(Oxford IB Diploma Programme) Sergey Bylikin, Gary Horner, Brian Murphy, David Tarcy - IB Chemistry Course Book The Only DP Resources A Developed with the IB-Oxford University Press (2014)
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O X F O 2 0 1 4 R D I B D I P L O M A P R O G R A M M E E D I T I O N CHEMISTRY C O U R S E C O M PA N I O N Sergey Bylikin Gary Horner Brian Murphy David Tarcy Library; p310: Shutterstock; 3 Education Lambert Great Clarendon Street, Oxford, OX2 6DP, United University Press is a department of the the University’s objective of University excellence of Oxford. in research, p365: education Oxford by publishing University Press worldwide. in the UK Oxford and in Oxford is a certain registered other University moral rights of Press the trade Images; F been All in published rights a the system, by law, reprographics the Oxford No rights scope of University this publication transmitted, in by of writing licence of or above Press, at in address be or agreed reproduced, by University Enquiries be may form terms should the any Oxford under organization. the must this same not circulate any Press, with concerning sent to the means, or the as Tech stored this on work any in without Clive expressly Data Library 7 9 10 8 other form and you must used made from conforms in in Publication Data in Library; p414: Photo Gabriel Library; Photo Dr p409: Morley Photo p391: Library; Andrew p404: Lambert Read/Science Library; of Library; p464: Photo Library; P482b: Photo p462: Photo Du Photo Library; p497: p495: Martyn F Birmingham Photo Cane Wilson/Science Photo TC Young/Virginia p487: Biosym p488b: Photo Library; Library; p491: Lambert Photo Photo Library; p517: Martyn Library; p506: David Consortium/Science F Photography/ Chillmaid/Science Library; High Brian Kletr/Shutterstock; Andrew Visions/Science Photo Jon p482a: Science Library; p488a: Technologies/Science Habbick Stefano Herve Torrione/Hemis/Alamy; p544a: Conge, Microeld p530: Scientic ISM/Science Photo Library; Pasieka/Science the production the grown in of this book sustainable environmental is a natural, forests. regulations of The the recyclable product manufacturing Photo p551b: Photo p559b: country of p559e: process origin. 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Science Library; p25: Vaughn Library; Photo REX/KPA/Zuma; Photo Library; p41: Picture One-Image p601a: Library; Photo Library; Photo Brown/Science p658a: Unlimited Science Library; Institution/Science Photo Getty p662: Library; Photo Royal Winters/Science Studio/ p10: Images; Images; The Winters/Science Library; Kenneth Shutterstock; Syred/Science Freeman, D Photo Photo Library; Charles Library Barrington p5a: and p600b: and p285b: Library; Technology; Chemistry; Winters/Science Photo by Six Red Marbles and OUP Science Photo Library; p780: Contents 1 Stoichiometric relationships Introduction of The matter mole Reacting to the and particulate chemical change 12 and Condensation Atomic structure (AHL) Electrons in atoms 291 volumes heavy 13 polymers Environmental 1 concept masses 12 nature 528 impact— metals 534 The periodic table—the B 20 Biochemistry transition metals (AHL) 2 Atomic structure The nuclear Electron atom First-row d-block Coloured complexes elements 50 14 to Proteins enzymes biochemistry 539 301 and 547 319 37 conguration Introduction Lipids 565 Carbohydrates 580 Chemical bonding and Vitamins 590 structure (AHL) Biochemistry 3 and the Periodicity Further aspects of covalent environment Periodic table bonding and structure 329 Proteins Periodic trends and enzymes 606 75 Hybridization 345 Nucleic acids Biological 4 597 67 619 pigments 629 Chemical bonding and 15 Energetics/thermochemistry Stereochemistry in biomolecules 641 structure (AHL) Ionic bonding and structure 93 Energy Covalent bonding cycles Entropy Covalent 357 C Energy 97 structures and spontaneity 364 Energy Fossil Intermolecular forces bonding 657 Nuclear Chemical kinetics (AHL) fusion and ssion 665 133 Rate expression and Solar reaction mechanism 5 653 fuels 122 16 Metallic sources 104 375 energy 674 Environmental impact—global Energetics/thermochemistry Activation Measuring energy changes energy warming 384 679 139 Electrochemistry, Hess’s Law 148 batteries 1 7 Bond enthalpies and Nuclear equilibrium law fusion Chemical kinetics solar 18 theory and 687 ssion and reaction cells 710 acids and bases 395 161 D involving Medicinal chemistry acids Pharmaceutical and bases drug pH Equilibrium products and 397 Equilibrium curves action 717 403 179 Aspirin and penicillin 725 Opiates 19 8 pH Electrochemical of Properties acids of and acids bases pH and bases scale and acids and bases deposition stomach 737 impact 745 of some Organic chemistry (AHL) medications of organic reactions 751 437 Taxol—a 200 routes chiral auxiliary 448 case 204 Stereoisomerism study 758 451 Nuclear Drug 9 the medications Environmental 195 Synthetic Acid of 413 197 weak regulation Anti-viral Types Strong cells 191 20 The 732 Redox processes (AHL) Acids and bases Theories 702 dye-sensitized Acids and bases (AHL) Calculations 7 cells nuclear rates Lewis of and 389 Photovoltaic Collision fuel Equilibrium (AHL) 152 The 6 rechargeable medicine detection 765 and analysis 775 Redox processes 2 1 Oxidation and reduction Measurement and analysis 209 (AHL) Electrochemical cells Internal Assessment 226 Spectroscopic organic 10 of compounds (with 461 thanks assistance to with Mark this Headlee chapter) for his 785 Organic chemistry Fundamentals of organic A chemistry Materials Functional Index 235 Materials group chemistry science introduction 791 471 248 Metals and plasma 11 identication inductively (ICP) coupled spectroscopy 475 Measurement and data Catalysts 484 processing Liquid Uncertainties and errors crystals 489 in Polymers measurement and results 494 261 Nanotechnology Graphical techniques Environmental Spectroscopic identication impact—plastics compounds 509 of Superconducting organic 501 272 metals and X-ray 277 crystallography 516 iii Course book denition The IB Diploma resource throughout course help is Programme materials of their study students expected designed two-year in a gain from course to the books support Diploma particular an The IB Learner Prole of subject. an IB The They of aim of all a will better and what learner subject while that illustrates reect the presenting the purpose philosophy and and content in and encourage a deep aims approach of the of by providing making understanding connections opportunities for to of wider critical books mirror in the terms IB of a issues use of a wide and of Diploma of the range of resources, learner Programme knowledge, action, IB service the core prole essay, issues book materials required a and and variety of be so the indeed, in and encouraged resources. and further and suggestions reading for conjunction students to of draw the theory creativity, are given how to in with IB for each extend to curiosity. conduct research and actively snow enjoy independence learning and in this will be sustained throughout their love lives. that They have explore local and concepts, global ideas, signicance. doing, they acquire understanding in-depth across a knowledge broad and and balanced of disciplines. Thinkers: thinking They skills exercise critically approach complex ethical initiative and in applying creatively problems, to and recognize make decisions. other Communicators: are conclusions Suggestions natural IB and used their necessary international and (CAS). can ten below. the reasoned, Each the through approach; requirements, extended skills Knowledgeable: viewing whole-course develop the They learning range mindedness, described create of and thinking. philosophy They acquire develop the person to aim each In curriculum as this work The IB. and The develop world. the of subject peaceful develop who a learning. IB to attributes, inquiry They is to people Diploma They way programmes minded more programme Inquirers: Programme IB internationally Programme understanding study are students from additional ideas of book research are and more They information than one in condently language communication. willingly understand and They in work collaboration with a and and express creatively variety of effectively in modes and others. provided. Principled: In addition, and the guidance requirements They are being course on the and specic on distinctive books provide course academic and with advice assessment honesty authoritative protocol. for a the action act sense dignity communities. own without They strong of the They and the with of integrity fairness, individual, take and justice groups responsibility consequences honesty, and and for that respect their accompany them. prescriptive. Open-minded: They understand and appreciate IB mission statement their The International inquiring, who help world Baccalaureate knowledgeable to create through a and better caring and intercultural aims more to develop young people peaceful understanding and respect. To this of to other end the organization and and works international challenging education points with schools, organizations programmes rigorous These programmes world to their of to individuals of international assessment. Caring: respect They to to view, personal and seeking and become who encourage active, understand differences, can students compassionate also that be other right. across and people, They show towards have act are values histories, and communities. and evaluating willing to and are traditions They a grow are range from of the to a empathy, needs personal make and the to a feelings commitment positive the compassion and of and others. to service, to the difference and lives of environment. the lifelong with Risk-takers: They and uncertainty and have new the roles, articulate iv and perspectives, accustomed others learners cultures the experience. governments develop own open and unfamiliar courage independence ideas, in approach with and of spirit strategies. defending their situations forethought, to They beliefs. explore are brave and Balanced: They intellectual, achieve understand physical personal and the importance emotional well-being for of balance themselves What constitutes malpractice? to Malpractice and result others. in, is advantage in Malpractice Reective: to their able to They own assess limitations personal give learning and in thoughtful and order to their support or any one that student or more includes results in, gaining an assessment plagiarism and or may unfair component. collusion. consideration experience. understand behaviour you They are strengths their Plagiarism and learning and ideas or is work following are dened of as the another some of representation person the ways as to your of own. avoid the The plagiarism: development. ● words one’s and ideas of arguments another must be person to support acknowledged A note on academic honesty ● It is of vital importance appropriately when After that all, credit to the owners information owners of acknowledge is used ideas of in and your (intellectual property rights. To have an that enclosed work, it must be based on work. authentic your original ideas with the work verbatim quotation must marks and CD-Roms, email messages, web sites on the piece and any other electronic media must individual be and quoted property) Internet of are within acknowledged ● have passages be information of others treated in the same way as books and fully journals acknowledged. Therefore, all assignments, written ● or oral, own completed language used or for and referred assessment expression. to, whether must Where in the use sources form of the sources of all illustrations, computer or paraphrase, appropriately such sources maps, audio-visual and programs, data, graphs, are similar material must be direct acknowledged quotation photographs, your must if they are not your own work be acknowledged. ● works of theatre art, arts whether or visual music, arts lm and dance, where the How do I acknowledge the work of others? creative The way that you acknowledge that you have ideas of footnotes other and people is through the use endnotes to be from (placed (placed provided another at at the the Collusion when bottom end you document, of a quote or of a page) document) or ● summarize provided in another artist is not that is need part denitions part of the to of provide a do ‘body not a of document. need assumed footnote for knowledge’. to be allowing for footnoted include as Other is, they are that that you used in your formal list of work. forms of that accepted involves into as works to nd how the should of a compulsory art) of (e.g. or the viewer of A supporting malpractice by by to be copied another or submitted student work for and/or different diploma assessment requirements. you of malpractice an unfair another student. unauthorized include advantage or Examples material into an any action affects the include, misconduct during an examination examination and ‘Formal’ a CAS record. several you use magazines, resources, full the acknowledged. usually that books, providing Extended the This resources information. in one internet-based and reader same the categories articles, of use presentation. separating different newspaper and you forms place, the falsifying means takes includes: work assessment gives room, resources be as This your duplicating results a work information That knowledge. should a You taking Bibliographies must dened student. components do of the ● information part or are paraphrase closely a of bibliographies. another Footnotes of used original the use CDs information your work bibliography can is Essay. v Using your IB Chemistry Online Resources What is Kerboodle? Kerboodle is subscription able to to an to access guide you online IB a learning Chemistry huge bank through this platform. Kerboodle of resources, If your Online school has Resources assessments, and a you will be presentations course. What is in your Kerboodle Online Resources? There are three planning, main resources, areas and for students on the IB Chemistry Kerboodle: assessment. Resources There a hundreds Kerboodle develop your Watch of Online. skills videos science in Hundreds extra You and and resources can use available these knowledge animations at as of home you on or the in progress experiments, IB the Chemistry classroom through difcult the to course. concepts, and action. of simulations, worksheets practice – read your articles, skills, or use perform your experiments knowledge to and answer questions. Look Find at galleries out answer more of by images looking questions, or do from at the book and recommended more see sites their on details the close up. Internet, research. Planning Be prepared extra resources Learn Plan and about and Learn vi for the on the practical the to skills that experiments analyse accurately. and Chemistry different prepare how IB work data of and your internal Kerboodle you need your draw assessment with online. to perform an investigation. own. conclusions successfully Assessment Click on the assessment examinations. style practice Here tests: auto-marked about attempts marks how a at are you as if it each one mark. The These an of knowledge interactive marks where and these until are you to quiz you or revise quizzes for and – go your exam- can be the back to printed You after there’s condent will every markbook, for every happy, come test. how so have one you two question. you can The see year. more these will the practice for comprehension, Evaluate feedback the are need use you your in automatically may practice: worksheets get reported throughout use questions check complete examination about Assessment written your lots sub-topic. then question tests: to every auto-marked were sure see check nd these for sub-topic, progress aren’t can use quiz automatically Summative there’s to will questions. Formative feel tab you your topic. and exams Work change then any submit reported in or the as revision, through the the test questions test for markbook, a you nal so you practice. practice across out answering when and you the are performed longer examined. as a timed test. Don't forget! You can also nd extra resources on our free website www.oxfordsecondary.co.uk/ib-chemistry Here you can nd answers to questions in the book . vii Introduction This book is a companion Baccalaureate Chemistry is one Programme. with is the often the Diploma It is of called the from a prerequisite and environmental being Programme knowledge IB With full the structure Subject Topics both 1 SL higher - 11 and Each as study and of the a two are its wide International closely skills. own as underpin right, A of years chemistry of an the nature study of additional IB of numeracy Diploma to develop nature scientic and is biological problem-solving, encouraged of both systems. medicine, range study Chemistry biological skills, of local Diploma academic critical-thinking, the level of all such understanding issues IB engineering. a skills, students develop on the principles in analytical During of combines and disciplines scientic book in detail courses. optional Materials, of science literacy global scientic (NOS), that concern, will with a perspective. follows has a the Topics Topics topics core 12 A, - B, the chemistry cover number of material 21 C explain and four Biochemistry, techniques, chemistry topics this of live fostering an that chemical we understanding act the subjects materials such and will material). option analytical of explain HL The Chemistry: of other Course, on to the programme in Guide. level options. All them as worthy skills. learners prepare in investigational communication focus understanding The skills, chemistry its and which involves Chemistry of in many and Chemistry science science, sciences, literacy Chemistry better IT of science laboratory subject for transferable scientic inquiry. a invariably data-handling, and of central also chemistry pivotal environment Apart students experimental acquisition physical generic, the an for Programme. of D common environmental cover the Energy that the the major and strands – is common AHL content domains Medicinal and of in the Applied Chemistry. quantitative perspectives to (additional aspects, integrated organic linkages. in the book include the following elements: Understandings The in specics detail. of the Concepts content are requirements presented in ways for each that sub-topic promote are covered enduring understanding. Applications and skills These a sections specic method help approach involving you illustrative key or to develop example, by your often considering laboratory understanding following a a by considering step-by-step particular chemical working experiment, techniques. Nature of science Here you issues, can endeavour. chemical world. viii explore theories, This is research NOS the methods hypotheses done that underpins and using led to each of laws science that carefully paradigm topic and are selected shifts presented some of associated in examples, our and the with knowledge scientic including understanding throughout the of book the there are a wide your chemical NOS is style questions an range of NOS understanding assessable are based and component integrated questions draw of the on and your programme throughout the exercises scientic and to challenge perspectives. sample NOS book. Theory of K nowledge These shor t sections have headings that are equivocal 'knowledge questions'. The text that follows often details one possible answer to the knowledge question. We encourage you to draw on these examples of knowledge issues in your TOK essays. Of course, much of the material elsewhere in the book , par ticularly in the NOS sections, can be used to prompt TOK discussions. TOK provides a space for you to engage in stimulating wider discussions about questions such as whether there should be ethical constraints on the pursuit of scientic knowledge. It also provides an oppor tunity for you to reect on scientic methodologies, and how these compare to the methodologies of other areas of knowledge. TOK is not formally assessed in the IB Chemistry programme, but it plays a pivotal role in the teaching of IB science. Activities and quick questions A variety of shor t topics or challenging questions are included with a focus on active learning. We encourage you to research these topics or problems yourselves using information readily available in textbooks or from the Internet. The aim is to promote an independent approach to learning. End -of-topic questions At the end of each (multiple-choice, problems Answers and can topic you data-base hypothesis be found at will nd exercises, style a wide range extended of questions response, NOS style questions). www.oxfordsecondary.co.uk/ib-chemistry ix Meet the authors Sergey Bylikin in 1998 and, in Chemistry. University, UK. since 2007 Gary since Head Hong in Gary and is where he to at Athlone been IB David Chemistry in USA, project of the this appreciation Brian an earlier devotion, for through age, x to my and wife have to IB had Kong. in and the in take of up he 1998 is science in IB School coordinator with the in IBO, India, curriculum course. to from the position UK at Gary and a University and UAE to senior involved in Ireland, University, associate Ireland currently was CAS Chemistry DP the Chemistry workshops Chemistry back and of the Kong. posts a in with review. taught that IB latest Medical International involvement Chemistry moved has his the Hong teaching where since on State University curriculum leading Inorganic to he In University Federation associated including advising and been signicant School PhD has Open Queensland, developing V years IB He the Chemistry of Department 8 cum has laude done is active and and possible of many the for degree work Science in sciences Asia question professor take up a lecturer. the post Brian design Workshop IB love has for Leader and and and of the He been has Field taught involved exam discussion and from information various education Chemistry support Education Australia. and setting chemistry their in Queensland, latest without families a Southeast and Chemistry design their in with graduate USA Europe, moderation, the been thank Mary, and Lorraine Joe and unwavering selessness input and and teaching support not like Teresa sister, their my patience, - would would and Programme in Russian at State Russian curriculum. writing, involved at Hong began Technology, the professor Canada Emirates the Northwest David has team a the positions, and Moscow of in boards boards, is and an Representative and curriculum. encouragement. patience. In To the particular, greatest special to: parents, vision friendship, Brian, goes Murphy buíochas!), size authors the he he with After with Australia, Diploma was of graduated technology IB of associated institutions. extent, Head Chemistry. and from Prize role University George Arab a latest postdoctoral United Institute curriculum A the Tarcy the State textbooks. Europe graduated College the of up leadership 2010, Chemistry assistant the the King Following Whitworth in In member at in 2000, across Japan. became Inorganic of in the took Switzerland Since teaching of latest a he several various Science. Cork. moved has held Murphy College he graduate PhD was involved Australia, and currently Brian a he of a received point author workshops Kong review 2009, was has of attending is and later, which an Horner, 1986 career and is awarded year Until after Sergey was one and for all (RIP) her family; support professional support, career for as their invaluable her who of love, Gary their expertise; well as the support, comments. understanding instilled in me Horner children's David many advice, an - my Tarcy unremitting parents happiness, friends and to and appreciation - to and my Tina of Dennis sister friendship; Sergey and Susan Walton, professional support (míle internationalisation my Myrtle for her brothers colleagues Bylikin - to I for from their eternal Gary have and met Natasha for her 1 S T O I C H I O M E T R I C R E L AT I O N S H I P S Introduction There is within and a a broad wide community variety approaching of their of people scientic inquiry working disciplines with reactions terminology and equations, Chemistry can be chemists, regarded as science, and mathematics the science. In this chapter we begin to many of the foundations on to of which of of chemistry matter to is the based. IUPAC the nomenclature of organic and and the representations number gas elicit stoichiometry, From Its is of denitions in of particles, universal mass and the understanding the the relative quantitative amounts of method of reactants the products in developed. a particular Treatment of chemical the gas reaction laws and the inorganic application compounds concept organization is of mole an and classication the importance. the a examining understanding the lay and down discusses chemistry. language volume of chapter the relation central of reasoning fundamental processes. this language common For methodology, by comprehensive of of volumetric analysis complete this chemical introductory chapter. 1.1 Itotio to t a ti at at o att a ia a Understandings Applications and skills ➔ Atoms of dierent elements combine in xed ➔ Deduction of chemical equations when ratios to form compounds, which have dierent reactants and products are specied. proper ties from their component elements. ➔ ➔ Application of the state symbols (s), (l), (g), Mixtures contain more than one element and/ and (aq) in equations. or compound that are not chemically bonded ➔ Explanation of obser v a ble changes in together and so retain their individual proper ties. ➔ physical proper ties changes of and tem pera tu re du rin g Mixtures are either homogeneous or state. heterogeneous. Nature of science ➔ Making quantitative measurements with replicates to ensure reliability – denite and multiple propor tions. 1 1 S TO I C H I O M E T R I C R E L AT I O N S H I P S The atomic theory A universally today is that However, the this matter has seventeenth was of accepted all a widely within it called was revealed the To been was that said released magnesium of so. explain phlogiston, of in to atoms. scientists During breaking theory conditions the a Over process be Some found burning gains The burning. rather the mass when phlogiston Scientists use instruments, to obtain it burns in oxygen, to feed for in do all not available made appetite stand those and the knowledge. have been understanding. test that to ground- primitive for hypotheses precision are the not often relatively and renewed theories was who their theories theories best today past, are of time. simple and facts. than The loses the discoveries with account metals used in time, tested re-like during fact commonly science phlogiston proposed investigations that of composed always belief. substances, Quantitative is century held combustion element not axiom atomic theory states that all matter is contradicting composed of atoms. These created destroyed, atoms cannot be theory. a wide and range of advanced evidence experimentation. through Much of computing the reactions. properties power observation or chemical methodologies, of matter arrangement and of and are Physical depend these rearranged and on during chemical the bonding and atoms. technology TOK States of matter Antoine often referred modern to Lavoisier as is well Lavoisier phosphorus they gained His that the Matter contribution In through when were is of documented. mass. contradicted “father discovered experimentation and the chemistry”. science 1772 to (1743–1794) sulfur is everywhere. surrounds form of planet seeks and matter and to us, the we which entire expand our We can we are see made and know universe is are up of touch there, made understanding of matter, many though up of matter we forms we and see belief results that particles – atoms, would be lost during combustion molecules, as phlogiston was released. or ions Could phlogiston mass? from Empirical Lavoisier ’s eventually of data the His a negative derived experiments accepted community. some have by work first the was scientific contained examples of particles are quantitative law of chemistry conser vation experiments present-day may of appear in the mass. standards ground-breaking and they occupies a MATTER motion His simple but their in constant by day. The discovery of oxygen by Joseph Priestly and Carl Scheele the phlogiston theory. invalidated This is an example of a paradigm shift. The has a mass dominant paradigm or belief is replaced by a new paradigm. Is this how scientic knowledge progresses? Figure 1 The characteristics of matter 2 volume in space were it. itsproperties. made up of mass it Air is a Our chemistry combusted These it, matter. cannot matter and consume of 1 . 1 The I n T r O d u c T I O n properties of T O the T h e three p A r T I c u l A T e states of n A T u r e matter are Soi xed volume ● xed ● xed shape ● no of ● cannot be attractive a xed the forces hold c h A n g e below. gas – be takes it the shape ● no xed volume ● no xed shape occupies occupy compressed ● can ● forces be the – space expands to available compressed forces between particles are between particles are between the close-packed particles shape container cannot particles than in solids taken as zero in arrangement ● ● c h e m I c A l volume weaker particles summarized A n d compressed ● ● m A T T e r liqi ● ● O f vibrate in particles vibrate, rotate, ● and particles vibrate, rotate, and xed translate (move around) translate faster than in a liquid positions The way the particles of matter move depends on the temperature. As SI (Système International) units are a the temperature increases the average kinetic energy of the particles set of standard units that are used in increases – the particles in a solid vibrate more. The particles in liquids science throughout the world. This will be and gases also vibrate, rotate, and translate more. discussed in great detail in sub-topic 1.2. Temperature When describing room temperature, we There are a number of different temperature scales. The most commonly might say ‘25 degrees Celsius (25 °C)’ or used are the Fahrenheit, Celsius, and Kelvin scales. All three are named ‘298 kelvin (298 K)’ (to the nearest kelvin). in honour of the scientist who developed them. Note that we use just the word kelvin, not The SI unit for temperature energetics calculations (see Absolute zero on is the topic kelvin (K). The Kelvin scale is used in degrees kelvin. The boiling point of water is 100 °C or 373 K, and the melting point of 5). water is 0 °C or 273 K. this is stops. even You 273 At in can is °C). It zero is the temperatures solid the Kelvin temperature greater than scale, at 0 K which absolute (on all the Celsius movement zero, all scale of particles particles vibrate, matter. convert temperatures from the Celsius scale to the the Kelvin evaporation scale using the algorithm: 100 (K) = temperature (°C) + C8 /erutarepmet temperature 273.15 Changes of state If you heat a block of ice in a beaker it will melt to form liquid water. condensation water melting If 0 you continue heating the water, it will boil to form water vapour. Figure2 freezing ice shows during a heating these temperature curve changes and the for of water state. kinetic – it We energy shows shall of how look at particles its temperature the changes relationship during these between changes of state. Figure 2 The heating curve for water 3 steam 1 S TO I C H I O M E T R I C R E L AT I O N S H I P S Atiity 1 What happens to the particles during Explain why the changes of state? temperature of a boiling liquid does not increase 2 ● As a sample despite energy being in constantly applied. until Deduce which would be ● the The solid it ice reaches ice begins more painful, scalding your Figure skin with water vapour or melting boiling water. lattice, 2 is at to –10 begin the shows melt the (263 point and All a is of to is heated, more. of The water at solid–liquid no the attractive molecules K) vibrate there occurring. the °C to melting that breaking allowing 3 of lattice change energy forces move 0°C (273 molecules being freely. used the The increases K). is temperature between more water equilibrium in is the temperature to set up. while disrupt molecules level of the and disorder Explain why you might feel increases. (The nature of the forces between molecules is discussed cold and shiver when you get in sub-topic 4.4.) out of the water at the beach on a very hot, windy day. ● Once all the molecules until the it ice has vibrate reaches water the starts melted, more to further and boiling move point heating faster. of makes The water at the water temperature 100 °C (373 rises K), and boil. Freeze-drying is a food ● At 100 °C a liquid–gas equilibrium is established as the waterboils. preservation technique Again the temperature does not change as energy isrequired to that uses the process of overcome the attractive forces between the moleculesin the liquid sbiatio. Foods that require water in order to free water molecules from theliquid toform a dehydration are rst frozen and gas. (Equilibrium The curve is covered in sub-topic 7.1.) then subjected to a reduced pressure. The frozen water ● in then sublimes directly to water temperature vapour, eectively dehydrating converted gure 2 remains to steam, shows that at °C. 100 the while Once temperature the all water the will is liquid increase boiling water above its has 100 been °C. the food. The process has ● Melting and boiling are endothermic processes. Energy must widespread applications in be transferred to the water of state. from the surroundings to bring about areas outside the food industry these changes The potential energy (stored energy) of the such as pharmaceuticals molecules increases – they vibrate more and move faster. (vaccines), document recovery for water-damaged books, and ● Cooling brings about condensation scientic research laboratories. freezing ● of is transferred changes they ● of less Vaporization during to a and potential move change boiling, or by form to heating liquid water, – the and the solid. exothermic surroundings The the form are processes to from energy the of processes. water the Energy during molecules these decreases – slower. of state from liquid evaporation at to gas which temperatures may below u b boiling the solid point. In sublimation matter changes state directly from e d m a il the gas phase without becoming a liquid. Deposition is the it to reverse process of sublimation – changing directly from a gas to a solid. n o it m o n le p it o n is g g n iz e e r f s happen is reverse vapour freezing the state. vibrate water and to the water liquid Condensation solid of vaporization Elements and compounds An condensation in element a xed contains ratio to rearrangements liquid atoms form of the of only compounds particles of one type. matter are of the of elements molecules or fundamental combine ions. These cornerstone gas of chemistry, represented in formulae and balanced Figure 3 Changes of state for water (Atoms 4 Atoms composed are covered in detail in sub-topic 2.1.) chemical equations. 1 . 1 I n T r O d u c T I O n Chemists the study many created in these important The and group oxidation Sodium abundant p A r T I c u l A T e elements chemical reactions, and and and n A T u r e compounds physical how they O f react with properties can be m A T T e r of used one the in A n d c h e m I c A l c h A n g e another, substances many sodium chloride, NaCl, is made up of the elements chlorine. 1 in is T h e applications. compound sodium The how different T O alkali air metal and stored violently under element sodium on oil the is a reacts to soft with prevent planet, metal water, these (2.26 % that undergoes creating reactions. by It rapid alkaline is the solutions. sixth most mass). Figure 5 The structure of sodium chloride. It consists of a crystalline Figure 4 Elemental sodium is a reactive alkali metal The halogen chlorine is a gas at room lattice of sodium ions (purple) and chloride ions (green) temperature. Chlorine, Cl , is 2 highly The the highly ionic table and irritating to the reactive and uses of constituent daily chloride and sodium compound consumed sodium skin, elements crystalline salt eyes, the and sodium in are the respiratory chlorine chloride, food very upper we to commonly eat. different combine tract. The from form called properties those of its elements. Mixtures A pure substance chemical include and the is matter physical elements that properties nitrogen, has are N a constant distinct and composition. and argon, Ar consistent. and Its Examples compounds such as 2 water, H O, table salt, NaCl, and glucose, C 2 Pure 6 substances example, sea substances such as H can water can be ltration, physically contains separated fractional combine mainly from to 12 form sodium the O 6 a mixture. chloride mixture distillation, . by and physical For water. Pure techniques orchromatography. The 5 1 S TO I C H I O M E T R I C R E L AT I O N S H I P S elements or compounds that make up a mixture are not chemically post’s aw o ostat oositio bound together. (1806) stated that compounds have Homogeneous mixtures have both uniform composition and uniform distinct properties and the same properties throughout the mixture. Examples include salt water or a elemental composition by mass. metal alloy such composition Examples as and hence include prawn soup) Figure 9 or brass. foods Irish and their properties such stew summarizes compounds, Heterogeneous the as (a tom vary yum mixture of classication mixtures throughout goong cubed of have (Thai meat matter non-uniform the hot and into a mixture. and sour vegetables). elements, mixtures. matter – any substance that occupies mixture – a combination pure substance – has a of two or more pure denite and constant substances that retain their composition individual properties homogeneous mixture – heterogeneous mixture – has both uniform has element – made up of non-uniform compound – made up atoms that each have of a combination of composition composition and the same atomic atoms or ions in a xed and properties varying properties, number, eg lead, Pb, ratio and having dierent eg salad dressing, mercury, Hg, throughout, eg salt water, paint, garden soil properties from the bromine, Br constituent elements, eg Figure 6 Chlorine reacts vigorously metal alloys water, H O, carbon dioxide, 2 with sodium metal CO , sodium chloride, NaCl 2 Figure 9 Elements, compounds, and mix tures The language of chemistry Chemistry enables the has wider symbols chloride, NaCl(s). It has very dierent properties and from those of its constituent elements a the to process Chemical symbols and in The International an nomenclature is to provide language country of or are a for of of and st u de n t s , with t ha t e l em e nt s each in a l lo w i ng bo rd er s an d oth e r. re qu i re s an d d is p la yed in expressing both Pure no ba la n c e d a nd t h e ir eq u at i on und er s t an d in g the Applied a inorganic naming to which that of elements inorganic of the Chemistry system of no ( IUPAC) resulting translation from The IUPAC Gold Book (http://goldbook .iupac.org/index.html) is IUPAC’s compendium of chemical terminology. is IUPAC’s another. investigate industrial dyes by separating them present standardized compounds. compounds, require are compounds. us rso 6 of C he m ic a l Figure 8 Paper chromatography is used to into their pure constituent components and c i t iz en s t r a ns l at i on . c om po un d s a and monitors and words language organic and and organic consistency culture’s in develops both symbols l a ng ua ge as t ra n sc e n ds e xa mined . way Union that a for i nf or ma ti on, b e i ng th a t l e ctur er s , communic at e are proportions, organization a nd a no the r of chemical which to s y mb ol s one wealth l anguag e te a che r s, equa ti o ns of relationship unlocks uni v e r sa l communi ty Knowledge Figure 7 T able salt is the compound sodium a scientists, in role a one 1 . 1 I n T r O d u c T I O n T O T h e p A r T I c u l A T e n A T u r e O f m A T T e r A n d c h e m I c A l na o TOK c h A n g e foa na o oyatoi io foa oyatoi io Language is a crucial component in the communication of + ammonium ion NH 3 phosphate(V) ion PO 4 4 knowledge and meaning. Does the language of chemistry with carbonate ion 2 3 phosphonate ion CO its equations, symbols, and units promote or restrict universal PO 3 hydrogencarbonate understanding? What role does linguistic determinism play? 3 2 sulfate(VI) ion HCO SO 3 4 ion For example, the concept of equilibrium is often initially hydroxide ion 2 sulfate(IV) ion OH misinterpreted. Preconceived ideas focus on a 50:50 balance SO 3 between reactants and products. It requires an understanding nitrate(V) ion 2 ethanedioate ion NO C 3 O 2 4 that equilibrium means that both the forward and reverse nitrate(III) ion 2 peroxide ion NO O 2 2 reactions are occurring at the same rate before we can see that an equilibrium reaction might favour the formation of products T able 1 Common polyatomic ions or reactions, or that such a reaction could be non-spontaneous. na o ai foa hydrochloric acid HCl Common combinations of elements: Background nitric(V) acid HNO 3 to writing equations phosphoric(V) acid H PO 3 An ion is positively a charged species. Anions are negatively charged and cations 4 are sulfuric(VI) acid charged. H SO 2 There the are a numb e r substances yo u of common wi l l s tudy p o ly a t om i c a nd w o rk io ns wit h . t ha t You ex is t ne e d to in m a ny be of ethanoic acid CH 4 COOH 3 fa m i li a r T able 2 Common acids with the names a nd fo r mula e of the s e io n s, s h own in t a bl e s 1 to 3. na o aio foa sulde ion ability to write equations is essential to chemistry and requires -ide S Writing and balancing equations An nai sx 2 2 sulfate(VI) ion -ate SO 4 a full understanding of the language of equations. At the most 2 sulfate(IV) ion -ate SO 3 fundamental hand side level, along formulae with their for state the reactants symbols (s), are (l), put (g), on the (aq), left- and those T able 3 Naming anions. The prex identies the for the products on the right-hand side. The arrow represents a element present and the sux the type of ion boundary deduced matter between by of reactants referring the to element the or and products. solubilities compound of at a State ionic given symbols salts and can the be state (eg element or polyatomic ion) of temperature. Worked example A reaction The may process equation ionic and of starts be described transforming with covalent the in terms these will starting words construction formulae of be of into a materials balanced chemical discussed in and chemical formulae. depth in products. Writing topic 4. Magnesium form a Write equations for the following chemical reactions, including state symbols. Refer to the working method on the next page on balancing equations if you need to. oxygen oxide. equation represent to state known Write a this to as chemical change, symbols. Solution The reactants magnesium, are a temperature, 1 in powder magnesium including Qik qstios burns white the solid and metal at the room diatomic Zinc metal reacts with hydrochloric acid to form the salt zinc chloride. Hydrogen molecule, oxygen, which is a gas is evolved. gas. 2 Hydrogen gas and oxygen gas react together to form water. 3 At a high temperature, calcium carbonate decomposes into calcium oxide and The magnesium, which carbon dioxide. product is a the oxide magnesium solid 2Mg(s) is + of oxide substance. O (g) → 2MgO(s) 2 7 1 S TO I C H I O M E T R I C R E L AT I O N S H I P S Working method: how to balance Step 1: Balance the metal Ca rst. chemical equations It The examples Figure 10 below reminds involve you that reactions metals of are is balanced. metals. below and Ca(s) + H O(l) → Ca(OH) 2 to the left of the metalloids Remember that change coefcient to balance in an the periodic equation of a formula (add a front of the formula). You do not change it (g) 2 2: occurs Balance in only O one next, formula on the each formula H number as in + you Step the (aq) 2 table. side. (H occurs in both products.) itself. Multiply O H by 2 to balance O. 2 Step 1: First balance each in side front of of the the the metallic element equation– symbol on add one a on number side if Ca(s) + 2H O(l) → Ca(OH) 2 necessary of atoms so of that this there is element the on same each 2: Balance any elements that occur You in has only one formula on the reactant can ions side. remain they can be Sometimes unchanged balanced been 3: Balance the remaining now reactions easily at this elements if balanced 5 6 7 8 9 C N O F Boron Carbon Nitrogen Oxygen Fluorine 13 14 15 16 17 Al Si P S Cl Aluminium Silicon Phosphorus Sulfur Chlorine 30 31 32 33 34 35 Zn Ga Ge As Se Br equation Always check to which make often sure. is now Potassium hydroxide neutralize the produce two Zinc Gallium Germanium Arsenic Selenium Bromine 48 49 50 51 52 53 Cd In Sn Sb Te I Cadmium Indium Tin Antimony Tellurium Iodine H SO 2 the Balance 80 81 82 83 84 85 Hg Tl Pb Bi Po At Mercury Thallium Lead Bismuth Polonium Astatine (aq) + a overall. soluble acid base sulfuric hydrogen Balance 1: is diprotic the ions following K by KOH(aq) → when Step 2: two Both O can Diprotic they equation. KOH on side. K SO 2 in that acid. doubling reactant 4 occur 116 balanced Example 2 Step 115 hydrogen 2, and dissociate. 114 step stage. acids 113 by 3: that necessary. B 112 see polyatomic in The Step (g) 2 and happens. products + H side. Step Step (aq) 2 number and compounds (aq) + H 4 O(l) 2 H on both 117 sides of the unchanged equation. in the The reaction sulfate and is ion is balanced, so metals the coefcient for H SO 2 will stay the same. 4 Figure 10 Metals are below and to the left of the metalloids in There are 4 H atoms on the reactant the periodic table side, so multiply H O by 2. 2 Example 1 The alkaline earth metal calcium reacts with SO H 2 water to produce an alkaline solution. Balance H SO 2 following + 2KOH (aq) → K SO 2 (aq) + 2KOH(aq) 4 → K SO 2 equation. The 8 (aq) 4 (aq) + H 4 O(l) 2 the equation is now balanced. (aq) 4 + 2H O(l) 2 1 . 1 I n T r O d u c T I O n T O T h e p A r T I c u l A T e n A T u r e O f m A T T e r A n d c h e m I c A l So tys o atio The names and symbols of the elements can be found in cobiatio or sytsis reactions involve the combination of two or more section 5 of the Data booklet substances to produce a single product: C(s) + O (g) → CO 2 c h A n g e (g) 2 doositio reactions involve a single reactant being broken down into two or more products: CaCO (s) → CaO(s) + CO 3 (g) 2 Si at reactions occur when one element replaces another in a compound. An example of this type of reaction is a redox reaction (topic 9): Mg(s) + 2HCl(aq) → MgCl (aq) + H 2 (g) 2 dob at reactions occur between ions in solution to form insoluble substances and weak or non-electrolytes, also termed tatsis reactions: HCl(aq) + NaOH(aq) → NaCl(aq) + H O(l) 2 This example is an acid-base reaction discussed fur ther in topic 8. Some applications and reactions of butane Fuels and refrigerants Butane, C H 4 create the variety of is in a the same that other hydrocarbons petroleum gas (LPG). such This is as propane used in a to wide applications. different CFCs with liqueed Methylpropane have mixed 10 fuel (also called chemical way. were isobutane) formula but Methylpropane previously used is for is an their used this isomer atoms as a of are butane. Isomers arranged refrigerant, structurally replacing the purpose. cfcs a t iat o si a tooy Ozone occurs Ozone lters Without many this forms naturally out most in of protection of life, the the the causing stratosphere, harmful the ultraviolet ultraviolet skin in radiation cancer in upper rays from would humans atmopshere. and be the harmful other The process of refrigeration sun. to problems. involves the energy changes of a condensation–evaporation cycle using volatile liquids. H Chlorouorocarbons (CFCs) H H were traditionally used O in refrigerators and air- H C C O conditioning units. They cause O depletion of the ozone layer H C in the atmosphere, which H C H H protects us from the harmful eects of ultraviolet radiation H in sunlight. H CFCs are now banned in many Figure 11 Ozone, O Figure 12 Methylpropane is used as a refrigerant countries, and non-halogenated 3 hydrocarbons such as propane CFCs undergo reactions with the ozone in the stratosphere, causing it are more commonly used to break down. The ‘ozone hole’ is a thinning of the ozone layer that instead. There is more about this appears over the polar regions of the Earth each spring. The use of in sub-topic 5.3. CFCs been has caused replaced by this depletion of the ozone layer, so they have now methylpropane. 9 1 S TO I C H I O M E T R I C R E L AT I O N S H I P S The combustion of hydrocarbons, C x H produces y carbon dioxide and water. Since 1997, taxis in Hong Kong have been powered by liqueed petroleum gas (LPG). Today there are over 18 000 LPG taxis and 500 LPG light buses operating there. LPG, consisting of butane and/or propane, undergoes combustion to release energy to power the vehicle. The reaction produces carbon dioxide and water (sub-topic 10.2). LPG burns much more cleanly than petrol or diesel. Figure 13 The ozone ‘hole’ was rst noticed in the 1970s and is monitored by scientists worldwide Balancing the equation for the combustion of butane The combustion C H 4 Step (g) O 10 1: atoms + of butane (g) → is CO 2 There rst by an (g) exothermic + H 2 are no O(l) 2 metal multiplying reaction. atoms CO by to balance, so balance the carbon 4. 2 C H 4 (g) Step 2: leave this the + O 10 (g) Oxygen right, → 4CO 2 until so (g) + H 2 is found last. in two Hydrogen multiply H O O(l) 2 by compounds has 10 atoms on on the the product left and side 2 so atoms on 5. 2 C H 4 Step To (g) 3: H Figure 14 Rush hour in Hong Kong 10 + 6.5O (g) are not by + 5H now (g) contain 6.5 → 13 oxygen molecules 4CO (g) + of 5H 2 used (see O(l) 2 2 in 15). are an odd number. required. O(l) 2 balanced topic atoms, oxygen We equations, therefore except when multiply the calculating whole 2. (g) + 13O 10 complex balancing 4CO equation enthalpy 4 The products the (g) equation → 2 10 Fractions 2C (g) 2 The H 4 lattice O 10 balance C + (g) → 8CO 2 coefcients equations on (g) + 10H 2 in this page 8 example is O(l) 2 more show efcient why than the method just trial of and error. 1 . 1 I n T r O d u c T I O n T O T h e p A r T I c u l A T e n A T u r e O f m A T T e r A n d c h e m I c A l c h A n g e The atom economy The global demand population, and dwindling need to must be fewer the To for nite conserve less toxic and resources resources. increasingly and goods developing services have led Synthetic efcient along economies, to emissions. to a an raw Sustainable and increasing levels heightened reactions preserve with increasing of awareness industrial materials world pollution, and development is of the processes produce the way of future. this Trost of rapidly end of the efciency atoms atom Stanford in of the economy University chemical reactants reactions with the Molecular percentage was developed Stanford, by CA, of mass atoms of Professor This comparing molecular mass by USA. looks the of Barry at the molecular useful useful level mass of compounds. products ____ = atom The atom which the × economy economy we will amount economy Molecular of of is discuss important later reactants 100% = would in mass in this the atoms of that in In an of ideal products no atoms 100% reactants discussion book. amounts suggest of Green produced. are Chemistry, chemical process So an atom wasted. Atiity a) Suggest why even if a chemical reaction has a yield close to 100%, the atom economy may be poor. Carry out some research into this aspect. b) Discuss some other ways a chemical process may be evaluated other than the atom economy, eg energy consumption etc. ) Deduce the percentage atom economy for the nucleophilic substitution reaction: CH (CH 3 ) 2 OH + NaBr + H 3 SO 2 → CH 4 (CH 3 ) 2 Br + H 3 O + NaHSO 2 4 Qik qstios Identify the type of reaction and then copy and balance 4 Al(s) + O (g) → Al 2 2 O (s) 3 the equation, using the smallest possible whole number 5 KClO (s) → KCl(s) + O 3 (g) 2 coecients. 6 1 SO (g) + H 3 O(l) → H 2 SO 2 7 2 NCl (g) → N 3 2 (g) + Cl CH 4 (g) + O 2 H (g) + O 8 2 (g) → CO (g) + H 2 O(g) 2 Ni(OH) (s) + HCl(aq) → NiCl 2 (g) (aq) + H 2 O(l) 2 2 8 3 C 3 (aq) 4 (g) → CO 2 (g) + H AgNO (aq) + Cu(s) → Cu(NO 3 O(g) 3 ) (aq) + Ag(s) 2 2 9 Ca(OH) (s) → CaO(s) + H 2 O(l) 2 11 1 S TO I c h I O m e T r I c r e l AT I O n S h I p S 1.2 T o o t Understandings Applications and skills ➔ The mole is a xed number of par ticles and Calculation of the molar masses of atoms, ions, ➔ refers to the amount, n, of substance. molecules and formula units. ➔ Masses of atoms are compared on a scale Solution of problems involving the ➔ 12 relative to C and are expressed as relative relationships between the number of par ticles, atomic mass (A ) and relative formula/ r the amount of substance in moles and the molecular mass (M ). r mass in grams. 1 ➔ Molar mass (M) has the units g mol ➔ The empirical formula and molecular formula Interconversion of the percentage composition ➔ by mass and the empirical formula. of a compound give the simplest ratio and the Determination of the molecular formula of ➔ actual number of atoms present in a molecule a compound from its empirical formula and respectively. molar mass. Obtaining and using experimental data for ➔ deriving empirical formulas from reactions involving mass changes. Nature of science ➔ Concepts – the concept of the mole developed from the related concept of ‘equivalent mass’ in the early 19th century. SI: the international system of measurement Throughout history measurement. so an internationally measurements Units of million tonnes do and desire the seven from for 12 the a of allows language e ss e ntia l US in do ll a rs, p r eci ous mi ll io n me tal s use s a of all the ran g e (p pm) of and us our to w a l ks of to another, understand of li fe . in m e a su r es a g en c i e s, pa r t ic u l at e Th e i n du s t ri es me a su r ed of forms culture culture. re s ourc e s a re standard international a that system Système base these in units different country us e o un c e s , inc l ud in g a mo n gs t m a t t e r. y i el d ot h e rs , W h ic h un i t s use? development cultures– are the of one e nv i r o nme nta l pr ot e c t io n partspe r chemists The (MT ) , of developed from set manuf a cturi ng hectare, about sp e a k s have vary agreed measureme nt agricultural talk may regardless nancialworld per societies These units seven of base International the SI units. set transcends system. of all units d’Unités All led languages other (SI). units to the and Ta b l e are 1 shows derived 1 . 2 T h e m O l e c O n c e p T Accuracy and SI units Continual in the improvements measurement physical constants Bureau of initials in applications of Aeronautics and are all precision have changed and Space in the units the from over instrumentation that time. use school Administration the The (known correct the of meant Measures monitors science, equivalent in SI have Weights French) of as of values units, laboratory (NASA), some International BIPM SI used of to SI from so the units its that US are in all National used and cases. po ty uit Sybo mass kilogram kg temperature kelvin K time second s amount mole mol electric current ampère A luminosity candela cd length metre m Figure 1 A platinum–iridium cylinder at the National Institute of Standards and Technology, Gaithersburg, MD, USA, represents the standard 1 kg mass Sty tis T able 1 The seven base units of the SI system Physical constants and unit Table 2 shows chemistry, used to two along convert quantities with SI their units to that units. a are used Table suitable 3 size throughout is a for list the of the study standard application of conversions are available in prexes you section 2 of the Data booklet. are The value of Avogadro’s measuring. constant (L or N ) will be A provided in Paper 1 questions, moa o o a ia as at 273 K a Aoao’s ostat (N and may be referred to in the ) Data booklet when completing A 10 0 kpa both Papers 2 and 3. 23 1 2 mol 6.02 × 10 2.27 × 10 3 m 1 mol 3 (= 22.7 dm 1 mol ) T able 2 Useful physical constants and unit conversions px Abbiatio Sa nano n micro µ milli m centi c deci d standard – kilo k 10 mega M 10 giga G 10 Amount of substance: The mole Chemists to need control processes involved and such in companies precise All of understand as use processing amounts of of of constituent aspects reaction. smelting food and medicines reacting substances all the electrolytic synthesizing chemical their to make are atoms, which a chemical From of up of vary drugs, is of to ability crucial the and in order 6 10 industrial industries pharmaceutical the elements in reaction large-scale aluminium beverages, and substances made of 9 10 to 3 10 2 10 measure 1 10 importance. that number are of composed 1 3 protons, 6 neutrons, and electrons (topic 2). Chemists use a system to measure 9 equal are, SI amounts which unit, of allows symbol different them mol, to elements calculate dened as a regardless reacting xed of how big quantities. amount, n, of a their The atoms mole substance. is an This T able 3 Useful prexes, their abbreviations and scales 13 1 S TO I C H I O M E T R I C R E L AT I O N S H I P S denition can be applied to atoms, molecules, formula units of ionic Stoiioty uses the compounds, and electrons in the process of electrolysis. quantitative relationships This xed amount is a number of particles called Avogadro’s constant between amounts of 23 (symbol L or N reactants and products in ) and it has a value of 6.02 × 1 10 mol . Avogadro’s A constant enables us to make comparisons between chemical species. A a chemical reaction. These mole of any chemical species always contains an identical number of relationships depend on the representative units. law of conservation of mass and denite propor tions. They allow chemists to calculate Relative atomic mass, relative formula mass, the propor tions of reactants to mix, and to work out expected yields, from the ratios of and molar mass Isotopes reactants and products of according to the balanced topic chemical equation. 2.1). relative occurs The a rati Atoi are protons a sample of atom an of of atom of atoms of is the nucleus Isotopes masses of atoms the abundance in single A in an of each isotope equals weighted is a with 12 average of the mass measure same neutrons of number (see numbers. the sub- The percentage that 4). one units. of have different (table compared that numbers have element carbon-12 a element different element the are same but the another The on a relative atomic masses scale in atomic of its which mass isotopes r Isoto aba ass and in their relative carbon having abundances. an A of The 12.01. existence The of relative different isotopes molecular mass results or r 35 Cl 35.0 75% relative formula mass M for a molecule or formula unit is determined r 37 Cl 37.0 25% by combining the A values of the individual atoms or ions. r have rati atoi ass A no units as A and M r they are both r ratios. 35.5 The molar mass is dened as the mass of one mole of a substance. It T able 4 The relative atomic mass of –1 has the unit of grams per mole, g mol (gure 2). chlorine is the weighted average of the atomic masses of its isotopes and their relative abundance Mg 24.31 g 58.44 g 23 6.02 × 10 H NaCl 18.02 g 23 6.02 × O 2 10 atoms formula units of Mg of NaCl 23 6.02 × 10 molecules of H O 2 Figure 2 The molar mass of a substance contains Avogadro’s number of representative par ticles (the par ticles may be atoms, molecules, or ions) TOK Scientic discoveries are the product of many dierent ways of knowing (WOK). To construct knowledge and understanding, scientists can use intuition, imagination, reasoning, and even emotion, as well as detailed investigation and analysis of large volumes of data that either suppor t or disprove observations and hypotheses. Sometimes it can just be a matter of serendipity. The scale of Avogadro’s constant (602 000 000 000 000 000 000 000) passes beyond the boundaries of our experience on Ear th. The population of the planet is dwarfed by this number. How does this experience limit our ability to be intuitive? 14 1 . 2 T h e m O l e c O n c e p T 13 Worked examples: A and M r r Al Example 1 26.98 State the relative atomic mass A of aluminium. r Figure 3 The element aluminium as represented in the periodic table Solution Figure A 3 shows (Al) = the periodic table entry for aluminium. 26.98 r Example 2 Calculate the molar mass M of sulfuric acid, H r SO 2 4 nati iis a its An ix or ow is a mathematical Solution notation that shows that a quantity Table 5 shows the data needed to answer this question. or physical unit is repeatedly multiplied by itself: et rati atoi nb o cobi atos ass/ 2 m × m = m ass A A ati ix shows a reciprocal: hydrogen 1.01 2 2.02 sulfur 32.07 1 32.07 oxygen 16.00 4 64.00 1 _ 1 = x x 1 _ 3 dm = 3 dm cotatio (oaity): units T able 5 3 may be written as mol dm , M, or 1 M (H r SO 2 ) = (2 × 1.01) + (1 × 32.07) + (4 × 16.00) mol L (US). 4 1 M (H r SO 2 ) = 98.09 g etay o taizatio: units mol 4 1 are kJ mol Example 3 Iitia at o atio: units are 3 Calculate M of copper(II) sulfate pentahydrate, CuSO r 5H 4 mol dm O. 1 s 2 Solution Many transition molecules metal bonded to complexes the central (sub-topic metal ion. 13.1) The contain formula water CuSO 5H 4 shows that 5 mol et of water combines with rati atoi ass A 1 mol of copper(II) O 2 sulfate. nb o cobi atos ass/ Sty tis • copper 63.55 1 63.55 sulfur 32.07 1 32.07 When adding and subtracting numbers, always express the nal answer to the same oxygen 16.00 oxygen 16.00 4 number of decimal places as 64.00 the least precise value used. 5 × 1 = 5 80.00 • hydrogen 1.01 5 × 2 = 10 When dividing or multiplying, 10.10 always express the answer to T able 6 Calculating the molar mass of copper(II) sulfate pentahydrate the same number of signicant gures as the least precise –1 M (CuSO r 5H 4 O) = 249.72 g mol value used. 2 15 1 S TO I C H I O M E T R I C R E L AT I O N S H I P S Qik qstio piay staas Calculate the molar mass of the A iay staa is any substance of very high purity and large molar mass, following substances and ions. which when dissolved in a known volume of solvent creates a primary standard solution. a) Mg(NO ) 3 2 Primary standard solutions are used in acid–base titrations to improve the b) Na CO 2 3 accuracy of the nal calculation. The concentration of a primary standard can be ) Fe (SO 2 ) ) 4 determined accurately. 3 S 8 ) Zn(OH) Mole calculations 2 ) Ca(HCO All ) 3 chemists, industries, ) whe the r in the s ci e nt i c c om m u n i t y, m an u fac t u r in g 2 or rese a r ch facil i ti es , wor k e ve r y da y wi t h r ea c t i n g I 2 quantities ) MgSO 7H 4 O [Al(H O) 2 stoichiometric P s ubs ta nce s calcul a ti ons . The a nd so n ee d r el a t i on sh i p to pe r for m be t wee n t he amou n t (in mol), number of p a r ti cl e s, a nd the mass of the s a mp le is s u m m ar i z ed ] 6 in j) chem i ca l 2 3+ i) of gure 4. O 2 5 × Avogadro’s constant, L × molar mass ÷ molar mass number of particles ÷ Avogadro’s constant, L Figure 4 The relationship between amount, mass, and number of par ticles Worked examples: mole calculations ➔ 1.50 mol of glucose contains 9 mol of C atoms. Example 1 number Calculate the amount (in mol) of carbon of atoms = n(CO ) in a sample of 1.50 × 10 amount (in mol) n × dioxide, Avogadro's 23 constant, L molecules. 2 23 = 9 mol = 5.42 × 6.02 × 10 1 mol Solution 24 number of × 10 C atoms particles ___ amount (in mol) n = Avogadro’s constant, L Sty ti Rearranging and substituting values: The answer is recorded to 3 signicant gures, as this is the precision of the data given by the examiner (1.50 mol). 23 1.50 × 10 __ n(CO ) = 2 23 6.02 = × 0.249 1 10 mol Example 3 mol Calculate Example 2 Calculate in 1.50 the mol the amount (in mol) of water molecules 22 in number of of glucose, carbon C H 6 atoms contained 3.01 × 10 ethanedioic formula acid, H C 2 units O 2 of 2H 4 hydrated O. 2 O 12 6 Solution Solution ➔ 1 ➔ molecule carbon, 12 of glucose atoms of contains hydrogen, 6 atoms and 6 of of every 16 1 formula 1 mol of a substance of particles. oxygen. number ➔ 1 mol of glucose contains 6 mol of C unit there are 2 molecules water. atoms ➔ of For atoms. contains Avogadro’s 1 . 2 Therefore, T h e m O l e c O n c e p T Example 6 number of particles ___ amount (in mol) n = Calculate Avogadro’s constant, the number of chlorine atoms in a L 6.00 mg sample of the anti-cancer drug cisplatin, 22 3.01 × 10 __ C n(H 2 O 2 2H 4 O) = O) = 2 × 0.0500 Pt(NH mol ) 3 Cl 2 2 23 6.02 n(H cis-diamminedichloroplatinum(II), = 2 0.0500 × mol 10 = 0.100 Solution mol 2 uits ➔ First ➔ Next the Amount of substance n has the units mol convert nd the the molar mass amount in in mg to mol g. by calculating mass. m ___ n = ➔ Finally remember that there are 2 mol of molar mass chlorine atoms in every mol of cisplatin. 1 Mass m has the units g; molar mass has the units g mol 3 6.00 mg n[Pt(NH = ) 3 6.00 Cl 2 × 10 g ] 2 3 Example 4 6.00 × 10 g ____ = Calculate dioxide, the amount (in mol) in 8.80 g of 195.08 carbon + 2(14.01) + 6(1.01) + 2(35.45) CO 5 2 = Solution 2.00 × 10 mol 5 n(Cl) = 2 × 2.00 × 10 5 mol = 4.00 × 10 mol m __ n(CO ) 5 = number 2 molar of atoms (Cl) = 4.00 × mol 10 × mass 23 8.80 6.02 × 2.41 × 10 1 mol g ___ = 19 = 1 12.01 = 0.200 + 2(16.00) g 10 mol mol H H Cl Example 5 Calculate the mass in g of 0.0120 mol of sulfuric N acid, H SO 2 4 Pt H H Solution Calculate the molar mass of H SO 2 into the and substitute N 4 equation: Cl H H mass (g) = n(H SO 2 = ) × M 4 0.0120 (H r mol × SO 2 ) 4 [2(1.01) + 32.07 + –1 4 (16.00)] g mol Figure 5 The anti-cancer drug cisplatin = 1.18 g Qik qstios 1 Calculate the amount (in mol) in each of the following masses: a) 8.09 g of aluminium b) 9.8 g of sulfuric acid ) 25.0 g of calcium carbonate ) 279.94 g of iron(III) sulfate. 17 1 S TO I C H I O M E T R I C R E L AT I O N S H I P S 2 Calculate the mass (in grams) in each of the following: a) 0.150 mol of nitrogen, N 2 b) 1.20 mol of sulfur dioxide, SO ) 0.710 mol of calcium phosphate, Ca 2 (PO 3 ) 0.600 mol of ethanoic acid, C H 2 3 ) 4 2 O 4 2 Calculate the number of par ticles present in the following: a) 2.00 mol of vanadium, V b) 0.200 mol of sodium chlorate(VII), NaClO 4 ) 72.99 g of iron(III) chloride, FeCl ) 4.60 g of nitrogen(IV) oxide. 3 Experimental empirical and molecular formula determination The term “empirical” observation and/or laboratories involved or food production compound in describes information investigation, in will medical often processes that using research carry may out be that is scientic and through Chemical development, analyses either derived methods. of the qualitative manufacturing, composition or of quantitative a in nature. Qualitative present in a analysis Quantitative of elements The of or elements formula in Therefore formula (in one the formula. Table for ionic the within the the shows compounds compound, the of Sbsta work each actual or some to is one is a of simplest present atoms of the is the relative a compound. amount ratio as of (in empirical the formula formula is the represents same the moa oa C water as H of ions eiia oa CH 3 O H H C C 6 HO 2 H 4 O 2 O 2 O 8 H 12 C 2 O 6 ions in its empirical formula. It is used in some 18 formula ratio 6 H hydrogen peroxide countries to enhance the health of teeth empirical the simplest 2 glucose of molecular 6). 2 Figure 6 Sodium uoride, NaF has a 1:1 ratio of mol) respectively. the the ratio The examples. ethane butanoic acid masses composition. compound, same are substance. whole-number in or the the exact whole-number formula of determine the mole elements purity their element number which the out compound of unit (gure verify chemists to formula the determining also empirical since structure a mol) molecular Sometimes 7 them structural formula. For is on could enables allows amount molecular It analysis which empirical atoms focuses compound. T able 7 Some examples of molecular and empirical formulae H 2 O 4 CH 2 O 1 . 2 T h e m O l e c O n c e p T Worked examples: percentage composition by mass You the can use molar your mass percentage by understanding of a compound mass of of to elements how to calculate in a calculate the Sometimes ratio to multiplication whole is needed to convert the numbers: compound. example 1 1:1.25 Multiply each 4(1):4(1.25) side ≈ by4: 4:5 Example 1 example Calculate the percentage by mass of sulfur 2 1:1.33 Multiply each 3(1):3(1.33) sulfuric acid, H side by3: in ≈ 3:4 SO 2 4 Sty ti Solution Empirical formulae are based on experimental A (S) r _ % sulfur = × M (H r SO 100% data; those for example 2 would likely have been ) 2 4 determined by a combustion reaction. The value 32.07 ___ = × 2(1.01) × (32.07) × 100% of 3.97 rather than 4 for hydrogen comes from 4(16.00) experimental error. = If you but have you mass the 32.69% of a compound know the of unknown percentage elements empirical molecular the present, formula and, in formula composition you can some Example 3 by calculate cases, Upon analysis, mass formula. 0.25 of g 194.13 of oxygen. Example 2 the compound that empirical formula of an hydrogen by 75 % carbon and mol hydrogen, of an was acid 8.0 g the found of sulfur, empirical 8.0 _ = mass. = 1.0 = n(C) to determine the ratio of the = 1 = 4 = 1 0.25 _ = 0.25 1.01 to g and 0.25 0.25 _ is 16.0 1.0 _ = Solution step molar 0.25 16.0 _ n(H) and formula 0.25 32.07 16.00 rst a contain 0.25 _ = 25% n(O) The with to formula. organic n(S) contains sample g Determine molecular Determine a –1 the 0.25 of n(H): Therefore the empirical formula is HSO . 4 % composition __ relative amount of substance = molar To mass calculate the molecular formula, calculate the 75 _ n(C) = = 6.24 empirical formula empirical formulae mass and determine how many 12.01 make up the molar mass. 25 _ n(H) = = 24.75 molar mass ___ 1.01 empirical Now the take the divisor to smallest quotient determine the (6.24). lowest Use this formula mass as 194.13 ___ whole-number = ratio of the 194.13 _ = 1.01 + 32.07 + = 2 97.08 4(16.00) elements: 6.24 _ carbon = 1 6.24 24.75 _ O percentage S 2 = 2 . This formula of compound is the acid called is 2(HSO ) or peroxodisulfuric 8 3.97 acid 6.24 the molecular H 4 hydrogen Because The composition (gure 7). O is O O experimentally determined it is acceptable H to O S round to the nearest whole number if the number S is close to a whole number. Therefore the simplest O H O O whole-number ratio of carbon to hydrogen is O 1:4 Figure 7 Molecular model of peroxodisulfuric acid and the empirical formula is CH . 4 19 1 S TO I c h I O m e T r I c r e l AT I O n S h I p S 1.3 rati asss a os Understandings Applications and skills ➔ Reactants can be either limiting or excess. ➔ The experimental yield can be dierent from ➔ quantities, limiting and excess reactants, the theoretical yield. ➔ Solution of problems relating to reacting theoretical, experimental, and percentage yields. Avogadro’s law enables the mole ratio of ➔ reacting gases to be determined from volumes Calculation of reacting volumes of gases using Avogadro’s law. of the gases. ➔ ➔ The molar volume of an ideal gas is a constant graphs involving the relationship between at specied temperature and pressure. ➔ temperature, pressure, and volume for a xed mass of an ideal gas. The molar concentration of a solution is determined by the amount of solute and the ➔ volume of solution. ➔ Solution of problems and analysis of Solution of problems relating to the ideal gas equation. A standard solution is one of known ➔ concentration. Explanation of the deviation of real gases from ideal behaviour at low temperature and high pressure. ➔ Obtaining and using experimental values to calculate the molar mass of a gas from the ideal gas equation. ➔ Solution of problems involving molar concentration, amount of solute, and volume of solution. ➔ Use of the experimental method of titration to calculate the concentration of a solution by reference to a standard solution. Nature of science ➔ Making careful obser vations and obtaining evidence for scientic theories – Avogadro’s initial hypothesis. Stoichiometry A balanced reactants and also chemical and the equations relative also include (see topic of examining the understanding chemical 20 to of relative this reactions, the success is provides their amounts may the linked are, reaction of of equation products of reactants specic 5). in and of the protability is reactants data the and processes percentage of the about their products. quantitative industrial particularly and symbols, Stoichiometry amounts vital information chemical on what state the of matter, Chemical the enthalpy quantitative products. where yield, organization. An the is method efciency directly 1 . 3 From a balanced chemical equation the coefcients r e A c T I n g can be m A S S e S A n d v O l u m e S interpreted TOK as the ratio equation of for the the amount, reaction in mol, used for of reactants the and products. manufacture of This ammonia is in the the When comparing the eight Haber process (see topic 7): areas of knowledge (AOK), N (g) + 3H 2 (g) ⇋ 2NH 2 (g) ∆H = −92.22 Mathematics involves kJ 3 knowledge and understanding It shows that one molecule of nitrogen gas and three molecules of of the highest certainty. The hydrogen gas combine in an exothermic reaction to produce two Nature of Science (NOS) molecules of ammonia. However, when setting up a reaction the reactants informs us that experimental may not always be mixed in this ratio – their amounts may vary from the data is often quantitative exact stoichiometric amounts shown in the balanced chemical equation. and mathematical analysis is required to enable precise descriptions, predictions The limiting reagent and, eventually, laws to be Experimental designers of industrial processes use the concept of a developed. Mathematics limiting reagent as a means of controlling the amount of products is an integral part of obtained. The limiting reagent, often the more expensive reactant, will scientic endeavours. The be completely consumed during the reaction. The remaining reactants use of numbers and an are present in amounts that exceed those required to react with the understanding of the mole limiting reagent. They are said to be in excess. concept have helped develop It is the Using reagent measured, specic here limiting is calculated amounts that identical to that of the determines amounts products to of be the amount limiting obtained. the experimental or the theoretical predicted or the actual yield of yield of reagent The formed. This is science. “Why is mathematics made achieved products. Chemistry into a physical enables assumption products of products so eective in describing the is natural world?” rarely the IB Diploma Chemistry Syllabus case. as Much this effort equates to is focused increased on improving prots and the yield efcient use of industrial of raw processes, materials. Worked example: determining the limiting reagent In the manufacture of phosphoric m _ acid, n(O ) = 2 molten and elemental then phosphorus hydrated according is to M oxidized the following 100.0 g __ = chemical = 2(16.00) P (l) + 5O 4 (g) + 6H 2 O(l) → 4H 2 PO 3 24.77 g of phosphorus reacts with g mol mol (aq) P 4 (l) + 4 If 3.125 1 equation: 100.0 g 5O (g) + 6H 2 O(l) → 4H 2 PO 3 (aq) 4 of ______________________________ oxygen and excess water, determine the limiting 1 M(g reagent, the amount in mol of phosphoric(V) mol ) 123.88 32.00 acid ______________________________ produced in g, of (the theoretical phosphoric yield) and the mass, acid. m/g 24.77 100.0 excess ______________________________ Solution n /mol 0.200 3.125 excess 0 i The amount in mol of phosphorus and oxygen ______________________________ is determined using the working method from n /mol sub-topic1.2: ______________________________ f To m _ n(P ) determine the amount of oxygen that will = 4 M react 24.77 with the phosphorus we can use a cross- g __ = multiplication = 0.2000 technique: mol 1 4(30.97) g mol 21 1 S TO I C H I O M E T R I C P : R E L AT I O N S H I P S O 4 P 2 1 : 5 0.200 : α (s) + 5O 4 (g) + 6H 2 → O(l) 4H 2 PO 3 (aq) 4 _________________________________ 1 M(g mol ) 123.88 32.00 _________________________________ 1 × α = 0.2000 × α = 0.2000 × 5 m/g 24.77 100.0 excess 5 _ _________________________________ 1 n /mol 0.2000 3.125 excess 0 i α = 1.000 mol _________________________________ Therefore 0.2000 mol of phosphorus requires n /mol 0.0 2.125 excess 0.8000 f 1.000 mol of oxygen to completely react. There is _________________________________ 3.125 and mol of oxygen phosphorus is available the limiting so this is in reagent. excess All the The mass of phosphoric acid, H PO 3 phosphorus will be consumed in the reaction produced be determined by multiplying n by M f 3.125 1.000 = 2.125 mol of oxygen will the reaction comes to : r remain m after can 4 and = M × n = [3(1.01) completion. 1 The limiting reagent dictates the amount × phosphoric to acid determine times the produced. the amount amount produced in compared mol The of of with mole product, ratio in the amount is acid of 30.97 + 4(16.00)] g mol 0.8000 mol = 78.40 g used mol. phosphoric + of Four will be phosphorus: This value represents phosphoric achieved in acid. the theoretical Theoretical yields yield are of rarely practice. Qik qstios 1 Butane lighters work by the release and combustion CO ) Calculate the mass, in g, of K ) Calculate the mass, in g, of O 2 of pressurized butane: produced. 3 produced. 2 2C H 4 (g) + 13O 10 (g) → 8CO 2 (g) + 10H 2 O(l) 4 A solution of 155 g of potassium iodide, KI is added to 2 a solution of 175 g of nitric acid, HNO Determine the limiting reagent in the following . The acid acts 3 as an oxidizing agent. reactions: a) 20 molecules of C H 4 6KI(aq) + 8HNO and 100 molecules of O 10 (aq) → 6KNO 3 + 3I b) 10 molecules of C H 4 ) 0.20 mol of C H 4 ) 8.72 g of C 4 (s) + 4H 2 and 91 molecules of O 10 O(I) 2 2 a) Deduce which reagent is in excess. b) Determine how many grams of this reactant will and 2.6 mol of O 10 H (aq) + 2NO(g) 3 2 2 and 28.8 g of O 10 2 remain unreacted. 2 Two aqueous solutions, one containing 5.3 g of sodium ) Determine how many grams of nitrogen carbonate and the other 7.0 g of calcium chloride, are monoxide, NO will be produced. mixed together. A precipitation reaction occurs: 5 Na 2 CO (aq) + CaCl 3 (aq) → 2NaCl(aq) + CaCO 2 Chlorine gas is produced by the reaction of (s) 3 hydrochloric acid, and the oxidizing agent Determine the limiting reagent and the mass, in g, of manganese(IV) oxide, MnO : 2 precipitate formed (the theoretical yield). MnO (s) + 4HCl(aq) → MnCl 2 3 (aq) + Cl 2 (g) + 2H 2 O(l) 2 The oxygen required in a submarine can be produced At 273.15 K and 100 kPa, 58.34 g of HCl reacts with by a chemical reaction. Potassium superoxide, KO 2 3 0.35 mol of MnO reacts with carbon dioxide, CO to produce oxygen and to produce 7.056 dm of chlorine gas. 2 2 potassium carbonate, K 2 . Write the balanced chemical equation for this reaction. b) 28.44 g of KO reacts with 22.00 g CO 2 a) Deduce the limiting reagent. b) Calculate the theoretical yield of chlorine. 3 a) the limiting reagent. 22 CO 2 . Deduce 1 . 3 r e A c T I n g m A S S e S A n d v O l u m e S Theoretical and experimental yields The balanced what is chemical theoretically carried out under equation possible ideal ● represents when conditions. a reaction It allows changes is amount of products to be reaction calculated – reverse of in industry reactions and work maximize to maximize prots. the the yield reactions conditions and However, especially in processes, many products. factors result in a These factors could of in loss ● impurity products from reaction of side-reactions due to the impurities. calculate made the percentage between the yield theoretical a comparison yield and the reduced amount produced in the process – the include: experimental ● of products under actual of consuming largeis yield as systems existence presence To experimental scale such yield. ● Scientists conditions, pressure the equilibrium theoretical and the ● expected in temperature yield: vessels experimental yield __ of % reactants yield = × theoretical 100% yield Worked example: determining theoretical yield Respirators concern are for being used workplace environmental increasingly safety pollution. and rising Iodine(V) with Step levels oxide, I of 2: 5 I O 2 with carbon monoxide, CO and can be remove I O 2 this (s) + 5CO(g) I g with of I (g) from + 5CO of : CO : 5 0.3000 (g) : α 1 × α = 0.3000 × α = 0.3000 × α = 1.500 5 5 33.6 5 _ g 1 mol carbon The reaction of 0.3000 mol of I O 2 and requires 1.50mol 5 given of an limiting 5 Calculate dioxide the 2 theoretical yield determine air: O 2 CO. gas 2 reacts the → 5 100.0 of poisonous ratios, used 1 to mole O 2 reacts Using reagent. CO for completion. However, only 1.20 mol of CO experimental is yield, 0.900 in mol, mol available; therefore this is the limiting reagent. of CO The , ratio of limiting reagent CO to product CO 2 2 calculate is the percentage 5:5 or 1:1. The number of mol of CO Figure 1 A chemist wearing a yield. that is 2 theoretically possible It that is therefore 1.2 mol. respirator for safety was found 0.90 mol or 39.61 g of CO was 2 Solution produced. Step 1: Calculate the initial amount in mol and determine the limiting is the experimental yield. of To reactants This determine the percentage yield of CO we rst 2 reagent: need to calculate the theoretical yield of CO : 2 m _ n(I O 2 ) = m = M × n 5 M 1 100.0 g = [12.01 = 52.8 + 2(16.00)] g × mol 1.20 mol _ _ _ = 1 2(126.90) = 0.2996 + 5(16.00) g mol mol g Then: m _ n(CO) experimental = yield __ % M yield × = theoretical 33.6 100% yield g _ _ _ = 1 12.01 + 16.00 g 39.61 mol g _ = × 52.8 = 100% = 75.0% g 1.20mol 23 1 S TO I C H I O M E T R I C R E L AT I O N S H I P S Qik qstios 1 Acetylsalicylic acid, also known as aspirin, C H 9 synthesized by reacting salicylic acid, C H 7 acetic anhydride, C H 4 C H 7 O 6 (s) + C 3 H 4 O 6 O 6 6 O 8 O is 3 3 9 H 8 O (s) + C 4 H 2 3 O 4 (l) (g) 2 produced. 2 3 can be produced in the following Sulfur trioxide, SO 3 H CH + O(l) + CO 2 If a 1.68 g sample of sodium hydrogen carbonate is O O O (s) + H 3 heated, calculate the mass, in g, of sodium carbonate (l) → C H CO 2 : 3 3 O (s) → Na 2NaHCO 4 with two-step reaction: O O + H O C 3 OH CH OH 4FeS O 3 (s) + 11O 2 (g) + O 2SO CH (g) → 2Fe 2 2 O 2 (g) → 2SO 2 (s) + 8SO 3 (g) 2 (g) 3 3 30.0 g of iron disulde (pyrite), FeS reacts in the 2 a) Calculate the theoretical yield, in g, of aspirin presence of excess oxygen to completion. when 3.0 g of salicylic acid is reacted with 4.0 g of a) Calculate the theoretical yield, in g, of sulfur acetic anhydride. trioxide. b) If the experimental yield of aspirin is 3.7 g, b) If an experimental yield of 28.0 g of sulfur calculate the percentage yield. trioxide is achieved, deduce the percentage 2 yield. The thermal decomposition of sodium hydrogen results in a 73.8% yield of sodium carbonate, NaHCO 3 carbonate, Na CO 2 : 3 Avogadro’s law and the molar volume of a gas The kinetic behaviour number to hold. 1 theory of of gases are distances. 2 a Gaseous gases or postulates made Most up of particles is a model microscopic postulates These Gases of at assumptions to The that explain theory must be is and predict based true for upon the the a theory are: of very the are used level. small volume particles, occupied constantly separated by moving in a gas is by large empty straight lines, space. but random directions. 3 Gaseous walls 4 SI unit of pressure particles the Gaseous Under The of particles conditions is obeys these Many theory. undergo container. of No exert no standard postulates and elastic loss of force collisions kinetic of attraction temperature the with energy equations and that each other and the occurs. on other pressure, follow gases. an from ideal the gas kinetic –2 the pascal other are units of N m . forces pressu re commonly different the (Pa), used countries, atmosphere high in inclu d ing gas At of degree of Hg), torr, of separation per and the and low gas they pressure, molecules act in a the is way signicance minimized that – adheres of any there to the is a ideal at high pressure and low temperature the particles of a gas mercury bar, more slowly and the distances between the particles decrease. an d Intermolecular pounds between model. move (mm temperature (atm ) , However, millimetres high attraction square attractions (sub-topic 4.4) become signicant and in ch eventually the gas can liquefy. These responses to changing conditions 5 (psi). The bar (10 Pa) is now mean widely used as a that behaviour unit , as it is very atmospheric clos e pressure, of can real depart from ideal gas behaviour and exhibit the gases. to 1 a tm . The early terms and 24 gases convenient by postulates scientists Joseph Louis of the such as kinetic Robert Gay-Lussac. theory Boyle, were Edme explained Mariotte, in quantitative Jacques Charles, 1 . 3 In 1806, Gay-Lus s a c volumes ratio of of reacting whole p r opo se d gases a nd tha t the the r e A c T I n g re l at i on s h ip pro duc t s c o ul d bet we en be m A S S e S A n d v O l u m e S the e xpr e ss e d as a Sty tis numb e r s . Physical There and are examine the many Avogadro’s gaseous behaviour property of particles Imagine gases a of taking and same (0 is are s imp le inate K) (gure as a and a pp ly. the force 2). equal balloon. an d th e ir the gas used i mpo rt a n t exerted by a conversions la w s be h a vi ou r m o de ls An pressure These to the Under particular the the and to e xp la i n gas as Data volume found p hys i c al of in can be booklet. an fo u nd The ideal section u nit in molar gas is 2. its and different will and STP Cl 2 1 of have pressure pressure, O 2 2 1 16.05 g mol of conditions balloons temperature N 1 mass same the temperature 4 4.00 g mol molar (100kPa) CH 1 2.02 g mol T he to standard He 2 r e a c t io ns unde rs t a n d numerically °C/273 H to and s ur fa ce . to known us pressure , a mass g as - phase qua nti ta ti vel y. are i ts each volume conditions s y s te m s w i th a using e na bl e gases gas collide temperature the imp o r ta nt la w constants 1 28.02 g mol 1 32.00 g mol 70.90 g mol Figure 2 The molar volume of any gas is identical at a given temperature and pressure Figure 3 Amedeo Avogadro (1 776–1856) proposed in 1811 3 At STP the balloo ns w i ll hav e id ent ic a l v ol u m e s of 22. 7 1 dm mol . that equal volumes of any gas This is the molar volume of an id ea l g as and it is constant at a at the same temperature and given temperature a nd p re s sur e . Eac h bal lo on c on t a in s 1 mol of the pressure contain the same number 23 gas so it contains 6.02 × 10 atoms or molecu l e s of the g a s. T his of molecules relationship measured number at The 3 H 8 the s a me law as Avogadro’s te mpe r a tur e simplies coefcients correspond C kno wn law : a nd equal pr es s ur e volumes c o n t ai n of the a ny gas s a me ofmolecul e s . Avogadro’s gases. is (g) to the + of ratio stoichiometric a of balanced volumes calculations chemical of the 5O 2 involving equation gases (gure reacting involving gases 4). (g) 2 + 4H O(l) 2 Figure 4 Volumes of gases obey Avogadro’s law 25 1 S TO I C H I O M E T R I C R E L AT I O N S H I P S Worked examples: Avogadro’s law Qik qstio Ammonium carbonate Example 1 decomposes readily when 3 Calculate n(O heated: ) found in a 6.73 dm sample of oxygen gas at STP . 2 3 (NH ) 4 2 CO (s) → 2NH 3 1 (g) mol O (g) + H 2 occupies 22.7 dm at STP 2 3 + CO O(l) 2 Solution Determine the volume, in 3 3 dm 6.73 dm _ , of the individual gases n(O ) = = 2 0.296 mol 3 22.7 produced on decomposition dm of 2.50 mol of ammonium carbonate. Example 2 The hydrogenation gas, H The product of ethyne, C H 2 in the presence of a involves reaction with hydrogen 2 nely divided nickel catalyst at 150 °C. 2 C H 2 (g) is + ethane, 2H 2 (g) C → H : 2 6 C H 2 2 (g) 6 3 When 100 3 cm of C H 2 volume and reacts with 250 cm of H 2 , determine the 2 composition of gases in the reaction vessel. Solution According to 2molecules Avogadro’s of law, hydrogen, for every 1molecule 1molecule of ethane of will ethyne be and formed. 3 Looking at the volumes reveals that only 200 cm of the hydrogen is 3 required, of gases and that contains 100 both cm of ethane ethane C and H 2 (g) will be formed. unreacted + 2H 2 (g) → The nal mixture hydrogen: C 2 H 2 (g) 6 3 initial volume, V /cm 100 250 0 0 50 100 i 3 nal volume, V /cm f 3 After reaction there will be 3 50cm 150 cm of gases in the vessel comprising 3 of H and 100 cm of C 2 H 2 6 The gas laws The gas mass of laws gas are in a series changing of relationships conditions of that predict temperature, the behaviour pressure, and of a xed volume. 3 You at have STP) seen is that Avogadro’s independent of the law states that composition of the the molar volume (22.7 dm gas. Boyle’s law Robert Boyle constant, Gases (1627–1691) inverse contained collisions The an with in the relationship discovered relationship smaller surface between volumes of the ∝ or V p 1 V 26 = 1 V p 2 2 when between will have container, pressure 1 _ p that exists p and so the temperature pressure an increased exert volume and V a number higher can be remains volume. of pressure. expressed as: 1 . 3 where V and p 1 the nal represent the initial volume and r e A c T I n g pressure and m A S S e S V 1 and and pressure, v O l u m e S p 2 volume A n d 2 respectively. aP/P ,erusserp aP/P ,erusserp 3 –3 volume, V/dm 1/V / dm Figure 5 Boyle’s law: the pressure of a gas is inversely propor tional to the volume at constant temperature Worked example: Boyle’s law A helium-lled weather balloon is designed to rise to altitudes as high 3 as 37 000 of 101 m. kPa is atmospheric A balloon released pressure with and is a volume rises 68 kPa. to an of 5.50 altitude Calculate dm of the and 3500 new a m pressure where the volume, 3 in dm . remain It is assumed that the temperature and amount, in mol, constant. Solution First make p = a 101 summary of the data: kPa 1 3 V = 5.50 dm 1 p = 68 kPa 2 3 V = α dm 2 Making V the subject of the expression: 2 p 1 _ V = V 2 × 1 p 2 101 kPa _ 3 = 5.50 dm × 68 kPa 3 = 8.17 dm Charles’s law Jacques the Charles (1746–1823) temperature mass of gas at a and volume constant investigated of a gas. pressure, the He the relationship discovered volume V of that the between for gas is a xed directly 27 1 S TO I C H I O M E T R I C R E L AT I O N S H I P S proportional Absot zo can be to the expressed absolute temperature T in kelvin. This relationship as: We saw in sub-topic 1.1 V V 1 2 _ V that absolute zero is zero ∝ T _ or = T T 1 2 on the kelvin scale, 0 K ( 273.15 °C). The idea of When an inated negative temperatures and (boiling point the existence of a minimum The possible temperature had frequency been widely investigated balloon by the scientic community room gaseous is balloon 196 °C), particles and then placed and removed the into average collide energy temperature is the it with the begins from the balloon a energy internal to deate liquid will container kinetic wall – of the nitrogen of of liquid the the balloon volume and nitrogen particles decreases. with reduces. allowed to If less the return to reinate. before Lord Kelvin’s time (1824–1907). Kelvin stated that absolute zero is the temperature at which molecular motion ceases. According to Charles’s law, if the temperature of a system was to double from 10 K to 20 K, the average kinetic energy of the par ticles would double and the volume would correspondingly double. Figure 6 Reducing the temperature reduces the average kinetic energy of the par ticles of a gas, and the volume reduces V ,emulov Worked example: Charles’s law 3 A glass gas ice-cold syringe water contains over the 76.4 outside cm of the of a gas gas at 27.0 syringe, °C. the After running temperature 3 of the gas reduces to 18.0 °C. Calculate temperature, T (K) occupied by the gas. Figure 7 Charles’s law: the volume of a gas is directly proportional to absolute Solution temperature at constant pressure 3 V = 76.4 cm = 27.0 + 1 T 273.15 = 300.15 K 273.15 = 291.15 K 1 3 V = α cm = 18.0 2 T + 2 V V 1 2 _ _ = T T 1 2 V × 1 T 2 _ = V 2 T 1 3 = 28 74.1 cm the new volume, in cm , 1 . 3 r e A c T I n g m A S S e S A n d v O l u m e S Gay-Lussac’s law Collaboration Having established proportional is directly laws volume proportional remaining constant to gas at to relationship stating that constant temperature temperature involves pressure at pressure is and constant and inversely that The volume pressure, temperature, is the scientic highly Evidence at to volume. collaborative. that (1778–1850) work with ideal gases led him to that when the volume of a gas is constant, the the gas is directly proportional to its absolute temperature. can be expressed by investigate _ or 8 K), no kinetic pressure. walls of increased As that energy the the general laws. of when the temperature container with the ideal temperature gas particles increases, increased the force reaches is zero particles and absolute and it collide frequency, zero exerts with P erusserp the the developing 2 demonstrates the of T 1 (0 to = T Figure new 2 _ 1 T scientists understanding as: p p ∝ other new possibility p often and The and relationship is tested, pressure develop of fundamental the utilized understanding is understanding challenged, Gay-Lussac’s community causing pressure. The combined gas law absolute The three gas laws, Charles’s law, Boyle’s law, and Gay-Lussac’s law, are zero, 0 K combined of gas, in the one law called relationship the combined between gas temperature, law. For pressure, a xed and amount volume is: temperature T p V p 1 2 Figure 8 Gay-Lussac’s law: the pressure of _ 1 _ V 2 = T T a gas is directly propor tional to absolute 2 1 temperature at constant volume The ideal gas equation The ideal volume, gas established both of a that pressure gas these equation temperature, and pressure and the describes and and volume amount the of a relationship amount, volume have gas a in are direct mol, between of gas inversely the ideal Having proportional relationship particles, pressure, particles. with gas the and that temperature equation combines interrelationships: pV = nRT T as ostat a t its o t ia as qatio –1 R is called the as ostat and it has a value of 8.31 J K –1 mol . This value is provided in section 2 of the Data booklet The inclusion of R in the ideal gas equation requires the following units: p (Pa), 3 V (m –3 ), and T (K). Note that 1 Pa = 1 J m ; this allows you to see how the units in the ideal gas equation are balanced: –3 p(J m 3 1 dm 3 ) × V(m –1 ) = n(mol) × R(J K –3 = 1 × 10 –1 mol ) × T (K) 3 m 29 1 S TO I C H I O M E T R I C R E L AT I O N S H I P S TOK Worked example: using the ideal gas The ideal gas equation is a equation to calculate volume model which is the product 3 Calculate the volume, in m , of a balloon lled with 0.400 mol of of a number of assumptions hydrogen gas at a temperature of 22.90 °C and a pressure of 1.20 Pa. about the ideal behaviour of gases. These have been Solution discussed earlier in the –1 topic. Scientic models Convert all data to SI units to enable the use of R as 8.31 J K –1 mol are developed to explain p = 1.20 Pa observed behaviour. In the 3 V = α m n = 0.400 R = 8.31 T = 22.90 development of models what role do imagination, sensory mol perception, intuition, or the 1 J K 1 mol acquisition of knowledge in the absence of reason play? + 273.15 = 296.05 K nRT _ V = p ra ass deviate from 1 1 0.400 mol × 8.31 J K mol × 296.05 K ____ ideal behaviour at very low = 1.20 Pa temperature and high pressure. 3 Under these conditions = 820 m the forces between the gas par ticles become signicant, and the gas gets closer to the Worked examples: determining the molar point where it will condense from gas to liquid. mass of a substance An organic hydrogen, compound and A oxygen containing was only the elements carbon, analysed. Example 1: Empirical formula A was being found to oxygen. contain 54.5% Determine the C and 9.1 % empirical H by formula mass, of the the remainder compound. [3] Solution 54.5 _ n(C) = 4.54 _ = 4.54 12.01 9.1 _ n(H) = 9.0 = 4 = 1 2.28 _ = 2.28 2.28 16.00 empirical ≈ 2.28 36.4 _ The 2 9.0 _ = 1.01 n(O) ≈ 2.28 formula is C H 2 O. 4 Example 2: Relative molecular mass 3 A 0.230 at95 30 °C g sample and 102 of A kPa. when vaporized Determine the had a relative volume of molecular 0.0785 mass of dm A. [3] 1 . 3 r e A c T I n g m A S S e S A n d v O l u m e S Solution pV = nRT n = m/M pV = mRT _ M 1 0.230 mRT _ M = g × 8.31 J 1 K mol × 368 K ____ = 1 = 3 pV 102 × 3 10 Pa × 0.0785 × 10 87.9 g mol 3 m Example 3: Molecular formula Figure 9 A homogeneous mix ture is characterized by a constant Determine the molecular formula of A using your answers composition throughout from parts (a) and (b). [1] Solution molar mass ___ 87.8 ___ = empirical formula 2(12.01) mass + 4(1.01) + (16.00) 87.8 _ ≈ 2 = 44.06 molecular formula = C H 4 IB, Nov O 8 2 2005 Concentration In a typical solution solutions A a of solution solvent. the laboratory rather known is a The solvent than is the in majority the gaseous reactions phase. carried Chemists out need are to in make up concentrations. homogenous solute water of is the mixture usually a solution of solid, is a solute but could described as that be an has a been liquid aqueous or dissolved gas. in When To make up solutions of known solution concentration, volumes must be measured accurately. The molar concentration of substance of a solution is dened as the amount (in mol) Apparatus used to do this 3 a dissolved in 1 dm 3 of solvent. 1 dm = 1litre (1 L). include burettes, pipettes and volumetric asks. amount of substance n/mol ___ 3 concentration c/mol dm = 3 volume of solution V/dm Par ts per million (ppm) is not an SI unit but is often used uits o otatio for very dilute concentrations such as when measuring Units of concentration include: pollutants (see sub-topic 9.1). –3 ● mass per unit volume, g dm ● mol per unit volume, mol dm –3 Concentration in mol dm may also be referred to as oaity, –3 and square brackets are sometimes used to denote molar –2 6 ● parts per million (ppm): one par t in 1 × 10 par ts. concentration, for example [MgCl ] = 4.87 × 10 –3 mol dm 2 3 1 ppm = 1 mg dm 31 1 S TO I C H I O M E T R I C R E L AT I O N S H I P S Worked examples: concentration calculations Example 1: Molarity of solution Solution 3 Calculate the concentration, in mol dm , of a n(C H 8 solution formed when 0.475g of O 5 K) = V × [C 4 H 8 O 5 K] 4 magnesium 3 1 dm _ 3 chloride, MgCl is completely dissolved in water = to 250 cm × –3 × 1.25 mol dm 3 2 1000 cm 3 make a solution with a volume of 100cm = 0.313 mol Solution m = H n(C 8 First calculate n(MgCl O 5 K) × M 4 ): 2 = 0.475 m _ n(MgCl ) 0.313 = mol × [8(12.01) + g 5(1.01) + –1 _ _ _ 4(16.00) + 39.10] g mol = 2 1 M 24.31 + 2(35.45) g mol = 63.9 g 3 = 4.99 × mol 10 Example 4: Concentration of standard 3 Convert the volume in 3 cm to dm : solution 3 1 dm _ 3 100 cm × A 3 = 0.1 standard solution is prepared by dissolving dm 3 1000 3 5.30g cm of sodium carbonate, Na CO 2 in 250 cm of 3 3 distilled Calculate the concentration of the sample n _ ] in a volumetric ask. A 10.0 cm solution: 3 [MgCl water of this pipetteand solution diluted is with removed water to by the bulb nal volume 4.99 × 10 mol __ = = 3 2 3 of V 0.1 0.100dm . Calculate the concentration, in dm –3 mol 2 = 4.99 × 10 dm , of the diluted solution. 3 mol dm Solution Example 2: Concentration of ions 3 First calculate n(Na CO 2 ) in a 10.0 cm sample of 3 3 Determine chloride the ions concentration, in example 1 in mol dm of the the standard solution: above. 3 10.0 cm _ m _ n(Na CO 2 ) = × 3 3 M 250 Solution cm 5.30 g ____ When solid MgCl is dissolved in water, the = 1 2 2(22.99) constituent ions are + 12.01 + 3(16.00) g mol liberated: 3 10.0 cm _ 2+ MgCl (s) → Mg (aq) + 2Cl (aq) × 2 3 250 cm 3 2(4.99 n _ [Cl ] = × 10 mol) __ = = 0.00200 mol 3 V 0.1 dm Finally 2 = 9.98 × 10 calculate the concentration 3 mol dm –3 solution in mol dm : Example 3: Mass of solute n _ [Na CO 2 Calculate the phthalate, C mass, H 8 O 5 K in (a g, of potassium primary ] = 3 V standard) 3 32 3 of a dm in = cm 0.100 4 3 250 0.00200 mol __ = 3 hydrogen 1.25 mol dm solution. 0.0200 mol dm of the diluted 1 . 3 r e A c T I n g m A S S e S A n d v O l u m e S Titrations Quantitative analysis includes a range of laboratory techniques used An to determine the amount or concentration of an analyte. The results aayt that expressed as numerical values with is involving of analysis two known a is until the monitored quantitative A concentration concentration progress is solutions. titration which is chemical through being a substance analysed by a units. given Volumetric is are technique involves added to reaction colour a is a used solution of complete. changes by standard using analytical proced u re. chemists solution unknown The reaction indicators (topic8). A staa sotio or iay Qik qstios 3 1 Calculate the mass, in g, of H SO 2 required to prepare 500 cm 3 sotio is prepared using a of a 2.0 mol dm 4 volumetric ask. Solvent is solution of sulfuric acid. added to a high purity sample 2 is to be used in the laboratory during an A solution of aluminium bromide, AlBr 3 until the level of the solution 3 electrolyte investigation. Calculate the total number of ions present in 2.5 dm of a reaches the mark on the ask. 3 1.6 mol dm solution of AlBr 3 Worked example: acid–alkali titration Sty ti When solving quantitative calculation problems involving 3 Calculate the volume, in dm 3 , of 0.390 mol dm potassium concentrations and volumes 3 hydroxide, KOH solution that will neutralize 25.0 cm of 0.350 mol of solutions, the focus is on –3 dm sulfuric acid, H SO 2 2KOH(aq) + H SO 2 the amount, in mol, of the 4 (aq) → K 4 SO 2 (aq) + 2H 4 substances reacting and their O(l) 2 relationship as shown by the Solution mole ratios in the balanced chemical equation. Step 1: Calculate the amount, in mol, of H SO 2 n(H SO 2 ) = c × = 0.350 = 8.75 : 4 V 4 –3 mol dm 3 × 0.0250 dm –3 × 10 mol –3 Step 2: The mole ratio of acid:alkali is 1:2. Therefore –3 of acid reacts with 2(8.75 × 10 8.75 × 10 mol –2 mol ) = 1.75 × 10 mol of KOH. not Step 3: Calculate the volume of KOH: In topic 9, we will introduce a n V general, simple-to-use formula. = c This formula can also be used for 2 the type of volumetric chemistry 1.75 × 10 mol __ V(KOH) = 3 = 0.0449 dm 3 0.390 mol dm question shown above. 33 1 S TO I C H I O M E T R I C R E L AT I O N S H I P S Questions 1 Epsom used salts as (magnesium bath salts. sulfate) However, the are commonly f) Determine anhydrous Fe(NH ) 4 form the of the water salt of is a drying hydration of agent. To Epsom determine salts, a 2.50 value of the evaporating Bunsen were salt dish burner was and gently ame observed. placed until Table 8 in a shows over further the changes May 3 The mol, of hence the 2 x. equation results. CO mass of evaporating basin 24.10 + of 3H 2 [2] that for is → reaction CH OH be occurring in the is: + H 3 the can a methanol 2 What xH in and 2008 synthesis mass/ 4 2 O a dsitio mass of evaporating basin + MgSO 4 amount, ·xH porcelain heated no 2 of ) g IB, sample the (SO O 2 maximum formed amount from 2 mol of of methanol carbon 26.60 O 2 dioxide mass of evaporating basin after heating and 3 mol of hydrogen? 25.32 A. 1 mol B. 2 mol C. 3 mol D. 5 mol IB, May T able 8 a) Calculate the evaporated b) Calculate of H mass, from the in the g, of water sample. amount amount, in [1] mol, 2006 O. 2 c) Calculate the mass, in g, d) Calculate the amount, of MgSO 4 in mol, of 4 MgSO Calcium carbonate decomposes on heating as 4 e) Calculate MgSO : the ratio amount of of amount H 4 value O and shown below. CaCO → of deduce the 2 of When f) State CaO the formula of the hydrated 2 g of 50 g of The value of x in Fe(NH ) 4 found mol of in the was ) 4 the xH 2 in added. water The O can oxide carbonate yield A. 7% B. 25% C. 50% D. 75% IB, November of are formed. are decomposed, calcium What is the oxide? 2 amount compound. wasdissolved (aq) (SO 2 determining sulfate sample BaCl by calcium calcium percentage be CO salt. 7 2 + 3 x A in 0.982 and g excess precipitate [1] of 2 BaSO was separated and dried and found to 2006 4 weigh a) 1.17 g. Calculate the amount, in mol, of BaSO 4 5 Ethyne, C H 2 in the 1.17 g of precipitate. the b) Calculate the amount, in , reacts with oxygen according mol, of sulfate in equation below. What volume 3 (in dm ) of 0.982 g sample of Fe(NH ) 4 (SO 2 ) 4 ·xH 2 with 0.40 dm of C H 2 2C Calculate 0.982 g the amount, sample of in Fe(NH mol, ) 4 d) Determine present Fe(NH ) 4 (i) e) in (SO 2 iron Use the the the your g ·xH 2 in g g, in of the iron in ·xH O. 2 (g) 2 A. 0.40 B. 0.80 + 5O (g) → 2 from of (d) water Fe(NH ) + 2H O(g) 2 [1] 2 following sulfate. to ) 4 [3] determine present (SO 2 (g) 2 C. 1.0 D. 2.0 IB, November of (iii) part 4CO the 2 of H 2 [1] O: mol sample ) 4 sample 4 34 (SO 2 ammonium answer amount 0.982 ) 4 (ii) mass, 0.982 of ? 2 O. [1] c) oxygen 3 reacts 2 the to 2 [2] in ·xH 2 the O. 2 [2] 2007 Q u e S T I O n S 6 A xed mass of an ideal gas has a volume of 10 A toxic gas, A, consists of 53.8 % nitrogen 3 800cm under certain conditions. The pressure and 46.2% carbon by mass. At 273 K and 5 (in kPa) What with is and the other temperature volume of conditions (in the K) gas are after remaining both these the doubled. 1.01 changes Determine same? × 3 10 Calculate Pa, 1.048 the the 200 cm B. 800 cm C. 1600 cm D. 3200 cm IB, May 2005 3 IB, specimen An oxide of empirical molar anddetermine 3 A. g its A occupies formula mass of the molecular paper of 462 cm . A. compound structure. [3] 2009 3 3 [1] 11 of of hydrogen Assuming complete reaction, what volume of were After gas apparatus 7 copper hydrogen. was had was reduced heating, the maintained cooled. The in a stream stream until of the following results obtained. –3 0.200 mol dm potassium hydroxide solution, 3 KOH(aq) is required to neutralize 25.0 –3 of 0.200 mol dm aqueous sulfuric Mass of Mass of SO 2 dish dish and = 13.80 contents g before heating = acid, 21.75 H empty cm g (aq)? 4 Mass of dish and contents after heating and 3 A. 12.5 cm B. 25.0 cm C. 50.0 cm D. 75.0 cm IB, May leaving to cool = 20.15 g 3 a) Explain why the stream of hydrogen gas was 3 maintained until the apparatus cooled. [1] 3 [1] b) 2007 Calculate the of using copper complete c) 8 Copper metal may be produced by the Write an empirical the reduction equation formula data of for of above, the the oxide assuming oxide. the [3] reaction that reaction occurred. of copper(I) oxide and copper(I) d) according to the below equation. State two O + Cu 2 A → 6Cu + kg of of IB, 2 10.0 kg copper(I) of copper(I) sulde was oxide reaction heated until Determine as that it would was be observed heated. [2] November 2004 no 0.502 g of an alkali metal sulfate is dissolved occurred. in a) tube and 12 further the SO 2 mixture 5.00 S changes [1] inside 2Cu [1] sulde the limiting reagent in water and excess barium chloride solution, this BaCl (aq) is added to precipitate all the 2 reaction, showing your working. [3] sulfate ions as barium sulfate, BaSO (s). The 4 b) Calculate that of the could maximum be obtained mass from of copper these precipitate masses reactants. 0.672 May Calculate organic of compound carbon, remainder 24.1% being A contains by mass of 62.0 % the amount Determine nitrogen, the c) Determine metal hydrogen. the percentage by mass hydrogen and the empirical formula (in mol) of barium Dene c) The the term relative molecular molecular mass of A of A. mass. is the November molecular formula (in mol) of the present. [1] the Deduce the molar and mass state identity of its of the alkali units. the alkali [2] your metal, workings. [2] [3] Write an equation for the precipitation [2] including state symbols. [2] 116. IB, Determine amount sulfate sulfate reaction, relative [2] of e) b) the metal showing IB, weighs by d) a) and formed. Determine alkali mass dried 2006 b) An and g. sulfate 9 ltered [2] a) IB, is of A. May 2007 [1] 2006 35 1 S TO I C H I O M E T R I C 13 Aspirin, the one world, equation of the can be given R E L AT I O N S H I P S most widely prepared used drugs according to in B. the A student times, below. and OH of out three obtained Mass OCOCH carried with the brass = this following 0.456 g ± brass three nails, results. 0.001 g + O H CH COOH 3 C COOH H Tit 1 CH O 3 Iitia o o aspirin excess solid aspirin ethanoic salicylic O 2 acid anhydride. 0.00 0.00 28.50 28.60 28.40 28.50 28.60 28.40 0.10 0 o 2– some 0.00 –3 ethanoic acid S reacted with 3 COOH ethanoic anhydride student 2 C C 3 salicylic acid A small 3 O A. experiment identical 3 (± 0.05 ) 3 fia o o Impure –3 was obtained by ltering 0.10 0 o the 2– reaction mixture. Pure aspirin was O S obtained 2 by recrystallization. Table 9 shows the 3 (± 0.05 ) 3 data vo a o recorded by the student. –3 0.10 0 o 2– O S Mass of salicylic acid used 3.15 ± 0.02 g Mass of pure aspirin obtained 2.50 ± 0.02 g 2 3 (± 0.10 ) 3 Aa o 28.50 a o 0.10 0 o –3 2– S Table 9 O 2 3 3 (± 0.10 i) Determine salicylic the acid, amount, C H 6 ii) Calculate the in mol, (OH)COOH, used. [2] Table 10 4 theoretical yield, in g, i) ofaspirin, C H 6 (OCOCH 4 )COOH. Calculate the [2] Determine the percentage yield of S O ii) the gures Calculate number associated of the aspirin with obtained, percentage this the mass Calculate and Another iv) Calculate uncertainty obtained 150%. an The repeated the 14 May Brass uses. checked found no is a the percentage of samples of analysis are is carried copper brass. [1] alloy out present The shown to in reactions with many determine three the identical involved in this below. + Step 1: Cu(s) + 2HNO (aq) + 2H (aq) → 3 2+ Cu (aq) + 2NO (g) + 2H 2 O(l) 2 2+ Step 2: 4I (aq) + 2Cu (aq) → 2CuI(s) + I (aq) 2 2 Step 3: I (aq) + 2S 2 36 O 2 of 3 mass of [1] copper the percentage in [1] by mass in the brass. [1] The manufacturers 2 (aq) → 2I of brass claim contains that the 44.2 % mass. Determine the copper percentage the result. copper-containing analysis the copper in errors. 2009 An mol, brass. error IB, in brass. yield IB, Commenton the experiment experimental teacher and in [2] student calculations amount, associated mass. by of the of sample and [2] calculate v) v) in 3. 3 present iii) of with step signicant the pure amount, in [1] copper State added of pureaspirin. iv) average 2 mol, 3 2 iii) ) of (aq) + S O 4 6 (aq) May 2010 the result. [1] 2 AT O M I C ST R U CT U R E Introduction Australian-born British Lawrence (1890–1971) 1915 Bragg Nobel Prize in physicist Physics with A William shared his wrong than the no theory theory is at always so much father, William Sir William crystal the structures development William on of Henry Bragg record, only 1969 is using of the having 25. Bragg In a Bragg, for their X-rays, X-ray analysis which led received Nobel the tape-recorded to prize at the Chemistry science” laureate interview Lawrence Bragg of crystallography. youngest better all. age in said: theory. terms is sometimes and at Every of the chemical atoms. In described centre this various theories current understanding and of reaction topic models of as the chemistry we can that the be shall “central lies atomic explained examine have led structure of to in the our the atom. 2.1 T to Understandings Applications and skills ➔ Atoms contain a positively charged dense A ➔ Use of the nuclear sy m bol notation X to Z nucleus composed of protons and neutrons deduce the number of proto ns , neu trons, an d (nucleons). electrons ➔ in atoms and ions . Negatively charged electrons occupy the space ➔ Calcu lations involving non- integer relative outside the nucleus. atomic ➔ The mass spectrometer is used to determine from masses given and data, a bu nd a nce inclu d ing of ma ss is otope s s pectra. the relative atomic mass of an element from its isotopic composition. Nature of science ➔ Evidence and improvements in instrumentation – alpha par ticles were used in the development of the nuclear model of the atom that was rst proposed by Rutherford. ➔ Paradigm shifts – the subatomic par ticle theory of matter represents a paradigm shift in science that occurred in the late 1800s. 37 2 ATO M I C S T R U C T U R E Background to atomic theory Two Greek philosophers, Democritus stated wascomposed of Leucippus around 440 indivisible BC contain and that particles by matter However, no concrete scientic The law of given to support accepted to any this hypothesis great degree by and the at so it 1808 the developed supported formed an the schoolteacher atomic model experimental origin of of matter data. atomic John This theory of destroyed. of a modern this a not only science. We shall that see scientic model over was progressively called comprise ● Postulate very Matter total mass cannot of be matter the is start equal of the to the total reaction. an to be accepted explanation but correctly outcomes should be able of to it should known the of future predict experiments. to deduce was Dalton used his theory another law: underpins in this The law of rened multiple proportions : If two topic X and Y combine in different ways and form more than one compound, the time. the matter summarized theory provide masses Dalton matter: reaction before observations to replaced chemical matter elements how of The Dalton model that For ● much oxygen thetime. English by parts was Dalton’s atomic theory In 3 scientic mass community conservation or following not and evidence created was sulfur termed ● atomos. 1part mass. as 1: small indivisible “atoms”. building Dalton’s blocks theory that can of be Y of can whole X be that combine expressed as with a a ratio xed of mass small numbers. follows. All matter particles (materials) called consists of Example: The law of multiple atoms propor tions ● Postulate one 2: type An element consists of atoms of only. Chemists as a an ● Postulate 3: Compounds consist of atoms example than one element and are formed atoms in whole-number of of the the time did amount law of not of use the mole substance. multiple As proportions, measuring the mass of carbon and oxygen by in combining Dalton’s of consider more in measure forming the two compounds carbon monoxide, ratios. CO(g), and carbon dioxide, CO (g). An experiment 2 ● Postulate 4: In a chemical reaction atoms are might not The created simple “laws wereknown 1800s and or to The by law a that 3 g of carbon combines with destroyed. of chemical the Dalton’s scientic theory combination” community explains a in the 4 g of oxygen 3 g of carbon to form have number to form carbon combines carbon combined with dioxide. in monoxide, 8 g Carbon different ratios of oxygen and to whereas oxygen give different compounds: ofthese. ● measure of denite French proportions: scientist, This Joseph was Proust, proposed in CO(g): C:O ratio = 3:4 CO C:O ratio = 3:8 1799. (g): 2 The law states that a compound always has The the same proportion of elements by mass. ratio with example, if you measure the mass of of sulfur the same in sulfur trioxide, SO it will masses mass of of oxygen carbon to that form combine the two and compounds oxygen the For is 1:2 (a simple ratio of whole numbers). always 3 Stdy tip In science, a w can be considered a summary of several observations. 38 2 . 1 T h e n u c l e a r TOK a T O m Qik qstio Deduce the ratio of the mass John Dalton was a brilliant scientist. He never married and said: “My head of oxygen per gram of sulfur in is too full of triangles, chemical proper ties, and electrical experiments to the compounds sulfur dioxide, think much of marriage!” He was an multidisciplinary scientist, who worked SO (g), and sulfur trioxide, 2 in the disciplines of physics, mathematics, biology, and philosophy, as well SO (g). 3 as chemistry. Do you think philosophy still has a place in modern scientic thinking? Debate this question in class and consider why scientists should always try to embrace an interdisciplinary approach in their thinking. Dalton was colour blind and saw himself as being dressed in grey clothes. His only known pastime was bowling. Could the wooden balls on the bowling green have inuenced his theories of the atom? How impor tant is the work–life balance for the scientic practitioner or indeed for society as a whole? atoms of element X atoms of element Y compound consisting of elements X and Y Figure 1 Schematic showing some of the principles of Dalton’s theory. Examine each of the four postulates and discuss each one in relation to the three representations shown here Thomson’s “plum-pudding” model of the atom Although Dalton’s theory did “What is not the 1808 answer atom postulates one had merit, fundamental composed his question: the atom eaten with of?” on was negatively embedded It was almost another 100 years before One to of physicist at the gather the rst J.J. evidence leaders Thomson Cavendish Cambridge, UK. to in eld this was (1856–1940), laboratory In answer the 1906 at the Thomson in a to a day charged positive plum in the pudding UK particles region and (like (the (a dessert Ireland), raisins) “pudding”) of scientists the began similar Christmas question. the who English worked University won atom. the + of Nobel + Prize in Physics for the discovery of the + electron. + + Thomson worked on cathode rays, which he “pudding” of positive + suggested consist of very small negatively + charged + charge spread over negatively particles called electrons. + + the entire sphere charged par ticles The term “electron” was originally proposed by the Irish scientist George Johnstone Stoney in 1891. Figure 2 Thomson’s “plum-pudding” model of the atom. In the analogy, raisins represent negatively charged par ticles embedded in a pudding of positive charge. Overall there is a balance between the positive and negative charges since the Thomson proposed what is now termed the atom is electrically neutral “plum-pudding” model of the atom – he said that 39 2 aTO m I c S T r u c T u r e Rutherford’s gold foil experiment Most gold Thomson’s model raised a number of of the foil matter is electrically neutral, of negatively charged particles in that very they must particles. The also contain search for for a New more detailed Zealand model physicist of Ernest the sli g ht l y some p a r ti cle s we re as de e c t e d back angle s a nd towar d s so me the e ve n so ur ce . bo un c e d Th e s e the had colli d e d hea d- on wi th wha t we particles know to be t he nucleus in the gold atom atom ( gure led th ro u gh positively these now and But large particles charged we nt de e cte d atoms straight implies were the by presence p a r ti cle s some questions. expected. Because alpha and 4) . Ruther f o r d d e scr ibe d thi s r es ult by Rutherford commenting: (1871–1937) foil and experiment experiment model by tested placing evacuated co-workers in 1909. to Thomson’s a thin chamber gold and conduct Published in the 1911, gold It this “plum-pudding” metal foil bombarding it in was as 15inch itcame an straight particles ( gure3). Alpha ( α) as shell back if at you a and had piece hit red of a paper and you! with Rutherford alpha incredible, artillery based his explanation on the fact that particles the gold foil consists of thousands of gold atoms. 2+ are high-energy, positively charged He ions When emitted from elements naturally such as occurring the beam of positively charged alpha radioactive particles bombarded particles passed the foil the majority of the radium. atom consists through mainly of unde ected, empty space. since the However, some alpha par ticles are deected at (scattered) at large angles the core of the atom lies a dense region of most alpha par ticles positive charge called the nucleus. When an are undeected alpha gold particle atom it when it hit initial path. came close de ected the to the through nucleus it nucleus a large re ected of a angle, back and along its par ticle deected thin gold foil beam of par ticles + + + deected par ticle + lead block containing circular zinc sulde a source of radioactive uorescent screen beam of alpha + + deected alpha par ticles par ticles par ticle + Figure 3 Rutherford’s experiment. The zinc sulde uorescent screen was used to detect alpha par ticles that had passed + through or been deected by the gold foil + + The results of Rutherford’s ground-breaking at the experiment time. Based were on Thomson’s + model alpha of the atom, particles Rutherford would have expected sufcient that energy + the to + pass directly mass that that the their 40 made alpha direction would his through involve results up the the particles on were gold a astonishing. He the of reected par ticle predicted decelerate through minor distribution atoms. would going only uniform gold deection. and gold atoms that foil However, Figure 4 Rutherford’s model, which explains his ndings in the gold foil experiment 2 . 1 T h e n u c l e a r a T O m T s i si The scale of the atom The word “nucleus” means the Rutherford’s work has formed the basis of much of our thinking on central and most important part the structure of the atom. Rutherford is rumoured to have said to of an object. The word is used hisstudents: in both chemistry (the nucleus All science is either physics or stamp-collecting! of an atom) and biology (the nucleus of a cell). The is vast hard space to in fully rugby-playing and on looking the centre represent The relative diameter the of an nucleus. sub-topic the the volume The of the centre stand. open is a pitch space in occupies a to the the atom of row at you and 4 the scale is a great of 6) seats the would nucleus. vast of nucleus (gure placed and times the top gure volume 100000 idea the the is in the Zealand were and of stadium grape model tiny New Park grape the atomic size from electron approximately return Eden small the tiny native at the the between between nucleus shall of If Rutherford’s atom We being distance of to Rutherford’s Imagine distance representation unrealistic. at of eld, the compared appreciate. tier the atom nation. down upper of the our is atom the of simple obviously and the diameter the atom of in Figure 5 The 100 New Zealand dollar note, issued in 1999, shows a picture of Lord 2.2. Rutherford, reecting his immense contribution Atoms themselves are extremely small. The diameter The unit of most atoms to science. Do any bank notes in your own 10 is in the range 1 × 10 10 to 5 × 10 m. used to describe the country have pictures of famous scientists? dimensions of atoms is the picometre, pm: 12 1 In is pm = X-ray the 10 m crystallography angstrom, a symbol commonly used unit for atomic dimensions Å: 10 1 For Å = 10 m example, the atomic radius of the uorine atom is quoted in 12 section toÅ 9 we given of the can Data use booklet as 60 dimensional × 10 m analysis, (60 using pm). the To convert conversion this factors above: 12 10 m _ 60 pm × 1 Å _ 1 × = 0.60 Å = 6.0 × 10 Å 10 1 pm 10 m Figure 6 Eden Park, Auckland, New Zealand Can we see atoms and are they real? All the models However, be seen. working for In 1981 at IBM tunnelling generates This The for have two in discussed people physicists, Zurich, microscope allowed scientists Prize in the ground-breaking are Gerd ability in an images to 1986 assumed only and invented electron of atoms when Heinrich the at the to real. can Rohrer, scanning that atomic individual awarded are they microscope surfaces observe was that “real” Binnig Switzerland (STM) , Physics have objects three-dimensional Nobel their we many atoms Binnig level. directly. and Rohrer work. 41 2 ATO M I C S T R U C T U R E TOK us so A simulation of Rutherford’s The American theoretical physicist Richard Feynman (1918–1988) said: gold foil experiment has been If ... all of scientic knowledge were to be destroyed, and only one developed by PhET at the sentence passed on to the next generation... I believe it is that all things University of Colarado, Boulder, are made of atoms. USA and is available on their Are the models and theories that scientists create accurate descriptions of the website. natural world, or are they primarily useful interpretations for the prediction, http://phet.colorado.edu/ explanation, and control of the natural world? No subatomic do we use to par ticles interpret ca n be ind ire ct directly obs er v ed. evidence, ga i ned Which ways throu g h t he of know ing use of technology? Subatomic particles and descriptions of the atom After Rutherford’s followed in having much a Atoms the consist more of ● the proton ● the neutron ● the electron. Section 4 of the coulombs, C, very and small these of masses experiment period Data types booklet each of 1909 picture of a number 1935, of the subatomic gives these atomic (table in approximately detailed three the to the mass unit, experiments structure of in the scientists atom. particle: mass, subatomic of culminating kg, and the particles. in The masses amu, is a charge convenient in given unit are for 1). –24 1 amu = 1.660539 Sbtoi × 10 g c mss/ +1 ∼ 1 lotio p ti proton neutron 0 electron nucleus nucleus ∼ 1 1 __ 1 outside the nucleus in the 1836 electron cloud T able 1 A comparison of the subatomic par ticles The neutron 1932 (gure Chadwick’s which was discovered by British physicist James Chadwick in 7). discovery beryllium, of Be, the neutron placed in a was vacuum based on chamber an experiment was in bombarded with 2+ alpha particles, to emit to prove He neutrons , emitted and categorically based that from on the polonium. Chadwick’s particles were Figure 7 British physicist Sir James Chadwick gamma rays as had been previously (1891–197 4), who was awarded the Nobel Prize 4 in Physics in 1935 for discovering the neutron α 2 42 9 + Be 4 12 → C 6 1 + n 0 The mass thought: in beryllium was calculations fact neutrons he found was and able not 2 . 1 The discovery jigsaw the of the neutron puzzle of atomic existence of the Chadwick’s was structure. neutron but at the time the Rutherford had no last always conclusive piece of T h e n u c l e a r a T O m the postulated evidence until discovery. The atomic number, Z The ● atoms The an of each atomic atom of element number an have is the element. an individual number Different of atomic protons elements in have number, the Z: nucleus different of atomic numbers. For a neutral protons, ● Z for for atom the number of electrons equals the number of example: oxygen, O, is 8. Therefore the oxygen atom has 8 protons and 8electrons. ● Z for 29 copper, Cu, is 29. Copper atoms have 29 protons and electrons. The mass number, A The mass both ● of the protons The mass neutrons For ● ● atom and is number, in concentrated in the nucleus, which contains neutrons. the A, is nucleus the of number an of protons + the number of atom. example: Z for 9 electrons. A uorine, for F , uorine-19 is 9. Therefore is 19. uorine Therefore has uorine-19 9 protons has 19 9 and = 10 neutrons. The nuclear symbol The for nuclear a symbol particular includes element X and both is A and Z For example, hydrogen has three isotopes: represented 3 H (tritium) 1 like this: 1 proton, 1 electron, 2 neutrons 1 neutron 0 neutrons A X 2 H Z (deuterium) 1 1 proton, 1 electron, Isotopes 1 H As you forms saw of atomic in the sub-topic same number, Z, 1.2, element but isotopes that different have mass are the different same numbers, 1 they have different numbers of neutrons 1 electron, nature most elements exist as mixtures of isotopes. in For their proton, A, In because (hydrogen) 1 example, boron contains the two naturally nuclei. occurring 19.9%) isotopes and boron-10 boron-11 (natural (natural abundance abundance 80.1%). 43 2 ATO M I C S T R U C T U R E Stdy tip Isotop it: n y d wpos An easy way to remember the Uranium found in nature consists of three isotopes with the relative abundances A X order of the nuclear symbol and atomic compositions shown in table 2. Z is “A to Z”, after the rst and last Isotop rtiv nb o nb o nb o bd potos tos tos U 0.0055% 92 protons 92 electrons 142 neutrons U 0.7200% 92 protons 92 electrons 143 neutrons U 99.2745% 92 protons 92 electrons 146 neutrons letters of the alphabet. 234 235 238 T able 2 Isotopes of uranium Uranium-235 (split ting) has be Stdy tips a much used the higher id uranium ● is with to into in nuclear release of a abundance increase enriched the reactors large of U-238 propor tion uranium and where amount than of of it u n d e r g o e s i ss i o energy. U-235 U-235. depleted so The Natural uranium uranium ore separation uranium is the of may natural p h y s i ca l Isotopes are often written process of i s o to p s p ti o . with just their mass number 37 Cl may be A. For example, Because they are the same element (same Z) isotopes have the same chemical 17 37 written as chlorine-37, ● Cl, proper ties but they show dierent physical proper ties due to their dierent mass or Cl-37. numbers, The atomic number, Z, can be The dierence in mass between U-235 and U-238 can be used to enrich a fuel obtained directly from the with U-235. In some nuclear reactors natural uranium is used as the fuel but periodic table (sub-topic 3.1; uranium used for nuclear weapons needs to be of higher grade and is usually section 6 of the Data booklet). enriched. rdiotiv isotops can occur naturally or can be A ativity 1 In class, discuss the pros and cons of nuclear energy and debate the issue of ar tically produced. Carbon-14 countries developing nuclear weapons programmes. is an example of a radioisotope 2 ) Deduce the number of protons, electrons, and neutrons in the isotopes that occurs naturally. 37 35 Cl and 17 Cl. 17 37 b) Deduce the number of protons, electrons, and neutrons in the ion, Cl 17 us so Radioisotopes The WebElements website, developed by Professor Mark Winter at the University of Sheeld, UK , contains lots of information about the elements. It includes a link to isotopes, showing the naturally As well as boron-10 radioisotopes boron-12, for and boron-11, boron-13. diagnostics, pharmaceutical archaeological and (radioactive boron isotopes). Radioisotopes treatment, research, and and are research, as also has Examples as “chemical used a are in tracers clocks” number of boron-8, nuclear in boron-9, medicine biochemical in geological and and dating. occurring isotopes and radioisotopes of the various Iodine radioisotopes as medical tracers elements of the periodic table. The http://www.webelements.com/ into thyroid the control 44 gland in the bloodstream. the body’s neck These growth releases thyroxine hormones and or metabolism. and chemical An triiodothyronine messengers overactive thyroid 2 . 1 gland the produces anxiety, Iodine goitre is and also iodine-131 example of in in and cancer the In image rays to is and such the gland) gland. in In as is gland levels of loss. (short- treatment whether contrast, a T O m accelerates high weight n u c l e a r radioisotope the patient thyroid this high-energy determine a and The are used hospital, of hormones symptoms which camera. brain to thyroid ( γ) Iodine-131 gamma and two thyroid diagnostics, an a these leading the gamma normally. using of body photons. functioning prostate the (swelling emits wavelength) cancer excess of concentrated iodine-131 is an metabolism T h e the given can be of thyroid thyroid gland radioactive obtained, iodine-125 is for used to treat tumours. Figure 8 A single-photon emission computed Positron emission images tracer tomography (PET) scanners give three-dimensional tomography scanner can be used to detect the of concentration in the body, and can be used to detect gamma rays from iodine131 cancers (see sub-topic tomography emitted from D.8). (SPECT) Single-photon imaging can be emission computed used to detect used to treat the gamma rays iodine-131. Cobalt-60 in radiotherapy us so Cobalt-60 also emits gamma rays and is cancer. The Nobel Prize in Chemistry is awarded annually by the Royal Carbon-14 in cosmic, geological, and archaeological dating Swedish Academy of Sciences, Radioisotopes cosmic, are often geological, and used as radioactive archaeological clocks matter. for The the dating American of Stockholm, Sweden. scientist The Nobel Prize website Professor Willard Libby won the Nobel Prize in Chemistry in 1960 for gives information about the his method that uses carbon-14 for age determination in archaeology, various Nobel Prize winners in geology, geophysics, and other branches of science. chemistry and other elds of Nitrogen is present nitrogen-14. The in the Earth’s atmosphere is atmosphere constantly as the isotope bombarded by science and medicine. highly Chemistry was deemed the penetrating cosmic rays from outer space and this neutron most impor tant science for the bombardment causes radioactive carbon-14 to form, along with work of Alfred Nobel. hydrogen, according to the nuclear equation: ● 14 1 N + 7 This 14 n → neutron H Chemistry Nobel laureate? 1 bombardment atmosphere, gas consists The + 6 the of half-life, What is the average age of a 1 C 0 as it 78% t is is of results in continuously the the Earth’s time it air takes a constant formed by for supply from of carbon-14 nitrogen-14. in Nitrogen volume. an ● When will this year ’s Nobel Prize in Chemistry be announced? amount of radioactive isotope 1/2 to decrease decay process Carbon-14 carbon other their to is can dioxide 5730 be carbon continually When a the is initial to form photosynthesis in their ratio bodies. and between essentially exchanged living its The or carbon and half-life they exhale carbon-12 the dies with at dioxide. assimilate Animals constant with organism atmosphere value. for the carbon-14 years. compounds, the atmosphere of oxidized for compounds organisms with one-half consume carbon and any other its in plants, dioxide. time, the carbon is plants carbon carbon-14 given atmosphere however, Living the taking In found since all in longer of in living the carbon processes no absorb into is life. exchanged organisms. 45 2 ATO M I C S T R U C T U R E The carbon-14 emitting beta 14 → N + result is tha t the r e to carb o n- 1 2 carbon-14 in the of measured. artefacts and undergo in the decay to form nitrogen, process: e carbon-14 be then -1 7 net may (electrons) 0 14 C 6 The isotope particles body of Scie nti s ts such as in is a g ra dual d ec re a se or g a ni sm ’s pl a nt ca n w o o d, a the use or a ni m a l this to p a py ru s , the The that met h o d p ai ntings, in bod y. ra tio of am ou n t wa s on c e d et er m i n e a nc i e n t of l iv i n g the ca n age man u s c ri pt s , scrolls. The Shroud of Turin The Shroud of one used wrap shows the many in at from on the is the of that Rome the Technology, dating Turin image believe Vatican based to a it a person cloth the of who All originated Christ of three of the Swiss Arizona, results and In to ad, and laboratories out the the the Institute carry that 1390 be cloth 1988 analytical Federal USA, to The traumatized Jesus. conrmed 1260 people death. independent UK, between many his physically crucixion three Oxford, by after appears the University Shroud. believed Jesus commissioned and the of represents University cloth linen body of carbon-14 samples suggesting taken that the Figure 9 The Shroud of Turin Shroud was not and debate and historians the about to burial the this cloth Shroud of Jesus. continues Nevertheless, amongst the controversy scientists, theologians, day. us so In July 2013 Giulio Fanti and co-workers from the University of Padua, An app (application) has Italy, published research in the journal Vibrational Spectroscopy which been developed by the shows a two-way relationship between age and a spectral property Diocese of Turin in Italy and of ancient ax textiles. The media reported their ndings worldwide, the International Centre of claiming that the results dated the Shroud of Turin between 300 bc and Sindonology (scientic study 400 ad, which could date from the time of Christ. of the Shroud). The app is named Shroud 2.0 and using this you can explore the ativity various images, scientic and In class, consider and debate the aspects of hypothesis, theory, technology, and theological interpretations. analytical evidence surrounding the Shroud of Turin. Relative atomic mass 1 ____ The mass of the electron is negligible amu ( ). The mass of the atom 1836 is concentrated the in mass of a section4 of the relative the ● single atomic unied mass are The atomic as is booklet, masses. unied nucleus atom Data dened dened the The mass tiny, and the as it seen is atomic unit protons in more mass 46 amu or mass one-twelfth 1 u table according neutrons. 1 of this convenient unit This unit of is unit the is mass used to a to (more IUPAC) non-SI of a = 1.6605402 × 10 kg. unit to However, sub-topic use a correctly and of carbon-12 express 27 1 and and system of termed relative atomic asfollows: atomic ground-state. in masses mass atom of and in atomic is its particles: 2 . 1 ● The relative atomic mass, A , is the ratio of the average mass T h e of n u c l e a r a T O m the r atom to the unied atomic mass Stdy tip unit. rtiv toi ss is a ratio As mentioned in sub-topic 1.2, the average mass of the atom is a weighted so it does not have units. average of the atomic masses of its isotopes and their relative abundances. The mass spectrometer The mass atomic spectrometer mass of an is element. an It instrument can also used show its to determine isotopic the relative composition. detector (stage 5) lighest par ticles positive ions are (deected most) accelerated in the electric eld (stage 3) heating lament to vaporize magnet (stage 4) sample (stage 1) inlet to inject heaviest par ticles sample (deected least) N electron beam to ionize sample (stage 2) S Figure 10 Schematic diagram of a mass spectrometer There ● are ve Stage 1 where stages in this process: (vaporization): it is heated and The sample vaporized, is injected producing into the gaseous instrument atoms or molecules. ● Stage 2 (ionization): energy electrons, X(g) e The gaseous generating atoms positively are bombarded charged by high- species: 100 + ● Stage 3 → X (g) + (acceleration): charged plates Stage (deection): eld the 2e 4 and 80.1 (area under peak) The positive accelerated perpendicular mass-to-charge The to the positive their ratio in path. (the ions attracted electric ions The m/z are are ratio). The negatively eld. deected degree to of by a magnetic deection species with depends the on ecnadnuba evitaleR ● + 50 19.9 (area under peak) smallest 0 mass, m, and the highest charge, z, will be deected the most. 0 Particles with no charge are not deected in the magnetic 2 4 6 8 10 eld. m/z ● Stage ratio. The (detection): ions instrument charge The 5 The are mass isotope) can spectrum the m/z the mass so a detects and deection therefore or relative detector counter adjusted The is The the be detected. versus indicates hit an that will plot only then of number, abundance of the species electrical positive depend relative A. The of a particular signal is ions of only respective of a on abundance height m/z generated. Figure 11 Mass spectrum of boron. The two peaks correspond to two isotopes single the (of each mass. each peak isotope. 47 12 2 ATO M I C S T R U C T U R E Stdy tips Worked examples: calculations involving In the periodic table of non-integer relative atomic masses and elements in section 6 of the Data booklet, the atomic abundances of isotopes number Z is given above the symbol for each element. The Example 1 number below the symbol Boron has two naturally occurring isotopes with the natural represents the relative atomic abundances mass, A shown in table 3. (gure 12). r Isotop nt bd/% 10 5 Z B 19.9 B 80.1 11 B Table 3 Isotopes of boron 10.81 A r Calculate Figure 12 Periodic table entry the relative atomic mass of boron. Solution for boron The ● of Don’t confuse the nuclear relative the atomic isotopes and mass their is the weighted relative average of the atomic masses abundance: 11 symbol, eg B for boron-11, 5 80.1 _ 19.9 _ relative atomic mass = 10 ( with the representation × ) + ( 11 × ) = 10.8 100 100 given in the periodic table. The nuclear symbol refers Example 2 to a par ticular isotope or Rubidium has a relative atomic mass of 85.47 and consists of two naturally id, and shows both 85 occurring isotopes, 87 Rb (u = 84.91) and Rb (u = 86.91). Calculate the atomic number Z and mass percentage composition of these isotopes in a naturally occurring sample number A, with A shown of rubidium. above Z. In the periodic table however, the relative atomic mass A Solution is given r along with Z, with Z above A . Note that in this example exact u values are given correct to two r decimal ● places so you need to use this information in your answer. Never round the relative In Example 1 no such precise information was given. atomic mass when answering a question. Always use the data given 85 Take a sample of 100 atoms. Let x = number of Rb 87 (100 x) = number of Rb atoms in the sample. in section 6 of the Data 84.91x + 86.91(100 x) ___ booklet and express values A = 85.47 = r 100 to two decimal places, eg A (H) = 1.01. cross-multiplying: r 84.91x + 86.91(100 84.91x + 8691 solve by making x x) 86.91x the 2.00x = -144 x = 72.00 = = subject 8547 8547 of the expression: 85 The 48 sample contains 72.00 % Rb 87 and 28.00% Rb. atoms and 2 . 1 T h e n u c l e a r a T O m 45 Example 3 Deduce the spectrum relative in gure atomic 13 and mass of identify the X element from the X from periodic its mass 40 table. 35 ● The ● In of mass spectrum theory atoms taken peak as the of area each an shows under two each isotope. In approximation heights are ● The naturally ● The total X-69 = occurring isotopes, peak is X-69 27 the isotopes the relative units X-71. proportional calculations of and and sum = 41 to the height numbers X-71 must peak to of number can be atoms. The units. ecnadnuba evitaler Solution 30 25 20 15 100 % 10 height of both peaks is 68 units. To deduce the relative 5 atomic mass of X we need to determine the relative abundance of 69 each 7 1 isotope: 0 0 20 40 60 80 100 27 _ X-69: ( ) × 100 = 40% m/z 68 Figure 13 Mass spectrum of X showing the 41 _ X-71: ( ) × 100 = relative abundances of its naturally occurring 60% 68 isotopes ● The relative procedure atomic from mass worked of X can example now be determined 40 _ relative atomic mass = ( 69 × ● From Ga (Z the = periodic 31), table which is in 70.2 (or section quoted as the 60 _ + ) ( 71 × 100 = using 1: 70 6 ) 100 of correct the having A to Data = 2 SF) booklet, 69.74. X must be The value of this calculation r 70.2 if from you the use this calculation peak calculations gures were heights will be expressed is closest instead 2 to SF 2 at of to this peak best, so value. areas, this is the the In precision reason of why all SF . 49 2 aTO m I c S T r u c T u r e 2.2 eto otio Understandings Applications and skills ➔ Emission spectra are produced when photons ➔ Description of the relationship between colour, are emitted from atoms as excited electrons wavelength, frequency, and energy across the return to a lower energy level. electromagnetic spectrum. ➔ The line emission spectrum of hydrogen ➔ Distinction between a continuous spectrum provides evidence for the existence of and a line spectrum. electrons in discrete energy levels, which ➔ Description of the emission spectrum of the converge at higher energies. hydrogen atom, including the relationships ➔ The main energy level or shell is given an between the lines and energy transitions to the integer number, n, and can hold a maximum rst, second, and third energy levels. 2 number of electrons, 2n ➔ ➔ Recognition of the shape of an s atomic orbital A more detailed model of the atom describes and the p , p x , and p y atomic orbitals. z the division of the main energy level into s, ➔ Application of the Aufbau principle, Hund’s p, d, and f sublevels of successively higher rule, and the Pauli exclusion principle to write energies. electron congurations for atoms and ions up ➔ Sublevels contain a xed number of orbitals, to Z = 36. regions of space where there is a high probability of nding an electron. ➔ Each orbital has a dened energy state for a given electron conguration and chemical environment and can hold two electrons of opposite spin. Nature of science ➔ Developments in scientic research follow improvements in apparatus – the use of electricity and magnetism in Thomson’s cathode rays. ➔ Theories being superseded – quantum mechanics is among the most current models of the atom. ➔ Use theories to explain natural phenomena – line spectra explained by the Bohr model of the atom. The electromagnetic spectrum What visions Samuel the The Beckett Nobel Prize light. in Visible light include light, is radio dark of one Literature that structure Visible the (1906–1989), developments electronic 50 in of the of Irish in led atom light type waves, have the light! we novelist, poet, and playwright who won 1969 to much have see, is of come full of our infrared experiments scientic electromagnetic microwaves, understanding from the intrigue. radiation . radiation of involving Other (IR), examples ultraviolet 2 . 2 radiation (UV), spectrum various X-rays, (EMS) types of is a and gamma spectrum of electromagnetic rays. The e l e c T r O n c O n f I g u r a T I O n electromagnetic wavelengths that comprise Stdy tip the radiation. The wavelengths of the dierent types of waves in the EMS are The energy, E, of electromagnetic radiation is inversely proportional to given in section 3 of the Data the wavelength, λ: booklet 1 _ E ∝ λ High-energy radiations wavelengths, and such as low-energy gamma rays radiations and such as X-rays radio have small waves have long wavelengths. Wavelength is related to the frequency of the radiation, ν, by the expression: c = νλ 8 where c The unit for SI is the of frequency speed of energy the light is hertz, the (3.00 joule, × J; 10 -1 m for s ). wavelength the metre, m; and Hz. absoptio, issio d otios spt A white-hot metal object such as an incandescent light bulb lament emits the full range of wavelengths, producing a otios spt including all the colours of the rainbow from red to violet. If a pure gaseous element such as hydrogen is subjected to an electrical discharge the gas will glow – it emits radiation. The resultant issio spt Figure 1 The aurora borealis in Lapland, Sweden. The aurora borealis (or Nor thern consists of a series of lines against a dark background. Lights) is a display of coloured light visible in If a cloud of a cold gas is placed between a hot metal and a detector, an the night sky at high latitudes. It occurs when bsoptio spt is observed. This consists of a pattern of dark lines against charged and energetic par ticles from the sun a coloured background. The gaseous atoms absorb cer tain wavelengths of light from the continuous spectrum. are drawn by the Ear th’s magnetic eld to the polar regions. Hundreds of kilometres up they collide with gaseous molecules and atoms, Absorption and emission spectra are widely used in astronomy to analyse light causing them to emit light from stars. Emission spectra and Bohr ’s theory of the hydrogen atom In the 1600s passed colours of all Isaac a generating merging continuous 400 to a prism (1642–1727) the continuous and into so the spectrum 700 Newton glass wavelengths colour from Sir through is as with a no rainbow. showed light spectrum. appears next a visible is This The if The series familiar wavelengths of sunlight into spectrum continuous gaps. that separated contains of is different light colours, example visible each of light a range nm. Figure 2 White light as perceived by the Many sources of radiation produce a line spectrum rather than a human eye consists of many colours or continuous spectrum. If a pure gaseous element is subjected to a high wavelengths of light. Shown here is the voltage colour under of reduced light. For pressure, example, the sodium gas will emits emit yellow a certain light. If characteristic this light is continuous spectrum of white light emitted by an incandescent light bulb lament 51 2 ATO M I C S T R U C T U R E passed through consists each a prism, black element identify emission to a the has the to its a resultant spectrum own of spectrum a small example, sodium nm is not number continuous of coloured but lines wavelength. characteristic For 589.0 (gure with characteristic element. wavelengths background the background corresponding Each to of line in spectrum the two distinct and 589.6 visible yellow nm, can which region lines, be can of the be used line corresponding seen on a black 3). Flame tests Figure 3 Line emission spectrum of sodium. Flame The spectrum looks like a single bright yellow The tests colour are of often the used ame in the varies laboratory for different to identify elements certain and can metals. be used line but at high resolution it is possible to see to identify unknown of electrons substances. The colours are due to the excitation two lines very close together corresponding to in the metals by the heat of the ame. As the electrons the wavelengths 589.0 nm and 589.6 nm lose the energy they have just gained, they emit photons oflight. Quantization of energy aoy The precise lines in the line emission of an element have specic You might think of a line wavelengths. Each characteristic wavelength corresponds to a discrete emission spectrum as being amount of energy. This is the basis of quantization, the idea that analogous to a barcode. Every electromagnetic radiation comes in discrete “parcels” or quanta. A product in a shop has its own photon is a quantum of radiation, and the wavelength, λ, and energy, unique barcode which gives E, of a photon are related by the equation: it an identity, and the same hc _ E is true of the line emission = hν = λ spectra of the elements. Each where: line emission spectrum is 34 dierent and is characteristic h = Planck’s constant = 6.63 × 10 J s of a specic element. ν = frequency of the radiation 8 c = speed This E is equation the 1913 proposed the way of s section the vice 1 of the greater Data the booklet. energy of It the shows that photon, the versa. Neils Bohr hydrogen based (1885–1962) atom. on Bohr classical examined proposed a the line theoretical mechanics. His model following: at atom its electron that consists centre, is balanced velocity in suggested energy its a the two by a positively which circular sun. charged negatively path or Although oppositely the a acceleration of orbit, there charged particle is particle similar an species, the called charged inherent this electron to force moving of at orbit. that of in orbit the of around moves planets attraction the m λ: spectrum between Bohr in to and the 1 10 physicist the hydrogen the × found attraction it: 52 for proton high ● be Danish the called the can 3.00 proportional spectrum explanation The = wavelength, the emission ● light inversely smaller In of the each orbit electron has a denite orbiting the energy positively associated charged with centre 2 . 2 in a particular electron in a orbit is xed particular or orbit quantized. is given by The the energy e l e c T r O n of c O n f I g u r a T I O n the aoy expression: Think about standing on the E = 1 _ –R ( H bottom step of a ight of stairs. ) 2 n You could jump to the second where: step, or you could jump higher 18 R = Rydberg constant = 2.18 × 10 to the third or four th step. J H Suppose you jump from the n = principal quantum number, with positive integer values 1, 2, 3, 4, ... rst step to the fth step. You depending on the orbit or energy level the electron occupies stay there for a few seconds ● When an electron subjecting level and it to in an stays in its ground-state electrical this is discharge), excited-state excited it for moves a (for to fraction a example, higher of a by energy and then jump back down. You might jump down to the rst step, or jump two steps down second. to the third step, or jump three ● When the electron falls back down from the excited-state to a lower steps down to the second step. energy level it emits a photon, a discrete amount of energy. This This is analogous to the way photon corresponds to a particular wavelength λ, depending on the excited electrons can jump energy difference between the two energy levels (gure 4). from a higher energy level to a lower one. n > 1 You always jump to a step, not to some place between steps. This shows the idea of quantization – each step is analogous to an energy level, n = 1 which has a denite, discrete ground-state excited-state e falls back down to a lower level and energy. Jumping up steps energy is emitted as a photon of light of wavelength, λ, corresponding to the energy dierence between the two energy levels requires an amount of energy, and jumping down steps releases discrete amounts Figure 4 Principles of the Bohr model of an atom when an electron is excited. n is the of energy. principal quantum number Note that current excited an electron level: to n in can gure =3, n = 4 4, be excited instead etc. The of to any being electron energy excited can level to also n fall = higher 2 it back than could down its be to any Wy t tiv si? lower energy level. The negative sign in the The difference in energy between the two energy levels can be expressed expression for E is an arbitrary as follows, where i represents the initial state and f represents the convention. It means that the nalstate: energy of the electron in the ΔE = E E f atom is less than its energy i if the electron was located an hc _ = hν = innite distance away from the λ nucleus. We can rearrange this expression noting that: Conventions are often used in E = chemistry. Another example 1 _ –R ( H 2 ) n ΔE = E of an arbitrary convention is E f = i 1 _ R H ( [ 2 n - [ )] 1 _ R f ( H 2 n i always placing the cathode )] on the right-hand side in a cell diagram (see topic 9). Can you 1 _ R = [ ( H 2 n 1 _ R )] [ H ( 2 n i )] think of any other conventions f that we use in chemistry? ΔE 1 _ R = [ ( H 2 n i hc _ 1 _ = 2 n )] hν = λ f 53 2 ATO M I C S T R U C T U R E The hydrogen different of the spectrum. series, upper line colours which energy Colour emission (violet, The series comprises levels spectrum blue, of lines back consists blue–green, lines shown associated down to the n in 2 a series red) in gure with = of and 5 is lines of visible called electronic energy of the region the Balmer transitions level. violet blue blue–green red 410 434 486 656 λ/nm n = 6 n = 5 to Transition from to n = 2 n = 3 to n = 2 n = 4 to n = 2 n = 2 n = 6 Stdy tip n = 5 You are not required to know the names n = 4 of the individual series of spectral lines. However, you are required to know which n = 3 transition corresponds to which region n = 2 of the EMS, eg the transition n = 4 to n = 1 will be seen in the UV n = 1 region, etc. outside the atom n = ∞ n = 5 continuum inside the atom n = 4 n = 3 Figure 5 Line emission spectrum of the hydrogen atom. Four lines are seen in the visible and ultraviolet regions of the spectrum; these make up the Balmer series n = 2 Other n = 3 series of energy infrared lines levels regions Sis of exist (table the n n = 1 Figure 6 Some transitions to the n = 1 level corresponding 1). These are to transitions observed in to the the EMS. n rio o emS i Lyman 1 2, 3, 4, 5, ... UV Balmer 2 3, 4, 5, 6, ... visible and UV Paschen 3 4, 5, 6, 7, ... IR from higher levels for the Lyman series (in the UV region) of spectral lines that occur in the emission spectrum of the hydrogen atom 54 n = 1 ultraviolet T able 1 Dierent series of lines in the hydrogen line emission spectrum and and 2 . 2 e l e c T r O n c O n f I g u r a T I O n Quantization and atomic structure The line emission of electrons to converge) merge, any in at forming energy; outside the it spectrum discrete higher a is hydrogen levels, energies. continuum. no atom. of energy longer Such the the electron get limit Beyond under an At provides which this of be this of for the together the referred the as a are the electron nucleus to existence (they convergence continuum inuence may evidence closer and free is said lines can have therefore electron. n = 2 n = 3 Models of the atom and electron arrangements n = The Bohr theory arrangements. electrons in the each atom electron shell or is a basis for orbit, for arrangement of H: 1 arrangement of P: 2, 8, 5 electron arrangement of Ca: 2, 8, 8, arrangements the the Bohr are chemical model of a very useful properties the atom gives electron the number of example: electron predicting writing arrangement electron Electron In of An 1 of the 2 tool an for explaining and element. energy levels are often drawn as Figure 7 Electron arrangement for phosphorus according to the Bohr model concentric circles, as shown in gure 7 for phosphorus. Limitations of the Bohr theory This of model has ● It assumes is incorrect; introduce It ● It ● is Bohr suggests were superseded and is associated with a number idea an positions orbits of an energy do of the not electron actually orbits exist (we are xed. shall This shortly orbital). levels incorrect that some limited elements Bohr ● the fact are circular or spherical in nature. incorrect. 2.1 hydrogen. the the scale atom fundamental for is the made atom – up mainly theoretical of remember problems from empty space. pertaining to the the ● It to containing that did more the to not just one explain than one electron is element, the line namely spectra of other electron. a subatomic particle orbiting Bohr made structure a and signicant in contribution particular, some of to the our understanding merits of his theory following: was that It calculations model nucleus. electronic are his The suggested Nevertheless, ● been model: Bohr ● in that also sub-topic There that the assumes This of now misconceptions: based on electrons the exist incorporated the fundamental in denite, idea of idea of discrete electrons quantization energy moving – the fact levels. from one energy level another. 55 2 ATO M I C S T R U C T U R E The quantum mechanical model of the atom The Bohr and in replaced of theory by the more quantum the provided particular electron. a sophisticated mechanics, Some Heisenberg’s rst approximation arrangement of the uncertainty electrons. mathematical which key of are principle the described states atomic has that structure, since theories incorporates ideas of It been from the wave-like eld nature of below. it is impossible to TOK determine accurately simultaneously both (topic the 12). momentum This means and that the the position more we of a know particle about Heisenberg’s uncertainty the position of an electron, the less we know about its momentum, and principle states that there vice versa. Although it is not possible to state precisely the location of is a theoretical limit to the an electron in an atom and its exact momentum along a trajectory at precision with which we can the same time, we can calculate the probability of nding an electron in know the momentum and the a given region of space within the atom. position of a particle. What are the implications of this for the limits of human knowledge? One aim of the physical sciences has been to give an exact picture of the material world. One Schrödinger’s physicist mathematical nature equation Erwin of the subsequent was Schrödinger equation electron. integrates This development Schrödinger received The to formulated the in (1887–1961). the dual 1926 His wave-like ground-breaking of the Nobel eld of Prize in by the Austrian sophisticated work quantum Physics and led to particle the mechanics. with Paul birth In and 1933 Dirac. achievement ... has been solution Schrödinger’s equation generated a series of to prove that this aim is mathematical functions called wavefunctions describing the electron unattainable. in Jacob Bronowski the can hydrogen occupy. atom Each and associated wavefunction is possible energy represented by the states the electron symbol, ψ. The 2 (1908–1974): Polish-born British square an of the electron wavefunction, in a region of ψ , space represents at a given the probability point a of distance, r, nding from 2 mathematician, biologist, scientic historian, inventor, the nucleus equations the basic of are the atom. very ψ is complex principles termed but at underpinning the this the probability level all we density. need to The consider are results. and poet. The What are the implications of wavefunctions of electrons in an atom are described by atomic orbitals: this claim for the aspirations An ● atomic orbital is a region in space where there is a high of the natural sciences in probability of nding an electron. par ticular and for knowledge in general? Any orbital types shape of can atomic and hold a orbital: associated maximum s, p, d, of and f, two etc. electrons. Each type There has a are several characteristic energy. Stdy tip aoy Atomic orbitals have dierent Imagine that you are a student in an IB chemistry class in Quito in Ecuador, waiting shapes. For SL you need to be for your teacher to arrive at 8.00 am. At 8.15 am there is no sign of your teacher and familiar with the shapes of the your class decide to go looking for him. You decide rst to dene the most probable s and p atomic orbitals, while places the teacher is likely to be. Suggestions from the class include: for HL you need to know the shapes of the s, p, and d atomic The teacher: orbitals. We shall return to the ● is possibly in the sta room, the chemistry laboratory, or the library ● may be in the school principal’s oce or in the school car park ● could be at his house in Quito shapes of the d orbitals in topic 13 when we discuss crystal eld theory. 56 2 . 2 ● could perhaps be at the airpor t ● might even have gone home to South Africa! e l e c T r O n c O n f I g u r a T I O n If the class went looking for the teacher they would most likely star t looking in the most probable locations closest to the classroom. But at 8.15 am they do not know with any degree of cer tainty precisely where the teacher is. A three-dimensional graph could be drawn with a cluster of dots showing areas where there is a high probability of nding him. This is the idea of an obit. A boundary surface could be drawn around this cluster of dots to dene a region of space where there is a 99% chance of nding the teacher. This might be the school perimeter, or Quito where he lives. If you were also asked to measure the distance from the classroom to the exact location where the teacher is you could not do this at 8.15 am, as you do Figure 8 An orbital is a three-dimensional not know his exact location with absolute cer tainty. graph with a cluster of dots showing the What aspects of quantum mechanics does this analogy capture? probability of nding the electron at dierent distances from the nucleus The s atomic orbital An s orbital is spherically boundary surface, chance probability or symmetrical. meaning of that nding The within an sphere the represents sphere electron there (gure is a a y 99 % 9). The p atomic orbital A p orbital is dumbbell shaped. There are three p atomic orbitals, p , p x and p , all with boundary surfaces conveying probable electron , y density z pointing in different directions along the three respective Cartesian x axes, x, y, and z (gure 10). Energy levels, sublevels, orbitals, and electron spin The by Bohr n, model which positive is integer increases, the introduced called the values mean the idea principal 1, 2, 3, position etc. of In an of a main quantum the energy quantum electron is level, number. described This can mechanical further from the z have model, as nucleus. n The Figure 9 The s atomic orbital is spherically energies of the orbitals also increase as n increases. Each main of electrons given energy level symmetrical 2 or shell can electron we in have the The hold capacity two energy maximum for n elements second, common a = 1 in is number 2, the for rst n = row 2 of is 8, the for n = 3 periodic is by 2n 18. table, . So That eight is the why elements y etc. levels types: s, are p, split d, up and f. into sublevels, Each sublevel of which contains a there are number y four of x orbitals, each of which can hold a maximum of 2 electrons (table 2). z x nb o obits i mxi b o sbv tos i sbv x x z z p Sbv y p p z y Figure 10 The three p atomic orbitals are dumbbell shaped, aligned along the x, y, and s 1 2 p 3 6 d 5 10 f 7 14 z axes T able 2 Sublevels of the main energy levels in the quantum mechanical model 57 2 ATO M I C S T R U C T U R E For convenience, an “arrow-in-box” notation called an orbital us so diagram is used to represent the electrons in these atomic orbitals The “Orbitron” website, (gure 11). We shall use orbital diagrams to represent electron developed by Professor Mark congurations. Winter at the University of Sheeld, UK is an excellent s sublevel (one box representing an s orbital) resource for exploring the shapes of the various atomic orbitals. It also provides information on the associated sophisticated mathematical p sublevel (three boxes representing the three p orbitals p , p x , and p y ) z wavefunctions. http://winter.group.shef.ac.uk/ orbitron/ d sublevel (ve boxes representing the ve d orbitals) f sublevel (seven boxes representing the seven f orbitals) Figure 11 Orbital diagrams are used to represent the electron congurations for atoms. Arrows are drawn in the boxes to represent electrons, a maximum of 2 electrons in each box (orbital) Two electrons in the same orbital have opposite values of the 1 N S magnetic quantum number, m . The sign of m s indicates electron. facing by in two the A orientation pair of opposite arrows in of the electrons in directions a box an (gure + s magnetic and ( eld orbital is - 2 as ) 2 generated behaves therefore spin 1 or by two commonly the magnets represented 12). Qt bs In this mathematical four qt S N N S represents qt number, model of the electronic structure bs. The rst is the piip the energy b, l, level. The describes the ti second the qt quantum sublevel, and b, m , the of the atom there are qt b, n, which number, the third atomic the zit quantum orbital. The four th l quantum number, the spi ti qt b, , describes the s magnet analogy spatial orientation examined of S in the levels, IB the electron Chemistry sublevels, spin. Diploma, atomic Quantum but you orbitals, numbers need and to are know electron not the formally principles spin. N half-arrows representing electrons energy of You might think of the four quantum numbers as an electronic postal of opposite spin address. The country represents the energy level, the province the in an orbital sublevels, the town the orbitals, and the street number or postal code the spin of the electron. Figure 12 Electron spin is represented by arrows in orbital diagrams 58 2 . 2 e l e c T r O n c O n f I g u r a T I O n Writing electron congurations We shall now develop congurations for these atoms ideas and further by writing electron ions. 3p 3s ygrene 2p 2s 1s Figure 13 This is the order of energy levels of the rst few sublevels There are electron 1 The three Aufbau orbital rst Up that few to with 3d principles that must be followed when representing congurations. principle is available energy Ca (Z = since that electrons Figure 13 ll shows the the lowest-energy sublevels for the levels. 20) the experimental level states rst. it is Aufbau data and lower in principle the 4s correlates level energy. The is lled precisely rst condensed before the electronic 2 conguration (Z = 21), level the the now 3d is for Ca two slightly lled is written levels are higher rst. The in as comparable energy correctly 3d in written sublevel. energy For than as Zn the [Ar]3d (Z = 3d 30), Zn is best written and as However, energy the 3d for level conguration Sc with the and for Sc 4s hence therefore 2 4s . the the This 4s trend level [Ar]3d continues now condensed 10 for . in than condensed 1 is [Ar]4s is much electron along the higher conguration 2 4s for this reason. This is Stdy tips consistent with experimental data which shows that when • the 3d-block elements are ionized, the electrons are For the IB Chemistry removed Diploma you need to be from the 4s before the 3d levels, which makes sense since the able to deduce the electron 4s is higher in energy than the 3d for this block of elements. congurations for the atoms The situation overall is quite complex as in the case of Sc the and ions of the elements up lling of the last three electrons does not continue in the 3d to and including level, and experimental data does not provide evidence for Z = 36 (Kr). an 3 [Ar]3d the 3d electron orbitals electrons conguration are entering more the for compact 3d orbitals Sc. The than will the reason 4s for orbitals experience a this and much is that hence greater • The periodic table showing atomic number, Z, is provided in section 6 of the Data booklet 59 2 ATO M I C S T R U C T U R E mutual repulsion. written by and E. In Scerri, Biochemistry, an excellent Department at the orbitals article of This Chemistry University USA Chemistry, 7th explained as and published November in 2013, singly before illustrated in occupying gure them in pairs. 14. of 1 California, is Education the reason 4 2 3 in is 2p feature can is that relieve repulsion, make follows: full although such slightly the additional different use “The of atoms this form unsettling relevant s orbital Figure 14 Electrons ll each orbital singly before electron-electron do of not occupying them in pairs always sheltering There because the situation is more be than is just that through set of the described. nuclear the atoms, interactions nucleus as One charge thing there between well as to increases and are is the between a we Pauli electrons the 1 full 2 condensed hold a principle maximum electron 3 orbital write of states two these electrons have opposite Hund’s rule of maximum that an that when lling of equal conguration “build the we successive orbitals described the according to the three above. periodic table to can the be shown four as sublevels four s, p, blocks d, orbitals ll all 15). the p block main-group elements 1 18 1s 1s 2 13 14 15 16 2p 2s d block 3p 3s 3 4 5 6 7 8 9 10 11 12 4s 3d 4p 5s 4d 5p 6s 5d 6p 7s 6d 4f f block 5f Figure 15 The blocks of the periodic table correspond to the sublevels s, p, d, and f 60 use electrons multiplicity electrons s block up” spin. degenerate energy), and any (gure (orbitals conguration representation. electron corresponding states electron electrons, The 3 conguration diagram table, principles and can electrons To exclusion can congurations and in orbital electron move complicated periodic The ways illustrated: consider as themselves”. 2 three complicated 1 7 and f 2 . 2 e l e c T r O n c O n f I g u r a T I O n Full electron congurations Table 3 shows the full electron congurations for some of the rst 36elements. et Z eto otio Period 1 elements: 1 H 1 1s He 2 1s 2 Period 2 elements: 2 Li 3 1s Be 4 1s B 5 1s C 6 1s N 7 1s O 8 1s 1 2s 2 2 2s 2 2 2s 2 2 2s 2 2 1s Ne 10 1s 3 2p 2 2s 2 9 2 2p 2s 2 F 1 2p 4 2p 2 2s 2 5 2p 2 2s 6 2p Period 3 elements: continue with the same lling pattern, for example: 2 Na 11 1s Mg 12 1s Al 13 1s Ar 18 1s 2 2s 2 6 2p 2 1 3s 6 2 2s 2p 3s 2 6 2 2 2s 2 2p 2 2s 3s 6 2p 1 3p 2 3s 6 So xptios: 3p coi d opp Period 4 elements: After Z = 30 the 4p sublevel is lled: 2 K 19 1s Ca 20 1s Sc 21 1s Ni 28 1s Zn 30 Ga 31 1s Br 35 1s Kr 36 1s 2 2s 2 Two of the rst 36 elements have 6 2p 2 2 3s 6 6 3p 2 1 electron congurations that dier 4s 6 from what you may predict. These 2 2s 2p 3s 3p 4s 2 6 2 6 1 two elements are Cr (Z = 24) and 2 2s 2p 3s 3p 3d 2 4s Cu (Z = 29): 2 2 2s 6 2p 2 3s 6 3p 8 3d 2 4s 2 2 1s 2 6 2 6 2s 2p 3s 3p 2 6 2 6 10 3d Cr 1s Cu 1s 2 2 2s 2p 3s 3p 10 3d 2 2s 6 2p 2 3s 6 3p 5 3d 1 4s 2 4s 2 4s 2 2s 6 2p 2 3s 6 3p 10 3d 1 4s 1 4p In these two elements electrons 2 2 2s 6 2p 2 3s 6 3p 10 3d 2 4s 5 4p go into the 3d orbitals before 2 2 2s 6 2p 2 3s 6 3p 10 3d 2 4s 6 4p completely lling the 4s orbital. Chromium has a half-lled 3d T able 3 Full electron congurations for some of the rst 36 elements sublevel of 5 electrons and copper has a completely lled 3d sublevel Condensed electron conguration of 10 electrons. Half-lled and You can see above that full electron congurations become quite completely lled 3d sublevels lengthy and cumbersome with increasing atomic number. An element’s reduce the overall potential chemistry is dictated by its outer valence electrons (as opposed to the energy of an atom, so the electron inner core electrons), and a more convenient way of representing 5 congurations 3d electron congurations is as the condensed electron 1 4s conguration : 10 and 3d 4 are more stable than 3d 9 [nearest noble gas core] + valence electrons 3d 1 4s 2 4s and 2 4s , respectively. 61 2 ATO M I C For S T R U C T U R E example: He [He] O [He]2s Ne [He]2s P [Ne]3s 2 4 2p 2 6 2p 2 or simply [Ne] 3 3p Orbital diagrams Orbital arrows boxes joined Orbital two make the steps types of orbital 1s 2 2s to of show the the core energy diagrams arrows-in-boxes boxes their all gas the and show noble showing 2 F may nearest use electrons together diagrams beyond have diagrams representing energy orbitals as levels that in or can notation representing in Degenerate gures orbitals 11 and are 13, with represented as in the full condensed may be used electron electron represented to represent on conguration, conguration. one line. For or just Orbital the orbitals diagrams example, gure may 16 shows uorine: 5 2p 1 4 2 5 3 1 5 or 2p 4 2 5 3 [He] 2 5 2s 2p ygrene 2 2s 2 1s Figure 16 Orbital diagrams showing the electron conguration for uorine The condensed diagrams for version the 5 Cr [Ar]3d is more elements convenient chromium, and cobalt, will and be used bromine in are this 1 2 3 4 5 4 5 [Ar] 1 5 4s 3d 7 Co [Ar]3d 2 4s 1 6 2 7 3 [Ar] 2 7 4s 3d 10 Br [Ar]3d 4s 2 5 4p 1 4 2 5 [Ar] 2 4s 62 10 3d 5 4p book. For represented 4s 1 by equivalence. the be described orbitals. 3 as example, follows: the orbital 2 . 2 e l e c T r O n c O n f I g u r a T I O n Worked examples: electron congurations Example 1 Deduce the Example 2 full 2+ Mg electron congurations for Mg, Deduce the condensed O, and 2 O S, S 2+ , Fe, Fe congurations of + , Cu, 2 From table 2 Mg 1s 2 2s ● 3: 6 2p S [Ne]3s Cu 2 4 3p 2- 3s For the S anion 2 To and Solution Solution ● electron 2 , write the electron conguration for the 2 S [Ne]3s we add 2 electrons: 6 3p or simply [Ar] 2+ Mg cation, 2 electrons must be removed. ● These are principal taken from quantum the orbital number n; of in The electron conguration this case, be deduced Fe (Z = 26) n; in can Fe [Ar]3d as: the 6 2 3s for highest orbital: 2 4s 2+ 2+ 2 Mg 1s 2 2s 6 For 2p the Fe removed ● From table 2 1s 2 2s the 2 electrons orbital of are highest 3: this O cation from case, the 4s orbital: 4 2p 2+ 6 Fe [Ar]3d 2 To write the electron conguration for the O ● anion, to the two electrons same must principles as be added The copper one of electron conguration is according the two exceptions that you must before: remember: 2 2 O 1s 2 2s 6 2p 10 Cu Notice that the electron congurations [Ar]3d 1 4s for + To 2+ the species Mg form the O are identical: the ion, again the electron is they removed contain Cu 2 and same number of electrons from the orbital of highest n; in this and case, the 4s level: + are said to be isoelectronic . Na , F , and + 2+ Ne are also However, number table isoelectronic each of of these protons with species (atomic 2 Mg and has a number O . Cu 10 [Ar]3d different Z 4): Spis atoi b, Z nb o (b o potos) tos 2 O F Ne 8 10 9 10 10 10 11 10 12 10 + Na 2+ Mg T able 4 Isoelectronic species 63 2 ATO M I C S T R U C T U R E Stdy tip Example 3 Warning: Do not be tempted 2+ Deduce the orbital diagrams for Ni, Ni , and Se. to rearrange the conguration 2+ for Fe 5 to [Ar]3d 1 4s as for Solution 5 chromium. The 3d 10 and 3d First write the condensed electron conguration for the species. congurations for chromium Then draw the orbital diagram, remembering that two electrons in and copper apply only to the same orbital have opposite spin quantum numbers: neutral atoms, not to ions. 8 Ni [Ar]3d 2 4s 1 6 2 7 3 8 4 5 4 5 [Ar] 2 8 4s 3d 2+ Ni 8 [Ar]3d 1 6 2 7 3 8 [Ar] 0 8 4s 3d 2+ Notice that electrons For in in the the orbital 4s diagram orbital – the for box the Ni should cation be left there are no blank. selenium: 10 Se [Ar]3d 2 4s 4 4p 1 4 2 3 [Ar] 2 10 4s 4 3d 4p Experimental evidence for electron congurations Direct be of evidence found from magnetism, paramagnetic can be eld. and In can number scientists 64 by contrast, be a a to in conguration measurements. at least magnetic eld. greater the diamagnetic by a determine the in and can types diamagnetism. greater of element different electron the number attraction has all its the past and in a A hence of magnetic electrons paired eld. research improvements The force an are unpaired material magnetic scientic one for There paramagnetism has the repelled of electron including electrons, Developments a the material attracted unpaired of magnetic over instrumentation number of unpaired 50 years which have have electrons in led to allowed an atom. Q u e S T I O n S Questions 1 What is the neutrons number in of protons, electrons, and 5 What with boron-11? A. 5 protons, 5 electrons, and 11 B. 5 protons, 5 electrons, and 10.81 C. 5 protons, 5 electrons, and 6 D. 11 is the the relative mass atomic spectrum mass shown of in an element gure 17? neutrons 100 neutrons neutrons 80 What is the 11 number 34 neutrons electrons, in of and 5 protons, neutrons electrons, ecnadnuba % 2 protons, and 2 S ? 60 40 16 A. 18 protons, 18 neutrons 16 16 protons, 34 neutrons 16 protons, 18 neutrons 16 protons, 18 neutrons electrons and 20 B. 18 electrons and 0 22 C. 18 electrons 23 24 25 and 26 27 28 29 30 mass/charge Figure 1 7 D. 3 Which 16 statements 35 chlorine, They about and the isotopes of A. 24 B 25 C. 26 D. 27 IB, May 37 Cl and 17 I. electrons Cl, are correct? 17 have the same chemical [1] 2009 properties. II. They have the same atomic number. III. They have the same physical 6 Which is correct electromagnetic for the following regions of the spectrum? properties. A. I and II B. I and III C. II and D. I, II utviot (uV) only III high shor t low low energy wavelength energy frequency A. only only high low low long energy frequency energy wavelength B. IB, and May Id (Ir) III [1] 2011 high shor t high long frequency wavelength energy wavelength C. 63 4 A sample of element X contains 69 % of X high long low low frequency wavelength frequency energy D. 65 and 31% of X. What is the relative atomic [1] mass of X in this sample? IB, A. 63.0 B. 63.6 C. 65.0 7 In May the 2009 emission electronic D. 69.0 IB, May spectrum transition of would hydrogen, produce a which line [1] in the visible region of the electromagnetic 2010 spectrum? A. n = 2 → n = 1 B n = 3 → n = 2 C. n = 2 → n = 3 D. n = ∞ IB, May → n = 1 [1] 2011 65 2 ATO M I C 8 Which S T R U C T U R E describes spectrumof the visible b) emission spectrum c) A. A series Distinguish hydrogen? of lines converging at The thinning between and of a the line a continuous spectrum. ozone layer [1] increases longer the amount the Earth’s of UV-B radiation that reaches wavelength B. A series of regularly C. A series of lines converging spaced at lower D. A series of lines converging at higher May (table 5). lines energy frequency IB, surface Typ o ditio Wvt / UV-A 320–380 UV-B 290–320 [1] 2010 T able 5 9 What is the order of increasing energy of the Based orbitals within a single energy on explain A. d B. s < C. p D. f IB, May < s < f p < d s < f < < d < < < p why dangerous p < the information in table 5 level? IB, f UV-B than Specimen rays are more UV-A. [3] Paper d < s [1] 13 a) Deduce the full and Mn electron conguration 2+ 2009 for b) Mn Deduce the condensed electron 2+ conguration 10 What is the condensed for Cu electron 2+ 3+ congurationfor 2 A. [Ar]4s c) ? [Ar]4s [Ar]3d D. [Ar]4s 14 series are often of IB, label an atom. of In lines its energy your in the emission labelling May level diagram diagram show for spheres. model. Developments in scientic improvements in apparatus. ultraviolet and each spectrum are series. research follow Discuss how statement with regard to the use of visible and magnetism in Thomson’s produced, with cathode rays. [4] 2010 the following radiation in wavelength In many textbooks order types of of electromagnetic widely increasing (shortest [Ar]4s I. Yellow II. Red . This the electronic vanadium is accepted However, rst). for is written as 3 3d electronic common by suggest the why practice chemical this conguration for way and community. of writing vanadium the may light be at odds with experimental evidence. light You III. Infrared radiation IV . Ultraviolet might like to read radiation the following article: http://www.rsc.org/eic/2013/11/aufbau- electron-conguration [1] your 66 As. this a 2 List as of as conguration a) and the 16 12 use representation experiments clearly drawn onthe electricity regions Co 5 this the for 3d 15 hydrogen Atoms Comment 1 and diagrams 4 3d C. Draw orbital 3d 6 11 Draw 7 2 B. Co answer. to guide you in 3 P E R I O D I C I T Y Introduction Science some of is resulted vast full the of factual greatest from amounts scientists of information. scientic data being and However, discoveries able deduce to have interpret clear important table from it. In 1869 the Russian Mendeleev arranged in recognized order that according if to (relative atomic mass), a the denite be led seen in the ultimately properties (after some of the a table topic theory) to the of the core periodic of physical it became properties clear of that the the elements function we shall of Z, the examine atomic the number. nature periodic table, establish what of information elements. renement development the pattern be extracted the from it, and explore how of repeated the developed and periodic this can This at atomic the could chemists, lies elements their In weight to which chemist are were available chemistry. chemical Dmitri tool elements, patterns As emerging of (periodic) patterns can be linked to the most properties of the elements. 3.1 P a Understandings Applications and skills ➔ The periodic table is arranged into four blocks ➔ Deduction of the electron conguration of associated with the four sublevels - s, p, d, an atom from the element ’s position on the and f. periodic table, and vice versa. ➔ The periodic table consists of groups (ver tical columns) and periods (horizontal rows). ➔ Nature of science The period number (n) is the outer energy level that is occupied by electrons. ➔ ➔ ➔ Obtain evidence for scientic theories by The number of the principal energy level and making and testing predictions based on the number of the valence electrons in an them – scientists organize subjects based atom can be deduced from its position on the on structure and function; the periodic periodic table. table is a key example of this. Early models The periodic table shows the positions of metals, non-metals and metalloids. of the periodic table from Mendeleev, and later Moseley, allowed for the prediction of proper ties of elements that had not yet been discovered. 67 3 P E R I O D I C I T Y The development of the periodic table Evidence for predictions theories. based table then Scientists on of scientic and structure elements theories testing often and is a try is obtained them to classify function, good against and example by their the of making proposed subject periodic scale. In when development over a from of number different each of of the years periodic and countries others’ work has and took involved on key the scientists modern the the it scientists table, be to It the of the German the chemist (1780–1849) elements calcium, in similar and (relative atomic of that mass strontium of mean calcium of elements other the and as sum of barium. and the a triads– triad. one or same more iodine, and and called law weights this also there and and of This triads . trio community at the attention to elements this into law triads a few elements. discovery time and was H o w e v e r, suggesting between atomic there did apply elements in the was an the of the each music. Thus nitrogen in and seven elements; fourteen; between between fourteen; and and bismuth, lastly, fourteen relationship term The Law of I propose to Octaves. Newlands, News, ‘a 12 letter (18th to the editor’, August 1865). of of octaves known applied to elements. only He a tried at principle the to time). the known However, elements they did not pay neatly t this type of pattern: highly reactive classication such as lithium, sodium, and caesium became potassium, to an grouped such as silver and with copper. very One Newlands had was to place two elements properties stepping periodic metals inherent the in one box of a periodic table, to take stone account development to extremities D ö b e r e i n e r ’s and important idea this 60 together, of of the was not limited was weight of some the idea link are peculiar number unreactive hypothesis by s u l f u r, rubidium, just stand as between arsenic, Chemical metals of Members antimony, J.A.R. all much or bromine, Surprisingly that numbers seven to horizontal of of (about scientic belonging same the by relation antimony This recognized chlorine, involving tellurium. their barium to the the that differ octaves phosphorus limited selenium, on of t o d a y ’s atomic Döbereiner another elements seen group, phosphorus Newlands’s and order transpositions, approximately classied involving This atomic using was the He be in slight that seven. provisionally the weight. the pattern) that also. terminology) few appear also the between weight of Johann discovered strontium, properties idea repeated octaves arranged a elements nitrogen arsenic had are with group will one the 1817 atomic of this a below. Döbereiner Döbereiner of is, development summarized other In (that law observed same line. as the elements will the foundations ideas. contributed periodic order as equivalents place multiples of in known analogous Four published elements this. table building Newlands of arranged became If The 1865 periodicity table of this. Newlands presented his law to the of Chemical Society in England but his ideas were elements. not accepted. Society of His this presentation work in 1866 to was the Chemical not published. Newlands As In 1864 the English chemist John a his (1837–1898) arranged to be of the in discovered order evidence of of a that atomic pattern when weight, with the elements repeated in there seven elements, such octaves that each of appeared or below analogy is similar it. of repeated This an at to the term octave eighth was in intervals music of – eight element based the on 1869, four rst at a and sugar returned plant. years mooted, after Newlands’s Russian ideas chemist (1834–1907) discovered, Dmitri like above on same the the had Newlands, that in order atomic if the of their elements were arranged the of weight, a repeated pattern note properties could be identied. This was musical termed 68 ridiculed consisting element named felt chemist Mendeleev Mendeleev properties chief properties were of Newlands position were In elements result Newlands the periodic law. The main difference to 3 . 1 between that Newlands’s Mendeleev elements only In very elements 1869 table of and Mendeleev’s considered carefully that had Mendeleev the and grouped similar published work properties was of the rst It soon improved the table over time atomic gaps for undiscovered fell into the elements, correct so the approach existence elements. to be J.W . version iodine placed in some of the table link of weight elements threes. in (1817)– atomic different groups of – did law. of For of was scientic elements problematic. Henry the elements atomic Moseley in number, weight. This table elements. the despite is the basis atomic appeared to pattern octaves seven be table summarizes over Dimitri (1865) were order weight 1 various scientists the Z, for the In in 1913 order the (1887–1915) periodic instead of table in atomic the modern of there evidence with the repeated consisting in The periodic Mendeleev – in order weight pattern a were of law the arranged atomic repeated of properties of the some of periodic their was Henry Moseley (1887–1915) when elements developed time. periodic (1869) contributions who (1834–1907) octaves in of Figure had elements arranged a weight the the obey Newlands properties ▲ not 126.90) when of to arranging predict tellurium, Law atomic Using (1837–1898) triads and to periodic weight after higher undiscovered J.A.R. Döbereiner between able of elements the (atomic (1780–1849) Law was properties However, Mendeleev’s example, Mendeleev and a that group. order this apparent that physicist arranged element had and British each tellurium became community of left that (127.60). Moseley periodic elements. Mendeleev fact weight t A b l e together properties. his the P e r i o d i c found. The law modern (1913) elements in order atomic (Z), – periodic when were of the arranged increasing numbers their recurred properties periodically. of elements. Figure 1 Scientists who contributed to the development of the periodic table of elements Avy In modern science do you think that theoretical research has a much greater chance of acceptance by the scientic community if it is suppor ted by empirical evidence? Discuss this in class. 69 3 P E R I O D I C I T Y toK A hyphss is a proposal that tries to explain particular What role did inductive and deductive reasoning play phenomena. A hy results from testing a hypothesis and in the development of the periodic table? What role may subsequently replace the hypothesis. A hypothesis do inductive and deductive reasoning have in science can therefore be considered a tentative explanation in general? that can be tested through investigation and exploration inductive reasoning (“bottom-up” approach): whereas a theory is an established array of ideas or 4. theory concepts which may then be used to make predictions. 3. hypothesis In science there are two ways of arriving at a particular 2. pattern conclusion – nuv asnng and uv 1. obser vation asnng (gure 2). Inductive reasoning is a “bottom-up” approach whereas deductive reasoning may be described deductive reasoning (“top-down” approach): as a “top-down” approach. With inductive reasoning 1. theory denite measurements and observations can lead scientists to establish the existence of possible trends 2. hypothesis or a pattern. From such a pattern a hypothesis can be 3. pattern formulated that can ultimately lead to a theory based on 4. obser vation certain conclusions. In deductive reasoning, the starting ▲ Figure 2 Inductive and deductive reasoning point involves the theories themselves. These are tested based on experimental (empirical) work. The periodic table today In the modern increasing physical called have periodic atomic properties groups. their The own table number, placed elements with are (table are elements underneath groups names the Z, each numbered arranged having other from 1 in to in order similar vertical 18; of chemical and columns certain groups 1). Gup num rmmn nam 1 alkali metals 2 alkaline ear th metals 15 pnictogens 16 chalcogens 17 halogens 18 noble gases Usfu su Much information on each ▲ T able 1 Names of groups recommended by IUPAC in the periodic table of elements element can be found on the “ WebElements” periodic table website. This resource was compiled by Professor Mark Winter at the University The current gure For 3. periodic Each example, there are group the known table is consists of characterized noble gases compounds in 118 by group a 18 containing elements number are very noble and of is shown distinct unreactive gases, such in properties. as (though XeF ). 4 of Sheeld, UK . http://www.webelements.com/ 70 Helium, the has many not typically lightest industrial form of the noble applications chemical gases, is because compounds used it is with for lling balloons non-ammable any elements. and and does 3 . 1 P e r i o d i c t A b l e 1 18 1 1 2 H He 2 13 14 15 16 1 7 5 6 7 8 9 10 B C N O F Ne 20.180 1.008 4.0026 3 2 4 Li Be 6.94 9.0122 10.81 12.011 14.007 15.999 18.998 11 12 13 14 15 16 1 7 Na Mg 22.990 24.305 3 Metalloids 3 4 6 7 8 9 10 11 18 Al Si P S Cl Ar 26.982 28.085 30.974 32.06 35.45 39.948 12 19 20 2 1 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr 39.098 40.078 44.956 47 .867 50.942 51.996 54.938 55.845 58.933 58.693 63.546 65.38 69.723 72.63 74.922 78.96 79.904 83.798 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 4 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe 85.468 87 .62 88.906 91.224 92.906 95.96 [97 .91] 101.07 102.91 106.42 107 .87 112.41 114.82 118.7 1 12 1.76 127 .60 126.90 131.29 5 55 56 Cs Ba 132.91 87 6 7 7 1 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 Lu Hf T a W Re Os Ir Pt Au Hg Ti Pb Bi Po At Rn 137 .33 1 74.97 1 78.49 180.95 183.84 186.2 1 190.23 192.22 195.08 196.97 200.59 204.38 207 .2 208.98 [208.98] [209.99] [222.02] 88 103 104 105 106 107 108 109 110 111 112 113 114 115 116 11 7 * Fr Ra [223.02] [226.03] 118 Lr Rf Db Sg Bh Hs Mt Ds Rg Cn Uut Fl Uup Lv Uus Uuo [262.11] [265.12] [268.13] [27 1.13] [270] [277 .15] [276.15] [281.16] [280.16] [285.1 7] [284.18] [289.19] [288.19] [293] [294] [294] ** 57 *lanthanoids 58 59 60 6 1 62 63 64 65 66 67 68 69 70 La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb 138.91 140.12 140.91 144.24 [144.91] 150.36 151.96 157 .25 158.93 162.50 164.93 167 .26 168.93 1 73.05 89 90 91 92 93 94 95 96 97 98 99 100 101 102 Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No [227 .03] 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] **actinoids ▲ 5 Figure 3 The modern periodic table of elements The horizontal periods. number, period. energy The n, of For rows of period the highest example, levels elements number with n 1, equal occupied calcium = numbered is 2, and Z = from the energy (Ca), 3, to level 20, is 1 to 7 principal in in the are termed quantum elements period 4 so has of the four 4. Metals, non-metals, and metalloids The periodic these are this line and the table is separated are the also by a metals non-metals lie split broadly stepped (excluding to the into diagonal metals line. and The non-metallic non-metals; elements hydrogen to the which is left a of gas) right. Metals: ● are good ● are malleable ● are ductile conductors of heat and electricity Quk qusn (capable of being hammered into thin sheets) Suggest w reasons why (capable of being drawn into wires) authorities in Sweden banned the use of mercury dental ● have lustre (they are shiny). llings since 2008. Mercury, Hg, solutions formed Hg can be sub-topic Z used = 80, in as a is a this liquid way lling and are for can called teeth. dissolve many amalgams; We shall for discuss other metals. example, metals Ag further The Sn in 4.5. Non-metals Non-metals metals gain metals lose are poor conductors electrons electrons in of chemical (they are heat and reactions electricity. (they are Typically reduced), non- whereas oxidized). Metalloids Some of the elements close metallicand non-metallic silicon, = Si, Z 14, to the stepped properties. germanium, Ge, Z The = diagonal elements 32, line have boron, arsenic, As, Z B, = both Z = 33, 5, antimony, 71 3 P E R I O D I C I T Y Sb, Z the = 51, tellurium, metalloids. semiconductors, dependent material Z = 52, to their as in astatine, such as which computers At, silicon intermediate, conductivity such and metalloids due electrical science, Te, Some has and Z and highly = 85 called are temperature- widespread smart are germanium applications in phones. Main group, transition elements, and s, p, d, and f blocks The ● periodic the the The on of be properties of position transition discussed The in periodic sublevels. also group 1 into two (excluding broad H), sections: group 2, and the in periodic elements. is groups main-group the detail table The elements: in split The this chemistry 13. into four of elements table; topic occupancy 3–11. blocks electrons for can is of often less the based each be true predicted for the transition on the sublevel s, is p, based properties elements d, and shown in will f table2. Maxmum num f Num f am ns n suv as n ah suv s 2 1 p 6 3 d 10 5 f 14 7 T able 2 Occupancy of electrons for each sublevel, and the related number of atomic orbitals group 1 (excluding H), group 2, and groups 13–18 tansn mns groups 3–11 (the f-block elements are sometimes described as the nn ansn mns) s-k mns groups 1 and 2 and He p-k mns groups 13–18 (excluding He) groups 3–12 (including Z and Z f-k mns = = 57 (La) and Z 90 (Th) to Z elements from Z = lanhans Ans 72 divided Man-gup mns -k mns ▲ further elements: Suv ▲ be 13–18 transition their the can main-group groups ● table T able 3 The elements in the blocks of the periodic table = = 89 (Ac), but excluding Z = 58 (Ce) to Z = 71 (Lu) 103 (Lr), which are classied as f-block elements 58 (Ce) to Z = elements from Z elements from Z 71 (Lu) and from Z = = 57 (La) to Z 89 (Ac) to Z = = = 90 (Th) to Z 71 (Lu) 103 (Lr) = 103 (Lr) 3 . 1 P e r i o d i c main-group elements s-block t A b l e p-block 1 18 1s 1s 13 2 2s 14 15 16 1 7 2p d-block 3s 3p 3 4 5 6 7 8 9 10 11 12 4s 3d 4p 5s 4d 5p 6s 5d 6p 7s 6d 4f f-block 5f ▲ Figure 4 The four blocks of the periodic table corresponding to the s, p, d, and f sublevels The number from the calcium so has 7 dropped of group is in valence number group valence from 2, so electrons of has electrons the group the 2 s- (outer-shell and p-block valence (note that number in electrons) elements. electrons. for the order to Fluorine p-block nd the can For is be found example, in elements, number group the of 1 17, is valence electrons). 73 3 P E R I O D I C I T Y Electron congurations and the periodic table Sub-topic 2.2 showed ● full ● condensed ● orbital For electron that the electron conguration of an element can be expressed in three ways: conguration electron conguration diagram. example, for uorine, F , Z = 9: 2 ● full ● condensed ● orbital electron conguration: 1s 2 2s 5 2p 2 electron conguration: [He]2s 5 2p diagram: [He] 2 2 2s 2 2p Figure 4 can periodic be table a 2p y powerful can 1 2p x be used z tool to when deduce writing the electron electron congurations: conguration, as the the position following of an worked element in example the shows. Worked example: deduction of the electron conguration from the element’s position in the periodic table 1 Consider the element number selenium, protons State the b) State in c) State the number of valence electrons d) State the number of protons and e) Deduce the full f) Deduce the condensed g) Draw group of the and has a) which of which the electrons periodic table in chemical in an atom selenium an symbol atom of Se. Se. belongs. of Se. 2 the electron orbital electrons conguration electron diagram for of in the anion, Se Se. conguration of Se. Se. Solution a) Z = 34, b) Se c) Group d) For is in so Se has group 16 16 34 protons (the elements and 34 electrons (atoms are neutral). chalcogens). have 6 valence electrons. 2 Se the number two negative charges e) The full f) The condensed g) The orbital of it protons has equals gained Z two for Se, 2 electron conguration for Se is namely electrons, 1s 2s 2 so 6 2p it 2 3s 34. has 6 3p diagram for conguration Se is given for Se is [Ar]3d However, total 10 3d 10 electron a 2 4s 2 4s 4 4p 4 4p below: [Ar] 2 4s 10 3d 2 4p 74 1 4p x 1 4p y z of since 36 it is an electrons. anion carrying 3 . 2 P e r i o d i c t r e N d S 3.2 P ns Understanding Applications and skills Ver tical and horizontal trends in the periodic ➔ Prediction and explanation of the metallic and ➔ table exist for atomic radius, ionic radius, non-metallic behaviour of an element based on ionization energy, electron anity, and its position in the periodic table. electronegativity. Discussion of the similarities and dierences in ➔ Trends in metallic and non-metallic behaviour ➔ the proper ties of elements in the same group, are due to the trends above. with reference to alkali metals (group 1) and Oxides change from basic through amphoteric ➔ halogens (group 17). to acidic across a period. Construction of equations to explain the pH ➔ changes for reactions of Na O, MgO, P 2 O 4 , and 10 the oxides of nitrogen and sulfur with water. Nature of science Looking for patterns – the position of an element in the periodic table allows scientists to make accurate ➔ predictions of its physical and chemical proper ties. This gives scientists the ability to synthesize new substances based on the expected reactivity of elements. Trends in physical and chemical properties Electron congurations explained through understand as atomic and radius, quantum aspects this of chemical topic, in patterns turn are with mirrored in mechanics, a and and such reactions new these heart The the show its periodic trends across position allows about and of properties table predictions the elements groups. periodic properties and – at chemical down afnity, understanding time, lie elements us described trends, chemical Patterns be help electron better same can properties properties, the patterns which atomic provide At 2), energy, These reactions. peppered of ionization electronegativity. in are many (topic of to behaviour therefore their periods an scientists in make of atomic and element in facilitate table in the accurate chemical the synthesis of compounds. properties. Atomic radius The radius to point of a circle, R , is the distance from the centre of the circle c a on the circumference. It is easily measured and has the core a denitevalue. In is the the Bohr model nucleus model it of while would the the appear hydrogen single that atom electron the radius (sub-topic lies of in the a 2.2) xed atom, orbit. R , can of Based also be the on atom this measured, e as according to Bohr the electron is in a xed position within a dened orbit. 75 3 P E R I O D I C I T Y However, the atom as is orbitals, nding xed, we which an so described highly cannot measure need to often that are the move surface probability of way is consider to is, of consider of the an overcoming an X space that that in of there When the looking quantum on as spheres orbital) in that problem non-metallic molecule. the nding is same way atoms we as we are know boundaries. a of not models, where represents of of atomic probability atomic xed fact chemically The in model in electron mechanics with in high the at Bohr located a model region and atoms is of atom Based the fact position Bohr electron diatomic are simplistic atomic this know where the radius circle. the now electrons represented (i.e. nding two a we the spheres. be boundary of means from as 2 and measure radius cannot One This away represented atoms topic regions electron. we in simplistic The 99 % space. the radius bonded distance of an atom together, between the that two 2 nuclei of the X atoms is given by d, and the bonding atomic radius , R , b is dened as: 1 R = d b This 2 is shown atomic radius in is gure 1, using sometimes the termed example the of iodine. covalent The bonding radius. 1 For metals the bonding atomic radius is d’ where d‘ now represents the 2 distance of the An between two atoms adjacent to each other in the crystal lattice metal. alternative atomic radius is the non-bonding atomic radius , R . nb d = 2R b ▲ Figure 1 The iodine diatomic molecule, I Consider a with another one densities. . The bonding atomic group Argon of gaseous there does is not argon very form atoms. little a When penetration diatomic two of species. argon their If atoms electron argon is collide cloud frozen in 2 radius, R , for iodine is 136 pm (d = the solid not be phase the atoms would touch each other (topic 1) but would b 12 272 pm), where 1 pm = 10 m chemically bonded. In this case the distance between the argon toK ● We saw in sub-topic 3.1 that Mendeleev examined the proper ties of elements in minute detail and grouped elements with similar proper ties Usfu su together. The “Periodic Table of Videos” hence website, developed by The predictive power of Mendeleev’s periodic table illustrates the “risk CBE and co-workers at the taking” nature of science. What is the distinction between scientic and University of Nottingham in the pseudoscientic claims? UK provides videos for all 118 a research professor and is also a pioneer in the eld of gn hmsy which is discussed at several points in the IB Chemistry Diploma programme. http://www.periodicvideos.com/ 76 elements fell into their correct groups. Mendeleev was therefore able to predict the proper ties of yet undiscovered elements at the time. Professor Mar tyn Poliako, elements. Professor Poliako is When Mendeleev published his rst periodic table of elements in 1869 he left gaps in the table for as yet undiscovered elements, and ● The periodic table is an excellent example of classication in science. It classies elements in several ways – metals, non-metals, and metalloids; main-group and transition elements; groups and periods; elements with acidic, basic, and amphoteric oxides; and s, p, d, and f sublevels. How do classication and categorization help and hinder the pursuit of knowledge? For example, scandium will be discussed fur ther in topic 13. Why is it incorrect to classify scandium as a non-transition element? 3 . 2 atoms could be measured and hence R could be found (gure 2). P e r i o d i c t r e N d S The nb non-bonding atomic radius is often termed the van der Waals’ radius. d = 2R nb ▲ Figure 2 Atoms of argon in the solid phase. The atoms are touching but not chemically bonded. The non-bonding atomic radius of argon R is 188 pm (d = 376 pm) nb Section of the the 9 mean over of the a range radius approximate For The bonding wide atomic from Data elements. their is atomic of atomic for provides term radius elements always bond example, booklet general length for from the two covalent is used atomic to Note that non-bonding elements can the atomic also be radii represent experimental compounds. than between the radius” obtained and smaller data “atomic data bonding radius. The estimated radii. the interhalogen compound BrF: Quk qusn atomic radius of bromine atomic radius of uorine = 117 pm Predict the bond lengths in: bond Compare length this of with Br the F = = 60 177 pm a) iodine monobromide, IBr ) trichloromethane pm experimental bond length of Br F in the (chloroform), CHCl gas 3 phase (176 nm). ev nua hag an snng In an atom the negatively charged electrons are the attracted to the positively charged nucleus. A valence electrons: charge, S, that is shielded or screened by the core or outer-shell electron is also repelled by the other Z = Z S e electrons in the atom. The ns in the inner where Z = actual nuclear charge (atomic number) and non-valence energy levels of the atom reduce the S = snng shng nsan positive nuclear charge experienced by a valence electron. This eect of reducing the nuclear charge Z experienced by an electron is termed snng or these rules in advanced textbooks on inorganic chemistry, can be worked out using Sa ’s us. You can read about e but you are not required to calculate Z shng. using Slater’s rules e as part of the IB Chemistry Diploma programme. You do need The net charge experienced by an electron is t er med to understand the principle of screening and for our purposes the ff v n ua hag , . Z This is the nuclear eff you can consider S as a parameter related to the number of charge, Z, (representing the atom ic nu mber) m inu s core electrons in an atom. 77 3 P E R I O D I C I T Y Worked example: estimating nuclear charge Estimate the effective nuclear charge (For experienced comparison, using Slater’s rules Z for eff by the valence electron in the alkali metal potassium potassium. need to is be calculated aware assumptions, of as the 2.2.) As chemists limitations equations, and of we many of our rules. Solution Potassium, 2 1s K 2 2s 6 2p has a 3s core does of (it 19 charge. 3p 4s of is the and 19 in protons 18 Z = conguration 19. electrons group 1). (gure experience The electron 1 electrons not the has 6 total electron 18 K 2 the that core and This 3). full one The electrons there valence force provide valence means of the are 19+ 19+ nuclear charge 18 18 electron attraction nuclear partially cancel core electrons this 4s valence electron positive charge approximately and the effective nuclear charge is 1: ▲ Z ≈ Z S = 19 18 = Figure 3 Shielding of the outer valence electron in the 1 eff potassium atom Periodic trends in atomic radius Across a period the increasing the period. from left effective to right, nuclear atomic charge, radii Z , decrease. going This from left is to because right of across eff nucleus, Down a period group away The the radii reason level of the from from nuclear trends atomic the the period. are of for top to bottom, nucleus. charge, Z, summarized the transition this is that the are atomic enter This in a electrons of gures number number, added 4 do of they radii new has because elements quantum electrons (outer-shell) closer to the radius. electrons the principal As valence atomic outer-shell increasing These pulls reducing the further This a increase. energy greater shielding and not 5. n, by Figure electrons in remains enter effect change the 5 1) are new located the core shows greatly the each so than the electrons. that across outermost almost (n In level constant rather a the period. energy across than the th n energy level. So the number of valence electrons and hence Z remain eff essentially constant, resulting in little variation in atomic radius. puorg a nwod esaercni iidar cimota atomic radii decrease across a period ▲ Figure 4 Trends in atomic radii. Some people think of these shapes as snowmen – going down a group the snowman is standing upright, while across a period the snowman is sleeping! 78 3 . 2 P e r i o d i c t r e N d S 2 0 0 K 300 L i N a R b 2 3 8 C s 9 M C a g S 2 0 6 S c Y 1 9 4 T i L a R a Z 2 0 1 )mp( suidar A H r B a 200 3 2 9 B e 2 4 2 F r 1 4 4 V r 1 5 6 c N H C r b f M o T a F e T c W R u 8 4 C o R e R h P d O s C P 7 5 Z A g Ir S n 7 1 N P S n 4 P O A S T i S S 6 2 A B r T e r K r I P o F C l e B i 0 1 6 0 1 0 0 s b b 6 4 1 0 4 G e t 1 4 C i G a d In A u B A l C u 100 X e A t 2 R n 3 4 5 6 7 8 9 10 11 12 13 14 15 16 1 7 18 ▲ Figure 5 Values of atomic radii of elements in pm. These data can be found in section 9 of the Data booklet Periodic trends in ionic radius The radii they The are of cations formed radii of in and the cations anions vary following are smaller from the parent atoms from which way. than those of their parent atoms; for ins example, the atomic radius of K is 200 pm while the ionic radius of + K is 138 electrons attracted The radii pm. in to of The the the reason cation so for the this is that valence there are electrons more are protons more than anions: nucleus. anions are An n is a charged species. Ions are either cations or strongly larger than those of their parent atoms; • for A an is positively + example, F is 133 greater the pm. atomic This repulsion is radius of because between F is the the 60 pm extra valence while electron the in electrons. ionic the radius anion charged, such as Na of results in • 2+ , Mg An ann is negatively 2 charged, such as Cl , O 79 3 P E R I O D I C I T Y Values for ionic radii are also given in section 9 of the Data booklet Suy p An easy way to remember the uorine potassium dierence in size of ionic radii is as follows: Aol America onine Anion larger + K F K F (It follows from this that atomic radius of potassium (K) = 200 pm atomic radius of uorine (F) = 60 pm cations are smaller.) + ionic radius of K = 138 pm ionic radius of F = 133 pm ▲ Figure 6 Atomic and ionic radii for potassium and uorine Ionization energy The an ionization electron The rst energy, from a IE, neutral ionization is the minimum gaseous energy, IE , atom of a in energy its gaseous required to remove ground-state. atom relates to the process: 1 + X(g) The → X second (g) + e ionization energy relates to the removal of a further electron + from the ion X (g), and the third ionization energy is associated with the 2+ removal of another electron from X (g). Values of ionization energy are 1 quoted The kJ values section as in 8 there of of is mol rst the an (per mole ionization Data input booklet. of of atoms). energies for Ionization energy in order the elements energy to values remove an are are provided always in positive, electron. Periodic trends in ionization energy Ionization left 1 to As energies right the vary ionization effective across energy nuclear the periodic values charge, table. increase Z , for Across the increases a period following from left to from reasons: right across eff Suy p a period the Trends in ionization energy valence attraction makes across a period and down a the it electrons between more difcult the to are pulled electrons remove and an closer the to the nucleus electron from nucleus, so increases. the This atom. group are the pps to the 2 Atomic radii decrease across a period – because the distance between trends in atomic radius. The the valence electrons and the nucleus decreases, it becomes more snowman diagram (gure 4) difcult to remove an electron from the atom. and its opposite (gure 7) will Going down a group from top to bottom ionization energy values help you remember both, and decrease for the following reasons: you need to know the reasons underlying these trends. 1 Atomic radii electron 80 increase from the down atom. a group, making it easier to remove an 3 . 2 P e r i o d i c t r e N d S puorg a nwod esaerced seigrene noitazinoi ionization energies increase across a period ▲ Figure 7 Trends in ionization energy are the opposite of the trends in atomic radius 2372 He 2081 Ne 2500 1681 F 1312 H 1402 1520 N 1314 Ar O 2000 1086 1251 C 1351 Cl Kr 900 1012 1 Be 1000 1140 801 P S Br B I E 1 lom Jk/ 1500 11 70 787 944 941 738 Xe Si As Mg Li 590 1008 I 869 Sb Ge Al 496 Na Se 944 762 578 1037 Te 579 Rn 920 709 Ca 1000 Ga At 812 Sn 419 558 549 K In Sr 7 16 Po 703 Pb Bi 589 403 Ti 503 y g r e n e Rb Ba 376 500 Cs atomic number Z 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 1 7 18 increasing ionization energy ▲ Figure 8 Trends in rst ionization energy, IE , for groups 1 2 and 13 18 of the periodic 1 table. IE values increase across a period and decrease down a group 1 2 The shielding nuclear and If a outer graph shown in increase is not topic effect charge, of electrons rst gure across smooth the in the ionization 9, a of core weakening the a energy but versus trend decrease period. increases attractive force faster between than the the nucleus atom. general period across electrons the The is atomic that down spikes rst a and number ionization group, dips is though will be plotted, energy the as values graph explained in 12. 81 3 P E R I O D I C I T Y 2500 He Ne 2000 1 Ar Kr Xe Rn I E 1 lom Jk/ 1500 Hg 1000 Zn Ga Al 500 Li Cd Tl ln Na K 10 Rb 20 30 Cs 40 50 Fr 60 70 80 90 100 Z ▲ Figure 9 Plot of rst ionization energy, IE , versus atomic number, Z. Notice the general 1 trend that IE increases across a period but decreases down a group, though the graph is 1 not smooth across a period Electron anity According to IUPAC, the electron afnity, E , is the energy required to ea detach This is X A an the (g) more the electron energy → X(g) common energy from the associated + singly with mol of and released neutral X(g) + Electron e → equivalent (E E afnity example, for atoms X the negative ion in the gas phase. process: e initial 1 charged denition ) when 1 is that mol of the electron electrons is afnity attached is to nal or molecules in the gas in section phase: (g) values are provided 8 of the Data booklet. For uorine: 1 F(g) + e → F (g) E = -328 kJ mol ea The negative process is sign indicates exothermic endothermic (in process). that energy contrast The more to is released ionization negative the during energies E value, this process: which the relate greater the to is an the ea attraction of the ion for the electron. However, gure 10 shows that the E ea values for some elements, for example group 1 1 H 60 3 Na 53 4 K 48 5 Rb 47 noble gases, are positive. He 2 Li the 18 13 14 15 16 1 7 73 2 18 >0 Be >0 Mg B 27 Al >0 42 Ca Ga 2 Sr 5 41 In 29 C N 122 >0 Si P 134 72 Ge As 119 78 Sn Sb 107 101 O F 141 Ne 328 S Cl 200 Ar 349 Se Br 195 I 190 >0 Kr 325 Te >0 >0 Xe 295 >0 1 ▲ Figure 10 Electron anities E , in kJ mol , for a selection of main-group elements. ea Notice that some of the elements have positive E values. The group 18 elements have ea theoretical, calculated values 82 3 . 2 P e r i o d i c t r e N d S Periodic trends in electron anity Trends in electron anity across a period Trends as the in electron trends across a afnity observed period from in for left the periodic atomic to right radius E table and values are not as ionization become well highlighted energy. more In negative general, (with ea some The exceptions). group 17 elements, the halogens, have the most negative E values: ea 1 for example, E (Cl) = -349 kJ mol . This is expected since on gaining Quk qusn ea an electron these elements attain the stable noble gas conguration. If Suggest why the E values for the ea you look across period 4 (n = 4 energy level) in gure 10 you can see group 2 elements are more positive 1 that from left to right E becomes more negative from 48 kJ mol forK ea than expected. 1 to 325 kJ energies, mol there for are Br. However, examples of within elements each that period, do not as for follow ionization this trend. For 1 example, arsenic, As, has E 78 kJ mol while you might expect this to ea 1 lie E between value 119 for As kJ can 1 mol be for Ge explained and by 195 kJ mol examining its for Se. electron The higher conguration ea 10 [Ar]3d 1 2 4s 4p 1 4p x already applies y contains for 1 4p : an electron is added it will enter a 4p orbital that z one other if electron, members of causing group repulsion. 15, in A similar particular for argument nitrogen where the Positive E values ea E value is positive. ea A positive afnity value for suggests electron that the anion Trends in electron anity down a group is In the case of the group 1 alkali metals, values of E generally become not stable, formed in so the it cannot gas phase. be For ea less negative going down the group (table 1). However, for the last three example, E for krypton is ea or four elements there is little difference between E 1 values. positive (41 kJ mol ), so Kr ea does not exist. Interestingly, 3 1 Gup 1 mn E the N anion the solid is well known in /kJ m a Li 60 sodium state (for nitride, example, Na N), in despite 3 Na the 53 fact that E for nitrogen ea 1 is K positive (20 kJ mol ). 48 3 In Rb 47 Cs 46 by crystals, the lattice (sub-topic provides Fr T able 1 Electron anity values for the group 1 elements The do patterns not show ionization is stabilized enthalpy 15.1), which sufcient energy 47 to ▲ N of electron the same energy, and afnity clear vary trends by electronegativity repulsion group, down a overcome so group electron as (discussed do afnity atomic in the the electron nitride anion. values radius, next). Electronegativity Electronegativity, atom In has 1932 for the the American electronegativity attract scales electrons but which the has symbol shared and to one the χ, pair scientist dened itself”. used symbol in χ . is of it Linus as There this 8 as in the a Pauling “the are section On dened electrons a of proposed power number the scale relative covalent of an of Data the atom the is concept in different booklet uorine, attraction that an bond. molecule to electronegativity the most a of Pauling scale, electronegative p element in the periodic table, has a value of electronegativity of 4.0. 83 3 P E R I O D I C I T Y Periodic trends in electronegativity As shown period the same Across the in and a gure down a reasons period effective 11, electronegativities group (see from nuclear that pages left to charge mirror show those for periodic trends ionization across energies, a for 80–81). right and electronegativity atomic radii both values increase increase because across a period. puorg a nwod esaerced seitivitagenortcele electronegativities increase across a period ▲ Figure 11 Trends in electronegativity are the same as those in ionization energy and the opposite to the trends in atomic radius Down a group because from atomic increases, its top radii effect to bottom increase is shielded and by electronegativity although the core the values nuclear decrease charge, Z, electrons. H B e N a K 4 M g S c T i R b r Y Z r X p B a F r L a N b f A N i R h T a R e F C o W R a O F e T c R u H N C B M o C s 2 V C r S C u A l P d O s c Ir P 0 Z A g t A u C n G a In A s i P S e S C l B r S n H g 3 S d 1 2 G e S b T e I T i 4 P b B i 5 P o 6 A t 7 8 9 10 11 12 13 14 15 16 1 7 ▲ Figure 12 Electronegativity values, χ , increase across a period from left to right and decrease down a group from top p to bottom. Fluorine is the most electronegative element in the table with a χ p 84 value of 4.0 on the Pauling scale 3 . 2 P e r i o d i c t r e N d S Science and peace Pauling Nobel in was the Prizes, 1962 for as his rst he person also to won opposition to win the two Nobel weapons Pauling unshared Peace of also vitamin Prize the mass is destruction. C common given in Pauling’s ● Do you know of any other scientists who peace through their scientic role peace in can scientists world play today? in the Discuss promotion this taking may be structure of the Data correct? large doses effective of ascorbic booklet.) Carry out of against acid Was this aspect, using the some library, the work? in literature, and an online search. Discuss of your the (The 35 suggestion into scientic What cold. section that acid) have research promoted suggested (ascorbic ndings in class. class. Periodic trends in metallic and non-metallic character As described classied topic in into sub-topic metals, 3.1, the elements non-metals, and in the metalloids periodic (see table gure 3 can in be sub- 3.1). Metallic shown character in gure decreases across a period and increases down a group, as 13. puorg a nwod sesaercni retcarahc cillatem metallic character decreases across a period ▲ Figure 13 Trends in metallic character in the periodic table As well as the properties metals also have to electrons low of metals ionization described energy previously values – they in sub-topic have a 3.1, tendency 1 lose oxidized. We during shall a chemical explore redox reaction, processes that is, they further in tend topic to 18 be 9. 1 2 13 14 15 + The properties addition, of non-metals non-metals show were highly also described negative in electron sub-topic afnities 3.1; – in they have 2 Li N to gain electrons during a chemical reaction, that is, they tend 3 Na 4 K reduced. 5 Rb 6 Cs 2+ Mg 14 metals. always In shows For 1+, topic 13 different the and we stable the charges cations for of the shall see of the some alkali alkaline that the common metals earth in metals transition ions group in of 1 group metals metals the 2 form it a and charge is non- is always number F 3+ Al 3 P 2 S Cl 2+ Ca + Figure O to + be 1 7 2 a + tendency 16 3 + 2 Se 2+ Sr Br 2 Te I 2+ Ba 2 +. of ▲ Figure 14 The charges of some common ions ions. 85 3 P E R I O D I C I T Y Trends in the proper ties of metal and non-metal oxides Suy p You can work out the formulas of the main-group metal oxides and hydroxides An oxide We make deduce oxide is the ion formed use of from the chemical to be 2 , the charge combination on formula for the of a metal metal of an cation oxide, element as with shown taking in the oxygen. gure charge 15 on to the example: using the periodic table. + ● 2 Na combines with O to form Na 2+ of the non-metals are less O 2 The corresponding oxides ● Ca ● Al 2 combines with O to form CaO straightforward and you 3+ 2 combines with O to form Al O 2 3 should memorize these for the elements carbon, Metal nitrogen, sulfur, phosphorus, hydroxides: oxides are basic: they react with water to form metal and halogens along with the CaO(s) + H O(l) → Ca(OH) 2 (aq) 2 corresponding acids formed. Na O(s) + H 2 In contrast, form oxides acidic CO (g) + H (l) + H (g) O 4 of the O(l) ⇋ non-metals H + O(l) H → H are acidic: they react with water to O(l) ⇋ + SO 6H 10 H (aq) SO → carbonic acid sulfuric acid 4 2 O(l) (aq) 3 2 2 (s) CO 2 2 2 P 2NaOH(aq) 2 3 SO → solutions: 2 SO O(l) 2 (aq) sulfurous acid 3 4H 2 PO 3 (aq) phosphoric acid 4 Namng xanns an as In naming oxoanions the following rules are useful: Students often struggle with the names of the oxoanions • If only one oxoanion is formed, the ending is “-ate”. • If two oxoanions are formed, the one with the smaller and their corresponding oxoacids. Table 2 summarizes some of these names. number of oxygens ends in “-ite” and the one with the greater number of oxygens ends in “-ate”. Fmua f xann Nn-sysma nam If • 2 there are four oxoanions , the one w ith the carbonate CO 3 smallest number of o xy gens ends in “ -ite” and is 2 C ethanedioate (oxalate) O 2 prefixed 4 ends in by “hypo”; “-ate”, and the the next one ends w ith in the “ -ite”; mos t the third oxyg en s nitrite NO 2 is nitrate NO prefixed oxoanions 3 this 2 by of system “per ” and chlorine, (table end s in bromine, “ -ate”. and T he four iodine follow 3) . sulte SO 3 2 sulfate SO Fmua Nn- Fmua Nn- f sysma f xa sysma xann nam 4 3 phosphite PO 3 nam 3 phosphate PO 4 ClO ClO hypochlorite HClO hypochlorous hypochlorite acid ClO chlorite ClO 2 chlorite chlorate ClO 3 chlorate ClO 4 86 perchlorate HClO 4 hydroxide ▲ ▲ chloric acid perchlorate 4 OH HClO 3 3 ClO chlorous acid 2 2 ClO HClO T able 2 The non-systematic names of some oxoanions T able 3 The oxoanions and acids of chlorine perchloric acid 3 . 2 P e r i o d i c t r e N d S Some interesting oxides ● Silicon dioxide, SiO , does not dissolve in water. However, it is 2 classied as hydroxide, an acidic NaOH to oxide form because sodium it can react silicate, Na with SiO 2 SiO (s) + 2NaOH(aq) → Na 2 ● Aluminium oxide, Al O 2 means SiO 2 it can react is (aq) + H 3 classied sodium (aq) and water: 3 O(l) 2 as an amphoteric oxide. This 3 both as an acid and as a base. See topic 8 for more information. Acting Al as O 2 an (s) acid: + 2NaOH(aq) + 3H 3 O(l) → 2NaAl(OH) 2 sodium Acting as O Al 2 a (aq) 4 aluminate base: (s) + 6HCl(aq) → 2AlCl 3 (aq) + 3H 3 O(l) 2 aluminium chloride Amph an amphp xs The terms amphoteric and amphiprotic are often mixed up. A par ticular type of amphoteric species is described • Amphiprotic species are described fur ther in sub- as amphp. These are species that are either + topic 8.1. proton (H ) donors or proton acceptors. Examples include self-ionizing solvents (such as water, H According to the IUPAC Gold Book, a chemical species • O and 2 methanol, CH that behaves both as an acid and as a base is termed OH), amino acids, and proteins. 3 amph. Aluminium oxide is classied as an amphoteric oxide. Table that 4 shows there across the is a how the trend period oxides from from Fmua f x left Na of basic to some period through ▲ elements to vary. acidic It shows oxides right. O(s) MgO(s) Al 2 Nau f x 3 amphoteric O 2 as as (s) SiO 3 (s) P 2 amph O 4 a (s) SO 10 (l) and SO 3 (g) 2 a a T able 4 Trend in the proper ties of the oxides of some period 3 elements Chemical properties within a group: Group 1, the alkali metals + The group 1 potassium, metals K, are lithium, rubidium, Rb, Li, sodium, caesium, Cs, Na Na, 1e → Na and 1 [Ne]3s francium, that Fr (see hydrogen metals – it is a is sub-topic not a 3.1, member non-metal and gure of a the 3). alkali On group 1 metals are descending and gas. of The the the having one valence group ionization alkali 1 the energy metals with atomic radius decreases. water The therefore increases reactions become characterized more by [Ne] Note electron; they vigorous further down the group. Less energy therefore is required to remove the valence electron from + form the ion M in ionic compounds by 1 potassium, K (IE = 419 kJ mol ) than from 1 losing this electron (they are oxidized, 1 sodium, Na (IE = 496 kJ mol ), for example. 1 topic 9). For example: 87 3 P E R I O D I C I T Y alkaline Reaction with water solution liberated The a group metal 1 metals hydroxide, react with MOH(aq), water to which in this (table 5). Hydrogen gas Li 2M(s) an + 2H O(l) → 2MOH(aq) + H + 2H (g) 2 ran wh wa 2Li(s) also form gives 2 Gup 1 ma is reaction: dspn O(l) → 2LiOH(aq) + H 2 Lithium reacts slowly and oats on the water (due (g) 2 to its low density). Bubbling is observed. Na 2Na(s) + 2H O(l) → 2NaOH(aq) + H 2 Sodium reacts vigorously. Heat is evolved and the (g) 2 sodium melts to form a ball of molten metal which whizzes around on the surface of the water. K 2K(s) + 2H O(l) → 2KOH(aq) + H 2 Potassium reacts more vigorously than sodium: the (g) 2 reaction is violent. It evolves enough heat to ignite the hydrogen, so bursts into ames instantly. A characteristic lilac-coloured ame is observed. Rb 2Rb(s) + 2H O(l) → 2RbOH(aq) + H 2 Both rubidium and caesium react explosively with (g) 2 water. Cs 2Cs(s) + 2H O(l) → 2CsOH(aq) + H 2 ▲ (g) 2 T able 5 Reactions of the alkali metals with water become progressively more violent as you descend the group Only two elements in the Chemical properties within a group: periodic table exist as liquids: bromine, Br Group 17, the halogens and mercury, Hg. 2 The group chlorine, gure 3). electrons, noble 17 Cl, gas elements, bromine, Their giving Cl + 2 The them a group e halogens, iodine, chemistry conguration [Ne]3s the Br, is 17 tendency (they are → to the non-metals astatine, gain by an reduced, At their (see seven electron topic 9). to For uorine, F , sub-topic3.1, valence attain the example: Cl 2 [Ne]3s elements are and characterized 5 3p I, exist 6 3p as or simply diatomic [Ar] molecules X . Fluorine and 2 chlorine room The gases, halogens structure the halogens highly group with atomic an the most radius electron. in liquid, me ta l of co v a l ent tho ug h decrease a compo und s the b o ndi ng form is iodine and astatine are solids at pressure. i o ni c w i th and reactive, this and form combining the for bromine temperature anion 88 are the i onic co mpou n ds . r e acti ve do wn t op ic ha logen be i n g g r ou p w it h 4 for Wi th Ha lo ge n s d ec r e as e s d es cendi ng the m e t a ls , (s e e co m p ou n ds ). r e activi ty r e a ctivi ty incr e a s e s wit h ca tion in g ro up m a ki n g it X de t a il s is of n on - m e t a ls g en e r al g oi ng uo ri n e . the the down The that l es s are the r ea s on the e a sy to ga i n 3 . 2 P e r i o d i c t r e N d S Reaction between halogens and alkali metals The halogens, X , react with the alkali the ionic metals, M(s) to form ionic alkali 2 metal halide salts, MX(s). In compound, MX(s), the cation is + M and the 2M(s) anion + X is (g) X → : 2MX(s) 2 For example: 2Na(s) + Cl (g) → 2NaCl(s) 2 Reactions between halogens and halides A solution of a more reactive halogen, X (aq), will react with a 2 solution of summary In table an of 6 complete ions, these the aqueous reactions bromide, of and (aq), are is of given can chlorine aqueous the formed also is Br as be potassium (aq) a reactive ionic to halogen. A 6. equations. written. a For is A example, colourless chloride colour (gure less table added yellow/orange bromine, in by represented equation solution colourless, formation X reactions balanced potassium is halide formed, observed is when solution due of which to the 15): 2 ▲ Figure 15 Gaseous chlorine, Cl (g), is bubbled 2 Cl (aq) + 2KBr(aq) → 2KCl(aq) + 2 Br (aq) 2 colourless through a solution of potassium bromide, which yellow/orange is initially colourless. On reaction, aqueous bromine is displaced from the potassium bromide solution and the yellow/orange colour of Br (aq) is obser ved 2 Suy p You can think of this displacement reaction as being a competition between the chlorine and the bromine for an extra electron. Remember that the atomic radius increases down a group (gure 4). The atomic radius of chlorine (100 pm) is smaller than that of bromine (117 pm) so chlorine has a stronger attraction for a valence electron than does bromine. Therefore chlorine forms the chloride anion, Cl more readily than bromine forms the bromide anion, Br . Going down group 17 the xzng ay, that is, the ability to gain an electron, decreases. X (aq) c (aq) b (aq) i (aq) 2 c (aq) no reaction 2 Cl (aq) + 2Br (aq) → 2Cl (aq) 2 + Br (aq) Cl 2 observation: yellow/orange solution due to formation of Br (aq) + 2I (aq) → 2Cl (aq) (aq) no reaction no reaction 2 + I (aq) 2 to formation of I 2 b (aq) 2 observation: dark red/brown solution due (aq) 2 Br (aq) + 2I (aq) → 2Br (aq) 2 + I (aq) 2 observation: dark red/brown solution due to formation of I (aq) 2 i (aq) no reaction no reaction no reaction 2 ▲ T able 6 Reactions between halogens X (aq) and halides X (aq) 2 89 3 P E R I O D I C I T Y Suy ps 1 The order of oxidizing ability for the group 17 elements follows the order of electronegativity: F χ : > Cl 4.0 > Br 3.2 > I 3.0 2.7 p oxidizing ability: F > Cl 2 2 > Br 2 > I 2 2 Be careful with the term s van when describing a chemical reaction. An observation is something that you directly witness, such as bubbles of a gas, the colour of a solution, or a precipitate forming. The formation of a gas is not in itself an observation. Worked example: explaining pH changes Construct the pH (see a balanced changes sub-topic for equation, the including reaction of state nitrogen symbols, dioxide to with explain water 8.1). Solution ● Nitrogen NO is a reacts non-metal with water and to therefore form a 1 :1 may form mixture of an acidic nitrous oxide. acid, HNO 2 and , 2 nitric acid, HNO . Nitrous acid is a weak acid and nitric acid is 3 a ● strong We acid. next 2NO + write H 2 ● Finally, 2NO O we (g) Because less 90 → balanced HNO 2 + 2 ● the + include H O(l) the → a 3 state HNO (see symbols: (aq) + HNO 2 mixture 7 of topic equation: HNO 2 2 than chemical acids 8). (aq) 3 is formed the pH of the solution will be Q U e S t i o N S Questions 1 What can A. 2 is the maximum occupy a d number of electrons that 6 C. 6 D. 10 Which of the following elements can 3 as Cl B. K C. Br D. F IB, May [1] 2010 What happens when sodium is added to water? I. A gas II. The is evolved. temperature of the water increases. Si III. A A. I and clear, II colourless B. I and III C. II and D. I, II, IB, November solution is formed. Te I and II B I and III C II and D I, II, How 2 B. 6 and valence electrons does Which earth and III [1] 2009 selenium 8 Which II. 34 only III P oxides to produce an acidic solution when water? O 4 D. III only I. 16 only only III many C. only only added A. II. radius? Al A I. largest metalloids? contain? 4 the be 7 classied III. has + 5 II. ion A. 2 B. I. Which sublevel? 10 MgO III. SO A. I and II B. I and III C. II and D. I, II, IB, May 3 of the following elements are only only alkaline III only metals? and III [1] Rb 2010 Sr III. Ba A. I and II 9 only Which 17 B. I C. II and and III D. I, II, statement about the elements in group correct? only III and is A. only Br will oxidize Cl 2 B. III F has the least tendency to be reduced. 2 C. Cl D. I will oxidize I 2 5 Which period property generally decreases IB, Atomic B. Electronegativity C. Atomic D. First IB, May a stronger 2 across 3? A. is May oxidizing agent than F . [1] 2 2011 number radius ionization energy [1] 2011 91 3 P E R I O D I C I T Y 10 How many of the following oxides 12 are Na O, MgO, Al 2 O 2 A. None B. 1 C. 2 D. 4 Describe , SiO 3 periodic table explain is what bubbled you through will a see if solution of: a) potassium iodide [2] b) potassium uoride. [1] IB, May The shows the a) relationship 2010 alkali periodic The gas 2 13 11 and chlorine amphoteric? metals table State the of are found in group 1 of the elements. full electron conguration + between electron properties of arrangement elements and is a and of the valuable tool predictions in and Identify the elements in property the Describe used periodic to arrange table. the c) [1] State rst Outline two what K you reasons ionization and energies down b) ion, understand by the chemistry. term a) its for b) making K explain of the group energy how alkali the rst metals ionization vary going 1. why + electronegativity period one 3 in reason assigned the increases periodic why noble across table gases electronegativity d) and are smaller not values. parent May 2010 why than atom, Suggest an why alkali working 92 the the ionic radius atomic of radius K of is the K. [3] e) IB, Explain you metal in the should with never your touch ngers laboratory. when C H E M I C A L B O N D I N G A N D 4 S T R U C T U R E Introduction At the very heart understanding structural chemical that this can atoms oppositely together in topic of chemistry chemical arrangements bond holds holds of the we be in together case of an in a ions and as the the covalent “glue” we different shall electron or used species) we In types of to ionic, covalent, as in how pair a also For look the at and metallic between covalent simple repulsion polarity and structure compounds. see determine shall such – differences A compound. three at the molecule, (charged ionic explore bonding our compounds. considered charged shall lies bonding model, (VSEPR) shape some of key – look and compounds valence theory, a and ionic shell can molecule, chemical intermolecular be and principles, forces. 4.1 Ioni ondin nd stt Understandings Applications and skills ➔ Positive ions (cations) form by metals losing ➔ Deduction of the formula and name of an ionic valence electrons. compound from its component ions, including ➔ Negative ions (anions) form by non-metals polyatomic ions. gaining electrons. ➔ ➔ The number of electrons lost or gained is determined by the electron conguration of the Explanation of the physical proper ties of ionic compounds (volatility, electrical conductivity, and solubility) in terms of their structure. atom. ➔ The ionic bond is due to electrostatic attraction between oppositely charged ions. ➔ Under normal conditions, ionic compounds are usually solids with lattice structures. Nature of science ➔ Use theories to explain natural phenomena – melting points of ionic compounds can be used molten ionic compounds conduct electricity but to explain obser vations. solid ionic compounds do not. The solubility and 93 4 C H E M I C A L B O N D I N G A N D S T R U C T U R E Ionic bonding Dnition of n ioni Ions are formed when one or more electrons are transferred from one ond atom An ioni ond refers to the to another. formation of a The noble driving gas force electron for this electron transfer is usually the conguration. electrostatic attraction For example, the electron conguration of sodium, Na is: experienced between the 1 [Ne]3s electric charges of a tion (positive ion) and where [Ne] is the noble gas core. A sodium an nion (negative ion). atom can lose its one valence + (outer-shell) electron to form the Na cation, [Ne]. That is: + Na We e say The → that Na sodium electron If a oxidized conguration 2 [Ne]3s is of in chlorine, Cl We + say Cl (it loses an electron). is: 3p atom gains an electron 2 gas process 5 chlorine noble this conguration, e → that [Ne]3s to form the or [Ar]. That Cl anion it will adopt a 6 3p is: Cl chlor i ne is reduced in this process (it g a i ns an electron). Hence, the formation Ionic but electron of the compounds note that between a that ionic are the by sodium is compound sodium chloride, lost generally strict cation is and formed denition an anion between involves (for gained by chlorine metals electrostatic example, the in the NaCl. and non-metals, attraction compound ammonium + chloride, NH Cl, which consists of the ammonium cation, NH 4 chloride Let us anion, take Cl , is another Magnesium , and the 4 is a ionic, but example group 2 of does an alkaline not ionic earth contain a metal). compound, metal, and magnesium so has two oxide. valence electrons: 2 [Ne]3s 2+ A magnesium also adopts atom the can [Ne] lose noble these gas two core. electrons That forming Mg , which is: 2+ Mg – 2e Magnesium Oxygen is electrons. → is in oxidized group The 2 [He]2s Mg 16, electron in the this process. chalcogen conguration group, of and oxygen so has six valence is: 4 2p 2 An oxygen adopts a atom noble can gas gain That or [He]2s electrons conguration: 2 [Ne] two 6 2p is: 2 O + Oxygen 94 2e is → O reduced in this process. to form the O anion, which 4 . 1 Hence, by the oxygen oxide, two in el e ctr o ns the that f o r ma tio n of are los t the by ion i c I O N I c ma gn e s iu m com p ou n d a re b O N D I N g a N D S T r u c T u r e ga i ne d ma g ne s iu m Stdy tips MgO. • Under normal lattice-type of positive conditions, structures and negative ionic that compounds consist ions of (gure are typically solids, three-dimensional and repeating have You should know the names of the various ions, their units formulas and charges, 1). including some oxonions (oxygen-containing anions, 2 + such as NO sodium ion Na , SO 3 , etc.) 4 chloride ion Cl (table 1). + • + It is incorrect to use the term “molecule” when referring to + ionic compounds. We never + say “molecules of sodium chloride”, but instead + “sodium chloride formula units” to indicate that ions + + are involved in the lattice + structure. + + + + Figure 1 Lattice structure of sodium chloride, which consists of sodium cations, Na chloride anions, Cl + Na 1 2 (102 × 10 , and . From the ionic radii given in section 9 of the Data booklet you can see that 1 2 m = 102 pm) is smaller than Cl (181 × 10 m = 181 pm) TOK Ion ● Nm General rules in chemistry (such as the + ammonium NH 4 octet rule) often have exceptions. How many hydroxide OH exceptions have to exist nitrate NO for a rule to cease to 3 being useful? hydrogencarbonate HCO 3 ● 2 carbonate CO What evidence do scientists have for the 3 existence of ions? What 2 sulfate SO is the dierence between 4 direct and indirect 3 phosphate PO 4 evidence? Topic 9 may ▲ help you when reecting T able 1 Names of various ions on this point. The octet rule The octet and can rule be a has useful its own starting place in point the in discussion trying to of chemical understand how bonding chemical Qik qstion bonds (that are is, undergo electrons rst two basis of formed. in rule states oxidation), order processes covalent The to acquire are the bonding, gain a that electrons noble basis of which elements gas ionic we tend (reduction), core electron bonding. shall to discuss The in lose or electrons Can you think of an example share conguration. third process sub-topic is The the in which the octet rule is not obeyed? 4.2. 95 4 C H E M I C A L B O N D I N G A N D S T R U C T U R E Worked example: deduction of the formula and name of an ionic compound Deduce and/or the formula polyatomic a) magnesium b) aluminium c) sodium and and and and name of the ionic compounds formed between the following pairs of elements species: uorine d) oxygen calcium e) and ammonium nitrate and phosphate. oxygen Solution comintion ) Fom Nm 2+ magnesium and uorine Mg is in group 2, so forms Mg magnesium uoride ; ; F is in group 17, so forms F fom is MF 2 ) 3+ aluminium and oxygen Al is in group 3, so forms Al aluminium oxide ; 2 ; O is in group 16, so forms O O fom is a 2 ) 3 + sodium and oxygen Na is in group 1, so forms Na sodium oxide ; 2 ; O is in group 16, so forms O O fom is N 2 d) 2+ calcium and nitrate Ca is in group 2, so forms Ca calcium nitrate ; ; nitrate is NO 3 ) fom is c(NO 3 ) 2 + ammonium and phosphate ammonium is NH ammonium phosphate ; 4 3 ; phosphate is PO 4 ) fom is (NH 4 ▲ (PO 3 ) 4 T able 2 Formulas and names of some ionic compounds from their component ions. In naming ionic binary compounds, ab, consisting of a metal and a non-metal, the ending will be-ide Physical properties of ionic compounds Q Melting and boiling points , and inversely proportional to the square 2 2 of Ionic compounds have high melting the distance high boiling points because of the of attraction , as given law of electrostatics from by physics: Q 1 forces r strong Q electrostatic them, points Coulomb’s and between between 2 _ the F ∝ 2 r ions in their melting lattice point point is there must 1413 of structures. NaCl °C. In is 801 order to For °C example, and melt its an the boiling ionic solid Hence, in charges the case of correspond magnesium to 2+ for the oxide, the 2+ apart the be a large input electrostatic of energy to break cation, these forces. Mg two , two magnesium 2 and 2 charges for are the oxide greater anion, than those O of . As 1+ + and The electrostatic force of attraction, F, is 1 in melting proportional to the interacting charges, Q and 1 96 the case of the Na and Cl ions, the directly point for MgO is higher, that is 2852 °C. 4 . 2 such Volatility as c O v a l e N T hexane. The b O N D I N g molecule of water is polar + and Volatility refers to the tendency of a has vaporize. For ionic compounds the of attraction are strong, and so itself, δ 2 on H and δ partial charges are attracted to the on ions in O. the electrostatic lattice forces charges substance These to partial the (for example, in the case of sodium chloride, volatility + the of such compounds is very δ on each H in the water molecule is attracted low. to a the negatively result charged individual ions chloride are pulled anion, out of Cl the ). As lattice Electrical conductivity and For an occupy ions ionic xed are solid compound not ionic positions free to contrast, to move in and the in do molten conduct the the move compounds In in in solid state lattice. the not Hence solid state, conduct state, the the ions the so electricity. ions are become case of a between the the solvent remain surrounded non-polar ions of the molecules, within the by solvent, water there ionic so the molecules. is no compound cations In the attraction and and anions lattice. free electricity. uss of ioni iqids Ionic liquids are ecient solvents and electrolytes, Solubility used in electric power sources and green industrial Ionic as compounds water, but do dissolve not in dissolve polar in solvents non-polar such processes. solvents 4.2 c ont ondin Understandings Applications and skills ➔ A covalent bond is formed by the electrostatic ➔ Deduction of the polar nature of a covalent attraction between a shared pair of electrons bond from electronegativity values. and the positively charged nuclei. ➔ Single, double, and triple covalent bonds involve one, two, and three shared pairs of electrons, respectively. ➔ Bond length decreases and bond strength increases as the number of shared electrons increases. ➔ Bond polarity results from the dierence in electronegativities of the bonded atoms. Nature of science ➔ Looking for trends and discrepancies – ➔ Use theories to explain natural phenomena – Lewis compounds that contain non-metals have introduced a class of compounds which share dierent proper ties from compounds that electrons. Pauling used the idea of electronegativity contain non-metals and metals. to explain unequal sharing of electrons. 97 4 C H E M I C A L B O N D I N G A N D S T R U C T U R E cont ondin Dnition of ont ond In A ont ond is formed by the electrostatic attraction between a shared pair of electrons and the positively charged nuclei. According to IUPAC (the ionic order bonding to attain chemical other in a bond order bonding is In to we saw noble exists, to atoms a can electron however, attain covalent how gas in noble which gas bonding, either atoms electron and lose or conguration. it gain A share occurs type electrons conguration. usually electrons second with This between in of each type of non-metals. International Union of Pure and Applied order look at this type of Lewis of bonding in detail, it is useful rst to Chemistry), a covalent bond is a region of introduce the idea a symbol, which is a simple and convenient relatively high electron density between method of representing the valence (outer shell) electrons of an element. nuclei that arises at least partly from In sub-topic 4.3 we shall develop this further into what we term the the sharing of electrons and gives rise Lewis (electron dot) structure of a compound, based on a system to an attractive force and characteristic devised by the US chemist, Gilbert N. Lewis (1875–1946). internuclear distance. In a Lewis number of element. N Let us symbol dots Some consider species, F , O 2 , representation, (or crosses), examples the N 2 , are presence and each which given of element represent in gure covalent is the surrounded valence by a electrons of the 1. bonding in four different HF . 2 Fluorine, F 2 ● Cl Fluorine is acquiring electron ● The in group one more 17, conguration Lewis symbol so has electron, for with seven valence uorine a would complete uorine octet electrons. attain of a Hence noble by gas electrons. is: F B ● If two uorine atoms share one electron each with each other, each Figure 1 Lewis symbols of three uorine atom gains one more electron to attain a complete octet of elements. Nitrogen has ve valence electrons, which results in the uorine atoms. This formation of a covalent bond between electrons, chlorine has seven the two covalent bond is a single bond and the valence electrons, and boron has shared pair can be represented by a line: three valence electrons + Stdy tip Remember, to deduce the nm of F F n tons of an element you can use the op nm from the periodic ● Note that in this Lewis structure of F there are a total of six non- 2 table of elements. For example, sodium (s-block) is in group 1, so has one valence bonding pairs bonding pair of of electrons (often called lone pairs) and one electrons electron; calcium (also s-block) is in group 2, so has two valence electrons. For the Oxygen, O 2 p-block elements you simply drop the ‘1’ ● Oxygen is in group 16, so has six valence electrons. Hence by in the group number to nd the number of acquiring two more electrons, oxygen would attain a noble gas valence electrons: silicon (p-block) is in electron conguration with a complete octet of electrons. group 14, so has four valence electrons. Fluorine (also p-block) is in group 17, so has seven valence electrons, and so on. ● If two oxygen electron covalent double 98 atoms each conguration bond bond between and the share can be the two two electrons achieved two and oxygen shared pairs with results atoms. can be each in This the other, this formation covalent represented bond by two of is a a lines. 4 . 2 c O v a l e N T b O N D I N g + O ● Note that in this Lewis O structure of O there are a total of four non- 2 bonding pairs of pairs of electrons (the lone pairs) and two bonding electrons Nitrogen, N 2 ● Nitrogen is acquiring electron ● If two group three 15, more atoms bond bond can between and the three a Note that valence nitrogen complete three two would octet and nitrogen shared electrons. pairs can be a by noble gas electrons. with results atoms. Hence achieve of electrons achieved in each the This other, this formation covalent represented by of a bond is a three lines: N N ● ve share be the + N has with each conguration covalent so electrons conguration nitrogen electron triple in in this Lewis N structure of N there are a total of two non- 2 bonding pairs of pairs of electrons (the lone pairs) and three bonding electrons Hydrogen uoride, HF ● Fluorine is acquiring in group one conguration 1, so has electron, ● Note is just more with one a in has octet electron. does nature, seven attain not and valence uorine complete would hydrogen historical so valence hydrogen that 17, electron, would of electrons. Hence the key by noble acquire the electrons. attain an to Hence noble Hydrogen acquiring gas just (the octet remember by gas is electron in one conguration octet point a of group more helium. rule here for In the Lewis structure of hydrogen is the formation of a noble gas electron conguration). a molecule, the electrons ● The Lewis symbols for hydrogen and uorine involved in the covalent bond are: are indistinguishable. x H F atiity For convenience electrons for the in two each we of use the different two symbols Lewis (a symbols cross to and signify a dot) for different the electrons Using a similar approach to that of the examples here, elements. deduce the Lewis structures ● To achieve noble gas congurations, uorine and hydrogen can of the molecules carbon each share one electron with each other, forming a covalent bond. dioxide, CO This covalent bond is a single bond and the shared pair can be , and water, H 2 O, 2 showing the steps involved in represented by a line. the formation of the covalent bonds in each case. x H + H H x F 99 4 C H E M I C A L B O N D I N G A N D S T R U C T U R E ● Note that in this non-bonding pair of Lewis pairs structure of of electrons HF there (the lone are a total pairs) of and three one bonding electrons bond stnth nd ond nth The examples above triplecovalent bond describe bonds. These molecules bonds with differ in single, both double, bond and strength and length. Bond strength anoy The trend in bond strength is: You can think of ond stnth ≡ > = > − in terms of windows – a That is, a triple bond is stronger than a double bond, which in turn is window that is triple-glazed is stronger than a single bond. stronger than a window that is double-glazed, which in turn is stronger than a window with a The bond Bond enthalpies enthalpies in will section be 11 discussed of in the Data booklet sub-topic show this (table 1). 5.3. single pane. Bond length This is − A a the > = single triple opposite > to bond strength: ≡ bond bond. illustrate trend this is longer The than covalent (table a double bond bond, lengths in which section in turn 10 of is longer the Data than booklet 1). bond nthpy cont ond nth / pm bond 1 (t 298 K) / kJ mo C≡C C=C C ▲ C 839 120 614 134 346 154 T able 1 Bond strengths (enthalpies) and bond lengths compison of ont onds nd ioni onds We now bonds. H F understand Table 2 the inherent summarizes some difference of these between ionic and covalent differences. etontiity We saw the single in the case of both uorine, F , and hydrogen uoride, HF , that 2 + δ - δ Figure 2 Dipole moment represented each covalent molecule. atoms in F , In there bond the is is case an made of equal up of identical sharing a shared atoms, of the pair such of as electrons electrons the in two the for uorine shared pair 2 by a vector in the polar molecule, HF between shared pair atoms. In two 100 the two is unequally fact, atomic atoms. you This shared might partners for is think the not the case, between of this shared the as pair! a In however, hydrogen in and “tug-of-war” reality, HF , and uorine between uorine the has a the much 4 . 2 c O v a l e N T Ioni ondin b O N D I N g cont ondin Usually formed between non-metals. Formed between a cation (usually metal) and an anion (usually + non-metal). Some cations (such as NH ) can be comprised of 4 ) can contain metals. non-metals and some anions (such as MnO 4 Formed from atoms either losing electrons (process of oxidation) Formed from atoms sharing electrons with each or gaining electrons (process of reduction) in order to attain a other in order to attain a noble gas electron noble gas electron conguration. conguration. Electrostatic attraction between oppositely charged ions, that is, a Electrostatic attraction between a shared pair of cation (positive ion) and an anion (negative ion). electrons and the positively charged nuclei. Ionic compounds have lattice structures. Covalent compounds consist of molecules.* Ionic compounds have higher melting points and boiling points. Covalent compounds have lower melting points and boiling points. Ionic compounds have low volatilities. Covalent compounds may be volatile. Ionic compounds tend to be soluble in water. Covalent compounds typically are insoluble in water. Ionic compounds conduct electricity because ions are free to Covalent compounds do not conduct electricity move in the molten state. They do not conduct electricity when because no ions are present to carry the charge. solid, however, as the ions are not free to move. T able 2 Dierences between ionic and covalent bonding ▲ *We shall discuss covalent network structures that involve lattices later. greater to a attraction what we partial for describe negative the as a shared polar charge, δ , pair than covalent and one hydrogen bond, atom does with adopting and one a this atom partial leads adopting positive + charge, the δ . shared negative In this pair case, of charge, since electrons δ , and uorine in the hydrogen has a covalent then greater bond, adopts pulling it the power acquires partial the for partial positive + charge, dipole δ . This separation moment, symbol of μ charge (gure can be represented vectorially by a Tnds in tontiitis 2). ● If the two atoms involved in the formation of the covalent bond are Going from left to right across a period, χ values P identical, covalent the bond bond is is said to non-polar be and a pure has no covalent dipole bond; moment. that is, Hence, the the F increase. F Reasons: bond in F is a non-polar covalent bond 2 i) decreasing atomic radii The US chemist Linus Pauling (1901–1994) introduced the idea of ii) increasing nuclear charge. electronegativity (χ ) as the relative attraction that an atom of an P element has for the shared pair of electrons in a covalent ● bond. Going down a group, χ P values decrease. Pauling in devised section 8 of a the electronegative scale of Data electronegativity booklet. element in the On the values, Pauling periodic table which scale, with a can uorine value of be is χ found the most =4.0 i) increasing atomic radii P (sub-topic 3.2). There certain Reasons: ii) primary screening have a are already trends seen for in the electronegativity ionization values energies across that a mirror period what and we down (shielding) eect of inner electrons. group. 101 4 C H E M I C A L B O N D I N G A N D S T R U C T U R E One Stdy tips ● of the main uses estimate, based on a ionic, pure of electronegativity electronegativity values value is that differences, we Δχ can , whether P bond is covalent (non-polar), or polar covalent. This Polar covalent bonds estimation is based on the rules in table 3 which you need to have an unsymmetrical remember. distribution of electron density and are bond typ represented by par tial Δχ P + charges, δ and δ ionic Δχ > 1.8 P ● For ionic bonds, we use full charges to represent the pure covalent (non-polar) Δχ = 0 P ions, eg + and in the case of NaF. These charges are polar covalent 0 < Δχ ≤ 1.8 P not par tial and hence we do not use the δ sign. ▲ T able 3 Rules to estimate whether a bond is ionic, pure covalent (non-polar), or polar covalent For example, χ F 2 (F) = from Δχ 4.0, P χ = 0, so F (F) = 4.0 has a pure covalent 2 bond and χ P NaF 3: P (non-polar) HF table (H) = 2.2, Δχ hence P polar covalent χ = (F) 4.0 = 1.8, so HF has a highly P bond and χ P (Na) = 0.9, hence Δχ P = 3.1, so NaF has ionic P bonding. chmisty in th kithn Miows are part of the electromagnetic spectrum (EMS), as seen from section 3 of the Data booklet. The reason why microwaves are so ecient at heating food relates to the interaction of the microwave radiation of wavelength, λ, (in the range 1 mm to 1 m, corresponding to frequencies, ν, in the range 300 GHz to 300 MHz) with the molecules of water, H O, present in food in the liquid state. Water consists of two - 2 O - - - H bonds, which are both polar, and water itself is a polar molecule (this is explained later). Hence the oxygen side of the water molecule is negatively charged and the hydrogen 2 2 δ 2 δ 2 δ δ side of the molecule is positively charged, resulting in a net + δ + O + δ + O δ + δ + O δ + δ O δ + δ dipole moment: 2 δ + δ O + δ + δ O + δ + δ O 2 δ In food, polar molecules of water absorb microwave radiation and constantly change their orientation (“ip”), ▲ 2 δ + δ δ+ O + δ 2 δ Figure 3 Interaction of electrical charges seen on a sinusoidal wave representation of microwaves (showing their oscillating capacity) interacting with the water molecules present in aligning themselves with the alternating electromagnetic food and causing them to ip eld created by microwaves (gure 3). The ipping and rotation of water molecules results in the dissipation of heat energy and increases the temperature of the food. 102 4 . 2 c O v a l e N T b O N D I N g Worked examples: compounds Example 1 Deduce Solution which of the following compounds are In order to deduce this we need to work out Δχ , P molecular: ● the electronegativity the various binary difference for compounds, each and bond in remember SO 2 that Δχ = 0 is indicative of a pure covalent P ● PCl (non-polar) 3 bond, 0 Δχ < ≤ 1.8 is indicative of P a ● Na polar covalent bond, and Δχ O > 1.8 is indicative P 2 of ● NH an ionic bond. NO 4 3 Δχ H 2 = 0, so H has a pure covalent (non- 2 P polar) bond. χ = Solution HCl SO and PCl 2 are molecules as they contain only (Cl) and no ions. In Na O, Δχ 2 is (H) = 2.2, hence = 2.5, so = ionic (since >1.8). Although 1.0, it consists with NH NO 4 is, in HCl has fact, ionic chlorine a polar having a covalent partial of negative non-metals, so P P bond, all χ and P Δχ non-metals it 3.2 P 3 charge, δ , and hydrogen having a because 3 + + it consists of an ammonium cation, NH , partial positive χ = charge, δ and 4 a nitrate oxoanion, NO . Remember that ionic 3 KBr (Br) 3.0 and χ P compounds have lattice structures. (K) = 0.8, hence P Δχ = 2.2, so KBr has ionic bonding with P potassium Example 2 Deduce which having of the bonds in the following CO χ (O) a = having 1 3.4 and χ P binary compounds are ionic, pure covalent (non- a 1+ charge and bromine charge. (C) = 2.6, hence P Δχ = 0.8, so CO has a polar covalent bond, P polar), or polar covalent: with and ● oxygen carbon having having a a partial partial negative positive charge, charge. H 2 ● HCl ● KBr ● CO Qik qstion Explain why, when you heat a refrigerated bowl of soup in a microwave cooker, the soup close to the outside of the bowl can appear warm but near the centre of the bowl the soup can often be cold. 103 4 c H e M I c a l b O N D I N g a N D S T r u c T u r e 4.3 c ont stts Understandings Applications and skills ➔ Lewis (electron dot) structures show all the ➔ Deduction of Lewis (electron dot) structure of valence electrons in a covalently bonded molecules and ions showing all valence electrons species. for up to four electron pairs on each atom. ➔ The “octet rule” refers to the tendency of atoms ➔ The use of VSEPR theory to predict the electron to gain a valence shell with a total of eight domain geometry and the molecular geometry electrons. for species with two, three, and four electron ➔ Some atoms, like Be and B, might form domains. stable compounds with incomplete octets of ➔ Prediction of bond angles from molecular electrons. geometry and presence of non-bonding pairs of ➔ Resonance structures occur when there is more electrons. than one possible position for a double bond in ➔ Prediction of molecular polarity from bond a molecule. polarity and molecular geometry. ➔ Shapes of species are determined by the ➔ Deduction of resonance structures, examples repulsion of electron pairs according to the 2 include but are not limited to C H 6 , CO 6 and O 3 3 valence shell electron pair repulsion (VSEPR) ➔ Explanation of the proper ties of covalent theory. network (giant covalent) compounds in terms ➔ Carbon and silicon form covalent network (giant of their structures. covalent) structures. Nature of science ➔ Scientists use models as representations of the real world – the development of the model of molecular shape (VSEPR) to explain observable properties. Nature of science But the a what and, on scholar must knowledge he as says for will what ultimately only he lives be that soon is be true, seeing long content what it is Scientists with false world in of exposed he can built if on in Ronald Nobel 104 Coase (Recipient Economic (1910–2013) laureate until his was death the on 2 shape certain of the Nobel Sciences oldest in a representations has been Every assumptions for a model, scientist its theory used to model – is one to as a of the explain in of science the and is major appreciate limitations, real model the whether Prize it will is one withstand the test of time. VSEPR theory 1991). living September of as VSEPR properties. considerations enough. Coase models example, molecular validity Ronald use for observable count accepted, – 2013. such limitations. model, although not without its 4 . 3 c O v a l e N T S T r u c T u r e S lwis (ton dot) stts Earlier which in this shows represented idea of of the by Lewis by two by a can dots, we i ntr od uced numb er eithe r bo nd be by of do ts (electron covalent electrons For top i c the in or two cr o ss e s in a ( or a of Fr om struc tu res , mol ecul e. r e p r e s ente d i de a a e le ct r on s cro ss e s . d ot ) a the v al e nce In a this an we based Le w i s nu m be r Lewis of of de ve l ope d on the of a ea c h wa ys dot the forma ti o n s t ru c t u r e, d iffe r en t com bi n a t ion symbol , e l em e n t a nd – pa ir of e ith e r a c ro ss ) , or line. example, phosphine, some PH , of the might ways be in which represented the are Lewis shown structure in gure of 1(a). 3 (a) x x - P P P H H H H H Cl (b) H Cl H C Cl H H Cl Figure 1 (a) Two dots, two crosses (or a combination of the two), or a line can be used to represent each pair of electrons in a Lewis (electron dot) structure. (b) Lewis (electron dot) structure of CCl . Remember the bond angles shown in a Lewis structure do not 4 necessarily represent the actual bond angles in the molecular geometry In ● such a representation bonding double, ● pairs or non-bonding which In the are Lewis electrons Similarly, of triple of the one of of to distinguish (showing the between: covalent bond as single, and electrons , electrons lone Lewis important electrons structure and is bonds) pairs pairs it not often involved phosphine there called the in bonding. are the three lone bonding pairs, pairs of pair. structures of carbon dioxide, CO , and carbon 2 monoxide, shown in CO, which gure contain multiple bonds, can be represented as 2. Figure 2 Lewis structures of CO In CO , each double bond represents two bonding electron pairs, and in 2 CO, the Lewis bond in triple structures (single, molecules. actual Lewis bond represents help double, us or H o w e v e r, shapes structure of of three understand triple Lewis molecules, a bonding molecule the bonds) and hence may be different and structures electron the tell the pairs. types existence us nothing of of covalent lone about representation drawn with a of pairs the the geometrical 105 4 C H E M I C A L B O N D I N G A N D S T R U C T U R E arrangement For example, that the differs Lewis completely structure of from its carbon real shape in tetrachloride, space. CCl , 4 is typically existence represented of 90 ° tetrachloride angles. very We C Cl molecule shall useful valence Cl as model shell for figure1b, bond is shortly in angles. tetrahedral see how predicting electron pair to which In fact with the this molecular suggest shape 109.5 ° deduce repulsion might Cl C shape of Cl (VSEPR) the carbon bond based geometries, the on called a the theory Lewis (electron dot) structures of cations and anions Stdy tip: us of and ionic compounds sq kts in Lewis structures can be written not only for neutral molecules but also lwis (ton dot) for stts an cations anion charged and anions. there ions, is an which In a compound electrostatic forms the containing attraction ionic bond. both between a the However, cation and oppositely the bonding The chemical formula of within the cation and anion separately may be covalent in nature; for ammonium nitrate is often example, written as NH NO 4 in ammonium nitrate, NH , but in NO 4 (gure 3(a)) the bonding in 3 3 + [NH reality it is made up of a cation, ] and in [NO 4 ] is covalent, even though the bonding between 3 the cation NH Cl and the anion is ionic. In the case of ammonium chloride, ammonium, and an oxoanion, nitrate. When you write Lewis (gure 3(b)) the Lewis structure of the chloride anion can be 4 represented with the chlorine surrounded by eight dots to represent structures of cations or anions, the eight valence electrons present in the anion. including oxoanions, you should always include square (a) H (b) + H brackets and the charge in the O representation. H N H H N H N H H covalent bonds covalent bonds covalent bonds in the cation in the cation in the anion Cl ionic bonds between the cation and the anion − so the compound overall is ionic H H N H Cl H ionic bonds between the cation and the anion − so the compound overall is ionic Figure 3 (a) Lewis structure of ammonium nitrate. (b) Lewis structure of ammonium chloride vn sh ton pi psion (vSePr) thoy Much of of the structure and as chemists dimensions. 106 core and understanding bonding. we need Every to of chemistry molecule have the involves has ability to a discussions particular always shape think in three 4 . 3 a) c O v a l e N T S T r u c T u r e S b) CH 3 O HO H C 3 N CH 3 Figure 4 (a) 2D representation of the drug tramadol, whose molecular formula is C H 16 NO 25 . 2 Tramadol is a centrally acting synthetic opioid analgesic used in treating severe pain. (b) Three-dimensional molecular space-lling model of tramadol. The atoms are represented as spheres and are colour coded: carbon (grey), hydrogen (white), nitrogen (blue), and oxygen (red) As mentioned previously, representations shell the electron shapes The basis charged apart In as to balloons is as of two-dimensional about theory shape. can be Valence used to deduce pairs of – since electrons electrons repel one are negatively another to be as far space. the examine two aligning are nothing (VSEPR) follows particles, maximum pair–electron Then case us molecules. theory in structures tell repulsion determine electron the pair this possible together. In of Lewis ultimately covalent subatomic order this of and pair the balloons at 180° spatial a to angle repulsion shape linear each that try the a be Think is of achieved number balloons geometry other. can tying of from balloons ultimately obtained, dividing with a adopt. the circle two up into 360 ___ halves: = 180° (gure 5). 2 Figure 5 Two balloons tied together showing a linear arrangement in space In the case of three balloons, a trigonal planar arrangement is generated, 360 ___ similar to taking a circle and slicing it into three segments: = 120°. 3 Hence the balloons each other Now consider you give might a what (the to on tying angle happens: might help the in taking a to the appreciate of be and two 109.5 ° of four on means – one at) Thinking dividing the space apart 360 ° balloons the try it! plane (gure in by of two 4, 90 °. balloons This at 120 ° dimensions, which This would is not maximize shape to 6). creates their a 7). tetrahedron the lie together. circle any (gure to chemistry three-dimensional geometry repulsion in balloons between imagine you themselves planar four arrangement tetrahedral You term visualize bond spatial arrange sitting in the three-dimensionality electron pairs (gure environment of this of geometry a cube based 8). 107 4 C H E M I C A L B O N D I N G A N D S T R U C T U R E y x z Figure 7 Four balloons tied together Figure 6 Three balloons tied together showing showing a tetrahedral arrangement Figure 8 A tetrahedron ts into a cube a trigonal planar arrangement in space in space in three-dimensional space The basic molecular in table of electrons might 1, like on the is to geometries basis of described imagine Nm of as two, as can therefore three, or occupying being a eld four an of be summarized, pairs of electron electron Mo omty as electrons. domain, shown Each which pair you density. bond n exmps of ton mos o ions domins hin this shp linear AB 2 two 180° BeCl , CO 2 2 trigonal planar AB 3 three 120° BF , [NO 3 ] 3 Figure 9 The Scottish scientist, engineer, and inventor, Alexander Graham Bell (1847–1922) sitting in his tetrahedral chair. Most famous tetrahedral for his invention of the telephone, Bell was also fascinated by the theory of engineering AB 4 structures and ight. He championed the four 109.5° + CH cause of tthd stts, frameworks , [NH 4 ] , [ClO 4 ] 4 based on a series of interlocked tetrahedra. He is seen here watching trials of his kite designs ▲ T able 1 Molecular geometries based on two, three, and four electron domains The set of three molecular geometries, AB (linear), AB extended to 2 planar), and AB (tetrahedral), can also be (trigonal 3 generate 4 additional than 108 the shapes number for of species electron that have domains fewer bonding present. In pairs such of cases, electrons the electron 4 . 3 domains not occupied non-bonding additional pairs of by the bonding electrons molecular (lone geometries are pairs of pairs). electrons In such generated: AB are cases, E c O v a l e N T lled S T r u c T u r e S by three (V-shaped or bent), 2 AB E (trigonal pyramidal), and AB 3 2 represents We ● can the a H The electrons domain domains this of distinguish molecular illustrate O. pair electron the To lone therefore electron ● E let us bent), where E 2 (table geometry geometry of or 2). between: predicted idea number (V-shaped from (based VSEPR (which take electron the gives on total theory); the example domains the of of and shape the predicted number of the water from molecule). molecule, VSEPR theory is 2 four (we electron formula shall learn domain we presence of occupied by molecular see how geometry that two to there bonding two lone geometry, deduce is only pairs pairs of two (not electrons. on an AB E This However, O four). 2 (table shortly). tetrahedral. are based this H This from bonds, The implies is two that the which other structure, means the chemical suggests domains that the the are actual V-shaped or bent 2 2). Nm of eton Mo ton domin omty domins omty bond n exmps of mos o ions hin this shp boon noy fo mo shp V-shaped trigonal planar three <120° AB E SO , [NO 2 (bent) ] 2 Returning to the balloon analogy, 2 you can see this in action if you tetrahedral again take four balloons and tie trigonal 2 four <109.5° AB E NH , [SO 3 pyramidal ] + , [H 3 O] 3 them together. This time have 3 two of the balloons blue and tetrahedral two of the balloons yellow, the V-shaped four <109.5° AB E 2 H O, [NH 2 (bent) ] 2 latter representing lone pairs of 2 electrons. Make the two yellow ▲ T able 2 Geometries involving lone pairs based on three and four electron domains balloons bigger than the two blue balloons (the text opposite explains why). To emphasize the Bond angles in molecular geometries: fact that the lone pairs are nonLone pairs of electrons affect the bond angles in a molecule. Lone pairs bonding pairs of electrons take a occupy more space than bonding pairs, so they decrease the bond angle black marker and mark two dots between bonding pairs. The degree of electron pair–electron pair on each yellow balloon. You still repulsion follows this order: have four electron domains, so LP|LP > LP|BP > the electron domain geometry BP|BP is designated as tetrahedral, but where LP represents lone pairs of electrons and BP represents bonding now it is made up of two bonding pairs of electrons. electron pairs and two non- Table 3 illustrates decreases the how bond repulsion between lone pairs of electrons bonding electron pairs. angles. 109 4 C H E M I C A L B O N D I N G A N D S T R U C T U R E Mo Nm of Mo omty bond n ton domins H C H H four CH 109.5° 4 H tetrahedral AB 4 N H H H NH four 107° 3 trigonal pyramidal AB E 3 O H O four 104.5° 2 V-shaped AB E 2 2 T able 3 Eect of lone pairs on bond angles ▲ Interpreting the VSEPR model Using the model of VSEPR theory it is not possible hydrogen sulde, S, H is V-shaped, based on an 2 to predict exact bond angles when lone pairs are AB E 2 present. will the be All angle However, students to determined is to bond (104.5°) be and account. make state angles and then molecular geometries a mistaken that than with the the basic repulsions the for angle from shape. should mistake that be many experimentally ammonia assume have (107 °) by feasible. bonds ethene, trigonal C=C and all V-shaped be interpretation of bond the angles. For example, phosphine, PH , the 92.1°. so Two H S the H H bond C predicted to arrangement H is be bond angle angles exact factors is occupy bond 121°, 120° about in angle even based each play and more example, is role are just molecule 117° a a multiple though on affected space, the is are predictions that differences also For angle The making other latter pairs). bond and both trigonal of the H would planar carbon: VSEPR H model. (the lone all these at factors, electronegativity like and lower many not but 2 much that geometries also bond predicted common learn molecular is less LP|LP A pyramidal This is associated LP|BP into water can expected bond taken you structure, also 12 1° H has 3 C an AB E structure and is trigonal pyramidal, 11 7° C but 3 H its H P H bond angle drops to 93.5 °. H Likewise, Working method to deduce both Lewis (electron dot) structures and electron domain and molecular geometries We can theory combine in a following structures method and geometries: 110 Lewis structures simple-to-use can be electron and working used to domain VSEPR method. deduce and The Lewis molecular 1 Draw a central ball-and-stick atom. electrons about draw in bond the Each the in at this any identifying represents covalent angles sticks diagram, stick bond. stage a Don’t – direction you to pair the of worry can commence 4 . 3 the process. the negative In the case of oxoanions, localize E AB 2 atoms; the charges on remaining any terminal bonds should oxygen other into anions double (not bonds. In oxoanions) (V-shaped) – all with associated bond- considerations. be the and S T r u c T u r e S 2 angle ● converted c O v a l e N T case Oxoanions. of cations Worked examples use square brackets and place the charge Example 1: Carbon tetrachloride, CCl outside these. For central 4 ● 2 the number in atom, the deduce periodic table from the its group number A ball-and-stick diagram for CCl : 4 of Cl valence 3 From electrons. the number of sticks, count the number Cl of single sigma bonds, (σ) which we shall designate C Cl as bonds. Cl 4 Add one electron for each negative charge ● (but not oxygen for localized atoms electron for in a charges oxoanions positive already in step charge. assigned 1 ). Delete Subtract C one one pi (π) four valence electrons (it is in group 14); four σ bonds; for so each has to the total number of valence electrons is eight; bond. 8 _ = 5 Combining by two which to steps 2, obtain equals 3 the the and 4, divide number number of of this number electron electron 4 so there are four electron domains. 2 pairs, Thus domains. (AB the electron domain geometry is tetrahedral ). 4 6 Based on deduce the the number electron of electron domain domains, ● geometry. There are are four present tetrahedral 7 Determine if the applicable, number and of deduce lone the pairs – C the and Cl bonds molecular the bond so no lone geometry angle will is pairs therefore be 109.5 ° present, molecular Cl geometry. Then draw an exact representation 109.5° of the structure, complete with predicted C Cl bond angles, electron-pair LP|LP > taking into account the order Cl repulsion: LP|BP > BP|BP ● Finally each 8 Finally, draw Cl of a Lewis representation excluding the octets hydrogen on all (which terminal will need Cl to in complete order to the octets generate on the by Lewis completing you terminal structure. atoms, already have Cl attained a noble gas electron conguration of 109.5° two). Remember to include square brackets C Cl for any cation or Cl anion. Cl 9 Draw page any 115) resonance where structures (explained on applicable. + Example 2: Ammonium cation, [NH ] 4 Let us put are three this working method to the test. There + ● Ball-and-stick diagram for [NH ] : 4 types of to work out: ● Basic shapes structure that you are required + H – AB (linear), AB 2 planar), and AB (trigonal 3 (tetrahedral). 4 H ● Species with lone pairs of electrons – AB E 2 (V-shaped), AB E 3 (trigonal pyramidal), and H 111 4 C H E M I C A L ● N has four ve σ B O N D I N G valence A N D electrons S T R U C T U R E (as it is in group 15); the lone the difference positive in the bond electronegativity angle is but likely also to play total role. charge; ● so inuencing bonds; a one pair number of valence electrons = 8; Finally, each 8 _ you need terminal F to in complete order to the octets generate on the Lewis structure: = 4 so 4 electron domains. 2 N Electron domain geometry is tetrahedral (AB ). F 4 F <109.5° ● There are are four present tetrahedral – N H the and bonds so molecular the bond no lone geometry angle will F pairs is therefore be 109.5 ° Example 4: Sulfur diuoride, SF 2 + ● Ball-and-stick diagram for SF H : 2 109.5° F N H H H ● The above structure ● structure as is hydrogen also is a valid by has two Lewis surrounded S which is the maximum number electrons (group 16); bonds; the total number of valence electrons of is electrons σ valence two so electrons, six eight; permissible. 8 _ = 4, so four electron domains. 2 Example 3: Nitrogen triuoride, NF Electron 3 domain geometry is tetrahedral (AB E 2 ● Ball-and-stick diagram for NF ). 2 : 3 ● There are must only be two two S F lone pairs covalent present, bonds. as Hence there the F molecular angle geometry will be less is V-shaped than and 109.5° due the to bond the F presence ● N has ve valence electrons (as it is in group much three σ bonds; LP|LP more is which so the total number of valence electrons of the two lone pairs, which occupy 15); space. greater is The than greater than repulsion that that between between a between a LP |BP a BP |BP , so is the bond angle is reduced signicantly from its eight; predicted value of 109.5 ° 8 _ = 4, so there are four electron domains. 2 S Electron domain geometry is tetrahedral (AB E). F 3 ● There are must only three molecular bond presence than – one N F geometry angle space) be will of the the that lone covalent is be trigonal less lone than pair repulsion between pair present, bonds. as Hence pyramidal 109.5° (which between there and due to the is from role more the greater F experimentally angle, the occupies LP |BP The the <109.5° which the of model, the LP|LP Finally you need terminal F F Lewis structure: F S determined F N F bond F angle, the 112 which model, is cannot 102.2°, of S showing in the F bond precisely signicant operation uorine will (also have an BP|BP . N experimentally 98°, F determined repulsion electronegativity each The is be inuence). ● F determined cannot be determined suggesting precisely that not only from is <109.5° F to in complete order to the octets generate on the 4 . 3 Example 5: Nitrite oxoanion, [NO This is an oxoanion, so in the S T r u c T u r e S ] 2 ● c O v a l e N T O xonions ball-and-stick In the case of oxoanions, we do not add an additional diagram we rst localize the one negative electron here for the negative charge on the nitrite charge on any one of the two oxygen atoms. oxoanion because this has already been accounted for Since oxygen has a valency of two, this means in the rst step. This is a very impor tant point and you the other nitrogen-to-oxygen bond must be a need to note this dierence for oxoanions. doublebond. ● There must be one lone pair present, as there O are only two σbonds. Hence V-shaped Note that in the case of an oxoanion, we with square brackets with the due outside, but we then the molecular bond angle geometry will be is to the presence of the less one than lone pair, negative which charge and the covalent rst 120° begin nitrogen-to-oxygen localize the occupies much more space. charge N on any one remove the stepwise of the two square working oxygen brackets atoms until and later in O our method. The ● N has ve two σ one pi valence O <120° electrons (as it is in experimentally which group15); cannot model, is Finally you be determined determined bond angle, precisely from the 115° bonds; ● ( π) bond, which counts as 1(it each important to remember this; see the need to complete the octets on is terminal O in order to generate the box Lewis structure: below); N so the total number of valence electrons is six; O O <120° 6 _ = 3 so there are three electron domains. 3 Note Electron domain geometry is trigonal planar (AB that for the oxygen containing the E). 2 double bond, addition oxygen of completing two with lone the the pairs, single octet entails whereas bond in for this the the structure, How to hnd π onds in vSePr thoy completion of the octet requires the addition π bonding involves o-axis bonding, as we will explain of in topic 14. Hence, as the shape of a molecule three lone pairs. The actual structure of is nitrite is a combination of two contributing controlled by the σ bonding framework along the Lewis structures. This is resonance, which we internuclear axis, in counting the valence electrons we shall return to shortly. Contributing resonance subtract 1 for each π bond present. For SL students, structures are represented by a double-headed you do not need to go into the reason behind this (the arrow. explanation is given at HL) but you do need to know N N the method involved. O <120° O <120° You can think of it like this – the shape is controlled by the geometrical arrangement along the The two nitrogen-to-oxygen bond lengths in internuclear axis, where the σ bonding framework nitrite are equivalent and intermediate in length lies. A double bond is described as a (σ + π) bond between a single and a double bond. The two and a triple bond is described as a (σ + 2π) bond. To contributing resonance structures therefore could reduce these back to σ bonds, you simply subtract be combined, each represented with square out the π components. brackets and the negative charge placed outside. 113 4 C H E M I C A L B O N D I N G A N D S T R U C T U R E S N N O O O <120° <109.5° <120° Note that for the oxygen containing the 2 Example 6: Sulte oxoanion, [SO ] double 3 ● This is stick also an oxoanion, diagram charges on oxygen other bond we two has a rst of the valency order to in three of sulfur-to-oxygen in so localize this ball-and- two oxygen two, bond satisfy the the this negative atoms. means must be valency a for Since the the addition the two bonds of octet the of two pairs. in lone this The octet the three involves whereas atoms structure, actual of the pairs, oxygen requires combination oxygen. completing terminal single lone double bond, with for the completion addition structure of of three sulte contributing is a resonance structures: 2 2 S O S has six valence electrons (as it is in group σ <109.5° π so is bond the total ( three length 1); number of valence equivalent between a and bond lengths single intermediate in and a double in bond. Inompt nd xpndd otts 8 _ 4, sulfur-to-oxygen are electrons eight; = <109.5° bonds; sulte one O 16); The three O O O <109.5° ● S O O O 2 S O O 2 S O O so there are four electron In most Lewis structures, the central atom will be domains. 2 surrounded by an octet of electrons. However, in some Electron domain geometry is tetrahedral (AB E). 3 ● There are must only be three one lone pair present, sulfur-to-oxygen as species, the central atom will have less than an octet there of valence electrons: these are inompt otts (for example, the linear molecule beryllium chloride, BeCl covalent , 2 σbonds. Hence pyramidal 109.5° and the the because of molecular bond the geometry angle will presence of be is trigonal less the than one lone which has the central beryllium atom surrounded by only four electrons, or the trigonal planar molecule , in which the central boron atom boron trichloride, BCl 3 pair, which occupies much more is surrounded by only six electrons): space. Cl S 120° 180° B O Cl <109.5° The experimentally which ● model, is Finally you each be determined determined bond angle, precisely from 106° Cl Cl O structure. to in complete order to the octets generate on the Cl In other species, an xpndd ott is possible. This is the discussed in topic 14. In such cases alternative Lewis structures involving octets may be used. need terminal Lewis 114 cannot Be 4 . 3 c O v a l e N T S T r u c T u r e S rsonn stts Dnitions As we Lewis saw in the structures arrangements of case to of have the the nitrite identical oxoanion, sometimes arrangements of atoms it is but possible for different ● rsonn involves using two or more Lewis structures to represent electrons. a par ticular molecule or ion. A The individual Lewis structures that contribute to the overall structure resonance structure is one of two or are called resonance forms. The actual electronic structure of the more alternative Lewis structures species is called a resonance hybrid of these resonance forms. In order for a molecule or ion that cannot to represent this idea of resonance, the contributing resonance forms be described fully with one Lewis are linked via a double-headed arrow. structure alone. One of the best known examples of resonance is the molecule benzene, ● C H 6 (gure In doiztion electrons are 10). 6 shared by more than two atoms in a molecule or ion as opposed to being localized between a pair of atoms. (a) (b) Figure 10 (a) Two Kekulé structures of benzene showing resonance. (b) Representation of benzene showing the delocalized nature of its π electrons C The two resonance forms represented here are termed the non-polar molecule Kekulé section 140 structures 10 pm, of the of Data intermediate benzene . booklet, In each between a benzene, as can carbon-to-carbon carbon-to-carbon be seen bond from length double is bond F (134 pm) structure where and of the further in a carbon-to-carbon benzene, circle topic therefore, represents the is single often bond (154 drawn as delocalization pm). in The Figure (which we 10(b), shall discuss B 10). F In topic 14 we shall discuss resonance in more detail, together with F π non-polar molecule electrons and bond order. Mo poity Earlier We in now this focus topic on we O discussed molecular the idea polarity , of that bond is, H polarity . whether the H molecule polar molecule itself is distinct may polar from have or non-polar the polar polarity bonds. In (gure of 11). individual order to The polarity bonds; deduce the a of molecules non-polar molecular is molecule polarity we N can follow a simple three-step working method described in the H box H below. H Wokin mthod to dd mo poity polar molecule 1 Using VSEPR theory, deduce the molecular geometry. 2 For each bond present, using electronegativity dierences, Δχ Figure 11 Examples of non-polar and polar molecules. The net dipole moment, μ, of a , deduce the P polar molecule can be represented vectorially. bond polarity for each bond present and draw the associated dipole moments; The dipole moment represents the non- these are best represented as vectors. symmetrical distribution of charge in a polar 3 Using vector addition, sum all the dipole moments present to establish molecule (compared with a symmetrical whether there is a net dipole moment, μ, for the molecule. If so, the molecule distribution of charge in a non-polar is polar. molecule). In the vector the head of the vector + represents δ and the tail represents δ 115 4 C H E M I C A L B O N D I N G A N D S T R U C T U R E Worked example: deducing molecular polarity › Deduce the molecular polarities of the following: › ν + › ν 1 = SF b) CO . This results in a net dipole net moment, a) ν 2 μ; the molecule is polar. 2 2 ν S ν 1 µ ν 2 ν 1 2 ν net Solution F F a) ν SF net 2 As seen on VSEPR of SF in is an earlier theory, worked the V-shaped. example molecular From section <109.5° geometry 8 of the b) CO 2 Data 2 booklet: Using χ (S) = 2.6 and χ P (F) = 4.0 to be VSEPR a linear theory, carbon molecule. dioxide From is section found 8 of the P Data Hence, sulfur uorine and the is S more F electronegative bond is polar with booklet: than the χ (C) = 2.6 and χ P following dipole (O) = 3.4 P moment: Hence, each C=O bond (Δχ = 0.8) is polar. P The S two opposite F other + that out, is, though deduce need to in are equal direction resulting and in no in magnitude hence net cancel dipole but each moment, - δ δ To vectors the sum molecular the two S F polarity, vectors. μ it = 0. has The two molecule polar is non-polar (even bonds). we The SF 2 molecule vectors (see is V-shaped using Study so we add the theparallelogram tip two C law below): µ ● Stdy tip = 0 A tug-of-war is a model that can be used to consider vectors on the same line (axis). In deducing molecular polarities based on molecular geometries you need to nd the vector sum of the individual dipole moments. The parallelogram law is a useful method. › ● The parallelogram law. If you have two vectors v and 1 › v , and both vectors star t from the same point, the 2 › sum of the two vectors, v, can be found by completing the parallelogram. The diagonal will give the resultant (the vector sum). → ν 1 Figure 12 Students in Montserrat in a tug-of-war. Both teams → are pulling along the same axis ν More polar bonds (resulting from a greater dierence in → ν 2 electronegativity, Δχ ) win the tug-of-war, provided the P pull is along the same axis. that is: › › v = v 1 116 › + v 2 4 . 3 c O v a l e N T S T r u c T u r e S aotops Dnition of otops Allotropes of the same element can vary in both physical and chemical As described by IUPAC, properties. otops are dierent Carbon is one and forms of the most fascinating elements in the periodic table, structural modications of the life on Earth are based on carbon. Carbon has a number of same element. allotropes: graphite, diamond, graphene, and C fullerene. 60 Covalent network solids ● Graphite, diamond, network solids . atoms held are dimensional well silicon by of a are examples network covalent structure example dioxide, graphene covalent together lattice known and A (in solid bonds large covalent is of one in a which giant networks network covalent in the three- orchains). solid is quartz, Another which is SiO 2 ● In contrast, C fullerene is molecular. 60 Graphite Graphite is an are layers are connected called example the pencils form geometry, of hexagonal by London andin in of of and weak forces, (the covalent network consisting intermolecular leading so-called graphite). is a rings the ‘lead’ Each covalently to in of our of In These attraction, is other as not adopts three graphite atoms. graphite pencils atom to solid. carbon forces use carbon bonded of a a which are lubricant lead but trigonal each there layers carbon planar carbon atoms Pop tis of ont at a bond angle of 120 °. The coordination number of each carbon nt wok soids is three within which can be solids, in the thestructure. sheets, allowsthe used as a graphite the Although London layers to lubricant is a good the forces slide past (gure between each 13). conductor covalent of bonds the other, Unlike layers and other electricity are as strong are thus covalent it has weak, ● graphite Melting points. Covalent network solids have high network melting points (typically delocalized greater than 1000 °C and πelectrons. much higher than the melting points of molecular substances). ● Electrical conductivity. Covalent network solids are poor electrical conductors (though graphite and graphene are clear exceptions – electrical conductivity is one of the characteristics that makes graphene remarkable). ● Solubility. They are typically insoluble in common solvents. ● Hardness. Generally, covalent network solids are hard, Figure 13 Graphite is a covalent network solid that consists of hexagonal layers of carbon atoms, which can slide past each other. The layers are connected by weak intermolecular though in graphite the layers can slide past one another. forces of attraction (London forces) 117 4 C H E M I C A L B O N D I N G A N D S T R U C T U R E Diamond Diamond is diamond, atoms in (gure four. a are and of is This the and means dental points single σ a C in lattice C to C structure four of of cutting 109.5° diamond because arrangement of carbon angle within known heavy-duty other bond carbon substances of tetrahedra. tools is this such For as drills. of diamond graphite, covalent diamond the each structural used Unlike of In bonded with number hardest often boiling the solid. covalently interlocking tools, in is arrangement, respectively). localized freely. one diamond melting network atom coordination is polishing 4827°C, covalent bonded reason The The Diamond saws, a carbon tetrahedral 14). covalently this also each does in are bonds, not very diamond and conduct high the (3550 and valence therefore electrons cannot move electricity. Figure 14 Diamond is an allotrope of carbon with a covalent network lattice structure. Strong covalent bonds in diamond make it is insoluble in all common Large crystals of diamond are mined for use solvents. as gemstones. Small crystals are used as an industrial abrasive. High-quality crystals of diamond are found in South Africa, Russia, Brazil, and Sierra Leone. Diamonds are forever? It Intntion pspti has ever been said that considered Unfortunately if “diamonds diamonds not! thermodynamically Under are last ambient unstable and a girl’s best friend”, but have you forever? conditions, eventually diamond turns into is another allotrope Throughout history diamonds of carbon, graphite. However, at room temperature this process is have often been a potential extremely slow, so diamond is said to be kinetically stable. At 1000 °C source of signicant global the conversion of diamond into graphite accelerates and at 1700 °C it conict. The term “blood completes within seconds. When we talk about stability in chemistry we diamond” has been coined need to consider both thermodynamic stability and kinetic stability to describe diamonds mined in regions of conict and subsequently sold to fund such conicts. What responsibilities Graphene – the super material! do nations and governments Graphene is materials, but not it only one Graphene is covalent of the thinnest and strongest of known have in the import of products is also the rst two-dimensional crystal ever discovered. such as gemstones and precious a network solid, but differs from graphite in that it metals? consists of a single (gure 15), carbon atom and coordination is is planar only one covalently number sheet of of atom in bonded each carbon atoms thickness. to three carbon in arranged As other in graphite, carbon graphene is hexagonally each atoms three. The so the carbon usf so atoms are Look at the history of the lattice is discovery of graphene and structure. densely actually packed planar, in a honeycomb which makes it crystalline remarkable lattice, as a but the crystalline current research developments The experimental evidence for the existence of graphene was obtained in using this material at the in 2004 by the Russian scientists Andre Geim and Konstantin Novoselov, University of Manchester, UK , who won the Nobel Prize in Physics in 2010 for their ground-breaking the university where Geim and experiments at the University of Manchester in the UK. Novoselov did their research to win the Nobel prize in Physics Graphene in 2010, http://www.graphene. more efcient manchester.ac.uk/story/ three million When 118 is an excellent than sheets graphite is thermal copper. of A graphene, prised apart and piece it of with electrical graphite one becomes conductor, 1 mm stacked thick on essentially top 300 times consists of of another. graphene. If a 4 . 3 graphene sheet When this, which looks Graphene electrical in like is rolled a a ● used high devices After ball a into as in especially the the aerospace nanotube it (sub-topic becomes a A.6). fullerene, below). exibility, “new because and its Some research composite industry of superb transparency. silicon”. following graphene–plastic the sphere (discussed material, lie carbon a S T r u c T u r e S of materials because of the future areas: to their replace low density strength due displays to the (LCD) and exibility, exible touch-screens transparency, and for electrical mobile conductivity graphene. the discovery materials BN forms up described of in it strength, graphene liquid-crystal of soccer been of development metals up, folded remarkable has applications and is conductivity, Graphene ● is turn, c O v a l e N T and have of graphene emerged, molybdenum in which 2004 include disulde, MoS . a whole the BN class single is an of layers two-dimensional of excellent boron nitride, lubricant and can Figure 15 Graphene 2 be used in ceramic a vacuum materials. so it MoS is is important also a very in space good research and is also used in lubricant. 2 atiity C fullerene 60 In gphn nnoions - th 1985 a new form of carbon allotrope called fullerene, with carbon ft ndmk in th d of atoms arranged in closed shells, was discovered by Robert F . Curl Jr phn! (working at Rice University at Sussex University at Rice in the USA), Sir Harold W . Kroto (working Find out about graphene in the UK), and Richard E. Smalley (also working nanoribbons (GNRs) by University). In 1996 these scientists were awarded the Nobel accessing the chemical Prize in Chemistry for their discovery of fullerenes. The number of literature or online and why are carbon atoms in the shell was found to vary, which led to the discovery material scientists so excited of several new carbon structures. Fullerenes were found to form when about their future development. vaporized of C and carbon C 70 being condensed were initially in an atmosphere synthesized, with of an more inert C 60 gas. than C 60 formed. The structure of each C molecule was Clusters clusters 70 found to consist of 60 a truncated spherically icosahedral symmetrical cage, C which has molecule the was shape unique of in a soccer nature ball. at its The time of 60 discovery. In the C polyhedron cage there are 20 hexagonal surfaces and 60 12pentagonal to three others arrangement The has is geodesic Fuller surfaces, for this the so that not a each carbon coordination (gure d e s i gned 196 7 of the planar dom e shape and Mo ntre a l s o ccer by is number covalently is three, bonded but the 16). the US Wor l d ba ll atom a nd ar c h i t ec t E xh i bi t io n he nc e C R. in has B u c km in s t er Ca n a da been named 60 buckminsterfull e re n e . referred to as These spherical ful l e r e ne s a re some ti me s Figure 16 Molecular structure of c buckyballs 60 fn (kminstfn), C fullerene is not a covalent network solid, and so is different from showing 60 carbon atoms arranged 60 graphite, diamond, and graphene. C is composed of individual 60 molecules between with the strong covalent molecules. bonds, but with weak London forces in a spherical structure that consists of interlinking hexagonal and pentagonal rings, like a soccer ball 119 4 C H E M I C A L B O N D I N G A N D S T R U C T U R E Fullerenes in some they and are black non-polar form coloured varies from solids that solvents solutions; red to do (for not the brown dissolve example, to colour in water, benzene). depends magenta. C , In on but the unlike can organic dissolve solvents solvent both graphite 60 and graphene, delocalized one C does not electrons, molecule to conduct the the electricity. electrons next do because not of Although have the the it does contain capacity symmetrical to move nature of 60 Many with 60 new compounds atoms inclusion other or small of the Fullerenes the enzymes Carbon of in have active and lie can unusual also fullerenes molecules complexes applications because forms from C of thereby nanotubes enclosed potentially the areas of ability the used synthesized the as properties inside the subsequently, fullerene gene superconductivity t human inhibit are to been within be electromagnetic the site have of and and cage. drug ferromagnetism inclusion hydrophobic immunodeciency Such carriers; virus complexes. cavity (HIV) that protease them. tube-shaped molecules, closely related to C , 60 and have copper many new electrical and are 100 electronic times that incorporated approximately stronger applications, compounds when conductivity allow within than replacing unstable the steel. ten They silicon, The are and substances tubes. times to in better also the of used in synthesis become diameter than such of stabilized tubes is 9 extremely and small, carbon chemical in the nanotubes research in order have of 10 mm, become materials a on the vibrant science, with nano and scale. dynamic numerous Buckyballs area of applications. Serendipitous discoveries The accidental discovery of the importance in the library discovered. famous in and What of of buckminsterfullerene serendipity online other to in nd chemistry. out examples how of is Carry a classic out example some research buckminsterfullerene serendipitous discoveries was are chemistry? Siion dioxid, SiO (q tz) 2 Silicon dioxide, SiO , often called silica, is found in its amorphous form 2 (that is, a solid crystalline with form it is three-dimensional no ordered called structure) quartz. covalent as Quartz network is solid. It sand. In another consists its most common example of arrays of a of SiO 4 tetrahedra arranged covalently to four covalently to two is bent on because each (1710°C) strong of oxygen and a covalent in a lattice oxygen silicon the high atoms atoms. presence atom. (gure Silicon boiling and The of each Si two dioxide point 17). O Each silicon oxygen Si atom (2230 both °C) is geometrical non-bonding has atom a due pairs high to is bonded arrangement of electrons melting the bonded point existence of bonds. Figure 1 7 Structure of quar tz, which is a Both crystalline form of silicon dioxide, SiO crystalline and amorphous dioxide are insoluble in water and solid . 2 Crystals of quar tz are used in optical and crystalline does not conduct electricity (since there are no delocalized 2 scientic instruments and in electronics, such electrons in as quar tz watches electricity 120 SiO present) orheat. however as Note electrons that are molten free to silicon move in dioxide the can molten conduct state. 4 . 3 c O v a l e N T S T r u c T u r e S coodint ont ondin The term oodintion ond We have just considered covalent network solids. Another type of is often used (based on covalent bonding covalent bond, is called coordinate covalent bonding . In a typical IUPAC recommendations) the shared pair of electrons originate from both atoms to designate a coordinate that form pair and the bond; one atom contributes one electron to the shared covalent bond. the covalent the A two second bonding, atoms; number of atom the this shared atom species contributes pair donates have of the second electrons both coordinate electron. comes electrons covalent to from the In coordinate only shared bonding. one of pair. Examples include: + ● [NH ] 4 + ● [H O] 3 ● CO ● Al Cl 2 ● 6 transition metal complexes (discussed in topic 13). + Ammonium cation, [NH ] 4 + When ammonia, NH , reacts with an acid, H , the lone pair on the 3 + nitrogen in NH combines with the proton, H , to form the ammonium 3 + cation, [NH ] : 4 + H N: + + H → [NH 3 ] 4 + H 109.5° N H H H + Hydronium cation, [H O] 3 + O H H H <109.5° The coordinate covalent bond is represented by an arrow Carbon monoxide, CO to signify the origin of the electrons in the bond. Once C O formed, however, all the bonds are equivalent (whether Dimer of aluminium chloride, Al Cl 2 coordinate covalent or normal 6 covalent). Previously, the Cl Cl Al was used for this type of Cl Cl bond but, based on IUPAC recommendations, this term is Al Cl term dative covalent bonding Cl now largely obsolete. 121 4 c H e M I c a l b O N D I N g a N D S T r u c T u r e In the solid state aluminium chloride is ionic. AlCl Cl is six-coordinate involving 3 an ionic lattice (but with signicant covalent characteristics). At atmospheric Cl pressure it sublimes at 180 °C. On increasing the pressure it melts. On melting Al Cl Al at 192.4 °C it forms the dimer, Al Cl 2 structure, aluminium three-dimensional is tetravalent structure of the , which has coordination bonding. In this 6 with a dimer coordination is shown in number gure 18. of four. The The bridging Cl chlorines are on a different plane compared to the terminal chlorines. Cl Al Cl Cl 2 this Figure 18 The structure of the Al Cl 2 molecules predominate in the gaseous state up to 400 °C. Above 6 temperature, it dissociates to form molecules of AlCl dimer in with a trigonal 3 6 the gaseous phase planar geometry (120° bond angles). Qik qstions rpsnttions of stts 1 Apar t from aluminium For tetrahedral structures it is common to use wd-nd-dsh nottion to show chloride, identify th other the various planes: substances that sublime ● a wedge indicates that the bond is in front of the dening plane ● a dash indicates that the bond is behind the dening plane ● a solid line indicates that the bond lies on the dening plane. readily. 2 Deduce whether both solid and molten aluminium chloride conducts For example, the tetrahedral structure of methane, CH , can be represented as 4 electricity. follows using this notation: H C H H H 4.4 Intmo fo s Understandings Applications and skills ➔ Intermolecular forces include London ➔ Deduction of the types of intermolecular force (dispersion) forces, dipole–dipole forces, and present in substances, based on their structure hydrogen bonding. and chemical formula. ➔ The relative strengths of these interactions are ➔ Explanation of the physical proper ties of London (dispersion) forces < dipole–dipole covalent compounds (volatility, electrical forces < hydrogen bonds. conductivity, and solubility) in terms of their structure and intermolecular forces. Nature of science ➔ Obtain evidence for scientic theories by making and testing predictions based on them – London (dispersion) forces and hydrogen bonding can be used to explain special interactions. For example, molecular covalent compounds can exist in the liquid and solid states. To explain this, there must be attractive forces between their particles that are signicantly greater than those that could be attributed to gravity. 122 4 . 4 I N T e r M O l e c u l a r F O r c e S Theories on intermolecular forces In sub-topics 4.2 intramolecular and 4.3 forces we of saw that there attraction ionic are that compound whereas hold the sodium melting chloride, point of NaCl(s), water, H is O(s), 801 °C, which is 2 the atoms covalent together bonding. molecular within Such geometries, a molecule, resulting intramolecular physical forces properties, in a covalent compound, is much lower, at 0 °C. affect and chmisty in th kithn reactivities could be of compounds. described as Intramolecular “bonding” forces forces of attraction. The next time you are having a drink with cubes of ice in it, take one cube of ice and try to break it with your Another type of attraction, intermolecular ngers. As you will discover this is vir tually impossible forces, are interactions between molecules within and the reason for this can be associated with the a compound (gure 1). Intermolecular forces are intermolecular forces of attraction in the vast network of largely responsible for the bulk properties of water molecules present in the ice. matter, melting that is, point its and physical properties such as boilingpoint. One of the relation + - δ + δ to assumptions the made in topic kinetic–molecular 6 in theory of gases - δ δ is that and that collisions another gaseous adhere polar every liquid covalent bond to are one species molecules of condensed a forces This can of covalent phase gaseous elastic. molecules is be temperature. intermolecular intermolecular or another. some one completely atoms gaseous at between do This not incorrect converted It is the attraction and suggests stick or because into a existence that compound (liquid particle to solid). of enable exist in Figure the 2 force of attraction demonstrates an example of this. Figure 1 Dierence between intramolecular and intermolecular forces of attraction for hydrogen uoride. The intramolecular forces result in covalent bonding As seen in topic 1, the par ticles in a solid or liquid are tightly packed together – that is why we use the term Intermolecular forces of attraction are much weaker ondnsd phs than covalent enthalpy ΔH bonds. change (the For of enthalpy example, the vaporization change standard of associated water, with the vap conversion boiling of point one at mole of standard pure liquid pressure, into 100 kPa) 1 kJ to gas is at its 44.02 1 mol (at break 298 the molecule from a K) two of O water section 11 whereas H polar (see of the the 926 kJ covalent bond Data mol is bonds enthalpies booklet: 2 × 463 required in a for = O H 926 kJ 1 mol are ). Since the relatively compound this example, weak, are reason, N intermolecular not many (g), molecules strongly covalent O 2 (for the held (g), CO 2 example, H forces of of a covalent together compounds (g) attraction and and, are CO(g)) for gases or (for liquids 2 O(l)). In contrast ionic compounds Figure 2 Sperm bank shipping containers being lled with liquid nitrogen to keep the sperm frozen. We think of nitrogen 2 have very between are the solids melting strong at electrostatic ions, room points. meaning forces that temperature For example, of ionic and the attraction as being in the gas phase, but all gases can be conver ted into liquids at some temperature because of the intermolecular compounds have melting forces of attraction between the molecules. The boiling point of high point of the liquid nitrogen is –195.8 °C at atmospheric pressure 123 4 C H E M I C A L In order to properties, and take The in of be need to account all question, types can understand we into attraction B O N D I N G three A N D S T R U C T U R E these bulk widen our intermolecular phases, therefore, intermolecular is solid, what forces the forces of and theories. of As that stated the science we obtain evidence for by making and testing predictions The The relative strengths of these can that can (dispersion) types be exist the in explain and special molecular liquid explanation forces signicantly forces intermolecular to how the attractive are of used example, above, of two and is covalent solid related between greater states. to their than those scientic that theories For existence particles In attraction compounds present? London bonding, interactions. gas. various attraction the hydrogen forces liquid, are of on physical discussion main discuss could be attributed to gravity. based three types of intermolecular forces of attraction that we shall are: interactions in general are: ● London london fos < dipo–dipo forces (also dipole-induced called dipole dispersion forces or instantaneous induced forces) fos < hydon onds ● dipole–dipole ● hydrogen forces However, we shall also consider bonding. examples where London forces are stronger than dipole–dipole forces, but the above order is what occurs often! Collectively induced by the dipole rst two forces) intermolecular are termed van forces der (as well Waals as dipole- forces , as specied IUPAC. London (dispersion) forces + dipole–dipole forces + dipole-induced dipole = The can strengths vary of intermolecular signicantly, considerably weaker but it than forces must ionic be or of attraction emphasized covalent n d Ws fos between that these molecules forces are bonds. london fos London (dispersion) recognized London by the all molecules. Such German–American forces theoretical physicist (1900–1954), London, the capital dispersion forces forces. origin The happens to a of or of exist hence the in the name United instantaneous the non-polar latter (which Kingdom!) term molecule has can such be as were Fritz nothing London induced forces forces to rst Wolfgang do are with also dipole–induced understood diatomic if we called dipole consider hydrogen, H , what when it is 2 approached Although the another moment), nuclei surrounded respect clouds to molecule, However, one part another the is if of an each by a you the the were In this on of is non-polar molecule located symmetrical, to take case, a moment adjacent a might as random have termed hydrogen in shown the in electrons. no net charged The distribution cloud at with throughout a given instant electron moment is instantaneous molecules. is the gure3. more dipole an there positively average snapshot slightly temporary is The is, of charged position. are (that consists negatively change electrons molecule dipole inuence cloud constantly time molecule. molecule hydrogen spherically part. temporary hydrogen hydrogen dipole electron 124 by of density time than generated. dipole and This has 4 . 4 H H H I N T e r M O l e c u l a r F O r c e S H Figure 3 The average distribution over time of the electrons in any H molecule is spherically 2 symmetrical Therefore, molecule nucleus of of higher one hydrogen has the At each the molecule acquired rst electron molecule. within if that a hydrogen density same molecule in molecule the time, will now short-lived the repel will electron two approaches be other second dipole, attracted cloud regions each a instantaneous of of on the to the second electron hydrogen the region hydrogen density approach, as both are negativelycharged. As described electron by cloud by the be dispersed, IUPAC, of a proximity minimized. of the polarizability molecular a charged repulsion Therefore, the probability of results in non-spherical orbital is is a nding pulled generated out that entity its results is between orbital regions of of distribution of electrostatic the space the of distortion eld (such mobile electron effectively symmetrically in ease electric Because (region can the an particle). electrons) of by electron spherical can is shape, cloud; shape. attractions there its A the that caused electrons density where change of as a can be high which that is, the temporary between the dipole partial + positive charge, charge, δ , of δ the , of one hydrogen neighbouring molecule hydrogen and the molecule partial (gure 4). + δ + δ H negative δ δ H H H London force Figure 4 At a given instant in time, a temporary dipole, the instantaneous dipole, is established This interaction such an is the arrangement different pattern of basis is of only induced a London temporary dipoles + force. – may in δ next however, instant of that time a + δ δ Note, emerge. + δ the δ δ δ + + δ δ δ δ + δ δ + + δ δ δ δ δ + + δ δ δ + δ δ + + δ δ δ δ + δ + δ δ + δ δ δ δ + δ δ + δ + δ + δ + δ Figure 5 Dierent arrangements of the interactions of the London forces of attraction between molecules, which result from interactions between an instantaneous dipole on one molecule and an induced dipole on an adjacent molecule 125 4 C H E M I C A L B O N D I N G A N D S T R U C T U R E What affects There are the magnitude three ● number ● size ● shapes of London forces? factors: electrons (volume) of of of the electron cloud molecules. Number of electrons The the greater electrons can the valence be number electrons to the nucleus polarized For example, Ne, and booklet more Kr, electrons, the will the nucleus. be larger The reduced the distance attraction and hence of the the between valence electron cloud easily. consider krypton, (table of and the boiling using the points of the information in two noble section 7 gases of the neon, Data 1). No s boiin point / °c Nm of tons 10 Ne (Z = 10) 246.0 36 Kr (Z = 36) 153.4 T able 1 Boiling points and number of electrons for neon and krypton 6 London forces decrease rapidly with increasing distance, r , based on the relationship: 1 _ V ∝ 6 r where V Hence, n = are 2 is in the the and Hence the krypton the in of level, much means great case energy located This potential neon but in further krypton higher the of associated eight case from the the in that outer of the in can are are the eight the n electrons cloud interactions. electrons nucleus krypton of with krypton outer electron forces than the the attraction London is energy be to = located outer 4 the energy nucleus polarized stronger, so the in the electrons more level. is not as easily. boiling point of neon. Size (volume) of the electron cloud As the number of dispersed The magnitude of the London forces will also depend on the size of the electrons can be linked to the electron cloud, that is its volume in space. In a large electron cloud, the molecular mass the greater the attraction of electrons to the nucleus will not be as great as in a smaller molecular mass, the greater electron cloud, and hence the electrons in a large electron cloud can be the number of London forces polarized more easily. present. For example, propane, CH consider CH 3 number 126 of CH 2 carbon results in octane than stronger for the and boiling octane, points CH 3 atoms 3 in London propane. octane forces of (CH is and the ) 2 two CH 6 greater hence alkanes (table 2). The 3 than a in higher propane, boiling which point for 4 . 4 akn H 3 octane (C ) 42.0 8 H 8 F O r c e S Sp-in mod boiin point / °c propane (C I N T e r M O l e c u l a r ) 125 18 T able 2 Boiling points of two alkanes Shapes of molecules The molecular shape is the third factor that inuences the magnitude the two Isom of London of C H 5 , forces. pentane, Let us CH 12 compare (CH 3 ) 2 CH 3 , the boiling and points of 2,2-dimethylpropane, boiin point / °c isomers (CH 3 ) 3 C 4 pentane 36.1 (table3). 2,2-dimethylpropane Both isomers contain the same number of electrons, but the 9.5 boiling T able 3 Boiling points of two isomers of pentane point of 2,2-dimethylpropane point of pentane. of pentane’s across the full interaction pentane. the In of length the of of contrast, is reason allows because molecules shape The shape the the the for considerably for this is that molecules molecule; better to that contact lower the is, (gure smaller there between because than the boiling straight-chain interact 2,2-dimethylpropane, considerably molecule is is the the of with a each large area molecules contact the nature other area almost of of for soccer-ball 6). In pentane, there is a large contact area across the entire molecule for In 2,2-dimethylpropane, there is a much smaller contact area for adjacent molecules to interact. adjacent molecules to interact. Figure 6 Space-lling models of pentane and 2,2-dimethylpropane showing areas of contact between adjacent molecules for London forces of attraction 127 4 C H E M I C A L B O N D I N G A N D S T R U C T U R E As the contact London which area forces results is much between in a the higher larger between molecules boiling will point for molecules have a of pentane, greater the magnitude, pentane. Final points on London forces – a warning on misinterpretation! ● It must be stressed molecules are molecules. almost small. is to zero, A molecule of has electrons magnitude ● London in He) H ), but they This greater to is exist with a when useful are explaining it means resulting in mass a greater forces have remember an is very London molecular but between molecules molecules marker, mass between attraction between greater polarized, that if number increase in the forces. attractive occur attraction individual molecular be of attraction of molecules able are also forces gravitational misinterpretation London forces and masses forces. a are of the that London to gravitational since alone London attributed fact, common state greater a In that not forces between between between non-polar polar atoms (for molecules molecules as well example (for (for example, example, 2 HCl). That non-polar is, every or molecule will experience London forces (whether polar). Dipo–dipo fos The second type which exist dipole moment, and CH in CHO. of all μ. In intermolecular polar molecules Examples this type of of forces with such a are dipole–dipole permanent molecules intermolecular ( not include force, forces, instantaneous) HF , there is ICl, an HCl, attraction 3 between end Let of us the positive another compare molar masses: halogen, end of one permanent the the bromine, boiling points interhalogen, Br permanent dipole (table of on two iodine an dipole and adjacent molecules that monochloride, the negative molecule have ICl, similar and the 4). 2 Typs of intmo Isom boiin point / °c fos psnt London forces + –1 97.4 ) ICl (M = 162.35 g mol dipole–dipole forces –1 Br (M = 159.80 g mol 58.8 ) only London forces 2 T able 4 Boiling points of ICl and Br 2 Since ICl dipole lead 128 to is highly forces a of higher polar, in attraction boiling addition between point. to the London ICl forces molecules it also (gure has 6), dipole– which 4 . 4 + I N T e r M O l e c u l a r F O r c e S + δ δ δ I δ CI I CI dipole–dipole force of attraction Figure 7 Dipole–dipole force of attraction between permanent dipoles on adjacent molecules of ICl. Note that iodine has a larger atomic radius compared to chlorine (see section 9 of the Data booklet), but chlorine is more electronegative (see section 8 of the Data booklet). Hydon ondin This and third is type one Hydrogen O H, or of of the intermolecular most bonding an N H can bond force important occur types between holds of a special place intermolecular molecules when in chemistry force. there is Dnition of th hydon ond a H F , an present. As recommended by IUPAC in 2011, a hydon ond is dened as an attractive A typical hydrogen bond may be depicted as: interaction between a hydrogen atom from a X - - Z molecule or a molecular fragment, X–H, in which X is more electronegative than H, and an atom or a group hydrogen bond of atoms in the same or a dierent molecule, in which where the: there is evidence of bond ● hydrogen ● acceptor Y ● Z in donor may which hydrogen a) water X an H formation. atom or to an Y is bonded bond is represented sometimes Hydrogen be is anion, Y, a fragment or a molecule Pure and Applied Chemistry, 83(8), (2011) pp1637-1641 Z by the three dots though dashes are used. bonds occur, molecules, H for example, between Ky point O 2 The H b) ammonia molecules, F, O H, and N H bonds NH are polar covalent bonds and 3 are not hydrogen bonds. c) hydrogen d) water uoride molecules molecules, and HF dimethyl ether molecules, (CH ) 3 a) b) O H H O 2 c) d) H O F H rpsnttion of H N hydon onds H H F H In this book we use dashes O H O H H H C 3 N H to represent hydrogen bonds CH 3 to distinguish them from lone H pairs of electrons. H Hydrogen bonding structures of often materials. has a large inuence on both the properties and 129 4 C H E M I C A L B O N D I N G A N D S T R U C T U R E exmp of th t of hydon ondin 150 Let us compare the boiling points of some hydrides of groups 14, 15, 16, and 17. H O 2 100 Figure 8 shows a plot of the boiling points of the series of hydrides versus period number. As you move down 50 a group, the boiling points increase within a par ticular hydride series, because of an increase in the number of C°/tniop gniliob electrons, resulting in a greater number of London forces. O, HF, and NH However, in the case of the hydrides H 2 , the 3 boiling points are considerably higher. This is because of the existence of hydrogen bonding in these compounds. Methane, CH HF H Te 2 period 0 1 2 3 4 number 6 SbH 3 H Se 2 H NH 50 HI S 2 3 SnH 4 AsH 3 , however, has a lower boiling point as PH 3 4 HCl GeH 4 100 expected, because it does not show hydrogen bonding. SiH 4 The strength of the hydrogen bond depends on the 150 electrostatic attraction between the lone pair of electrons CH 4 of the electronegative atom and the nucleus of the proton. Hence the hydrogen bonding in HF is stronger the hydrogen bonding in H 200 than O because uorine is more Figure 8 Boiling points for the series of hydrides (HX, H 2 P (O) = 3.4] and XH P is more random, compared water on did Ear th hydrogen (ice) state. This molecule show be has is molecules, in a gaseous means ice f loats leads lat tice bonds that to on 9). cavities In with The in all In the in ice, results in a higher density of liquid each water water ordered presence the which ice. water the adjacent very to addition, phase water water. regular, (figure creates solid than bonds a state. the density hydrogen which the hydrogen-bonding, the lower why forms hydrogen in bonding water network not would , 3 ) from groups 14, 15, 16, and 1 7 4 Hydon ondin nd wt If X, XH 2 (F) = 4.0, χ electronegative than oxygen [χ of lat tice. the In Figure 9 Open cavity structure in the lattice structure of ice contrast , in the liquid phase the hydrogen bonding Hydrogen bonding is also present in biomolecules such as in the double TOK helix structure of DNA. (sub-topic B.8) Both theoretical and empirical evidence are used for the –1 existence of hydrogen bonding. Typ of intmo fo rti stnth / kJ mo The nature of the hydrogen London forces bond is a topic of much weak (1–10) – this can increase with number of electrons, size (volume) of discussion and the current electron cloud, and shape of molecule IUPAC guidelines contain six criteria that should be used as dipole–dipole forces weak to moderate (3–25) hydrogen bonds moderate to strong (10–40) evidence for the occurrence of hydrogen bonding. How does a specialized vocabulary help and hinder T able 5 Comparison of the various relative strengths of intermolecular forces between molecules the growth of knowledge? 130 4 . 4 I N T e r M O l e c u l a r F O r c e S Worked examples Example 1 Identify ● the intermolecular forces in the following substances: ● He (CH ) 3 ● CH (CH 3 ● ) 2 ● CH 4 CH 3 O 2 F 3 ● NF CH 3 CH 3 OH 2 Solution Sstn Intmo fos psnt He London only CH (CH 3 ) 2 CH 4 commnt Non-polar molecule so London only 3 Since F is more electronegative than N [χ P (F) = 4.0, χ (N) = P 3.0], this trigonal-pyramidal molecule has a net dipole moment and therefore is polar: Polar molecule so London + N NF 3 F dipole–dipole F F This molecule contains no H atoms, so no hydrogen bonding is possible. Even though the highly electronegative element oxygen is present, there is no O–H bond so therefore no hydrogen bonding is possible. Polar molecule so London + ) (CH 3 O 2 O dipole–dipole H CH C 3 3 This molecule is tetrahedral; F is more electronegative than H and C [χ P (F) = 4.0, χ (C) = 2.6, χ P (H) = 2.0] so there is a net P dipole moment present making the molecule polar: F Polar molecule so London + CH C F 3 dipole–dipole H H H Even though the highly electronegative element uorine is present, there is no H–F bond so therefore no hydrogen bonding is possible. The molecule is polar and an O–H O CH CH 2 3 bond is present, so: H London + dipole–dipole + CH 3 CH OH 2 hydrogen bonding O CH CH 2 3 H 131 4 C H E M I C A L B O N D I N G A N D S T R U C T U R E Example 2 State and species Example 3 explain can molecules: form which of the hydrogen ammonia, NH , following bonds with propane, CH 3 ethanoic acid, CH As water CH 3 CH 2 a general London forces CH 2 does not contain OH, NH or on hydrogen basing the It forces that cannot dipole–dipole on intermolecular boiling point of carbon HF form hydrogen bonds with CCl water is 76.72 °C, COOH have N–H and O–H bonds, 3 they the boiling 4 have the ability to form of uoromethane, CH F is 78.2 °C. Solution hydrogen ● with whereas 3 CH 3 therefore bonds < answer the molecules. and of order: 3 point NH strengths the bonds your fact tetrachloride, bonds. follow (dispersion) < Comment, forces, CH relative , 3 Solution 3 the forces COOH. 3 CH rule intermolecular We rst work out the types of intermolecular water: forces CCl : of attraction only present London forces, in each because compound. this is a non- 4 H polar CH molecule F: London with no forces net and dipole moment. dipole–dipole forces, 3 O because this is a polar molecule. H ● On the rules, N H point H basis we of of this would the forces opposite is following that uoromethane intermolecular Infact, and expect the our general boiling with its additional should be much the higher. case! H ● H The the of reason for strength CH F , the this of the must be associated London number of forces. valence In with the case electrons is 3 considerably O fewer than in CCl . In CCl 4 presence H greater of more valence polarizability of 4 electrons the the leads electron to cloud. a This CH 3 results O C which The O the H in signicantly outweigh key point syllabus challenged is here the is that relative based on stronger the and the above every data forces, forces. order example cited must in be provided! Qik qstion atiity Suggest a second way in which Researchers have seen hydrogen bonds for the rst time! ethanoic acid CH London dipole–dipole COOH can 3 Researchers in China recently used tomi fo miosopy (aFM) to produce hydrogen bond with water. the rst high quality images of hydrogen bonds that exist between molecules of 8-hydroxyquinoline. Find out more about this from the chemical literature or online. What is especially surprising about the atoms involved in this type hydrogen bond (hint – consider the involvement of carbon!)? 132 4 . 5 M e T a l l I c b O N D I N g 4.5 Mti ondin Understandings Applications and skills ➔ A metallic bond is the electrostatic attraction Explanation of electrical conductivity and ➔ between a lattice of positive ions and malleability in metals. delocalized electrons. Explanation of trends in melting points of ➔ ➔ The strength of a metallic bond depends on the metals. charge of the ions and the radius of the metal ion. Explanation of the proper ties of alloys in terms ➔ ➔ Alloys usually contain more than one metal and of non-directional bonding. have enhanced proper ties. Nature of science ➔ Use theories to explain natural phenomena – the properties of metals are dierent from covalent and ionic substances and this is due to the formation of non-directional bonds with a “sea” of delocalized electrons. Mti ondin In topic 3, periodic section we table 8 of delocalized gure 1, is saw of the that metals elements. Data booklet), throughout a surrounded regular by a the giant “sea” of lie to Metals so the have left low valence metal. lattice The that delocalized of the stepped ionization (outer-shell) structure consists of of a line energies electrons metal, positive in the (see can shown ions be in ( cations) electrons. Dnition of mti ond A mti ond is the electrostatic attraction between a lattice of positive ions (cations) and delocalized electrons. + + + + + + + + + + + + + + + + + + Figure 1 Structure of a metal showing an array of positive ions (cations) surrounded by a “sea” of delocalized electrons 133 4 C H E M I C A L B O N D I N G A N D S T R U C T U R E Delocalized a metal, lattice The but electrons instead forming strength a of “sea” a ● the number ● the charge ● the ionic are of radius associated move the with a throughout particular the entire nucleus of crystalline electrons. depends electrons metal of to bond valence the not mobile metallic of of are free on that three can factors: become delocalized ion metallic positive ion (cation). aoys Dnition of n oy An An oy can be dened alloy metal as a metallic material, or homogeneous on a more and macroscopic scale, is (or the mixture non-metal metallic hardness, elements that consists combined non-metals strength, consisting of two or more a metals) (for (table either an example, carbon). and with cast Alloys durability of two alloying iron have or consists enhanced which more element differ metals, of the those of metal properties, from or composed of of a one iron such the as parent 1). elements so combined that they cannot be readily separated by composition of oy mix ts mechanical means. Alloys (mjo mnt: a metal) + (oyin mnt: can be metal or non-metal) are to be considered as mixtures for the purpose of aoy composition uss brass copper and zinc door handles, window classication. United Nations (2011) ttings, screws 4th Edition. steel Iron, carbon, and other bridges, buildings metals such as tungsten dental amalgam mercury, silver, and tin used by dentist for teeth llings T able 1 Examples of alloys Explanation of electrical conductivity and malleability in metals Electrical conductivity Metals the is applied move carry in are a to the the electrical wiring can metal, the electric metal conductors delocalized through an through 134 good mobile a high electricity and presence movement Hence degree in an copper of because a of of purity. can Metals of or a are solid other metals impurities (cations) electrons in in electrical to a malleable. to be shape hence increase used of potential electrons structure The the When mobile resulting resistance. needs the metallic metal, of electrons. current. restrict Malleability have without this can dened bonds or past of within direction non-directional the the (they as is one they that The the another, shape lattice are is the hammered breaking. property slide rearrangement metallic Malleability pounded often act in of do ability into a positive which the not why ions leads solid. The have described every sheet reason any as direction 4 . 5 about is the applied over one metallic xed by immobile pounding, another bonding but cations). the there (gure Thus cations is no may if pressure slide disruption The melting strength to the 2). M e T a l l I c positive of ions electrons. than point the the within The a metal attractive the melting melting following of b O N D I N g “sea” point point depends forces of that of of on the hold the delocalized calcium potassium is for higher the reasons: applied force under pressure ● Calcium atom, has delocalized + + + + + two whereas electron ions + + + + ● and 1+ + + + + atom. per one Therefore, between delocalized the the electrons positive will be + ● From a 2+ Therefore the electrons + calcium. forms ion. between + attraction the in Calcium a + only + greater + per electrons has + electrostatic + delocalized potassium ion the positive will also section 9 ions be of whereas and greater the potassium electrostatic Data the in forms attraction delocalized calcium. booklet + we see that + the size of the ionic radius of K is 138 pm, 2+ whereas the size of the Ca ionic radius is structure of metal after being pounded into a sheet smaller Figure 2 Metallic bonding remains intact even after a metal is at 100pm, delocalized which electrons hammered into a sheet or other object without breaking. This will implies be that more the strongly 2+ attracted to the Ca ion. illustrates the proper ty of malleability This Aluminium this foil, property of often used to malleability wrap for the food, shows variation observed metals. metal As increases aluminium. going expntion of tnds in mtin in melting descending we saw in will a attractive the group, decrease forces points group topic (remember down points the on and with 3, can 1, the radius diagram) the melting decreasing (table be alkali ionic snowman hence also the strength of 3). points of mts Metallic bonds are very strong and Mt therefore Ioni dis of Mtin point / °c + metals often have Forexample, 3414 occur Table °C, metal, as potassium calcium some 38.8 compares (section melting melting although such 2 the high °C the of low for the the of tungsten melting is points also / pm lithium (Li) 76 180.5 sodium (Na) 102 97.8 mercury. melting and 7 point M points. points alkaline Data of the earth alkali potassium (K) 138 63.5 rubidium (Rb) 152 39.3 caesium (Cs) 167 28.5 metal booklet). T able 3 Melting points for the group 1 metals Mt Mtin point / °c When comparing metals, potassium (K) the number melting of points delocalized of the valence alkali electrons 63.5 per atom does calcium (Ca) the not (one) and change the within charge the on the group, so cation the (1+) only factor 842 thatinuences T able 2 Melting points of potassium and calcium the melting point comparisonis the + size of the ionic radius ofM 135 4 C H E M I C A L B O N D I N G A N D S T R U C T U R E Even expntion of th poptis of adding element oys in tms of non-dition ondin an alloy As seen the Alloys can have a number of improved to the parent metallic greater greater strength to enhanced magnetic ductility for a metal example into a to (a mechanical deform being as those the of alloying the the metallic non-directional can be modied if (cations) different network then becomes property to the slide the properties parent bonds shape of metal. within of by force a because of can slide atoms positive are past one the present ions is another. the disturbed more difcult for the and positive able to under tensile metal stretch the past each (gure other 3). This and is change why the alloys shape are that generally, allows 3, an properties of greater with of change corrosion ions ● gure are ions However, resistance amount element: it ● in metal regular ● small properties positive ● a dramatically compared lattice pure compared can much stronger than pure metals. stress, metal wire). applied force under pressure applied force under pressure + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + structure of metal after being pounded into a sheet Figure 3 The presence of dierent atoms in an alloy disturbs the regular lattice and hinders the movement of positive ions past one another Mts – n intntion dimnsion In this sub-topic we explored some key aspects of metals. The availability of metal resources and the means to extract them varies greatly in dierent countries, and is a factor in determining national wealth. As technologies develop, the demands for dierent metals change and careful strategies are needed to manage the supply of these nite resources. Discuss this in class in the context of your own country and the regional economy. 136 Q u e S T I O N S Qstions 1 What is the name of K SO 2 ? 7 What the A. potassium calcium C. potassium D. calcium the electr o n molecular d o ma in geo me tr y, a nd g e ome try, the Cl – P– Cl sulte bond B. is 4 angle for the mo l ecul e phos phor u s sulte trichloride, PCl ? 3 sulfate eton sulfate Mo c–P–c ond domin omty n / ° omty 2 What is the A. MgO B. MgO formula of magnesium oxide? A. tetrahedral B. tetrahedral tetrahedral 109.5 trigonal 109.5 2 pyramidal C. MnO D. MnO trigonal tetrahedral C. 100.3 2 pyramidal 3 What is the A. NaNO B. NaNO C. Na formula of sodium nitrate? trigonal trigonal pyramidal pyramidal 100.3 D. 2 3 N 8 3 D. 4 A. is NaCN Which of Na Which the following species is molecular? O following KBr C. NH graphite B. graphene C. C D. diamond 9 What are the intermolecular forces present 4 in the molecule CH F 2 Which molecule oxygen bond A. CH B. (CH C. (CH CH 3 has the shortest carbon-to- A. London forces B. London forces C. London forces, 3 2 and dipole–dipole forces OH O hydrogen 2 ) ? length? 2 ) 3 dipole–dipole forces, and bonding CO 2 D. only hydrogen bonding CO 10 6 carbon 3 O 2 D. of NO 4 5 allotropes 60 B. N the A. 2 D. of molecular? The electronegativities, χ , for four elements Which statement best describes metallic are P bonding? given in table emnt 4. H c O A. Electrostatic B. Electrostatic charged 2.2 χ 2.6 3.4 attractions between oppositely between a c ions. attractions lattice of 3.2 P positive ▲ ions and delocalized electrons. T able 4 C. Which bond is the most negative polar? D. A. C Electrostatic attractions ions Electrostatic and between delocalized attractions a lattice of protons. between protons H and B. O H C. H Cl D. C O IB, electrons. May [1] 2009 137 4 C H E M I C A L 11 Consider B O N D I N G the following A N D S T R U C T U R E species 15 Compare and (diamond, BF Cl NCl 2 contrast graphite, the allotropes graphene, and ) carbon in 60 OF 3 of C 2 termsof: For a) b) each species ● structure ● bonding ● intermolecular ● melting ● electrical deduce: (i) its electron (ii) its molecular (iii) its bond (iv) its molecular draw an domain geometry forces geometry points angle(s) polarity appropriate Lewis (electron conductivity. dot) structure. 16 In chemistry can 12 Consider the following ] [ClO 3 both lead to terminology certain ] [BF 3 ] COF 4 Suggest is why incorrect the term based on each “macromolecular” IUPAC for covalent network species solids a) models 2 recommendations For and assumptions. species: a) [NO often such as graphite. deduce: b) The O–Cl–O bond angle in OCl is 110.9°. 2 (i) its (ii) electron its domain molecular geometry Discuss with geometry the (iii) b) its draw bond an whether predictions model of this of bond bond VSEPR angle angles agrees based on theory. angle(s) appropriate Lewis (electron dot) 17 structure. Suggest work why for the VSEPR theory majority of does not transition metal 2 complexes, such as [FeCl ] , but 4 13 Deduce the intermolecular forces present in few complexes, such as [MnO ] 4 each of ● Ar ● CH the CH 3 ● CH following CH 2 species: OH 2 Cl 3 ● CH CH 3 14 Deduce form ● which CH CH ● NH ● C 3 H 2 4 PH 3 138 3 of the following bonds OCH 2 CH 2 hydrogen 3 ● OCH 2 CH 2 3 with species water may molecules: does for a 5 E N E R G ET I C S AND T H E R M O C H E M I ST RY Introduction Chemistry involves elements of energy a is the study periodic use models, and scientic with reactions to of Conservation principle empirical terminology associated of table. fundamental The changes of the of data, central of examine science. mathematics explain chemical the the energy reactions is to the the chemistry function Hess’s the nature and energy. enthalpy, Law of science. relationship and of We a will greater bond In this exists investigate gain applications that topic introduce the we between the state applications understanding of of enthalpies. 5.1 M Understandings Applications and skills ➔ Heat is a form of energy. ➔ Temperature is a measure of the average kinetic ➔ using Total energy is conser ved in chemical reactions. ➔ ➔ A q = the of a hea t pure cha nge when s u bs tance is the changed mc ∆T calorimetry reaction Chemical reactions that involve transfer of heat of temperature energy of the par ticles. ➔ Calculation experiment should be f or covered an and enthalpy the of resu lts evaluated. between the system and the surroundings are described as endothermic or exothermic. ➔ The enthalpy change (ΔH) for chemical 1 reactions is indicated in kJ mol ➔ . ΔH values are usually expressed under standard conditions, known as ΔH , including standard states. Nature of science ➔ Fundamental principle – conser vation of energy is a fundamental principle of science. ➔ Making careful obser vations – measurable energy transfers between systems and surroundings. 139 5 E N E R G E T I C S A N D T H E R M O C H E M I S T R Y What is thermodynamics? The utilization Agricultural, consume of energy industrial, vast amounts is central and of to our domestic energy will lives. activities remain law all is the study of how it is interconverted. The can only states that rst energy law can from one form to be of neither law created converted is often energy and nor between called states the that destroyed; different it another that for a given system we forms. can of for and quantify all the energy changes. be This converted be means account thermodynamics This energy This and constant. conservation energy daily. can Thermodynamics of and is one of the most fundamental principles that ofscience. the total amount of energy for a given system Chemical potential energy, heat, and entropy In a chemical energy the is Heat, of q, cooler the is a to as conduction, When as heat of a total the of reacting ions, and energy result thermal of and transferred to is conserved. bonds of mixture is is a It an function transferred can be object, and potential products, of the while kinetic present. temperature radiation. Chemical reactants molecules that the energy. convection, is energy chemical the atoms, form body, referred system + in temperature energy universe = reaction stored from transferred Heat the a gradient. has result by the is warmer Heat is the processes ability an body to a sometimes to do increase of work. in the surroundings average kinetic temperature At absolute or reaches its kelvin) is matter. As a zero, theoretically surroundings energy change 0 stops K ( minimum its in particles the °C), all entropy possible to and therefore an increase in its phase. 273.15 and proportional the of the temperature value. motion S (see The average the absolute kinetic increases, of the particles sub-topic 15.2) a temperature energy kinetic of of the energy system (in particles of the of particles increases. system (contents of ask) Chemical energy When examining the energy changes involved in a chemical reaction, Figure 1 The universe is the combination of the system and its surroundings we divide chemical think of the universe reaction the is system into taking as two being solvents. The reaction, thermometers parts: place, surroundings all and the include the its system reactants, the in which surroundings. products, apparatus that the You and can any contains the In an op tm the transfer or other measuring devices, the laboratory, and of matter and energy is possible everything external to the reacting substances. across its boundary (eg matter can be added to a beaker, and When energy can be transferred rearranged to through its sides). A od are and tm allows no transfer of Energy matter, though energy may be endothermic transferred across the boundary. energy In an otd tm, matter released can neither enter nor exit, but exothermic can only move around inside. and a broken is the changes 140 chemical reaction create new required and it can is an This energy reaction. for bonds transfer important the are is are type of the reactants are in the reactants form the products. bond the made: energy of to bonds: termed of bonds made each of part atoms Chemical chemical quantied The place, bonds the chemical process. system a break be new takes products. chemical to process. when in new breaking bond bond. bond is an dissociation Energy making is is an between the surroundings understanding the energy 5 . 1 M e a s u r i n g e n e r g y c h a n g e s Enthalpy and thermochemistry A omt is any apparatus Enthalpy is an example of a state function. For a state function any used to measure the amount change in value is independent of the pathway between the initial and of heat being exchanged nal measurements. Other examples of state function include volume, with the surroundings. In an temperature and pressure. exothermic reaction heat is For example, pool early in afternoon a simple and the (nal uctuations is taking value), one. temperature morning that cooling the may does have However, which (the has initial not tell does occurred the water value) the occurred it of and whole give any throughout a then story throughout not in the again of in any day. indication the generated which is transferred swimming temperature The of to the surroundings. In an the calculation the heating endothermic reaction heat is consumed. In the school laboratory, experiments focus on the change in temperature day. of the reaction solvent, which ∆T = T T (nal) in most cases is water. (initial) Thermochemistry chemical is reactions. dened reaction. as The commonly unit of the is heat term used the At of “change in by ∆H changes a is the the closed enthalpy” describing change heat pressure, transferred when enthalpy study constant or that change system “heat occur in during enthalpy during of a reaction” thermodynamics of a ∆H chemical is reaction. The exothermic reaction kJ. ygrene laitnetop Endothermic and exothermic reactions A chemical to the reacti o n in surrounding s is which he a t d ene d as is an t r an s fe r re d e xoth e r m i c from t he r e act i on sy s t em wi t h a a reactants ∆H products negative their ∆H. In contrast, surrounding s positive are chemi ca l de ned as r e a ctio ns tha t e n dot h e r m i c a bso rb r e ac t ion s he a t wi t h fro m a reaction pathway ∆H Figure 2 In an exothermic reaction the enthalpy of the products is lower than that The calculations involved in investigating the energetics of the reaction of the reactants. The products are described between zinc and aqueous copper(II) sulfate, CuSO are described 4 as being energetically more stable than the later in the topic. Measured quantities of copper(II) sulfate solution reactants. (We shall study the activation and zinc are mixed in a calorimeter. The mixture is stirred and the energy of a reaction in topic 6) change in temperature measured using a thermometer or data-logging equipment. endothermic reaction Zn(s) + CuSO (aq) → Cu(s) + ZnSO 4 result involving of calculating the exothermic displacement reaction Ammonium ΔT nitrate, shows of (gure NH this single copper(II) ion replacement by the metal reaction zinc is an 2). NO 4 the that is an important component of fertilizers. 3 ygrene laitnetop The (aq) 4 products a ∆H reactants When the nitrate solid ions, dissolves the reaction in water requires to form heat to aqueous proceed. ammonium This heat is and absorbed reaction pathway by the system from the surroundings, resulting in a decrease in the Figure 3 In an endothermic reaction the temperature of the surroundings as recorded by a thermometer. The enthalpy of the products is greater than that apparatus of an containing endothermic the reaction reaction feels (gure 3). cold to touch. This is an example of the reactants. The products are described as being energetically less stable than the reactants + NH NO 4 (s) 3 → NH (aq) 4 + NO (aq) 3 141 5 E N E R G E T I C S A N D T H E R M O C H E M I S T R Y Calculating enthalpy changes in aqueous solutions We stated earlier that the change in enthalpy ∆H or a copper 1 K. For example, the 1 is dened as the heat transferred by closed is 0.385 1 system ∆H for heat a a a chemical reaction change. pure an during we When substance therefore of as the To need calculating such understanding reaction. the water, to nd heat we physical calculate the change need to quantity, 2.44 of of have a J K heat capacity, specic heat capacity of 1 K . The while lower substance, achieved transferred that of ethanol is for to the the the the specic higher same heat the amount capacity temperature of heat sample. the Specic specic g 1 g given rise J heat capacity is an intensive property c that does not vary in magnitude with the size of 3 the 1 The units The specific for specific hea t ca pacity are kJ kg system being described. For example, a 10 cm 1 K . sample of copper has the same specic heat 1 heat cap a city of w ater is 4 . 18 kJ capacity kg as a 1 tonne block. 1 K and this can be found in s ection 2 of the Specic Data q of a q The specic dened the heat capacity is used to calculate the heat booklet as heat the capacity amount temperature of 1 g of a pure of heat of the substance needed to substance 1 = using the relationship: mcΔT is raise by system °C where m is mass in kg and ∆T is the change in temperature. Wokd xmp : t tp Example 1 Example 2 When a 1.15 g sample of anhydrous lithium chloride, LiCl 180.0 was added to 25.0 g of water in a coee-cup calorimeter, iron, resulting a temperature rise of 3.80 K was recorded. Calculate the 26.0 °C. J of heat is in trans f erred a Calculate to tempera ture the specif ic a 1 00 . 0 ris e hea t g from s a mp le 2 2.0 °C capa city of of to iron . enthalpy change of solution for 1 mol of lithium chloride. Solution Solution ∆T = (299 295) K = 4 K . q = mc∆T Make c the subject of the equation and solve: 1 = 0.025 kg × 4.18 kJ kg q __ 1 K × 3.80 K c = m∆T = 0.397 kJ 0 180 kJ ___ = Conver t to energy gained for 1 mol of LiCl. 0.100 g × 4 K 0.397kJ / 1.15g LiCl × 42.394 g/mol = 14.6 kJ/mol LiCl 1 = 0.450 kJ K 1 ∆H = -q = -14.6 kJ mol thermometer co-p omt Performing reactions in a polystyrene coee cup to measure the enthalpy change is a convenient experimental procedure. The methodology introduces glass stirrer cork stopper systematic errors that can be analysed and the eect of their directionality assessed. stmt o are a consequence of the experimental procedure. Their eect on empirical data is constant and always in the same direction. With the two polystyrene cups nested coee-cup calorimeter, the measured change in enthalpy for a reaction will together containing reactants always be lower than the actual value, as heat will be transferred between the in solution contents and the surroundings in every experiment. Figure 4 A coee-cup calorimeter 142 5 . 1 M e a s u r i n g e n e r g y c h a n g e s Investigation to nd the molar enthalpy change for a reaction Earlier we looked displacement at the reaction exothermic between zinc ● metal and The copper(II) specic solution is heat the capacity same as of that an of aqueous water. sulfate: Loss Zn(s) + CuSO (aq) → Cu(s) + ZnSO 4 is (aq) following heat main from method is used to calculate that is enthalpy change for this reaction the this surroundings experiment to quantify. The change and in the = mc∆T ∆T systematic means that will lower be calculated Experimental method to determine ΔT calculated or The effect the from directional Using an electronic balance, a graph error. will This include loss CuSO the mass solution. of 25 Subtract from solution following to the the coffee-cup mass of the the of a 1.0 mass mol of heat the of the cylinder An transfer of the + accepted or record a temperature the errors in of curve the probe temperature of until a every 30 constant seconds for temperature the than making the actual procedure is on important in the value. the result of considering up is to 3 in experimental procedures. to data of calculating compensate is to after this look the back at for the the maximum systematic cooling reaction to the is section complete, point of and introduction and the zinc, as shown in gure 5. A more accurate the value solution lower in method temperature solution calorimeter. thermometer software, q value, recorded dm of related of actual calculations extrapolate Using the errors improvements 3 cm 4 cylinder than temperature accurately 3 measure maximum value of subsequent 2 to in equation: q 1 system error from a the of difcult temperature molar the source 4 one The of the minutes, for ΔT can then be determined. or achieved. Calculation of molar enthalpy change 3 At 3 minutes, (between and 4 1.3 g and commence Continue to introduce 5 to g, zinc previously weighed) M of opp(ii) ft oto/ 28.8 M of z/ 1.37 stirring. take minutes 1.4 powdered temperature after the readings maximum for up c tmpt T() 39.0 T(t)/°c temperature T able 1 Sample results has 5 been Produce reached. a temperature determine the change versus in time graph to temperature. Taking molar q the results enthalpy = mc∆T = 0.0288 = 4.69 in table change as 1 we can calculate the follows: Assumptions and errors 1 A number this ● of assumptions The when heat released coffee loss coffee is cup to cup difcult also to polystyrene The acts the transferred be from transferred as an reaction the has it a insulator heat from quantify cup, so it the is water. heat and It to would be reached the heat capacity assumed temperature against representation of kJ kg 1 K × 39.0 K kJ is of ∆T a zero. an 0 accurate 4.18 is However, capacity the × water. surroundings. to maximum the to kg using C°/erutarepmet The heat ● made method: completely ● are the heat 2 4 6 8 10 evolved time/min during the reaction. Figure 5 Determination of the change in temperature in calorimetry experiments 143 5 E N E R G E T I C S A N D T H E R M O C H E M I S T R Y 1 __ amount of CuSO = 1.37 g The × 4 shape of the graph and the change in 1 65.38 g mol temperature = 0.0210 from a lower to a higher value lead mol to the conclusion that the reaction is exothermic: 1 4.69 kJ __ molar enthalpy change ΔH = -223 kJ mol = 0.0210 mol 1 = 223 kJ mol TOK Tmpt In theory of knowledge there The SI unit of temperature is the kelvin (K). Note that a change in temperature ΔT are eight specic ways of calculated from experimental data in Celsius will be identical to the value of ΔT knowing. These are: language, calculated in kelvin. sense perception, emotion, Throughout the world, the majority of countries use the Celsius scale for the reasoning, imagination, everyday description of temperature. As the Celsius and kelvin scales are linked, faith, intuition, and you will often see both scales being used in an IB question. The USA uses a memory. Scientists perform mixture of metric and imperial units of measurement. For example, the Fahrenheit experiments and process scale of temperature is used in the USA. the raw data to enable us to draw conclusions. We compare experimental and °F theoretical values. What 100 criteria do we use when 80 making these comparisons? 60 Are our judgments subjective 40 or objective? When analysing 20 and appraising experimental 0 20 limitations and making theoretical assumptions, which of the ways of knowing Figure 6 We use SI units in science, but a mix ture of imperial and metric systems of are we utilizing? measurement is used in dierent countries The tdd tp of to ΔH is determined Enthalpy change of formation 298 at temperature 25 °C/298 K and The change in enthalpy during a reaction can be determined using the pressure 100 kPa with all species following equation: in their standard state. stdd ΔH reaction = ∑(ΔH odto are denoted by the symbol . products) – ∑(ΔH f ΔH is the standard reactants) f enthalpy change of formation of a substance. f This is from the its existing energy change constituent enthalpy upon elements of the in formation formation their data of standard to 1 mol state. calculate the of We a substance can enthalpy use of std tp reaction. The value and sign of the calculated enthalpy of formation Section 12 in the Data booklet informs us about the energetics of the reaction. gives the standard enthalpy of formation for a large number For example, the standard enthalpy change of formation of common compounds. In for methane is: 1 C(s) + 2H (g) → CH 2 (g) ∆H 4 = -74.9 kJ mol f examinations, questions will provide any other values not included in the Data booklet It is in to note their that the standard elements states. carbon Equations and for hydrogen ∆H must are represent f the 144 important represented formation of 1 mol of a substance. In some cases, such as the 5 . 1 M e a s u r i n g e n e r g y c h a n g e s 1 formation of phenol shown below, this results in mol of diatomic 2 Qk qto oxygen O appearing on the reactant side: 2 Write equations to describe 1 _ 6C(s) + 3H (g) + 1 O 2 (g) → C 2 H 6 OH(s) ∆H 5 = -165.0 kJ the standard enthalpy change mol f 2 of formation for the following compounds and state the Enthalpy change of combustion enthalpy value by referring to The standard enthalpy change of combustion ∆H is the heat the Data booklet c evolved The are upon the enthalpies values classied of complete combustion derived as combustion under liqueed found standard petroleum in of 1 mol section conditions. gas (LPG) is of 13 substance. of the Butane, highly Data one of ) propane b) chloromethane ) ethanol d) benzoic acid ) carbon monoxide booklet the gases ammable: 13 _ C H 4 (g) + O 10 (g) → 4CO 2 (g) + 5H 2 O(l) 2 2 1 ∆H = -2878 kJ f) mol methylamine c This thermochemical enthalpy of equation combustion value can also included in be written the with equation. the The negative 1 compod ∆H /kJ mo f enthalpy change indicates an exothermic reaction so the value would C H 6 be included on the product (l) +49.0 6 side: CO (g) 393.5 2 13 _ C H 4 (g) + 1 O 10 (g) → 4CO 2 (g) + 5H 2 O(l) + 2878 kJ mol O(l) H 2 2 285.8 2 T able 2 Standard enthalpy changes Working method of formation. Benzene, 2C C H 6 H is 6 6 (l) + highly 15O 6 ∆H (g) ammable, → 12CO 2 reaction = (g) producing + 6H 2 ∑(∆H [12 × ( = ( = -6535 393.5) 4722 + products) 6 sooty ame: O(l) 2 - ∑(∆H f = a reactants) f × 1714.8 ( 285.8) 98.0) 2 × (+49.0) 15 × 0] kJ kJ kJ Investigation to nd the enthalpy change of combustion 3 The enthalpy change of combustion of common 2 Accurately determine the mass of 30 cm of 3 alcohols From a can be determined homologous series in of the laboratory. alcohols, water enthalpy change of combustion values can and subsequently Place the beaker with the Using either temperature following spirit of in burners the procedure standard are alcohols butan-1-ol 1 a and Determine using an utilizes school required, methanol, initial electronic cm beaker. or metal spirit calorimeter burner on a beneath. each Five containing ethanol, mass of balance. temperature determine of the probe and or record a the initial water. equipment laboratory. one propan-1-ol, pentan-1-ol. the a thermometer, Experimental method available 250 analysed. 4 The a be tripod determined in patterns 3 in contained the spirit burners 5 A spirit the burner alcohol period over in of a) one allow is is lit burnt which two under to it heat burns different each alcohol temperature the calorimeter the can water. be and The monitored ways: to change burn of until 30 °C is a achieved 145 5 E N E R G E T I C S A N D T H E R M O C H E M I S T R Y b) Qk qto Write equations to describe the 6 standard enthalpy change of allow After cap, this each time re-weigh alcohol period, each to burn for extinguish one and a period each record the spirit of 2 minutes. burner change in by mass replacing of the the alcohol. combustion for the following compounds, and state the Calculation of enthalpy of combustion enthalpy value by referring to aoo ∆m of oo/ ∆T/°c 0.348 30.0 M of wt/ section 134 of the Data booklet methanol ) octane, C 31.2 H 8 18 T able 3 Sample results b) chloroethane, C H 2 Cl (hint: 5 q a corrosive strong acid is = mc∆T = 0.0312 = 3.91 one of the products) ) cyclohexanol, C H d) methanoic acid, CH 6 1 kg × 4.18 kJ kg 1 K × 12 0.348 g __ 2 amount of CH OH = = 3 6 H 12 0.0109 mol 1 32.05 glucose, C K kJ O 2 ) 30.0 O g mol O 6 3.91 kJ __ molar energy change = 0.0109 mol 1 = 359 kJ mol As in all investigations, rst 1 ΔH = –359 kJ mol determine the dependent and independent variables and the variables that will be controlled. Assumptions ● Heat loss to signicant ● All the the but environment cannot alcohols are be pure is negligible (in reality, it is quantied). and undergo complete combustion. Dt o can be used to record temperature changes accurately and the associated software to perform data Obt d t ott of food analysis and graphing. The The world increase in obesity use of data-logging equipment Obesity, eating disorders, and unhealthy diet are serious health issues facing many demonstrates the practical cultures throughout the world, as many societies become more auent and food is application of technology in the readily available. Obesity is generally dened as an excessive accumulation of fat laboratory. that can lead to health problems. The bod m dx (BMi) is found by taking a person’s mass (in kilograms) and dividing it by the square of their height (in metres). An adult with a BMI above 25 is considered overweight while one with a BMI greater than 30 is obese. The World Health Organization (WHO) has been Energetics experiments monitoring the eect of changes in diet on dierent nations for decades. Their provide a useful set of research has found the following: raw data and involve ● In 2013 the occurrence of obesity worldwide was more than double the level experimental procedures in 1980. that can be evaluated for random and systematic errors ● (topic 11). The identication In 2008 over 1.4 billion adults worldwide were overweight, with approximately 200 000 000 men and 300 000 000 women being classied as obese. of the systematic errors ● 65% of the world’s population reside in countries where more people die from and examination of their obesity-related causes than from being underweight. directionality is an essential ● More than 40 000 000 children under the age of 5 years were overweight in 2010. ● Obesity is a preventable disease. aspect of the analysis of experimental results. 146 5 . 1 M e a s u r i n g e n e r g y c h a n g e s In China, the rapid increase in auence and the globalization of the economy has seen an unprecedented expansion in the fast-food industry and of nutritional choices. With these has come a signicant increase in the number of children who are overweight and obese. Type 2 diabetes is normally associated with adults, but the rise in the prevalence of the disease amongst children in China and in other countries is seen as a signicant threat to the wellbeing of future generations. Food labelling and determination of energy content Governments throughout the world have a responsibility to their citizens to provide leadership, education, and guidance in health and nutrition. Linked to the globalization of economies and free-trade agreements has been the standardization of labelling of food products to include an analysis of the contents including energy content. To determine their energy content, foods were traditionally placed in a calorimeter surrounded with water and completely burnt, causing the water to rise in temperature. This temperature change was then used to calculate the energy content (sometimes referred to as “caloric value”) of the food. Today the preferred method of calculation is using the Atwater system. This system relies on average energy values for proteins, carbohydrates, fats, and alcohol being applied to foods of a known composition. The National Data Laboratory (NDL) in the USA holds information on the energy content of over 6000 foods. Figure 7 Nutritional information displayed on food packaging 147 5 e n e r g e T i c s a n D T h e r M O c h e M i s T r y 5.2 h’ w Understandings Applications and skills ➔ The enthalpy change for a reaction that is Application of Hess’s law to calculate enthalpy ➔ carried out in a series of steps is equal to changes. the sum of the enthalpy changes for the Calculation of ∆H reactions using ∆H ➔ data. f individual steps. Determination of the enthalpy change of a ➔ reaction that is the sum of multiple reactions with known enthalpy changes. Nature of science ➔ Hypotheses – based on the conser vation of energy and atomic theory, scientists can test the hypothesis that if the same products are formed from the same initial reactants then the energy change should be the same regardless of the number of steps. TOK Testing hypotheses In TOK , a primary focus is on Experimental questions about knowledge hypothesis. which are open ended atomic with multiple perspectives that and expressed without change using subject-specic and language. Hess’s law can be can evidence Based theory, when in the be on enables the scientists products enthalpy number analysed of are are able formed should be chemical and scientists principles used as to of test from to or identical reactions disprove of same set regardless involved. for of of a energy experimentally the evidence prove conservation the and hypothesis reactants, the route Quantitative the taken data thishypothesis. considered an application of the law of conservation of energy. What are the Overall and net reactions challenges in applying If you have travelled to New York, Tokyo, London, Hong Kong, Paris, general principles of a law Beijing, Berlin, or Seoul, you will have experienced the subways that to something as specic as criss-cross these enormous cities and transport millions of people every Hess’s law? day. In any between often The point same idea shows reactions, Hess’s law Regardless always Figure be 2 can is of can the an the then B. For which be net is true by that in a the applied of to – eld it is than the all each, chemistry. summary result reaction and the nal in of A a to half travel the of an enthalpy chemical number overall energy proceeds, states in of an equations, results the way traveller, conservation chemical one fun is route. of a together nd more adventurous initial summing is fastest the of which magnitude be the added providing there the reaction when route shows network application same and and out which the direction 148 A working usually law transport the the taking change enthalpy are into change the the will same. account equation. for different reaction. law: system overall equation of the Hess’s reaction. 5 . 2 h e s s ’ s l a W A H H y x C H z B Figure 2 Hess’s law Figure 1 You can take many alternative routes on the Paris Metro Worked example : calculating enthalphy change Using Hess’s calculate law, the and the enthalpy following change ΔH reversing information, for the change reaction: the the chemical sign of the equation enthalpy we must value: 4 1 _ 1 _ C + 2H + CO O 2 → CH 2 OH ∆H 3 (4) + 2H 2 O → CH 2 OH + 1 O 3 2 2 4 2 ∆H 1 _ CH OH + 1 = +676 kJ 1 O 3 → CO 2 + 2H 2 O 2 2 ● ∆H = -676 kJ Because oxygen is found in all three (1) 1 equations, should C + O → CO 2 ∆H 2 = -394 kJ be O → H 2 point hydrogen. We of focus require 2 mol of (2) 1 _ + 2 next 2 hydrogen H the O ∆H 2 = -242 kJ 3 (3) can be as a used reactant. in the Therefore, direction as equation written but 3 2 with IB, May, that 2006 double the ● of at the overall formation of equation 4 for the enthalpy From Now equations 1–3, the reactant carbon product focus ● For of methanol your carbon, should be the require a 2H can and O add ∆H three species summing be = This means doubled: -484 kJ 3 the those moles. must 2 the in table equations common enthalpy to together, both values as 1. main rtt reaction of and methodology. we → value 2 we shown the O eliminating methanol. sides ● + 2 ● Look number enthalpy 2H Solution the that Podt C + uses → etp ΔH = –394 kJ 2 1 mol of carbon as a reactant. Carbon is a _ + O → CH OH + 1 ΔH 3 reactant used as in equation 2, so this equation can = +676 kJ 1 be 2 1 _ written: 2H + O 2 → ΔH 2 2 = –484 kJ 3 2 C + O → CO 2 ∆H 2 = -394 kJ 2 C + 2H 2 ● For methanol we need to use equation → CH 3 OH ΔH = –202 kJ 4 1 _ 1, + O 2 but we need methanol is to a reverse product the not equation a so reactant. that 2 When T able 1 149 5 E N E R G E T I C S The combination A N D of T H E R M O C H E M I S T R Y the se re a ctio ns can a l so 1 C + 2H + O 2 be represented gure You diag r a mma tical l y as sh o wn CH 2 OH 3 2 in 3. will notice that the enthalpy cycle diagram 1 O O 2 1 2 O 2 2 shows the reversed, combustion in summation the of same of methanol way equations it was equation during the method. CO 2 + 2H O 2 Figure 3 Enthalpy cycle for the formation of methanol T t of With the world population exceeding 7 billion people, phosphor dust, and other hazardous substances. Guiyu in and increasing international concern over the world’s China has become a centre for e-waste disposal and the resources, recycling of materials has become mainstream rapid expansion of recycling processes in villages around in many countries. Recycling developed from a desire to Guiyu has resulted in heavy metal contamination of the use raw materials more eciently, to reduce energy use groundwater and soil, and air pollution from the burning of in the production of goods, to protect the environment plastics within the waste. from excessive pollution, and to utilize waste materials Despite good intentions, the end result of recycling can and thereby reduce landll waste disposal. However, be extremely negative. What is more, the eciency economic and other pressures to recycle can lead of recycling processes in energy terms varies widely. to potentially harmful impacts on the environment. Many countries and environmental organizations are For example, electronic or e-waste is an escalating investigating how we can address the long-term problem throughout the world. This waste contains eects of recycling programmes on the environment heavy metals such as lead and cadmium, highly toxic and communities. Figure 4 Technology waste recycling in Guiyu, China 150 5 . 2 h e s s ’ s l a W Worked example Determine the enthalpy of formation C the enthalpy of combustion H 2 using of ethane, data In summary: in 6 ■ section 13 of the Data The equation is ■ equation hydrogen Write a balanced chemical equation for of 1 combustion of for the combustion of combustion of carbon mol of is tripled. the ■ formation the doubled. The Solution ● for booklet The ethane: is ∆H equation for the ethane reversed. f 2C(s) + 3H (g) C 2 H 2 (g) 6 2C(s) + 3H (g) → C 2 ● Write equations hydrogen, and for the ethane combustion and of determine H 2 (g) 6 carbon, ∆H their = –84 kJ c enthalpy 1....C(s) values + O from (g) → the CO 2 ∆H = Data booklet: (g) Figure 5 can found be shows how using the an enthalpy enthalpy of formation cycle diagram. 2 -393.5 kJ c ∆H 1 _ 2....H (g) + O 2 (g) → H 2 2C(s) O(l) + 3H 2 f 2 2 H 6 (g) 2 ∆H = -286 kJ c 7 _ 3....C H 2 (g) + O 6 (g)→ 2CO 2 (g) + 3H 2 O(l) 2 2 3 7 O ∆H = -1561 kJ c ● Multiply 2O or ● Now for reverse each the sign combine enthalpy of 1....(×2): equation of the the 2C(s) to for = -787 → the 156 1 net (g) 2CO 2 (g) + 3H 2 O(l) kJ Figure 5 Alternative method: enthalpy cycle to nd the c __ 2....(×3): 3H (g) = -858 enthalpy of formation of ethane + 2 ∆H + ethane: 2 ∆H 3(-286) 2 enthalpy form equation + 2 accordingly. equations formation 2 2 2(-394) change O 2 → 2 2 2 kJ c 3....(reversed): 2CO (g) 2 + → 2 2 _ (g) + 2 ∆H = +1561 kJ c std tp You may nd the summation of equations method easier when working out the direction of the equations and the mole coecients. During examinations you may be asked to use the summation of equations method and/or construct an enthalpy cycle. Often candidates make simple arithmetical errors when calculating the enthalpy of reaction. It is advisable to clearly show your full working rather than simply recording the nal answer. This gives the examiner the oppor tunity to assign par t marks where applicable. 151 5 e n e r g e T i c s a n D T h e r M O c h e M i s T r y 5.3 Bod tp Understandings Applications and skills ➔ Bond forming releases energy and bond ➔ Calculation of the enthalpy changes from breaking requires energy. known bond enthalpy values and comparison ➔ Average bond enthalpy is the energy needed of these with experimentally measured values. to break 1 mol of a bond in a gaseous molecule ➔ Sketching and evaluation of potential energy averaged over similar compounds. proles in determining whether reactants or products are more stable and if the reaction is exothermic or endothermic. ➔ Discussion of the bond strength in ozone relative to oxygen in its impor tance to the atmosphere. Nature of science ➔ Models and theories – measured energy changes can be explained based on the model of bonds broken and bonds formed. Since these explanations are based on a model, agreement with empirical data depends on the sophistication of the model and data obtained can be used to modify theories where appropriate. Modelling energy changes Scienti c certain directly. of the models processes Based on processes, evidence inform in are d e v el ope d tha t a ca nnot the or etica l s uch suppo r t of modi cati o ns mod el s the to to be e x pla i n models und e r st a n di n g can degree p r odu c e the or i es, the changes ob se r ve d or the or i es . the can in of of empirical accuracy ca n br e a k i ng agreeme nt dependent E n er g y reactio ns bond data on of the the be unde rs too d a nd b ond b e twe e n o b ta i ne d v a l i di ty the s e in of the the u s in g m a ki ng. m ode l s The an d l a bor a t or y mode l and is t he da ta . Bond enthalpy The breaking requires example) gaseous Bond of energy. is covalent for hydrogen The dened breaking value, the is bond as the energy molecules an molecule enthalpy into a to standard process individual bond required under endothermic of (the break hydrogen H 1 H mol bond of atoms in this bonds and has a positive enthalpy example: 1 H (g) → 2H(g) ∆H = +436 kJ mol 2 Bond 152 enthalpy and selected and in is also values table1. referred are These to provided are as in average bond dissociation section values in conditions. 11 and of the are enthalpy , Data booklet therefore only an 5 . 3 approximation. They are derived from experimental data B O n D e n T h a l P y involving Bod the breaking of For example, the same bond found in a wide variety of av bod compounds. 1 tp/kJ mo series If a as the the would upon be of signicant for each of enthalpy of time, at as successive the in the that the vary the a the of These bond in is take are an the account not H O O 436 one 144 bond of O H C H C C 463 the considered value 498 O=O changes accepted H enthalpy Additionally, not limitations enthalpy changes. which environment does alkane dissociation atoms. state the bonds steps bond chemical hydrogen exist. through individual series time, gaseous average reaction a will of underwent removed forces and enthalpy environment molecules intermolecular be was removal for bond methane atom different the enthalpy H chemical molecule hydrogen C to to 414 use 346 calculations. 614 C=C Bond length Consider the molecules hydrogen uoride H F , hydrogen chloride C H Cl, hydrogen discussed with how 839 C≡C bromide as increasing you H Br, move atomic and down number, hydrogen group Z. The 17 iodide the H I. atomic consequence of In topic radius this is 3 increases that 358 O we 804 C=O bond 615 C=N length you increases, move and down bond group 17 strength (table decreases, in the hydrogen halides as N Bod H F H 890 C≡N 2). Cl H Br H 158 N I 470 N=N Bod t/pm 92 128 141 160 945 N≡N T able 2 Bond lengths of the hydrogen halides Bond strength The bond enthalpy reects the strength of the covalent bond. As Cl Cl Br Br the from bond to increases shortening carbon single of bond the of double resulting bond the to triple in length. an increase This homologous bonds trend series of the in can number of electrostatic be seen alkanes, in electrons forces the alkenes, 151 I in and a carbon– and 193 we I move 242 T able 1 Average bond enthalpies at 298 K alkynes. Bond polarity The polarity of a electronegativity bond of can the be described bonded atom etotvt H 2.2 F 4.0 Cl 3.2 Br 3.0 atoms Bod by the (tables difference 3 and in 4). ∆(etotvt) Bod tp/ 1 kJ mo H F H Cl H Br 1.8 567 1.0 431 0.8 366 δ δ+ T able 3 Electronegativity values T able 4 Bond polarity and bond enthalpy at 298 K H Fluorine The has polarity atom. attract The one the of highest the bond is H F said another, electronegativity bond to results have increasing in ionic the a value partial of charge character . strength of any The the F element. on each partial bond charges (gure 1). Figure 1 The polar H F bond 153 5 E N E R G E T I C S A N D T H E R M O C H E M I S T R Y Worked example: using bond enthalpies to nd the enthalpy change of reaction Using the change form data for from the section 11 electrophilic of the addition Data of booklet hydrogen nd the enthalpy bromide to ethene to bromoethane. C H 2 (g) + HBr (g) → C 4 H 2 ∆H = ∑(BE ∆H = [4BE bonds + broken) BE C–H ∆H = [(4 × ∆H = 2636 + + - (g) ∑(BE BE C=C 414) Br 5 ] bonds [5BE H–Br 614 + formed) + BE C–H 366] [(5 × + BE C–C 414) + 346 ] C–Br + 285] 1 The bond using ∆H 2701 enthalpy = -65 kJ calculated reaction = ∑∆H mol will vary signicantly (products) - ∑∆H f ( 105.7kJ) when do because liquids not take are into bond account in the the calculation f enthalpies involved from (reactants) the are average reaction, bond intermolecular values. Additionally, enthalpy forces within calculations the liquids. std tp A frequent error made by candidates is to confuse the dierent equations for the calculation of a change in enthalpy. For bond enthalpy, think in terms of bond breaking and formation: ∆H = ∑(BE bonds broken) - ∑(BE bonds formed) For enthalpy of formation, think in terms of products and reactants: ∆H reaction = ∑(∆H products) - ∑(∆H f reactants) f Bond enthalpy values and enthalpies of combustion Gasoline used to or witnessed developed The petrol power is produced various signicant of the fractional of transport. changes in its economies enthalpy by modes is being combustion markets overtaken of The as by enthalpy values from octane, section 11 the that C H 8 bond distillation of of automotive demand of the can petroleum industry for automobiles developing Data be calculated using booklet: 25 _ C H 8 (g) + O 18 (g) → 8CO 2 (g) + 9H 2 O(g) 2 2 ∆H = ∑(BE bonds broken) - ∑(BE bonds formed) c 25 _ = (18BE + 7BE C–H + BE C–C ) (16BE O=O + C=O 18BE ) O–H 2 25 _ = (18(414) + 7(346) + (498)) (16(804) 2 1 = 154 16 099 21 198 = -5099 kJ mol + in economies. 18 the and has 18(463)) 5 . 3 In comparison, change of the experimentally combustion ∆H for determined octane (section value 13 of for the the B O n D e n T h a l P y enthalpy Data booklet) is Qk qto c 1 5470 kJ mol 1 The reason change takes for using place However, octane difference bond in the the and the enthalpy gaseous is in their when values, state, experimentally water that it with assumed no derived standard is calculating the that intermolecular enthalpy states, of forces liquid. with hydrogen gas reaction combustion namely chlorine gas is combined enthalpy the In the chlor-alkali industry to produce hydrogen involved. chloride: involves Additionally, Cl 2 as mentioned above, all bond dissociation enthalpy values are (g) + H (g) → 2HCl(g) 2 averaged If the enthalpy change for across a wide range of related compounds so they represent only an his reaction is approximation of the true 185 kJ, value. calculate the bond enthalpy for the H–Cl bond. Ozone 2 Ozone, O is both created and d e s tr oy ed in the st r a t os ph e ri c l a ye r Earth’s Methane and chlorine react of to produce chloromethane 3 a tmosphere . Ul tr av iol e t (UV ) r ays fro m the s un ar e a bso rbe d and hydrogen chloride. by oxygen, O , splitting the mol e cul e into s ingl e ox yge n a t om s . 2 ) These oxygen ato ms ca n the n comb in e w it h ox yg e n m ol ec u l es Write the balanced to chemical equation for form ozone: this reaction. UV b) O (g) → O (g) + O Using bond enthalpy (g) 2 values from the Data O (g) + O (g) → O 2 Ozone UV (g) booklet, determine 3 is very effective radiation. molecular This oxygen at absorbing absorption and a harmful breaks single down oxygen long- the atom. and ozone molecule Without the enthalpy change short-wavelength the to for this reaction. reform presence ) of ozone in the stratosphere, life on Earth would change forever, Deduce whether this as reaction is exothermic harmful UV radiation would damage cells in both plants and animals. or endothermic. 1 The bond dissociation enthalpy of an oxygen molecule is 498 kJ mol . In d) comparison, the energy required to break an oxygen–oxygen bond Which are more within energetically stable, 1 an ozone molecule is 364 kJ mol . The consequence of this is that an the reactants or the ozone molecule is decomposed by UV rays more readily than an oxygen products? molecule. reaction The with potential atom molecule. exist at a the energy Examination oxygen ozone of required prole this have The photolysis a for energy this energy of is from reaction that energy are above the shown reveals combined this described coming reaction prole greater products higher reaction UV the to an the The 2. oxygen be endothermic radiation. ingure than said is molecule reactant less stable, and ozone as they energy. ygrene laitnetop O + O 2 products a ∆H O 3 reactants reaction pathway Figure 2 Endothermic energy prole for ozone photolysis 155 5 E N E R G E T I C S A N D T H E R M O C H E M I S T R Y Dpto of t ozo Since the early 1980s scientists have been monitoring ozone depletion globally, par ticularly the giant holes in the ozone layer which have appeared above the Arctic and Antarctic polar icecaps (gure 3). Figure 3 ERS-2 satellite map of Antarctic ozone hole in 2010 Chlorouorocarbons (CFCs) are a type of hydrocarbon containing carbon, hydrogen, and halogen atoms. The compounds themselves are considered to be non-toxic and have a low level of both ammability and reactivity. First used in industry in the late 1920s, a CFC called Freon gas became an industry standard refrigerant in domestic refrigerators throughout the world. Over the following decades the use of CFCs in commercial air conditioning and the automobile industry, as an aerosol propellant, and as a solvent saw the demand for this class of compounds increase rapidly. Scientific close these of to research the Ear th’s compounds chlorine. The destruction directly of the ClO in in w hen this tha t exposed chemical releas ed the then lay er of reaction, de s tru ction (g) → ClO 3 discovered un d er w ent ozone for Cl(g) + O 1973 su rf a ce, chlorine consumed responsible in of (g) + O a CFCs rema i ned ra d iation in resulting ca talytic atm osp her e. s mall large UV reaction s had the while to in effect As the the am ou nts of CFCs qu an tities of ozon e. the on h armless s tratosp her e rel ea se the chlor ine wer e is fou nd not to be (g) 2 (g) + O 3 (g) → Cl(g) + 2O (g) 2 Science has made many advances that have improved people’s daily lives and extended life expectancy and quality. As a consequence the world’s population is increasing. The use of CFCs has had a massive economic (positive) and environmental (negative) impact on the world. Whether the blame for the environmental consequences lies with multinational companies who utilize technology or the scientists who invent it is the focus of many discussions. 156 Q u e s T i O n s Questions 1 Consider the following specic metals heat (table capacity of the 5 4). Use data each 1 385 Ag 234 Au 130 Pt 134 these section 12 enthalpy of the change chemical Data of booklet reaction to for reactions: 1 a) Cu of the K sp t pt/J k Mt from calculate H C 2 b) (g) + H 4 CH (g) + H 4 6 a) → C H 2 O(g) → (g) 6 CO(g) + 3H 2 Dene T able 4 (g) 2 the formation, (g) 2 term ΔH standard enthalpy change of . [2] f Which metal will temperature to a 0.001 initial A. show increase kg sample the if 50 of greatest J of each b) heat metal is at (i) Use supplied the same information the complete temperature? C C Au H 4 Ag D. Pt (g) to + May the 6O 8 table of → to for the but-1-ene following (g) 5 change 4CO 2 equation. (g) + 4H 2 O(g) 2 [1] compod IB, in enthalpy combustion according Cu B. the calculate C H 4 2007 (g) CO 8 (g) H 2 O(g) 2 1 ∆H +1 /kJ mo 394 42 f [3] 2 When solid 40 H O joules at of 16.0 heat °C are the added to a temperature sample of increases T able 5 to 2 (ii) 8.0 °C. What is the mass of the solid H O Deduce, giving a reason, whether sample? 2 1 Specic heat capacity of H O(s) = 2.0 J g thereactants 1 or the products are K 2 morestable. A. 2.5 g C. 10 B. 5.0 g D. 160 (iii) IB, Nov g Predict, [1] giving enthalpy based The temperature of a 2.0 g sample of of heat from 25 energy °C to were 30 °C. added? How many (Specic 1 capacity of aluminium = 0.90 J g 0.36 C. 9.0 B. 2.3 D. 11 of with on for the how but-2-ene that average of the complete would but-1-ene bond enthalpies. [1] May 2007 joules heat 1 K .) 7 The of A. reason, aluminium IB, increases a change combustion 2007 compare 3 [2] g ∆H values nitrogen are for the given formation of two oxides below. 1 _ 1 N [1] (g) + O 2 (g) → NO 2 (g) ∆H = –57 ∆H = +9 kJ mol 2 2 IB, May 1 2003 N (g) + 2O 2 Use these following 4 What is the energy temperature of 20 change g of (in water kJ) when increases (g) → B. 20 × 10 × 20 × 283 values to reaction (g) calculate (in kJ mol 4 ∆H for the kJ): the by 10 °C? 4.18 × O 2 2NO (g) → N 2 A. N 2 a) 4.18 b) O 2 (g) 4 105 –48 c) +66 d) +123 20 × 10 × 4.18 __ IB, C. November 2007 1000 20 × 283 × 4.18 __ D. [1] 1000 IB, November 2003 157 5 E N E R G E T I C S 8 The standard combustion A N D T H E R M O C H E M I S T R Y enthalpy reactions change is given of Reaction three below in II kJ. 1 _ SO (g) + O 2 (g) ⇋ SO 2 (g) ∆H = –92 kJ 3 2 2C H 2 (g) + 7O 6 (g) → 4CO 2 (g) + 6H 2 O(l) 2 ∆H = a) –3120 State with 2H (g) + O 2 C (g) → 2H 2 H 2 (g) + O(l) ∆H = 4 (g) → name reason, accompanied 2CO 2 (g) + 2H 2 of the term whether ∆H . reaction State, I would be –572 2 3O the a by a decrease or increase in temperature. O(l) [3] 2 ∆H = –1411 b) At room temperature sulfur trioxide, SO , is 3 Based the on the standard following above information, change in a calculate enthalpy, ∆H , for solid. ∆H the Deduce, value negative reaction. if with a reason, would be SO instead (s) more of SO 3 formed C H 2 (g) → C 6 H 2 (g) + H 4 (g) November (g) or the less were 3 reaction II. [2] [4] 2 c) IB, in whether negative Deduce the ∆H value of this reaction: [1] 2009 1 _ S(s) + 1 O (g) → SO 2 (g) 3 2 IB, 9 Approximate values of the average November 2005 bond 1 enthalpies, shown in in kJ table mol , of three substances are 6. 12 H carbon 430 H But-1-ene the F following and in oxygen water to vapour produce according to equation. C H 4 + 6O 8 → 4CO 2 + 4H 2 O 2 565 F a) T able 6 What burns 155 F H gas dioxide Use of is the enthalpy change, in kJ, for this the ∆H data for in the Bod C table 7 to calculate combustion C C=C C of the value but-1-ene. H [3] O=O C=O 496 743 O H reaction? av 2HF → H + 2 348 612 412 463 bod F 2 tp/ A. +545 IB, May B. +20 C. 20 D. 545 1 kJ mo 2006 T able 7 b) 10 The is reaction between ethene and hydrogen gas State and above is explain whether endothermic or the reaction exothermic. [1] exothermic. IB, a) Write b) Deduce an equation the for relative this reaction. stabilities of Explain, by the reactants and 2006 [1] and 13 energies May products. Given the following data: [2] 1 c) the referring molecules, why to the the bonds reaction C(s) in + 2F (g) → CF 2 (g); ∆H 4 = –680 kJ mol = +158 kJ mol = +715 kJ mol 1 is 1 F (g) → 2F(g); ∆H 2 exothermic. 2 [2] 1 C(s) → C(g); ∆H 3 IB, November 2007 1 calculate for 11 Two reactions occurring in the sulfuric Reaction I S(s) (g) + O 2 158 acid ⇋ are SO shown (g) 2 below: ∆H C F average bond bond. manufacture IB, of the the = –297 kJ November 2003 enthalpy (in kJ mol ) Q u e s T i O n s 14 Methanol is made in large quantities as c) it The Data booklet value for the enthalpy of 1 is used in in the production of polymers combustion and methanol is 726 kJ mol fuels. Suggest values The be of enthalpy of determined combustion theoretically of or methanol why this value calculated in differs parts a) from and the b) can experimentally. (i) Part a) [1] (ii) Part b) [1] 1 _ CH OH(l) + 1 O 3 (g) → CO 2 (g) + 2H 2 O(g) 2 2 a) Using of the the IB, information Data booklet, from section determine enthalpy of 11 One important property of a rocket fuel methanol. the large volume of gaseous products provide thrust. Hydrazine, N H 2 The e nthalpy of co mb us ti on of also be determi ne d e xp er ime nta ll y used school laborator y. A b ur ne r water in gure 4. a was we i g he d test tub e as a nd as a rocket fuel. The 4 combustion is represented by the equation of below. containi n g N methanol is in hydrazine a , m e t h a no l often can formed [3] which b) mixture combustion is of 2011 the 15 theoretical May us e d to i ll us tr a te d in he a t H 2 (g) + O 4 (g) → N 2 (g) + 2H 2 O(g) 2 1 ∆H = –585 kJ mol c a) Hydrazine nitrogen reacts and with uorine hydrogen to uoride, produce all in the thermometer gaseous state. State an equation for the reaction. [2] b) Draw the Lewis and nitrogen. c) Use the structures for hydrazine [2] test tube with water stand section the shield 11 of a) Based bond the enthalpy part d) average Data change enthalpies booklet for the to given in determine reaction in above. on [3] your answers to parts a) and c), burner suggest and whether uorine mixture of is a a mixture better hydrazine of hydrazine rocket and fuel than a oxygen. [2] Figure 4 IB, The data shown in table 8 were May 2010 collected. Initial mass of burner and methanol/g 80.557 16 Two students from Final mass of burner and methanol/g the were Data asked booklet to to use information calculate a value for the 80.034 enthalpy Mass of water in test tube/g 20.000 C H 2 26.4 Final temperature of water/°C hydrogenation of ethene to form ethane. 21.5 Initial temperature of water/°C of John (g) H (g) → C 2 used section + 4 11. the average Marit H 2 used (g) 6 bond the enthalpies values of from enthalpies T able 8 of i) Calculate the methanol ii) Calculate by iii) the amount, in mol, of burned. the heat absorbed, in kJ of 1 mol the mole , for enthalpy of the section b) change, Marit section value of bond for 12. the ethene enthalpy obtained enthalpies given of using in 11. [2] arranged section combustion methanol. the theaverage [3] 1 in Calculate from hydrogenation kJ, water. Determine a) [2] combustion 12 into the an values energy she found cycle in (gure 5). [2] 159 5 E N E R G E T I C S A N D T H E R M O C H E M I S T R Y ∆H C H 2 + (g) H 4 (hydrogenation) (g) C 2 1 1 lom Jk 682 O 1 2 O3 2 lom Jk 1141 2 - l o m J k 0 6 5 1 - 2CO + (g) 3H 2 O(l) 2 Figure 5 Calculate of the value hydrogenation for of the enthalpy ethene from the energycycle. c) Suggest one isslightly [1] reason less why accurate John’s than answer Marit’s answer. d) John [1] then enthalpy to decided of produce C H 6 (i) + H (g) the average a value of of John’s was hypothesis ofaverage less accurate bond for forethene. is moreaccurate 160 May 2009 was [1] between the for for that Determine use to of enthalpies greater it information enthalpy (average and is equationshown IB, enthalpies the difference enthalpies same. (l) 12 cyclohexenethan was the cyclohexene cyclohexene. methods combustion) bethe of H bond for percentage thesetwo bond C 6 hydrogenation The → 2 deduce (ii) determine cyclohexane. (l) Use to hydrogenation 10 H 2 ethene. it why would the enthalpies is cyclohexene above, Deduce needed answer. than what to it extra provide a [2] (g) 6 6 C H E M I CA L K I N ET I C S Introduction In this topic kinetics. the of will proper razor also in learn discuss the form a the key science the topic part a the the of as a exploring gases and guide the the of to aspect the of reaction. of this a In and namely reaction. outline molecules addition of collide we numerical branch reactions, chemical the We chemical will then importance of describe the principle theories. that and chemical of theory developing graphical treatment of rate collision probability rate analysis our important by greater higher of very chapter theory that orientation Throughout will will begin kinetic-molecular Occam’s We we We data physical with consider sufcient the obtained effect from of energy a and catalyst rate on rate. experiments chemistry. 6.1 C oo o o co Understandings Applications and skills ➔ Species react as a result of collisions of ➔ Description of the kin etic theory in terms of sufficient energy and proper orientation. the ➔ The rate of reaction is expressed as the change movement kinetic in concentration of a par ticular reactant/ in energy of is par ticles w hose prop or tional to average tempera ture kelvin. product per unit time. ➔ ➔ Concentration changes in a reaction can be Analysis rate of graphical a nd nu merical d ata from experiments. followed indirectly by monitoring changes in ➔ Explanation of the eff ects of temperatu re, mass, volume, and colour. pressure/concentration, ➔ Activation energy (E ) is the minimum energy rate a of a nd par ticle s ize on reaction. that colliding molecules need in order to have ➔ Construction of Maxw ell -Boltzmann successful collisions leading to a reaction. energy ➔ By decreasing E , a catalyst increases the rate the a of a chemical reaction, without itself being probability factors permanently chemically changed. a ➔ distribution of affecting to succes s f u l thes e, accou nt f or collisions inclu ding and the effec t of the resu lts. energy profiles catalyst . Investigation of experimentally ➔ cu r ves Sketching with and and rates and of reaction eva lu a tion explan a tion without of of cata ly s ts . Nature of science ➔ The principle of Occam’s razor is used as a guide to developing a theory - although we cannot directly see reactions taking place at the molecular level, we can theorize based on the current atomic models. Collision theory is a good example of this principle. 161 6 C H E M I C A L K I N E T I C S Studying reaction rates When to a chemical reaction takes 1 What occurs 2 What is 3 Does 4 What the the This the of which is given four information involved sufcient can be questions are of interest discussed in terms 3 This is the of the study This is the energy or nothing One of the whether much be eld heat about main it point in a period of of the in how a take from out a airbags in or a reaction chemical (and (topic is 5), reaction. their states), a given if reactions while for it it to takes occur others sub-topic a the branch focus which is of 7.1). of this the reaction a useful. 150 very years as of tells us place. reaction There for quickly, such takes chemical be topic. study Thermodynamics examining enough reaction reactions, which slowly when cars, balanced products equilibrium chemical chemical chemical could 1.1). of fast and following: the reactants, kinetics, quickly place the chemical rate reaction time? thermodynamics ow carrying of of chemical Many ination the considerations will formed! rapid over called in slowly? involves (sub-topic is reaction? or obtained identies This 4 rapidly questions stoichiometry chemistry chemical happen 2 to four reaction? the transfer occur these equation and extent energy Answering 1 during reaction potentially is place, chemists: the such rusting is not product as take the place years. Rate of reaction The rate of reactants reaction or is products 3 Units: mol dm dened per unit as 1 s , the change 3 mol in concentration of time. dm 1 min , etc. Experimental measurements of reaction rates Δc, the change property that Examples concentration, change when 1 change in pH 2 change in conductivity 3 change in mass 4 change in colour In order is to of (for or acid base (for volume (for determine (or plotted. gradient be measured reactants are by monitoring converted into a products. reactions) reactions (for reactions involving reactions electrolytes) involving involving solids transition or metals gases) or other compounds). concentration time can the include: coloured 162 in will the the The the rate of property rate of tangential reaction, associated reaction line at is time then t. at time with t, a graph of concentration) determined from versus the slope or 6 . 1 Let us consider (limestone), the following CaCO and reaction hydrochloric C O l l i s i O n between acid, t h e O r y calcium a n d r a t e s O f r e a C t i O n carbonate HCl: 3 CaCO (s) + 2HCl(aq) → CaCl 3 In this reaction, volume, rate of set-up V,of gaseous carbon reaction for (aq) + CO 2 this can carbon dioxide be dioxide gas (g) + H 2 is one produced determined. Figure is 1 O(l) 2 of the products. recorded shows over the If time the t, the experimental reaction: clamp delivery tube measuring cylinder carbon dioxide ask trough hydrochloric acid water calcium carbonate Figure 1 Experimental set-up for measuring the rate at which carbon dioxide is produced in the reaction between calcium carbonate and hydrochloric acid Table 1 shows the t/ data recorded during this experiment. 0.0 10.0 20.0 30.0 40.0 50.0 60.0 70.0 80.0 90.0 100.0 0.0 19.0 30.0 37.5 45.0 50.0 52.0 53.0 53.0 53.0 53.0 3 V(CO )/cm 2 T able 1 Volume, V, of carbon dioxide evolved at time t in the reaction between calcium carbonate and hydrochloric acid 60 Figure 2 shows a plot of V against t 50 The rate of reaction can be expressed in three ways: average mc/) 3 ● rate instantaneous ● initial OC(V 2 ● rate rate. 40 30 20 10 0 0 Average rate 20 40 60 80 100 t/s The average reactant or rate is product a measure in a given of the time change interval, in concentration of Figure 2 Plot of volume of carbon dioxide gas t evolved V(CO ) against time t in the reaction 2 between calcium carbonate and hydrochloric Mathematically, average rate can be expressed as: acid Δc _ average rate = Δt where: Δc = change Δt = time For gases, in concentration interval the over average which reaction of reactant the change rate can be or in product concentration also expressed as was the measured change in volume: ΔV _ average rate = Δt where during ΔV the is the change in volume of the gas produced or consumed reaction. 163 6 C H E M I C A L K I N E T I C S Hence in this example, as the reaction 53.0 _ average rate is complete 1 = = 7.57× 10 3 cm at 70.0 s, then: 1 s 70.0 Instantaneous rate The instantaneous Δc _ lim rate is given by: dc _ = Δt → 0 dt Δt Initial rate When t = t , the instantaneous rate = the initial rate. 0 In order drawn gives to to the deduce the the curve initial at rate initial 0 s, rate and (gure of the reaction slope or at t = gradient 0 s, of a tangent the is tangential line 3). 60 (x , y 2 ) = (20, 50) 2 50 (x , y 3 ) = (50, 50) 3 40 , y 4 2 mc/) 3 (x ) = (20, 39) 4 30 OC(V 20 10 (x , y 1 ) = (0, 0) 1 0 0 20 40 60 100 80 120 t/s Figure 3 Plot of volume of carbon dioxide evolved versus time showing tangents used to measure both initial rate at t = 0 s and instantaneous rate at t = 50 s (x , y 1 (x ) = (0.0, 0.0) = (20.0, 1 , y 2 ) 50.0) 2 Δy 50.0 0.0 _ _ Initial rate = = So at the start of the 3 = 20.0 Δx 2.50 cm 1 s 0.0 reaction, carbon dioxide gas is being generated at a 3 rate In of 2.50 order tangent to is tangential (x , y 3 (x , 4 ) cm per deduce drawn line second. the to gives instantaneous the the curve at instantaneous = (50.0,50.0) = (20.0,39.0) t, of and rate reaction, the at slope this at t or time = 50.0 s, gradient (gure y ) 3). 4 Δy 50.0 39.0 __ _ rate = = Δx 1 = 50.0 20.0 3.67 × 10 3 cm 1 s a of 3 instantaneous 164 rate point the 6 . 1 C O l l i s i O n t h e O r y a n d r a t e s O f r e a C t i O n Rate equation A rate equation rate expressed A B + → C + in D, is a mathematical terms a rate d[A] = equation = In the general = xA where + yB x, y, → q, qC and = + as which the shows reaction follows: _ = + dt dt p pD are the stoichiometric d[B] 1 _ _ - = x written in where: d[A] 1 _ rate expression, example, d[D] + dt case be _ - dt can For d[C] _ - differential concentration. d[B] _ rate of - d[C] 1 _ _ · = y dt coefcients: · = q d[D] 1 _ _ + dt + _ · p dt dt Kinetic–molecular theory of gases The the ideal gas equation (sub-topic 1.3) is given in by At pV = kelvin. a given equation does not order gases, to a of describes explain why understand model Clausius how gases the devised the kinetic more simply gases act as physical was called gases, in behave, they mo v e me nt due to instant by the as of the of gases. The theory can be the ir in of a to ms the r mal ti me , than energy temperature, theory have tr a v e l l ing the temperatur e kinetic Rudolf molecular will velocities In properties 1857 termed but do. so me or ene r g y. the wi l l the same o the r at of t he pa rticl es . i ncr ea s es be g re a te r the hig he r. p a r ti cl es ave r a g e of At all ki ne tic H oweve r, ave ra g e a gi ven g ase s wil l e ne r gy. kinetic Hence, theory is nRT particles The The molecules expression: be the kinetic molecular theory can offer summarized an explanation of temperature and pressure at asfollows: the 1 Gases consist which are of a large moving at number high of particles, velocities in molecular molecular equation level. theory can be Starting postulates, with the the ideal kinetic gas derived. randomdirections. 2 The In size this a gaseo us ther e even though their volumes can the be or are the is co ns id er e d at to ha ve or the that, m a s s, ne gl igib l e. no rma l a to m s compar e d neg l ig ibl e. a ss ump ti o n pa r ti cl e s since , between large p a rticle is gaseo us justied space very 3 of model mole cul es si ze Th is pr e s su r e, of th e is a t om molecule. Collisions and between another particles are collide, energy energy one gaseous completely there is is no simply particle elastic . net loss When in transferred. A Figure 4 In the kinetic useful analogy here is what happens molecular theory of gases, a sample when of gas is modelled as consisting of a collection of par ticles two snooker balls collide. The total kinetic moving at high velocities in random directions. The sizes of energy is equivalent before and after the the par ticles are negligible. The par ticles collide with each collision of the snooker balls. other and with the walls of the container containing the gaseous sample. All the collisions are c. The pressure 4 The average kineti c e ne rg y of the pa r t ic l es exer ted by the gas results from collisions of its par ticles is proportional to the a b so lute tem pe r at u r e with the walls of the container 165 6 C H E M I C A L K I N E T I C S Collision theory Occam’s razor is a principle fourteenth-century theologian, though was and many known states logician suggest earlier to Franciscan William references much attributed English than of of the a Ockham, that this. the The quantum razor friar, is theory can a taking formulate models. that: as guide although reactions principle principle mechanics. used we place principle the the is directly molecular based theory of a Occam’s development cannot at theories Collision The in on level, current good of see we atomic example of this principle. “Entities should not be multiplied For unnecessarily.” a chemical reacting Scientists have formulated the principle like reaction particles, a to occur number of between two conditions must this: befullled: “When two competing theories exist that 1 explain observed facts, both giving The two that the same two be predictions, theories used is until the then the optimum more simpler one evidence particles must collide with each other, essentially that of there must be physical contact. the 2 should transpires is, The colliding mutual to particles must have the correct orientations. proveotherwise.” 3 The of British the theoretical University of physicist Cambridge Stephen and The author History of Time of could laws for that some observe without of us the that determines present disturbing universe mortals. the imagine supernatural the principle all still It are it. features as of there events being, state of of to Occam’s the who the helps set three must initiate conditions is that cannot on new out lead the and on form have reaction. the basis why a rates of hydrogen slowly principle is often encountered widely For justication of example, the the only reacting a taking particles formation. H In the , and The at idea it of has been used uncertainty in iodide, room as of place. bounce Let us In off resulting gaseous iodine, I , For fraction of most one in no reaction to form 2 the reaction temperature, collisions proceeds since only a very results in a small reaction: a the eld H (g) + I 2 do o cvo HI, model theory. small unchanged, that in number chemistry. of model chemical molecular reaction hydrogen, a temperature. kinetic to is 2 observed.” The sufcient the theory reactions, remain product between be the chemical another cut depend collisions, to This understand collisions the and based most models interest employ us theory. reactions could such razor theory a universe much better is completely However, not seems known to said: collision “We particles energy of These ABrief reacting kinetic Hawking consider each of (g) ⇋ 2HI(g) 2 these three conditions in the box above separately: g, E 1 The rst, that physical contact is necessary for a reaction to occur, is The activation energy is the relatively obvious to appreciate. minimum energy that colliding 2 Figure 5 illustrates the second condition. In (a) the orientation of par ticles need for a reaction the colliding particles is not favourable, so no reaction will occur. In to occur. (b) the particles effective have 166 and sufcient have a a suitable result in kinetic energy . orientation, chemical reaction, so the provided collision the can reacting be particles 6 . 1 C O l l i s i O n t h e O r y a n d r a t e s O f r e a C t i O n (a) ineective collision No reaction occurs, since + orientation is not favourable. (b) eective collision + +2 Orientation is favourable, so may result in reaction if there is sucient kinetic energy. x Figure 5 The possibility of a collision leading to a reaction depends on the orientation of the par ticles a y A useful analogy somebody trying in understanding to push a large activation rock over a energy hill is to (gure imagine available 6). energy release Initially, there must be a minimum input of energy in order for the person z to shift the rock over the hill. Once over the hill, the rock can fall to point z Figure 6 Analogy of activation energy, E a Catalysts The analogy (gure just described can be conveyed in a potential energy prole 7). (a) (b) transition state transition state ygrene laitnetop ygrene laitnetop a reactants ∆H a products ∆H reactants products reaction coordinate reaction coordinate Figure 7 Potential energy prole for (a) an exothermic reaction with ΔH<0; (b) an endothermic reaction with ΔH>0 The arrangement transition state of atoms or activated at the crest of the energy prole is termed the complex do o c Catalysts may be either homogeneous or heterogeneous. A catalyst is a substance that increases the rate of a Homogeneous catalysts chemical reaction, but is not A homogeneous catalyst is in the same physical phase or state as the consumed in the reaction reactants. The destruction of gaseous ozone, O , by chlorine atoms is an 3 example of homogeneous catalysis, since itself. A catalyst provides chlorine atoms (which act as an alternative pathway for the catalyst) have the same state as the gaseous reactants. the reaction and lowers the In the stratosphere (upper atmosphere), ozone in the ozone layer activation energy, E a absorbs over protecting us 95% of from the this UV radiation harmful reaching Earth from the sun, radiation. 167 6 C H E M I C A L K I N E T I C S hν O (g) → O 3 O (g) + O → O (g) + O (g) 2 Thus, With (g) 2 (g) + heat 3 there the is a net energy progressive radiation can increased reach risk of conversion depletion the skin of Earth’s cancers the from ozone surface. This (melanomas) UV to layer can and heat energy. (gure be 8), more associated UV with an cataracts. Uu rouc: Ozo ho Wc - images, data and information for both the southern and nor thern hemispheres can be found on this website hosted by NASA - Goddard Space Flight Centre: http://ozonewatch.gsfc.nasa.gov Figure 8 The largest ozone hole to date seen here shown in purple was recorded on September 24, 2006 for the Antarctic Hemisphere Chlorine (CFC) atoms with UV refrigerators, are produced light. and CFCs aerosols. in the were reaction previously Freon, CF Cl 2 presence is of greater), UV and light, the radicals weaker are C of a used (g), chlorouorocarbon in air is one is broken conditioning example of a units, CFC. In the 2 Cl produced. bond The chlorine (the radicals C F bond then strength attack ozone. hν CF Cl 2 rst step: Cl (g) → CF 2 (g) Cl (g) + + O (g) → ClO (g) + O 3 second step: ClO·(g) reaction: 2O + O (g) → Cl·(g) acts represent in gure the as a catalyst potential (g) and energy (g) + 2O (g) 2 → 3O 3 Chlorine (g) 2 3 overall Cl 2 (g) 2 is regenerated prole for this in the second reaction step. scheme We as can shown 9. transition states ygrene laitnetop a Cl•(g) + 0 (g) 3 + O•(g) (uncatalysed) a (catalysed) (catalysed) ∆H Cl•(g) + 20 (g) 2 (catalysed) reaction coordinate Figure 9 Potential energy prole for the catalytic destruction of ozone, showing both the catalysed and uncatalysed pathways 168 6 . 1 As can for be the and, in prole seen from catalysed general, the more the catalysed state Although ozone harmful for to may the in of formed typical the are in for uncatalysed pathway. acts as UV-a a of as gaseous of is a (the shield just one 10. from high atmosphere) emphysema. nitrogen oxides, r e a C t i O n energy gure radiation, lower and O f states shows in r a t e s reaction potential shown UV-b a n d transition particular protective and asthma t h e O r y pathways This troposphere such reaction two this representation and the actually specic stratosphere problems in is high-energy ozone respiratory be of there catalysed the effects concentrations lead 9, This showing transition the gure pathway. C O l l i s i O n can Ozone NO and NO , 2 from in car the exhaust gases troposphere with also acts VOCs as a (volatile organic greenhouse compounds). Ozone gas transition states at crests ygrene laitnetop (uncatalysed) a reactants (catalysed) a ∆H products reaction coordinate Figure 10 Typical potential energy prole showing catalysed and uncatalysed pathways Heterogeneous catalysts acv A heterogeneous catalyst is in a different phase or state from the Can you name ve countries reactants. Typically, the heterogeneous catalyst is in the solid phase around the world that still and the reactants are in the liquid or gaseous states. An example of a continue to use leaded petrol? heterogeneous system CO, of a catalyst car. nitrogen In a is car the catalytic engine monoxide, NO, converter pollutants and such unburned used as in the carbon exhaust monoxide, hydrocarbons, C H x produced. of The carbon hydrocarbon atmospheric car. The fuels. nitrogen catalytic substances, monoxide Nitrogen and oxygen converter namely comes carbon from monoxide at converts dioxide, is high , incomplete temperature these CO the produced in substances water, H 2 from O, into and are combustion the the , y reaction engine less of of the harmful nitrogen, N 2 : 2 catalyst 2NO(g) N (g) + O 2 (g) 2 catalyst 2NO (g) N 2 (g) + 2O 2 (g) 2 catalyst 2CO(g) + O (g) 2CO 2 (g) 2 catalyst CH CH 3 CH 2 propane Examples + 5O catalysts and vanadium(V) Cr O 2 if a car poison (g) + 4H O(g) 2 used oxide, V O . Leaded in and , catalytic converters transition copper(II) metal oxide, are oxides CuO, platinum, such and as chromium(III) 5 petrol (leaded gasoline) is not used in modern cars – 3 tted the 3CO 2 rhodium, 2 oxide, (g) 2 fuel of palladium, (g) 3 with a catalytic converter used leaded petrol, the lead would catalyst. 169 6 C H E M I C A L K I N E T I C S Maxwell-Boltzmann energy distribution and temperature Rates of reaction at in the the gas phase molecular can level be using different approaches: 1 the collision 2 Maxwell theory Boltzmann energy distribution curve 3 temperature We have model. effects already We shall on kinetic explored now the look at energies. collision theory approaches 2 and 3. ygrene citenik htiw selcit rap fo noitcarf interpreted area A a for catalysed reaction area B kinetic energy Maxwell–Boltzmann energy distribution Figure 11 Maxwell curve The Boltzmann energy distribution cur ve showing the activation energy for a catalysed reaction. Area kinetic-molecular theory says that A shows the fraction of particles in the sample that do not particles have sucient energy to react. Area B shows the fraction of gas move randomly in different directions at of particles in the sample that have the minimum energy high velocities. and, since However, these velocities differ required to initiate a reaction with the use of a catalyst with other changes this particles the of scenario. curve. with a of kinetic with of is This is value a needs of energy. of representation the curve is is the seen particle discuss particle is in the particles . by the distribution (that energy be to described fraction energy can the of the considered energy the kinetic As be of gaseous gaseous to colliding sides realistic velocities plot the single velocities kinetic of a not Boltzmann given that It constantly individual What distribution Maxwell of an are and velocity constantly. velocity distribution The particles of is particles the probability occurring) from asymmetric. versus gure The ygrene citenik htiw selcit rap fo noitcarf container, the the a area A the total number area of reaction E a for uncatalysed reaction area B area C kinetic energy 11, under Figure 1 2 Maxwell represents for catalysed Boltzmann energy distribution cur ve gaseous showing the activation energy for an uncatalysed reaction. particles the in the majority kinetic time, energy some either high sample. of of or gaseous near the the At certain mean gaseous low a particles value. particles velocities, temperature, will but have At will the a a not have sucient energy to react. Area B + area C shows the fraction of par ticles that have the minimum energy given required in order to initiate a reaction using a catalyst. Area have majority Area A shows the fraction of par ticles in the sample that do will C shows the fraction of par ticles in the sample that have the minimum energy required to initiate a reaction without the have velocities and hence kinetic energies close to use of a catalyst the 170 mean kinetic energy. 6 . 1 C O l l i s i O n t h e O r y a n d r a t e s O f r e a C t i O n Temperature eects on kinetic energies increasing of particles to overcome increase. the that As greater in atter and Therefore, frequency be T more now activation increases, the proportion with a higher increasing collisions successful more at particles curve be of an having the Maxwell becomes temperature. increase. collisions, which will temperature, will will velocity particles result, energy barrier there of a proportion kinetic mean so distribution broader of As the energy the and energy. energy T, sufcient increases kinetic Boltzmann have the particles increase temperature since have ygrene citenik htiw selcit rap fo noitcarf With T 1 T 2 a the There there sufcient kinetic energy will are kinetic Figure 1 3 Maxwell Boltzmann energy distribution cur ves for two temperatures, T > T 2 energy to overcome the activation energy barrier. . Notice that, at the higher 1 temperature, the energy distribution is broader and the mean kinetic energy is greater. The proportion of particles that have This results in an increase in the rate of reaction. with an increase in temperature sucient thermal energy to overcome the activation energy Typically of barrier has increased. The area under both curves is the same as 10°C, the reaction rate will double. this signies the total number of gaseous particles in the sample tOK su p In a potential energy prole, Temperature can be considered as a measure of the average amount of kinetic the y-axis is the potential energy of par ticles, which is the energy due to motion. As the temperature energy and the x-axis is the increases, so does the kinetic energy of the par ticles. The lowest temperature reaction coordinate, which that can be attained theoretically is bou zo, which is 273.15 °C. This is represents the progress of taken as the zero point on the Kelvin scale. At 0 K , all thermal energy has been the reaction. In a Maxwell removed from a substance and the motion of par ticles has eectively ceased. Boltzmann energy distribution However, if you consider Heisenberg’s uncer tainty principle, all molecular cur ve, the y-axis is the movement may not cease entirely at 0 K . Consider why this may be the case. fraction of par ticles with a The Kelvin scale gives a natural measure of the kinetic energy of a gas, and cer tain kinetic energy and the is independent of physical proper ties, whereas the ar ticial Celsius scale is x-axis is kinetic energy. Don’t based on the physical proper ties of water. The Celsius scale is dened about an arbitrary zero point and hence negative °C values occur. Are physical proper ties mix up the two axis labels in these two representations. such as temperature invented or discovered? Could Anders Celsius, the Swedish astronomer credited with the Celsius scale, have chosen the melting and boiling points of another substance other than water to devise the ar ticial Celsius scale? 171 6 C H E M I C A L K I N E T I C S Factors that aect the rate of a chemical reaction There are four factors that can increase the rate of 4 Decreasing the par ticle size of a chemical reaction: reactants in the solid phase 1 increasing the temperature reaction conducted at which the In is a 2 addition of a catalyst a increasing the concentration of the decreasing the particle size of reactants in us the examine each of these the for 1 Increasing the temperature at which surface just discussed this factor. A the surface rate have example involves of the the above milk to down 0 °C of involving rate area up of of into a gas reaction the solid smaller is (or will increased pieces. is that of the the reaction takes The place only turn bacterial if in can the a of milk. sour temperature on At solid a This is liquid is of reactant. reacting the solid reactant. used, and particles of liquid So, there will will a for be be for example, greater For only direct As the contact solid increase reaction. increase. solid, divided an a a have nely will will available reaction If there with in the number of a the result, grain can combust example, explosively. nely This has led rate to a in grain number of major explosions in conned spaces room over reactions. a good temperature refrigeration temperature, due effect of area divided slowed solid particles powder solid the reaction is conducted time surface the this case surface individually. or of the phase. with We the reaction solid, the in Let a reactants on solid if breaking reason 4 and increase by 3 heterogeneous liquid) factories. period process brought to is just refrigerator. acv Find out some examples from dierent locations around the world where this has happened. 2 Addition of a catalyst We have also already discussed this in detail. Measuring the rate of a chemical reaction 3 Increasing the concentration of the We reactants If in a xed reactant be volume species corresponding collisions. the concentr a ti on incr e a s e s , incr e a s e Hence the r e in ther e the wil l wi l l an look to stated at a number monitor the of rate techniques of a that chemical can reaction previously: a fre que nc y be as of be shall used 1 Change in pH (for acid-base of inc re a s e in the reactions) number of success f ul the of reactio n col li si ons , a nd t he r efor e In rate will incr ea s e. An a reaction where + H of this involves the d e s tr ucti on of st a t u es m a de O limestone (calc i um car bo na te , hydronium cations, + (or Ca C O ). If simply H ) or hydroxide ions, OH , are 3 present of either ex am pl e as either the reactant or product species, the 3 the concentration of the p o ll uta nt s ulfur change probe SO , in the atmosphe r e i ncr e as e s, in pH can be monitored using a pH di ox ide , the ra te and meter of 2 destruction of the l i me stone statue s i n c re a se s . 2 Change in conductivity (for reactions 2CaCO (s) + 3 2SO (g) + 2 O (g) → 2CaSO 2 (s) 4 involving electrolytes) + 2CO (g) 2 Consider The solubility of calcium sulfate is greater the following reaction: than + that of calcium carbonate, leading to erosion of IO (aq) + 5I (aq) + 6H (aq) → 3I 3 the In this in an the 172 (aq) + 3H 2 O(l) 2 limestone. reaction acidic of iodate medium concentration of ions there ions is with a from net a iodide ions decrease total of 12 in on 6 . 1 the reactant This be side decrease in monitored meter. If to zero the using the net on the product concentration a of conductivity number of C O l l i s i O n ions t h e O r y a n d r a t e s O f r e a C t i O n side. ions can probe absorbance and decreases photoelectric light during the reaction, then the total electrical cell source coloured lter conductivity will also decrease and vice versa. meter solution Figure 1 4 Schematic diagram of a colorimeter that records 3 Change in mass or volume (for absorbance. According to Beer ’s law the absorbance A is directly proportional to the concentration c, that is A ∝ c or A = cεl, reactions involving gases) where l is the path length and ε is the extinction coecient. We have seen an example of this physical method earlier. In is colorimetry, passed light through monitored. The of the a certain coloured colorimeter or wavelength solution being spectrophotometer 4 Change in colour (for reactions then measures the intensity of the transmitted light. involving transition metals or other The coloured compounds) absorbance absorbed curve Colorimetry (gure 14) is used to by can the be reactions thathave a colour in or a coloured intensity product One can be example product. corresponding concentration of the of this to the a (aq) + 5C 4 (aq) the + light calibration versus of the standard coloured solution 15). change or colorimeter. ethanedioate + O 2 A absorbance of in reaction: 2 2MnO change reactant using involves mixture. amount ecnabrosba manganate(VII) of the monitor The coloured monitored reaction of coloured (gure reactant indicative plotted concentration chemical is 16H (aq) 4 purple 2+ → 2Mn (aq) + 10CO (g) + 8H 2 pale O(l) 2 pink 3 concentration/mol dm Figure 15 Sketch of a typical calibration cur ve Worked examples: rates of reaction 1 b) M(CO Example 1 ) = 12.01 + 2 × 16.00 0.25 _ n(CO In the = 44.01 g mol 2 chemical reaction of calcium carbonate ) = 3 = 5.7 × 10 mol 2 44.01 with hydrochloric acid, 0.25 g of carbon dioxide 3 5.7 × 10 _ Average was a) generated Deduce in the 60.0 rate = s. balanced 5 = 9.5 × 10 1 mol s 60.0 chemical equation su p forthe reaction, including state symbols. Be careful with g c gu in a question like -1 b) Calculate the average rate, in mol s this. Since division is involved the answer should be expressed with the smallest number of signi cant Solution gures from the experimental data, which in this a) CaCO (s) 3 + 2HCl(aq) → CaCl (aq) 2 + CO (g) + case will be two. 2 H O(l) 2 173 6 C H E M I C A L K I N E T I C S Example 2 Figure 16 dioxide shows formed hydrochloric calcium how the varies acid volume with solution carbonate in a time, is of carbon when added to Then a as the gradient excess dioxide ask. of acid The is proceeds as generated the slows with amount as the time, of the carbon concentration decreases. plot acid reaction decreases eventually attens as hydrochloric consumed. ii) OC( V 2 ) OC( V 2 ) 0 time 0 Figure 16 time i) Explain ii) Copy the shape of the curve. Figure 1 7 [3] Decreasing the graph and sketch the rate youwould obtain if double the hydrochloric acid solution of concentration, as in the less gas instead, with all other constant shape of from the thecurve original.Explain is Outline one other way in will less this reaction which can be Sketch a be the selected variable studied graph in to There wouldchange the state between term one activation reason calcium takes slower. dioxide are a why is volume time. halved, Since collisions the rate of will reaction there is the same amount acid, will the be same volume of produced. to of possible measure the answers rate at here. which as place carbonate at a and reasonably the rate gas is the beginning of [2] graph. the produced quickest, hydrochloric acid is as reaction, the carbon contents versus time dioxide concentration greatest. of given mass off. of time (gure or a plot of This ask mass 18). ssol ssam of plot at 2010 gradient a hydrochloric fast temperature. = involve reaction Figure 18 174 is hence Since decreases simply time is longer energy the Solution At and number stnetnoc + ksa fo ssam room Rate frequent mass versus i) a [2] Dene May maximum over with + IB, same concentration method would acid the evolved a the and the (blue) illustrate time. iv) but be hydrochloric One how curve the iii) schoollaboratory. decreases the [4] carbon rateof 17 why different. of iii) steep, acid will the gure variables be kept In example the above,isused concentration reaction. half of the of volume is of the curve of time loss 6 . 1 Other the possibilities pH or the might include C O l l i s i O n monitoring t h e O r y b) a n d Draw a versus pressure. r a t e s graph O f of r e a C t i O n total volume of oxygen time. 3 c) iv) Activation energy is the minimum energy Calculate correct colliding particles need to have in order the average rate, in 1 cm min , the to one decimal place. for 3 a reaction calcium takes to carbonate place at temperature barrier occur. is a The and between hydrochloric reasonably because quite reaction the fast rate activation d) acid at Deduce, oxygen in to s, how be long it took for 40 cm of collected. room 3 e) Determine the initial rate, in f) Determine the instantaneous at = cm 1 min energy rate, in low. 3 1 cm g) min Explain , t whether 4 min. the catalyst used is Example 3 homogeneous The data in table decomposition manganese(IV) 2 of were recorded hydrogen oxide for peroxide catalyst. The using total a Solution 1 H O 2 oxygen gas collected was heterogeneous. volume a) of or the measured (aq) → H 2 O(l) + O 2 (g) 2 2 at differenttimes. b) 70 tm/m to voum o ox g g 3 coc/cm 0 1 18 2 32 3 42 4 50 5 56 6 61 7 64 8 64 9 64 negyxo fo emulov latot 0 60 50 40 30 20 10 0 0 2 4 6 10 8 12 time/min Figure 19 Plot of total volume of oxygen given o versus time 3 64 cm _ c) Average rate = 10 64 T able 2 d) 2.8 min e) (x y , 1 ) = = 2.8 × (2,50); 3 = 7 9.1 60 (x 1 s , 2 = y ) 168 Initial Deduce the the balanced reaction, chemical including equationfor rate = (0,0); 2 3 = = 0 1 min s 0 50 _ a) cm min 25 cm 1 min 2 statesymbols. 175 6 C H E M I C A L f) (x , 3 y ) = K I N E T I C S (6, 64); (x 3 , y 4 ) = (2, 36); g) Manganese(IV) a Instantaneous rate at t = 4 different 3 = 2 7cm phase to is solid, aqueous so is in hydrogen min: peroxide. 36 64 _ oxide 4 Manganese(IV) oxide is acting as 1 min heterogeneous 6 catalyst in this reaction. su p There are a number of points that you have to be careful 2 about in graphical questions: 1 Graphs should have a title which involves a plot of y versus x (not the other way round!). 176 Both the x- and y-axes must be labelled and units included. 3 When finding choose two the slope points as of far a tangential apar t as line, possible. try to a Q U e s t i O n s Questions 1 Consider and I the gaseous (g) + H 2 reaction between gaseous iodine 4 hydrogen. (g) ⇋ 2HI(g) ∆H = -9 Equal masses of powdered carbonatewere added of acid. hydrochloric to calcium separate The calcium solutions carbonate kJ 2 was Why do some hydrogen collisions not result in between the iodine formation of and in excess. the intervals. product? The Which represent against A. I and H 2 molecules do not The dioxideproduced the time volume was of carbon measured at curves in gure evolution of carbon forthe acid regular 20 solutions best dioxide shown in have 2 table sufcient 3? energy. system The temperature is in equilibrium. D. The activation of the energy system for this is too high. reaction is mc/)g( The C. 3 B. I II IB, low. May [1] 2011 3 2 At 25 OC fo emulov 2 very °C, 200 cm III IV 3 of 1.0 mol dm nitric acid time/s is added to 5.0 experiment of is in the of magnesium repeated magnesium result g powder, same using the which initial powder. same If mass conditions reaction the ▲ Figure 20 will 3 rate? 25 cm 3 o 50 cm 3 o 3 2 mo m Voum Coco o 25 cm 3 1 mo m 1mo m hC hC hC A. I III IV B. I IV III C. I II III D. II I III tmpu / °C o 3 3 / o hnO hnO 3 /mo m 3 3 cm 200 A. 2.0 25 B. 200 1.0 50 C. 100 2.0 25 D. 100 1.0 25 ▲ T able 3 [1] [1] IB, IB, May 5 3 Which for of rate A. s B. min C. cm May 2009 2011 the of following reaction? is an appropriate unit Hydrochloric of calcium acid repeated using does change and this the is reacted carbonate; the calcium collision affect large carbonate the is pieces then powder. activation How energy frequency? acvo g 3 with reaction Coo s quc 3 D. mol dm 1 min A. increases increases B. stays constant increases C. increases stays constant D. stays constant stays constant [1] IB, November 2009 177 6 C H E M I C A L 6 Which K I N E T I C S factors can affect the rate of a chemical 9 a) A solution of hydrogen peroxide, H O 2 reaction? added to acidied I. The concentration of the a solution with of sodium hydrochloric acid, , is 2 iodide, NaI, HCl. The reactants. yellow colour of the iodine, I , can be used 2 II. The temperature at which the reaction to takes determine H III. The the rate of reaction. place. physical state of the O 2 reactants. (aq) + 2NaI(aq) + 2HCl(aq) 2 → 2NaCl(aq) + I (aq) + 2H 2 A. I and II B. I and III The experiment II D. I, and II, is repeated with III only and of the III to the reaction changes reason, i) its The that effect conditions. follow, on the concentration predict, rate of H of O 2 at 7 In an ethyl acid-catalysed ethanoate, hydrolysis the reaction in 3.5 from min, 1.50 concentration at a mol given dm of the ii) ester The 3 to 0.35 temperature, mol T . constant For each stating a reaction. is increased 2 temperature. [2] of 3 changes some only changes C. O(l) 2 only a dm solution ne of powder NaI is prepared instead of from from large crystals. Which [2] 1 of the following statements are correct? b) Explain when I. The average rate of the 3 0.33mol reaction If the rate of a of reaction the increases system increases. [3] min IB, II. the temperature is 1 dm why the reaction temperature, is T carried , the out at reaction a November 2009 higher rate will 2 begreater. 10 III. The products of the reaction will Models the ethanoic acid and I and II B. I and III C. II I, II, and the Factors and that reaction affect and Dene the rate particle the of a size, chemical Design in concentration temperature of the an rate the term rate of a an of appropriate reaction alkaline the Design an rate three characteristic properties hydrolysis to measure reaction particles reaction as theory. 2011 of the ester methyl ethanoate medium. appropriate experiment to measure of reaction using that described affect by the the rate involving the the “clock reaction of reaction” magnesium of with 178 experiment a [1] reactant May of chemical technique IB, theory. of the List collision reaction. reaction. ii) Discuss molecular III 12 i) chemistry. only include reactants, in kinetic only III (saponication) 8 vital the only the D. of ethanol. 11 and prove principles theory A. can be of collision [3] dilute hydrochloric acid solution. 7 E Q U I L I B R I U M Introduction An understanding of reactions that are equilibrium, in examine the equilibrium constant K , c equilibrium the to and equilibrium science and how is of to control the fundamental society. The Haber position of importance process, used examine the the effects the value We will information of of changing K it conveys and experimental applying Le discuss conditions Châtelier’s on principle. c for an the equilibrium impact we large-scale on discuss the manufacture system history how that of has the reactions of ammonia, had World. can be in a profound In a is this state chapter of Q, also which products not in a is introduce a and state measure the of reactants of term the reaction relative present in a quotient, amounts reaction of that is equilibrium. 7 .1 E Understandings Applications and skills ➔ A state of equilibrium is reached in a closed ➔ The characteristics of chemical and physical system when the rates of the for ward and systems in a state of equilibrium. reverse reactions are equal. ➔ ➔ Deduction of the equilibrium constant The equilibrium law describes how the expression (K ) from an equation for a c equilibrium constant (K ) can be determined for c homogeneous reaction. a par ticular chemical equation. ➔ ➔ Determination of the relationship between The magnitude of the equilibrium constant dierent equilibrium constants (K ) for the c indicates the extent of a reaction at equilibrium same reaction at the same temperature. and its temperature dependence. ➔ ➔ Application of Le Châtelier ’s principle to The reaction quotient (Q) measures the relative predict the qualitative eects of changes of amount of products and reactants present temperature, pressure, and concentration on during a reaction at a par ticular point in time. the position of equilibrium and on the value of Q is the equilibrium constant expression with the equilibrium constant. non-equilibrium concentrations. The position of the equilibrium changes with changes in concentration, pressure, and temperature. ➔ A catalyst has no eect on the position of equilibrium or the equilibrium constant. Nature of science ➔ Obtaining evidence for scientic theories – ➔ Common language across dierent disciplines – isotopic labelling and its use in dening the term dynamic equilibrium is used in other equilibrium. contexts, but not necessarily with the chemistry denition in mind. 179 7 E Q U I L I B R I U M Equilibrium reactions in chemistry Many state important of occurring control being reaction synthesized understanding Systems that example, with a leave An the Water liquid evaporation. energy in have molecules through the and same collision of and in the enter time, and the liquid liquid the enter in in that of being enables the chemists desired relies everyday a storage product heavily liquid life. on in water phase For container sufcient phase gaseous the a constantly Industry have gaseous some yield in reactions reactions. common hot exist equilibrium reactions. equilibrium are and reverse reactants maximize some and chemical equilibrium equilibrium you reversible forward products and controlling phase At with are the understanding involving imagine lid. both conditions and are reactions with simultaneously interconverted. to chemical equilibrium, the process molecules in the sealed energy will of will process lose of condensation. As evaporation saturated particles with resulting evaporation further gas will change present The continues, water in forward constantly described in the of being as a atmosphere and the condensation equal and concentration in the vapour the the rate amount closed of gas and and is change, between the the at at two the rate will of be amount no of equilibrium. equal but becomes between rates. molecules phases. This The are is equilibrium H O(l) ⇋ H 2 O(g) 2 liquid phase ▲ system there then occurring not closed collisions Eventually present system are does interconverted dynamic increases. The reactions and the of condensation liquid system. reverse liquid of in chance gaseous phase Figure 2 In a dynamic equilibrium the for ward and reverse reactions occur at equal rates Heterogeneous equilibrium and solubility ▲ Figure 1 Bromine in a closed A system will establish a dynamic equilibrium equilibrium between the liquid and gaseous phases. Eventually saturated chloride, The soluti o n if NaCl ther e and concentration in is a clo se d e xce s s wa ter, of i ons sy st e m s ol id initia ll y pr e se nt w il l p re s e nt . the in s ol id the e st a bli sh If you wil l a mi x beg i n a que ou s dyn a m i c so li d to so di um di ss ol ve. s ol u t io n wi l l the colour stops changing as the increase. Some solution. At aq ue o us io ns wi ll r e c o mbin e and pr e c ip it a t e ou t of equilibrium concentration of vapour this is reached precipitation. 180 s ta g e the ra te of dis so lut i on is fas t e r t h an the r ate of 7. 1 Eventually, when the solution becomes saturated, the rate of E q u i l i b r i u m dissolving TOK will equal the rate of precipitation. A dynamic equilibrium has been established. Scientists use specialized terminology to precipitation + NaCl(s) ⇋ Na (aq) + Cl facilitate unambiguous (aq) communication of meaning dissolution and understanding. They initial recognize the need for a equilibrium common language. For example, a closed system is one in which no matter is exchanged with the surroundings. The language of science and its internationally agreed terminology and symbols aid communication and understanding. However, when meaning is inexible Cl + (aq) (aq) Na + + Na Cl does language also play a (aq) Na (aq) (aq) (aq) Cl (aq) Cl + Na role in forming boundaries (aq) to our experience of new knowledge? ▲ Figure 3 A saturated solution is in dynamic equilibrium Chemical systems When describing “reactants” and completion, a state of depend chemical producing equilibrium. on chemical the and on give new The physical conditions products), reactions “products” (a the the in an substances. relative in presence In rates conditions change open impression of system, that reality of of a many forward temperature concentration the of terms reactions and and proceed reactions reverse exist in reactions pressure, reactants to on Stdy tp and catalyst. For non-reversible reactions, Nitrogen(IV) oxide or nitrogen dioxide, NO is a toxic brown gas that the use of a single arrow → in a 2 exists in equilibrium with colourless dinitrogen tetroxide, N O 2 chemical equation represents : 4 the complete conversion of N O 2 When (g) ⇋ 2NO 4 a (g) 2 sample of reactants to products. Reversible colourless frozen N O 2 which is then sealed, initially N O 2 colour is is is placed into a reactions and systems in container 4 the only substance present and no equilibrium are designated by 4 the use of a double-headed observed. arrow: ⇋ At room temperature N O 2 to brown signals the begins to decompose and a colour change 4 presence of NO . Your understanding of kinetic decomposition of N 2 theory will tell you that initially the rate of O 2 be greatest, when the concentration [N O 2 passes, sufcient NO molecules will be ] is at a maximum. As will 4 time 4 present in the closed system for 2 successful collisions to reform N O 2 The concentration of N O 2 concentration of NO . 4 decreases progressively while the 4 increases. The system approaches a dynamic 2 equilibrium become as the rates of the forward and reverse reactions equal. 181 7 E Q U I L I B R I U M At equilibrium: NO 2 noitartnecnoc N ● The forward ● There ● There ● The is no and reverse change in reactions the are occurring concentrations of at equal reactants rates. and products. O 2 4 is no change equilibrium in can macroscopic be properties approached from such either as colour the and forward density. or equilibrium achieved reverse direction. 0 time ▲ The ● time for the reaction N O 2 (g) ⇋ 2NO 4 equilibrium continue Figure 4 Change in concentration versus but products (g) is no dynamic. overall The change forward in the and reverse concentration reactions of reactants and occurs. 2 approaching and achieving equilibrium Any ● changes pressure, or in the reaction concentration conditions, of reactants such or as temperature, products, can affect the rate of for ward equilibrium, demonstrating its dynamic nature. reaction qck eston etar Which statement is always true for a chemical reaction that has reached equilibrium achieved equilibrium? (rates are equal) rate of reverse reaction 0 a) The yield of product(s) is greater than 50% ) The rate of the forward reaction is greater than the rate of the reverse time reaction. ▲ Figure 5 At equilibrium, the rates of the for ward and reverse reactions remain constant c) The amounts of reactants and products do not change. d) Both forward and reverse reactions have stopped. [1] ib, Nov 20 05 Animated computer simulations are available in The equilibrium law which the user can decide When a reaction system has established equilibrium, the forward and how to change the conditions reverse reactions are occurring at equal rates and there is no change of a reaction to illustrate in the concentration of reactants and products. To make best use of the concept of dynamic the reaction we need to understand the position of the equilibrium , equilibrium. that is, whether manipulate of this reactants to or products maximize the are yield of favoured, products and and how the we can protability industry. TOK The Scientists have a common terminology and a common reasoning process, which involves using deductive the law ratio molar of of chemical of the coefcients) their molar equilibrium equilibrium concentration to the states products concentration coefcients) constant, of K is . a of constant. The that a given to reactants This subscript at (raised ‘c’ the temperature power (raised constant indicates is to the called that of values for products and reactants are being concentration used. through analogies and For example: generalizations. They share O mathematics, the language (g) + 4HCl(g) ⇋ 2 2H O(g) + 2Cl 2 [ H O [ ] (g) 2 2 2 of science, as a powerful Cl 2 ] 2 __ K tool. Indeed, some scientic = c 4 [ O ] [ HCl ] 2 explanations exist only in mathematical form. The value reaction of at a the equilibrium given constant temperature. is Values specic of K c 182 for have each no power the c logic and induction their chemical units. 7. 1 The magnitude of K tells you about the position of the E q u i l i b r i u m equilibrium. c If K is a very large number, K c 1, this indicates that at a given c temperature, of >> K , the products greater the are favoured proportion of over reactants. products that The exists larger the value compared with c reactants at equilibrium. Conversely, a very small value of K ( K c << 1 ) c In a hoogeneos e indicates that the reaction is unfavourable at this given temperature. all the reactants and products are present in one phase. The most common example is Writing equilibrium constant expressions reversible reactions that occur When constructing the equilibrium constant expression for a in the gaseous phase. homogeneous reaction, the following need to be considered: In a heteogeneos 1 For an aqueous reaction the concentration of the solvent water e the reactants or does not appear in the equilibrium constant expression, as its products exist in more than concentration does not change during the reaction. one phase, such as gaseous 2 If the reaction esterication takes place discussed in in a non-aqueous sub-topic 10.2), solution water (such must be as the and solid, liquid and solid. included in The equilibrium constant the K expression as any other reactant or product. c Table 1 expressions derived here are gives different some examples of equilibrium constant expressions for homogeneous equilibria. for reactions. Checa eaton E constant expesson 2+ 3+ Fe ] [ Fe(SCN) ___ 2+ (aq) + SCN (aq) ⇋ Fe(SCN) (aq) K = c 3+ 1 [ Fe [ C ][ SCN H 3 ] COOCH 7 ] 3 __ CH OH(aq) + C 3 H 3 COOH(aq) ⇋ C 7 3 H COOCH 7 (aq) + H 3 O(l) K 2 = c [ CH OH ][ C 3 H 3 COOH ] 7 2 [ NH ] 3 __ N 2 (g) + 3H (g) ⇋ 2NH 2 (g) K 3 = c 3 [ ▲ N 2 ][ H 2 ] T able 1 Equilibrium constant expressions for some chemical reactions TOK Scientists approach their understanding of the universe from varying perspectives. Observations made using our senses examine phenomena at the macroscopic level. The models and theories developed by the scientic community focus on our understanding of the microscopic world. Which of the ways of knowing (WOK) qck eston enable us to make the transition from the macroscopic to Deduce the equilibrium constant expression for each of the microscopic? the following homogeneous equilibrium reactions. macoscopc properties of substances can be identied 3 1 a) using our senses, and can be directly determined by measurements. Examples include colour, texture, and N 2 ) (g) + 2 H 2 ClNO (g) ⇋ NH 2 (g) 3 (g) + NO(g) ⇋ NO 2 (g) + ClNO(g) 2 density. mcoscopc properties exist at an atomic level c) and can be determined only indirectly. 4NH 3 (g) + 5O 2 (g) ⇋ 4NO(g) + 6H O(g) 2 183 7 E Q U I L I B R I U M Worked example: deducing equilibrium constant expressions For the homogeneous 0.300 × 0.300 __ equilibrium: = 2 (3.00) H (g) + I 2 (g) ⇋ 2HI(g) 2 2 = 1.00 × 10 3 the equilibrium concentrations (in mol dm ) are Note: as follows: 1 _ K = c(1) [ H (g) = ] 0.300, [ I 2 Deduce (g) ] = 0.300, [HI(g)] = K 3.00 c(2) 2 the equilibrium constant expression, The K , value of K and determine the value of K for the forward also depends on how the chemical c c equilibrium and reaction is balanced: c reverse 1 _ reactions. 1 _ H (g) + I 2 (g) ⇋ HI(g) 2 2 2 Solution [HI] _ = K 1 c(3) For the forward reaction: 1 2 H [ ] 2 I [ 2 ] 2 2 [HI] 3.00 __ _ K = K c(1) = c(3) [ H I ][ 2 1 1 ] 2 2 ( 0.300 ) 2 ( 0.300 ) 2 ( ) 3.00 __ = = 0.300 × 10.0 0.300 ___ 2 = 1.00 × 10 = K c(3) For the reverse ⇋ H (g)+ 2 I by (g) 2 last ][ c(1) shows gives a that dividing new the equation equilibrium constant 2 I 2 example throughout which [ H K reaction: This 2HI(g) √ is equal to the square root of the original ] 2 _ K equilibrium = c(2) constant. 2 [HI] qck estons 1 2 The equilibrium constant for the reaction between The equilibrium constant for the equilibrium between 4 hydrogen and chlorine gas to produce hydrogen N O 2 and NO 4 is 7.7 × 10 at 273 K: 2 33 chloride gas is 2.40 × 10 at 298 K . N O 2 H (g) + Cl 2 (g) ⇋ 2NO 4 (g) 2 (g) ⇋ 2HCl(g) 2 Calculate the equilibrium constant for 1 Calculate the equilibrium constant for the reverse N 2 O 2 (g) ⇋ NO 4 (g) at 273 K . 2 reaction, namely the decomposition of HCl. Combining equilibrium constants Table and 2 summarizes equilibrium the ways constants that can be the equilibrium constant expressions combined. Change n eacton eaton E constant expesson reverse the reaction inverse of the expression E constant 1 _ K (reverse) = (forward) 1 _ c K or K c c K c __ halve the coecients square root of the expression √ K c 2 double the coecients square the expression K c sum equations product of the expressions K c ▲ T able 2 The equilibrium constant K 184 K c1 × K × … c2 for the same reaction at the same temperature can be c expressed in a number of ways = 7. 1 E q u i l i b r i u m The eect of changing experimental conditions on the equilibrium constant When equilibrium constant change provided in is established, that the experimental However, the value of the conditions K position temperature remains can and of affect constant the equilibrium pressure the do not equilibrium unless the remains change. A position. temperature c changes (table 3). Change n condton E poston K c concentration of changes in response to a change in product or reactant [reactants] or [products] no change pressure in a reaction with gaseous reactants or no change products, the pressure can aect the equilibrium position temperature usually changes: the direction of change unless exothermic or endothermic ΔH = 0 catalyst ▲ changes, depends on whether the reaction is no change no change T able 3 The eect of changing conditions on the equilibrium position and the value of K c Le Châtelier ’s principle Le Châtelier’s principle changing conditions If is At of a change the forward the system a given made and to will to a reactants or a useful on system reverse tool the that reactions for predicting equilibrium is in will equilibrium, shift to the effect that position: offset the this balance change between and return equilibrium. temperature, equilibrium is have products constant K table does . The 3 shows not that result in equilibrium changing a change position the in of concentration the the value of reaction will c change in response to the change in concentration so as to return K c to its original value. For example, gure 6 illustrates the chromate– ▲ dichromate Figure 6 The chromate–dichromate equilibrium. equilibrium: 2 Aqueous dichromate Cr O 2 2 Cr 2 (aq) + H 7 O(l) ⇋ 2CrO 2 orange (aq) + 2H aqueous chromate CrO (aq) yellow (aq) is yellow. The 4 4 dichromate (aq) is orange while 2 + 2 O 7 colour of the solution gives an indication of the chromate position of equilibrium If the tells concentration us that change. the of a reactant forward Decreasing forward reaction. mixture results the The is reaction increased, will be concentration addition of Le Châtelier’s favoured of a to product hydroxide ions, principle counteract will OH , also to this favour the the reaction + to form orange The water. in a The dichromate reverse increased reduction equilibrium reacts reaction or the in is to concentration mixture form favoured concentration + H if of becomes yellow the a a as OH paler reacts colour with as H more chromate. concentration reactant is of a product decreased. is The addition HCl results + of H the ions in reverse observed. the form reaction In both of concentrated being cases, favoured the value hydrochloric and of a deeper equilibrium acid, orange colour constant K in being remains c unchanged as long as the temperature remains the same. 185 7 E Q U I L I B R I U M Le Châtelier’s conditions an to principle equilibrium ammonia N (g) uses + the reaction. the 3H 2 The allows maximize (g) The yield nitrogen is ⇋ 2NH equilibrium of chemists the to desired process for manipulate product the reaction formed manufacture in of reaction: (g) 3 increased and Haber following 2 industrial amount by hydrogen, using and high concentrations removing the product of reactant ammonia gases from the mixture. Cause and eect “Cause and situation direct where develop or a second They which is this the rst and (the test that the collection either to the effect) event (the and suggests factors to refers concept that then used event the between false. disprove of hypothesis experimentation data, “causation” understand causation true a or consequence Scientists or effect” a a relationship predict of the not empirical modify, or of a effects demonstrate a changing change of in helping changes reaction and However, explanation the tool concentration constant. an on useful and equilibrium changes resultant hypothesis. an provide between is qualitative pressure, equilibrium not these by principle the temperature, does be hypothesis support, to Châtelier’s position often may of is cause). Le for equilibrium in on the the the the value principle effects reaction: it of does cause-and-effect relationship these and in the conditions equilibrium the system. The Haber process Over the past century, of agriculture have of the land and Earth’s the over 7 employment and being population billion food of people. sector security many, including World Trade advances resulted in in cultivated the planet developing remains of with is crops a to major countries great international science one-third increasing Agriculture in the over concern bodies such a process a large one of (WTO), the the the world. in fertilizers, manufacture to as Food Organization Organization (FAO), and the demand decade for demands for endeavours fertilizers. and of food increases More will and the the not water elds than onset with plastics, of the production exponentially nitric of 50 % be able without to meet HNO scientic biotechnology of current and used (aq) on nitrate Haber + this will the use of was 1918 (1868–1934) awarded for his was a German ammonia, the work Nobel on the Prize in NH from NO century wor l d w ar, mi li ta r y to H a be r a dva n c e NH its at elements. the A Ammon ia (g) → ammo nium re a c t s NH also NO 4 as a (aq) n i t ra t e , which is 3 fe rtil ize r. Ammo ni u m und e r g o ex pl os ive whe n d e to na te d: → N O(g) + 2H 2 O(g) 2 is sometimes warfare”, Haber referred to advancing as the the “father research of into utilization rst world of many war. He poisonous was not gases alone in during making shortage beginning prompted fo r m of of advances to in science with military the during this time. Gustav Hertz research and 186 rs t e x p l o s ive s . to day (s) applications twentieth the 3 signicant fertilizers and Chemistry synthesis 3 natural explosives, chemist the of the fertilizers. and in in food chemical who processes application used bres, Ge r m an of acid decomposition Haber Fritz also 3 to 4 relies is industrial major World NH production its on now (WHO). we food in as ammonia 3 every well is the with Global widespread As ammonia manufacture and the Health manufacture pharmaceuticals. worked Agriculture could Ammonia most in With Organization that scale. James Franck were physicists who devised 7. 1 investigations the atom mechanics see the in atomic an (for the Hahn 1944 his the for on course Bohr’s the in topic of Later in life range of awards and the founder was the he They 1925, in for of his on Prize. Franck, Haber’s and research developing were serve to while Their research illustrate understanding ethical to their receive work, and several discoveries. is how were who the while the Nature of goes the of on members Their of questions results utilize (NOS). scientic of intellectual to on stories Science advanced raising concept society all worked weapons. considerations scientic He team signicantly of heavy Hahn chemical Physics continued Peace atomic in ssion nominated Nobel of 12). Prize the Hertz, of quantum mechanics, Physics Nobel model for quantum book, Prize discovery nuclei. impressive Niels groundwork more Nobel achievements as the received for occasions support lay Physics awarded Otto to and E q u i l i b r i u m about some the of property scientic regarded age. The eect of pressure on reactions in the Stdy tp gas phase We have effect of position rather seen a When considering the eect how change of Le in Châtelier’s equilibrium. than the principle concentration A aqueous system phase of a that will can involves be be product affected used or to predict reactant substances by changes on of changes in pressure on a the reaction, you must refer to the the in the gaseous in applied moles of gaseous reactants or products in the reaction. For example: pressure: 4HCl(g) + O (g) ⇋ 2H 2 O(g) + 2Cl 2 C(s) + H (g) O(g) ⇌ H 2 2 (g) + CO(g) 2 In this heterogeneous reaction, In this reaction there are 5 moles of gas on the reactant side and 4 moles the solid carbon is not included on the product side. A change in pressure applied to the system will when considering the eect of result in a shift in the equilibrium position. If the pressure is increased, changes in pressure. Le Châtelier’s change. of the the N principle forward system. reverse In The In reaction Haber (g) + the 3H are decrease Such a same being 4 (g) side. ⇋ of A way, 2NH gas a favoured decrease in will to shift to reduce pressure reverse the will this pressure result in the (g) in on high pressure change equilibrium equilibrium 3 moles the the becomes favoured. 2 product that process: 2 there the says reaction the of the reactant pressure the K , if side and favour the only 2 moles forward of gas reaction, on to system. equilibrium constant, will the position will temperature not affect remains the value of constant. c Temperature and the equilibrium constant An understanding considering constant. N (g) the For + this the of thermodynamics changing the of a reaction temperature on is the required when equilibrium example: 3H 2 In of effect (g) ⇋ 2NH 2 reaction (g) ΔH = –92 kJ 3 energy can be considered a product and is released to the surroundings. 187 7 E Q U I L I B R I U M exothermic reaction TOK The work of Fritz Haber a had mixed outcomes for society. While his method of ygrene laitnetop ammonia production had a signicant and long-lasting eect on increasing food production for humanity, reactants ∆H his work with the German military in the manufacture products of explosives and the use of chlorine in chemical warfare brought into question reaction progress the role of scientists in ▲ Figure 7 The potential energy prole of an exothermic reaction society. The outcomes of scientic endeavours The reverse reaction is endothermic – it requires energy from the often have signicant surroundings. ethical implications. At equilibrium the forward and reverse reactions occur at equal rates Should scientists be held and there is no net change in energy. responsible for the way in which society utilizes their discoveries? For the exothermic equilibrium accordance move to change; This in with the the results the left, reaction, direction Le Châtelier’s favouring concentrations in a an decrease in increase that in the reactants, nitrogen the temperature consume principle, the of will to and equilibrium the extra will equilibrium minimize hydrogen constant the . effect the In position will K shift energy. will of the increase. Conversely, c a decrease in equilibrium the temperature to the concentration right, of for the exothermic favouring ammonia, the NH . reaction forward This will reaction results in an shift and increase 3 Table 4 summarizes the effects on the increasing in K . c the equilibrium system of changing temperature. Type of Change n eacton tepeate E poston E constant K c increase moves to the left, decreases favouring reactants exothec decrease moves to the right, increases favouring products increase moves to the right, increases favouring products endothec decrease moves to the left, favouring reactants ▲ 188 T able 4 The eects on the equilibrium system of a change in temperature decreases 7. 1 E q u i l i b r i u m The eect of a catalyst on equilibrium reactions The addition lowering a catalyst energy In a of the means to on catalyst that overcome reversible effect a activation both forward and position of greater the reaction, the the a energy lowered reactions alternative reaction proportion and equilibrium constant an In activation forward reverse the equilibrium a provides energy. increase will not of by to an and The equal and for have has rates is reaction, sufcient an of amount. there a completion, become energy reactions. change goes reactants barrier activation reverse pathway that no products. equal the The effect on the K c qck eston PCl (g) ⇋ PCl 5 (g) + Cl 3 (g) ∆H = +92.5 kJ 2 Predict and explain any shift in the equilibrium position when: a) the temperature of the system is decreased ) additional chlorine gas is injected into the system c) the pressure applied to the system is increased d) a catalyst is added. Reaction quotient If a system product to has not reached reactants will equilibrium, not equal K . the This ratio ratio of is concentration called the of reaction c quotient as it Q moves favoured to and this toward helps you equilibrium establish to determine and equilibrium the (table the direction progress of the of the reaction reaction that is 5). The concentration of products is greater than at equilibrium and the Q > K c reverse reaction is favoured until equilibrium is reached. The concentration of reactants is greater than at equilibrium and the Q < K c forward reaction is favoured until equilibrium is reached. The system is at equilibrium and the forward and reverse reactions Q = K c ▲ occur at equal rates. T able 5 The relationship between the reaction quotient Q and the equilibrium constant K c 189 7 E Q U I L I B R I U M Questions 1 The equation industry shown to for a reversible convert methane reaction to used hydrogen in Poston of is Vae of e e constant shifts towards the decreases below. A. CH (g) + H 4 O(g) ⇋ CO(g) + 3H 2 Δ = reactants (g) 2 +210 kJ shifts towards the increases B. reactants Which statement is always correct about shifts towards the thisreaction when equilibrium has decreases been C. products reached? shifts towards the A. The concentrations of methane and increases D. carbon products monoxide are equal. IB, B. The rate than C. The of the the rate amount forward of of the reaction reverse hydrogen is is of three times the The equation The value of ΔH for the reverse reaction IB, May kJ. oxides one of reversible nitrogen is reaction shown below: is N 210 for methane. involving D. 2003 reaction. 4 amount November greater O 2 [1] (g) ⇋ 2NO 4 Experimental 2006 (g) ΔH = +58 kJ 2 represented data on for the this reaction following can graph be (gure 8). 1.0 2 Sulfur dioxide and oxygen react to form sulfur product according to the 3 trioxide equilibrium: 0.8 (g) + O 2 How (g) ⇋ 2SO 2 are the md lom/noitartnecnoc 2SO (g) 3 amount of SO and the value 2 of the equilibrium affected A. The by an constant increase amount of SO in for the reaction pressure? and the value of 0.6 reactant 0.4 0.2 the 3 equilibrium constant both increase. 0 0 B. The amount of SO and the value of 2 4 6 8 10 the 3 time/min equilibrium constant both decrease. ▲ C. The amount of SO increases but Figure 8 the 3 valueof the equilibrium constant a) Write an expression for the equilibrium decreases. constant, K , for the reaction. Explain c D. The amount of SO increases but the 3 value of the equilibrium constant the signicance of the horizontal the lines graph. parts of does notchange. on the State what can be [1] deduced about the magnitude of K for the c IB, November 2007 reaction, b) 3 What and will the happen value of to the the position equilibrium of equilibrium temperature is increased in constant the Use Le and explain Use + Cl Le (g) ⇋ 2BrCl(g) ΔH = +14 kJ d) the the of to predict increasing position of the equilibrium. [2] Châtelier’s the effect principle of to predict increasing the and pressure State position and of explain equilibrium. the effects of [2] a catalyst 2 onthe the forward position value of K of and IB, . November reverse equilibrium, reactions, and on on the [6] c 190 principle effect [1] on (g) on [4] when reaction? 2 the reason. following explain Br a Châtelier’s temperature c) the giving 2005 8 A C I D S A N D B A S E S Introduction The theories applications and of the of Nature scientists and of laid development acids acids of bases. bases as bases the The This of that donors for work a and the acids acceptors, of are acid The of measure acids of examines properties ions the describes and respectively. their scientists de nitions chapter theory of early foundations range proton and work Science. the a Brønsted–Lowry and and exemplify of the all general acids the and a the topic. reactions bases, characteristics explored threat the concentration deposition, age and to a in of of of scale as hydronium strong detail. product the and pH The the and weak chemistry industrial environment, concludes 8.1 To o Understandings Applications and skills + ➔ A Brønsted–Lowry acid is a proton/H donor ➔ Deduction of the Brønsted–Lowry acid and + and a Brønsted–Lowry base is a proton/H base in a chemical reaction. acceptor. ➔ ➔ Deduction of the conjugate acid or conjugate Amphiprotic species can act as both base in a chemical reaction. Brønsted–Lowry acids and bases. ➔ A pair of species diering by a single proton is called a conjugate acid–base pair. Nature of science ➔ Falsication of theories – HCN altering the theory of that oxygen was the element which gave a false. compound its acidic proper ties allowed for other acid–base theories to develop. ➔ Theories theory of being super s ed ed acidity derived – f rom ➔ a sour taste, but this ha s been proven Public understanding of science – outside of the arena of chemistry, decisions are sometimes one the ea rly referred to as “acid test” or “litmus test”. s ensation 191 8 A C I D S A N D B A S E S The role of acids and bases The process of to Acids and bases are familiar in our everyday lives and have a signicant refers to the heating of materials role in chemistry. For many hundreds of years scientists have been to very high temperatures in investigating acid-base reactions and developing a range of denitions air in order to bring about their and theories that will be discussed in this topic and topic 18. Each of thermal decomposition (in the these theories has its strengths and weaknesses and some of the earliest case of limestone), the removal ideas about acids and bases have now been disproved. of water from a hydrated compound (for bauxite), or the It removal of a volatile matter from The has long minerals and ores. associate word a (ethanoic acid). A derived been acid understood is sour derived taste acid), with lemon base that from the is that from many juice soluble Arabic acids the or in word and Latin bases acidus substances grapefruit water is al-qaly behave that (citric called as meaning are meaning acidic acid), an and alkali. opposites. sour sour The calcined and such we as milk word still vinegar (lactic alkali is ashes. Early theories about acids Antoine believe Lavoisier’s that reactions, source of oxygene that was their had present acidic previously Lavoisier’s disproving However, acidic oxygen integral in the his in been work that led acids and He Greek) was the subsequently by theory even the from Joseph developed time presence the by inuenced by differences explain The being by and a served the of new laws, on and to used to propose leads Modern trying Evidence is theories disproved, theory. focus scientic observation these or bases, important and community supported, bases and phenomenon. and testing cultures acids an curiosity generalize scientic to about over to or existing even theories explain on why are they sometimes the replaced globe to theories, theories of disproved, generating hypotheses. 1.1). disproved. around in specic theories experimentation acids Theories later develop in vocabulary Early endeavour characteristic the if purpose name element (sub-topic to of languages. the fundamental due meaning to was gave to properties were him combustion discovered phlogiston belief compounds was all to properties. (“acid-forming” Priestly. of investigations oxygen, in react in the way they do. the Arrhenius’s theory of acids and bases Svante August Arrhenius Chemistry in 1903 for an as a substance alkali, a soluble (1859–1927) his work in the was eld awarded of acids the and Nobel bases. Prize He in dened + acid An of an the acid and base combination of that base, is in produces well the ionizes known hydrogen water to hydroxide as a ion produce ions, OH neutralization and the hydrogen . The H involving ion. + H (aq) + OH (aq) → H O(l) 2 An example acid in the of this type stomach of with neutralization aluminium is the hydroxide reaction of contained hydrochloric in an antacid tablet: 3HCl(aq) + Al(OH) (s) → AlCl 3 Arrhenius’s Figure 1 The reaction between the base vapours of concentrated ammonia explained, theory ammonia as and had its limitations. hydrogen ammonia does chloride not and hydrogen chloride solutions NH produces solid ammonium chloride, visible as a white smoke 192 (g) 3 + HCl(g) → (aq) + 3H 3 NH Cl(s) 4 O(l) 2 The gas contain reaction (gure between 1) hydroxide could ions. . combination reaction hydroxide ions, the not weak be 8 . 1 T h e O r i e s O f a c i d s a n d b a s e s TOK Science in society Terminology community, often have general the say serves is as often life, is life as the test add of test” Public decisions of meaning by a or specic “litmus event ability to is within everyday or a to of as understanding of scientic associated at nal science ndings vital hypotheses is obtained in the teacher laboratory through observation in data in and experimentation. One Science everyday assumption in this process that is that scientists are able to uninuenced is understanding. Evidence to test with a university.” and to develop knowledge and in examination situations and variety of methodologies can example, providing experimentation, about is the study perceived For Scientists employ a wide scientic phenomenon: in myriad the understanding, life. test” performance their process in some credibility supported bias. informed “acid student’s scientic rigorous, human “the to meaning certainty litmus used accepted communicating term the that the an different of a has clearly a testing might that recreate conditions in the by laboratory that accurately making represent what is occurring in issues. the universe outside. How then is this methodology used in chemistry dierent from the Brønsted–Lowry acids and bases Scientists of often information subject of the sometimes idea work and ideas research. discovering simultaneously. Thomas Lowry Arrhenius’s Referring could be collaboratively, that In and This leads other developed participating to a cases better a the case denition in work theorizing when of an open understanding scientists subsequently was methodologies employed in and of the independently, about Johannes acids other areas of knowledge? exchange the same Brønsted bases that and broadened theory. to a hydrogen dened as a ion as proton a proton, donor and they a proposed base as a that proton an acid acceptor. + H In an aqueous solution a proton can be represented + hydrogen ion, as either O the + H or as the hydronium ion, H O . The hydronium O 3 H ionis formed when a water molecule forms a coordinate bond H with + H H aproton (gure 2). For H example: + + HCl(aq) + H O(l) → H 2 Common acids hydrochloric are often acid), phosphoric acid). phosphoric acid O (aq) + Cl Figure 2 The hydronium ion, H (aq) referred diprotic (such Hydrochloric is a weak O 3 3 to as as and being sulfuric sulfuric monoprotic acid), acids or (such triprotic are strong as (such acids as while acid. + HCl(aq) → H (aq) + Cl (aq) + H SO 2 (aq) → 2H 2 (aq) + SO 4 (aq) 4 + H PO 3 (aq) ⇋ 3H 3 (aq) + PO 4 Ethanoic (aq) 4 acid, CH COOH is also a weak acid: 3 + CH COOH(aq) ⇋ CH 3 COO (aq) + H (aq) 3 or + CH COOH(aq) + H 3 In the and last water reaction, O(l) ⇋ CH 2 equilibrium is the acting as ethanoic a ethanoate COO + H acid CH O (aq) 3 is acting Brønsted–Lowry ion, (aq) 3 COO is as base. a Brønsted–Lowry Focusing acting as a on proton the acid reverse acceptor and 3 193 8 A C I D S A N D B A S E S the hydronium A to is assumed to base undergo complete dissociation ion of is the the ion as a proton Brønsted–Lowry conjugate acid donor. acid of The ethanoate (ethanoic another acid), ion while Brønsted–Lowry is the the conjugate hydronium base, water. in water (sub-topic 8.4). The conjugate acid and base differ from one another by a single proton. This For example, in hydrogen is termed a conjugate acid–base pair. Figure 3 shows another example. chloride, HCl the hydrogen ion has almost no anity for the conjugate base 1 chloride ion. A wk undergoes only par tial dissociation in water, - - H CO 2 establishing an equilibrium, (aq) + OH (aq) ⇋ HCO 3 (aq) + H 3 O(l) 2 and a solution of a weak acid is only a weak electrolyte. conjugate acid 2 Figure 3 Conjugate acid–base pairs in the neutralization of carbonic acid (Brønsted–Lowry acid) with a hydroxide ion (Brønsted–Lowry base) Quk quto 1 Identify a) H the bases of the SO 2 b) conjugate following d) C 4 H 6 HNO acids. OH 5 e) OH f) H 3 c) C H 2 2 OH 5 Identify O 2 the conjugate acids of the following bases. 2 a) OH d) CO b) H e) HNO c) NH f) C 3 O 2 3 3 3 H 2 Identify the conjugate acid–base pairs NH 5 2 HCO (aq) + 2 in the following equations: 2 S (aq) ⇋ HS (aq) + CO 3 (aq) 3 2 CH COOH(aq) + HPO 3 (aq) ⇋ CH 4 COO (aq) + H 3 PO 2 (aq) 4 Amphiprotic species Some a substances part. These molecule Lowry acid. N 3 are donate It can the base species can Polyprotic + H have Brønsted–Lowry said a also species are ability to depending to be proton accept in a act on as the either amphiprotic. a reaction, proton, frequently a Brønsted–Lowry reaction involved For thus acting in in which example, acting as a they as a the in or taking water Brønsted– Brønsted–Lowry reactions acid are which base. they - CH COO behave amphiprotically for example: 2 H PO 2 (aq) + OH (aq) ⇋ HPO 4 (aq) + H 4 O(l) 2 + H PO 2 R Amino acids (aq) + H 4 O (aq) ⇋ H 3 acids (sub-topic contain a PO 3 weakly B.2) also acidic (aq) + 4 act as carboxyl H O(l) 2 amphiprotic group and a species. weakly All 2-amino basic amino zwitterion group. Figure 4 A 2-amino acid is amphiprotic as an In acid presence 194 the in of ionized the a form presence strong acid (a of it zwitterion, a strong acts as a gure base, base 4) the donating and accepts a a compound proton. proton. In acts the 8 . 2 p r O p e r T i e s O f a c i d s a n d b a s e s 8.2 po t o Understandings Applications and skills ➔ Most acids have obser vable characteristic Balancing chemical equations for the reactions ➔ chemical reactions with reactive metals, metal of acids. oxides, metal hydroxides, hydrogencarbonates, Identication of the acid and base needed to ➔ and carbonates. make dierent salts. ➔ Salt and water are produced in exothermic Candidates should have experience of acid– ➔ neutralization reactions. base titrations with dierent indicators. Nature of science ➔ Obtaining evidence for theories – obser vable proper ties of acids and bases have led to the modication of acid–base theories. Acid–base theories For scientists, provide refute sense experimentation evidence the of that theories our can we and either have observation support formulated has or to come and make world. have about been the data, rationalizing reactions modi ed over of time acids as and more observed Scientists quantitative de ning Theories from bases. properties analyse establishing discrepancies of acids qualitative patterns with the and and goal of relationships. bases evidence Properties of acids and bases Acids and bases (concentrated block domestic remover, perform sodium and changing many useful hydroxide) commercial iron(III) functions dissolves drains. oxide (Fe . Ammonia is used as a general daily and Phosphoric O 2 FePO in grease , rust) acid into life. oil is Caustic deposits an iron(III) soda that effective can rust phosphate, 3 cleaner while mild acids such as 4 vinegar Table ▲ 1 are sometimes shows some put on wasp properties of stings, acids which and are alkaline. bases. a b taste sour taste bitter pH < 7.0 pH > 7.0 litmus is red litmus is blue phenolphthalein is colourless phenolphthalein is pink methyl orange is red methyl orange is yellow T able 1 General proper ties of acids and bases and their effects on some common indicators 195 8 A C I D S A N D B A S E S The reactions of acids with metals, bases, Tt o yo and carbonates Putting a lighted splint in Most acids react with metals, metal oxides, hydroxides, hydrogencarbonates, hydrogen gas results in a and carbonates. combustion reaction, and a “squeaky pop” is heard. Hydrogen gas is highly ammable. 2H (g) + O 2 All as these reactions “common compound (g) → 2H 2 salt” produce but this composed of a is an salt. just Sodium one anion chloride example and of a is referred salt, which to is a cation. O(g) 2 It is important to understand how a salt is derived from an acid and a + energy base. The following reactions illustrate the formation of a wide variety ofsalts. Metals react that with are found acids to above form acid + 2HCl(aq) a hydrogen salt and metal + → Zn(s) in salt → the hydrogen + ZnCl activity SO 2 (aq) 2CH (aq) + Fe(s) → + H These COOH(l) The salt + Figure 1 Hydrogen gas with of give the 2Na(s) off metal produced example, (aq) + H ⇋ 2CH (g) 2 COONa(l) + H 3 reactions reactivity (g) 4 3 hydrogen and the depends magnesium hydrochloric on gas the is at acid a (g) 2 strength chloride 9.1) 2 FeSO 4 (sub-topic hydrogen 2 H series gas: different and from chloride rates according concentration which salt it was of the to the acid. produced, produced in a for reaction acid. explodes upon ignition The standard change reaction with between conditions. an enthalpy associated This exothermic a change the strong enthalpy process of neutralization formation acid and change (sub-topic a has of 1 mol strong a of base negative is water under value the energy from the standard – neutralization 5.1): + H (aq) + OH (aq) → H O(l) 2 The salt from produced the include which base and metal is a in neutralization an anion hydroxides, weak from metal reactions the acid. oxides, 2HCl(aq) + SO 2 + base Ca(OH) (aq) → salt → CaCl (aq) COOH(aq) dissolves + CaO(s) → in a cation of bases hydroxide, water (aq) + 2H + NH OH(aq) ⇋ CH not to (s) + H directly an COONH with alkaline O(l) 2 3 react create O(l) 2 4 does water + CaSO 4 oxide of examples ammonium 2 4 3 Calcium Common and 2 CH composed base: acid H is (aq) + H 4 aqueous solution of O(l) 2 acids. This calcium base hydroxide, Figure 2 Zinc reacting with which neutralizes the acid: hydrochloric acid CaO(s) + H O(l) → Ca(OH) 2 Many metal oxides act as bases Calcium an hydroxide alkali. Many (aq) 2 is slightly other bases, soluble in such iron(II) in aqueous solutions whereas hydroxide, most non-metal oxides are acidic (sub-topic 3.2). 196 are insoluble in water. as water. A soluble hydroxide base or is called aluminium is 8 . 3 Metal carbonates carbon acid dioxide + metal carbon and and hydrogencarbonates with acids to + p h s c a l e produce water: Tt o o ox carbonate/metal dioxide react T h e hydrogencarbonate → salt Passing carbon dioxide gas + through limewater (calcium water hydroxide) results in a cloudy 2HCl(aq) + Na CO 2 HCl(aq) + (s) → 2NaCl(aq) + CO 3 NaHCO (g) + H 2 (s) → NaCl(aq) + CO 3 O(l) 2 (g) + H 2 (milky) suspension of insoluble O(l) calcium carbonate, CaCO 2 : 3 Ca(OH) (aq) + CO 2 (g) → CaCO 2 (s) 3 Acid–base titrations + H O(l) 2 A titration involves with a a a progress loggers into of point the the to 1.3) a volumetric between a substance solution solution reaction combined plotted colour reaction is standardized burette be (sub-topic with produce change as (sub-topic the a being can a pH pH (the be of titrant). monitored curve. can An be in titrant small using used and that concentration is delivered increments. several to acid–base approaches technique unknown The analysed, probe titration analysis techniques. collect indicator reaches from The data that undergoes the Data can a equivalence 18.3). Figure 3 Limewater provides a test for carbon dioxide The colour changes of dierent indicators can be found in section 22 of the Data booklet 8.3 T h Understandings Applications and skills + ➔ ➔ pH = -log [H + (aq)] and [H pH ] = 10 + ➔ Solving problems involving pH, [H ], and [OH ➔ Students should be familiar with the use of a ]. A change of one pH unit represents a 10-fold + change in the hydrogen ion concentration [H ]. pH meter and universal indicator. ➔ pH values distinguish between acidic, neutral, and alkaline solutions. + ➔ The ionic product constant, K = [H ][OH Nature of science ] = w 14 10 at 298 K . ➔ Occam’s razor – the pH scale is an attempt to scale the relative acidity over a wide range of + H concentrations into a very simple number. 197 8 A C I D S A N D B A S E S Occam’s razor Scientic theories can be comprehensive models works. from Derived of complex how while and the analysis, and hypothesis, they elegant and also high capacity for gaining pH scale is a very effective method of can representing be a experimentation, The observations, maintaining understanding. universe a continuous range of hydrogen ion imposing. + concentration The for of principle the Occam’s development elds theory of of of knowledge. should remain razor is theories Its as a in blueprint a number philosophy simple as is ease is that and a of ] as simple interpretation the general distinguishes possible [H for basic/alkaline both population. between numbers students The acids, that pH of scale neutral enable science clearly solutions, and solutions. The pH scale The pH scale is a simple and effective way of representing the + concentration of hydrogen ions, [H ] in a solution. This concentration + is often very low; for example, in water [H 7 ] = 1.0 × 10 3 mol dm . + Comparing result in values ratios that + [H ∶ [H 1 Scientists : 0.000 oven employ treatment and pH for different difcult to substances to water can comprehend: of a cleaner a number to ion of of is and dened use results valid by mathematical The concentrations non-scientists solution different data. the for = -log [H of a in a following approaches logarithmic simple scientists + pH relative 001 presentation hydrogen accessible The are ] ] water is [H + ] display of visual (gure two scale to the to scale that 1). expressions: + (aq)] or pH = -log [H O (aq)] 3 + [H pH ] = 10 neutral acidic pH 0 1 2 alkaline 3 4 5 6 7 gastric pure juice water 8 9 10 11 12 13 14 oven detergent cleaner Figure 1 The pH scale h + [h ] As 1 the pH scale equivalent 2 7 10 14 a logarithmic 10-fold change scale in to the base 10, hydrogen a change ion of 1 pH unit concentration. A is small in the pH of a solution represents a large change in the hydrogen 5 ion 1 × 10 concentration (table 1). Note that the pH scale has no units. 7 1 × 10 10 1 × 10 Calculating pH 14 1 × 10 T able 1 pH values and their corresponding hydrogen ion concentrations 198 a 2 Strong acids aqueous ▲ to 1 × 10 change 5 is 1 1 × 10 and strong solutions. bases Therefore are the assumed to completely concentration of a ionize strong in monoprotic 8 . 3 acid will mol dm be the same as the concentration of the hydrogen ion. A T h e p h s c a l e 0.1 TOK 3 + solution of hydrochloric acid equates to [H 3 ] = 0.1 mol dm The language of mathematics is integral to the development Ionization of water of scientic theories based on The pH scale covers both the acidic and alkaline regions of aqueous the analysis of quantitative systems. When we to considering solutions involving strong and weak bases, data. Chemists use the need examine the relationship between hydrogen ion and language of mathematics as a hydroxide ion concentrations. Water undergo means of communicating their can equilibrium auto-ionization, expression (sub-topic according to the following ndings. It is well accepted 7.1): that language plays a pivotal role in the communication of + H O(l) ⇋ H (aq) + OH (aq) 2 knowledge. Language not only + [H ][OH ] _ K allows us to form descriptions = c [H O] of our experiences, it can also 2 + as [H O] is constant, K 2 = [H 14 ][OH ] = 1.0 × 10 at 298 inuence the way we view K. w and subsequently understand This expression is the ion product constant for water. In pure water, our interactions with the _________ + [H 14 ] = [OH ] √ = 1.0 × 7 10 = 1.0 × universe. For what reasons is 10 it important to have only one “scientic” language? Worked examples: calculating pH Example 1 + A solution of fresh milk has a pH of 6.70. Calculate [H ] and [OH ]. Solution + [H pH ] = 10 6.70 = 7 10 = 2.0 × 10 + K = [H × 10 3 mol dm 14 ][OH ] = 1.0 w 14 14 1.0 × 10 __ [OH ] = 1.0 × 10 __ = + 7 [H ] × 10 2.0 × 8 = 5.0 10 3 mol dm Example 2 2 Calculate the pH of a 1.0 × 3 10 mol dm solution of sodium hydroxide. Solution Sodium hydroxide is a strong base that completely ionizes in water: + NaOH (aq) → Na (aq) + OH 2 [OH ] = 1.0 × 10 3 mol dm + K = [H (aq) 14 ][OH ] = 1.0 × 10 w 14 1.0 × 10 __ + [H ] = [OH ] 14 1.0 × 10 __ + [H ] = 12 = 1.0 × 10 3 mol dm 2 1.0 × 10 + pH = -log [H (aq)] = -log (1.0 12 × 10 ) = 12.00 199 8 a c i d s a n d b a s e s pH and acid–base titrations The analytical laboratory by the for technique the addition illustrates the of past an features data be (gure can titration years. indicator progress characteristic that of 200 of of an the collected (sub-topic Traditionally (sub-topic acid–base titration. using a pH 1.3) a 18.3). titration These probe been Plotting and curves and has titration its is a pH enables are used in the monitored curve analysis generated associated of from software 2). Figure 2 A pH probe can be used to collect data during an acid–base titration 8.4 sto wk Understandings Applications and skills ➔ Strong and weak acids and bases dier in the ➔ Distinction between strong and weak acids and extent of ionization. bases in terms of the rates of their reactions ➔ Strong acids and bases of equal concentrations have higher conductivities than weak acids and bases. ➔ A strong acid is a good proton donor and has a with metals, metal oxides, metal hydroxides, metal hydrogencarbonates, and metal carbonates, and their electrical conductivities for solutions of equal concentrations. weak conjugate base. ➔ A strong base is a good proton acceptor and has a weak conjugate acid. Nature of science ➔ Improved instrumentation – the use of advanced analytical techniques has allowed the relative strengths of dierent acids and bases to be quantied. ➔ Looking for trends and discrepancies – patterns and anomalies in relative strengths of acids and bases can be explained at the molecular level. ➔ The outcomes of experiments or models may be used as fur ther evidence for a claim – data for a par ticular type of reaction suppor ts the idea that weak acids exist in equilibrium. 200 8 . 4 s T r O n g a n d W e a K a c i d s a n d b a s e s Predictions, patterns, and anomalies Advances in techniques computing (including between research this looking data gain greater of acids to be strengths institutes for bases with of allows high different chemical behaviour evidence also Science trends and recognizes modern certainty. Science strengthened be by developed idea there an that are in allow at networking Scientists patterns on the and chemical anomalies molecular acids provides human exist in all data to strengths behaviour in the of relative their level. in analyse anomalies relative explanations the weak analytical and data. their and limitations inherently advances of data about bases of meters), wealth Patterns instrumentation is pH Quantitative and the that a predictions acids to development as discrepancies, reliability. supports however, the such provide understanding. and made power, sensors Empirical equilibrium. measurements; that is endeavour close which to is technology. Strengths of acids and bases Sulfuric acid is a strong acid. The strength of an acid or base depends on the degree to which it ionizes However, the rst ionization or dissociates in water. A strong acid is an effective proton donor is complete but the second that is assumed to completely dissociate in water. Examples include dissociation is incomplete so hydrochloric acid, HCl, sulfuric acid, H SO 2 , and nitric acid, HNO 4 : 3 an equilibrium sign is used: + HCl(aq) + H O(l) → H 2 O (aq) + Cl (aq) H 3 SO 2 (aq) + H 4 O(l) → 2 + HSO + H SO 2 (aq) + H 4 O(l) → H 2 O (aq) + HSO 3 (aq) (aq) + H 4 O (aq) 3 4 + HNO (aq) + H 3 O(l) → H 2 O (aq) + NO 3 HSO (aq) (aq) + H 4 3 O(l) ⇋ 2 2 SO These to go base. reactions to represented completion. For which are The hydrochloric has almost no by chemical conjugate acid, the afnity base of conjugate for a proton a equations strong base in is that acid the aqueous is are a very chloride + (aq) + H 4 assumed O (aq) 3 weak ion, Cl solution. stuy t A weak donor. acid The dissociation equilibrium acid dissociates (sub-topic molecules higher have afnity for a only of a partially weak 7.1). At is water; a than The only conjugate does the it is reversible equilibrium dissociated. proton acid in a a poor reaction small base conjugate proton of that reaches proportion a weak base of a acid of the has strong a acid. The terms “ionization” and “dissociation” are interchangeable and are both equally acceptable in examination answers. + CH COOH(aq) + H 3 O(l) ⇋ H 2 O (aq) + CH 3 COO (aq) 3 + H CO 2 In the reactions accepting The (aq) terms distinct + H 3 a O(l) ⇋ H 2 of both (aq) + HCO strong and (aq) 3 weak acids, water is acting as a base, proton. “strong” from and “weak” “concentrated” when and cott sto O 3 applied “dilute”. to Table acids 1 bases some are quite examples. dut 3 6 mol dm and gives 3 HCl 0.5 mol dm HNO 3 3 Wk 10 mol dm 3 CH 3 COOH 0.1 mol dm H 2 CO 3 T able 1 Some examples of strong, weak, dilute, and concentrated acid solutions 201 8 A C I D S A N D B A S E S A amot: Species that strong base hydroxides are also all completely soluble in dissociates water and in are water. good The group examples of 1 metal strong bases: behaves both as an acid and a + NaOH(aq) → Na (aq) + OH (aq) base eg aluminium hydroxide: + KOH(aq) → K (aq) + OH (aq) Acting as a base: A Al(OH) metal hydroxide does not act as a Brønsted–Lowry base because it (s) + 3HCl(aq) → 3 does AlCl (aq) + 3H 3 not have the capacity to accept a proton. However, in solution the O(l) 2 hydroxide ion acts as a Brønsted–Lowry base, accepting a proton: Acting as an acid: + OH (aq) + H O (aq) → 2H 3 Al(OH) O(l) 2 (s) + NaOH(aq) → 3 Ammonia Na[Al(OH) is an example of a weak base. In the reaction with water, ](aq) 4 ammonia accepts a proton and effectively undergoes ionization. amot: One type of + NH amphoteric species are amphiprotic molecules. These can act as Brønsted-Lowry acids (aq) + H 3 In this O(l) ⇋ NH 2 reaction water Brønsted–Lowry (aq) + OH (aq) 4 displays acid, its donating amphiprotic a nature by acting as a proton. (proton donors) or Brønsted- Lowry bases (proton acceptors) eg water and amino acids. Experimental determination of the strength of acids and bases A number of techniques concentration, one so that can they be can used be to compare assigned an acids order of and bases strength of equal relative to another. Conductivity All acids and conductivity present. and graphite can current passing they the acids a be be through and a degree in a connected identical bases higher to solution measured for of are of water to an on and the experiment ammeter solution so solutions using (gure that of create ions. The concentration any different a 1). of power The voltage difference acids or ions pack in bases ions. strong conductivity comparison in depends simple each aqueous concentration display example, dissociate aqueous electrodes must Strong strong acid. The lamp gives a qualitative This an applied reects Figure 1 Testing the conductivity of a bases of the electrolytes than weak conductivity (sub-topic acids of and 19.1), bases. equimolar so For solutions reading of current; an ammeter would of hydrochloric acid and ethanoic acid would demonstrate that the provide quantitative readings hydrochloric higher weak acid degree of gives a higher dissociation ammeter than reading ethanoic acid, and so which has is a a acid. Energy changes on neutralization Neutralization reaction is exothermic neutralization The strong enthalpy water 202 for a neutralization reaction, A occurs so or (ΔH strong the base is consideration from hydrogen < an 0, acid reaction driving acid when acid sub-topic is almost removes reaction to completely in and this and a base 5.1). react The identical ionized to species together. enthalpy that for from change a the The weak of acid. dissociation completion. dissociated reaction hydroxide is the ions. in solution exothermic so the only formation of 8 . 4 For a neutralization reaction involving a weak s T r O n g acid or a n d base W e a K there a c i d s a n d b a s e s are TOK other enthalpy undissociated considerations. forms in These aqueous species solution. exist The predominantly ionization of a in weak their acid We use our senses – sight, or base is mildly endothermic. Therefore the enthalpy of neutralization touch, smell, taste, and for a strong base–weak acid reaction will be slightly less exothermic than hearing – to perceive the that for a strong base–strong acid reaction. The weaker the acid, the world and develop our more endothermic the dissociation reaction becomes and thus the lower understanding. Technology the enthalpy change of neutralization. allows us to extend our HCl(aq) + NaOH(aq) → NaCl(aq) + H senses, unveiling new O(l) 2 1 ΔH = −57.1 kJ knowledge and experiences mol neutralization that may challenge the boundaries of our current HCl(aq) + NH (aq) ⇋ NH 3 Cl(aq) 4 understanding. How does 1 ΔH = −53.4 kJ mol neutralization this increased sensory perception aect our view CH COOH(aq) + NaOH(aq) → CH 3 COONa(aq) + H 3 O(l) 2 of the world? How might 1 ΔH = −56.1 kJ mol neutralization reasoning based on evidence and discussion help us to CH COOH(aq) + NH 3 (aq) ⇋ CH 3 COONH 3 (aq) decide if this new knowledge 4 1 ΔH = −50.4 kJ mol changes or reinforces our neutralization view of the world? Monitoring the rate of a reaction 3 The reactions metals, metal carbonates all of strong produce (topic through observation and weak acids hydrogencarbonates, reaction analysis and 6) can a be gas. The and rate determined (gure quantitatively 2) by The with metal of of the by which gas is evolved monitoring through loss of hydrogencarbonate of the mass. reaction shown of such experiments enables a strong distinguished from a weak 1 mol ethanoic sodium acids strong solutions with carbonate, demonstrate by dm and the zinc and sodium different weak rates acids. rate acid 3) the reactions enables mass on data an to electronic be balance collected over to time. be of A (gure series and powdered Performing at reactions hydrochloric granules, qualitatively followed individual Graphing these results illustrates the differences acid. in the initial rate of reaction (sub-topic 16.1). Figure 2 Obser vation provides qualitative Figure 3 Monitoring the loss of mass provides data for a reaction that evolves a gas quantitative data for a reaction that evolves a gas 203 8 a c i d s a n d b a s e s 8.5 a oto Understandings Applications and skills ➔ Rain is naturally acidic because of dissolved Balancing the equations that describe the ➔ CO and has a pH of 5.6. Acid deposition has a 2 combustion of sulfur and nitrogen to their pH below 5.6. oxides and the subsequent formation of H SO 2 ➔ Acid deposition is formed when nitrogen or H SO 2 sulfur oxides dissolve in water to form HNO , H 2 ➔ SO 2 and H 4 4 , 3 and HNO 2 3 , 3 HNO , HNO Distinction between the pre-combustion and ➔ SO 2 3 post-combustion methods of reducing sulfur Sources of the oxides of sulfur and nitrogen oxides emissions. and the eects of acid deposition should be Deduction of acid deposition equations for acid ➔ covered. deposition with reactive metals and carbonates. Nature of science ➔ Risks and problems – oxides of metals and non-metals can be characterized by their acid–base proper ties. ➔ Acid deposition is a topic that can be discussed from dierent perspectives. ➔ Chemistry allows us to understand and reduce the environmental impact of human activities. Acid deposition Acid are deposition deposited on is the the economicdevelopment increasing emissions acidrain, Acid the deposition deforestation, acid levels plants, of most toxic shing in metals industry marble, ofthe the and in by many the and of of and of rivers, pH the of metal in other world ledto that ways. soils of and rapidly cause health, from can soil by uptake affect corrosive and include: elevated increased which and bridges, to minerals systems, life These leading toxic marine people’s oxides many river buildings, have deposition. from and pollutants industrialization sulfur uptake lake and the acid minerals ultimately and parts form acid-forming Increased environment shellsh limestone, which nitrogen leaching lakes reduction in by surface. prevalent affects the process Earth’s the effects on vehicles. Acid rain Pure water presence H CO 2 . A has of a pH of dissolved typical pH 7.0. Rainwater carbon value of dioxide is naturally which rainwater is forms 5.6. 3 CO (g) + H 2 O(l) ⇋ H 2 CO 2 (aq) 3 + H CO 2 (aq) ⇋ H (aq) + HCO 3 HCO (aq) 3 204 (aq) 3 + ⇋ H (aq) 2 + CO (aq) 3 acidic weak due to the carbonic acid, 8 . 5 Acid rain has deposition a are pH less sulfur than 5.6. dioxide, The SO major and pollutants nitrogen that oxides, cause NO and and are the products of natural decomposition of d e p O s i T i O n acid NO 2 These a c i d . a oto: a 2 occurrences vegetation, as such well as as volcanic eruptions man-made o om primary Acid deposition, a secondary pollutants from the combustion of fossil fuels containing high levels pollutant, can take many of sulfur impurities (optionC.2). Acid rain results principally from the dierent forms including rain, formation of two strong acids, nitric acid, HNO , and sulfuric acid, H 3 SO 2 4 snow, fog and dry dust. The and can be considered as a major global environmental problem. components of acid deposition For a example, car or oxide, a jet at high engine, nitrogen(II) temperature nitrogen oxide gas in the reacts (nitrogen internal with combustion oxygen gas to engine form of the (the primary pollutants) may be generated in one country monoxide): and depending on climate patterns may be deposited High Temperature in neighbouring countries or N (g) + O 2 (g) → 2NO(g) even dierent continents. 2 There are no boundaries On reaction with oxygen, the oxide, nitrogen(IV) oxide (nitrogen for acid deposition. For dioxide) can form: example in Europe industrial 2NO(g) + O (g) → 2NO 2 Nitrogen(IV) air pollution USA and (g) conurbations in countries 2 oxide causes which Mexico is the often brown observed colour in of cities smog, such as a common Los type Angeles in of the such as Germany and the UK may act as the source of acid rain but due to factors City. such as prevailing winds, The reaction and nitrous between water and nitrogen(IV) oxide produces nitric acid acid deposition may occur in acid: Scandinavian countries fur ther 2NO (g) + H 2 O(l) → HNO 2 (aq) + HNO 3 nor th such as Norway and (aq) 2 Sweden. Hence the eects Another oxide of nitrogen, NO, is easily oxidized to nitrogen(IV) oxide of acid rain may occur away by atmospheric oxygen: from the actual source leading 2NO(g) + O (g) → 2NO 2 (g) to widespread deforestation 2 and pollution of lakes and Nitrous acid can be also oxidized by atmospheric oxygen: river systems. National and 2HNO (aq) + O 2 Therefore, (g) → 2HNO 2 all oxides (g) regional environmental 3 of nitrogen eventually produce nitric acid, protection agencies throughout HNO 3 the world collaborate in an Sulfur dioxide combines with water to form sulfurous acid: eor t to better understand SO (g) + H 2 O(l) ⇋ H 2 SO 2 and control acid deposition. (aq) 3 The US Environmental + H SO 2 (aq) + H 3 O(l) ⇋ HSO 2 (aq) + H 3 O (aq) 3 Protection Agency and the Acid Some coal can contain almost 3% sulfur. On combustion, sulfur dioxide forms: Deposition Monitoring Network in East Asia (EANET) websites provide data that can be used S(s) + O (g) → SO 2 (g) 2 in the discussion of secondary On subsequent sulfur trioxide 2SO (g) + reaction is O 2 Sulfur SO oxygen in the atmosphere, the oxide, generated: (g) ⇋ 2SO 2 trioxide sulfuric with can pollutants and their political implications. (g) 3 then react with rain in the atmosphere, to form acid: (g) 3 + H O(l) 2 → H SO 2 (aq) 4 205 8 A C I D S A N D B A S E S Pre- and post-combustion technologies Pre-combustion techniques used methods on fuels mineral beneciation reduces the different of amounts technologies dioxins and CaO(s) to their sulfur and methods emissions combustion. crushing sulfur methods remove from atmosphere. dioxide of reduce coal, other followed impurities. result in the refer Physical by to cleaning otation Combinations removal of up to or that of 80–90 % sulfur. Post-combustion and involves pre-combustion inorganic to before For the + SO (g) it → ue CaSO 2 several complementary nitrogen gases calcium from on dioxide, combustion example, remove focus sulfur before oxide or oxides, they lime are will heavy metals released react with into the sulfur gases: (s) 3 The eects of acid rain on buildings Limestone are of and commonly signicant the world. marble used in cultural Both are building monuments importance contain calcium materials and that buildings throughout When calcium neutralization gradually eroded, (s) + H 3 only in their structural is exposed occurs causing and to the signicant acid rain, building a is damage. carbonate, CaCO differing carbonate reaction SO 2 (aq) → CaSO 4 (s) + 4 CO (g) + H 2 O(l) 2 composition. The role of chemists in studying acid deposition Science, to and chemistry understand occurs and environment. is wide interactions 206 Acid – which its of cross G: particular, in of study and option Environmental 5.8: ways extent The ranging Geography Global the the in acid impact this on and The societies interest in reects the deposition about the evidence HL– change; and sub-topic this acid decisions subject to reduction of deposition. on economic needed concern that deposition (IB environments; environmental systems us phenomenon disciplinary Urban deposition). enables informs how on to the of acid wider Scientic community research discussions reduce the between success in deposition. which impact environment. cooperation achieve the of Political nations the provides lead and is control to acid also and Q U e s T i O n s Questions 1 Consider the equilibrium 6 below: An aqueous reacts CH CH 3 COOH(aq) + H 2 O(l) ⇋ CH 2 CH 3 COO with solution of magnesium which of the following metal? (aq) 2 + + H O (aq) A. Ammonia B. Hydrogen C. Potassium D. Sodium 3 Which A. species CH CH 3 B. H represent COOH and a conjugate H 2 O acid–base chloride pair? hydroxide O 2 and CH 2 CH 3 hydrogencarbonate [1] COO 2 IB, May 2003 + C. H O and H 3 O 2 + D. CH CH 3 IB, COO and H 2 May O [1] 3 7 2011 Which property aqueous A. 2 Which is not a conjugate acid–base HNO and CH C. H of acids in Acids react with ammonia hydrogen gas solution and a to salt. NO 3 B. characteristic pair? produce A. is solution? 3 COOH B. and CH 3 Acids COO react oxygen 3 with gas, a metal salt, oxides and to produce water. + O and OH C. 3 Acids 2 D. HSO and IB, May SO 4 react hydrogen [1] D. Acids a Which species behave as Brønsted–Lowry the a metals to produce salt. react following reversible salt, with metal hydrogen and carbonates to gas, water. [1] acids IB, in reactive and 4 2011 produce 3 with gas May 2010 reaction? 2 H PO 2 (aq) + CN (aq) ⇋ HCN(aq) + HPO 4 (aq) 4 8 A. HCN and A solution of acid A has a pH of 1 and a CN solution of acid B has a pH of 2. Which 2 B. HCN and HPO statement must be correct? 4 2 C. H PO 2 D. and HPO 4 HCN 4 and H PO 2 IB, May A. Acid B. [A] C. The A is stronger than acid B [1] 4 > [B] 2010 + concentration than in of H of H ions in A is higher B is twice B + D. 4 Explain, using the Brønsted–Lowry theory, The concentration ions in how + the water case IB, can act identify May either the as an acid conjugate or acid a or base. base In concentration of H ions in A [1] each IB, formed. November 2010 2011 3 9 100 cm of a NaOH solution of pH 12 is mixed 3 with 5 Which of the following is/are formed when 900 oxide I) A reacts metal Water III) Hydrogen I only B. I and C. II and D. I, II a dilute IB, is the pH of the solution? A. 1 B. 3 C. 11 D. 13 gas May [1] 2009 only III and What acid? IB, II water. salt II) A. with of a resulting metal cm only III November [1] 2003 207 8 A C I D S 10 Black a pH A N D coffee of 8. B A S E S has a pH Identify of 5 which and is toothpaste more acidic has III) Use and each battery solution and lamp in a and circuit see with how a bright the + deduce 11 how the more IB, May many acidic times the [H is greater product. 100 the pH of the solution of 0.50 mol dm and II only b) I and III c) II and d) I, II, with IB, Specimen only III only and III [1] HCl(aq) 200 3 cm of 0.10 mol paper dm [5] 16 May I 3 cm NaOH(aq). IB, a) resulting 3 mixed glows. [2] 3 is lamp in 2011 Determine when ] Describe two different properties that could be used 2011 3 to distinguish of a between a 1.00 mol dm solution 3 strong monoprotic acid and a 1.00 mol dm 3 12 Which 0.10 mol dm solution would have the solution highest of a weak monoprotic acid. [2] conductivity? IB, A. HCl B. NH C. CH 3 17 May 2011 Ethanoic acid, CH COOH, is a weak acid. 3 COOH 3 a) D. H CO 2 IB, Dene May the term weak acid and state the [1] 3 equation for the reaction of ethanoic acid 2011 with b) water. Vinegar, [2] which contains ethanoic acid, 3 13 A student sodium has equal hydroxide volumes and of 1.0 ammonia mol dm can solutions. be used carbonate kettles. Which statement about the solutions is Sodium hydroxide has a lower B. Sodium than hydroxide concentration Sodium ethanoic May has than a of calcium of electric acid equation with for calcium the reaction carbonate. [2] 2009 higher hydrogen The equations hydroxide has of two acid–base reactions are ammonia. given C. the deposits elements ammonia. 18 ion State the electrical IB, conductivity clean correct? of A. to from a higher pH below. than Reaction A ammonia. + NH D. Sodium ion hydroxide concentration has than a higher hydroxide ammonia. (aq) May reaction reactants the Which list contains only strong CH B. HCl, COOH, H 3 CO 2 , H , H OH (aq) mixture in 3 3 2 A consists mainly HNO , H lies to B 4 (aq) + H O(l) ⇋ NH 2 , H (aq) + OH (aq) 3 reaction in B consists mainly of 4 SO 2 mixture SO 2 products 3 equilibrium 3 3 HNO the PO The COOH, HCl, + CO 3 D. (aq) 4 because 2 CH NH left. NH HNO 3 C. ⇋ acids? Reaction A. O(l) 2 2010 of 14 H [1] The IB, + 3 because the equilibrium lies to the [1] 4« right. IB, May 2009 a) For each whether 15 Which methods will distinguish solutions of a strong base and water reactions is acting A as and an B, acid deduce explain your or a base answer. [2] a b) strong the between and equimolar of In reaction B, identify the stronger base, acid? NH or OH and explain your answer. [2] 2 I) Add magnesium to each solution and c) look for the formation of gas In reactions A and B, identify the stronger bubbles. + acid, NH or 4 II) Add aqueous sodium hydroxide to your solution and measure the change. (underlined) and explain 3 answer. temperature IB, 208 NH each November 2009 [2] 9 R E D O X P R O C E S S E S Introduction Redox lie at reactions the chemical centre and applications. different be on many reduction everyday biochemical) In ways considered based of this both and and chapter we oxidation introduce and have the (both numerous explore and oxidation processes the reduction idea of state, which oxidation be problems will examine also chemical can can titration of and will be useful tool volumetric the energy electrical electrochemistry. cells a in solving conversion energy, Both in which voltaic redox chemistry. and is We between the basis electrolytic introduced. 9.1 O ao a o Understandings Applications and skills ➔ Oxidation and reduction can be considered in ➔ Deduction of the oxidation states of an atom in terms of oxygen gain/hydrogen loss, electron an ion or a compound. transfer, or change in oxidation number. ➔ ➔ Deduction of the name of a transition metal An oxidizing agent is reduced and a reducing compound from a given formula, applying agent is oxidized. oxidation numbers represented by Roman ➔ Variable oxidation numbers exist for transition numerals. metals and for most main-group non-metals. ➔ ➔ The activity series ranks metals according to reduced and the oxidizing and reducing agents the ease with which they undergo oxidation. ➔ The Winkler method can be used to measure Identication of the species oxidized and in redox reactions. ➔ biochemical oxygen demand (BOD), used as a Deduction of redox reactions using half- equations in acidic or neutral solutions. measure of the degree of pollution in a water ➔ Deduction of the feasibility of a redox reaction sample. from the activity series or reaction data. ➔ Solution of a range of redox titration problems. ➔ Application of the Winkler method to calculate BOD. Nature of science ➔ How evidence is used changes in the denition oxidation numbers is a good example of the of oxidation and reduction from one involving way that scientists broaden similarities to specic elements (oxygen and hydrogen), to general principles. one involving electron transfer, to one invoking 209 9 R E D O X P R O C E S S E S Redox reactions Three of the main ● acid–base ● precipitation ● redox A redox Both of reaction that occur in chemistry are: reactions reactions reactions. reaction reduction different The types ways, different involves and and ways two oxidation all of three processes, can in terms of specic ● in terms of electron ● in terms of oxidation these elements reduction considered descriptions describing ● be have and a merit processes oxygen in and oxidation . number in their of own right. are: hydrogen transfer number. Oxidation: Combining with oxygen At a the simplest substance 2Mg(s) level oxidation combines + O (g) with → can be oxygen. considered Examples as a reaction in which include: 2MgO(s) 2 2CH OH(l) + 3O 3 ▲ (g) → 2CO 2 (g) + 4H 2 O(l) 2 Figure 1 The Statue of Liber ty, New York, USA. It took many years after restoration in 1986 4Fe(s) + 3O (g) → 2Fe 2 O 2 (s) 3 for the copper to be oxidized and for the statue Fe to reform the green patina O 2 (s), iron(III) oxide, is rust. Rusting is an example of the process of 3 corrosion Global examples of corrosion The deterioration electrochemical described as of metals processes caused (redox used, by reactions) is the addition polymer polytetrauoroethene (PTFE): corrosion. F F C C F F Oxidation of iron and copper: The Statue of Liber ty n The Statue restored in of Liberty 1986 as it in New was York, found USA that was corrosion PTFE had occurred between the wrought is iron structural support Shellac resin and the outer copper , secreted by the female lac on trees in Thailand and India) inserted and the as iron an but insulator over between time the an failed the and the renovation iron work supports a rusted. different As its brand as a name non-stick cooking pans. in outer the Statue green of Liberty coating called oxidized the to patina. the restoration work on the Statue of was completed the statue was brown in part insulator It has taken many years for it to was fully 210 by used the colour. of also insulation Liberty had for copper When copper known also was form originally is bug The found which skin. coating (a commonly ® Teon and reform the patina (gure 1). oxidize 9 . 1 O x i d A t i O n substances zarda), of a in added betel person. mouth cause nut, is of to r e d u c t i O n it lime tobacco of the the tobacco found and (shada/ damage with have corrosive corrosion as can Paan Scientists highly the such and Chewing cancer. Paan A n d this that is health can cause the the lime primary bridge. Avy Can you think of some initiatives that governments and city councils can adopt to inform the general ▲ public of the health eects related to chewing Paan in Figure 2 Howrah bridge in Calcutta, India countries where this is par ticularly prevalent? Steel and Paan: The Howrah bridge Another Howrah 2010 interesting bridge the bridge corrosion is a India. of example Calcutta, was caused mixture (calcium in in found by an betel The leaf to be areca chewed itself corrosion unusual leaf, hydroxide) of India is not (gure the nut, in Paan. and harmful, Paan slaked millions of but In contrast, digestive In undergoing agent, by is 2). lime people other many and of is these a that substance down useful used the as a aphrodisiac, research properties nature slows also recent antioxidant A Reduction: Removal of oxygen or India Paan stimulant, of rate the at mnemonic may Paan. delays has for due to tonic that the antioxidant onset which nerve suggested be An post-meal and of oxidation oxidation remembering is or occurs. this is OILRIG: addition of hydrogen Oxidation Reduction oxygen, may for NiO(s) be considered as the removal Reduction example: + C(s) → Ni(s) + Let CO(g) us with In this carbon reaction to give nickel(II) metallic oxide is reduced may also be return oxygen Magnesium considered as the Loss of electrons Is to Gain the gas to of electrons reaction form of magnesium magnesium metal oxide. by nickel. earth Reduction Is of is a metals), member and has of the group 2 electron (alkaline conguration addition 2 [Ne]3s of hydrogen. An example of such a reaction the WO (s) + 3H 3 (g) → W(s) + 3H 2 . It mirrors the its previous two valence electrons to attain noble gas core conguration, [Ne]: O(g) 2+ 2 Mg This loses is: interpretation → Mg + 2e of 2 [Ne]3s reduction, oxide in as the oxygen is removed from [Ne] tungsten(VI) Oxygen process. is a is a member non-metal. 2 [He]2s It has of group the 16 (chalcogens) electron and conguration 4 2p and gains two electrons to attain the Oxidation and reduction in terms of noble gas conguration [Ne]: electron transfer 2 O In terms of electron transfer, oxidation + and 2e 2 [He]2s reduction can be dened as Oxidation involves the loss of electrons involves the 2 2p [Ne] or [He]2s 6 2p magnesium gain of is oxidized (loses electrons) and and oxygen reduction O follows: Hence ● → 4 is reduced (gains electrons). electrons. 211 9 R E D O X The P R O C E S S E S overall reaction is: example, in 2Mg(s) + O (g) → the complete form of combustion coal) in oxygen of solid yields carbon carbon (eg dioxide: 2MgO(s) 2 C(s) We can consider the reaction as being However, covalent 2+ → 2Mg + O (g) → CO 2 processes: 2Mg + two (g) 2 carbon bonds, dioxide so no is molecular, ionic bonds are with formed. We 4e cannot describe this combustion reaction as a redox 2 O + 4e → 2O process 2 in electrons Considering redox processes in terms of is a common interpretation must approach; be applied however with are of lost electron or gained transfer and as in carbon theory dioxide no is a electron neutral transfer terms species! The original denition of oxidation this caution. as the as carbon addition of oxygen is more appropriate here For is clearly oxidized in this process. An application of redox chemistry from optometry Optometrists often prescribe glasses with The chlorine atoms formed by the exposure + photochromic lenses. These lenses darken in the light are reduced by the Cu The 2+ presence of ultraviolet light (from sunlight); this are oxidized to to + ions. Cu ions 2+ Cu ions. These Cu ions then + change is based on a redox reaction. oxidize silver atoms to 2+ Ordinary glass photochromic is composed lenses of contain silicates while copper(I) Cu and silver chloride, ions: + + Ag → Cu + + Ag chloride, The CuCl, Ag lenses then become transparent again and the AgCl. + silver The on chloride exposure ions to are oxidized ultraviolet to light chlorine atoms and and Cl chlorine atoms return to the initial Ag species. ( hν). hν Cl → Electron cation Cl + e transfer to be then reduced takes to place metallic causing silver the silver atoms. + Ag The + silver making The e → atoms the inhibit lenses darkening chloride Ag process allowing transparent turn the the transmittance of light, dark. is reversed lenses to by copper(I) become again. ▲ When the following lenses are reaction takes + Cu removed from the light, Figure 3 Photochromic lenses the place: 2+ + Cl → Cu + Cl Electron book-keeping Chemists developed which reactants and the denition specic that can to of one scientists be products an used (oxygen invoking broaden electron to during oxidation elements transfer, 212 have reactions a and and track book-keeping the number chemical reduction process. from hydrogen), oxidation similarities states to of to is a general a model The example principles. of involving involving good redox in development denition one for electrons electron of the way 9 . 1 O x i d A t i O n A n d r e d u c t i O n Oxidation and reduction in terms of oxidation states The a oxidation molecule, In terms ● of or state an oxidation Oxidation and is the apparent charge of an atom in a free element, ion. state: describes reduction a process describes a in which process in the which oxidation the state oxidation increases state decreases. rs fo assgg oao sas 1 The oxidation state of an atom in a free element is 0; , O for example, S 8 , P 2 5 The oxidation state of uorine is , and Na all have atoms with an 4 , and LiF. For the other 2 oxidation state of 0. group 17 halogen elements the oxidation state is usually 2 1 in all its compounds, for example HF, OF 1 in binary compounds (HI, NaCl, KBr) but in Group 1 metals always have a +1 oxidation state in combination with oxygen in oxoanions and oxoacids their ions and compounds. the oxidation state is positive (for example, in HClO 4 Group 2 elements always have a +2 oxidation state in chlorine has a +7 oxidation state). their ions and compounds. 6 In Aluminium, which is a member of group 3, has an a neutral states of oxidation state of +3 in the majority of its compounds. the sum equals 3 molecule all of the the the the atoms is oxidation overall sum z ero. of In the a states of cha rge of the oxidat ion poly ato mic all ion. the For ion at om s ex amp le, The oxidation state of hydrogen is +1 when hydrogen in is bonded to a non-metal, such as in HCl and HNO the NH . oxidation state of nitrogen is 3 and 3 3 hydrogen is +1; the sum of the oxida tion states is However, when hydrogen is bonded to a metal, 3 + (3 × +1) = 0, which equals the net cha rge for example in a metal hydride such as NaH, the on oxidation number of hydrogen is the ammonia molecu le. 1. + In 4 The oxidation state of oxygen is usually the ammonium O and H 2 SO 2 , NH the individual 4 oxidation in H cation, 2, such as states of nitrogen and hy drogen are the . The main exception is in a peroxide 4 same (a species with an O O O 2 and NH the sum of the oxid a tion states 3 is 3 + (4 × +1) = +1 which equals the net 1. A typical example of such a charge compound is H in linkage); here the oxidation now state of oxygen is as on the ammoniu m ca tion. , hydrogen peroxide. 2 An application of redox chemistry at the hair salon Proteins have a number of functions, one International directives of which involves having a structural role in In the body. Proteins are polymers composed the European peroxide monomeric units called amino acids (see B.2). The protein molecules in hair restricted thiol groups, and hydrogen peroxide these to sulfonic acid groups, skin, (EU) and the oral use of hydrogen hygiene to maximum products concentrations of 12 %, and 0.1%, respectively. As stipulated by the can EU oxidize hair, contain 4%, SH in sub- is topic Union of SO H. Cosmetics Directive such products must be This 3 labelled: oxidation of the of the proteins thiol and groups hair can changes become the “Contains hydrogen peroxide. Avoid contact structure more with eyes. Rinse with them.” immediately if product comes in contact brittle. Avy The Directive Do people with bleached hair use particular conditioners? products Find out what a suitable conditioner might be for heavily should states containing be that when hydrogen using hair peroxide gloves worn. bleached hair. What might its chemical components be? 213 9 R E D O X P R O C E S S E S Variable oxidation states d b w As mentioned in the r ul es for a ss i gn i n g ox ida ti on s t a t es ab o ve , oao mb a although oao sa many compounds, group “Oxidation number ” and exist “oxidation state” are used the interchangeably in many 2 such alkaline for many transition variable textbooks. metals. el e me nts as e a r th me ta ls elements range s ta te s of xe d g ro up ma i n- g r oup oxidation The the ha ve 1 ox id at i on al ka li (e g +2 for non-met a l s (also ar e called a met a l s Ca), a nd the (e g +1 for f or and o x i d a ti on of the ions Na) fo r the and the s ta te s most metals). p ro pe rt y s t at e s t h e ir pa r t ic ul a r transition ox ida t io n in variable in cha r ac t er is t i c d i ff er e nt sta t e s, of Indeed tr a ns i t ion d- b l oc k Note however IUPAC elements is shown in g ur e 4, whic h is g ive n in s ec t ion 14 of the recommends the use of Data booklet. IUPAC describes tr a ns i ti on e le me nts as el e me n t s wh o se Roman numerals for oxidation atoms have an inco mp le te d- subs he ll or wh i c h ca n g i ve ri se to c at i ons numbers. Interestingly, with an incomple te d - s ubs he ll . In t he r st - r ow d- blo c k el e m e n t s, oxidation states in some the transition eleme nts ar e Sc to Cu in c l u si ve (but n ot Z n, wh i c h is compounds may not have explained in topic 13). integer values; they may be fractional, for example the oxidation state of oxygen in the S t V c M F co n c Z +1 +1 +1 +1 +1 +1 +1 1 is – superoxide anions, O 2 . 2 +2 +2 +2 +2 +2 +2 +2 +2 +2 +3 +3 +3 +3 +3 +3 +3 +3 +3 +4 +4 +4 +4 +4 +4 +4 +5 +5 +5 +6 +6 +2 O zg a g ags +6 An ozg ag causes another species to be oxidized, +7 and is itself reduced in the process. A g ag type A: type B: type C: causes another species to be Sc, Ti, and V Cr and Mn Fe, Co, Ni, Cu, and Zn reduced, and is itself oxidized in the process. ▲ Figure 4 Oxidation states of the rst-row d-block metals. The most stable oxidation states are marked in green Worked example Deduce the Solution oxidation states of each x a) K Cr 2 O 2 7 x atom (marked ) in each of the following 2(+1) + 2x + 7(–2) species: x a) K x x Cr O 2 2 b) Mn 7 = +6 O 4 x x c) Mg x N 3 d) [NH ] 4 214 [Fe 2 Mn O 4 8 x e) b) S 2 (H O) 2 ][SO 6 x + 4( x = +7 ] 4 2 2) = –1 = 0 9 . 1 O x i d A t i O n A n d x c) Mg x N e) [NH 2 3 3(+2) ] 4 + 2x = 0 To [Fe = (H 2 O) 2 answer this knowledge x r e d u c t i O n of ][SO 6 question, the ] 4 2 you charges of should use your ammonium ( +1), –3 water (0) and sulfate ( 2) species 2( 2) = (see sub- x d) S topic 8 x = 0, since this is a free element. 4.1). 2(+1) x = + x + 6(0) + 0 +2 Qk qsos x 1 Deduce the oxidation states of each atom (marked with an ) in each of the following species: x a) C x O b) HCl O 2 4 x ) Na P x O 3 ) O 4 3 x ) P x H f) I Cl 3 x g) x (SO Fe 2 4 ) h) H 3 C 2 O 2 4 x ) N O 3 2 In the following balanced equation: Cl (aq) + 2KI(aq) → 2KCl(aq) + I 2 (aq) 2 a) Deduce the oxidation states of chlorine and iodine in the reactants and products. b) State which element is oxidized and which element is reduced. ) Identify the oxidizing agent and the reducing agent. Oxidation states and the nomenclature of transition metal compounds As is stated that previously they can Traditionally, used to the name numbers. The one have of Roman such the variable characteristics oxidation numeral compounds system is called states system and the this of of transition in their nomenclature system Stock is elements compounds. based on nomenclature has been oxidation system . In Sy ps the 1 Stock system, Roman numerals (I, II, III etc.) are used to indicate Remember when writing the oxidation states the oxidation number. charge goes before the number and not after it. For example, the The Stock nomenclature system oxidation number of hydrogen in HBr is +1 In KMnO , often called by its old name potassium permanganate by 4 and not 1+. many chemists, manganese has an oxidation state of +7. However, 7+ from is a purely highly Stock electrostatic improbable. nomenclature Its perspective name system. is the presence potassium of an Mn manganate(VII) cation using the 2 The oxidizing and reducing agents are always the reactants. 215 9 R E D O X P R O C E S S E S in -ate, and the one with the greatest number of Sy ps oxygens will be prexed by “per ” and end in -ate. The 1 When deducing the name of a transition metal four oxoanions of chlorine, bromine, and iodine follow compound using the Stock system, do not be this system. tempted to use the subscript representing the number of atoms of the other element in the Foma of ooao no-sysma am compound to write the oxidation number in Roman 2 carbonate CO 3 is iron(II) numerals. For example, the name of FeCl 2 2 C chloride because iron is deduced to have the +2 oxidation state [x + 2( ethanedioate (oxalate) O 2 4 1) = 0, nitrite NO 2 so x = +2]. It is coincidental that 2 matches the nitrate NO 3 number of chlorine atoms in the formula. In the 2 sulte SO compound FeO this becomes clearer: the correct 3 name is iron(II) oxide [x + ( 2) = 0, so x = +2]. 2 sulfate SO 4 2 In working out the names of many transition metal 3 phosphite PO 3 compounds, knowledge of the non-systematic 3 phosphate PO 4 names and charges of the various oxoanions can be useful (table 1). hypochlorite ClO chlorite ClO In naming oxoanions, a good rule of thumb is as follows: 2 chlorate ClO If there is only one oxoanion, the ending will be -ate. ● 3 If there are two oxoanions, the one with the smaller ● perchlorate ClO 4 number of oxygens will end in -ite and the one with hydroxide OH the greater number of oxygens will end in -ate. 4 or thosilicate SiO 4 If there are four oxanions, the one with the lowest ● T able 1 Formulas and non-systematic names of some ▲ number of oxygens will end in -ite and be prexed oxoanions by “hypo”, the next will end in -ite, the third will end tOK Qk qso Chemistry has developed Using the Stock nomenclature system, deduce the name of each of the a systematic language that following transition metal compounds: has resulted in older names a) CoF ) Cu(OH) 3 ) Cu 2 O 2 becoming obsolete. What b) V has been lost and gained in O 2 ) MnO 3 2 this process? Nomenclature In theory the as names stated one new is not all set of inorganic state, many such In guidelines and the Although systematic include required. oxoanions oxoacids. 216 of could previously oxidation state of one additive the compounds. elements as the IUPAC the new names, system, has huge in only +1, so published systematic corresponding state However, have potassium, 2005 for oxidation the a naming inorganic based merit, on IUPAC does completely such as acid, etc. called be recognize eliminate carbonate, In this that old system, ), are acid, ), nitrate, and formed? acid nitric Can you to names nitric would nitrate carbonic hydroxidodioxidonitrogen. names unrealistic carbonate trioxidocarbonate(2 trioxidonitrate(1 is non-systematic carbonic dihydroxidooxidocarbon, these it be would would acid be would suggest be how 9 . 1 O x i d A t i O n A n d r e d u c t i O n Expressing redox reactions using half-equations in acidic or neutral solutions Half-equations complex redox represents processes. the can be separate The very reactions. useful Each in balancing oxidation following general balance redox and reduction working Step the be used to a reaction State and states. neutral to In the balance an IB syllabus equation you in are acidic 4: the reactant oxidation and product states for each for process. Balance of Step 5: write the these electrons half-equations lost equals the so that number the of gained. Add Step 6: atom the two overall Check product Assign oxidation only media. 1: the or Working method Step for half-equation method electrons required half-equation corresponding involving number oxidation the the reduction Step can 3: process half-equation half-equations redox the together to reaction. total charge on the reactant and sides. in species. + Step 7: Balance the charge by adding H and H O 2 to Step 2: which Deduce species which is species is oxidized the appropriate sides. and reduced. Worked example 1 Iron The tablets iron in are often the prescribed tablets is to patients. commonly Step The as anhydrous iron(II) sulfate, FeSO . to determine the oxidation 2+ in of iron in such tablets percentage involves of Fe changes from +2 3+ to Fe +3 a in Fe , so the oxidation state by increases, mass state An 4 experiment 2: present indicative of oxidation. The oxidation redox state of Mn changes from +7 in MnO to 4 reaction, shown in the following unbalanced 2+ +2 in Mn , indicative of so the oxidation state decreases, equation: 2+ Fe 3+ (aq) + MnO (aq) → Fe (aq) Step 4 a) Deduce the balanced redox 3: equation Oxidation in acid and identify the reduction. 2+ (aq) + Mn oxidizing (loss of 2+ Fe reducing 3+ (aq) → Fe (aq) + e agents. Reduction b) electrons): and Consider the oxidation state (gain of electrons): 2+ of MnO (aq) + 5e → Mn (aq) 4 manganese in the permanganate anion, Step MnO . Comment on the 4: following 4 Oxidation (loss of electrons): statement: 2+ 5Fe “If oxidation apparent element state charge has in is considered that an an ion, atom then as of the 3+ (aq) → 5Fe (aq) + 5e the Reduction an oxidation (gain of electrons): 2+ MnO (aq) + 5e → Mn (aq) 4 state of manganese here must signify the Step presence of the corresponding 5: ion!” 2+ Oxidation: 5Fe 3+ (aq) → 5Fe (aq) + Solution 5e 2+ Reduction: MnO (aq) + 5e → Mn (aq) 4 a) Step 1: 2+ Overall: 5Fe (aq) + MnO (aq) → 4 2+ Fe : Fe, MnO : x = Mn, 3+ +2 x + 5Fe 4( 4 O, x = 2) = –1, so x = +7; Step 2+ (aq) + Mn (aq) 6: –2 Total charge on reactant Total charge on product side = 9+ 3+ Fe : Fe, x = +3 side = 17+ 2+ Mn : Mn, x = +2 217 9 R E D O X Step P R O C E S S E S 7: b) Possible a) + To balance this equation 8H must be the response oxidation the reactant However, + MnO (aq) + 8H (aq) 3+ Water need can whichever part in anion was found to be the +7. oxidation be to balance included side of the at the + Mn (aq) state very equation it last is of ionic hydrogens. the different states meanings, and and ionic an charges oxidation 2+ (aq) stage +7 are required: does charge properties on of of Based not 7+. ions, theoretical real. 2+ 5Fe in → have 5Fe we question: manganese + (aq) 4 Next of side: 2+ 5Fe NOS inserted permanganate on to state on signify Ionic a corresponding charges whereas constructs and electrostatic are real oxidation are states not considerations + (aq) + MnO (aq) + 8H (aq) → 4 the 3+ 5Fe oxidizing agent of a 7+ ionic charge is most 2+ (aq) + Mn (aq) + 4H O(l) unlikely. 2 The presence is MnO (aq) and the bonds Oxidation are ionic. states assume However, in that MnO 4 the 4 2+ reducing agent is Fe (aq). manganese–oxygen bonds are covalent in nature. The activity series The activity they those lower series is it is a most series undergo down primarily non-metal. reactive (table oxidation. from based The metals 2) ranks Metals solutions on are em of metals, series is metals higher at their according in in the the is section top of the of ease series salts. often 25 the to activity respective hydrogen given found up Data which displace Although included the with can the even booklet. though The series. dasg avy eas of oao ass lithium potassium sodium magnesium aluminium manganese zinc iron lead hydrogen copper silver mercury gold ▲ T able 2 The activity series Let’s consider some examples: 2+ ● Zn(s) Zinc + Cu metal is 2+ (aq) → above Zn (aq) copper + metal Cu(s) in the series, so therefore it is more 2+ reactive 218 and can displace the Cu ions in solution to form copper metal. 9 . 1 ● Zn(s) + 2HCl(aq) → ZnCl (aq) + H 2 O x i d A t i O n A n d r e d u c t i O n (g) 2 Qk qsos Zinc the metal is above hydrogen ions hydrogen in in the hydrochloric series. acid to Therefore, form it can hydrogen displace 1 gas. Deduce the oxidizing and reducing agents in the reaction of ● 2Al(s) + Fe O 2 Aluminium (s) → 2Fe(l) + Al 3 O 2 metal is above iron (s) 3 in potassium bromide with chlorine. the series. Hence molten iron can 2 form according to the reaction Table 4 shows reactions involving above. aqueous solutions of halogens ● 2Na(s) + 2H O(l) → 2NaOH(aq) + H 2 with aqueous potassium iodide (g) 2 solution. Copy and complete the In this reaction, hydrogen is displaced from water by the very table, and in each case: reactive alkali metal, sodium, to liberate hydrogen gas in the process. a) A reactivity series can also be written for the group 17 state whether a reaction will elements, occur or not uorine, chlorine, bromine, and iodine (table 3). b) identify the colour of the halide solution after reaction Gop 17 m Aom as eogavy (pm) iasg (Pag sa) χ ) avy deduce the balanced p equation for any reaction that uorine 60 4.0 chlorine 100 3.2 occurs. bromine 117 3.0 iodine 136 2.7 Haog c (aq) 2 B (aq) 2 a) rao wh T able 3 Reactivity series for group 1 7 elements ▲ Ki(aq) Again the more 2KBr(aq) + reactive Cl (aq) elements → are 2KCl(aq) found + Br 2 The atomic chlorine a is greater is by oxidation in the state losing states. oxidation of of bromine chlorine is smaller electronegative. gaining in the series. (aq) 2 attraction reduced, oxidized radius more higher for an an an electron electron electron Chlorine state to to from than form of 1 than a does the 0 so bromine, bromine, chloride to so is so the an b) coo has of ha chlorine Bromine change there process. there so therefore ion. Note 1, reduction 0, of nucleus bromine. from to that chlorine form changes indicative changes The is a The soo is in decrease oxidation increase in the ) Baa oxidation state indicative of an oxidation process. qao In the laboratory, potassium colourless the solution formation reactivity when bromide of series chlorine there of a can gas is bubbled corresponding potassium aqueous so is bromide bromine. displace to a solution change from yellow/orange, Chlorine bromide through colour ions is higher from up in of the indicating the potassium bromide to ▲ form bromine (see topic T able 4 3). 219 9 R E D O X P R O C E S S E S Uses of chlorine in everyday life Chlorine widely is a used Calcium powerful as a oxidizing disinfectant hypochlorite, agent and Ca(OCl) , is and is antiseptic. often used in 2 hospitals their hands NaOCl, homes The is as healthcare (gure another is the which a can syndrome that of professionals Sodium and T he US Preve nti o n disinfection of used syr i ng es f acto r a cq uir ed (AIDS) . and often in the imm uno de cie ncy to disinfect in our bleach. ne e d l e s lead to hypochlorite, disinfectant, contributo r y human Control 5). household sharing users of by am on g vi ru s i mmune C ente rs (C DC) s y r i ng e s and ( H I V) , d e c i en c y for ha s d rug t r an s m i ss i on D i se a se r e por t e d ne ed le s ▲ with household b l e a ch ma y go s om e way Figure 5 A solution of calcium hypochlorite acts to as both a disinfectant and an antiseptic. Can you alleviating this ris k . explain the dierence between these two terms? Use of chlorine and ozone as disinfectants H in drinking water Access to United over C a supply Nations one billion fundamental of as a clean drinking fundamental people resource worldwide to water human do mankind. has been right, not yet have Water the supplies recognized it is luxury are by estimated of the that such disinfected a using Cl Cl strong oxidizing agents such as chlorine, Cl or ozone, O 2 pathogens. be added In in the USA, three chlorine forms: is chlorine used gas, for Cl ; to kill microbial 3 this purpose. sodium Chlorine hypochlorite, can NaOCl; 2 Cl and calcium hypochlorite, Ca(OCl) . All three of these solutions yield 2 hypochlorous The use object of to chlorine acid, chlorine the can taste also HOCl, can and react trichloromethane, which cause with , the problems general CHCl is odour other of antibacterial for the general residual chemicals commonly to known agent. form as public. chlorine in toxic Some water. products chloroform people Residual (gure such as 6). 3 In Europe, water Nice supplies. in 1906 chlorine ▲ France for was The for rst this water one of the rst industrial purpose. countries ozonation Table 5 to plant compares use ozone was the to disinfect established use of ozone in and treatment. Figure 6 The structure of trichloromethane, CHCl 3 Avy Ozo cho can be used to treat viruses cannot be used to treat viruses leaves no unpleasant residual leaves a residual taste and unpleasant odour taste or odour In your country, nd out fewer toxic by-products can form toxic by-products, often carcinogenic more expensive cheaper whether chlorine or ozone is used to disinfect municipal water supplies. ▲ T able 5 Advantages water supplies 220 and disadvantages of using ozone and chlorine in the treatment of 9 . 1 O x i d A t i O n A n d r e d u c t i O n Redox titration reactions Moay In topic eld of 1 titrations volumetric titrations, those were introduced, analysis. involving In and addition redox these to play reactions reactions are also a pivotal role involving extremely in acid the base Note that for convenience concentration is often useful. termed the “moay” In order to solve to recall some titration questions involving redox reactions, you need (the unit is sometimes of the formulae used in volumetric analysis: abbreviated to M), but it is 1 Amount of substance (in mol) = best practice to use the unit n 3 mol dm m _ n in calculations. = M 1 where m = mass in g; M = molar mass in g mol 3 2 n = volume (in dm 3 ) × concentration (in mol dm 3 volume (in cm ) 3 ) × concentration (in mol dm ) _____ = 1000 3 because 1 = 1000 cm 1 _ 1 _ (n 3 3 dm ) = (n A ν ) and hence B ν A B 1 _ 1 _ (V × c A ν ) = (V A × c B ν A ) B B 3 V = volume of reactant A (in dm ) A 3 c = concentration of reactant A (in mol dm ) A 3 V = volume of reactant B (in dm ) B 3 c = concentration of reactant B (in mol dm ) B ν and ν A are the stoichiometry coefcients B Working method Step 1: Deduce Step 2: From given from the V , c A to be the balanced redox information , V A , and c B determined. equation, given, and state identify using which the oxidation three fourth pieces variable states. of data that B Identify the stoichiometry coefcients ν and ν A from the Step 3: balanced Set up following expression and ll × c A ) = A (V 4: × B ν A the known data: c ) B B Solve for the unknown variable (V , A Step in 1 _ (V ν Step B equation. the 1 _ are needs 5: Answer concentration in any riders particular to the question c , V A , or B (such c ). B as expressing a units). Worked example Sy p 1 Consider the following balanced equation for the reaction of This potassium manganate(VII) with ammonium iron(II) type of frequently 2+ 5Fe (aq) + + MnO (aq) + 8H 3+ (aq) → 5Fe a titration to + Mn (aq) + 4H the in O(l) 2 determine appears 2+ (aq) 4 In question sulfate. concentration of a Question 1 of Paper 2. potassium 3 manganate(VII) solution, 28.0 cm of the potassium 221 9 R E D O X P R O C E S S E S manganate(VII) Sy p solution solution 3 25.0 reacted completely with 3 cm of a 0.0100 mol dm solution of ammonium iron(II) Notice that all variables 3 sulfate. Determine the concentration, in g dm , of the potassium are given correct to three manganate(VII) solution. signicant gures, hence the nal answer for the Solution concentration should also Step 1: Deduce the balanced redox equation, using oxidation states. be expressed correct to three signicant gures. This step Step are 2: is not From given required the from V to be as information , c A needs here , V A , and equation given, c B determined. the and state is given which identify the in the three question. pieces fourth of variable data that B Identify the stoichiometry coefcients, ν and A ν Sy p , from the balanced equation. B 2+ Although molar masses are A represents Fe and B represents MnO 4 given correct to two decimal 2+ V = volume of Fe 3 = 0.0250 dm A places in the Data booklet, 2+ c the number of signicant = concentration of Fe 3 = 0.0100 mol dm A gures in the nal answer 3 V = volume of MnO B = 0.0280 dm 4 is determined from the c experimental data given = concentration of MnO B : this is what must be calculated 4 in the question and not ν = 5 = 1 A from published data. The ν B same would apply to any constants used to answer Step 3: Set up the following expression and ll in the known data: questions. 1 _ 1 _ (0.0250 × 0.0100) = (0.0280 × c ) B 5 Step 1 4: Solve for the unknown variable (V , c A , V A , or B c ). B 3 c (concentration of MnO B Step ) = 0.00179 mol dm 4 5: Answer concentration any in riders to particular the question (eg expressing a units). 3 To calculate 1 mol the of concentration KMnO ≡ in (39.10) g + dm , we (54.94) + use dimensional 4(16.00) ≡ analysis. 158.04 g 4 So: 0.00179 158.04 mol __ g _ × 3 = 0.283 g dm 3 1 1 dm mol An environmental application of redox chemistry: The Winkler method Aquatic dioxide life and depends oxygen on gases dissolved such in as the carbon water The in solubility dependent. of At oxygen 273 K (0° in water C) the is temperature solubility is 3 order to survive. Oxygen, O , is a non-polar 14. 6 mg 7.6 mg dm (or 14.6 ppm), compared with just 2 3 molecule, but water, H O, is polar. Therefore the dm (7.6 ppm) at 293 K (20 °C). Clearly 2 solubility of oxygen in water will be very low. as the gas 222 temperature decreases. increases, the solubility of the 9 . 1 The O x i d A t i O n degree water can oxygen dipole moment as the of be organic demand of matter in over units environmental or a measured In pollution by the BOD. oxygen temperature in r e d u c t i O n measured amount organic A n d period of a This is of to water of sample of dened required sample a in biochemical 5 oxidize at days. a denite BOD is ppm. science, ppm is often used as O the standard maximum toxic H or unit of concentration allowable carcinogenic upper limit to of a indicate the potentially (cancer-causing) substance. For H example, according to recommendations from the Figure 7 Water is a polar molecule. The vectorial sum ▲ World Health Organization (WHO) the maximum of the two individual OH polar bonds results in a net 2+ allowed concentration of lead(II) cations, Pb (aq), in dipole moment for the molecule 3 drinking water is 0.001 mg dm or 0.001 ppm. coaos pa s p mo Avy The concentration of very dilute solutions is often measured in pa s p mo, ppm Go to the WHO website (http://www.who.int/en/) and try to nd data about the maximum allowed concentration in ppm concentrations of other metals in drinking water. mass of component in solution _______ = 6 Compare this data with the limits set by the × 10 total mass of solution government of the country where you live or by mass of solute in mg _____ directives set by a wider union of countries (eg the = 3 volume of solution in dm The a amount of barometer water. The to dissolved indicate Winkler oxygen the is quality method, European Union). often of based a used body on as of Typical values of BOD redox Pure reactions, is one technique that can be used water Water measure the amount of dissolved oxygen in from general, a high concentration of indicates a low level of ▲ be considered from a river considered a BOD has a less BOD than of 1 1 ppm. ppm very with a clean. BOD However, of 20 ppm water would poor quality (table 6). BOd (ppm) pure water less than 1 untreated domestic sewage 350 euent from a brewery 500 water from an abbatoir 3000 T able 6 Typical biological oxygen demands for water samples When organic water, it material Figure 8 The WHO is the directing and coordinating authority of eamp so present. for health within the United Nations has that pollution. be ▲ river dissolved taken oxygen a water. would In generally to matter provides The water The carbon hydrogen in is is is discharged source bacteria into and a break compounds a series of oxidized oxidized of down such as carbon water, into for the a any body of bacteria organic carbon oxidation to to food dioxide reactions. dioxide, and any the nitrogen 223 9 R E D O X present is P R O C E S S E S oxidized to nitrate, NO . The The bacteria procedure developed by Winkler is an indirect 3 multiply more and their dissolved processes. If increased oxygen the is uptake levels used of for mean these oxygen by one, that oxidation the react faster than the rate at which dissolved with replaced from the atmosphere and the depleted body of of water oxygen. will Under the hydrogen bacteria sulde, H will S, produce NH 2 amines), and PH . the gas commonly then the in Hydrogen left dark. to with rotten eggs; it is also often was for a The measure the water water is taken from discharged a into location a rst saturated with marina oxygen period of Winkler 5 days method at 293 was K carried the dissolved sample before oxygen and content after the 5-day sulde incubation associated of efuent UAE, the period. The following is the series of odour reactions from sample treated such 3 is reagent. (and 3 phosphine, directly anaerobic products ammonia, not eventually such out as cm Dubai, in conditions redox does from and become the oxygen 3 50.0 in photosynthesis, dissolved oxygen where is the bacteria A is as liberated related to the method: from 2+ volcanoes. Due odour its and to its characteristic potential source H Mn unpleasant S is often (aq) + 2OH (aq) sewer (s) referred 2Mn(OH) (s) + O 2 as Mn(OH) 2 2 to → (g) → 2MnO(OH) 2 (s) 2 gas. + MnO(OH) (s) + 4H (aq) + 2I (aq) → 2 em Sbsa Sbsa po po aaob aob oos oos 2+ Mn (aq) + I (aq) + 3H 2 2 I (aq) + 2S 2 carbon CO CH 2 O 2 O(l) 2 2 (aq) → 2I (aq) + S 3 O 4 (methane, (aq) 6 3 It 4 was found that 5.25 3 cm of a 0.00500 mol dm commonly known as solution of sodium thiosulfate, Na S 2 O 2 (aq) was 3 marsh gas) required hydrogen H O CH 2 , NH 4 , H 3 S, and H 2 NH NO 2 sulfur H SO , amines the iodine produced. Determine the oxygen, ppm, in concentration in the of sample dissolved of water. S (hydrogen b) 2 4 with O 3 3 react 2 a) nitrogen to Deduce the BOD, in ppm, of the water sample, sulde) assuming 3 phosphorus PH PO oxygen in the the maximum water is 9.00 solubility ppm at of 293 K. 3 4 ▲ (phosphine) that T able 7 Substances produced by bacteria under aerobic and c) Comment on the BOD value obtained. anaerobic conditions Solution The the reduction depletion in of d i s s o l ve d s h oxy ge n s to cks . If the c an B OD re s u lt is in Step 1: Deduce The than the dissolved co nte nt in the wa t er, cannot surviv e . Ty p i ca ll y s h re qu i re 3 ppm of diss o l v e d ox yg en in series w a t e r, in sustain a healthy a q ua tic of dissolve d e nv ir onme n t o x yg en in wat e r below 6 equation. redox The equations important is point the correct stoichiometric is ratio a nd oxygen and thiosulfate; this is needed the s h ou ld the calculation. Careful examination of the not three fall balanced question. determine for content of the between to redox at to least balanced aq u at i c given life the g r e at e r reactions gives the following ratio: ppm. 1 mol O (g) → 2 mol MnO(OH) 2 (s) → 2 2 4 mol S O 2 (aq) 3 Worked example: measuring BOD Step using the Winkler method three 2: From pieces the of information data are given given, from V state , A In the Winkler method, an iodine/thiosulfate c and identify the fourth variable that c , A which V and B needs to be B redox titration dissolved is oxygen carried out present in to a measure water the sample. determined. ν A 224 and ν , B Identify from the the stoichiometry balanced equation. coefcients, 9 . 1 O x i d A t i O n A n d r e d u c t i O n 2 A represents S O 2 and B represents = volume of S A 5: Answer any riders to the question. 2 O 2 3 3 2 V Step O 3 = 0.00525 In dm order to calculate the concentration in g dm , 3 we use dimensional analysis. 2 c = concentration of S A O 2 3 3 = 0.00500 mol 1 dm mol of O ≡ 2(16.00) ≡ 32.00 g 2 3 V = volume of O B = 0.0500 So: dm 2 4 c = concentration of O B ; this must be 1.31 calculated × 10 32.00 mol __ 2 g _ × 3 1 ν = 4 = 1 1 dm mol A 3 ν = 4.19 × 10 = 4.19 mg 3 g dm B 3 Step the 3: Set known up the following expression and ll dm = Hence data: the oxygen (BOD) = 9.00 1 _ 4.19 ppm in used by the bacteria 4.19 = 4.81 ppm. 3 (0.00525 dm × 0.00500) This BOD value shows reasonable water 4 quality 1 _ = (0.0500 × c for discharge effective Step 4: Solve for the unknown variable (V , c A , taken at the point in sewage Dubai, efuent suggesting treatment plan that must be an in V A B place. c sample ) A 1 or the Typically untreated domestic sewage has ). B a 4 c (concentration of O B ) = 1.31 × 10 BOD in the range 100–400 ppm. 3 mol dm 2 Activity Chemistry Discuss is this full of commenting on ionic and charge, domain abstract statement may be with aspects a concepts, theories, reference to such negative as oxidation charge preferable the term centres. to and assumptions. thiosulfate numbers, Suggest negative oxoanion, formal why charge charge, electron centre in this context. 225 9 r e d O x P r O c e S S e S 9.2 eohma s Understandings Applications and skills Voltaic (Galvanic) cells: Construction and annotation of both types of ➔ ➔ Voltaic cells conver t energy from spontaneous, electrochemical cells. exothermic chemical processes to electrical Explanation of how a redox reaction is used ➔ energy. to produce electricity in a voltaic cell and how ➔ Oxidation occurs at the anode (negative current is conducted in an electrolytic cell. electrode) and reduction occurs at the cathode Distinction between electron and ion ow in ➔ (positive electrode) in a voltaic cell. both electrochemical cells. Electrolytic cells: Performance of laboratory experiments ➔ ➔ Electrolytic cells conver t electrical energy involving a typical voltaic cell using two metal/ to chemical energy, by bringing about non- metal-ion half-cells. spontaneous processes. Deduction of the products of the electrolysis of ➔ ➔ Oxidation occurs at the anode (positive a molten salt. electrode) and reduction occurs at the cathode (negative electrode) in an electrolytic cell. Nature of science ➔ Ethical implications of research – the desire to produce energy can be driven by social needs or prot. Energy Energy The created There due is law or are to the of capacity destroyed many motion), energy, to do conservation heat but is different The energy converted forms potential energy, work. of of energy nuclear SI from energy, (stored energy, unit states of that one form such or sound energy energy as to is energy, joule (J). be another. kinetic positional the cannot energy energy), chemical (energy light energy, and Figure 1 Voltaic cells conver t energy electrical energy. from spontaneous exothermic chemical processes to electrical energy Electrochemical cells In an take electrochemical place, which electrochemical 1 Voltaic (or electrical cell go chemical in either energy–electrical direction. There energy are two galvanic) chemical Electrolytic cells Voltaic cells – – cells these processes these convert convert to electrical convert chemical energy from energy to chemical energy, bringing about a non-spontaneous process 226 types bringing about a to energy. electrical non-spontaneous energy spontaneous, energy Figure 2 Electrolytic cells convert electrical energy, conversions main cell: energy. exothermic 2 can process. to chemical of 9 . 2 e l e c t r O c H e M i c A l c e l l S Sy p Early ideas about electricity An easy way to remember Electrochemistry explores energy conversions between chemical and the energy conversion in an electrical energy. The Italian physician, physicist, and philosopher Luigi electrolytic cell Galvani origins, (1737– 1798) whereas the considered Italian electricity physicist essentially Alessandro Volta biological in is “ee”: its (1745–1827) electrolytic cell – electrical to did not. The initial discovery of electrochemical cells resulted from chemical. serendipitous driven by demand observations potential for of prots energy has scientists, from driven but improved innovation recent developments technology. in devices The and have been increasing processes. Sy p An easy way to remember Electrodes Electrons are conductor circuit, cell In of such both of electricity as contains redox processes at electrodes carriers the two voltaic electric used solution to in electrodes, and charge make a cell the electrolytic in metals. contact (the anode An with a electrode non-metallic electrolyte). and the is An a is the mnemomic “crOA”: part of a electrochemical cathode – reduction and Oxidation – Anode. cathode. cells: ● oxidation always takes place at the anode ● reduction always takes place at the cathode. Sy p An easy way to remember The In a polarity of voltaic the electrodes differs in the different types of cell. the polarities at electrodes is “cnAP”: for the electrolytic cell: cell, cathode–negative; ● the cathode ● the anode is the positive electrode Anode –Positive. If you can is the negative electrode. recall this mnemonic for the In an electrolytic cell: electrolytic cell, you know that ● the cathode ● the anode is is the the negative positive electrode the opposite is true for polarities in a voltaic cell. electrode. The voltaic cell A voltaic (the cell anode) consists and are different ion electrodes, ion electrode. required to types of metal For be of two reduction half-cells. occurs electrode ions the IB familiar in at used two with in other voltaic different Chemistry only Oxidation the cells, at as states, programme, metal/metal-ion one (the such oxidation Diploma the occurs half-cell half-cell cathode). There metal/metal- and SL the gas- students are electrode. The metal/metal-ion electrode A metal/metal-ion solution type of electrode containing electrode cations consists of the of same a bar of metal. metal Typical dipped into examples a of this include: 2+ ● Fe(s)|Fe (aq) (gure 3) 2+ ● Zn(s)|Zn (aq) bar of metallic iron, Fe(s) 2+ ● Cu(s)|Cu (aq). 2+ solution of Fe In this notation, the vertical line represents a phase boundary (aq), or called the electrolyte junction In a voltaic cell the two half-cells are separated – if the solutions were 2+ Figure 3 The Fe(s)|Fe allowed to mix in a single container, a spontaneous reaction would (aq) electrode occur 227 9 r e d O x P r O c e S S e S but there circuit, would and be hence no no movement current. of The electrons two through electrodes are the in external electrical contact Activity via Discuss the origins of of from a It electrochemistry Nature perspective, of some of the the eld working to the It provides of with produce has a number of functions: reduction of the cathode processes, and anode preventing and mixing hence of the continuity – a path for the migration of the positive (the cations) and the negative ions (the anions) in the cell. It reduces the liquid-junction p ot e nt ial . This is the voltag e ethical when tw o d i ffe r ent so lut i on s co m e in t o con t a c t wi t h current our global more separation and electrical other, which o ccur s due to un e qu a l ca t io n and a n i on ever- migration increasing This in each research, bridge. solutions. generated implications salt original ● this a discoveries ions scientists called physical oxidation two ● of junction allows the Science from serendipitous liquid the ● eld a desire across the j unction. to energy. A salt high bridge contains concentration a of concentrated ions in the solution salt bridge of a strong allows ions electrolyte. to diffuse The out of 2+ it. For example, the Daniell voltaic cell consists of the Cu(s)|Cu (aq) 2+ and for Zn(s)|Zn this cell (aq) could electrodes. be sodium Typical sulfate, compounds Na tOK SO 2 KCl(aq). Is energy real, or just an react The with ions the used other in ions the in salt (aq) used or in the potassium salt bridge chloride, 4 bridge must be inert – they should not thesolution. abstract concept used to justify why cer tain types The Daniell voltaic cell of changes are always In the Daniell cell (gure 4), the following half-equations show the associated with each other? redox ● processes Anode occurring. (negative electrode): oxidation. 2+ Zn(s) ● → Cathode (aq) Zn + (positive 2e electrode): reduction. 2+ Cu ● (aq) Overall + cell 2e → Cu(s) reaction: 2+ Cu Once the of the as it 2+ (aq) + cell side Zn sulfate coated drawing right-hand → connected, copper(II) becomes When is Zn(s) in voltaic as in (aq) as + the solution more cells, gure Cu(s) redox fades. copper, by processes The and occur copper the convention zinc the bar bar the blue increases gets cathode is colour in size thinner. drawn on the 4. e e V 2 SO 4 + Na Zn anode Cu cathode salt bridge ( ) (+) 2 2 SO 4 SO 4 cotton wool 2+ Zn 2 SO ZnSO CuSO 4 4 4 2 SO 2+ Cu 4 2+ Zn(s) → Zn 2+ (aq) + 2e Cu (aq) + movement of cations movement of anions 2+ Figure 4 The Daniell cell: a cell consisting of Zn(s)|Zn 228 2+ (aq) and Cu(s)|Cu (aq) half-cells 2e → Cu(s) 9 . 2 How will can be For the more you determine reduced Daniell easily in a cell, which voltaic zinc oxidized – is the metal cell? The higher zinc up will be answer in half-cell the acts oxidized lies in series as and e l e c t r O c H e M i c A l which metal the activity series. than copper, so the c e l l S it is anode. Cell diagrams Cell cell. on diagrams By the For are convention right. the The Daniell used the salt cell a bridge the 2+ Zn(s)|Zn as anode cell convenient is is always shorthand written represented diagram would by be on to the two represent left and parallel written a the voltaic cathode vertical lines. as: 2+ (aq) || Cu (aq)|Cu(s) Sy p When answering questions about voltaic cells, make sure you know the direction of ow of the electrons and ions (gure 5). e e V + Zn anode Cu cathode salt bridge ( ) (+) + + anode cathode (oxidation) (reduction) Figure 5 The direction of ow of electrons, positive ions and negative ions in a voltaic cell Qk qso 2+ For a voltaic cell consisting of a Zn(s)|Zn 2+ (aq) half-cell and an Fe(s)|Fe (aq) half-cell: a) State the cell diagram for the cell. b) Write half-equations for the reactions occurring at the cathode and the anode. ) Identify a suitable compound that may be used in the salt bridge. ) Identify the direction of the movement of electrons and ion ow, both in Ky pma wok solution and in the salt bridge. In this topic it is impor tant to ) Explain why the cation and anion of the salt used in the salt bridge should have have carried out laboratory approximately the same size and charge. Identify using section 9 of the Data experiments or seen videos or booklet the ionic radii of the cation and anion of the compound given in (c). simulations involving a typical f) Cotton wool is often used at the tips of the salt bridge. Suggest the function of voltaic cell using two metal/ this. metal-ion half-cells. 229 9 R E D O X P R O C E S S E S The global energy perspective One type of fuel cell is the biological fuel cell, Fuel cells which The combustion of fuels such as oil, uses natural gas releases heat energy which into electrical energy in an energy waste plant. The energy loss in this process can or greater. Fuel cells however can an 70% electrical energy. based redox on A of the fuel energy cell processes. is a The in a fuel cell, the so common hydrogen–oxygen reaction between hydrogen fuel and fuel is a the (g) + O 2 Under (g) → 2H 2 acidic methane or fuel alternative cell to is the non-polluting internal engine, the cell uses problem. storage The of hydrogen methanol liquid methanol rather than fuel hydrogen, type cell , is much easier to transport. Methanol uses be produced from biomass as a carbon- oxygen: fuel (it does not contribute to the O(l) 2 conditions, hydrogen major neutral 2H as is can the such into cell which of chemicals convert fuel voltaic most in materials. efcient combustion approximately from be and 67% electricity electric Although power generate is organic converted to coal, chemical or bacteria greenhouse the following effect). reactions Fuel cells and the International take place at the anode and cathode: Space Station ● anode (negative electrode): oxidation The hydrogen–oxygen fuel cell can be used as an + 2H (g) → 4H + 4e 2 energy ● cathode (positive electrode): reduction Space ve + O (g) + 4H + 4e → 2H 2 cells the only cell so it are they of C.6). are One by lifetime expensive of cell cells is space (ISS) agencies is a The International collaborative representing 15 product nations, of as are the added. of 6 that the very the not are cells and fuel produce. is in requires does Fuel of November inhabited by and humans since 2000. fuel need continuous (gure to cells purication it a is continuously Another reactants disadvantage fuel Water been conventional fuels). that provide impurities or (unlike fossil expensive of devices. hydrogen–oxygen applications very poisoning the because disadvantage their fuel reactants number topic a Fuel electricity more of reactions of recharging. a Station spacecraft. O(l) efcient non-polluting, advantage up, highly product is combustion of in 2 has Fuel source have sub- cells is that Another they fuel, supply used are prone which complex to reduces and fuel. Figure 7 The Japanese pressurized experiment module for the International Space Station, shown here at its manufacturing facility in Nagoya, Japan. The module, called Kibo or “hope” in Japanese, is Japan’s rst human space facility. Experiments in Kibo focus on space medicine, biology, Ear th obser vations, material science, biotechnology, and communications research Figure 6 This bus in Reyjavik, Iceland, is powered by a fuel cell that runs on hydrogen 230 9 . 2 e l e c t r O c H e M i c A l c e l l S Avy 1 2 Find out the cathode and anode half-equations and with wind turbines. Consider a number of the overall cell reaction for the direct methanol fuel countries worldwide where wind farms are located cell. Compare and contrast the methanol fuel cell with and compare and contrast any government the hydrogen–oxygen fuel cell from an environmental regulations that may be in place regarding their perspective. construction. a) Discuss some aspects of what is commonly termed the “hydrogen economy”. Your answer might address aspects such as the advantages and problems of using hydrogen as a fuel in motor cars, the various methods for generating hydrogen, and the use of renewable energy sources such as wind and solar energy. b) Suggest how wind farms may be assisting developing countries in dealing with their energy needs, and so driving the global “hydrogen economy”. Explore what problems wind turbines may pose for rural communities. Research and Figure 8 Example of an on-shore wind farm in Kilmore, Co. discuss any possible health eects associated Wexford, Republic of Ireland Electrolytic cells Electrolysis energy is is used the to process drive a by which ● electrical cathode (negative electrode): reduction non-spontaneous 2+ Pb chemical reaction. An electrolytic cell is used ● this purpose, two which electrodes (the consists cathode of a and single the overall container, anode), (l) + 2e → Pb(l) for cell reaction: a PbBr (l) → Pb(l) + Br 2 solution be (the electrolyte), considered There cell are but at as many SL an electron different you and will be a battery which (g) 2 can pump. types only of electrolytic assessed on the e e electrolysis of a molten salt. Electrolysis of a molten salt such as + graphite electrode graphite electrode lead(II) bromide e e In the PbBr electrolysis (l), inert of molten graphite lead(II) electrodes bromide, are e dipped 2 2+ into the PbBr (l) electrolyte. The following Br half- Pb 2 equations show the processes that take place at the e Br e electrodes: 2+ Pb Br ● anode (positive electrode): oxidation PbBr (l) 2 2Br → Br (g) + 2e 2 Figure 9 Electrolysis of molten lead bromide, PbBr (l) 2 231 9 R E D O X P R O C E S S E S Working method for the electrolysis of a molten salt Step 1: Identify all species present. Step 4: show Step 2: Identify which species are attracted to (negative attracted to the electrode) anode and (positive which and annotate direction of the the electrolytic movement of cell and electrons the and cathode Draw the species the direction of ion ow. are electrode). Step 5: State what would be observed at each electrode. Step 3: Deduce place at the cell the cathode two and half-equations anode and taking the overall reaction. Worked example Describe the electrolysis of molten sodium chloride. Step 4: Solution e Step e 1: battery iner t electrode iner t electrode + NaCl → Na + Cl Cl (g) 2 + So Na(l) and Cl(l) ions are present. porous separator Step 2: NaCl(l) + Cathode (negative electrode): Na + (l) Cl Anode (positive electrode): Cl cathode Step (+) 3: Anode (positive electrode): ( ) oxidation: Figure 10 Electrolytic cell for molten sodium chloride, NaCl(l). This experimental set-up is used commercially in the dows 2Cl(l) → Cl (g) + 2e 2 for the electrolysis of sodium chloride. The liquid sodium Cathode (negative electrode): reduction: metal is less dense than the molten sodium chloride, so it oats on the surface and is collected + Na(l) This + needs number of e to → be Na(l) multiplied electrons from by the 2 to balance anode the Step 5: equation: At the anode (positive electrode): bubbles of + 2Na(l) Overall + cell 2e → 2Na(l) chlorine reaction: At + 2Cl(l) + 2Na(l) → the liquid 2Na(l) + Cl gas are cathode sodium observed. (negative electrode): a pool of forms. (g) 2 Qk qsos 1 Explain why solid lead(II) bromide does not conduct ) State a suitable material for each electrode. ) Identify the direction of movement of electrons electricity. 2 a) Construct and annotate a diagram of the and ion ow. electrolytic cell for the electrolysis of molten ) State what would be observed at each electrode. f) Discuss, with reference to dierences in aluminium oxide. b) Identify the half-equations occurring at the proper ties, why aluminium is used to replace iron cathode and at the anode. in many applications. 232 Q u e S t i O n S Questions 1 Which species could be reduced to form NO ? 5 Which statement about the electrolysis of 2 molten A. N sodium chloride is correct? O 2 B. A. A B. A yellow-green gas is produced at the NO 3 negative C. electrode. HNO 2 D. NO silvery metal is produced at the positive [1] electrode. IB, May 2011 C. Chloride ions electrode 2 Consider voltaic the overall reaction taking place in D. a cell. O(s) + Zn(s) + H 2 O(l) → 2Ag(s) + Zn(OH) 2 What is the role of The positive May are are and attracted undergo attracted undergo to the positive oxidation. to the negative oxidation. [1] 2011 (s) 2 zinc in the cell? 6 A. ions electrode IB, Ag Sodium and electrode and the oxidizing What is the reducing agent in the reaction agent. below? B. The C. The positive electrode and the reducing agent. 2MnO negative electrode and the oxidizing (aq) + Br (aq) + H 4 O(l) → 2MnO 2 agent. D. The (s) + 2 BrO + 2OH (aq) 3 negative electrode and the reducing agent. [1] A. Br B. BrO C. MnO D. MnO IB, May 3 IB, May 2011 4 [1] 2 3 What BrO happens , are to bromine converted to when bromine bromate ions, molecules, Br 3 A. ? 2 It undergoes state B. 2012 It undergoes state reduction changes from 1 and to oxidation changes from 1 and to its oxidation 7 0. Which changes electrode its oxidation could (cathode) take in a place voltaic at the positive cell? 2+ I. Zn II. Cl III. Mg(s) A. I and II B. I and III (aq) to Zn(s) 0. (g) to Cl to Mg (aq) 2 C. It undergoes reduction and its oxidation 2+ state D. It undergoes state 4 changes the unknown from + +5 following metals (aq) +5 to oxidation changes Consider 2XNO from X, Y(s) Y, → and to its 2X(s) of Y(NO 3 Y(NO 2XNO (aq) + Z(s) → no the II and D. I, II, IB, May III and only III [1] 2010 (aq) 2 reaction 2 (aq) + Z(s) → 2X(s) + Z(NO 3 What ) 3 ) 3 C. only three Z. + only oxidation 0. reactions and (aq) 0. ) 3 is the metals order (least A. X < Y < Z B. X < Z < Y C. Z < Y < X D. Y < Z < X IB, May of increasing reactive (aq) 2 reactivity of rst)? [1] 2011 233 9 R E D O X 8 Metal P R O C E S S E S A voltaic is more cell is reactive made as than shown metal B. (gure A standard 10 Describe how the concentration 11). a) V a car factory river voltmeter b) A a in after farmer dissolved a river releases using puts it decrease warm for large oxygen would water if: into the cooling [1] quantities of a fertilizer B salt bridge on IB, 11 a eld May how phosphates solution containing solution containing 2+ A IB, of to the river. [1] 2009 Describe value next a the to addition water water November can of nitrates increase the or BOD sample. [2] 2009 2+ (aq) B (aq) Figure 11 12 Which statement is correct? The nd Winkler the method uses concentration of redox reactions oxygen in to water. 3 100 A. Electrons ow in the external circuit cm of analysed Ato using was this taken from method. The a river and reactions B. taking B. water from Positive ions ow through the salt place are summarized below. bridge 2+ from A to Step B. 1: 2Mn (aq) + 4OH (aq) + O (aq) → 2 2MnO C. Positive from B ions to ow in the external circuit (s) + 2H 2 O(l) 2 + A. Step 2: MnO (s) + 2I (aq) + 4H (aq) → 2 2+ D. Electrons ow through the salt bridge Mn from (aq) + I (aq) + 2H 2 B to A. [1] 2 Step 3: 2S O 2 IB, November O(l) 2 2 (aq) + I 3 (aq) → S 2 O 4 (aq) + 2I (aq) 6 2010 a) State what happened to the O in step 1 in 2 terms 9 A 0.1337 XIO , g was sample of dissolved an in alkali water, metal b) iodate, acidied, State of electrons. the change manganese and in [1] in step oxidation number for 2. [1] 3 an excess of potassium iodide, KI added. The c) 0.0002 moles of I were formed in step 3. 3 resulting the iodine sodium solution thiosulfate required 36.64 pentahydrate cm of Calculate solution, O , the dissolved amount, in the 2 3 Na S 2 O 2 .5H 3 titration the starch relative identify half-equations 234 (25.49 g dm ) for complete IB, using Calculate hence O 2 the solution atomic metal. involved. as an mass Deduce indicator. of all X and relevant November 2009 in moles, water. of oxygen, [1] 10 O R G A N I C C H E M I S T R Y Introduction Organic chemistry containing array From from of is compounds biological foods pesticides the compounds to and and systems fuels, chemistry and their to from fertilizers, of fundamental of our of of a carbon- organic the to dyes will is expansion material we world. be the In and their alkenes, and develop system of of understanding organic rules focus, in reactions. will The of the The nomenclature to functional chemistry of halogenoalkanes, be of compounds. addition important alcohols, benzene an of IUPAC main identication to chemistry the topic, application biotechnology, and this classication vast reactions. paints importance understanding studies the groups alkanes, polymers explored. 10.1 Fnnts of ognc chst Understandings Applications and skills ➔ A homologous series is a series of compounds ➔ Explanation of the trends in boiling points of of the same family, with the same general members of a homologous series. formula, which dier from each other by a ➔ Distinction between empirical, molecular, and common structural unit. structural formulas. ➔ Structural formulas can be represented in full ➔ Identication of dierent classes: alkanes, and condensed format. alkynes, halogenoalkanes, alcohols, ethers, ➔ Structural isomers are compounds with aldehydes, ketones, esters, carboxylic acids, the same molecular formula but dierent amines, amides, nitriles, and arenes. arrangements of atoms. ➔ ➔ Identication of typical functional groups in Functional groups are the reactive par ts of molecules eg phenyl, hydroxyl, carbonyl, molecules. carboxamide, aldehyde, ester, ether, amine, ➔ Saturated compounds contain single bonds nitrile, alkyl, alkenyl and alkynyl. only and unsaturated compounds contain ➔ Construction of 3-D models (real or vir tual) of double or triple bonds. organic molecules. ➔ Benzene is an aromatic, unsaturated ➔ Application of IUPAC rules in the nomenclature hydrocarbon. of straight-chain and branched-chain isomers. ➔ Identication of primary, secondary, and ter tiary carbon atoms in halogenoalkanes, and alcohols and primary, secondary, and ter tiary nitrogen atoms in amines. ➔ Discussion of the structure of benzene using physical and chemical evidence. Nature of science ➔ Serendipity and scientic discoveries – PTFE and superglue. ➔ Ethical implications – drugs, additives, and pesticides can have harmful eects on both people and the environment. 235 10 O R G A N I C C H E M I S T R Y Introduction to organic chemistry TOK Organic The theory of “Vitalism” was based on the belief that a vital force was involved in the chemistry of living organisms. Indeed the word chemistry compounds. form four process bonds, bonds by to which is the eld atoms other many producing chemistry is Carbon of have atoms. identical a wide atoms can are varied studies electrons undergo joined branched, and that valence Carbon straight-chain, therefore chemistry four or eld carbon-based so they together cyclic can catenation, by the covalent structures. Organic ofstudy. “organic” originated from Understanding natural and synthetic organic compounds requires the scientists’ understanding study of chemical bonding and nomenclature, chemical structure, at that time, that organic stoichiometric relationships, functional groups, and reaction mechanisms. compounds could only be The energetics and the of reactions and their role in industry, chemical kinetics, synthesized within living impact of synthetic medicines and drugs on the health of society organisms. This belief are some of the points of focus of organic chemistry. remained until the German chemist Friedrich Wöhler ar ticially synthesized Homologous series urea from the inorganic Classication is a common human activity. Just as biology uses compound ammonium scientic cyanate, NH taxonomy to classify organisms on the basis of shared OCN. 4 characteristics, Are there other examples in group science where vocabulary of and chemists name utilize compounds a unique that share system of important nomenclature features and to patterns reactions. has developed from a A homologous series is a series of compounds that can be grouped misunderstanding that together based on similarities in their structure and reactions. A was the product of the homologous series has the same general formula which varies from one technology of the day? member to another by one CH (methylene) group. 2 Language plays a vital role in the communication The alkane of knowledge and its are subsequent understanding. alkenes Therefore should language that series has the general formula C H n hydrocarbons and contain (they alkynes contain are two carbon–carbon carbon more double and (table 2n + hydrocarbon and triple 1). The alkanes 2 hydrogen only). The homologous bonds, series respectively. be universal so that Homologous series that contain functional groups can also be misnomers arising from described by a general formula and also show similar physical and misconceptions may be chemical properties within the series. The functional groups are the eliminated? reactive as parts oxygen the of and the nitrogen. carbon–carbon functional groups homologous molecules double of series In the that and the and commonly alkene triple series. and contain alkyne bonds Table2 contain elements homologous respectively shows the make structures such series up of the three oxygen. Physical properties of a homologous series The physical gradually boiling as points apparatus point molar rises 236 length of with mass), as in an is the the gure a of in the 1. the table members carbon series: liquid. an number 1. of This a chain alkane Such increasing seen within pentane of of members shown temperature while properties the series can experiment of can butane homologous increases. is carbon be a seen gas at be atoms the room change example, measured shows by series For that (or the using the increasing state at the boiling room temperature, 10 . 1 N Fo F u N d a m e N T a l S O F Conns stct fo O r G a N i C C H e m i S T r y Stct fo Bong ont / °C H methane C CH CH 4 161 4 H ethane C H 2 propane C C 4 pentane C 5 hexane C 6 ▲ CH 8 H CH 10 H CH 12 CH 3 H CH 14 CH 3 CH 2 C H H H 89 2 H CH 2 CH 2 H H H C C C H H H H 42 H H H H C C C C H H H H H 0.5 3 CH 2 H CH 2 CH 2 C 3 CH 2 H CH 2 CH 3 H 3 CH 3 H CH 3 H 3 butane CH 6 H H H H H H C C C C C H H H H H H 36 3 CH 2 H H H H H H C C C C C C H H H H H H H CH H 69 3 T able 1 The homologous series of alkanes Hooogos cohos hs ktons ss Gn C H n fo C OH H H H n 2n+1 H O C 2n H H H n O 2n H O H C C C O H C C C C H OH C C H C H 3 H H H H H H H H H H H H H H H H O H H C C C C H H O H C C C C C H OH C C C H C H 4 H H H H H H H H H H H H H H H H H H H H O H H C C C C C H H H H O C H C C C C C OH H C C C C H C H 5 H H ▲ H H H H H H H H T able 2 The general formula and structural formulae of the homologous series of alcohols, aldehydes, and ketones 237 10 O R G A N I C C H E M I S T R Y This trend forces becomes chain oil is in boiling (London longer. length a chain. are mixture Trends well of boiling in point. fractions A in points to by that is density the vary in the as and viscosity the the length separation mixture apparatus non-volatile strong 4.4) petrochemical physical separate distillation long-chain, a increasingly sub-topic increasing understood simple from forces, distillation boiling from results hydrocarbons Fractional differences point (dispersion) can intermolecular carbon of their process into in carbon industry. a Crude carbon that fractions effectively compounds chain with uses of separate school similar volatile laboratory. thermometer condenser Homologous series have similar chemical proper ties due to the presence of the water same functional group; this is to sink responsible for their overall water chemical reactivity and distillate from faucet the types of characteristic reactions they undergo. ▲ Figure 1 Distillation apparatus incorporating a temperature probe Qck qstons 1 Alkenes are impor tant star ting materials for a variety b) Applying IUPAC rules, state the names of of products. ) isomers B and C. State and explain the trend of the boiling [2] iB, Nov 20 09 points of the rst ve members of the alkene homologous series. b) [3] Describe two features of a homologous series.[2] iB, m 2011 2 The boiling points of the isomers of pentane, C 5 H , 12 A shown in gure 2 are 10 °C, 28 °C, and 36 °C, but not necessarily in that order. ) Identify the boiling point for each of the isomers a, B, and C in a copy of table 1 and state a reason for your answer. iso [3] a B C Bong ont B ▲ 238 T able 3 ▲ C Figure 2 10 . 1 F u N d a m e N T a l S O F O r G a N i C C H e m i S T r y Chemical formulae of organic compounds The structure different In of ways sub-topic organic providing 4.3 structures. an we These compound varying examined are useful levels the to may use of of be represented the H (electron-dot) valence electrons present molecular compounds and polyatomic H C C H H H in ▲ simple H several information. Lewis visualize in Figure 3 Lewis structure ions. of ethane, C H 2 Empirical formulae atoms present actual number formula offer in a of (sub-topic molecule. atoms little or The present no 1.2) represent molecular in the information the formula molecule. about simplest the Both ratio describes these possible the types structure O of of H larger, more Structural complex 6 of C O H molecules. formulae take three forms: full, condensed, and ▲ skeletal. Figure 4 Lewis structure of methanoic acid, HCOOH ● Full structural showing another all in the a formulae atoms and are two-dimensional bonds, and their representations positions relative to one compound. In structural formulae a ● In a condensed structural formula all the atoms and their relative covalent bond between two positions are represented but the bonds are omitted. atoms is represented by a ● A skeletal formula the end atoms table formula where of each present the the carbon line in is and most and each functional basic representation hydrogen vertex groups atoms represents are also are a of not the shown carbon included as structural single line that describes two but atom. shown bonding electrons. For a double The bond two lines are used and in for a triple bond, three lines 4. N propane (sub-topic 4.2). F stct fo H H H H C C C Conns stct fo H CH CH 3 H H H H H H H C C C CH 2 3 H OH propan-2-ol CH CH(OH)CH 3 H O Skt fo 3 H H H H C C O O propanal H C CH CH 3 CHO 2 H H H H O H C C C O propanone H H CH C(O)CH 3 H 3 H H H C propene H CH H C H ▲ C 3 CH=CH 2 or CH 3 CHCH 2 H T able 4 Full, condensed, and skeletal structural formulae can all be used to represent organic compounds 239 10 O R G A N I C C H E M I S T R Y Nomenclature of organic compounds The International world authority substance of needs compound substituents parts The that Union on to which form the the and Applied nomenclature. enough the functional describe alkanes Pure provide from and of chemical information chemical groups compound backbone is the name to derived, present. (gure of Chemistry The The (IUPAC) of a signpost the including name has is the chemical a class any number of 5). IUPAC rules for naming organic compounds. The longest carbon chain that includes the principal group or the most complex cyclic Sux indicating or heterocyclic system the principal group -ane a b c -ene Principal (parent) chain -yne Prexes Suxes indicating in alphabetical order saturation or unsaturation of the principal chain ▲ lngth of Figure 5 Outline of the nomenclature of organic compounds Nomenclature of alkanes N cbon chn 1 Examine the continuous 1 structure carbon of the chain. compound This provides and the determine root name the for longest the alkane meth- (table 2 eth- 3 prop- 4 but- 5 pent- 6 hex- 2 If 5). alkyl the substituents branch sufx will will be change are present, determined from creating by “-ane” to the branched number of chains, carbons the name (table 5). for The “-yl”. H methyl H ▲ T able 5 The IUPAC root names for the C substituent H H H H C C C C H H H H alkane series 3 Sbsttnt Conns n fo When numbering substituent must numbering from on carbon 2. substituent the be longest the left to right Numbering on carbon 5. CH methyl 3 H CH ethyl CH 2 3 H propyl CH CH 2 C H CH 2 3 H butyl CH 2 CH CH 2 2 CH 3 H H C C 1 H ▲ T able 6 Naming alkyl substituents 240 C 2 H H C 3 H H H H C 4 H H C 5 6 H carbon lowest results from chain, numbered right in the to left the position carbon. methyl would In this of any example, substituent incorrectly being have the 10 . 1 4 When there are alphabetical several order different prior to the F u N d a m e N T a l S substituents, root arrange O F them O r G a N i C C H e m i S T r y in name. H ethyl H H H C H C H H H C H C 2 C H C 4 5 H H H C 3 H H C C 1 H substituent 3-ethyl-2-methylhexane 6 H H H methyl substituent H alphabetically, ethyl comes before methyl. 5 Use a comma 6 Use a hyphen 7 The number prexes 8 to to of in words table are 1 mono 2 di numbers. separate multiple shown Successive separate numbers and substituents of letters. the same type is indicated 3 tri 4 tetra 5 penta by 7. merged into one word. ▲ T able 7 Numerical multipliers in the IUPAC nomenclature system To or demonstrate structural compounds formula. 1 the isomers that have Isomers Begin by application have drawing carbon–carbon of the the of rules hydrocarbon same unique the these chemical physical molecule we shall hexane. formula and with Structural but chemical the examine a the isomers isomers different are structural properties. longest straight chain of isos atoms. Isomers may dier from one another in their physical H C CH CH 3 2 CH 2 CH 2 CH 2 3 proper ties. The ability of hexane molecules of one isomer to pack closer together will result 2 Reduce to act so this of the as the a longest methyl derivative chain lowest chain numbered one substituent. of must by hexane result in is carbon The a the and longest chain substituted methyl use the is now pentane. group removed ve The branching carbon carbons numbering off at the in increased intermolecular forces and therefore an increase in the boiling point. Hexane molecules (boiling carbon. point 69 °C) can approach each other more closely than those CH 3 H C of the branched-chain hexane CH 3 CH CH 2 1 2 2 3 derivative 2,3-dimethylbutane CH 3 4 5 (boiling point 58 °C). 2-methylpentane 3 Examine another is the not the isomer carbon. last exist of as If the it is it to is see if substituted the the moved same as to methyl C3 pentane substituent another isomers isomer can is be moved formed. to This (4-methylpentane does 2-methylpentane). CH 3 H C 3 CH CH CH 2 1 2 CH 2 3 4 3 5 3-methylpentane 241 10 O r G a N i C C H e m i S T r y 4 In a similar way, substituted Qck qston now butane remove another compounds from carbon the atom original and create hexane: Applying IUPAC nomenclature CH CH 3 CH 3 3 rules, state the name of each of the H molecules shown in gure 6. C C 3 a) CH 3 CH CH 2 CH CH 2 2 CH 2 CH H CH 2 1 C CH 3 3 3 4 CH CH 3 1 2 3 4 CH 3 3 2,2-dimethylbutane CH 2,3-dimethylbutane 3 CH 3 b) CH 3 CH C CH 2 CH 3 CH CH 3 2 CH 3 CH Saturated and unsaturated hydrocarbons Hydrocarbons 3 CH c) CH 2 CH 3 2 organic atoms only. bonds single bonds. or CH are hydrogen are triple In a compounds saturated Unsaturated carbon–carbon bonds. The consisting compound of all compounds simplest carbon the contain example of and carbon–carbon a double and/ saturated 3 hydrocarbon is methane, CH , a member of the alkane family. Alkanes 4 CH 3 CH CH 3 are CH 2 CH aliphatic The CH 3 This 3 ▲ naturally compounds occurring (gure 7). mixture is extracted from hydrocarbons beneath the come Earth’s from crude surface, re ned, CH C 3 and CH of straight-chain 3 oil. d) majority or CH 2 separated petroleum, by fractional butane, and distillation into useful substances such as kerosene. Figure 6 The of mixture mainly of Cycloalkanes (gure hydrocarbons alkanes, 8), are ring whereas consisting of that cycloalkanes structures aromatic alternating makes and that up contain hydrocarbons single and crude aromatic double oil single or is a combination hydrocarbons . carbon–carbon arenes are ring carbon–carbon bonds structures bonds. Functional groups Tens is of millions constantly companies and organic as new chemical products of Natural compounds organisms. reactions All compounds ▲ of rising these compounds exist compounds are industries. involving found containing Synthetic both in substances natural plants are specic in the world synthesized and man-made animals functional into are number are the compounds. synthesized classes groups the pharmaceutical compounds and organized and by (table of by organic 7). Figure 7 Computer-generated 3D models When naming compounds that contain a functional group, the position of methane, ethane, and propane, the of the group is identied by giving the number of the carbon atom to first three members of the alkane series which it groups is attached. take priority When over numbering substituents the and carbon atoms, carbon–carbon functional multiple bonds. Unsaturated hydrocarbons The or primary triple left to right, double Figure 8 Computer-generated model of the bond numbering ▲ chain in unsaturated carbon–carbon methyl would is from be bond. H H H H C C C C C hydrocarbons the substituent at right H If C3. to molecule would However, left as must below branch the off double include was from bond the double numbered C2 takes and from the priority shown. 4-methylpent-2-ene. Note the use of hyphens and the fact cyclohexane, a cycloalkane H 5 H 4 3 CH 3 242 2 H 1 H that the substituent is named before the functional group so 10 . 1 Css Fncton go Sx alkanes ― -ane F u N d a m e N T a l S O F O r G a N i C Gn fo C H e m i S T r y ex Constructing 3-D models C Propane C H n H 3 2n + 2 8 of organic compounds is an excellent interactive but-2-ene alkenes C alkenyl C -ene H C n technique, enhancing 2n CH=CHCH CH 3 3 visualization of the molecule C C and mutual orientation of but-2-yne alkynes -yne C H n 2n 2 alkynyl individual atoms. Models can C≡CCH CH 3 3 enhance understanding of a benzene arenes phenyl ― C variety of concepts from the H n 2n 6 H C 6 naming of organic molecules, 6 visualizing stereoisomers 2-chlorobutane ―X halogenoalkanes ― C H n (including optical isomers) X 2n + 1 (X = F, Cl, Br, I) CH CH(Cl)CH 3 CH 2 3 to complex reactions OH mechanisms (sub- topic 20.1). butan-2-ol alcohols -ol ROH CH(OH)CH CH hydroxyl 3 CH 2 3 O ethanal aldehydes aldehyde -al C RCHO CH CHO 3 H C propanone ketones -one RC(O)R′ O C(O)CH CH 3 3 carbonyl O -oic C carboxylic acids ethanoic acid RCOOH acid OH COOH CH 3 carboxyl O methyl ethanoate C -oate esters RCOOR′ COOCH CH O 3 3 ester O ethoxyethane ethers ― ROR′ CH ether CH 3 OCH 2 CH 2 3 H N propan-1-amine RNH 2 CH H CH 3 amines -amine N CH 2 NH 2 2 RNHR′ N-methylethanamine N RN(R′)R′′ H CH CH 3 NHCH 2 3 amino O C ethanamide amides -amide RCONH 2 CONH CH NH 2 3 2 amido C N nitriles propanenitrile -nitrile RCN CH cyano ▲ 3 CH CN 2 T able 7 A summary of classes of organic compounds showing their functional groups 243 10 O R G A N I C C H E M I S T R Y Aliphatic compounds H H H H H H H H H H Cl H H H C C C C C H Cl H H H O H C C C H C H C C C C H C C C H O H H H H H H H OH H H H butan-2-ol Number chain so that group the from the is numbered butanal carbon right to The left for functional on the H carbon atom. there this group aldehyde denition lowest carbon functional an propanone on the atom is no is has the state be general on In For this group multi-substituted halogenoalkanes, compound functional only a RC(O)R ′. three-carbon so to ketone formula terminal (C1), need The by 2,2-dichloropentane carbon can along C2. number with signify number. the halogens in H H H H C C H H O C C C H O H halogens they are H H C C H H are listed order. H H C C H H C C O H of If H H H to number alphabetical O H used prex present. different present, a the is H C O H H H butanoic acid Note that carbon the atom functional By convention, the functional counted group of the carboxylic acids, carbon aldehydes, esters and amides functional are positioned at the right-hand the end of the structural formulae there of organic compounds. include Esters in the group in the chain. group is The hydrogen the functional COOH. The atom are The on is O R ′ the group group C1 in is no need the the position, substituent to rst, number the name. is OCH alkyl followed name of group alkoxy so CH ethoxy named as alkanes. called 2 terminal named substituted alkyl replaces the in Ethers acid substituent the is are derivatives. longest As ethoxyethane methyl propanoate is the 3 group. by the acid anion. PTFE, a fortunate discovery PTFE or polytetrauoroethene polymer that commercial polymer is more name, which is commonly Teon. means it It is can a be a synthetic known by its thermoplastic moulded This in uoropolymer 1938, a product polymerization retaining its new shape upon a properties low coefcient electrical surface all of and PTFE is of a chemical friction, and solvents. as high thermal qualities, known implements. 244 include high very The non-stick low most resistance, melting insulation, point, non-stick solubility common surface on discovered iron-catalysed tetrauoroethene gas. describes fortunate accidental cooling. discoveries Its accidently the when Serendipity heated, of was of in use cooking stems and from within ashes serendipity. “superglue” discovery. risk is of good.” the (IB of Methyl and scientic Scientic inspiration, another “Ethical assessment parts science. cyanoacrylate example discussions, the Chemistry of of an or accidental risk-benet precautionary way endeavour imagination, addressing syllabus.) analyses, principle the are common all H 10 . 1 F u N d a m e N T a l S O F O r G a N i C C H e m i S T r y Classifying molecules: primary, secondary, C o x tcton on and tertiary compounds th wo We shall look at three classes of compound to understand the effect A large majority of the world’s oil on their chemical reactions of changing the position of the functional reserves are controlled by a small group on the carbon chain. We shall look at the carbon attached to the number of countries. Countries functional group: a primary carbon atom is bonded to one other carbon with oil reserves in decreasing atom, a secondary carbon atom to two other carbon atoms, and a order of their magnitude include tertiary carbon atom to three other carbon atoms (gure 9). Saudi Arabia, Venezuela, Canada, Iran, Kuwait, United Arab Emirates, Classifying halogenoalkanes Alkanes undergo (sub-topic known 10.2) as and the reaction. is This formed idea Whether be reactions with mono-substituted depends will Russia, Libya, Kazakhstan, Qatar, substitution resulting halogenoalkanes. halogenoalkane the free-radical a on primary, the developed alkanes secondary, conditions in and sub-topic USA, China, Brazil, Algeria, and halogens Mexico. These countries dier are or widely in their economic, social, tertiary mechanism of and political demographics. 20.1. It is instructive to compare countries that are net expor ters H 3°carbon with those that are net impor ters 1°carbon C H 2°carbon H of crude oil. Nt xo ts produce H H H C C C H H H H H H C C C H Cl H H H more barrels of oil than they H Cl H H H C C C H H Cl H consume, while nt o ts consume more barrels of oil than they produce. For many years 1 -chloropropane 2-chloropropane 2-chloro-2-methylpropane the USA has been the largest net ▲ Figure 9 Primary (1°), secondary (2°), and ter tiary (3°) halogenoalkanes impor ter of crude oil, but in 2013 it was over taken by China. China’s wish to reduce its heavy reliance Classifying alcohols on non-renewable energy sources Alcohols The can position when the be classied of the alcohol potassium in the hydroxyl group undergoes dichromate(VI) same way halogenoalkanes determines oxidation or as in potassium the the (gure products presence of manganate(VII) 10). formed is reected in its domestic and international policies – China acidied is a world leader in investment (sub-topic 10.2). in renewable energy sources, par ticularly wind energy. H A country’s reliance on oil 1°carbon H 2°carbon C H 3°carbon shapes its global policies. H H H C C C H H H H H H C C C H OH H H H Relationships with countries that H OH H H H C C C H OH H H are net expor ters are of paramount propan1 -ol propan-2-ol impor tance and the need to secure energy supplies for a nation can 2-methylpropan-2-ol result in governments overlooking ▲ Figure 10 Primary (1°), secondary (2°), and ter tiary (3°) alcohols perceived political dierences of trading par tners. Biofuels are in Alcohols plants. alternative large-scale it to substances fuels to and other petrol production traditional whose fossil of energy biofuels (gasoline) ethanol fuels. Fossil is are and from fuels derived from increasingly diesel. sugar Brazil cane remain the for carbon being has xation used as undertaken decades, primary St t adding source of When classifying compounds as energy on a global scale, but the complex mixture that makes up crude primary, secondary, or ter tiary, oil contains components that can be used for the synthesis of various focus on the carbon that is bonded products – from dyes and cosmetics to pesticides and polymers. The ever- to the functional group. increasing in the combustion depletion materials for a not vast of only array valuable of of the non-renewable fuels themselves, substances that are a fossil but part fuels also of of our could result valuable daily raw lives. 245 10 O r G a N i C C H e m i S T r y Classifying amines Octn nbs of fs An amine is classied as a primary, secondary or tertiary amine The fractional distillation of crude depending on the number of alkyl groups bonded to the nitrogen atom oil or petroleum separates this of the functional group (unlike halogenoalkanes and alcohols, which mixture of hydrocarbons into consider the carbon atom next to the functional group).When naming various fuels including natural amines the root loses the -e and is replaced by “amine”. “ N” signies that gas, gasoline or petrol, kerosene the substituent, namely the methyl group in gure 11, is bonded to the used in the aviation industry, nitrogen atom rather than the carbon atom. diesel fuel used in transpor t, construction, and agriculture, 1°amine 2°amine 3°amine and fuel oil for heating. H H H H The octn nb is a standard H H H H H H C C C H H H CH CH 3 H C C C H H H N H C C C H H H 3 N H N method of describing the H H CH 3 performance of fuels used in cars and aircraft. The octane number propan1 -amine N-methylpropan- N, N-dimethylpropan- 1 -amine 1 -amine is not indicative of the energy content of the fuel but is rather ▲ Figure 11 Primary (1°), secondary (2°), and ter tiary (3°) amines a way of describing its ability to combust in a controlled manner without causing excessive Aromatic hydrocarbons ngn knockng. In simple The alkanes, alkenes, and alkynes are differentiated by the presence of terms, engine knocking is the single, double, or triple bonds respectively. Aromatic hydrocarbons result of uncontrolled detonation are characterized by the presence of the benzene ring. The German of the air–fuel mixture in a scientist August Kekulé (1829–96) proposed a ring structure for benzene combustion engine. This is more C H 6 composed of six carbons bonded together by alternating double and 6 common in fuels that have a low single bonds. This structure would result in an unsymmetrical molecule octane number. with carbon–carbon bonds of different bond lengths. Automobile manufacturers Benzene crystallizes upon cooling and analysis of X-ray diffraction are multinational companies patterns generated from the crystalline substance revealed that all six that produce and sell their carbon–carbon bonds in its molecule have identical bond lengths of products globally. The octane 140 pm. It is now understood that the carbon–carbon bonds in benzene rating system is expressed are intermediate in length between single (154 pm) and double (134 pm) dierently in dierent countries, carbon–carbon bonds and thus have a bond order of 1.5. Electrostatic making comparisons difficult. potential mapping of benzene (gure 12) conrms that all the carbon The sch octn nb atoms have equal electron density, so the molecule (RON) is the most common method of determining the octane number, though the oto octn nb (MON) is also used either in isolation or in combination with the RON. Thus methods of repor ting fuel octane numbers at the pump can vary from country to country, leading to inconsistency in communication to consumers. ▲ 246 Figure 12 An electrostatic potential map of benzene is symmetrical. 10 . 1 F u N d a m e N T a l S O F O r G a N i C C H e m i S T r y 2 Benzene bonded contains to one six sp another hybridized and each carbon carbon is atoms (sub-topic bonded to a 14.2) single hydrogen 2 atom by atoms and sigma overlap below electrons one the over structures of bonds. p another, plane the The of the orbitals forming six a carbon six carbon nuclei benzene shown in H of the sp hybridized continuous atoms. can gure six be π The bond carbon that lies delocalization represented by the above of the π resonance 13. The Kekulé structure of H H H H H benzene was drawn as a series of alternating double H H H H and single bonds. H The circle H within the six-member ring H structure represents a system H H of delocalized pi electrons that H H are evenly distributed between H ▲ Figure 13 Resonance structures of benzene Hydrogenation in the presence carbon–carbon is of benzene of hydrogenation is as in the the fact would much enhanced benzene topic be less readily 20.1) aliphatic decolorize benzene but and does cyclic than π the of this, or the and A do; for one the difference this reactions as isomers suggested reactions other for if by from of the stability benzene exists that known evidence resulting example, two as is change show energy, energy, of three hydrogenation enthalpy in substitution isomer be bonds of hydrocarbons contained experiments molecule addition would double unsaturated benzene consequence electrophilic Only there this If times However, benzene show water. three to change delocalization electrons. not 10.2). enthalpy and compounds compounds; single hydrogen approximately undergoes bromine alternating bonds, energy the of (sub-topic cyclohexene. stability of addition catalyst of resonance delocalization the a double of it each of the six carbons. is that (sub- unsaturated does not 1,2-disubstituted the benzene ring had Kekulé. TOK August Kekulé is best known for his discovery of the structure of benzene. While many scientic discoveries are the product of reasoning suppor ted by evidence obtained through observation and experimentation, some discoveries are born from moments of inspiration or ashes of intuition, as well as a healthy imagination with a high degree of creativity. Kekulé is said to have visualized the cyclic structure of benzene in a dream. While the impor tance of evidence is universally accepted in the scientic community, there is an understanding of the role of less analytical ways of knowledge in the acquisition of scientic knowledge. To what degree do these ways of knowing play a par t in the acquisition of new knowledge, and can you think of some recent examples? ▲ Figure 14 August Kekulé 247 10 O r G a N i C C H e m i S T r y 10.2 Fncton go chst Understandings Applications and skills Alkanes: Alkanes: ➔ Alkanes have low reactivity and undergo free➔ Writing equations for the complete and radical substitution reactions. incomplete combustion of hydrocarbons. Alkenes: ➔ ➔ Alkenes are more reactive than alkanes and Explanation of the reaction of methane and ethane with halogens in terms of a free- undergo addition reactions. Bromine water can radical substitution mechanism involving be used to distinguish between alkenes and photochemical homolytic ssion. alkanes. Alkenes: Alcohols: ➔ ➔ Alcohols undergo nucleophilic substitution Writing equations for the reactions of alkenes with hydrogen and halogens and of reactions with acids (also called esterication symmetrical alkenes with hydrogen halides or condensation) and some undergo oxidation and water. reactions. ➔ Halogenoalkanes: ➔ Halogenoalkanes are more reactive than Outline of the addition polymerization of alkenes. ➔ alkanes. They can undergo (nucleophilic) Relationship between the structure of the monomer to the polymer and repeating unit. substitution reactions. A nucleophile is an Alcohols: electron-rich species containing a lone pair that ➔ Writing equations for the complete combustion it donates to an electron-decient carbon. of alcohols. Polymers: ➔ ➔ Writing equations for the oxidation reactions Addition polymers consist of a wide range of of primary and secondary alcohols (using monomers and form the basis of the plastics acidied potassium dichromate(VI) or industry. potassium manganate(VII) as oxidizing Benzene: ➔ agents). Explanation of distillation and reux Benzene does not readily undergo addition in the isolation of the aldehyde and carboxylic reactions but does undergo electrophilic acid products. substitution reactions. ➔ Writing the equation for the condensation reaction of an alcohol with a carboxylic acid, in the presence of a catalyst (eg concentrated Nature of science ➔ sulfuric acid) to form an ester. Use of data – much of the progress that has Halogenoalkanes: been made to date in the developments and ➔ Writing the equation for the substitution applications of scientic research can be reactions of halogenoalkanes with aqueous mapped back to key organic chemical reactions sodium hydroxide. involving functional group interconversions. 248 10 . 2 F u N C T i O N a l G r O u p C H e m i S T r y Converting one functional group to another Chemical form the research the reactions basis into of natural in a to both advance the improve their to to the A of has their and and of so on the they to utilize can various food supplies populations, develop be of S o c i e t y ’s ensure global to involves chemistry molecules. communities, research that the chemists organic developing drives subsequently data Scientific production synthesis of interconversions compounds. their enabled new health life and wealth create group organic structure achieve developed quality of compounds groups pathways functional synthesis l a b o r a t o r y. functional reaction desire of pathways produced organic involving the determination design for for new and organic compounds. Alkanes Alkanes are the simplest hydrocarbons. Wi t h low bond polarity and 1 strong covalent carbon–carbon bonds (bond energy 346 kJ mol ) 1 and are carbon–hydrogen relatively inert. bonds (bond H o w e v e r, energy alkanes do 414 kJ undergo mol ), they some important reactions. The combustion of alkanes Alkanes in are commonly combustion liquid to Alkanes and gas) decreases used as fuels as fuels, Volatility as the tend to releasing (the length be large tendency of the to carbon short-chain amounts change chain molecules of energy state from increases. such as butane octane. Alkanes undergo oxygen. This and used reactions. water. complete highly Carbon contributing to hydrocarbons dioxide global with combustion exothermic has a warming. octane reaction in signicant Petrol present in or the the presence produces carbon of environmental gasoline highest is a excess dioxide impact, mixture of proportion. 1 _ C H 8 (l) + 12 O 18 (g) → 8CO 2 (g) + 9H 2 O(g) ∆H = -5470 kJ 2 2 Qck qston Incomplete combustion When oxygen combustion. gas, is is in In produced. reducing its limited this It supply reaction irreversibly oxygen-carrying alkanes carbon binds undergo monoxide, to Deduce balanced equations for incomplete which hemoglobin in is a the poisonous blood the complete combustion of: thus ) propane b) pentane c) hexane. capacity. 1 C H 5 (l) 12 + 5 O 2 (g) 2 → 5CO(g) + 6H O(g) ∆H = -1830 kJ 2 249 10 O R G A N I C C H E M I S T R Y Wtng ognc chnss Gob o ts to c gnhos gs ssons “Curly arrows” are used to The work of environmental organizations has long focused on the ght illustrate the movement of to reduce greenhouse gas emissions and slow the rate of pollution that electrons in organic reaction accompanies economic development throughout the world. The US mechanisms as bonds are Environmental Protection Agency estimates that globally 80 million tonnes of broken and made. A sh- methane annually (28% of global methane emissions) can be attributed to hook arrow is used to show ruminant livestock . Countries such as Brazil, Uruguay, Argentina, Australia, and hootc sson, breaking a New Zealand contribute dispropor tionally large amounts of greenhouse gases bond to produce two par ticles for their levels of population and economic development, due to the scale of that both have a single their livestock industries. unpaired electron, a radical. Landll in developed countries contains an increasing amount of organic “green” The half arrow represents the waste and domestic kitchen waste. In anaerobic conditions common in landll movement of a single electron sites, microbes produce methane in vast quantities. This form of anaerobic as the bond breaks: respiration is known as thnognss. Governments and environmental agencies are developing technologies to reduce these emissions, using the gas to generate electricity for domestic power grids through methane capture systems. Governments and local councils in some countries are collecting green A full arrowhead shows the waste to compost, avoiding the waste going to landll and contributing to the movement of a pair of electrons production of methane. during htotc sson, when both electrons move together to form a new bond: The halogenation of alkanes Alkanes to When drawing mechanisms are increase the relatively their inert, reactivity. and One chemists way of often work achieving this to is activate to alkanes halogenate alkane. using curly arrows: Common ● addition, originate from the exact individual location of the electrons In being moved. single an from ● and atoms addition with in other reaction molecule, the studied elimination. while two organic chemistry Substitution single atoms molecules elimination is is or are the include the substitution, replacement with added a small together removal of two of group to of atoms. produce a substituents molecule. The arrowhead must accurately nish at the exact destination of the electrons. ● reactions The base of the arrow must Alkanes can unsaturated three types undergo alkenes of free-radical and reaction alkynes. listed substitution Alkenes and and elimination alkynes can to form undergo all above. The arrow commences at an electron-rich region and Free-radical substitution ends at an electron-poor An example of free-radical substitution is the reaction between methane region of the molecule. and chlorine The term in the presence of UV light. Students are not required to free-radical refers to a species that is formed when a molecule make the distinction between undergoes homolytic ssion: the two electrons of a covalent bond are sh-hook arrows and full split evenly between two arrowheads when drawing have a single electron: reaction mechanisms. In this text, only full arrowheads will be used in mechanisms. 250 A B A + B atoms resulting in two free-radicals that each 10 . 2 Heterolytic electrons ssion involved of in a bond creates the bond are a cation unevenly and split an F u N C T i O N a l anion, between as the G r O u p C H e m i S T r y the two atoms: + A + A B When methane B reacts halogenoalkane with chlorine chloromethane H is in the presence of UV light the produced: H hv H C H + H C H Cl + H Cl H methane There chlorine are three reactions: chloromethane stages initiation, involved in such propagation, free-radical and substitution termination, described below. Initiation The homolytic light produces ssion two of the chlorine chlorine radicals molecule that have in a the short presence of UV lifespan. hv Cl Cl 2 Cl initiation Propagation The rst propagation chlorine free-radical. Cl + H CH H stage involves Cl a reaction between methane and a CH 3 3 propagation 1 The production chain reaction producing chlorine CH + the the set methyl up. desired radical Cl of is that The allows radical halogenoalkane, can Cl radical methyl Cl now CH 3 take + part the reaction reacts with a chloromethane, in the rst to continue chlorine along with propagation as a molecule a reaction. Cl 3 propagation 2 Termination A termination mixture. concentration radicals, Cl Cl step 3 C the reactions concentration become of the hydrocarbon slowing the rate of Cl Cl Cl CH Cl CH 3 H reduces Termination CH 3 more begins reaction and to of radicals prevalent decrease. eventually in the when They reaction the “mop stopping it up” the completely. 3 H 3 C CH 3 termination reactions 251 10 O R G A N I C C H E M I S T R Y Alkenes Alkenes are unsaturated carbon–carbon alkenes more undergo double reactive addition hydrocarbons bond. than The the that presence contain of the corresponding at least double saturated one bond makes alkanes. Alkenes reactions. Test for unsaturation The presence using the of a double addition of bond bromine in a hydrocarbon water, Br (aq). A can be mixture demonstrated of the alkene 2 and bromine water will undergo a colour change from brown to colourless: C H 2 (g) + Br 4 (aq) → 2 mechanism change with absence of of Br 4 (aq) 2 colourless this bromine the H 2 brown The C reaction water is this carbon–carbon is shown a in gure negative double 1. result, If there is no indicating the CH CH bond. + CH CH 2 CH 2 CH 2 2 2 2 δ+ Br Br Br Br Br induced dipole Br δ ▲ ▲ Figure 1 Testing for the presence of a C=C bond by the addition of bromine water Figure 2 Bromine water is decolorized by gaseous ethene 252 colour 10 . 2 F u N C T i O N a l G r O u p C H e m i S T r y Addition of hydrogen: hydrogenation ethn n nng ft Large quantities product of ethene thermal hydrocarbons, organic of are used decomposition ethene is an in or the chemical catalytic important raw industry. cracking material in of The Ethene is used in the food long-chain the production industry to accelerate the of ripening of fruit. Fruits are polymers. generally picked while they In the presence of 150 to produce of a nely divided nickel catalyst at a temperature are still unripe to enable °C, ethene will undergo an addition reaction with hydrogen gas them to be transpor ted from the saturated alkane ethane: farm to supermarket without Ni C H 2 (g) + H 4 → (g) 2 C H 2 ∆ becoming damaged and (g) 6 looking unappealing to the This type of reaction is very important in the food industry. The addition consumer. Ethene is a natural of hydrogen to unsaturated fats and oils occurs in the manufacture of par t of food ripening, as it is margarine. Removing the carbon–carbon double bonds increases the released by ripening fruit. melting point, making a substance that is solid rather than liquid at Exposure of the fruit to ethene room temperature. increases the rate of ripening, The partial results solid oils be in at as hydrogenation an elevation room they stored saturated a of Partial past the in the health double work decade been bonds of the have introduced consumer (see diet also trans-fats to avoid Cis- and t increase which Raised results length for the are levels uses the in the of the LDLs negative trans- fats many of of these content producers time a are in the is a hydrogenated product can However, concentration conversion dangers use of while preventing the build- which up of ethene around fruit by introducing carbon dioxide to the container holding the fruit slows the rate of ripening. of transport associated with B.3). community, the partially blood therefore margarine involved of oils consumption. trans-carbon–carbon aware food the vegetable creating industry life, sub-topic into point, remain (LDLs) reduce more from share. shelf recognized to food and scientic multinational market The the blood. risks large and in hydrogenation carbon the oils lipoproteins cholesterol increased of and melting prolonged supermarket fats low-density a polyunsaturated the temperature. have in of of have are trans-fats of cis-carbon– bonds. governments harmful and and discussed resulting in detail a result the legislation foods. distanced As over products processed also publicity of double and has make Many themselves loss in of sales sub-topics and B.3 B.10. Halogenation of alkenes The electrophilic addition of halogenation elemental of halogens symmetrical such as alkenes chlorine, Cl or 2 resulting C H 4 H in (g) a dihalogenated + Br 8 (g) → 2 H H H H C C C C H + H but-2-ene H 4 Br 2 H C involves the bromine, Br , 2 alkane: Br 8 (l) 2 H H H H H C C C C H Br Br H H 2,3-dibromobutane 253 10 O R G A N I C C H E M I S T R Y The addition in single a two are of a alternative discussed products in sub-topic 20.1). C + H 4 H hydrogen halide, mono-halogenated (g) this are topic HBr(g) → possible; C H 4 H H H C C C C H H + and steam a C H in (g) petrol, + H of H a 6–7 symmetrical are covered at alkenes HL in H H H H C C C C H H Br H H a ethanol of a is achieved catalyst, by reacting phosphoric(V) acid, ethene at 300 °C MPa: O(g) → C H 2 variety creating of presence 2 has only alkenes results alkene Br(l) HBr production the 4 Ethanol alkene unsymmetrical 2-bromobutane pressure 2 symmetrical an 9 H large-scale with a however, but-2-ene The to With (unsymmetrical 8 H HX, alkane. of OH(g) 5 uses biofuel including (sub-topic as an additive to gasoline or 10.1). Polymerization of alkenes The plastics world, variety they of can have a a The monomer H 2 one of → the range Polymers negative ethene largest of impact is on the industry, the the quality of bodies used our in widely lives the for a although environment. of bond, supplied undergoes polymers reaction double monomer, manufacturing addition improve carbon–carbon petrochemical nC is broad polymerization contain polymer. the a purposes. Addition that industry producing to addition many linking the small monomers together plastics to form industry polymerization to a by the produce polyethene: [ CH 4 CH 2 ] 2 n St t C C When drawing diagrams to C C C represent polymerization, it is C C impor tant to draw continuation C bonds through the brackets. C C C C C ▲ 254 C C C C Figure 3 The polymerization of ethene C C C C C C C C C C C 10 . 2 Any monomer that polymerization. structure by a single adjacent of of contains The the monomer bond, a carbon–carbon repeating and monomers. the For structural that formed electrons example, double unit it, with released gure of 4 the the bond can polymer double forming shows F u N C T i O N a l new the G r O u p C H e m i S T r y undergo reects bond the replaced bonds to the polymerization propene. n CH CH CH 2 CH 2 CH CH 3 3 n ▲ Figure 4 The polymerization of propene Alcohols Alcohols form applications Alcohols a diverse and can play a undergo group of compounds signicant complete part in that have synthetic combustion, a wide range of reactions. releasing carbon dioxide ▲ and Figure 5 Colour change during the water: 2 reduction of orange Cr O (aq) 7 2 3+ C H 2 OH(l) + 3O 5 (g) → 2CO 2 (g) + 3H 2 O(g) ∆H = -1367 kJ ions to green Cr (aq) ions 2 Soc ctons of coho Oxidation of alcohols conston Acidied potassium manganate(VII) equations to for construct dichromate(VI) can these these be used for oxidizing equations the (gure or oxidation agents was 5) are as of potassium alcohols. follows. developed in (The sub-topic The half- working method Excessive alcohol consumption is a growing problem in many countries. Bng nkng is 9.2.) sometimes dened as drinking + MnO (aq) + 8H 2+ (aq) + 5e → Mn (aq) + 4H 4 2 Cr O O(l) + (aq) + 14H 3+ (aq) + 6e → 2Cr (aq) + 7H 7 2 5 (for a man) or 4 (for a woman) 2 alcohol units over a 2-hour period O(l) 2 and increasing the blood alcohol The oxidation products of alcohols depend on the type of alcohols content above 0.08% by volume. involved. An alcohol unit varies from country to country. In general, one alcohol unit is equivalent Primary alcohols 3 to approximately 10 cm The oxidation of a primary alcohol is a two-stage process that of rst pure alcohol. Having reached produces an aldehyde followed by a carboxylic acid. Potassium alarming proportions amongst dichromate(VI), K Cr 2 manganate(VII), O 2 is a milder oxidizing agent than potassium adolescents and young adults in 7 KMnO . When the primary alcohol ethanol, 4 is heated with acidied This K aldehyde H 2 Cr 2 produced. C O 2 can , the aldehyde ethanal, 7 be CH CHO OH 5 further oxidized to the many western societies, binge is drinking is having a signicant 3 carboxylic acid impact on economies, social ethanoic acid, CH COOH. structure, law and order, and 3 ultimately health systems. H H Associated health risks include H H O O physical and psychological H C C H C C C O H H H H C H dependence on alcohol, liver and H brain damage, elevated risk of ethanol ethanal ethanoic acid cancer of the throat, mouth, and esophagus, depression, anxiety and social problems at work and within the family. 255 10 O R G A N I C C H E M I S T R Y The aldehyde Carboxylic acids are capable preventing of forming dimers, paired point molecules held together by forces: hydrogen bonds (gure 6). The carboxylic increased size of the molecule have can its than the recovered acids have have boiling by oxidation. carboxylic aldehydes higher be further acid weak the The due to process distillation , has differences dipole–dipole stronger of aldehyde a in lower the intermolecular intermolecular boiling intermolecular hydrogen forces bonds while and so points. leads to stronger van der Waals’ If the carboxylic acid is the desired product, the aldehyde must remain forces and a higher boiling in the reaction of time. mixture with the oxidizing agent for a longer period point. Instead Reuxing δ O 3 a of the distillation technique that apparatus involves the a reux cyclic column is evaporation used. and δ+ H condensation O does CH is C C not of a volatile reaction mixture, preserving the solvent as it evaporate. CH 3 H δ+ δ Secondary alcohols ▲ Figure 6 A dimer of ethanoic acid The oxidation formation of H of a a secondary alcohol such as propan-2-ol results in the ketone: H H H H 2H H C C H H C O H removed C C H H H C O H H oxidation Upon as formation the carbon attachedto of the atom of ketone, the no further functional oxidation group has no is possible hydrogens it. Condensation reaction of an alcohol and a carboxylic acid Esters are derived applications and from ranging carboxylic from acids avouring acid and an is a reversible alcohol concentrated are sulfuric reaction heated in the a variety of medications to solvents CH 3 COOH(l) IUPAC carboxylic 7). occurs of CH OH(l) SO acid names the CH The acid esters name second (in a carboxylic catalyst, this of CH 3 methanol of when a normally (conc) 2 4 ________ → 3 from “methyl”). (gure + 2 propanoic derived that presence acid: H CH 256 have and explosives. Esterication The and agents consist the word case, of alkyl is COOCH 2 two words. The rst in the alcohol composed of the root followed by + H O(l) 2 propanoate chain “prop”), (l) 3 methyl word (in name the sufx this of is example, the “anoate” 10 . 2 F u N C T i O N a l G r O u p C H e m i S T r y methyl is derived -anoate is the from the alcohol, methyl proponate sux for esters methanol -prop- signies three carbons present in the carboxylic acid parent molecule ▲ Figure 7 Naming esters ▲ Figure 8 3D computer-generated image of methylpropanoate. Grey = carbon, white = hydrogen, red = oxygen Nucleophilic substitution reactions: An introduction Once form an a types of bond, unreactive reaction. C X, the electron-rich a lone pair aqueous polarized X There being are of These An of the distinct of aqueous (ethanol) and CH 3 now carry atom a reaction part polar open attack Nucleophiles full in to other carbon–halogen to negative NaOH(aq), charge by a take by contain charge. 18.1). hydroxide, susceptible halogen reaction will The to on the attack the sodium releases + mechanisms mechanism present, be namely discussed nucleophilic Cl(g) 2 it is (sub-topic positive can contain carbon sometimes sodium makes substitution nucleophiles. bases partial reaction. mechanisms CH of The halogenoalkane example with . and as a molecule by contains carbon the nucleophile the atom in hydroxide will result the ion. in an formed. two substitution class known Lewis bond substitution alcohol as solution :OH The undergone resulting halogenoalkanes electrons act nucleophile C Since has the electron-decient species of Nucleophiles An alkane halogenoalkane, OH chloride (aq) → This ion, CH primary, is the , the nucleophilic depends secondary, of produces leaving OH(aq) on or the tertiary. 20.1. reaction reaction Cl CH 3 this occurs sub-topic substitution hydroxide. a in for that + Cl chloroethane an alcohol group. (aq) 2 257 10 O R G A N I C C H E M I S T R Y Electrophilic substitution reactions: An introduction As discussed addition reactions. accepting While the in An an than illustrating pair. In unique of an acts as requires a Lewis are a 20.1 of of the reaction the is species undergo of substitution capable (sub-topic to the π will be of 18.1). electrons leads mechanism to of in substitution the nitration developed, benzene. the takes base readily ring benzene properties Lewis acid attracted stability substitution This not electrophilic electron-poor reaction reaction does undergo sub-topic bromine. and is the electrophilic elemental environment ring, substitution) the It benzene will electrophiles benzene addition. example with it electrophile electron-poor (electrophilic 10.1, Instead electron aromatic rather An sub-topic reactions. reaction place (FeBr or in of an AlBr 3 ) benzene anhydrous as the catalyst. 3 FeBr 3 C H 6 One + ___ → Br 6 C 2 bromine H 6 atom Br + HBr 5 from replaces Br a hydrogen atom in benzene; 2 the remaining product, an 258 hydrogen organic products bromine and bromide. (non-aqueous) are omitted. hydrogen Please atoms note environment, that so form this the the inorganic reaction states of takes by- place reactants and in Q u e S T i O N S Questions 1 Which be a A. three compounds homologous CH OH, CH 3 can be considered 5 to Alkenes CH 3 OH, are important series? CH 2 CH 3 CH 2 a) OH The an reaction CH CH 3 OH, CH 2 CHO, CH 3 CH CH 3 CH(OH)CH 2 ) 3 , CH 3 CH 3 CH 2 CH 2 CH CH 2 CH 2 OH, CH 2 CH 3 OCH 2 CH 2 a test ) CH 2 May Deduce the colour in the change water is added to when chloroethene. [1] the Lewis CHO identify the structure formula of of chloroethene the repeating [1] 2 of the polymer poly(chloroethene). [2] 2009 Besides uses the IUPAC name for CH CH 3 A. 1,1-dimethylpropane B. 2-ethylpropane C. unsaturation , c) is for Describe 3 unit What water COH CH 3 2 bromine OH, and IB, with 3 3 (CH chemically 2 b) D. alkenes and compounds. COOH bromine (CH of organic 3 laboratory. C. of 2 provides B. economically family CH(CH 2 )CH 3 IB, ? May 6 State IB, May reactions of state two commercial alkenes. [2] 2010 and explain molecules tertiary 3-methylbutane polymerization, the 3 2-methylbutane D. of (gure whether 9) are the following primary, secondary, or halogenoalkanes. [4] [1] 2009 H C 3 Which conditions are required to obtain is yield of oxidized a carboxylic using acid potassium when H H C C C H H H H H H C C C Cl H H C H Cl H Cr 2 O 2 H C C H H H H C H dichromate(VI), H C Add sulfuric II. Heat III. Distil the reaction mixture under product as the oxidizing I and II B. I and III C. II and D. I, II, Figure 9 IB, May 2011 only only III 7 Consider only the reaction following 1 RCH 3 and May III sequence reaction _____ → RCH IB, c) agent added A. b) reux ▲ the H acid a) is H (aq)? 7 H I. H C ethanol H K H a H good H _____ → Br of reactions: 2 RCH 2 OH 2 [1] reaction 3 _____ → RCOOH 2009 RCH is an unknown alkane in which R 3 represents 4 Alkenes are important starting materials for of State and points alkene b) of explain the rst the ve homologous Describe two series. The trend of the alkane May contains members of the series. features of a Determine showing boiling b) [3] Equal [2] your 81.7 % by mass of of alkane mass, signicant its formula, [3] carbon are gures, at dioxide found measured pressure. empirical working. volumes unknown same homologous and IB, group. products. carbon. a) alkyl a a) variety an to the Deduce an to have the the accuracy same the and of two temperature molecular 2011 formula of the alkane. [1] 259 10 O R G A N I C c) (i) C H E M I S T R Y State for (ii) the reagent reaction State the and conditions needed 1. 10 [2] reagent(s) and Alkenes for reaction 3. Reaction 1 Describe giving involves the a free-radical stepwise equations propagation to and economically of mechanism. mechanism, The represent the termination in by reaction of the provides initiation, steps. when test and chemically compounds. with for bromine unsaturation Describe bromine the water colour is added to chloroethene. [1] [4] Deduce the Lewis chloroethene November alkenes a laboratory. change b) IB, organic [2] water d) an family conditions a) needed are important structure and identify of the formula 2010 of the repeating unit of the polymer poly(chloroethene). 8 a) Identify A E, the formulas formed in the of the organic reactions, I products, CH (CH 3 II. ) 2 (CH ) 3 OH + K 8 Cr 2 + CBr + NaOH of H __ → O 2 Besides polymerization, commercial + H __ I. c) IV: → A [2] uses of the state two reactions alkenes. [2] B 7 IB, __ → May 2010 C 3 + H __ III. (CH ) 3 IV . H CHOH + K 2 Cr 2 C=CH 2 + __ → Br 2 → O 2 D 7 11 E Chloroethene, H C=CH 2 can react to form a polymer. of polymer and draw the of a section of this consisting of three repeating units. Draw the Lewis predict c) compounds, C H 4 Draw a A and D, each have the with a is reacted hydroxide formula of to C to C with produce H 4 O. potassium Compound sodium D is of C H 4 react with Deduce B B is then C with resists a manganate(VII) formula further of oxidation by the May manganate(VII). to O. 2011 dilute aqueous compound Compound potassium structural A, with produce E does E with not 10 acidied compounds 260 compound Compound potassium reacted hydroxide formula IB, aqueous 10 compound C dilute 8 acidied a O. acidied Compound with produce H 4 oxidized of B, C, manganate(VII). formulas D, and E for [5] six why structure H – C of organic polymer – for Cl chloroethene bond angle. [2] poly(chloroethene) of May 2010 carbon the economic disposal IB, sodium Outline is Cl. A the section 9 Compound the paper containing formula important [2] b) Specimen Two an polymer and 9 is manufacture structural a) IB, Cl, 3 to poly(chloroethene). type formula used Name 2 this H 2 compound b) C [5] 2 atoms. [1] polymerization importance plastics is a and of alkenes why problem. the [2] M E A S U R E M E N T A N D D ATA 11 P R O C E S S I N G Introduction Analytical techniques chemistry. As appreciate the we must limit of taken power realize that precision into lie chemists, the very only analysis, any and account of at not core we but in measurement accuracy, when do and to addition has this evaluating that of need we can graphical be experimental the means, analytical (IR), uncertainty effectively identication a must underpin how mass of and in measurement, represent examine organic of infrared (MS), and by spectroscopic compounds, techniques spectrometry data the looking at spectroscopy proton nuclear 1 results. In this topic we will explore the principles magnetic resonance spectroscopy ( H NMR). 11.1 Uc as a os masum a sus Understandings Applications and skills ➔ Qualitative data includes all non-numerical ➔ Distinction between random errors and information obtained from obser vations not systematic errors. from measurement. ➔ ➔ Record uncer tainties in all measurements as a Quantitative data are obtained from range (+) to an appropriate precision. measurements, and are always associated with ➔ Discussion of ways to reduce uncer tainties in random errors/uncer tainties, determined by the an experiment. apparatus, and by human limitations such as ➔ Propagation of uncer tainties in processed data, reaction times. including the use of percentage uncer tainties. ➔ Propagation of random errors in data processing ➔ Discussion of systematic errors in all shows the impact of the uncer tainties on the experimental work , their impact on the results, nal result. and how they can be reduced. ➔ Experimental design and procedure usually ➔ Estimation of whether a par ticular source of lead to systematic errors in measurement, error is likely to have a major or minor eect on which cause a deviation in a par ticular the nal result. direction. ➔ ➔ Calculation of percentage error when the Repeat trials and measurements will reduce experimental result can be compared with a random errors but not systematic errors. theoretical or accepted result. ➔ Distinction between accuracy and precision in evaluating results. Nature of science ➔ Making quantitative measurements with replicates to ensure reliability – precision, accuracy, systematic, and random errors must be interpreted through replication. 261 11 M E A S U R E M E N T A N D D ATA P R O C E S S I N G Qualitative and quantitative analysis According to IUPAC, qualitative and quantitative analysis can be The analytical Analysis distinguished as: Quaav aayss Substances are identied can chemist be ● qualitative ● quantitative of is two often described as the chemical detective. types: analysis analysis. or classied on the basis of their chemical or physical proper ties, such as chemical reactivity, solubility, molar Uncertainty in measurement In science numerical data can be divided into two types: mass, melting point, radiative ● data involving exact numbers (that is, the values are known exactly – proper ties (emission, there is no uncertainty) absorption), mass spectra, ● nuclear half-life, etc. data involving degree of inexact numbers (for these types of numbers there is a uncertainty). Quaav aayss As scientists, when we carry out a particular experiment involving The amount or concentration of measurement, there will always be some uncertainty associated with the an analyte may be determined measured data, that is the data will involve inexact numbers to some (estimated) and expressed degree. Such uncertainty may be associated with factors such as the as a numerical value in instruments used in the laboratory. For example, the mass of a sample of appropriate units. potassium bromide, KBr(s), will depend on the type of balance used. In ‘Nomenclature in evaluation of a typical school laboratory, top-pan balances often read to 0.01 g, but an analytical methods including analytical balance, to 0.0001 used in more precise analytical experiments, can read detection and quantication at least g or better) (gure 1). Uncertainty may also depend capabilities’ Pure and Applied human error. Chemistry, 67(1699), (1995), p1701 Data may also be classied as qualitative or quantitative data: Quaav aa Qualitative data includes all non-numerical information obtained from observations not from measurement. Quaav aa Quantitative data are obtained from measurements, and are always associated with random errors/uncer tainties (dened shor tly) determined by the apparatus and by human Figure 1 (a) A top-pan balance used to measure mass in a typical school laboratory limitations, such as reaction times. can read to 0.01 g. (b) An analytical balance used to measure mass to a high degree of precision can often read to 0.0001 g. The shutters on the balance should be closed to reduce both air ow and dust collecting which can both aect the reading 262 on 11 . 1 U n C e r t A i n t i e S A n d e r r O r S i n M e A S U r e M e n t A n d r e S U lt S An example of a awed experiment in science The OPERA experiment – a case-study involving CERN and LNGS Neutrinos neutral and are that are one make as unlike electrical OPERA of up considered but, Gran a greater The In betwe e n that international the do L a b ora tor i in tha n Gra n to ve loci ty all ove r of of This in wor ld , of that the nd i ngs the in wa s and the re In in of the imp o r ta nce of the it by that in fa c t , whi ch a ws l ed v el ocity li g ht . the ex pe ri me n t al the in ar e , of te s ti n g to s et r e por te d ne utri nos und e r s ta nd ing there or i g in a l wa s of Albe rt S ub s eq ue n t l y, wer e , 2 0 12 wi th not hi n g s tau n c h ly s a id e q uip ment, pr o b l e ms a bou t tha t pos tula t e d e x p e ri me nt. val i d i ty and conce rned as velocities the and science l i g ht, tha t consistent shows v e ry consi de r ed ho w e ve r, th e timing uncertainty but their in set-up reliability li g ht . than reported indeed, at was measurements. Nature ma de the the of aw possible po ur I t a ly) tr a vel no was in Na zi on al i d is co ve ry hea d l i ne s it t he Ge n eva , Sa ss o, a ppe a r the in was were it OPERA, defensive an as faster Einstein. r e se a rch Euro p é e n ( C ERN), scie nti c c ar r y fr om physicists ndings travels be e le ct r on not r e sults Co ns e il (L NGS ) m ig h t inte r na ti ona l ne utr i no s velocity proposed the to the r e ac t ion s pa r t ic l es T he y the y 2 0 11, (an Nuc l é a i r e Sasso suggested s i m il ar many e le ct r ic a l ly nucl ea r f und ame nta l ele ctr o n, and in uni v e r se . bein g charge. Switzerland, del the experime nt Recherche s ma ll , p r o d uce d the the collaboration la extr e me l y particles id ea r e su l t s of me a s ur e me nt. A xamp of mpac of os fom sac spac Cas of Mas Cma Ob Spaccaf Dr. Edward Weiler, NASA’s Associate Administrator for Space Science stated the following: In 1998, NASA launched the Mars Climate Orbiter, a space probe designed to examine the climate on the People sometimes make errors. The problem here was not planet Mars. However in 1999, the spacecraft crashed the error, it was the failure of NASA’s systems engineering, and completely disintegrated as it approached Mars at and the checks and balances in our processes to detect an incorrect altitude (http://mars.jpl.nasa.gov/msp98/ the error. That’s why we lost the spacecraft. orbiter/). The reason for the crash related to an error in It was subsequently repor ted that one team used imperial the transfer of information between the spacecraft team units while the other team used SI units for a pivotal based in Colorado, and the mission navigation team based operation for the space probe, which led to the incorrect in California, USA. trajectory required to place the probe on Mars. Dierence between precision and accuracy In order need to to understand consider the the idea difference of uncertainty between in measured precision Precision According of to under by we accuracy. Accuracy IUPAC, agreement obtained and values, precision between applying stipulated random part of shortly) which the the closeness test experimental conditions. the is independent The experimental results procedure smaller errors the According of the to IUPAC, agreement measurement (which is the accuracy between and a true particular the results, the more value of quantity closeness of the to a measurand be measured). for the presentation of results precise of the the result (dened ‘Nomenclature affect is the chemical analysis’ Pure and Applied Chemistry , procedure. 66(595), (1994), p598 263 11 M E A S U R E M E N T A N D D ATA P R O C E S S I N G Aaogy Case 3: high precision and high accuracy The analogy of the dar tboard is useful in appreciating the dierence between precision and accuracy (cases 1–3). Case 1: high precision and low accuracy As scientists we always our experimental not only do need to work . strive That to is, replicate for any Case 3 in experiment Case 2: low precision and low accuracy we we be need able each time In experiment an and low with calibration are of a to the be same pH level we errors often accurate, reproduce where accuracy, instrument to of have the in addition experiment consistency. high associated common but (for precision with the example, poor meter). Signicant gures Signicant a given greater the quantity. a gures certainty In order measurement, notation SF to it is the to the The number greater the of digits number about the numerical know the number useful (sometimes represent For refer measurement. to called number of express of the signicant the signicant measured exponential signicant of value of the reecting precision gures, measured gures For calculated associated parameter notation). or in of the with scientic convenience, let gures: example: Masum Scc oao numb of SFs 2 135.680 g 1.35680 × 10 3 0.00620 dm 6.00 kg 3 6.20 × 10 3 six 3 dm 6.00 kg 2.0600 m g three three 3 2.0600 m ve 1 0.2 mg mg 2 × 10 one 2 300 kg* 3 × 10 kg* one *If a number is expressed with no decimal point, for example 300 kg, then it is assumed that the zeros are not signicant. Hence 300 kg has just one SF. 264 11 . 1 In numerical gures ● For always two an result ● calculations should employing be simple be number For operation be number of decimal e r r O r S measured account. i n M e A S U r e M e n t quantities, These multiplication based signicant are A n d r e S U lt S signicant handled based on the or division: measurement with The the gures involving expressed of with into involving expressed smallest should dealing taken A n d rules: operation should an U n C e r t A i n t i e S on addition the or subtraction: measurement with the The result smallest places Worked example: using signicant gures The mass of 35.4200g. a sample The mass bottle of the and a piece empty of sample aluminium bottle is metal 28.9200 is g. If 3 the aluminium displaces 2.41 cm of water, calculate the density of 3 aluminium, in g cm Solution d = m V density = = mass of of volume aluminium aluminium of aluminium. m _ d = roug V Let m = mass of empty sample bottle = 28.9200 g = mass of empty sample bottle + aluminium In calculations in chemistry 1 we often have to round o Let m = 35.4200 g 2 numbers. If the digit to be m = m m 2 = (35.4200 g 28.9200 g) = 6.5000 g removed for rounding purposes 1 is less than the number ve, In nding m, the operation is subtraction, so the value of m is the digit immediately before it expressed to the smallest number of decimal places, which is four for will not be changed. However, both m and m 1 2 if the digit is equal to ve or 3 d = (6.5000 g)/(2.41 cm ) greater than ve, the number immediately before it will be increased by one. ve SF three SF 3 For example 3.42 cm rounded 3 Hence, the result should be correctly reported as d = 2.70 g cm , 3 to two SFs would be 3.4 cm as the operation involves division and to nd the result from the 3 while 23.46 cm quotient we used the smallest number of signicant gures which , 3 or 23.45 cm will rounded to three SFs would be be three from the volume, V 3 23.5 cm Signicant carefully as the gures in power composed associated calculations. of to which two The the with log base logarithms of can a (logs) number be raised parts: to is need dened get that to be handled mathematically number. A log is Suy p You can ignore SFs associated with physical constants given in the Data booklet for the ● The characteristic ● The mantissa – the integer part purposes of calculations – only – the decimal part consider SFs for measured data. 265 11 M E A S U R E M E N T A N D D ATA P R O C E S S I N G The number signicant base that the 10 and digits logs involve natural equation. of gures to base log, For in for the base logs is mantissa logarithmic the 10 ln, a e, are that is used, in pH, for the (this natural common frequently example, indicates entity number covers logs, both ln). whereas example in of logs to the Calculations in kinetics the Arrhenius in: 1 log (2.7) = 0.43 = 4.3 × 10 10 two the number with the SFs 2.7 mantissa contains mantissa (the two SFs, decimal so the part) (decimal answer having part) should two SFs. two be SF reported Here is another example: ln(6.28) three the number with the 6.28 SFs 1.837 mantissa contains mantissa = (the three decimal SFs, (decimal so part) the part) answer having three three SFs should be reported SFs. Experimental errors As stated already, associated types Sysmac o of with every it. This experimental ● systematic ● random single is measurement termed has a experimental degree error. of uncertainty There are two error: error error. According to IUPAC, systematic error is the mean that would Systematic errors result from an innite number Systematic of measurements of the same design measurand carried out under the or errors with measured are the associated with instrumentation quantity will always a aw used. be in the actual Systematic greater or less experimental errors than imply the that true value. repeatability conditions minus Systematic errors can be further classied into three types: a true value of the measurand. ● instrumentation ● experimental ● personal errors raom o methodology errors According to IUPAC, random error is the result of a errors. measurement minus the Examples of systematic errors: mean that would result from an innite number of ● Faulty ● Errors gas syringes that have associated leakage ( instrumentation error). measurements of the same in the readings taken from a pH meter due to faulty measurand carried out under calibration of the instrument ( instrumentation error). repeatability conditions. ● Poorly insulated calorimeter in a thermochemistry experiment International Vocabulary (experimental methodology error). of Basic and General Terms in Metrology, Second Edition, ISO, 1993. 266 ● Measuring or burette the volume from the top of of a colourless the liquid meniscus in a instead graduated of from the cylinder bottom. 11 . 1 Such a systematic manipulation ● Evaporation methodology ● of error data would lead (experimental volatile liquids on A n d to e r r O r S greater a M e A S U r e M e n t volumes methodology heating i n in A n d r e S U lt S the error ). sample ( experimental error). Occurrence being of U n C e r t A i n t i e S of side-reactions measured which (experimental can interfere methodology with the parameter error ). Usfu souc ● The ● parallax exact colour of a solution at its end point ( personal error). The BPIM document Evaluation of error associated with reading a graduated cylinder Measured Data – Guide to the Expression incorrectly of Uncer tainty in Measurement can be accessed at http://www.bipm.org/en/ publications/guides/gum.html. BiPM The mission of BIPM (Bureau International des Poids et Mesures, correct reading of volume at whose headquar ters are based in bottom of meniscus of water Paris, France) is to ensure worldwide uniformity of measurements and their traceability to the iaoa Sysm of Us (SI). With the authority of the Convention of the Metre, a diplomatic treaty between 55 nations, BIPM functions through a series of consultative Systematic errors can often be reduced by adopting greater care committees, whose members are the to the experimental design . Such errors are consistent and can be national metrology laboratories of the detected and ultimately corrected. signatory states, and through its own Systematic errors will affect the accuracy of the results. experimental programmes. BIPM carries out measurement-related research. Par t Random errors of the work of BIPM involves looking at international comparisons of national Random errors occur because of uncontrolled variables in an measurement standards as well as experiment and hence cannot be eliminated . They can, however, performing calibrations for its member be reduced by repeated measurements. Random errors affect the states. As a result of collaboration precision of the results. between seven international Examples of random errors: organizations, including IUPAC and the ● estimating a quantity particular instrument apparatus (for which eg example, lies with a a between marked graduations spectrophotometer) or of a measuring International Organization for Standards (ISO), the Evaluation of Measurement Data – Guide to to the Expression of burette). Uncer tainty in Measurement was rst ● not being able to read an instrument due to uctuations in readings published in 1995. This was revised in that occur during measurements due to changes in changes in the 2008 and has been widely adopted in surroundings (for example, temperature variations, airow, changes most countries; it has been translated in pressure) into several languages. ● reaction time. 267 11 M E A S U R E M E N T A N D D ATA P R O C E S S I N G Absolute and relative uncertainty Uc ay of masum According to IUPAC, uncertainty of measurement is a parameter, associated with the result of a measurement, that Suppose m, in of a the in an experiment sample of laboratory. aspirin Your you use a top-pan (2-acetoxybenzoic recorded mass was balance acid) m = that 3.56 to measure you have the mass, synthesized g. characterizes the dispersion of the values As stated, there will be a degree of uncertainty associated with every that could reasonably be attributed to the single measurement. So here there is an uncertainty associated with the measurand. instrument used, International Vocabulary of Basic and been between General Terms in Metrology, mass should in 3.55 be this g case and reported the 3.57 as g. top-pan Hence, follows in balance. the your The mass uncertainty laboratory is could 0.01 g have and the notebook: Second Edition, ISO, 1993. m Any = (3.56 ± 0.01) experimental experimental Where the A An uncertainty ● Absolute the ● result result represents uncertainty result Relative absolute result, g in = the A can (A strictly a given is or the ∆A, to the experimental speaking the in form: unit the result and absolute ∆A represents uncertainty relative. margin of measurement. uncertainty uncertainty, is reported ∆A) absolute uncertainty from ± be measured or, be should ratio the uncertainty Its symbol comparing size of the is the associated with ∆A. size measured of the experimental A ∆A _ relative uncertainty = A Example 3 A calibrated burette has an absolute uncertainty of ±0.02 cm . During Pcag (%) av uc ay = 3 a titration, the volume of a 0.15 mol dm solution of hydrochloric acid ______ ∆A relative uncer tainty ( ) × 100 % 3 at the end point was recorded as 12.25 cm . Calculate the associated A relative uncertainty. Solution 3 absolute uncertainty (∆A) = ±0.02 cm 3 measured experimental result ( A) = (12.25 ± 0.02) cm 3 ∆A _ relative uncertainty _ = 3 = ru 1 2 × 10 3 A When adding or subtracting measurements, the absolute uncertainty associated with the Note that each other relative uncertainty is dimensionless, since the units cancel out! net measured parameter is the square root The relative uncertainty is often expressed as the percentage relative of the sum of the squares of the absolute _____ uncertainties that is √ 2 uncertainty, so in this example the percentage relative uncertainty Σ∆A would be 0.2% ru 2 When multiplying or dividing measurements, Propagation of uncer tainty the relative uncertainty associated with the net measured parameter is the square root of the sum of the squares of the relative uncertainties. After identifying quantities, combine the to give propagation 268 the next the of uncertainties step is to associated gure resultant out how uncertainty . uncertainties, and in with experimentally these This order is to different what do is this measured uncertainties termed two the rules are applied. 11 . 1 Percentage U n C e r t A i n t i e S A n d e r r O r S i n M e A S U r e M e n t A n d r e S U lt S error Pcag o = Percentage relative uncertainty is different to percentage error. literature value experimental value ________ × 100% For example, change CaCO of the the (s), literature value decomposition was found to be for the reaction +178.1 of standard calcium enthalpy literature value carbonate, kJ: 3 CaCO (s) → CaO(s) + CO 3 The (g) ∆H = +178.1 kJ 2 experimental percentage error value is was given found to be +172.0 kJ. Hence the by: 178.1 172.0 __ percentage error = ǀ × 100% 100% = 3.4% ǀ 178.1 6.1 _ = ǀ ǀ × 178.1 tOK The vertical lines, |...|, used in the expression for percentage error represent the idea of a modulus mathematically, that is any negative value is considered positive. This is also used in describing the modulus of a complex number in mathematics, |z|, which is the distance the complex number is, in the form z = x + iy, from the origin. ______ 2 |z| is expressed as √ (x 2 + y ) . This entity has to be positive. tOK The same symbols are often used to represent alternative meanings in dierent Science has been described scientic disciplines. For example, in chemistry we use ver tical lines in cell as a self-correcting and diagram notations to represent dierent phase boundaries. In both physics and communal public endeavour. chemistry, ver tical lines of dierent lengths represent a cell, used in a battery, To what extent do these that is | | where the shor ter ver tical line represents the negative pole and the characteristics also apply longer ver tical line represents the positive pole. Equally, even in chemistry we to the other areas of often use the square brackets, [ ], symbol for dierent purposes, for example knowledge? concentration, idea of a complex, etc. Worked examples Example 1 The mass of a sample bottle metal is 33.2901 g. The bottle is 26.3505 g. If 4.506 g and mass of a piece the of empty titanium V is the sample density of titanium at 298 K, calculate the volume of titanium. = V is 3 cm of m _ d the volume The mass (m) of titanium sample is: 3 water, in cm , displaced at this temperature by m = 33.2901 g 26.3505 g = 6.9396 g themetal. (this operation express the involves reported subtraction result based on so you the Solution smallest ● d is the density of number of decimal places, which is titanium; four). m is the mass of titanium; 269 11 M E A S U R E M E N T D ATA 6.9396 m _ V A N D P R O C E S S I N G g __ = Solution 3 = = 1.540 cm 3 d 4.506 g cm + pH = -log [H ] = 3.75 10 (this operation express the involves reported division result so based you on The mantissa here contains two SFs. the + [H 3.75 ] = anti-log of 3.75 = e 10 smallest number of SF , which is four). 4 = expressed Example 2 State the number of signicant gures 1.8 as × two 10 3 mol dm SFs. associated Example 5 with the following: A a) 0.00390 kg of calibrated burette has an absolute uncertainty Cu(s) 3 of ±0.02 cm . During a titration, the volume of a 3 b) 136.250 g of NaCl(s) 0.10 mol dm solution of hydrochloric acid at the 3 end was percentage Solution First point express both masses in scientic recorded relative as 22.18 cm . Calculate its uncertainty. notation Solution 3 3.90 × 10 kg of Cu(s) 3 absolute uncertainty (∆A) = ±0.02 cm 2 1.36250 × 10 g of NaCl(s) measured a) three b) six experimental result ( A) = (22.18 ± 3 SFs 0.02) cm 3 SFs (note the last zero is ∆A _ signicant). relative uncertainty ( _ ) = ( 3 ) A 4 = 9 × 10 Example 3 percentage (%) relative uncertainty = 3 Calculate the perchloric pH acid, of a 0.020 HClO mol dm solution of 4 (9 × 10 ) × 100% = 0.09% (aq). 4 Example 6 Solution During Perchloric acid completely is a strong dissociated in acid, so is assumed to be solution: a titration the following titres were 3 recorded for a hydrochloric 0.10 acid mol from dm a solution of burette: + HClO (aq) + H 4 O(l) → H 2 O (aq) + ClO 3 3 (aq) initial 4 titre = (5.00 ± 0.02) cm (21.35 ± 0.02) cm 3 + pH = -log [H 10 O ] = -log 3 nal (0.020) titre = 10 3 3 0.020 mol dm 2 = 2.0 × 3 10 mol Calculate dm the uncertainty which involves pH = two SFs. volume of this delivered, in cm , and the volume. Therefore: Solution 1.70 3 volume (since the mantissa must contain two In Also note also there are no units for pH since based on a logarithmic = 21.35 5.00 = 16.35 cm order to obtain the uncertainty in this volume it we is delivered SFs). need to use the expression: expression. _____ 2 √ ΣΔA Example 4 since The pH of a carton of orange juice was found to a subtractive operation is involved. _______________ be 2 uncertainty 3.75. Calculate the + [H 270 ], hydrogen ion = √ [(0.02) 2 + (0.02) ] = 0.03 concentration, 3 in mol dm 3 The volume would be reported as (16.35 ± 0.03) cm 11 . 1 U n C e r t A i n t i e S A n d e r r O r S i n M e A S U r e M e n t percentage Example 7 A n d relative r e S U lt S uncertainty in concentration = ______________ (13.3 ± 0.1) g of sodium chloride salt, 2 NaCl(s), is dissolved of water, of sodium in H a ask O(l). containing Calculate the (2.0 ± 0.1) 2 10 _ √[ ( 3 dm 10 _ ) + ( ) 13.3 = ] 5% 2.0 concentration 3 2 concentration = 6.7 g dm (±5%) 3 and the chloride, percentage concentration. dissolved in in g relative Assume the dm , in the solution uncertainty that the salt is of The this absolute uncertainty subsequently fully from can also be found this: solution. 5 × 6.7 _ ∆A = 3 = ±0.3 g dm 100 Solution 3 concentration 13.3 = (6.7 ± 0.3) g dm g _ concentration = 3 = 6.7 g dm 3 2.0 Notice that because is this division required on dm answer is is expressed involved calculation. and Check as two rounding this using Example 8 SFs up The your literature change of value for combustion the of standard methanol, enthalpy CH OH(l), 3 1 calculator! When was multiplying or dividing found to be 726.0 kJ mol : measurements, 3 _ CH the percentage the sum of the uncertainty squares of is the the square root percentage of OH(l) + O 3 (g) → CO 2 (g) + 2H 2 O(l) 2 2 relative 1 ∆H = -726.0 kJ mol c uncertainties: The percentage relative experimental value was found to uncertainty 1 be in mass 680.0kJmol = Calculate the percentage error, correct to two 0.1 _ ∆A _ × 100% = A × 100% decimal places. 13.3 percentage relative uncertainty Solution in volume = ( 726.0) ( 680.0) __ percentage error = ǀ ( ǀ 726.0) × 100% 0.1 _ ∆A _ × 100% = A × 100% 46.0 _ 2.0 = ǀ ǀ × 100% = 6.34% 726.0 Writing thermochemical equations Thermochemical equations can be written with non-integer 3 stoichiometric coefcients (for example, O ratios. molecules, only of when considering 2CH molecule of an are equation used, oxygen molecules it as being is we is considered would broken better to in not a write in of terms consider reaction! the of two- Hence, equation for this as: OH(l) 3 when integers thirds reaction a However, indicative 2 2 mole (g)) + 3O (g) 2 → 2CO (g) 2 + 4H O(l) 2 271 11 M e A S U r e M e n t A n d d AtA P r O C e S S i n G 11.2 Gapca cqus Understandings Applications and skills Graphical techniques are an eective means ➔ Drawing graphs of experimental results, ➔ of communicating the eect of an independent including the correct choice of axes and scale. variable on a dependent variable, and can lead Interpretation of graphs in terms of the relationships ➔ to determination of physical quantities. of dependent and independent variables. Sketched graphs have labelled but unscaled ➔ Production and interpretation of best-t lines ➔ axes, and are used to show qualitative trends, or cur ves through data points, including an such as variables that are propor tional or assessment of when these can and cannot be inversely propor tional. considered as a linear function. Drawn graphs have labelled and scaled axes, ➔ Calculation of quantities from graphs by ➔ and are used in quantitative measurements. measuring slope (gradient) and intercept, including appropriate units. Nature of science The idea of correlation can be tested in experiments, the results of which can be displayed graphically. ➔ Graphs and correlation In science a graph representing interpreted. statistical or of is, versus the two X, We as an plotted considered two variables. on the (that way is, sets In a the already joy of discovery described detective. chemist that (hence the Part to analytical of data (that one is chemist role of statistics, all about within detective a and the set of can be described as a and technique that 272 direction of the This is extent one where parallel is of the to one second idea experiments can represented coefcient a is be deduced measure relationship between represented by various a the the with two one in which variable correlation whose results can be is possible a on a useful linear of two scatter points by from the the plots to correlation strength variables. graph. way the symbol, that The Data show This the are the often scatter correlation quantify relationship r. of the between extent the two statistical indicates relationship while coefcient, of in correlation versa. of change graphically. coefcient analogy). in measure decrease increases vice a Correlations the between in thedata set. The degree from and or is correlation or negative tested variables measure A can it variables positive variable be is, two increase is of Correlation A decreases plotted that the the statistical and embedded chemical the explore chemistry patterns the is involve analytical of which another. graph effect) variables; to variables data and of another. any of variable x-axis of be displayed chemical relationships data is useful subsequently independent variable analytical the is very y-axis. have the a can between random the cause) the be which Dependence dependent on can relationship between Y data, two sets 1 to+1 (gure 1): value of r can range 11 . 2 y G r A P h i C A l y t e C h n i Q U e S y x x r = +1 x r = 0 r = -1 Figure 1 Sketches of various scatter plots showing dierent correlation coecients, r. The independent variable is the caus, represented on the x-axis. The dependent variable is the c, represented by the y-axis ● r = +1, is indicative relationship (all of a points perfect lie on a positive straight lead linear and ● r = 0, no linear complete relationship scatter of exists (there to predictions, government line) r = -1, is and of a perfect negative variable increases, the graphs, barriers, – all the data points will lie on line but the gradient will be used in representations diverse analysis, Interpretation ● Graphical effect lead ● of to the ● a have only graphs appropriate are a an by have of units Units ● the slope ● the intercept, ● the idea or of a as of health largely facilitate transcend communication worldwide. Chile, where At the a research language presenters is of of benet of the but the Spanish, variable, would pictorially and tongue is presentation research communicating dependent native a chemist clearly Arabic from given in may the understand ndings. This represented not Spanish graphical shows the scientic data! the can quantities. unscaled variables generally axes, that would and are are used proportional not need to be to or shown variables. and scaled Drawn axes, graphs and are always based display on the variables. features gradient but as often means a physical such labelled for on in whose language, data can effective which can scientists understand population trends variable measurements. number of the as widely modelling. labelled trends, proportional. quantitative There are are such statistical determination sketch, Drawn elds independent graphs data climate these qualitative inversely on of and techniques an Sketched show subject nance, of setting negative). Oman Graphical the such a used straight areas, other conference decreases underpins linear between (one so many education. Charts points) indicative relationship in is language ● and policies of a that line, you are required to know for graphs: m c “best-t” line The slope or gradient of a line, m Mathematically, angle, the θ, that x-axis. In the the slope line order to of makes nd a line, with the m, the slope is the tangent positive of a line, of direction you need the of to 273 11 M E A S U R E M E N T A N D D ATA P R O C E S S I N G choose two points on the line, ideally well separated from each y other, (x , y 1 ) and 1 of the 2 1 = x ∆x the line negative -x 2 there x-axis, ). _ = Where y -y 2 _ c , 2 y ∆y m (x is 1 an incline gradient in the will of the have positive a line in the positive direction of positive value; the direction where x-axis, the there of is a gradient the decline will have a value. x Figure 2 The intercept, c, is the point where the line cuts the y-axis at x = 0 The intercept, c The intercept, c, is the point where the line cuts the y-axis at x = 0 y (gure The 2). intercept can be found ● using extrapolation ● using the by two methods: c equation Sometimes when graph appropriate with you of plot a line, a y graph, scales = it where mx is + c more the convenient x-axis scale to begins draw at a the point x greater than zero as the data points may not be located at zero on the Figure 3 The method of x-axis. If this is the case, by extrapolation you can simply extend the ex trapolation involves ex tending a line back to the y-axis to ndc (gure 3). line back to the y-axis to nd c Alternatively, you could choose some point on the line, (x , y c ), and use c y the equation c = y of the line y = mx + c to nd c, as long as you know m: mx c c The idea of a “best-t ” line When that, lie you plot although exactly on data there the obtained is line. a from linear For this an experiment relationship, purpose, it is not you all best to may the nd data draw a points line of x best t (gure 4). Remember, this line may not necessarily contain all Figure 4 Line of “best t” experimental data points. Worked examples: using graphs 3 Example 1 Beer’s A law = where is The based A the relationship: 0.1002 0.130 0.2008 0.270 is the c is absorbance, the ε is the concentration, 0.2819 0.380 0.4000 0.540 0.5082 0.685 0.6000 0.810 Unknown sample 0.460 extinction and l is the length. following standard sulfate data solutions concentration 274 on Absobac, A εcl coefcient, path Cocao, c/mo m using of was in an atomic recorded order to using six determine unknown sample absorption of the copper(II) spectroscopy. the 11 . 2 a) Explain what you understand by a standard G r A P h i C A l t e C h n i Q U e S Example 2 solution. The b) Draw a suitable relationship Beer’s at plot low showing the linear concentrations that graph reactant proves law. the A table → below A is versus for a a plot time of concentration based on decomposition the data reaction of given of in reagent products. 3 c) Calculate the concentration, in mol dm , of 3 tm t/s the unknown copper(II) sulfate [A]/mo m solution. 2 d) Calculate the slope, m, of the line, and 3 4.00 × 10 state 2.30 × 10 itsunits. 3 3 1.00 × 10 2.00 × 10 3 a) A 3 2.00 × 10 Solution standard solution concentration is is one known 1.50 × 10 3 whose 3 3.00 × 10 1.00 × 10 exactly. 3 4 4.00 × 10 b) Based on on the graph data paper a graph (never is now drawn on just white The paper!) or using a such Microsoft computer graph have a title and Excel. all the The axes can be expressed mathematically as: programme [A] as 5.0 × 10 drawn graph must = -kt + [A] must be o labelled where k represents the rate constant and [ A] the o with the scaled. appropriate A “best-t” units line is and appropriately then plotted. initial In a) this case all the data points lie exactly concentration. Calculate its the line, so absorbance proving is Beer’s directly law; that proportion to is, slope, m, of the graph and state units. the the b) Determine concentration. state Plot of the on its the intercept, c, of the graph and units. c) Calculate the rate d) Deduce the state units. constant, k, and state its units. A versus c 0.9 initial concentration, [A] , and o (x , y 2 ) 2 0.8 its 0.7 Plot of [A] versus t 2.5 0.6 (x 3 A 0.4 , y 1 md lom 0.5 ) 1 2.0 (x , y c ) c 1.5 3 01 0.3 × /]A[ 0.2 (x , y ) 1 1 0.1 1.0 (x 0.5 , y 2 ) 2 0 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0 0.7 1 2 3 4 5 3 3 t/ × c/mol dm 10 s 3 c) When d) (x , y 1 (x , ) A is C = 0.34 = (0.1002, 0.130) = (0.6000, 0.810) mol Solution dm a) and (x , y 1 ) = (0.400, = (4.00, 2.30) and 1 1 y 2 ) (x 2 ∆y = 0.810 0.130 __ = ∆x , y 2 ___ m 0.460, ∆y 0.680 _ 0.1002 = 0.4998 1.36 mol = 1.80 _ = 4.00 ∆x 1 = 0.50) 0.500 2.30 __ _ m = 0.6000 ) 2 0.400 = -0.500. 3.60 3 dm Next we need 3 10 to take 3 mol dm /10 the units 3 of m into 6 s = 10 account: 3 mol dm 1 s 275 11 M E A S U R E M E N T A N D D ATA P R O C E S S I N G 6 Therefore m = -0.500 1 = 5.00 × × 10 × 10 10 3 mol 3 10 mol 7 5.00 × 6 dm 3 mol 1 dm s Suy p 1 s = 1 dm s Mathematically, any propor tionality sign, “∝”, in a mathematical expression can be replaced by an b) Let (x , y c ) = (2.00, 1.50) c equality expression plus a constant, for example “= k”. y = mx c c + c c = y = 2.50 mx c = 1.50 ( 0.500 × 2.00) c 3 × This 3 10 mol is in the form y = mx + c, so a suitable linear dm 1 __ sketch would be to plot p on the y-axis and ( ) on V the x-axis. m would equate to the constant k. As Suy p 1 __ there is no term to the right of the k( ) part of the V Note that the units of c will always correspond to the expression, mathematically the intercept, c, will units of the y-axis variable, which in this case is [A]. bezero. c) The expression [A] = -kt + [A] is in the Plot of p versus 1/V (n, T constant) form o of y So = m mx = + c -k, and hence: 7 k = -m = -( 5.00 × 10 3 mol dm 1 s ) p 7 = 5.00 × 10 3 mol 1 dm s 3 d) c = [A] , so [A] o = 2.50 × 10 3 mol dm o For HL students only. This data is based on a zero-order 1/V reaction, discussed fur ther in topic 16. Suy p Example 3 Sketched graphs have labelled but unscaled axes, and are used to show qualitative trends, such as variables Boyle’s law is one of the gas laws and states that that are propor tional or inversely propor tional. Units with the temperature, T, and the amount of gas, generally, therefore, do not have to be included in n, constant, to the the pressure is inversely proportional sketches. In the IB programme you should know the volume. State how you would represent dierence between the command terms draw and this law mathematically as an expression and sketch. You should always use graph paper when suggest the type of linear graph you would sketch drawing a graph (complete with a title, labelled to illustrate this relationship. and scaled axes, and units). For a sketch, you can represent the graph on white paper (including a title Solution and labelled, but unscaled, axes). The key phrase here is inverse proportionality. Hence: 1 _ p ∝ V 1 _ p = k ( ) V where 276 k is a constant of proportionality. 11 . 3 S P e C t r O S C O P i C i d e n t i F i C A t i O n O r O r G A n i C C O M P O U n d S 11.3 Spcoscopc cao o ogac compous Understandings Applications and skills The degree of unsaturation or index of hydrogen ➔ Determination of the IHD from a molecular ➔ deciency (IHD) can be used to determine from formula. a molecular formula the number of rings or Deduction of information about the structural ➔ multiple bonds in a molecule. features of a compound from percentage Mass spectrometry (MS), proton nuclear ➔ 1 composition data, MS, H NMR, or IR. 1 magnetic resonance spectroscopy ( H NMR), and infrared spectroscopy (IR) are techniques that can be used to help identify compounds and to determine their structure. Nature of science Improvements in instrumentation – mass ➔ ➔ Models are developed to explain cer tain spectrometry, proton nuclear magnetic phenomena that may not be obser vable – resonance and infrared spectroscopy have made for example, spectra are based on the bond identication and structural determination of vibration model. compounds routine. Degree of unsaturation or index of hydrogen deciency (IHD) The can or degree be of used multiple The degree and π to bonds of bonds ● a double ● a triple ● a ring ● an The unsaturation determine bond is is can be ring in a index of hydrogen molecular is as used structure, counted is a formula deciency the number (IHD) of rings molecule. counted counted aromatic IHD a unsaturation present bond is in or from one as as degree worked ● from the structure ● from the molecular out as two calculate the number of rings where: one two counted to degree degrees of of of unsaturation unsaturation unsaturation four degrees of unsaturation. ways: formula. 277 11 M E A S U R E M E N T A N D D ATA P R O C E S S I N G From the structure Compou Sucu numb of gs, ihd oub bos, a p bos benzene • one ring 4 • three double bonds (in Kekulé structure) cyclobutane • one ring 1 cyclohexane • one ring 1 • one ring 3 • two double bonds • one ring • ve double bonds • one triple bond (chair conformation) cyclopentadiene 2-acetoxybenzoic 6 OH O acid C (aspirin) CH O 3 C O ethyne 2 From the molecular formula In order C H c use N h to O n X o the = for = IHD is a for the halogen generic atom = 8 n = 0 o = 2 x = 0 IHD = Therefore molecular Cl, Br, or formula I), we can C expression: (0.5)(2c H O 8 + 2 - h - x + n) : 2 (0.5)(8 the There + 2 8 molecule are several 0 + contains isomers 0) = illustration: 1 either of C H 4 278 (F , 4 h ring. X x 4 c the where following IHD Hence deduce , one O 8 . 2 double Here are bond just or one three for 11 . 3 isom of C h 4 S P e C t r O S C O P i C O 8 i d e n t i F i C A t i O n O r Sucu O r G A n i C C O M P O U n d S ihd 2 methyl propionate 1 O H O C CH 3 2 3 1 ethyl ethanoate O H C O CH 3 2 3 tetrahydro-3-furanol 1 OH O Let us take deduce four their compounds IHD using the with different molecular formulas and formula: Mocua fomua ihd Acvy Using the ChemSpider RSC C H 17 NO 21 (cocaine) 8 4 database (www.chemspider. C H 27 O (cholesterol) com) look at the structures of 5 46 these molecules and check to C H 6 N (aniline) 4 see if the above calculations 7 agree with what you expect the C H 15 ClN 10 O 3 (clonazepam) 12 3 IHD to be from the respective structures. Electromagnetic spectrum (EMS) The electromagnetic spectrum (gure 1) is given in section 3 of the Databooklet energy 16 14 10 12 10 10 10 10 8 6 10 γ UV V I 4 10 2 10 10 0 2 10 10 IR B G Y O 400 4 10 6 10 8 10 wavelength/m radio waves R 700 wavelength/nm Figure 1 The electromagnetic spectrum The of energy the of electromagnetic radiation by Planck’s radiation, E, is related to the frequency, ν, equation: hc _ E = hν = λ where: 34 h = Planck’s Data constant = 6.63 × 10 J s (given in section 2 of the booklet ) 279 11 M E A S U R E M E N T A N D D ATA P R O C E S S I N G E = energy of radiation (measured in J) tOK ν = frequency of radiation (measured in Hz) Electromagnetic waves 8 c = speed of light = 3.00 × 1 10 m s (given in section 2 of Data booklet) can transmit information beyond that of our sense λ = wavelength (measured in m). perceptions. What are From this relationship it can be seen that the energy is directly the limitations of sense proportional to the frequency and inversely proportional to the perception as a way of wavelength, that is: knowing? 1 _ E ∝ ν and E ∝ λ The various regions spectroscopy radiation) of the (which and is various EMS the are study techniques the of basis the are of way used different matter to types of interacts identify the with structures of substances: ● X-rays from to – the as information structure ● Visible and and and hence electronic UV-vis ● the ● of high, the of these of X-ray (UV) light an atom electrons Diffraction distances spectroscopy in cause atoms. bond basis levels radiation and to be patterns bond removed can angles in lead a crystallography give rise gives or to electronic information molecule. This transitions about is the the basis of causes stretch and groups cause certain bend) present. molecular bonds and This as is in such the rotations a molecule provides basis and of can IR to vibrate information on spectroscopy give information on lengths. Radiowaves because can spin states environments the now cause radiowaves spectroscopy Let’s as the type energy Microwaves on is levels ultraviolet this functional their such form example, bond energy energy spectroscopy Infrared (for ● their inner to is of consider ● infrared ● proton ● mass of the (IR) nuclear be change. based atoms connectivity cornerstone can of three on Nuclear this can the transitions absorbed be and different in a certain strong nuclei, magnetic which present types of in a on leads of to eld causes (NMR) different chemical information molecule. spectroscopy identication magnetic which resonance information deduced, atoms spectroscopic by that organic form the compounds: spectroscopy 1 nuclear magnetic spectrometry resonance ( H NMR) spectroscopy (MS). x Infrared spectroscopy Unlike UV energy F to spring model used to understand the vibration of molecules 280 various in which about electronic their is bonds. the IR the Hence, can be but vibration using identied correspond spring radiation transitions, from groups transitions spectroscopy radiation, result functional vibrational IR visible result vibrations, molecules Figure 2 Hooke’s law is the basis of the and to can of in not a have cause certain this denite model does type molecular groups of of spectroscopy molecule. energy sufcient The levels. The basis of 11 . 3 In a the spring spring or can twisted, vibration F That ∝ is, exerted be a covalent (both a bond is considered symmetrically distortion. law to the of directly the with by = where on to expansion will applied F rise based the expressed The giving is every stretched i d e n t i F i C A t i O n from The physics and force called as a O r O r G A n i C spring. Such asymmetrically), required to Hooke’s cause C O M P O U n d S bent, the law: x position load model, be S P e C t r O S C O P i C a the length of proportional spring (gure negative sign 2). a to spring the By where F ( x) force from ( F) convention represents its equilibrium caused the the law by is the typically restoring force spring: -kx k is a constant fundamental obeying Hooke’s of proportionality, frequency law can be of the called vibration, related to the the ν, spring based mass, m, on by constant a the system expression: __ k _ 1 _ ν = √ 2π Hence, for vibrate at the vibration higher frequencies, and m triple ν. of an atom frequencies, The bonds). same If you ν, it can be and heavier for multiple applies imagine two seen that atoms bonds atoms lighter will (for connected atoms vibrate at example by a will lower double spring H Cl the Figure 3 Stretching in HCl molecule stronger and For of the bond therefore a more diatomic molecular connecting energy molecule, vibration is is the two atoms required such as to stretch hydrogen possible, that is the tighter the string will be it. chloride, stretching HCl, only (gure one form 3). O If we HCl compare has the the frequencies smallest mass and of HCl, HBr, greatest and bond HI we enthalpy nd (see that because section 11 of H the Data booklet) it will have the greatest H frequency. symmetric stretch 1 Mocu H Cl H Br 1 Bo apy/kJ mo Wavumb/cm 431 2886 366 2559 O H H 298 I H 2230 asymmetric stretch T able 1 Bond enthalpies and wavenumbers for selected HX molecules Different required As can molecules to be execute seen absorb a from at different vibration table 1, IR will frequencies depend absorptions on are the because bond typically the energy enthalpy. cited as the O 1 reciprocal of the wavelength ( ) . This is the wavenumber and has λ 1 units of cm H For polyatomic For example, species, the water there may molecule be has several three 1 ● a ● an ● a symmetric stretch (3652 cm stretch (3756 cm of modes of vibration vibration. (gure 4): symmetric bend ) 1 asymmetric different modes H Figure 4 Modes of vibration of the ) water molecule. All three modes of 1 symmetric bend (1595 cm ). vibration are IR active 281 11 M E A S U R E M E N T A N D D ATA O P R O C E S S I N G However, for a in change vibration a covalent the bond molecular mode. Let us to absorb dipole take some IR radiation moment there associated must with be the examples: symmetric stretch Mocu Poay of mocu A bsopo of ir aao non-polar no (IR inactive) non-polar no (IR inactive) non-polar symmetric stretch – no (IR H O 2 asymmetric stretch O 2 CO 2 (gure 5) inactive) asymmetric stretch – yes (IR O 1 ) active) (2349 cm symmetric bend – yes (IR active) 1 (667 cm ) symmetric bend Figure 5 Modes of vibration of the The CO absorbance, A, of a sample can be related to the transmittance by molecule. The asymmetric 2 the expression: stretch and the symmetric bend are IR active because of a change in A = -log T 10 the molecular dipole moment that results from the molecular vibration An IR spectrum is a plot of the percentage transmittance, %T, versus the 1 wavenumber spectrum, for the and of IR (in ), functional for where groups absorptions different booklet; cm of the can various functional example, %T be from identied. bonds groups C=C ranges are in The different given absorption 0% in in to 100%. In an characteristic classes section alkenes 26 of of ranges molecules the typically IR Data occurs in 1 the range 1620–1680 cm , etc. 100 )%( ecnattimsnart 50 0 4000 3000 2000 1500 1000 500 1 wavenumber/cm Figure 6 IR spectrum (in liquid lm) of butanoic acid In gure 6, note the following absorptions: 1 ● strong, O 282 H broad bond of peak a in the range carboxylic acid 2500–3000 cm characteristic of the 11 . 3 S P e C t r O S C O P i C i d e n t i F i C A t i O n O r O r G A n i C C O M P O U n d S 1 ● strong peak in the range 1700–1750 cm characteristic of the C=O group 1 ● peak the in C the H range 2850–3090 cm characteristic of bond. 1 Typically can be in the identied. IR spectrum. IR is often is a it of on chemistry. 300–1400 is termed supporting provides bonds little technique the IR or her the analytical to structural some journey is often into also of the organic only physics at for in the sometimes, for of of groups However, (and, point work an absence decisions starting in the or functional detective used as of information. key an the vibrations region presence different elucidation is complex technique the with making and spectroscopy more ngerprint relates other in structural compound his cm associated provides inorganic) chemist a really powerful beginning an it chemical molecules– it This termed information certain region the organic analytical heat sensors and remotesensing. 1 Proton nuclear magnetic resonance ( H NMR) spectroscopy 1 H NMR spectroscopy environments the most chemist. states. As eld orientation. two may with the the align nuclear of they can with energy external anti-parallel the two the applied the difference hydrogen as are The have is of corresponds magnetic depends on eld a the and available exist is to in the two energy same a magnetic or against alignment lower the it. of different is in random chemical spin results nuclear region of in spin to energy, will spin no the compared radiowave ∆E and eld, This difference increased, there the energy organic possible the in chemical possibly When This to is different magnets. eld parallel eld the can placed conguration. levels on molecule tiny magnetic magnetic alignment EMS. energy the a atoms states nuclei levels. between As behave spin the in technique hydrogen two When information atoms structural nuclei such gives hydrogen important The magnetic states of the ∆E, the increase. The environments of atoms. 1 In a H NMR standard δ, is spectrum, expressed assigned in as 0 environments Data the position (tetramethylsilane, parts ppm. have per TMS) million Hydrogen different of is the NMR termed (ppm), nuclei chemical of in the the the shifts signal proton. different (see relative chemical δ for to a shift, TMS chemical section 27 of the booklet) O 1 Therefore the the number number of of different signals on chemical a H NMR spectrum environments in shows which the 1 hydrogen of atoms methanoic the two are acid, different found. HCCOH, chemical For example, two signals in are environments of the H found, the NMR spectrum which two C shows A atoms, A and H hydrogen B B. 283 11 M E A S U R E M E N T A N D D ATA P R O C E S S I N G typ of poo Cmca sf (ppm) CH 0.9–1.0 3 CH R 1.3–1.4 2 R CH 1.5 2 O 2.0–2.5 C RO CH 2 O 2.2–2.7 C R CH 2 CH 3 H CH Hal 2.5–3.5 1.8–3.1 3.5–4.4 2 CH 3.3–3.7 2 O 3.7–4.8 C R O CH 2 O 9.0–13.0 C R O HC H H 1.0–6.0 CH 4.5–6.0 2 OH 4.0–12.0 H 6.9–9.0 O 9.4–10.0 C R H T able 2 Typical proton chemical shift values (δ) relative to tetramethylsilane (TMS). R represents an alkyl group, and Hal represents F , Cl, Br, or I. These values may vary in dierent solvents and conditions 284 11 . 3 The signals (gure 7) can typ of poo S P e C t r O S C O P i C be assigned as i d e n t i F i C A t i O n O r O r G A n i C C O M P O U n d S follows: Pc δ/ppm Acua δ/ppm (fom sco 27 of 1 (fom h nMr spcum) Data booklet) hCO 9.4–10.0 8.06 COOh 9.0–13.0 10.99 A: B: 12 1 Another useful feature of a H NMR spectrum is that it contains 10 8 6 4 2 0 2 an ppm integration trace that shows the relative number of hydrogen atoms 1 Figure 7 1 present. In the case of the H NMR spectrum for methanoic acid H NMR spectrum of methanoic acid, this HCOOH. will be 1:1. 1 An important the fact that detected by the protons magnetic dimensional topic application view of of in H NMR water resonance organs in spectroscopy molecules the imaging human is within (MRI), body associated human which (discussed with cells can gives a be three- further in 21). Mass spectrometry (MS) In topic 2 we introduced the principles of mass spectrometry (MS). When a + gaseous ion molecule peak in compound. mass of which spectrum a masses. ionized the the be are molecular energetic molecule ions. 28 The can, in of the to M , the fact, on is formed. molecular ionization break fragmentation information some ion, corresponds highly further Section Here its spectrum to will provides molecule. their mass Owing spectrometer, some in a is Data booklet lists into pattern certain mass process up of molecular of the involved smaller groups these in a fragments, observed functional some The in a mass present fragments and examples: + ● (M 15) results from the loss of CH results from the loss of OH results from the loss of CHO r 3 + ● (M 17) r + ● (M 29) or r the loss of CH CH 2 3 + ● (M 31) results from the loss of OCH r 3 + ● (M 45) results from the loss of COOH. r 100 80 ytisnetni evitaler 60 Figure 8 A nuclear magnetic resonance (NMR) spectrometer. NMR spectroscopy measures 40 the resonance between an applied magnetic eld and the magnetic moment of a molecule’s 20 atoms. It allows identication of molecules in a sample. This 400 MHz Agilent Unity Inova 0 12 15 20 25 30 35 40 45 50 55 60 NMR Spectrometer is located at the Magnetic m/z Resonance Facility of the National Renewable Figure 9 MS of propan1 -ol Energy Laboratory (NREL). The NREL is based in Golden, Colorado, USA 285 11 M E A S U R E M E N T A N D D ATA P R O C E S S I N G Some of the peaks for propan-1-ol (gure m/z 8) can be assigned as follows: Pc δ / ppm (fom sco 27 of Data booklet) + 60 molecular ion peak , CH CH 3 CH 2 OH 2 + 31 (high relative intensity) (M 29) from the loss of CH r CH 2 3 + 29 (M 31) from the loss of CH r OH 2 Worked examples Example 1 Deduce the Solution index of hydrogen deciency of: For (CH ) 3 a) Zanamivir caused by Section (an the 37 of inhibitor inuenza the used A to and treat B infections viruses) a) 2 b) 9:1 COH: 3 using Data booklet For CH COOCH 3 b) Carbolic C H 6 acid which has the molecular : 3 formula O. a) 2 b) 1:1 c) m/z 6 Solution + = 74 molecular ion peak (CH ) 3 a) Zanamivir has four ring, IHD double bonds and COH 3 one + m/z so its = = 57 (M 5. 17) from loss of OH. from loss of CH r + m/z b) For C H 6 IHD = 59 (M O: 15) r 3 6 = (0.5)(12 + 2 6 0 + 0) = 4 Suy p In MS, don’t forget the + sign for species that remain Carbolic acid is, in fact, phenol which OH after a fragment has been lost. has the Kekulé structure: Example 3 An unknown compound, X, of molecular formula 1 C H 2 Example 2 For each of IR the following two O, has the following IR and H NMR spectra. 4 spectrum (in liquid lm): compounds: 100 (CH ) CH COH 3 COOCH 3 a) State 3 how many signals each compound shows 1 in b) its State H NMR what spectrum. you expect the integration trace )% ( ecnattimsnart 3 50 to 1 be for each H NMR. 0 4000 c) For the MS of (CH ) 3 values, 286 giving a COH, state possible 3 reason 3000 2000 1500 m/z for your answer. 1000 1 wavenumber/cm 500 11 . 3 S P e C t r O S C O P i C i d e n t i F i C A t i O n O r O r G A n i C C O M P O U n d S 1 H NMR spectrum (in CDCl ): absorption for C=O in the wavenumber range 3 1 1700–1750 Data cm booklet. , based Indeed, on there section is a 26 strong of the peak at 1 approximately presence ● If C=O an of is a 1727cm C=O present, aldehyde or a , which suggests the bond. then X might ketone. An be either aldehydic 1 proton is quite spectrum, with 9.4–10.0ppm, Data booklet. single ● 10 9 8 7 6 5 4 3 2 1 If X an CHO. atoms a CH of X showed peaks at m/z values 15, 29, Deduce given the and the and Data 44 structure any other booklet. spectroscopic structure (other of For peaks of X were using additional each the as found). δ assign possible shift, from there = 9.8 that must , portion the δ, does H in section NMR the 27 appear of to range the be a ppm. of means contain this which the we now molecule, remaining one suggests number carbon indicates have a that is of and methyl X that three group, ethanal, CHO. 3 information information spectrum information also fact, the 3 CH of asseen with As hydrogens, spectrum chemical aldehyde identied is MS a in 0 ppm The In signal is characteristic O H from as based much on C H the X. H A B Solution ● ● As the molecular formula of X is given, Let’s on is worth rst nding out the IHD, now test this proposed structure based it the spectroscopic data gained from the which 1 H indicates the index of hydrogen NMR atoms or degree of spectrum. Two types of hydrogen deciency are present in different chemical unsaturation environments, For the generic molecular formula C H c IHD = (0.5)(2c + 2 h N h O n X o A and B. : x typ of Pc igao Acua poo δ/ppm ac δ/ppm x + n) (fom (fom For C H 2 O the IHD is: Sco 1 4 h nMr 27 of Data (0.5)(4 + 2 4 0 + 0) = spcum) 1 booklet) Therefore double ● We the bond note molecule or that, one contains either one ring. based on A: the molecular 9.4–10.0 ChO 9.8 1:3 formula, B: 2.2–2.7 X contains just one oxygen atom. The 2.2 classes COCh 3 for (C X could CO be C), an an ether (C aldehyde O (C C), a CHO), ketone or an ● alcohol (C Having is ● Based on spectrum established the structure of X, it OH). the and above see we now whether examine there is a the IR strong IR to worth returning conrm range for the the to the additional infrared IR spectrum characteristic absorption due to 287 11 M E A S U R E M E N T the CH bonds in A N D the D ATA P R O C E S S I N G wavenumber as range follows (using section 28 of the Data 1 2850–3090cm IR spectrum, . As there can are, be seen indeed, from booklet): the absorptions ■ m/z = 15, signies the presence of CH , 3 within this range. which indicates loss of CHO from molecule + ● F i n a l l y, consider the MS. There should X, that is (M 29) r be a molecular ion peak at m/z = 44, + ■ corresponding to the relative molar m/z = 29, signies the presence of CHO , mass which indicates loss of CH from molecule 3 of C H 2 O, calculated as 44.06. This, indeed, + 4 X, that is (M 15) r is present. In addition, the other dominant ● m/z values in the MS can be assigned This conrms compound X to be ethanal. Usfu soucs ● Spectral Database for Organic Compounds, SDBS, hosted by National Institute of Advanced Industrial Science and Technology (AIST), Japan. http://sdbs. db.aist.go.jp/sdbs/cgi-bin/cre_index.cgi ● EURACHEM – a network of organisations in Europe having the objective of establishing a system for the international traceability of chemical measurements and the promotion of good quality practices. There is an excellent guide on uncer tainty in measurement which might be useful for your IA and other laboratory work . http://www.eurachem.org/ ● NIST – National Institute of Standards and Technology, USA. The chemistry por tal is wor th accessing for spectroscopic data etc. http://www.nist.gov/ chemistry-por tal.cfm 288 Q U e S t i O n S Questions 1 How many 0.0200 signicant gures are D. in g? with IB A. 1 B. 2 Measuring 5 a May Which gas 2 3 D. 5 A burette reading is recorded as 27.70 ± volume of a gas produced [1] 2010 are likely experiment C. the syringe is to be reduced repeated A. Random B. Systematic C. Both D. Neither a when number of an times? errors errors random and systematic errors 3 0.05cm . Which of the following could bethe random nor systematic errors [1] actualvalue? IB November 2009 3 I. 27.68 cm II. 27.78 cm III. 27.74 cm 3 6 Deduce the IHD for codeine using section 37 3 ofthe A. I and II B. I and III C. II booklet only only 7 and Data III Deduce the IHD formula C H 5 D. I, IB II, and May for a molecule of molecular only III N 10 2 [1] 2011 1 8 The H NMR formula C H 3 3 A piece of metallic aluminium with a mass spectrum O is of shown X with molecular below. 6 of 3 10.044 A g was student to found carried determine the to out have the a volume following of 3.70cm calculation density: 3 2 10.044 _ 3 density (g cm ) 1 = 3.70 What is the reportfor best the value density the of student could aluminium? 3 A. 2.715 g B. 2.7 C. 2.71 D. 2.7146 cm 3 g cm 3 g cm 10 9 8 7 6 3 g cm 5 4 3 2 1 0 chemical shift/ppm [1] Source: SDBSWeb, http://sdbs.riodb.aist.go.jp (National IB May 2011 Institute of Advanced Industrial Science and Technology) a) 4 Which experimental procedure is Deduce which compoundsis likelytolead to a large systematic of the following most X and explain your error? answer. A. Determining the concentration of an alkali CH [2] CO CH 3 by titration with a CH =CH CH 2 B. Measuring the volume of a solution using CH CH 3 CHO; 2 OH 2 a b) volumetric ; 3 burette Deduce which one of the signals in pipette 1 the C. Determining the neutralization in enthalpy a change HNMR spectrum of occurinthe spectrum of X would also of one of the beaker otherisomers, giving your reasoning. [2] 289 11 M E A S U R E M E N T c) The infrared werealso (i) Apart C C A N D and D ATA mass P R O C E S S I N G spectra for X recorded. from absorptions andC H absorption,in wouldbe bonds, due to suggest wavenumbers, present in the one that infrared spectrum. (ii) Apart C C [1] from absorptions andC H absorption,in infrared inone of shownin d) IB 290 Suggest of two in the May the 2011 to but one absent present compounds a). formulas species mass spectrum, other part due suggest wavenumbers, inthis the bonds, that [1] and m/z would spectrum. be values detected [2] 12 AT O M I C ST R U CT U R E (AHL) Introduction The is quantized related atoms to and nature the of energy molecules. energy states In of this transitions empirical electrons spectra topic we in see how for the evidence has been gained used existence of to from line provide energy emission strong evidence levels. 12.1 Eeto to Understandings Applications and skills ➔ In an emission spectrum, the limit of ➔ Solving problems using E = hv ➔ Calculation of the value of the rst ionization convergence at higher frequency corresponds to the rst ionization energy. energy from spectral data which gives the ➔ Trends in rst ionization energy across periods wavelength or frequency of the convergence account for the existence of main energy levels limit. and sublevels in atoms. ➔ ➔ Deduction of the grou p of an element f rom i ts Successive ionization energy data for an successive ionization energy da ta. element give information that shows relations ➔ Explanation of the trends and discontinuities in to electron congurations. data on rst ionization energy across a period. Nature of science ➔ Experimental evidence to suppor t theories – emission spectra provide evidence for the existence of energy levels. 291 12 ATO M I C S T R U C T U R E ( A H L ) CERN Scientic theories experimental (Conseil Européen Nucléaire) Nuclear near and is Geneva France the on has 20 several of its are universe, set most and up in in the CERN it the Europe. has 1911 and CERN The origins of. Our focus of When second at the CERN world war on understanding signicantly nucleus (sub-topic an main concentrated the is Scientists progressed of involves international the the time atom. then project made most instruments physics. after Switzerland and Europe, research. is for located fundamental extensive discovery Rutherford largest discover the laboratory, across outside to at since the study of Organization between the CERN Recherche scientic The 1954 inside matter to what research beyond of particle trying exploring of border scientic is la CERN states example research CERN was used matter. collaborative of The by accepted. European the countries supported be pour some member excellent be to sophisticated world, particles over the Research. advanced in must evidence by Figure 1 The Large Hadron Collider (LHC) at Ernest CERN is the world’s largest and most powerful 2.1). par ticle accelerator. It consists of a 27 km ring of CERN was also the birthplace of the World Wide superconducting magnets Web (the scientist internet), Tim invented Berners-Lee; it by the was British originally In developed so that scientists and 1964 a number theoretical around the globe could instantly and 2013 the the preliminary Large CERN Hadron suggested Higgs by a the Standard universe in the is Collider in Model , from In gure existence people is at of the is an termed how the everything plants to stars was is considered to be composed of just a no until who 2013. blocks. These are the particles of boson particles Higgs themselves boson is one of are 17 governed has (another are sub-topic 2.2). is the constituents others of (including Some are 292 in photon, Higgs responsible nature, main for matter particle for excluding its is Peter the described for named Higgs, existence for was problem that its after one of originally existence as the linking there existence British six in of the greatest theory scientists 1964. Higgs scientic with evidence. Nobel Prize in Physics 2013 was awarded few to François Englert, Université Libre de matter. by Belgium, and Peter W . Higgs, University forces. Edinburgh, of a Scotland, for all the gravity. mechanism the that theoretical contributes example, our understanding of the origin of mass of particles subatomic particles, conrmed through and which was recently whereas and the discovery of the predicted the forces particle, by the that experiments occur the years evidence 100years, fundamental photon) a existence fundamental fundamental everyday the 50 particle been in experimental to see the to discovery particles The evidence discovery of The However, almost Professor Bruxelles, Such for proposed jointly building boson. experimental Empirical The trees Higgs scientists physicist predicted physics model to 1) boson describes this from existence Higgs particle which constructed. universe, the The whose model results (LHC, for particle. particle theoretical experimental evidence boson elementary suggested outlining data. for In physicists mechanism share of information of universities at CERN’s LHC. ATLAS and CMS 12 . 1 E l E c T r O n s i n a T O m s TOK In topic 2 we discussed a number of key discoveries Theoretical scientists often work in elds where associated with the structure of the atom. In this topic we the application of their research to real life may continue to look at historical developments in this eld. be dicult to predict, or dicult to comprehend by non-scientists. de Broglie: Wave proper ties of electrons In 1924 the French scientist Louis de Broglie brought the wave–par ticle dual theory of the electron to the fore in Should governments and funding bodies fund basic theoretical research, or should they concentrate on applied or strategic research where the end the de Boge equto: application may be more economically tangible? Can h _ λ = p you think of examples of scientic discoveries resulting where: from basic research that have resulted in unforeseen applications? λ = wavelength h = Planck’s constant Heisenberg’s uncer tainty principle can be expressed p = momentum = mv = mass × velocity mathematically as follows: In this equation, wavelength is inversely propor tional h _ to momentum. ∆p × ∆q ≥ 4π You might nd that interpretation of this expression may where: warrant an exploratory journey of how we consider the ∆p = uncer tainty of momentum measurement movement of particles. Suppose you have a particle of large mass, such as a tennis ball. If the mass is large then so will be the momentum, ∆q = uncer tainty of position measurement p. The wavelength, λ, associated h = Planck’s constant with the tennis ball when moving at high velocity becomes Suppose you wish to measure the location of a moving negligible. This agrees with what we observe: you don’t see electron. If the position is measured with high accuracy, a tennis ball moving in a wave pattern across a tennis court! then ∆q will approach zero. What then happens However, if the mass of the par ticle is tiny, s uch as simultaneously to ∆p? To explore this we can rearrange 31 the electron with mas s m = 9.109 × 10 kg, then the equation: e the wavelength will be la rge, s ugges ting th at a w av e h _ motion can be associa ted w ith the electron in an atom. ∆p ≥ ( 1 _ ) 4π So de wave Broglie suggested proper ties, but so that d oes not only does li g ht Δq hav e ______ 1 m at ter! as ∆q → 0, then → ∞ Δq The de Broglie equation shows that macroscopic par ticles so ∆p → ∞ have too shor t a wavelength for their wave proper ties to be observed. that is, as the uncer tainty of position measurement approaches zero, the uncer tainty of momentum Is it meaningful to talk of proper ties that cannot be measurement approaches innity so the momentum observed with our senses? becomes eectively undened. Heisenberg’s uncer tainty principle In 1927 another theoretical physicist, Werner Heisenberg from Germany, published the ground-breaking theory termed Heebeg’ ue tty ppe. Professor Heisenberg was one of the pioneers in the eld of qutu eh and the basis of his principle is as follows: The more precisely the position is determined, the less precisely the momentum is known in this instant, and vice versa. Heisenberg, uncer tainty paper, 1927. Heisenberg said: “ What we observe is not nature itself, but nature exposed to our method of questioning.” Can our senses give us objective knowledge about the world? The idea of uncer tainty lying at the hear t of Heisenberg’s thinking is an example of a historical journey of discovery that embraces not just physics but the persona of an individual as well. Find out more about Heisenberg and consider what is meant by this statement. 293 12 ATO M I C S T R U C T U R E ( A H L ) The Schrödinger wave equation The dual wave–particle nature of the electron has been one shödge wve equto. See sub-topic 2.2 for more of the great discussions in the history of subatomic particles. information on the Schrödinger wave equation. The Austrian physicist Erwin Schrödinger (1887–1961) Schrödinger ’s wave equation accurately predicted the was an advocate of wve eh, expressed in the energy levels of atoms. Emission spectra and ionization The values of some rst 1 ionization energies, in kJ mol , are given in section 8 of the Data booklet In sub-topic Emission atomic In 2.2 energy sub-topic required in its the spectra to line emission provide spectrum experimental of hydrogen evidence for the was introduced. existence of levels. 3.2 ionization remove ground-state. an energy electron The rst was from dened a as neutral ionization the minimum gaseous energy ( IE ) atom of a energy or molecule gaseous atom 1 Uefu eoue is related to the process: Chemsoc Timeline – This is + X(g) a visual exploration of key → X (g) + e events in the history of science Successive ionizations are also possible; for example, the second with par ticular emphasis on ionization energy (IE ) is associated with the process: 2 chemistry. It was developed + X 2+ (g) → X (g) + e by Murray Roberston in collaboration with ChemSoc, th The n ionization energy (IE ) relates to the process: n the chemical network of the (n 1)+ X n+ (g) → X (g) + e Royal Society of Chemistry (RSC). You can even make For a given element the IE increases for successive ionizations, in the order: predictions for inventions or IE < IE 1 < IE 2 < IE 3 < IE 4 . . . 5 discoveries that you think will be made in years to come! This is because removed from with an each successive increasingly ionization positive species, an electron and hence is being more energy is http://www.rsc.org/chemsoc/ required. For example, for magnesium, Mg: timeline/pages/timeline.html 1 IE = 737.7 kJ mol = 1450.7 kJ mol = 7732.7 kJ mol = 10542.5 = 13630 1 1 IE 2 1 IE 3 1 IE kJ mol 4 1 IE kJ mol 5 In sub-topic hydrogen of we inuence of occurred). shown The the in the increase frequency of can 2 the be in the merge, can have the in at any principal radiation to the in higher forming electron represents used emission converge lines nucleus: gure convergence that lines electron the The saw the convergence continuum ∞ 2.2 atom, a spectrum of energies. At quantum the so outside process IE . of In is no the spectrum the of at Lyman the under the (ionization from ionization limit Beyond longer atom number emission determine the continuum. energy, is the n = the the series 1 to has n = atom. limit for of the 1 hydrogen relates to returns 294 atom the to (UV energy the region), given the out ground-state, n frequency when = 1, an at the electron asshown in limit falls gure of convergence from 6 of n = ∞ and sub-topic 2.2. 12 . 1 E l E c T r O n s i n a T O m s outside the atom continuum n = ∞ inside the atom n = 5 n = 4 n = 3 n = 2 n = 1 Figure 2 The ionization process for the hydrogen atom Determining the wavelengths of lines in spectra: The Rydberg equation hc _ The Rydberg equation can be used to nd IE = E E ∞ = hν = 1 λ the wavelengths of all the spectral lines in the (since emission spectrum of hydrogen, and is given c = νλ) by 8 the c expression: = speed of light = 2.998 × 10 1 m s 1 1 _ = 1 _ R 1 _ From the Rydberg equation, can be inserted into λ ( H λ 2 2 n n ) the expression for IE and rearranged to: f i where: IE λ = = Rydberg = constant = 1.097 × 10 2 = nal ) 2 ∞ 1 34 (6.626 × 8 10 state J 7 (1.097 f ( ) m = initial 1 ___ 1 __ ( 1 H i hcR H 7 R = wavelength × s)(2.998 × 10 1 m s ) 1 10 m )(1 0) state 18 = n = principal quantum 2.179 × 10 J number –1 The Note that n is greater than f energy in kJ mol is found by: n i 18 IE The IE can then be determined as = (2.179 = 1.312 = 1312 × 10 23 J)(6.022 6 ΔE = × 10 1 mol ) follows: × 10 1 J mol hν 1 kJ mol where: This value calculated for the rst ionization 34 h = Planck’s constant = 6.626 × 10 J s energy (IE ) for hydrogen is given in section 8 of 1 ν = frequency the Data booklet 295 12 ATO M I C S T R U C T U R E ( A H L ) study tp ● Worked examples: determining energy In any calculation you Example 1 should use the data given in the question or otherwise Determine the energy, in J, of a photon of red light, correct to four data from the Data booklet. signicant gures, In example 1 you should given that the wavelength 34 h = 6.626 × 8 10 J s; c = 2.998 × 10 λ = 650.0 nm. 1 m s use the values of h and c provided. Note that the Solution question requires an hc _ ΔE = hν = answer to a given number λ of signicant gures. 34 8 1 6.626 × 10 J s × 2.998 × 10 m s ____ ΔE 19 = = 3.056 × 10 J 9 ● Always read carefu lly. the Also, 650.0 question make × 10 m s u re Example 2 you include the units 1 Calculate throughout the that stages of your the rst ionization energy, in its shortest-wavelength line in the 34 the to will correct the final answer. you help In must conver t you units for hydrogen series is 91.16 given nm. this = 6.626 × 8 10 J s; c = 2.998 × 10 1 m s 23 ; N = 6.022 × 1 10 mol A related Solution questi on remember to , Lyman obta in numerical nm mol answer. h This kJ variou s The to shortest-wavelength transition of m. n = ∞ to n line = in the Lyman series corresponds to a 1. hc _ IE = hν = 1 λ 34 8 1 6.626 × 10 J s × 2.998 × 10 m s ____ IE 18 = = 1 2.179 × 10 J 9 91.16 × 10 m × (6.022 1 expressed in kJ mol : 18 IE = (2.179 = 1312 × 23 10 J) × 10 1 mol 6 ) = 1.312 × 1 10 J mol 1 1 kJ mol Periodic trends in ionization energies 1 Figure group metal from 3 2 shows ten alkaline titanium, IE to IE 2 ; successive earth Ti. In the metal the third ionization calcium, case of Ca ionization Ca energies, and there is energy the a in kJ group mol 4 signicant corresponds , for the transition jump to the going removal 3+ an electron do not from the fully occupied 3p sublevel. As a result Ca species Ca Ti occur. This supports observations that for the group 2 metals 2+ 20 000 there is one contrast, Ti stable oxidation exhibits state, oxidation +2 states (forming of +2, +3, 2+ ions, and +4 eg Ca (see ). topic In 13 1 lom Jk/I 15 000 and Ti section is +4. In 14 of the gure3 Data there booklet). is a The large most jump in stable IE for oxidation Ti going state from of IE to 4 IE 10 000 , corresponding to the removal of the fth electron, supporting the 5 observation that species with Ti in the +5 oxidation state do not occur. 5000 The electron deduced as congurations follows: 0 2 0 1 2 3 4 5 6 7 8 9 10 11 Ca: [Ar]4s I No. 2+ Ca : [Ar] Figure 3 Ten successive ionization energies for calcium and titanium 2 Ti: [Ar]3d 4+ Ti 296 of 3 25 000 : [Ar] 4s 2 for the most stable ions of Ca and Ti are 12 . 1 It is interesting to note that in the case of Ti, the ionization E l E c T r O n s i n a T O m s energies 1 IE ube increase the 3p 3d and and The more 4s 4s gradually orbitals, than which for are Ca as electrons much closer in are being energy removed compared to 1 alkali metal potassium, K (Z = 19) has the 1 418.8 2 3052 3 4420 4 5877 5 7975 6 9590 7 11343 8 14944 9 16963.7 10 48610 11 54490 12 60730 13 68950 14 75900 15 83080 16 93400 the orbitals. group IE/kJ o from electron conguration: 2 K: 1s 2 2s From 6 2p this 2 3s 6 3p 1 4s conguration we might expect a large jump going from IE 1 to IE , from IE 2 to IE 9 associated principal with , and from IE 10 the removal quantum to IE 17 number of n. electrons Table 1 . These signicant jumps are 18 from shows energy the rst levels 19 of IEs different for K. TOK In plotting ionization energies, a ogth e allows all data points to be plotted on a single graph. 1 The dierence between IE and IE 1 for K is 2633.2 kJ mol , while the dierence 2 1 between IE 18 and IE is 31191 kJ mol . Therefore it is dicult to represent all 19 19 ionization energies for K on a linear scale. Look at the unreasonably long y-axis when comparing the plot of IE versus IE No. to the plot of log 17 99710 18 444870 19 476061 IE versus 10 IE No. in gure 4 (a) and (b). Can you think of examples in chemistry or other sciences that present data in a par ticular way to suppor t the scientist’s postulates, theories, and hypotheses? T able 1 Values of ionization energy Where else in chemistry do we use logarithmic plots? Do you know the dierence between log 10 and log (IE) for the rst 19 ionization (ln) and can you give examples of where energies for potassium. Signicant e jumps are evident between IEs 1 each type of log is used in chemistry? and 2, 9 and 10, and 1 7 and 18 Plot of log Plot of I versus I No. for K I versus I No. for K 10 6 500000 450000 5.5 400000 5 350000 4.5 300000 1 I gol lom Jk/I 250000 4 200000 3.5 150000 3 100000 2.5 50000 0 2 0 5 10 15 20 0 5 I No. 10 15 20 I No. Figure 4 (a) Plot of IE versus IE No. for potassium; (b) Plot of log IE versus IE No. for potassium 10 297 12 ATO M I C S T R U C T U R E ( A H L ) Worked examples Example 1 Values for element group X of expect successive are the to study tp the given in periodic nd X. IEs table table State for the 2. of an unknown Deduce in elements name of you this When writing electron congurations, electrons in which individual orbitals should be presented as superscript: 2 would 1s group. 2 2s 6 2p 1 3s rather than 1s2, 2s2, 2p6, 3s1. In addition, always take note of the type of electron conguration requested – this question asks for the 1 IE ube IE/kJ o full electron conguration so you should not write a 1 condensed electron conguration such as [Ne]3s IE 899 1 IE 1757 2 Solution 14850 IE 2 3 Na: 1s 2 2s 6 2p 1 3s 21005 IE 4 The the plot in easiest gure to 5 shows remove. that This is the rst because electron it is is furthest Table 2 Ionization energies (IEs) for X. from the nucleus, occupying Solution is a large the being the outermost increase going n valence = from 3 electron energy IE to IE 1 The largest jump in IE occurs between IE level. There because 2 and 2 the next electron is removed from the n = 2 level. 1 IE , corresponding to ΔIE = 13093 kJ mol . This 3 However, mustcorrespond to a change in energy X must be in group 2, the the next seven electrons the small, the level; gradual therefore for increase in IE reects the fact that all eight alkaline electrons occupy the same n = 2 energy level. There earthmetals. is another large jump in IE going from IE to IE 9 associated with the removal of an , 10 electron from the Example 2 n Figure 5 represents the successive = 1 energy nucleus energies of sodium. The vertical axis plots and energy) instead of ionization so This will be electron very is closest difcult to energy electron also comes from the 1s data unreasonably to be long represented vertical without The sublevel, IE shows only a small increase over IE 11 the the to so allow to remove. log eleventh (ionization level. ionization using . 10 an axis. Example 3 6.0 Figure 6 energies shows for the the variation second-row in rst ionization elements in the 5.5 periodic table from Li to Ne. 5.0 He I gol 4.5 2400 Ne 4.0 2000 3.5 F 1600 N 1 3.0 1 2 3 4 5 6 7 8 9 10 11 number of electrons removed lom Jk/I 2.5 O H 1200 C Be 800 Figure 5 Successive ionization energies of sodium B Li State and the full explain electron how the conguration successive of 400 sodium ionization 0 energy data for sodium are related to its 0 electron conguration. 2 4 6 8 10 [4] Z IB, 298 May 2010 Figure 6 Ionization energies for the rst 10 elements 12 . 1 a) Explain why as you go across a period, E l E c T r O n s i n a T O m s IEs increase. b) Although there a general increase in IE 1 across the second period as expected, there 2 is 1 2s 1 2p evidence of some discontinuity. This is to as a dog-teeth plot. 2p y z often For referred 1 2p x Explain nitrogen the most loosely bound electron is why: any one of the three 3p electrons which are all 1 i) IE for oxygen (1314 kJ mol ) is lower are 1 degenerate (have the same energy). 1 than IE for nitrogen (1402 kJ mol ) 1 ii) Again start by drawing orbital diagrams for Be 1 ii) IE for boron (801 kJ mol ) is lower than 1 and B: 1 IE for beryllium (900 kJ mol ). 1 2 Be: [He]2s Solution a) IEs increase across ● decreasing a period atomic radii for two across reasons: a period 2 from left to 0 2s right increasing nuclear charge, B: Z [He]2s 1 2p b) i) First consider elements 2 O: the oxygen [He]2s 2 2p orbital and 1 2p x diagrams of 2p y 0 2p x 0 2p x 2 ● 0 2p z 0 2p y z the nitrogen: 1 2p y z 2 1 2s 0 2p The the most 2p loosely orbital 2p y bound while 0 2p x for z electron Be it is for one B of occupies the two x 2 2 2s 1 2p 1 2p x electrons 2p y in the 2s level. It will be easier to z remove the electron from the 2p orbital in B x The rst electron to be removed from a neutral since gaseous atom will come from the highest 2p is of the highest energy level. In in energy than 2s. This criterion occupied overrides sublevel higher this any consideration of paired electrons case in 1 an orbital. Hence IE for B (801 kJ mol ) is lower 1 this is the 2p sublevel which is higher than 2s 1 than IE for Be (900 kJ mol ). 1 in energy. obvious This from represented of the (sub-topic an orbital with horizontally. order difference the energy diagram outermost Remember energies in of that the may which not is energy be often levels just shown within an energy level sublevels is s < p < d the < f 2.2). aogy Suppose you have a two-story building and you need to remove one oor in order to meet new height regulations. Which oor would you remove? Obviously the top oor (oor 2) – the building would collapse if So an electron will be removed from the 2p you removed the ground oor (oor 1)! It is the same sublevel. Which is the most loosely bound electron? when removing electrons from energy levels and There will be maximum repulsion in an orbital that sublevels – electrons are removed from the energy contains paired electrons, so the most loosely bound level of highest principal quantum number n rst, and electron will be a 2p electron as circled in the orbital x from the sublevel with the greatest energy, within that diagram for oxygen above. This is the reason why energy level. IE is lower for oxygen than for nitrogen, whose 1 orbital diagram 2 N: [He]2s is 1 2p shown 1 2p x on the here: 1 2p y z 299 12 ATO M I C S T R U C T U R E ( A H L ) Questions 1 Figure 7 represents remove nine atom an have of been the energy electrons, element. one Not all at of needed a time, the 4 to from The an graph against electrons of the atomic elements rst ionization number shows for the periodicity energy rst (gure plotted twenty 8). removed. 2500 1 lom Jk/ygrene noitazinoi tsr I gol number of electrons removed ▲ 2000 1500 1000 500 0 Figure 7 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 1 7 18 19 20 atomic number Which element could this be? ▲ A. C B. Si Figure 8 i) Dene state C. P D. S IB, May the what term is rst meant ionization by the energy and term periodicity. ii) Explain of which ionization energies of main there be the greatest energy State A. Between 1st B. Between 2nd C. Between 3rd and 4th ionization energies D. Between 4th and 5th ionization energies IB, November Which of A. > IE the 2nd ionization and 3rd ionization energies [1] 2009 following is correct? 4 B. Molar C. The ionization third energies ionization are energy measured represents in kJ. the process: 2+ D. → X Ionization going 300 3+ (g) (g) + e energies from left to levels is meant decrease right. energy. energies IE 3 X this and the existence sublevels [4] by the term second difference? IB, 3 from for atoms. what ionization and evidence boron iii) will information provides within Between how 2010 graph 2 [2] [1] across a period [1] May 2009 [1] T H E P E R I O D I C TA B L E – T H E 13 T R A N S I T I O N M E TA LS ( A H L ) Introduction Transition elements incomplete consider d one complexes of have sublevels simple of characteristic such theory, transition metals. crystal metal are properties. In eld often this These topic theory we which properties explore can help can these us in be associated properties and understanding with in the particular why the coloured. 13.1 Fi- -c Understandings Applications and skills ➔ Transition elements have variable oxidation ➔ Explanation of the ability of transition metals to states, form complex ions with ligands, have form variable oxidation states from successive coloured compounds, and display catalytic and ionization energies. magnetic proper ties. ➔ ➔ Explanation of the nature of the coordinate Zn is not considered to be a transition element bond within a complex ion. as it does not form ions with incomplete d ➔ Deduction of the total charge given the formula orbitals. of the ion and ligands present. ➔ Transition elements show an oxidation state of ➔ Explanation of the magnetic proper ties in +2 when the s- electrons are removed. transition metals in terms of unpaired electrons. Nature of science ➔ Looking for trends and discrepancies – transition elements follow cer tain patterns of behaviour. The elements Zn, Cr, and Cu do not follow these patterns and are therefore considered anomalous in the rst-row d-block . 301 13 T H E P E R I O D I C TA B L E – T H E T R A N S I T I O N At IUPAC ii f a the family aii centre of chemical of M E TA L S the elements, ( A H L ) periodic called properties often table the of elements transition play a key role lies a very elements, in many important whose facets of physical and everyday life. According to IUPAC, a aii In the periodic table (gure 1) the rst-row transition elements are the is an element that has elements in period elements below 4 from scandium (Sc) to copper (Cu) inclusive. The an atom with an incomplete these elements in periods 5, 6, and 7 are also described d-sublevel or that gives rise as transition elements. to cations with an incomplete The lanthanoids are the elements from Z = 57 to Z = 71 and the d-sublevel. These refer to actinoids are the elements from Z = 89 to Z = 103. La (Z = 57) and elements in groups 3–11. 1 Ac (Z = 89) respectively all the have (so other do not The transition congurations contain lanthanoids congurations. inner electron and f-electrons actinoids f-block of in [Xe]5d their contain elements are 2 1 6s outer and [Rn]6d energy f-electrons sometimes in levels), their described 2 7s electron as the elements 1 18 2 1 He H hydrogen [1.007; 1.009] 3 , but helium 13 2 14 5 4 15 6 16 1 7 8 9 7 4.003 10 Li Be B C N O F Ne lithium beryllium boron carbon nitroge n oxygen uorine neon [6.938; 6.997] 9.012 [10.80; 10.83] [12.00; 12.02] [14.00; 14.01] [15.99; 16.00] 19.00 20.18 11 12 13 14 15 16 1 7 Na Mg Al Si P S Cl Ar sodium magnesium aluminium silicon phosphorus sulfur chlorine argon 22.99 24.31 26.98 [28.08; 28.09] 30.97 [32.05; 32.08] [35.44; 35.46] 39.95 3 4 5 6 7 8 9 10 19 20 2 1 22 23 24 25 26 27 28 29 30 31 32 33 34 35 K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr potassium calcium scandium titanium vanadium chromium manganese iron cobalt nickel copper zinc gallium germanium arsenic selenium bromine krypton 39.10 40.08 44.96 47 .87 50.94 52.00 54.94 55.85 58.93 58.69 63.55 65.38(2) 69.72 72.63 74.92 78.96(3) 79.90 83.80 37 38 39 40 41 42 43 44 45 11 18 46 12 47 48 49 50 51 52 36 53 54 Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe rubidium strontium yttrium zirconium niobium molybdenum technetium ruthenium rhodium palladium silver cadmium indium tin antimony tellurium iodine xenon 85.47 87 .62 88.91 91.22 92.91 95.96(2) 101.1 102.9 106.4 107 .9 112.4 114.8 118.7 12 1.8 127 .6 126.9 131.3 57-7 1 55 56 Cs Ba caesium barium Cs 132.9 87 Fr francium lanthanoids 74 75 76 77 78 79 80 81 82 83 84 85 86 W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn hafnium polonium astatine radon tantalum tungsten rhenium osmium iridium platinum gold mercury thallium lead bismuth 1 78.5 180.9 183.8 186.2 190.2 192.2 195.1 197 .0 200.6 [204.3; 204.4] 207 .2 209.0 104 105 106 107 108 109 110 111 112 Rf Db Sg Bh Hs Mt actinoids rutherfordium dubnium seaborgium bohrium hassium meitnerium Ra radium 73 T a 89103 137 .3 88 72 Hf 57 59 58 60 6 1 62 Ds darmstadtium 63 114 116 Rg Cn Fl Lv roentgenium copernicium erovium livermorium 64 65 66 67 68 69 70 7 1 La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu lanthanum cerium praseodymium neodymium promethium samarium europium gadolinium terbium dysprosium holmium erbium thulium ytterbium lutetium 138.9 140.1 140.9 144.2 150.4 152.0 157 .3 158.9 162.5 164.9 167 .3 168.9 1 73.1 1 75.0 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr actinium thorium protactinium uranium neptunium plutonium americium curium berkelium californium einsteinium fermium mendelevium nobelium lawrencium 232.0 231.0 238.0 ▲ Figure 1 IUPAC periodic table of the elements The as elements transition of group elements 12, that is according Zn, to Cd, Hg, IUPAC as and all Cn, four are not classied elements have 10 d-sublevels scandium Ufu uc containing and yttrium ten are d-electrons classied as (for transition 1 an incomplete d-sublevel 3+ The electron congurations of only all the elements in the periodic to table can be found by accessing (Sc http://www.webelements.com/, of compiled by Professor Mark metal Winter at the University of and Sheeld, UK. a Sc be [Ar]3d 1 ; Y , [Ar]3d elements 2 4s Zn: [Kr]4d as they full 2 4s ). Both have 2 5s ). In 1920, when 3+ and Y compounds non-transition 3+ , (Sc, example, elements were known, because they their ions were widely contained considered no d-electrons 3+ [Ar], these Y , [Kr]). elements bonding. Since have For then been example, many lower synthesized, scandium oxidation most can of exist which in the state compounds involve +2 metal– oxidation state, 1 because transition compound its electron element in which conguration according scandium to is the in is [Ar]3d IUPAC the +2 , scandium denition. oxidation An state is considered example is CsScCl 3 302 of a 13 . 1 Collectively, Ac) are comprise These the referred the in as f-block elements f-block elements to the are the in are those formal periodic groups d-block in 3–12 which members table inclusive elements. (gure of the The 4f group 3 F I r s t- r o w (including elements and but 5f La e l e m e n t s and which orbitals they d - b l o C k form are a lled. separate 2). 1 18 1 1 2 H He 2 13 14 15 16 1 7 2 3 4 3 4 5 6 7 8 9 2 1 22 23 24 25 26 27 10 28 11 29 12 30 Sc Ti V Cr Mn Fe Co Ni Cu Zn 39 40 41 42 43 44 45 46 47 48 s-block p-block elements 5 elements Y Zr Nb Mo Tc Ru Rh Pd Ag 57 72 73 74 75 76 77 78 79 Cd 80 6 La* Hf T a W Re Os Ir Pt Au Hg 89 104 105 106 107 108 109 110 111 112 7 Ac* Rf Db Sg Bh Hs Mt Ds Rg Cn transition elements d-block elements 58 59 60 6 1 62 63 64 65 66 67 68 69 70 7 1 Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 90 91 92 93 94 95 96 97 98 99 100 101 102 103 Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr lanthanoids (*also includes La) actinoids (*also includes Ac) f-block elements (sometimes called the inner transition elements) main-group elements group 1 (excluding H), group 2 and groups 13–18 transition elements groups 3–11 (the f-block elements are sometimes described as the inner transition elements) s-block elements groups 1 and 2 and He p-block elements groups 13–18 (excluding He) d-block elements groups 3–12 [including Z = 57 (La) and Z = 89 (Ac), but excluding Z = 58 (Ce) to Z = 71 (Lu) and Z = 90 (Th) to Z = 103 (Lr), which are classied as f-block elements] f-block elements elements from Z = 58 (Ce) to Z = 71 (Lu) and from Z = 90 (Th) to Z = 103 (Lr) lanthanoids elements from Z = 57 (La) to Z = 71 (Lu) actinoids ▲ elements from Z = 89 (Ac) to Z = 103 (Lr) Figure 2 Periodic table of the elements showing the main-group elements, the transition elements, the s-, p-, d-, and f-block elements, the lanthanoids and the actinoids The metallic described as nature the of the transition transition elements means they are often metals. 303 13 T H E P E R I O D I C TA B L E – T H E T R A N S I T I O N M E TA L S ( A H L ) Electron congurations of rst-row d-block elements suy ip and their ions You need to be able to write the The electron congurations (both full and condensed) and the orbital diagrams for atoms and following are some congurations of the corresponding orbital examples rst-row of full d-block and condensed elements, their electron ions, and their diagrams: ions up to Z = 36, that is to Kr. ● For vanadium, 2 1s 2 2s 6 2p 3s 3 [Ar]3d The 2 V (Z 6 3p = 3 3d 23): 2 4s (full electron conguration) 2 4s orbital (condensed diagram electron conguration) is: [Ar] 2 3 4s In the rst 3d orbital before diagram, lling multiplicity, ● For 2 1s Hu’ u f aiu uipiciy see nickel, 2 2s Ni 6 2p 8 [Ar]3d in three pairs, sub-topic (Z 2 3s the them = 6 3p ll the Hund’s 3d orbitals rule of singly maximum 2.2. 28): 8 3d d-electrons following 2 4s (full electron conguration) 2 4s (condensed electron conguration) Hu’ u f aiu The orbital diagram is: uipiciy states that when lling degenerate orbitals (that is, of the same energy) [Ar] electrons ll the orbitals singly before lling them in pairs. 2 8 4s 3d 2+ ● For Ni 2 1s (Z 2 2s 6 2p = 28): 2 3s 6 3p 8 3d (full electron conguration) suy ip 8 [Ar]3d (condensed electron conguration) When removing electrons to The orbital diagram is: form cations (positive ions) electrons are always removed from the level of highest [Ar] principal quantum number, n. In the case of the rst-row 0 8 4s 3d d-block elements this will be the 4s level. 2+ In the the 3d case of Ni the electrons are removed from the 4s level before level. Transition metals and the Auf bau principle In an and article D. Chemical relative 304 by L.G. Devoghel Education energies Vanquickenborne, published (71, of the in the (1994), 3d and K. Journal Pierloot, of p469-471), 4s orbitals the are discussed respect to and the ground-state is, why an explanation lling and electrons in of these is transition are given orbitals removed metal from with in both ions the the (that 4s level 13 . 1 before this is the 3d for beyond article is regards to rst-row). current interesting misconceptions with the the to that IB read in appear what scope some to An of but relation in orbital The syllabus, more some need an to earlier based the sources energies F I r s t- r o w on e l e m e n t s explanation proposed electron–electron schematic introductory electron be d - b l o C k in nature level in by R.L. Rich, interactions, and can be was useful understanding at d-block congurations. considered. An orbital However, is assumed when two to have one electrons energy occupy an level. orbital, Electron congurations involving because exceptions be In the rst-row in transition terms of metals, electron there are have to be careful with: Cr (Z = 24) = 29). In the case of Cr, you may be and write nuclear is the electron incorrect. conguration Chromium has a as conguration of [Ar]3d charge of the occurs for Cu. One (Z) (both additional pairing increases, electrons: from [Ar]3d orbitals. 2 4s As a the result are factor energy, P. doing 1 4s . A would there is greater d orbitals nucleus to the are not same shielded extent electrons will occupy the as orbitals, previously which in is earlier what we topics in have been writing For example, vanadium (Z electron = 23) to be [Ar]3d 10 electron conguration is [Ar]3d a 2 4s . The attempts are often made to already . At a terms of the extra stability of the fully-lled (d ) d-sublevel. half-lled (d is far too simplistic explanation (in the However, and a much previously stated, cited point after (note that Chem Educ., paper and references the effect of of increasing the 4s and 3d nuclear levels lines electrons the explanation that occupy energies of involves the all nding electrons with Rich’s work, vanadium, crossover that on in line A strong correlation experimental data and on advanced same the as energy is a crossover to onto chromium copper, point). This process there produces the subshell. energies As Rich of the points individual out, is followed until the subshell is the half-lled; the upper the line outer is also used. has electrons are Hence, simply for given orbital. sum their lowest energies available. As can be interpreted of from the lled (d diagram, this does not lead exactly to half- respective been theoretical computational on leading represent each element 10 ) or fully-lled (d 0 ) (or empty d ) subshells. found these often occur because some additional data energy based the the However, between from there leading nickel, 5 interactions. seen interactions the This be relates every between can more therein) charge and from a thereafter, energies as mentioned lower to but after another electrons J 2 4s this two detailed [Ar]3d ) is approach of this 10 and conguration simplistic rationalize 5 in electron correct 1 4s condensed levels level has the 3 conguration s lowest similar expect 9 electron to As . congurations. anomaly an condensed 5 electron the repulsion tempted available This termed is Cu 4 to there that (screened) (Z electrostatic two congurations attraction you their charged), considered, the exceptions of negatively is required to go beyond them. calculations. 1 + 2 3d number of electrons 1 2 in half-subshell electron 1 1 + 1 spin 2 + + Sc 2 2 + Ti + V + Cr E 1 + Mn + Fe 3 1 + Co + Ni 2 4s 1 1 + Cu + Zn 1 2 1 2 1 5 1 1 1 1 1 1 5 1 1 1 1 5 1 Sc 2 3 5 1 4s 1 1 4 Ti 3 1 V Cr 5 5 Mn 1 4 1 Fe 1 Co 5 5 5 5 Ni 1 + 5 5 Cu Zn 5 5 5 (a) 1 1 1 (b) 2 3d 5 1 5 ▲ 2 Figure 3 Schematic representation of Rich’s interpretation of electron congurations for transition elements in terms of intra-orbital repulsion and trends in subshell energies. In the rst diagram, the order in which the levels are occupied is presented. In both diagrams key crossover points feature and in the second diagram one sees how an electron is removed from the 4s level before one from the 3d level 305 13 T H E In on this approach, the lled P E R I O D I C model and of TA B L E the a invalid. T H E commonly perceived fully-lled somewhat – held extra d-sublevels In the case T R A N S I T I O N rationale stability could of the be of based the half- considered formation M E TA L S of A ( A H L ) more found detailed in Inorganic G.L. account Miessler, Chemistry (5th of P .J. this discussion Fischer Edition), and 2013, can D.A. be Tarr., Prentice Hall. n+ the transition removed, the metal cation, overall M electron , when electrons repulsion is are Reproduced with permission from R.L. Rich decreased (Periodic Correlations, W.A. Benjamin, Inc., 1965). and the greater energy extent of the d orbitals compared to that is lowered of the s to a orbitals. Aciviy suy ip On closer examination of the For the IB Chemistry Diploma programme, you are only required to know electron congurations of the entire anomalous electron congurations of the elements chromium and copper from periodic table, other elements also the rst-row transition metals. These congurations are: Cr, [Ar]3d 5 10 1 4s , and Cu, 1 4s . All other rst-row transition metals will have electron congurations convey deviations from expected [Ar]3d patterns. Look at the webelements as predicted based on their position in the periodic table. website and try to nd four other Note the following point, however. Once deduced, do not be tempted to modify electron d-block elements with electron 5 congurations of cations further to follow this 3d congurations that dier from what 6 has an electron conguration of [Ar]3d 4s is expected. 10 and 3d pattern. For example, Fe 2 2+ . However, the electron conguration of Fe 6 is [Ar]3d 5 [Ar]3d . Do not then be inclined to rearrange this electron conguration further to 1 4s 2+ . This is an incorrect electron conguration for Fe . In summary, just note the two exceptions of Cr and Cu for the IB Diploma Chemistry programme. Quic qui 1 Deduce the full electron tok congurations of: 3+ a) Co; ) Zn; c) Ti The medical symbols for the female and male genders originate from the symbols used for copper and iron by the alchemists. These symbols have and explain why Zn is not been used since Renaissance times (see the Royal Society of Chemistry (RSC) described as a transition Visual Elements Periodic Table, www.rsc.org/periodic-table/alchemy). element, according to IUPAC recommendations. 2 Deduce the condensed electron congurations of: 2+ a) 3 V; ) Mn; c) Mn Deduce the orbital diagrams of: 3+ a) Co 3+ ; ) Cr + ; c) Cu Mars symbol – symbolizes a male Alchemist’s symbol for iron organism Venus symbol – symbolizes a female organism 306 Alchemist’s symbol for copper 13 . 1 F I r s t- r o w d - b l o C k e l e m e n t s r by Iron and copper are two of the seven metals of alchemy (gold, silver, mercury, copper, lead, iron and tin). Alchemists are often considered as the rst chemists. Rober t Boyle, who was born in Lismore Alchemists developed a unique language to describe not only chemical in Ireland in 1627, is often described reactions, but also philosophical doctrines. Some commentators claim that the as “ The Father of Chemistry” pseudoscience of alchemy has played a key role in the development of modern (see www.rober tboyle.ie/). medicine and chemistry. Alchemists made a signicant contribution to the chemical industries of that period, in areas such as the metallurgical industry, the dye industry and the glass-manufacturing industry. Alchemists extracted metals from their ores and tried to arrange the information known at that time of the various substances. The original idea of a periodic table might therefore in par t be attributed to the alchemists. During the early days of alchemy the astronomical signs of the planets were used as alchemical symbols. Alchemical symbols were used to represent some elements and their compounds until the 18th century. Boyle was not only a devotee of “natural philosophy”, an advocate of the Characteristics of transition elements As mentioned the of a nuclear these increase lower In in electrons the the the can case rst shown to be enter electrons greater going the of IE rst and across of the the the valence to right atomic energy elements, period, the fact the orbital, shell that effect than radii ( IE) rate of experimental sciences, but also a key the will there valence is increase a is a table, result across gradual elements. electrons. This the elements, electrons This have founder of the Royal Society in England. Boyle proved the inverse relationship between the volume of a gas and its pressure, known as by’ a (sub- topic 1.3). Although Boyle made the much elements, main-group Inner-shell As increase transition for periodic decrease. main-group for whereas orbital. across although corresponding to (screening) left ionization transition that a from increases inner-shell enter gure 3, attributed an shielding in Z, factors, compared difference topic charge, two period. in a trend is 4. transition to modern science, much of his thinking centred around what is termed “chaici”, which has its basis as the extension of knowledge by reasoning and inference. All these attributes are still important to the modern day chemical practitioner. Have any of the principles of the earlier alchemists also been carried 2500 He through the ages to modern day scientic methodology and chemical practice? You Ne might wish to reect on the importance of 2000 1 hypothesis and observation in chemical lom Jk/ygrene noitazinoi tsr experiments you carry out in the laboratory. Ar 1500 Kr tok Xe Rn Hg Zn 1000 Rober t Boyle was a scientist and a Cd philosopher. In many countries such as France the study of philosophy is Al Ga 500 Li mandatory at high school. In France Ti In Ni the philosophy curriculum aims K Rh Cs Fr at producing enlightened citizens capable of intelligent criticism. 10 20 30 40 50 60 70 80 90 100 Find out what other countries prescribe the teaching of philosophy as mandatory atomic number at school and discuss the role and ▲ Figure 4 Trends in the rst IE for main-group and transition elements. Notice that the rate of importance of taking a philosophical increase in the rst IE across the period is much more gradual for the transition elements compared to that for the main-group elements view in scientic discourse. 307 13 T H E P E R I O D I C TA B L E – T H E T R A N S I T I O N Transition M E TA L S metals have a ( A H L ) number of key characteristics: laay ip • ● they ● compounds ● transition metals form ● transition metals are ● magnetic have variable oxidation states When reading the meniscus for potassium permanganate, the top of of transition elements and their ions are often coloured the meniscus should always be read complexes with ligands (B in gure 6), because it is convex often used as catalysts upwards. This is because the deep colour makes it very dicult to read the meniscus. This is in contrast to normal states properties and of transition coordination metals depend sodium, where on their oxidation number. practice for most clear solutions (A) where the meniscus is read from the Variable oxidation states bottom, that is concave upwards. In ▲ contrast to an alkali metal such as the oxidation state Figure 6 How to read a meniscus: A, for is always +1 in its ion and compounds, transition metals are often found clear solutions; B, for KMnO 4 with the different rst-row oxidation d-block states. elements The (see range of sections different 9 and oxidation 14 of the states Data for booklet) A can be B be seen split from the according sc to ti diagram their shown oxidation in gure states into 5. The three d-block types C – elements A, B, V C m F ni +1 +1 +1 +1 +1 +1 +1 and Cu +2 +2 +2 +2 +2 +2 +2 +2 +2 +3 +3 +3 +3 +3 +3 +3 +3 +3 +4 +4 +4 +4 +4 +4 +4 +5 +5 +5 +6 +6 can C. Z +2 +6 +7 • In carrying out a redox titration type involving potassium permanganate Sc, the colour change is typically from purple to colourless (with a ▲ faint pink tinge, signifying the +7 A: Ti, type and V Cr B: and type Mn Fe, C: Co, Ni, Cu, and Figure 5 Range of oxidation states of the rst-row d-block metals. The most common oxidation states are marked in green to +2 oxidation state change for manganese.). If, however, the colour The characteristics of Type A are dominated by: changes from purple to brown, this ● stable high oxidation states (for example, V is +5 in VO ) 3 would signify the formation of the 4+ intermediate ion of manganese, Mn ● , unstable low oxidation states. with an associated oxidation state of The characteristics of Type B are dominated by: +4, which is also a stable oxidation ● stable high oxidation state. This may occur if there is insucient acid in the conical ask . 308 states (for example, Mn is +7 in MnO , 4 2 is +6 in Cr O 2 ) 7 Cr Zn 13 . 1 F I r s t- r o w d - b l o C k e l e m e n t s 2+ ● stable low oxidation states (for example, Mn is +2 in [Mn(H O) 2 ] , 6 3+ Cr is +3 in [Cr(H O) 2 The characteristics ● unstable ● stable high of ] ). 6 Type C oxidation are dominated by: states 2+ low oxidation states (for example, Fe is +2 in [Fe(H O) 2 Manganese +7. In the is characterized chemical permanganate, in manganate(VII) practice deep in oxidation +2, the state which is redox though +7 almost is you In reduced often (more compound colour. states may workplace!) (purple) of oxidation titrations this chemical burgundy by laboratory This named named reagent is reactions, manganese from reagent +1 potassium this way in characterized an to potassium manganese with ). by with oxidation a an state of colourless: 7+ 2+ Mn oxidation rarely range the correctly is redox to that use ] 6 state: + 5e → +7 Mn +2 2+ species: [MnO ] [Mn(H 4 Another states. transition In yellow its metal, highest compounds, chromium in a +3 chromium, oxidation which state, can oxidation be can +6, oxidation The state: oxidation alcohol is of rst oxidized also exist to green in various forms oxidation orange complexes and with state. 3+ + 3e → +6 primary ] 6 chromium reduced 6+ Cr O) 2 Cr +3 alcohols into an is a two-step aldehyde, process. which in turn A is primary oxidized laay ip further into the corresponding carboxylic acid. • Primary alcohols can be oxidized by strong oxidizing agents In the oxidation of a primary such alcohol, the aldehyde can as potassium dichromate(VI), K Cr 2 O 2 , in sulfuric acid, H 7 SO 2 , to 4 be isolated by iiig it o form the corresponding carboxylic acid, under reux (as discussed in as it forms. Distillation is a sub-topic 10.2): technique used to separate K Cr 2 CH CH 3 O 2 K 7 Cr 2 OH liquids that have dierent O 2 7 CH 2 COOH 3 3 + + H boiling points (boiling point H of ethanal is 20.2°C; that of ethanol ethanal ethanoic acid (primary alcohol) (aldehyde) (carboxylic acid) ethanoic acid is 118°C). • Alternatively, if a milder Oxidation of a primary alcohol oxidizing agent is used, such as pyridinium chlorochromate Secondary alcohols can also be oxidized by potassium dichromate(VI) in (PCC), with an organic solvent sulfuric acid to form the corresponding ketone: such as tetrahydrofuran (THF), K Cr 2 CH CH 3 CH(CH 2 O 2 the aldehyde forms as the nal 7 )OH CH 3 CH 3 C(O)CH 2 3 + product of the reaction. H PCC butan-2-ol butan-2-one CH 3 (secondary alcohol) As outlined oxidation in and +6 to OH topic 9, each reduction. the In redox the chromium process case is of CH CHO 3 THF involves this reduced → 2 (ketone) dichromate(VI), of CH two reaction from an ethanol ethanal (primary alcohol) (aldehyde) half-reactions, with potassium oxidation state +3. 309 13 T H E P E R I O D I C TA B L E – T H E T R A N S I T I O N Oxidation M E TA L S ( A H L ) half-reaction: + CH CH 3 OH(aq) + H 2 Reduction O 2 → CH COOH(aq) 4H + (aq) + 14H (aq) + 4e 3+ (aq) + 6e → 2Cr (aq) + 7H 7 Overall O(l) 2 equation: 2 3CH + 3 half-reaction: 2 Cr O(l) 2 CH 3 OH(aq) + 2Cr 2 O 2 + (aq) + 16H (aq) → 3CH 7 COOH(aq) + 3 3+ 4Cr (aq) + 11H O(l) 2 bahay The redox reaction involving potassium dichromate(VI) is the basis of the ahay used by police forces worldwide to determine if a driver of a vehicle has consumed alcohol. In this test, crystals of potassium dichromate(VI), which are orange/yellow in colour, change 3+ to green, which signies the formation of the Cr species. Since 2012, it has been required by French law that all vehicles need to be equipped with a breathalyser. As seen in gure 7, the simple version of this on-board vehicle breathalyser test kit involves the colour change from orange/yellow to green. This type of breathalyser does not record the ach ccai (BAC), which is the concentration of ethanol in a person’s blood. BAC is the 3 mass, in milligrams, of ethanol per 100 cm of blood. • In order to measure the BAC three devices can be used: • semiconductor oxide-based sensor • fuel-cell sensor • intoximeter, which is an IR spectrometer; this type If the crystals are all yellow/orange, the result is zero – you are clear to go! • If the crystals are green below the line, according to the tube, you are under the maximum limit. But you do have alcohol in your blood and your judgement and reaction times will almost cer tainly of technology is often used in large, table-top be aected – you should consider waiting a while breathalysers found at police stations. and retesting sicuc i • These are relatively new to the market and have a If the crystals are green beyond the line – you are denitely over the limit. do not drIVe! number of advantages, such as their low cost, low power ▲ Figure 7 Example of a simple breathalyser test kit used in consumption, and por tability. The disadvantage of this France. Notice the orange/yellow to green colour change, type of breathalyser is that their sensors need to be 6+ which signies the Cr 3+ + 3e → Cr reduction caused calibrated more frequently than fuel-cell based testers. by ethanol Incorrect calibration can result in systematic errors. suppor t a legal case in a cour t of justice. For this reason, Fu-c positive tests obtained by preliminary screening need to be Another type of breathalyser is based on the fu c conrmed by more advanced analytical techniques, such Ethanol is oxidized initially into ethanoic acid and then as gas liquid chromatography(GLC), in which a sample into carbon dioxide and water. The fuel cell conver ts is sent to a forensic science laboratory and the exact chemical energy generated from the oxidation process concentration of ethanol in the blood is determined. GLC is into electrical energy. The electric potential is used to used to analyse volatile substances. determine the concentration of ethanol. Ii This type of fuel cell can also be quite basic and the results typically determined may not be suciently accurate to 310 A third type of breathalyser is the intoximeter based on IR spectroscopy. This is discussed in detail in option D. 13 . 1 F I r s t- r o w d - b l o C k e l e m e n t s Explanation of the ability of transition metals to form variable oxidation states from successive ionization energies As stated they metals, an already exhibit which alkaline have earth compounds. In states +3, of types +2, of one variable metal of the only metal and one and contrast, is key oxidation xed occurs the +4. related This the is reason a for this in transition stark state. +2 metal patterns of in oxidation with transition The to characteristics states. For titanium to state occurs difference in that s-block calcium its ion with between successive is the example, oxidation the metals contrast is and oxidation the ionization two energies. Coloured compounds of transition metals and their ions Transition metal compounds and KMnO ions are often burgundy coloured, for example: (purple) 4 2+ [Mn(H O) 2 K Cr 2 ] almost O 2 colourless, with a faint pink tinge 6 orange 7 3+ [Cr(H O) 2 ] CuSO 5H 4 [NH ] 4 green 6 O blue 2 [Fe(H 2 O) 2 Crystalline ][SO 6 ] 4 hydrated pale green 2 copper(II) sulfate, CuSO ·5H 4 blue in colour (gure ofcrystallization, 8). and the Upon heating solid, the anhydrous O, is Mediterranean 2 compound CuSO loses forms, its water which is a 4 white powder. O O O O O O Cia ig S H S H In coordinate bonding the O pair of electrons comes from O O OH H H 2 O O the same atom, unlike typical OH 2 2 2 covalent bonding where Cu Cu the shared pair consists of electrons that originate from OH O H 2 H 2 O OH O 2 O 2 both atoms, A and B, which O form the covalent bond, as H H S S O O discussed in topic 4. The older O O name for coordinate bonding O O was dative covalent bonding. 2+ ▲ Figure 8 Structure of CuSO ·5H 4 O. Note the presence of hydrogen bonding and that Cu The use of this older name is 2 has an octahedral stereochemistry, which may not be obvious from the formula no longer recommended by IUPAC. As stated previously, zinc is not classied as a transition element, as it IUPAC recommends the term 10 has a complete d-sublevel, [Ar]3d 2 4s 2+ . Its ion, Zn , has the electron coordination bonding but in 10 conguration, [Ar]3d this text we will use the more Compounds (explained atoms which of zinc(II) below) responsible can absorb in for in are the the the usually complex colourless, have absorption visible of region a unless the chromophore electromagnetic of the widely used term coordinate ligands (group of radiation), electromagnetic bonding as applied in the IB Chemistry guide. spectrum. 311 13 T H E P E R I O D I C TA B L E – T H E T R A N S I T I O N M E TA L S ( A H L ) Complexes of transition metals Compounds that contain transition elements and in which the central n+ metal ion, M molecules or complexes. , is bonded, via ions (termed the Such compounds, compounds to signify the coordinate ligands) are often bonding, are termed described coordinate bonding as to a group transition of metal coordination present between the dcipi f a iga ligand(s) and the central metal ion. A ligand is an atom, molecule, Examples of species with coordinate bonding: or ion that contains a lone pair of electrons (non-bonding + ● hydronium cation, [H O] ● carbon ● a monoxide, CO 3 pair) that coordinates, through cia ig, to a + central transition metal ion to O H H form a cp. H The term piga is often + used to describe a species ● ammonium cation, [NH ] transition metal complex, for 4 2+ that has the ability to act as a example [Ni(NH ) 3 ] 6 ligand in a complex, but is not + + 2 O, yet coordinated. Hence, H 2 H NH 3 because of its two lone pairs, is H NH N 3 3 a neutral proligand, but in the N 3+ complex [Cr(H O) 2 ] Ni H , water H 6 acts as a ligand. H H N NH 3 3 NH 3 ) a) Bonding models of transition + 1 + 3 OH 2 + 1 + 1 OH 2 H metal complexes H O electroneutrality OH principle is 2 Fe OH 2 Pauling’s O 2 3 2 an Fe H OH O 2 2 approximate method of estimating how H charge O + 1 OH 2 2 + 1 OH is distributed of in The basis any individual a this molecule principle atom in or is the complex that the 2 ion. charge molecule or + 1 OH 2 on ion is c) ) 1 OH + 2 restricted to a range between 1 to 1+ and 2 ideally H the charge should be close to 2 zero. 2 2 2 H 3+ 9 shows various representations of O OH 2 Fe Figure 2 Fe the H + 2 + OH 1 OH 1 O 0 O 2 OH 3+ cationic complex, [Fe(H O) 2 ] 2 . 6 O H 1 OH 2 2 1 + OH + OH 2 2 2 2 1 + ● In gure cationic 9(a) a typical complex is representation given. As the of lone the pair 2 of ▲ electrons on each water ligand contributes to Figure 9 Various representation and bonding models for 3+ the cationic complex, [Fe(H O) 2 ] . (a) Conventional 6 3+ the coordinate bond, an arrow is used instead representation of the cationic complex [Fe(H O) 2 of a straight ] . The lone 6 pair on each water ligand forms the coordinate bond with the line. 3+ central Fe ● If we were to adopt the model proposed ion. Square brackets here represent the complex, in which has an octahedral stereochemistry. The overall charge gure 9(b), it would mean a net transfer of on the complex is 3+. (b) Charge distribution in the cationic charge from each water ligand to the metal 3+ complex [Fe(H O) 2 centre. The charge distribution that results ] based on a 100% covalent bonding 6 3+ model. (c) Charge distribution in [Fe(H O) 2 from this 100% covalent bonding model ] based on a 6 100% ionic bonding model. (d) Approximate charge distribution 3+ would confer 3 on Fe and 1+ on each water. in [Fe(H O) 2 312 ] 6 based on Pauling’s electroneutrality principle 13 . 1 (b) is not charges a valid residing model, on however, metals is as F I r s t- r o w a negative target In gure 9(c), a 100% ionic bonding protein shown. ion and The the neutral. 3+ charge water This resides molecules theoretical on stay model the results have the body, quantum performed which interact on with the the mechanical atoms target in drug classical mechanics was used to simulate iron the remainder the press of the protein. As outlined in effectively is also invalid, shown the release given by the The Royal Swedish as Academy experimental in were e l e m e n t s model and is protein calculations atypical. the ● d - b l o C k of Sciences on the 2013 prize they remark existence that “Todaythe computer is just as important a 3+ of the [Fe(H O) 2 ] species in aqueous solution; 6 tool that is, it remains a single unit in for chemists realistic In ● gure 9(d), principle is distribution iron, the are total Pauling’s applied and means central electroneutrality the that metal, approximate now the should as the test tube . Simulations are so solution. net be predict the outcome of traditional experiments” charge charge zero. thatthey As Activity on there Potassium permanganate, KMnO , is frequently 4 a of six water ligands in the compound, used as an oxidizing agent. The manganate(VII) 3+ the Fe cation needs, effectively, three electrons ion has the formula [MnO ] . 4 to confer on it a net zero charge. The charge (i) distribution, then, on each water ligand will Comment, VSEPR 1 three electrons/six ligands = +. Hence, giving a reason, whether or not deduce the be theory could be used to in 2 geometry 3+ this model, coordinate bonds in [Fe(H O) 2 be 50% covalent and 50% is a good to try example to of evaluating understand the structure of If the bonding terms of a Prize in Chemistry 2013 jointly to Martin Karplus in this 100% what France and the and Harvard Michael of Levitt Medicine, this (Stanford USA) and of Southern charge for the The California, Los development chemical used model American of multi-scale systems. Chemists models ranging sophisticated further the for may from known spheres and computational structures proteins. was of remarkable the their of understand Their 2013 about Noble models both research the Prize combined traditional quantum chemical of invalid properties two simulating For how of the the Linus Pauling, is well development but of Pauling the is scale of perhaps for his electroneutrality electroneutrality principle. principle to the manganate(VII) ion, was suggest the be approximate on each charge oxygen if distribution manganese help in a net charge of 1 +. Determine on What this basis and the the percentage covalent character recipients was percentage ionic character. that approaches mechanics example a for programmes Chemistry the be molecules, processes. work classical mechanics. of in the explain and resulted chemists on ion. chemist, his Pauling’s might complexes be and models what explore would have applied to model, Angeles, If to considered Warshel less complex sticks was bonding atoms electronegativities, always anion ionic University Arieh known (University for the angles). University, (iii) USA) identify bond oxygen manganate(VII) School and the (Université why Strasbourg, USA), ion, was manganese de the idea. Nobel awarded of (including a deduce The Draw various nature in scientic ion. ] (ii) models manganate(VII) ionic. geometry This the 6 the would of drug in and their interacted with Classication of ligands The ion number depends pairs) in the coordinate known as of on coordinate the ligand. bond of formed donor Monodentate with chelate bonds number a metal ligands) ion can by ligands while form one centres ligand (atoms are able to polydentate two or more with with a metal lone form ligands such electron only one (also bonds. 313 13 T H E P E R I O D I C TA B L E – T H E T R A N S I T I O N M E TA L S ( A H L ) Monodentate ligands O N Monodentate ligands contain a single donor atom and have one lone H H H H pair contributing to the coordinate bond in a complex. Typical examples H include ▲ water, ammonia, and the halides such as Cl etc. Figure 10 Notice that the water proligand contains two lone pairs of electrons, but Cl only one contributes to the coordinate bond in a transition metal complex Polydentate (chelate) ligands These are ligands coordinate A f caui! examples which bonds are with given have a two or transition more metal donor centre. atoms Some that form common below. Monodentate does not refer to the number of lone pairs present 1,2-ethanediamine (en), H NCH 2 CH 2 NH 2 2 in the proligand, but refers to the number of lone pairs from the ligand that actually are H NH N 2 2 involved in the formation of the coordinate bond. Monodentate M ligands are often described as “one-toothed ligands”. The structure bidentate transition its older N centre in in it a section has two 16 of donor complex. en is the Data atoms still booklet. that en is a coordinate sometimes referred to to the by ethylenediamine. ligands language, are often meaning described crab-claw) as as chelate the ligands ligands look like (coming from they grabbing are the N the N N metal between two or more donor atoms, just like a crab can grab N your Fe toes on a beach! The complexes formed from chelate ligands are very Fe stable. N N given 3 Greek N is because + 3 N en metal name, Polydentate + of ligand N An example of a chelate complex is [Fe(en) N number N ]Cl 3 optical of the isomers iron is (two six as each en ligand non-superimposable is . bidentate. mirror The coordination 3 The images, complex gure has 11). 2 Ethanedioate ▲ Figure 11 Optical isomers of [Fe(en) (ox), (C O 2 ]Cl 3 ) 4 3 - - M Ethanedioate, dianionic often referred to by its older name, oxalate, ligand. 4 Ethylenediaminetetraacetate, (EDTA) O O - - O C CH 2 O - 314 C CH 2 CH C O CH C O 2 N N 2 - is a bidentate, 13 . 1 F I r s t- r o w d - b l o C k e l e m e n t s 4 (EDTA) bonds. in an is It a polydentate has the octahedral [Co(EDTA)] , ability ligand to complex. the EDTA that wrap For acts can itself example, as a form around in up a the hexadentate to six coordinate transition anionic metal centre O complex O N Co ligand. O EDTA ● is used Removal as its with of use heavy in other N O in: metals. the The treatment metal ions ligand of lead present in has a number poisoning. blood. of applications, EDTA When Na can such ▲ Figure 12 Structure of [Co(EDTA)] ▲ Figure 13 Use of EDTA as a preser vative coordinate [Ca(EDTA)] is 2 administered to a patient, lead can displace calcium to form the 2 anionic complex [Pb(EDTA)] 2 : 2+ [Ca(EDTA)] + 2 Pb → 2+ [Pb(EDTA)] + Ca 2 Once ● formed, Chelation therapy. potential use considered of the the lled from plaque, Water free The Food the is loss perceived in by rancidity and chains double acids of are of a lipids oxidized bond in to are be Restorative sculpture. a oxygen have catalysed for sculptured metal of can pieces. the risk has of been (“hardening reduces removes reduce cholesterol- cardiovascular therapy softening (which added has shown a to food down or lipid. food that no with to fats and “gone off” because In across the ions. It is of hydrolytic the fatty fatty acid carbon-to-carbon aldehydes The oils. components, rancidity, Volatile (for leading in their metal ensure products appearence. odours. or to precipitate reactions occurs has into added can reason. catalyse oxidative light can urine. its bloodstream chelation same can noxious by the is therapy effectively This water taste, is EDTA the process EDTA and carboxylic involves acts as a ions. EDTA artwork coating In the Rancidity odour, of into disease. the when unsaturated can such ions broken of in often colour. bad and the which reactions develop or in and remain for kidneys atherosclerosis tissue. used is ions use ions Metal senses scavenger of also for EDTA heart shampoos taste the the for the Chelation reduces the propan-1,2,3-triol. form, radical is Ca-EDTA development acids benet mayonnaise). rancidity, date magnesium used of calcium potentially EDTA or preservation. example presence by application treatment free to passed surgery. atherosclerotic which softening. It of be medical by-pass potential limited calcium soaps). ● a However, somewhat the as can Another heart arteries”). problems. ● in concentration calcium ● [Pb(EDTA)] also Old be used brass insoluble solid or in the copper complex, restoration sculptures brochantite, 2 CuSO 3Cu(OH) 4 form, ● which Cosmetics. . Upon the addition of EDTA, [Cu(EDTA)] can 2 is soluble EDTA is and easily sometimes removed. used as a preservative in cosmetics. in cosmetics 315 13 T H E P E R I O D I C TA B L E – T H E T R A N S I T I O N M E TA L S ( A H L ) Coordination numbers The six majority of (octahedral transition metal geometry) schiy or complexes four or coordination square planar numbers 90 (and 180) eap 6 [Fe(H O) 2 tetrahedral 109.5 4 square planar 90 (and 180) 4 K K ]Cl 6 theory ions. The of X-ray available. distances Many the and transplatin, of deduce such complexes the features, testicular no can complexes geometries. Cisplatin, cancer, is can if such structure, geometry be of be all of crystals the bond elucidated 4 in transition the planar. by the angles using Its the metal metal structural complex and this of bond technique. number treatment ] 4 transition of coordination used of the determined single as square anticancer the d-sublevels crystallography in shows to incomplete platinum(II) planar bladder used the Structural present of square be of structures technique are cannot because ] [Ni(CN) 2 complexes 2 [CoCl 2 VSEPR of geometries). Ciai u b ag/° octahedral have (tetrahedral four have ovarian, geometrical isomer, activity. Transition metals as catalysts Transition are some sections ● metals examples of Haber N (g) the often of reactions + 3H catalyst: (g) ⇋ 2NH catalysts that you in may chemical be reactions. familiar with Here from other (g) 3 Fe(s) decomposition of 2H O(l) O as process: 2 2 used programme. 2 ● are (aq) → 2H 2 hydrogen + O 2 catalyst: MnO peroxide: (g) 2 (s) 2 ● hydrogenation H C=CH 2 (g) + of H 2 alkenes: (g) → Ni(s), Pd(s), hydrogenation RCH=CHR’ + H of or Pt(s) oils (g) → RCH 2 catalyst: product a is ● oils has The controlled. Mono- be This R’ 2 hydrogenated, is greater texture The and saturated 316 can liquid. also oxidation. CH 2 Ni(s) Unsaturated of (g) 3 ethane catalyst: instead CH 3 ethene ● CH 2 main chemical (that is, its form for stability fats of are a semi-solid cooking due hardness disadvantages polyunsaturated fats. to advantageous to and a reduced plasticity) hydrogenation healthier (or purposes. for solid) The rate of the of product are: the heart than 13 . 1 ● Trans fatty acids metabolize tissues of the lipoprotein which of the can with body. (LDL) may formed Trans and in This in partial therefore fatty cholesterol result arteries. be difculty acids be hydrogenation. may increase (colloquially cardiovascular will the levels known further as in of of e l e m e n t s the fatty low-density “bad because in d - b l o C k These accumulate problems discussed F I r s t- r o w cholesterol”), the sub-topic narrowing B.3. Catalytic conver ters in cars In a running temperature N (g) + car O 2 When engine, conditions (g) → gaseous (1500 nitrogen °C) to form and oxygen nitrogen react under high- monoxide: 2NO(g) 2 NO(g) is released into the atmosphere, it combines with O (g) to 2 form nitrogen dioxide NO (g): 2 2NO(g) + O (g) → 2NO 2 Nitrogen the dioxide brown (g) 2 is colour a of secondary pollutant photochemical that smog. is primarily Nitrogen responsible dioxide is toxic for and ▲ can result in respiratory Figure 14 Catalytic conver ter on the underside problems. of a car. Three- way catalysts conver t oxides of Carbon also monoxide, emitted Most from modern reduce NO(g) CO(g), the a highly exhaust cars now and NO are (g) of a toxic, car, equipped to N and H 2 (g) odourless, as well with while as and colourless unburned catalytic oxidizing is hydrocarbons. converters CO(g) gas, and nitrogen, carbon monoxide and hydrocarbons into nitrogen, carbon dioxide and water. However, unleaded fuel has to be used in that unburned vehicles tted with catalytic conver ters. If 2 leaded fuel is used (that is, fuel containing hydrocarbons to CO (g) 2 O(g), which are less harmful substances: 2 added lead compounds used as anti-knocking 2NO(g) + 2CO(g) → N (g) + 2CO 2 CH CH 3 Ethane In one Rh and (g) propane of CO(g) often so CuO Cr NO(g) the of there or in and (g) → 3CO O a , (g) + 4H 2 exhausts catalytic exhaust second which O(g) 2 can result converter unburned the is 2 reducing 5O agents) the catalyst can be poisoned 2 2 temperature NO(g) + 3 chamber oxidize the of CH 2 (g) in ozone (gure hydrocarbons. gases and chamber operates at a 14) of it Pt, a lower Pd, and increases additional contains much beads However, produces that formation. amounts different catalyst, temperature, 3 to N (g). 2 Hgu a hgu caay Catalysts in green chemistry Hgu caay Catalysts American and play an important Chemical Society , implementation eliminate health the and use the of and role green in green chemistry chemical chemistry. is products generation of the and According design, the development, processes substances to to hazardous reduce to or A homogeneous catalyst is one that is in the same phase or physical state as the substances involved in the reaction that human it is catalysing. environment. Hgu caay Biological catalysts A heterogeneous catalyst is one that is An enzyme is a biological enzyme-catalysed metals. One hemoglobin oxygen subunit in of reactions example (Hb). the catalyst. is that heme The hemoglobin occur (gure Hemoglobin blood. In contains red an in human cells 15), (gure vibrant the is colour atom and which 16) of body the of is involve the to in a dierent phase to the substances many centre that stems which are involved in the reaction that it is transition iron protein blood iron, there catalysing. Industrial catalysts that of involve transition metals are usually transports from heme. oxygen Each heterogeneous catalysts. binds. 317 13 T H E H P E R I O D I C TA B L E C 2 – T H E T R A N S I T I O N M E TA L S ( A H L ) CH 3 CH 2 O H C 3 N N O rest of porphyrin not shown N Fe N H N Fe N N C 3 N CH 3 heme N protein (globin) N O HO O ▲ HO Figure 15 Structure of heme. In heme, iron has ▲ a coordination number of four and a square Figure 16 Structure of human oxyhemoglobin. Oxygen is carried through the blood stream by the formation of a weak bond with 2+ planar geometry. The Fe 2+ ion is at the centre heme. The O Fe bond is then broken relatively easily. When 2 2+ of a large nitrogenous heterocyclic ring called O bonds to Fe 2+ , O 2 functions as a monodentate ligand and Fe 2 a pphyi. Each Hb molecule contains four then adopts an octahedral stereochemistry, with a coordination heme groups. The iron can bind to one O number of six, as heme is linked to the protein (the globin) via an 2 molecule and therefore a single Hb molecule can transpor t up to four O additional Fe N bond. molecules 2 Magnetic properties of transition metals Magnetic many number, contain by an not properties factors, and external magnetic transition the geometry electrons magnetic unpaired elds. compounds 318 the unpaired contain of including are Para- of the that eld. further complex. as contrast, and in their of the complexes metal, its Paramagnetic tiny magnets are properties sub-topic A.2. are by metals, on materials attracted materials repelled of depend coordination and diamagnetic therefore diamagnetic discussed and state behave In electrons and metals oxidation do external ions, and 13 . 2 C o l o U r e d C o m P l e x e s 13.2 C u cp Understandings Applications and skills The d-sublevel splits into two sets of orbitals of ➔ Explanation of the eect of the identity of the ➔ dierent energy in a complex ion. metal ion, the oxidation state of the metal Complexes of d-block elements are coloured, ➔ and the identity of the ligand on the colour of as light is absorbed when an electron is excited transition metal ion complexes. between the d orbitals. Explanation of the eect of dierent ligands on ➔ The colour absorbed is complementary to the ➔ the splitting of the d-orbitals in transition metal colour obser ved. complexes and colour obser ved using the spectrochemical series. Nature of science Models and theories – the colour of transition ➔ Transdisciplinary – colour linked to symmetry ➔ metal complexes can be explained through can be explored in the sciences, architecture, the use of models and theories based on how and the ar ts. electrons are distributed in d-orbitals. In an ion, isolated they atom, split into these sublevels light, which d leads are orbitals two to have sublevels. absorption responsible for the The and the same energy electronic emission colour of but in a transitions of the complex between photons of visible complex. Theories on complexes A number proposed in of to different explain complexes. theories the These have bonding theories are of been transition d-block listed metals below Valence bond developed which had theory is Ligand by theory Linus (VBT). Pauling hybridization rarely used as in its VBT the is was often 1930s, basis. not This and today. eld Crystal an eld theory electrostatic limitations, order (this ● of be Molecular theory, for (CFT). model. example, ligands will theory CFT in the is orbital it does the ligands are (LFT). cannot theory explain (MOT). an a models. with discussed in differs LFT is from an electrostatic model. combination The LFT is terms bonding more of CFT of the as it LFT is CFT description detailed electronic and can energy the series ● Angular relative in In between involving frontier orbitals*. its later). interactions on but based have spectrochemical considered covalent CFT CFT, considered MOT levels on of based associated be ● and order: extension ● centre in ● chronological metal considered. a overlap sizes of molecular model. orbital orbital In this energies (MO) model, are the estimated calculation. this the 319 13 T H E ● P E R I O D I C These theories many of the complexes, and TA B L E and – models characteristics such magnetic as colour, properties. T H E help of T R A N S I T I O N us explain transition electronic The M E TA L S ( A H L ) details metal of spectra, the this comprehensive of IB these book, explain models Chemistry we the shall colour are beyond Diploma use of only the scope programme. the transition CFT In model metal to complexes. *As outlined in the IUPAC Gold Book (http://goldbook.iupac.org/), fi ia refer to the highest-energy occupied molecular orbital (HOMO) (lled or par tly lled) and the lowest-energy unoccupied molecular orbital (LUMO) (completely or par tly vacant) of a molecular entity. The IUPAC Gold Book is an invaluable source for chemists. Crystal eld theory (CFT) The d-sublevel and d consists of ve d-orbitals (gure 1) d , d xy , d yz , d xz , 2 x 2 y 2 z As can be electron seen from density gure pointing 1 at three 45 ° to of these the orbitals Cartesian have axes their (d , d xy contrast, electron the remaining density two pointing orbitals along the ( d , 2 and 2 x d y 2 z Cartesian have ) axes. , lobes d yz ). of In xz their However, lobes in the of free n+ metal ion, d-orbitals CFT is M , are based with ligands (L) at an innite distance away, these ve degenerate. on an electrostatic model, where the ligands are considered n+ as point eld charges created by symmetrical), Ufu uc will increase Look at the Orbitron website to created see the shapes of the d orbitals: will by split the the in the into that surround ligand point energies energy ligand two of of metal charges the d uniformly. point sets the is will however, is degenerate M . isotropic orbitals If, charges cation, the the is, electrostatic spherically degenerate electrostatic then t the (that remain octahedral, energy, If set the and d orbitals the e 2g http://winter.group.shef.ac.uk/ Three orbitron/ are of the orbitals (the t set) will decrease in but eld set. g energy (that is, they 2g stabilized) and two of the orbitals (the e set) will increase in energy g (that is, they are destabilized). y The stabilized orbitals that z z x y x d d xy d xz yz third car tesian axis in each case is or thogonal (90°) to the 2D plane y z x z 2 x 320 x y d ▲ comprise Figure 1 Five d-orbitals 2 y d 2 z the 13 . 2 t set are the d 2g is , d xy associated electron , and d yz with density orbitals. The reason for this C o l o U r e d stabilization xz the fact lying at that 45 ° these to the three orbitals Cartesian have axes. In their lobes contrast, the of d 2 2 x and d orbitals 2 (e z C o m P l e x e s ) are destabilized because their lobes of y electron g density are between directed the two along split the Cartesian degenerate sets axes. of The orbitals energy is separation dened as Δ , the o crystal eld splitting energy n For the rst three 1 d electron will occupy the t set of and will the orbitals congurations, degenerate orbitals d in 2 , d an 3 , and d , the octahedral electrons crystal eld, 2g ll singly before lling them in pairs, following 3 Hund’s rule electron of has a maximum choice – it multiplicity. can either However, occupy the after d , the destabilized fourth e level or g else occupy the stabilized t level. Although the electron would enter 2g a stabilized pair the electron additional So energy what with energy are parameter, the Δ ? level, is to do termed factors First so another the that of all, would electron pairing affect it is the require in an additional already energy, crystal important to lled to This P eld stress splitting that Δ o experimental energy orbital. energy is an o quantity. The following are the factors that affect the size of Δ : o ● identity ● oxidation ● nature ● geometry of the metal state of the of of ion the metal ion ligands the complex ion. Symmetry Science which of is peppered form science. The historically never just should part of with the origin of interesting. accept always symbolic universal such As symbols try to representations symbols chemists at grasp their the ● language can we face origin be should and Part of the IB learner to to a a triply The symbols one orbital term for a as IB learners we strive to be prole t and e 2g notations used in b are used of set if orbitals. of there Ungerade is e orbitals. is the only German odd. The number 2 is used if the sign of the is changes upon rotation about inquirers. the The or set degenerate involved. wavefunction that degenerate doubly such ● representations. refers refers value of t an let octahedral Cartesian us look at axes (gure what 2). happens For to example, the sign of the g crystal eld energy splitting diagram have their wavefunction with respect to the d orbital on xy origin ● g in symmetry: comes the from rotation the wavefunction upon inversion; parity of the term does that orbital. is, u gerade, not meaning change there comes is no from that about the x-axis: y sign change the in change of sign from + to - term x on rotation around x-axis ungerade, changes is u a meaning sign change are only that upon in the used if the wavefunction inversion; parity a of that the is, there orbital. geometrical entity g d and has xy a ▲ centre of of inversion. inversion (but the not in in a German the Hence, octahedral tetrahedron), term as for g there is a stereochemistry is used. Figure 2 Explanation of the 2 symmetry label centre Gerade is As is the sign changes from + to , the number 2 used. even. 321 13 T H E P E R I O D I C TA B L E 1 Gup 9 cp Δ – T H E T R A N S I T I O N M E TA L S ( A H L ) Identity of the metal ion / c The identity of the metal ion can inuence the extent of the crystal eld 3+ [Co(NH ) 3 22900 ] splitting. 6 In general, Δ increases descending a group with the metal in o the same oxidation state. 3+ [Rh(NH ) 3 34100 ] 6 3+ [Ir(NH ) 3 41100 ] Oxidation state of the metal ion 6 For a given metal, Δ increases as the oxidation state increases. Since the o metal–ligand 1 Cp Δ / c the metal interaction increases, the is partly electrostatic distances between in the nature, metal as and the charge ligands on decrease resulting in a better overlap between the metal orbitals and the ligand orbitals. 2+ [Co(NH ) 3 10200 ] 6 Nature of the ligands 3+ [Co(NH ) 3 22900 ] 6 Ligands ligand, may NH have , has a different greater charge charge densities. density For example, compared to the water, 3 hence the ammonia H O, and 2 crystal eld splitting caused by ammonia will be greater. 1 Cp Δ / c 3+ [Co(H O) 2 18200 ] 6 spcchica i 3+ [Co(NH ) 3 22900 ] - 6 I - < Br - < Cl 2 < F < [C O 2 ] 4 ≈ H O < NH 2 < en < bpy < phen < NO 3 < CN ≈ CO 2 weak-eld ligands strong-eld ligands increasing ∆ 0 In the case of weak-eld ligands, the conguration adopted involves a pi-f conguration (gure 3), whereas in the case of strong-eld ligands, such as CN , the conguration adopted involves a pi-pai arrangement (gure 4). The following is a guideline when considering whether a conguration involves a spin-paired or a spin-free e ygrene latibro 2+ • m : d g arrangement: : In the spectrochemical series, for complexes that 2 2 x and d y 2 z orbitals destabilized 3 ∆ 0 5 barycentre 2 ∆ 0 5 : d t 2g , d xy , d xz yz 2+ , ligands to the involve M orbitals stabilized right of NO are designated 2 as g- iga (and hence adopt a spin-paired 2+ conguration), whereas ▲ Figure 3 Crystal eld splitting for [Fe(H O) 2 4 ligand. The t complexes with ligands to the ] , which involves the H 6 O weak-eld 2 2 e 2g conguration adopted is a pi-f conguration g are designated as left of NO 2 a- iga (and hence adopt a spin-free conguration). e 3+ m series, for complexes that 3+ involve M of H : d g : In the spectrochemical ygrene latibro • , ligands to the right O are designated as strong- 2 2 2 x and d y 2 z orbitals destabilized 3 ∆ 5 0 barycentre 2 ∆ 0 5 t : d 2g , d xy , d xz yz eld ligands whereas complexes orbitals stabilized with ligands to the left of H O 2 are designated as weak-eld ligands. 3 ▲ Figure 4 Crystal eld splitting for [Fe(CN) ] , which involves the CN strong-eld 6 5 ligand. The t 2g 322 0 e conguration adopted is a pi-pai conguration g 13 . 2 A Japanese arranged Δ . The chemist, into a R. Tsuchida, suggested spectrochemical spectrochemical series, series , which is that based given ligands on in could order section of 15 C o l o U r e d C o m P l e x e s be increasing of the Data o booklet is based on empirical evidence. The geometry of the complex ion The geometry of the complex ion can also inuence the crystal eld 4 splitting parameter. For example, Δ for a tetrahedral complex is ∼ t It must is empirical values be of emphasized, in the nature, crystal however, as a eld model energy Δ o 9 that because CFT cannot splitting the spectrochemical account for the series relative parameters. Explanation of the colour of transition metal complexes The colour of transition metal ions is associated 2+ d orbitals. For example, conguration, so its d has sublevel an is partially condensed lled and thus electron one would 2+ Cu example, [Ar]3d incomplete, 2+ expect with 9 Cu ions to crystals be of coloured. CuSO · 5H 4 Cu O ions have a are often dominant blue in colour. Mediterranean For blue 2 2+ colour. In contrast, conguration, and ions so of the zinc’s d-block d-sublevel metal is zinc, fully Zn lled. 10 , As have a an result, [Ar]3d ions of 2+ Zn are White metal be typically light consists complexes transmitted. the colour colourless. of The the of all absorb the colour light colours some of of these wheel transmitted, the visible colours, (gure that is 5) can the spectrum. allowing be Transition other used to colours to determine complementary colour 3+ of the absorbed light. For example, [Ti(H O) 2 light. The complementary colour to colour wheel. ] absorbs yellow–green 6 yellow-green is red-violet, which lies 3+ at the opposite side of the Therefore, [Ti(H O) 2 a appear in red-violet. transition metal will 6 Let us now complexes. i v t le io v – e lu b b lu e e g n ra o – w llo e y b l u n e e rg – w o lle y rg e e n w o l l e ▲ l o n absorption o ra n g e y and ions t e r e d g light e r of colour te lo iv – d e r e nature d the complementary – examine the o r transmit ] e – g r e e n Figure 5 The colour wheel 323 13 T H E P E R I O D I C TA B L E – T H E T R A N S I T I O N As outlined are split M E TA L S previously, into two sets the of ( A H L ) ve d-orbitals degenerate in orbitals an – octahedral the crystal stabilized t set eld and the 2g destabilized e set. If the d orbitals are partially lled, d-to-d electronic g transitions energy t can set of occur. In orbitals other to the words, an electron higher-energy e 2g set can of jump from orbitals. This the lower- d-to-d g electronic transition is the origin of the colour of transition metal complexes. 3+ In the case of [Ti(H O) 2 yellow-green absorbed is a light ] , such an electronic transition requires a photon of 6 to be measure absorbed. of Δ . Since The frequency of the ΔE represents the yellow–green energy change, light the o frequency of the light, v (or f ), is related to ΔE via the expression: hc _ ΔE = hv = λ where: 34 h = Planck’s constant = 6.63 × 10 J s 8 c = λ speed = of light wavelength, Since ΔE is in in related a vacuum = 3.00 × 10 1 m s m to Δ , the actual colour of any complex will depend o on all the factors oxidation state geometry of described of the the previously, metal complex ion, the including nature of the the identity ligands, and and the ion. a) e : d g 2 2 x and d y 2 z ygrene latibro orbitals destabilized 3 ∆ 5 0 barycentre 2 ∆ 5 0 : d t 2g , d xy , d xz yz orbitals stabilized b) d-to-d electronic transition hν e : d g 2 2 x and d y 2 z ygrene latibro orbitals destabilized 3 ∆ 5 0 barycentre 2 ∆ 5 0 t : d 2g , d xy , d xz yz orbitals stabilized 3+ ▲ Figure 6 (a) Crystal eld splitting for ground-state [Ti(H O) 2 ] 1 , which involves a t 6 conguration. (b) Crystal eld splitting for excited-state [Ti(H O) 2 conguration 324 0 e 2g 3+ ] 6 g 0 involving a t 2g 1 e g 13 . 2 C o l o U r e d C o m P l e x e s Worked examples base. Each water molecule acts as a monodentate Example 1 2+ ligand, For the complex K [Fe(ONO) 3 ], forming a coordinate bond with Ni . As deduce: 6 there are six water ligands involved, the geometry 2+ a) The oxidation state of the transition metal in of the cationic complex, [Ni(H O) 2 ] , is octahedral, 6 2+ the with complex. Ni having perchlorate b) The condensed electron conguration of metal in this oxidation coordination are in the number lattice and do of 6. not The form the part transition ions a of the cationic complex. state. OH 2 c) The coordination number of the metal in the H OH O 2 2 complex. Ni d) The stereochemistry e) The charge (geometry) of the [ClO 90° ] 4 2 complex. H OH O 2 2 OH 2 on the complex. Example 3 Solution Consider a) Let x = the oxidation state of iron in the complex [Ni(NH the ) 3 ]Cl 6 2 complex. a) Deduce of 3(+1) + x + 6( 1) = 0, so x = the the transition Fe is c) Six, [Ar]3d 2 4s 3+ , so Fe a in its conguration associated state in this complex. 5 is [Ar]3d b) assuming metal electron +3. oxidation 6 b) condensed monodentate nitrito State the geometry of the transition metal ligand, complex and draw a diagram of the complex. (ONO) c) d) Octahedral, assuming a monodentate Identify the ligand, the nature of the bonding between nitrito ligand and the transition metal ion in the (ONO) complex. e) Each potassium has a charge of 1+, so be 3+, meaning the net d) State the e) Draw denticity of the ammonia ligand. + charge that for the three charge K ions on the will anionic complex a diagram d-sublevel. (that is nested in the square brackets) showing the splitting of the part will be Label the orbitals involved and 3 populate f) Explain each of the orbitals with whether the complex is electrons. paramagnetic suy ip or diamagnetic. Oxidation states are written with the sign rst and then the number (for example, here iron has an oxidation Solution state of +3); ions are written with the number rst and 3+ a) ion). then the charge (for example, Fe Let x = the oxidation state of nickel in the complex. x + 6(0) + 2( 1) = 0, so x = +2; Example 2 8 Ni Ni(ClO ) 4 reacts with water to form the O) 2 of ][ClO 6 an ] 4 is 2+ , Ni 8 is [Ar]3d . Explain this reaction Octahedral, acid–base formed CN = 6. in 2 theory, and outline how the NH 2+ bond 2 4s 0 [Ni(H terms [Ar]3d 2 b) ion is complex between Ni and H 3 O. 2 H NH N 3 3 Ni Cl 90° 2 Solution A Lewis acid is an electron-pair acceptor and a H NH N 3 3 NH 3 2+ Lewis acts base as the is an electron-pair Lewis acid and H donor. O acts Hence as the Ni Lewis 2 325 13 T H E P E R I O D I C TA B L E – T H E T R A N S I T I O N M E TA L S ( A H L ) 6 c) Coordinate bonding. e) t 2 e 2g d) NH has one lone pair involved in the f) g Paramagnetic since there are two unpaired 3 coordinate bond to Ni so it is monodentate. electrons. e : d g 2 and d 2 x y 2 z orbitals destabilized 3 ygrene latibro ∆ 0 5 barycentre 2 ∆ 0 5 t : d 2g , d xy , d xz yz orbitals stabilized 326 Q U e s t I o n s Questions 1 Which of the transition following elements is not a 7 What of the following can act as a ligand? element? I. PH II. H III. NO A. I and II B. I and III C. II and D. I, II, 3 A. Fe O 2 B. Cu 2 C. D. 2 Sc only Zn What is the condensed electron only III only conguration and III 2+ of Co ? 2 A. [Ar]4s B. [Ar]4s 7 3d 8 2 Which electron transitions are responsible for 5 3d the colours of transition metal compounds? 7 C. [Ar]3d D. [Ar]4d A. Between B. Among d orbitals and s orbitals 7 3 What is the condensed electron the C. From D. Between the IB, May attached metal ion ligands to the attached ligands conguration d orbitals [1] 2+ of Fe ? 2 A. [Ar]4s B. [Ar]4s 1 5 3d 9 6 C. 2009 6 3d Which salts dissolved 2 D. form coloured solutions when [Ar]3d [Ar]4s in water? 4 3d I. ZnCl II. FeBr III. Co(NO 2 4 What [NH is ] 4 the ligand [Fe(H 2 O) 2 in the ][SO 6 ] 4 2 complex ? 2+ A. Fe B. [SO ) 3 3 2 2 A. I and II only B. I and III only ] 4 C. H C. II and D. I, II III only O 2 + D. [NH and III ] 4 10 5 What is the oxidation state of iron in Which the complex Na[Fe(EDTA)] 3H of the following statements is correct for the complex [Cr(H O) 2 O? ]Cl 6 ? 3 2 A. A. It is paramagnetic. B. It is diamagnetic. C. The +1 B. +2 C. +3 ion D. coordination is number of the chromium 3. +6 D. H O acts as a bidentate ligand in the 2 complex. 6 What is the total charge, n, in the following n complex of Ni(II), [Ni(NH ) 3 ] ? 6 11 A. 0 B. 1+ Explain, energies, by why states,but C. 2+ D. 3+ referring Ca Ti to forms only successive variable occurs in ionization oxidation the +2 oxidationstate. 327 13 T H E 12 P E R I O D I C Explain why TA B L E [Ni(H O) 2 – T H E ][BF 6 ] 4 T R A N S I T I O N is coloured. M E TA L S 14 ( A H L ) In an article Journal 13 Consider the complex, K written by W .B. Jensen in the 2 [Fe(C 4 O 2 ) 4 ]. 3 of Chemical p1182-3), it quantities of was Education reported HgF have (85, that been 9, (2008), minute detected, 4 using a) Deduce the condensed matrix of the transition metal extreme State why, the metal geometry of the complex Identify in and draw a diagram the ligand the nature and the of the bonding transition metal State Draw the and the denticity a diagram d-sublevel. populate Explain of the the be rst considered metal. In the publication, Label each of the the or the a however, challenges this claim. counterargument Explore may have why the splitting orbitals orbitals complex diamagnetic. is merit, in view of conventional ion ethanedioato showing whether paramagnetic 328 in now thinking on what complex. electrons. f) basis, might between ligand. e) this of considers d) K complex. Jensen’s the 4 transition Jensen c) on mercury transition the at conditions. complex. instance b) non-equilibrium in Suggest this techniques, electron under conguration isolation of involved with as a transition element. IUPAC C H E M I C A L B O N D I N G 14 A N D S T R U C T U R E ( A H L ) Introduction More and in-depth closer often require theories topic we of and of more bonding expand introduced ve explanations analysis six in bonding to 4 electron be to concepts considered. principles of explore domains systems arrangements sophisticated the topic of structural In and this VSEPR species and Theory involving discuss the roles play that in bonding. as a formal such a charge Hybridization mathematical hybridization examination around a and treatment the central is and can be number interior delocalization structure also model schemes of of and introduced we see how deduced of atom from electron in a an domains species. 14.1 Ft sts of ot o stt Understandings Applications and skills ➔ Covalent bonds result from the overlap of atomic ➔ Prediction whether sigma (σ) or pi (π) bonds orbitals. A sigma bond (σ) is formed by the direct are formed from the linear combination of head-on/end-to-end overlap of atomic orbitals, atomic orbitals. resulting in electron density concentrated between ➔ Deduction of the Lewis (electron dot) the nuclei of the bonding atoms. A pi bond (π) is structures of molecules and ions showing all formed by the sideways overlap of atomic orbitals, valence electrons for up to six electron pairs on resulting in electron density above and below the each atom. plane of the nuclei of the bonding atoms. ➔ ➔ Application of FC to ascer tain which Lewis Formal charge (FC) can be used to decide which (electron dot) structure is preferred from Lewis (electron dot) structure is preferred from dierent Lewis (electron dot) structures. several. The FC is the charge an atom would ➔ Deduction using VSEPR theory of the electron have if all atoms in the molecule had the same domain geometry and molecular geometry with electronegativity. FC = (Number of valence ___ 1 electrons) - ve and six electron domains and associated (Number of bonding electrons) – 2 bond angles. (Number of non-bonding electrons). The Lewis (electron dot) structure with the atoms having ➔ to dissociate oxygen and ozone. FC values closest to zero is preferred. ➔ Exceptions to the octet rule include some species Explanation of the wavelength of light required ➔ Description of the mechanism of the catalysis of ozone depletion when catalysed by CFCs having incomplete octets and expanded octets. and NO ➔ Delocalization involves electrons that are shared x by more than two atoms in a molecule or ion as opposed to being localized between a pair of atoms. ➔ Nature of science Resonance involves using two or more Lewis (electron dot) structures to represent ➔ Principle of Occam’s razor – bonding theories a par ticular molecule or ion. A resonance have been modied over time. Newer theories structure is one of two or more alternative need to remain as simple as possible while Lewis (electron dot) structures for a molecule maximizing explanatory power, for example the or ion that cannot be described fully with one idea of formal charge. Lewis (electron dot) structure alone. 329 14 C H E M I C A L B O N D I N G A N D S T R U C T U R E ( A H L ) Theories of bonding and structure To study large structures explanations of sophisticated concepts, bonding we look than at and bonding we have species consider systems models, met based so on and far. ve this and six should more theories In underpinning in-depth requires maintaining of and consider (and challenge) models and theories used in for capacity and chemical theories theories as for is is that a theory possible while maximum discovery of application. have need to been modied remain as over simple as time. possible bonding maximizing their explanatory power. In this structure. chapter The principle simple of while and as electron some New the a understanding topic Bonding domains, this remain principle the of Occam’s development different elds of of razor theories knowledge. is a in The and blueprint a number several bonding associated of models are and theories presented, each of with structure its pros and limitations. philosophy Formal charge TOK In As we shall see throughout this topic, covalent bonding can be described using valence bond theory are as obey having alternative ways of describing the same (FC) can of be a the structure theory. To what extent is 4.3 a is the we in a the considered of A as a idea molecular useful in the process worked of showing is to (electron the or a structures in or valence the ion. be ion. drawn which formal The electronic dot) electrons polyatomic can deciding determine molecule as Lewis species Lewis involving out a how approach appropriate present charge the of different rule. most atoms way covalent number octet hypothetical introduced convenient distributed Sometimes all or molecular orbital sub-topic structure that Lewis charge calculation of book-keeping; it FC is a follows: phenomena a strength or a 1 _ FC = (number of valence electrons) - weakness? (number of bonding 2 electrons) For example, in the - (number molecule of non-bonding tetrachloromethane, electrons) (gure CCl 1) the 4 Cl formal charge on each atom is calculated as follows: 1 _ FC(C) C Cl = (4) (8) Cl 0 = 0 2 1 _ FC(Cl) = (7) (2) 6 = 0 2 Cl 2 In the case of the carbonate anion, CO (gure 2), the FCs on the 3 ▲ Figure 1 Lewis structure of carbon and oxygen atoms are: tetrachloromethane, CCl 4 1 _ FC(C) = (4) (8) O 0 = 0 2 A 1 _ FC(O - 2 ) = (6) (2) 6 = -1 (4) 4 = 0 A 2 O 1 _ FC(O ) = (6) B 2 C If O O there rule, ● are the the a number most one of possible reasonable with FC one Lewis will difference structures ( ΔFC = FC FC max O B ▲ to O 0, all obey ) closest min and A Figure 2 Lewis structure of the ● the one that has the negative 2 carbonate anion, CO 3 330 that be: electronegative atoms. charges located on the most the octet 14 . 1 For BF example, . For gure structure 3 F u r T h e r shows two a S p e c T S Lewis O F c O v a l e n T structures for boron b O n d i n g a n d S T r u c T u r e triuoride, St t (a): 3 To check if you have calculated formal 1 FC(B) = (3) (6) 0 = 0 2 charges correctly, nd the sum of the 1 FC(F) = (7) (2) 6 = 0 FCs for the molecule or ion. For a neutral 2 molecule the sum of the FCs = 0 (for ΔFC = 0 example, for CCl : 0 + [4 × 0] = 0). For 4 For structure (b): a polyatomic ion the sum of the 1 FC(B) = (3) (8) 0 = FCs = charge on the ion (for example, for -1 2 2 CO 1 FC(F ) = (7) A : 0 + 2[ 1] + 0 = 2 ). 3 (2) 6 = 0 (4) 4 = +1 2 1 FC(F ) = (7) B ΔFC 2 = FC FC max Since ΔFC is = (+1) ( 1) = +2 min closest to zero for Lewis structure (a), this is the F most A reasonable representation of BF . 3 Although preferred Lewis based incomplete Species than see 8 can in FC octet also valence later structure on be of obeys electrons found electrons this (b) the considerations with octet even (fewer than expanded surrounding the rule, though 8 structure boron valence octets central of (a) has is F F B B an electrons). electrons atom), as (more we shall topic. F F F F Dierent interpretations of “charge” F (a) The idea state, and of charge formal we has charge, need to many ionic interpret connotations charge, the partial intended in chemistry charge, meaning total (b) F A B (oxidation charge), depending ▲ Figure 3 Two possible Lewis structures for boron triuoride, BF on 3 the To context. distinguish charge, and the ionic terms oxidation charge, consider state, the formal charge, hydrogen partial uoride molecule, H HF (gure F 4). ▲ Figure 4 Lewis structure for hydrogen Oxidation states uoride, HF hydrogen: +1 uorine: 1 St t Formal charges 1 FC(H) = (1) (2) 0 = In writing oxidation states, the sign goes 0 2 before the number (eg +2). For charges, 1 FC(F) = (7) (2) (6) = 0 2 ΔFC = FC the sign goes after the number (3 FC max = 0 ). Formal charges do not represent the actual min charge on the ion so hence signs are put Par tial charges From H = on section 2.2 and uorine hydrogen. 8 of for to F be One before the number (eg +1). Oxidation the = Data 4.0, more rather booklet , so you negative the electronegativity would than approximate expect the but the partial simple for partial charge way of charge on calculating numbers are represented by Roman numerals (eg I, II) whereas oxidation states are represented by Arabic numerals (eg +1, +2). 331 14 c h e M i c a l b O n d i n g H a n d S T r u c T u r e the F as + partial charge based - electronegativity values is (2.2) __ H: = (2.2 + on follows: 0.30 0.30 ( a h l ) + 0.35 of the charge of the bonding pair 4.0) - δ δ (4.0) __ ▲ F: Figure 5 The par tial charge in a = (2.2 + 0.65 of the charge of the bonding pair 4.0) molecule of hydrogen uoride, HF, based on simple approximations Hence, since electrons, uorine neutral on HF atom uorine The and 6. the F The is 1.86 The length 1.30e. to charge a on of 0.30 is than partial hydrogen. which two Therefore charge experimentally shows the nding of = consists negative moment, shows of 2e HF HF a charge is represented calculated a polar by the dipole D. region region in equates partial 5. × more electrostatic method bond This dipole bond 0.65 0.30e 0.30+ gure red experimentally have have net in HF blue Another H a a molecular gure covalent uorine. and with for will will of shown moment single uorine in molecule vector the 92 the area area partial pm calculated potential of lowest 10 of HF involves Data 1.86 is shown electron electron the moment for greatest charges (section dipole (MEP) of density. taking booklet) the and the D: F μ _ Q = H r where: Q = apparent μ = dipole r ▲ Figure 6 The molecular electrostatic potential We = H can F use charge on each end of the molecule moment bond the length following conversion: (MEP) on the van der Waals surface of the 30 1 D = 3.34 × 10 C m hydrogen uoride molecule hence: 30 (1.86 × 3.34 × 10 C m) ___ Q = 12 (92 × 10 m) 20 = 6.75 × C 10 19 The the charge Data of an booklet), isolated electron therefore the = 1.60 partial × 10 charge δ C is (section given 4 of by: 20 (6.75 × 10 C) __ δ = = 0.42 19 (1.60 Computer accurately, has δ + = × 10 programs and from 0.54+ and C) can the F calculate MEP has δ- partial model = 0.54 Net charge on the HF molecule The 332 net charge on the HF molecule is charges shown 0. in much gure in density 6, more H 14 . 1 F u r T h e r a S p e c T S O F c O v a l e n T b O n d i n g a n d S T r u c T u r e Molecular geometries based on ve and St t six electron domains In topic and six 4 four we electron electron using examined the domains. domains procedure eqto and x positions molecular are geometries Molecular summarized outlined in topic based geometries in table 1, on two, based and on can ve be occur only for geometries three, based on ve electron and domains, not those based deduced on six electron domains. We 4. usually do not refer to axial For molecular geometries based on ve electron domains, remember or equatorial positions for an that lone pairs rst instance. (non-bonding pairs) occupy equatorial positions in the octahedral geometry. order of We = shall is based interactions LP|LP (LP This > LP|BP lone > pair; examine is as the follows fact that (topic in terms of repulsion, the 4): BP|BP BP the on = bonding reason for pair) this in nm of eto om to oms omt a worked example later in this topic. Mo omt nots trigonal bipyramidal equatorial F position B a F AB B 5 e 120° B A P e 5 trigonal bipyramidal 5 BPs B e 90° F example: PF B a 5 F axial position F see-saw F B a AB F E 4 B e 5 trigonal bipyramidal less than 120° A 4 BPs and 1 LP S B e example: SF 4 F B a less than 90° F T-shaped F B a AB 3 B E 2 A e 5 trigonal bipyramidal 3 BPs and 2 LPs Cl F B a example: ClF less than 3 90° F 333 14 C H E M I C A L B O N D I N G A N D S T R U C T U R E nm of eto om to oms omt ( A H L ) Mo omt nots Linear I B a AB E 2 5 3 180° A trigonal bipyramidal 2 BPs and 3 LPs I example: I B a 3 I octahedral F F F B AB S B 6 6 B octahedral A 6 BPs 90° B B example: SF 6 B F F F square-based pyramidal F AB 6 B octahedral E 5 B F B 5 BPs and 1 LP F Br A B example: BrF B 5 F F square planar F AB E 4 B 2 B F 6 octahedral A 4 BPs and 2 LPs 90° B B F example: XeF 4 Xe F ▲ T able 1 Electron domain geometries and molecular geometries based on ve and six electron domains. B = axial substituent; B a = e equatorial substituent; E (in formula) = lone pair of electrons; BP = bonding pair of electrons; LP = lone pair of electrons Overlap of atomic orbitals: Sigma (σ) and pi (π) bonding In topic and a bond The A 4 discussed consists single double between 334 we multiple bond of bond, two difference as a atoms double electrons represented covalent two the such bond A and shared by a of triple a is a four single bond. between stick, consists B. between or two sigma A covalent single atoms bond electrons, two A bond covalent and B: (σ). pairs, shared 14 . 1 F u r T h e r a S p e c T S O F c O v a l e n T b O n d i n g a n d S T r u c T u r e σ The A double triple bond, represented covalent between two bond atoms A by two consists and sticks, of six is a sigma electrons, plus or pi three bond pairs, (σ + π). A B shared B. σ + π A The triple bonds A (σ Lewis bond, + represented by three sticks, is a sigma plus 2 two 2π). structure is a simple model showing how the valence (outer- σ shell) B pi electrons are distributed in a molecule or polyatomic ion. + 2π In A sub-topic 14.2 quantum mechanics helpful in we shall look called visualizing the at a more sophisticated molecular difference orbital between a theory theory sigma based (MOT), bond and which a pi is bond. ▲ For atomic relatively be orbitals close identical. X in to overlap energy, atomic and orbitals and the form molecular symmetry combine to of form orbitals the x atomic new they must orbitals molecular B on Figure 7 Covalent bonds be must orbitals. i o of o t: There are three possible outcomes: A 1 bonding orbital: 2 anti-bonding sigma (σ) or pi (π) B > A=B > A≡B orbital i o of o stt: orbital: sigma star ( σ*) or pi star (π*) orbital A≡B > A=B > A 3 non-bonding Table 2 shows a B situation. number of combinations of atomic comto of tom ots orbitals. Mo ots fom T σ + bonding s + s σ* + anti-bonding σ + bonding s + p x σ* + anti-bonding + NB + NB s + p non-bonding y s + p non-bonding z σ + bonding p x + p x σ* + anti-bonding 335 14 C H E M I C A L B O N D I N G A N D S T R U C T U R E comto of tom ots ( A H L ) Mo ots fom T π + bonding p + p y y π* + anti-bonding π + bonding p + p z z π* + anti-bonding p + p x p + NB + NB + p x NB + p y ▲ non-bonding z + p non-bonding y non-bonding z T able 2 Combination of atomic orbitals (LCAOs). Black represents the positive wavefunction, Ψ , and white represents the negative + wavefunction, Ψ Delocalization and resonance dsto of sm Figure 8 shows two Lewis structures for the nitrite oxoanion, NO that 2 o have In the formation of a sm o identical arrangements of atoms but different arrangements electrons. there is a direct head-on overlap - of the atomic orbitals along the - internuclear axis and the electron N N density is located along this axis. In the formation of a o there is a sideways overlap of the atomic orbitals and the ▲ Figure 8 Lewis structures of the nitrite oxoanion, NO 2 electron density is located above and below the internuclear axis. Each Lewis double two 336 structure bond, but structures. the An shows one position ion with of one N O the of single double these bond bond and is structures one N=O different would in have the of 14 . 1 one In shorter fact the N=O bond (114 experimentally intermediate between F u r T h e r a pm) and measured single a S p e c T S and one longer bond double O F N lengths NO c O v a l e n T O are bond. bond both Both b O n d i n g (136 125 a n d pm). rso pm, Lewis S T r u c T u r e structures, In so, two or more called resonance called a forms, contribute to the electronic structure which is Lewis structures can represent resonance hybrid. This idea of resonance is shown by linking a par ticular molecule or ion. the contributing resonance forms by a double-headed arrow (gure 9). A so stt is one of two or more alternative N Lewis structures for a molecule N O or ion that cannot be described O fully with one Lewis structure alone. Figure 9 Resonance in the nitrite oxoanion, NO ▲ 2 Note that resonance structures The resonance in the NO ion can also be represented as shown in are purely hypothetical 2 gure 10. The dashed curve conveys delocalization. As specied by species that do not actually IUPAC, delocalization is a quantum mechanical concept used to describe exist. the A pi bonding conjugated represented as in a conjugated system a is system a of system. molecular alternating entity single whose and structure multiple can bonds. be Conjugation is the interaction bond. bond A a NO one conjugated and a conjugated has of p “fractional oxoanion is may containing not double the orbital system orbital system p with also a bond bond form lone localized another of between character” order the pair of each intervening interaction atoms bond O an electrons. two or N across but order. bond is of The 1.5, a sigma double bonding instead In the since in each case both of a N link the bonds are 2 equivalent and intermediate between a single and a double bond: ▲ total number of NO bonding pairs 3 _ ____ bond order in NO = = = 1.5 Figure 10 An alternative representation of resonance in the NO oxoanion 2 2 total Another oxoanion that has number resonance of NO structures is the carbonate 2 CO 2 positions oxoanion, 2 . Three resonance structures can be written for CO 3 (gure 11). 3 2 2 O O doto doto involves C C electrons that are shared by O O O O more than two atoms in a molecule or ion as opposed to being localized between a pair of atoms. 2 O C O O 2 ▲ Figure 11 Resonance in the carbonate oxoanion, CO 3 St t 2 In CO the three C O bond lengths are equivalent (129 Covalent bond lengths (both pm), 3 intermediate bond length between (122 a pm). single The C bond O bond order is (143 pm) worked and out as a double C =O single and multiple) are provided in section 10 of the follows: Data booklet total order in CO number of CO bonding pairs 4 _ ____ 2 bond = = = 1.33 3 total number of CO positions 3 337 14 C H E M I C A L B O N D I N G A N D S T R U C T U R E ( A H L ) Worked examples (iv) To draw the Lewis structure, complete the Example 1 octets Consider a) the following BrF b) six d) c) e) the (ii) the F ] [ICl ] 4 each (i) atoms: 2 4 For [ICl 5 SOF uorine species: IF 3 on species, deduce: electron domain molecular Br F geometry geometry F (iii) the (iv) a approximate valid Lewis bond angle(s) (electron dot) b) structure. Ball-and-stick diagram of the IF molecule 5 (ignoring bond angles): Solution F a) Ball-and-stick diagram of the BrF molecule 3 (ignoring the bond F angles): F I F F F Br For F For F I: number of valence number of sigma electrons bonds = 7 = 5 = 12 = 6 Br: total number of valence number of sigma electrons = 7 = 3 = 10 number number bonds (i) total number of valence electrons of of valence electron From table electrons domains 1, electron domain geometry: octahedral. number of electron domains = 5 (ii) This is an example of an AB E system, so 5 (i) From table trigonal (ii) This is 1, electron domain geometry: the bipyramidal. an example six of an AB E 3 one lone pair equivalent is located in any of the positions. system, 2 F so the two equatorial lone pairs are located in the positions: F F I F F Br F Molecular (iii) less than 90° F geometry: (iii) angles: Bond experimental cannot 338 be than value predicted is angles: angles T-shaped. less geometry: Bond of Molecular F 90 °. angles (The 86.2 °, from but F Br F this VSEPR theory.) are these is square-based six less F than I 90°. experimental 80.9°.) F F pyramidal. bond (Each I F one bond 14 . 1 (iv) To draw octets the on F u r T h e r Lewis uorine a S p e c T S structure, O F complete c O v a l e n T b O n d i n g (iv) the To draw octets atoms: a n d the on S T r u c T u r e Lewis chlorine structure, complete the atoms: - F Cl F F I I F F Cl c) Ball-and-stick diagram of the [ICl ] anion 2 (ignoring bond angles): St t - When drawing the Lewis structure of an anion or cation, Cl include square brackets and the respective charge. Don’t forget to complete the octets on the terminal atoms in a Lewis structure, unless the terminal atom is I hydrogen which has only the two electrons in a bonded Cl pair, so attaining the noble gas conguration of helium. For I: d) Ball-and-stick diagram of the SOF molecule 4 number of valence number of sigma electrons = 7 = 2 (ignoring bond maximum bonds cannot ve charge = 1 member S total number number of of valence electron electrons = domains be must valency the of be angles); period the of central 3, note oxygen atom. can central that is in so its From table trigonal (ii) This is 1, electron domain 5 F the an example three equatorial lone O of an pairs AB positions bipyramidal of electron E so F F O S system, located the trigonal S F F 3 are domain octet, geometry: bipyramidal. 2 so a SOF F (i) oxygen as 4 10 = 2 Sulfur, expand atom the in F F geometry: For S: number of valence number of sigma electrons = 6 = 5 = -1 = 10 = 5 Cl 1 pi bonds bond I total number number of of valence electron electrons domains Cl p os vSepr to Molecular geometry: linear. Pi bonding is o-axis bonding. Because the shape of a (iii) The bond pairs are angle remains arranged 180 °, as the symmetrically. lone molecule is controlled by the sigma-bonding framework along the internuclear axis, when counting the valence electrons we subtract 1 for each pi bond present. 339 14 C H E M I C A L (i) (ii) From B O N D I N G table 1, A N D electron geometry: trigonal Molecular geometry: S T R U C T U R E ( A H L ) (i) domain From bipyramidal. trigonal table 1, electron-domain geometry: octahedral. bipyramidal: (ii) This is an example of an AB E 4 the F two lone as possible to minimize pairs from are each other system, so 2 located as (i.e. far at apart 180 °) repulsion: F less than 120° - S O F Cl greater than 90° Cl 90° I F Cl (iii) Bond angles: less than 120 ° for F S F eq greater than 90° double bond for O=S F . ; eq The axial F S F 120°. for bond The F To is F ; 90.7° the on Molecular space decreased for values O=S so (iii) from are geometry: square planar. the Bond pairs 110 ° so F eq draw octets angle more experimental S eq (iv) occupies Cl angle(s): present all bond 90°. are The two arranged angles remain sets of lone symmetrically 90 ° axial Lewis uorine structure, and complete oxygen (iv) the To draw octets atoms: the on Lewis chlorine structure, complete the atoms: - F Cl Cl F I S O Cl F Cl F Example 2 e) Ball-and-stick diagram of the ICl anion 4 Deduce (ignoring bond the molecular polarities of the following: angles): a) SF 6 - b) cisplatin, Pt(NH ) 3 Cl Cl 2 2 Cl c) transplatin, Pt(NH ) 3 Cl 2 2 I Solution Cl Cl a) Since SF has a symmetrical octahedral 6 structure For I: of individual and number of valence number of sigma electrons = 7 = 4 = 1 (see the table S F molecule 1), the bonds as a dipole cancel whole moments one another, remains non-polar. bonds b) Cisplatin, [Pt(NH ) 3 ve charge From section 8 of electronegativities total number of valence electrons = of electron domains = = 3.0. Hence are the is square Data planar. Pt booklet, = 2.2, Cl the Pt Cl bond = 3.2, (difference in 6 electronegativity Pt 340 ] 2 12 N number Cl 2 the N bond = 1.0) (difference is = more 0.8). polar than the 14 . 1 To deduce the polarity of F u r T h e r the a S p e c T S complex O F c O v a l e n T b O n d i n g a n d S T r u c T u r e we cst s ttmt consider the moments. is direction In cisplatin approximately using This the 90°, in a cisplatin the the so parallelogram results below; of Cl we dipole is polar Pt Pt add law net a two Cl Cl bond the from dipole two angle vectors mathematics. moment as shown complex: Cisplatin is used to treat ovarian and testicular cancer. In the body the two Cl ligands are replaced by the DNA base guanine, and the geometrical arrangement of the resulting complex is a perfect t to the two double helix strands in DNA . Transplatin is ineective in the treatment of cancer due to the trans conguration of the Cl ligands. The two complexes can be H N Cl 3 distinguished by their dierent molecular polarities. Pt Example 3 H N Cl 3 a) Below the are two dinitrogen alternative oxide Lewis molecule, structures N for O: 2 St t In deducing molecular polarities based on molecular N O A B geometries you should nd the vectorial sum of Based the individual dipole moments using the Lewis parallelogram law. › For two vectors on formal structure charge is considerations, which preferred? › v and v 1 both star ting from the same b) Below are three alternative Lewis structures 2 › point, the sum of the two vectors v for can be found the cyanate anion, OCN by drawing and completing the parallelogram. The › diagonal gives › v › = v - v (gure 12): › C + v 1 2 A → v 1 → v - → C v 2 B ▲ Figure 12 The parallelogram law for adding vectors - c) Transplatin, Pt(NH ) 3 Cl 2 is also square planar 2 C but the Cl vectorial is 0 since sum of and bond the are along complex. moment Cl they directions the Pt two equal the Cl Therefore hence H angle Pt Cl but Pt is in Cl there dipole The moments C opposite linear is transplatin N 180°. no is axis net in dipole non-polar: Based which on Lewis charge structure considerations, is deduce preferred. Cl 3 Solution Pt a) Table Lewis Cl formal 3 shows the structures formal A and charges (FC) for B. NH 3 341 14 C H E M I C A L B O N D I N G A N D S T R U C T U R E ( A H L ) on atom Fc = ΔFc = Fc the more electronegative element, in this case - Fc mx m oxygen. The preferred Lewis structure is therefore C. Lewis structure A Example 4 1 terminal N (4) (5) (4) = -1 Explain 2 why bromine triuoride, BrF has its lone 3 pairs of electrons located in equatorial positions. 1 central N (5) (8) (0) = +1 (+1) ( 1) = +2 2 Solution 1 O (6) (4) (4) = 0 2 In example 1(a) the molecular geometry of BrF 3 Lewis structure B was arrangement 1 terminal N (4) (5) deduced to is be T-shaped. based on an The AB geometrical E 3 (4) = -1 system. Below 2 2 are (+2) ( (6) structures for BrF with the lone 3 pairs (8) Lewis 1) = +3 1 O three surrounding Br located in different positions: (0) = +2 2 F ▲ T able 3 Formal charges (FCs) for the dinitrogen oxide F molecule, N O: Lewis structures A and B Br F 2 F F Br Since ΔFC preferred b) Table B, for A Lewis 4 and is shows C less than structure for the the ΔFC will FCs for cyanate be for B, the Lewis anion, structures Fc = F A B C both lone pairs one lone pair equatorial both lone pairs equatorial and one axial axial A, OCN Table atom F F A. Br F ΔFc = Fc - Fc mx 5 shows possible Lewis the interactions for these three structures. m Lewis structure A ittos Lewis Lewis Lewis structure structure structure A B C posto of o both equatorial both axial s equatorial and axial 120° 90° 180° 6 × 90° 1 O (6) (6) (2) = +1 2 1 C (4) (8) (0) = 0 = (+1) ( 2) = +3 2 1 N (5) (2) (6) = -2 2 a t w Lewis structure B lp|lp 1 O (4) (6) C (4) = 0 2 a t w 1 lp|bp (4) (8) (0) = 0 (0) ( 1) = +1 4 × 90° 3 × 90° 2 × 120° 2 × 120° 2 × 90° 1 × 120° 2 a t w 1 N (5) (4) 3 × 120° (4) = -1 2 bp|bp 2 × 90° Lewis structure C ▲ T able 5 Interactions for three Lewis structures A, B, and C for 1 O (2) (6) (6) = -1 bromine triuoride, BrF 2 . (LP = lone pair; BP = bonding pair) 3 1 C (4) (8) (0) = 0 (6) (2) = 0 = (0) ( 1) = +1 Remember that: 2 1 N (5) 2 ▲ LP|LP T able 4 Formal charges (FCs) for the cyanate anion, OCN > LP|BP > BP|BP : The greatest bond angle for LP |LP interactions Lewis structures A, B, and C occurs Based are B atom 2.6). on and C. the two However, preferred O is (electronegativities: Since second 342 ΔFC both criterion B is and that C the O = have the Lewis most 3.4; the N structures electronegative = 3.0; same negative FC C ΔFC, = the should reside Lewis in structures structure interaction: while will both the A C gives preferred occupying and Next structure structure be B. A C, so we considering gives only 6 4 structure, equatorial × × can discard LP |BP 90° 90° with positions. bond angles angles, the so lone A pairs 14 . 1 F u r T h e r a S p e c T S O F c O v a l e n T b O n d i n g a n d S T r u c T u r e An environmental perspective: Catalysis of ozone depletion Ozone, O is a V-shaped (bent) molecule with where: 3 a bond angle of 116.8° and its two O O bond 34 lengths are resonance shown in equal forms gure (128 can pm). be Two written h = Planck’s for ozone, as ν = frequency = 6.63 × of the 10 J s radiation 13. 8 c = λ + constant contributing O O speed = of light wavelength = of 3.00 the × 1 10 m s radiation + Worked example ▲ The average 362 kJ bond enthalpy in ozone is Figure 13 Resonance forms of ozone, O 3 1 mol section The bond order for the O O bond in ozone as and bond order in section required number of O O bonding oxygen pairs enthalpy given data in given 11 of the Data = number wavelength, booklet, in nm, calculate of the UV the radiation of O O to break and the the O O =O O double bond in bond in ozone. 3 _ _ _ _ _ = total bond relationships O 3 total the the follows: maximum each Using is in calculated 1 . = 1.5 2 positions Solution The wavelength of light required to Oxygen: The bond enthalpy for O =O is 1 498 dissociate oxygen and ozone kJ mol photons In topic 6 we mentioned that in the ozone radiation layer from absorbs the over 95 % J. Next First calculate convert the kJ to energy J and of then stratosphere, use the in . of harmful Avogadro’s constant, L: UV 1 sun: 498 × 1000 J mol __ E = 19 = 23 6.02 × 8.27 × 10 J 1 10 mol hν O (g) → O 3 (g) + O•(g) hc __ 2 E = , so by cross-multiplication Eλ = hc. We λ O (g) + O•(g) → O 2 (g) + then heat make λ the subject of the equation: 3 34 There is a net energy conversion from UV radiation to heat (6.63 hc _ λ = × 8 10 J s)(3.00 × 10 1 m s ) ____ = 19 energy. E 8.27 × 10 J 7 = The progressive allows more surface, cancers depletion of the radiation to reach UV resulting in an (melanomas) ozone increased and the risk bonds radiation in ozone ( hv). The can 10 m 9 of 1 nm = 7 10 m, so λ = (2.41 × 10 m) × 9 skin (1 nm/10 m) = 241 nm cataracts. be bond × layer Earth’s Ozone: The 2.41 broken order in by ozone The bond enthalpy for the O O bond in 1 UV ozone is 362 kJ mol (1.5) 1 is lower than the bond order in oxygen (2), 362 × 1000 J mol __ so E = the O=O double Radiation of bond shorter in oxygen wavelength is is 6.02 stronger. break the stronger bond in × 10 oxygen. = is that the energy, E, of a photon of 10 J (6.63 × 10 8 J s)(3.00 × 10 1 m s ) ____ = 19 E reason × mol 34 λ The 6.01 1 required hc _ to 19 = 23 6.01 × 10 J light 7 = is inversely proportional to the 3.31 × 10 m wavelength 7 λ, so the greater wavelength and the vice energy, the shorter the λ = The hc _ E = hν (3.31 × 10 9 m) × (1 nm/10 m) = 331 nm versa. depletion greater detail of in the ozone topic layer was discussed in 6. = λ 343 14 C H E M I C A L B O N D I N G A N D S T R U C T U R E ( A H L ) Models and theories of structure and bonding You come she to knocks nature them with all all her theories, and at. Another of Pierre-Auguste approach structure Renoir (1841–1919), with is the S to simply having a draw a Lewis complete octet electrons: French impressionist artist. 2 O Chemical models bonding and and theories structure which can are riddled often be with interpreted O in specic should ways be used to suit with our requirements. caution in science. Models Let us take O an example and we its of a Lewis associated mean by structure molecular of an geometry oxoanion to see what Here this. of It is the theory that decides what we can O although shown physicist services Einstein who to the (1879–1955), Nobel theoretical Prize physics, in German Physics especially in for theoretical 1921 his for his at the the law of the obeys the octet rule, However, as the S valency is same by four electron geometry for domains the we oxoanion, arrive namely tetrahedral. discovery Both of S incorrect. observe. surrounded Albert is photoelectric models result in each S O bond being effect. equivalent, a single S intermediate O bond. The between question a double then is and which A question of approach: Expanded octets, Lewis structure is the most valid. octets, or formal charge considerations If In topic 4 we looked at one method we we determining the shape of an oxoanion the sulfate oxoanion, get the now to formal results shown charge in considerations table 6. such 2 as turn of Table SO 6 shows that in both Lewis structures 4 the more negative FC resides on the more O electronegative oxygen atom, so this is not a 109.5° factor here. ΔFC is closer to zero for the Lewis S O O structure in octet, this so which sulfur supports has the an expanded suggestion that this is O the Using this method correctly the sulfate is can represented molecular worked out as geometry preferred structure. of tetrahedral and atom Fc = ΔFc = Fc - Fc mx be structures. Try equivalent in to by write six contributing these! All S O m resonance bonds are Lewis structure: expanded octet length. 1 The nal step involves completing the octets S on (6) (12) (0) = 0 2 the four terminal oxygen atoms, in order to write 1 six valid Lewis structures for sulfate. Here are O two (6) (2) (6) = -1 (4) (4) = 0 (0) ( 1) = +1 2 of these as an example: 1 O= (6) 2 O O S S Lewis structure: octet O O etc. O O 1 S O (6) (8) (0) = +2 (2) (6) = -1 2 O (+2) ( 1) = +3 1 O (6) 2 This of model shows electrons: in that each of S has the an six expanded Lewis octet structures, ▲ T able 6 Formal charges (FC) for the Lewis structures of the 2 the central S atom is surrounded by 12 electrons. sulfate oxoanion, SO 4 344 14 . 2 A similar analysis with the phosphate oxoanion, evidence points h y b r i d i z a T i O n to two shorter S =O bonds, 3 [PO ] reveals that the Lewis structure in which supporting a Lewis structure for the sulfate 4 phosphorus based on has FC an expanded calculations. octet is However, favourable oxoanion with an expanded octet: theoretical H quantum mechanical calculations suggest that the O most with favoured the single octet of Lewis electrons structure and is many the one H chemists S argue that Lewis structures for the obeyed when Lewis choosing for structure rather than between oxoanions, the in it which alternative is better the expanded octet octet to rule O is alternative. As we said theories If we go one molecular step further geometry of and examine sulfuric acid, O O opt at with the an outset, open approach scientic models and mind! the experimental 14.2 h to Understandings Applications and skills ➔ A hybrid orbital results from the mixing of 3 Explanation of the formation of sp ➔ 2 , sp , and sp dierent types of atomic orbitals on the same hybrid orbitals in methane, ethene, and ethyne. atom. Identication and explanation of the ➔ relationships between Lewis (electron dot) structures, electron domains, molecular geometries, and types of hybridization. Nature of science ➔ The need to regard theories as uncer tain – Quantum mechanics involves several theories hybridization in valence bond theory can help explaining the same phenomena, depending on explain molecular geometries, but is limited. specic requirements. Models, theories, assumptions, and deductions What look we see depends mainly on what and we theories critical for. Theories Sir John Lubbock (1834–1913), biologist, banker and model each theory the our merits in chemistry has of a its own We its own limitations. particular perspective. has theory perhaps should merits and depend treat as not with Samuel be seen the same Beckett in black degree (gure and of 1). we should also look for white shades of – grey politician. Sometimes may should scientists and Each chemistry FRS, as English in perspective on models question with a every experimental combination to see the possible of bigger with assumption, evidence, models picture. this and challenge and consider theories “Real to science” theories using allow us becomes perspective! 345 14 C H E M I C A L In topic dot) 4 we B O N D I N G saw structure in the A N D value providing of a a S T R U C T U R E Lewis simple ( A H L ) (electron model Valence bond theory (VBT) which The shows how valence or outer-shell electrons VBT each distributed in a molecule. Sub-topics 4.3 and other how the number of electron domains from deduced from a Lewis structure, considers enabling form a that atoms molecule. The approach bond the overlap of atomic formed orbitals can resulting be to 14.1 results showed model are in a bonding orbital with the electrons the localized between assumed that their own Figure atoms 2 and A form the H , two the respective shows H the when atoms. atoms atomic VBT it is they retain orbitals. interaction each In interact having of two one hydrogen electron, to B the diatomic molecule H . The two atoms 2 approach each interaction there to is the a other between decrease interaction, and subsequently their in a the electrons energy chemical of bond there and the is is an nuclei. system If due formed. + H H A ▲ H H B Figure 2 Formation of the H diatomic molecule based on VBT. 2 The two electrons are localized between the two atoms and each atom involved in the bonding retains its own respective ▲ Figure 1 The Irish playwright Samuel Beckett. John Calder, atomic orbital author of The Theology of Samuel Beckett, said to Beckett “It We is a ne day,” to which Beckett replied “So far!” can represent potential prediction of the ultimately the electron-domain molecular geometry geometry using and the two energy atomic this bonding versus nuclei the model distance, (gure by d, a sketch of separating 3). VSEPR Z theory. VSEPR of the bonds Neither theory) detailed or even of these models however offers electronic why (Lewis an + Y explanation structure chemical and of bonds X covalent exist. To 0 we explain need to look phenomena scale, with sub-atomic forms the the can formation to quantum be models of explored that particles. chemical explain the the Schrödinger’s launch-pad for the bonds mechanics. on eld ygrene laitnetop fully Physical microscopic behaviour wave of of equation bond energy quantum bond length mechanics. bonding shapes Two It and of helps us provides understand an chemical understanding of the internuclear distance, d molecules. theories in quantum mechanics that aid an ▲ understanding of molecular geometries Figure 3 Interaction energy diagram showing potential energy are: versus internuclear distance, d, for the H molecule 2 ● valence ● molecular We 346 shall bond theory orbital discuss each (VBT) theory one On gure ● At 3: (MOT). separately. X the the potential hydrogen energy atoms are is too essentially far apart zero, to as interact. 14 . 2 ● At Y The the hydrogen electron on atoms atom A, approach e , is each attracted other. to the orbitals the h y b r i d i z a T i O n overlapping, formation of but new the overlap orbitals called results in molecular A nucleus on atom B, H . Simultaneously, e B e repel each other nuclei H B and H A each other. energy There going is from a X and orbitals. decrease to Y as molecular than the electrons orbitals, are and assigned associated to these with the whole B in the repel molecule potential attraction rather than individual atoms. is Figure greater The A and 4 shows the formation of the molecular repulsion. orbitals for orbitals on the H molecule. The two 1s atomic 2 ● At Z the minimum achieved. of H . This The H potential represents H bond energy the length most is 74 is stable pm state (see two new which is the hydrogen molecular additive, atoms orbitals. results in combine One a to form combination, bonding 2 section 10 of the Data booklet). molecular energy, orbital 1 Explain and difference, Question why signicantly the to potential the left of energy point Z orbital the other results sigma sigma in star (σ), which combination, an is which is of lower which anti-bonding (σ*), of is a molecular higher energy. rises on gure 3. + St t bonding orbital σ Remember from chapter 5 that bond breaking is an endothermic process (ΔH positive) and bond node formation is an exothermic process (ΔH negative). In VBT, the greater the degree of orbital overlap anti-bonding orbital σ* (interaction), the stronger the bond will be. ▲ Figure 4 Molecular orbital theory (MOT) model for the H 2 VBT can be applied to other homonuclear molecule. There is a build-up of electron density in the bonding diatomic molecules such as F and to molecular orbital, σ, between the two nuclei. If we think of a 2 heteronuclear or HF . One model the is inherent the a of of the be the used the chemical applied the in VBT not that together, the electron density is the negative “glue” that holds the to the two positively charged nuclei together. In the anti-bonding molecular orbital, σ*, there is a nodal plane (node) between the explain bonds. occur two nuclei, which means that there is zero electron density here VBT on We can represent this orbital diagram as We shall For hybridization, bonds that an atom explain can form the and molecular conditions the of these discussing The hybridization note that VBT its The limitations and has been superseded as by molecular 5. orbitals must be to be formed two met. atomic for orbitals effective must be relatively close in overlap. orbital symmetry the sign of of the the atomic orbitals, wavefunction Ψ, that must a be model gure also is, has molecular bonds. 2 Before in a spatial energy orientation shown by number 1 of model electronic molecules. to covalent bond as the “glue” that holds atoms in a molecule bond. to model HCl considered covalent energy as structure are does between in such Lewis bonds model polyatomic expand concept of changes also structure shortly so molecules covalent differences formation can all and considers VBT limitation that same diatomic identical. For example, the signs of both theory. Ψ could be positive (both Ψ ), and sum to + form the σ bonding molecular orbital. In Molecular orbital theory (MOT) contrast VBT assumes that when bonds are formed the subtractive combination would from have one wavefunction positive, Ψ , and one + the overlap of atomic orbitals, the original nature wavefunction of the atoms is retained. MOT involves negative, Ψ , resulting in the σ* atomic anti-bonding molecular orbital. 347 14 C H E M I C A L B O N D I N G A N D S T R U C T U R E ( A H L ) In MOT x atomic orbitals combine to form x new 1 molecular orbitals; for example, two 1s atomic σ* orbitals combine orbitals, Many ygrene each σ and theories theory mechanics to form two new molecular σ*. have has its provides same their own advantages constraints. several explain the specic requirements. theories phenomena, but equally Quantum that can depending on σ 1 H : 1s ▲ 1 H H : 1s 2 Figure 5 Molecular orbital diagram for the H molecule 2 Hybridization hto VBT uses the concept of hybridization to provide an electronic ots description of polyatomic molecules such as CH and 4 NH , but it can also 3 hto is a term used to account for the geometries of molecules. describe the mixing of atomic We shall now consider three different types of hybridization: orbitals to generate a set of new hybrid orbitals that are 3 1 sp 2 sp 3 sp as seen in methane as seen in ethene equivalent. Hybridization is a 2 mathematical procedure. as seen in ethyne. A ot results from the mixing of dierent types of atomic orbital on the same atom. The hybridization domains around type the can be central determined from the number of electron atom. TOK hto is a mathematical tool that allows us to relate the bonding in a molecule to its symmetry. What is the relationship between the natural sciences, mathematics, and the natural world? Smmt is key in several dierent areas of knowledge. In the eld of mathematics, for example, there is great emphasis on the concept of symmetry. Early Greek mathematicians such as Pythagoras and Euclid used the idea of the symmetry of specic shapes and objects to develop their theories. The structure of DNA is an excellent example of the symmetry of the helix, and it was this symmetry that ultimately led Crick and Watson to unravel the structure of DNA. Ssmmt (SUSY) is an impor tant theory met in par ticle physics. What is meant by supersymmetry, why is it discussed so much, and to which elementary par ticles does it relate? How are architects, musicians, ar tists, and dancers inuenced by symmetry in their creativity? 348 14 . 2 h y b r i d i z a T i O n 3 The formation of sp Methane, CH is a hydrocarbon hybrid orbitals in methane and is one of the shaped) combined with 25 % of the characteristics 4 3 major the components electron of natural conguration of gas. Let’s consider carbon: electron conguration: 1s 2 electron spherical 2s conguration: orbital s is orbital. shown orbital in shape gure of an sp 6. 2p [He]2s 2 2p + ground-state The 2 2 condensed the hybrid 2 full of = diagram: 3 ▲ [He] Figure 6 One of four sp orbitals which each have 75% p character and 25% s character. This hybrid orbital is aligned in the z 2 1 1 2p 2s direction 0 2p x 2p y z 3 The 1 Hydrogen has the electron conguration 1s . a might expect electrons in that its since 2p carbon sublevel, it has two would four sp hybrid orbitals point to the corners of You tetrahedron (gure 7). unpaired form two 3 sp bonds with hydrogen in its ground-state, hybrid orbitals forming 109.5° CH rather than CH 2 extremely The . Although CH 4 does exist it is central atom 2 unstable. tetrahedral hybridization nature of its of one methane 2s and involves three 2p the orbitals on central carbon the C atom in an excited-state as follows. 3 ▲ Figure 7 (a) The tetrahedral orientation of the four sp Step 1 orbitals on carbon. Each of these sp One of the electrons ground-state to the hybrid 3 the 2s orbital of con guration of carbon is vacant 2p in orbital, to form an hybrid orbitals contains one electron. (b) For simplicity, the smaler of the two lobes is the often omitted on diagrams promoted excited-state: z Step 3 [He] The nal step involves overlap of each carbon 3 sp orbital with a hydrogen 1s atomic orbital. 1 Hydrogen This arrangement gives four potential C H has a 1s electron conguration and the bonds. 1 s However, we know from topic 4 that methane orbital is spherically symmetrical. Each 1s atomic has 3 orbital a tetrahedral molecular geometry with H C the bond angles of 109.5 °. the excited-state the geometry orbitals are is at So accounts although for incompatible 90° to each the this four because other (topic model C H the on hydrogen and each sp hybrid orbital on H central carbon contains one electron (gure 8). for bonds, three H 2p 2). central 109.5° carbon atom 3 sp Step 2 The four atomic orbitals 2s, 2p , 2p x , and hybrid 2p y z H combine to form The new a set of four new hybrid orbitals. 3 four equivalent sp and hybrid each one orbitals are consists of H entirely 25 % s 1 H character and 75% p character, since they 1s atomic were orbitals formed from one s orbital and three p orbitals. 3 ▲ Figure 8 In methane, each carbon sp orbital overlaps with a 3 The shape 75% of of the each sp hybrid characteristics of orbital a p will orbital have hydrogen 1s orbital to form a tetrahedral geometry (dumbbell 349 14 C H E M I C A L Any molecule geometry on B O N D I N G with its a A N D S T R U C T U R E tetrahedral interior central electron atom ( A H L ) domain (based Step 2 on To four electron domains) would be predicted account for the 120° approximate bond angles to (based on three electron domains), the next step 3 have sp hybridization. involves 2s, 2p hybridization and 2p x . These of three of combine the to atomic form a set orbitals of three y 2 St t new sp hybrid orbitals. The 2p orbital remains z unhybridized. Hybridization is deduced from the electron domain geometry rather than the molecular geometry. For 2 The has a tetrahedral electron example, ammonia, NH three atoms 3 domain geometry, corresponding to four electron are 33.3% domains (three bonding pairs and one non-bonding point or lone pair). Its molecular geometry is trigonal s sp entirely character to (gure new the hybrid orbitals equivalent and corners of 66.7% a on and p the each carbon one character. trigonal planar has They system 10). pyramidal (with a bond angle of 107° due to the p repulsion between the lone pair and bonding pairs, z which reduces the bond angle from 109.5°). The 2 3 sp since it is based on hybridization of ammonia is sp the four electron domains. 2 sp C 2 The formation of sp hybrid orbitals 2 sp in ethene Ethene, C H 2 an alkene , is another hydrocarbon. Ethene is 4 2 Figure 10 Each of the three sp ▲ with one C=C double bond. hybrid orbitals contains one Around electron. The 2p unhybridized atomic orbital also contains z each carbon there are three electron domains so one electron. As before, for simplicity of representation, the the electron domain geometry (and molecular smaller of the two lobes is omitted geometry) angle is (gure is trigonal 121.3° and planar. the H The C H H C=C bond bond angle is 117° Step 3 9). The 12 1.3° H next bond H step along involves the the formation internuclear axis by of the a sigma overlap of 2 two C sp orbitals, one on each carbon atom. 11 7° C A pi the H bond two H p is formed from unhybridized the sideways atomic overlap orbitals, with of the z overlap ▲ hybrid regions above and below the internuclear Figure 9 The molecular geometry of ethene. The double axis bond occupies more space so the H C (gure 11). H bond angle is reduced from the predicted 120° for a trigonal planar p p z z pi geometry 2 sp 2 sp To deduce the hybridization scheme of the sigma carbon orbital for atom in diagram ethene, of we carbon start as with shown the previously C C methane. 2 2 sp sp 2 sp 2 sp Step 1 One of the electrons in the 2s orbital of conguration of carbon is the ▲ ground-state Figure 11 The formation of the C=C double bond: a sigma bond promoted 2 forms from the overlap of two sp to the vacant 2p orbital z to form an excited-state, the overlap of two p z as 350 for methane above. orbitals orbitals, and a pi bond from 14 . 2 and Step 4 2p . These h y b r i d i z a T i O n combine to form a set of two hybrid orbitals. The remaining 2p and 2p y The new x sp nal step involves overlap of each z remaining orbitals remain unhybridized. 2 sp orbital orbital on carbon (gure with a hydrogen 1s atomic The 12). two atoms new are sp hybrid entirely orbitals equivalent on and the carbon each one pi consists H H 2 of They point Each of 50% in s character opposite and 50% directions p character. (gure 14). 2 sp sp 2 these sp hybrid orbitals contains one 2 sp sp electron. C The remaining orbitals are the 2p C and y 2 2 sp sp 2p unhybridized atomic orbitals, also containing z sigma one H electron each. H p p z z 2 ▲ Figure 12 In ethene, four carbon sp orbitals overlap with four hydrogen 1s orbitals to form a trigonal planar geometry on sp C sp each carbon atom Any molecule with a trigonal-planar electron domain p y 2 geometry will be predicted to have sp hybridization. p z The formation of sp hybrid orbitals in ▲ Figure 14 Each of the two sp hybrid orbitals contains one electron. The 2p and 2p y unhybridized atomic orbitals also z ethyne contain one electron each Ethyne, C H 2 bond. As , is an alkyne and has one C≡C triple 2 seen in sub-topic 14.1, around each carbon Step 3 atom there are two electron domains so the electron The domain geometry (and molecular geometry) next sigma linear, with a 180° bond angle (gure step involves the formation of a is bond along the internuclear axis by the 13). overlap of two sp hybrid orbitals, one on each 180° carbon H the atom. sideways Two pi bonds overlap of are the formed two p and from two y unhybridized ▲ atomic orbitals, p z with the overlap Figure 13 The molecule of ethyne is linear regions To deduce atom in the hybridization ethyne, again we scheme start with of the the carbon (gure above and below the internuclear axis 15). orbital pi diagram of carbon as shown for methane. sp sp Step 1 One of the electrons ground-state to the as for vacant the 2s orbital of conguration of carbon is 2p in orbital to form an the sigma promoted excited-state, z pi p p y y methane. p p z Step 2 ▲ z Figure 15 The formation of the C≡C triple bond in ethyne: a sigma bond forms from the overlap of two sp orbitals, a pi To account for the 180° bond angle (based on bond from the overlap of two p two electron domains), the next step from the overlap of two p hybridization of two of the atomic orbitals 2s orbitals, and another pi bond y involves orbitals z 351 14 C H E M I C A L B O N D I N G A N D S T R U C T U R E ( A H L ) H Step 4 The sp nal step orbital orbital on involves carbon (gure overlap with a of each hydrogen H O H remaining 1s atomic H C O H 16). A H H pi H B ▲ sp Figure 1 7 Methyl propanoate sp C H C C H Solution sigma pi p ito tom A B C nm of 4 3 4 tetrahedral trigonal tetrahedral p y y p p z z to oms ▲ Figure 16 In ethyne, two carbon sp orbitals overlap with two eto om hydrogen 1s orbitals to create a linear geometry on each omt carbon atom Any molecule geometry on predicted to with its a linear interior have sp electron central domain atom would planar Mo be tetrahedral trigonal V-shaped planar (bent) omt hybridization. bo (s) H C H: C C 109.5° Summary O: C 120° O C: less than 109.5° nm of eto om to oms omt 4 tetrahedral hto 3 hto 2 3 sp sp sp 3 sp ▲ T able 2 2 3 trigonal planar 2 linear sp Example 3 sp The ▲ T able 1 Electron-domain geometry and hybridization condensed (traditional structural name formula aniline) is C H 6 a) Using Worked examples the Example 1 Deduce the interior atom VSEPR domain and carbon theory, phenylamine . 2 deduce the molecular and of NH 5 electron geometries nitrogen atoms phenylamine. hybridization of the central nitrogen b) in A model of the molecule is shown in ammonia. Solution As seen in electron topic 4, domains because in there ammonia, are the four hybridization 3 must be sp Example 2 For each molecule deduce interior of the molecular atom methyl electron A, B, and propanoate domain geometry, the C in a (gure geometry, bond angles, 17), the and the ▲ hybridization 352 of in state. Figure 18 Molecular model of phenylamine gure 18. 14 . 2 In this model nitrogen of the the appear two to be hydrogen atoms above horizontal the attached to For h y b r i d i z a T i O n C: plane number of valence number of sigma electrons = 4 = 3 = -1 = 6 = 3 molecule. Figure 19 shows phenylamine, electrostatic a space-lling which is potential the map model basis of one an showing pi bond electron total charge bonds of number of valence electrons density. number electron of domain molecular bond electron domains geometry: geometry: angle: trigonal trigonal planar planar 120° predicted hybridization domains: sp based on 3 electron 2 Ball-and-stick bond diagram for nitrogen (ignoring angles): C H For N: number of valence number of bonds total ▲ number of electrons valence electrons = 5 = 3 = 8 = 4 Figure 19 Space-lling model of phenylamine number Deduce gure two c) the 18 hybridization and from of gure the 19 nitrogen and electron from comment on of electron domain domains geometry: tetrahedral the predicted molecular predicted H geometry: trigonal pyramidal models. A of theoretical study phenylamine of the found electronic the H N H N H bond angle: less than 109.5 ° structure bond predicted hybridization domains: sp based on 4 electron 3 angle is in very phenylamine close to the to be 112.79 °, experimental which value from See gas-phase microwave studies. Discuss gure 20. what H you may conclude about the H molecular 3 nitrogen sp N geometry around the nitrogen in the NH 2 group in the structure of phenylamine, H and H 2 deduce its hybridization state on this carbon sp basis. H H Solution H a) Using VSEPR theory to consider the geometry 3 ▲ at carbon, start by drawing the Figure 20 VSEPR theory suggests sp hybridization at nitrogen ball-and-stick in phenylamine diagram (ignoring bond angles): b) C C The model in gure 18 shows a trigonal C 3 pyramidal H at geometry, nitrogen as suggesting outlined above, sp hybridization while that in 353 14 C H E M I C A L gure 19 around B O N D I N G shows a nitrogen A N D trigonal with a S T R U C T U R E planar 120° ( A H L ) on geometry bond pi angle, N and system increases on the the electron aromatic ring. density A more of the planar 2 suggesting sp geometry hybridization. is therefore adopted around the NH 2 group. Although VSEPR theory is a useful model We structure, in some cases it does such the In with cases the it geometry cannot be found used to experimentally; accurately the electron domain around N in one of these three dipolar structures: in deduce number of valence number of sigma electrons = 5 = 3 = -1 = -1 = 6 = 3 hybridization. phenylamine the NH functional group type attened geometry interactions resulting about with the N, in due a trigonal to aromatic resonance ring. 1 pi bond 1 +ve planar- The p interacts with the pi system on the charge orbital total N number resulting electron pair structure of in delocalization (lone pair). The phenylamine is of the most of in Lewis gure with three dipolar resonance electron + NH 2 domains domain geometry: trigonal planar geometry: trigonal planar is consistent with the planar geometry + NH 2 electron structures. This + NH electrons 21 molecular along valence non-bonding stable shown of aromatic number ring bonds is 2 slightly on out not resonance agree work for geometry predicting can NH 2 2 2 from the space-lling model, suggesting sp hybridization. c) The experimental suggests ▲ a H geometry trigonal pyramidal trigonal planar N H bond somewhere (109.5 ° bond angle 113° between angle) and Figure 2 1 Lewis structures for phenylamine (120°), and a hybridization 2 state Delocalization reduces the of the electron non-bonding density in the pair p on somewhere N orbital usf so The “ChemEd DL” website contains numerous digital resources. It is a collaboration with the Journal of Chemical Education, the education division of the American Chemical Society, and the ChemCollective Project, initiated through the National Science Foundation in the USA. It has an excellent library of three- dimensional models. 354 of between sp 3 and sp . Q u e S T i O n S Questions 1 What are the formal charges on P and O in the 5 What is the molecular geometry of [PF ] ? 6 Lewis (electron oxoanion dot) structure represented in of gure the phosphate A. Trigonal planar B. Trigonal bipyramidal C. Square D. Octahedral 22? 3 pyramidal O O P O 6 Which of the following molecules is O non-polar? ▲ Figure 22 A. SF B. ClF C. BrCl D. SeF 4 A. B. P P is 1 is and +5 C. P is 0 D. Both O and and O O are is is is 3 0 5 2 6 0 3 7 Which of orbitals 2 What are oxygen ΔFC, formal atoms in carbon the the and Lewis dioxide charges the formal (electron shown on in the carbon charge dot) gure following in combinations gure 24 results in of a atomic sigma bond? and difference, structure the shown I. + of 23? II. + O ▲ C O III. Figure 23 + FC(C) FC(O) ΔFC ▲ 3 A. +4 2 B. +4 +4 Figure 24 +6 0 A. I and II I and III C. +4 +2 +2 B. D. 0 0 0 C. II and D. I, II, What is the electron domain geometry only only III and only III of 2 the sulte oxoanion, [SO ] ? 8 3 What in A. Trigonal planar B. Trigonal pyramidal C. D. is the hybridization of the oxygen hybridization of the carbon atom ethanol? A. sp Tetrahedral B. sp V-shaped C. sp D. dsp 2 3 (bent) 3 4 What is the molecular geometry of BrF ? 5 9 A. Octahedral B. Square C. T-shaped D. Square What atom is in the methanal? planar A. sp B. sp C. sp D. dsp 2 pyramidal 3 3 355 14 C H E M I C A L 10 How many molecule B O N D I N G sigma of and propyne, pi H A N D S T R U C T U R E bonds are present ( A H L ) in a 12 Consider the following + 3 [NO ] [SiF 3 sigma pi A. 5 3 B. 6 2 C. 7 1 D. 8 0 IB, May For a) each i) its electron the following NH 4 ii) its molecular iii) the Draw geometry geometry bond angle(s). an appropriate each and Lewis calculate the (electron formal dot) charge atom. 2 The following reactions take place in the ozone species, by the absorption of ultraviolet light. Deduce: I O → O 3 i) its electron-domain ii) its molecular iii) bond O → and hybridization state of of central Draw molecular an structure on each O• explain, the by reference reactions, wavelength. atom IB, its + polarity. appropriate and Lewis calculate atom. the I or II, to the bonding, requires a the shorter v) O• angle which the O• 2 geometry State iv) + 2 geometry II 356 domain CS 3 layer b) 4 species: 13 a) SCl [1] on each ] 4 2011 Consider For [IF species: structure CCl ] 6 Deduce: b) 11 species: CCCH? (electron formal dot) charge May 2011 [2] E N E R G E T I C S A N D 15 T H E R M O C H E M I S T R Y ( A H L ) Introduction In sub-topic compounds lattice cycles, the 4.1 structures. derived calculation determined While we form some examined and This from of their topic Hess’s energy directly from reactions go how begins Law values with (topic that empirical to reach ionic three-dimensional energy 5) be evidence. completion, others as associated disorder, for cannot equilibrium products, behind Gibbs with can and of 7. and Entropy, randomness as chemical enables the reactants topic considered energy spontaneity in molecular be physical free between discussed the or driving changes chemists to force and assess the reaction. 15.1 Ee les Understandings Applications and skills + ➔ Representative equations (eg M (g) → ➔ Construction of Born– Ha ber cy cles for + M (aq)) can be used for enthalpy/energy of group 1 and 2 oxides a nd chlorid es . hydration, ionization, atomization, electron ➔ Construction of energy cy cl es from hy dration , anity, lattice, covalent bond, and solution. lat tice, ➔ and solution e ntha lpy. For example, Enthalpy of solution hydration enthalpy and dissolution of solid Na OH or NH Cl in water. 4 lattice enthalpy are related in an energy cycle. ➔ Calcu lation Haber Nature of science ➔ Relate or enthalpy dissolution size hydration ➔ of and cha nges energy charge of from Born– cycles. ions to lat tice and enthalpies. Making quantitative measurements with replicates to ensure reliability – energy cycles allow for the calculation of values that cannot be determined directly. ➔ Perform include laboratory single aqueous ex periments replacem ent which reactions cou ld in solutions. 357 15 E N E R G E T I C S A N D T H E R M O C H E M I S T R Y ( A H L ) TOK Models for nding enthalpy changes Subject-specic terminology of reaction with precise denitions We summarize many chemical and biological processes in a single plays a central role within chemical equation representing one pathway from reactants to the areas of knowledge: products. These processes in reality often take place involving several the natural sciences, reactions with alternative pathways. We can use these alternative mathematics, history, pathways to determine changes in enthalpy values that cannot be human sciences, the ar ts, measured directly. ethics, and religious and indigenous knowledge In systems. Communication taking using this specic models terminology may shape reactants understanding of concepts Using within these areas of methodology doing this place are we develop within in a models system. agreement that are empirical broken data in to that We with and can represent assess empirical bonds conrm or data that are modify the the energy extent by to focusing formed proposed to changes which on bonds make models these in products. is a central science. knowledge. How impor tant is this terminology and what role does it play? The Born–Haber cycle and enthalpy of formation The standard represented enthalpy by a single of formation of an ionic compound can be equation: Stu ti 1 Na(s) + You need to know the individual steps of the Born–Haber cycle in order to apply the method to a variety of situations. However, 1 Cl An a application series of (g) → of ΔH of an = -411 kJ mol f Hess’s reactions formation NaCl(s) 2 2 law that ionic (sub-topic can be 5.2), combined to the Born–Haber determine the cycle, enthalpy is of compound. rather than writing these steps for specic elements and compounds it is permissible to Constructing a Born–Haber cycle write representative equations The when answering questions, for several Born–Haber steps in cycle the combines formation of the an enthalpy ionic changes compound, associated dened with below. + example, M(g) → M (g) + e . This approach will be used in Lattice enthalpy this chapter. The lattice occurs on enthalpy the is formation dened of 1 as mol the of standard gaseous enthalpy ions from the change solid that lattice: + MX(s) → M (g) + X (g) ∆H > 0 lat Lattie ethalies are often quoted as negative values The process 298K can is be endothermic. found in Experimental section 18 of the values Data of lattice enthalpy at booklet. that represent the exothermic formation of the lattice from Enthalpy of atomization gaseous ions. However, in this The enthalpy of atomization ΔH is the standard enthalpy change at textbook we shall consider the that occurs on the formation of 1 opposite process (endothermic element in its standard state: formation of gaseous ions from M(s) → M(g) ΔH lattice), which is consistent with the denition given in the X 2 Data booklet 358 > 0 > 0 at 1 (g) 2 → X(g) ΔH at mol of separate gaseous atoms of an 15 . 1 E n E r g y c y c L E S Ionization energy As introduced in topic 3, the ionization energy, ΔH , is the standard IE enthalpy 1mol ions of change atoms with third that or positively multiple ionization occurs valence energies on the removal charged ions electrons are the in of 1 the rst, mol of gaseous second, electrons phase. and from For metal sometimes dened. + IE : M(g) → M (g) + e ∆H + IE > 0 > 0 IE 1 : M 2+ (g) → M (g) + e ∆H IE 2 2 Electron anity The electron afnity, ∆H , is the standard enthalpy change EA on the addition gaseous of 1 mol of electrons to 1 mol of atoms in the phase: X(g) + e → X (g) ∆H < 0 EA As discussed are in exceptions, topic such 3, as electron the afnity electron is typically afnity for negative, but there helium. Constructing the Born–Haber cycle The lattice and the and nd enthalpy, electron the Born–Haber energy afnity enthalpy cycle relationships each the of enthalpy are 1) the atomization, combined formation (gure between of of focuses on individual to an the construct ionic the steps ionization the Born–Haber compound. processes rather The involved than energy, the cycle standard and the magnitude of change. + M ∆H (g) + X(g) +e (X) at ∆H (X) EA Stu ti 1 + M X (g) 2 (g) e 2 Values for lattice enthalpies can be found in the Data + ∆H M (M) + (g) X (g) booklet (section 18), along IE with enthalpies of aqueous solutions (section 19), and 1 + M(g) X 2 (g) 2 enthalpies of hydration ∆H lat (section 20) which will be used ∆H later in this topic. The Data (M) at booklet will be available during ∆H f 1 + X 2 the examination, except in (g) 2 Paper 1. Figure 1 A generalized Born–Haber cycle 359 15 E N E R G E T I C S A N D T H E R M O C H E M I S T R Y ( A H L ) Worked examples Example 1 Write an Example 2 equation formation of Born–Haber for the potassium cycle to enthalpy bromide. calculate change Construct the lattice of Use a the the Born–Haber enthalpy of cycle in formation gure for 3 to calculate magnesium oxide. enthalpy 2+ Mg for this + (g) O(g) +2e compound. Solution 1 ∆H (O) +249 kJ mol at ∆H K(s) + 1 (O) 141 kJ mol EA(1) 1 Br (l) → KBr(s) 2 2 1 2+ Mg O (g) 2e (g) 2 2 + K + (g) Br(g) +e 2+ Mg + (g) O (g) 1 ∆H (Mg) +1451 kJ mol IE(2) 1 ∆H (Br) ∆H +97 kJ mol 1 ∆H (Br) 325 kJ mol +798 kJ mol 1 + Mg + (g) O EA +e (g) 2 2 1 + K 1 (O) EA(2) at Br (g) e (l) 2 2 2+ 1 ∆H (Mg) Mg +738 kJ mol 2 + (g) O (g) IE(1) + 1 ∆H (K) K +419 kJ mol + (g) Br (g) IE 1 + Mg(g) O (g) 2 2 1 ∆H 1 (MgO) +3795 kJ mol lat + K(g) Br (l) 2 2 1 ∆H x kJ mol lat 1 ∆H (Mg) +148 kJ mol at 1 ∆H (K) +89 kJ mol ∆H at (MgO) f 1 + O (g) 2 2 1 x kJ mol ∆H f 1 + Br (l) 2 2 1 Figure 3 Born–Haber cycle to calculate the enthalpy of 392 kJ mol formation for magnesium oxide Figure 2 Born–Haber cycle to calculate the lattice enthalpy for potassium bromide Solution To for determine the potassium (endothermic) bromide, follow lattice the enthalpy pathway ∆H (MgO) = ∆H f (Mg) + ∆H at + on (Mg) IE(1) ∆H (Mg) + ∆H IE(2) gure 2 and add up the values for the enthalpy + ∆H - ∆H (O) + ∆H EA(1) changes shown, taking note of their sign. (O) at (O) EA(2) (MgO) lat ∆H (KBr) = ∆H f (K) + ∆H at + (K) = IE ∆H (Br) + ∆H at (Br) + ∆H EA (148 + (KBr) ( + 738 141) + + 1451 798 + 249 3795) lat 1 - 392 = 89 = - + 419 + 97 + [ 325] + = x - 552 kJ mol 1 672 kJ mol Figure 4 The lattice structure of crystalline magnesium oxide 2+ Grey: Mg 360 2 Red: O 15 . 1 E n E r g y c y c L E S collaboatio i the sieti ommuit “ Why is it impor tant for countries to collaborate to combat global problems like global warming?” Chemistry Syllabus sub-topic 15.1 The scientic community brings together various scientic disciplines and also other elds such as engineering, technology, and mathematics. Some notable examples of international collaboration include the Human Genome Project, CERN, and the Manhattan project. Scientists look to the past and the future to understand the patterns in the Ear th’s climate. The Intergovernmental Panel on Climate Change (IPCC) was established in 1988 by the United Nations Environment Programme (UNEP) in conjunction with the World Meteorological Organization (WMO), to coordinate data collection, independently analyse it, and publish repor ts. Complex models make projections about the Ear th’s future climate using indirect indicators of global warming. These include data from ice cores, cores from ancient coral formations, ocean and lake sediments, borehole temperatures, evaporation and precipitation cycles, glacial recession patterns, and receding polar ice caps. More accurate data results from linking repeated measurements on a global scale. Global collaboration is essential to investigate the causes and eects of global warming. Variations in lattice enthalpy values The by magnitude both An the increase oppositely the For ionic the 17 lattice on the ionic is, the from this is and the lattice radius of the by in anion iodine. a compound a greater of the directly attraction required (table affected the to as in you distance electrostatic decrease break between apart 1). increases As is radii. energy enthalpy strength reected a ionic result the the to for the will increasing uorine increases, and enthalpy ions charge ions, that halides, ions decreases (table in group bonded the charged lattice, metal down of charge the move between the attraction lattice enthalpy 2). ∆H (theoetial value)/ Vaiatio i hae a ioi lat 1 aius omae with nacl kJ mol MgO 3795 greater ionic charge NaCl 769 — KBr 671 increased ionic radius T able 1 Variations in the lattice enthalpy with ionic radius and ionic charge 1 ∆H (theoetial value)/ kJ mol lat NaF 910 NaCl 769 NaBr 732 NaI 682 T able 2 Lattice enthalpy of metal halides 361 15 E N E R G E T I C S A N D T H E R M O C H E M I S T R Y ( A H L ) Enthalpy changes in solution Many reactions therefore between the The to the lattice studied consider in the enthalpy chemistry enthalpy change of take of place in solution solution, the solution. and the hydration It is useful relationship enthalpy, and enthalpy. standard enthalpy change of solution , ∆H , is the change in sol Enthalpy changes of solution enthalpy when 1 mol of a substance is dissolved in a large excess of a can have either positive or pure solvent: negative values. 1 + NH Cl(s) → NH 4 (aq) + Cl (aq) ∆H = +14.78 kJ mol = -48.83 kJ mol sol 4 + LiBr(s) → 1 (aq) Li + Br (aq) ∆H sol It is or possible by ionic using solid to an and calculate energy the the cycle enthalpy that subsequent change involves hydration the of solution lattice enthalpy empirically, enthalpy of the of the gaseous ions produced. The enthalpy change of hydration , ∆H , for an ion is the enthalpy hyd The enthalpy change of change when 1 mol of the gaseous ion is added to water to form a dilute hydration always has a negative solution. The term solvation is used in place of hydration for solvents value. other than water. + M + (g) → M 1 (aq) ∆H = - kJ mol hyd 1 X (g) → X (aq) ∆H = - kJ mol hyd Worked example Find the enthalpy change of solution ∆H for sodium hydroxide sol using the enthalpy cycle in gure 5. + Na (g) + OH (g) 1 ∆H = ( 424) + ( 519) kJ mol hyd 1 ∆H = +900 kJ mol lat + Na (aq) + OH (aq) 1 ∆H = x kJ mol sol NaOH(s) Figure 5 Enthalpy cycle to calculate the enthalpy change of solution for sodium hydroxide Solution + ∆H = sol ∆H (NaOH) + ∆H lat (Na ) + ∆H hyd 1 = 900 = -43 + ( 424) -1 362 kJ mol + ( 519) (OH hyd kJ mol ) 15 . 1 E n E r g y c y c L E S Solvation, dissolution, and hydration + δ + δ Three terms commonly used when describing the interactions between + solvents briey and solutes and the subsequent solutions formed, are described δ δ here: + δ δ Solvation is interaction described of a solute by the and IUPAC the Gold solvent or Book a as “any similar stabilizing interaction of + solvent δ δ with groups of an insoluble material. Such interactions generally involve + δ + electrostatic specic forces effects and such as van der Waals’ hydrogen forces, bond as well as chemically δ more + formation.” δ + δ Water is oxygen a polar and solvent. hydrogen, molecule (bent) electrons on due the negative on hydrogen the their partial shell, also (gure6). new The repulsive atom, on Water as a phase is the in oxygen cations polar shell and is known lone the solvent with a charges so that solvation is water liquid solution. positively charged sodium ion of having themselves the (solvent) molecules surround a pairs positive forming mixed as the Figure 6 Solvation shell: the water water molecule anions, when between the partial orientate and of between this hydroxide formed geometry atom molecules hydration sodium electronegativity forces result the surround solid with in water, This is a the dissolution enthalpy released of known When of the atoms. homogeneous process to charges charges difference combined oxygen partial The of during hydration is hydration the is process inuenced a of by way of quantifying solvation. the charge 1 catio ∆H the amount The magnitude and size of the of ion energy enthalpy (table 3). 1 Aio /kJ mol of the ∆H h /kJ mol h + Li 538 F 504 424 Cl 359 1963 Br 328 + Na 2+ Mg 3+ Al 4741 I 287 T able 3 Enthalpies of hydration (more data is available in section 20 of the Data booklet) As you move down hydration decreases hydration enthalpy group 17. decrease For in a group as in cations, size results the in group an in the ionic 1 while increase a periodic radius in table the increases. uorine charge signicantly has on larger enthalpy Lithium the the has highest ion of the combined enthalpy of greatest value in with a hydration. Quik questio Calculate the enthalpy change of solution of barium chloride given the following data: 2+ BaCl (s) → Ba 1 (g) + 2Cl 2 2+ Ba (g) ∆H = +2069 kJ mol lat 2+ (g) → Ba 1 (aq) ∆H = -1346 kJ mol hyd 1 Cl (g) → Cl (aq) ∆H = -359 kJ mol hyd 363 15 E n E r g E T I c S A n d T H E r M O c H E M I S T r y ( A H L ) 15.2 Eto a sotaeit Understandings Applications and skills ➔ Entropy (S) refers to the distribution of Prediction of whether a change will result in an ➔ available energy among the par ticles. The more increase or decrease in entropy, by considering ways the energy can be distributed the higher the states of the reactants and products. the entropy. Calculation of entropy changes (ΔS) from ➔ ➔ Gibbs free energy (G) relates the energy that standard entropy values (S ). can be obtained from a chemical reaction to the Application of ΔG ➔ = ∆H - T∆S in change in enthalpy (ΔH), change in entropy predicting spontaneity and calculation of (ΔS), and absolute temperature (T). various conditions of enthalpy and temperature ➔ Entropy of gas > liquid > solid under the same that will aect this. conditions. Relation of ΔG to position of equilibrium. ➔ Nature of science ➔ Theories can be superseded – the idea of entropy has evolved through the years as a result of developments in statistics and probability. Sotaeous haes Chemists work reactions will systems A to reaction is said or intervention. may place be a enthalpy just one The and in conservation of energy to the system. the will of the (S) and movement us of to it set chemical control moves of chemical can exothermic. are occur be external different that do rates not take non-spontaneous positive examining usually either without at Reactions to whether when are said towards conditions or the negative, is spontaneity spontaneous thermodynamics focuses Spontaneous which and but there of a are rule. concerns to when spontaneous or applications is under modify given reaction, a of on The of original reactions predict the greater to the shift and the of the gain total the law, the of the total higher in an chemical being the the the of chance entropy total reaction. change of of entropy understanding entropy of second available energy lower law The spontaneity increase can fundamental world. from particles, an of rst distribution we direction are The physical and state If quantify the the the the lead surroundings. so in entropy amongst their and chemistry. energy measure widespread system allow a particles. returning a conditions laws energy, being particles this of are reactions this practical Entropy the freedom 364 of second between localized within to can outcomes. under that of thermodynamics reactions. desired considered Exothermic exceptions importance law be conditions they spontaneous set change to the that endothermic given aspect rst be Reactions The many to the so equilibrium either under reaction. understand achieve completion and to proceed, for a of this system, 15 . 2 E n T r O p y A n d S p O n T A n E I T y Changes in entropy Figure 1 a containing glass shows difference and the that glass) condensation iced exists and water. on between the the The outside that of gas temperature the system surroundings (iced water under is greater than the system) results in thermal energy from to and the surrounding glass equilibrium. entropy of its With the contents, this until thermal water/ice they energy mixture of a of a the liquid entropy which in of a turn is solid. S is is a state function, determined by the so a change difference in between atmosphere its the that conditions, that being entropy transferred same than (everything Entropy, outside the greater will reach nal and initial values: an transfer, increase ∆ the (reaction) = ∑∆ (products) 298 298 while ∑∆ (reactants) 298 the entropy of the surroundings will decrease as The energy is transferred from it. The condensed conditions entropy on the surface of the glass is lower in entropy water vapour in the change, be and specied the for subscript a particular “298” refers to than a the must water temperature of 298 K. atmosphere. The second chemical in the law reactions entropy When the of the of rather in which it ∆S is = in the than is of in ∆S an are the + overall that increase spontaneous. universe is found a describes (table says equilibrium. describing opposite written in universe this the total result universe entropy non-spontaneous, spontaneous that system entropy negative, thermodynamics the overall unchanged, overall of remains If to reaction a reaction direction to the be that is that the is way 1). ∆S system surroundings Figure 1 Changes in entropy are associated with every chemical and physical process ∆S spontaneous > 0 total ∆S Predicting changes in entropy Simple representations of particles in the equilibrium = 0 total ∆S different non-spontaneous < 0 total states of matter show an increasing entropy as T able 1 The second law of thermodynamics allows us to the particles gain more freedom of movement predict the direction of a reaction and more particles ways move of distributing from solids the energy through as liquids the to gases. An increase in the system the particles, heat will energy result leading in to (enthalpy) increased greater within movement disorder and of an increase in entropy (S) increase liquid the the spontaneity gas in entropy changes ● solid in the both Exothermic of of a reactions as this enthalpy greater are system. and chemical spontaneous, and the enthalpy affect reaction. more leads Therefore, entropy to a stability likely to be reduction of the in reaction products. Figure 2 Entropy increases from the solid through to the liquid ● An increase in entropy makes reactions more to the gaseous phase likely leads Achieving a change of state from solid to liquid to to is sometimes absorbed particles which described results increasing. In in in terms the terms of kinetic of energy energy entropy we spontaneous, more uniform as greater distribution disorder of energy to within gas be the system. being of can the say This will be revisited in greater depth later in thistopic. 365 15 E N E R G E T I C S A N D T H E R M O C H E M I S T R Y ( A H L ) reteati laies: Iteatioal ollaboatio The impor tance of being able to obtain data from around the globe is highly signicant. Glacier recession is an indirect indicator of global warming. Why is it impor tant for countries to collaborate to combat global problems like global warming? Chemistry Syllabus sub-topic 15.1 Glaciers around the world are retreating to higher altitudes as global temperatures steadily rise. Ice cores from glaciers provide vital information to climate scientists, enabling them to build up a picture of the Ear th’s climate and signicant events, such as volcanic eruptions, over thousands of years. Studies into the consequences of the melting of glaciers have stimulated widespread discussions within the media, government, and scientic organizations. Figure 3 Jostedalsbreen glacier, Nor way. calulati eto haes The entropy change thermodynamic The standard temperature To calculate nd the total and of of a which entropy standard to ● be can provided values, entropy between the S , be in calculated section relate to 12 from of the standard Data booklet. conditions (reaction) the change total associated entropy of = ∑∆S (products) - performing the ∑∆S 298 with a reaction products and entropy (reactants) 298 change calculations the following that values for entropy are specic for points different 1 matter, for example, 70.0 The J K S (H O(g)) = 188.8 J K mol states of while S (H O(l)) 2 1 mol coefcients molar 366 need 1 2 1 ● we the considered: Remember = of reactants: 298 When system is pressure. difference entropy ∆S molar the ΔS data entropy used values to balance when the equation calculating the must overall be applied entropy to change. 15 . 2 Examine ● reaction degree be the to of used chemical have disorder to reaction positive check in or the your and predict negative products nal whether entropy and you change reactants. E n T r O p y expect based This on A n d S p O n T A n E I T y the the prediction can calculation. Stu ti Worked example Standard molar entropy has the unit Calculate the standard entropy change for the following reactions: 1 J K 1 a) H (g) 1 + O 2 (g) → H 2 2 S O(l) (H 298 ) 130.7 J for standard enthalpy of formation. These 1 b) NH Cl(s) → NH 4 (g) + S HCl(g) (O 298 1 ; compare this with kJ mol mol 2 2 1 mol 1 K ) 205.1 J K 1 values are combined in Gibbs free energy mol 2 3 calculations (see later in this topic). When combining these quantities be sure to Solution convert units appropriately. 1 a) H (g) + O 2 (g) → H 2 2 O(l) 2 1 ∆S = [∆S 298 (H 298 O)] [∆S (H 298 2 ) + 2 ∆S (O 298 2 )] 2 1 = [70.0] [130.7 + × 205.1] 2 1 = The -163.3 negative J entropy K change associated with this chemical reaction 1 indicates a decrease in disorder (greater order), with 1 mol of gas in large 2 changing b) NH into Cl(s) 1 mol → NH 4 of (g) a liquid. + HCl(g) 3 ∆S = [∆S 298 (NH 298 ) + ∆S = [192.5 + = +284.55 (HCl)] - [∆S 298 3 186.9] (NH 298 Cl)] 4 [94.85] 1 Transforming increase data in can 1 mol of disorder, be found J K a solid hence in into the section 2 mol large 12 of of a positive the Data gas results entropy. a Thermodynamic Booklet. Quik questios 1 Predict whether the following reactions will have a positive or negative entropy change, ∆S a) NH NO 4 (s) → N 3 O(g) + 2H 2 . Table 2 shows the standard entropy values of the substances in the reaction above. O(g) Substae 2 CO (g) 2 b) N (g) + 3H 2 ) N O 2 ) (g) → 2NH 2 (g) → 2NO 4 CaCO 2C H 2 2 1 S 214 131 (g) H 4 186 O(g) 2 189 (g) Calculate the standard entropy change for the 2 6 1 mol CH Table 2 (s) → CaO(s) + CO (g) + 7O /J K (g) 2 (g) 2 3 e) (g) 3 H (g) → 4CO 2 (g) + 6H 2 O(l) reaction. [3] 2 The equation for the reaction between carbon dioxide Explain how the sign can be predicted from the equation for the reaction. and hydrogen is shown below. [2] IB, nov 20 07 CO 2 (g) + 4H 2 (g) → CH 4 (g) + 2H O(g) 2 367 15 E N E R G E T I C S A N D T H E R M O C H E M I S T R Y ( A H L ) Sustaiable ee “Sustainable energy is a UN initiative with a goal of doubling of global sustainable energy resources by 2030.” Chemistry Syllabus sub-topic 15.1 “Sustainable energy for all” is a United Nations (UN) initiative that aims to reduce the inequalities that exist in the provision of clean and ecient energy services, improve energy eciency thereby reducing energy demand, and increase the proportion of energy that comes from renewable resources (http://www.sustainableenergyforall. org/objectives/universal-access). The project has three main objectives: 1 Ensure universal access to modern energy services which focuses on improving the lives and economic conditions of people throughout the world. Approximately one-fth of the world’s population do not have electricity in their home and almost 40 per cent utilize fuel sources such as animal waste, charcoal, and wood to provide heat for cooking. Toxic products from this form of combustion result in the deaths of over 2 million people annually, mainly women and children. “Electricity enables children to study after dark. It enables water to be pumped for crops, and foods and medicines to be refrigerated.” 2 Energy eciency is the par t of the project that looks at countries, including the way in which we use power. From industry to households, oce and accommodation buildings to transpor tation, lighting to electrical appliances, a variety of people, agencies, and governments are being encouraged to both educate and legislate, with the aim of decreasing the global electricity demand. Energy-saving light bulbs, energy-ecient televisions, buildings that require less energy to heat and cool, and the use of information technology in industry to better manage power usage are all examples of how the global community is reducing the demand for power. This ultimately saves governments, individuals, and businesses money and lessens the impact of coal-red power stations on the environment. 3 Renewable renewable The cost of energy energy the development appreciably over governments, renewable – the has of can dams in play set to a target global renewable decades businesses, energy hydroelectric UN contributed and and a Brazil energy now role generate doubling in sources Where power 83% of the production represents individuals. major of energy a has share by decreased viable option resources generation. the of 2030. country’s are For for available, example, electricity. Figure 4 Itaipu dam, built between Brazil and Paraguay, is the second largest hydroelectric power plant in the world 368 15 . 2 E n T r O p y A n d S p O n T A n E I T y gibbs fee ee The H, Gibbs free entropy S, importance now look of at a G of is a absolute entropy the temperature For energy and in state dening relationship the spontaneous function, temperature the T. along with Having spontaneity between total enthalpy established of entropy, a the reaction, enthalpy, we and shall the system. Reactions that are reaction: spontaneous and are ∆S = ∆S + total ∆S > sys 0 surroundings therefore thermodynamically favourable can sometimes be A chemical reaction may be either exothermic or endothermic: the kinetically improbable, due transfer of heat across the system/surroundings boundary is directionally to the existence of very high dependent on the change in enthalpy. For an exothermic reaction in an activation energies (see open system, heat is transferred from the system to the surroundings. sub-topic 16.2). This results The the impact an that low and energy to melt in a as each the has will of to G = H = Gibbs reaction - a a change at high 60 less state of a the reaction two The a the level transfer water marked entropy, function as the a on the the the in The entropy the will increases, ice the so at of ice amount ice begin water additional level of entropy. temperature of system Gibbs free at of resulting hot the of system. one block same However, with and called of molecules effect on systems, such effect. entropy. compared has existing separate different of surroundings. conditions temperature, °C. of the of the into have enthalpy, new on disorder much of at will in entropy energy energy change energy, can be G: TS ∆H free - T∆S energy system spontaneity. have one signicant dene ∆G heat water kinetic a the enthalpy system have in dependent and combination used The bowl signicant energy The a into already is transferring temperature °C increase the surroundings Imagine 0 in at For negative a provides constant reaction value ( ∆G an effective temperature to be < 0). way and spontaneous ∆H ∆S positive (> 0): positive (> 0): endothermic more disorder of focusing pressure the to Gibbs on a determine free energy its must ∆G Sotaeit Exlaatio negative at high T dependent on spontaneous only at high positive at low T temperature temperatures when T∆S > H positive (> 0): negative (< 0): endothermic more order always positive never reverse reaction > 0 spontaneous spontaneous at all temperatures negative (< 0): positive (> 0): exothermic more disorder always negative always forward reaction < 0 spontaneous spontaneous at all negative at low T dependent on spontaneous only at low positive at high T temperature temperatures when temperatures negative (< 0): negative (< 0): exothermic more order T∆S < H T able 3 Factors aecting ∆G and the spontaneity of a reaction 369 15 E N E R G E T I C S A N D T H E R M O C H E M I S T R Y It is be not always possible spontaneous increasing not with to will of 3). always whether be > 0. The chemical reactions spontaneous, increasing the a Exothermic of ∆G temperature predict (table reactions spontaneous, the or disorder endothermic on ( A H L ) order will spontaneity with ∆G always of reaction that other be will involve < 0. Similarly, non- reactions depends system. gibbs fee ee hae of fomatio The Gibbs free energy change of formation , ΔG , represents the free f energy under change when standard ΔG = 1 mol conditions ΣΔG r of of (products) a compound 298 - ΣΔG Gibbs free energy change of a formed pressure from of 100 its elements kPa: (reactants) f from ΔG r the and f Worked example: nding ΔG Calculate K is values f reaction, Solution ΔG for the combustion of ethanol, C r H 2 give CO (g) and H 2 OH to 5 C O(g). H 2 OH(l) + 3O 5 (g) ΔG = ΣΔG r Substae C H 2 OH(l) H 5 O(g) CO 2CO (g) + 3H 2 O(g) 2 2 (products) - ΣΔG f (reactants) f (g) 2 = [2ΔG (CO f ∆g → 2 2 ) + 3ΔG (H f 2 O)] 2 values/ [ΔG f (C f 175 228.6 H 2 394.4 OH)] 5 1 kJ mol = [2 × -394.4 = -1299.6 + 3 × -228.6] [ 175] 1 Substae kJ mol 1 ∆g /kJ mol f Quik questio SO (g) 371.1 3 Calculate the Gibbs free energy change for the following reactions. Values for H SO 2 (l) 690.0 4 can be found in section 12 of the Data booklet; additional data is listed in ∆G f NH Cl(s) 202.9 table 4. 4 CaCO (s) 1129.1 a) SO CaO(s) 604.0 (g) + H 3 3 b) O(l) → H 2 2NH SO 2 (l) 4 Cl(s) + CaO(s) → CaCl 4 CaCl (s) 748.1 ) 2 NH C H 2 (g) (s) + H 2 (g) + H 4 O(l) → C 2 H 2 O(l) + 2NH 2 (g) 3 OH(l) 5 16.5 3 T able 4 ∆G values not found in f section 12 of the Data booklet calulati the gibbs fee ee hae of a eatio fom ethal a eto ata To determine need to standard calculate spontaneity the conditions formation 370 the calculate Gibbs (298 of free K and of reactants and/or the enthalpy and a reaction energy 100 kPa). products entropy from change If are ∆G for the = the Gibbs unknown, changes for the ∆H reaction free we T∆S energies need reaction. , we under rst of to 15 . 2 Worked example: calculating ∆G Standard 2C enthalpy H 2 (g) H + O (g) combustion 4CO (g) + → 2H on 3O the (g) → 6H O(l) are from ∆H given H 2 (g) = -3120 ∆H = -572 2CO (g) + 2H ΔH = -1411 kJ kJ information, change in O(l) C 6 H 2 (g) kJ 2 b) calculate enthalpy, ∆H , for The an the sign + H 4 for increase number reaction: → the of the in stating disorder moles or c) for the is gas entropy evident is as increases positive: the from 1 to 2 (g) a reason, whether the sign ∆S = [∆S (C 298 of [∆S above reaction would be H 2 (C 298 ∆S of in reaction. 298 Predict, change 2 c) b) - T∆S below: ∆H in C S p O n T A n E I T y 2 2 above standard following + O(l) 2 Based (g) 2 reactions A n d 2 4 the → 2 2 a) (g) of 2 2 C 7O 6 (g) 2H + change E n T r O p y )(g) + ∆S 2 (H 298 4 H )(g)] 2 )(g)] 6 positive = [220 = 120 + 131] [230] negative. Calculate the standard entropy change for J K 1 the reaction. d) ∆G = ∆H T∆S = +137 120 _ d) Determine the value of ∆G for the reaction ( 298 × ) 1000 at 298 K. = e) Determine the temperature at which In reaction will occur +101 this calculation, Nov 2009 by 1000. Gibbs is Solution Rearrange the three combustion reactions the standard The rst change in The free joules entropy positive energy examining the to value kilojoules is value indicates This for positive the that can by the be value dividing change in reaction predicted for change to in nd from non-spontaneous. by a) the spontaneously. converted IB, kJ this enthalpy (endothermic) and the low enthalpy. temperature. equation will occur in the same e) direction but only half of the stoichiometry To determine reaction needed so halve the enthalpy the temperature at which this is will occur spontaneously, we make value: the assumption that the value for Gibbs free 1 C H 2 (g) + 3 6 O (g) → 2CO 2 2 (g) + 3H 2 O(l) energy 2 = ∆H -1560 second equation needs to be reversed zero and solve for T. kJ ∆G The is = ∆H T∆S 0 = ∆H T∆S T = and halved: 137 __ ∆H _ 1 H O(l) → H 2 The (g) + 2 third O equation (g) ∆H = +286 needs to be (g) + 2H 2 O(l) → C 2 H 2 standard ∆H C H 2 = these -1560 (g) 6 of enthalpy (g) + → C 1142 K 10 reaction becomes temperatures (g) greater spontaneous than 1142 at K. 2 = +1411 determines kJ the change: + H 2 3O 4 equations × reversed: ∆H Summation = 3 120 ∆S The 2CO = kJ 2 2 286 (g) 4 + + 1411 H (g) = ∆H + = 137 kJ +137 kJ 2 371 15 E N E R G E T I C S A N D T H E R M O C H E M I S T R Y ( A H L ) gibbs fee ee a hemial equilibium We and have established pressure reversible the free As towards the From is Gibbs which 5 we time reverse reaches a can the As is in between constant ∆G the as Gibbs that a the as the its reaction the will same Gibbs be In way as reactants at to the (A) the the Gibbs the change detail in products moves point energy point of of a topic (B). reaction. G(reactants) ∆G r ygrene eerf sbbiG A G(products) B equilibrium 0 mol reactant 1 mol product Figure 5 How the Gibbs free energy changes as the reaction proceeds 372 The again 17. 0 mol product free equilibrium reaction 1 mol reactant 5). during energy forward of forward increases, free (for (gure Gibbs the non-spontaneous the during in At a equilibrium, equilibrium free then when reaction or region temperature time reaches proceeds and energy examined it the Gibbs continues, and the decreases. this becomes free constant reactions) minimum reaction of and reaction spontaneous the ratio energy the at From where minimum minimum. a 0. increases free reached place < point non-reversible reaches then to products see forward minimum equilibrium has energy reaction relationship the taking when changing of towards favoured. free is (for system decreases reaction the the gure energy energy amount reactions) equilibrium reactions commences completion reversible that spontaneous reaction Gibbs alters. are and The the Q U E S T I O n S Questios 1 The be in lattice enthalpy calculated gure from of magnesium the chloride Born–Haber cycle II) can The shown of ionic charge of lithium is less than that calcium. 6. A. I only B. II C. I D. Neither IB, May 2+ Mg III = +738 + 1451 (g) + 2e kJ + 2Cl(g) IV 2+ Mg(g) + 2Cl(g) Mg(g) + Cl Mg (g) + 2Cl (g) only and II I nor II [1] 2004 II H 3 Which reaction occurs with the largest increase (g) 2 ∆H I = +148 (MgCl lat kJ in entropy? A. Pb(NO ) 2 ) 3 Mg(s) + (s) + 2KI(s) → PbI 2 (s) + 2KNO 2 (s) 3 (g) Cl 2 B. CaCO C. 3H D. H (s) → CaO(s) + CO 3 V = -642 (g) 2 kJ (g) + N 2 MgCl (g) → 2NH 2 (g) 3 (s) 2 (g) + I 2 IB, Figure 6 a) Identify by I the and V enthalpy in the changes Use the [2] energies given in above and further data 4 The booklet lattice c) The calculate of theoretically lattice is to enthalpy enthalpy +2326 kJ. a from value magnesium calculated of and Explain the for value theoretically for [4] value the The the is reaction correct spontaneity of this reaction at temperatures? be spontaneous at all B. It will be spontaneous at high but not at low temperatures. temperatures temperatures. between and It will be spontaneous at low temperatures the value. experimental certain will not at high temperatures. [2] D. d) a statement It but experimental for A. chloride difference calculated values Which the chloride. magnesium ∆S positive. the C. the 2004 ∆H both different Data [1] the about cycle 2HI(g) labelled cycle. ionization → 2 May are b) (g) lattice enthalpy It will not be spontaneous at any of temperature. magnesium oxide is given in section 18 IB, the Data oxide booklet. has a magnesium IB, November Explain higher why lattice [1] of May 2004 magnesium enthalpy than chloride. [2] 5 The following temperatures 2010 CaCO (s) → reaction above CaO(s) 850° + 3 Which 2 The lattice enthalpy values for lithium calcium uoride are shown ∆H = +1022 kJ only at C. CO (g) 2 combination 1000 is correct for this reaction °C? below. 1 LiF(s) spontaneous uoride at and is ΔG ΔH ΔS A. - - - B. + + + + + - - mol 1 CaF (s) ∆H = +2602 kJ mol 2 Which of explain less the why than following the that for value statements for calcium lithium help(s) uoride to is C. uoride? D. + [1] I) The ionic radius of lithium is less than that IB, of May 2007 calcium. 373 15 E N E R G E T I C S 6 Explain which can in terms both 7 May of ΔH sometimes sometimes IB, A N D be Dene the formation spontaneous question, Data use reaction are for positive and of undergoes enthalpy your state nitric relevant booklet standard illustrate including formation Propyne the term and equation, C a values 2004 this as why [4] from b) , ΔS ( A H L ) not. information the ΔG and Throughout a) T H E R M O C H E M I S T R Y change answer symbols, of with an for acid. [4] complete combustion follows: H 3 (g) + 4O 4 (g) → 3CO 2 Calculate the given following the (g) + 2H 2 enthalpy O(l) 2 change additional of this reaction, values: [4] 1 ΔH of CO f (g) = -394 kJ mol 2 1 ΔH of H f c) Predict ΔS for and the negative, IB, 374 May O(l) = -286 kJ mol 2 explain close 2005 whether reaction to in zero, part or the (b) value would positive. of be [3] 16 C H E M I CA L K I N ET I C S (AHL) Introduction In this topic we mathematical rate can of a only many explore equations chemical be cases limited various that reaction. determined are the relate Rate the the and slowest a in step reaction. reaction equations empirically by the to reaction in of terms and the A detailed mechanism and of understanding allows optimize yield, the reaction environmental chemists reaction time, of to the control conditions product cost impact. 16.1 Ra rn an racn can Understandings Applications and skills ➔ Reactions may occur by more than one step ➔ Deduction of the rate eq u a tion from and the slowest step determines the rate of experimental data and s olvi ng problems reaction (rate determining step/RDS). involving ➔ the rate equ a tion. The molecularity of an elementary step is the ➔ Sketching, identifying, a nd a na lys ing number of reactant par ticles taking par t in graphical representations for zero, first , and that step. second ➔ order reactions . The order of a reaction can be either integer or ➔ Evalu ation of proposed rea ction mecha nis ms fractional in nature. The order of a reaction can to be consistent with kinetic and describe, with respect to a reactant, the number stoichiometric data. of par ticles taking par t in the rate-determining step. ➔ experimentally. ➔ ➔ Nature of science Rate equations can only be determined ➔ Principle of Occam’s razor – newer theories The value of the rate constant (k) is aected by need to remain as simple as possible while temperature and its units are determined from maximizing explanatory power. The low the overall order of the reaction. probability of three-molecule collisions means Catalysts alter a reaction mechanism, stepwise reaction mechanisms are more likely. introducing a step with lower activation energy. 375 16 C H E M I C A L K I N E T I C S ( A H L ) Rate equation In topic 6 we differential example, consider xA + yB where x, y, The introduced expression rate → q, = Ra quan m rate = k[A] rate p is the d[A] [A] rate ∝ [B] rate ∝ [A][B] rate = k[A][B] equation in terms as of the mathematical concentration. For 1 _ = d[C] = q on 1 _ _ + dt depends coefcients follows: _ y reaction as - dt ∝ rate rate d[B] 1 _ _ rate a stoichiometry expressed = a of pD - of idea expresses reaction: are equation x The + and 1 _ rate the qC the that the d[D] _ + p dt dt concentrations of the reactants: n [B] where: k = rate constant This [A] = concentration of reactant A is the rate general, can account the equation be (sometimes expressed exponents as m shown and n, in the called the the box orders rate to the with law) left, respect and, in taking to into each [B] = concentration of reactant B reactant, which convey how sensitive the rate of reaction is to changes m = exponent in rate equation in the concentrations of A and B. described as the rr The overall and n order of the reaction is then dened as the sum of the m w rc racan A exponents: n = exponent in rate equation overall reaction order = m + n described as the rr w rc racan B Rate equations orders Let’s can take the (g) NO N: can only + be only be determined deduced following CO(g) → The and n) cann be worked carbon because the reaction: NO(g) + CO 2 The orders (for example, m experimentally empirically. (g) 2 rate equation for the reaction of nitrogen dioxide, NO (g), with 2 monoxide, out from the stoichiometry CO(g), has been found experimentally to be: 2 rate = k[NO ] 2 coecients (for example, x m Hence the rate equation rate = k[A] n 2 [B] corresponds to rate = k[NO and y) of a par ticular reaction. ] . 2 This means that m = 2 and n = 0; that is, the order with respect to NO (g) 2 is m two and of and n, are which One a order not are However, implies the with deduced respect from to the CO(g) is zero. stoichiometry Notice how coefcients the x orders, and y, both 1. the overall order second-order method of of the reaction, given equation is by m + n = 2 + 0 = 2, reaction. deducing the rate to use the method of Typically orders with respect initial rates, the principle of which we introduced in topic 6. The to reactants are either two, one value of the rate constant, k, is affected by temperature and its units are or zero order, but orders can determined from the overall order of the reaction. in fact be fractional or even negative! In this book , only the Catalysts reactions with whole-number As discussed in topic 6, a catalyst is a substance that increases the rate orders will be discussed. of a chemical catalyst reaction, provides activation an energy, is not alternative E (gure a 376 but 10 consumed pathway in for sub-topic in the the reaction reaction 6.1). itself. and A lowers the 16 . 1 R A t e e x p R e s s i o N A N d R e A C t i o N m e C h A N i s m The contact process Uful rurc The rst catalyst used in industry was for the production of sulfuric acid. The American Chemical Council (ACC) ● In this process, called the contact process, elemental sulfur, S(s), is has developed C AB, a Ccal Ac rst reacted with oxygen gas, O (g), to form sulfur dioxide gas, SO 2 S(s) + O (g) → SO 2 (g): Barr, which is an economic 2 indicator that predicts peaks and (g) 2 troughs in the overall economy in the ● Sulfur dioxide then reacts with oxygen gas to produce sulfur trioxide, USA and highlights potential trends in SO (g): 3 other industries. The barometer ser ves V O 2 2SO (g) + O 2 (s) 5 (g) 2SO 2 as a pivotal tool in predicting broader (g) 3 economic health in the USA . CAB is The catalyst used is vanadium(V) oxide, O V 2 heterogeneous (s), which is a 5 a leading index of overall industrial catalyst. production and has a number of dierent indicators including chemical Can you see why this is a heterogeneous catalyst? company stock data, etc. (For example, have a look at the video on their website ● Sulfur trioxide is next absorbed into concentrated sulfuric acid, (http://www.americanchemistry.com/ H SO 2 (l). This produces oleum, H 4 S 2 O 2 (l). Oleum reacts with water to 7 Jobs/CAB) for a greater insight into the produce aqueous sulfuric acid, H SO 2 (aq). 4 impor tance of chemistry to the global SO (g) + H 3 H The 2 S 2 O 2 for a war, O(l) S → the 2 2H oleum trioxide For the (l) economy). CAB was the rst of its kind 7 SO formed is rst too mirrored during the decreased, reversed developed globally. (aq) 4 water closely acid sulfuric is and example trend O 2 sulfuric of H 2 production of production H why time. production + sulfur acid → 2 reason long world (l) (l) 4 7 between Sulfuric SO and a there that the country’s rst but is direct reaction vigourous. and economic second immediately was a world after dramatic health wars the the second increase in the acid. What are some current indicators of a country’s economic health? Molecularity and rate-determining step (slow step) of a reaction The sequence reactants to of the reaction kinetics. This reaction mechanism or sequence elementary step is classied molecules or steps formation of any its atoms events unimolecular: ● bimolecular: is or single two is the a step is elementary as molecule molecules or the reaction process. in as In the an of turn, the in In elementary an a step elementary number of reaction: elementary involved chemical mechanism. elementary in from aspect an represents involved atoms pathway important described which reactants reaction very termed molecularity, involved ● outlining products individual reaction by of collision step in an elementarystep ● termolecular: elementary Each molecules or atoms involved in collision in an step. elementary energy, three step has its own rate constant, k, and its own activation E a Let’s return NO (g) 2 to + the reaction CO(g) → of NO(g) nitrogen + CO dioxide with carbon monoxide: (g) 2 377 16 C H E M I C A L K I N E T I C S ( A H L ) The Rar f trlcular Racn reaction events elementary Termolecular reactions are step mechanism leading from representing reactants to the products sequence is of actually molecular composed of two steps: 1: NO (g) + (g) 2 very rare as it is very unlikely → NO(g) + (g) 2 step 1 is 3 bimolecular that three par ticles would (g) + CO(g) → (g) 3 collide simultaneously with + CO 2 (g) 2 each other in the correct orientation. For example, have overall reaction: NO (g) + CO(g) → NO(g) + CO 2 (g) 2 you ever seen three snooker In this mechanism, NO (g) is described as a reaction intermediate , as 3 balls colliding at the same it is formed in step 1 and then is consumed subsequently in step 2. time when watching a World Championship snooker match Therefore, on TV? step reactions determines may the rate-determining occur rate of step the by more than reaction. one The step slow and step is the slow termed the (RDS). Deduction of a rate equation from a proposed reaction mechanism In order 1 to deduce Decide 2 on equal to From (1) the which the rate deduce rate step of equation is this the the RDS. slow rate from a The proposed rate of reaction the overall mechanism: reaction is step. equation for the RDS. Anal For temperatures the reaction less than 498 K, the experimental rate equation for Passengers having arrived just discussed has been found to be: through passpor t control at 2 rate an airpor t have to follow a = k[NO ] 2 sequence involving two stages ● In effect, the reaction mechanism is essentially a hypothesis of the in order to exit the airpor t. The sequence of events that has led to the overall reaction converting the rst step involves collecting reactants into products. There might, therefore, be a number of possible luggage at the baggage carousel reaction mechanisms that equate with the experimental rate equation. and the second step involves For the example just discussed, here is a proposed reaction mechanism. exiting the arrivals area. The step that determines the rate at Consider step 1 as the slow step (so is which the passengers can be the RDS) and step 2 as the fast step: k __ → 1 step 1: NO (g) + (g) 2 on their way out of the airpor t NO(g) + (g) 2 (slow) 3 k is determined by the rate at (g) + CO(g) 2 __ → (g) 3 + CO 2 (g) (fast) 2 which their luggage arrives on overall reaction: NO (g) + CO(g) → NO(g) + CO 2 (g) 2 the carousel. This is the ra- rnn of the two- Hence: step sequence of events and rate of overall is analogous to the idea of the reaction = rate of the slow step (in this case step 1) 2 = k[NO ] 2 lw in chemical kinetics. where k represents mechanism If step 1 is the slow step, the ● activation energy for this step, In is the rate consistent contrast, at rate equation rate = constant with the temperatures for the for the overall experimentally greater reaction just than 498 discussed reaction. This determined K, has the rate proposed equation. experimental been found to be: (1) will be large. If step 2 is E a k[NO ][CO] 2 (2) will be the fast step, then E a small . A proposed reaction mechanism here might be a single-step bimolecular process: single step: NO (g) 2 378 + CO(g) k _ _ → NO(g) + CO (g) 2 (slow) 16 . 1 R A t e e x p R e s s i o N A N d R e A C t i o N m e C h A N i s m Hence: rate of step) overall = k[NO reaction = rate of the slow step (in this case the single ][CO] 2 This proposed rate equation. mechanism is consistent with the experimentally determined toK A reaction mechanism can be suppor ted by indirect Cancer research, for example, is all about identifying evidence. What is the role of empirical (experimental) mechanisms for carcinogens as well as for cancer- evidence in the formulation of scientic theories? Can killing agents and inhibitors. we ever be cer tain in science? Worked example: deduction of the rate equation from experimental data and solving problems involving the rate equation 1 Consider the balanced stoichiometry equation, coefcients of and the note reactants the 5 Deduce the overall products. For + yB reaction → qC + y, q, and p are the stoichiometry Determine these coefcients 2 Write the order the rate rate equation, where m and (1, values deduce down of the reaction: = m + constant, k, n for each pD experiment x, order example, 6 xA overall and the to 2, 3, etc.). give the appropriate Find mean units the mean value for of k of and k n Example 1 represent the orders with respect to each Consider reactant: m rate 3 From = k[A] the ratios (as A(g) data for deduce each each of of Based the the 1) (rate 2) (rate _ , 3) for the , 2) (rate 3) , (rate pairs of concentration rate does data not Deduce → C(g) orders and the from D(g) with overall i niti al one where change experiment respect the each orders ratio to m obtained and n. Use logs may be tools with helpful Deduce the rate equation. in one of them [B()] inal ra another. in of step some respect here, each order. the 3, 3 / l for 3 / l 1 deduce fundamental 2 mathematical to reaction 3 From d at a etc. / l 4 rate 4) [A()] going + expe r i me nta l the reactant _ ● Look on B(g) appropriate): _ (rate + below: following ● (rate reac ti o n: n [B] given experiments, the to indices Experiment 1 1.00 × 10 Experiment 2 2.00 × 10 Experiment 3 2.00 × 10 2 1.00 × 10 3 4.20 × 10 and example: 0 x log = 2 1 (XY) = log X + log 2 1.00 × 10 3 8.40 × 10 Y X _ log ( Y ) = log X log Y 2 2 2.00 × 10 2 3.36 × 10 p log X = plog X 379 16 C H E M I C A L ● Calculate for the K I N E T I C S the value reaction of from ( A H L ) the rate constant, experiment 3 and k, Then substituting the data state from experiment 2 (3.36 × 3 10 mol 3: 1 dm s ) _____ its k units. = 2 (2.00 ● Determine the rate of the reaction = 3.00 × × 10 mol dm and 3 [B(g)] 4.20 × dm )(2.00 2 10 6 mol × 10 3 mol dm 1 dm The dm units were worked s out as follows: 3 1 mol dm s ____ units of k = 3 Solution In order to working from mol solve thi s method the to method q ue s ti on d e duce of i ni tia l we the ca n ra te u se dm × 2 t he units of k = 3 mol mol 6 3 dm × mol dm 1 dm s eq u at i on ra tes : The orders may also be deduced by inspection. By ● There are equation two so reactants the rate m rate = k[A] in the equation chemical is given keeping [B] constant in experiments 1 and 2 and by: doubling [A], the initial rate is seen to double. Hence the order with respect to A will be one. Likewise, by n [B] keeping [A] constant in experiments 2 and 3, [B] ● You to next use. data not have In in order which change down to to – just choose to decide one this one of the helps the this, reduce order.For k(0.010) for the ratios pairs doubles. However, this time the initial rate is seen of to increase by a factor of four, meaning that the does order with respect to B is two. This is a quick way of problem deducing the orders, but with more dicult numbers example: nding the orders by this method might be quite n tricky – following the working method using ratios (0.010) 0.00420 _ __ = = m rate look concentrations m rate 1 _ appropriate 2 will always allow you to nd t
