorides, Cl−
Three tests are available
■ Add AgNO3(aq) followed by NH3(aq): a white
precipitate of AgCl(s) is formed which is soluble in NH3
to form Ag(NH3)2+(aq).
Ag+(aq) + Cl−(aq) → AgCl(s)
■ Add conc. H2SO4: the pungent, colourless hydrogen
chloride gas is evolved.
H2SO4(l) + Cl−(s) → HCl(g) + HSO4−(aq)
■ Add Pb2+(aq): a white precipitate of PbCl2(s) is formed
which dissolves on heating and re-precipitates on
cooling.
Pb2+(aq) + 2Cl−(aq) → PbCl2(s)
Bromides, Br−
Three tests are available
■ Add AgNO3(aq) followed by NH3(aq): a white or cream
precipitate of AgBr(s) is formed which is partially
soluble in NH3.
cream precipitate of AgI(s) is formed which is insoluble
in NH3.
Ag+(aq) + I−(aq) → AgI(s)
■ Add conc. H2SO4: iodine is formed as a black or violet
precipitate. On warming, violet vapours of iodine are
evolved. HI is initially formed, but is oxidized to iodine.
H2SO4(l) + I−(s) → HI(g) + HSO4−(aq)
2HI(g) + [O] → I2(s) + H2O(l)
■ Add Pb2+(aq): a yellow precipitate of PbI2(s) is formed
which is soluble in excess of the iodide solution.
Pb2+(aq) + 2I−(aq) → PbI2(s)
PbI2(s) + 2I−(aq) → [PbI4]2−(aq)
Chromate(VI), CrO42−
Three tests are available.
■ Add AgNO3(aq): a red-brown precipitate of Ag2CrO4(s)
is formed which is soluble in NH3.
2Ag+(aq) + CrO42−(aq) → Ag2CrO4(s)
■ Add Ba2+(aq): a pale yellow precipitate of BaCrO4(s) is
formed which is soluble in strong acids.
Ba2+(aq) + CrO42−(aq) → BaCrO4(s)
■ Add Pb2+(aq): a yellow precipitate of PbCrO4(s) is
formed.
Pb2+(aq) + CrO42−(aq) → PbCrO4(s)
Flow charts
The identification of anions is provided in a different form
in Figures 18.3–18.6.
Ag+(aq) + Br−(aq) → AgBr(s)
■ Add conc. H2SO4: red-brown vapours of Br2(g) and
HBr(g) are seen. The HBr(g) that is formed is oxidized
to Br2(g).
H2SO4(l) + Br−(s) → HBr(g) + HSO4−(aq)
■ Add Pb2+(aq): a white precipitate of PbBr2(s) is formed.
This precipitate dissolves on heating and re-precipitates
on cooling.
Pb2+(aq) + 2Br−(aq) → PbBr2(s)
ITQ 3 Two solutions, labelled X and Y, contain either chloride
or bromide ions. Describe how AgNO3 followed by NH3(aq) can
be used to identify the ion in each solution. Write the balanced
equations for the reactions occurring.
ITQ 4
(a) Describe what you would see when an aqueous solution of
silver nitrate is added to a solution containing iodide ions
followed by aqueous ammonia.
(b) Write balanced equations for the reactions occurring in part (a).
(c) List and explain the observations when conc. H2SO4 is added
to solid potassium iodide.
Chapter 18 Qualitative inorganic analysis
heat the solid:
is gas given off?
CO2
SO2
NO2
O2
CO32–
SO32–
NO3–
(Na + )NO3–
no
gas
Cl – Br– I – SO42–
Figure 18.3 Testing for anions; heating the solid.
dilute acid on the solid:
is gas given off?
CO2
SO2
CO32–
SO32–
Testing for gases
As we have seen in several of the reactions encountered,
gases can be produced in some tests. Confirmatory tests for
these gases should be performed when they are generated
in the test. When a gas needs to be tested with materials
such as moist litmus paper and splints, the material should
be placed at the mouth of the test tube where the gas is
escaping.
Carbon dioxide, CO2
■ Colourless, odourless.
■ When bubbled through lime water (Ca(OH)2(aq)), the
Figure 18.4 Testing for anions; adding dilute acid.
mixture turns milky. The milky appearance is due to
the formation of a solid precipitate of CaCO3.
Ca(OH)2(aq) + CO2(g) → CaCO3(s) + H2O(l)
acidified (HNO3) aqueous silver nitrate with
a solution of the sample
white
precipitate
darkens in
the light
cream
precipitate
Cl –
Br –
yellow
precipitate
I–
Figure 18.5 Testing for anions; adding AgNO3(aq)/H+(aq).
Hydrogen, H2
■ Colourless, odourless.
■ ‘Pops’ with lighted splint or may burn with a blue
flame. H2(g) is recognized by the ‘pop’ when it burns.
This ‘pop’ is the sound of a small explosion since H2 is
extremely flammable.
Hydrogen chloride, HCl
■ Colourless, pungent.
acidified (HCl) aqueous barium chloride with
a solution of the sample
■ HCl(g) is an acidic gas; it dissolves in water to form
hydrochloric acid, HCl(aq).
white precipitate
SO4
2–
■ Fumes in moist air; turns moist blue litmus red; fumes
with NH3(g). The fumes are due to the formation of
NH4Cl.
HCl(g) + NH3(g) → NH4Cl(g)
Figure 18.6 Testing for anions; adding BaCl2(aq)/H+(aq).
Hydrogen sulfide, H2S
■ Colourless, odour of rotten eggs.
■ Turns Pb2+ ions black due to the formation of PbS.
ITQ 5
Ammonia, NH3
Two calcium salts D and G were heated and the following results
were obtained:
■ Colourless, pungent.
Compound D decomposed to give a gas which formed a white
ppt when bubbled into Ca(OH)2(aq).
■ Ammonia is an alkaline gas.
■ Turns moist red litmus blue; fumes with HCl(g).
Compound G decomposed to give two gases. One was brown
and turned blue litmus red and the other rekindled a glowing
splint.
Nitrogen dioxide, NO2
(a) Identify the gases evolved in heating compounds D and G.
■ Turns moist blue litmus red.
(b) Deduce the molecular formulae of D and G.
■ Red-brown, pungent.
177
178
Unit 1 Module 3 Chemistry of the elements
Oxygen, O2
■ Colourless, odourless.
■ Relights a glowing splint.
■ Oxygen is the only gas which supports burning and
hence will relight a glowing splint.
Sulfur dioxide, SO2
■ Colourless, choking odour.
■ Turns moist blue litmus red.
■ Turns KMnO4/H+ colourless.
■ Turns K2Cr2O7/H+ green.
■ SO2(g) is an acidic, reducing gas.
Chlorine, Cl2
■ Pale yellow-green, pungent, choking odour.
■ Bleaches moist blue litmus.
■ Cl2(g) has a bleaching effect.
Bromine, Br2/HBr
■ Red-brown, pungent.
■ Moist blue litmus turns red then bleached; fumes in
moist air.
■ Br2(g) has a bleaching effect.
Iodine, I2
■ Violet-black solid and violet vapours.
■ Bleaches moist litmus.
■ Turns starch/iodide paper blue-black.
Chapter 18 Qualitative inorganic analysis
Review questions
1
Write the formula of the coloured species present in
each of the following:
(a) ammonia solution is added to aqueous copper(II)
sulfate(VI) and a light blue precipitate formed;
(b) excess ammonia solution is added to part (a) and a
deep-blue solution formed.
2
Describe the product of each of the following reactions
and write an equation for each reaction:
(a) NaOH(aq) added to iron(II) chloride;
(b) NaOH(aq) added to iron(III) chloride.
3
When an acidified solution of manganate(VII) ions
is added to an aqueous solution of iron(II) sulfate, a
yellow solution is formed.
(a) Write the two half-equations to explain the above
observations.
(b) Deduce the change in oxidation state occurring in
these reactions.
4
Describe how you would perform a laboratory test to
detect each of the following gases:
(a) carbon dioxide;
(b) ammonia;
(c) sulfur dioxide;
(d) nitrogen dioxide;
(e) chlorine.
5
6
Identify each of the substances Q, R and S, indicated
below, by explaining the reactions described and
writing ionic equations where applicable.
(a) Substance Q is a brown solid which effervesces
with dilute acid to produce a colourless, odourless
gas which does not support combustion. The
resulting solution formed a brown precipitate on
the addition of an alkali.
(b) Substance R is a white solid, which when treated
with dilute acid, gives off a reducing gas with a
pungent odour. The remaining solution, when
treated with aqueous ammonia, formed a white
gelatinous precipitate which is soluble in excess.
(c) Substance S is a white powder which is soluble
in water to give a blue solution. When aqueous
barium chloride is added to a solution of S, a
white precipitate is formed which is insoluble in
dilute hydrochloric acid.
The following tests were done on a sample T which
consisted of pale green crystals. All gases evolved were
tested. Complete Table 18.4 by filling in observations
and inferences where applicable. Tests (b) and (c)
were performed on separate portions of an aqueous
solution.
Table 18.4
Test
Observations
Inferences
T was heated gently Droplets of a colourless
in a hard glass test liquid collected on the top
of the tube turned cobalt
tube.
chloride paper pink.
(b) (i) Excess NaOH(aq)
A pale green precipitate
was added.
was formed which turned
dirty green on standing.
(ii) The mixture from
part (b) (i) was
warmed.
(c) (i) Pb(NO3)2(aq) was
Cl− or SO42− ions
added.
are present in
sample T.
Sample T
(ii) The suspension was
contains SO42−
heated to boiling
then allowed to
ions.
cool.
(a)
(d) Write an ionic equation for the reaction occurring
in test (b) (ii).
7
(a) Complete Table 18.5 by deducing the cations and
anions present in the salt being tested.
Table 18.5
(i)
Experiment
Observations
Heat the salt in a dry test
tube.
Colourless gas; choking
odour. Turns blue litmus
red; decolorizes KMnO4/
H+. Red-brown residue
after heating.
A pale green solution is
produced.
A dirty green ppt.
(ii) Make an aqueous solution
of the salt.
(iii) Add NaOH(aq) to part (ii)
and warm.
(iv) Add NH3(aq) to part (ii).
(v) To part (ii) add dilute HNO3
and then Ba(NO3)2(aq).
Cations and
anions present
Dirty green ppt insoluble
in excess NH3(aq).
White ppt insoluble in
dilute acid.
(b) (i) State two reagents that can be used to confirm
the identity of the cations in part (a).
(ii) State an alternative reagent for the
identification of the gas liberated in part (a) (i).
(c) In part (a) (v), why is HNO3 added prior to the
Ba(NO3)2(aq)?
179
180
Unit 1 Module 3 Chemistry of the elements
8
(a) Figure 18.7 shows part of the periodic table.
Answers to ITQs
1
Mg
K
Al
Cl
Fe
I
Ba
or Cu2+(aq) + 4NH4OH(aq) →
Cu(NH3)4]2+(aq) + 4H2O(l)
Pb
Figure 18.7
Which of the ions shown in Figure 18.7 will react
with each of the following substances?
(i) KOH(aq) to produce a red-brown precipitate;
(ii) Na2CO3(aq) to give a white precipitate;
(iii) AgNO3(aq)/H+(aq) to form a white precipitate;
(iv) Pb(NO3)2(aq) to produce a yellow precipitate.
(b) D is a powdered mixture containing a soluble
and an insoluble salt. A sample of D is treated as
follows:
I Water is added to D and the mixture filtered.
II The residue reacts completely with dilute
HNO3(aq) and a colourless gas is given
off which forms a white precipitate with
Ca(OH)2(aq). The resulting solution reacts
with both NaOH(aq) and NH3(g) to form a
white precipitate which does not dissolve in
excess of the reagents.
III One sample of the filtrate reacts with
BaCl2(aq)/H+(aq) to form a white precipitate.
Another sample reacts with NH3(aq) and
NaOH(aq) to form a white precipitate which is
soluble in excess of the reagents.
(i) Using the information provided, deduce the
possible ions in the residue as well as in the
filtrate.
(ii) Write a balanced ionic equation for the
reaction between nitric acid and the residue.
(a) A cloudy blue-green ppt is seen which dissolves
to give a deep purple/blue solution. (Note that
the question asked ‘what is observed …’ and not
‘what is produced …’.)
(b) Cu(NO3)2 + 4NH4OH → [Cu(NH3)4](NO3)2 + 4H2O
2
(a) Add NaOH solution. If there is no ppt, then you
have Ca2+ ions.
If a ppt forms you have either Al3+ ions or Zn2+
ions.
Filter and wash the ppt. Add a little to some
NH4OH(aq). If the ppt dissolves you have
Zn2+ ions. If it does not dissolve you have Al3+ ions.
3
Add a solution of silver nitrate acidified with a little
nitric acid. Cl– ions produce a pure white ppt. Br– ions
produce a cream coloured ppt.
The colours can be hard to distinguish. Filter and
wash the ppt. Add a little to a solution of ammonia. If
it dissolves, the ppt is AgCl and the original solution
contained the Cl– ion. If only some of the ppt dissolves
it is AgBr and the original contained the Br– ion.
Cl–(aq) + Ag+(aq) → AgCl(s)
Br–(aq) + Ag+(aq) → AgBr(s)
4
(a) You would see a yellow ppt. Adding aqueous
ammonia has no effect, the ppt does not dissolve.
(b) I–(aq) + Ag+(aq) → AgI(s)
(c) H2SO4(l) + I–(aq) → HI(g) + HSO4–(aq)
then the HI gas is oxidized by the acid:
2HI(g) + [O] → I2(s) + H2O(l)
5
(a) The gas from compound D was carbon dioxide.
The brown gas from compound G was nitrogen
dioxide NO2.
The other gas from compound G was oxygen.
(b) D and G are calcium salts. D is CaCO3 and G is
Ca(NO3)2.
Answers to Reviews questions
5
(a) Q is Fe2(CO3)3
(b) R is ZnSO3
(c) S is anhydrous CuSO4
6
T is FeSO4.xH2O
8
(b) D contains CaCO3 and ZnSO4
Chapter 1 Atomic structure
Unit 2
Chemical principles and
applications II
181
182
Module 1
The chemistry of carbon
compounds
Chapter 19
Alkanes
Learning objectives
■ Explain why carbon forms compounds comprised of carbon chains and rings.
■ Describe in detail the bonding in saturated alkanes and cycloalkanes.
■ Describe and account for the three-dimensional shape of methane.
■ Define the terms homologous series, structural isomerism, sp3 hybrid orbital and substituent.
■ Systematically name alkanes and cycloalkanes.
■ Write or draw, from molecular formulae or systematic names, the structures of alkanes and
cycloalkanes.
■ Describe the physical properties, sources and uses of C1 to C10 n-alkanes.
■ Describe in outline the processes which occur when alkanes are subjected to combustion, thermal and
catalytic cracking and bromination.
Introduction to carbon compounds
Organic chemistry is the chemistry of compounds of carbon.
Living organisms, to a large extent, consist of carbon-based
compounds. Cellulose, a very large molecule built up from
carbon, hydrogen and oxygen, provides structural tissue
in plants. Animals use proteins, also very large molecules
containing nitrogen in addition to carbon, hydrogen and
oxygen. Living organisms also produce many other organic
compounds, some of which are very useful to us.
■ Quinine, C20H24N2O2, is found in the bark of trees
of the genus Cinchona. For many years quinine was
the most important drug for treatment of malaria
and many modern anti-malarial drugs are similar in
structure.
■ Table sugar is sucrose, C12H22O11. The production of
sugar from sugar cane was one of the driving forces
behind the transatlantic slave trade and the eventual
formation of modern West Indian societies. One could
say that we are here because of sucrose!
■ Cholesterol, C27H46O, is found in animals. An
accumulation of cholesterol in the walls of blood
vessels may lead to many illnesses, including strokes
and heart attacks. Cholesterol, however, is the starting
material for the formation of other important natural
compounds such as sex hormones.
Many complex molecules which occur in living organisms
(known as natural products) can be made in the laboratory
(synthesized) from simple compounds. Synthesis is one of
the important areas of organic chemistry, and each year
chemists succeed in synthesizing many natural compounds
with very complicated structures. In principle, we should
be able to synthesize any natural organic compound – but
some are extremely large and complex macromolecules
that probably no one will synthesize. The speed and
efficiency with which living organisms synthesize complex
molecules from simple starting materials (carbon dioxide
and water in plants, glucose in animals) is truly amazing.
Biosynthesis means synthesis carried out by living
organisms.
Chapter 19 Alkanes
Period
1
1
2
3
2
H
4
Li
5
Be
B
6
7
C
N
8
O
9
He
10
F
Figure 19.1 The first two periods
of the periodic table.
Ne
Bonding in carbon compounds
What is so special about carbon, just one element from
the periodic table, that we now have a major division of
chemistry devoted to it? An enormous number of carbon
compounds, probably about 20 million, have already been
described. No other element comes close to matching this.
If we look at the periodic table we can see that carbon is
centrally situated in the second row (period 2, see Figure
19.1). Carbon is midway between the alkali metal lithium,
which forms ionic compounds as the cation Li+, and
fluorine, which forms compounds as the fluoride anion F−.
Ionic and covalent compounds
Ionic compounds such as NaCl are held together by the
attraction between the positively charged Na+ cations
and the negatively charged Cl− anions. In solution all
of the cations are, in effect, associated with all of the
anions, and there are no bonds between specific pairs
of ions. In contrast, the covalent bond in a compound
X–Y, where the connecting line represents the covalent
bond with a shared pair of electrons, keeps X joined to
Y permanently (or until the compound decomposes).
There is more about chemical bonding in Chapter 4.
In contrast to lithium and fluorine, carbon forms covalent
bonds – bonds that result from sharing a pair of electrons
between two atomic nuclei. Carbon can form stable bonds
between two carbon atoms and then continue forming
further C–C bonds almost without limit, leading to chains
and networks. An example is
C–C–C–C–C–C–C–C–C–C–C–C–C–C–C–C
which is a C16 chain, found in palmitic acid; palmitic acid
occurs in many natural oils and fats and is used in making
soaps. The C27 framework of cholesterol (Figure 19.2) is more
complicated but still small by comparison with many others.
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
Table 19.1 Selected bond energies
Bond Bond energy / kJ mol−1
H–H
431.0
C–H
414.2
C–C
347.3
C–O
334.7
C–N
284.5
C–S
272.0
S–S
225.9
N–N
163.2
O–O
146.4
A carbon atom has six electrons, four of which are involved
with making four covalent bonds. Carbon is tetravalent. We
can make a simple model of a carbon atom in a molecule
such as CH4 by using a ball as the carbon atom and four
‘arms’ (springs or sticks) to represent the bonds to the
hydrogen atoms (Figure 19.3).
Each carbon–hydrogen bond is formed from one electron
provided by C and one electron provided by H. The
H
These H atoms
are in the plane
of the paper
C
C
H
H
C
C
C
Figure 19.2 The framework of cholesterol contains 27 carbon
atoms.
This H atom is below
the plane of the paper
H
C
C
C
C
C
Look at Table 19.1 for some selected bond energies. Carbon
also readily forms covalent bonds with hydrogen – there
are very few organic compounds that do not contain
hydrogen. Organic compounds frequently contain C–O
bonds and C–N bonds. Bonds to other elements such as
halogens can be formed, but natural organic compounds
most commonly contain the elements C, H, O and N.
Bonds such as S–S, N–N and O–O are all weaker than the
bonds to carbon.
C
C
C
Other elements do not form stable bonds of this sort. For
example, oxygen can form an O–O bond, but it breaks
easily. As a consequence, peroxides such as hydrogen
peroxide, H–O–O–H, are reactive oxidizing agents.
This H atom is above
the plane of the paper
Figure 19.3 Ball-and-spring representation of methane, CH4. The
angle between any two of the bonds radiating from the carbon
atom is 109.5°.
183
Unit 2 Module 1 The chemistry of carbon compounds
electrons in each bond are paired; they have opposite
spins. Negatively charged electrons provide the force
that holds the positively charged atomic nuclei together.
The electron pairs in one bond repel the electrons in a
neighbouring bond. Consequently, the four bonds radiating
from the carbon atom get as far from each other as they
can. This leads to a three-dimensional arrangement with
angles of 109.5° between any two of the bonds. If we draw
lines between the H atoms we find we have outlined a
tetrahedron; the bond angles are, consequently, said to
be tetrahedral (Figure 19.4).
the general formula CnH2n+2 (where n is any integer). They
are said to be saturated hydrocarbons because they each
contain the greatest number of hydrogen atoms for the
given number of carbon atoms. The only bonds between
carbon atoms are single bonds.
Writing formulae
Formulae such as CH3CH3 and CH3CH2CH3 are condensed
formulae.
In displayed formulae the symbols of all atoms are
written and all bonds are drawn as plain lines.
H
H
H
C
H
H
H
Figure 19.4 The tetrahedral shape of
methane, CH4, revealed by drawing
lines between the H atoms.
Methane, CH4, is the simplest hydrocarbon. ‘Hydrocarbon’
means a compound that contains carbon and hydrogen
only.
If we were to join two carbon atoms together with a
covalent bond, using one electron from each carbon, and
then attach hydrogen atoms to all other valence arms, we
would make ethane, H3C–CH3.
H
C
H
H
H
H
C
C
H
H
H
H
If we started with three carbon atoms we would make
propane, H3C–CH2–CH3.
C
H
H
H
H
C
H
H
H
H
H
H
H
H
H
C
H
+
C
H
H
H2
C
C
H
H
ethene, C2H4
C
C
C
The double bond produces unsaturation in the molecule.
If we were to add H2 across the double bond in ethene, we
would produce ethane.
H
H
C
carbon–carbon
double bond
C
H
H
C
H
H
H
H
H
H
C
C
H
H
The simplest unsaturated hydrocarbon is called ethene
because it is derived from ethane. The molecular formula
of ethene is C2H4, corresponding to ethane with two H
atoms removed. We have such good reason to believe that
carbon is normally tetravalent that we look for a way of
forming four bonds to each carbon in ethene. If we join the
two carbon atoms together using two bonds we can write
the formula for ethene as H2C=CH2.
H
H
H
H
Methane, ethane and propane are the first three members
of a class of hydrocarbons, the alkanes. The alkanes have
In the drawings above, ethane and propane are drawn
using the flying wedge convention. The carbon atoms
and the hydrogen atoms bonded by plain lines (–) are
in the plane of the paper. Hydrogen atoms bonded by
) are below the plane of the paper
dashed lines (
) are above the
and the H atoms bonded by wedges (
plane of the paper.
C
184
H
ethane, C2H6
This gives us a satisfactory picture of the simplest member
of a class of unsaturated hydrocarbons called alkenes, of
general formula CnH2n. Each member of the series has one
double bond between two adjacent carbon atoms.
ITQ 1 Answer this question by using these compounds:
(i) C6H14 (ii) C7H15 (iii) C8H16 (iv) C9H22 (v) C10H22 (vi) C11H20
(a) Which are alkanes?
(b) Which are alkenes?
(c) Which are alkynes?
(d) Which can’t exist as a hydrocarbon molecule?
Chapter 19 Alkanes
We can also make a triple bond between two adjacent
carbon atoms. The simplest member of this class of
compounds is ethyne, HC≡CH, corresponding to ethane
with two hydrogen atoms removed from each carbon
atom. The common name of ethyne is acetylene.
H
C
C
more carbon atoms can have a straight chain or a branched
chain. The C4 and C5 alkanes will shows us how this works.
H3C
H
H
H
C
C
H
H
carbon–carbon
triple bond
C
C
+
H
H3C
H2
ethyne, C2H2
H
H
H
C
C
H
H
H2
H
C
C
H
H
H
ethene, C2H4
H
ethane, C2H6
Figure 19.5 Sequential addition of two equivalents of hydrogen
to ethyne gives ethene and then ethane.
Ethyne is more highly unsaturated than ethene. It is the
first member of the alkynes. The general formula of the
alkynes is CnH2n−2.
Alkanes
Structural isomerism
Alkanes have the general formula CnH2n+2. Many alkanes
can be generated by making a chain of CH2 groups and
terminating the chain by adding one more H to each
terminal CH2.
H
H
C
H
m
H
H
C
C
H
H
CH3
H
terminal groups
CH3(CH2)mCH3
The compounds described here are members of a
homologous series, a series of compounds of the same
chemical class that can be formed by adding CH2 units step
by step to the first member of the series.
A carbon atom is not limited to having two attached
carbon atoms and can be linked to three or four other
carbon atoms. Instead of a straight-chain alkane, we then
have molecules with branched chains. A C3 compound
can have only a straight chain, but an alkane with four or
H3C
H
C
C
H
H
H3C
butane, C4H10
C
CH3
H
1
H3C
If we add one molecule of H2 across the triple bond of
ethyne we produce ethene. Adding another molecule of H2
would convert ethene to ethane (Figure 19.5).
H
CH3
H
H
H
C
C
C
H
H
H
CH3
2
pentane, C5H12
3
CH3
CH3
H3C
4
C
CH3
CH3
5
We can see that there are two alkanes with the formula
C4H10. Compound 1 has a straight chain and 2 has a
branched chain. These two compounds are said to be
isomers. Isomers are compounds that have the same
molecular formula but differ in some way. There are several
ways in which compounds may differ, leading to isomerism
of several types. Here, the two isomers of C4H10 differ in
structure (the way the atoms are connected) so they are
said to exhibit structural isomerism. There are three
structural isomers of formula C5H12. There is the straightchain compound 3, the branched-chain compound 4 and
compound 5 which is doubly branched.
The first two isomers of C5H12, 3 and 4, could be constructed
by replacing one H in 1 with a CH3 group. If you examine 1
carefully, you will see that it contains two distinct types of
hydrogen atoms. There are the six that are part of the –CH3
groups and the four that are part of the –CH2– groups. All
six CH3 hydrogen atoms are equivalent, and replacement of
any one of them by CH3 leads to compound 3. In the same
way, the four CH2 hydrogen atoms are all equivalent, and
replacement of any one of them leads to compound 4.
In isomer 2 all nine CH3 hydrogen atoms are equivalent, and
replacement of any one of them by CH3 leads to 4. There is
one unique H in 2, and replacement of that by CH3 leads to 5.
Nomenclature of alkanes
When alkanes such as those we looked at above were
discovered, they were assigned names that seemed logical
at the time.
■ Compound 1 was called butane and its isomer 2 was
named isobutane.
■ Compound 3 was named pentane and, since it had
two isomers, two different prefixes were required to
distinguish them: 4 was called isopentane and 5 was
called neopentane.
185
186
Unit 2 Module 1 The chemistry of carbon compounds
by the suffix -ane. The prefix ‘n-’, as an abbreviation for
‘normal’ is often used with the names of straight-chain
alkanes containing four or more carbon atoms.
Table 19.2 C1 to C10 straight-chain alkanes
Number of C atoms
Alkane
Condensed formula
1
methane
CH4
2
ethane
CH3CH3
3
propane
CH3CH2CH3
4
butane
CH3CH2CH2CH3
5
pentane
CH3(CH2)3CH3
6
hexane
CH3(CH2)4CH3
7
heptane
CH3(CH2)5CH3
8
octane
CH3(CH2)6CH3
9
nonane
CH3(CH2)7CH3
10
decane
CH3(CH2)8CH3
1
methyl
CH3–
Me
Alkyl groups (also known as radicals) are formed by
removing one hydrogen atom (H) from the alkane. These
alkyl groups normally appear attached to another group
in derivatives of the alkane. The name given to an alkyl
group formed by removing a terminal H from a straightchain alkane is that of the alkane with -ane replaced with
-yl (Table 19.3). Thus, methane gives methyl, ethane gives
ethyl, propane gives propyl, and so on. These names apply
only when the H is removed from a terminal position and
the alkane has a straight chain. The abbreviations Me, Et,
Pr and Bu are commonly and conveniently used for the
first four alkyl groups.
2
ethyl
CH3CH2–
Et
Branched-chain alkanes are treated differently.
3
propyl
CH3CH2CH2–
Pr
4
butyl
CH3CH2CH2CH2–
Bu
5
pentyl
CH3(CH2)3CH2–
–
6
hexyl
CH3(CH2)4CH2–
–
7
heptyl
CH3(CH2)5CH2–
–
8
octyl
CH3(CH2)6CH2–
–
9
nonyl
CH3(CH2)7CH2–
–
10
decyl
CH3(CH2)8CH2–
–
Table 19.3 C1 to C10 straight-chain alkyl groups
Number of C atoms
Alkyl
Alkyl condensed formula
Abbreviation
When we look at alkanes with more than five carbon atoms,
the number of isomers increases much more rapidly than
the number of C atoms. Distinguishing isomers by prefixes
becomes much too cumbersome. A more systematic naming
system had to be found, so we shall look at that next.
Before we do, we can observe that, like people, organic
compounds are often known by two names: a formal or
systematic name, and a common nickname which we can
use when we are talking to friends. Names like ‘isobutane’
are now effectively nicknames.
The International Union of Pure and Applied Chemistry
(IUPAC) assigned the responsibility for designing an
unambiguous nomenclature system to a committee that
produced the IUPAC system of nomenclature that is widely
accepted today.
Straight-chain alkanes were assigned the names that had
been used for them for many years. The first four names,
methane, ethane, propane and butane, had been used
from very early times. The names of alkanes with five or
more carbon atoms were based on the Greek or Latin word
for the number of carbons present (Table 19.2) followed
ITQ 2 Draw structures of as many isomers as you can with the
formula C7H16 and label them with their systematic names.
1 Identify the longest chain of carbon atoms in the
molecule. Name the compound as a derivative of this
straight-chain alkane. Thus, for
H
H3CCH2CH2
C
CH2CH3
CH3
the longest chain is six carbons and there is a methyl
group. This compound is named as a derivative of
hexane. Butane and pentane chains can also be seen,
but the longest chain, hexane, is chosen.
2 Identify the group or groups attached to the longest
or parent chain; in a formal sense hydrogen atoms on
the parent hydrocarbon are regarded as having been
substituted by these groups, so the groups are known
as substituents. The substituent here is a methyl
group, so this alkane is a methylhexane. This name is
ambiguous because the position of the methyl group
is undefined. We remove the ambiguity by assigning a
position, or locant, to the substituent by numbering the
parent chain starting from one end. We number the
carbon atoms so that the position of the substituent is
assigned the smallest number possible. Numbering this
compound from the right gives the locant 3 (not 4):
6 5
4
3
2
1
H
H3CCH2CH2
C
CH2CH3
CH3
1 2
3
4
5
6
The systematic name for this alkane is 3-methylhexane.
3 When the longest chain carries more than one
substituent, we number the chain in the way that gives
one of the substituents the smallest possible locant,
Chapter 19 Alkanes
and the numbers assigned to the other substituents
follow. Thus in
CH3
2
4
H
H
C
CH2
C
CH3
Bonding in alkanes: a detailed description
We have already looked briefly at how and why the
covalent bonds of the carbon atoms in alkanes form with
the maximum angular distance between them, leading to
tetrahedral geometry. We know that the atomic number of
carbon is six, and the electronic configuration of an isolated
carbon atom in the ground state is 1s2 2s2 2p2.
CH2CH3
CH2CH3
there is a methyl group at position 2 and an ethyl
group at position 4 (not 3 and 5, respectively). The
name is written with the substituents placed in
alphabetical order (not numerical order), so the name
is 4-ethyl-2-methylhexane.
Spectroscopic results tell us that an isolated carbon atom
has two kinds of atomic orbitals in its valence shell. There
is an s orbital that forms a sphere around the nucleus and
has no directional characteristics. There are also three p
orbitals, shaped like dumbbells directed at right angles to
each other along x, y and z axes (Figure 19.6).
If two substituents are attached to the same carbon
atom, the locant is shown for each substituent. Thus:
CH3
CH3CH2CH2
C
z
CH2CH3
is 3-ethyl-3-methylhexane.
CH2CH3
y
If the same alkyl substituent occurs twice, the prefix
‘di-’ is used to form the name, and the locant for each
substituent is shown as before.
H
CH3
x
H
C
CH2
s orbital
C
CH3
CH2CH3
px , py and pz orbitals
is 2,4-dimethylhexane.
CH3
Figure 19.6 s and p orbitals: shapes and directionality.
CH3
CH3CH2CH2
C
p orbital
In the isolated atom, two of the four valence electrons of
carbon should first fill the 2s orbital, and the remaining
two electrons should populate 2p orbitals, which are at
higher energy. From this we might expect the carbon atom
to form two bonds at right angles to each other. However,
CH2CH3
is 3,3-dimethylhexane.
CH3
ITQ 3 Assign IUPAC names to each of the structures a to f.
What is the relationship between these compounds?
Are they structural isomers or are they identical?
a
H
H
b
H
H
H
H
H
H
H
H
C
C
C
C
C
C
C
C
H
H
H
H
H
H
H
H
H
H
c
H
H H
C
H H
H
H
C
C
C
C
C
C
H
H
H
H
H
C
H
H
H
H
H
H
H
H
H
H
C
C
C
C
C
C
H
H
H
H
H
H
C
C
H
H
H
H
H
H
d
H
H
H
C
C
H
e
H
H
H
H
H
C
C
C
C
C
H
H
H
H
C
H
H
H
H
H
H
H
C
C
H
H H
H
C
H
H
H
H
C
C
C
C
H
C
H H
H
H
H
H
f
H
H
H
H
H
H
C
H
H
C
C
C
H
H
H
C
H
H
C
C
C
H
H
H
H
H
H
187
188
Unit 2 Module 1 The chemistry of carbon compounds
z
Energy
these sp3 orbitals are
in the plane of the paper
y
this sp3 orbital is behind
the plane of the paper
x
hybridization
this sp3 orbital is in front
of the plane of the paper
2px , 2py , 2pz orbitals
four sp3 hybrid orbitals
with tetrahedral angles of
109.5˚ between them
2s orbital
we know that this is not the case. Four tetrahedral bonds
are formed, so they must be formed from four orbitals with
tetrahedral angles. If we imagine that one of the 2s electrons
is promoted to a 2p orbital we now have four unpaired
electrons occupying atomic orbitals in the valence shell.
When bonds are formed in compounds of carbon, these
atomic orbitals are hybridized to give molecular orbitals.
In alkanes, the tetrahedral molecular orbitals of carbon
are formed by mixing (hybridizing) the non-directional
s orbital with the three directional p orbitals to give four
hybrid orbitals with both shape and direction (Figure 19.7).
The carbon orbitals are said to form an sp3 (tetrahedral)
hybrid. When a C atom and an H atom combine, their
orbitals overlap and form a molecular orbital which
contains the two bonding electrons. It is these electrons
which make up the bond. In the alkanes it is the end of a
p orbital of the C atom which overlaps with the s orbital
of the H atom, head-on (Figure 19.8). Such bonds are
described as sigma (σ) bonds, because their symmetry is
related to that of an s atomic orbital (σ is Greek for ‘s’). In
an s orbital the electron density forms a sphere around the
atomic nucleus. In a σ molecular orbital electron density is
highest between the two nuclei and is symmetrical around
the line joining them. In structural drawings, the bond is
represented by a line.
H
space occupied by electrons
of opposite spin; one electron can
be thought to originate from H and
one from C
C
H
H
H
Figure 19.8 Structure of methane: the electrons comprising the
vertical C–H sigma (σ) bond are shown; the remaining three C–H
sigma (σ) bonds are represented by lines.
Figure 19.7 Hybridization of
2s with 2px, 2py and 2pz orbitals.
capped by H atoms. Now imagine that, instead of adding
those terminal H atoms, we were to join the two ends of
the chain together to form a ring. The type of compound
we end up with can be represented by cyclohexane.
The structure of cyclohexane can be given in a shorthand
version as
Each corner of the structure is a carbon atom and carries
sufficient hydrogen atoms to satisfy the valence of the
carbon atom. In cyclohexane, each corner represents a
–CH2– group. The formula of cyclohexane is C6H12.
Chemically, these cyclic products are very similar to the
straight-chain alkanes.
Physical properties, sources and uses of
alkanes
The straight-chain alkanes from methane to decane,
C1–C10, which we have discussed constitute a homologous
series. The individual members of a homologous series are
known as homologues. The homologues in a series differ
in the number of –CH2– groups between the terminal –CH3
groups, so the molecular weights increase by 14 amu as
we move up the series. This increase in relative molecular
mass (RMM) and molecular size is reflected in changes in
physical properties; examples are the physical state and
higher boiling points as well as increases in density and
viscosity.
Methane
■ CH4, RMM = 16.
■ Colourless gas at RTP, boiling point is −164 °C.
Cycloalkanes
When we discussed straight-chain alkanes we described
them as chains of –CH2– groups with the terminal groups
■ Found in natural gas, coal gas, gases from oil wells,
cracked petroleum, decay of organic material in
swamps and marshes, fermented sewage sludge.
Chapter 19 Alkanes
■ Used as fuel, synthesis of ‘carbon black’ for use in
printing, rubber industry, starting material in chemical
synthesis.
Ethane
■ CH3CH3, RMM = 30.
■ Colourless gas at RTP, boiling point is −89 °C.
■ Found in natural gas, gases from oil wells, cracked
petroleum.
■ Used as fuel.
Propane
■ CH3CH2CH3, RMM = 44.
■ Colourless gas at RTP, boiling point is −44.5 °C.
■ Found in natural gas, gases from oil wells, cracked
petroleum.
■ Used as fuel.
Butane
■ CH3CH2CH2CH3, RMM = 58.
■ Colourless gas at RTP, boiling point is −0.5 °C.
■ Found in natural gas, gases from oil wells, cracked
petroleum.
■ Used as fuel.
Pentane
■ CH3(CH2)3CH3, RMM = 72.
■ Colourless liquid at RTP, boiling point is 36 °C.
■ Found in natural gas, gases from oil wells, cracked
petroleum.
■ Used as solvent, starting material in chemical synthesis.
Hexane
■ CH3(CH2)4CH3, RMM = 86.
■ Colourless liquid at RTP, boiling point is 69 °C.
■ Found in natural gas, cracked petroleum.
■ Used as solvent, especially for extraction of edible oils
from seed crops.
Heptane
■ CH3(CH2)5CH3, RMM = 100.
■ Colourless liquid at RTP, boiling point is 98.4 °C.
■ Found in petroleum fractions.
■ Used as solvent.
Octane
■ CH3(CH2)6CH3, RMM = 114.
■ Colourless liquid at RTP, boiling point is 125.5 °C.
■ Found in petroleum fractions.
■ Used as fuel component, solvent.
Nonane
■ CH3(CH2)7CH3, RMM = 128.
■ Colourless liquid at RTP, boiling point is 151 °C.
■ Found in petroleum fractions.
■ Used as gasoline and jet fuel component.
Decane
■ CH3(CH2)8CH3, RMM = 142.
■ Colourless liquid at RTP, boiling point is 174 °C.
■ Found in petroleum fractions.
■ Used as fuel component.
Solubility of alkanes
‘Like dissolves like’ is a useful guideline in predicting
solubility. Alkanes are non-polar compounds and water is
a polar solvent, so alkanes, and hydrocarbons in general,
are insoluble in water. Alkanes are soluble in each other
and in less polar solvents such as ethanol and ether. Their
solubility decreases with increasing molecular weight.
Reactions of alkanes: an introduction
Alkanes react with very few of the reagents used in organic
chemistry.
Combustion in oxygen
In older days alkanes were called paraffins, a name derived
from Latin (parum affinis) and meaning that they had
little affinity or reactivity. Our everyday use of the name
‘paraffin’ describes a fluid that is used to produce heat by
burning it. Alkanes are used as fuels. They react vigorously
and exothermically with oxygen when ignited. In coal
mines this property of methane causes violent explosions.
The products of complete combustion of hydrocarbons
are carbon dioxide and water. For methane, ethane and
propane the balanced equations are:
CH4 + 2O2 → CO2 + 2H2O + heat
2C2H6 + 7O2 → 4CO2 + 6H2O + heat
C3H8 + 5O2 → 3CO2 + 4H2O + heat
189
190
Unit 2 Module 1 The chemistry of carbon compounds
In complete combustion all the bonds in the hydrocarbon
are broken. The heat evolved is therefore related to the
differences between strengths of the bonds in the starting
material and the strengths of the bonds in the products.
Heats of combustion can therefore provide valuable
information about molecular structure.
If not enough oxygen is available to completely oxidize the
carbon to CO2, then the toxic gas carbon monoxide (CO) is
formed. This is shown for propane:
C3H8 + 4O2 → CO2 + 2CO + 4H2O + heat
Almost all organic compounds will produce carbon dioxide
and water on complete combustion. This is the basis of
an old analytical method, known as combustion analysis,
for determining the empirical formula of a molecule.
In combustion analysis a known weight of a compound
is subjected to complete combustion and the quantities
of carbon dioxide and water which are produced are
separated and the weights obtained using an automated
instrument called a C,H analyser. The weights of carbon
and hydrogen in the sample are calculated, and from these
the percentages of carbon and hydrogen in the compound
are obtained. You have seen how to do this in Chapter 6
(page 60).
Cracking
Cracking is the process in which large hydrocarbon
molecules are broken down into mixtures of smaller, more
useful, alkanes and alkenes. Cracking can be carried out at
high temperatures and pressures without catalysts (thermal
cracking), or with catalysts at lower temperatures and
pressures (catalytic cracking). In the petroleum industry
cracking of heavier fractions obtained from distillation of
crude oil yields diesel oils, gasoline and kerosene.
In thermal cracking a C–C bond of an alkane may break to
give two fragments in which one electron of the electron
pair comprising the covalent bond is located on each
fragment. This is known as homolytic bond cleavage,
and the products are alkyl radicals (Figure 19.9).
H
H
C
C
H
heat
The alkyl radicals then lose a hydrogen atom (H•) and may
rearrange to produce hydrocarbons of smaller molecular
weights.
Substances known as zeolites are commonly used as
catalysts in catalytic cracking (Figure 19.10). Zeolites are
microporous materials comprised of aluminium, silicon and
oxygen (aluminosilicates) and incorporating cations such
as Na+, K+, Ca2+ and Mg2+. Many zeolites occur naturally
in mineral deposits; the zeolites used in the petrochemical
industry are synthetic.
H
H
C
C
H
H
zeolite
C
H
H
C
loss of H
rearrangement
H
two alkyl radicals
Figure 19.9 Thermal cracking of an alkane.
C
C
+
H
+
loss of H
fragmentation
rearrangement
lower RMM
hydrocarbons
Figure 19.10 Catalytic cracking of an alkane.
Halogenation
If a bromine molecule, Br2, dissociates into two Br atoms,
reaction with an alkane can be initiated. The dissociation
step requires energy, provided by heat (a thermal reaction)
or light (a photochemical reaction). One Br atom then
removes a hydrogen atom from the alkane to form one
molecule of HBr and an alkyl radical, which then reacts
with another Br atom to form a bromoalkane (Figure
19.11).
In practice this reaction is difficult to control. We can not
be sure which H will be replaced by Br in more complex
alkanes, and the reaction may replace more than one
hydrogen by Br, leading to a mixture of products. For us
at present, the reaction serves simply to introduce us to
derivatives of alkanes and to substitution reactions.
Br
Br
H
H
C
C
H
H
HBr
lower RMM
hydrocarbons
+
H
carbocation
energy
+
part of an alkane
H
H-
part of an alkane
part of an alkane
CnH2n+2
H
zeolite
H
+
2 Br
Br
H
H
C
C
Br
H
an alkyl radical
CnH2n+1
Figure 19.11 Free-radical bromination of an alkane.
H
H
C
C
H
Br
alkyl bromide
CnH2n+1 Br
Chapter 19 Alkanes
Summary
✓ Carbon is tetravalent and, due to its central
position in the second row of the periodic table,
forms stable covalent carbon–carbon bonds
(single and multiple) and compounds consisting
of carbon chains and rings.
✓ Acyclic alkanes have the general formula CnH2n+2;
the general formula of alkenes is CnH2n; the
general formula of alkynes is CnH2n−2.
✓ Alkanes are known as saturated hydrocarbons
because they contain the greatest number of H
atoms for a given number of C atoms. Alkenes
and alkynes are unsaturated hydrocarbons.
✓ There are structural isomers of alkanes with four
and more carbon atoms.
✓ Alkanes, and all organic compounds, are named
unambiguously using a system of nomenclature
developed by the International Union of Pure
and Applied Chemistry (IUPAC rules).
✓ Carbon–carbon single bonds and carbon–
hydrogen bonds are known as sigma (σ) bonds.
These bonds represent molecular orbitals which
are combinations of atomic orbitals.
✓ In alkanes the four σ bonds from each carbon
atom point toward the corners of a regular
tetrahedron.
✓ In alkanes each carbon atom has four sp3 hybrid
orbitals which are formed by mixing one 2s with
three 2p orbitals.
✓ Cyclic saturated alkanes are known as
cycloalkanes and are named using the IUPAC
system.
✓ The lower homologues in the linear alkane series
occur in natural gas and petroleum fractions and
are used mainly for fuel and as solvents.
✓ The products of complete combustion of
hydrocarbons are CO2 and H2O and the quantities
of CO2 and H2O produced by burning a given
mass of a hydrocarbon can be used to calculate
the empirical formula.
✓ Large hydrocarbon molecules can be broken
down into smaller alkanes and alkenes by
thermal or catalytic cracking.
✓ Alkanes are unreactive compounds, but can
be made to react with halogens, e.g. bromine,
to form bromoalkanes in which H atoms are
replaced by Br.
191
192
Unit 2 Module 1 The chemistry of carbon compounds
Review questions
Answers to ITQs
1
Which types of orbitals overlap to form:
(a) the C–C bond in ethane?
(b) the C–H bond in ethane?
1
2
Write displayed structures and give systematic names
for the three structural isomers of C5H12.
3
Write condensed structural formulae for each of the
following structures.
(a)
H
(b)
H
2
(a)
(b)
(c)
(d)
C6H14 and C10H22 are alkanes.
C8H16 is an alkene.
C11H20 is an alkyne.
C7H15 can’t exist as a hydrocarbon (needs an even
number of H); C9H22 can’t exist as a hydrocarbon
(has too many H).
i
ii
n-heptane
H
H
H
H
C
C
C
C
H
H
H
H
H
H
H
H
H
C
C
C
C
C
H
H H
C
H H
H
H
iii
2-methylhexane
iv
2,2-dimethylpentane
3-methylhexane
H
v
vi
H
(c)
H
H
H
C
C
H
4
H
2,4-dimethylpentane
2,3-dimethylpentane
H
N
vii
viii
H
3,3-dimethylpentane
For each of the following condensed formulae:
3-ethylpentane
CH3CH(CH3)CHBrCH3 and CH3C≡CCH2CH=C(CH3)2
ix
(a) write the displayed formula;
(b) draw the line formula.
5
Convert each of the following line drawings to a
displayed formula, showing each atom and each bond.
(a)
(b)
(c)
O
6
What can you deduce about the possible structures
represented by each of the following molecular
formulae?
(a) C14H30
(b) C12H24
(c) C3H7
7
Draw structures for:
(a) n-butylcyclopentane
(b) 1,3,5-triethylcyclohexane
(c) 1,1,2-trimethylcyclobutane
8
Ascorbic acid (vitamin C) contains carbon, hydrogen
and oxygen. Combustion analysis of 0.528 g of
ascorbic acid gave 0.792 g of CO2 and 0.216 g of H2O.
(a) Calculate the empirical formula of ascorbic acid.
(b) If the molecular weight of ascorbic acid is
176 amu, what is the molecular formula?
2,2,3-trimethylbutane
3
a
b
c
d
e
f
n-octane
2,4-dimethylhexane
3-ethylhexane
2,5-dimethylhexane
2,4-dimethylhexane
2,2,4-trimethylpentane
b and e are identical; all are structural isomers (all are
C8H18).
193
Chapter 20
Alkenes and alkynes
Learning objectives
■ Describe in detail the bonding in alkenes and alkynes.
■ Define the terms sp2 hybrid orbital, sp hybrid orbital, stereoisomer, geometric isomer.
■ Describe and account for the trigonal and linear shapes of alkenes and alkynes.
■ Account for the rigidity and reactivity of carbon–carbon multiple bonds.
■ Systematically name alkenes, cycloalkenes and alkynes.
■ Write or draw, from molecular formulae or systematic names, the structures of alkenes, cycloalkenes
and alkynes using the following formats: displayed structures, condensed formulae, line drawings.
■ Describe the physical properties, sources and uses of C1 to C4 alkenes and alkynes.
■ Describe the outcome of addition of X2 and H2 to alkenes and alkynes.
Introduction
Alkenes
Carbon forms compounds with double and triple bonds
between carbon atoms, known as alkenes and alkynes
respectively. Alkenes are sometimes called olefins; olefin
is an old name for ethene, H2C=CH2, which also used to be
known as ethylene. Alkynes are also known as acetylenes;
acetylene is the trivial name of the simplest alkyne, HC≡CH.
Bonding in alkenes
The ground-state electronic configuration of carbon is
1s2 2s2 2p2. As we did when we were describing bonding
in alkanes (Chapter 19), we will first uncouple the 2s2
electrons and promote one to a 2p orbital. The electronic
configuration of carbon is now 1s2 2s1 2px1 2py1 2pz1. We can
Compounds containing multiply bonded carbon atoms are now hybridize the atomic s orbital of carbon with two p
said to be unsaturated compounds because other atoms can orbitals (instead of the three in sp3), and we will choose the
be added to them across the carbon–carbon bonds. In this px and py orbitals for this purpose. The hybridized orbitals
chapter we will look closely at some simple unsaturated will also be in the xy plane and, since we started with three
atomic orbitals, we will get three hybridized orbitals (Figure
hydrocarbons.
20.1). These orbitals are said to be sp2 hybridized; they
lie in the xy plane with equal
z
Energy
angles of 120° between them,
and they are described as
y
trigonal because they possess
node
three-fold symmetry about
y
x
the z axis. This means that you
hybridization
2px , 2py orbitals
unhybridized
can rotate the orbitals about
x
2pz orbital
that axis and they will come
three sp 2 hybrid orbitals directed towards the
into an identical position three
corners of an equilateral triangle in the x, y plane;
2s orbital
the angles between the orbitals are 120˚
times in one revolution. The pz
orbital has not been used yet,
Figure 20.1 Hybridization of 2s with 2px and 2py orbitals to give three sp2 atomic orbitals in a
so we shall save it for use later.
carbon atom.
194
Unit 2 Module 1 The chemistry of carbon compounds
Each sp2 orbital can be used to produce a σ bond by
overlapping head-on with an orbital on another atom.
In alkenes, we start by making a σ bond between two
carbon atoms with sp2-hybridized orbitals. Then we use the
remaining four hybridized orbitals, two on each carbon, to
make σ bonds to hydrogen atoms (Figure 20.2).
H
H
H
C
H
C
C
C
H
H
nodal
plane
H
atomic orbitals the
pz orbital on each C atom
is above and below the plane
H
overlap atomic orbitals
to give molecular orbitals
H
H
H
H
H
H
H
H
overlap p orbitals sideways
to give / molecular orbital
redraw and rotate to show
m bonds and pz orbitals
Figure 20.2 Bonding in ethene: there is a σ bond between
the two carbon atoms and the so-far unused pz lobes have
overlapped sideways to form a π bond
Figure 20.2 shows the two carbon atoms with sp2-hybridized
orbitals that overlap to form the σ bonded framework of
H2C–CH2. The framework is shown again with the bonds
represented by lines (to simplify the picture) and rotated
so that we can see the two pz orbitals side by side, at right
angles to the plane of the molecule. When these atomic
orbitals (AOs) overlap sideways they lead to a π molecular
orbital. The electrons are concentrated above and below the
C–C sigma bond because the sp2 hybrids, like the atomic p
orbitals from which they were formed, have a central point
(the node) at which the electron density is zero.
bond is spread above and below the plane of the molecule,
as illustrated for ethene in Figure 20.2. Electrons in a π
bond are more easily pushed or pulled than the electrons
in a σ bond. We can say that the π bond is more polarizable
than the σ bond. The chemical consequence of this is that
the π bond is more reactive than the σ bond.
Rotation about σ and π bonds
In ethane, H3C–CH3, rotation about the single C–C σ bond
meets little resistance, and at room temperature the C–C
bond rotates freely. However, in ethene, H2C=CH2, the
C=C bond resists rotation. This is because the two atomic
pz orbitals must be side by side and pointing in the same
direction for the π molecular bond to form. Rotation
would, therefore, break the π bond. A considerable
amount of energy would be required to do this. So at
room temperature the C=C bond is rigid, and the relative
positions of the groups attached to it are fixed.
Isomerism and nomenclature of alkenes
We can construct other alkenes by replacing one or more
of the H atoms of H2C=CH2 with alkyl groups. If we replace
just one H with an alkyl group, it does not matter which H
is replaced because all four hydrogen atoms in ethene are
the same (they are equivalent). So, for example, there is
only one compound CH3CH=CH2. This compound contains
three carbon atoms, so it is a derivative of propane, and is
given the name propene. In naming an alkene, we replace
-ane of the alkane with -ene.
If we replace an H of H2C=CH2 with ethyl instead of methyl,
the product, CH3CH2CH=CH2, contains four carbon atoms
and is a butene. If we replace one H on each of the carbons
A node is a region of space where the mathematical
sign of the wave function which describes the electron
density changes. We have used a change of colour to
show the change of sign. The significance for us is that
the electron density is zero at the node.
these H’s are on one side of the C
H
H
C
120˚
One is a liquid at room temperature and is similar in behaviour
to most alkenes. One has been prepared only at very low
temperature. At −100 °C it undergoes a spontaneous change.
(b) Which is which? Suggest an explanation.
120˚
H
these H’s are on the other
side of the C C
ethene; the C
C does not rotate
H
H
(a) Draw the structures of cyclopentene and cyclopropene.
C
H
The carbon–carbon double bond is made up from a σ bond
and π bond. These two bonds are different in character.
Most of the electron density in the σ bond is concentrated
between the two C atoms; the electron density in the π
ITQ 1
C
C
H3C
H
H3C
C
CH3
one isomer of 2-butene is derived
from ethene by replacing both
hydrogens from one side of the
C C by CH3 groups
Figure 20.3 Isomers of 2-butene.
C
H
C
CH3
another isomer of 2-butene is
derived from ethene by replacing
one H from each side of the
C C by CH3 groups
Chapter 20 Alkenes and alkynes
z
Energy
y
x
hybridization
2px orbital
of ethene with a methyl group the product, CH3CH=CHCH3,
also contains four carbon atoms and is also a butene. So the
name ‘butene’ is ambiguous; there are isomeric butenes.
Isomers are compounds containing the same atoms,
differently arranged. The double bond can be at the end
of the four-carbon chain or in the middle, CH3CH2CH=CH2
or CH3CH=CHCH3. We need to use a number to describe
where the double bond is located in each of these isomers.
The number is assigned by numbering the carbon chain in
the way that gives the carbon atom at one end of the double
bond the smallest possible number. So CH3CH2CH=CH2 is
1-butene. Its isomer, CH3CH=CHCH3, is 2-butene.
However, there are two different ways of making the
replacement of H by methyl, and both give a 2-butene, so
this name is still ambiguous. The second methyl substituent
can be on the same side of the double bond as the first, or it
can be on the opposite side (Figure 20.3).
We cannot convert one of these isomers to the other by
a rotation because the double bond is rigid and resists
rotation. They are different compounds. To name these
isomers we use the prefix cis, which is Latin for ‘on the same
side’, to describe the first, and trans, which is Latin for ‘on
the opposite side’, for the second (Figure 20.4). These two
isomers have the same bond structure. In these isomers the
atoms are connected in the same sequence. The difference
that makes these compounds isomers is the arrangement
of the atoms in space, i.e. the spatial arrangement of their
atoms.
H
C
cis-2-butene
CH3
H
C
H3C
C
CH3
Figure 20.6 Hybridization of 2s
and 2px orbitals to give two sp
atomic orbitals in carbon.
two sp hybrid orbitals directed in
the x and -x directions; the angle
between the orbitals is 180˚
2s orbital
H
unhybridized
2py and 2pz orbital
x
H3C
C
H
position 1; the number isn’t always needed though (Figure
20.5). If a substituent is present, the numbering chosen is
that which gives the substituent the lower number.
cyclohexene
Figure 20.5 Naming cycloalkenes.
Alkynes
Bonding in alkynes
An alkyne is characterized by the presence of a carbon–
carbon triple bond in the molecule. In ethyne, the first
member of the series, both carbon atoms are sp hybridized.
If we assume that it is the px atomic orbital that mixes with
the s orbital, two hybridized orbitals result, one pointing
in the x direction and one pointing in the opposite (−x)
direction (Figure 20.6).
An sp hybridized orbital on one C atom overlaps with an sp
hybridized orbital on the another C to form a C–C σ bond.
Each remaining sp hybridized orbital forms a σ bond to H.
The H–C–C–H framework produced is linear. The py and
pz orbitals on the two C atoms are then able to overlap to
produce two π orbitals at right angles to each other (Figure
20.7).
z
x
C C m bond formed
by overlap of two sp
hybrid orbitals, one
from each C atom
y
H
H
C
trans-2-butene
C
Figure 20.4 Named isomers of 2-butene.
C
H
H
Cycloalkenes
Double-bonded derivatives of cycloalkanes are called
cycloalkenes and are named systematically in the same way
as other alkenes, with one end of the double bond being
C H m bond formed by
overlap of sp hybrid orbital
of C and s orbital of H
Figure 20.7 Bonding in alkynes.
two C C / bonds formed by
overlap of lobes of unhybridized
2py and 2pz orbitals
195
196
Unit 2 Module 1 The chemistry of carbon compounds
Higher alkynes have one or both of the H atoms of ethyne
replaced with alkyl groups. The carbon atoms of the triple
bond and the atoms, of whatever element, directly attached
to the sp hybridized carbon always form a straight line.
Nomenclature of alkynes
The nomenclature of alkynes follows the same pattern as
used for alkanes and alkenes, except that the -ane or -ene
ending is replaced by -yne. Unlike alkenes, though, we do
not have to describe geometric isomers. Just as with the
alkenes, the numbering system assigns the smallest possible
number to one of the sp hybridized carbons (Figure 20.8).
HC
CCH2CH3
H3CC
1-butyne
CCH3
2-butyne
H3CC
CCH2CH2CH3
2-hexyne
Figure 20.8 Naming alkynes.
Physical properties, sources and uses of
alkenes and alkynes
As with alkanes, the lower homologues in the alkene
and alkyne homologous series are gases. As we move up
the series the physical states of the compounds change to
liquids and then to solids. C2–C4 alkenes are gases, C5–C17
alkenes are liquids and alkenes with 18 and more carbon
atoms are solids. A summary of some properties, sources
and uses of C2–C4 alkenes is given in Table 20.1 and in Table
20.2 for the C2–C4 alkynes.
Ethene
Alkenes
Alkenes contain carbon–carbon double (C=C) bonds.
The double bond is reactive because the electrons in the
π bond are more mobile than those in a σ bond, so they
interact more readily with chemical reagents. The most
characteristic reaction of an alkene is addition. If bromine,
Br2, is added to a solution of an alkene in an inert solvent, the
characteristic orange colour of bromine disappears rapidly,
and the alkene is converted to a dibromoalkane. Br2 has
added across the double bond. The two Br atoms are found
to have added to opposite sides of the double bond. The two
Br atoms are added one after the other, and the second Br
comes from the opposite side to avoid the first. This is most
readily seen if we carry out the addition to a cycloalkene.
Figure 20.9 shows the addition of bromine to cyclohexene.
The product is trans-1,2-dibromocyclohexane.
bond is above the page
Br
H
Br
+
Br
H
Br
the cyclohexene ring
is in the plane of
the paper
bond is below
the page
trans-1,2-dibromocyclohexane
Br
Br
+
Br
the cyclohexene ring
is perpendicular
to the plane of
the paper
Br
trans-1,2-dibromocyclohexane
Figure 20.9 Addition of bromine to cyclohexene.
Table 20.1 Properties, sources and uses of C2–C4 alkenes
Name
An introduction to the reactions of
alkenes and alkynes
Propene
But-1-ene
But-2-ene
Condensed formula CH2=CH2
CH3CH=CH2
CH3CH2CH=CH2
CH3CH=CHCH3
RMM
28
42
56
56
Physical state
Colourless gas
Colourless gas
Colourless gas
Colourless gas
Boiling point / °C
−105
−48
−6
1
Sources
Natural gas, coal gas, gases from oil wells, cracked
petroleum
Fruit ripening, anaesthetic, manufacture of plastics,
synthesis of solvents
Natural gas, gases from oil
wells, cracked petroleum
Manufacture of plastics,
synthesis of solvents
Natural gas, gases from oil
wells, cracked petroleum
Synthesis of solvents
Natural gas, gases from oil
wells, cracked petroleum
Synthesis of solvents
Uses
Table 20.2 Properties, sources and uses of C2–C4 alkynes
Name
Ethyne
Propyne
But-1-yne
But-2-yne
Condensed formula HC≡CH
CH3C≡CH
CH3CH2C≡CH
CH3C≡CCH3
RMM
26
40
54
54
Physical state
Colourless gas
Colourless gas
Colourless gas
Colourless liquid
Boiling point / °C
−84
−23
8
27
Sources
Natural gas, cracking of methane/ethane mixtures,
Cracking of propane
from calcium carbide CaC2 + 2H2O → C2H2 + Ca(OH)2
Used in welding (oxy-acetylene blowpipe), starting
Rocket fuel, starting material
material in chemical synthesis
in chemical synthesis
Synthesized from ethyne
Synthesized from ethyne
Starting material in chemical
synthesis
Starting material in chemical
synthesis
Uses
Chapter 20 Alkenes and alkynes
CH3
CH
CH2
+
H2
CH3
propene
CH2
CH3
propane
bond is above the page
CH3
H
catalyst
+
H
CH3
H
bond is below
the page
H
CH3
CH3
the cyclopentene ring
is in the plane of
the paper
cis-1,2-dimethylcyclopentane
Alkynes
In alkynes the π bonds react in a way similar to the π bonds
of alkenes. If we count the double bond of an alkene as
one unit of unsaturation then an alkyne has two units of
unsaturation. Alkynes undergo addition reactions just as
alkenes do, but if we add X2 (for example) to the triple
bond, the product is an alkene that has a double bond, and
can react again to add another X2 molecule (Figure 20.10).
It is often very difficult to stop the reaction after the first
X2 has added.
CH3
C
C
CH3
+
Br
Br
CH3
Br
CH3
CH3
C
C
further
reaction
CH3
Br
Br
Br
C
C
Br
Br
H
H
C
C
H
H
CH3
CH3
+
H
H
(with catalyst)
H3C
H
CH3
further
reaction
H
CH3
CH3
Figure 20.10 Addition reactions of 2-butyne.
ITQ 2 What two products would be possible from the reaction of
hydrogen chloride with 1-butene: H3C–CH2–CH=CH2.
ITQ 3 What product or products would you expect from the
reaction of:
(a) hydrogen and a catalyst with 2-hexyne?
(b) bromine with 2-hexyne?
197
198
Unit 2 Module 1 The chemistry of carbon compounds
Summary
Review questions
1
✓ Alkenes, cycloalkenes and alkynes are named
State how each carbon atom in compound A is
hybridized: sp, sp2 or sp3.
unambiguously using the IUPAC system.
Br
✓ In alkenes, one bond of the C=C is a σ bond and
C
✓ In alkenes, the three σ bonds from each carbon
atom point toward the corners of an equilateral
triangle.
✓ In alkenes, each carbon atom has three sp2
hybrid orbitals which are formed by mixing one
2s with two 2p orbitals.
✓ In alkenes, the π bond is formed by overlap
of coplanar unhybridized 2p orbitals on each
carbon atom of the C=C.
✓ The C=C of alkenes is rigid and this gives rise to
geometric isomerism.
✓ In alkynes, one bond of the C≡C is a σ bond and
two are π bonds. Each carbon of the C≡C also
forms one additional σ bond to another carbon
atom or to hydrogen.
✓ In alkynes, the two σ bonds from each carbon
atom point in opposite directions (180° apart).
✓ In alkynes, each carbon atom has two sp hybrid
orbitals which are formed by mixing one 2s with
one 2p orbital.
✓ In alkynes, the two π bonds of the C≡C are
formed by overlap of coplanar unhybridized 2p
orbitals on each carbon atom of the C≡C.
✓ The loosely held electrons which form the
π bonds of alkenes and alkynes cause these
compounds to undergo addition reactions
readily; two atoms or groups add to each π bond.
C
H3C
one bond is a π bond. Each carbon of the C=C also
forms two additional σ bonds to other carbon
atoms or to hydrogen atoms.
C
C
H
CH3
A
2
Give the IUPAC names of the following compounds:
a
b
CH3CH2CH2CCH2CH3
CH2
c
Chapter 20 Alkenes and alkynes
Answers to ITQs
1
(a)
ne
cyclopentene
cyclopropene
(b) Cyclopentene is a liquid at room temperature
(boiling point is 44 °C) and is relatively stable. The
internal angles of a pentagon are 108°, which is
the same as the angle between sp2 hybrid orbitals
and not very different from the angle between sp3
hybrid orbitals (109.5°). So there is relatively little
ring strain in cyclopentene.
Cyclopropene is very strained. The internal angles
of an equilateral triangle are 60°, which is vastly
different from the angle between sp2 hybrid
orbitals (120°) and the angle between sp3 hybrid
orbitals (109.5°). Cyclopropene is difficult to
prepare and is very unstable.
2
CH3–CH2–CH2–CH2Cl and CH3–CH2–CHCl–CH3
(You will see in Chapter 23 which is more likely.)
3
(a)
CH3CH2CH2C
CCH3
H2
CH3CH2CH2
CH3
C
catalyst
C
H
2-hexyne
CCH3
Br2
CH3CH2CH2
2-hexyne
n-hexene
Br
C
Br
CH3CH2CH2CH2CH2CH3
H
cis-2-hexene
(b)
CH3CH2CH2C
H2
catalyst
C
Br2
CH3CH2CH2CBr2CBr2CH3
CH3
trans-2,3-dibromo-2-hexene
2,2,3,3-tetrabromohexane
199
200
Chapter 21
Alcohols and amines
Learning objectives
■ Define the term functional group and provide examples of functional groups.
■ Classify and systematically name simple alcohols and amines.
■ Describe the bonding to oxygen atoms and nitrogen atoms in alcohols and amines and account for the
■
■
■
■
presence of lone pairs of electrons on these atoms.
Provide explanations for the relatively high boiling points and solubility in water of the lower alcohols.
Describe the oxidation reactions of alcohols and apply these reactions and the colour changes observed
in the oxidizing agents to qualitative analysis.
Describe the reactions which convert alcohols to esters, ethers and haloalkanes.
Explain why amines are basic, derive Kb and pKb, and relate their values to basicity.
Introduction
Alkanes do not react with most laboratory reagents, but
alkenes undergo addition reactions with reagents such as
bromine and HBr. These addition reactions take place at
the carbon–carbon double bond, We call the double bond
a functional group. Many derivatives of alkanes can be
described by the formula R–X, where R is an alkyl group,
and X is a functional group.
We can study this area of chemistry systematically by
describing the preparation, properties and reactions of
classes of compounds, defining them by the functional
groups they carry. There are many millions of organic
compounds, so it would be impossible to describe the
chemistry of each separately. Fortunately, we can make a
good first approximation by writing the general equation
for the conversion of R–X to R–Y, and expect that what we
know to be true of R–X, where R is one alkyl group, will
also be true when R is a different alkyl group. Later we will
look at the reactions in more detail, and consider how and
why the exact nature of R can influence the course of the
reaction, whether it is faster or slower, and so on.
Haloalkanes
Haloalkanes were introduced in Chapter 19 as derivatives
of alkanes. We can consider all haloalkanes to be members
of a class of compounds and expect them all to show similar
properties and behaviour. If we consider how readily a
haloalkane undergoes a particular reaction, we find that
it depends on the nature of both the alkyl group and the
halogen. For example, the conversion of a haloalkane, R–X,
to an alcohol, R–OH, can be carried out using hydroxide
ions from NaOH in a suitable solvent (Figure 21.1).
R
X
+
OH –
R
OH
+
X–
Figure 21.1 Conversion of a haloalkane, R–X, to an alcohol,
R–OH, using hydroxide ions from aqueous sodium hydroxide.
The rate of this reaction varies very considerably, depending
on the nature of the alkyl group R and on whether X is I,
Br, Cl or F.
Alcohols
Some of the most important organic functional groups
contain oxygen. They are important in organic chemistry
and in biochemistry; among these are the alcohols, R–OH.
In alcohols the functional group is the hydroxyl group
and R is an alkyl group. We can see that an alcohol is an
alkane, R–H, with one H replaced by OH. Another way of
looking at an alcohol, R–O–H, is as a monoalkyl derivative
of water, H–O–H, derived by replacing one of the hydrogen
atoms by an alkyl group, R. The chemistry of the alcohols
illustrates both views: the reactivity can be traced to the
high electronegativity of the oxygen atom.
Chapter 21 Alcohols and amines
The simplest alcohol, CH3OH, is a liquid with a low boiling
point (65 °C) while C18H37OH is a waxy solid with physical
properties similar to those of the alkane C18H38. Both these
alcohols undergo similar chemical reactions, although they
show differences in, for example, the rate of the reaction.
gives the smallest possible number to the C that carries the
OH. Some common names are still used for alcohols such
as isopropyl alcohol and tert-butyl alcohol and for complex
alcohols such as cholesterol, where the systematic name
would be very cumbersome.
Many alcohols occur naturally. The well-known liquid
commonly called ‘alcohol’, which gives the class of
compounds its name, is ethanol, written as CH3CH2OH
(or C2H5OH). Ethanol occurs naturally through the
fermentation of sugar, a process that our early ancestors
learned how to use to produce alcoholic drinks. Cholesterol,
C27H45OH (Figure 21.2), is a solid alcohol produced in
animal metabolism. It has had a bad press in recent years,
but it is, nevertheless, of great biological importance.
Alcohols are classified according to the number of alkyl
groups present on the carbon to which the OH is attached;
this position is called the ‘alpha’ (α) position. A primary
alcohol has one alkyl substituent at the α position, a
secondary alcohol has two and a tertiary alcohol has three.
The notations 1°, 2° and 3° are sometimes used to indicate
primary, secondary and tertiary respectively. Methanol is
unique because it has no alkyl substituents on the α carbon.
In most ways, the properties of methanol are similar to
those of a primary alcohol.
Table 21.1 gives details of some simple alcohols.
Bonding in alcohols
We will examine bonding in alcohols using methanol,
CH3OH, as an example.
HO
H
cholesterol
Figure 21.2 The structure of cholesterol.
H
Nomenclature and classification of alcohols
Ordinary alcohols are easily named as derivatives of
alkanes. We simply replace the final -e of the alkane
with -ol: CH3OH is methanol and C2H5OH is ethanol. If
necessary, the chain is numbered starting from the end that
Table 21.1 Details of some simple alcohols
Formula
Name
_
Old name
methanol
CH3C OH
_
CH3CH2
Type
OH
_
CH3CH2CH2
OH
ethanol
primary
1-propanol
primary
2-propanol
secondary
alcohol
CH3
_ CH
OH
isopropyl
alcohol
CH3
_
CH3CH2CH2CH2
OH 1-butanol
CH3CH2
C
OH
2-butanol
C
CH3
OH
2-methyl-2propanol
tertiary
H
This is the displayed formula of methanol. All the bonds
shown are σ bonds. A formula such as this does not display
the bond angles accurately.
The three C–H bonds of methanol are σ (sigma) bonds;
each represents a molecular orbital formed by combination
of an sp3 hybrid atomic orbital of carbon and an s orbital of
hydrogen, as described for methane in Chapter 19.
The atomic number of oxygen is eight and the electronic
configuration is 1s2 2s2 2p4. There are three 2p orbitals,
which are dumbbell-shaped and directed at right angles
to each other along x, y and z coordinates. Each orbital
accommodates up to a pair of electrons, so the electronic
configuration of oxygen can be written more precisely as
1s2 2s2 2px2 2py1 2pz1.
■ two are completely filled (contain two electrons each).
CH3
_
H
■ two are half-filled (contain one electron each);
secondary
CH3
CH3
O
As with carbon, the 2s and the three 2p orbitals hybridize to
four sp3 orbitals. Of these four sp3 oxygen atomic orbitals:
primary
H
_
C
tert-butyl
alcohol
One of the half-filled oxygen atomic sp3 orbitals combines
with the fourth carbon atomic sp3 orbital (which also
contains one electron) to form a σ molecular orbital with
the electrons paired; the second half-filled oxygen atomic
sp3 orbital combines with a singly occupied hydrogen
201
202
Unit 2 Module 1 The chemistry of carbon compounds
atomic s orbital to form a second σ molecular orbital in
which the electrons are paired. The oxygen atom, which
is now bonded to carbon and to hydrogen, therefore has
two pairs of electrons which are not involved in bonding.
These electron pairs are lone pairs of electrons, and are
represented as pairs of dots next to the symbol for oxygen.
Because the oxygen is sp3 hybridized the angle between the
groups bonded to O is approximately 109° (Figure 21.3).
lone pairs of
electrons occupying
sp3(O) orbitals
C
H
O
H
H
m bond from sp3(O)
- s(H) overlap
CH3CH2CH2OH; RMM = 60; boiling point = +97 °C; miscible
in all proportions with water. Main source is reduction
of propanal. Main uses are solvent; starting material in
synthesis.
Figure 21.3 Bonding and electron lone pairs in methanol,
indicating the C–O–H bond angle.
General properties of alcohols
Alcohols with one O–H group are monohydric alcohols
and form a homologous series with the general formula
CnH2n+1OH, often written as ROH. The lower members of
the series are liquids with pungent smells and a distinctive
taste; higher homologues are solids with almost no smell.
An important characteristic of alcohols is the ability to form
intermolecular hydrogen bonds. Oxygen is much more
electronegative than hydrogen, so the electrons which
comprise the O–H bond are slightly displaced toward
oxygen, giving rise to a partial negative charge (δ−) on
oxygen and a partial positive charge (δ+) on hydrogen. We
can say that the O–H bond is polarized.
b–
b+
O
H
CH2CH2OH; RMM = 46; boiling point = +78 °C; miscible
in all proportions with water. Main source is fermentation
of sugar. Main uses are alcoholic beverages; solvent; fuel;
starting material in chemical synthesis.
1-Propanol
m bond from sp3(C) - sp3(O) overlap
R
CH3OH; RMM = 32; boiling point = +65 °C; miscible in all
proportions with water. Main sources are from heating
synthesis gas (CO2 + CO + H2) under pressure over metal
catalysts; synthesis gas produced from CH4 in natural gas.
Main uses are as solvent; starting material in synthesis;
antifreeze; fuel; denaturant for ethanol in ‘methylated’ spirits.
Ethanol
H
m bonds from sp3(C)
- s(H) overlap
Methanol
This polarization causes hydrogen bonds to form between
alcohol molecules. In water, H–O–H, the O–H bonds are
similarly polarized and hydrogen bonds can form between
water molecules and alcohol molecules (Figures 21.4 and
21.5).
Hydrogen bonding between alcohol molecules causes
alcohols of low molecular weight to have relatively high
boiling points. For example, the boiling point of propane
(RMM 44) is −42 °C, while that of ethanol (RMM 46) is
+78 °C.
CH3
2-Propanol
CH
1-Butanol
CH3CH2CH2CH2OH; RMM = 74; boiling point = +118 °C;
solubility in water is 7.7 g per 100 mL. Main source is
reduction of butanal; obtained by addition of H2 and CO
to propene. Main uses are solvent; perfume component;
synthesis of butyl esters and ethers which have many
applications.
2-Butanol
H
CH3CH2
C
OH
CH3
; RMM = 74, boiling point = +100 °C;
solubility in water is 12.5 g per 100 mL. Main source is
addition of H2O to 1-butene or 2-butene. Main uses are
solvent; starting material for oxidation to the ketone,
butanone, an important solvent.
b+
R
O
b–
R
H
b+
O
b–
R
H
b+
O
b–
R
H
b+
O
b–
R
H
b+
O
b–
Figure 21.4 Hydrogen bonding between alcohol molecules.
R
H
b+
OH
Also known as isopropyl alcohol. CH3
; RMM = 60;
boiling point = +82 °C; miscible in all proportions with
water. Main source is addition of H2O to propene. Main
uses as solvent; cleaning fluid; fuel additive.
O
b–
b+
H
H
b+
O
b–
R
H
b+
O
b–
H
H
b+
O
b–
R
H
b+
O
b–
H
b+
Figure 21.5 Hydrogen bonding between alcohol molecules and
H2O.
Chapter 21 Alcohols and amines
2-Methyl-1-propanol
CH3
CH2OH
Oxidation of alcohols
C
Oxidation is a reaction that is of great importance in both
the laboratory and in living organisms. A considerable range
of oxidizing agents is available, some specially designed for
special cases. Chromic acid is a common, though rather
brutal, oxidizing agent that sees frequent use. Its formula
corresponds to H2CrO4 in H2SO4. Dichromate ion Cr2O72−,
from the potassium or sodium salt, K2Cr2O7 or Na2Cr2O7, is
another popular chromium-based oxidizing agent.
H
Also known as isobutyl alcohol.
CH3 ; RMM = 74;
boiling point = +108 °C; solubility in water is 9.5 g per 100
mL. Main source is reduction of the aldehyde obtained from
propene + CO. Main uses are solvent; starting material for
preparation of esters, which are also important solvents.
CH3
2-Methyl-2-propanol
CH3
C
OH
Also known as tert-butyl alcohol.
; RMM = 74;
CH3
boiling point = +83 °C; miscible with water. Main source is
addition of H2O to methylpropene. Main uses are solvent;
starting material for preparation of esters and ethers which
are used as flavours, gasoline additives.
Summary
Figure 21.6 shows the boiling points of some alcohols and
alkanes with comparable values of relative molecular mass.
Hydrogen bonding between alcohol molecules and water
causes lower alcohols to be very soluble in water (water is
a polar solvent). As the size of the alkyl group, R, increases,
the solubility of the alcohol decreases because the size
of the non-polar hydrocarbon portion of the molecule
becomes large relative to the polar (O–H) portion (recall
that ‘like dissolves like’).
Alcohols do not protonate water, therefore aqueous
solutions of alcohols are neutral. You may recall that the
defining characteristic of an acid is its ability to protonate
water, that is, to ionize and donate a proton to the water
molecule.
120
alcohols
150
alkanes
Boiling point / ˚C
100
50
0
-50
In chromic acid and dichromate, the chromium is at
the oxidation level CrVI; it gains electrons as it carries
out the oxidation, and is reduced to CrIII in the process.
CrVI compounds are orange-yellow in colour and CrIII
compounds are green. For the oxidation that concerns
us here, the alcohol must have at least one H on the α
carbon. The oxidation converts the alcohol to a carbonyl
compound, that is one that has a C=O functional group
(Figure 21.7).
carbonyl group
R
OH
C
R’
R
oxidizing
C
agent
H
O
R’
if R and R’ are alkyl groups, the compound is a ketone
Figure 21.7 Oxidation of alcohols to aldehydes and ketones.
When a solution of the CrVI oxidizing agent (orange) is
added dropwise to a cold solution of the oxidizable alcohol
the reaction mixture becomes green, which is characteristic
of the CrIII oxidation state. Addition of the oxidizing agent
is continued until the reaction mixture is slightly orange;
this indicates that CrVI is no longer being reduced and all
the alcohol present has been oxidized.
The permanganate ion, MnO4−, which is deep purple,
also oxidizes alcohols. The products are the same as those
obtained by oxidation with CrVI compounds. In KMnO4
the oxidation state of Mn is +7; when KMnO4 effects an
oxidation the MnVII is eventually reduced to MnII and a
colour change from purple to brown (MnIV) or, in acid
solution, to colourless.
A secondary alcohol undergoes oxidation to a ketone; that
is a carbonyl compound where R and R' are both alkyl
groups. This is looked at in Chapter 24.
-100
-150
-200
0
20
40
60
80
100
Relative molecular mass
120
140
Figure 21.6 Boiling points of alcohols and alkanes. The higher
boiling points for the alcohols is due to hydrogen bonding.
ITQ 1 Draw the structures of compounds you choose to illustrate
the following reaction sequences, and label each reaction either
oxidation or reduction. (Reagents need not be shown.)
(a) alcohol → aldehyde → carboxylic acid
(b) ketone → alcohol
203
204
Unit 2 Module 1 The chemistry of carbon compounds
There is a test for secondary alcohols that have a methyl
group CH3 adjacent to the –OH group.
■ Add aqueous NaOH is added to the alcohol.
■ Add a solution of I2 in aqueous NaOH.
■ Warm the reaction mixture.
■ Dilute the solution with cold water. A pale yellow
precipitate indicates the presence of a secondary
alcohol – see Chapter 23 (page 219) for more details.
■ The precipitate is iodoform, HCI3.
Note that a tertiary alcohol can not undergo this oxidation
as it has no α hydrogen atom.
Aldehydes and ketones often undergo the same kinds of
reaction, but early organic chemists saw enough difference
between them to justify giving them different names. A
very important difference is that, while further oxidation of
a ketone is difficult, aldehydes undergo further oxidation
so easily that it is difficult to stop the oxidation of a primary
alcohol at the aldehyde stage, and further oxidation of the
aldehyde to a carboxylic acid is common (Figure 21.8).
O
R
chromic
acid
CH2OH
R
primary alcohol
C
O
H
chromic
acid
aldehyde
(not isolated)
R
C
OH
carboxylic acid
Figure 21.8 Oxidation of primary alcohols with chromic acid,
HCrO4, a strong oxidizing agent.
In the conversion of an alcohol to a carbonyl compound,
the oxidation leads to the removal of two H atoms from the
alcohol but, in the conversion of an aldehyde to a carboxylic
acid, the oxidation leads to the addition of one O atom to
the aldehyde. In organic chemistry, oxidations very often
follow this pattern: the loss of H, or the gain of O, or some
combination of the two. Ultimately, if we want to examine
oxidation in great detail, we do have to be concerned about
the movement of electrons in the manner we discussed for
the conversion of CrVI to CrIII, but the simple description of
oxidation as ‘gain of O or loss of H’ serves us well for most
organic reactions. The reverse of oxidation is reduction,
which we can then consider as ‘gain of H or loss of O’.
A ketone can be reduced to a secondary alcohol, and
an aldehyde can be reduced to a primary alcohol by the
addition of H atoms. This can, in fact, be carried out by
adding H2 to the C=O in presence of a catalyst under very
forcing conditions or, more conveniently, by chemical
reduction where a reagent such as sodium borohydride,
NaBH4, provides the H atoms. The reduction of a carboxylic
acid, RCOOH, to a primary alcohol, RCH2OH, can be also
be carried out by catalytic hydrogenation or by chemical
reduction. It is possible, but more difficult, to reduce the
carboxylic acid only as far as the aldehyde, RCHO.
Other reactions of alcohols
We have looked first at the oxidation of alcohols because
it illustrated differences in the chemistry of primary,
secondary and tertiary alcohols, and because it introduced
us to three classes of compounds, ketones, aldehydes and
carboxylic acids. Clearly, oxidation involves the O–H group
of the alcohol and an H on the α carbon atom. Now we shall
look at typical reactions that involve just the O–H group.
First we shall look at those reactions which break the O–H
bond and so replace H with something else.
Preparation of esters from alcohols
In an esterification reaction, an acyl group replaces the
hydrogen attached to the oxygen of an alcohol. A solution
containing an alcohol (ROH) and a carboxylic acid
(R'COOH) is heated with a catalytic amount of a mineral
acid, setting up the equilibrium shown in Figure 21.9.
ITQ 2
(a) Provide the IUPAC name for compounds a to f.
(b) What product or products, if any, would you obtain if you treated each with chromic acid.
a
H
H
H
H
H
H
C
C
C
C
C
H
H
H
H
H
d
C
C
H
H
c
H
CH3CH2
OH
C
CH2CH3
OH
CH3CH2
C
f
H
CH3
C
CH2CH3
H
C
H
OH
OH
H
H
CH3
CH3
CH3
OH
e
CH3 OH
CH3
b
H
C
C
H
H
C
C
C
H
H
H
H
Chapter 21 Alcohols and amines
O
R’
C
O
OH
+
H
carboxylic acid
O
HCl
R
R’
alcohol
C
OR
ester
(+ HOH)
water
Figure 21.9 The equilibrium reaction between carboxylic acids
and alcohols, catalysed by mineral acid.
If we cause this equilibrium to move from left to right (by
removing H2O, for example) the reaction is an esterification
(producing the ester R'COOR). This is a condensation reaction,
a reaction in which two molecules, in this case the carboxylic
acid and the alcohol, combine to form a product, the ester,
and a small molecule (in this case, water) (Figure 21.10).
O
R’
C
O
OH
+
H
carboxylic acid
O
HCl
R
R’
alcohol
C
OR
(+ HOH)
ester
Figure 21.10 If water is removed, the equilibrium shifts to the right.
Alternatively, we could cause the reaction to move from
right to left by treating the ester with a large excess of H2O
in presence of the acid catalyst. That reaction would be a
hydrolysis, converting the ester to a carboxylic acid and
an alcohol. Hydrolysis of the ester results in breakage of the
C–O single bond and addition of the elements of water to the
fragments (Figure 21.11). The suffix -lysis means breaking
or cleavage and hydro, in this context, refers to water.
O
R’
C
(haloalkane). There are several ways of accomplishing this
change; the method chosen depends on which alcohol we
start with.
■ The alcohol is treated with concentrated HCl or HBr
(Figure 21.12). This method works well only for
tertiary alcohols. For primary or secondary alcohols we
would choose another method.
CH3
CH3
C
CH3
OH
+
HCl (conc.)
CH3
+
carboxylic acid
H
O
HCl
R
alcohol
R’
C
C
Cl
(+ HOH)
CH3
2-methyl-2-propanol
(tert-butanol)
a 3˚ alcohol
2-chloro-2-methylpropane
(tert-butyl chloride)
Figure 21.12 Conversion of a tertiary alcohol to a chloroalkane
with HCl.
■ The alcohol is treated with thionyl chloride, SOCl2
(usually with a base present) (Figure 21.13).
OH
CH3
C
O
+
CH3
Cl
S
(base)
Cl
thionyl chloride
H
2-propanol
Cl
CH3
C
CH3
(+ SO2 + HCl)
H
O
OH
CH3
2-chloropropane
OR
ester
+ HOH
large excess
Figure 21.11 If water is present in excess, the equilibrium shifts
to the left.
Naming esters
An ester is named as the alkyl derivative of the carboxylic
acid. The alkyl group originates from the alcohol and is the
first part of the name.
gaseous by-products
Figure 21.13 Conversion of a secondary alcohol to a
chloroalkane with SOCl2.
■ The alcohol is treated with phosphorus tribromide,
PBr3. All three Br atoms are delivered stepwise to three
molecules of ROH (Figure 21.14).
3 CH3CH2CH2
OH
+
PBr3
1-propanol
3 CH3CH2CH2
O
Br
( + H3PO3 )
1-bromopropane
CH3
C
OCH2CH3
from ethanoic acid
from ethanol
Ethyl ethanoate is the ester formed from ethanol and
ethanoic acid.
O
CH3CH2CH2CH2CH2
C
from hexanoic acid
OCH2CH2CH3
from propanol
Propyl hexanoate is the ester formed from propanol and
hexanoic acid.
Preparation of haloalkanes
The entire OH group of an alcohol R–OH can be replaced
by another group (X) to produce R–X. Examples are found
in reactions that convert an alcohol to an alkyl halide
Figure 21.14 Conversion of a primary alcohol to a bromoalkane
with PBr3.
Amines – RNH2
Ammonia, NH3, is the hydride of trivalent N in the same way
as H2O is the hydride of divalent O. Remember that alcohols,
R–OH, can be viewed as monoalkyl derivatives of H2O.
Amines are the alkyl derivatives of NH3 (Figure 21.15). Since
ammonia has three H atoms that could be replaced by alkyl
groups, we have to deal with a greater number of variables
than we do with derivatives of H2O. In monoalkyl derivatives
of NH3, RNH2, the –NH2 group is known as the amino group.
The CAPE syllabus only looks at primary amines.
205
206
Unit 2 Module 1 The chemistry of carbon compounds
H
O
H
H
N
H
H
water is the hydride of divalent O
ammonia is the hydride of trivalent N
amino group
R
O
R
H
N
H
H
a monoalkyl derivative
of H2O is an alcohol
a monoalkyl derivative of NH3 is an
amine; this amine is a primary (1˚ ) amine
Figure 21.15 The relationship between H2O and alcohols and
between NH3 and primary amines.
Compounds of nitrogen are major constituents of plants
and animals. Proteins are fundamentally important to all
forms of life. They are constructed from very large numbers
of amino acids that contain, as their name implies, amino
(–NH2) groups and carboxylic
H
O
acid (–COOH) groups. Amino
one form of
N
C
an amino acid
acids in proteins are joined
H
OH
C
together by amide units that
H
R
link the –COOH of one amino
acid to the NH2 of another.
Alkaloids are constituents of plants; most are amines,
and some have strong physiological effects on animals
(including human beings). Sometimes the physiological
effect is detrimental, as it is with poisons such as strychnine.
However, the effect can be beneficial, as it is with quinine,
which is used medicinally. Sometimes the beneficial medical
effect can be abused, as it is with morphine and cocaine.
The three C–H bonds of the methyl group are σ bonds,
representing σ molecular orbitals formed by combining sp3
hybrid atomic orbitals of carbon and s orbitals of hydrogen.
This is as described for methane in Chapter 19 and for
methanol on page 202 of this chapter.
The atomic number of nitrogen is seven and its electronic
configuration is 1s2 2s2 2p3 or, more specifically, 1s2 2s2 2px1
2py1 2pz1. As with carbon and oxygen, the 2s and the three
2p orbitals of nitrogen hybridize to form four sp3 atomic
orbitals. Of these four sp3 nitrogen atomic orbitals:
■ three are half-filled (contain one electron each);
■ one is completely filled (contains two electrons).
One of the half-filled nitrogen atomic sp3 orbitals combines
with the fourth carbon atomic sp3 orbital (also half-filled)
to form a σ molecular orbital with paired electrons (N–C
bond). The other two half-filled nitrogen atomic sp3 orbitals
each combine with a singly occupied hydrogen s orbital to
form two more σ molecular orbitals with paired electrons
(two N–H bonds) (Figure 21.16). The nitrogen atom which
is now bonded to carbon and two hydrogens has a pair of
electrons which is not involved in bonding. This lone pair
or non-bonded electrons are represented as a pair of dots
above the symbol for nitrogen.
3
m bonds from sp (C)
- s(H) overlap
R
NH2
methylamine
(methanamine)
CH3CH2
N
ethylamine
(ethanamine)
H
m bond from sp3(C) - sp3(N) overlap
H
Figure 21.16 Bonding and electron lone pair in methylamine.
H
NH2
m bonds from sp3(N)
- s(H) overlap
N
H
RNH2 is a primary (1°) amine. In primary amines the
nitrogen is bonded to one alkyl group. Primary amines
are named as alkylamines in common nomenclature. In
IUPAC nomenclature they are named by replacing the ‘-e’
of the parent alkane with ‘–amine’.
CH3
H
C
H
Classification and nomenclature of amines
The terms primary, secondary and tertiary as
applied to amines depend on the number of
alkyl groups attached to N.
lone pair of electrons
occupying sp3(N) orbital
H
NH2
cyclohexylamine
(cyclohexanamine)
The three groups bonded to nitrogen and the lone pair of
electrons point toward the corners of a tetrahedron, so the
angle between the groups is approximately 109°.
General properties of amines
The very low molecular weight amines are gases, but
most aliphatic amines are liquids with strong, somewhat
unpleasant, fishy odours. Nitrogen is more electronegative
than hydrogen, so the N–H bonds in primary amines are
polarized.
b–
b–
b+
Bonding in amines
Methylamine, CH3NH2, is the simplest
primary amine.
Note the lone pair on the nitrogen atom.
This can accept protons and therefore
makes the amine a base.
R
H b+
H
H
H
C
N
N
H
H
methylamine
H
b+
R
N
H
R
Primary amines can therefore hydrogen bond to each
other. This intermolecular H-bonding is illustrated in Figure
21.17. The O–H bonds in water are polarized, so primary
amines also form hydrogen bonds to water.
Chapter 21 Alcohols and amines
b–
R
N
H b+
b–
R
b–
R
b+
H
N
b+
N
electronegative than O, so electrons on N are more readily
available to be donated to form a two-electron bond with
an atom having a vacant orbital, and N is more able to
carry the positive charge that results. In aqueous solution,
an amine is a weak base and accepts a proton from water in
a reversible acid/base reaction (Figure 21.19).
b+
H
H b+
H
H b+
b–
R
H
b+
H
N
R
N
H b+
H
+
H
O
H
H
R
+
N
H
+
–
OH
H
Figure 21.17 Intermolecular hydrogen bonding in a primary amine.
Figure 21.19 Deprotonation of H2O by a primary amine.
Methylamine
The equilibrium constant for the process shown in Figure
21.19 is known as a basicity constant, Kb, and is defined as:
CH3NH2; RMM = 31; gas at room temperature, boiling point
= −6 °C; very soluble in water. pKb = 3.36; pKa of conjugate
acid = 10.64.
CH3CH2NH2; RMM = 45; gas at room temperature, boiling
point = +17 °C; very soluble in water. pKb = 3.25; pKa of
conjugate acid = 10.75.
Propylamine
CH3CH2CH2NH2; RMM = 59; liquid at room temperature,
boiling point = +49 °C; very soluble in water. pKb = 3.33;
pKa of conjugate acid = 10.67.
Amides
Just as an ester can be seen as the product of the elimination
of a molecule of water between a carboxylic acid and an
alcohol, so an amide results if we link a carboxylic acid to
an amine with the elimination of H2O (Figure 21.18).
O
N
H
[RN+H3][OH−]
[RNH2]
We can measure Kb and so can calculate pKb of an amine:
Ethylamine
R
Kb =
+
HO
C
O
R’
H
amine
R
N
C
R’
(+ HOH)
H
carboxylic acid
amide
water
Figure 21.18 Formation of an amide from a primary amine and a
carboxylic acid.
pKb = –log10 Kb
Remember that when water is present in large excess its
concentration is assumed to be constant and is not included
in the equilibrium constant.
■ A low numerical value of pKb means that the
equilibrium lies to the right and the basicity of the
amine is high.
■ A high numerical value of pKb means that the
equilibrium lies to the left and the basicity of the
amine is low.
An alternative and more widely used way of defining
basicity of amines is based on the ability of the conjugate
acid of the amine, RN+H3, to protonate water in the
following equilibrium reaction:
RN+H3 + H2O ҡ RNH2 + H3O+
The equilibrium constant for this reaction is an acidity
constant, Ka:
Ka =
[RNH2][H3O+]
[RN+H3]
and pKa = –log10 Ka
Basicity of amines
The amines are bases, and this is the property that dominates
their chemistry and makes them distinctly different from
oxygen-containing compounds. Oxygen compounds such
as alcohols, ROH, have two sets of lone-pair electrons,
while amines have only one set. However, N is less
ITQ 3 Draw the structures of compounds 1 and 2.
1
pentylamine
2
3-aminocyclohexanol
■ A low pKa for the conjugate acid means that the
equilibrium lies to the right and the basicity of the
amine is low.
■ A high value of pKa for the conjugate acid means that
the equilibrium lies to the left and the basicity of the
amine is high.
207
208
Unit 2 Module 1 The chemistry of carbon compounds
Summary
Review questions
1
✓ Alcohols, R–OH, contain the O–H functional
group and are classified as primary, secondary
and tertiary on the basis of the number of
carbons attached to the carbon atom bearing the
OH group.
✓ The oxygen atom in alcohols is sp3 hybridized
Match the boiling points given with compounds drawn
below. Give reasons for your answer.
Boiling points: 97 °C; 197 °C; 49 °C
CH3CH2CH2NH2
CH3CH2CH2OH
propylamine RMM 59
propanol RMM 60
2
alcohols to form hydrogen bonds with other
alcohol molecules and with water.
✓ Primary alcohols are oxidized to aldehydes
which are then rapidly oxidized to carboxylic
acids. Secondary alcohols are oxidized to
ketones. The oxidizing agents contain transition
metal ions which change colour when they are
reduced in the reaction.
✓ Two important reactions of the O–H group in
alcohols are:
–
–
conversion to esters, RCOOR′
conversion to haloalkanes, R–X
✓ Amines are derivatives of NH3. RNH2 is classified
as a primary amine.
✓ The N atom in amines is sp3 hybridized and
carries one lone pair of electrons.
✓ The N–H bonds in primary amines are polarized
and these compounds form hydrogen bonds
with other amine molecules and with water.
✓ Amines are basic compounds.
H
C
C
H
H
OH
ethylene glycol RMM 62
and carries two lone pairs of electrons.
✓ Polarization of the O–H bond in alcohols causes
HO
H
How would you prepare ethyl ethanoate using ethanol,
CH3CH2OH as the only organic starting material?
H
H
O
C
C
H
O
H
H
C
C
H
H
H
ethyl ethanoate
3
Deduce the structures of compounds M and N from the
scheme below.
O
N
C3H7Cl
SOCl2
M
C3H7OH
H2CrO4
H3C
C
CH3
Chapter 21 Alcohols and amines
Answers to ITQs
1
(a)
H
H
H
C
C
H
H
H
oxidation
OH
H
O
C
C
O
H
C
C
C
reduction
H
H
H
H
a
H
C
C
OH
ethanoic acid
carboxylic acid
H
H
O H
C
C
C
H
H
H
H
C
C
C
C
C
H
H
H
H
H
H
OH
H
H
H
H
H
H2CrO4
1-pentanol, a primary alcohol
b
O
2-propanol
alcohol
H
H
H
H
H
H
propanone
ketone
2
H
ethanal
aldehyde
(b)
H
H
H
ethanol
alcohol
H
oxidation
C
C
C
C
H
H
H
H
O
H2CrO4
C
H
H
pentanal, an aldehyde (not isolated)
H
H
H
H
C
C
C
C
H
H
H
H
O
C
OH
pentanoic acid, a carboxylic acid
H
CH3CH2
H2CrO4
CH2CH3
C
CH3CH2
C
CH2CH3
OH
O
3-pentanol, a secondary alcohol
3-pentanone, a ketone
c
d
OH
CH3CH2
C
H2CrO4
CH3
CH3 OH
CH3
NO REACTION
CH3
H
H
CH3
C
CH2CH3
H
C
H
H2CrO4
C
H
H
CH3
CH3
C
f
HO
H
H
O
C
CH3
C
3-methylbutanone,
a ketone
O
H
C
C
C
CH3
OH
H
H
CH2CH3
CH3 O
H
3-methyl-2-butanol,
a secondary alcohol
2-methyl-2-butanol, a tertiary alcohol
e
C
H2CrO4
C
C
C
H
H
H
H2CrO4
H
H
H
C
C
H
H
C
C
C
H
H
OH
2-methylbutanoic acid,
a carboxylic acid
2-methyl-1-butanol,
a primary alcohol
3
1
2
H
CH3CH2CH2CH2CH2
cyclopentanol,
a secondary alcohol
OH
1
N
H
3
pentylamine,
a primary amine
NH2
3-aminocyclohexanol,
a primary amine
cyclopentanone,
a ketone
H
H
209
210
Chapter 22
Stereochemistry
Learning objectives
■ Explain the meaning of structural isomerism.
■ Describe and give examples of chain, functional group and positional isomers.
■ Explain the term geometrical isomerism.
■ Describe and give examples of cis/trans isomers.
■ Explain the origins of chirality and optical isomerism.
■ Give examples of chiral compounds.
Introduction
Stereochemistry is the study of the shapes of molecules
and the relationships between their component atoms and
groups in two and three dimensions. These sub-molecular
relationships determine the course of many vital life processes
and drug interactions which involve large and small organic
compounds such as proteins, enzymes and drugs.
You will find it easier to understand stereochemistry if you
use molecular models. They are often quite cheap to buy (try
your local bookshop) and you can share them with friends.
We saw in Chapter 20 (page 195) that the double bond in
butene could be placed between carbons 1 and 2, to give
–C=C–C–C (1-butene), or between carbons 2 and 3, to give
–C–C=C–C– (2-butene). These two substances are structural
isomers of each other. The two isomers contain different
groups. For example, 1-butene contains a –CH2 group whilst
2-butene does not. For more about 2-butene, see below.
If we place two methyl substituents on the same carbon of
H2C=CH2, we get a third compound that is another isomer
of each of the previous two.
C
2
Some alkenes (for example, 2-butene) can exist as
distinguishable stereoisomers. This is because the carbon–
carbon double bond, >C=C<, does not allow rotation about
the C–C axis. Groups which are bonded to sp2 hybridized
carbons are therefore fixed in space relative to each other.
So 2-butene, for example, exists as two chemically distinct
isomers, cis-2-butene and trans-2-butene, that cannot easily
be interconverted by rotation about the double bond.
H3C
CH3
C
H
CH2
1
2-methylpropene
Its older name is isobutene but, because this isomer does
not have four carbons in a continuous chain, systematic
nomenclature does not consider it a butene. It is a
derivative of propene. The position of the double bond
H3C
C
H
C
H
cis-2-butene
CH3
3
Geometric isomers
cis/trans isomerism
Structural isomers
CH3
(at position 1) determines the numbering system, so this
isomer is 2-methylpropene. Because there is no doubt
where the double bond is located in the propene chain
there is no need to add the locant ‘-1-’ to the name. Again,
this structural isomer contains different chemical groups.
For example, 2-methylpropene contains a carbon atom
with no attached hydrogens.
H
C
CH3
trans-2-butene
This form of stereoisomerism is called geometric
isomerism. Geometric isomerism results from rigidity in
molecules and occurs in alkenes and cyclic compounds.
The geometric isomers of 2-butene are not readily
interconvertible because rotation about the double bond
is unlikely, on energetic grounds. Geometric isomers of
alkenes are generally illustrated using displayed formulae.
Chapter 22 Stereochemistry
Isomerism about a carbon–carbon single bond
Ethane, H3C–CH3, is the simplest compound with a C–C
single or sigma (σ) bond. The bond angles about each C
are all tetrahedral, so the molecule has a three-dimensional
structure. Atoms or groups connected by σ bonds can
rotate around the σ bond. In simple alkanes this rotation
occurs readily at room temperature. Atoms or groups in a
molecule can therefore assume different arrangements in
space relative to each other. These arrangements are known
as conformations. There are several ways of representing
conformations and some of these are illustrated, for ethane,
in Figure 22.1.
H H
C
H
H
H
H
H
C
H
H H
flying wedge
H
H
H
H
H
H
H
H
sawhorse
ball-and-stick
H
H
H
H
H
H
Newman projection:
an end-on view of
two carbon atoms
Figure 22.1 Four ways of representing a single conformation of
ethane, H3C–CH3.
The flying wedge convention for representing threedimensional structures was introduced in Chapter 19. Groups
or atoms attached to bonds drawn as wedges are above the
plane of the paper; groups or atoms attached to bonds drawn
as hashed lines are below the plane of the paper. Ball-andstick and sawhorse representations are perspective views.
A Newman projection is an end-on view of two connected
carbon atoms; the bond which connects the two carbon
atoms is hidden. In the Newman projection of ethane the
methyl groups are like two wheels, each with three spokes
(the C–H bonds), connected by an axle (the C–C bond),
and each wheel can rotate independently about the axle.
In the Newman projection of ethane in Figure 22.1, the
C–H bonds on the carbon atom to the front (blue) are as
far away as possible from the C–H bonds (shown in black)
on the carbon atom to the rear (which we cannot see). In
this conformation the two sets of C–H bonds are said to be
staggered. If we keep the carbon atom at the front fixed
and rotate the C–C bond, the angular separation between
the two sets of C–H bonds decreases. When the C–C bond
has been rotated by 60° the two sets of C–H bonds are
exactly superimposed; in this conformation the C–H bonds
are said to be eclipsed (Figure 22.2). After a rotation of
120°, the C–H bonds are again staggered, and the model
looks exactly as it did at the start.
The H atoms of one CH3 repel the H atoms of the other CH3.
In the eclipsed conformation, with the smallest distance
between the H atoms, the repulsive interaction is highest. In
the staggered conformation, the distance between H atoms is
greatest, and the repulsive interaction is lowest. If we could
take a snapshot picture of a sample of ethane we would see
some molecules at every rotation angle, but the vast majority
of the molecules would be clustered at or near the staggered
conformation (lowest energy). Molecules in the staggered
and eclipsed conformations are different, so they are isomers.
They have the same structure, but they differ in the spatial
arrangement of their atoms, so they are stereoisomers. More
precisely, they are conformational isomers.
Conformational isomers can be readily interconverted, in
simple cases by rotation about a single bond. The energy
barrier between these isomers is too low for us to isolate a
single isomer at room temperature. In principle, the isomers
could be separated and isolated at a very low temperature
where there is not enough thermal energy available to cause
one isomer to climb the energy barrier separating it from
another. Usually, however, we depend on spectroscopy to
show us that a sample contains conformational isomers.
Isomerism at a single carbon
All four H atoms of methane, CH4, are identical. When one
H atom is replaced with Cl to make chloromethane it does
not matter which H is replaced, as there is only one possible
CH3Cl. In CH3Cl all three H atoms are identical, so replacing
one with Br yields only one possible CH2BrCl. If we
continue this process and replace another H with I, we find
H
H
C
C
H
H
H
Cl
methane
chloromethane
one structure possible
H
H
Newman
H
H
HH
H
staggered
rotate the
rear C atom
by 60˚
H
H
H
H
H
eclipsed
rotate the
rear C atom
by 60˚
H
H
H
H
H
staggered
Figure 22.2 Interconversion of the staggered and eclipsed
conformations of ethane by rotation about the single C–C bond.
Cl
B
H
Cl
bromochloromethane
one structure possible
Br
H
Br
H
H
A
I
C
pro-A
C
H
H
H
C
Br
I
Cl
bromochloroiodomethane
two isomers
Figure 22.3 Replacing three hydrogens of methane with Cl, Br
and I leads to two isomers of bromochloroiodomethane.
211
212
Unit 2 Module 1 The chemistry of carbon compounds
Optical activity
A beam of light is made up of waves, corresponding to
oscillating electric and magnetic vectors moving at the
speed of light. If we could see the waves moving along the
x-axis (in an xyz frame), the waves would be oscillating
at every possible angle on the yz plane. It is possible to
use a transparent mineral such as Iceland spar to select
waves oscillating in only one orientation; let us choose
the orientation parallel to the z-axis to illustrate. Such
light is said to be plane polarized (Figure 22.4).
If the beam of polarized light is passed through substances
with chiral properties the plane of polarization is rotated
in either a clockwise or a counterclockwise direction.
Molecules with the ability to rotate the beam of polarized
light are said to be optically active, and retain this property
in solution. Clearly this optical activity is the property of
chiral molecules.
an important difference. Now two distinguishable products
CHBrClI can be formed. The two H atoms of CH2BrCl are
not identical in every respect; if we replace one of them we
get A, while replacing the other gives us B (Figure 22.3).
Note that it is best to construct molecular models to follow
this discussion.
It is impossible to superimpose isomer A on isomer
B, no matter how they are rotated. Molecule B is the
non-superimposable mirror image of molecule A. Any
two figures (not just molecules) with this relationship are
described as enantiomorphs, so A and B are enantiomorphic
isomers, a term that has been contracted to enantiomers.
This relationship may be more easily seen if A is rotated
and placed beside B.
Br
H
H
C
C
I
Br
I
Cl
Cl
A
mirror
B
Pairs of enantiomers were first recognized because of a
property usually observed in solution: they rotate the
plane of polarization of plane-polarized light in opposite
directions. The enantiomer that rotates the plane to the right
(clockwise) is called dextrorotatory, and the enantiomer
that rotates the plane to the left (counterclockwise) is
called laevorotatory. The terms are derived from the Latin
words for right and left. This is known as optical activity
and the compounds are called optical isomers.
A 1:1 mixture of two enantiomers is known as a racemic
mixture. A racemic mixture will not rotate the plane
vertically polarized
light waves
z
x
x
y
incident beam
(unpolarized)
z
z
polarizing lens
y
Figure 22.4 Light that oscillates in all directions can be plane
polarized. The light transmitted by the polarizing lens only
oscillates in one direction.
of plane-polarized light because the rotation on one
enantiomer cancels the rotation of the other enantiomer.
A racemic mixture is optically inactive.
Chirality
Molecules that exist in enantiomeric pairs are said to be
chiral. A chiral object cannot be superimposed on its mirror
image. An object that is not chiral is achiral. Your hands
are chiral; your left hand is the mirror image of your right
hand, but they are not identical. If you try to superimpose
one on the other, you will find that you can not do so; if
you place them palm-to-palm, one is effectively ‘back-tofront’, but if you place one over the other, the thumbs
point in opposite directions. The word ‘chiral’ is derived
from the Greek word (χειρ – cheir) for hand.
Chiral objects lack certain symmetry properties. The
presence or absence of a plane of symmetry is recognized
most easily, and the absence of a plane of symmetry is
usually sufficient to make an object chiral. A plane of
symmetry is an imaginary plane drawn through an object
and serving as a mirror so that one side of the object is an
exact reflection of the other side. Figure 22.5 shows two
very similar molecules, both with a plane of symmetry.
Cl
H
Cl
H
C
H
chloromethane
C
H
H
Br
bromochloromethane
Figure 22.5 Chloromethane has a plane of symmetry, as does
bromochloromethane.
Chapter 22 Stereochemistry
Many of the molecular building blocks, such as the amino
acids that are used to make proteins, molecules essential
for life, are chiral (Figure 22.6). There is an absolute
requirement that only one enantiomer can be used.
Chirality is a property of the whole molecule. Any carbon
atom carrying four different substituents is at the centre
of a chiral array, and is sometimes called a stereogenic
centre. When there is only one such carbon, the whole
molecule is chiral. Here are a few examples of common
chiral compounds.
CH3
COOH
H
C
R
NH2
H
COOH
R
C
H
C
COOH
OH
H
C
OH
CH2CH3
CH3
2–butanol
lactic acid
CHO
H
C
OH
CH2OH
glyceraldehyde
NH2
Figure 22.6 Amino acids are chiral.
For a similar reason, only one enantiomer of some
medicinal compounds has a beneficial effect; the other
enantiomer is inactive or, in some cases, detrimental. The
most famous of these is thalidomide, which when used to
control morning sickness in pregnant women, resulted in
the birth of deformed children. Only one enantiomer of
the drug is responsible for this effect (Figure 22.7). Giving
the stereopure compound doesn’t help as it converts to the
racemic mixture in the body, so there is always a supply
of the drug that causes the birth defects. However, when
used carefully, thalidomide has beneficial effects in the
treatment of some cancers.
Figure 22.7 The two enantiomers of thalidomide. Can you see
that the molecule contains no planes of symmetry?
ITQ 1 Many of the things we use quite frequently are
dissymmetric, though sometimes we have to look closely
to recognize the dissymmetry. If you are left-handed you
have probably already recognized this because the bulk of
dissymmetric manufactured products have been designed for
use by right-handed people, and you would prefer to have the
other enantiomorph.
Identify the dissymmetric objects in the lists below:
■ Sports equipment: cricket bat; hockey stick; golf club; tennis
racquet
■ Musical instruments: guitar; violin; flute; clarinet; saxophone;
drum; keyboard.
■ At home: scissors; computer mouse; mug; can opener; knife.
ITQ 2 Which of the following compounds are chiral? Draw both
enantiomers of the chiral molecules in a manner that makes clear
that they are mirror images of each other. Use the flying wedge
convention.
(a) 1-bromobutane
(b) 2-bromobutane
(c) 1-bromo-1-chlorobutane
(d) 1-bromo-2-chlorobutane
(e) 2-bromo-2-chlorobutane
213
214
Unit 2 Module 1 The chemistry of carbon compounds
Answers to ITQs
Summary
✓ Carbon atoms connected by single bonds rotate.
The distance between the groups on neighbouring
sp3 hybridized carbon atoms can therefore vary.
✓ The various arrangements resulting from rotation
around C–C single bonds are called conformations.
The most stable conformations are generally those
in which the groups on neighbouring C atoms are
furthest from each other.
✓ There are two isomers of any compound with
one sp3 carbon centre bonded to four different
groups. These isomers are non-superimposible
mirror images and are called enantiomers.
✓ One enantiomer of an enantiomeric pair rotates
the plane of plane-polarized light to the left;
this is the laevorotatory enantiomer. The other
enantiomer is the dextrorotatory enantiomer,
and rotates the plane of plane-polarized light to
the right.
1
For discussion.
2
(a) 1-bromobutane is not chiral
(b) 2-bromobutane is chiral
H
H
H
C
C
C
H
H
Br
H
1
2
C
*
H
Et
Br
C
H
Me
Br
of a given compound is a racemic mixture and
does not rotate the plane of plane-polarized light.
mirror
(c) 1-bromo-1-chlorobutane is chiral
H
H
H
C
C
C
C
Cl
H
H
H
Br
1
H
*
Br
C
H
Cl
H
S
C
C
H
NH2
O
OH
c
CH3
(d) 1-bromo-2-chlorobutane
Br
1
H
H
C
C
C
H
H
Cl
2
C
*
H
H
Et
Cl
Cl
C
C
Et
BrH2C
Et
CH2Br
H
OH
H
mirror
(e) 2-bromo-2-chlorobutane
f
CH3
OH
1
(b) Choose one of the isomers of C3H7OBr which has a
stereogenic centre and draw the two stereoisomers
using the flying wedge representation.
H
C
C
C
Cl
H
2
C
*
HO
(a) (i) Draw the three structural isomers of C3H7OBr.
(ii) Which of the three structural isomers in part
(i) are chiral? Give a reason for your answer
and mark the stereogenic centres with *.
H
Br
H
H
2
H
Cl
OH
C
e
d
Br
C
mirror
H
b
Pr
Pr
Mark the stereogenic centres, if any, in the following
compounds.
H
H
Pr
H
Review questions
H
Me
✓ A mixture of equal parts of the two enantiomers
H
C
Et
Et
Br
optical rotation; all other physical properties are
identical.
a
H
Me
✓ Enantiomers differ only in the direction of
1
H
H
Me
H
Et
Br
Br
C
Me
H
Et
Et
Cl
C
Cl
mirror
Me
215
Chapter 23
Aldehydes and ketones
Learning objectives
■ Systematically name simple aldehydes and ketones.
■ Define the terms canonical form and resonance hybrid.
■ Describe the fundamental features of the carbonyl group: bonding, polarization, canonical forms,
■
■
■
■
■
■
resonance hybrids.
Apply known oxidation reactions of alkenes and alcohols to the preparation of aldehydes and ketones.
Describe the reduction of aldehydes and ketones to alcohols with NaBH4 and LiAlH4.
Demonstrate the use of the following oxidation reactions of aldehydes in qualitative analysis: reaction
with Cr2O72−, MnO4−, Fehling’s or Benedict’s solution, Tollens’ reagent.
Describe the iodoform reaction of methyl ketones and of secondary alcohols with an adjacent methyl
group.
Draw and explain the mechanism for the reaction of aldehydes or ketones with HCN.
Outline the condensation reaction between aldehydes or ketones and compounds with –NH2 groups
and apply the formation of derivatives of 2,4-dinitrophenylhydrazine to qualitative analysis.
Introduction
Aldehydes and ketones contain the carbonyl or oxo
functional group. In aldehydes, the carbon of the carbonyl
group is always bonded to one hydrogen and to a carbon
substituent, so an aldehyde group always occurs at the end
of a chain. In ketones, the carbonyl carbon is bonded to
two carbon substituents, so the keto group occurs within a
chain or a ring.
a
b
C
c
R
O
C
H
carbonyl
R
O
aldehyde
reduced to the corresponding alcohol, vitamin A. Geranial
is a fragrant aldehyde which occurs in lemongrass oil and
is used in perfumes and flavourings. The open-chain forms
of sugars contain the aldehyde or ketone functional group,
as well as many –OH groups.
O
O
C
O
O
progesterone
testosterone
H
R’
ketone
Many ketones and a few important aldehydes occur
in plants and animals (Figure 23.1). Testosterone and
progesterone are important hormones that contain
ketone carbonyl groups as well as other functional groups.
Carbonyl groups are often reduced to alcohols in metabolic
processes (which may also be responsible for the reverse
reaction that oxidizes the alcohol).
Simple aldehydes are less common as natural products
because they are easily oxidized or reduced. Retinal is an
aldehyde that is important to us because it is part of the
mechanism of vision in which the eye absorbs light and
converts it to a signal carried by nerves to the brain; it can be
OH
H
O
O
retinal
geranial
O
H
CH2OH
C
H
C
OH
OH
C
H
H
C
H
C
C
O
HO
C
H
OH
H
C
OH
OH
H
C
OH
CH2OH
glucose (open chain form)
CH2OH
fructose (open chain form)
Figure 23.1 Some naturally occurring ketones and aldehydes.
216
Unit 2 Module 1 The chemistry of carbon compounds
Nomenclature of aldehydes and ketones
Bonding in the carbonyl group
Aldehydes are named systematically by replacing the final
-e in the name of the parent alkane with -al to give alkanal
(Figure 23.2). A locant is not needed since –CHO must be
at the end of a chain; it defines position 1 if a locant is
needed for any substituents that may be present on the
chain. The first two members of the series are sometimes
given their common names of formaldehyde (methanal)
and acetaldehyde (ethanal).
Using methanal, H2C=O, as an example, we will examine
bonding in the >C=O group (Figure 23.4).
CH3
O
H
Cl
H
O
C
O
2-chloropropanal
H
C
C
ethanal
(acetaldehyde)
O
O
H
CH3
H
CH3
trans-2-butenal
O
H C
C
H
CH3 H
C
CH3CH2
H
H
CH3
H
methanal
(formaldehyde)
C
C
propanal
C
C
H
CH3 H
C
CH3
propanone (acetone)
C
C
The carbonyl carbon is sp2 hybridized, so there are three
sp2 atomic orbitals and one p atomic orbital from carbon.
This total of four orbitals corresponds to four outer-shell
electrons of carbon. The oxygen atom is also sp2 hybridized,
and there are three sp2 atomic orbitals and one p atomic
orbital from oxygen. Note, however, that oxygen has six
outer shell electrons. The three carbon sp2 atomic orbitals
are used to produce three sigma (σ) bonds:
overlap;
■ the carbon and oxygen p atomic orbitals form a π
bond.
Four bonding molecular orbitals, three σ and one π, result
from combination of the atomic orbitals. Each of these
four molecular orbitals is occupied by two paired electrons,
giving a total of eight electrons involved in bonding:
■ two of the bonding electrons originate from H;
■ four electrons were the outer-shell electrons from C;
Oxygen, therefore, has four outer-shell electrons which
are not involved in bonding. These form two lone pairs
which are represented by pairs of dots above and below the
symbol for oxygen.
CH2CH3
butanone
O
CH3CH2
CH2CH2CH3
sp2 O
■ two electrons originate from O.
O
CH3
H
■ one σ bond is formed with oxygen by sp2(C)–sp2(O)
O
CH3
m bonds
lone pairs or
non-bonded
electrons
O
overlap;
Ketones are named systematically as derivatives of the
corresponding alkane by replacing the final -e with -one
to give alkanone (Figure 23.3). The position of the C=O
is shown by a locant at the start of the name; the carbon
skeleton is numbered to give the locant the smallest possible
number. The first two ketones, propanone and butanone,
do not need a locant; the names are unambiguous since the
C=O must be at position 2 in both molecules. The common
name ‘acetone’ is often used for propanone. The name
pentanone is ambiguous and a locant is required.
CH3
C
■ two σ bonds are formed with two H atoms by sp2(C)–s(H)
3,4-dimethylpentanal
Figure 23.2 Structures and names of some aldehydes.
O
/ bond
H
Figure 23.4 The bonding in methanal.
H
C
C
sp2 C
2-pentanone
C
CH2CH3
3-pentanone
Figure 23.3 Structures and names of some ketones.
The carbonyl group, >C=O, is planar because the carbon
atom is sp2 hybridized. As in alkenes, the angles between
the bonds radiating from the sp2 carbon are 120°.
ITQ 1
(a) Provide systematic names for the following compounds.
i
H
H
CH3
C
C
H
H
O
ii
H
C
H
Br
H
H
C
C
C
H
H
H
O
C
H
iii
H
H
O
H
H
C
C
C
C
H
H
H
iv
Br
O
H
H
H
H
C
C
C
C
C
H
H
H
Br
(b) Draw the structures of: (i) 2-bromopropanal; (ii) 1-penten-3-one; (iii) 2,4-dimethylpentan-3-one
H
H
Chapter 23 Aldehydes and ketones
General properties of aldehydes and
ketones
Oxygen is more electronegative than carbon, so the O of
the >C=O pulls more electron density towards itself. The
Pauling electronegativity scale values are O = 3.5 and
C = 2.1 (see page 23, Chapter 2). This polarization has a
greater effect on the electrons in the polarizable π bond,
leading an increase in electron density on O and a decrease
on C. Organic chemists often describe this using a resonance
representation, as shown in Figure 23.5.
+
C
O
–
O
C
Figure 23.5 Canonical forms of the carbonyl group.
The diagram in Figure 23.5 tells us that neither the left-hand
structure nor the right-hand charge-separated structure
describes the carbonyl group completely; the true structure
is a blend of both. This existence of a structure in two
forms which differ only in their electron arrangement is
called resonance. The concept of resonance was developed
by organic chemists who realized that their traditional
representation of molecules as arrays of atoms joined by
bonds (shown as lines) that served very well for many
molecules was inadequate in some cases. The structures we
have drawn here are called canonical forms; each on its
own is inadequate, but together they form a resonance
hybrid that more closely describes the true structure. It
is important to distinguish the double-headed resonance
arrow (↔) from the arrow used to show an equilibrium (ҡ).
A
B
equilibrium
C
D
resonance
A and B are two different molecules that can interconvert
in the equilibrium. C and D are two canonical forms of the
same molecule. There is only one molecule. The resonance
arrow is our statement that C and D separately are not
adequate representations of the molecule, which has
characteristics of C and D, but we cannot draw a single
structure that shows both together.
The carbonyl group is exceptionally stable, and its chemistry
is dominated by this stability. An extension of resonance
theory tells us that a resonance hybrid must be more stable
than either of its canonical forms, but it is not intuitively
obvious why the carbonyl group should be as stable as it
is. The polarization of the group is the other feature that
has a dominant effect on its chemistry. It is sometimes
convenient to indicate the polarization of the C=O by
placing a partial negative charge (δ−) on the oxygen atom
and a partial positive charge (δ+) on the carbon atom.
b+
b–
C
O
You will recall that δ+ and δ− are used to show the
polarization of the O–H bond in alcohols and in water.
Water forms hydrogen bonds to the oxygen of carbonyl
compounds. Low molecular weight carbonyl compounds
are therefore soluble in water, and the resulting solutions
are neutral.
Low molecular weight carbonyl compounds are extremely
flammable, with low flash points.
O
H
C
Methanal
H
Also known as formaldehyde; RMM = 30. colourless gas,
boiling point = −21 °C. Used as a solution in H2O. Very
high solubility in water; forms a hydrate with H2O. Formed
from oxidation of methanol, CH3OH. Used in synthesis of
polymers; the aqueous solution, formalin, is a preservative
and disinfectant.
OH
H
C
methanal hydrate
OH
H
O
CH3
Ethanal
C
H
Also known as acetaldehyde; RMM = 44. Colourless
liquid, boiling point = +20 °C; fruity odour. Soluble in
all proportions with water. Formed by the hydration of
HC≡CH or the oxidation of H2C=CH2 (Wacker process).
Used in the synthesis of ethanoic acid, various esters and
other compounds.
O
CH3CH2
ITQ 2 Draw the resonance showing the two canonical forms of
propanal using Figure 23.5 as the model.
H
H
H
C
C
H
H
propanal
O
C
H
Propanal
C
H
RMM = 58. Colourless liquid, boiling point = +47 °C;
pungent odour. Solubility in water of 20 g/100 cm3 at
20 °C. Formed by hydroformylation:
CO + H2 + C2H4 → CH3CH2CHO
Used in the synthesis of alkyd resins and small molecules.
217
218
Unit 2 Module 1 The chemistry of carbon compounds
O
special reagents oxidize primary alcohols only as far as the
aldehyde.
C
CH3CH2CH2
Butanal
2–
H
RMM = 72. Colourless liquid, boiling point = +76 °C;
sweat-like odour. Solubility in water of 7.6 g/100 cm3 at
20 °C. Formed by hydroformylation:
CH3CH2CH2
i Cr2O7
OH
1-propanol,
a primary alcohol
H
CO + H2 + C3H6 → CH3CH2CH2CHO
C
CH3CH2
Used as starting material in chemical synthesis.
CH3
O
C
CH2CH3
Also known as methyl ethyl ketone (MEK); RMM = 72.
Colourless liquid, boiling point = +80 °C. Solubility in
water = 27.5 g/100 cm3 at 20 °C. Formed by the oxidation
of 2-butanol or of 2-butene. Used as an important solvent.
Preparation of aldehydes and ketones
Aldehydes and ketones can be prepared by the oxidation
of alcohols (see page 203). Oxidation in this context means
removal of two H atoms. Ketones are obtained by oxidation
of secondary alcohols with a variety of oxidizing agents.
The most commonly used oxidants are compounds of CrVI
(orange solutions) which, after oxidizing the alcohol, are
reduced to CrIII (green). Widely used CrVI oxidizing agents
are dichromate, Cr2O72−, as the potassium or sodium salt,
and a solution of CrO3 in H2SO4.
OH
H3C
C
H
2-propanol,
a secondary alcohol
The carbon of the >C=O bears a partial positive charge
(δ+) which is balanced by a partial negative charge (δ−) on
the oxygen. The carbonyl carbon is therefore electrophilic
(electron seeking) and is subject to the addition of
nucleophiles (species which seek positive charges) (Figure
23.6). This feature dominates the chemistry of aldehydes
and ketones.
electron pair
from nucleophile
forms a bond
to carbonyl C
b+
C
b–
–
O
C
O
Nu
Nu
–
nucleophile
seeks positive
charge
electron pair comprising
one of the C–O bonds
moves on to O so
that C remains
tetravalent
intermediate product
of addition of
Nu – to C O
Figure 23.6 Addition of a nucleophile, Nu:− to >C=O.
Aldehydes are much more susceptible to nucleophilic
addition than ketones. One reason for this is that the
carbonyl carbon of aldehydes (with one H and one alkyl
group) is less sterically hindered than the carbonyl carbon
of ketones (Figure 23.7). Another reason is that the partial
positive charge (δ+) on the carbonyl carbon of ketones is
stabilized by the electrons which form the bonds of the two
attached alkyl groups.
O
2–
CH3
O
propanoic acid,
a carboxylic acid
Reactions of aldehydes and ketones
Also known as acetone; RMM = 58. Colourless liquid,
boiling point = +57 °C. Soluble in water in all proportions.
Synthesized from propene, CH=CH2CH3, via various
processes. Is an important solvent; starting material for
synthesis of many compounds and polymers, including
polymethylmethacrylate.
Butanone
C
CH3CH2
C
CH3
Propanone
CH3
O
propanal,
cannot be isolated
O
OH
2–
ii Cr2O7
Cr2O7
C
H3C
CH3
propanone,
a ketone
Aldehydes cannot be isolated from the oxidation reaction
of primary alcohols because, once formed, they rapidly
undergo further oxidation to carboxylic acids. Some
propanone
propanal
Figure 23.7 Space-filling models of propanal and propanone
show that the >C=O of propanal is more sterically accessible.
ITQ 3 Which alcohol would you need to use to prepare (a) 2-pentanone and (b) 3-pentanone?
Chapter 23 Aldehydes and ketones
Reduction of aldehydes and ketones
An aldehyde can be reduced to a primary alcohol and a
ketone can be reduced to a secondary alcohol, both by the
addition of H atoms.
H
CH3CH2
C
+
O
2H
CH3CH2
H
C
OH
H
propanal, an aldehyde
1-propanol, a primary alcohol
H
CH3
C
+
O
2H
CH3
CH3
C
OH
CH3
propanone, a ketone
2-propanol, a secondary alcohol
These reductions can, in fact, be carried out by adding H2
to the carbonyl compound in the presence of a catalyst.
However, this method of reducing aldehydes and ketones
is not commonly used on a laboratory scale.
Aldehydes and ketones are easily reduced to alcohols by
reagents which deliver hydride, ‘H:−’. Sodium borohydride,
NaBH4, and lithium aluminium hydride, LiAlH4, are the
most widely used hydride reducing agents.
H
H
B
H
H
+
–
H
Al –
Na
H
H
Li
+
H
This reducing property of aldehydes is utilized in the
Fehling’s and Benedict’s reactions and in the Tollens’ silver
mirror reaction which are used in qualitative analysis.
Fehling’s and Benedict’s reagents contain CuII ions, which
are blue. Addition of an aldehyde to either of these solutions
causes reduction of the CuII to CuI and formation of Cu2O,
which is a red-brown precipitate.
Tollens’ reagent is ammoniacal silver nitrate solution. It
is prepared by adding a small amount of aqueous NaOH
to a solution of AgNO3. Silver oxide, Ag2O, is precipitated
and aqueous ammonia is added to the mixture until the
precipitate disappears. On addition of an aldehyde to
Tollens’ reagent the Ag+ in the complex cation, Ag(NH3)2+,
is reduced to metallic silver which forms a mirror on the
walls of the (clean) reaction vessel.
Ketones are oxidized with dichromate, Cr2O72−, or
permanganate, MnO4−, under forcing conditions of high
temperature and acidity. This oxidation causes C–C bond
rupture and is of little synthetic or analytical use and will
not be discussed further. Ketones do not reduce CuII or
AgI, so do not react with Fehling’s, Benedict’s or Tollens’
reagents.
O
An important oxidation reaction of ketones
is the iodoform reaction of methyl ketones, CH3 C
The
group is known as the acetyl group.
O
Sodium borohydride is very much less reactive than lithium
aluminium hydride.
Reductions with NaBH4 can be carried out in alcoholic
and even aqueous solutions, but strictly non-protic and
anhydrous conditions must be maintained for reductions
with LiAlH4. Both NaBH4 and LiAlH4 reduce >C=O groups
by addition of H:− (a nucleophile) to the electrophilic
carbonyl C.
Oxidation of aldehydes and ketones
Oxidation of aldehydes with common laboratory oxidizing
agents such as potassium permanganate or dichromate
proceeds rapidly to yield carboxylic acids in good yield.
H
Iodoform is triiodomethane, HCI3, a pale yellow antiseptic
smelling solid, melting point = 123 °C, which is insoluble in
water. The iodoform reaction is used as a qualitative test for
O
.
methyl ketones,
CH3
O
C
The equation for the iodoform reaction is:
O
+
C
CH3
4OH –
+
3I2
R
O
HCl3
+
+
C
O
78% yield
The reaction of aldehydes with CrVI reagents can be used to
detect aldehydes; the colour of the solution changes from
clear orange (CrVI) to cloudy green (CrIII) as the aldehyde is
oxidized and the CrVI is reduced by the aldehyde.
R
Secondary alcohols that form methyl ketones on oxidation
also undergo the iodoform reaction because
OH
they are oxidized to the methyl ketones under
CH3
C R
the conditions of the iodoform reaction (see
H
Chapter 21, page 204).
H2SO4, H2O 15-20 ˚C
OH
C
CH3
KMnO4
O
R
–
3I – +
3H2O
R
To carry out the iodoform reaction, follow this sequence.
■ The methyl ketone (or methyl alcohol) is dissolved in
the solvent dioxane.
219
220
Unit 2 Module 1 The chemistry of carbon compounds
■ Dilute aqueous NaOH is added to the dioxane solution.
■ A solution of I2 in aqueous NaOH is then added.
CH3
■ The reaction mixture is warmed.
the reason for adding H2O is to cause the HCI3 to
precipitate.
Nucleophilic addition
Hydrogen cyanide, displayed formula H–C≡N, adds to
aldehydes and to some ketones to form cyanohydrins.
CH3
OH
+
C
H
C
CH3
N
H
C
N
H
ethanal
cyanohydrin
b
OH
O
CH3CH2
C
+
C
H
C
CH3CH2
N
CH3
C
C
cyanohydrin
Hydrogen cyanide (H–C≡N) is sometimes called hydrocyanic
acid and is a weak acid. A strong base will remove the
proton from H–C≡N to produce nitrile ions.
B
base
+
H
C
N
B
H
+
–
C
N
nitrile or cyano
ion, a very strong
nucleophile
–
CH3
O
C
C
N
ii H +
CH3
H
alkoxide
O
H
C
C
N
H
cyanohydrin
(ii) alkoxide intermediate is protonated
Figure 23.8 Mechanism of cyanohydrin formation.
The cyanide ion, −:C≡N, is lethally toxic to human beings
because it inhibits important enzymes which are involved
in the production of adenosine triphosphate.
Derivatives of cyanohydrins occur in a number of plants,
including bitter almonds and some beans and tubers,
making them potentially toxic. However, some of these
plants are used as a source of food. A noteworthy example
is the cassava root, which people in several parts of the
world, including many Caribbean countries, use in their
daily diet. The grated cassava root must first be carefully
treated to hydrolyse the toxic components (linamarin and
lotaustralin) and remove the cyanide.
Addition/elimination reactions
The C≡N group (the nitrile or cyanide group) adds to the
carbonyl carbon and H adds to the oxygen. A new C–C
bond is formed, so this reaction is very important in organic
synthesis as the construction of carbon frameworks is a key
process. The –C≡N of a cyanohydrin can be converted to
a carboxyl group (COOH) by hydrolysis or to a primary
amine (-CH2NH2) by reduction.
–
C
N
CH3
butanone
–
(i) nucleophilic nitrile ion adds to electrophilic carbonyl C
Analogous reactions occur between acetyl groups and
bromine or chlorine to produce bromoform, HCBr3
(tribromomethane), or chloroform, HCCl3 (trichloromethane). Chloroform and bromoform are liquids at
room temperature, while iodoform is a nicely crystalline
compound that is easily observed, so the iodoform reaction
is used for analysis. The general reaction is known as the
haloform reaction.
O
i
+
C
H
■ H2O is added, and the iodoform, HCI3, precipitates;
a
O
N
nitrile or cyano ion
a very strong
nucleophile
The cyanide ion, −:C≡N, is a very strong nucleophile
and readily adds to the electrophilic carbonyl carbons of
aldehydes and some ketones. The intermediate alkoxide
which is formed is then protonated (Figure 23.8).
Aldehydes and ketones undergo a condensation reaction
with most compounds containing a primary amino group,
–NH2. If the –NH2 group is part of a primary amine, as
shown in Figure 23.9, the product is an imine, also called
a Schiff base.
Derivatives of hydrazine, H2N–NH2 (of which
2,4-dinitrophenylhydrazine,
H
NO2
H
abbreviated
2,4-DNP
and
N N
sometimes
called
Brady’s
reagent, is very important) react H
NO2
with aldehydes and ketones.
Aldehydes and ketones also react with hydroxylamine,
H2N–OH.
When an aldehyde or a ketone condenses with
2,4-dinitrophenylhydrazine the product is a yellow or
orange-coloured crystalline solid 2,4-dinitrophenylhydrazone (a 2,4-DNP derivative) (Figure 23.10).
The formation of 2,4-dinitrophenylhydrazones is a useful
qualitative test for aldehydes and ketones. An alcohol
solution of 2,4-dinitrophenylhydrazine is added dropwise to
a solution of the test compound. The formation of an orange
Chapter 23 Aldehydes and ketones
or yellow precipitate of the 2,4-dinitrophenylhydrazone
confirms that the test compound is an aldehyde or a ketone.
Before spectroscopic methods were used routinely for
identification of organic compounds, the solid 2,4-DNP
derivative of an aldehyde or ketone of unknown structure
would be prepared. It would be purified by recrystallization
and its melting point determined. The aldehyde or ketone
could then be identified by comparison of the melting
point of its 2,4-DNP derivative with melting points of
2,4-dinitrophenylhydrazones of known structure.
The condensation of hydroxylamine, H2N–OH, with
aldehydes or ketones yields oximes (Figure 23.11). Many
oximes are crystalline solids and this reaction, like the
reaction with 2,4-DNP, can be used in qualitative analysis
for identification of carbonyl compounds.
CH3
O
C
H
+
N
CH3
CH2CH3
H
H
C
CH2CH3
N
+
H2O
H
ethanal
ethylamine
imine or Schiff base
Figure 23.9 The reaction between ethanal and ethylamine.
CH3
C
H
H
O
+
N
H
NO2
N
H
+
N
H
NO2
ethanal, colourless
liquid bp 20 ˚C
N
C
CH3
H
NO2
2,4-dinitrophenylhydrazine,
orange solid used as a
solution in methanol or ethanol
NO2
2,4-dinitrophenylhydrazone of ethanal,
orange crystalline solid
precipitates from solution
mp 165 ˚C
Figure 23.10 The reaction between ethanal and 2,4-dinitrophenylhydrazine.
CH3
O
C
H
+
CH3
propanone (acetone),
colourless liquid
bp 56 ˚C
N
OH
CH3
H
C
N
OH
CH3
hydroxylamine,
colourless solid used
as a solution in
methanol or ethanol
oxime of acetone,
white crystalline solid
precipitates from solution
mp 62 ˚C
Figure 23.11 The reaction between propanone and hydroxylamine.
ITQ 4 Compounds E, F and G are isomers with the formula
C4H8O; all are derivatives of butane. From the data in the table
below deduce the structure of each of these compounds and of
the derivative produced when a test is positive.
Compound Iodoform test Fehling’s test 2,4-DNP test
NaBH4 treatment
E
negative
positive
positive
C4H10O formed
F
positive
negative
positive
C4H10O formed
G
positive
negative
negative
no reaction
+
H2O
H2O
221
222
Unit 2 Module 1 The chemistry of carbon compounds
Review questions
Summary
1
Cyclopentanone, shown below, is a cyclic ketone.
✓ Aldehydes and ketones contain the carbonyl,
O
>C=O, group, which, due to the difference in
electronegativity between C and O, is polarized.
Polarization leads to an increase in electron
density on O and a decrease on C.
cyclopentanone
(a) Show how the carbonyl group in cyclopentanone
is polarized; use δ+ and δ−.
(b) Show the reaction of cyclopentanone with cyanide
ion (−:C≡N) to give an alkoxide. Use two curly
arrows to illustrate the mechanism. Refer back to
page 220 if you need.
✓ The >C=O group is a hybrid of two canonical
forms.
+
C
O
C
–
O
✓ Aldehydes and ketones are prepared by
oxidation of alcohols, and are converted to
alcohols by reduction.
2
✓ Aldehydes are easily oxidized to carboxylic acids
and, in the process, reduce the oxidizing agent.
This reducing property of aldehydes is used
analytically to detect them by observation of the
following changes:
CrVI (orange) → CrIII (green)
MnVII (purple) → MnII (colourless)
CuII (blue) → CuI (brick-red)
AgI (colourless) → Ag0 (gray)
✓ Methyl ketones and secondary alcohols which
(a) Write the structures of three straight-chain
compounds with molecular formula C5H10O which
contain a carbonyl group.
(b) Name each of the compounds which you have
drawn in part (a).
(c) How would you distinguish between these C5H10O
isomers by subjecting each of them to the same
two chemical tests? Negative evidence is just as
important as positive evidence.
Answers to ITQs
1
form methy ketones on oxidation methyl groups
can be detected by the iodoform reaction.
✓ Secondary alcohols with adjacent methyl groups
(a) (i) 2-methylpropanal
(ii) 4-bromobutanal
(iii) 1-bromobutanone
(iv) 1-bromopentan-2-one
(b)
are oxidized to methyl ketones under the
reaction conditions of the iodoform reaction (in
situ) and give a positive iodoform test.
i
H
✓ The carbonyl C of aldehydes and ketones is
H
Br
C
C
H
H
O
H
C
C
C
H
C
H
2-bromopropanal
electrophilic, and nucleophiles add to this C. An
important example of nucleophilic addition to
carbonyl compounds is the reaction with H–C≡N
to form cyanohydrins.
CH2CH3
H
1-penten-3-one
iii
H
✓ Aldehydes and ketones condense with
compounds containing –NH2 groups. The
condensation of aldehydes and ketones with
2,4-dinitrophenylhydrazine to give yelloworange crystalline 2,4-dinitrophenylhydrazones
is used in the detection and identification of
aldehydes and ketones.
O
ii
CH3 O
CH3
C
C
C
H
CH3
CH3
2,4-dimethylpentan-3-one
2
H
H
H
C
C
H
H
C
H
propanal
3
H
O
(a) 2-pentanol
(b) 3-pentanol
H
–
H
C
C
H
H
+
C
H
O
Chapter 23 Aldehydes and ketones
4
Compound E
O
O
Fehling’s test
C
H
CH2CH2CH3
+
C
(2Cu2+ + 5OH – )
blue solution
–
O
CH2CH2CH3
Cu2O
+
3H2O
+
H2O
brick-red ppt
aldehyde,
butanal
2,4-DNP test
NaBH4
NO2 H
H
N
OH
NO2 H
N
N
H
H
C
CH2CH2CH3
N
O2N
H
CH2CH2CH3
C
orange solution
in MeOH or EtOH
O2N
H
2,4-dinitrophenylhydrazone of butanal
orange ppt
Compound F
O
iodoform test
O
CH3
C
CH2CH3
+
HCI3
(4OH – + 3I2 )
+
C
–
iodoform
yellow ppt
O
3I
–
+
3H2O
CH2CH3
methyl ketone,
butanone
2,4-DNP test
NaBH4
NO2 H
N
OH
H
NO2 H
N
N
H
H
C
CH2CH3
O2N
C
orange solution
in MeOH or EtOH
CH3
+
N
O2N
H2O
CH2CH3
CH3
2,4-dinitrophenylhydrazone of butanone
orange ppt
Compound G
OH
CH3
C
HH
O
H
C
C
H
2˚ alcohol with an
adjacent methyl group,
2-butenol
O
CH3
C
O
H
C
H
methyl ketone
C
H
iodoform test
HCl3
(4OH – + 3I2 )
iodoform
yellow ppt
+
–
H
C
C
O
H
C
H
+
3I
–
+
3H2O
223
224
Chapter 24
Carboxylic acids and derivatives
Learning objectives
■ Systematically name simple carboxylic acids, esters, acyl chlorides and amides.
■ Explain the consequences of polarization of the –COOH group.
■ Describe three methods for preparing carboxylic acids.
■ Explain the relationship between Ka, pKa and acidity of carboxylic acids, and the effect of
■
■
■
■
electronegative substituents on the acidity of carboxylic acids.
Describe the general features of amino acids.
Write equations for the reactions of carboxylic acids with various bases and for carboxylate salts with
mineral acid.
Demonstrate the relationships between carboxylic acids and esters, carboxylic acids and acyl chlorides,
and carboxylic acids and amides.
Suggest methods for the preparation and hydrolysis of esters, acyl chlorides and amides.
Introduction
O
Carboxylic acids contain the
C O H functional
group, known as the carboxyl group. The carbon of the
carboxyl group is sp2 hybridized and is doubly bonded to
oxygen and singly bonded to a hydroxyl (O–H) group. The
carboxyl carbon cannot undergo further oxidation without
fragmenting the molecule.
Carboxylic acids with saturated carbon chains have the
general formula CnH2n+1COOH. They are sometimes called
fatty acids because some higher members of the series
occur in natural fats. Examples of acids that occur in
natural fats are stearic acid, C17H35COOH, and palmitic acid,
C15H31COOH. You can recognize where stearic acid was
originally discovered if you know that the word ‘stearic’ is
derived from the Greek for animal fat (tallow).
Nomenclature
In IUPAC nomenclature the name of a carboxylic acid is
obtained by changing the -e of the parent alkane to -oic
acid, giving alkanoic acid. The carbon chain includes the C
of the –COOH, which must be at the end of the chain and
is considered to be position 1. The name does not need this
locant to be unambiguous, but the locants for substituents
on the chain must be shown. The first few members of
the series are sometimes called by their trivial names; these
names are derived from the original natural sources of the
acid. Figure 24.1 gives some examples.
O
H
C
O
OH
CH3
methanoic acid
or formic acid
C
OH
CH3
O
CH3CH2
C
ethanoic acid
or acetic acid
OH
CH3CH2
propanoic acid
or propionic acid
O
C
CH
H
Br
C
OH
2-bromo-3-methylpentanoic acid
Figure 24.1 Structures and names of some carboxylic acids.
General properties
The C1–C3 carboxylic acids are pungent-smelling liquids,
the C4–C9 compounds are rank-smelling oils and the acids
with ten and more carbons are odourless solids.
The presence of two electronegative oxygen atoms in the
carboxyl group causes it to be polarized, as shown here for
ethanoic acid.
b–
b+
CH3
O
C
O
b–
H
b+
Chapter 24 Carboxylic acids and derivatives
One molecule of a carboxylic acid can form two hydrogen
bonds to a second molecule of the same acid, as shown in
Figure 24.2, to produce a dimeric structure.
CH3
b–
b+
O
H
O
C
CH3
C
H
O
O
b–
b+
Figure 24.2 Hydrogen bonding between two molecules of
ethanoic acid.
These dimers occur in the vapour phase. In liquid carboxylic
acids, i.e. those members of the homologous series with
1–9 carbon atoms, many molecules associate by hydrogen
bonding. As a result, the boiling points of carboxylic acids
are relatively high. Carboxylic acids also form hydrogen
bonds with water, as illustrated in Figure 24.3. One
consequence of this hydrogen bonding with water is that
carboxylic acids with up to four carbons are miscible with
water in all proportions.
CH3
b–
b+
O
H
C
b– O
O
b–
Also known as valeric acid; CH3(CH2)3COOH; RMM = 102.
Boiling point = +187 °C; solubility in water = 4.97 g/100
mL at 25 °C. pKa = 4.82. Natural source is valerian root.
Hexanoic acid
Also known as caproic acid; CH3(CH2)4COOH; RMM = 116.
Boiling point = +205 °C; solubility in water = 1.08 g/100
mL at 25 °C. pKa = 4.54. Natural source is goats’ cheese.
Preparation of carboxylic acids
Oxidation reactions of alcohols, aldehydes and alkenes
can be used to prepare carboxylic acids. Primary alcohols
are oxidized to aldehydes and, with strong oxidizing
agents such as acidified potassium dichromate or acidified
potassium permanganate, the aldehydes are rapidly
oxidized to carboxylic acids (Figure 24.4). Reaction 2 is
faster than reaction 1, so it is not possible to isolate the
intermediate aldehyde.
b–
H
O
H
b+
b+
H
H
O
b+
1 K2Cr2O7
CH3CH2OH
C
b+
H
Pentanoic acid
O
CH3
C
CH3
O
O
2 K2Cr2O7
CH3
ethanol
OH
ethanal
ethanoic acid
(acetic acid)
b–
b–
b+
Figure 24.3 Hydrogen bonding between ethanoic acid and H2O.
Methanoic acid
Also known as formic acid; HCOOH; RMM = 46. Boiling
point = +101 °C; soluble in water in all proportions. pKa =
3.75. Natural source is ants.
C
H
Figure 24.4 Oxidation of ethanol to ethanoic acid with acidified
potassium dichromate.
Aldehydes can also be used as starting materials for the
preparation of carboxylic acids (Figure 24.5).
O
R
O
1 [oxidation]
C
2 H+
H
aldehyde
R
C
OH
carboxylic acid
Ethanoic acid
Figure 24.5 General reaction for the oxidation of aldehydes to
carboxylic acids.
Also known as acetic acid; CH3COOH; RMM = 60. Boiling
point = +118 °C; soluble in water in all proportions. pKa =
4.76. Natural source is vinegar.
Acidity of carboxylic acids
Propanoic acid
The acid strength of a compound is the extent to which it
gives up a proton to a dipolar solvent in which it is dissolved.
Carboxylic acids ionize in water, which is a dipolar solvent.
Also known as propionic acid; CH3CH2COOH; RMM = 74.
Boiling point = +141 °C; soluble in water in all proportions.
pKa = 4.87. Natural source is milk, butter, cheese.
RCOOH
+
H2O
RCOO
–
+
+
H3O
Butanoic acid
ITQ 1 How would you prepare butanoic acid, CH3CH2CH2COOH,
from each of the following compounds?
Also known as butyric acid; CH3(CH2)2COOH; RMM = 88.
Boiling point = +164 °C; soluble in water in all proportions.
pKa = 4.81. Natural source is rancid butter.
(a) CH3CH2CH2CH2OH, butanol
O , butanal
(b)
CH3CH2CH2
C
H
225
226
Unit 2 Module 1 The chemistry of carbon compounds
The equilibrium constant for this reaction is known as an
acidity constant, Ka.
Cl
Cl
[RCOO−][H3O+]
Ka =
[RCOOH]
C
O
–
Cl
A simple equilibrium constant would have [H2O] in the
bottom line. However, because the equilibrium is far to
the left for this reaction, [H2O] is effectively constant and
is, in effect, incorporated into Ka. The numerical value of
Ka is very small so acidity is generally expressed as pKa,
which is −log10 Ka, just as pH = −log10 [H+].
large Ka → small pKa → stronger acid
small Ka → large pKa → weaker acid
–
+
H3O
H
C
b+
O
b–
ethanoic acid
pKa 2.86
pKa 4.76
The influence of the inductive effect on acid strength
decreases with increasing distance between the –COOH
group and the electronegative group (Cl); this is shown in
Figure 24.7.
Cl
CH3CH2
C
Cl
CH3
COOH
H
Chlorine is more electronegative than carbon, so the
polarization of the Cl–C bond by the inductive effect results
in a partial negative charge (δ−) on chlorine and a partial
positive charge (δ+) on the carbon. This carbon is adjacent
to the carbonyl C, which also carries a δ+ charge. The two
δ+ charges further polarize the O–H bond and ionization
is enhanced. This effect increases with increasing Cl
substitution, as shown in Figure 24.6.
CH2COOH
H
3-chlorobutanoic acid
pKa 4.0
H
C
CH2CH2COOH
CH3CH2CH2COOH
butanoic acid
H
pKa 4.8
4-chlorobutanoic acid
pKa 4.5
Figure 24.7 pKa values of butanoic acid and isomeric
monochlorobutanoic acids.
Amino acids
Carbon atoms bonded to C=O groups are known as α
(‘alpha’) carbons, and the atoms or groups attached to α
carbons are also designated α. Carboxylic acids with an
–NH2 (amino) group bonded to the C next to the –COOH
are α-amino acids.
H
b+
adjacent b charges
C
Cl
+
O
COOH
chloroethanoic acid
pKa 2.9
b–
b+
C
b+
C
H
H
Ethanoic acid, CH3COOH, is a weak acid (pKa in H2O = 4.76).
Chloroethanoic acid, ClCH2COOH, with pKa = 2.81 in H2O,
is almost one hundred times as strong as ethanoic acid.
(The pKa scale is a logarithmic scale, so a decrease of one
unit represents a tenfold increase in Ka.) Chloroethanoic
acid is polarized:
H
H
COOH
2-chlorobutanoic acid
any factor which stabilizes RCOO− relative to RCOOH will
shift the equilibrium to the right and increase the acidity
of the carboxylic acid. When an acid releases a proton the
anion formed is known as the conjugate base of the acid.
So RCOO− is the conjugate base of RCOOH.
Cl
C
Figure 24.6 pKa values of ethanoic acid and chloro-substituted
ethanoic acids.
For the ionization of a carboxylic acid in water:
RCOO
H
pKa 1.48
H
Carboxylic acids protonate water only to a small extent, so
the equilibrium shown above lies far to the left.
b–
H
pKa 0.70
O
H2O
COOH
dichloroethanoic acid
O
O
C
Cl
trichloroethanoic acid
C
C
+
Cl
COOH
Cl
If the H atom is lost the remaining anion has two forms
that resonate. The anion is stabilized, which enhances the
ionization.
–
RCOOH
H
R
_
C
NH2
O
C
OH
α-Amino acids are the structural units of proteins. In the
animal kingdom, to which we all belong, proteins are
essential components of structural tissue (muscles, bones,
tendons), enzymes (which control metabolic processes)
and many other materials necessary for our maintenance
and preservation (e.g. blood, hair, hoofs, feathers).
Chapter 24 Carboxylic acids and derivatives
We know that the carboxyl group, –COOH, is acidic and
the –NH2 group is basic. The –NH2 group removes a proton
from –COOH, converting it to –COO− and becoming –NH3+.
Therefore, the general structure shown above is, strictly
speaking, incorrect, although you will sometimes see amino
acids drawn in that way. The structure of an α-amino acid
is correctly represented as shown here.
H
R
_
C
Table 24.1 Some essential amino acids
Name and
Structure
Abbreviation
glycine, Gly
CH2
H3N
NH2
297
117
315
131
295
131
284
165
283
the side chain
contains an
aromatic ring
105
228
the side chain
contains a
hydroxyl group
121
not
the side chain
crystalline contains a thiol
(SH) group
133
270
CH2 CH2CH2CH2NH2 146
224
C
–
NH3
CH(CH3 )2
O
Amino acids are always called by their trivial names. The
trivial name of each of the 20 essential amino acids is
assigned a three-letter abbreviation which is used when
describing and discussing the long sequences of amino
acids in peptides and proteins. Ten of the 20 essential
amino acids are listed in Table 24.1.
COO
+
valine, Val
The proteins and peptides which are fundamental to life
on Earth are constructed from a pool of 20 essential amino
acids. The simplest essential amino acid is glycine (Table
24.1). In glycine, the α carbon has two H atoms attached,
so this C atom is not a stereogenic centre. In all the other
19 essential amino acids, the α carbon has four different
substituents, so these amino acids are chiral. In these
19 essential amino acids the groups are attached to the
stereogenic centre, as shown in Table 24.1.
–
CH3
O
Such a structure is known as a zwitterion. A zwitterion
is a dipolar ion: a chemical species with both a positive
and a negative ionic charge. The zwitterionic structure
of amino acids causes them to have distinctive salt-like
properties. Amino acids are very polar; they are therefore
soluble in water and insoluble in non-polar solvents. They
are generally solids with high melting points, in contrast
to other organic compounds of comparable molecular
weight which are liquids at room temperature. The –NH3+
of amino acids is weakly acidic and will release a proton
to a strong base; the –COO− group is a weak base and will
accept a proton from a strong acid.
O
C
89
alanine, Ala
–
+
O
+
H
C
RMM Melting
Comments
point / °C
75
290
C
H
leucine, Leu
COO
+
–
NH3
CH2CH(CH3 )2
C
H
isoleucine, Ile
COO
+
–
NH3
H3C
CH2CH3
CH
C
H
phenylalanine,
Phe
COO
+
–
NH3
CH2
C
H
COO
+
serine, Ser
–
NH3
CH2OH
COO –
C
H
+
cysteine, Cys
NH3
CH2 SH
COO –
C
H
aspartic acid,
Asp
+
NH3
CH2 COOH
C
H
lysine, Lys
C
H
+
+
COO
an acidic
amino acid:
the side chain
contains a –
COOH group
a basic amino
acid: the side
chain contains
an –NH2 group
–
NH3
COO
–
NH3
ITQ 2
(a) The pKa values of four carboxylic acids, A, B, C and D are
given below. Calculate the pKa value of each and compare the
acid strength of that carboxylic acid with that of ethanoic acid
(pKa 4.76).
A Ka = 1.36 × 10−3
B Ka = 1.77 × 10−5
C Ka = 1.48 × 10−5
D Ka = 9.30 × 10−6
(b) Assign each of the structures below as A, B, C or D.
O
O
CH3CH2CH2
C
ClCH2
OH
CH3
OH
CH3
C
CH3
C
O
C
O
H
OH
C
OH
227
228
Unit 2 Module 1 The chemistry of carbon compounds
Hypoglycin A – an unusual amino acid in ackee.
COO
H
+
–
NH3
hypoglycin A
The fruit of ackee (Blighia sapida) produces a toxic amino
acid called hypoglycin A. This unusual structure, which
contains a cyclopropane ring, was determined in 1958
by chemists at the University of the West Indies, Mona
Campus, in Jamaica. Blighia sapida is native to West Africa
where its poisonous properties were well known. A few
seeds found their way to Jamaica on a slave ship and
when they germinated the tree flourished. Consumption
of unripe ackee fruits results in ‘Jamaican vomiting
sickness’, which can be fatal. Jamaicans have discovered
that the amount of toxin decreases as the fruit ripens,
and it is edible if properly prepared. The ackee fruit has
become a distinctive component of Jamaican cuisine
and is a part of the national dish, ackee and saltfish.
When the base is carbonate or hydrogencarbonate, CO2 is
evolved; this is a useful qualitative test for carboxylic acids.
Carboxylate salts formed from C1–C5 carboxylic acids are
completely water-soluble. Carboxylate salts of carboxylic
acids with very long carbon chains, e.g. sodium stearate,
CH3(CH2)16COO−Na+, are soaps. The long, non-polar carbon
chain (the ‘tail’) of the soap is insoluble in water and
the polar, ionic ‘head’ is water-soluble. The difference in
polarity between the two sections of long-chain carboxylate
salts enable soaps to serve as cleaning agents by forming
micelles. Micelles are globular clusters of carboxylate salts,
‘tails’ in and ‘heads’ out (Figure 24.9); when they surround
particles of dirt or grease the dirt or grease can be washed
away.
O
H
hydrophilic,
polar head group
H
H
KOH
CH3(CH2)4C
hexanoic acid
+
+
H2O
potassium hexanoate
O
O
hydrophobic,
non-polar
hydrocarbon chain
O
O
H
H
H
H
+
O
+
2 CH3(CH2)3C
+
H2O
+
–
CO2
O Na
H
H
O
H
H
H
H
O
H
CH3(CH2)3C
O
+
–
+
HCl
+
CH3(CH2)3C
NaCl
OH
O Na
pentanoic acid
The carboxylate ion is a weak base, and readily accepts a
proton from the strong mineral acid.
■ carboxylate salt + mineral acid →
carboxylic acid + inorganic salt
RCOO Na + HCl → RCOOH + NaCl
NaHCO3
O
OH
CH3(CH2)6C
octanoic acid
+
+
O–Na
+
H2O
sodium octanoate
O
decanoic acid
H
O
−
+
O
+
OH
O
H
H
O
When carboxylate salts are treated with mineral acids,
(HCl, H2SO4, HNO3) carboxylic acids are regenerated.
sodium pentanoate
O
CH3(CH2)8C
H
O
H
RCOOH + NaOH → RCOO− Na+ + H2O
O
OH
CH3(CH2)6C
H
O
■ carboxylic acid + base → carboxylate salt + water
Na2CO3
pentanoic acid
H
O
H
The two reactions described above:
O
2 CH3(CH2 )3C
H
H
H
H
sodium pentanoate
O–K
H
O
Figure 24.9 An illustration of a micelle.
O
+
H
H
H
Carboxylic acids undergo typical acid/base reactions with
basic compounds such as NaOH, Na2CO3 and NaHCO3 and
with amines or ammonia, RNH2, to form carboxylate salts
and water (Figure 24.8).
OH
O
H
H
O
Salts
CH3(CH2)4C
H
O
O
H
O
H
O
H
Reactions of carboxylic acids and their
derivatives
H
H
NH3
CH3(CH2)8C
+
O–NH4
ammonium decanoate
Figure 24.8 Reactions of carboxylic acids with bases.
CO2
+
can be used to separate mixtures consisting of carboxylic
acids and neutral water-insoluble compounds. The sodium
salt is always water soluble, so the mixture can be treated
with sodium hydroxide solution to dissolve the acid and
then filtered to remove any water-insoluble compounds.
The filtrate can then be treated with a mineral acid to
regenerate the water-insoluble carboxylic acid.
Chapter 24 Carboxylic acids and derivatives
Table 24.2 Properties of some esters
Esters
Name
Formation of esters
An ester is a derivative of a carboxylic acid in which the –H
of the –COOH group is replaced by an alkyl group, R′.
O
R
ethyl
butanoate
C
R
CH3CH2CH2
O
carboxylic acid
2-methylpropyl
propanoate
R’
ester
H of COOH is replaced by R’
O
O
CH3CH2
O
ethanoate
CH2CH2CH3
hexanoate
CH2CH3
pentyl
ethanoate
146
130
banana
149
172
orange
211
102
apple
103
C
O
CH3
propyl
C
O(CH2 )4CH3
octyl ethanoate
O
CH3
name: propyl hexanoate
ethyl
rum
CH3
CH3CH2CH2CH2CH2 C
CH3 C
130
C
OCH2CHCH3
An ester is named as the alkyl derivative of the carboxylic
acid.
O
116
Boiling
point /
°C
pineapple 121
OCH2CH3
C
H
RMM Odour/
flavour
O
O
O
O
Structure
C
O(CH2 )7CH3
name: ethyl ethanoate
Carboxylic acids condense with alcohols to form esters and
water.
O
H2SO4
CH3CH2CH2CH2CH2 C
CH3
+
O
H
H
hexanoic acid
methanol
O
CH3CH2CH2CH2CH2 C
methyl hexanoate
+
O
O
CH3CH2CH2
H2O
CH3
This condensation reaction is catalysed by mineral acid,
and the reaction mixture is usually heated at the boiling
point of the alcohol over a period of time, a process known
as refluxing. The alcohol is both a reactant and the solvent,
and is therefore present in large excess.
Esters of low molecular weight are generally pleasant
smelling liquids which are soluble in most organic solvents
(Table 24.2). Volatile esters occur in many fruits, giving
them both flavour and aroma, and are used in beverages
C
OCH3
(catalytic)
reflux
O
methyl
butanoate
and perfumes. A natural flavour can be a complex mixture
of several esters, often with one dominant, and other
organic compounds. Our senses of taste and smell are quite
subtle, and we can usually recognize quite small differences
in these blends. Artificial flavours are mixtures of synthetic
esters and other compounds that have been made in an
attempt to match the natural flavour, but in many cases we
still find that they taste artificial.
Fats, oils and waxes, all of which occur in plants and
animals, are also esters. Fats and oils are formed from
long-chain carboxylic acids and glycerol. Glycerol is a C3
compound with three –OH groups on adjacent carbons; it
is a triol. Fats and oils are therefore called triglycerides.
Trimyristin, found in nutmeg, is a typical triglyceride
(Figure 24.10).
O
CH3(CH2 )12
HO
O
3 CH3(CH2 )12
myrisitc acid
CH2
+
C
+
O
C
H
HO
CH
HO
CH2
glycerol
H
CH3(CH2 )12
CH3(CH2 )12
O
O
CH2
O
O
CH
O
CH2
C
C
trimyristin, a triglyceride
ITQ 3 Provide structures and names for the esters formed
between the each of carboxylic acids and each of the alcohols
shown below. You should end up with four esters.
Figure 24.10 Formation of trimyristin, a
triglyceride, from myristic acid and glycerol.
O
H
O
OH
CH3OH
methanoic acid
methanol
C
CH3CH 2
C
propanoic acid
OH
CH3C H2CH2OH
propanol
229
230
Unit 2 Module 1 The chemistry of carbon compounds
In triglycerides the portion of the ester derived from the
carboxylic acid consists of a C14, C16, C18 or C20 carbon chain
which may or may not contain double bonds.
Waxes are simple esters in which long-chain saturated
RCOO– groups are linked to long-chain alkyl groups. An
example is the main component of beeswax – its structure
is shown below.
O
CH3 (CH2 )24
O
R
O
NaOH(aq)
+
C
O
R
C
+
(CH2 )27CH3
ester
H2SO4
C
R’
+
O
H
H
carboxylic acid
O
alcohol
Figure 24.13 Ester hydrolysis, catalysed by base (saponification).
R
The formation of an ester from a carboxylic acid and an
alcohol is a reversible reaction (Figure 24.11). Removal of
H2O (catalysed by the mineral acid) causes the equilibrium
to shift to the right.
R
O
R’
sodium carboxylate
O
Hydrolysis of esters
O
H
O– Na+
R’
The carboxylic acid is formed when the reaction mixture is
acidified with mineral acid (Figure 24.14).
C
O
under basic conditions, the initial product is the sodium
carboxylate salt (Figure 24.13).
O
+
C
–
R
C
O Na
+
O
sodium carboxylate
NaCl
H
carboxylic acid
Figure 24.14 Conversion of a carboxylate salt to a carboxylic acid.
Acyl chlorides
(catalytic)
HCl
+
O
R
C
An acyl chloride,
Cl is derived from a carboxylic
acid by replacement of the –OH of the carboxyl group with
–Cl. Acyl chlorides are also known as acid chlorides.
reflux
alcohol
O
R
O
+
C
O
H2O
O
C
R
R’
R
OH
ester
C
Cl
carboxylic acid
acyl chloride
OH of COOH is replaced by Cl
Figure 24.11 General reaction for esterification.
The reverse reaction, the conversion of the ester to a
carboxylic acid and an alcohol, will occur if the ester is
treated with a large excess of H2O in the presence of the
acid catalyst (Figure 24.12). This reaction is hydrolysis
(hydro – water; lysis – cleavage).
O
R
H2SO4
+
C
O
R’
H2O
Acyl chlorides are important as intermediates in organic
synthesis; they are not generally themselves synthetic targets.
Acyl chlorides are very reactive, very much more reactive
than carboxylic acids, and are easily converted to other
derivatives of carboxylic acids such as esters and amides.
In the IUPAC system acyl chlorides are named by replacing
the -ic acid portion of the name of the carboxylic acid with
–yl chloride.
(catalytic)
reflux
(in excess)
O
ester
CH3
O
R
CH3CH2CH2
C
Cl
C
+
O
C
O
H
carboxylic acid
H
O
ethanoyl chloride
R’
alcohol
Figure 24.12 Ester hydrolysis, catalysed by acid.
Bases such as NaOH and KOH can also catalyse hydrolysis of
esters. A very important difference is that the base-catalysed
reaction is irreversible. So, in practice, we usually choose
to hydrolyse esters under alkaline conditions. The name
for base-catalysed hydrolysis of esters is saponification.
This terms arises because the process was originally used
to produce soap (Latin: sapo, saponis – soap) from the
triglycerides in animal fat. When an ester is hydrolysed
O
CH3(CH2 )4
C
Cl
butanoyl chloride
Cl
hexanoyl chloride
Acyl chlorides are prepared by the reaction of carboxylic
acids with thionyl chloride, SOCl2, or with phosphorus
trichloride, PCl3, or phosphorus pentachloride, PCl5 (Figure
24.15).
We saw previously that an ester can be prepared directly
from a carboxylic acid by heating with an alcohol and a
catalytic amount of mineral acid. The preparation of esters
via acyl chlorides is an alternative route. The acyl chloride
method is used when the alcohol is unstable, and cannot
withstand the high temperatures and strongly acidic
conditions otherwise needed (Figure 24.16).
Chapter 24 Carboxylic acids and derivatives
Amides
O
+
C
CH3CH2CH2
heat
SOCl2
OH
butanoic acid
thionyl chloride
O
( + SO2
C
CH3CH2CH2
+
HCl)
Cl
butanoyl chloride
An amide is derived from a carboxylic by replacement of
the –OH of the –COOH with –NR2. Nitrogen is trivalent,
so the N of the amide is linked to two other groups, which
may be H or alkyl.
O
+
C
3 CH3(CH2)3
C
R
O
carboxylic acid
( + H3PO3 )
C
3 CH3(CH2 )3
Cl
pentanoyl chloride
O
+
C
OH
heat
PCl5
phosphorus
pentachloride
ethanoic acid
amide: R’ and R’’ may be H or alkyl
OH of COOH replaced by NR’R’’
The acyl group, RC=O, of the amide has a strong effect on
the N, which is no longer appreciably basic. We can use
resonance theory to explain this difference. When we
draw the two principal canonical forms of an amide, we
see that one of them no longer has a lone-pair of electrons
on the N.
O
–
(+
C
CH3
POCl3
+
O
HCl)
C
R
Cl
N
ethanoyl chloride
Figure 24.15 Preparation of acyl chlorides.
CH3
O
CH3CH2
+
C
HO
Cl
C
+
CH3
N
CH3CH2
CH3
C
O
C
CH2CH3
temp
triethylamine
O
CH3CH2
room
CH3CH2
CH3
+
CH3
O
H
N
CH3CH2
+
Cl
–
CH2CH3
N+
Figure 24.16 Reaction of propanoyl chloride with tert-butanol to
give tert-butyl propanoate.
Peptides and proteins are polymers of amino acids which
are connected via amide linkages. The amide functional
group is therefore of crucial importance to life on Earth.
Amides are the least reactive of the carboxylic acid
derivatives covered in this chapter:
R
O
C
R
N
COOH
glutathione
(a) Identify the amide groups in glutathione.
(b) Show the three products of hydrolysis of glutathione in
aqueous HCl or aqueous NaOH.
Use Table 24.2 to identify two of the three hydrolysis products.
C
R’
N
ester
amide
R’’
Hydrolysis of amides is much slower and much difficult
to achieve than hydrolysis of esters. Amide hydrolysis
requires prolonged heating in concentrated aqueous acid
or base.
The reduction of an amide to an amine with lithium
aluminium hydride, LiAlH4, is a useful synthetic reaction.
O
O
R
OR’
N
H
O
C
Cl
acyl chloride
ITQ 4 Glutathione occurs in muscle and other animal tissues.
This compound is a tripeptide, a small protein that yields three
amino acids upon hydrolysis.
SH
H
O
R’
R’’
O
triethylammonium chloride
HOOC
C
These electrons have been delocalized, and are partly on N
and partly on O, so they are no longer as readily available
to be donated to a proton as they are in basic amines.
CH3
tert-butyl propanoate
NH2
R
R’
R’’
CH3CH2
tert-butanol
propanoyl chloride
R’
N
O
CH3
C
R’’
phosphorus
trichloride
pentanoic acid
R
OH
heat
PCl3
OH
O
CH3
C
N
CH3
LiAlH4
CH3
H
C
CH3
H
N
CH3
CH3
231
232
Unit 2 Module 1 The chemistry of carbon compounds
Review questions
Summary
1
(a) Draw the structure of the following carboxylic
acids:
(i) 2-bromoethanoic acid
(ii) 2-chloroethanoic acid
(iii) 2-fluoroethanoic acid
(b) Assign each of the pKa values below to one of the
carboxylic acids in part (a).
pKa: 2.81, 2.66, 2.87
Give reasons for your assignments.
2
Provide structures for the compounds F to K in the
following reaction sequences.
✓ Carboxylic acids contain the carboxyl group,
–COOH, which is very polar.
✓ Carboxylic acid molecules form hydrogen bonds
with each other and with H2O. This causes liquid
carboxylic acids to have high boiling points and
lower members of the homologous series to be
very soluble in H2O.
✓ Carboxylic acids can be prepared by oxidation of
primary alcohols or of aldehydes and hydrolysis
of nitriles.
F
✓ Carboxylic acids are weak acids and protonate
NaOH, H2O
C7H14O2
G
C3H5ONa
H2O in the equilibrium reaction:
✓ Polar substituents on R cause the equilibrium
to shift to the right, and increase the acidity of
RCOOH.
C
3
(a) Explain what is meant by the acid strength of a
compound.
(b) Define Ka and pKa.
4
(a) Draw the structures of the following carboxylic
acids:
(i) ethanoic acid
(ii) dichloroethanoic acid
(iii) propanoic acid
(iv) 2-chloropropanoic acid
(b) Place the acids in order with the strongest acid at
the top and assign each acid its pKa value from this
list: pKa values: 1.48, 2.83, 4.76, 4.87
(c) Give reasons for the order in which you have
placed the acids.
5
Compound A can be converted to compound B as
outlined below.
O
C
O
✓ Amino acids are polar, salt-like substances.
✓ Carboxylic acids react with bases to form
carboxylate salts.
✓ Carboxylate salts are reconverted to carboxylic
acids by treatment with mineral acids.
✓ Carboxylic acids condense with alcohols to give
esters.
✓ Hydrolysis of esters to carboxylic acids is
catalysed by acid or by base.
✓ Acyl chlorides are formed from carboxylic acids
J
C3H8O
C3H5OCl
–
NH3
I
C3H6O2
K
The structures of α-amino acids are correctly
written as zwitterions.
+
H2CrO 4
SOCl2
✓ Amino acids are the structural units of proteins.
R
C4H10O
HCl, H2O
RCOOH + H2O ҡ RCOO− + H3O+
H
H
+
O
by reaction with SOCl2, PCl3 or PCl5.
CH3CH2CH2
reaction (i)
C
OCH2CH3
A
product
O
reaction (ii)
CH3CH2CH2
C
Cl
B
(a) What are the functional groups present in A and B?
(b) Give the systematic names of compounds A and B.
Chapter 24 Carboxylic acids and derivatives
4
(c) Identify the reagents and, where possible,
the reaction conditions necessary to carry out
reactions (i) and (ii).
(d) Draw and name the product of reaction (i).
(e) How would you convert compound B to
compound A?
(a)
amide groups
COOH
O
glutathione
O
K2Cr2O7 or KMnO4
CH3CH2CH2
(oxidation)
C
OH
(b)
O
SH
HOOC
H
OH
OH
N
O
CH3CH2CH2
O
(i) oxidation
C
CH3CH2CH2
+
(ii) H
H
(b)
OH
O
CICH2
O
C
H
C
OH
OH
A pKa = 2.87
B pKa = 3.75
CH3
O
CH3CH2CH2
CH3
C
OH
C pKa = 4.83
3
C
O
C
OH
CH3
D pKa = 5.03
O
H
C
O
OCH3
methyl methanoate
CH3CH2
OCH3
methyl propanoate
O
C
C
O
OCH2CH2CH3
propyl methanoate
CH3CH2
C
OCH2CH2CH3
propyl propanoate
NH2
H
H
glutamic acid
C
(a) Carboxylic acid A: Ka = 1.36 × 10−3; pKa = 2.87
A is a stronger acid than ethanoic acid.
Carboxylic acid B: Ka = 1.77 × 10−4; pKa = 3.75
B is a stronger acid than ethanoic acid.
Carboxylic acid C: Ka = 1.48 × 10−5; pKa = 4.83
C is a weaker acid than ethanoic acid.
Carboxylic acid D: Ka = 9.30 × 10−6; pKa = 5.03
D is a weaker acid than ethanoic acid.
H
H
NH2
(b)
2
N
N
(a)
CH3CH2CH2CH2OH
H
HOOC
Answers to ITQs
1
SH
O
N
COOH
cysteine
H
glycine
Glycine and cysteine are listed in Table 24.1.
O
233
234
Chapter 25
Aromatic compounds
Learning objectives
■ Identify and differentiate between aliphatic, alicyclic, conjugated, non-conjugated and aromatic
■
■
■
■
■
■
■
compounds.
Describe the bonding in benzene.
Show the general mechanism of electrophilic aromatic substitution and the mechanisms for
bromination and nitration of benzene.
Define the terms canonical form, resonance hybrid, resonance stabilization, and draw the canonical forms
and resonance hybrids of benzene and of the cationic intermediate in electrophilic aromatic substitution.
Name derivatives of benzene.
Describe the properties, uses and main reactions of nitrobenzene and aniline.
Explain why phenol is acidic and describe its main reactions – formation of trihalo- derivatives, esters
and ethers.
Explain what is meant by an azo compound.
Introduction
In the early days of organic chemistry, chemists recognized
that there was a class of compounds that had distinctive
properties that were different from those of most other
compounds. These compounds were seen to burn with a
smoky flame, producing soot; this was an indication that
they had a high carbon content, i.e. a high carbon : hydrogen
ratio. Some of these compounds had a distinctive aroma
(smell), so they were classified as aromatic compounds. It
later turned out that the smell was not the most important
property that distinguished these compounds. In fact, some
had no smell at all, but the name ‘aromatic’ stuck and is
still used today.
Organic compounds can be classified as aliphatic, alicyclic
or aromatic (Figure 25.1):
■ aliphatic compounds are open chain compounds;
■ alicyclic compounds are cyclic and saturated or not
highly unsaturated.
Benzene was the key compound that led to an understanding
of the nature of aromaticity. Benzene has the formula C6H6,
so it is unsaturated since it has eight H atoms fewer than
the saturated straight-chain C6 compound hexane, C6H14.
Aromatic compounds include benzene and its derivatives,
and compounds that resemble benzene in chemical
O
CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2
H
C
OH
palmitic acid, C16H32O2
aliphatic
H
C
C
CH3CH2CH2CH2CH2CH3
hexane, C6H14
aliphatic
CH3CH2CH
CHCH2CH3
hex-3-ene, C6H12
aliphatic
H
C
C
Br
H
H
C
C
C
H
benzene
C6H6
H
H
C
C
H
H
H
C
C
C
H
cyclohexane, C6H12
alicyclic
cyclohexene, C6H10
alicyclic
Figure 25.1 Examples of aliphatic and alicyclic compounds.
C
C
C
C
H
H
C
C
H
C
C
C
H
C
bromobenzene
C6H5Br
H
H
naphthalene
C10H8
Figure 25.2 Examples of aromatic compounds.
H
Chapter 25 Aromatic compounds
H
H
C
H
C
H
C
H
C
C
H
C
H
electrophile
H
C
The structures of aromatic compounds are particularly
stable. This stability would be disrupted by addition of H2
or Br2 or H2O, etc. However, in the substitution product
the stable aromatic structure is preserved (see Figure 25.3).
Conjugation
A compound with alternating double and single bonds is
said to be conjugated. In a conjugated diene there are two
C=C units separated by a single bond and in a conjugated
triene three C=C units alternate with single bonds (Figure
25.4). Conjugation can also occur within cyclic structures.
H
H
H
H
H
C
C
H
H
butadiene, conjugated
C
H
H
C
C
C
H
H
C
line formula
Figure 25.5 Displayed and line formulae of ‘cyclohexatriene’
(benzene) C6H6.
Stability
C
H
H
Characteristics of aromatic compounds
C
C
displayed formula
Benzene and other aromatic compounds undergo
substitution reactions with electrophiles such as +NO2 and
+
Br; this is known as electrophilic aromatic substitution
(Figure 25.3). In this chapter, E+ will be used to represent a
generalized electrophile.
C
C
H
C
H
structure and some aspects of electronic configuration.
Figure 25.2 shows some examples of aromatic compounds.
H
C
H
E
Figure 25.3 Electrophilic aromatic substitution in benzene.
H
C
C
+
substitution product
H
benzene
C6H6
H
H
C
+
E
C
C
C
+
+
C
H
H
H
hexatriene, conjugated
Figure 25.4 Examples of conjugated systems.
Benzene can be regarded as the cyclic conjugated system
‘cyclohexatriene’, i.e. a ring of six carbons with three C=C
alternating with C–C. This structure is shown in Figure
25.5, along with a simplified representation. In this line
formula benzene is written as a hexagon without the
CH units explicitly shown. You assume that a ‘corner’ is
a carbon atom and then add hydrogen atoms until the
carbon atom has four bonds.
Canonical forms
The structure of Figure 25.5 can be written in two ways:
A
B
1
6
5
1
2
6
3
5
4
2
3
4
Structures A and B can be interconverted by the electron
movement shown with the curly arrows. Note that benzene
is neither structure A nor structure B, but a combination
of the two. Structures A and B are canonical forms and
benzene is a resonance hybrid of its canonical forms A
and B. Note carefully that the double-headed ‘resonance’
arrow (↔) between the two formulae is different from the
double arrow (ҡ) used to show a reaction equilibrium; it
tells us that our representation of the structure by either
one of the canonical forms is inadequate. Benzene is not A
some of the time and B some of the time; it is always the
resonance hybrid of the two.
Canonical forms are hypothetical structures which
contribute to the actual structure of a compound or ion.
Canonical forms are interconvertible by movement of
electrons. A single canonical form cannot fully explain the
properties of a compound or ion.
A resonance hybrid is the combination of canonical forms
that describes the actual structure of a compound or ion.
The internal energy of a resonance hybrid is lower than
that of any of the contributing canonical forms.
The resonance hybrid of benzene can be
represented as a circle within a hexagon.
However, the structure of benzene is often represented by
one of the canonical forms.
The stability of benzene
Canonical forms A and B (see above) differ only in the
positions of the double bonds; they are equivalent, so they
contribute equally to the hybrid. The internal energy of the
resonance hybrid is much lower than the energy of A or B.
235
Unit 2 Module 1 The chemistry of carbon compounds
A conjugated diene is more stable than an isomeric
unconjugated diene. Similarly, a conjugated triene is more
stable than an unconjugated isomer. However, benzene is
much more stable than we would expect ‘cyclohexatriene’
to be, so it is clear that there is something very special
about benzene.
The electron structure of benzene
The six carbon atoms in benzene are sp2 hybridized. Each of
these carbon atoms is σ bonded to two other carbon atoms
(sp2–sp2 overlap) and to a hydrogen atom (sp2–s overlap)
(Figure 25.6). The six carbon atoms link to form a flat,
planar hexagon. This is a very stable geometric arrangement
because the angle in a hexagon (120°) is identical to the
angle between sp2 orbitals. The lobes of the p orbitals above
the plane of the ring overlap, and so do the lobes below the
plane of the ring. So there is a cloud of electron density,
known as a π cloud, above and below the ring (Figure 25.7).
three sp2 orbitals in a plane
perpendicular to the paper;
angle between orbitals is 120˚
lobes of unhybridized
p orbital perpendicular
to sp2 orbitals
The six π electrons from the six sp2 carbons in benzene
are not localized in three distinct –C=C– bonds, but are
spread out (delocalized) over the entire ring. All the
carbon–carbon bonds in benzene are 139 pm long, which is
intermediate between the length of a carbon–carbon single
bond between sp2 carbons (147 pm) and a carbon–carbon
double bond (133 pm). This delocalization of π electrons
is a stabilizing feature, which is known as resonance
stabilization or resonance energy.
Figure 25.8 shows the heats of reaction for the conversion
of cyclohexene (1), cyclohexadiene (2), cyclohexatriene (3)
(calculated) and benzene (4) to form cyclohexane C6H12. If
hydrogenation of each C=C makes the same contribution
to the heat of reaction, then the heat of reaction of
‘cyclohexatriene’ should be 3 × 120 kJ mol–1. However,
the heat of hydrogenation of benzene is 208 kJ mol–1, so
benzene is approximately 150 kJ mol−1 more stable than
the imaginary ‘cyclohexatriene’. Benzene liberates less
heat than cyclohexatriene would, and so must have had
less internal energy than expected.
+H2
H
H
C
H
H
C
C
C
C
C
H
C
H
H
C
H
H
H
C
C
C
C
C
H
+3H2
+3H2
1
2
3
4
-50
C
lobes of p orbitals overlap
H
H
C
H
m-bonds (lines) and p orbitals (lobes)
0
C
C
120˚
H
+2H2
H
H
C
H
/cloud above and below the ring
Heat of reaction / kJ mol –1
236
-100
-150
-120
-200
-208
-250
-238
resonance energy
-300
-350
-360
Figure 25.6 Bonding in benzene.
-400
/ cloud
Figure 25.8 Heats of hydrogenation of cyclohexene,
cyclohexadiene and benzene.
Resonance stabilization or resonance energy is the difference
between the observed enthalpy of formation of a resonance
hybrid and the calculated enthalpy of formation of the
canonical form with the lowest internal energy. The greater
the difference, the more stable the resonance hybrid.
Aromaticity
Figure 25.7 The delocalized π electrons in the structure of benzene.
ITQ 1 Which compound in each of the following pairs of isomers
is conjugated? Which is more stable?
a
(i)
(ii)
■ it must be cyclic and it must be planar (flat);
■ it must be conjugated all the way round the ring;
b
or
A compound must satisfy the following criteria to be
classified as aromatic:
(i)
(ii)
or
■ it must have 4n + 2 π electrons in the ring, where n is
zero or any integer.
Chapter 25 Aromatic compounds
Nomenclature of benzene derivatives
Here are some example of the relative positions of the
substituents being designated by numbers only.
There are a few compounds that are almost always called
by their trivial names. These trivial names are not related to
the name of the attached substituent group. The following
are important.
NH2
OH
CH3
CH3
CH3
1
O2N
5
phenol
(not usually called
hydroxybenzene)
toluene
(not usually called
methylbenzene)
I
Many derivatives of benzene are named by attaching
‘-benzene’ to the name of the substituent group. Examples
are the halobenzenes and nitrobenzene.
Br
bromobenzene
Cl
I
NO2
chlorobenzene
iodobenzene
nitrobenzene
NO2
4
COOH
benzoic acid
2
6
3
I
NO2
3,5-diiodotoluene
aniline
(not usually called
aminobenzene)
1
2,4,6-trinitrotoluene (TNT)
In naming derivatives of benzene, substituent groups
are placed in alphabetical order and assigned the lowest
combination of numbers.
Properties and uses of aromatic
compounds
Benzene itself is a carcinogen but many aromatic
compounds which incorporate one or more benzene
rings possess useful physical, spectroscopic and biological
properties. Benzene and its derivatives therefore find
application in many areas.
Further rules are needed when the benzene ring has two
substituents attached.
■ Benzene, nitrobenzene and toluene are important
■ If the substituents are the same, the prefix di- is used.
■ Among pharmaceuticals, aspirin is a widely used
■ If the substituents are different, the names of the
substituents are combined and used as a prefix to
‘-benzene’, or the name of the substituent is used as a
prefix to the trivial name.
■ The relative positions of the substituents are designated
either by numbers or by using the prefixes ortho (o),
meta (m) or para (p). Note that ortho, meta and para, or
the letters o, m and p, are italicized in the compound
names. ortho means 1,2-, meta means 1,3- and para
means 1,4-.
Br
1
Br
2
Br
NO2
4
NO2
CH3
1
2
1-bromo-3-nitrobenzene
meta-bromonitrobenzene
m-bromonitrobenzene
NH2
Br
analgesic, antipyretic and anticoagulant.
■ Sulfanilamide is an antibacterial agent.
■ Eugenol, the compound responsible for the smell of
pimento and cloves, has a simple benzenoid aromatic
structure.
■ Butylated hydroxytoluene (BHT) is a common
preservative for foods and cosmetics.
■ The laboratory indicator methyl orange has the
structure of a typical azo dye.
Cl
1
1
3
1,2-dibromobenzene
ortho-dibromobenzene
o-dibromobenzene
solvents and starting materials for chemical synthesis.
(a) Name the compounds drawn below. Both are used as
fungicides.
OH
i
1
OH
ii
Cl
1-chloro-4-nitrobenzene
para-chloronitrobenzene
p-chloronitrobenzene
COOH
1
ITQ 2
Cl
Cl
Cl
Cl
Cl
(b) Draw the structures of the following compounds:
3
NO2
4
NO2
2-bromotoluene
ortho-bromotoluene
o-bromotoluene
3-nitroaniline
meta-nitroanaline
m-nitroaniline
4-nitrobenzoic acid
para-nitrobenzoic acid
p-nitrobenzoic acid
(i) 2,4,6-trinitrophenol, also known as ‘picric acid’ (used as a
stain and is highly explosive);
(ii) 2,4-dihydroxybenzoic acid, used as an analgesic;
(iii) p-nitroaniline (4-nitroaniline), starting material for the
synthesis of many dyes.
237
238
Unit 2 Module 1 The chemistry of carbon compounds
■ Styrene and terephthalic acid are the monomeric units
OH
of the polymers polystyrene and Dacron™.
■ Naphthalene has insecticidal properties and is the
main component of mothballs. Naphthalene is an
example of a fused bicyclic aromatic hydrocarbon.
Naphthalene is said to be fused because it contains
rings with at least two carbon atoms in common.
naphthalene
fused bicyclic aromatic hydrocarbon
fumigant
Benzene
Phenol
C6H6O, RMM = 94. Pinkish-white crystalline solid; boiling
point = +182 °C; melting point = +41 °C; antiseptic smell.
Sources: oxidation of benzene; reduction of benzoic acid;
coal oxidation; cumene process. Uses: antiseptic; starting
material for synthesis of drugs, herbicides, synthetic resins
and various cosmetics. Also known as carbolic acid; acidic,
pKa 9.95; causes burns and is toxic and carcinogenic.
COOH
Benzoic acid
C6H6, RMM = 78. Colourless liquid; boiling point = +80 °C;
melting point = +5.5 °C; characteristic odour. Sources: coal
tar, crude oil. Uses: important industrial solvent; starting
material for preparation of drugs, plastics, synthetic rubber,
dyes. Toxic and carcinogenic; forms an azeotrope with H2O;
first isolated and identified by Faraday in 1825.
CH3
C7H6O2, RMM = 122. White crystalline solid; boiling point
= +249 °C; melting point = +122 °C. Sources: oxidation
of toluene; occurs naturally in many plants and animals.
Uses: food preservative; industrial feedstock.
Reactions of benzene
Benzene will react with an electrophile, E+, to give a product
in which one H on the ring is replaced by E. This is known
as electrophilic aromatic substitution (Figure 25.9).
Toluene
C7H8, RMM = 92. Colourless liquid; boiling point = +110 °C;
melting point = −93 °C. Sources: crude oil, petroleum
cracking. Uses: industrial solvent; octane booster in gasoline;
industrial feedstock. Toxic, but less so than benzene.
NO2
H
H
C
C
H
H
H
C
+
C
C
H
C
+
C
H
H
H
Nitrobenzene
C6H5NO2, RMM = 123. Pale yellow oil; boiling point =
+211 °C; melting point = +93 °C; almond-like odour.
Sources: nitration of benzene. Uses: industrial solvent;
synthesis of aniline, rubber chemicals, dyes, explosives,
pharmaceuticals; used in shoe polish; perfume in soaps.
Toxic; readily absorbed through the skin.
H
C
+
E
C
C
C
C
+
H
E
H
Figure 25.9 Substitution reaction of benzene with E+, a
generalized electrophile.
Alkenes, non-aromatic compounds with C=C, undergo
addition reactions, initiated by addition of an electrophile.
This is shown in Figure 25.10 for 2,3-dimethylbutene. See
Chapter 27 (page 262) for a full discussion.
H3C
NH2
C
CH3
C
H3C
b+
b–
E
A
CH3
alkene
Aniline
C6H7N, RMM = 93. Colourless liquid, but the normal
appearance is red-brown due to the presence of aerial
oxidation products; boiling point = +184 °C; melting point
= +6.3 °C; sharp, rotting-fish smell. Sources: reduction of
nitrobenzene. Uses: manufacture of dyes and polyurethane.
Toxic and carcinogenic; basic.
H3C
+
C
H3C
CH3
–
+
C
H3C
E
CH3
carbocation
A
A
C
C
H3C
CH3
E
CH3
addition product
Figure 25.10 Addition of a generalized reagent E-A to
2,3-dimethylbutene, in which E is the electrophilic portion of the
reagent.
Chapter 25 Aromatic compounds
The reagent E–A is polarized. As E
is added, the E–A bond breaks and
both electrons in the bond remain
with A, which leaves as :A−. :A−
then reacts with the carbocation to
give the addition product.
The electrons in the π orbitals
of benzene resist reaction, so
the electrophiles have to be
more reactive than those used
for reactions with alkenes. In
electrophilic aromatic substitution,
the strong electrophile E+ pulls the
π electrons from the benzene ring
and forms a two-electron σ bond
to C.
The cationic intermediate (I) formed
is similar to the carbocation in the
alkene reaction (see Figure 25.10),
but the charge is delocalized across
the remaining five C atoms, which
are still linked by π bonds. The
partially stabilized cation (I) very
readily loses a proton (H+) from the
C to which E+ has added, to recover
full aromatic stability (Figure
25.11). Any weak base present is
able to remove the proton, and the
overall reaction is the substitution
of one H of benzene with E.
H
H
C
C
H
H
H
C
+
C
H
C
C
E
C
slow, r.d.s.
+
C
H
H
C
C
H
+
C
E
C
H
m bond
H
H
benzene
cationic intermediate (I)
+
Step 1: E pulls electrons from the benzene ring and a m bond is formed to C
H
H
C
C
H
C
C
H
H
H
H
+
C
+
C
C
E
C
C
C
H
H
H
Ia
H
H
C
C
E
C
C
H
H
C
C
+
H
H
Ib
Ic
H
C
E
C
H
The three canonical forms of the cationic intermediate
H
H
C
C
H
C
+
C
H
C
+
E
E
C
H
H
H
The resonance hybrid of the cationic intermediate
H
H
C
C
H
C
C
H
+
H
H
C
C
fast
E
C
H
–
A
C
H
C
+
C
H
C
C
H
A
E
H
H
cationic intermediate (I)
a generalized weak base
aromatic substitution product
+
Step 2: H is removed from the cationic intermediate by any weak base present in the reaction mixture, and the
In
summary,
the
aromatic aromatic
substitution product is formed
stabilization of benzene makes
Figure 25.11 The mechanism of electrophilic aromatic substitution.
it resistant to reaction. When a
strong electrophile does react, the
intermediate formed loses a proton in order to recover A Lewis acid is a chemical species (cation or molecule)
aromatic stability, and the net reaction is substitution.
which can accept a pair of electrons. Examples are H+, Al3+,
Cu2+, BF3, AlBr3, AlCl3, FeCl3, FeBr3, CO2, SO2 and SO3.
Bromination of benzene
■ Bromine, Br2, first forms a complex with the FeBr3.
Molecular bromine, Br2, does not react with benzene in the
same way in which it adds to alkenes (see Chapter 20). For
bromine to react with benzene the electrophile Br+ must be
generated, and this is usually done by treating Br2 with a
Lewis acid catalyst such as FeBr3.
+
Br2
FeBr3
+
heat
Br
bromobenzene
75% yield
HBr
■ This complex dissociates to give Br+ and FeBr4−.
■ The electrophile, Br+, then reacts with benzene to give
the cationic intermediate.
Nitration of benzene
Nitration is an aromatic substitution reaction of great
usefulness. In aliphatic chemistry, in contrast, it is rarely
used. The electrophile is the nitronium ion, NO2+, which
is usually produced from nitric acid, HNO3, by the action
of the stronger acid sulfuric acid, H2SO4. A mixture of
239
240
Unit 2 Module 1 The chemistry of carbon compounds
concentrated HNO3 and concentrated H2SO4 is used to carry
out aromatic nitration. This mixture is known as nitrating
mixture. It is usually prepared at low temperature and
extreme care must be used in its preparation and use. Do
not try to prepare or use nitrating mixture without expert
supervision.
O
–
N
O
+
O
+
H from H2SO4
–
OH
O
N
O
+
O+ H
nitric acid, HNO3
+
N+
H2O
O
nitronium ion
H
The nitronium ion then adds to the benzene ring in an
electrophilic substitution reaction.
O
+
N+
+
O
O
N
+
nitronium ion
nitrobenzene
H
+
–
O
The overall reaction is the substitution of an –NO2 group
for an –H atom.
+
H2SO4
HNO3
+
50 ˚C
H2O
NO2
The nitro group, –NO2, makes the aromatic ring less
nucleophilic, so –NO2 is said to be a deactivating
substituent, and this is of considerable practical importance.
Nitrobenzene is therefore less reactive than benzene, so it
is possible to convert all of the benzene to nitrobenzene
and stop the reaction there.
Trinitrobenzene is a very powerful explosive. Toluene
(methylbenzene) is easier to nitrate than benzene because
the methyl group is an activating substituent. Consequently,
2,4,6-trinitrotoluene (TNT), being easier to prepare, is a
much better known explosive.
CH3
nitrating
nitrating
mixture
mixture
An activating substituent is a group which causes the
aromatic ring to be more reactive in electrophilic aromatic
substitution. Examples are methyl (–CH3), amino (–NH2)
and hydroxyl (–OH) groups. These groups release electron
density or electrons into the aromatic ring, making it more
nucleophilic.
Chemists do, sometimes, carry out nitrations because
they want the aromatic nitro compound for its own sake.
However, the nitro compound is often prepared so that
it can be reduced to an amine (Figure 25.12). There are
several ways of carrying out the reduction; treatment with
tin, iron or zinc in hydrochloric acid is a classical method.
nitrate
R
aromatic
compound
Properties and reactions of aniline
Aniline (aminobenzene) is an important industrial
chemical. It is the starting material in the manufacture of
polyurethane, which is a widely used polymer, and of many
dyes used to colour fabrics, ceramics, plastics and even food.
[H]
+
NO2
CH3
O2N
NO2
mixture
NO2
amino
derivative
reduction
CH3
2,4-dinitrotoluene
R
There is no electrophile that corresponds to +NH2, so we
can not introduce –NH2 directly by aromatic substitution.
Aniline (aminobenzene) and other aromatic amines are
important and are used in many ways. They are usually
prepared by nitration followed by reduction.
nitrobenzene
nitrating
NH2
Aniline is prepared by the reduction of nitrobenzene with
metal (Sn, Fe or Zn) and acid (usually HCl) or by catalytic
hydrogenation.
para-nitrotoluene
NO2
R
Sn/HCl
nitro
derivative
NO2
toluene
NO2
Figure 25.12 Formation of aromatic amines by nitration followed
by reduction.
nitrobenzene
85% yield
CH3
A deactivating substituent is a group which causes the
aromatic ring to be less reactive in electrophilic aromatic
substitution. Deactivating substituents pull electrons away
from the aromatic ring, making it less nucleophilic. A prime
example is the nitro group, –NO2.
NO2
2,4,6-trinitrotoluene (TNT)
2H2O
NH2
aniline
Pure aniline is a colourless liquid, boiling point +184 °C,
with a strong, rank odour. It is easily oxidized by air and
the oxidation products are highly coloured, so the normal
appearance of aniline is a red-brown liquid. It is toxic and
carcinogenic, so inhalation and contact with the skin must
be avoided.
Chapter 25 Aromatic compounds
Aniline is a primary amine; it consists of an –NH2 group
bonded to an aromatic ring. The aromatic ring can be
regarded as a phenyl substituent, abbreviated Ph. So a
convenient shorthand way of writing aniline is PhNH2, and
yet another name for aniline is phenylamine.
Aniline is a base, like all amines, and it is protonated by
acids to give the anilinium ion.
H
+
+
Cl
Diazonium ions react with phenols to form highly coloured
azo compounds, many of which are used as dyes or
indicators.
R
+
+
N
N X
OH
–
R’
aromatic dizonium salt
phenolic compound
–
+
NH2
R
–
NH3Cl
aniline (liquid)
N
anilinium chloride, or aniline
hydrochloride (crystalline solid)
Aniline is a weaker base than aliphatic amines of comparable
molecular weight, e.g. pentylamine, hexylamine,
dipropylamine and triethylamine (Table 25.1). This means
that the equilibrium for the deprotonation of H2O by
aniline lies further to the left than for the aliphatic amines.
The pKb of aniline (9.42) is therefore higher than the pKb of
the aliphatic amines.
H
–
NH2
+
H
O
NH2
H
+
+
O
H
anilinium ion
The lone-pair electrons on the nitrogen of aniline are
somewhat delocalized over the aromatic ring. This
delocalization causes these electrons to be less available for
bonding to H+ than the electrons on the nitrogen atoms of
the saturated amines.
N
OH
azo compound
(highly coloured)
R’
Azo compounds are compounds with the general structure
Ar–N=N–Ar; each aryl group is attached to the nitrogen
atom by a carbon atom. Azo compounds are coloured.
Properties and reactions of phenol
Phenol (hydroxybenzene), is also known as carbolic acid.
As this name suggests, phenol is acidic. Comparison of
the pKa of phenol with pKa values for water, alcohols and
carboxylic acids reveals that phenol is more acidic than
water and alcohols and less acidic than carboxylic acids
(Table 25.2).
Table 25.2 pKa values of phenol, water, ethanol and some
alcohols and carboxylic acids
Name and molecular formula
Structure
phenol, C6H6O
OH
RMM
pKa
94
9.89
18
15.7
122
4.19
Table 25.1 Basicity of selected amines
Name
Formula
RMM pKb
Aniline (aminobenzene, C6H5NH2 (PhNH2) 93
phenylamine)
Pentylamine
CH3(CH2)3CH2NH2 87
pKa of conjugate acid
9.42 4.63
water, H2O
3.37 10.63
O
H
benzoic acid, C7H6O2
Diazotization
Aniline and other aromatic amines also serve as key
intermediates in further synthetic sequences. Diazotization,
the treatment of aromatic amines, Ar–NH2, with cold
nitrous acid, HNO2, converts the –NH2 group to –N2+, a
diazonium ion. Diazonium ions are very reactive and are
not usually isolated.
HNO2
R
NH2
(prepared in situ from
NaNO2 + acid) 0–5 ˚C
X
R
+
N
N
aromatic amine
aromatic diazonium salt
–
H
COOH
hexanoic acid, C6H12O2
CH3(CH2)4COOH
116
4.85
ethanoic acid, C2H4O2
CH3COOH
60
4.76
ethanol, C2H6O
CH3CH2OH
46
15.9
241
242
Unit 2 Module 1 The chemistry of carbon compounds
The H of the –OH group of phenol is quite easily removed
by H2O, i.e. phenol readily protonates water; the products
of this reaction are the phenoxide and hydroxonium ions.
O
H
O
+
–
H2O
+
H3O
+
phenol
phenoxide ion
hydroxonium ion
The –H of the –OH group of phenol can be replaced by an
alkyl group when phenol is made to react with a haloalkane
in the presence of a strong base. The phenoxide ion is first
formed. This is a nucleophile, which displaces the halide
ion from the haloalkane in a nucleophilic substitution
reaction (Figure 25.13). The product of this reaction is a
phenyl ether.
Step 1: formation of phenoxide by reaction with base
O
[phenoxide ion][H3O+]
Ka=
= 10−9.89
[phenol]
H
O
+
pKa = −log10 Ka = 9.89
The phenoxide ion is resonance stabilized and this causes
the equilibrium shown above to lie somewhat to the right.
When phenol is treated with bromine water or chlorine
water, 2,4,6-tribromophenol or 2,4,6-trichlorophenol
precipitates immediately.
O
H
+
Br
H2O
3Br2
3HBr
–
+
H2O
Step 2: phenoxide (nucleophile) displaces halide from the haloalkane
O
–
O
H
K
+
+
potassium
phenoxide
+
Br
K
OH
phenol
H
O
+
K
–
H
C
b
+
b
CH3
–
+
I
K
+
I
–
+
H
potassium
phenoxide
iodomethane
methyl phenyl
ether
Figure 25.13 Methylation of phenol.
phenol
Br
2,4,6-tribromophenol
fluffy white precipitate
It is thought that it is the phenoxide ion which actually
reacts. This reaction is a qualitative test for phenol and may
also be used for quantitative measurement.
Phenol reacts with acyl chlorides to form phenyl esters
(Figure 25.14). In this respect, phenol is similar to other
alcohols and other compounds with –OH groups.
O
O
O
H
C
CH3
O
Phenol is sufficiently acidic to cause burns to the skin and
therefore must be handled with caution. It will dissolve in
strong bases, e.g. NaOH, but is not acidic enough to react
with carbonates or hydrogencarbonates.
O
H
O
+
+
Na
OH
–
+
phenol
+
C
CH3
HCl
Cl
ethanoyl chloride
(acetyl chloride)
phenyl ethanoate
(phenyl acetate)
an ester
+
Na
–
Figure 25.14 Esterification of phenol.
+
H2O
sodium phenoxide
phenol
ITQ 3
(a) Name compounds G and H as derivatives of aniline.
O2N
G
H3C
NH2
H
NH2
(b) Which is the more basic of the two? G or H? Explain your
answer.
(c) Assign the pKb values below to G and H.
(i) pKb 13.00
(ii) pKb 8.92
(d) Which compound is more basic than aniline?
(e) Which compound is a weaker base than aniline?
Chapter 25 Aromatic compounds
Summary
Review questions
1
(a) Draw structures of:
(i) 4-bromoaniline
(ii) 3-iodophenol
(iii) 2-chlorobenzoic acid
(b) Name the following compounds.
✓ Aromatic compounds are very stable, cyclic,
planar, unsaturated compounds with continuous
conjugated –C=C–C=C–C=C– systems. They burn
with a smoky flame due to the high C:H ratio.
Br
✓ Benzene, C6H6,
, is the prototypical
aromatic compound. The stability of benzene
and aromatic compounds is due to the
continuous overlap of the electrons which
originate from the p atomic orbitals of the sp2
hybridized carbon atoms.
i
✓ Benzene is nucleophilic (seeks a positive
charge in reactions) and reacts with strong
electrophiles (E+, species which seek electrons).
The intermediate formed in this reaction loses
a proton in order to recover aromatic stability
and the net reaction is electrophilic aromatic
substitution.
✓ Aniline (aminobenzene/phenylamine), which
can be prepared by reduction of nitrobenzene,
is used to form diazonium salts. Diazonium salts
can be coupled with phenols to form coloured
azo compounds, or can be converted to a number
of other derivatives.
✓ Phenol (hydroxybenzene) has properties of both
a weak acid (deprotonates H2O, pKa 9.89, and
reacts with strong base) and an alcohol (reacts
with acyl chlorides to form esters).
iii
I
NO2
2
CH3
ii
I
COOH
Compound K can be converted to O as shown below.
K
L
(i) Sn-HCl
✓ The structures of benzene and of aromatic
compounds are hybrids (resonance hybrids) of
contributing structures which are known as
canonical forms.
CH3
C6H7N
(ii) HNO2
NO2
M
N
O
C6H6O
+
N
N
X
N
N
OH
–
(a) Name compound K.
(b) Draw and name compound L.
(c) Describe the conditions necessary to carry out
reaction (ii).
(d) What type of compound is M? Give a brief
description of the properties and uses of these
compounds.
(e) Draw and name compound N.
(f) What type of compound is O? Briefly describe the
properties and uses of these compounds.
243
244
Unit 2 Module 1 The chemistry of carbon compounds
Answers to ITQs
1
(a) Structure (ii) is conjugated and more stable.
(b) Structure (ii) is conjugated and more stable.
2
(a) (i) 2,4,5-trichlorophenol
(ii) 2,4,6-trichlorophenol
(b)
i
O2N
COOH
NO2
NO2
3
ii
OH
iii
NH2
OH
OH
NO2
(a) G is 4-nitroaniline or p-nitroaniline;
H is 4-methylaniline or p-methylaniline.
(b) Compound H is more basic than compound G:
the CH3 group pushes electron density across
the aromatic ring and enhances the ability of the
lone-pair of electrons on the nitrogen of the NH2
group to remove a proton from H2O.
Compound G is less basic than compound H:
the NO2 group pulls electron density across the
aromatic ring and reduces the availability of the
lone-pair of electrons on the NH2 group for bond
formation to a proton from H2O.
(c) Compound G: pKb = 13.00;
compound H: pKb = 8.92
(d) Compound H (pKb 8.92) is more basic than aniline
(pKb 9.42).
(e) Compound G (pKb 13.00) is a weaker base than
aniline (pKb 9.42).
245
Chapter 26
Macromolecules
Learning objectives
■ Define the terms macromolecule, polymer and monomer and provide naturally occurring and synthetic
■
■
■
■
■
■
■
examples of each.
Describe the key features of addition polymerization and condensation polymerization.
Predict whether a given monomer or pair of monomers will polymerize by addition or condensation.
Draw the structure of the repeating unit of a polymer formed from a given monomer or pair of
monomers.
Recognize the repeating units in polymer chains and determine the structures of the monomers.
Discuss, using specific examples, the uses and advantages of synthetic polymers.
Describe aspects of the impact of plastic on the environment.
Outline measures for minimizing and managing plastic waste.
Introduction
Polymerization
Life on Earth is based on the structure and properties of
macromolecules such as nucleic acids, proteins, enzymes
and cellulose.
Polymerization is the process whereby many small
molecules (monomers) combine to produce a polymer.
The term macromolecule simply means ‘very large
molecule’. A macromolecule may contain thousands or
even hundreds of thousands of atoms. The term polymer
is applied to a macromolecule composed of many smaller
identical or similar molecular units called monomers,
which are often arranged in a highly organized order. The
term oligomer is applied to molecules of intermediate size,
built up from a relatively small number of monomers.
The development of synthetic polymers has resulted in
tremendous improvement in the quality of life. Try to
imagine life without plastics (e.g. tanks for water storage,
piping), synthetic fibres (fabrics for clothing) or synthetic
rubbers (car tyres). These are, in fact, the three groups
into which polymers are classified: plastics, fibres and
elastomers.
■ In addition polymerization the monomers combine
(add) without the loss of atoms, so no by-products are
formed.
■ In condensation polymerization, combination of
the monomers is accompanied by the formation of
small molecules such as H2O or HX (X = Cl, Br).
Addition polymerization
Addition polymerization via radicals
Addition polymerization is most easily illustrated by the
polymerization of ethene to produce poly(ethene), by a
radical process. To learn more about radical reactions, you
need to look at Chapter 27 (page 259).
Chain initiation starts the process. We shall call the initiator
the radical R•, which then reacts with a molecule of ethene
to form a new radical (Figure 26.1).
H
ITQ 1 Correlate each of the prefixes in (a) with one of the
meanings in (b).
(a) Prefixes: di; macro; mono; oligo; poly; tri.
(b) Meanings: one; two; three; few (two to eight); very large; many.
R
H
C
C
H
radical
R
H
ethene
H
H
C
C
H
H
new radical
Figure 26.1 Chain initiation in the polymerization of ethene.
246
Unit 2 Module 1 The chemistry of carbon compounds
R
H
H
C
C
H
C
C
H
H
H
R
H
R
H
H
H
H
H
C
C
C
C
H
H
H
H
H
H
H
C
C
C
C
C
C
H
H
H
H
H
H
R
H
H
C
C
H
H
H
H
H
H
H
H
H
H
H
H
H
C
C
C
C
C
C
C
C
H
H
H
H
H
H
H
H
n
repeating monomeric units
Figure 26.2 Chain propagation in the polymerization of ethene.
R
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
n
R
n
R
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
C
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
n
n
R
OR
R
H
H
H
H
H
H
H
H
C
C
C
C
C
C
C
C
H
H
H
H
H
H
H
H
n
R
R
H
H
H
H
H
H
H
H
C
C
C
C
C
C
C
C
H
H
H
H
H
H
H
H
n
R
Figure 26.3 Chain termination in the polymerization of ethene.
Chain propagation follows, as the new radical forms a bond
to a second molecule of ethene, and the process is repeated
many times as the chain grows (Figure 26.2).
In other cases the initiator is a peroxide, R–O–O–R. The
O–O bond in the peroxide breaks easily to produce two
RO• (alkoxy) radicals.
Chain termination ends the process. This is when the
growing radical chain reacts with another radical, which
can be R• or a second growing radical chain (Figure 26.3).
In practice, technical details of initiator, temperature and
pressure are used to control polymer chain length and
other features. There are many types of poly(ethene)
that are marketed: the most familiar, sometimes called
polythene, is made into a film that is used for packaging
and is produced at high temperature and pressure with O2
as the initiator.
H
H
C
H
C
H
ethene (ethylene)
(monomer)
O2, heat
pressure
or catalyst
H
H
H
H
H
H
C
C
C
C
C
C
H
H
H
H
H
H
poly(ethene)
(polymer)
R
O
O
R
R
H
O
O
+
R
H
C
C
X are used to make a large
Substituted alkenes, H
number of polymers with chains in which the substituent
X appears on alternate carbon atoms.
H
H
C
H
C
X
substituted ethene
(monomer)
H
H
H
H
H
H
C
C
C
C
C
C
X
H
X
H
X
initiator
H
substituted poly(ethene)
(polymer)
Many common plastics and some fibres are polymers
of substituted ethenes. Plastics are polymers that can be
extruded into sheets or moulded into shapes of various
sorts.
Chapter 26 Macromolecules
■ Poly(vinylchloride) (PVC), where X = Cl, is used to
make plastic pipes and many other products.
■ Polystyrene, where X = C6H5 (Ph), is a thermoplastic
that can be melted by heat and moulded into shapes.
H
H
C
C
■ Derivatives of acrylic acid, H
COOH , are used to
make a wide variety of polymers.
H
H
C
■ Acrylonitrile, H
C
C
technical developments have made it a superior method of
controlling the outcome of some polymerization processes.
The manufacture of polymers, some very common and
some with very special properties, is a major industry.
There is the technology to modify the polymerization
process to produce materials with predictable properties.
Co-polymers are polymers made from two or more
monomers, such as acrylonitrile and styrene, where the
properties of the polymer can be varied by varying the ratio
of the monomers incorporated.
N , gives us polyacrylonitrile,
which has many uses: the fibre with the trade name
Orlon™ is polyacrylonitrile.
Addition polymerization via ions
We have discussed chain-reaction polymerization, where
the growing chain is a radical and the process is initiated
by a free radical. Polymerization can be initiated in other
ways. In ionic polymerization the initiator is an acid or
a base which reacts with an alkene to produce a cationic
or anionic intermediate that reacts with further alkene
molecules to produce a growing ionic chain (Figure 26.4).
Natural addition polymers
Rubber is a natural polymer which is produced from the
latex of several different species of tree, particularly Hevea
brasilensis, a large tree found in the Amazonian rainforest.
Rubber is formally derived from addition polymerization of
2-methyl-1,3-butadiene (isoprene) (Figure 26.5).
isoprene
etc.
n
rubber (all cis-isoprene polymer)
H2C
X
H
+
+
H2C
CH
R
H3C
+ X
–
+
H2C
–
H2C
CH
R
H2C
cationic
chain
R
–
CH
R
Figure 26.5 The derivation of the rubber polymer from 2-methyl1,3-butadiene.
R
CH
B
B
CH
CH
R
anionic
chain
Figure 26.4 Addition polymerization via cationic and anionic
intermediates.
Chain termination comes when a growing cationic chain
reacts with an anion or an anionic chain reacts with a
cation. Two growing ionic chains can not react together in
a termination step. Acidic initiators may be Brønsted acids
such as H2SO4 or Lewis acids such as BF3; basic initiators are
inevitably very strong bases such as n-BuLi.
In outline, ionic polymerization looks very similar to
radical polymerization, but in practice the products of
the different methods may differ in important ways. For
example, polymer chain lengths may be very different, the
amount of chain branching and cross-linking may differ,
stereochemistry may differ, and so on. A relatively recent
method of polymerization using complexes of aluminium
and titanium as the initiators (Ziegler–Natta catalysts)
has some similarity to radical or ionic polymerization, but
All of the double bonds in the rubber polymer are cis.
This somewhat disordered molecular structure causes the
material to be soft and elastic.
Gutta percha is another natural polymer formed from the
latex of other trees. Gutta percha is also a poly(isoprene),
but with trans double bonds and a more ordered molecular
structure.
CH3
H
CH3
H
CH3
H
n
gutta percha (all trans-isoprene polymer)
Gutta percha does not have the elasticity of rubber, but its
hardness and crystallinity make it useful as an insulator, as
a building and dental material and in the manufacture of
golf balls.
In the preparation of rubber the watery latex of the tree is
tapped, and it sets to a sticky elastic solid as it loses water.
The Amerindians have known about rubber for thousands
of years, and they used it to make balls for games they
played. When Europeans first saw these balls they were
amazed by how well they bounced, but it was their discovery
that the material could be used to erase or rub out pencil
247
248
Unit 2 Module 1 The chemistry of carbon compounds
marks that gave us the English name rubber. Raw rubber
is of limited value, and it must be given further treatment
to enhance its useful characteristics of elasticity, resilience
and impermeability to water and air. The most significant
advance was made when the process of vulcanization was
discovered: when rubber is heated with sulfur, it becomes
tougher, or even hard. In the vulcanization process C–S
bonds are formed at double bond sites in adjacent polymer
chains, forming C–S–C cross-links between the two chains.
As more cross-links form, the rubber becomes harder and
stronger, and technology controls the degree of crosslinking needed to produce rubber with the characteristics
required for a particular purpose.
presented a major problem since the synthetic polymer
had both cis and trans double bonds, not the all-cis
stereochemistry of natural rubber. Polymerization of
2-chloro-1,3-butadiene (called chloroprene) led to a
poly(chloroprene), given the trade name Neoprene®.
Many years of investigation and technological advance
were required before superior synthetic rubbers became
available. With improvements in technology, synthetic
rubbers are now designed to have properties suitable for
special purposes, not just the manufacture of superior tyres.
Condensation polymerization
The two most important reactions in condensation
polymerization are:
Synthetic rubber
The invention and development of the automobile
equipped with pneumatic tyres caused a huge increase in
the demand for rubber and, with some difficulty, plantations
of H. brasiliensis were established in a number of tropical
countries. The recognition that the supply of rubber could
become severely limited, especially in wartime, led to
the invention of synthetic rubbers. Almost all tyres, for
example, are now made of synthetic rubber.
■ formation of an amide (+ H2O) from an amine and a
carboxylic acid;
■ reaction between an alcohol and a carboxylic acid to
give an ester and H2O.
These reactions were discussed in Chapter 24 and so are
shown in outline only in Figure 26.6.
1
O
O
A logical approach to the synthesis of a substitute for
natural rubber was to polymerize derivatives of butadiene.
An early attempt used butadiene itself and the polymerizing
agent was sodium (Na), so the product was named Buna
rubber. It was much inferior to natural rubber, but later
technological advances led to a range of Buna rubbers with
specialized properties.
R
Polymerization of isoprene itself was carried out, but the
product was also inferior to natural rubber. Stereochemistry
carboxylic acid
C
R’
+
O
H
H
carboxylic acid
R
C
O
O
alcohol
+
H2O
+
H2O
R’
ester
1
O
R
O
C
R’
+
O
H
H
R
C
N
N
R’’
amine
amide
R’
R’’
Figure 26.6 Important reactions in condensation polymerization.
ITQ 2 Each of the following monomers forms an addition polymer
for which the name is given. Draw the structure of the polymer.
H
a
H
C
H
H
polyvinyl
chloride
C
b
Cl
H
C
C
polystyrene
H
chloroethene
(vinyl chloride)
phenylethene
(styrene)
F
c
F
C
F
C
Teflon
Polymers from amides
The general term peptide is used to describe the amide
formed by linking the –COOH of one amino acid with the
–NH2 group of a second amino acid (Figure 26.7).
H
H2N
O
C
R
+
C
1
OH
H
H
N
H
F
tetraflouroethene
H2N
C
R
H
C
R1
ITQ 3 Using the scheme for the polymerization
of isoprene as a template (Figure 26.4),
show how 2-chloro-1,3-butadiene, in the
Cl
form drawn below, can polymerize to give
2-chloro-1,3-butadiene
Neoprene™, which is an all-trans polymer.
O
can react further
C
OH
2
O
C
H
H
N
C
R2
dipeptide
O
+
H2O
C
OH
can react further
Figure 26.7 Formation of a dipeptide by condensation of two
amino acids.
Chapter 26 Macromolecules
The dipeptide so formed
still has a free –COOH group
and a free –NH2 group, so
the peptide-forming reaction
can be repeated by linking a
third amino acid at either the
–COOH or the –NH2 to form
one of two possible tripeptides
(Figure 26.8).
The process can be repeated
over and over again to give
polypeptides
of
various
lengths
and
various
combinations of amino acids.
H
H
C
N
H
H
N
C
H
H
O
O
C
R1
H
H
N
C
R
C
C
N
H
condensation at the
–NH2 of the dipeptide
H
H
N
C
H
C
N
R3
H
H
H
C
O
H
C
C
R
+
N
H
N
R1
tripeptide
O
C
C
OH
R3
condensation at the
–COOH of the dipeptide
O
H
H
H
OH
2
dipeptide
OH
R3
O
O
H
C
O
C
H
C
N
R1
H
O
C
R2
OH
2
C
H
H
N
C
O
C
R3
H2O
tripeptide
+
OH
H2O
Glutathione (see ITQ 4, Figure 26.8 Formation of two possible tripeptides from a dipeptide.
Chapter 24) is a common
natural tripeptide formed from the amino acids glutamic ■ When we watch a spider making a web we are
impressed by the strength of the slender filament that
acid, cysteine and glycine; it occurs in muscle tissue, yeasts
the spider spins. The filament contains a protein that
and elsewhere.
from glutamic acid
from cysteine
from glycine
provides the strength.
SH
O
HOOC
N
N
caterpillar, produces another protein filament called
fibroin. Fibroin provides us with silk – the luxury
fabric that originated in China many thousands of
years ago, and over centuries was the basis of the trade
between China and the rest of the world.
COOH
O
H
NH2
■ Another insect, the moth Bombyx mori, when it is a
H
glutathione
Two examples of nonapeptides (nine amino acids) are
vasopressin, which controls blood pressure, and oxytocin,
which stimulates uterine contraction. Insulin, important in
the control of carbohydrate metabolism, contains 51 amino
acid units and differs in detail from one species to another.
■ Another important natural fibre is wool, which
consists of the protein keratin.
The polypeptide chain of these protein fibres led chemists
to think of synthesizing a polyamide chain. Chemists
at the Dupont company achieved practical success with
the product they gave the commercial name Nylon. A
monomer with two –NH2 groups, 1,6-diaminohexane, was
made to react with a monomer with two –COOH groups,
adipic acid (Figure 26.9).
Proteins are polypeptides that may be very large indeed;
they are also more complex than the simple polypeptide
chains we have described above. Proteins provide animals
with muscles, skin, hair, feathers and other structural tissue.
H
O
H
N
CH2CH2CH2CH2CH2CH2 N
H
H
O
CH2CH2CH2CH2CH2CH2 N
C
H
N
C
+
CH2CH2CH2CH2
HO
H
1,4-diaminohexane
O
C
OH
adipic acid
O
CH2CH2CH2CH2
+
C
H2O
OH
H
O
H
C
N
CH2CH2CH2CH2CH2CH2
H
O
N
C
CH2CH2CH2CH2
nylon-6,6
O
H
C
N
H
CH2CH2CH2CH2CH2CH2
N
x
Figure 26.9 Formation
of nylon-6,6 from
1,6-diaminohexane and
adipic acid.
ITQ 4 Which two of the following terms apply to glutathione? monomer; oligomer; oligopeptide; polymer; polypeptide; protein
249
250
Unit 2 Module 1 The chemistry of carbon compounds
HO
O
O
C
C
OH
+
HO
HO
OH
HO
O
C
C
OH
O
O
C
C
O
O
C
C
O
O
C
C
OCH2CH2
OCH2CH2
OCH2CH2
TM
O
O
O
+
O
O
C
C
OH
+
H2O
O
O
C
C
O
+
H2O
O
O
C
C
O
+
H2O
TM
Terylene /Dacron
CH2CH2
HO
O
C
C
OH
OCH2CH2OH
ethylene glycol
terephthalic acid
HO
CH2CH2
O
O
H2O
OH
CH2CH2OH
CH2CH2O
x
Figure 26.10 Formation of TeryleneTM/DacronTM from terephthalic and ethylene glycol.
Since both monomers have six carbon atoms, this polymer
was called nylon-6,6. There is now a family of nylons, each
one a polyamide.
Polyesters
Polyesters are synthetic fibres prepared using an
esterification reaction. Polyesters are synthesized by
condensing a diol monomer with a dicarboxylic acid
monomer. Chemists at the company ICI in the UK used
ethylene glycol and terephthalic acid as the monomers to
make the polyester to which they gave the trade name
Terylene®; Dupont called it Dacron® (Figure 26.10).
We have briefly surveyed polymerization and polymers
without going deeply into details. Of course, polymers have
become very familiar to us in everyday life. We use many
ITQ 5 Polymerization of 6-aminohexanoic acid gives nylon-6.
H
N
CH2CH2CH2CH2CH2
(a) Draw the zwitterionic form of 6-aminohexanoic acid.
(b) Draw a short segment of nylon-6.
O
H
kinds of polymers in such things as plastic bags, plastic
water tanks, synthetic fabrics, spectacles, contact lenses,
various forms of rubber, and so on. We have suggested that
some polymerization methods are ‘superior’ to others, but
we recognize that polymers are used in so many different
ways that different polymerization processes will be needed
to produce a product with just the properties we want for
a particular purpose. Take the example of pneumatic tyres,
which were all made at first from natural rubber. When
synthetic rubbers were developed, it was important to use
a process that made a tough, hard-wearing rubber for the
tyres, but a different type of rubber, elastic and impermeable
to air under pressure was needed to make air-tight inner
tubes. Now most automobile tyres do not need an inner
tube because the synthetic rubber used is not just tough, it
is also air-tight.
C
OH
6-aminohexanoic acid
ITQ 6 A short segment of the polymer Kodel™ (a fibre) is shown below.
O
O
O
O
C
C
O
CH2
CH2
O
C
C
O
CH2
n
Kodel
(a) What functional group is present in Kodel™?
(b) Draw the structures of the monomeric units from which Kodel™ could be made.
(c) Is Kodel™ an addition polymer or a condensation polymer?
CH2
Chapter 26 Macromolecules
Nowadays not only the structure of a car tyre, but also the
characteristics of the rubber itself can be changed to meet
the tyre’s intended use. For example in Formula 1 racing,
tyres must provide maximum grip but need not last for
many miles. For road use, mileage is of greater importance.
In F1, teams can choose two from four different hardnesses
of rubber, varying from very hard to soft. They pick
tyres according to the particular track and the expected
temperature. The different rubbers are manufactured by
varying the amount of carbon, sulfur and oil mixed with
the rubber.
Carbohydrates
Carbohydrates form a very large class of natural compounds
of central importance to life in plants and animals. The
general formula of many simple carbohydrates is Cn(H2O)n
where n is typically 4, 5 or 6 (but it occasionally can have
other values). Carbohydrates appear to be ‘hydrates of
carbon’, although they do not actually contain water. The
name is now applied to a much wider family of compounds.
Carbohydrates are the most abundant compounds in the
natural world, making up more than 50% of the dry weight
of the Earth’s biomass.
Carbohydrates are synthesized in plants by photosynthesis.
Light from the Sun is absorbed by chlorophyll and this
is converted to the energy necessary to biosynthesize
carbohydrates.
nCO2 + nH2O + solar energy → Cn(H2O)n + nO2
carbohydrate
Carbohydrates act as a repository of solar energy. The
energy is released when animals or plants metabolize
carbohydrates.
Cn(H2O)n + nO2 → nCO2 + nH2O + energy
Much of the energy released by oxidation of glucose is
trapped in the molecule adenosine triphosphate (ATP).
Monosaccharides are simple sugars and are the building
blocks for disaccharides, oligosaccharides and
polysaccharides. Hydrolysis of the higher saccharides
converts them to monosaccharides. The names of sugars
generally contain the suffx ‘-ose’. Glucose and fructose
are examples of monosaccharides with n = 6, so they are
termed hexoses. Glucose contains an aldehyde function,
so it is an aldohexose; its isomer fructose is a ketone, so
it is a ketohexose (Figure 26.11). Glucose is the most
important hexose; fructose is an isomer of glucose and
occurs widely in fruits.
1
1
CHO
2
H
OH
HO
3
H
4
H
5
CH2OH
O
2
HO
3
H
OH
H
4
OH
OH
H
5
OH
H
6
6
CH2OH
CH2OH
D-(+)-glucose
D-(+)-fructose
Figure 26.11 Open-chain structures of glucose and fructose
drawn using the Fischer convention. Carbons 2–5 are not shown
explicitly; bonds drawn vertically are in or behind the plane of
the paper and horizontal bonds are above the plane; the prefix D
shows that the stereochemistry of these compounds at C-5 is the
same as D-(+)-glyceraldehyde.
The open-chain structures of sugars contain >C=O
(carbonyl) groups; in glucose the carbonyl group is part of
the aldehyde at C-1 and in fructose the carbonyl group is
a ketone at C-2.
Open-chain sugars form rings (cyclize) when the oxygen of
one of the –OH groups adds to the C of the >C=O and the
H of the same –OH adds to the O of the >C=O. In glucose
the –OH group at C-5 adds readily to the aldehyde >C=O at
C-1 to give a six-membered ring; the reverse reaction also
occurs readily – the six-membered ring opens to give the
open-chain form of glucose – so a solution of glucose exists
as an equilibrium mixture of the open-chain and cyclic
forms (Figure 26.12).
OH
1
CHO
H
OH
2
HO
3
H
1
H C
H 2 OH
HO
3
H
4
OH
H
4
H
5
OH
H
5
6
CH2OH
D-(+)-glucose
open-chain form
H
6
CH2OH
5
O
OH
6
CH2OH
=
4
O
OH
OH
1
OH
OH
D-(+)-glucose
cyclic (pyranose) form
Figure 26.12 Equilibrium between the open-chain and cyclic
forms of glucose; the wavy line indicates that the –OH group at
C-1 can be above or below the plane of the six-membered ring.
Sugars with ring structures consisting of five C atoms and
one O atom are pyranoses. Sugars with ring structures
consisting of four C atoms and one O atom are furanoses.
In fructose the C-5 –OH adds to the ketone at C-2 to give
a five-membered ring and this cyclic form of fructose is in
equilibrium with the open-chain form (Figure 26.13).
The functional group which results from the addition of
an –OH group to the >C=O of an aldehyde or ketone is a
hemiacetal, so the cyclic forms of glucose and fructose are
hemiacetals (Figure 26.14). The equilibrium between the
251
252
Unit 2 Module 1 The chemistry of carbon compounds
1
1
CH2OH
2
O
HO
3
H
4
H
5
a
CH2OH
R
R
HO
C
2
HO
3
OH
H
4
OH
H
5
H
6
HOCH2 O
H
OH
=
O
HO
CH2OH
C
H
aldehyde
b
OH
CH2OH
D-(+)-fructose
open-chain form
R’’OH
H
OR’’
alcohol
OH
hemiacetal
OH
6
CH2OH
O
+
R
R
C
O
+
R’’OH
R’
D-(+)-fructose
cyclic (furanose) form
ketone
C
R’
alcohol
OR’’
OH
hemiacetal
Figure 26.13 Equilibrium between the open-chain and cyclic
forms of fructose; the wavy lines indicate that the groups at C-2
can be above or below the plane of the five-membered ring.
Figure 26.14 The formation of hemiacetals from (a) aldehydes
and (b) ketones; the hemiacetal functional group contains sp3 C
bonded to two O atoms, one of which is –OH.
open-chain and cyclic (hemiacetal) structures described for
glucose and fructose is typical of monosaccharides.
Maltose (malt sugar) is derived from barley and is used in
brewing beer; it is formed by linking C-1 of one glucose unit
to the –OH at C-4 of another glucose unit. Lactose (milk
sugar) is found in the milk of mammals. Lactose is formed
by joining C-1 of galactose (an isomer of glucose) to C-4
of glucose. Some individuals lack the enzyme needed to
hydrolyse lactose and must keep milk and dairy products
out of their diet.
The substance that is commonly called ‘sugar’ is the
disaccharide sucrose. Sucrose is formed from the cyclic
forms of glucose and fructose (shown in Figures 26.12 and
26.13). Note the link between the C-1 –OH of glucose and
C-2 of fructose in the structure of sucrose.
6
CH2OH
5
Oligosaccharides are carbohydrates of intermediate
molecular weight and consist of three to nine
monosaccharide units. They occur most abundantly in
plants but some are also of importance in animals. The
factors that determine our ABO blood group contain an
oligosaccharide component, and it is a small variation in this
component that decides our blood group and determines
which type of blood we can safely receive by transfusion.
O
4
OH
OH
3
1
2
O
OH
6’
HOCH2
O
HO
2’
1’
CH2OH
OH
Maltose and lactose are important natural disaccharides.
CH2OH
a
CH2OH
O
OH
O
1
4’
OH
OH
O
OH
OH
OH
maltose
CH2OH
b
O
CH2OH
OH
4
5
O
OH
OH
4’
OH
O
OH
1
from galactose
from glucose
lactose
OH
Starch and cellulose are described as polysaccharides, but
neither of them is a single chemical compound. They are
both polymers of glucose but one important difference
between them is the stereochemistry of the links between
the glucose units.
In starch, the bond linking C-1 of one glucose unit to O
and C-4 of the other glucose unit is below the plane of the
six-membered rings, as shown in maltose; such links are
described as α (‘alpha’). Hydrolysis of starch gives maltose.
In cellulose, the bond linking C-1 of one glucose unit to
O and C-4 of the other glucose unit is above the plane of
the first six-membered ring; such links are described as β
(‘beta’). Hydrolysis of cellulose gives cellobiose, which is an
ITQ 7
ITQ 8
(a) What is hydrolysis?
(a) What product or products would be obtained if (i) maltose and
(ii) lactose are subjected to hydrolysis?
(b) Is sucrose an addition or a condensation product of glucose
and fructose? (Hint: hydrolysis of sucrose gives glucose and
fructose).
(c) Identify the hemiacetal carbons in glucose and fructose.
(d) Is sucrose a hemiacetal?
(b) Suggest reaction conditions under which hydrolysis can take
place.
Chapter 26 Macromolecules
CH2OH
a
CH2OH
O
CH2OH
O
OH
4
OH
4
O
OH
CH2OH
O
O
1
OH
OH
CH2OH
O
b
OH
CH2OH
cellobiose
O
4
OH
O
1
OH
1
OH
O
1
OH
1
O
OH
4
OH
4
OH
4
OH
a section of the cellulose polymer
isomer of maltose. Structures of cellobiose and a section of
the cellulose polymer are shown in Figure 26.15.
Cellulose, the major constituent of plant fibres, contains
chains of a thousand or more glucose units, linked as shown
in Figure 26.15. Cellulose is insoluble in water, and is not
hydrolysed by the enzymes that digest starch. It occurs in
great abundance in the structural tissue of plants. It also
provides us with fibres we use every day; cotton thread,
for example, is almost pure cellulose. Most mammals can
not digest cellulose, but it is present in their diet and is
beneficial because it provides bulk or roughage that helps
the digestive system. Some bacteria produce enzymes
that can digest cellulose. Cows and other ruminants can
digest cellulose because there are such bacteria in their
rumens. Cellulose can be converted chemically to glucose
by treatment with aqueous mineral acids; this process is
now used to produce ethanol by fermentation of glucose
obtained from fibrous plant material.
Starch occurs as granules in plants, and the nature of the
starch varies from one species to another. The starch in
cereal plants provides food for human beings and other
animals. Closer examination shows that it is made up of
two components: amylose and amylopectin.
■ Amylose consists of chains of glucose units, all joined
by α glycosidic links from C-1 in one unit to C-4 in the
next; the chains may contain hundreds or thousands
of glucose units, so amylose is not a single chemical
compound.
■ Amylopectin is also built up from chains of glucose
units with α glycosidic links as in amylose, but it
differs in having many cross-links that connect the C-1
terminal of one chain by a link to C-6 of a glucose unit
in another chain. Amylopectin may contain hundreds
of thousands or millions of glucose units in a highly
branched network connecting hundreds of relatively
short chains containing dozens of glucose units.
Figure 26.15 Structures of
(a) cellobiose and (b) a section
of the cellulose polymer.
Herbivorous and omnivorous mammals, which include
humans, produce enzymes that can cleave the links of
amylose and amylopectin in the digestive process, and
starch is the major component of the diet of many animals.
Pectins are polysaccharides built up from several different
monosaccharide monomers; they occur in plant cell walls
and function as a sort of cement or glue, and we use them
in the kitchen as thickening agents when we make jellies
and jams.
There are many other important natural polysaccharides,
but our discussion has served to describe those that are the
most abundant and important to us.
Plastics in the environment
Synthetic polymers, particularly plastics, possess many
very useful properties. Some of these properties are:
■ strength
■ flexibility
■ resistance to chemicals
■ impenetrability
■ inexpensiveness
■ lightness.
However, many of these properties have also made plastics
an environmental hazard. Since the 1950s, one billion
tonnes of plastic have been discarded, and will persist in the
environment for hundreds of years. The estimated lifetime
of plastic shopping bags is between 500 and 1000 years.
Discarded plastic bags find their way into the environment
where they decay very slowly or not at all. We can see them
‘decorating’ trees where they have been blown by the wind.
Plastics can block drains and sewers, causing flooding after
heavy rainfall. Ocean currents cause the plastic discarded
in the seas to collect into large floating islands of plastic
debris, known as ocean gyres. The plastic that is visible
at the ocean surface is a fraction of the plastic discarded in
253
254
Unit 2 Module 1 The chemistry of carbon compounds
the ocean, as more than half of all thermoplastics will sink
in seawater.
Incineration of hydrocarbon plastics at appropriately high
temperatures (1500 °C) produces carbon dioxide and water.
Chemists have tackled the challenge of producing
biodegradable polymers that decay at a suitable rate. The
development and widespread use of these materials will lead
to accumulation of less plastic waste in the environment.
One approach has been to incorporate a number of lightsensitive monomers into the polymer. Sunlight causes
a photochemical reaction in these monomer units that
breaks the polymer chain; when the polymer has been
broken down to small enough units, bacteria can take over
and complete the degradation.
An interesting spin-off of this approach has been that sheets
of plastic have been made that undergo degradation at a
predictable rate in sunlight. These can be used to protect
beds of newly planted seeds for just the length of time it
takes for the seedlings to begin to sprout; then the plastic
crumbles and lets the plant grow in the normal way.
The Society of the Plastics Industry (SPI) has instituted
a classification scheme in which polymers are identified
using a number from 1 to 6, with 7 being used for otherwise
unspecified materials (Table 26.1). A plastic object is
usually stamped with a rounded triangle of three clockwise
arrows (the universal recycling symbol) surrounding the
identification code of the polymer of which it is made
(Figure 26.16). Resin identification codes make it easier to
sort plastics for recycling.
Table 26.1 Resin identification codes
Code Polymer
1
2
3
Use
PET or PETE (polyethylene
thermoplastic used in plastic soft drink
terephthalate)
bottles and rigid containers
HDPE (high-density polyethene) milk and water bottles; the base of 2 L
soda bottles
PVC (polyvinyl chloride)
pipes; window and door frames among
other uses
4
LDPE (low-density polyethene)
cellophane wrap; diaper liners
5
PP (polypropylene)
6
PS (polystyrene)
light thermoplastic resin used in
packaging
food containers, insulators (foam); plastic
cutlery, lab equipment (extruded)
7
others
Waste management
Plastic is an integral part of modern life, and most of the
widely used plastics are not biodegradable. The problems
caused by discarding plastic in the environment make it
necessary for us to minimize, or even avoid, generating
plastic waste.
There are well developed systems for managing such waste.
The three R’s – Reduce, Reuse, Recycle – classify waste
management methods for plastic and other materials.
Retailers in several countries encourage reduced use and
reuse of plastic shopping bags by making customers pay
extra for new shopping bags. Recycling of thermoplastics
requires that they be melted and remoulded. However,
there are a number of problems associated with recycling
of plastic. These include:
■ lack of profitability, because the value of the material is
low;
■ technical challenges to separating the additives in the
plastic from the polymer itself;
■ the necessity to sort plastic waste into the various
polymer types prior to melting; sorting is largely
manual, and therefore labour intensive.
1
2
3
4
PETE
HDPE
V
LDPE
5
6
7
PP
PS
OTHER
Figure 26.16 Resin identification codes as stamped on plastic
objects.
Chapter 26 Macromolecules
Summary
✓ Polymers are very large molecules formed by
addition or condensation of small molecules
(monomers).
✓ Natural polymers (nucleic acids, proteins,
cellulose) play important roles in living systems.
✓ Synthetic polymers provide plastics, fibres and
elastomers and improve our quality of life.
✓ Alkenes undergo addition polymerization.
Addition polymerization commonly proceeds
via a radical mechanism, requiring an initiator
for the chain initiation step and proceeding
through chain propagation with the formation of
new radicals to chain termination, which is the
combination of radicals.
✓ Condensation polymers are commonly formed
by the reaction between amines and carboxylic
acids to yield polyamides or the reaction
between alcohols and carboxylic acids to give
polyesters.
✓ Plastics decompose very slowly. Discarded plastic
causes litter and will persist in the environment
for hundreds of years. It also blocks waterways
and is mistaken by marine animals for food,
filling their digestive systems and causing them
to starve.
✓ Reduce, Reuse and Recycle (the three Rs) are the
most important waste management methods for
plastics and other materials.
✓ The number inside a rounded triangle of three
clockwise arrows stamped on a plastic object
is the resin identification code; this number
identifies the polymer of which the object is
made.
255
256
Unit 2 Module 1 The chemistry of carbon compounds
Review questions
1
Draw the structure of the monomer from which each of the following
addition polymers is formed.
a
H
C
C
C
H
H
Orion
2
N
b
n
TM
H
CH3
C
C
H
COOCH3
n
polymethyl methacrylate
A short segment of the polymer Kevlar® is shown below.
O
O
H
H
O
O
C
C
N
N
C
C
H
H
N
N
n
Kevlar
(a)
(b)
(c)
(d)
3
What functional group is present in Kevlar®?
Draw the structures of the monomeric units from which Kevlar® is made.
Is Kevlar® an addition polymer or a condensation polymer?
Kevlar® is a super fibre with a tensile strength greater than that of steel
and it is stable at very high temperatures. It is used for making army
helmets, bullet-proof vests and protective clothing for firefighters. Suggest
a reason for the strength and inertness of Kevlar®.
(a) Show the condensation reaction between one molecule of phosgene
and two molecules of bisphenol A to yield two molecules of HCl and an
organic product B.
CH3
O
Cl
C
Cl
HO
C
OH
CH3
phosgene
bisphenol A
(b) Show the reaction between compound B, phosgene and another
molecule of bisphenol A.
(c) Draw a short segment of the polymer that is eventually formed.
(This polymer is a polycarbonate; polycarbonates are strong, transparent
polymers which are widely used in the automobile industry and to make
compact discs. The example which you have just drawn is Lexan™, used
for bullet-proof windows and traffic light lenses.)
4
(a) Why are the temperature settings on our clothes irons for synthetic
fabrics lower than those for linen and cotton?
(b) What would happen if you tried to remove a stain from your silk shirt
using ordinary household chlorine bleach? (Don’t answer this question
by carrying out an experiment!)
(c) What would happen if you accidentally spilled aqueous NaOH on your
polyester trousers?
Chapter 26 Macromolecules
Answers to ITQs
1
2
mono = one; di = two; tri = three; oligo = few (two to eight); macro = very large; poly = many.
H
a
H
C
C
H
Cl
H
H
H
H
H
H
C
C
C
C
C
C
H
Cl
H
Cl
H
Cl
chloroethene
(vinyl chloride)
H
b
polyvinyl chloride
H
C
C
H
H
H
H
H
H
H
C
C
C
C
C
C
H
H
phenylethene
(styrene)
F
c
polystyrene
F
C
C
F
H
F
F
F
F
F
F
F
C
C
C
C
C
C
F
F
F
F
F
F
tetraflouroethene
Teflon
3
Cl
Cl
Cl
oligomer, oligopeptide.
5
(a)
H
N
CH2CH2CH2CH2CH2
H
N
6
C
O–
H
(b)
n
O
+
H
Cl
Cl
Cl
4
etc.
CH2CH2CH2CH2CH2
(a) ester
(b)
O
O
HO
C
C
OH
O
H
C
N
CH2CH2CH2CH2CH2
H
O
CH2
O
H
C
N
CH2
O
CH2CH2CH2CH2CH2
O
C
H
(c) Condensation polymer
7
(a) Breaking of a bond with simultaneous addition of the elements of water (H2O) to the
fragments formed: e.g. R–O–R’ + H2O → R–O–H + H–O–R’; hydrolysis is the reverse of a
condensation reaction which produces H2O.
(b) Sucrose is a condensation product of glucose and fructose.
(c) C-1 in glucose and C-2 in fructose.
(d) No.
8
(a) (i) Maltose would give glucose.
(ii) Lactose would give galactose and glucose.
(b) Generally aqueous acid or aqueous base.
257
258
Chapter 27
Reaction mechanisms
Learning objectives
■ Define the following terms: reaction mechanism, homolytic cleavage, heterolytic cleavage, nucleophile,
■
■
■
■
■
electrophile, leaving group, chain reaction, chain initiation, chain propagation, chain termination,
substrate, solvolysis, SN1 reaction, SN2 reaction.
Illustrate electron movement in bond cleavage and bond formation using singly barbed/fish hook
arrows for single electrons and doubly barbed/curly arrows for pairs of electrons.
Show and explain the reaction mechanism for the free radical chlorination of methane.
Show and explain the reaction mechanism for the addition of Br2 to an alkene.
Predict and explain the outcome of addition of H–X to an unsymmetrical alkene.
Describe the main features of nucleophilic substitution (SN1 and SN2) reactions.
Introduction
Chemists of the nineteenth century established that the
atoms in organic molecules are held together by covalent
bonds, which are directional. Because they are directional,
they give the molecules structure and stereochemistry. This
important advance was followed in the twentieth century
by the recognition that when organic compounds undergo
a reaction, the mechanism of that reaction can be described
and explained. For a reaction such as the conversion of
bromoethane to ethanol (Figure 27.1):
■ we can describe how the bonds in the starting
materials break, how the bonds form in the products
and how the atoms move as the reaction proceeds;
■ we can provide a rational explanation for these
changes.
The mechanisms of many reactions have been thoroughly
studied and are understood in great detail. An important
tool has been the study of reaction rates, i.e. kinetics. By
measuring how the rate of a reaction depends on variables
such as temperature and the concentrations of reacting
molecules we can learn:
■ how much energy is needed to drive the reaction
forward;
■ how the reacting molecules come together;
■ how bonds are broken and made as the reaction
proceeds.
H
H
H
C
C
H
H
bromoethane
Br + NaOH
H
H
H
C
C
H
H
OH + NaBr
ethanol
Figure 27.1 Conversion of bromoethane to ethanol.
From these studies, organic chemists have been able
to propose logical and self-consistent explanations, i.e.
mechanisms, for the course of many types of related
reactions without carrying out detailed studies of every
one of them.
Homolytic and heterolytic cleavage
In Br2 the two atoms are held together by a two-electron
covalent bond which can break in one of two ways during
a reaction. In homolytic cleavage (homolysis) the two
Br atoms separate, and each retains one electron from
the bonding pair to form two identical fragments (Figure
27.2). The Br fragments are said to be radicals because
they each have one unpaired electron in the valence shell.
That valence shell is one electron short of a closed shell.
In reactions involving homolytic cleavage and reactions of
radicals, singly barbed (fish hook) arrows are used to show
the movement of single electrons.
Chapter 27 Reaction mechanisms
a nucleophile because it seeks out positive charge. Many
nucleophiles are negatively charged so we will use X:− as
a symbol.
singly barbed or fish hook arrows; each
shows the movement of a single electron
Br
Br
+
Br
covalent bond = a pair of electrons
Br
The bromine species accepting those electrons is called an
electrophile. The heterolysis of Br2 assisted by X:− is shown
in Figure 27.4.
bromine radicals
Figure 27.2 Homolytic cleavage of molecular bromine, Br2.
In the reverse of this reaction, two radicals can combine to
form a covalent bond (Figure 27.3).
singly barbed or fish hook arrows; each
shows the movement of a single electron
doubly barbed or curly arrows; each
shows the movement of a pair of electrons
X
–
+
nucleophile
Br
Br
Br
bromine radicals
Br
Br
electrophile
Br
covalent bond = a pair of electrons
Figure 27.3 Combination of bromine radicals to form molecular
bromine, Br2.
In heterolytic cleavage (heterolysis), both electrons which
form the covalent bond remain with one fragment. This
fragment then has an even number of electrons and one
unit of negative charge because of the extra electron.
Heterolysis of Br2 produces the bromide anion, :Br−, so we
might expect that the other fragment would be the cation
Br+. This is an unstable species with two gaps in the valency
shell, but it can be stabilized in the presence of another
species with two electrons to spare. Such a species is called
X
Br
+
product formed by combination
of nucleophile and electrophile
Br
–
bromide ion
Figure 27.4 Heterolytic cleavage of bromine, Br2, assisted by a
nucleophile, X:−.
Figure 27.4 shows the use of doubly barbed arrows to show
the movement of pairs of electrons. These arrows show
the direction of movement of electron pairs. Do not use
these arrows to describe the movement of atoms. You may
argue that, in the sequence above, the X:− moves forward
and :Br− moves away. That is true, but the purpose of the
arrows is to show what is happening to the electrons.
Curly arrows can begin only in areas of high electron
density. These are:
■ in the middle of covalent bonds;
■ at negative charges;
ITQ 1
■ at lone pairs of electrons.
Draw the required number of curly arrows to show the
mechanism of each of the following reactions. The covalent
bonds formed or broken are shown in colour.
Curly arrows can end at the following positions, with the
results described:
CH3
a
H3C
■ at an initially uncharged bonded atom or group,
CH3
C
Br
H3C
CH3
+
+
C
H3C
C
Br
(one arrow)
CH3
CH3
+
–
+
OH
H3C
C
CH3
OH
(one arrow)
H
O
d
+
+
H
H3C
O
(one arrow)
H
H
+
H3C
H
+
O
H
+
(two arrows)
–
OH
H3C
O
H
e
CH3CH2
+
H
H
N
CH2CH3
CH2CH3
+
H
+
CH3CH2
bond formation or bond breakage and neutralization of
the positive charge.
Homolysis and radical reactions
CH3
c
H3C
resulting in bond breakage and formation of a negative
charge on the atom or group at which the arrow ends;
■ on a positively charged atom or group, resulting in
CH3
b
–
N
OH
(one arrow)
+
CH2CH3
CH2CH3
The halogens fluorine, chlorine, bromine and iodine can all
react with alkanes to form halogenated alkanes. Fluorine
reacts so vigorously that an explosion occurs if special care
is not taken, chlorine reacts rapidly, bromine less rapidly
and iodine, although it can react, does so too slowly to be
of practical use (Table 27.1).
A curly, or curved, arrow in the context of a reaction mechanism shows
movement of a pair of electrons. Covalent bonds consist of paired
electrons, therefore curly arrows illustrate breakage and formation of
covalent bonds.
259
260
Unit 2 Module 1 The chemistry of carbon compounds
Table 27.1 Enthalpy change (ΔH ) for the reaction:
CH4 + X2 → CH3X + HX
/ kJ mol−1
X
ΔH
F
−432
explosive!
Cl
−101
useful
Br
−26
useful
I
+53
very slow reaction
H3C
Comment
If an alkane such as methane is present, the Cl• can react
at a C–H bond to break it homolytically to produce H–Cl
and an alkyl radical such as •CH3. That •CH3 can react with
an unchanged Cl2 molecule by homolysis to give Cl–CH3
and Cl•, which can react with unchanged CH4 in a chain
reaction (Figure 27.5).
energy
Cl
Cl
thermal or
photochemical
H
H
H
C
H
C
H
H
H
H
C
H
+
Cl
chain initiation
H
H Cl
Cl
Cl
H
H3C
CH3
Cl
Cl
Cl
Cl
H3C
Cl
H3C
Cl
chain termination
Figure 27.6 Chain termination steps in the free radical
chlorination of methane.
Consider chlorination, a reaction of some practical value.
Homolytic cleavage (homolysis) of Cl2 requires energy,
which can be provided by heat or light. For clarity, we can
write the chlorine atoms produced as Cl• to emphasize that
they are radicals.
Cl
CH3
C
+
H
Cl
chain
propagation
Cl
+
Cl
H
Figure 27.5 Chain initiation and propagation steps in the free
radical chlorination of methane.
In principle, the chain initiation step needs to take place
only once. Each of the two Cl• fragments can initiate a
chain propagation sequence by removing H• from CH4
to give H–Cl and •CH3; •CH3 can then react with Cl2 to
give CH3Cl and regenerate Cl•, which continues the chain
propagation. This process could then continue until all the
reactant molecules had been consumed. In practice, some
chain termination occurs when two radicals meet and
the lone electrons pair together to form a new molecular
bond (Figure 27.6).
In the early stages of the reaction, Cl• has the greatest
probability of meeting a CH4 molecule to produce CH3Cl.
As the CH4 is removed and the amount of CH3Cl grows, the
probability increases that Cl• will meet a CH3Cl molecule
and react to form CH2Cl2. Eventually, all the H atoms of
CH4 can be replaced by Cl.
Refrigerators, CFCs and the ozone layer
Refrigerators and air-conditioners are inventions that
make life more pleasant for us by letting us store food
longer and cooling our buildings in hot weather. They
both use an engine that depends on a so-called working
fluid or refrigerant to move heat in the required direction.
The engine uses a cycle in which the fluid cools by
evaporation and expansion, and is then compressed and
condensed by a piston, requiring work to be done. The
fluid is then returned to the evaporator. The details of
thermodynamics and technology need not concern us
here; our present interest is in the nature of the working
fluid. Ammonia, boiling point = −33 °C, has excellent
physical properties, but it causes severe problems if the
refrigerator develops a leak. In about 1930, freons such as
dichlorodifluoromethane, CF2Cl2, boiling point = −30 °C,
were introduced. They are ideal refrigerants because they
are chemically inert. Numerous other chlorofluorocarbons
(CFCs) were later developed commercially.
Because of its inertness, a CFC lasts for a very long time.
However, CFCs are gases which can rise into the stratosphere,
where they can undergo photochemical degradation by the
high-energy photons of ultraviolet light from the Sun. This
leads to the formation of chlorine atoms, Cl•, which initiate
a chain reaction which destroys ozone, O3.
Cl
F
ITQ 2 Tetrachloromethane, commonly called carbon tetrachloride
or just carbon tet, has been used as a fire extinguisher and
as a solvent, especially in dry cleaning. However, it is very
toxic, and its use has been discouraged in recent years. Use
the conventions illustrated in Figure 27.5 to show the reaction
sequence that could be used to prepare CCl4 from CH4.
Cl
C
hi
Cl
F
Cl
F
+
C
Cl
initiation step
F
+
O3
ClO
+
Cl
+
O2
propagation steps
Cl
O
O
O2
Chapter 27 Reaction mechanisms
The same reaction can be used to chlorinate (or brominate)
other alkanes. With alkanes such as CH3CH2CH3 and
(CH3)3CH, the H atoms in each molecule are not all identical,
as they are in CH4, so isomeric monochloro derivatives can
be formed. The number of isomers increases with further
chlorination. Alternatively, exhaustive chlorination can be
used to produce a product with all H atoms replaced by Cl.
Bromination of alkenes
Alkenes react rapidly with Br2 to give an addition product,
the dibromoalkane.
H
R
C
R
+
C
R
Br2
H
a trans alkene
red
Ozone in the stratosphere protects us from the destructive
effect of these high-energy photons. The photons cause
the degradation of O3 into O2 and O•, but the chemical
process is reversible, providing a steady supply of ozone
in the stratosphere to protect us.
O3
O
+
hi
O2
O2
O
+
O2
+
O
O3
hi
O
In contrast, the chain reaction initiated by Cl• is irreversible
and continues to degrade other molecules of O3, leading
to the depletion of ozone in the stratosphere and the
weakening of our protective shield.
Don’t confuse the two places where ozone occurs in the
atmosphere. Ozone is destructive when formed at sea
level as it forms part of photochemical smog. However, in
the stratosphere, ozone serves to protect us.
C
C
H
H
R
The disappearance of the bromine colour is used as a
laboratory test for alkenes.
Laboratory test for alkenes and alkynes
Bromine, Br2, is a red-brown liquid (boiling point =
+59 °C). A dilute solution of Br2 in dichloromethane is
also red-brown in colour and is used as a reagent to test
for the presence of C=C and C≡C in a compound.
When an alkene is mixed with a Br2/CH2Cl2 solution, the
Br2 from the solution adds to the alkene. The product of
the reaction is a dibromo compound, which is colourless.
The reaction starts by the electrophilic (electron-loving)
attack of Br2 on the π bond of the alkene. The double bond
in the alkene is a region of high electron density. As the
Br2 molecule approaches the C=C, the electrons which
form the Br–Br covalent bond are repelled away from the
Br atom closer to the C=C. The Br–Br bond is polarized
and the Br atom carrying the δ+ charge is therefore
electrophilic (Figure 27.7). The Br–Br bond does not break
spontaneously, and free Br+, a high-energy species, is not
formed.
this bond is polarized: one Br
atom becomes electrophilic
The effect of CFCs on ozone in the stratosphere was
discovered in about 1980 by the chemists Frank Rowland,
Mario Molina and Paul Crutzen. They were awarded the
1995 Nobel Prize for Chemistry for this work.
After the detrimental effect of CFCs on the ozone shield
was discovered, many governments began to restrict their
use or even ban them, while scientists looked for safer
refrigerants. An international agreement known as the
Montreal Protocol on Substances that Deplete the Ozone
Layer came into effect in 1989. Under this agreement
production and use of all CFCs will cease in 2040. If this
schedule is kept, then the ozone layer is expected to
recover by 2050.
Br
a dibromoalkane
colourless
Heterolysis and ionic reactions
By far the greatest number of organic reactions used in the
laboratory fall into the category termed ionic reactions. We
shall examine the halogenation of alkenes. In so doing we
shall see the contrast between these ionic reactions which
entail heterolytic cleavage of halogen molecules and the
radical reactions that occur in the halogenation of alkanes.
Br
an area of
high electron
density
C
b+
b–
Br
Br
C
electrophilic Br atom
Figure 27.7 Polarization of Br2 on approach to an alkene.
The electrons of the π bond move towards the electrophilic
Br atom and a C–Br bond forms as the Br–Br bond breaks
(Figure 27.8).
Br
+
C
C
b+
C
Br
b–
Br
C
Br +
Br
–
C
C
Br
cationic intermediate
Figure 27.8 The addition of Br2 to an alkene.
The reaction is completed by the addition of :Br− to C+ to
give the dibromoalkane.
261
262
Unit 2 Module 1 The chemistry of carbon compounds
Addition to unsymmetrical alkenes
H
H
H
If we add H–X to an unsymmetrical alkene such as
Me2C=CH2, two products are possible:
H
■ Me2CH–CH2X
■ Me2CX–CH3
In the middle of the nineteenth century, the Russian chemist
Markovnikov established a rule based on his observation of
a large number of reactions in which a hydrogen halide
H–X (X = Cl, Br or I) adds to an unsymmetrically substituted
alkene. The rule is the H adds to the C that already has the
greater number of attached H atoms, and the X adds to
the C with the fewest H substituents. Some examples are
shown in Figure 27.9.
Me
Me H
H
C
+
C
H
Cl
HCl
H
Me
+
C
Me
C
H
H
H
Me H
H
C
C
Cl
HCl
H
C
C
Me H
The explanation for Markovnikov’s rule is simple in
principle. The addition reaction begins by the protonation
of the alkene by the H of H–X. The process is an electrophilic
attack on the π electrons of C=C. This step is similar to that
in the reaction with Br2, but with the difference that, in an
ionizing solvent, the H–X bond has already been broken, so
the electrophile is H+(solvated) (Figure 27.10).
e.g. water
+
H (aq)
proton
electrophile
+
+
C
C
H
H
C
H
H
C
+
C
C
H
H
H
H
H
C
C
H
H
+
H
secondary (2˚) carbocation:
the C atom bearing the
positive charge has
tertiary (3˚) carbocation:
tertiary (1˚) carbocation:
2 alkyl groups
the C atom bearing the
the C atom bearing the
positive charge has
positive charge has
3 alkyl groups
1 alkyl group
H
Figure 27.11 Relative stability of carbocations.
A tertiary cation is more stable than a secondary cation,
which is in turn more stable than a primary cation: Me3C+
> Me2CH+ >> MeCH2+. The empirical rule is explained,
because addition of the proton to the C already having the
greatest number of hydrogens generates the cationic centre
at the C which has the greatest number of alkyl substituents
(Figure 27.12).
Me
R
C
+
C
Me
H
+
C
H
nucleophile
H
Me
+
Me
electrophile
C
H
H
stable tertiary carbocation forms
Me
Me
H
C
C+
H
H
unstable primary
carbocation does
not form
Figure 27.12 Protonation of an unsymmetrical alkene.
The addition reaction is completed when the nucleophilic
Cl:− reacts with the cationic carbon (Figure 27.13).
Stability of cationic intermediates
(carbocations)
ionizing solvent
H
H
X
Markovnikov’s rule was empirical. It was a generalization
based on many experimental observations, but there was
no explanation for it. Subsequently it was found to apply
to the addition of many compounds of general formula
H–X, not just hydrogen halides.
Cl
H
C
H
Figure 27.9 Addition of HCl to alkenes following Markovnikov’s
rule. Note that the extra hydrogen adds to the carbon that already
had the greater number of hydrogens before the addition.
H
H
–
Cl (aq)
chloride ion
nucleophile
Figure 27.10 Ionization of HCl.
The proton is added to the alkene to generate the more stable
cationic intermediate. Ions with positive charges on carbon
are called carbocations. The more stable carbocation is
that which has the greater number of alkyl substituents
(Figure 27.11).
H
Me
Cl
–
+
+
C
Me
chloride ion
nucleophile
C
Me H
H
Cl
H
C
C
H
Me H
stable tertiary carbocation
electrophile
Figure 27.13 Addition of chloride ion to a carbocation.
Acid-catalysed addition to alkenes
A reaction closely related to the addition of H–X takes place
when an alkene is treated with an aqueous mineral acid
such as dilute sulfuric acid.
H
H3C
C
H
C
H
H
H3C
+
H2O
HO
H
C
C
H
H
Chapter 27 Reaction mechanisms
The reaction begins when the acid protonates the alkene,
generating the more stable carbocation.
H3C
H3C
H
C
+
C
H
H
Very many organic reactions can be described by the
simplified equation shown in Figure 27.15.
H
+C
H
C
H
H
R
H
stable tertiary carbocation forms
The most abundant nucleophile present is water. H2O is not
an anion, but a molecular nucleophile. One of the electron
lone-pairs on O is easily donated to C+. The immediate
product is a protonated alcohol, which then loses a proton
to give the final product, the alcohol with the –OH group at
the C which had been the most substituted C on the C=C.
Again this is Markovnikov addition.
H
H
O
H3C
H CH3
H
+C
C
+
H
H
O
H
H
H
C
C
H
H
H
stable tertiary carbocation
protonated alcohol
CH3
+
–H
H
Nucleophilic substitution reactions
O
C
H
C
H
H
H
X
+
substrate:
molecule on which
the reaction is
occurring
Y
R
nucleophile
H
C
C
H
+
H
OH
O
S
sulfuric acid
O
H
Me
C
+
Me
O
C
H
H
Me
C
H
Me
+
3˚ carbocation
–
OH
O
H
H
OSO3
H
an alkyl hydrogen
sulfate
S
O
C
O
Figure 27.14 Addition of cold concentrated sulfuric acid to an
alkene.
X
leaving group:
the group
which leaves
Figure 27.15 A nucleophilic substitution reaction in outline.
This is a substitution reaction in which Y replaces X
in the substrate molecule R–X. X and Y can describe a
very large number of groups, but most substituents X are
more electronegative than carbon. Consequently many
C–X bonds are polarized δ+C–Xδ−. If the bond is to break
heterolytically, we can expect the leaving group X to
leave with the bonding electrons and appear as the ion
:X−. The group Y attacks the electron-deficient Cδ+, so it
is a nucleophile and the reaction is termed a nucleophilic
substitution. The nucleophile is often an anion, :Y−. It is
useful to focus on the C–X bond that is being replaced by
the C–Y bond and write the equation as shown in Figure
27.16. The dashed lines (- -) represent bonds that are
partially formed or partially broken.
–
A similar reaction takes place with concentrated sulfuric
acid. The product, an alkyl hydrogen sulfate (Figure 27.14),
can then be hydrolysed by heating with water, and the net
result of this reaction is the Markovnikov addition of H2O
to the alkene.
Me
+
substitution
product
alcohol
Me
Y
b–
Y–
b+
C
X
b–
Y
C
X
Y
C
b+
+
X
–
Figure 27.16 A nucleophilic substitution reaction in more detail.
We shall see later that the charges on X and Y can vary
(we have already seen that a nucleophile need not carry
a negative charge), but it is easier and clearer to use the
simple case shown to describe the general reaction.
We can now logically ask about the timing of the
bond-breaking and bond-making processes. Does the C–X
bond break before the C–Y bond forms, or does the C–Y
bond ‘grow in’ as the C-X bond is breaking? We cannot
have five complete bonds to C, so complete addition of Y
to C cannot occur first, but a partial C–Y bond could be
ITQ 3 What are the products of the reaction of 1-butene with
each of the following reagents?
(a) liquid Br2
(b) HCl
(c) water or dilute H2SO4
(d) cold concentrated H2SO4
ITQ 4 Using Figure 27.16 as a template, show
the mechanism of the reaction of each of the
nucleophiles listed below with bromomethane.
(a) :I−
(c) :−C≡N
(e) :−S–H
(b) :−OH
(d) :−O–CH3
H
H
b
C
b–
Br
H
bromomethane
263
Unit 2 Module 1 The chemistry of carbon compounds
forming while the C–X bond is in the process of breaking.
Organic chemists studying the mechanism of a great
number of nucleophilic substitution reactions concluded
that there is a spectrum of possibilities ranging from:
SN1 mechanism
■ complete rupture of the C–X bond before any C–Y
SN 2 mechanism
C
critical
ratedetermining
step
X
C
+
+
X
–
Y
C
fast step
critical
rate-determining
■ simultaneous formation of the C–Y bond as the C–X
–
Y
intermediate carbocation
bond forms, to
Y
C
X
step
Y
C
+
X
–
bond breaks.
Kinetic studies provide a method of distinguishing between
the two possibilities.
When the C–X bond breaks before any C–Y bond forms, the
rate of the reaction, measured by the disappearance of R–X,
will depend only on the concentration of substrate R–X. The
critical energy barrier that must be crossed by the reaction
involves only one molecule, R–X. This ‘unimolecular’
nucleophilic substitution has been designated the SN1
mechanism.
In the alternative mechanism, both nucleophile and
substrate take part in the critical step that determines
the rate of the reaction. This ‘bimolecular’ nucleophilic
substitution has been designated the SN2 mechanism.
These two mechanisms are shown in Figure 27.17.
We can see that an SN1 reaction takes place in two stages
or steps, but only the substrate molecule is involved in the
critical rate-determining step. The reaction profile (showing
Figure 27.17 The SN1 and SN2 mechanisms showing the ratedetermining steps.
energy against progress of reaction) has two humps (Figure
27.18). The first hump shows the ionization step leading to
the carbocation, which has a finite lifetime, but eventually
crosses the lower second hump when it meets a :Y− anion
and reacts to give the substitution product. Compounds
which will form the relatively stable secondary and tertiary
carbocations will undergo substitution reactions via the
SN1 mechanism.
An SN2 reaction takes place in one step, and both the
substrate molecule and the nucleophile are required
for that critical rate-determining step. The reaction
profile consequently has but one hump (Figure 27.19).
Compounds with two H atoms on the C bonded to X will
undergo substitution reactions via the SN2 mechanism.
Examples are primary alkyl halides.
Potential energy, E
energy of carbocation
C
–
X + Y
activation energy (E act )
for reaction with –
C+
activation
energy (E act )
for formation
of carbocation
Y
–
Y + X
C
6H for the
reaction
energy of reactants
energy of products
Figure 27.18 Reaction profile
for an SN1 reaction.
Progress of the reaction
energy of transition state (t-s)
Potential energy, E
264
C
–
X + Y
energy of reactants
activation
energy (E act )
for formation
of transition state
6H for the
reaction
Progress of the reaction
C
–
Y + X
energy of products
Figure 27.19 Reaction profile
for an SN2 reaction.
Chapter 27 Reaction mechanisms
Summary
✓ Reaction mechanisms describe the way in which
✓ The addition of Br2 to an alkene is an ionic
reaction which begins by electrophilic addition
of ‘Br+’ to the C=C and is completed by addition
of :Br− to give the dibromoalkene.
bonds break and form in a chemical reaction.
✓ Bond breakage and bond formation in reactions
of organic compounds are illustrated by curved
single headed and double headed arrows which
represent movement of single electrons and
electron pairs respectively.
✓ The addition of H–X to an unsymmetrical
alkene proceeds via formation of the most stable
carbocation intermediate.
✓ A nucleophilic substitution reaction:
✓ Bond breakage can be homolytic, leading to
R–X + :Y− → R–Y + :X−
can proceed via the SN1 mechanism in which R+
is formed, or via the SN2 mechanism in which
displacement of :X− and addition of :Y− are
simultaneous.
radicals and radical reactions or, more commonly,
heterolytic, leading to ions and ionic reactions.
✓ The formation of radicals often leads to chain
reactions. The chlorination of methane is an
example of a radical chain reaction.
Review questions
1
Insert the single-electron (fish-hook) arrows required
to complete the mechanism of the intramolecular free
radical reaction shown here.
H
H
CH2
H2C
C
C
H
homolysis
– Br
CH2
CH2
(2 arrows)
CH2
CH2
(3 arrows)
H
This question is about the alkenes K, L and M.
H
L
H
C
C
H
H
H
C
C
C
H
CH3
H
H3C
(a) Draw the four structural isomers of C4H9Br.
(b) Giving reasons, state which isomer would be most
reactive towards substitution via:
(i) the SN1 mechanism;
(ii) the SN2 mechanism.
5
This question is based on the chemistry of the alkene,
N.
CH3
reduction
+H
H
CH3
C
C
H
C
H
CH2CH3
H3C
M HC
3
C
H
CH2 Br
4
CH2
H2C
(2 arrows)
K
CH3
Br
bond
formation
CH
CH2
2
Identify which of the two molecules below will
undergo substitution via the SN1 mechanism.
H
CH
Br
3
H
(a) Show the mechanism of the addition of a proton,
H+, to the C=C to form the more stable carbocation.
(b) Show the mechanism of the addition of water
(H2O is a molecular nucleophile) to each of the
carbocations.
(c) Show how a proton is lost from each of the
products of the reactions in part (b) to form
alcohols. Which, if any, of the alcohols formed are
identical?
C
CH3
(a) Give the structure of the product formed when
compound N is treated with each of the following
reagents.
(i) HI
(ii) cold alkaline KMnO4
(iii) H2 with a Pt catalyst
(iv) Br2
(b) Write a detailed annotated mechanism for the
reaction of compound N with HI.
(c) What will occur if the product of the reaction of
compound N with HI is stirred with methanol
(CH3OH) at room temperature?
265
266
Unit 2 Module 1 The chemistry of carbon compounds
Answers to ITQs
1
2
CH3
a
H3C
CH3
C
Br
H3C
CH3
C
+
+
H3C
C
–
+
OH
H
H3C
C
CH3
OH
(one arrow)
CH3
c
H
H
O
H3C
d
H
H3C
O
(one arrow)
O
H
+
OH
H3C
O
H
C
H
+
H
CH3CH2
H
N
CH2CH3
CH2CH3
+
+
H
Cl
H
H
H
H
H
C
Cl
Cl
H
C
CH3CH2
N
(one arrow)
+
H
CH2CH3
CH2CH3
H
C
C
Cl
H
Cl
+
Cl
Cl
Cl
H
C
Cl
+
H
Cl
H
Cl
Cl
Cl
H
C
H
H
Cl
Cl
C
+
H
H
OH
C
H
Cl
e
Cl
H
H
H
(two arrows)
–
+
Cl
H
+
H
+
H3C
H
+
+
hi or heat
H
CH3
+
Cl
(one arrow)
CH3
CH3
b
Br
Cl
–
Cl
H
C
Cl
+
Cl
+
Cl
+
Cl
+
H
Cl
Cl
+
Cl
Cl
H
Cl
H
Cl
Cl
H
C
Cl
Cl
Cl
H
C
Cl
Cl
C
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
Cl
H
C
C
Cl
Cl
Cl
C
Cl
Chapter 27 Reaction mechanisms
3
H
a
C
b
CH2CH3
H
H
H
CH2CH3
H
H
H
CH2CH3
H
H
C
H2O
H+
H
+ H2SO4
C
H
H
HCl
+
C
H
Br2
+
C
C
d
+
C
H
C
c
Br H
H
H
CH2CH3
C
C
H
Br
H
H
C
C
H
Cl
H
H
C
C
H
OH
H
H
C
C
H
OSO3H
CH2CH3
CH2CH3
CH2CH3
(conc.)
CH2CH3
4
H
a
H
–
I
H
C
I
Br
C
H
H
–
HO
H
C
HO
Br
N
C
H
Br
N
C
–
O
H
C
Br
CH3O
e
HO
C
–
Br
N
C
–
H
H
H
+
Br
C
H
+
Br
S
H
C
H
Br
HS
Br
H
+
Br
H
+
Br
H
–
H3CO
C
–
H
–
H
C
–
H
H
–
H
–
H
H
H
Br
H
C
H
+
H
H
H
H
H
H
H
CH3
–
Br
C
H
d
C
H
H
H
C
I
H
H
–
Br
C
H
c
H
H
H
b
–
H
Br
HS
C
H
–
267
268
Module 2
Analytical methods and
separation techniques
Chapter 28
Measurement in chemical analysis
Learning objectives
■ Define the terms mean and standard deviation and calculate the mean and standard deviation of data
■
■
■
■
values provided.
Explain the meanings of the terms accuracy, precision, systematic error and random error.
Define uncertainty in measurement and include values for the uncertainty in reported data for
temperature, volume, mass and length.
Report data and the results of calculations using the correct numbers of significant figures and digits
after the decimal place.
For a given experiment, choose the correct glassware for measurement of volume and the correct
balance for measurement of mass.
Introduction
Accuracy
In the experimental physical sciences, of which chemistry
is one, descriptions of substances and phenomena are based
on data collected by measurement of mass, volume, length,
temperature, time and other parameters. In this chapter
the important terms used in statistical analysis of data are
explained. Also discussed are the factors which can affect
the validity of data obtained in chemical analysis.
Defining some terms
Mean
The mean (the average or arithmetic mean) of a group
of values is obtained by dividing the sum of values by the
number of values in the group.
Mathematically this is:
x=
x1 + x2 + x3 + ...xn−1 + xn
n
(28.1)
where x1 … xn are the actual values and n is the number of
values.
The accuracy of a measurement is the closeness of the
measurement to the true or generally accepted value of the
quantity being measured.
Precision
The precision of a series of measurements of the same
quantity obtained in the same way is the closeness of the
values to each other. Precision is therefore an indication of
reproducibility.
The precision of a group of measurements can be quantified
by calculating the standard deviation (SD).
ITQ 1 A sample of sodium chloride which is known to contain
exactly 60.66% Cl is analysed first by method A and then by
method B.
■ Average value of Cl obtained by method A = 60.73%
■ Average value of Cl obtained by method B = 60.81%
Which method provides more accurate data?
Chapter 28 Measurement in chemical analysis
Standard deviation is calculated using the formula:
s=
(x
− x ) + ( x2 − x ) +...+ ( xn − x )
n −1
s=
∑( x − x )
− 2
1
− 2
−
2
(28.2)
− 2
(28.3)
n −1
Worked example 28.1
Q
Calculate the standard deviation for the values obtained in
method A for the % Cl in the sample in ITQ 2.
A
(i) First calculate the mean:
60.85 + 60.59 + 60.75
= 60.73%
3
(ii) Use the formula in Equation 28.2:
Systematic errors result from limitations in experimental
techniques, poor choice of method, faults in apparatus
and instruments and inaccuracies in concentrations of
reagents. Each systematic error is due to a positive or a
negative factor in the experimental procedure. Systematic
errors affect the accuracy of a result.
Random errors can change from positive to negative in going
from one determination to the next. The experimentalist
has no control over random errors. Random errors affect
precision.
There are two details to bear in mind.
2
2
2
s = (60.85 − 60.73) + (60.59 − 60.73) + (60.75 − 60.73)
3−1
2
2
2
s = (0.12) + (−0.14) + (0.02)
2
If the errors in a set of measurements are truly random (for
example, as might occur when you repeat an experiment
as carefully as you can a number of times) then the ‘real’
result has a 70% chance of lying in the range mean ± SD and
a 95% chance of lying in the range mean ± 2SD. Random
errors are errors which cause unknown and unpredictable
changes in the measured result.
There can be large systematic errors present in
measurements which are precise. These are errors built
in to a measuring device. Examples are a pipette which
does not deliver its stated volume of liquid or an electronic
balance which does not accurately return to zero.
s = 0.0144 + 0.0196 + 0.0004
2
s = 0.0344
2
=0.13
Usually you would use a calculator to do the hard work. Many
calculators have ‘calculation of SD’ as one of their functions.
The standard deviation is usually included in the statement
of the mean of a series of measurements. The mean values
for the chloride percentages given in ITQ 2 are stated as:
■ chloride content determined using method A = (60.73
± 0.13)%
■ chloride content determined using method B = (60.81
± 0.05)%
Uncertainty in single determinations
Despite the increasing popularity of instruments with
digital displays, many laboratory devices which you will
encounter in chemistry are graduated. Instruments such
as rulers, burettes, measuring cylinders, some pipettes,
syringes, thermometers and some balances are marked
with a scale: i.e. a line divided into equal spaces (scale
units) which are numbered at regular intervals.
■ On a 50 cm3 burette the scale unit is usually 0.1 cm3
■ On a 100 cm3 measuring cylinder the scale unit is 1 cm3
■ On a 10 cm3 measuring pipette the scale unit is 0.1 cm3
Errors
Experimental results are subject to errors which are classified
into two main groups – systematic errors and random errors.
ITQ 2 For the analysis of Cl in sodium chloride, three replicate
samples were analysed by method A and another three replicate
samples by method B (see ITQ 1). Values for the percent Cl in
each sample are given below:
% chloride
method A
60.85
69.59
60.75
method B
60.76
60.85
60.82
Which of the two sets of values are more precise – those from
method A or those from method B?
■ On a standard laboratory mercury thermometer the
scale unit is 1 °C
When reading a value from one of these pieces of apparatus
you must estimate the value to one decimal place more
than the level of graduation. If the reading coincides with
ITQ 3
(a) Calculate the standard deviation in the values obtained via
method B for the % Cl in the sample in ITQ 2.
(b) Is this value lower or higher than the standard deviation in the
values obtained for method A?
269
270
Unit 2 Module 2 Analytical methods and separation techniques
a
a
b
b
16
16
60
60
50
50
16.25 cm3
16.60 cm3
17
17
42.5 ˚C
18
18
40
40
37 ˚C
19
19
30
30
Figure 28.1 Burette readings of (a) 16.60 cm3 and (b) 16.25 cm3.
Figure 28.2 Thermometer readings of (a) 37.0 °C and (b) 42.5 °C.
a scale unit there is a zero in the last decimal place of the
reading. If the reading does not coincide with a scale unit,
the digit in the last decimal place is the scale unit divided by
2. The burette reading illustrated in Figure 28.1(a) would
therefore be 16.60 cm3 and the burette reading in Figure
28.1(b) is 16.25 cm3. Recall that burette readings are taken
at the lowest point of the meniscus.
Uncertainty in addition and subtraction
The thermometer reading illustrated in Figure 28.2(a) is
37.0 °C and in Figure 28.2(b) it is 42.5 °C.
The estimated digit in the burette reading of 16.25 cm3 and
the thermometer reading of 42.5 °C is the uncertainty in
the reading. These readings are therefore correctly written
as (16.25 ± 0.05) cm3 and (42.5 ± 0.5) °C. Similarly, the
readings which coincide with the scale units are written
correctly as (16.60 ± 0.05) cm3 and (37.0 ± 0.5) °C.
Uncertainty in this context is defined as the range which
a data value may have because the last digit which has
been read from the instrument scale has been estimated
(interpolated).
Most modern balances have digital displays, and a number
of factors contribute to the uncertainty in the values of the
masses measured. The uncertainty, in grams, of weighing
with digital balances is usually written on the balance itself.
If the uncertainty is not written on the balance it can be
assumed to be equal to the smallest increment displayed.
This also applies to other digital devices.
The uncertainty in measured values affects the calculations
in which these values are used. We are concerned here
only with addition and subtraction, to which a very simple
rule applies: the uncertainty in the result of adding or
subtracting two numbers is the square root of the sum of
the squares of the uncertainties in the two numbers.
For A (± a) − B (± b) = C (± c) and A (± a) + B (± b) = C (± c),
the uncertainty in the answer (c) is:
c = a2 + b2
Worked example 28.2
Q
At the start of a titration the reading on a 50 cm3 burette was
3.55 cm3. At the end point of the titration the reading was
13.80 cm3. The scale unit on the burette is 0.1 cm3. What
amount of solution was used?
A
(a) State each burette reading correctly (include the
uncertainty).
initial reading = (3.55 ± 0.05) cm3
final reading = (13.80 ± 0.05) cm3
(b) Give a correct statement of the volume of the solution
delivered from the burette (include the uncertainty).
final reading = (13.80 ± 0.05) cm3
initial reading = (3.55 ± 0.05) cm3
volume of solution delivered
= (10.25 ± 0.052 +0.052 ) cm3
ITQ 4 Name two digital devices, other than balances, which you
might encounter in a modern laboratory.
= (10.25 ± 0.07) cm3
Chapter 28 Measurement in chemical analysis
Significant figures
The number of digits in a measurement is called the
number of significant figures. A burette reading of
16.25 cm3 has four significant figures and a thermometer
reading of 42.5 °C has three significant figures.
■ Zeros in the middle of a number are significant: 3.708
has four significant figures.
■ Zeros at the beginning of a number are not significant:
0.003888 has four significant figures and 0.03708 also
has four significant figures.
■ Zeros at the end of a number and after a decimal place
are significant: 37.0 has three significant figures and
3.850 has four significant figures.
If we add or subtract two values with different numbers of
significant figures, the answer must not have more digits
after the decimal place than either of the original numbers:
1.95
cm3 ← three significant figures;
two decimal places
3
+ 0.04137 cm
= 1.99
← four significant figures;
five decimal places
cm3 ← three significant figures;
two decimal places
In carrying out multiplication or division of values with
different numbers of significant figures the answer cannot
have more significant figures than either of the original
values.
÷
=
355
13.53
26.2
← three significant figures
← four significant figures
← three significant figures
The guiding principle here is that the number of significant
figures in a calculated result depends on the term with the
largest percentage uncertainty.
When we multiply or divide numbers using calculators, the
answers contain up to eight digits after the decimal point.
These answers must therefore be rounded to give the
correct number of significant figures and digits after the
decimal point.
When the number of digits to be kept has been decided, the
rules for rounding numbers are applied.
■ If the first digit to be removed is less than 5, round
down by dropping it and all following digits.
3.763 541 becomes 3.76 when rounded to three
significant figures.
■ If the first digit to be removed is greater than or equal
to 6, round up by adding 1 to the digit on the left.
3.763 541 becomes 3.8 when rounded to two
significant figures.
■ If the first digit to be removed is 5 and there are more
non zero digits following, round up by adding 1 to the
digit on the left.
3.763 541 becomes 3.764 when rounded to four
significant figures.
■ If the first digit to be removed is 5 and nothing follows,
round down.
3.763 5 becomes 3.763 when rounded to three
significant figures.
Glassware used for measuring volume
Measuring cylinders, volumetric flasks, pipettes, burettes
and syringes are all used to measure volume.
■ Measuring cylinders are the least accurate of these
measuring devices and so are not used to measure
liquids for quantitative chemical analysis.
■ Volumetric flasks are used to prepare solutions of
known concentration. A volumetric flask is designated
To-Contain (TC) a specific volume of liquid.
■ Volumetric pipettes are used to transfer a specific
volume of a solution from one container to another.
A volumetric pipette is designated To-Deliver (TD)
the volume stated. After the liquid has been delivered
from the pipette, some of it remains in the tip and on
the inner surface; the stated TD volume takes account
of this, so the liquid in the tip and on the inner surface
should not be blown out.
■ Burettes are used to accurately deliver a variable
amount of liquid, as required in titrations.
Volumetric glassware must be properly calibrated to avoid
introducing systematic errors into the measurements for
which they are used. Calibration is accomplished by very
carefully measuring the mass of water that is contained
in or delivered by the volumetric flask, pipette or burette.
The mass is then converted to volume, using the density of
water at the temperature at which the mass was measured.
volume =
mass
density
Figure 28.3 shows items of glassware used for measuring
volume.
271
Unit 2 Module 2 Analytical methods and separation techniques
(b)
(c)
(d)
(e)
10cm
3
100
(f)
5
90
6
7
8
9
3
70
10
11
12
13
14
60
50
50 ml
80
15
30
20
20
21
22
23
24
20
25
250 ml
Figure 28.3 (a) Measuring cylinder, 100 cm3,
(b) volumetric flask, 250 cm3, and (c) volumetric
pipette, 25 cm3, (d) graduated pipette, 10 cm3,
(e) burette, 50 cm3, and (f) gas syringe, 50 cm3.
30 40
16
17
18
19
40
10
0
1
2
3
4
10
(a)
25cm
272
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
(a)
(b)
(c)
Figure 28.4 (a) Mechanical single
pan beam balance, (b) top loading
electronic balance and (c) analytical
electronic balance.
Measuring mass
The terms ‘mass’ and ‘weight’ are often used interchangeably,
but they are different properties.
■ Mass is the measurement of the amount of matter in
an object.
■ Weight is a force, the pull of gravity on an object by
the Earth or some other celestial body.
Weight varies with location, but at the same location objects
with identical masses have identical weights. The mass of
an object is measured by comparing (balancing) its weight
with the weight of a reference standard of known mass.
Your school laboratory is probably equipped with at least
one mechanical single pan beam balance. The object
whose mass is to be determined is placed in the pan and
the sliding counterweights on the beams are moved until
the object in the pan is balanced. The capacity of such
balances is usually about 600 g and the smallest scale unit is
0.1 g. Increasingly, masses are determined using electronic
balances with digital displays. Top-loading electronic
balances generally have a capacity of around 300 g and are
used to determine masses to ± 0.01 g. Analytical electronic
balances have smaller capacities (around 100 g) and some
are sensitive to ± 0.00001 g.
Figure 28.4 shows three types of balance.
Vibrations of the surface on which they are placed and air
currents affect all balances and cause the measurements
to fluctuate. The effect of these factors increases with the
sensitivity of the balance. In the mechanical single pan and
the top-loading electronic balances the balance pan is open
to the air, so these balances must not be placed in draughts
or air currents. In an analytical balance the balance pan is
enclosed in a balance chamber with doors which are closed
when masses are being determined. Placing objects which
are above or below room temperature on the balance pan
sets up convection currents in the chamber, causing the
measurement to fluctuate, so objects must be at room
temperature when placed on an analytical balance.
ITQ 5 Which of the piece of glassware illustrated in Figure 28.3 would you use for each of the following operations?
Choose from: measuring cylinder, volumetric flask, pipette, burette, syringe.
(a) Preparing 250 cm3 of a 1 M NaOH solution.
(b) Transferring 25 cm3 of 1 M NaOH solution from a volumetric flask to a conical flask for a titration.
(c) Introducing 1.4 g of butanal (a liquid of density 0.80 g cm−3) into a reaction mixture.
(d) Measuring the water to prepare 100 cm3 of an approximately 5% weight volume solution of sodium hydroxide.
Chapter 28 Measurement in chemical analysis
Summary
4
✓ Measurement of mass, volume, temperature
Initial temperature = 16.0 °C
and time is fundamental to chemistry and is
particularly important in chemical analysis.
Final temperature = 42.5 °C
The scale unit on the thermometer used was 1 °C.
✓ Two important terms in statistical analysis of
(a) Write the initial and final temperatures to include
the uncertainty in the measurement.
(b) Calculate and state the increase in temperature
and the uncertainty in this value.
data are the mean and standard deviation.
✓ Key concepts in measurement and reporting
of experimental data are accuracy, precision,
systematic error, random error and uncertainty.
✓ The number of significant figures and/or decimal
5
Round off each of the following quantities to the
number of significant figures indicated in brackets.
(a) 2.2046 kg (3)
(b) 453.59 g (4)
(c) 1.6093 km (3)
6
Carry out each of the following calculations and
express each result with the correct number of
significant figures. Do the calculation first using all
the figures, whether they are significant or not, then
round off the final answer.
(a) 0.489 cm3 + 21.2 cm3 − 2.4924 cm3= ??? cm3
(b) 5.7853 g ÷ 0.00260 dm3 = ??? g dm−3
(c) 36.791 g + 0.09855 g = ??? g
places in a data value is determined by the
magnitude of the scale unit on the measuring
device from which the data has been read.
Review questions
1
Measurements were made of the weight of an active
drug in two sets of prescription tablets of different
weights. Both sets of tablets contain 80.0% by weight
of the active drug. The tablets in set A each weighed
25.0 mg and the tablets in set B each weighed
100.0 mg. The data are given below.
Weight of active drug / mg
Set A (n = 5) 19.8
20.3
20.6
19.2
19.7
Set B (n = 4) 81.1
79.3
80.4
79.7
Calculate the mean and standard deviation in each set
of measurements.
2
Seven analyses for the phosphorus content of a
fertilizer gave values of 16.2, 17.5, 15.4, 15.9, 16.8,
16.3 and 17.1%. Calculate the mean and standard
deviation of these analyses.
3
Two methods of analysis of the % iron in a sample
gave the following data:
Method E: (36.27 ± 0.16)%
Method F: (36.34 ± 0.22)%
The true value is 36.32%
Which set of data is more accurate?
The initial and final temperatures of a water bath
were measured in an experiment to determine the
heat of combustion of a substance.
Answers to ITQs
1
Method A
2
Method B
3
(a) 0.05
(b) lower
4
thermometer, stopclock
5
(a) 250 cm3 volumetric flask
(b) 25 cm3 TD pipette
(c) syringe – use the density of the liquid to calculate
the volume of the mass given
(d) 100 cm3 measuring cylinder
273
274
Chapter 29
Gravimetric analysis
Learning objectives
■ Define the terms gravimetric analysis, precipitation gravimetry and
volatilization gravimetry.
■ Given the necessary data, calculate the percentage composition of a salt and
the experimental percentage and number of moles of water of crystallization
in a hydrated salt.
■ Give examples of precipitates which are useful in precipitation gravimetry
and describe their properties.
■ Describe a simple experiment to determine the composition of a salt by
precipitation gravimetry.
■ Determine the moisture content of foodstuffs and of soils and the amount of
water of crystallization of hydrated salts.
Introduction
In analytical chemistry we both identify materials and
determine the composition of substances. Identification is
a qualitative exercise. For example, consider the following
set of experimental observations.
A blue-coloured solid compound is dissolved in water. When
aqueous AgNO3 is added to this solution a white precipitate is
observed. An aqueous solution of the solid is also blue in colour,
and produces a blue precipitate on addition of aqueous NaOH.
This solid can be identified qualitatively as copper(II)
chloride, CuCl2. However, these qualitative tests do not
give us information about the purity of the sample: the
percentages that consist of copper ions, chloride ions or
water of crystallization. Gravimetric analysis, which is
quantitative analysis by weight, is a classical and relatively
simple method of determining this information.
The precipitation method
Gravimetric analysis of anions and metal cations often
entails the formation and precipitation and weighing of
known, insoluble and very stable compounds which contain
the anion or metal cation. This is known as precipitation
gravimetry. We could, therefore, determine the percentage
of chlorine in our sample of CuCl2 by dissolving a known
mass (approximately 0.2 g) of the sample in H2O, acidifying
the solution with HNO3 and adding AgNO3 to precipitate
AgCl as shown in this equation:
CuCl2(aq) + 2AgNO3(aq) → 2AgCl(s) + Cu(NO3)2(aq)
The AgCl could then be isolated by filtration, dried and
weighed.
Worked example 29.1 illustrates the separation of an ion
(Cl−) from the sample of CuCl2 by its transformation into the
insoluble stable AgCl. The AgCl can then be precipitated,
isolated and weighed.
The substances precipitated in gravimetric analysis must
possess the following properties.
■ They must be stable, of definite chemical composition
and precipitated in a pure form.
■ They must be highly insoluble in the solutions in
which they are formed so that virtually none of the
ions being analysed remain in solution and none of the
precipitate is dissolved in washing.
■ The precipitate must be granular; the size of the
particles must be sufficiently large so that they do not
pass through the filter.
Qualitative analysis of a number of anions and cations
relies on the formation of insoluble precipitates. These
precipitates include BaSO4, AgCl, AgBr and AgI, all of
Chapter 29 Gravimetric analysis
H3C
Worked example 29.1
Q
A
N
OH
N
OH
C
We have a mass of 0.2212 g of the solid which we know to
contain CuCl2. We obtain 0.3528 g of AgCl from it, as described
above. Calculate:
(a) the percentage of Cl in the sample;
(b) the percentage of CuCl2 in the sample (assuming that all the
Cl present is CuCl2).
(a) Calculating the percentage of Cl in the sample:
number of moles AgCl obtained:
mass AgCl
0.3528 g
=
= 2.4616 × 10–3 mol
FW AgCl
143.32 g mol−1
–
C
H3C
dimethylglyoxime
(DMG)
Figure 29.1 Structure of an organic reagent used for gravimetric
analysis of cations.
Dimethylglyoxime (DMG) is commonly used for
gravimetric analysis of nickel (Ni2+), with which it forms a
reddish precipitate (Ni(DMG)2).
H
+
1 mol Cl is chemically equivalent to 1 mol Ag
∴ number of moles Cl– = number of moles AgCl
= 2.4616 × 10–3 mol
–
mass of Cl in AgCl = number of moles Cl– × FW Cl
= 2.4616 × 10–3 mol × 35.45 g mol–1
= 0.08726 g
mass Cl
% Cl =
× 100%
mass sample
H3C
N
OH
H3C
+
C
H3C
N
Ni 2+
OH
(b) Calculating the percentage CuCl2 in the sample:
2 mol Cl– is chemically equivalent to 1 mol Cu2+
1 mol Cl– is chemically equivalent to 0.5 mol Cu2+
number of moles Cl–
∴ number of moles Cu2+ =
2
=
H3C
2.4616 × 10–3 mol
2
N
CH3
C
Ni
N
C
N
O
Ni(DMG)2
× 100%
N
C
dimethylglyoxime (DMG)
–
0.08726 g
=
0.2212 g
= 39.45%
O
C
C
2
O
CH3
O
H
The formation of precipitates for gravimetric analysis
must be carried out under carefully controlled conditions
of temperature, pH, concentration/dilution and ionic
strength. Specific conditions are required for the formation
of each specific precipitate. Other experimental aspects of
successful gravimetric analysis are:
■ accurate and precise weighing of substances to
be analysed (analytes), of precipitates and of the
containers in which they are placed;
■ quantitative (100%) transfer of analytes to reaction
vessels and of precipitates to weighing vessels.
–3
= 1.2308 × 10 mol
mass of Cu2+ = number of moles Cu2+ × FW Cu2+
= 1.2308 × 10–3 mol × 63.55 g mol–1
= 0.07822 g
mass of CuCl2 in sample = mass of Cu2+ + mass of Cl–
= 0.07822 g + 0.08726 g
= 0.16548 g
0.16548 g
% CuCl2 in sample =
× 100%
0.2212 g
= 74.81%
which are suitable for gravimetric analysis of the anions
and cations in salts of Ba2+, SO42−, Ag+, Cl−, Br− and I−.
Some metal ions form complexes with organic reagents
(see Chapter 17, page 163). These metal ion complexes are
usually coloured and some are sufficiently insoluble for use
in gravimetric analysis. The structure of dimethylglyoxime
(DMG) is shown in Figure 29.1.
Apparatus and glassware for gravimetric
analysis
■ Analytical balance: a balance for determination of
weights to four decimal places.
■ Laboratory oven: for drying and conditioning
glassware and drying reagents and precipitates.
■ Desiccators: a large tightly covered glass vessel in
which a drying agent (self-indicating silica gel or
CaCl2) is placed. The top of the desiccator bears a
stopcock that can be attached to a vacuum source.
Solid compounds, glassware and apparatus are kept
dry or cooled under dry conditions in a desiccator.
ITQ 1 An impure sample of NaCl (0.2063 g) produced 0.3735 g of
AgCl when treated with excess AgNO3. Calculate the percentage
of NaCl in the sample.
275
276
Unit 2 Module 2 Analytical methods and separation techniques
suspension to be filtered
sintered glass crucible
filter cone
Volatilization methods
to vacuum
filter flask
assembly for filtration using sintered glass
crucible, rubber cone and filter flask
Figure 29.2 Apparatus for filtration.
(a)
Figure 29.2 shows the apparatus for filtration and Figure
29.3 shows other pieces of apparatus used in gravimetric
analysis.
(b)
The amount of water of crystallization in a hydrated salt and
the percentage of moisture in soils and in foodstuffs can be
determined by heating the sample at a suitable temperature
until the weight of the sample is constant. This is an example
of volatilization gravimetry. If samples are to be heated
at very high temperatures they are placed in porcelain or
silica crucibles which can withstand temperatures of up
to 1000 °C. If we are doing this, then the substance itself
must not decompose at a high temperature. Aluminium
pans are suitable containers if samples are to be heated at
temperatures up to 130 °C. Crucibles and pans should be
conditioned by heating and cooling to constant mass.
Water of crystallization
(c)
(d)
Figure 29.3 (a) Drying oven, (b) desiccator, (c) sintered glass
crucible and (d) tongs.
■ Weighing paper: squares of paper with a very smooth
surface, sold commercially for weighing solids.
■ Source of vacuum: this can be a mechanical vacuum
pump or a water aspirator pump.
■ Sintered glass crucibles, porosity #4: cylindrical glass
funnel used for both filtration and weighing. The base
consists of a porous disc of ground glass that has been
subjected to heat and pressure (sintered). The porosity
of the disc is given by a number: 0 is the coarsest and
5 is the finest. For use in gravimetric analysis the
crucibles must be conditioned by heating and cooling
until there is no change in mass.
■ Filter cone: rubber or silicone cone in which the
sintered glass crucible is placed and which fits on to the
top of the filter flask.
Many metal salts occur as hydrates in the solid state. A
hydrate is a solid salt which consists of cations and anions
which are surrounded by weakly bonded water molecules.
The water molecules are important in maintaining the
crystalline structure of the solid, and each metal hydrate
(also known as a hydrated salt) is associated with a specific
number of water molecules. These water molecules are
known as water of crystallization (sometimes called water
of hydration).
Examples of hydrated salts are:
■ BaCl2.2H2O – barium chloride dihydrate
■ CuSO4.5H2O – copper sulfate pentahydrate
■ MgSO4.7H2O – magnesium sulfate heptahydrate
■ Na2SO4.10H2O – sodium sulfate decahydrate
Barium chloride occurs as a dihydrate (BaCl2.2H2O). The
relative formula mass of BaCl2.2H2O is:
Ba
+ 2 × Cl
+ 2 × H2O
=
137.33
70.90
36.04
244.27
The theoretical percentage of water in BaCl2.2H2O is:
36.04
× 100% = 14.75%
244.27
■ Filter flask: conical thick-walled glass flask with a side
arm used for filtration under vacuum.
■ Tongs: for transferring hot glassware from the oven to
the desiccator.
ITQ 2 How would you use the pieces of apparatus described in
this chapter in the gravimetric analysis of the impure sample of
NaCl for which the data is given in ITQ 1?
Chapter 29 Gravimetric analysis
Procedure
crucible and lid
ring clamp
clay triangle
Bunsen
flame
Figure 29.4 Experimental assembly for the determination of
water of crystallization in a hydrated salt.
When heated, many hydrated salts decompose to produce
the anhydrous salt and water. Barium chloride dihydrate
gives up all its water of crystallization, as shown in the
following equation:
1 Two porcelain crucibles with lids are washed and dried.
Each crucible and lid is placed on the clay triangle and
heated by a direct flame for 5 minutes. The crucible
and lid are cooled in a desiccator and weighed. Heating
and cooling are repeated until the weight of the
crucible and lid is constant.
2 Approximately 1 g of BaCl2.2H2O is placed in each
crucible and the crucible, lid and contents are weighed
accurately. The mass of each sample of BaCl2.2H2O is
calculated by subtraction.
3 One crucible is placed on the clay triangle with the lid
partially covering the crucible. The crucible is heated
with a low Bunsen flame for about 3 minutes and
then with a high flame for 10 minutes. The crucible
is allowed to cool in air for a few minutes and then
transferred to a desiccator for final cooling. The crucible
with anhydrous BaCl2 and the lid are then weighed.
4 The procedure in step 3 above is repeated with the
second sample. The percentage water in each sample is
calculated and the average of the two values is taken.
heat
BaCl2.2H2O(s) ⎯→ BaCl2(s) + 2H2O(g)
By weighing the sample before and after heating it is
possible to determine the amount of water of crystallization
in a hydrated salt. When expressed as a percentage this is
known as the experimental percentage of water.
ITQ 3 Here are the experimental procedure and data for the
determination of the percentage moisture of skimmed milk,
carried out in duplicate. Calculate the percentage moisture in the
sample.
Procedure
Here is the experimental procedure for determing the
percentage water in BaCl2.2H2O. This procedure can be
applied to determining water of crystallization in other
hydrates that readily give up water on heating.
Duplicate samples of approximately 1 g of skimmed milk were
weighed into covered pre-conditioned aluminium pans. The pans
containing the samples were dried in a ventilated laboratory oven
at 100–105 °C for 2 hours. The pans with samples were removed
from the oven, cooled in a desiccator and weighed. The pans
with samples were heated for a further two hours as before,
cooled and re-weighed. The second weights were taken as final.
Apparatus
Results
Experimental method
Sample A
■ Analytical balance
■ 2 × porcelain crucibles
■ Clay triangle
■ Bunsen burner
■ Tongs
■ Ring clamp
Sample B
Weight of aluminium pan + cover
16.2280 g
15.2294 g
Weight of aluminium pan + cover + sample
17.3534 g
16.2838 g
17.2671 g
16.2016 g
17.2668 g
16.2015 g
Weight of aluminium pan + cover + sample
after heating for 2 h
Weight of aluminium pan + cover + sample
after heating for 4 h
■ Desiccator
Chemicals
ITQ 4 Calculate the theoretical percentages water of
crystallization in the following salts:
BaCl2.2H2O (approximately 2 g)
(a) CuSO4.5H2O
(b) MgSO4.7H2O
(c) Na2SO4.10H2O
277
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Unit 2 Module 2 Analytical methods and separation techniques
Worked example 28.2
Q
A
The experimental percentage of water in an unknown hydrate
was found to be 20.59% using the method described above.
The formula weight of the anhydrous salt is 144.45 g mol−1.
Calculate the number of moles of water of crystallization per
mole of the anhydrous salt (AS).
Unknown hydrate AS.xH2O is 20.59% H2O
% AS in the unknown hydrate = (100 − 20.59)% = 79.41%
100 g of AS.xH2O contains 20.59 g H2O and 79.41 g AS
20.59 g H2O =
79.41 g AS =
20.59 g
H O = 1.14 mol H2O
18.02 g mol−1 2
79.41 g
AS = 0.55 mol AS
144.45 g mol−1
Number of moles water of crystallization per mole AS
moles H2O in 100 g hydrate 1.14
=
= 2.07
moles AS in 100 g hydrate
0.55
The number of moles of water of crystallization is generally
a whole number, so x = 2, and the formula of the unknown
hydrate is AS.2H2O.
=
Purity of carbonates
The purity of carbonates and hydrogencarbonates can
be determined by volatilization gravimetry. A weighed
sample of the carbonate or hydrogencarbonate is treated
with acid. This releases CO2, as shown below for sodium
hydrogencarbonate.
NaHCO3(aq) + H2SO4(aq) → CO2(g) + H2O(l) + NaHSO4(aq)
The CO2 evolved is made to react with NaOH which has
been pre-absorbed on to an inert support; the inert support
with the pre-absorbed NaOH (absorbent) is sold as a
commercial product.
2NaOH(s) + CO2(g) → Na2CO3(s) + H2O(l)
The mass of the absorbent (NaOH + inert support) is
determined before and after the reaction with CO2 and the
mass of the CO2 is obtained by subtraction.
Applications of gravimetric analysis
Modern instrumental techniques have, to some extent,
superseded gravimetric analysis. However, gravimetric
methods are still widely used.
1 To analyse and establish the purity of compounds
used as standards in other types of analysis and for
calibration of instruments.
2 To determine the concentrations/percentages of
specific substances in mixtures and materials such as
foodstuffs, drugs and soils.
3 To maintain quality control of fertilizers; the three
numbers displayed on the label of a fertilizer indicate
the percentages of nitrogen (N), phosphorus (P) and
potassium (K); phosphorus is present in the fertilizer
as phosphate ion (PO43−); the percentage of phosphate
ion in the fertilizer can be verified by converting
the phosphate in a known quantity of fertilizer to
the insoluble compound magnesium ammonium
phosphate hexahydrate (MgNH4PO4.6H2O).
4 To monitor the concentrations of particles in air; many
air quality monitors contain filters made of Teflon or
fibreglass to trap particles; the filters must be carefully
conditioned before use and weighed to a high degree
of accuracy both before and after use in the monitor;
the difference in the weight of the filter before and
after use gives the mass of the particulate matter
collected.
The classical method for the determination of the empirical
formula of a hydrocarbon by combustion analysis, described
in Chapter 19 (page 190), is a gravimetric method. You
will recall that in combustion analysis a known mass of an
organic compound is burnt in an excess of pure O2 and the
masses of the CO2(g) and H2O vapour which are produced
are found.
The balanced equation for the combustion of ethane is:
2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)
The CO2 and H2O formed in this reaction are trapped by
suitable absorbents and their masses obtained from the
increase in the weights of the absorbents.
ITQ 5 A sample of BaCl2.2H2O (1.0573 g) was heated in a
crucible as described in the experiment above. After heating, the
mass of the sample was 0.8982 g. Calculate the experimental
percentage of water in the sample.
ITQ 6 The experimental percentage of water in FeSO4.xH2O was
found to be 45.35%. The formula weight of anhydrous FeSO4 is
151.91 g mol−1. What is the value of x in FeSO4.xH2O?
Chapter 29 Gravimetric analysis
Summary
Review questions
1
A nickel-containing salt (0.3327 g) was treated with
the dimethylglyoxime (DMG) reagent to produce
the red Ni(DMG)2 complex, molecular formula
Ni(C4H7N2O2)2 with a final weight of 0.2422 g.
Calculate the percentage of nickel in the salt.
2
A sulfate-containing salt (0.2517 g) was treated with
an excess of BaCl2 solution to give BaSO4 with a final
weight of 0.2968 g. Calculate the percentage of sulfate
in the salt.
3
Antacid tablets which contain NaHCO3 and inert
ingredients were ground to a fine powder and a
sample of the powder (0.5248 g) was treated with
dilute H2SO4. Carbon dioxide was liberated:
✓ Gravimetric analysis is quantitative analysis by
weight. Gravimetric analysis utilizes classical
and relatively simple experimental methods
to determine the percentage composition of
samples.
✓ In precipitation gravimetry, one component of
a salt or mixture is transformed into a stable
insoluble precipitate which is isolated and
weighed. Precipitation gravimetry is used for
analysis of anions and cations.
✓ In volatalization gravimetry, H2O(g), CO2 or other
volatile compounds are generated from a sample.
The percentage moisture in a sample and the
amount of water of crystallization in hydrated
salts can be determined by volatalization
gravimetry.
NaHCO3(aq) + H2SO4(aq) →
CO2(g) + H2O(l) + NaHSO4(aq)
The CO2 was trapped by reacting it with NaOH, as
described on page 278, to form Na2CO3:
2NaOH(s) + CO2(g) → Na2CO3(s) + H2O(l)
The mass of Na2CO3 formed was found to be
0.6438 g. Calculate the percentage NaHCO3 in the
antacid tablets.
4
An alloy is known to consist of silver and copper,
both of which are non-reactive metals. However, both
silver and copper are oxidized with nitric acid, HNO3.
Ag(s) → Ag+(aq) + e−
Cu(s) → Cu2+(aq) + 2e−
4H+(aq) + NO3−(aq) + 3e− → NO(g) + 2H2O(l)
overall:
Ag(s) + Cu(s) + 4H+(aq) + NO3−(aq) →
Ag+(aq) + Cu2+(aq) + NO(g) + 2H2O(l)
You are provided with:
■ a sample of the alloy
■ nitric acid, HNO3 (6 M)
■ sodium chloride, NaCl(s)
■ distilled water
■ the apparatus necessary to carry out precipitation
gravimetric analysis (see page 274)
Describe in outline how you would determine the
percentage silver in the alloy.
279
280
Unit 2 Module 2 Analytical methods and separation techniques
Answers to ITQs
1
75.27%
2
You need to do the determination in duplicate. Discuss
the reasons for this in your class.
(a) Place a square of weighing paper on the analytical
balance and zero the balance.
(b) Use a spatula to place solid NaCl on the weighing
paper until the weight displayed on the balance
is close to 0.2 g. Record the weight of the NaCl to
four decimal places.
(c) Transfer the NaCl to a 250 cm3 beaker and add
water (150 cm3) to dissolve the NaCl.
(d) Add 0.1 mol dm−3 AgNO3 to the NaCl solution
slowly, with stirring to produce insoluble AgCl as
a precipitate. You need a slight excess of AgNO3
(why?). To ensure that excess AgNO3 is present,
allow the suspension in the beaker to settle and
add a few drops of AgNO3. If no more precipitation
is observed, the AgNO3 is present in excess.
(e) Allow the reaction mixture to stand for a while,
then filter the AgCl through a pre-conditioned,
pre-weighed sintered glass crucible placed in
a filter flask equipped with a filter cone and
connected to a vacuum (as illustrated in Figure
29.2). It is important to ensure that all the AgCl
is transferred from the beaker to the crucible. The
AgCl in the crucible can be washed by filtering
through additional water.
(f) Place the crucible with the AgCl to dry in an oven
heated to 130 °C.
(g) Remove the crucible with the AgCl from the oven,
allow it to cool in a desiccator, weigh and obtain
the weight of the AgCl by difference.
3
Sample A contains 7.70% H2O; Sample B contains
7.81% H2O; the average percentage H2O in the two
samples is 7.76.
4
(a) CuSO4.5H2O is 36.08% H2O
(b) MgSO4.7H2O is 51.17% H2O
(c) Na2SO4.10H2O is 55.92% H2O
5
15.05%
6
x = 7; the formula is FeSO4.7H2O
281
Chapter 30
Titrimetric analysis
Learning objectives
■ Explain the meanings of the following terms as used in titrimetric analysis: analyte, end-point,
■
■
■
■
equivalence point, indicator, primary standard, standard, titrant, titration error.
Provide a general description of the process which occurs in a titration and explain how titrations
are used in chemical analysis.
Calculate the concentrations of acids and bases using data obtained from direct acid/base titrations
and back titrations involving acid/base reactions.
Describe the principles of end-point detection by use of indicators, potentiometry, conductimetry
and thermometry.
Calculate concentrations of oxidizing and reducing agents using data obtained from direct redox
titrations and back titrations involving redox reactions.
Introduction
Titration is a very commonly used method in analytical
chemistry. In a titration the concentration (unknown) of
a chemical species (an ion or a compound) in a solution is
determined by reacting the ion or compound completely
with a reagent that is in a solution of known concentration.
We use the stoichiometry of the reaction between the
analyte and the standard and the concentration of the
standard in the titrant to calculate the number of moles of
the analyte in solution.
■ The reagent of known concentration is the standard.
In Chapters 11 and 12, acid/base titrations and redox
titrations were used to determine mole ratios and molar and
mass concentration. The most important terms associated
with acid/base equilibria are:
■ The solution containing the standard in known
■ Ka, the acidity constant; pKa = −log10 Ka
■ The chemical species present in the solution of
unknown concentration is termed the analyte.
concentration is the titrant.
■ The precise point at which all the analyte has reacted
with the standard is the equivalence point.
The attainment of the equivalence point is detected by (a) a
visual or colour change or (b) a physical method. The point
at which the visual, colour or physical change is observed
is the end-point of the titration. Ideally, the end-point
should be observed at the equivalence point; however,
these do not always coincide and the difference is known
as the titration error.
■ The visual or colour change may result from the
reaction which takes place between the analyte and
the standard or it may be produced by an indicator
which is an auxiliary reagent added to one of the
solutions.
■ The physical change may be a change in temperature,
pH or conductivity.
■ Kb, the basicity constant; pKb = −log10 Kb
■ Kw, the ion product of H2O ([H+][OH−]);
pKw = −log10 Kw
■ pH = −log10 [H+]
Acid/base titrations
Think about a simple acid/base titration: an example would
be determining the concentration of an aqueous solution
of NaOH (the analyte) by neutralizing it with a standard
solution of HCl (the titrant). The experimental set-up is
illustrated in Figure 30.1, and consists of a burette which
contains the NaOH solution and a flask containing the HCl
solution. The reaction which takes place is
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
At the equivalence point the HCl in the flask has been
completely neutralized by the NaOH.
Unit 2 Module 2 Analytical methods and separation techniques
D
A
Conductivity
burette containing
NaOH solution
equivalence point
B
C
Volume of NaOH added
Figure 30.2 Variation of conductivity with volume of NaOH added
in the reaction NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
■ We can monitor the temperature of the reaction
flask containing
HCl solution
Figure 30.1 Apparatus for titration.
The pH of the NaOH solution is in the range 11.5–13.5,
and the pH of the HCl solution is in the range 0.0–2.0; the
precise pH of each solution depends on the concentration.
At the equivalence point the pH of the solution is 7.0. We
can make a fairly accurate determination of when the
equivalence point is reached.
mixture as NaOH is added to HCl. The neutralization
reaction is exothermic, so the temperature of the
reaction mixture will increase to a maximum at
the equivalence point and will decrease on further
addition of NaOH. These temperature changes are
small but measurable, and such titrations are known
as thermometric titrations. A stylized graph of
temperature versus volume of NaOH added is shown
in Figure 30.3.
■ We can add an indicator to the solution in the flask. A
Temperature
282
suitable indicator must undergo a colour change at or
close to the pH of the solution at the equivalence point.
equivalence point
■ We can measure the actual pH of the solution with a
pH meter. This is a type of potentiometric titration.
■ We can measure the conductance of the solution.
The conductance will decrease to a minimum at
the equivalence point as H+ and OH−, from HCl and
NaOH respectively, combine to produce H2O which is
dissociated to a very small extent. Titrations monitored
by conductance are known as conductimetric
titrations. A graph of conductivity versus volume of
NaOH added is shown in Figure 30.2.
Volume of NaOH added
Figure 30.3 Variation of temperature with volume of NaOH added
in the reaction NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
The three important acid/base titrations are:
■ strong acid with strong base (usually a mineral acid
with a metal hydroxide);
■ weak acid with strong base (usually a carboxylic acid
with a metal hydroxide);
■ strong acid with a weak base (a mineral acid with, for
example, ammonium hydroxide).
ITQ 1
(a) 25 cm3 of 1 M HCl neutralized 15 cm3 of NaOH. Find:
(i) the molarity of the NaOH solution
(ii) the concentration of NaOH in g cm−3
(iii) the percentage concentration of NaOH (weight/volume =
g/100 cm3)
(b) 50 cm3 of 1 M KOH neutralized 25 cm3 of H2SO4. What is the
molarity of the H2SO4?
(c) If a solution contains 1.5 g of NaOH in 250 cm3, what volume
of 0.1 M HCl would be required to react with 20 cm3 of this
solution?
These titrations and the associated titration curves were
discussed in detail and illustrated in Chapter 11. The pH at
the equivalence points in examples of these titrations are
given in ITQ 4.
ITQ 2
(a) What ions are in solution at points A, B, C and D in the graph
in Figure 30.2?
(b) Why does the conductivity increase after point C?
Chapter 30 Titrimetric analysis
When we examine the titration curve for the neutralization
of HCl with NaOH (Figure 30.4) we see that:
■ just before the equivalence point the pH of the
solution is between 3 and 5;
■ at the equivalence point the pH has increased sharply
to 7;
■ the addition of even a very small volume of NaOH
after the equivalence point has been reached results in
a very large increase in pH.
pH
14
equivalence point
7
0
Volume of base
Figure 30.4 Titration curve for the neutralization of HCl with NaOH.
In classical titrimetric analysis, colour indicators are used to
signal such sharp pH changes. Colour indicators are highly
conjugated, structurally complex organic compounds. We
do not need to know their structures to understand how
they work. Examples of colour indicators are litmus, methyl
orange, methyl red and phenolphthalein. Indicators are
weak acids, and in aqueous solution they ionize to varying
extents and are in equilibrium with their conjugate bases.
The acid form of the indicator has a different colour from
the conjugate base. Using H-Ind as a general term for an
indicator and Ind− for the conjugate base of the indicator,
the equilibrium in aqueous solution can be shown as:
is visible to the human eye; when it is 1:10, only colour X
is visible. When [Ind−] and [H-Ind] are equal, the colour of
the solution is a composite of colour X and colour Y.
The equilibrium constant for the dissociation of an indicator
is designated Kind:
Kind =
[H+] [Ind−]
[H-Ind]
The pH at which [H-Ind] and [Ind−] are equal depends, in a
somewhat complex way, on this equilibrium constant which,
in turn, is determined by the actual chemical structure and
properties of the indicator molecule. This value is termed
pK’ind the apparent indicator constant. The pH at which
the protonated and deprotonated forms of litmus are equal
(pK’ind) is 6.5, for methyl orange pK’ind is 3.7, for methyl
red pK’ind is 5.0 and for phenolphthalein pK’ind is 9.6. The
visible change from colour Y (where [Ind−]:[H-Ind] = 10)
to colour X (where [Ind−]:[H-Ind] = 0.1) commonly occurs
over approximately two pH units. This range is designated
the colour change interval (pH range) of the indicator and it
spans pK’ind. In Table 30.1, pH ranges, colours and pK’ind for
selected indicators are shown. The indicator chosen for an
acid/base titration should undergo a colour change at a pH
as close as possible to the equivalence point.
Table 30.1 Selected indicators: pH ranges, colour changes
and pK ’ind.
Indicator
pH range
Colour in acid
solution
Colour in basic
solution
pK ϶ind
bromophenol blue
2.8–4.6
yellow
blue
4.1
methyl orange
2.9–4.6
red
orange
3.7
bromocresol green
3.6–5.2
yellow
blue
4.7
methyl red
4.2–6.3
red
yellow
5.0
litmus
5.0–8.0
red
blue
6.5
phenolphthalein
8.3–10.0
colourless
red
9.6
thymolphthalein
9.3–10.5
colourless
blue
9.3
H-Ind(aq) ҡ H+(aq) + Ind−(aq)
colour X
colour Y
In acidic solution the equilibrium is shifted to the left and
colour X dominates.
In basic solution H+ is removed, the equilibrium is shifted
to the right and colour Y dominates.
The observed colour of the solution depends on the ratio
of [Ind−] to [H-Ind]. When this ratio is 10:1 only colour Y
ITQ 3 Explain why there is a sharp increase in pH if a very small
volume of NaOH is added after the HCl has been neutralized in
the reaction NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
ITQ 4
(a) For the two titrations listed below choose one or more suitable
indicators from Table 30.1. Give the colour of the indicator at
the start of the titration and the colour at the end-point.
Solution in Solution in
burette
flask
pH at
Initial Colour at
equivalence Indicator
colour end-point
point
(i) NaOH(aq)
CH3COOH(aq) 8.72
(ii) NH4OH(aq)
HCl(aq)
5.28
(b) Write equations for reactions (i) and (ii) in part (a).
283
284
Unit 2 Module 2 Analytical methods and separation techniques
Apparatus, experimental procedure and technique
for titrations
Apparatus
■ burette
■ pipette
■ pipette filler
■ volumetric flasks
■ conical flasks
■ indicator bottle
■ funnel
■ wash bottle
A practical application of acid/base titrimetric analysis is
the determination of the concentration of ethanoic acid in
commercial samples of vinegar. Vinegar contains a dilute
solution of ethanoic acid (CH3COOH). Ethanoic acid is a
carboxylic acid; it reacts with sodium hydroxide to form
the carboxylate salt (sodium ethanoate) and water. The
equation for this reaction is:
O
CH3
O
+ NaOH
C
CH3
+
C
OH
–
+
H2O
O Na
sodium ethanoate
Procedure and technique
1 The solutions to be titrated are usually prepared in volumetric
flasks. These come in different sizes, and each size measures one
specific volume accurately.
2 The burette and the pipette should each be rinsed with the solution
which it will be measuring. If they are rinsed with water the
concentrations of the solutions will change, leading to inaccurate
results.
3 Solutions are transferred to the conical flasks using the pipette
and a pipette filler. Pipettes, like volumetric flasks, measure one
specific volume accurately.
4 Is an indicator being used in the titration? If so, about two drops of
indicator solution are added from an indicator bottle to the solution
in the conical flask.
5 Burettes also deliver accurate volumes. Burettes are graduated in
divisions of 0.1 cm3 and volumes can be read to halfway between
the divisions, i.e. to 0.05 cm3. Solutions are added to the burette
using a funnel.
6 The solution in the burette is added to the solution in the conical
flask until the end-point is observed.
7 A rough titration is carried out first: the solution from the burette is
quickly added to the solution in the conical flask; even if the endpoint is passed, this gives the approximate volume of the solution
in the burette needed to reach the end-point.
8 A number of accurate titrations are then carried out. About 90% of
the solution needed to reach the end-point can be added quickly
from the burette; close to the end-point the solution is added dropwise. The average of two or three results which are within
0.10 cm3 of each other (concordant results) is calculated and
taken as the titre of the solution in the burette.
9 During addition of the solution from the burette the conical flask
should be swirled (carefully) to ensure mixing and complete reaction
of the standard and the analyte. Solutions adhering to the tip of the
burette and the sides of the conical flask can be rinsed into the main
solution in the conical flask using distilled water from a wash bottle.
Ethanoic acid is a weak acid and sodium hydroxide is a
strong base. For such a weak acid/strong base titration the
pH at the equivalence point is 8.72 and the indicator used
for detection of the end-point is phenolphthalein.
ITQ 5 Why does the addition of distilled water from the wash
bottle (point 9 above) not affect the results?
ITQ 6 A 10 cm3 sample of vinegar was pipetted into a conical
flask and two drops of phenolphthalein indicator were added to
the solution. A 0.50 M solution of NaOH was added from a burette
until the contents of the flask were very pale pink. This procedure
was repeated until two concordant results were obtained. The
average of two concordant results was 13.75 cm3 of NaOH. What
is the concentration of ethanoic acid in grams per 100 cm3 of
vinegar? (FW ethanoic acid, CH3COOH = 60.05 g mol−1.)
Back titrations in acid/base titrimetric
analysis
Back titration (also known as indirect titration) is a method
whereby an analyte is added to a solution that contains
an excess of a standard reagent. The amount of unreacted
standard is then determined by titration against a second
standard. The quantity of the first standard which has
reacted is obtained by subtraction, and the quantity of the
analyte can be calculated. Back titration is used when:
■ the analyte is insoluble in water;
■ the analyte is volatile;
■ the reaction between the analyte and the standard is slow;
■ a direct titration would be a weak acid/weak base
titration; the end-points of such titrations are not sharp.
The structure of the painkiller aspirin is
HO
carboxylic acid
O
C
O
O
C
ester
CH3
The chemical name of this compound is acetylsalicylic acid
(ASA). There are two functional groups in this molecule: it
is both a carboxylic acid and an ester.
Chapter 30 Titrimetric analysis
Worked example 30.1
Q
A
Example of a back titration
It is claimed that antacid tablets of a certain brand each
contain 0.5 g of CaCO3. The following back titration was used to
investigate this claim.
■ The insoluble CaCO3 was reacted with a known amount of
HCl (which was an excess).
■ The amount of unreacted HCl was determined by titration
with NaOH solution.
■ The amount of acid which reacted with the CaCO3 was
obtained by subtraction.
■ The quantity of CaCO3 in each tablet was calculated.
This experiment is based on the following reactions:
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
Procedure
1 An antacid tablet was crushed in a mortar, using a pestle,
and the solid was transferred to a 250 cm3 volumetric flask.
2 25 cm3 of 1 M HCl was added to the flask, and water was
added to make the solution up to 250 cm3. This solution is
the stock solution.
3 25 cm3 portions (aliquots) of the stock solution were titrated
against 0.05 M NaOH using methyl red as the indicator. The
average of three concordant titrations was 30.85 cm3 NaOH.
Calculation
Find the number of mols of HCl remaining in the volumetric
flask after the reaction with the antacid tablet. The equation
between NaOH and HCl gives a 1:1 ratio.
30.85
moles of HCl in 25 cm3 = 0.05 mol dm−3 ×
dm3
1000
= 0.00015425 mol
This quantity of HCl remains from a 25 cm3 aliquot taken from
the 250 cm3 of the original stock solution.
250
moles of HCl in 250 cm3 = 0.00015425 ×
25
= 0.00154 mol
Number of moles HCl present in 25 cm3 of a 1 M solution
(before reaction with the antacid tablet):
25
=
= 0.025
1000
Number of moles HCl that reacted with the antacid
= 0.025 − 0.0154 = 0.0096
From the equation:
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)
2 moles HCl react with 1 mole CaCO3
∴ number of moles CaCO3 in antacid tablets
= 0.5 × number of moles HCl that reacted with the antacid
= 0.0096 × 0.5
= 0.0048
mass of CaCO3 in antacid tablet
= number of moles CaCO3 × FW CaCO3
= 0.0048 × 100.0 g
= 0.480 g
The carboxylic acid group in acetylsalicylic acid undergoes
a fast reaction with a strong base such as sodium hydroxide
to produce the sodium salt of the acid and water.
HO
+
O
–
O
Na O
C
+ H2O
C
O
O
+ NaOH
C
O
O
fast
C
CH3
CH3
sodium salt of acetylsalicylic acid
The ester group of acetylsalicylic acid also reacts with base;
it slowly undergoes hydrolysis to form the sodium salt of
ethanoic acid (sodium ethanoate) and the phenol.
+
–
+
O
Na O
–
O
Na O
C
phenol
C
O
O
C
+ NaOH
OH
slow
CH3
O
+
CH3
C
–
+
O Na
sodium ethanoate
So one molecule of acetylsalicylic acid reacts with two
molecules of base: the first molecule of base is consumed
very quickly in a neutralization reaction with the carboxylic
acid group; the second molecule of base reacts more slowly,
to hydrolyse the ester group.
Due to this difference in the rates of the two reactions
between acetylsalicylic acid (ASA) and base the quantity of
ASA in aspirin tablets is best determined by back titration. A
weighed sample of an aspirin tablet is dissolved in ethanol
(ASA is an organic compound and is not very soluble in
water) and this solution (in a conical flask) is titrated against
sodium hydroxide solution of known concentration (added
from a burette) using phenolphthalein as the indicator.
After the end-point of the titration has been reached (let
us say this required x cm3 of sodium hydroxide solution)
an additional volume of sodium hydroxide (x + 10 cm3) is
added to the solution in the flask. The contents of the flask
are warmed for about 15 minutes, then allowed to cool to
room temperature. The contents of the flask (which are
now basic) are then titrated against hydrochloric acid of
known concentration. The number of moles of hydrochloric
acid used in the titration is equal to the number of moles
of base which are not consumed in the hydrolysis reaction.
This is subtracted from the number of moles of base added
for the hydrolysis (number of moles of base in x + 10 cm3);
the difference is equal to the number of moles of base
required to hydrolyse the ester which, in turn, is equal to
the number of moles of ester present and the number of
moles of ASA in the sample.
285
Unit 2 Module 2 Analytical methods and separation techniques
Titrations monitored by measurement of
pH (potentiometric titrations)
It is possible, using a pH meter, to determine the end-point
of an acid/base titration by measuring the pH of the solution
being titrated. A pH meter consists of a voltmeter which
is calibrated in pH units and which is connected to two
electrodes – a reference electrode and an indicator electrode.
The two electrodes are often built into one cylindrical probe.
The reference electrode is usually a calomel electrode
which consists of a glass tube containing a paste of Hg,
Hg2Cl2 and KCl surrounded by an outer tube containing
saturated aqueous KCl. There is a small hole in the bottom
of the inner tube and the bottom of the outer tube is made
of a porous material (sintered glass or ceramic material) so
that contact is maintained between the electrode and the
solution in which it is placed. The indicator electrode, also
known as the glass electrode, consists of a silver wire with
an AgCl tip suspended in a KCl solution kept in a bulb of
special glass which is coated with metal salts (Figure 30.5).
pH meter converts voltage (potential
difference) into pH reading
glass electrode,
coated with
metal salts and
containing KCl
reference
electrode
+
+
H
+
H
+
+
H
+
H
+
H
+
+
+
H
+
H
H
+
H
+
H
H
In conducting acid/base titrations using a pH meter, the pH
of a solution is measured as a function of the volume of the
titrant added. The pH changes slowly until the end-point,
at which it changes sharply. A graph is plotted (x-axis is
volume of titrant; y-axis is pH) and from the graph the
precise volume of titrant required to cause the sharp change
in pH is determined. Small constant increments (e.g. 2 cm3
or 1 cm3 or 0.5 cm3) of titrant are added throughout the
titration. It is not necessary to add the titrant slowly when
the end-point is close, as is done with a visual indicator.
The graph which is obtained by plotting pH versus volume
of titrant is a titration curve (see ITQ 8).
solution being
tested
Ag wire with
AgCl tip
H
When the probe with the electrodes is placed in a solution
containing hydrogen ions (H+), the hydrogen ions move
towards the glass electrode and displace some of the metal
ions in the glass coating. This creates a potential difference
across the glass which is picked up by the silver wire and
detected by the voltmeter. Greater [H+] gives a higher
voltage, which is displayed as a lower pH; lower [H+] gives
a lower voltage which is displayed as a higher pH. A pH
meter must be calibrated before use by placing the probe in
buffer solutions of known pH and adjusting the display as
necessary. Measurements of pH made using pH meters are
sensitive to temperature, so must be carried out at constant
temperature or corrected for temperature variations. Some
pH meters have built in thermometers and automatically
adjust the pH measurements for temperature variations.
H
+
H
+
H
+
H
+
H
+
H
hydgrogen ions interacting with glass electrode
ITQ 8 A 21.649 g sample of a household cleaner which contains
ammonium hydroxide (NH4OH) was dissolved in distilled H2O.
The resulting solution was titrated against 0.1 M HCl, and the
titration was monitored using a pH meter. A plot of pH versus
volume of HCl is shown below (FW NH4OH = 35.05 g mol−1).
12
Figure 30.5 Diagram of a pH meter with the probe placed in a
solution containing hydrogen ions.
10
8
ITQ 7 A 0.3316 g sample of aspirin dissolved in ethanol (15 cm3)
and treated with two drops phenolphthalein indicator solution
required 15.25 cm3 of 0.0950 M NaOH for neutralization. An
additional 25.25 cm3 of 0.0950 M NaOH was added and the
sample was heated to hydrolyse the ester group. The reaction
mixture was cooled to room temperature and the excess base
was back titrated with 10.15 cm3 of 0.1030 M HCl.
(a) How many grams of acetylsalicylic acid are in the sample?
(b) What is the percentage of acetylsalicylic acid in the sample?
(i.e. what is the purity of the sample?)
FW acetylsalicylic acid = 180.2 g mol−1
pH
286
6
pH 5.3
4
2
10.5 cm3
0
0
2
4
6
8
10
12
14
16
Volume of 0.1 M HCl / cm3
Calculate the percentage of ammonium hydroxide in the sample
(g NH4OH per 100 g of cleaner).
Chapter 30 Titrimetric analysis
Thermometric and conductimetric
titrations
A very simple thermometric titration can be carried out
using the experimental set-up shown in Figure 30.6.
The polystyrene cup contains a solution of NaOH of
known concentration. As small constant volumes of
HCl (concentration unknown) are added to the cup the
temperature of the solution in the cup is read from the
thermometer.
A graph is plotted (x-axis is volume of HCl; y-axis is
temperature) and enough readings are taken to determine
the maximum temperature reached during the experiment.
The end-point, obtained from the graph, is the volume of
HCl added at the maximum temperature.
A more sophisticated way of obtaining thermometric titration
data utilizes a titration calorimeter comprised of a water
bath surrounding a reaction vessel. The temperature of the
water bath is electronically controlled and the temperature
of the solution is monitored with a temperature sensor,
which may be a thermally sensitive resistor (thermistor) or
a thermocouple.
Automatic titrators used in modern analytical chemistry
laboratories can utilize conductivity sensors which measure
the electrical conductivity of the solution in the flask
throughout the titration. The conductivity of the solution is a
function of the concentration and the relative conductivities
of the ions in the solution. In general, the equivalence point
corresponds to the minimum in the graph of volume of
titrant (x-axis) versus conductivity (y-axis).
Primary standards
The concentrations of all solutions used to analyse
substances by titrimetric analysis must be accurately
known. The concentrations of the hydrochloric acid and
the sodium hydroxide solutions used in the analysis of the
vinegar, antacid tablets and aspirin described above would
have been established by titrating these solutions against
solutions of primary standards. A compound used as a
primary standard must meet the following criteria.
■ The compound must be stable, so that it can be easily
weighed under normal atmospheric conditions and not
change during storage.
■ The compound must be easy to obtain in highly
purified form.
■ It should dissolve readily in the titration medium in
which it is to be used.
■ It should have a high formula weight so that weighing
errors are minimized.
■ The reaction between the primary standard and the
compound being standardized should be fast and
quantitative (go to completion).
■ The compound should not be expensive.
Chemical suppliers sell compounds in highly purified form
for use specifically as primary standards. These primary
standards have been analysed by the suppliers and the
composition is printed on the label of the container. In
general, primary standard reagents are of about 99.99%
purity.
The commonly used primary standard reagents for the
standardization of acids are:
■ sodium carbonate – Na2CO3
■ sodium hydrogencarbonate – NaHCO3
burette
Commonly used primary standard reagents for the
standardization of permanganate solutions are:
■ oxalic acid – (COOH)2
hydrochloric acid
■ sodium oxalate – (COONa)2
clamp
cork
thermometer
polystyrene cup
containing NaOH
solution
Figure 30.6 Apparatus for a simple thermometric titration.
Commonly used primary standard reagent for the
standardization of thiosulfate solutions is:
■ potassium iodate – KIO3
Redox titrations
In redox reactions one reactant is oxidized (the element at
the reacting centre loses electrons and its oxidation number
increases) and the other reactant is reduced (the element at
the reacting centre gains electrons and its oxidation number
287
288
Unit 2 Module 2 Analytical methods and separation techniques
Worked example 30.2
Q
Permanganate and oxalate
Preparation of 0.02 M KMnO4 solution
1 Approximately 3.2 g of KMnO4 is weighed on a watch glass and transferred to a 1500 cm3 beaker.
2 Water (1000 cm3) is added, the beaker is covered and the solution is boiled for 15 minutes. This is to ensure that all reducing impurities
are completely oxidized.
3 The solution is allowed to cool to room temperature and is then filtered through a funnel with a plug of glass wool in the stem or
through a sintered glass crucible. Filter paper should not be used as the cellulose of which it is made can be oxidized by MnO4−.
4 The filtered solution is stored in the dark in a clean (preferably brown) glass bottle with a glass stopper. Permanganate solutions
decompose in bright light.
Standardization of KMnO4 with sodium oxalate
Sodium oxalate is the sodium salt of the dibasic carboxylic acid oxalic acid.
H
O
O
+
Na
–
O
C
H
O
oxalic acid
O
C
C
O
+
Na
–
O
C
O
sodium oxalate
Sodium oxalate is a primary standard, as it is a stable solid and readily available in pure form. In the acidic solutions in which
permanganate titrations are carried out sodium oxalate is converted to oxalic acid:
C2O42− + 2H+ ҡ H2C2O4
Oxalic acid reduces permanganate in acid to produce carbon dioxide and water:
2MnO4− + 5H2C2O4 + 6H+ → 2Mn2+ + 10CO2 + 8H2O
1 Weigh about 1.7 g of sodium oxalate accurately. Transfer all the solid to a 250 cm3 volumetric flask and make the solution up to mark.
Shake the flask to ensure that the contents are thoroughly mixed. This solution should not be stored for more than 4 or 5 days, as
oxalate solutions attack glass.
2 Using a pipette, transfer 25 cm3 of this solution to a 250 cm3 conical flask and add 75 cm3 of 2 M H2SO4 to the flask.
3 Titrate this solution at room temperature with the KMnO4 (which is in a burette). Add the KMnO4 at a fast rate until the solution in the
conical flask is faintly pink. Leave the flask to stand at room temperature until the solution becomes colourless. Warm the solution to
50–60 °C and add KMnO4 from the burette until the warm solution in the flask has a permanent pale pink colour.
A
Calculation
Sodium oxalate (1.6853 g) was weighed accurately and a 250 cm3 solution was prepared as described above. Aliquots (25 cm3) of this
solution were acidified with H2SO4 and titrated against a solution of KMnO4 (nominally 0.02 M). The average of three concordant titrations
was 26.45 cm3 KMnO4.
2MnO4− + 5H2C2O4 + 6H+ → 2Mn2+ + 10CO2 + 8H2O
FW Na2C2O4 = 134.00 g mol−1
1.6853 g × 25 cm3
moles Na2C2O4 in 25 cm3 solution =
= 1.258 × 10−3 mol
134.00 g mol−1 × 250 cm3
From the equation:
5 mol of Na2C2O4 reacts with 2 mol of MnO4−
2
mol of MnO4−
∴ 1 mol of Na2C2O4 reacts with
5
2
and 1.258 × 10−3 mol of Na2C2O4 reacts with × 1.258 × 10−3 mol of MnO4−
5
2
26.45 cm3 KMnO4 solution contains
× 1.258 × 10−3 mol of MnO4−
5
2
1000
1000 cm3 KMnO4 solution contains × 1.258 × 10−3 ×
mol MnO4−
5
26.45
Molarity of the KMnO4 solution is 0.0190 M
Chapter 30 Titrimetric analysis
decreases). Analytes which are readily oxidized or reduced
can be analysed quantitatively by titrimetric analysis in
which the standard is an oxidizing agent or a reducing
agent. Such titrations are called redox titrations.
Important oxidizing agents used in redox titrations are
potassium permanganate (KMnO4), potassium dichromate
(K2Cr2O7), iodine (I2), potassium iodate (KIO3) and
potassium bromate (KBrO3).
Reducing agents are iron(II) compounds, tin(II) compounds
and sodium thiosulfate (Na2S2O3).
Potassium permanganate (manganate (VII))
The permanganate ion (MnO4−) in acid solution is a strong
oxidizing agent; manganese(VII) accepts five electrons and
is itself reduced to manganese(II). The equation for this
process is:
MnO4− + 8H+ + 5e− ҡ Mn2+ + 4H2O
The acid is usually sulfuric acid, as the chloride ions from
hydrochloric acid are oxidizible and can consume the
permanganate. The permanganate ion is deep purple, and
a minute quantity of a KMnO4 solution (0.01 cm3 of a
0.02 M solution) causes 100 cm3 of water to be pale pink.
Titration between potassium permanganate and solutions
of analytes which are colourless or faintly coloured
therefore do not require indicators; these titrations are said
to be self-indicating. The end-point of a titration with
potassium permanganate (usually added from the burette)
is taken as the appearance of a permanent pale pink colour
of the solution in the flask.
It is difficult to obtain potassium permanganate of
high purity. In both the solid form and in solution it is
contaminated with small quantities of manganese dioxide
(MnO2) which is a very insoluble brown solid. For this
reason potassium permanganate is not a primary standard,
and solutions must be standardized, usually with sodium
oxalate (Na2C2O4). Set out below are procedures for
preparation of a 0.02 M solution of potassium permanganate
and standardization of this solution with sodium oxalate.
Redox titrations with iodine: iodimetric titrations
Iodine (I2) is an oxidizing agent and readily takes two
electrons from a reducing agent to form two iodide (I−)
ions. The half-cell reaction is:
I2 + 2e− ҡ 2I−
Iodine is not very soluble in water (solubility = 0.0335 g/100
cm3) but dissolves readily in fairly concentrated solutions
of potassium iodide. The solubility of iodine in the presence
of I− is due to a reversible reaction in which the tri-iodide
ion (I3−) is formed:
I2(aq) + I− ҡ I3−
The tri-iodide ion is also an oxidizing agent, and is reduced
by two electrons to form three iodide ions:
I3− + 2e− ҡ 3I−
Equations for redox reactions which involve iodine can
be written using either the tri-iodide ion (I3−) or elemental
iodine (I2). The two redox equations shown below involving
iodine/tri-iodide and thiosulfate (S2O32−) describe the same
process, although the second equation is more correct than
the first.
I2 + 2S2O32− → 2I− + S4O62−
I3− + 2S2O32− → 3I− + S4O62−
A notable property of elemental iodine (I2) is that it
sublimes, i.e. it vaporizes directly from the solid. Because
of this property, aqueous solutions of iodine have a
high vapour pressure, and as the iodine evaporates the
concentration in solution decreases. This problem is
removed when the iodine is dissolved in potassium iodide
and the tri-iodide ion forms.
Starch is the commonly used indicator for titrations which
involve iodine. A solution of I2 in KI is red. This solution
turns deep blue when it is added to a dilute starch solution.
The starch solution used as an indicator is prepared by
making a paste of 0.1 g of soluble starch with about 15
cm3 of water. The paste is then poured, with stirring, into
100 cm3 of boiling water and the resulting mixture is
boiled for 1 minute. The starch is allowed to cool to room
temperature and 2.5 g of potassium iodide is added.
Solutions of iodine are commonly standardized with
sodium thiosulfate. The thiosulfate ion is a reducing agent,
releasing two electrons and forming the tetrathionate ion
in the half-cell reaction:
2S2O32− ҡ 2e− + S4O62−
Thiosulfate and iodine react quantitatively:
I2 + 2S2O32− → 2I− + S4O62−
The thiosulfate solution (colourless) is added from the
burette to the iodine/tri-iodide solution (red) in the conical
flask. When the solution in the flask becomes pale yellow,
starch solution (approximately 2 cm3) is added and the
solution turns blue. Addition of the thiosulfate is continued
slowly until the solution in the flask is just colourless.
Sodium thiosulfate is not a primary standard. Crystalline
sodium thiosulfate is nominally a pentahydrate
289
290
Unit 2 Module 2 Analytical methods and separation techniques
Iodine/thiosulfate titration
A: Preparation of 0.05 M iodine
Care: iodine sublimes, and the vapour solidifies on cold surfaces, so iodine should never be weighed on an analytical balance as I2 deposits will
form on and in the balance.
1 Pure potassium iodide (20 g) is dissolved in distilled water (30–40 cm3) and the solution is poured into a 1000 cm3 glass-stoppered
volumetric flask.
2 About 12.7 g of resublimed iodine is weighed in an appropriate container using a top-loading balance and transferred via a powder funnel to
the flask containing the concentrated KI solution.
3 The flask is stoppered and shaken until the I2 has dissolved. The reaction I2 + I− ҡ I3− is exothermic so the solution becomes warm. The
solution is allowed to cool to room temperature and made up to mark with distilled water. The solution should be stored in a cool, dark place.
B: Preparation of 0.1 M sodium thiosulfate
Ordinary distilled water contains carbon dioxide, which reacts with the water to form carbonic acid:
H2O + CO2 ҡ H2CO3
Carbonic acid is a weak acid:
H2CO3 ҡ HCO3− + H+
This causes distilled water to be marginally acidic.
Low concentrations of acid lead to slow decomposition of thiosulfate to bisulfite and sulfur:
S2O32− + H+ → HSO3− + S
When water is boiled the carbon dioxide is removed. Solutions of sodium thiosulfate must be prepared using previously boiled distilled water.
About 25 g of solid sodium thiosulfate (Na2S2O3.5H2O) is weighed and dissolved in previously boiled and cooled distilled water. The solution is
transferred to a 1000 cm3 volumetric flask and made up to mark with boiled distilled water.
C: Standardization of sodium thiosulfate with potassium iodate
1 Pure dry potassium iodate (close to 4.28 g) is weighed accurately, transferred to a 1000 cm3 volumetric flask and the solution made up to
mark with previously boiled and cooled distilled water.
2 A 25 cm3 aliquot of this solution is pipetted into a 250 cm3 conical flask and iodate-free potassium iodide (1 g) and 1 M sulfuric acid (3 cm3)
are added to this solution.
3 The liberated iodine in the flask is titrated with the sodium thiosulfate (added from the burette). The contents of the flask are shaken
throughout the titration. When the colour of the liquid in the flask becomes pale yellow the solution is diluted to approximately 200 cm3 with
distilled water, and starch solution (2 cm3) is added (the solution turns blue). Addition of thiosulfate from the burette is continued slowly until
the solution in the flask is just colourless.
4 Steps 2 and 3 are repeated until two concordant results are obtained.
To test for iodate in KI: add dilute sulfuric acid to a solution of the KI; the solution should not turn yellow, nor should a blue colour appear
when starch is added.
25
5 Molarity of Na2S2O3 solution = 6 × mass KIO3 ×
× titre Na2S2O3
214.00
D: Standardization of the iodine solution with the sodium thiosulfate solution
1 A 25 cm3 aliquot of the iodine solution prepared in part A (above) is pipetted into a 250 cm3 conical flask and diluted with previously boiled
and cooled distilled water (75 cm3).
2 Sodium thiosulfate solution (recently standardized as described in part C) is added from a burette until the solution in the flask is pale yellow.
Starch solution (2 cm3) is then added (the solution in the flask turns blue) and addition of thiosulfate from the burette is continued slowly until
the solution in the flask is just colourless.
3 Steps 1 and 2 are repeated until two concordant results are obtained.
titre S2O32– solution
4 Molarity of I2 solution = 2 × molarity S2O32– solution ×
25
Chapter 30 Titrimetric analysis
(Na2S2O3.5H2O), but it is efflorescent, so the composition
of a given sample cannot be accurately known. Solutions
of sodium thiosulfate must, therefore, be standardized
using a primary standard. The primary standard for the
standardization of thiosulfate is potassium iodate (KIO3).
In acid solution, iodate ions (IO3−) react with iodide ions
(I−) to liberate iodine (I2):
IO3− + 5I− + 6H+ → 3I2 + 3H2O
Practical applications of redox titrations
Analysis of ascorbic acid in vitamin C tablets
Ascorbic acid (vitamin C) is an essential nutrient and an
anti-oxidant (reducing agent) which protects the body
against oxidative stress. It occurs in many fruits and is widely
available, mainly in tablet form, as a nutritional supplement.
In the presence of an oxidizing agent, ascorbic acid is
converted to dehydroascorbic acid. The half-cell reaction
for the oxidation of ascorbic acid is:
C6H8O6
ascorbic acid
ҡ
C6H6O6 + 2H+ + 2e−
dehydroascorbic acid
O
HO
O
OH
OH
OH
O
OH
OH
O
O
ascorbic acid
O
dehydroascorbic acid
Iodine (or tri-iodide) rapidly oxidizes ascorbic acid:
C6H8O6 + I2 → C6H6O6 + 2I− + 2H+
Ascorbic acid is very soluble in water, so it is possible to
quantify the ascorbic acid in vitamin tablets and fruits by
titration with iodine/iodide solutions. However, in practice
an excess of the iodine required to oxidize the ascorbic acid
is produced by the reaction between a known amount of
potassium iodate and potassium iodide in acid.
KIO3 + 5KI + 3H2SO4 → 3I2 + 3H2O + 3K2SO4
The ascorbic acid is present in the solution in which the
iodine is formed and immediately reacts with it.
The excess iodine is then titrated with sodium thiosulfate.
This is another example of a back titration. Potassium
iodate is a primary standard, so the same potassium iodate
solution can be used to standardize the sodium thiosulfate.
Worked example 30.3
Q
Analysis of vitamin C tablets
Note that all the distilled water used will need to be boiled
before use.
A: Preparation of the KIO3 solution and standardization
of the Na2S2O3
1 A solution of potassium iodate was prepared by
dissolving 4.2813 g of the solid in a small volume
(around 40 cm3) of distilled water and transferring the
solution quantitatively to a 1000 cm3 volumetric flask
and making up to mark with distilled water.
2 A 25 cm3 aliquot of this solution was pipetted into a
conical flask and 5 cm3 of 1 M potassium iodide and
3 cm3 of 1 M sulfuric acid were added.
3 Sodium thiosulfate was added from a burette to the
conical flask until the contents of the flask were pale
yellow. Starch solution (2 cm3) was then added to the
flask and the addition of the thiosulfate solution was
continued until the contents of the flask were just
colourless. The average of two concordant titrations
was 14.70 cm3.
A
Calculations
FW of KIO3 = 214.00 g mol−1
molarity of the KIO3 solution =
25 cm3 of KIO3 contains
4.2813 g
214.00 g mol−1
4.2813 25
×
mol
214.00 1000
From the equation:
KIO3 + 5KI + 3H2SO4 → 3I2 + 3H2O + 3K2SO4
1 mol KIO3 generates 3 mol I2
4.2813 25
4.2813 25
×
mol KIO3 generates 3×
×
mol I2
214.00 1000
214.00 1000
From the equation:
I2 + 2S2O32− → 2I− + S4O62−
1 mol I2 oxidizes 2 mol Na2S2O3
4.2813 25
4.2813
25
3×
×
mol I2 oxidizes 2 × 3 ×
×
mol Na2S2O3
214.00 1000
214.00 1000
14.70 cm3 Na2S2O3 contains 2 × 3 ×
4.2813
25
×
mol
214.00 1000
4.2813
25
1000
×
×
mol
214.00 1000 14.70
= 0.2041 mol
Molarity of Na2S2O3 = 0.2041 M
1000 cm3 Na2S2O3 contains 2 × 3 ×
291
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Unit 2 Module 2 Analytical methods and separation techniques
Worked example 30.3 (continued)
Q
B: Analysis of vitamin C tablets
1 A vitamin C tablet was weighed and placed in
a 250 cm3 beaker, 75 cm3 of 1M H2SO4 was
added to the beaker and the contents of the
beaker stirred with a glass rod until the tablet
dissolved.
2 The solution in the beaker was poured into
a 250 cm3 volumetric flask. The beaker was
rinsed twice with distilled water and the water
used for rinsing added to the volumetric flask.
The flask was made up to mark with distilled
water.
3 A 25 cm3 aliquot of this solution was pipetted
into a 250 cm3 conical flask and to this was
added:
(a) 5 cm3 of 1 M potassium iodide;
(b) 3 cm3 of 1 M H2SO4;
(c) 25 cm3 of the standard potassium iodate
solution, measured accurately with a pipette or
a burette.
4 Sodium thiosulfate was added from a burette
to the conical flask until the contents of the
flask were pale yellow. Starch solution (2 cm3)
was then added to the flask and the addition
of the thiosulfate solution was continued until
the contents of the flask were just colourless.
The average of two concordant titrations was
12.15 cm3.
A
Calculations
4.2813 g
M
214.00 g mol−1
4.2813
25
25 cm3 of KIO3 contains
×
mol
214.00 1000
From the equation:
KIO3 + 5KI + 3H2SO4 → 3I2 + 3H2O + 3K2SO4
1 mol KIO3 generates 3 mol I2
4.2813
25
4.2813
25
×
mol KIO3 generates 3 ×
×
mol I2
214.00 1000
214.00 1000
= 1.50 × 10−3 mol I2
This is the number of mols of I2 formed initially; some of this I2 reacts with
ascorbic acid:
C6H8O6 + I2 → C6H6O6 + 2I− + 2H+
The I2 which remains after all the ascorbic acid has been oxidized is titrated
against Na2S2O3.
To calculate the number of mols I2 remaining in the flask after the reaction:
1000 cm3 Na2S2O3 contains 0.2041 mol
0.2041
1 cm3 Na2S2O3 contains
mol
1000
0.2041
12.15 cm3 Na2S2O3 contains
× 12.15 mol
1000
From the equation:
I2 + 2S2O32− → 2I− + S4O62−
2 mol Na2S2O3 oxidizes 1 mol I2
1
1 mol Na2S2O3 oxidizes mol I2
2
0.2041
1 0.2041
× 12.15 mol Na2S2O3 oxidizes ×
× 12.15 mol I2
1000
2
1000
= 1.24 × 10−3 mol I2
Number of mols of I2 which have oxidized the ascorbic acid in 25 cm3 of solution
= 1.50 × 10−3 − 1.24 × 10−3 mol
= 2.6 × 10−4 mol
From the equation:
C6H8O6 + I2 → C6H6O6 + 2I− + 2H+
I mol I2 oxidizes 1 mol ascorbic acid
∴ number of mols ascorbic acid in 25 cm3 solution = 2.6 × 10−4 mol
and number of mols ascorbic acid in 250 cm3 solution = 2.60 × 10−3 mol
FW of ascorbic acid = 176.14 g mol−1
mass of ascorbic acid in 250 cm3 solution
= number of mols × FW
= 2.60 × 10−3 mol × 176.14 g mol−1
= 0.4580 g
Molarity of the KIO3 solution =
Chapter 30 Titrimetric analysis
Worked example 30.4
Q
Analysis of hydrogen peroxide
1 The hydrogen peroxide to be analysed was
diluted to about 0.35% H2O2, so 15 cm3
of ‘20-volume’ (nominally 6%) solution of
hydrogen peroxide was transferred, using a
burette, to a 250 cm3 volumetric flask and the
solution made up to mark with distilled water.
2 A 25 cm3 aliquot of this H2O2 solution was
added slowly, with stirring, to a solution of
1 g potassium iodide in 100 cm3 of 1 M H2SO4
kept in a stoppered flask. After the addition
was complete the contents of the flask were
left to stand for 15 minutes.
3 The liberated iodine was titrated with
standardized sodium thiosulfate (0.204 M)
until the contents of the flask were pale
yellow. Starch solution (2 cm3) was then
added to the flask and the addition of the
thiosulfate solution was continued until the
contents of the flask were just colourless. The
average of two concordant titrations was
23.75 cm3.
A
Calculations
1000 cm3 of the standardized Na2S2O3 solution contains 0.204 mol
0.204
1 cm3 contains
mol
1000
23.75 cm3 contains
0.204
× 23.75 mol
1000
From the equations:
H2O2 + 2H+ + 2I− → I2 + 2H2O
I2 + 2S2O32− → 2I− + S4O62−
1 mol of H2O2 generates 1 mol of I2
and 2 mol of Na2S2O3 reduces 1 mol of I2
∴ 2 mol of Na2S2O3 ≡ 1 mol H2O2
1
× mol H2O2
1 mol of Na2S2O3 ≡
2
0.204
1 0.204
× 23.75 mol H2O2
× 23.75 mol Na2S2O3 ≡ ×
1000
2 1000
= number of mol H2O2 in 25 cm3 of the solution
250 1 0.204
no. of mol H2O2 in 250 cm3 of the solution =
× ×
× 23.75 mol
25 2 1000
−1
FW H2O2 = 34.02 g mol
250 1 0.204
mass H2O2 in 250 cm3 solution =
× ×
mol × 23.75 × 34.02 g mol−1
25 2 1000
This is the mass of H2O2 in 15 cm3 of the undiluted solution.
∴ mass H2O2 in 100 cm3 of the undiluted solution
250 1 0.204
100
=
× ×
× 23.75 × 34.02 ×
25 2 1000
15
= 5.494 g
100 cm3 of the undiluted solution of H2O2 contains 5.494 g
The actual concentration of the H2O2 is 5.494% w/v.
Analysis of hydrogen peroxide
Aqueous solutions of hydrogen peroxide (H2O2) are sold
as bleaching agents and disinfectants. Bottles are labelled
as either ’20-volume’, ‘40-volume’ or ‘100-volume’
hydrogen peroxide; this means that 1 cm3 of these solutions
liberate respectively 20 cm3, 40 cm3 and 100 cm3 of oxygen
(measured at s.t.p.) when decomposed by boiling. The
concentration of hydrogen peroxide in the ’20-volume’
solution is 6 per cent, in the ’40-volume’ solution it is 12
per cent and in the ‘100-volume’ it is 30 per cent. ‘Per cent’
in this context means weight/volume (w/v); so ‘6% w/v’
means that 100 cm3 of the solution contains 6 g of H2O2.
Hydrogen peroxide is an oxidizing agent. Under acidic
conditions it will oxidize iodide ions (I−) to iodine and
will itself be reduced to water. The equation for this redox
reaction is:
H2O2 + 2H+ + 2I− → I2 + 2H2O
This is not a very fast reaction, but it is fast enough to be
used for practical analysis of hydrogen peroxide.
293
294
Unit 2 Module 2 Analytical methods and separation techniques
Summary
✓ A titration is an analytical method by which
the concentration of an ion or compound (the
analyte) is determined by reacting the ion
or compound with a reagent (the standard)
in a solution of known concentration.
The stoichiometry of the reaction and the
concentration of the standard are used to
calculate the concentration of the analyte.
✓ The completion of the reaction between the
analyte and the standard can be detected by: (a)
use of a colour indicator, (b) measurement of pH,
(c) measurement of conductivity or
(d) measurement of temperature.
✓ The point at which the colour, pH, conductivity
or temperature change is detected is the endpoint of the titration. The end-point may differ
slightly from the precise point at which the
analyte has reacted with the standard, i.e. the
equivalence point.
✓ The concentrations of analytes which are acids
or bases are determined by acid/base titrations.
Strong acid/strong base and weak acid/strong
base reactions are commonly used in titrimetric
analysis; the weak base/strong acid titration
is less common and the weak acid/weak base
titration is not analytically useful because the
end-point is not sharp.
✓ The concentrations of analytes which are
reducing agents are determined by titrations
with standards which are oxidizing agents;
concentrations of analytes which are oxidizing
agents are determined by titrations with
standards which are reducing agents. These
titrations are termed redox titrations.
✓ Commonly used standards which are oxidizing
agents are permanganate ion (MnO4−), iodine
(I2 in solution with iodide, I−, as the tri-iodide
ion, I3−). Thiosulfate ion (S2O32−) is the most
commonly used reducing standard.
✓ The concentrations of the standard solutions
in titrimetric analysis must be determined by
titration against solutions of primary standards.
A primary standard is a very pure stable solid.
An accurate weight of a primary standard is used
to prepare a solution of known concentration;
this solution is used to standardize solutions for
titrimetric analysis.
✓ Primary standards for solutions used in acid/
base titrations are sodium carbonate and sodium
hydrogencarbonate. Primary standards for
solutions used in redox titrations are oxalic acid,
sodium oxalate and potassium iodate.
Chapter 30 Titrimetric analysis
Review questions
Answers to ITQs
1
How would you prepare the following solutions?
(a) 500 cm3 of approximately 0.0500 M KMnO4
(b) 2000 cm3 of approximately 0.2500 M KIO3
(c) 500 cm3 of approximately 0.0500 M Na2CO3
1
2
Describe how the concentrations of the NaOH solution
(0.0150 M) and the HCl solution (0.1030 M) used
in the aspirin analysis in ITQ 7 would have been
determined using sodium carbonate (Na2CO3) as a
primary standard.
(a) (i) 1.667 M
(ii) 0.06668 g
(iii) 6.668 g
(b) 1 M
(c) 30 cm3
2
(a) A H3O+, Cl−
B H3O+, Cl−, Na+, OH−
C Na+, Cl−
D Na+, OH−, Cl−
(b) After the end-point the solution contains excess
NaOH, giving Na+ and OH– ions. Both Na+ and OH–
have high conductivity.
3
Hydroxide ions are present in excess. Even with
0.1 cm3 excess of 1 M NaOH in 200.1 cm3 of
solution the hydroxide ion concentration [OH−]
0.1
= 200.1 = 5.00 × 10−4; pOH = 3.3; pH = 10.7
4
(a) (i)
(ii)
(b) (i)
(ii)
5
Addition of water to the flask has no effect on the
initial quantity of the material in the solution placed
in the flask.
6
4.128 g
7
(a) 0.2439 g
(b) 73.54%
8
0.17%
3
4
5
A solution of sodium thiosulfate was standardized
by dissolving 0.1327 g KIO3 (FW 214.00 g mol–1) in
water, adding excess KI and acidifying with H2SO4.
The iodine liberated from this reaction required
39.45 cm3 of the thiosulfate solution for reduction.
Calculate the molarity of the Na2S2O3.
Fumaric acid is a carboxylic acid that occurs in many
plants and is essential to respiration in animal and
plant tissues. A 0.3313 g sample of fumaric acid was
dissolved in 250 cm3 of water; 25 cm3 aliquots of this
solution were titrated against a 0.052 M solution of
NaOH. The average of two concordant titrations was
10.85 cm3.
(a) How many acidic hydrogens are there in one
molecule of fumaric acid?
(b) Suggest a suitable indicator for this titration.
Ferrous ions (Fe2+) are oxidized to ferric ions (Fe3+) by
permanganate (MnO4−) in acidic solution.
(a) Given that the half-cell reaction for the reduction
of MnO4− in acidic solution is
MnO4− + 8H+ + 5e− ҡ Mn2+ + 4H2O
write a balanced equation for the redox reaction
between ferrous ions and permanganate ions.
(b) What volume of 0.018 M KMnO4 is required
to oxidize 25 cm3 of 0.103 M FeSO4 in acidic
solution?
phenolphthalein; colourless; pink
methyl red; red; yellow
NaOH + CH3COOH → CH3COONa + H2O
NH4OH + HCl → NH4Cl + H2O
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Chapter 31
Introduction to spectroscopy
Learning objectives
■ Define the key features of wave motion: wavelength (λ), frequency (ν) and amplitude (A).
λν and as particles
with energy E = hν.
■ List the types of radiation comprising the electromagnetic spectrum in order of increasing or
decreasing energy.
■ Explain the meaning of the term ‘quantized energy levels’ and provide a relevant illustration.
■ Describe the changes which occur in atoms and molecules as a result of the absorption of
ultraviolet and visible radiation and of infrared radiation.
■ Describe electromagnetic radiation as waves with velocity 3.0 × 1010 cm s−1 =
Introduction to spectroscopy: resonance
same frequency and if the two systems are in phase with
each other (i.e. their extremes happen at the same time).
Imagine you are pushing a child’s swing. Each time the
swing returns to you it has undergone one oscillation, the
distance moved by the swing is related to the amplitude,
and the number of times per minute that the swing returns
to you is the frequency. If you push the swing each time it
comes to rest at the end of one oscillation, the amplitude of
the swing increases. Push at any other time and the motion
of the swing is hindered, or at worst, stopped altogether.
When these conditions are met the two systems are in
resonance with each other.
As an illustration, find two tuning forks with the same
frequency and a third tuning fork with a slightly different
frequency. Strike one of the pair and hold it on a solid
surface. Get a friend to hold the other two forks on the same
surface. Only the one with the same natural frequency as the
sounding fork will take energy from it and begin to sound.
Energy can only be transferred efficiently from one
oscillating system to another if the two systems have the
Penetrates
earth
atmosphere?
yes
Wavelength
(metres)
radio
10
buildings
yes
no
microwave
3
10
humans
infrared
–2
10
honey bee
–5
pinpoint
visible
0.5 x 10
–6
protozoans
no
ultraviolet
10
–8
X-ray
10
molecules
gamma ray
–10
atoms
10
–12
atomic nuclei
Frequency
(Hz)
10
4
10
8
12
10
IR analysis
15
10
16
10
UV analysis
18
10
10
20
Figure 31.1 The
electromagnetic spectrum.
Chapter 31 Introduction to spectroscopy
depending on what is most suitable for each part of the
electromagnetic spectrum.
Electromagnetic radiation
The electromagnetic force is one of the four main forces in
Nature; the other three are the gravitational force, the strong
nuclear force and the weak nuclear force. Spectroscopy is
the study of the interaction of matter (atoms, molecules, ions)
with electromagnetic radiation. Light is the most familiar
form of electromagnetic radiation, and visible light is only
a tiny part of the electromagnetic spectrum (Figure 31.1).
When we see a rainbow or when we pass white light
through a prism, we recognize that white light is composed
of a spectrum of colours. When we examine the colour of
a compound or carry out a colour test such as adding an
acid/base indicator to a sample we carry out a very simple
spectroscopic experiment. Some components of white light
have been absorbed by the sample, and the remaining part
that reaches our eye is interpreted by the brain as a colour.
In the laboratory a spectrometer (sometimes called a
spectrophotometer) serves a similar function and tells us
what ‘colour’ reaches it after the light has passed through
the sample.
Experiments have shown that electromagnetic radiation
has all the properties of a wave motion. For example, no
matter what its wavelength, electromagnetic radiation can
be diffracted. But electromagnetic radiation also behaves as
though it is a stream of particles, which we call photons.
The energy of a photon is fixed by the wavelength of its
associated wave. This dual nature (something being a wave
and a particle at one and the same time) is not something
we meet in everyday life, so it takes a little getting used to!
Characteristics of electromagnetic radiation
Electromagnetic radiation can be described as a train of
waves, characterized by wavelength (Figure 31.2).
nodes
points in
phase
h
h
Frequency is a related measure; it is defined as the number
of waves that pass a reference point in a unit of time. In
spectroscopy the unit of time is the second (s), so the unit of
frequency is the reciprocal second (s–1), which is now given
the name hertz (pronounced ‘herts’) (Hz). The symbol for
frequency is ν (‘nu’).
Frequency and wavelength are inversely proportional:
ν∝
1
λ
The amplitude (A) is the maximum displacement from
the horizontal axis. Amplitude is measured in units of
length. Waves of electromagnetic radiation travel at the
velocity of light, c, where c = 3.0 × 108 m s–1, and this is
the constant that connects frequency and wavelength of
electromagnetic waves:
ν=
c
and c = λν
λ
Electromagnetic radiation of long wavelength has
low frequency, and electromagnetic radiation of short
wavelength has high frequency.
Electromagnetic radiation can be described as consisting of
photons. A photon is a quantum (particle/packet, plural
‘quanta’) of light, comprising a specific quantity of energy.
You know from Chapters 1 and 2 that the energy of one
photon is given by the Bohr equation:
E = hν
where h is the Planck constant, the value of which is
6.63 × 10–34 J s.
Using the equation
ν=
c
λ
we can find that the energy, E, for a photon is given by:
A
line of
propagation
h
nodes
Figure 31.2 A wave, showing the line of propagation, amplitude,
wavelength and frequency.
One wavelength is the distance between two successive
nodes on a wave, or the distance measured along the line
of propagation between two points that are in phase on
adjacent waves. The symbol for wavelength is λ (‘lambda’)
and it is measured in units of length. We can measure
wavelength in metres (m) or sub-units of metres, such as
centimetres (cm), micrometres (μm) or nanometres (nm),
E=
hc
λ
Because electromagnetic radiation can be described either
as waves or as quanta, it is said to possess wave–particle
duality.
ITQ 1 Show that the units of Planck’s constant are joule-seconds,
J s.
ITQ 2
(a) If a radiation has a wavelength of 3 × 10−5 m, what is the
energy of its photons?
(b) What is this energy in J mol−1?
297
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Unit 2 Module 2 Analytical methods and separation techniques
Regions of the electromagnetic spectrum
Light can be divided into infrared (IR), visible and ultraviolet
(UV) light, in order of increasing energy. Microwaves are
lower in energy than IR radiation and X-rays are higher
in energy than UV light. Radiofrequency (rf) radiation is
lowest in energy and is placed below microwaves. These
are the five regions of the electromagnetic spectrum that
are important in spectroscopy (Table 31.1).
Table 31.1 The regions of the electromagnetic spectrum that are
important in spectroscopy
Region
Wavelength range / m
X-ray
10−12 to 10−8
−8
10 to 10
IR
10−6 to 10−3
■ any photons whose energy corresponds exactly to
the difference in two electron energy levels will be
absorbed and the electron will be promoted into its
higher level;
Microwave 10−3 to 1
Radio
An atom or molecule at its lowest possible energy level
is said to be in the ground state. After absorption of
electromagnetic radiation the atom or molecule is said to
be in an excited state.
When photons with a range of energies fall on a material:
−6
UV/VIS
When an atom or a molecule is subjected to electromagnetic
radiation (irradiated), the energy level of the atom or
molecule will increase if the energy of the photons (hν)
with which it is irradiated is equal to the difference (ΔE)
between two of its energy levels. The atom or molecule
absorbs the energy of the photons when, and only when,
ΔE = hν.
1 to 105
■ any photons whose frequency corresponds exactly to
The interaction of electromagnetic
radiation with atoms and molecules
The energy levels in atoms and molecules are quantized:
they have exact values. The differences between any two
energy levels are therefore quantized. This means that:
■ the electrons in atoms and molecules can have only
certain specific energies;
■ if a bond connects two atoms, that bond behaves as
the frequency of oscillation of a bond between atoms
will be absorbed and the amplitude of the oscillation
will increase.
In both of these cases, the photons are resonating with a
part of the system. A graph of energy transmitted against
frequency of the incoming radiation will show a dip at the
resonant frequency. Each resonance will be characteristic
of the particular atom or bond that is stimulated. This
makes resonant absorption a very powerful analytical tool.
a spring with a fixed natural frequency of oscillation
so molecules can vibrate and rotate only with specific
frequencies.
These energies can be predicted by quantum theory.
In atomic and molecular spectroscopy, energy levels are
represented as horizontal lines arranged along a vertical
axis: horizontal lines low on the vertical axis represent low
energy levels and the lines higher up on the axis represent
higher energy levels (Figure 31.3).
energy level of the
second excited state
Energy
(increasing
along this
axis)
energy level of the
first excited state
6E
difference in energy
between the groundstate and the first
excited state
ground-state energy
level; occupied before
irradiation by emr when the molecule is
in the ground state
higher energy
levels, vacant
in the ground
state; occupied
after irradiation
ITQ 3 From the equations E = hν and c = λν, derive the equation
which relates energy (E) to wavelength (λ).
ITQ 4
(a) What is the wavelength of electromagnetic radiation with
frequency 3.0 × 1015 s−1?
(b) In which region of the electromagnetic spectrum does this
radiation fall?
(c) What is the energy of the photons which correspond to
radiation of this frequency?
ITQ 5 In what region of the electromagnetic spectrum is radiation
with the following wavelengths found?
(a) 250 nm
(b) 6.2 μm
(c) 8 nm
(d) 3 cm
Figure 31.3 Quantized energy levels.
Chapter 31 Introduction to spectroscopy
Effects of irradiation
In a molecule in the electronic ground state, bonding
molecular orbitals are occupied by pairs of electrons, and
antibonding molecular orbitals are vacant. Read Chapter
19 if you need to revise the concept of bonding molecular
orbitals. The energy differences (ΔE) between bonding
molecular orbitals and antibonding molecular orbitals are
the largest energy differences (ΔE values) encountered in
spectroscopy.
Eantibonding MO − Ebonding MO = ΔE (large)
Remember that ΔE = hν, so high frequency electromagnetic
radiation is required to bring about a large energy
change (ΔE). To promote an electron from a bonding
molecular orbital to an antibonding molecular orbital
requires radiation in the visible and ultraviolet range.
If a molecule is irradiated with very short wavelength
(high frequency) UV light or with X-rays, an electron in
a bonding molecular orbital can be promoted beyond the
level of the corresponding antibonding molecular orbital
and hence out of the molecule, resulting in loss of the
electron and ionization of the molecule. Short wavelength/
high frequency UV light and X-rays are therefore called
ionizing radiation. Excessive exposure to ionizing radiation
is harmful to living organisms (including human beings)
because the ions which are formed can damage to DNA,
leading to mutations and cancer in animals.
The differences in vibrational energy levels of molecules
are small, so irradiation with infrared light, which is of
relatively low energy, causes increases in vibrational
energy.
In this chapter we have described the principles of
absorption spectroscopy. For the sake of completeness,
it should be made clear that there are other types of
spectroscopy. Emission spectroscopy was of the greatest
importance in the investigation of the structure of atoms.
Energy is transferred, by heat for example, to the atom,
and the excited atom then emits energy as light at specific
frequencies. The yellow colour produced when sodium
salts are placed in a Bunsen flame is a familiar example.
Our knowledge of the electronic structure of atoms is
derived from these studies, as is the terminology we use
to describe atomic orbitals (s, p, d, f, etc.). In fluorescence
spectroscopy, energy from light of one frequency is
absorbed by the sample molecule, the energy is partially
dissipated by the molecule, which later emits light of a
lower frequency.
299
300
Unit 2 Module 2 Analytical methods and separation techniques
Summary
✓ Waves are characterized by wavelength (λ),
✓ Energy levels in molecules and atoms are
quantized. This means that molecules can vibrate
only with specific frequencies; atomic and
molecular orbitals and the electrons which they
hold have specific energies; and the rotational
energies of molecules are specific.
which is the distance between corresponding
points on adjacent waves, and by frequency
(ν), which is the number of waves that pass a
reference point in a unit of time.
✓ The electromagnetic spectrum consists of X-rays,
ultraviolet (UV), visible and infrared (IR) light,
microwaves and radiofrequency radiation, in
order of decreasing energy.
✓ A photon of electromagnetic radiation of energy
hν will be absorbed by an atom or a molecule if
the energy of the photon is equal to the energy
difference between two energy levels of the
atom or molecule, i.e. if ΔE = hν.
✓ Electromagnetic radiation can be treated as
waves of velocity c = νλ or photons of energy
E = hν.
Review questions
1
2
3
Answers to ITQs
Arrange the following regions of the electromagnetic
spectrum in order of increasing energy:
1
E = hν so h =
E
ν
radio frequency, X-ray, ultraviolet, visible, infrared
units of energy are joules, J
(a) Find the frequency of light which corresponds to
photons of energy 5.0 × 10−12 J.
(b) What is the wavelength of this light?
(c) Is this radiation harmful? Give a reason for your
answer.
frequency = number per second, so units are s−1
J
therefore units of h are −1 = J s
s
hc
(a) E =
λ
6.63 × 10−34 J s × 3 × 108 m s−1
E=
3 × 10−5 m
2
The energy diagram shown below represents the four
lowest energy levels of an atom.
E = 6.63 × 10−21 J
E4
E3
Energy
units are
E2
(J s)(m s−1) i.e. J, which is correct since
E is an energy.
m
(b) 1 mole = 6.02 × 1023 particles
Energy of 1 mole of photons would be
E1
6.63 × 10−21 J × 6.02 × 1023 = 4 × 103 J
(a) Which level represents the energy of the atom if it
is in its lowest energy state?
(b) How does the energy of the atom change as it is
excited from E1 to E2 to E3 to E4?
(c) What must take place before the atom can be
excited from E1 to E2?
(d) On the diagram above, draw arrows to show the
transitions that the atom can undergo, starting
from its lowest energy level.
(e) Which of the transitions shown in the answer to
(d) above would require electromagnetic radiation
of (i) the longest wavelength? (ii) the highest
frequency?
c
hc
;E=
λ
λ
3
E = hν; ν =
4
(a) 1 × 10−7 m
(b) visible
(c) 2.0 × 10−18 J
5
(a)
(b)
(c)
(d)
UV
IR
X-ray
microwave
301
Chapter 32
Ultraviolet–visible spectroscopy
Learning objectives
■ Describe what can occur when a molecule with covalent bonds is irradiated with UV–visible light.
■ Define the terms lambda max (λmax), chromophore, UV–visible spectrum, absorbance (A), molar
■
■
■
■
extinction coefficient (ε), standard curve and chromophoric reagent.
In general terms, relate the value of λmax to the structure of a compound and explain why there is
no observable absorption of UV radiation by some compounds.
Describe the features of a UV–visible spectrophotometer and outline the procedure for obtaining
UV–visible spectra.
Use the Beer–Lambert law (A = εcl) to calculate the concentration of a given analyte in solution.
Explain how a standard curve is generated and how the concentration of an analyte can be
determined from a standard curve.
m› (anti-bonding) molecular orbital;
vacant in the ground-state
Molecular orbitals in covalent molecules
In a covalent molecule the atoms are held together by bonds
that consist of pairs of electrons. These electrons populate
molecular orbitals that are formed by the interaction of
atomic orbitals of the constituent atoms. Two s atomic
orbitals interact to form two molecular orbitals. One of
the molecular orbitals is of lower energy than the atomic
orbitals – this is the sigma bonding (σ) molecular orbital,
which is occupied by two electrons. The other molecular
orbital is of higher energy than the atomic orbitals – this is
the sigma anti-bonding (σ*) molecular orbital, which can
theoretically accommodate two electrons, but it is vacant
when the molecule is in the ground state (Figure 32.1).
The overlap of two adjacent p orbitals can form a pi bonding
(π) molecular orbital and a pi anti-bonding (π*) molecular
orbital. When a molecule contains atoms such as oxygen,
nitrogen, chlorine (and other halogens) on which there are
non-bonded electrons, these electrons occupy molecular
orbitals known as n orbitals; n electrons are not involved
in bonding, so there are no n anti-bonding orbitals. The
relative energy levels of σ, π, n, π* and σ* molecular orbitals
are shown in Figure 32.2.
A molecule with several atoms and different types of bonds
will have molecular orbitals at several energy levels; some
of the molecular orbitals will be occupied and some will be
vacant. In ethene, H2C=CH2, the molecular orbitals are σ,
π, π*and σ*.
Energy
electron in
s atomic orbital
electron in
s atomic orbital
m (bonding) molecular orbital;
occupied by two paired electrons
Figure 32.1 Interaction of two s atomic orbitals to form σ and σ*
molecular orbitals.
m›
/›
Energy
n
/
m
Figure 32.2 Relative energy levels of σ, π, n, π* and σ* molecular
orbits.
ITQ 1 How many of each type of molecular orbital, (σ, π, π* and
σ*) are there in ethene?
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Unit 2 Module 2 Analytical methods and separation techniques
Absorption of energy by electrons in
molecular orbitals
We can irradiate a covalent molecule with electromagnetic
radiation and, if the energy of the photon (hν) matches the
energy difference (ΔE) between an occupied and a vacant
orbital, the energy of the photon may be transferred to an
electron and lift it into the vacant orbital. The molecule
carrying this extra energy is said to be in an excited state.
We can describe the process as σ → σ*, n → σ* or π →
π*, depending on whether σ, n or π molecular orbitals
are involved. Excitation of electrons from bonding or
non-bonding to anti-bonding molecular orbitals requires
electromagnetic radiation in the ultraviolet and visible
regions to satisfy the equation ΔE = hν.
It is more convenient to use units of wavelength (rather
than frequency) in this part of the spectrum. The UV
region with wavelengths between 200 nm and 400
nm is called the near-ultraviolet. Experimentally, it is
relatively easy to produce electromagnetic radiation in this
wavelength range and observe its absorption by molecules.
For wavelengths shorter than 200 nm special equipment
is required because many common materials absorb light
at these wavelengths; this region is called the vacuum
ultraviolet since air absorbs at these wavelengths and
must be excluded from the apparatus if absorption in this
range is to be observed. Consequently, attention has been
generally confined to the near-UV and visible regions of
the spectrum for this type of spectroscopy.
The ΔE for σ → σ* transitions is relatively large, and
absorption to cause this type of transition requires
electromagnetic radiation in the vacuum UV region that is
not easily studied. The energy differences between n and
σ*, n and π* and between π and π* orbitals are smaller,
ITQ 2 When a molecule of ethene, H2C=CH2, absorbs UV
radiation of wavelength 170 nm, ΔE (as in Figure 32.3a) can be
calculated.
ΔE = hν =
hc
λ
where h is the Planck constant = 6.63 × 10−34 J s;
c = 3.0 × 108 m s−1.
ΔE =
6.63 × 10−34 J s × 3.0 × 108 m s−1
170 × 10−9 m
= 1.17 × 10−18 J
(a) Repeat the above calculation to find the energy absorbed
when a molecule of butadiene absorbs UV radiation of
wavelength 217 nm (ΔE in Figure 32.3b).
(b) Comment on the difference in ΔE between ethene and
butadiene.
and transitions between these types of orbitals result from
absorption in the near-UV region. Electronic transitions
which give useful information generally involve π orbitals.
The simplest example of a C–C π bond is in ethene, H2C=CH2,
but even here the π → π* transition requires electromagnetic
radiation of wavelength 170 nm, which is outside of the
convenient range. However, when two double bonds are
conjugated, ΔE between the π and π*orbitals is reduced,
so the absorption occurs at lower frequency (ν) and longer
wavelength (λ). 1,3-Butadiene, H2C=CH–CH=CH2, absorbs
UV at 217 nm. The π molecular orbitals of the conjugated
double bonds are derived from the four p atomic orbitals of
the four sp2 hybridized carbon atoms. This generates four
π molecular orbitals: two low-energy orbitals, each with
a pair of electrons, and two vacant π* orbitals of higher
energy (Figure 32.3).
1,3-butadiene
H
H
C
C
H
C
H
C
H
H
H
ethene
C
C
H
H
H
/›
4
/›
/›
3
6E
6E
Energy
/2
/
/1
a
b
Figure 32.3 π molecular orbitals in (a) ethene and (b) 1,3-butadiene.
ΔE between the occupied molecular orbital of highest energy and
the unoccupied molecular orbital of lowest energy is much less in
1,3-butadiene which has two conjugated double bonds.
The important change for our purposes is that the energy
difference (ΔE) between the highest occupied π orbital and
the lowest unoccupied π* orbital has become smaller. In
a compound with three conjugated double bonds, as in
1,3,5-hexatriene, H2C=CH–CH=CH–CH=CH2, there are
three occupied π orbitals and three unoccupied π* orbitals.
The energy difference between the highest occupied and
the lowest unoccupied orbitals is even smaller; smaller ΔE
corresponds to smaller ν and therefore longer wavelength λ.
As the number of conjugated double bonds in a compound
increases, the wavelength of the light which the compound
absorbs also increases.
The visible region of the spectrum starts near 400 nm and
extends to 800 nm. The principal wavelength of light which
a compound absorbs is designated λmax (‘lambda max’).
As λmax increases it gets closer to the visible region of the
spectrum. If a substance reflects (does not absorb) visible
light it will appear white. If it absorbs light with a wavelength
Chapter 32 Ultraviolet–visible spectroscopy
Table 32.1 Colour and wavelength of light absorbed and
observed colour
Studying absorption of UV–visible
radiation
Colour of light absorbed Wavelength absorbed / nm Observed colour
violet
400
yellow
blue
450
orange
blue-green
500
red
yellow
550
violet
orange
600
blue-green
red
700
green
in the visible region of the spectrum it will appear coloured.
The light in the visible region of the spectrum is coloured.
Violet light is at the short wavelength end of this range, and
as wavelength increases the colour changes to indigo, then
blue, green, yellow, orange and finally red. (One mnemonic
for R O Y G B I V – the first letter of the colours starting with
‘red’ – is Richard Of York Gave Battle In Vain.)
If we remove a little of the short-wavelength violet-blue
part of white light, the eye perceives a yellowish colour.
Many organic compounds appear to be cream-coloured
even though their λmax is not in the visible region. As the
λmax value increases and moves into the visible region, the
compound becomes strongly coloured. The eye sees the
part of the spectrum that is not absorbed and perceives the
colour that is complementary to the absorbed light. If blue
and green are absorbed the sample appears red to the eye
(Table 32.1).
The compound β-carotene, which is found in carrots, has
11 conjugated double bonds.
CH3
CH3
CH3
Analysis of samples by UV–visible spectroscopy entails
accurate measurement of the wavelength of maximum
absorption (λmax) and of the intensity of absorption (how
much light is absorbed). Absorption of UV–visible radiation
is determined using very dilute solutions, so the sample
to be analysed must be dissolved in a solvent which does
not absorb light at the same wavelength as the sample.
Solvents commonly used in UV–visible spectroscopy are
95% ethanol, water and hexane. The sample to be analysed
is weighed accurately and dissolved in a known volume
of a suitable solvent; the solution is usually prepared in a
volumetric flask. A portion of the solution is placed in a
cuvette (sometimes called a UV–visible cell). A cuvette is
tubular container, usually 1 cm square in cross-section and
with rectangular sides 4–5 cm high (Figure 32.4).
path length
=10 mm
45 mm
10 mm
10 mm
H3C
CH3
Figure 32.4 A cuvette.
CH3
CH3
CH3
CH3
CH3
`-carotene
The λmax for β-carotene is at 497 nm (the blue-green
part of the spectrum) and this gives carrots their familiar
red-orange colour.
The term chromophore describes a molecular feature that
is responsible for the absorption of light of any wavelength.
In β-carotene, the chromophore consists of 11 conjugated
carbon–carbon double bonds. The chromophore in
butadiene consists of two conjugated carbon–carbon
double bonds.
Cuvettes can be made of plastic, glass or quartz. Plastic
and glass absorb UV radiation, so cuvettes made of these
materials can only be used to measure absorption in the
visible region. Quartz is transparent to both UV and visible
radiation, so quartz cuvettes can be used for the entire UV–
visible range.
In a double-beam spectrophotometer, two cuvettes are
placed in the spectrophotometer. One cuvette contains
the solution of the analyte (the sample cell) and the other
contains pure solvent (the reference cell).
In some UV–visible spectrophotometers there are two
sources of light:
ITQ 3 Use the information in Table 32.1 to predict the colour of
the compounds for which the absorption maxima (λmax) are given
below.
■ a UV source, usually a hydrogen discharge lamp,
(a) 460 nm, a commonly used indicator
■ a source of visible light, which can be a tungsten
(b) 545 nm, a plant pigment
(c) 617 nm, a synthetic dye.
emitting light of wavelength 180– 400 nm;
filament lamp, which produces light of wavelength
400–800 nm.
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Unit 2 Module 2 Analytical methods and separation techniques
In some other spectrophotometers a high-intensity xenon
flash lamp is used for the source of both UV and visible
light.
Radiation from the source is split by mirrors into two
beams. One beam (the sample beam) passes through
the sample and the other (the reference beam) passes
through the reference cell. If, for example, the sample
under study absorbs radiation of wavelength 300 nm,
then the intensity of the light of wavelength 300 nm after
it has passed through the sample (transmitted light) will
be less than it was before it passed through the sample
(incident light). In the spectrophotometer the intensities
of the two transmitted beams are compared over the
range of wavelengths generated by the sources. The
spectrophotometer electronically subtracts the absorption
of solvent in the reference beam from the absorption of the
solution in the sample beam, minimizing the effects due to
absorption of light by the solvent.
Many modern spectrophotometers are single-beam
instruments. A cuvette containing solvent only is placed
in the spectrophotometer, the entire wavelength range is
scanned and the absorbance/transmittance data is stored
digitally. The solvent in the cuvette is replaced with the
solution of the analyte, the wavelength range is again
scanned and the absorbance/transmittance data for the
solution is stored. The stored data for the solvent is then
subtracted from the solution data.
Many UV–visible spectrophotometers automatically plot
the intensity of the transmitted or absorbed light on the
vertical axis (y-axis) versus wavelength on the horizontal
axis (x-axis). This plot is a spectrum (plural: spectra) (Figure
32.5).
Lambert’s law states that the fraction of incident light
absorbed by a homogeneous liquid depends on the nature
of the liquid, but is independent of the intensity of the
incident light.
Beer’s law states that the absorption of light is proportional
to the number of absorbing molecules.
Combined, these give us the Beer–Lambert law:
absorbance (A) = log10
I0
=εcl
I
where I0 is the intensity of the incident light, I is the intensity
of the transmitted light, ε (‘epsilon’) is the molar extinction
coefficient (the units are 1000 cm2 mol−1 but these are, by
convention, not expressed), c is the concentration in mol
dm−3 and l is the path length of the absorbing solution in
centimetres.
I
log10 0 is the absorbance (sometimes called the optical
I
density); absorbance is usually designated A, and
this is the parameter on the vertical axis of a UV–visible
spectrum. The value of A depends on the concentration of
the solution; an acceptable maximum is 2.
Epsilon (ε), the molar extinction coefficient (sometimes
referred to as the molar absorptivity or the extinction
coefficient), is an intrinsic molecular property of the
sample in solution and is related to the probability of a
given transition. The value of ε is a constant for a given
transition (λmax) in a specific molecule; the value of ε can
be very large, so log10 ε is sometimes quoted.
The path length (l) is the distance travelled by the light
through the solution under study. When the solution is
held in a cuvette of cross-section 1 cm square, the path
length is 1 cm.
1.00
Applications of ultraviolet–visible
spectroscopy
0.75
Absorbance
304
0.50
The Beer–Lambert law
In the previous section we discussed absorption of
electromagnetic radiation in the UV–visible region by
covalent molecules In the past, UV–visible spectroscopy
was widely used in organic chemistry. Many inorganic ions
and compounds also absorb UV–visible radiation. These
include transition metal ions in solution and the anions
nitrate (NO32−), nitrite (NO2−), chromate (CrO42−) and
permanganate (MnO4−).
There are two empirical laws of light absorption which
enable us to use experimental data to obtain useful
information.
Substances which do not absorb very strongly in the UV–
visible region can be made to react with reagents to produce
highly coloured derivatives which can be easily analysed
0.25
0
200
250
300
350
400
h / nm
Figure 32.5 A simple absorbance spectrum.
Chapter 32 Ultraviolet–visible spectroscopy
Worked example 32.1
Q
Worked example 32.2
For 2,4,6-octatriene (1) the wavelength of maximum absorption
(λmax) is 274.5 nm. The value of ε for this band is 30 000.
Q
The extinction coefficient (ε) of the Fe2+–1,10-phenanthroline
complex (shown in Figure 32.6) at the λmax of 508 nm is
11 100.
(a) What is the concentration of Fe2+, in mol dm−3, in a solution
which gives an absorbance (A) of 0.18 using a cell of path
length of 1 cm?
(b) Convert the concentration obtained in part (a) to parts per
million (ppm = milligrams per dm3) of Fe2+.
A
(a) Use the Beer–Lambert law:
A=εcl
A
c=
εl
A = 0.18; ε = 11 100; l = 1
0.18
c=
= 1.62 × 10−5 mol dm−3
11100 × 1
This is the concentration of the
Fe2+–1,10-phenanthroline complex.
Each mol of the complex contains one mol of Fe2+
the concentration of Fe2+
= concentration of the complex
= 1.62 × 10−5 mol dm−3
(b) Atomic weight of iron = 55.8 g
1 dm3 of a solution containing 1.62 × 10−5 mol
gives A = 0.18
This solution contains 1.62 × 10−5 × 55.8 g of Fe2+
= 9.05 × 10−4 g = 0.905 mg
The concentration of Fe2+ in this solution is 0.9 ppm
(1) C8H12
h max 274.5 nm (¡ 30,000)
(a) Calculate the concentration of the solution (in mol dm−3)
that would be required to give an absorbance (A) of 1, using
a cell of path length 1 cm.
(b) Convert the concentration obtained in part (a) to g/100 cm3.
A
(a) Use the Beer–Lambert law:
A=εcl
A
c=
εl
A = 1; ε = 30 000; l = 1
1
c=
= 3.33 × 10−5 mol dm−3
30000 × 1
(b) FW of 2,4,6-octatriene (C8H12) = 108 g mol−1
1 dm3 of a solution containing 3.33 × 10−5 mol dm−3
gives A = 1
1 dm3 of a solution containing 3.33 × 10−5 × 108 g
gives A = 1
100 cm3 of a solution contains
100
3.33 × 10−5 × 108 ×
g gives A = 1
1000
Mass needed = 3.60 × 10−4 g (which is 0.36 mg)
by UV–visible spectroscopy. The reagents which produce
the coloured derivatives are known as chromophoric
reagents. Two examples are shown in Figure 32.6:
a
N
+
3
■ 1,10-phenanthroline, which forms a red complex with
N
Fe2+
2+
Fe
N
ferrous ion (Fe2+);
N
■ dimethylglyoxime (DMG), which forms a red complex
with nickel ion (Ni2+); this complex is also used for
gravimetric analysis of nickel and was discussed in
Chapter 29 (page 275); when DMG is dissolved in an
organic solvent the nickel–DMG complex can be used
for the quantitative analysis of nickel by UV–visible
spectroscopy.
3
1,10-phenanthroline (-Phen)
max
H
H3C
N
OH
H3C
O
O
N
N
C
C
+
C
H3C
ITQ 4The fragrant naturally occurring compound
β-phellandrene (RMM 136) has λmax = 231 nm, and
the value of ε for this band is 21 000. A solution
of β-phellandrene shows absorbance, A, = 1.5
with a 1-cm cell. Calculate the concentration of
β-phellandrene in this solution, in g dm−3.
= 508 nm
b
2
CH2
Fe( -Phen)32+
N
OH
Ni 2+
Ni
C
H3C
C
N
N
O
dimethylglyoxime (DMG)
Ni(DMG)2
max
`-phellandrene
CH3
C
CH3
O
H
= 445 nm
Figure 32.6 Two chromophoric reagents used in quantitative
analysis of metal ions by UV–visible spectroscopy:
(a) 1,10-phenanthroline; (b) dimethylglyoxime.
305
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Unit 2 Module 2 Analytical methods and separation techniques
Ultraviolet–visible spectroscopy is applied to the analysis
of numerous organic, inorganic and biological substances.
The majority of the analyses in clinical laboratories utilize
UV–visible spectroscopy. The two worked examples shown
above demonstrate that UV–visible spectroscopy is a very
highly sensitive analytical method with detection limits in
the range of 10−5 mol dm−3.
The Beer–Lambert law (A = ε c l) is obeyed when absorbance
(A) is ≤ 2. When the law is obeyed absorbance (A) and
concentration (c) are directly proportional, because ε and
l are constant. A plot of absorbance (usually on the y-axis)
versus concentration (x-axis) gives a straight line. The
absorbance values, at a given wavelength, of solutions of
different and known concentrations can be used to generate
such a plot, which is known as a standard curve (also
known as a calibration curve or a reference curve). The
absorbance value of a solution of unknown concentration
(of the same substance) is then determined at the same
wavelength and the concentration of that solution can be
read from the standard curve.
Summary
✓ When a molecule with covalent bonds absorbs
UV–visible radiation, electrons are promoted
from occupied molecular orbitals to anti-bonding
molecular orbitals.
✓ The wavelength of light (λmax) which a molecule
absorbs depends on the energy difference (ΔE)
between the occupied molecular orbital of
highest energy and the anti-bonding molecular
orbital of lowest energy:
ΔE = hν =
hc
hc
.
;λ=
ΔE
λ
✓ Compounds which are highly conjugated
(contain several alternating double and single
bonds, –C=C–C=C–C=C–) absorb light in the visible
range (400–800 nm) and can be used to make
derivatives of non-absorbing species for analysis.
✓ Absorption of UV–visible radiation is governed
by the Beer–Lambert law: A = ε c l, where
A = absorbance, ε = extinction coefficient,
c = molar concentration, l = path length.
✓ Important terms used in UV–visible spectroscopy
are: spectrum (plural – spectra, a plot of
absorbance versus wavelength), chromophore
(molecular feature which absorbs light),
standard curve (a plot of absorbance versus
concentration), chromophoric reagent (a
compound used to make a coloured derivative of
a non-absorbing species).
✓ To determine the UV–visible spectrum of a
compound, a dilute solution is prepared using a
solvent which does not absorb in the UV–visible
range. A portion of the solution is placed in a
cuvette and irradiated through the required
wavelength range. The spectrophotometer
compares the intensities of the incident and
transmitted light, corrects for the effects of
solvent and, in most cases, plots the spectrum.
Chapter 32 Ultraviolet–visible spectroscopy
Review questions
1
Answers to ITQs
(a) How many of each type of molecular
orbital (σ, π, n, π*, σ*) are there in
methanal?
H
C
H
methanal
(b) Which is the occupied molecular orbital of highest
energy?
(c) Which is the vacant molecular orbital of lowest
energy?
(d) Which electronic transition from an occupied
molecular orbital to a vacant molecular orbital
requires the least energy?
2
A chemist prepared 1,3,5-hexatriene and
1,3,5,7-octatetraene, placed them in separate flasks, but
forgot to label the flasks. How could UV spectroscopy
be used to determine which compound is in each flask?
1,3,5-hexatriene
1,3,5,7-octatetraene
3
Calculate the extinction coefficient, ε, for a compound
in a solution of 1.5 × 10−4 mol dm−3 for which the
absorbance, A, is 1.2 when measured in a cell of path
length 1 cm. What would the value of A be for a
solution of half this concentration?
4
Match the structures of the naturally occurring
coloured substances shown in (a) with the absorption
maxima given in (b).
(a)
OH
O
CH3
glucose
COOH
OH
HO
OH
O
cochineal, a red food colouring extracted
from females of the insect Coccus cacti
i
O
N
H
Br
H
N
Br
O
Tyrian purple, a blue-purple dye from
mollusks of the genus Murex
ii
CH3
CH3
OH
O
O
O
sugar
iii
CH3
CH3
crocin, a component of the yellow food colouring saffron
(b) λmax = 440 nm; λmax = 500 nm; λmax = 580 nm.
1
σ (bonding) = 5; π (bonding) = 1; π* (anti-bonding) = 1;
σ* (anti-bonding) = 5.
2
(a) 9.17 × 10−19 J
(b) This energy difference is about an order of
magnitude less than ΔE between the π and π*
orbitals of ethene.
3
(a) orange (it is methyl orange)
(b) blue (it is cyanidin chloride)
(c) green (it is malachite green)
4
9.7 × 10−3 g dm−3
O
307
308
Chapter 33
Infrared spectroscopy
Learning objectives
■ Describe what can occur when a molecule with covalent bonds is irradiated with
■
■
■
■
■
infrared light.
_
Define the term wavenumber v (‘nu bar’), and convert frequencies and wavelengths to
wavenumbers.
_
In general terms, relate the value of v for the stretching frequency of a bond between
two atoms to the strength of the bond and the combined mass of the atoms.
Recognize the absorption peaks of important functional groups in infrared spectra.
Describe how gas, liquid, solution and solid samples are prepared for infrared analysis.
Describe how infrared absorption affects the Earth’s climate.
Introduction
The term ‘infrared’ describes a region of the electromagnetic
spectrum with frequencies lower than visible light. The
infrared region spans the wavelength range 0.7–100 μm
(μm is the abbreviation for micrometre; 1 μm = 10−4 cm
= 10−6 m). Chemists have found it convenient to use
wavenumbers with cm−1 units (reciprocal centimetres),
rather than frequency or wavelength, in this region of the
_
spectrum. The symbol for wavenumber is v (‘nu bar’) and
the conversion of wavelength to wavenumbers is shown
below.
balls of various masses (with the balls representing the
atoms) connected by springs of various strengths (with
the springs representing the chemical bonds). This ball and
spring model is relevant to infrared spectroscopy because
it indicates that there is always some vibration motion
associated with a molecule. Hooke’s law, from physics,
tells us that for two balls held together by one spring,
the vibration frequency, ν, depends on the strength of
the spring and on the masses of the balls. The stronger
the spring the higher the frequency of vibration, and the
greater the masses the lower the frequency.
How organic molecules absorb infrared
radiation
For a molecular vibration to interact with and absorb
infrared radiation, the motion of the atoms must produce
a change in the dipole moment of the molecule. In
simple terms this means that the vibration must cause
the distance or angle between bonded atoms of different
electronegativity to alternately increase and decrease (i.e.
to oscillate). Molecules such as hydrogen (H2), oxygen
(O2) and nitrogen (N2) do not absorb infrared radiation
because there is no change in the dipole as the H–H, O–O
and N–N bond is elongated and compressed. When the
oscillating dipole moment of the molecule interacts with
the electrical vector of infrared radiation (νvibration = νIR)
Infrared spectroscopy is the measurement of the absorption
of infrared radiation by compounds. A molecule is not a
rigid assembly of atoms. A simple analogy is a system of
ITQ 1 Convert the limits of the infrared range (0.7 μm and 100
μm) to wavenumbers.
_
v=
1
104
=
λ (in cm) λ (in μm)
Wavenumbers are directly proportional to energy.
Radiation of high wavenumber is of high frequency. In
the range 2.5–16 μm (4000–625 cm−1), which is known as
the ‘mid-IR’, the frequency of the radiation matches the
vibration frequencies of organic molecules.
Chapter 33 Infrared spectroscopy
the molecule absorbs the radiation. Absorption of this
energy results in increased amplitude of the molecular
vibration (in the excited state). As the molecule reverts to
the ground state the absorbed energy is released as heat;
in older spectrometers this is detected by very sensitive
thermocouples.
More on molecular vibrations
There are basically two types of molecular vibration –
stretching and bending. These are illustrated in Figure 33.1
for a group consisting of three atoms.
Stretching vibrations
symmetric
specify a narrower range depending on which atoms are
attached to the >C=O and generate correlation tables that
tell us where to expect a particular kind of carbonyl group
(ketone, aldehyde, ester, amide, etc.) to absorb.
Interpreting infrared spectra
An infrared spectrum is a chart showing downward
peaks (% transmittance) on the y-axis, plotted against
wavenumber (cm−1) on the x-axis. Peak intensity is
described qualitatively as either strong (s), medium
(m), weak (w) or variable (v). The infrared spectrum of
cyclohexanone is shown in Figure 33.2.
asymmetric
Bending vibrations
In-plane bending vibrations
scissoring
rocking
Out-of-plane bending vibrations
+
wagging
+
+
twisting
–
Transmittance / %
100
50
0
4000
3000
2000
1500
1000
500
Wavenumber / cm –1
Figure 33.1 Stretching and bending vibrations: + and − indicate
vibrations perpendicular to the paper.
Stretching vibrations generally occur at higher frequency
(higher wavenumber) than bending vibrations. In most
organic molecules there are many atoms of various masses
and bonds of various strengths, so the number of vibration
modes is very large.
The frequency of a vibration is directly proportional to the
strength of the bond, so triple bonds will vibrate with higher
frequencies than either double bonds or single bonds.
■ C≡C and C≡N; ν max = 2600–2100 cm−1
■ C=C, C=N, C=O; ν max = 1800–1500 cm−1
■ C–C, C–N, C–O; ν max = 1300–800 cm−1
The vibration frequency of a bond is inversely proportional
to the masses of the atoms which are bonded, so a C–C
stretch will occur at lower frequency than a C–H stretch.
Because vibration frequencies depend on the combined mass
of the two connected atoms and the strength of the bond
joining them, we can assign group frequencies to functional
groups or other definable structural features. For example,
we can expect all carbonyl stretching vibrations to absorb
in a region near 1700 cm−1, regardless of what is attached
to the >C=O group. It is possible to take a further step and
Figure 33.2 An infrared spectrum of cyclohexanone.
An infrared spectrum can contain more than 30 absorption
bands. The vast majority of the bands are not interpreted;
these result from combinations of different stretching and
bending modes in the molecule. In general, the identifiable
group vibrations are found at higher wavenumbers (4000–
1500 cm−1) and the absorptions associated with complex
coupled vibrations are found at lower wavenumbers in
what is called the fingerprint region (1300–600 cm−1). As
in detective stories, the molecular fingerprint is a criterion
of identity.
A survey of some of the absorption bands that we look for
in an infrared spectrum follows.
O–H and N–H (and C–H) stretching
These bonds are found in the 3600–2700 cm−1 region.
Recalling Hooke’s law, we expect these absorptions to be at
the high-frequency end of the spectrum since the combined
mass of the atoms is low.
The C–H bond is not very polar, so the vibration produces
very weak absorption, but organic compounds usually have
many C–H bonds, so their combined absorptions produce a
group of moderately strong peaks near 3000 cm−1.
309
310
Unit 2 Module 2 Analytical methods and separation techniques
The O–H bond absorbs in the same region of the spectrum
as the C–H bond, but it has two important differences.
■ The bond is more polar so the absorption is stronger.
■ The effective strength of the O–H bond is variable.
Alcohols and other compounds containing hydroxyl
groups can form hydrogen bonds between two OH
units, and these may form extensive chains.
O
H
O
H
O
H
In simple terms, if the hydrogen bond (dotted) is stronger,
the corresponding O–H bond (shown as a solid line) is
weaker. In a dilute solution where hydrogen bonding is
minimal, the O–H stretching absorption is a sharp peak
near 3600 cm−1; in solutions where hydrogen bonding
occurs, the absorption moves to about 3300 cm−1 and
becomes broader.
Organic chemists often dissolve solid samples before
measuring their spectra. Spectra of solid samples can be
obtained, but crystal effects can distort spectra. (We shall
return to experimental procedures later.)
Carboxylic acids, R-COOH, are very extensively hydrogen
bonded, and their O–H stretching absorption appears as a
distinctive very broad peak.
N–H stretching vibrations give peaks in the same region,
but they are not as affected by hydrogen bonding.
C≡N and C≡C stretching
These bonds are found in the 2600–2100 cm−1 region.
The stretching vibrations of these triply bonded groups
occur in this region.
The absorption due to C≡C is very weak, as there is little
change in dipole moment when this bond stretches.
a
R
R
4
3
1
R
C
C
2
R
C
O
b
R
R
4
3
c
R
1
C
2
R
C+
1
+
R
C
3
–
O
R
4
R
C
C
2
R
C
–
O
Figure 33.3 Resonance forms of a conjugated unsaturated
ketone; if (a), (b) and (c) contribute equally to the hybrid, the C–O
bond has about 67% single bond character.
A simple ketone, such as (CH3)2C=O, gives rise to a strong
peak near 1700 cm−1.
The peak for the corresponding aldehyde (CH3CHO in this
case) appears close by, less than 15 cm−1 higher.
The stretching frequency is almost entirely the property
of the C=O, and is only very slightly affected by the mass
of the substituents. If, however, we examine a conjugated
unsaturated ketone, we see a shift of about 35 cm−1 to
a lower wavenumber. The movement of the electrons
through conjugation significantly changes the character
of the C=O, making it less double-bonded in character.
Effectively, the ‘spring’ is weaker so the vibration frequency
OCH2CH3
is lower (Figure 33.3).
, the
In an ester such as ethyl ethanoate, CH3 C O
OCH2CH3 substituent has the opposite effect, increasing the
vibration frequency; esters absorb near 1735 cm−1.
We might expect carboxylic acid peaks to appear near the
ester peaks, but carboxylic acids are extensively hydrogen
bonded, affecting the C=O stretching as well as the O–H
stretching, lowering the carbonyl frequency and bringing it
back to about the same place as that of a ketone.
The absorption due to C≡N is of medium intensity because
this bond is more polar than the triple bond of alkynes.
C=C, C=N and N=O stretching
Compounds of the type X=Y=Z, such as CO2 and SO2, also
absorb in this region. Many spectra of compounds which do
not contain triple bonds show a weak band in this region
due to CO2 absorbed in the solvent.
The C=C stretching of unsymmetrically substituted alkenes
gives infrared absorption peaks in the range 1666–1640
cm−1. These peaks are of weak (w) to medium (m) intensity.
C=O stretching
These bonds are found in the 1800–1650 cm−1 region.
The polar carbonyl groups give rise to some of the strongest
peaks in the infrared spectrum because of the large
dipole moment of the carbon–oxygen double bond, and
these peaks often provide structural information of great
importance.
These bonds are found in the 1670–1500 cm−1 region.
Symmetrically substituted alkenes do not absorb infrared
because there is no change in dipole when the C=C
stretches.
Aromatic compounds produce three to four bands of
medium intensity in this region.
The fingerprint region
The fingerprint region is found around 1300 to 600 cm−1.
Chapter 33 Infrared spectroscopy
Gases
Gas samples are introduced into gas cells, 10 cm long, with
NaCl windows (Figure 33.4) and the gas cell is mounted
in the sample beam of the instrument. At atmospheric
pressure the concentration of a gas sample is low, so the
path length of the cell must be long. Gas cells are used for IR
analysis of air samples in the monitoring of environmental
pollution.
Liquids
Figure 33.4 Gas cell for infrared analysis of gas samples.
Many of the absorption peaks in this region of the spectrum
cannot be easily interpreted. There is one important
application of absorption in this region of the spectrum,
which is to prove that two samples are identical. Spectra
of unknown compounds can be overlaid with spectra
of known substances. Perfect matching of all peaks in
the infrared spectra of two compounds is proof that the
compounds are identical.
A liquid sample can be examined as a film between two
circular plates of NaCl. Plates are placed in a plate holder,
which is placed in the IR beam.
Solutions
Solutions (1–5% weight/volume) are prepared using
alcohol-free CHCl3 or CCl4, and the solution is transferred
using a syringe to a solution cell with NaCl windows spaced
between 0.1 and 1 mm apart. The solution cell is then
placed in the IR beam. Absorption due to the solvent can
be subtracted electronically.
Solids
Obtaining infrared spectra
Sample preparation
The container in which a sample is placed for collection of
infrared data must not absorb infrared radiation. Commonly
used IR-transparent materials are NaCl and KBr. Infrared
spectra can be obtained for gases, liquid compounds, solids
or liquids in solution – the usual solvents are chloroform
(CHCl3) and tetrachloromethane (CCl4) – and solids in the
solid state.
ITQ 2
(a) List the functional groups in each of these C5H8O isomers:
O
O
H
H
H
C
C
H
C
C
H
C
CH3CH2
CH3CH2C C C H
H
C
C
H
OH
C
H
H
H H H
i
ii
iii
(b) Match each of the groups of IR absorption bands to structure
i, ii or iii.
3300, 2155 cm−1
1685, 1619 cm−1
1740 cm−1
Solids are dispersed in KBr, as KBr discs, by grinding the
solid with 10–100 times excess KBr using a small mortar and
pestle (which is usually made of agate). The solid mixture
is pelletized with a special mould and hydraulic press to
produce the KBr disc, which is a thin tablet approximately
0.5 cm in diameter. The KBr disc is supported in the IR
beam on a disc holder.
Solids can also be dispersed in Nujol, as Nujol mulls. Nujol
is mineral oil, a hydrocarbon of very high boiling point. A
Nujol mull is a paste which is prepared by grinding about
1 mg of the solid with a few drops of Nujol using a mortar
and pestle.
In the more modern Fourier transform infrared (FT-IR)
spectrometers it is not necessary to prepare samples as
described above. Liquid or solid samples are simply placed
on a crystal of diamond, zinc selenide or germanium.
How data from infrared spectra are used
Infrared spectra provide us with information on the
presence or absence of functional groups in a molecule.
Negative evidence can be just as important as positive
evidence, so the absence of an infrared absorption band
characteristic of a particular functional group can be used
in deducing the structure of a compound. Compounds
with several functional groups will usually show separate
IR absorption bands for each functional group. Infrared
311
312
Unit 2 Module 2 Analytical methods and separation techniques
data is used in conjunction with other data in determining
the structures of compounds. This other data can include
elemental composition, molecular formula, boiling or
melting point and other types of spectra. Apart from when
IR spectra are overlaid (as described on page 311) it is not
possible to determine the structure of a compound solely
by interpreting its infrared spectrum.
Infrared absorption in climate and the
environment
Radiation from the Sun which reaches the Earth’s
atmosphere is approximately 3% UV light, 44% visible light
and 53% infrared light. The visible radiation is absorbed
by the Earth’s surface. This absorption of energy warms
the surface, and the surface then emits lower energy IR
radiation back towards space.
The gases nitrogen, oxygen and argon make up more than
99% of the Earth’s atmosphere. These gases do not absorb
either visible or infrared radiation. However, the small
percentages of carbon dioxide and water vapour present
in the atmosphere absorb the infrared radiation from the
Earth’s surface. This causes heating of the atmosphere, and
also makes the surface of the Earth warmer. This heating is
known as the greenhouse effect. Without the greenhouse
effect, the Earth’s surface would probably be frozen, like
the surface of Mars. The average temperature of the Earth’s
surface is +17 °C; it is estimated that without greenhouse
gases this temperature would be −17 °C.
So-called ‘greenhouse gases’ include carbon dioxide,
methane and nitrous oxide (N2O). Human activity, such
as the burning of fossil fuels (coal, oil), causes the amounts
of these gases in the atmosphere to increase and this, in
turn, leads to further warming of the atmosphere. This
global warming will gradually warm the oceans. It has
been calculated that the heat resulting from increases
in greenhouse gases (excluding water vapour) since
preindustrial times is equivalent to about 1% of the solar
radiation absorbed by the Earth’s surface.
Summary
✓ Covalent bonds vibrate with frequencies which
correspond to frequencies in the infrared
range of the electromagnetic spectrum. When
molecules are irradiated with infrared light (νIR =
νvibration) they absorb energy and the amplitude of
the vibration increases.
✓ Infrared radiation is described in wavenumbers,
with cm−1 units (reciprocal centimetres).
The symbol for wavenumber is v̄ (‘nu bar’).
Wavenumbers are proportional to frequency and
energy.
✓ Vibrational frequency is directly proportional
to the strength of a bond and inversely
proportional to the combined mass of the
bonded atoms. Functional groups, therefore,
absorb IR radiation within characteristic ranges.
✓ Important infrared absorption ranges are:
3600–2700 cm−1 for O–H and N–H (and C–H)
stretching;
2600–2100 cm−1 for C≡N and C≡C stretching;
1800–1650 cm−1 for >C=O stretching;
1670–1500 cm−1 for C=C, C=N and N=O stretching;
1300–600 cm−1, the fingerprint region.
✓ Infrared spectra can be determined for gases,
liquid compounds, solutions and solids.
✓ The greenhouse effect which warms the Earth’s
atmosphere and surface is the result of the
absorption of infrared radiation by carbon
dioxide and water vapour.
Chapter 33 Infrared spectroscopy
Review questions
1
Three compounds and three wavenumbers are shown in each of the following
parts. In each case, identify a wavenumber that describes a significant peak in
the infrared spectrum of each compound.
(a)
CH3
i
N
ii
H
O
iii
CH3CH2CH2
C
C
N
CH3
CH3CH2
CH2CH3
1712 cm –1 (strong)
3400 cm –1 (medium)
2250 cm –1 (weak to medium)
(b)
i
ii
CH3CH2CH2
C
C
CH3
H
H
C
H
C
H
H
CH3
1600 cm –1 (medium)
H
iii
3350 cm –1 (strong)
H
H
OH
C
C
H
C
H
H
H
C
C
2120 cm –1 (weak)
2
Three unlabelled bottles, each containing a different household chemical, are
discovered. The chemicals are thought to be nail polish remover, kerosene oil
and rubbing alcohol. Using infrared spectroscopy how would you determine
which is which? Nail polish remover consists mainly of the ketone propanone
(acetone); kerosene oil is a petroleum fraction consisting of C6–C16 hydrocarbons;
rubbing alcohol can be either ethanol or propan-2-ol (isopropanol).
3
Describe the features in the infrared spectra that you would use to distinguish
between the pairs of isomers shown below. The presence of certain absorption
peaks would be important, but note that the absence of particular peaks can also
provide useful information.
(a)
i
H
H
H
(b)
H
ii
H
C
OH
C
C
C
H
C
H
H
H
i
C
CH3CH2
ii
CH3 O
C
i
CH2CH3
O
H
C
C
H
(c)
O
CH3
H
CH3
ii
O
O
C
(d)
C
OCH2CH3
CH3
i
CH3CH2CH2
OH
Answers to ITQs
ii
H
CH3CH2CH2
C
N
H
C
C
CH2CH2N
H
(e)
i
O
ii
CH3
C
C
CH3
OH
C
C
H
CH3CH2
H
C
H
C
H
1
14 286 cm−1, 100 cm−1
2
(a) i >C=O, C=C
ii >C=O
iii O–H, C≡C
(b) iii = 3300, 2155 cm−1
i = 1685, 1619 cm−1
ii = 1740 cm−1
313
314
Chapter 34
Mass spectrometry
Learning objectives
■ Explain how atoms and molecules are made to form ions and ion radicals which can
■
■
■
■
■
■
be detected by mass spectrometry.
Define the terms base peak, fragment ion, ion radical, mass : charge ratio, molecular
ion and relative abundance.
Calculate the relative atomic mass of an element from its mass spectrum.
Describe in outline how a mass spectrum is obtained.
Deduce the number of carbon atoms in a compound from the relative abundances of
the M+• and the M + 1 peaks.
Recognize the presence of bromine and chlorine atoms in a compound from the
relative abundances of the M+• and M + 2 peaks.
Identify simple compounds from their mass spectral fragmentation patterns.
Introduction
The unit on the x-axis is atomic mass units divided by the
charge on the ion (z), which is usually +1. The unit on the
y-axis of the spectrum is percentage relative abundance.
The strongest peak (the base peak) is assigned the value
of 100%. The percentages of all the other ions are quoted
relative to that base peak. Detection depends on the
particle carrying an electric charge. Neutral particles are
not detected.
100
Relative abundance / %
In UV–visible spectroscopy, absorption of radiation in the
UV and visible regions of the electromagnetic spectrum
causes electrons in atoms and molecules to move to
higher energy levels. In infrared spectroscopy, absorption
of infrared radiation results in changes in the vibrational
energy of molecules. A mass spectrum is produced by
exciting an atom or molecule in the gas phase with enough
energy to cause it to ionize. The gaseous ions formed are
usually positively charged (+1). The mass : charge ratio
(m/z) and relative abundance, expressed as a percentage,
of each of the gaseous ions are displayed on the horizontal
and vertical axes of the mass spectrum. Figure 34.1 gives a
simple example.
50
0
0
2
4
6
8
10
12
m/z
Figure 34.1 The mass spectrum of natural boron.
Mass spectra of atoms
The mass spectrum of a pure sample of an element
which has more than one stable isotope will show peaks
of mass : charge ratio corresponding to the mass of each
isotope. The most abundant isotope will give the highest
peak which, by convention, is assigned a relative abundance
of 100%. This peak is called the base peak.
Chapter 34 Mass spectrometry
–e–
Worked example 34.1
O
Figure 34.1 shows that there are two stable isotopes of boron.
They are 10B and 11B. The more abundant isotope of boron is
11
B, so the relative abundance of the 11B+ peak is taken as
100%. The abundance of the 10B+ ion is 23.0%. Find the relative
atomic mass of natural boron..
Q
1 Add the relative abundances from the mass spectrum:
100 + 23 = 123
2 Take the relative abundance of the ion from each isotope
and express it as a percentage of the total:
100
% of 11B =
× 100% = 81.3%
123
A
23
× 100% = 18.7%
123
The relative atomic mass of an element is a weighted average
of the relative abundances of its stable isotopes. So the relative
atomic mass of a naturally occurring sample of boron is
81.3
18.7
× 11 +
× 10 = 8.94 + 1.87
100
100
= 10.8 (to 3 significant figures)
% of 10B =
C
O
O
C
O+
FW = 44
m/z 44
molecular CO2
16 valence shell electrons
molecular ion derived from CO2
15 valence shell electrons
Figure 34.2 Removal of an electron from carbon dioxide.
A simple example of the mass spectrum of a molecule is
the mass spectrum of carbon dioxide. The most easily lost
electron would be one of the lone-pair electrons on oxygen
(Figure 34.2).
Some of the molecular ions, CO2+•, can fragment to give an
oxygen ion at m/z = 16 and carbon monoxide (CO). The
carbon monoxide has no charge and so is not detected.
Other CO2+• molecular ions can fragment to a carbon
monoxide ion CO+ (m/z = 28) and an oxygen atom. The
oxygen atom has no charge and so is not detected.
These fragmentations and the mass spectrum of carbon
dioxide are shown in Figure 34.3.
Figure 34.3 shows three peaks in the mass spectrum of
carbon dioxide.
Mass spectra of molecules
■ m/z = 44, which is the molecular ion, CO2+•
When an electron is removed from a covalent molecule
(M) in the gas phase the positively charged gaseous ion
formed is known as the molecular ion, M+•; another term
for M+• is the parent ion. This process is
■ m/z = 16, which is the oxygen ion, O+
M
→
M
+
e
molecule in the gas phase
molecular ion
an electron
+•
−
The molecular ion, M+, is the molecule minus an electron;
M+ is a radical cation (sometimes called a radical ion)
containing one unpaired electron. Because M+• is lacking
an electron in its molecular framework, it is very unstable.
The unstable M+• falls apart (undergoes fragmentation)
to form ions of lower mass (and hence lower m/z) called
fragment ions. Also formed are neutral radicals or neutral
molecules. In some text books, fragment ions are called
daughter ions. The m/z values of the fragment ions can
provide information about the structure of M+•.
Relative abundance / %
■ m/z = 28, which is the carbon monoxide ion, CO+
100
80
60
40
20
0
15
20
25
30
35
40
45
50
m/z
+
O
m/z = 16
+
C
O
O
C
O+
m/z = 44
molecular ion
(15 valence shell electrons)
C
O+ +
O
m/z = 28
Figure 34.3 Fragmentation and mass spectrum of carbon dioxide
(ignoring the tiny contributions from 13C).
ITQ 1 The mass spectrum of the element silicon (Si) is shown below, with a listing of the m/z and relative abundance of each ion.
Relative abundance / %
100
75
50
25
0
0
10
20
30
m/z Relative abundance
(a) Identify the base peak in this mass spectrum.
28
100
29
5.1
(b) Calculate the percentage of each isotope in a natural
sample of silicon.
30
3.4
(c) Calculate the relative atomic mass of silicon to three
significant figures.
315
316
Unit 2 Module 2 Analytical methods and separation techniques
neutral molecules, neutral
fragments and negative ions
electron beam
to vacuum pump
filament
sample
analyser tube
magnet
magnet
neutral molecules
ion exit slit
positively charged
repeller plate
negatively charged
accelerating and
focusing plates
positively charged
ions (deflected
according to m/z)
recorder
collector
Figure 34.4 Schematic diagram
of a typical commercial mass
spectrometer.
How mass spectra are obtained
Mass spectra are determined using mass spectrometers,
which are complex instruments that operate under high
vacuum and require trained operators. A schematic
diagram of a typical commercial spectrometer is shown in
Figure 34.4 and the components are described below.
■ The inlet system is a metal chamber into which a
very small amount of a sample can be introduced and
vaporized under high vacuum at room temperature
or slightly higher. Samples must be vaporized so that
the molecules are separated from each other before
ionization.
■ The ion source is a source of energy which causes
the sample molecules to undergo ionization and
fragmentation. The most common method of
ionization is bombardment of the vaporized sample
with high-energy electrons. This is known as electron
impact ionization.
■ An exit slit: the positive ions produced in the ion
source are made to pass through this opening by
application of a repelling (positive) potential.
■ Acceleration zone: here a very strong negative electric
potential (6000–8000 V) is applied. The ions are both
accelerated and focused as they pass through.
■ Separator: the application of a combination of
magnetic and electrical fields causes ions to follow a
curved path. The important equation is
m/z =
B2r2
2V
where B is the magnetic field strength, V is the electrical
field strength and r is the radius of the curved path
followed. The value of B is usually fixed, so V must be
varied so that ions of different m/z can follow a path of
the same radius r. For a given B and V, ions of only one
m/z will follow the curved path of radius r and reach the
collector; the other ions will strike the sides of the tube.
■ Collector and detector: positive ions that strike a
collector plate produce a flow of neutralizing electrons
which is proportional to the ion abundance. This
current is amplified and measured very accurately.
Measurements are recorded and displayed electronically.
■ A computer attached to the spectrometer processes
the data from the detector and presents a calibrated
spectrum: the x-axis shows m/z values and the y-axis
shows ion intensity (a measure of the number of
ions produced). The units on the vertical axis of the
spectrum are percentages, with the strongest peak (the
base peak) assigned the value of 100%.
Applications of mass spectrometry
Deducing the number of carbon atoms in an
organic compound
The most abundant isotope of carbon is 12C (98.9%) and
the natural abundance of 13C is 1.1%. We can say that
there is a 1.1% probability of each carbon atom being 13C.
H
H
Benzene (C6H6, FW = 78)
compound.
H
H
C
C
C
C
C
C
is a very stable
H
H
When benzene is analysed by mass spectrometry the
molecular ion C6H6+• (M+• ) is also very stable. In the mass
spectrum of benzene the molecular ion is the base peak,
i.e. the peak of 100% relative abundance. Since there are
six carbon atoms in benzene, each with a 1.1% probability
of being 13C, the relative abundance of the M + 1 peak is
Chapter 34 Mass spectrometry
6 × 1.1% as high as the M+• peak. The actual mass spectrum
of benzene (Figure 34.5) shows this prediction to be true: the
relative abundance of the M + 1 peak is approximately 6.6%.
Relative abundance / %
100
80
The number of carbon atoms in a molecule can therefore
be deduced in this way.
60
number of carbon atoms =
40
relative abundance of the M + 1 peak
100
×
+•
relative abundance of the M peak
1.1
20
0
10
15
20
25
30
35
40
45
50
55
60
65
70
75
80
m/z
Figure 34.5 Mass spectrum of benzene. The peak at m/z = 79 is
about 6.6% the height of the peak at m/z = 78.
43
CH3CHCH3
80
Relative abundance / %
There are some elements with significant amounts of two
or more isotopes. These include:
■ sulfur:
100
60
32
S 95%; 34S 4.2%
■ chlorine:
35
■ bromine:
79
■ silicon:
Cl
Cl 75.5%; 37Cl 24.5%
Br 50.5%; 81Br 49.5%
28
Si 92.2%; 29Si 4.7%; 30Si 3.1%
The presence of these elements in compounds can be
readily recognized by the distinctive mass spectral patterns
that they produce.
40
63
41
20
78
65
Chlorine
80
0
0
10
20
40
30
50
60
70
80
90
100
m/z
Figure 34.6 The mass spectrum of 2-chloropropane, showing the
molecular ion, M+•, at m/z = 78 (CH3CH35ClCH3)+• and an ion two
mass units greater at m/z = 80 (CH3CH37ClCH3)+• that has one
third the intensity.
100
43
CH3CH2CH2Br
80
Relative abundance / %
Recognizing the presence of bromine and
chlorine atoms in an organic compound
Chlorine consists of the 35Cl and 37Cl isotopes in a ratio of
about 3:1. If one chlorine atom is present in a compound,
the molecular ion, M+•, is accompanied by an ion two mass
units greater (M + 2)+•, but one third the intensity.
In the mass spectrum of 2-chloropropane, shown in Figure
34.6, the relative abundance of the molecular ion at
m/z = 78 is about 22% while that of the (M + 2)+• ion at
m/z = 80 is 7.5%, which is about one third the intensity of
the M+• ion.
Bromine
60
27
40
41
122 124
20
15
0
10
20
30
40
50
60
70
80
90
100
110
120
130
m/z
Figure 34.7 The mass spectrum of 1-bromopropane, showing
the molecular ion, M+•, at m/z = 122 (CH3CH2CH279Br)+• and an ion
two mass units greater at m/z = 124 (CH3CH2CH281Br)+• of almost
equal intensity.
ITQ 2 How many atoms of 13C does one mole of elemental
carbon contain?
Avogadro’s number is 6.0221 × 1023. The percentage abundance
of 13C is 1.1%.
Bromine also has two isotopes, 79Br and 81Br, in almost equal
abundance. The 79Br isotope is very slightly more abundant
and is therefore considered to form the molecular ion. If one
Br is present in a compound the mass spectrum has M+• and
(M + 2)+• of almost equal intensity. In the mass spectrum of
1-bromopropane (Figure 34.7), the molecular ion at m/z =
122 (CH3CH2CH279Br)+• is accompanied by a peak two mass
units higher at m/z = 124 which is due to (CH3CH2CH281Br)+•.
ITQ 3 Data from the mass spectrum of a hydrocarbon are given
below.
■ M+•
m/z = 86; relative abundance 30%
■ M + 1 m/z = 87; relative abundance 2%
(a) How many carbon atoms are there in each molecule of the
hydrocarbon?
(b) What is the molecular formula of the hydrocarbon?
317
Unit 2 Module 2 Analytical methods and separation techniques
Determining the elemental composition of
molecular and fragment ions
One advance in the design of mass spectrometers was
especially important for interpreting the spectra of organic
molecules. Double focusing mass spectrometers use a
combination of magnetic and electrostatic fields to produce
high-resolution mass spectra. Older instruments could
resolve ions differing by one mass unit. For example, they
could tell the difference between m/z = 85 and m/z = 86.
However, there are many possibilities for the elemental
composition of a peak of any given integral mass. For
example, a peak at m/z = 86 could be C6H14, C5H10O2, C3H6N2O
or C4H6O2. Using a high-resolution mass spectrometer the
exact masses of each of these peaks can be determined.
Formula Exact mass
C6H14
86.1092
C5H10O
86.0729
C3H6N2O 86.0479
C4H6O2
86.0366
The non-integral values arise because only 12C has an
integral mass (exactly 12.0000 by definition). The mass
of any other atom is slightly different from integral. The
values for all important isotopes have been determined
with great precision, so it is possible to determine the
elemental composition of molecular ions or fragment ions
by exact-mass measurement. Molecular formulae and
structures of quite complex new compounds have been
determined entirely on the basis of their high-resolution
mass spectra.
■
12
■
1
■
16
■
14
C = 12.0000 (by definition)
H = 1.0078
O = 15.9949
N = 14.0031
+
H
H
H
H
C
C
C
H
H
H
H
H
In alkanes, all bonds are σ bonds and there are no lone-pair
electrons, so the electron removed to form the molecular
ion comes from a σ sigma bond (Figure 34.8).
ITQ 4 Suggest a molecular formula
for the compound which gives the
following mass spectral data:
m/z
Relative abundance
94
100
95
1.1
96
96
97
1.1
H
C
C
C
H
H
H
H
+
e–
The molecular ion from an alkane undergoes fragmentation
by breaking of a C–C bond and the formation of two alkyl
fragments, one a cation and the other a radical. C–H bonds
do not break readily because both hydrogen radicals (H•) and
hydrogen ions (H+) are high-energy species in the gas phase.
Alkanes with long unbranched chains give spectra with a
weak molecular ion (CnH2n+2)+• and a series of peaks starting
with M − 15 (i.e. M+• − CH3) separated by 14 mass units.
Each of these peaks results from an ion formed by breakage
of a C–C bond in the chain. The mass spectrum for pentane is
shown in Figure 34.9. The molecular ion, M+•, gives a peak at
m/z = 72, which corresponds to CH3CH2CH2CH2CH3⎤+ •. This
ion loses a methyl radical, •CH3, to produce a cation at m/z =
57, which corresponds to CH3CH2CH2CH2⎤+. The molecular
ion can also lose an ethyl radical (•CH2CH3) to produce a
base peak
100
43 (M - 29 )
CH3CH2CH2CH2CH3
80
60
27
40
molecular ion (M+ )
29
(M - 15)
57
20
72
0
0
20
40
60
m/z
m/z
Alkanes
H
Figure 34.8 Removal of an electron from an alkane; the example
shown here is propane.
15
Fragmentation pathways of common classes
of organic compounds
H
M+ derived from
propane is missing
an electron from
a C C bond
propane
Relative abundance / %
318
73
72
71
57
43
42
41
39
29
28
27
15
14
relative abundance
0.52
18.56
4.32
11.20
100.00
55.27
37.93
12.44
26.65
17.75
31.22
4.22
2.56
Figure 34.9 The mass spectrum of pentane.
80
100
Chapter 34 Mass spectrometry
propyl cation, CH3CH2CH2⎤+, at m/z = 43. The ion at m/z =
29 is produced by loss of a propyl radical (•CH2CH2CH3) from
the molecular ion, and the ion at m/z = 15 is formed when
the molecular ion loses a butyl radical (•CH2CH2CH2CH3).
Alkanes with branched chains give mass spectra which
differ only slightly from those of their unbranched isomers.
Breakage at the point of branching leads to secondary
carbocations that are energetically slightly favoured.
Alkenes
The electrons in the π bond of an alkene are at a higher
energy level than those in a σ bond, so the molecular ion is
formed by ejection of a π electron (Figure 34.10).
this bond cleaves subsequently
H
H
H
H
C
C R’
H
H
Derivatives of benzene which have alkyl groups attached
(alkylbenzenes) form the benzyl cation, C6H5CH2+ which
has m/z = 91. The benzyl cation is very stable because the
positive charge can be delocalized through resonance into
the aromatic ring.
Functional groups
Fragmentation patterns of compounds containing common
functional groups are summarized in Table 34.1.
C
C
R
C
_
H
C
`
H
C R’
+
e–
ITQ 5 Below are two mass spectra. One is obtained from
2-pentanone (A) and the other from 3-pentanone (B). Use the
information given in Table 34.1 on fragmentation of ketones to
determine which spectrum corresponds to 2-pentanone and
which corresponds to 3-pentanone.
H
M + derived from the
alkene is missing an
electron from the C C
alkene
Figure 34.10 Removal of an electron from an alkene.
a
The molecular ion usually fragments by cleavage of the
C–C single bond β to the C=C. However, in acyclic systems
the double bond in the molecular ion formed from an
alkene tends to migrate, making the fragmentation pattern
unpredictable.
Table 34.1 Fragmentation patterns of compounds with common
functional groups
Class of
compound
Alkyl halides,
ethers, amines
Molecular
ion
R
1
+
2
R
C
R
Main fragment
ion
R
3
⎯⎯→
1
R
Radical or
molecule
+
2
C
R
3
+
X•
H
C
H
O
C
H
H
b
H
C
C
C
H
H
H
H
H
H
H
C
H
2-pentanone
H
C
O
C
H
H
H
C
C
H
H
3-pentanone
100
Relative abundance / %
C
R
The spectra of aromatic compounds such as benzene and
naphthalene show the molecular ion as the base peak; the
molecular is stable and has no low-energy fragmentation
pathways.
+
H
H
Benzene and alkylbenzenes
80
60
40
20
X
X = halogen, OR, NR2
R
0
+
2
R
C
R
3
⎯⎯→ C
OH
Ketones
1
R
2
R
C
⎯⎯→
R
3
+
2
C
+
acylium ion
Cyclohexene
derivatives
+
+
⎯⎯→
30
40
50
60
70
80
60
70
80
100
R1•
+
O
O
20
m/z
OH
+
R
10
+
2
R1•
Relative abundance / %
Alcohols, ethers
(replace OH
1
with OR)
R
80
60
40
20
+
0
10
20
30
40
50
m/z
H
319
Unit 2 Module 2 Analytical methods and separation techniques
Summary
Review questions
1
✓ A mass spectrum of an atom or molecule is
produced by causing a gaseous sample to lose an
electron to form an ion-radical:
M → M+• + e−
Classify each of the following species as an atom, a
radical cation, a radical, a molecule or an ion. Say
whether or not it will give a signal in a mass spectrum.
(b) H
(c)
(a) H
H
H
H
+
C
sometimes be deduced by using the ratio of the
intensities of the M + 1 and M+• peaks in the mass
spectrum.
✓ In compounds which contain one chlorine
atom the ratio of the M+• to the M + 2 peak is
3 : 1 because the ratio of 35Cl to 37Cl is 3 : 1. For
compounds with one bromine atom M+• : M + 2 is
about 1 : 1 because 79Br and 81Br occur in almost
equal abundance.
✓ High-resolution mass spectrometry is used
to determine the elemental composition of
molecular and fragment ions by exact-mass
measurement; this is based on the fact that the
masses of all atoms other than carbon-12 are not
integral.
✓ Compounds within a given class of compounds
often give predictable fragmentation patterns
which can be used in deducing structure.
H
C
+
O
C
H
(e) C
O
(g) O
(j)
H
Br+
(f)
C
H
C
H
H
2
O
+
O
(h) Br
+
(i)
✓ Removal of an electron from an organic molecule
✓ The number of carbon atoms in a molecule can
N
H
H
(d) C
✓ A mass spectrum is a plot of mass : charge ratio
M produces an ion radical M+•, the molecular
ion. M+• undergoes fragmentation to form ions
of lower m/z (fragment ions) which can provide
information about the structure of M+•.
C
H
the molecule with high-energy electrons
(electron impact ionization).
(m/z) on the x-axis versus relative abundance on
the y-axis. The unit on the x-axis is atomic mass
units and charge is usually +1. The units on the
vertical axis of the spectrum are percentages; the
strongest peak (base peak) is assigned the value
of 100%. The percentages of all the other ions are
quoted relative to the base peak.
C
H
✓ Ionization is often accomplished by bombarding
(k) H
H
+C
C
H
H
H
H
H
The mass spectrum of 2-methylbutane (C5H12, RMM
72) is shown below. 2-Methylbutane is an isomer of
pentane; the mass spectrum of pentane is given in
Figure 34.9 and is discussed in the text.
100
43
CH3CHCH2CH3
80
Relative abundance / %
320
CH3
57
60
29
40
27
20
72
15
0
0
10
20
30
40
50
60
70
80
90
100
m/z
(a) Suggest structures for the ions which give signals
at m/z = 57, m/z = 43, m/z = 29 and m/z = 15 in the
mass spectrum of 2-methylbutane.
(b) Why is the peak at m/z = 57 in the spectrum of
2-methylbutane much stronger than it is in the
spectrum of pentane (Figure 34.9)?
Chapter 34 Mass spectrometry
3
Which of the isomers of C9H10O, C or D, gives the mass
spectrum shown below? Use the information in the text
on fragmentation of aromatic compounds and in Table
34.1 on fragmentation of ketones in your deduction.
O
H
H
C
C
C
H
C
(b) 28Si =
O
Relative abundance / %
5.1
× 100% = 4.7%
108.5
30
3.4
× 100% = 3.1%
108.5
H
Si =
D
100
× 100% = 92.2%
108.5
29
Si =
H
(c) Relative atomic mass of silicon
= (0.922 × 28) + (0.047 × 29) + (0.031 × 30)
= 28.1
100
80
60
2
6.6243 × 1021
3
(a) 6
(b) C6H14
4
CH3Br
5
The first spectrum is that of 3-pentanone. The base
peak at m/z = 57 is due to the acylium ion E. The peak
at m/z = 29 corresponds to the ethyl cation F.
40
20
0
25
50
75
100
125
m/z
4
(a) The base peak is m/z = 28.
C
H
H
H
1
H
C
C
H
Answers to ITQs
Refer to the mass spectrum of 2-chloropropane
(Figure 34.6).
(a) Suggest a structure for the peak at m/z = 43.
(b) The peak at m/z = 65 is approximately one-third
the height of the peak at m/z = 63.
(i) What do the relative heights of these peaks
indicate about the elemental composition of
the ions?
(ii) Suggest structures for the ions which give
these two peaks.
E
H
F
H
H
H
C
C
H
H
+
H
C
C
H
H
C
O
H
(C3H5O m/z = 57)
+
(C2H5 m/z = 29)
The second spectrum is that of 2-pentanone. The base
peak at m/z = 43 is due mainly to the acylium ion G;
it is possible that the propyl cation H (m/z = 43) also
contributes to this signal.
G
H
H
H
H
H
C
C
C
H
H
H
+
H
C
C
O
H
(C2H3O m/z = 43)
H
(C3H7 m/z = 43)
+
321
322
Chapter 35
Phase separations
Learning objectives
■ Discuss the chemical principles upon which simple distillation and fractional distillation are based.
■ Discuss the advantages of carrying out distillation processes under reduced pressures.
■ Discuss the chemical principles and use of steam distillation.
■ Discuss the principles upon which solvent extraction is based.
■ Select appropriate methods of separation, given the physical and chemical properties of the
components of a mixture.
■ Perform distillation experiments.
■ Carry out simple separation experiments based on solute partitioning between two immiscible solvents.
■ Cite examples of the applications of the distillation methods used in various industries.
Introduction
The word ‘phase’ is synonymous with ‘state of matter’.
Matter may be defined as the substance that makes up the
physical world. The three fundamental states of matter
are solid, liquid and gas. Hence, states of matter are the
characteristic forms that are adopted by the different phases
of matter. The concept of phase separation is associated
with how different types of mixtures can be separated;
most of these separation techniques involve a change of
state, i.e. melting, subliming, evaporating or dissolving.
■ solvent extraction
Mixtures are impure substances and often we need to
separate a mixture in order to obtain its pure components.
The act of separating mixtures takes place all the time.
■ liquid solutions of substances with widely differing
■ In the home, water is separated from pasta with the
use of a colander; tap water is filtered to remove
impurities.
■ In the laboratory, chemists prepare compounds and
then need to purify them.
■ In the petroleum industry, the components of crude oil
are separated and then selectively purified.
A variety of separation techniques exist and we shall
discuss the following during this chapter:
■ simple distillation
■ fractional distillation
■ vacuum distillation
■ steam distillation
■ separating funnel use.
The separation technique chosen depends on the differences
in the physical and chemical properties of the components
of the mixtures.
Simple distillation
Simple distillation is used to separate:
boiling points (more than 25 °C difference) – in
this instance, both components of the mixture are
obtained;
■ liquids from involatile solids or oils – in this case, the
solvent is collected rather than the solute.
Distillation is based on the differences in volatilities of the
components of the mixture.
During simple distillation, when the solution in the flask
boils, pure vapour rises upwards and enters the inner tube
of the Liebig condenser. The outer jacket of the condenser
contains cool, running water that cools the pure vapours in
the inner tube and allows it to condense. This condensed
liquid, called the distillate, is collected at the end of the
Liebig condenser. Figure 35.1 shows the usual laboratory
apparatus for simple distillation.
Chapter 35 Phase separations
thermometer
thermometer
adaptor for thermometer
clamp
stand
rubber tubing
rubber tubing
adaptor for
thermometer
condenser
still
head
condenser
adaptor
roundbottomed
flask
boiling
chip
salt solution
wire
gauze
heat
fractionating
column with
short lengths of
glass rod inside
cold
water out
cold
water out
cold
water in
cold
water in
conical flask
round-bottomed
flask
tripod
Figure 35.1 Distillation apparatus.
conical flask
wire
gauze
liquid mixture
distillate
Fractional distillation
Fractional distillation is used to separate miscible liquids
where the components of the mixture have boiling points
that are close together (usually less than 25 °C from each
other) or where the components are chemically similar
(such as ethanol and water). Fractional distillation is similar
to simple distillation except that a fractionating column
is added between the flask and the condenser (Figure 35.2).
A fractionating column is a vertical tube packed with inert
fragments that provide a large cool surface for the vapours
to cool, condense and vaporize again. Glass beads are often
used in a laboratory column.
Once in equilibrium the column is hotter at the base than
at the top, so there is a temperature gradient along the
column.
When the mixture boils, vapours travel up the fractionating
column and condense. At any point in the column the
vapour, and hence the condensing liquid, is richer in the
more volatile component than was the case lower down.
Most of the condensing liquid runs back down the column.
You can imagine a column with a series of ‘condensing
traps’ inside it. At each trap the condensing liquid contains
less of the more volatile component, because some of it
has passed further up the column. If the column is long
enough the vapour at the top comprises only the most
volatile component of the mixture.
Fractional distillation is often used in the petroleum refinery.
After the crude oil has been extracted, it is transported to
the refinery for separation into fractions using fractional
distillation.
tripod
heat
heat-proof
mat
support
Figure 35.2 Fractional distillation.
Raoult’s law
A problem when dealing with distillation of liquid mixtures
is that sometimes the components do not interact with each
other in any way, but sometimes the various molecules
interact (but without chemical change). This is especially
noticeable when the molecules are chemically similar
in some way. Francois-Marie Raoult, a French chemist
(1830–1901), investigated the behaviour of liquid mixtures
when heated. He found that, provided the intermolecular
forces between unlike molecules are no different from
those between like molecules, the total vapour pressure of
the mixture at any temperature is equal to the sum of the
vapour pressures of each component at that temperature,
each one multiplied by its mole fraction in the mixture.
For a pure liquid in a closed container, at any given
temperature, liquid particles continually enter the vapour
phase because they travel to the surface with enough
kinetic energy to escape from the attractive forces of
other molecules in that surface. Likewise, vapour particles
heading for the surface are recaptured by it and re-enter
the liquid. Equilibrium is reached when the concentration
ITQ 1 During simple distillation, water to cool the vapour enters
the condenser from the end closer to the receiver, and leaves
from the end close to the distilling flask. Why is this specific
connection advisable and not the opposite?
323
324
Unit 2 Module 2 Analytical methods and separation techniques
vapour particle
PA,pure
P
Figure 35.4 The relationship
between vapour pressure and
mole fraction for component A in
a mixture.
vapour
closed container
0
xA
1
PA,pure
liquid particles
liquid
Figure 35.3 The vapour pressure of a liquid.
of molecules in the vapour is high enough that the rate at
which they re-enter the liquid is equal to the rate at which
they are replenished from the liquid. At this stage, the pure
liquid has established an equilibrium with its vapour:
liquid ҡ vapour
The equilibrium vapour is said to be saturated and the
pressure is the saturated vapour pressure of the liquid
at that temperature (Figure 35.3).
If two miscible liquids (A and B) are mixed and have
reached equilibrium, the vapour pressure of the mixture
obeys Raoult’s law.
If the mixture contains a moles of substance A and b moles
of substance B, then there are (a + b) moles altogether.
The mole fraction of A is
a
b
and that of B is
a+b
a+b
If the saturated vapour pressure of A alone is pA and that of
B is pB, the saturated vapour pressure of A in the mixture
a
b
and that of B is PB
.
(P) is PA
a+b
a+b
nA
xA =
nA + nB
where nA and nB are the number of moles of A and B
respectively in the mixture.
n
From the expression xA = n +A n , it can be seen that
A
B
when xA = 1, liquid A is pure and the vapour pressure is
entirely due to liquid A. For gases, pressure p is proportional
to the number of moles n, thus the mole fraction can be
expressed in terms of pressure:
xA =
PA
PA + PB
If a graph of the vapour pressure of each liquid component
is plotted against its mole fraction, a straight line passing
through the origin would result (Figure 35.4).
From Figure 35.4, we can see that the right-hand side of
the graph intersects y-axis at the point that shows pure
PB,pure
0
1
xA
xB
1
0
Figure 35.5 Vapour pressure
composition graph of A and B.
vapour pressure. This re-establishes the point that was
made earlier; when xA = 1, liquid A is pure and the vapour
pressure is entirely due to liquid A.
Let us now consider the vapour pressure composition graph
using the previous example of a solution of two liquids, A
and B, that have reached equilibrium; this graph would
resemble that in Figure 35.5.
Here are some points to note from Figure 35.5.
■ The plots are on the same axes.
■ The mole fractions xA and xB run opposite each other.
Thus, as the mole fraction of A increases, that of B
decreases.
■ The blue and red lines show the partial pressures of the
liquids A and B respectively.
■ In this example, the vapour pressure of pure A is
higher than that of pure B. Therefore, the molecules
escape more easily from the surface of A than of B; A
is the more volatile liquid.
■ The blue and red lines combine to give the purple line
which represents the total vapour pressure of liquids A
and B. Thus, the total vapour pressure of the mixture
(ptotal) is equal to the sum of the individual partial
vapour pressures, i.e.
ptotal = pA + pB
The solution of liquids A and B in this example obeys Raoult’s
law and is described as an ideal solution. However, ideal
solutions are rare and this may be explained on the basis of
intermolecular attractions. In two separate pure liquids (A
and B), some of the more energetic liquid molecules have
enough energy to overcome the intermolecular attractions,
leave from the surface and enter the vapour phase. When
these two pure liquids are mixed, the tendency of the two
different sorts of molecules to escape is unchanged. Therefore,
this solution can only be deemed ideal if the intermolecular
forces between two liquid A molecules (which we will call
A–A) are identical to the intermolecular forces between a
Chapter 35 Phase separations
vapour
pressure
vapour
pressure
vp of
pure A
actual vapour
pressure
ideal vapour
pressure
vp of
pure B
A–A, B–B and A–B are quite different. As such, the
tendency for the molecules to escape is not the same in the
mixture as it is in the pure liquid. The dissimilar nature of
A and B results in the solution deviating from Raoult’s law
of ideality; such solutions are called non-ideal solutions.
Countless pairs of liquids deviate from Raoult’s law; they
either show a positive or negative deviation.
Positive deviation from Raoult’s law
0A
1.0 B
mole fractions
1.0 A
0B
Figure 35.6 Vapour pressure composition curve of non-ideal
solutions.
liquid A molecule and a liquid B molecule (A–B). Similarly,
the intermolecular forces between two liquid B molecules
(B–B), must be identical to the intermolecular forces
between A–B for ideality. Thus, for an ideal solution, the
intermolecular forces and interaction energies between
A–A, B–B and A–B must be equal so that there is very small
energy change when the substances are mixed. However, no
two solutions are identical and therefore a true ideal solution
does not exist. Some solutions get close to ideal behaviour if
their molecules are almost identical chemically. Examples of
solutions which get fairly close to being ideal include:
If the vapour pressure of a mixture is higher than you
would expect from an ideal mixture, there is said to be a
positive deviation. This is evidence that the molecules
are escaping more easily than they do in the pure liquids,
sometimes owing to dissimilarities of polarity. In such an
instance, the liquid pair expands on mixing and absorbs
heat which causes strong intermolecular attractions within
one or both of the components to disrupt. As a result, the
A–B intermolecular forces become weaker than the A–A
or B–B forces. Since 
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