Chemistry For CAPE®️ Examinations

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, stored in a retrieval system, transmitted in any form, or by any means, electronic, mechanical, photocopying, recording, or otherwise, 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. These materials may contain links for third party websites. We have no control over, and are not responsible for, the contents of such third party websites. Please use care when accessing them. Although we have tried to trace and contact copyright holders before publication, in some cases this has not been possible. If contacted we will be pleased to rectify any errors or omissions at the earliest opportunity. 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 50 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 51 52 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 60 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. 61 62 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. 63 64 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 66 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 75 76 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 78 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 79 80 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 81 82 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 277 278 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 × 291 292 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 295 296 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? 297 298 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? 302 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. 303 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. 305 306 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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