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Chemistry Course Companion for IB Diploma Programme
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Oxford Resources for IB Diploma Programme s s 2 0 2 3 E D I T I O N y n l y p o C n o i t a D avid Tarcy u a E Elisa Jimenez Grant O t i s r e v i n U d l v Gary Horner e CO M PA N I O N r o f x O Sergey Bylikin r y CO U R S E P C H E M I S T RY s s e r O n l y P y t i s y p r e C o v i n n U o i d t a r o u l f x a E v O Oxford Resources for IB Diploma Programme s s 2 0 2 3 E D I T I O N y n l y p o C n o i t a a D avid Tarcy u l v E Elisa Jimenez Grant O t i s r e v i n U d r o f x O Gary Horner e CO M PA N I O N Sergey Bylikin r y CO U R S E P C H E M I ST RY Commons; p69:Wirestock Creators/Shutterstock; p76:Sergey Bylikin; p90:DAVID PARKER / SCIENCE PHOTO LIBRARY; p94:Pi-Lens/Shutterstock; p96(l):KKStock/ Getty Images; p96(m): M arina Kryuchina/Shutterstock; p96(r):Nneirda/ Shutterstock; p100:MARTYN F . 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CHILLMAID / SCIENCE PHOTO LIBRARY; p338:SCIENCE PHOTO LIBRARY; p339:TED KINSMAN / SCIENCE PHOTO LIBRARY; p341(t):Andrea v Printed Obzerova / Alamy Stock Photo; p341(b):M ark Lorch/Shutterstock; p342:Reprinted in Italy by Rotolito SpA (adapted) with permission from J. support for IB teaching and IB curriculum and learning. The authors have the following acknowledgements and thanks: U I would colleagues for their support To Miljan. profes- permission from J. Chem. Society;p460:Richard Image Partnership Liu; p439:REUTERS / photos by Thomas D. Schneider Ltd Educ. 2012, 89, 5, 675–677 . Copyright (adapted) with 2012 Americ an Chemic al Liu; p468:sciencephotos / Alamy Stock Photo; p470(l):Heritage / Alamy Stock Photo; p470(r):The Granger Collection / Alamy Stock Photo; p477(l):MAXIMILIAN STOCK LTD / SCIENCE PHOTO LIBRARY;p477(r):ASHLEY COOPER / SCIENCE PHOTO LIBRARY; p481:nagelestock. expertise, Ilias Liakatas, Astha Acharya, and feedback: Nerissa Puntawe, com / Alamy Stock Photo; p484:ANDREW LAMBERT PHOTOGRAPHY / SCIENCE PHOTO LIBRARY; p497:Kenneth Brown / EyeEm; p499:dpa picture alliance / Alamy Stock Photo; p509:bluesnote/Shutterstock; p513:CHARLES D. WINTERS / SCIENCE Natalie Parker, Síle-C aitríona O'C allaghan. PHOTO LIBRARY; p520:ANDREW LAMBERT PHOTOGRAPHY / SCIENCE PHOTO LIBRARY;p526:S ankei Archive via Getty Images; p536:STEVE HORRELL / SCIENCE PHOTO LIBRARY; p538:CHARLES D. WINTERS / SCIENCE PHOTO LIBRARY; p543:SCI- Vicki Boyd, Chandan Bhosale, ENCE PHOTO LIBRARY; p553:ANDREW LAMBERT PHOTOGRAPHY / SCIENCE PHOTO Deepa u f x S athya and Photo;p431:Xinhua / Alamy Stock Photo; p434:Richard my t a r o K athryn Russell, Cendrella Kettaneh, Shutterstock; p426:tab1962/Getty Images; p427:R K awka / Alamy Stock like to thank the following members of the DP Science 2023 Research Panel for sharing their insights, S andy Hawkinson, Tina Reid/Getty Images; p388(tr): Dr Ajay Kumar WINTERS / SCIENCE PHOTO LIBRARY; p408:ilyankou/ Shutterstock; p452:SCIENCE PHOTO LIBRARY; p453:Reprinted c are and i In memory of David Tarcy University CHILLMAID / (https://alum.mit.edu/www/toms/images/boltzmann); p443(b):M ariyana M/ I dedic ate this book in d their loving memory. The Publisher would Dennis: Getty Images; p388(tl): John and stock;p390:CHARLES D. Alamy Stock Photo; p443(t): Boltzmann›s Tomb throughout my sister Susan for her unwavering friendship, my inspirational parents Myrtle and Elisa Jimenez Grant: 1249–1252 Copyright Singh/Shutterstock; p388(b):Valery Lisin/Shutterstock; p389:keerati/Shutter- o sional advice, 7, SCIENCE PHOTO LIBRARY; p352(br): photong/Shutterstock; p386:timandtim/ n To my friends and teaching c areer, 93, Press;p352(tr): chromatos/Shutterstock; p352(bl): MARTYN F . oers like to thank Dr Natalia K alashnikova for her support and suggestions. Gary Horner: 2016, Shutterstock; p351(b):Ron Kloberdanz/Shutterstock; p352(tl): Oxford in this textbook has been C high-quality guidance and fully aligns with current i n logo signies the content by the IB to ensure it Sergey Bylikin: Educ. 2016 Americ an Chemic al Society; p343:Stu Shaw/Shutterstock; p351(t):Julien_N/ The “In cooperation with IB” reviewed Chem. o Acknowledgements LIBRARY;p554:CHARLES D. Photolibrary, Inc. WINTERS/SCIENCE PHOTO LIBRARY; p555:D avid R. Frazier / Alamy Stock Photo; p557:SCIENCE PHOTO LIBRARY; p571:ANDREW LAMBERT PHOTOGRAPHY / SCIENCE PHOTO LIBRARY; p581:Judy Kennamer/ Shutterstock; p583:MARTYN F . CHILLMAID / SCIENCE PHOTO LIBRARY; p591:MARTYN The Publisher wishes to thank the International Bacc alaureate Organization for F . l permission to reproduce their intellectual property. use photographs and LIBRARY;p599:Dorling Kindersley/Getty Images; p608(l):© a authors would Corbis; p608(r): ggw/ Shutterstock; p619:REUTERS / Alamy Stock Photo; p632:Pictorial Press Ltd / Alamy Stock like to thank the following for permission to Photo;p637:ANDREW LAMBERT PHOTOGRAPHY / SCIENCE PHOTO LIBRARY . Artwork by Q2A Media, Aptara Inc., other copyright material: v O The publisher and CHILLMAID / SCIENCE PHOTO LIBRARY; p592:Colin Hawkins/Getty Images;p594(l):SCIENCE PHOTO LIBRARY; p594(r): SHEILA TERRY / SCIENCE PHOTO Cover:www.ickr/GettyImages. 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M arbles, Tech-Set Ltd, Trystan Mitchell, Russell Walker, Paul Gamble, Wearset Clive Clive Sergey Index by James Helling. / SCIENCE PHOTO LIBRARY; p7(r):stockcreations/Shutterstock; p12:Andrea Izzotti/Shutterstock; p13(t):Jan Halaska / Alamy Stock Photo; p13(b):KPG_Payless/ Shutterstock; p14:Alexey Kljatov/Shutterstock; p15:Wayne E astep/Getty Images; p16:© BIPM;p22:one-image photography / Alamy Stock Photo; p23:IBM RESEARCH / SCIENCE PHOTO LIBRARY; p26:Artiom Photo/Shutterstock; p28(t):PATRICK LANDMANN / SCIENCE PHOTO LIBRARY; p28(b):Science History Images / Alamy Stock Photo; p31:FRANCOIS GUILLOT/AFP via Getty Images; p35:Denis Belitsky/ Shutterstock; p36:Fotomaton / Alamy Stock Photo; p37(t):H.S. Photos / Alamy Stock Photo; p37(b):Zern Liew/Shutterstock; p63:CHARLES D. WINTERS / SCIENCE PHOTO LIBRARY; p66:Nneirda/Shutterstock; p67(l):Ashley Cooper/ Getty Images; p67(r):Ludovic C aritey / 500px/Getty Images; p68:Mysid/Wikimedia Although we have made every eort public ation this has not errors or omissions at Links to third Oxford to trace and contact been possible in all c ases. the earliest all copyright If notied, party websites are provided by Oxford in good faith and disclaims any responsibility for the materials contained referenced in this work. holders before the publisher will rectify any opportunity. for information only. in any third party website Contents Structure 1. Models of the particulate nature of matter 2 s s Structure 1.1 Structure 1.2 Structure 1.3 Structure 1.5 r Structure 2. Models of bonding and structure 94 Structure 2.2 y Structure 2.4 t i s 228 Structure 3.1 Structure 3.2 Tool 1: Reactivity 1. What drives chemic al reactions? Reactivity 1.2 Reactivity 1.4 n U Reactivity 1.3 Reactivity 2. How much, how fast and how far? i t a r o Reactivity 3. What are the mechanisms of chemical change? Reactivity 3.2 Reactivity 3.3 l Reactivity 3.4 460 536 u f x Reactivity 3.1 386 o Reactivity 2.2 d Reactivity 2.1 C Reactivity 1.1 o i n v Tool 3: p Tool 2: 308 a O Cross-topic exam-style questions 652 The inquiry process 655 v (authored by Maria Muñiz Valcárcel) The internal assessment (IA) E Index (authored by Maria Muñiz Valcárcel) Periodic Table Answers: y r e Tools for chemistry O Structure 3. Classic ation of matter n l Structure 2.3 y P Structure 2.1 Reactivity 2.3 e Structure 1.4 668 686 708 www.oxfordsecondary.com/ib-science-support iii Introduction The diploma programme (DP) chemistry course is aimed at students in the 16 The curriculum seeks to develop of the nature of science, chemistry and a conceptual understanding working knowledge of fundamental principles of practic al skills that c an be applied in familiar and unfamiliar As with all the components of the DP, this course fosters the IB learner prole attributes (see page viii) in the members of the school community. r denitely true. throughout the book and NOS features but they c annot This has led science throughout are encouraged to come up p Observations and experiments contain assumptions or unrealistic simplic ations, but lead to serendipitous results. U Measurements • all data are prone to error. It is important to takes observed These predictions may conrm a theory (within observable limitations) or may falsify it. • t a r o i o d know the limitations of your data. Evidence explanation that patterns and hypotheses and uses them to generate predictions. • to reduce its limitations. A theory is a broad Measurements c an be qualitative or quantitative, but and Theories n • and of science is to increase the complexity of the model, Sometimes the observations in experiments are unexpected the aim C i n • models as simplied explanations of their observations. Models oen o aspects: v into the following eleven to paradigm shis in history. Models Scientists construct NOS c an be organized be proved to be y • the programme. r e with further examples of your own as you work through O evidence, entire course. suggested n F alsic ation Hypotheses c an be proved false using other are specic to science. NOS is a central theme that is present across the You will nd l y • is concerned with methods, outcomes that t i s purposes and Science as a shared activity Scientists learn to be sceptic al about their observations and be fully supported by evidence. Patterns and trends u f x • Recognition of a pattern or trend part Patterns lead on a convention for clear forms an • Global impact of science Scientists are responsible to society for the consequences of their work, environmental, to a possible explanation. The E hypothesis is this provisional view and further veric ation. iv such as peer review of work before public ation or agreement communic ation. a v Hypotheses collaboration, of the scientist’s work whatever the science. • l O important Scientic activities are oen c arried out in they require their knowledge to it requires knowledge must and fairly. whether ethic al, economic or social. Scientic be shared with the public clearly y P Nature of science Nature of science (NOS) e contexts. s s to 19 age group. Syllabus structure Topics are organized into two main concepts: structure and reactivity. This is shown in the syllabus roadmap The skills in the study of chemistry are overarching experimental, skills that are integrated experimental work, into the course. inquiries and Chemistry is a practic al subject, technologic al, mathematic al and inquiry so these skills will be developed through investigations. Reactivity why simple to more complex forms chemic al reactions occur l Structure determines reactivity, which in turn transforms structure What enthalpy changes chemic al Structure 1.2 — The nuclear atom of matter O particulate nature Reactivity 1.1 — Measuring drives Reactivity 1.2 — Energy cycles in reactions? reactions Structure 1.3 — Electron Reactivity 1.3 — Energy from fuels particles by mass: The mole Reactivity 2. How much, how fast and U Structure 2.2 — The covalent structure Structure 2.3 — The metallic d Structure 2.4 — From models to i matter extent of chemic al change Reactivity 3. Reactivity 3.1 — Proton transfer What reactions Classic ation of elements are the mechanisms Reactivity 3.2 — Electron transfer u of chemic al reactions change? Structure 3.2 — Functional groups: Reactivity 3.3 — Electron sharing Classic ation of organic reactions a compounds v E roadmap above, Reactivity 3.4 — Electron-pair sharing reactions For example, “Structure determines reactivity, You are therefore encouraged help Reactivity 2.3 — How far? The table: Chemistry concepts are thoroughly interlinked. new and of chemic al change Reactivity 2.2 — How fast? The Structure 3.1 — The periodic l O f x t a Classic ation of r o materials amount rate of chemic al change o model Structure 3. Reactivity 2.1 — How much? The how far? n model spontaneity (Additional higher o bonding and Reactivity 1.4 — Entropy and level) C i n Structure 2.1 — The ionic model Models of v Structure 1.5 — Ideal gases y Structure 1.4 — Counting p r e congurations Structure 2. n Reactivity 1. the particulate nature of matter y Structure 1.1 — Introduction to Models of the t i s Structure 1. y P Reactivity refers to how and r Structure Structure refers to the nature of matter e Skills in the study of chemistry from s s below. as shown in the which in turn transforms structure”. to continuously reect on the connections between prior knowledge as you progress through the course. you explore those connections. identify and In assessment apply the links between dierent examples of DP-style exam questions that tasks, topics. Linking questions will you will be expected to On page 652, link several dierent there are three topics in the course. v How to use this book The aim of this development book and Feature boxes by through and for conceptual opportunities to understanding, aid in skills cement knowledge and practice. sections signposting opportunities develop throughout content practice. relating This is an the to book are particular overview of designed to support these e aims, to s s understanding is provide ideas and concepts, as well as these features: r boxes to studying a topic, condently or fully, them you might but Hence, you n U These illustrate NOS using issues from both modern science and science o d history, and show how the ways of doing science have evolved over the centuries. There is a detailed description of what is meant by NOS and the i r o dierent aspects of NOS on the previous page. The headings of NOS feature t a boxes show which of the eleven aspects they highlight. u f x Theory of knowledge an features issues. E LHA of in this IB how book Diploma we pose Parts of the book have a coloured question. This indic ates that Chemistry Higher Level. vi the a TOK part understanding v O The important and l is course. arrive at questions our for dierent may p other parts of the course. Nature of science thinking a It focuses on knowledge you that of critic al the highlight world. these bar on the edge of the page or next to a the material is for students studying at DP AHL means “additional higher level”. also prior o the connections between C there and where explored y r e v i n Linking questions within each topic highlight This is come back to them when you revise your Linking questions discussed book consider these as you work understanding. content parts you will be able to answer you should the of you the When you be able to answer these questions by studying that topic, with increasing depth. through the topic and not other in direct a a concept O start you thinking. t i s E ach topic begins with a guiding question to get will n y margin further direct need, dierent about or context. you knowledge will l These Guiding questions or way or in They to a skill give to think something. y P Developing conceptual understanding Developing skills Chemistry skills Approaches to learning ATL ATL scientists features have communic ation, about and how the of skills, develop and your how ATL self-management, social to examples experimental or inquiry skills, of experiments and research, prompt own These contain ways to develop famous skills you to c an be used think your mathematic al, especially through practic al work. Some of these as springboards for your Internal e thinking give demonstrated Assessment. strategies. r section Internal more book on how to Assessment. information. Links for use Flick in the are DP the to full of reference Chemistry, inquiry this margin process section material details as in your throughout on the working the book and the you rest of work the towards it as through book for too. y p o you these aware to of experiments answer questions the questions to prepare possibilities and to give you further practice at using your chemistry knowledge and n you These are designed o make Use i to requires data. day-to-day to and chemistry, check your own understanding and for your for Activity that for data to allow you to progress. are this. based These They give you opportunity acquisition your IA. to an apply chemistry knowledge oen in a and skills, practic al way. u l O f x t a for direct C You U designed analysis of d interpretation also and assessment r o are nal through will preferably Data-based questions your subject r e review these examples c arefully, aer attempting the question yourself. of the mathematic al modelling examples of how to answer questions or how to complete c alculations. the of essential and v i n These are step-by-step on the a ss e ss m e n t Practice questions Worked examples Part all analysis study Practicing should for data internal n guidance the and O your as of required p r o c e ss t i s well tools inquiry y experimental the y three c h e m i s t r y, l These for P To o l s s s These Use these questions at v a End-of-topic questions the end of each topic to draw together concepts from that topic and to practise answering exam-style questions. E vii Course book denition The IB Learner Prole The IB Diploma Programme course books are resource The aim of all IB programmes to develop internationally materials designed minded people who work to create a better and to support students throughout more peaceful world. in a particular subject. They will help students gain an develop understanding of what described is expected an IB Diploma Programme subject from the study of this person through ten learner attributes, as They develop e in a way that below. while presenting Inquirers: content The aim of the programme is to s s their two-year Diploma Programme course of study their natural curiosity. They illustrates the purpose and aims acquire the skills necessary to conduct inquiry and of the IB. They reect the philosophy and approach of encourage a deep snow independence in learning. They understanding of each this love of learning will be sustained throughout their lives. They explore concepts, l Knowledgeable: ideas and The books mirror the IB philosophy of viewing the have loc al and in terms of a whole-course approach; doing, global signic ance. In so they acquire in-depth knowledge and understanding across a broad the IB learner prole and the IB Diploma disciplines. Programme core requirements, the extended essay, and theory of knowledge, creativity, activity, service Thinkers: and encouraged students of the IB are required to draw conclusions from a variety v and indeed, Communic ators: Suggestions for additional and further ideas and In addition, more than one language and communic ation. approach ethic al express creatively in in a variety of modes of They work eectively and willingly in collaboration with others. the course companions provide advice U and suggestions for research are provided. and information condently and C how to extend i n reading are given in each book and They understand o of resources. to make reasoned, p decisions. materials and, develop range of creatively to recognize and complex problems, y skills critic ally and E ach book c an be used in conjunction with other balanced They exercise initiative in applying thinking r e (CAS). and O t i s the use of a wide range of resources, international mindedness, n curriculum y issues that guidance on the specic course assessment Principled: on ac ademic honesty protocol. They are distinctive and They act with integrity and n requirements and a strong sense of fairness, authoritative without being dignity of the individual, o i t a r o d prescriptive. justice and honesty, with respect for the groups and communities. They take responsibility for their own action and the consequences that Open-minded: accompany them. They understand and appreciate their IB mission statement The International Bacc alaureate aims to develop knowledgeable and who help c aring young people to create a better and u f x inquiring, more peaceful world through intercultural understanding and l the organization works with schools, governments and international organizations to challenging programmes of international educ ation and v develop a O To this end, respect. rigorous assessment. E to become active, learners who understand dierences, viii to the perspectives, individuals and seeking and lifelong that other people, with their c an also be right. values and communities. are willing to grow from C aring: and are open traditions of other They are accustomed to evaluating a range of points of view, and the experience. They show empathy, towards the needs and compassion and feelings of others. a personal commitment environment. compassionate and personal histories, respect They have to service, and to act to make a positive dierence to the lives of others and to the These programmes encourage students across the world own cultures and y by making connections to wider issues and providing opportunities for critic al thinking. P actively enjoy learning and subject r research and the IB and Risk-takers: the independence of spirit ideas and ‘Formal’ means that They approach unfamiliar situations and uncertainty with courage and strategies. forethought, and accepted have separating the resources that to explore new roles, c ategories (e.g. They are brave and articulate in and use one of the several books, internet-based This usually involves you use into dierent magazines, newspaper resources, and works of art) providing full information as to how a reader or They understand the importance of viewer of your work c an nd intellectual, physic al and the same information. A emotional ballance to achieve Essay. others. What constitutes malpractice? They give thoughtful consideration to their M alpractice is behaviour that own learning and experience. their learning and limitations in order M alpractice includes personal development. as the representation of the ideas or work of another person as your own. The following are some of the ways to avoid plagiarism: integrity ● It words and the owners of information when ● that information is used in your work. Aer all, passages that owners are quoted one’s y r e appropriately credit ideas of another person to support arguments must be acknowledged is of vital importance to acknowledge and O t i s A note on ac ademic n is dened y Plagiarism l plagiarism and collusion. y their strengths and gaining an unfair advantage in one or more assessment component. to support or may result in, P understand results in, They are able to assess you or any student and r Reective: e bibliography is compulsory in the Extended personal wellbeing for themselves and s s articles, defending their beliefs. Balanced: you should forms of presentation. verbatim must be enclosed within quotation marks and p of ideas (intellectual property) have property rights. acknowledged it must be based original ideas with the work of Therefore, ● all assignments, i n others fully acknowledged. email messages, must be treated written or oral, completed for assessment must use your or referred to, expression. ● whether in the form of direct quotation such sources must U or paraphrase, n not i o d t a r o you acknowledge that ● data, maps, illustrations, graphs, audio-visual and be acknowledged if they are your own work when referring to works of art, whether music, lm dance, theatre arts or visual arts creative use of a part original artist you have used the any other electronic media the sources of all photographs, similar material must How do I acknowledge the work of others? and in the same way as books and computer programs, be appropriately acknowledged. The way that journals Where sources are used C own language and o on your individual and v To have an authentic piece of work, must and where the of a work takes place, the be acknowledged. ideas of other people is through the use of footnotes and bibliographies. u f x Footnotes (placed at (placed of a document) are to be provided at the end the bottom of a page) or endnotes Collusion is dened as supporting malpractice by another student. This includes: ● allowing your work to be copied O l assessment or submitted for by another student when you quote or paraphrase from another document a ● or closely summarize the information provided in another document. You do not need to provide a duplic ating work for dierent assessment components and/or diploma requirements. v footnote for information that is part of a ‘body of knowledge’. E footnoted That is, Other forms of malpractice include any action that denitions do not you an unfair advantage or aects the results of another as they are part Bibliographies should of the assumed knowledge. student. Examples include, taking unauthorized material into an examination room, misconduct during include a formal list of the an examination and resources that gives need to be you used falsifying a CAS record. in your work. ix Experience the technology with for DP education Oxford’s digital Science but have you tried our digital course on Kerboodle? designed for the next generation of Oxford’s DP Science oer brings together the IB curriculum digital components for the best blended the digital to oΎine Course Book C o v i n n U o i d t a r o u l f x a E v O Encourage of motivation engaging interactive exercises, x y to and p access resources mobile- access r e student with onscreen learning experience. O anywhere optimized beyond. Use both print teaching and t i s Learn enabling success in DP and n future-facing functionality, and y and l teachers, content activities, with a variety including vocabulary animations, and videos y in cooperation with the IB and students and P Developed e resources, r You’re already using our print of s s oÏer future Embrace independent progression provides can a with personalized self-assign real-time learning adaptive journey auto-marked results and and are Deepen technology so that students assessments, oΊered and next understanding extension repetition, get up steps where questions intervals to with support, on and students are completed encourage intervention spaced asked topics knowledge at follow- regular retention s s e r O n l y P y t i s y p r e C o v i n n U o i d t a r o data reporting collected responsive u l f x individual to with teaching and rich support class at an level a E v O Recommend For Enhance more Oxford’s digital information oΊer and to for sign DP up Science for free to your trial school access, go today! to: www.oxfordsecondary.com/ib/dpscience xi s s e r O n l y P y t i s y p o the particulate C m a tt e r n of v of U nature i n Models 1 r e Structure o i d t a r o u l f x a E v O Introduction to the Structure 1.1 particulate nature of matter s s r e How c an we model the particulate nature of matter? atoms came from experimental evidence that could only be predictive power and could not account for the great variety of chemical compounds, so it was eventually abandoned. The systematic study of chemical changes led to the proportions suggested the existence of atoms. It was this experimentation which led to the modern atomic theory which Elements c annot be broken to down into c an be ratio, in more than which separated are by dierent elements a xed one not o so contain xed together of ratio. element or i no atoms chemic ally physic al Cornell University c aptured the What do models show us that The physic al gases) and kinetic properties changes of Structure 1.1.3 — average energy (E kinetic molecular of matter theory is a model (solids, liquids, state. Temperature (in K) is a measure of ) of particles. k compound t a in bonded of r o Mixtures consist d chemic ally explain and simpler substances. Compounds scientists at picture of atoms to date. Structure 1.1.2 — are the primary constituents chemic ally n matter, In 2021, C i n of 1 detailed microscope images c annot? U Structure 1.1.1 — Figure most o v way of processing knowledge through observation and p elements could only combine with one another in xed y broken down into simpler substances. The fact that these r e discovery of many chemical elements that could not be O t i s of earth, air , re, and water . However , this theory lacked n y Early classical theory suggested that all matter was composed Understandings l explained if matter were made of particles. y P The universally accepted idea that all matter is composed of bonded and methods. u f x The composition of matter (Structure 1.1.1) and a re touch to of matter In contrast, are considered study of matt e r see of i t. we in energy is energy are a and T he its c o mp o s i t i o n . c o n s u me ma tt e r. u n de rs t a n di n g shown as of ma tte r, fo r m s c annot our and produce up m a ny we ex pa n d M atter th e E th o u g h is made a We l and energy C h e mi s t r y v O M atter Air is a u n i ve rs e of i t, of made and M a tte r s u r ro u n ds fo r m is ma tte r it its ma tte r of us, th a t m a tt e r we and pro pe r t i e s . is e ve r y w he re. and we c an kn ow is c he mi s tr y The see t h e re, seeks characteristics gure 2. anything closely property of that exists but associated matter, such does with as not each the have other, ability to these and properties. energy is oen perform work or heat. 3 Structure 1 Models of the particulate nature of matter Although Chemic al reactions are nuclear in mass and energy c an be converted into one another (for example, in introduced reactors or inside stars), chemistry studies only those transformations of Reactivity 1.1. matter where both products have the from form to one mass same and energy mass another as rather are conserved. In starting than materials, created or chemic al reactions, the and the energy is transformed destroyed. s s made up of e particles – atoms, r molecules, or ions space y p r e C o v i n U has a mass n o i d u l a E v O f x t a r o The 2 famous are Einstein reactions 8 (3.00 × 10 changes This is eect without is E = However, relatively mc the small , shows energy while that mass (m) released the or and energy absorbed in speed of light (c) is very large –1 m s ). As a result, the loss or gain in mass c aused by chemic al negligible. example What equation, interconvertible. chemic al the volume in Thinking skills ATL (E) occupies a MATTER motion n in constant O particles are 4 l The characteristics of matter of a demonstrates certain factor compromising other examples inchemistry? the of the is importance minor, nal it c an of approximation in science: if oen be ignored in c alculations result. negligible eects have you encountered y P y Figure 2 t i s u Structure 1.1 Introduction to the particulate nature of matter The atomic theory The law always of of conservation combine elements. could form not water, consumed of be c arbon broken the monoxide, the chemic ally. of that water with with observation that certain substances led to elements showed mass react and that down experiments would and proportions theorized equalled c arbon mass of of idea Hydrogen the mass formed. 1.33 g 2.66 g the that combined Other oxygen oxygen of to to and was oxygen hydrogen c an and experiments through form matter composed form other substances react to oxygen showed that combustion c arbon to form dioxide. r It was proposed that elements, such as hydrogen, oxygen or carbon, are the internal structure and primary constituents of matter, and they cannot be chemically broken down into characteristics of atoms will be simpler substances. The idea of denite proportions suggested that particles of one discussed in Structure 1.2 y simple ratio, and that atoms of one element have a dierent mass than atoms of a dierent element. This, and other experimental evidence, led to the atomic theory. be reactions. states or Physic al that all matter destroyed, and but chemic al is composed of atoms. These atoms they are properties rearranged of during arrangement of these atoms. that 8th of that one smaller the is parts E What on to into their must What is is snap as to “kana”. said a to have seashell further. atomic D alton. to dierent powder “atomos”, “not any of John experiments evidence theories? broken development classied based be particles producing known a be until units, credited knowledge evidence. atomic in the Democritus successively not conservation “elements”, Scientic could later, could c alled l mass atoms could v O from stage years He BCE, indivisible that due proposed that “particles too u next 2000 of f x splittable”, The century increasingly composed are i into 5th wor ld parti cles. t a observed Āruni natu ra l matte r They mass together into the substances and experience”. in the r o Similarly, BCE, seen in these that pa rticl es. o objects be chan ges betwe en century to re a sone d indi visible d small tiny, s age n interactions Leuci ppus, of In dian U postu late d In and up the philos ophers theory, D alton over drew propose that types known as masses. be supported was used evidence? to Is by veriable develop these evidence shaped by our y made them Gre ek o w as the C Democ ritus among and i n Ā runi v atomists, Uddāl ak a p Evidence Ancie nt chemic al matter depend on the bonding r e and theory created O atomic c annot t i s The n l element, called atoms, would combine with atoms of another element in a xed, y P The e 1.0 g was and of denite s s but It in Figure 3 Top: Āruni lived in what is now modern day perspective? Northern India, depicted by the Ganges river. Bottom: Democritus is in a Renaissance-era painting 5 Structure 1 Models of the particulate nature of matter Chemic al symbols In modern which example, the chemic al of one atoms or chemic al symbol and two symbol for elements letters iron is for Fe and are are hydrogen (the rst represented derived is two H from (the letters rst of by the the letter the same element of L atin symbols, names. ferrum “iron”). Name Common chemic al elements and their symbols are listed in table1; the full list is given in the data booklet and in the periodic table at the end of this book. H hydrogen oxygen Na sodium Atoms are the properties. and form smallest While iron is another magnesium Figure 4 bound atomic formula magnesium is a species, one type composed Mg sulde. of of atom, sulfur compound, and as S Mg. atoms it Similarly, sulfur only. consists of (gure4). MgS is the p C sulfur (middle) and magnesium sulde (right) single more elements In contrast, two dierent, chemic al o v i n (le), n U M agnesium or example, magnesium metal is an only chemic al contain atoms of a two y (MgS) chemic ally of substance For of O Common chemic al elements sulde forces. contains atoms t i s Table 1 elementary it contain r e as chemic al Pure substances and mixtures o Figure 5 i E 6 t a v sulfur (S) classied as a pure substance or a mixture, depending on the type (gure 5). – any substance that definite and mixture u e.g., a atoms, be arrangement matter composition l O of magnesium (Mg), c an particle compound – composed of one kind of occupies space and has mass – has a uniform chemical element d r o f x pure substance M atter of two or atoms – composed more kinds of in a fixed ratio, e.g., magnesium sulfide (MgS), How matter is classied – a combination of two or more pure substances that retain their indiidual properties water (H ) homogeneous – has uniform composition and properties throughout, water, according to the arrangement of particles e.g., metal sea alloy heterogeneous – has nonuniform composition and e.g., arying paint, properties, salad dressing y Fe (S) chemic al certain n chlorine by substance, possess Elementary substances chemic al compounds together elementary still l Cl bound that individually, they tend to combine together P sulfur matter exist y magnesium S of c an chemic al substances. element, while Mg units atoms e c arbon O r C For hydrogen), and s s the Symbol chemistry, consist Structure 1.1 Introduction to the particulate nature of matter Pure substances cannot be separated into individual constituents without a chemical reaction, which alters their physical properties. In contrast, mixtures can be separated into individual components that retain their respective physical properties. A student had two pure substances, observations were A made and and B. They were heated in separate crucibles and some qualitative and recorded in table 2. r e quantitative s s Data-based questions B were n l and a to O is y A ice p 4. change qualitative changes y P y Melting the v 3. the a E State of M ass of crucible crucible Change and substance contents in Observations mass / g heating aer aer / g heating / g 26.12 ± 0.02 26.62 ± 0.02 Black colour 27.05 ± 0.02 25.76 ± 0.02 Black colour heating substances A and B a C alculate 2. u Results from l 1. Table 2 colour o C M ass and heating colour Green O f x Red n i t a r o before o Appearance aer heating each of the two substances Observations Substance B Substance B v i n U d A t i s Substance A r e in mass for substances A and B. observation physic al change substances both pure from A the while and B experiment rusting iron represented substances, not a mixtures. is performed on A and B. a chemic al physic al Discuss change. change or whether a the Explain, chemic al using the observations, whether change. experiment shows that A and B are elements. 5. Both A same and B turned black on heating. C an it be concluded that the heating of these two substances produced the substance? 7 Structure 1 Models of the particulate nature of matter Melting point determination Melting purity point of points, t e m p e ra t u r e v a l u e. over a The its c an which that presence melting used of to they matches impurities point a ss e ss substances me ans closely t e m p e ra t u r e and melt the in c auses a Method the h av e at a sharp (Your teacher will provide specic instructions, depending on specific the oretic al substance melting to occur 2. Prepare 3. being two samples of each record sucient amounts relevant relevant Following 4. qualitative Prepare, your of in a two third solids B and the the melting organic solids, for Record t i s 1. known relevant example, appropriate aspirin and salol (phenyl 2-hydroxybenzoate) 2. Comment S afety points • Note that • You protection. the teacher melting will give 3. point you apparatus further gets safety very hot. this environmentally salol and aspirin point of a substance not are contain more chemic ally Mixtures discussed in the c an be than If the the particles the top, mixture, is structural are which extent the to and evenly reveals of its could a , which of where the melting of an samples quantitative data in an impure substances. dierence in their point organic compound the air distributed, that is and with separated in by particles of is data be used to synthesis? no xed ratio, which physic al methods. are evenly distributed. milk is a as in a its and 80% nitrogen and 20% mixture of two solids, then Natural milk heterogeneous maintains explosive, roughly sampled. such heterogeneous. mixture hydrogen, H or be composition regardless not c an in three formulas of A and B and use explain success element so your sample of the oxygen, and small amounts of other gases. Air is a referred to as component example, nitrogen, consistent mixture u f x to is i oxygen one and homogeneous, of o mixture homogeneous E ach The most common homogeneous a t a r o d chapter. is what analyse bonded Tools for chemistry Air the C for determining the are melting To of small results, comparing the melting substances information n U Mixtures Methods 4. are irritants hazardous). the o v example, qualitative melting points. i n and (for on pure Research prec autions, depending on the identity of the solids being analysed a y eye for separate format. p Wear of r e • tube, point mixture). apparatus two two together. c apillary Questions of B) n S amples in O • point and y C apillary tubes (A l Determine (A, Melting solid mixture of the two solids. quantitative data 5. • solids teacher ’s instructions, mix small the Materials • analysed.) organic physic al and oxygen, O l 2 will have the cream rise mixture. chemic al properties. For , supports combustion. 2 a O mixtures, aqueous solutions, will When these substances are present in a mixture, their properties stay the same. be discussed in Reactivity 3.1, and In contrast, water, H O, is not a mixture of hydrogen and oxygen but a chemic al 2 the properties of metal alloys in E 8 v Structure 2.4. compound The new gas, is with its water formed substance not by explosive, own bonding has and properties without a none it and chemic al of two the does the hydrogen properties not support hydrogen reaction, which atoms of with combustion. and oxygen creates one hydrogen or It is c annot oxygen atom. oxygen. It is not a a pure substance be separated new substances. from y Inquiry and solids of P • observations the analysis. skills and of samples c apillary tubes. Tool 1: Melting point determination Identify identity Obtain ra n g e. • 2: the 1. r Relevant be Pure e l ow e rs data s u b s t a n c e. s s melting a Structure 1.1 Introduction to the particulate nature of matter Separating mixtures Mixtures mixture c an has separated of using property magnetic the to does components c an c an be it usually is be a a of magnetic them. sugar sand if we and s u ga r It sulfur is powders not. iron(II) maintains pure This sulde, none of c an be dierence FeS, is not the properties substance. understand sugar and placed each component of the sulfur compound sugar is and while bec ause between bec ause iron smell. individual separated from The sulfurous new, attractions m i x tu re is of their will intermolecular dissolve in forces. water, due to water. in water and t he s u ga r Intermolecular di ss o l ve s . The solution c an th e n be po u re d t hro u g h fi l t e r paper placed i n s i de p a ss th e by will and p a ss e v a p o ra te s e v a p o ra t i n g th ro u gh l e avi n g th e f i l te r re m a i n on t hro u gh w a te r pa p e r. th e t he behind f ro m S u ga r (fi gu re fi l t e r f i l te r th e t he 6 ). p a p e r, p a p e r. pu re fil t ra t e c r ys ta l s The — the fo r m sand w he re a s Th e sand. will l a rg e wet Th e t he sand s u ga r solution in t hi s p a r ti c l e s sugar is be which pa ss e d pro c e ss a residue p (We define discussed o C i n residue that remains aer n U as a substance evaporation, similar or any process) i filtrate u Filtration apparatus a O l f x t a r o d filtration o distillation, evaporating th e v filter funnel are y filter paper Figure 6 and forces Structure 2.2 o bt a i n e d c r yst al li za t i on 7 ). d i ss o l ve d dried, c an will r e (fi gu re the fil t ra t i on O w a te r th ro u gh w a te r c alled t i s in p ro c e ss n not a y funnel, l in a y solid have means mixture P Th e as A Iron separate not separated intermolecular physic al r the magnet. used and solids S and is a by properties. e Two separated s s in be unique sugar solution basin E v solution from evaporating basin cold tile leave for a few days t Figure 7 The crystallization heat for sugar to crystallize process 9 Structure 1 Models of the particulate nature of matter Distillation can be used to separate miscible liquids with dierent boiling points, such as ethanol and water. Ethanol has a lower boiling point and will evaporate rst. Once the vapours rise up a cooling column, they can be condensed to a liquid. As shown in gure 8, cold water surrounds the condenser and allows the vapours to condense to liquid ethanol. The water remains mostly in the distillation ask. s s r e thermometer water out Distillation apparatus flask water water in y distillate p heat r e and O t i s ethanol n y condenser l Figure 8 (mostly ethanol) o v i n chromatography will C Paper chromatography can be used to separate substances such as components Paper in inks. A piece of chromatography paper is spotted with the mixture. The bottom be discussed in more detail in of the paper, below the spot, is placed in a suitable solvent as in gure 9(a). substances in the n The U Structure 2.2 mixture phase) and the paper (the dierent anities forces of attraction between and or 9(c) ve the solvent o intermolecular for the solvent (the mobile paper. Figure the shows a pure substances mixture that was in the mixture composed of pure substances. i (a) the (b) (c) paper some some u l a E v O f x t a The stages in 2D paper chromatography 10 d Figure 9 r o u have stationary phase). The anity depends on the hours hours later later solvent drop of mixture turn and paper use a 90° clockwise different solvent Data-based questions Look at gure 9. 1. Which colour dot had the strongest anity for both solvent 1 and solvent 2? 2. Which colour 3. Which had a dots had stronger a stronger anity for anity solvent for 2 solvent than 1 than solvent 1? solvent 2? y P distillation u Structure Table 3 shows a summary of the separation techniques 1.1 Introduction to the particulate nature of matter discussed. Components Technique Description removed ltration is le s s mixture poured through a paper lter or solid(s) soluble insoluble substance(s) substance(s) solvent, the solution more soluble less soluble cools substance(s) substance(s) porous material or an is added organic mixture is to water solvent dissolved in e mixture (solvation) r dissolution P Activity hot water or an organic Suggest a suitable method down, and the heated up solid(s) and/ volatile one or more of its or distillation non-volatile liquid(s) components vaporize(s) liquid(s) placed on a. salt and pepper b. several c. sugar d. For less soluble a piece of paper; one more soluble side of the paper is chromatography submerged component(s) water or move(s) faster solvent; components 3 Summary of separation techniques Figure 10 An advanced each describe technique and component is the outline isolated. place n o i t a a by fossil fuels. mixture, u l v E provided d r o f x O ltration technique c alled water for millions of people. how slower stay(s) in C Ta b l e U i n move along the paper or dyes water o a move(s) v in water-soluble and separation component(s) p paper following iron and copper lings each y is r e mixture the mixtures: evaporation or until of O is are ltration t i s mixture by each n formed isolated y crystals separating for l crystallization y liquid(s) other However, Why might it reverse osmosis extracts salt from seawater, this process requires vast amounts of energy, be important most providing fresh of which is currently to consider alternative energy sources? 11 Structure 1 Models of the particulate nature of matter Planning experiments and risk assessments Relevant skills Tool 1: Separation • Tool 1: Addressing of mixtures safety of self, others and the environment s s • Instructions 1. Using the c alcium you Determine • Identify suitable time, try methodology the separation. the masses are before mix all and the them dry, iron you physic al and hazards and complete to mass of c alculate component. your teacher method and school’ s by c alcium health should validate comparing c arbonate) the mass before and each component prior to mixing c arry measure your with beforehand. your together, and aer of lings, C i n components salt, Measure Then the eectiveness (sand, that assessment o together. Remember risk aligned p the them out! and v aer evaluate component the allow powdered y could each identify methods r e of the you: control measures disposal it and risk relevant Extension You consider hazards level of would four substances. method, which safety policies. have must that lings O you your a in you these iron n • on protocol so, of method salt, l Assess If each a sand, y Identify the • the doing of decided • devise t i s 3. have assessment and In properties chapter, containing the the out your mass of separation, each percentage again. recovery make sure Compare of each n U Linking questions factors are i components t a How c an How do l u are contain of considered a mixture? products of intermolecular between Why the o d r o f x a E v O 12 What two a alloys generally choosing reaction forces substances? metallic in be method to separate the purified? influence the type (Tool 1) of mixture that forms (Structure 2.2) considered bonding? a (Tool 1) (Structure to be 2.3 mixtures, and even though they often Structure 2.4) y risk this P a in mixture r Once a c arbonate. chemic al 2. ideas separate e to Structure 1.1 Introduction to the particulate nature of matter States of matter (Structure 1.1.2) Solids, liquids and gases Matter is composed of particles. The types of interactions between these particles determine the state of matter of a substance: solid, liquid or gas. All substances s s can exist in these three states, depending on the temperature and pressure. The states of matter for of solid, substances (l) is a solid Water is a liquid below • Water is a gas are and 0 °C: H shown (g) for by gas. letters For in brackets aer the example: O(s) 2 0 and O(l) 100 °C: H P between 2 above 100 °C: H O(g). 2 For example, the expression “NaCl(aq)” tells us that sodium chloride is dissolved t i s in water while “NaCl(s)” refers to the pure compound (solid sodium chloride). The solid liquid gas • forces • around Steam, liquid but reached. A further c arbon water This states it the of until vaporizes of conditions, change CO of (s), move around liquid ice are the three states of water matter heated. ice vibrate, and in a a as The they (changes accelerates and absorb particles temperature melts temperature changes dioxide, is violently, point, in E the these certain melting. this increase eventually reverses At more as v is their energy rotate, move water and a positions, Under change absorb negligible particles around faster than l will are • u f x O ice forces rotate, and Changes of state Substances vibrate, i Figure 11 weaker than particles attractive between particles t a move • c an be • those in solids r o not forces o are vibrate in fixed positions but do attractive volume compressed between particles strong particles fixed shape • d are fixed no compressed between particles • no • n attractive U • c annot be • y c annot be compressed volume fixed shape o no C fixed • • volume p r e • fixed shape v fixed • i n • O properties of the three states of matter are summarized in gure 11. n l y A special symbol, (aq), is used for molecules or other species in aqueous solutions. y Water • liquid r • for e formula: (s) becomes a release to energy. Solid vibrate in xed known as the melting point its the or continue state from solid to liquid). movement of particles, and gas. The decrease in temperature state. solid state, substances known as gure12), c an turn into sublimation, which is is commonly gases directly, without typic al used for for dry ice (solid refrigerating ice 2 cream and biologic al samples. Figure 12 Sublimation of dry ice 13 Structure 1 Models of the particulate nature of matter The process water opposite vapour in the to air sublimation solidies and is c alled forms deposition. snowakes of At low temperatures, various shapes and sizes (gure13). When when are a substance energy a solid particles or a gas, from and the when from energy or to a solid, the a to less the forces and condensed become when a You becomes of state occurring in these will maize known as non-Newtonian uids, typic al liquids. The varies make starch depending a viscosity on the 2. force a non-Newtonian uid commonly slime or “oobleck”, and 3. and record relevant eye • 250 • Powdered • Water large spatula 3 cm t h re e to the fo u r h e a pe d b e a ke r. N o te i ts of a condensed molecular solid. the each of The process of water • the maize Suppose you of in gure 14. starch and mix. the by mixture adding achieves more maize needed. exploring the harden if properties of tapped, and ow slowly. the following: maize starch starch–water were relating Consider shown properties and identify the state of powdered • as are maize until Adjust should stirred matter question asked to a mixture. to maize develop a research starch–water mixture. possible independent and dependent variables. 3. Research non-Newtonian uids and identify other examples of these substances. 4. How has about this states experience of matter changed and their the way you think properties? Reect on ma i z e this, a pp e a ra n c e completing the following sentence starters: and • I • Now, I think... used to think... E v Linking questions Why are some conditions? Why are substances solid while others are fluid under standard (Structure 2.4) some (Structure 14 2. a O c o n s i s t e n c y. spoons l s ta rc h or starch u Ad d maize f x Method beaker t a r o or o Spoon i d • if the water, It Describe • protection. Materials 1. 1. qualitative to water time mixture. smoothly n S afety more some Questions U observations Wear more a o Identify or C 2: your water adding consistency. Spend v Inquiry i n variables thick starch explore its Inquiry 1: Identify dependent and independent add Continue applied Relevant skills • Slowly of non- properties. • a process. transformations p as for n them. known condensed O to uids to and, y Newtonian like less becomes a gas. These state y changes r e behave liquid exothermic t i s substances, a stronger. This happens when a gas liquid surroundings is an Non-Newtonian uids not to the product of The do a surroundings deposition of water Some state surroundings. This happens 2, changes of state Reactivity 1.2) endothermic and some exothermic? y energy condensed l A snowake, changes lose intermolecular liquid more P releasing the a liquid a particles processes. substance becomes a the e a the substance, Figure 13 becomes from by r state, changes absorbed endothermic When is s s state, Structure 1.1 Introduction to the particulate nature of matter gas (g) s s Endothermic and Figure 15 Orange growers spray their fruit exothermic with water on cold nights. t i s Freezing of water is an exothermic process that heat) to the fruit, protecting it against cold Kelvin temperature sc ale (Structure 1.1.3) the particles of a faster, a gas heated, changes added energy to is between a they there liquid used the to is and lattice move no increase. particles. more, As particles Temperature substances in a faster. liquid is a absorb vibrate energy, more and temperature change during the periods when when disrupt of the a liquid solid changes lattice molecules in the liquid. and to a gas (gure 16). The overcome the intermolecular n U forces is in in particles o water solid while vibrate of energy C a solid energies kinetic i n When the v move rises, average p of y temperature measure r e As releases energy (in the form of O changes of state n liquid (l) Figure 14 l y solid (s) y P melting e r freezing vaporization m c + d water steam s om t a u K A a Figure 17 are kilogram length, energy input The seven base SI units (kg) for mass, second (s) for time, electric current, meter (m) for ampere (A) for kelvin (K) for temperature, Graph of the heating curve for water E were accepted l freezing ice Figure 16 There water v O f x erutarepmet 0 + water melting kg condensation i / r o C° ice steam o d 100 many mole (mol) for amount of substance, and c andela (cd) for luminous intensity. All units attempts to measure relative temperature, but the rst widely of measurement temperature sc ale was introduced by the Polish-born Dutch c an be derived from these physicist seven base units D aniel Gabriel F ahrenheit. You The kelvin is the base unit of temperature of Units measurements (SI). c an There be are will learn more about the mole measurement in the International in System Structure 1.4. seven base units, and all other units of derived from these (gure 17). 15 Structure 1 Models of the particulate nature of matter Measurement 3 M aking, recording, and communic ating volume (m measurements 2 greatly benets from the Bureau French mesures), agreed of Weights in the and Measures = (BIPM, rene late 19th which –3 ), density (kg m ), energy ( joule, J, where The century, is an base ) and units including so on, are several are dened that Boltzmann constant, standards. derived from the seven base constant, N ; and you k; the according will to seven constants, recognize, such as the speed of light, Plank constant, c; the Avogadro h A International System of Units (SI, from the r The French The use of universal and precisely dened units is very Système international d’unités) is the most commonly seven units: (second, amount of current substance m), mass blocks important, as it allows scientists from dierent countries are the to understand one another and share the results of their (kilogram, kg), (ampere, A), studies. What other advantages are there to internationally temperature shared and continuously updated measurement systems in (mole, mol) and luminous the natural sciences? You might want to look up the Mars (c andela, cd). All other units, such as those of Climate Orbiter. each several with universal dierent temperature reference points. v summarized in table 4. n O boiling point = 212° 2 H O freezing point = 150° 2 o H O boiling point = 0° 2 1700s O H freezing point = 0° 2 i H O boiling point = 100° 2 t a l u CGPM 1950s Triple BIPM 2018 Kelvin Table 4 Examples a Temperature unit kg, and for m historic al is energy, and and practic al of related s. It v arious to has practic al reasons ( J), been water = 273.16 K k. energy which are decided reasons”. could of dened in terms of the Boltzmann t e m p e ra t u r e thermal joules zero = 0 point constant, the temperature = 12° to What sc ales and in as turn keep do you such it could be expressed dened in terms of the base kelvin think as an some SI of base these unit “for historic al be? bec ame obsolete in Look c arefully at table 4 above. Identify one thing you see, one thing it makes when the kilogram and all other SI units were redened think about, as exact quantities your 16 1700s H Absolute in points body 1800s are freezing point = 32° Kelvin you based O been these freezing point = 0° 2 H have of 2 Celsius units cylinder to dene a kilogram of This standard 2019, v mass. E A platinum–iridium in the US was used d r o f x O Figure 18 O Human 1700s Delisle Reference H sc ales Some o U F ahrenheit D ate 1700s C Newton i n Sc ale y history, developed, p Throughout r e Thinking skills ATL O t i s intensity electric (metre, building n K), s), length Its y (kelvin, measurement. l time base of on physic al constants class. and one thing it makes you wonder. Share your ideas with y system P used 1 J –2 s units. seeks to set up and measurement 1 kg m e organisation continuously sc ales. The Bureau international des poids et established international upon s s International from Structure Kelvin is temperature considered Absolute any an Under c annot normal get in proportional any that at collisions. colder. temperature pressure, 373.15 K. zero this average kinetic temperature M atter An of water Absolute the the particulate nature of matter energy of particles and sc ale. implies on to to at increase 1 degree boils on at the absolute in the particles zero c annot temperature Celsius. 0 °C is of 1 c annot lose kelvin equal to transfer heat and is equivalent 273.15 K. 100 °C, so that makes the boiling point of Celsius sc ale is –273.15 °C. Figure 19 The Celsius and Kelvin rounded to whole numbers) 400 K water l 373 K boils 273 K freezes 40 °C v solid CO 2 more about the C o i n 100 K U liquid 191 °C learn energy of particles in Reactivity 2.2. 150 K 150 °C will kinetic 195 K 100 °C 82 K air 50 K 250 °C o 0 K zero i Kelvin Celsius u l a v O f x t a r o d absolute n 200 °C 273 °C You p 200 K dry ice 78 °C y r e 250 K 50 °C O t i s 300 K water 0 °C n y 350 K 50 °C y P sc ales for temperature (all values are 100 °C r t e water (0 K) energy increase is absolute Introduction s s to zero kinetic hence an 1.1 Linking questions is sample E What the graphic al at fixed a distribution temperature? of kinetic energy values of particles in a (Reactivity2.2) What must happen to particles for a chemical reaction to occur? (Reactivity 2.2) 17 Structure 1 Models of the particulate nature of matter End-of-topic questions 5. Which changes of state are opposite to each other? Topic review 1. Using your answer the knowledge guiding from the question as Structure 1.1 fully as melting and s s A. condensation topic, B. vaporization and deposition C. deposition and sublimation D. sublimation possible: e How can we model the particulate nature of matter? Which of the and freezing following statements is r 6. incorrect? Exam-style questions A. solids and liquids are almost incompressible 2. Which of the following are examples of particles liquids D. particles in both and solids gases and have no liquids are mobile l B. C. homogeneous xed shape mixtures? elements Steel III. Aqueous KMnO potassium manganate(VII), A. oxygen B. hydrogen (aq). I and II only D. I, II and III B exothermic CO C endothermic CO (s) → CO (s) → C(g) + O 2 (s) 2 following sodium could I and II only C. I and III only E B. a I only v O distillation be u ltration I, II and III A. decrease B. increase C. decrease D. increase (g) Explain why by by temperature increase in on the Celsius temperature by sc ale is 20 K? 20 °C 20 °C by by 293.15 °C 293.15 °C the proportional Celsius to Kelvin temperature average temperature temperature is kinetic not, increment is even the is directly energy but the though same in a 1-degree each sc ale? [2] used to 10. Ionic salts c an unbalanced lead(II) be ionic broken down equation for in electrolysis. The the electrolysis of molten bromide is: 2+ Pb a. + Br One → of Pb + X the formula b. Balance c. The of products product the point and of is lead, Pb. State the X. [1] equation. electrolysis c arriedout matter 18 in the from a solution of sodium l evaporation II. C(g) + O 2 water? I. III. D. chloride → (g) 2 2 methods f x in (s) 9. 2 CO i solid chloride the CO → (g) t a obtain A. r o endothermic of to oxygen in dry ice Extended-response questions (g) 2 d 2 Which change equivalent n CO o process exothermic U endothermic? D magnesium and sulfur in magnesium sulde Equation describing the A water by o Exothermic or in other D. describes the sublimation of dry ice dioxide)? oxygen vibrate c arbon Which and c an each C. C (c arbon i n correctly 8. from y III only C. gases nitrogen in air and p B. and r e II only v A. and separated methods? 4 4. liquids be O physic al c an n Which t i s II. What solids, Air 7 . 3. in y I. at of molten 380 °C. boiling each of Write equation gave data, (b). bromide is reference to melting species state in lead(II) With point the temperature. you [1] deduce the state of in the symbols equation at this in the balanced [2] y P Multiple-choice questions Structure 11. The kinetic mass × energy the square of of particles the is c. equal to half of their Once to 2 mv = . Determine how much the the obtain Introduction excess removed, velocity of the particles: 1 E 1.1 the the copper(II) student pure to particulate oxide needed crystals of had to nature matter been gure copper(II) of out sulfate how from speed of k 2 the molecules in a pure gaseous substance will solution. couldfollow the Kelvin temperature is doubled. to obtain pure, dry copper(II) s s when Describe a method the student increase [2] sulfatecrystals. 12. Pure c aeine is a white [3] powder with melting point 14. Study the gure below. 235 °C. e vaporization the melting chemist is point of c aeine investigating the in kelvin. ec acy of [1] three r State A C° a. b. 100 condensation all once and melting 0.960 g. collects point of the the She uses following each data for the product: water ice + Method 3 0.229 0.094 0.380 t i s Method 2 O freezing ice Mass of c aeine obtained / g Melting point of 188–201 a. Explain why, C alculate ii. C alculate the mean and range of the mass constant c aeine student word a product. minimize the random oxide acid is insoluble equation for this sulfate with in solution excess water. reaction is by copper(II) as f x acid powder could was was be chemic al for this a The balanced heated, added v O b. a symbols, l state observed E use to a the iodine sample of energy remains time. chloride atmospheric in [2] 100.0 g of temperature and properties: –3 °C similar to the curve one for a in gure 16 sample of this solution. exists temperature solid cold that When surfaces pressure, as [2] diatomic molecules, I . At and readily cooled, without solid iodine pressure, forms it is violet a lustrous purple- fumes when heated gaseous iodine deposits on condensing. melts at Under 114 °C to increased form a deep- equation, including violet liquid. [2] copper(II) Formulate was in equations that represent all changes of oxide mentioned above. [3] excess and b. State the melting c. Suggest how gaseous iodine. point of iodine in kelvin. [1] suspended in the solution, reason, a method the student could remove period sodium heating black gently. quickly sinking to the bottom of the beaker. Suggest, giving increasing the 101 °C graph room state it point: point: the a following sodiumchloride Elemental a. then of standard the a show for the of 2 follows: reaction. until boiling to 15. u Write melting of oxide. sulfuric acid + copper(II) oxide → copper(II) sulfate + water a. has Sketch 0 °C 5.00 g b. at (at [1] [1] copper(II) sulfuric pressure) t a The c aeine r o Copper(II) [2] experiment. prepares dilute to of water reason, which method i reacting way solution pure o A one this purest a d 13. in gures. giving A appropriate number n error an U Suggest the to spite o signic ant Determine, gave c. answer C of iii. your [2] percentage yield of Method 1. i n Give obtained. the v of in temperature p input,the i. energy input y 229–233 r e 190–220 caeine product / °C steam water melting 0 Method 1 + n and is water y yield c ases theoretic al l method three The y in methods. erutarepmet yield extraction P / c aeine excess copper(II) oxide. liquid iodine c an be obtained from [1] [2] 19 The nuclear atom Structure 1.2 Structure 1.2.1 The answer to this question was obtained by nucleus years of brilliant how we know the late in the atoms idea that electrons that Structure 1.2.2 matter dierent was chemic al Isotopes of are atoms gaining relative — M ass atomic spectra masses of are electricity study the and popularity. The radioactivity structure used to determine elements of the allowed atom from their isotopic itself. O n of to same element with t i s scientists the (known as composition. discovery of neutrons. l was — numbers y theory) Negatively occupy the space outside the nucleus were indivisible and reactions the atomic charged, dense (nucleons). more fascinating than Structure 1.2.3 the neutrons LHA of rearranged positively The structure of the atom (Structure 1.2.1) atom contains neutrons occupy electrons are very 2. It is a It an has highly a the which Atoms nucleus. nucleus structure to the The contains contain protons, protons electrons, neutrons and atom containing itself. virtually all the mass of the atom. charge. designed particles made are: itself also are by Ernest were given in red Rutherford toward a in 1911, sheet of positively gold charged foil. The main gure 1. i Rutherford’ s explanation detector source + + u l a E v O large of t a f x alpha outside comparison dense alpha observations movable in o d r o beam of the positive experiment radioactive alpha particles of small n In factors U 3. nucleus, nucleons). o is region as C It vast v key 1. the charged known known as subatomic particles. i n The positively p which a (collectively y and r e An Most alpha particles are undeflected atom + gold foil Some alpha vacuum + particles are deflected slightly A few alpha undeflected + particles slight bounce + deflection deflection off nucleus 20 Figure 1 Rutherford’s gold foil experiment y composed is what is known 1800s, a and P In contain protons r the question of of research. Sometimes, the charged question of Atoms composed e 100 — over s s Understandings How do nuclei of atoms dier? Structure 1.2 The nuclear atom F alsic ation experiment vulnerable The plum-pudding claim present throughout. red at the undeected. model, this gold were foil the would Rutherford’ s paving the way that blob atom was that never electrons be stands proven c ase, all alpha knowledge have degree results for the with the gone through its of new model of the atom. is falsiable. up to true means C an a Determine in the which gold foil experiment and the observation is explained Property nucleus has a positive Occ asionally, some of the alpha The nucleus is particles comparison to the size of the atom. straight through the gold foil. repelled The nucleus virtually all nucleus. is very the dense, containing mass of the atom. model, of the the in the solar entire atom. by way electrostatic electrons planets the orbit atomic instead of the by l a E Figure 2 experiments orbit Sun. by claim? proposing the Rutherford model the Just positively as contains gravity, steer it in the the over Sun contains 99.9% electrons (gure2). charged atomic are of held the 99.8% mass of around the u – his nucleus attraction. v O as mass, However, f x nucleus same system’s charged of t a nucleus negatively results known as the i this the also o summarized r o In c an n U Rutherford planetary model of the atom, d 1911, evidence a by a nature of scientic o are approaching the C particles closely very small in i n alpha straight back. v bounced charge. p The c an y r e Observation In by property. when accompanied falsify but y observations are n the nucleus. they l show the Nearly all the alpha particles went The further strong O which is y below of that certainty. Scientic provisional t i s lists properties always counterexample Activity The testing absolute The that means P directions. This contradicts them. A scientic severe therefore new single that with uncertainty. knowledge contradicted the development of a are evidence r existing If suggested charged to e particles atoms model positively the atomic model s s preceded it, namely the “plum-pudding model”. amorphous falsied claims gold that an foil Scientic The + – electron – proton – neutron + + – nucleus The Rutherford model of the atom 21 Structure 1 Models of the particulate nature of matter Models Scientists use models to represent Atoms natural phenomena. All themselves are extremely small. The diameter of –10 models have 2. The useful size Consider of model the of which the nucleus the should depiction is be of the exaggerated most identied and atom but it in gure unit serves as a atoms used to is in the range 1 × 10 –10 to 5 × 10 m. The describe the dimensions of atoms is the picometre, pm: –12 nuclear atom. 1 pm = 10 In the nucleus is hard to fully appreciate. Rutherford’ s native X-ray m crystallography dimensions the a commonly angstrom, used unit –10 1 Å = 10 m the atomic radius of the uorine atom is the centre of the pitch from the top row of seats. If a golf –12 60 × 10 m (60 pm). To convert this toÅ ball were placed at the centre of the eld, the distance using the we conversion c an use factors between you and the golf ball would represent the above: –12 10 relative volume of open space in the atom is vast, and m 60 pm × 1 Å t i s The × = –1 0.60 Å = 6.0 × 10 Å –10 our simple representation of Rutherford’s atomic model In in gure 2 is obviously spite a tiny of its of much volume of the atom and the diameter of an atom is Rutherford 100 000 times the diameter of the nucleus. is of our Rutherford’ s thinking rumoured to on have the work has formed the structure of the atom. said to his students: y r e approximately m limitations, unrealistic. The nucleus occupies basis 10 O 1 pm n y given distance between the electron and the nucleus. l dimensional analysis, All science is either physics or stamp collecting! p C o v i n n U o i t a u l a 22 Eden Park, v E Figure 3 of the eld d r o f x O Auckland, New Zealand. If the atom were the size of the stadium, the nucleus would look like a golf ball in the centre y example, P being at Eden Park stadium (gure 3) and looking down at For for atomic symbol Å: r New Zealand is a great rugby-playing nation. Imagine is e The vast space in the atom compared to the tiny size of s s understood. limitations, Structure 1.2 The nuclear atom TOK All the could we argued physicists, Switzerland Binnig that gave Prize in the scientists in and assume only that “real” Heinrich sc anning generates Physics are atoms when Rohrer, tunnelling are they real. c an working microscope However, it be at seen. In 1981 IBM in Zurich, (STM), an electron three-dimensional images of surfaces at the atomic the ability 1986 was to observe awarded individual to Binnig atoms and directly. The Rohrer for their r groundbreaking work. c an nd an atomic sc ale lm created by IBM c alled A Boy and his Atom on the internet. are material tools of Relative nucleus same mass particles also as are l O charge and it relative charge a negligible –1 value particles are a of single electron approximately is known as the 1.602 × 10 of a the charge proton as +e. electrons and of The commonly protons an expressed electron symbol have charges for electron elementary The C. The in elementary actual e is c an oen charges of be represented as –e, omitted, –1 and so +1, it is and customary masses and charges of charges of charge units. the to particles c an be found in the For data example, neutron and outside nucleus these subatomic Relative masses and nucleus –19 has Table 1 Loc ation +1 0 by neutral subatomic The the proton, 1 c arried a proton. shown in table 1. Relative a (e) E electric v electron contains the t mass 1 neutron charge the subatomic proton The the u Particle that nearly the production or acquisition of o charges shown with observations of the i and have neutron, f x masses the the t a experiments particle, r o Other in d knowledge? O important make n How to y c apacity p human’ s o extended C technology natural world? n l y t i s r e A still from A Boy and his Atom U Has v Figure 4 i n y P You e This Nobel Gerd discussed objects invented microscope level. have that s s two models be booklet. charge say that respectively. 23 Structure 1 Models of the particulate nature of matter How small is small? Relevant skills Tool 3: Apply • Tool 3: Use and and use SI prexes and units interpret scientic notation s s • Instructions 1. A variety objects of small but in lengths rather order of are based size, shown in table 2. Without looking at their on what from you smallest stop, atom, 267 pm diameter 0.30 mm diameter fullerene, C 250 µm length 150 pm diameter 0.71 nm 60 y Lengths of various small items p 2. Table 2 r e O full bond, t i s c arbon 0.10 mm n iodine-iodine printed thickness diameter Convert the length values into metres and state them in standard form to v two signicant gures. Refer to the following conversion factors: micro, µ: 10 • nano, n: 10 • pico, p: 10 • femto, f: 10 the 4. Provide ATL a web than the i 5. question Conduct smaller –15 values following the full your 1 in table values to school’s nd given reference t a u Atomic number 2 in order of increasing size. Was the list you correct? search o d r o f x for three in your citing more table and 2, values to add to the list: one one larger, information referencing and the nuclear and one intermediate. sources in question 4, system. symbol As of 2023, there are 118 known elements, given atomic numbers 1 to 118. The atomic l a E v O 24 for –12 n gave length –6 –9 C List i n • U 3. milli, m: 10 o –3 • number of an element is also the number of protons in the nucleus of that atom. Gold, atomic number 79, has 79 protons, while carbon, atomic number 6, has 6 protons. As all the relative mass is in the nucleus, the dierence between the atomic number and mass number is the number of neutrons in the element. Gold has atomic number 79 and mass number 197 . Therefore, it has 197 – 79 = 118 neutrons. Each element is neutral, with no charge, so the number of electrons in a neutral atom must equal the number ofprotons. y paper, cell, 0.84 fm l of Length radius y onion each item, list these P sheet charge about largest. r Item proton, know to e lengths, Structure 1.2 The nuclear atom Activity Determine Atomic the missing symbol values Atomic from the table. number M ass number Protons Neutrons Electrons s s O 8 13 27 85 37 35 27 32 r 120 100 use nuclear symbol notation, X, to denote the number of isotope, for protons Z is the example, and electrons in an atom. atomic with mass number, number and 197 X is and A the represents chemic al atomic the mass number of the symbol number 79, (gure 5). Gold, would number of chemic al where neutrons compounds are protons. For the compound ionic example, the ion nucleus is sharing magnesium a 2+ two greater protons + also number: 12 displayed E (12 oxygen react with the number number the or loses of of nuclear As fewer oxygen M agnesium the in electrons. more to produce electrons positively negatively 2 to charged charged notation 24 24 result, these atoms two symbol a electrons than electrons below: + charge: 2+ (12 with a atom 2– protons – 10 electrons) Mg 12 chemic al element: Mg protons) gains the (magnesium) two electrons lost by magnesium to produce an 16 ion form protons electrons) atomic number: 12 The as than a mass (12 is having atoms charge, l charge v O resulting transferring oxide. remaining (10). The or neutral, magnesium with (12) by longer u f x a magnesium in no i form sometimes t a atoms r o Atoms The nuclear symbol notation o Figure 5 = protons d of n U atomic number C i n X Z number symbol for the element o = N v N + Z nuclear p mass number = a y Au. 79 A have r e 197 symbol notation of O t i s Z neutrons, n frequently l y A Chemists y 207 69 80 P Pb e 80 negative charge. The nuclear symbol for the oxide ion is oxide 2– O . 8 25 Structure 1 Models of the particulate nature of matter The overall chemical equation for the reaction between magnesium and oxygen is 1 2+ O Mg + Ionic in bonding is → Mg 2– + O 2 discussed further 2 Structure 2.1 2+ 2– Mg + O ions Ionic is more result bonds in a hold commonly force the of ions written attraction together as MgO, between to form as the them solid opposite charges on the known as an magnesium s s two ionic bond. oxide. e Activity Linking questions the 21 an ion with electrons, What determines the different chemical properties of atoms? (Structure 1.3) 24 and 28 How does the atomic number relate to the position of an element in the neutrons. periodic table? (Structure 3.1) l n y t i s Isotopes (Structure 1.2.2) O Isotopes are dierent atoms of the same element with a dierent number of neutrons. As a result, they have dierent mass numbers, A, but the same atomic number , Z. 35 37 Cl, and one 17 y r e Chlorine, for example, has two isotopes: one with mass number 35, Cl. They have similar chemical properties, as they are both with mass number 37 , 17 p chlorine atoms with the same number of electrons, but dierent physical properties, such as density, because atoms of one isotope are heavier than atoms of the other . is hydrogen-2 radioactive, consists of (deuterium). so it does two The U C (gure6), hydrogen i n and not o occurring (protium) v Naturally stable third occur in isotopes, isotope nature in of hydrogen-1 hydrogen, tritium signic ant quantities. Activity the table below n Copy and complete it by deducing the nuclear symbols and/ Isotope i hydrogen-1 Z symbol N A (protium) t a H 1 (deuterium) hydrogen-3 (tritium) u l a A portable tritium light Nuclear 1 hydrogen-2 Atomic v E Figure 6 d r o f x O o or composition of these isotopes. numbers of 1 3 isotopes are oen omitted in nuclear symbol notation. For 37 example, ‘Cl’ 17 , tells so you isotope the including written A , the with listed a for of chlorine isotope the chlorine atomic hyphen, each is number such element with as on mass and is therefore not 37 must necessary. chlorine-37, the number periodic or c an is not written as Cl. have an atomic number of These Cl-37 . The table be a isotopes c an also be relative atomic mass, whole number bec ause it r is the weighted average of all isotopes of that element. Natural abundance (NA) of an isotope is the percentage of its atoms among source. The radioactive dec ay of tritium produces high-energy electrons (beta particles). all atoms of the abundances given for all element isotopes of found an on our element, planet. we c an If we know c alculate the the natural average A of that r These electrons hit a uorescent material element. The opposite task (c alculation of natural abundances from A ) r and 26 make it glow in the dark only if the element is composed of two known isotopes. is possible y for protons, symbol P notation nuclear r Deduce Structure 1.2 The nuclear atom Worked example 1 C alculate the A r for iron Isotope using N atural the values in the following table. abundance (NA)/ % 54 5.845 Fe 91.754 s s Fe 56 57 Fe 2.119 Fe 0.282 We know A r = average abundance of the natural values add abundance to 57 × 100% of so each we isotope divide by multiplied 100 to by obtain their the mass numbers. average. r × 5.845 + 56 × 91.754 + 2.119 + 58 × y 54 A 0.282 = = 55.91 100 two stable of isotopes each of chlorine: isotope given Cl-35 that A and for Cl-37. chlorine C alculate is × NA of isotope 1) + (A of isotope 2 (35 × NA of Cl-35) + (37 × NA of Cl-37) = x = NA of Substituting Cl-35, then 100 in the above equation 35.45 the brackets 35.45 100 rearrange in terms of 3700 x = x: 22.5. x terms gives: Therefore, the natural abundance of Cl-35 is 77.5% and Cl-37 22.5%. a actual x = l 77 .5 and 100 v O The resolving the 3545 2 x = gives: u f x Then and Cl-37. t a 2x = r o 3700 of i 100 Expanding NA o x) = = d 35x + 37(100 x of isotope 2) n Let 35.45 U 100 NA C Therefore: × i n 100 o of isotope 1 v (A = r natural p Solution A the 35.45. r y (NA) r e are abundance O t i s Worked example 2 There l Therefore: n up y natural P The r Solution e 58 35 natural abundances of 37 Cl and Cl are 75.8 and 24.2%, respectively. Average A values for all elements r The results of our calculations are slightly dierent because we used mass numbers, are E which rounded are 35 values for the actual masses of the given in the data booklet and in 37 Cl and Cl atoms. the periodic table at the end of this book. 27 Structure 1 Models of the particulate nature of matter s s at e Density Melting Boiling point / °C point / °C Compound –3 r 4 °C / g cm 1 H 0.00 1.106 3.82 H 2 Figure 7 A pellet of enriched properties of uranium 235 used (increase as fuel in nuclear reactors nuclear in only 0.72% one type of of and in a U) at particular progress of t a (gure8). Her bomb stating “I will to work with is (Mt), a technology as such well Frisch In later being up is U. The enrichment (gure7), as most while c an This the natural also make oen have nuclei as the aer uranium it possible referred to as to years, the in large It from atoms releasing led tothe bomb. the discovery the has Meitner, Meitner developed ethic al, consequences. of atomic Lise doctorate led may economic development. named physics Otto 1939. is U, substance and splitting one fuel 235 applic ations cultural energy, receive Nature in atomic you nuclear history published in C an of their involves energy, o in of meitnerium i woman 109, which C amounts and social, n U ssion, colossal Element l u Austrian-Swedish physicist Lise Meitner in 1906 a E v O 28 d r o Figure 8 f x Nuclear development science politic al, for the o i n in environmental, nuclear reactions. Global impact of science Developments of least 3% of v isotope labelling. over with isotope. isotope mechanisms 238 U and used 238 U uranium this are p the require isotopes y Enriching track proportion of these r e contains to the reactors 235 uranium consists of two main isotopes, physic al O in n occurring dierences heavy water t i s Naturally Physic al properties of normal and 101.4 l Table 3 y 100.0 was US. the second University of Vienna of nuclear ssion, invited to work on She declined, famously have nothing to do with a bomb!” think of other ethic alimplic ations? scientic developments that have had important y 1.000 O P O 2 2 Structure 1.2 The nuclear atom LHA Practice questions Linking question 1. State the nuclear symbols numbers of 2. Naturally occurring protons and for potassium-39 neutrons in the and copper-65. nucleus of each Deduce the How isotope. c an provide sulfur has abundances: C alculate 33 S(95.02%), the isotopes average A with the following 34 S(0.75%), value for mechanism? 36 S(4.21%) and isotope tracers evidence for a reaction natural s s 32 four (Reactivity 3.4) S(0.02%). sulfur. r 3. The actual A value of sulfur is 32.07. Suggest why your answer to the question diers from this value. r (gure in a 9) is an instrument used to detect the relative sample. positive ions are (stage 5) (deflected most) accelerated to (stage 3) vaporize magnet sample are atoms c ations. into then lose For some Cu of (g) + 2e their and with vaporized high-energy electrons copper l atoms to form c an be (stage 1). The atoms electrons positively ionized as (stage 2). As charged ions, follows: a O → instrument example, + Cu(g) + e the bombarded least) o the known as injected u result, is sample f x a the t a sample within of a mass spectrometer r o The Schematic diagram S i Figure 9 (stage 2) d beam to sample (deflected n ionize U electron heaviest particles C N o inject sample v to i n inlet (stage 4) (stage 1) y filament electric p heating the r e field in O t i s detector lightest particles n isotopes y of l mass spectrometer abundance y P M ass spectrometry (Structure 1.2.3) The e r previous The resulting ions are then accelerated by an electric eld (stage 3) and deected by v a magnetic eld (stage 4). The degree of deection depends on the mass to charge ratio (m/z ratio). Particles with no charge are not aected by the magnetic eld and E therefore never reach the detector . The species with the lowest m and highest z will be deected the most. When ions hit the detector (stage 5), their m/z values are determined and passed to a computer . The computer generates the mass spectrum of the sample, in which relative abundances of all detected ions are plotted against their m/z ratios(gure 10). 29 Structure LHA u 1 Models Figure 10 of the particulate nature of matter M ass spectrum of a sample of copper 100 80 s s ytisnetni 60 r e evitaler 40 62 64 66 68 t i s m/z operational examination details of of 11 shows boron a from mass this spectrum mass from a sample of boron. C alculate spectrum. relative 80.1 then derive of 11, a which c alculate A 10 12 of boron the has o number c an to which i We 10, E mass need of u we number 8 a Solution First, 6 m/z M ass spectrum mass, 19.9 l Figure 11 4 v O 2 assessed in t a r o f x 0 d evitaler 0 be n U ytisnetni 50 atomic not o C i n 100 the v r will y , spectrometer p A mass r e Worked example 3 Figure the papers. O The information relative has by a from graph. abundance relative finding the of abundance sum of The 19.9%. the of peak at The m/z peak at = 10 m/z represents = 11 an isotope with a mass represents an isotope with a 80.1%. relative abundance of each isotope multiplied by its mass r number. The relative abundance values add up to 100%, so we divide the result by 100 to obtain the average. 11 × 80.1 + 10 × 19.9 = 100 30 10.8 n 60 l 0 y 0 y P 20 Structure 1.2 The nuclear atom LHA Data-based questions 1. Estimate atomic the relative mass, A , for abundance this of element each and isotope identify from the gure 12. Use your estimates to c alculate the relative element. r Figure 12 s s t M ass spectrum of unknown element 6 e 5 r 204 203 206 205 207 208 209 p m/z spectrometry of is used cosmic for discovering origin. For (gure14). and nickel of have these similar two properties metals are (gure13). they elements are relative c an in common atomic easily be masses. of iron by meteorites the mass isotopic spectrometry n 100 ytisnetni evitaler The samples, However, distinguished cobalt 60 40 20 0 58 60 0 62 58 60 62 m/z m/z the relative atomic abundance mass, A and of hence each isotope deduce for whether nickel. cobalt Use or your nickel estimates has the actual A value for nickel to c alculate larger r 3. geologic al components M ass spectra of cobalt (le) and nickel (right) Estimate relative identic al so o Figure 13 E i 0 specic nickel a v 0 of and 80 u 20 nearly dierent, nickel l O f x ytisnetni evitaler 60 cobalt t a r o 80 40 d 100 and very U compositions presence C Cobalt the example, o those i n including v M ass y r e 0 O 1 n y t i s 2 2. l evitaler 3 y P ytisnetni 4 its A r is 58.69. Suggest why your result in question 2 is dierent. r 31 1 Models of the particulate nature of matter LHA Structure s s e r to c an be practice in various databases average i n Relevant skills Tool 2: Identify • Tool 3: Percentages and extract U C • data atomic on the mass internet, values giving you from authentic o data. found c alculating y spectra chance v a in 1864 in the S ahara Desert p r e Mass spectra M ass found n l iron meteorite, O Tamentit from databases Instructions Using a database n 1. of your choice, search for the mass spectra of three dierent elements. From the mass element. i 3. Compare t a booklet. o d r o u l f x a E v O 32 2. your spectra, c alculate c alculated Comment on the relative any relative atomic dierences atomic mass you to mass that of each stated in the data observe. Linking question How does the fragmentation pattern of a compound in the mass spectrometer help in the determination of its structure? (Structure 3.2) y P y Figure 14 t i s Structure 1.2 The nuclear atom End-of-topic questions 5. Which of the following statements are correct? Topic review 1. Using your knowledge from the Structure 1.2 Nearly all mass of the atom is s s I. contained within topic, its nucleus. answer the guiding question as fully as possible: II. The mass number shows the number of protons in an atomic nucleus Isotopes of the same element numbers correct for 2+ Cu A. I and II only B. I and III only C. II and III only D. I, II and III ? Neutrons Electrons 34 27 B 29 34 31 6. Which of 34 63 31 34 29 27 A. values are the same for both 2 H and H 2 of combustion IV. density C. II and III only D. I, II and III Li. Which shows abundances for the correct lithium approximate lithium? u a 10 l 35 D Li 75 v O C The at gold gold page foil foil. experiment This involved ring alpha particles experiment is depicted in gure 1 on 20. An alpha particle is a helium nucleus. State the nuclear b. symbol Suggest that the would for an results have of been alpha the particle. gold foil observed in [1] experiment are following Li and each of the alternative scenarios: percentage i. Atoms of ii. are positive Atomic instead hard, dense, solid balls charge. nuclei are [1] instead negatively 7 abundance of Li charged. [1] 25 39 8. There are two stable isotopes of potassium: K and 50 41 K. The A of potassium is 39.10. Use this information LHA f x 6 50 7 . Percentage abundance of B cobalt-58 and cobalt-59 6 of Percentage A cobalt-59 and nickel-58 D. t a 7 isotopes C. a. i occurring numbers Extended-response questions o I and IV only cobalt-58 and nickel-59 n U I and III only B. naturally protons d A. The of equal o number contain cobalt-58 and nickel-58 B. C ∆H III. i n II. r o 4. boiling point v I. ? 2 species p Which following r e 1 3. the neutrons in their nuclei? y C D of O 29 t i s A n Protons y 29 y is equal l 63 What protons. P Multiple-choice questions of have r III. Exam-style questions 2. e How do nuclei of atoms dier? r 65 to determine isotopes 90 and the relative abundances of the two sketch the mass spectrum of potassium metal. E 9. “Dutch 14% [3] metal” zinc. oen used Dutch mass is This for metal an alloy alloy making c an be spectrometry. composed of 86% copper and closely resembles gold, so it is costume jewellery. distinguished Explain how from gold using [2] 33 Electron congurations Structure 1.3 s s How c an we model the energy states of electrons in atoms? This question is complex with many layers. What are electrons? How do we know they exist in energy states? What various e models about these energy states are there? of modern these no views, analogues clouds electrons in depend our on are quantum everyday the life, energies of objects we c an that behave visualize electrons, as both electrons which c an in have particles atoms only as certain, energy are produced electrons in by excited states Structure 1.3.5 — state electron for The line emission spectrum of provides evidence for the space existence of E ach c an there is has two a xed a high in discrete energy levels, which converge at Structure 1.3.6 — The main energy level is given an of 2 integer number, n, and c an hold a maximum of 2n convergence higher C i n Structure 1.3.4 — A more detailed model of the atom Structure 1.3.7 for an element describes the division of the main energy level into s, p, d an orbitals, regions emission spectrum, the limit frequency corresponds to — Successive give ionization information about energy data its electron conguration. n U o i Much of our understanding studies that involving sunlight prism. This c an u in which be l example of a a A gaseous pure glow — prism, E as an it 400 nm in to the within spectrum a prism and 1c) the is subjected series a continuous (gure into and spectrum words, a into congurations In the 1600s, dierent next, the as and atoms Isaac has come Newton from showed coloured components using a (gure appears in Sir a no rainbow. 1a). This type continuous gaps The are of series visible. spectrum of The colours, classic wavelength of visible light 700 nm. emission spectrum between down merges element other produces light. continuous spectrum continuous from electron with wavelengths, colour ranges will broken generates a each of interaction contains light of all v O f x t a r o d Emission spectra (Structure 1.3.1) lines 34 In at ionization. electrons. and f sublevels of successively higher energies. — of probability of nding an o v Structure 1.3.3 energy chemic al it of will lines (gure source to emit a high light. against 1b) of spectrum In dark contrast, visible will a voltage When light appear. under this light reduced passes background. when of all This a cold This gas pressure through a is is known placed wavelengths, a series of dark is known as an absorption LHA energies. and electrons of opposite p higher dened number electron. electrons a conguration hold contain where orbital y of hydrogen and Sublevels r e — given environment, levels. spin. Structure 1.3.2 a n lower spectra when and v alues. l to Emission photons such shapes O return — Although The y emitting t i s atoms waves. clouds. predened Understandings Structure 1.3.1 and fuzzy y sizes to has P behaviour r According Structure 1.3 Electron congurations a continuous spectrum s s spectrum r hot gas e emission b gas (c) visible light of all wavelengths C n U o i t a u l a v E Figure 2 d r o f x O y of all wavelengths (b) a heated o from (a) visible light p r e The spectra generated v Figure 1 passing through a cold gas i n spectrum n l absorption O cold gas y P y t i s c The aurora borealis (Northern Lights) in Lapland, drawn by the E arth’s magnetic eld to the polar regions, Sweden. Charged high-energy particles from where they excite atoms and the Sun are molecules of atmospheric gases, c ausing them to emit light 35 Structure 1 Models of the particulate nature of matter Emission spectra Emission spectra handheld be lamps when a observed by contain voltage through a simple holding it up low-pressure is to a gases light Method source. which 1. Observe natural light through the spectroscope. Note 2. O bs e r ve are down the details of the spectrum you observe. applied. L E D. Relevant a r ti fi c i a l No te d ow n l i g ht t he f ro m a details c o m pu te r of th e s c re e n s pe c tr u m Inquiry 3: Construct graphs and draw lines of best t 3. 2: Identify and record relevant Observe sucient relevant the Identify and describe patterns, 2: colours, Q uestions Assess accuracy Sketch 2. Describe the spectra each as you a observed. continuous, protection. 3. lamps will get very hot. Look the with up the Handle them c are. theoretic al • Further safety prec autions will be given by your on teacher, depending on the exact emission discharge the lamps and number, nature of the n eye discharge emission or spectra of the elements in you O The spectrum. t i s Wear • observe, y absorption you l 1. S afety • lines wavelengths and number of lines. observed. observed Compare the emission lines, commenting colours and positions of the emissionlines. Next, you will wavelengths of Discharge lamps • Handheld spectroscope orange light has its with Like i emission own on the characteristic For example, wavelengths colour c an be in of observed through accuracy a used shop to line a graph wavelength. your data. of the in and a that sodium observed test be of used chemic al which atoms 589.6 nm ame c an identify t a the line emission spectrum of sodium (right) spectrum, excited 589.0 appe ars a streetlights (le) and t observed wavelength data. barcodes spectra best vs and Construct Comment o substance. of wavelength lines. 6. element. yellow-orange line theoretic al Comment on the relationship shown in your graph. u l Sodium v E 36 Figure 3 d r o f x O same the a the emission o element identify Draw the 5. n to theoretic al C i n U E ach v • compare of y 4. M aterials p r e discharge lamps. c an emit (gure3, any to be used yellow- right). The sodium-containing identify elements. products, line y Inquiry discharge lamps. Note emission trends and relationships • the P 2: various of quantitative data. including Inquiry from details r and light qualitative down observations e o bs e r ve. Tool • or yo u skills • • s s Discharge ionized c an spectroscope Structure 1.3 Electron congurations Observations Chemists oen properties through instruments. boundaries of our through in the Sodium seen lamps is spectrum region also of of is in expand the to vapour lamps a strong spectrum. The light the more naked complex eye but the Figure 4 to a higher energy level by the heat of the ame. When • they fall back to a lower energy level, photons of certain • p of salts (e.g. LiCl, NaCl, o A metal ) 2 Clean the end of the ame test wire by dipping it into the HCl solution and placing it in a non-luminous Bunsen burner ame. Repeat until no ame colour is observed. 2. n • , CuCl 2 Method 1. heatproof mat various Dip the end Bunsen metal of and burner in the the ame place it in ame, salt and test the noting the wire into one of the salt edge of the non-luminous down the identity of the colour(s) you observe. o Dilute , SrCl nichrome) qualitative protection. prec autions 3. Clean the wire again and repeat with other salt samples. 4. Clear around open ames. up as instructed by your teacher. hydrochloric acid is an irritant. E • relevant a eye suitable v Take i record l O Wear and or hydrochloric acid t a and S afety • dilute of 2 C i n U u Identify observations • burner C aCl (platinum of samples samples, d r o f x 2: Small wire colours for dierent elements Relevant skills Inquiry Bunsen KCl, v visible region of the spectrum. r e wavelengths are emitted. Some of these photons are in the test portion n Small O Flame • y • ame tests is atomic emission. Electrons are promoted • l y Materials identify the presence of some metals. The principle behind Flame test emission spectrum t i s Flame testing is an analytical technique that can be used to Figure 5 Helium (gure 4). Flame tests natural gure 3, observing reveals the orange helium a revealing otherwise spectroscope yellow observing (oen sight), or with technology detail. As between directly and with the aid of an instrument? y helium emission a or light. dierence P from features the e light in is phenomenon c an be made senses observations, orange-yellow emission human What from observing the r the the data Observations Advancements imperceptible emit generate matter. s s directly of variety which • Dispose • Further of dierent are of chloride salts will be used, some Q uestions: irritants — all avoid contact with the skin. 1. substances Look up the emission spectra of the metals you tested. appropriately. Compare these to the colours you observed. Comment safety prec autions will be given by your on any similarities and dierences. teacher, depending on the identity of the salts being 2. Explain why the dierent metals show dierent analysed. amecolours. 37 Structure 1 Models of the particulate nature of matter TOK One of the ways knowledge up”: they take involves specic developed drawing observations is through conclusions and build reasoning. from general inductive Reasoning experimental principles reasoning c an be deductive observations. or Inductive (“bottom-up” a red ame test. gives a red ame test. to apply are your “top make down”: scientic the conclusion they infer knowledge in what grounds might On what grounds might we we classied of into gives each these a type two the following of claim reached through inductive claim reached through deductive E Visible light is one type of microwaves, The energy 1 E of ∝ λ the of premises: reasoning? a part existing observation types? a all approach): red ame test. doubt are You do this all the time pattern doubt light, 38 includes bromide disadvantages neatly v On and lithium premises. o be a always o i advantages that general 4. t a propose red ame tests. n U red ame tests. could reasoning knowledge salt. give from (“top-down” 3. u the give scientic lithium conclusions hypothesis d are salts you your a l C an O What this, is f x Lithium From suppose bromide r o example, Lithium salts context. reasoning theory 2. For new lithium C 1. specic a all i n deductive that y asked c an v when you p Deductive arguments salts: r e observations, lithium reasoning? reasoning? electromagnetic (EM) radiation. In addition to visible infrared radiation electromagnetic the radiation is (IR), ultraviolet (UV), X-rays and gamma spectrum. inversely proportional to the wavelength, λ: rays y gives iodide about O sulfate observations red ame test. n following Lithium these a the Lithium From gives l make chloride e might P you pattern y example, theory hypothesis observation Lithium “bottom approach): t i s For are r 3. 1. arguments from them. 4. 2. inductive. s s Inductive reasoning is Structure Electromagnetic waves all travel at the speed of light, 8 of light is approximately frequency of the equal radiation, f, to by 3.00 × 10 the c, in a vacuum. The 1.3 Electron congurations speed –1 m s following . Wavelength is related to the equation: c = f × λ energy EM waves, such as gamma rays, have short s s High wavelengths and high frequencies while low energy waves, such as microwaves, have long wavelengths and low frequencies. λm r 1 10 4 10 14 10 (γ rays) 10 ultraviolet 10 00 1 10 14 10 infrared 00 (IR) 4 10 1 10 v microwaves 10 10 p 10 Activity 00 waves o radio U C 10 10 i n 0 10 10 y ygrene (UV) r e elbisiv 10 O t i s 10 n y 400 10 X-rays 1 l λnm 1 10 0 10 y rays P gamma 10 e f Compare green in the colours red and gure 6. Determine which colour has: a. the highest wavelength b. the highest frequency c. the highest energy 4 10 4 Figure 6 The wavelength (λ) of electromagnetic radiation is inversely the spectra below. Explain how o at radiation i Data-based questions energy of that we know that stars are partly composed of hydrogen. u l a E v O f x t a r o d proportional to both frequency and Look n 10 3900 4000 Figure 7 7600 4500 5000 5500 The hydrogen emission spectrum (top) and 6000 6500 the absorption spectrum generated 7000 7500 from the Sun (bottom) 39 Structure 1 Models of the particulate nature of matter The line emission spectrum of hydrogen (Structure 1.3.2 and 1.3.3) E ach line which A idea that photon radiation = emission is as to a spectrum specic electromagnetic a quantum of of an element amount radiation energy, of comes which is has energy. in a specic This is proportional to the the specic energy possessed by the r E = photon, expressed in joules, J = constant, frequency of the 6.63 × 10 radiation, J s expressed The electron When that for a theory only the of the orbit returns rst to the a stationary orbits the lowest moves lower energy attempt to U making any i spectra. of the to of of model transitions the a energy based on its around the nucleus. level absorbs a photon energy level and remains level, it emits a photon of light. dierence between the main Classic al the two levels. problem of the electrodynamics predicted energy and quickly fall into the nucleus, atoms of energy higher overcome when wavelengths, measuring energies energy Bohr their specic By impossible. Bohr postulated that staying in stationary orbits. the atom could have only certain, well- between stationary orbits could absorb or emit producing wavelengths of characteristic lines in the atomic these lines, it was possible to c alculate electrons in stationary orbits. u l a E v O f x hydrogen atom, the electron energy (E ) in joules could be related to the n energy level number (n) by a simple equation: 1 E = –R n H 2 n –18 where R ≈ 2.18 × 10 J is the Rydberg constant. This equation clearly represents H the quantum only nature discrete, of the quantized half-integer parameters, number (n) c an numbers 40 in radiate existence radiate energies, photons a not electrons dened t a r o d Since would prolonged did o electrons electrons n orbiting atom were: Rutherford model of the atom (Structure 1.2). that inverse discrete energy levels with it hydrogen theory time. represents was in certain energy, short electron photon the his C Bohr ’s the in amount level of of o When This electron right a p at an exist model postulates associated with v the i n 3. of c an are or y 2. orbits main r e These proposed The Hz, O 1. Bohr spectra. hertz, n Niels t i s 1913, emission in y –1 seconds, s l f Planck’ s mean take values. where These known as only higher atom, positive energy. the energy values are of an electron characterized quantum numbers. The integer values (1, 2, 3, c an by have integer or principal quantum …), where greater y P h = For frequency of the follows: –34 the quantization: discrete packets, or quanta. h × f Where In wavelength, c alled e E the s s the in corresponds Structure The most electron stable has ground state c alled return state the of lowest of the atom. excited states. to the the ground hydrogen atom is the state at possible In contrast, Atoms state energy. by in the excited emitting This energy energy states of = is unstable specic 1, Electron congurations where the known as the levels with are photons n level 1.3 n = and 2, 3, … are spontaneously wavelengths (gure8). s s +energy e + + p e + p p dec ay r between amount electron c an level. of rungs amount of be of of ladders between a ladder. energy, with varying energy Jumping and levels, up distances between the much how each jumping rung down a like or rung level or you c annot requires a level releases energy. excited Electrons hydrogen to any energy returning to n = 2 level, will n, and return to any lower produce distinct lines in the visible (gure 9). v line. The energy of the photon released is lower when an electron falls from n = 3 to i n two of the allowable energy states of the electron in the hydrogen atom. 486 n = 6 n = 5 n C cyan 434 n blue 410 = 4 i u l n = 3 = 6 n = 5 n = 4 n = 3 n = 2 n red 656 = 1 a E v O f x t a r o n o from violet d transition U colour wavelength / nm o n = 2, than from n = 6 to n = 2. In both cases, it represents the dierence between p Note that the red line has a longer wavelength and lower frequency than the violet y spectrum exist r e energy the discrete same resemble c annot O An atoms t i s specic, the in n stand levels Electrons y rungs. l Energy Electrons returning to lower energy levels emit a photon of light, hf y Figure 8 hf P e excitation ◂ Figure 9 The visible lines in the emission spectrum of hydrogen show electrons returning from higher energy levels to energy level n = 2 41 Structure 1 Models of the particulate nature of matter n = 7 Electron transitions to the ground state, n = 1, release higher energy, shorter n = 6 wavelength ultraviolet light, while electrons returning to n = 3 produce lines in n = 5 the infrared region It important of the electromagnetic spectrum (gure 10). n = 4 is required IR to note that electrons between will allowable absorb energy or rele ase states. Any only the excess exact will not not = 2 and if an insucient amount of energy is supplied the electrons move. visible light levels closer to the nucleus hold fewer e Energy holds 3 has a level, electrons, at maximum of 18 n n, is 2n = 2 . For there example, could electrons, and n = be 4 a has the energy maximum a radiation 32 us convey concepts, our ideas we more sometimes Prepare a written explanation of atomic n = 2 is much smaller than that any diagrams. Exchange it with a electrons. Order • Whether have to diagram whether or are important shared make to voc abulary ideas each it other ’ s improvements accompany not given concepts adds to your the to are missing other feedback, from the explanation. feedback, spend some time using the your work. explanation. Finally, Discuss choose why a graph, image you chose it and explanation. n U C or any i n feedback which o you scientic in v When of each p Use • Give y r e • graphs or images to emission that does not include partner. n = 1 concentrating on: diagrams, clearly. The energy dierence between between n = 2 and use O help Linking questions qualitative i such as gas t a from o d r o f x gaseous How l u How does period and quantitative discharge do an tubes elements different a E v O 42 What emission and element’s in the data c an prisms in be the collected study of from instruments emission spectra from light? (Inquiry 2) spectra elements? number and provide evidence for the existence of (Structure 1.2) highest periodic = n explaining t i s When the electron moves further away from the y Communic ation skills ATL Notice how the allowable energy levels get closer together when n = 3 and of n occupied table? main (Structure energy 3.1) level 1 electrons, Electron transitions for the hydrogen atom. nucleus. level with eight l Figure 10 of maximum 1 UV = two energy relate to its y = to any P n n up in r electrons will electrons. The maximum number 2 of energy be radiation absorbed, n to move s s n = 3 Structure 1.3 Electron congurations The quantum mechanic al model of the atom (Structure 1.3.4) The Bohr model atoms. of was levels. narrow The model It assumed the the Bohr, incorrect levels. electron It of states electrons spectra these However, lines this of of electrons in existed in hydrogen discrete consisted corresponded to the model was limited by assumptions: predict electron. energy that emission wavelengths energy not the the idea the was was a emission only spectra of elements containing successful subatomic with particle in the a hydrogen atom. xed orbit about the not account for the eect of y could electric and magnetic elds on the spectral lines of atoms and ions. Heisenberg’s could not explain molecular bonding and geometry. The uncertainty principle states that it is impossible to principles the loc ation and momentum of an electron simultaneously. and stated that electrons exhibited xed momentum in specic these limitations, quantum the Bohr theory has been quantum the mechanics combines the less more we it the about know about is possible not trajectory c alculate the its of an probability f x region know of simultaneously. the position momentum, to pinpoint electron of nding in the an an of and This an electron, vice versa. we in u l boundaries are E What of the limits of Wave–particle duality subatomic species to is the The of released even whole atoms and interference (bending through or nature. around obstacles), all (combination obstacles) and of which are waves. Einstein (1879–1955) described by the in equation duality of the Schrödinger 1926 by the electron is equation, Austrian quantitatively which was physicist Erwin (1887–1961). Solutions to the Schrödinger give a series known as states and of three-dimensional wave functions, energies of mathematic al which describe the electrons in atoms. knowledge? ability behave characteristics to wave–particle as of both The concept that objects these of wave–particle duality illustrates the fact electrons and other particles and species, such as of study do not always fall neatly into the waves. discrete Certain diraction and of absorbed phenomena of light, but together they do. possible human c apable be particulate separately neither of them fully explains the functions, knowledge? electrons, are to their We have two contradictory pictures of reality; Schrödinger (1908–1974) implic ations of this uncertainty principle on tendency suggest (passing characteristic formulated a the v O the Bronowski are the photons, waveforms), Albert has been to prove that this aim is unattainable. and entities molecules, tunnelling each exact picture of the material world. One achievement … Jacob of c an One aim of the physical sciences has been to give an What small loc ation or atom, electron space. means t a predict particle r o Although we a However, states that it is accurately both the momentum i the of discrete n that position momentum as o the determine d and to idea of following key principles. Heisenberg’s uncertainty principle impossible the U with by o modern quantization superseded C i n TOK The eventually mechanic al model of the atom. are Structure 2.2 y of modern v the p Bec ause r e circularorbits. molecular geometry Bohr ’ s explained in model behind precisely bonding know O It 5. t i s 4. l nucleus. y 3. could one to the n It than and explain P more 2. bec ause allowable problems to r 1. attempt quantization: e several in an on According lines dierences was based s s energy It c ategories we have developed. What is the role mass, of c ategorisation in the construction of knowledge? 43 Structure 1 Models of the particulate nature of matter Schrödinger ’s probability wave density, electrons are path, theory this of region in There four space space are two uncertain. Instead gives at several electrons. atomic a types order of a electrons electron probability orbitals, has an follow will from the nucleus. An high atomic orbital are a that that and of increasing theoretic al, and energy these orbital shape are are a dened found an energy. labelled labelled number, These n, energy introduced levels are by split Bohr alphabetic ally model sublevels represents the comprised of and 3s. As n For example, increases, the s for n = orbitals 1, 2 are and 3, further the to the 12 shows away nucleus that, from for 1s, there is The is a There is zero the is true for nucleus with the two the high of that of zero probability an nding highest regions probability reaches highest electron the of nding when is at an electrons close move further be further away, found closer to the between even the two peaks. greater distance o C n 50 i o pm t a u l 50 100 pm average radius a E v O f x 0 3s 50 100 pm Figure 12 are 1s, 2s probability. 0 0 we somewhat could electron probability zero v U d r o 2s 2s, probability probability 3s, and i n 1s For small a never p from nucleus. there same probability orbitals from the nucleus. y nucleus. the this r e although and atomic O Figure t i s The s orbital c an hold two electrons s distanced n orbitals. y atomic where there is a 99% probability of nding 44 rst An s orbital is spheric al. The sphere represents the boundary space an electron. the into The s, p, d, and f. The plots of the wavefunctions for the rst three s orbitals 150 y Figure 11 quantum levels. is a l energy travel specic c an hold a maximum and P principal main a electron. (g, h, i, k and so on). The in atomic orbital nding each characteristic be r orbitals in of is electrons in atoms in terms of their idea that the momentum and position of saying distance there E ach of probability certain orbitals, Subsequent the where the e of describe Heisenberg’s s s region functions using Structure Imagine that 8.00am. they At could you are 8.15am, be. a Some possibly students no from might perhaps be at the airport • might even the have exact be school region where perimeter, the the town region of or teacher the is is where could a town 99% with be a the it drawn your library possible is a nding is of draw this cluster them. an a probability This loc ated, Similarly, probability to high around of school airport. high is there chance where includes space unknown, areas surface there that the c ar park Pole! showing boundary space around represents or a atomic nding an (gure13). t y y all describe There and z are axes boundaries the orbitals, These each are described with labelled p , p x highest and y probability of nding t Figure 14 The three p atomic orbitals are dumbbell shaped, aligned along the u z z x, y and z axes. There is zero probability of nding the electron at the intersection of the axes between the two lobes of the x x a x orbital l p v O f x z y with p 14). t a electrons in these orbitals. x three (gure y shapes y o These z x, i . shaped r o p dumbbell parallel to the o is x a sphere that encloses 99% n U p orbital orientations d A of the dots (right) C i n z z Representation of a 1s atomic orbital as a cluster of dots (le) and p r e v x Figure 13 O electron A the or school centre North dots the t i s orbital the of of in n certain teacher. region the laboratory, where teacher: y might the a of town or begin at y dene to loc ation cluster the oce the to wonder l nding to in gone chemistry that lesson you P three-dimensional of house suggest so r Although their class chemistry teacher, congurations e may • at the principal’ s DP your Electron s s • be room, your of your could school for sign is the sta waiting still • in the is • be in student there 1.3 dumbbell. E ach of the p orbitals c an hold two electrons y y p y orbital p z orbital E 45 Structure 1 Models of the particulate nature of matter Theories and models Current before. natural theories are theory are world. are evolved Contrary substantiated amassed, from previous comprehensive to by documented the vast and systems use of the amounts models, of ideas word of each that “theory” observations communic ated by a superseding model in and everyday and the one explain tested an that c ame aspect of s s the atomic Theories language, scientic hypotheses, which large number of scientists. r + + + + + + kana 1913 ohr model 1930 uantum pudding” model y o C n theories c an you are found of probabilit model in regions of high probability i of think u l f x a E v O 46 regions t a examples r o other and The atomic theory has seen the idea of atoms evolve from indestructible spheres to the quantum mechanic al model where electrons have specic energies and What 1926 Heisenberg’s uncertaint o Figure 15 d 1912 Rutherford’ s model model p r e “billiard ball” v i n U 1803 D alton’s mechanic al model O Democritus’ atomos 1897 Thomson’ s “plum t i s Āruņi’ s y 800–400 BCE n l + + of ? Linking question What is the periodic relationship table? between (Structure 3.1) energy sublevels and the block nature of the y + + P + e + + Structure 1.3 Electron congurations Electron congurations (Structure 1.3.5) E ach atomic spheric al oriented it has type the dierently. in energy has There than s a characteristic lowest or are possible ve d shape energy. orbitals and There and energy. The s orbital is are seven f three p orbitals, orbitals, and each these are s s higher orbital and p. z r x s z y For the s, p, Type d, principal only n = the 3, and there of u f x per 2 v E 3 p a s l O s three Total orbitals sublevel 1 are For types: n s, number orbitals per energy x = n, 2, p, p x and d. n x f 2 orbitals need c an hold there y f 1 number, exists. f. Number of (n) quantum orbital of quantum number s t a Principal p. 1, y f Only the shapes of s and O and by = p s types: n f orbitals. i four For and z o orbital: are dened 1). d C level (table p, x 0 2 z y f o of The shapes of the s, r o there –1 x d z n energy orbitals types x f d of i n x x 1 y U E ach z y –2 y d v z f Figure 16 x 0 y x y d –1 z y x d z –3 y r e x f z y –2 y 1 z y z p 0 z d t i s z x p –1 n x p y x y l y y z P z e y 3 to be known types are two For n = 4, M aximum number electrons of within type 1 2 level (n 2 ) energy level 1 2 4 8 9 18 (2n ) 1 3 s 1 p 3 d 5 t Table 1 c an hold 2n sublevels, s 1 p 3 d 5 to n. (s, 4 E ach energy level, dened by 2 n, 16 p, electrons. The number of or atomic orbital types, is equal For n = 4 there are four types of orbitals d, and f ) with 16 atomic orbitals in total 32 2 occupied by a maximum of 2(4) = 32 total electrons f 7 47 Structure 1 Models of the particulate nature of matter Activity State the following for the energy level with sublevel types b. the number c. the total number of atomic orbitals d. the of maximum atomic orbitals number of in each electrons at n = 5: sublevel that energy level. r convention, represent s sublevel (one box in notation arranged orbitals is Arrows are drawn in the boxes to represent so each box has a of two “arrows” representing the three p orbitals p , p x , and p y ) z five d orbitals) n o sublevel boxes i d (seven boxes representing the seven f orbitals) u Atomic l a E v O f x t a r o f the o U (five representing C sublevel boxes used c alled electron conguration v d (three i n sublevel is (gure 17). The y of two electrons p orbital diagram orbitals p maximum r e A maximum c an occupy each orbital, c alled an atomic representing an s orbital) shows the number of orbitals for each sublevel. electrons. in n each box This diagram electrons box” are O In orbital diagrams, in t i s Figure 17 represents an orbital. of “arrow electrons y arrangement u an how l to orbitals electrons. two electrons mechanics pair of regions are should solves electrons directions. one are Electrons Hence downwards not this with of space charged be able problem opposite each orbital half-arrow, where there negatively, to by occupy using spins box is (gure and a ± the is a like high same spin probability of nding charges region notation for repel of each space. each other, so Quantum electron. A behave like magnets facing in opposite shown 18). This with is an upwards known as the half-arrow, , and Pauli exclusion principle: Only two electrons c an occupy the same atomic orbital and those electrons must have opposite spins. 48 y P Orbital diagrams For e the s s a. Structure N t S Figure 18 1.3 Electron congurations Electron spin is represented by an arrow pointing up (positive spin) or down (negative spin) s s N in an orbital what of our shows the mathematics extent does limits as the atomic mathematics the orbitals same of the energy life degenerate 3s electrons in 2p Unfortunately, neither c an and basis: around its be c an visualized in any way, be 2s expressed only in degenerate of human support same are perception type of and, at the same 1s time, science. 1 knowledge 2 3 development in the n Figure 19 The three 2p orbitals are degenerate as they have the same energy. in one referred to as sublevel are degenerate of equal orbitals energy. (gure 19). These three degenerate atomic orbitals have lower energy than the three 3p orbitals a E v O with physic al rotate. electrons language sciences? the no everyday u of Orbitals in l E ach analogues f x natural behaviour has c annot electron form. This lack of visualization does not undermine the quantum rather of wave the i To but power a of t a the wave-like and r o theory interpretation waves, rotation 3p ygrene no mathematic al the o have the this as 3d degenerate o they nor like interpreted C However, behave spin oen d as is n axis. atoms the spin U Electron own v i n TOK n of opposite spin y representing p electrons r e half-arrows O t i s magnet analogy y S l S e r N P N y S 49 Structure 1 Models of the particulate nature of matter An atom of boron (B) has ve electrons, and its orbital diagram is drawn as follows: 1 2p s s 2 2s boron (B) e 2 1s single equal show lemost that every occupy any orbitals are equivalence. it is a matter of personal occupied same with spin an in each electron of of that them opposite gure which of A. the the reason 1s 2s 1s 2s i 1s f x u l doubly occupied for the 2s 2s four orbitals orbitals below Hund’s drawn are are fully and incorrect the a any correct by the c an electrons. 2p orbitals? electron exclusion being orbital (gure 20). before Pauli diagrams spin occupied lled represents rule three p orbitals must principle. wrong. 2p 2p 2s 2p 1s 2s 2p 1s 2s 2p t a 2p across the three degenerate and and diagrams on is D. The electrons are evenly orbitals in nitrogen before an orbital is o C. 1s 1s n U 2s Figure 20 distributed 2p r o 1s 2p d 2s the based B. 1s The C i n conguration State 20. think o State you v 2. at do p Why half-arrow y r e Look have all electrons in singly the before Practice questions 1. the they joined n the as boxes O doubly with by preference. and that means t i s become electron orbitals, degenerate orbital in a sublevel is singly occupied orbitals have the same spin. This one three Traditionally, of occupied before any orbital is doubly occupied have the represented l states energy although 2p y Hund’ s rule c an degenerate their box, boron E. a E v O The Aufbau principle states that as electrons are added to atoms, the lowest available energy orbitals ll before higher energy orbitals do. The third and fourth energy levels contain d and f orbitals (gure 21). These orbitals are typically lled aer the s orbitals of the following levels because they are higher in energy. As shown in gure 21, the 3d sublevel has a higher energy than 4s but lower than 4p, so 4s is lled with electrons rst, followed by 3d and nally 4p. For the same reason, 4d orbitals are lled aer 5s, and 4f orbitals are lled only aer 6s. 50 y to The in P the electron energies. together in 2p r The Structure t Figure 21 energy and 1.3 Electron congurations The 4s sublevel has a lower will ll before the 3d sublevel 4f 6s 5p s s 4d 5s ygrene 4p 3d e 4s 3p r 3s Ca with have experimental electrons in the data 4s that show that potassium, K, and sublevel, not in 3d. t Figure 22 Potassium orbital lling diagram electron in the 4s orbital bec ause 3d 3d order is o following i the observed: Activity u f x t a Generally, d r o 1s o U 2s orbitals are higher in energy n ygrene 2p C i n 3s v 3p p 4s showing the outermost y r e 4p n consistent O is c alcium, t i s This l y 1s y P 2p 2s 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p … Copy the orbital diagram from O l In the IB Diploma Programme, only the electron congurations of atoms and ions gure21 and complete it for the up to Z=36 will be assessed. Electrons in these species can ll sublevels up to 4p. sharing There 1. Full so it are is important three E ion. electron transfer v reactions, and a Electron ways are to to following know show the the to electron electron understanding Condensed Orbital The orbital ground a. aluminium, Al b. chlorine, Cl c. iron, conguration of an atom or an Fe conguration electron lling lling their conguration: Refer to the periodic table at the of this diagram diagram book to deduce the conguration number 3. in chemic al back 2. elements states: fundamental (“arrows for in of electrons in each atom. boxes” notation) potassium is given in gure 22. 51 Structure 1 Models of the particulate nature of matter Full electron congurations To write the a full electron electrons Rule and the in conguration, successive Pauli orbitals exclusion we use the according periodic to the table, Aufbau and “build principle, principle. s s Worked example 1 Determine the full up” Hund’s electron conguration for the c alcium atom. e Solution r The Aufbau principle states that as electrons are added to atoms, the lowest exclusion principle, we know that each orbital will have a maximum of two l electrons. n y The atomic number of calcium is 20. Let’ s split the 20 electrons evenly across each orbital, starting with the lowest energy rst. When writing electron O t i s congurations, the number of electrons within each sublevel is given in superscript, next to the sublevel: 2 2s orbital • The three • The 3s • The three has orbital 2p has conguration full brings electron next So, for for the phosphorus us up lowest have three 18 electrons, electrons with two 2 the full electron 6 each, six in total: 2p 2 electrons: 3s energy, 4s: 4s c alcium, atom. to electrons U C the two o the i n Determine 2 electrons: 2s v This 3. two have two orbitals Practice question electrons: 1s has orbitals orbital 3p two also y 1s The le p The • r e • 6 each, six in total: 3p over to go into the orbital with 2 conguration is 1s 2 2s 6 2p 2 3s 6 3p 2 4s Condensed electron congurations the gets mostly electron c an congurations core group 18 an electrons as is element by their inner to the the full electron conguration write. The chemistry of atoms and valence electrons, that is, the outermost the same A valence electron more convenient electrons and conguration way of writing represent the as the previous noble gases) element in the periodic table: u Condensed electron conguration = [previous noble gas] + valence electrons Table 2 shows some more examples of full and condensed electron congurations. l a Examples of full and condensed Condensed Atomic Element Full electron conguration electron E v number conguration 2 O 8 1s Ne 10 1s Mn 25 1s 2 2s 2 35 1s 2 4 2p [Ne] 6 2p 2 2s 2 [He] 2s 6 2p 2s 2 Br 4 2p 2 2s 2 52 to core electrons. highlight having (known as the increases, time-consuming determined t a inner of be rather than the electron congurations for selected elements it o is electrons, number and i d r o Table 2 f x O u ions atomic longer n As 2 3s 6 2p 6 3p 2 3s 2 4s 6 3p 5 4s 2 3d 2 [Ar] 4s 10 3d 5 4p 5 3d 2 [Ar] 4s 10 3d 5 4p y P available energy orbitals ll before higher energy orbitals do. From the Pauli Structure the condensed electron conguration for the c alcium Determine electron Solution the condensed conguration example 1, 2 c alcium to be 1s we 2 2s determined 6 2p 2 6 3s the full electron conguration of 2 3p 4s . The previous noble gas in the periodic table is argon, which has an atomic 2 2s 6 2p 2 3s e 2 number of 18. Argon has an electron conguration of 1s 6 3p , and we 2 also conguration. sometimes The be condensed shortened orbital by using diagrams for a condensed periodic oxygen and manganese are shown t i s in 2p 3d the Write key ve answer of the should that test key, concepts how to than a and skills: electron sheet of A4 from the quantum congurations. paper. is discussed further chapter. know your then write try from this chapter. understanding them out on of the ideas in this o an to longer from you questions M ake no takeaways voc abulary brief range one i t a r o d chapter. summary, key a atom, This n List chapter span the C three of i n a Write topic U Write this model sublevel o v Self-management skills in. of electrons of elements y 4s in structured Structure 3.1. p r e manganese: [Ar] ideas is type O 2s mechanic al the n y valence appear oxygen: [He] The table below. that ATL to l according y c an The electron P diagrams . r can therefore write the condensed electron conguration of calcium as [Ar] 4s Orbital for the phosphorus atom. s s worked congurations atom. 4. In Electron Practice question Worked example 2 Determine 1.3 of your peers. Exceptions to the Aufbau principle The Aufbau elements. the for Mn, with These 3 3d at . with are conguration the (Sc) valence periodic to oxidation states are 3d order of lling electrons to of atomic form ions, some table copper 2 [Ar] 4s are orbitals the lost for electrons rst. 5 3d , the 4s electrons will be lost 2+ electrons known as the 3d E look sc andium There the lose 5 , with electron conguration [Ar] 3d , a 2 elements you electron gives the manganese ion, Mn v O This ions. atoms sublevel with the highest principal quantum number ( n) not [Ar] 4s All predicts when l So, rst. correctly However, u in principle f x most (Cu). at tend to lose two 4s transition elements, or the back These of this transition book, metals electrons to form 2+ transition metals. If these c an elements also have are from variable in compounds. exceptions. With only one electron in its 3d orbital, sc andium 3+ readily forms only Sc ions, by losing this 3d electron and the two 4s electrons. 53 Structure 1 Models of the particulate nature of matter The Ionization and oxidation ground those covered in Structure 2.1 state congurations of copper and chromium are also dierent from are predicted by Aufbau principle. and 2 Structure 3.1 The predicted electron conguration of copper is [Ar]4s 9 3d , as the Aufbau principle suggests that the lower-energy 4s orbital should be lled rst. However , the observed 2 chromium, the predicted conguration is [Ar]4s 10 3d s s 1 ground-state electron conguration for copper is [Ar]4s (gure 23). For 4 1 3d and the observed is [Ar]4s Activity 5 3d (gure 23). In each case, promoting a 4s electron to a 3d level leads to a more stable e electron conguration. In the case of copper , this gives a full d sublevel, and in the Deduce the electron conguration case of chromium, there are no paired electrons but rather six half-occupied orbitals, 2+ of the Cu c ation. r each containing an electron with the same spin. The expected Cu (Z = 29) Cr ( Z = 24) electron l observed expected [Ar] configuration chromium 9 2 3d 4s 2 4s O [Ar] [Ar] 10 1 are that lled you of need chromium to with (Cr) and copper (Cu) electrons according to the general 1 4s are the only know. In all other elements up to Z=36, the order. i n o v sublevels congurations y exceptions 3d p electron two r e The 5 4s 3d LHA 4 3d t i s observed configuration n y [Ar] congurations of copper and The quantum mechanic al the minimum molecule X(g) The periods. 18. The electrons for p, the the atom helps to to eject an electron explain the elements. out of a trends and Ionization energy neutral atom or → X (g) + e f table the c an be (gure element u energy (IE ) are known periodic shown 24). are as The also table, four as groups, the groups blocks sublevels and are the rows corresponding holding are numbered the to known from 1 to the outermost four valence shown. generally decreases down the groups of the periodic 1 l and a Going increases down electrons energy are a shielding group, and the number from (so-c alled therefore the “inner less of pull sublevels of the electrons”). energy is increases. The outermost nucleus by The required the more to electrons sublevels, remove in the the electrons lower greater from the E sublevel. ac ross a outermost nucle ar across the periods. shielded sublevels outermost Going period, the ele ctrons charge. At constantbe c ause 54 of state. periodic and each required across table d, ionization table the in model ionization energies (IE) of + energy periodic s, the energy Going sublevels First v O f x t a Structure 3.1 as + columns discussed further in rst itsground i are in the o energy and periodic trends r o table d Ionization in n is U discontinuities C Ionization energy (Structure 1.3.6 and 1.3.7) the the number are held s ame time, number of of protons closer the inner to the in the nucleus nucleus shielding ele ctrons by ee ct does the inc re ases, inc re ase d remains not ne arly change. so y Figure 23 and P u Structure 1.3 Electron congurations LHA s-block 1 18 transition elements 1s 1s 2s 14 15 4 5 6 7 8 4f 5f trend period is re quire d of ac ross decreasing shown remove ionization gure 25: 2 Period outermost period. energy down a ele ctrons, group 3 Period 4 2500 Xe Al Na Rb K 10 2 Kr a E 0 u l Li v sriF O o i a i o i f x gree B 500 i O Be 5 Ar N H 1000 Period t a r o lo 1500 increasing o 1 Ne 2000 d He and so n Period in to the U a is inc re ases p general across energy o more energy p, d and f C ionization to the sublevels s, i n Therefore, The blocks of the periodic table correspond v Figure 24 y r e f-block 7p y 6d 6p n 7s l 5d O 6s 4p r 4d 12 P 5s 11 y 3d 10 t i s 4s 9 e 3p 3 The 17 2p d-block 3s 16 s s 13 2 18 36 54 Aoi ber Figure 25 Plot of rst ionization energy against atomic number for the elements from hydrogen to xenon 55 1 Models of the particulate nature of matter LHA Structure There are two clear discontinuities 1. Between the group 2 The valence The paired 2s across the period: and group 3 elements 2 electron conguration of 2 beryllium is 2s while for boron it is 2s 1 2p . 2 the electron shield slightly the easier single to 2p electron in boron from the nucleus, s s making electrons remove. e + Be Be 0 0 2p 2p 1 2s r 2 2s B 0 2p presence results that of to and The — not t in be any trend period. aluminium, the allows s ame c an be observe d in comparing group 2 to group 2 collect. For so example, the rst the 3s ionization ele ctrons energy magnesium. of further 3 elements 1 shield the aluminium is lone 3p lower ele ctron than that in of y conclusions patterns discrepancies do pattern for they r e overall out data n — look the O The in y trends 2s t i s Scientists 2 2s Patterns and trends l 1 2p 2 Suppose you have a two-story building and you need to remove one oor to meet drawn. new height regulations. Which oor would you remove? Obviously, it will be the p What can be inferred from the top oor, as the building would collapse otherwise! The same reasoning can be v patterns in successive ionization applied to the ionization of atoms — electrons are removed rst from the highest o energies? occupied energy level, and from the highest energy sublevel within that level. Between the group 15 and 16 elements group 15 conguration Nitrogen has a same p three and (gure region t a the is stable o nitrogen there more sublevel, i the p 16 also 2 of electrons space in 2s the drop This and 2p more is in ionization energy. The 3 2p electron therefore 26). a 2 nitrogen is 1s 2 while energy is the increased orbitals oxygen it is 1s conguration bec ause have for do not than required paired to remove electrons into in close N 3 2 2p 2p 2 2 2s 2s + O O 4 3 2p 2p 2 2s 2 2s electron in occupy nitrogen, proximity. + N an oxygen However, sublevel is sublevels with 2 or 4 4 2p oxygen as it has a half- repulsion. come electron 2 2s a E v O 56 from u A half-lled l electrons d r o f x Figure 26 more stable than p lled to of n U From u C i n 2. y P + B Structure 1.3 Electron congurations 6 most stable p orbital conguration is p , a completely lled p LHA The sublevel, 3 followed by p , a half-lled 10 example, d why sublevel. This is generally true for other sublevels. For 5 and d chromium and electron copper congurations do not obey the are also Aufbau stable, which principle partly explains (gure 23). s s C alculating ionization energy from spectral data As the principal lines the quantum levels converging in number converges the to hydrogen of a energy levels continuum. emission increases, the distance This c an spectrum, be observed shown in by gure spectral 27. r 5 level 4 Figure 27 Ultraviolet and visible light transitions in hydrogen and O high energy low energy p visible light C i n o light v ultraviolet y r e level 1 n 2 y level t i s 3 the resulting l emission spectrum level y t ∞ level P level e between –8 The spectral lines in the hydrogen emission spectrum converge at 9.12 × 10 m, or 912 Å (gure 28). This represents the wavelength of light at which the hydrogen wavelength c an be used to energy of hydrogen. o i a 1000 ionization t Figure 28 Hydrogen is ionized at the wavelength where the spectral lines converge in the emission spectrum u l 950 rst t a r o f x E v O 900 the d 912 Å c alculate n This U atom is ionized. 1050 1100 1150 1200 1250 Wavelength / Å 57 1 Models of the particulate nature of matter LHA Structure Worked example 3 8 Spectral lines converge at 9.12 × 10 m in the emission spectrum of the 1 hydrogen atom. C alculate the rst ionization energy of hydrogen in kJ mol s s Solution First, c alculate the frequency of radiation using to 3.00 8 of light, approximately 1 × 10 m s = c alculate f the = 2.18 × 10 × (s 9.12 × 10 energy × using 3.29 Planck’s constant and the s values of the J two steps c an be merged into The constant are one h Avogadro’ s λ This given in the data mole and represents absorbed Avogadro’s level, discussed further in or in the energy of a single photon of light which would be exciting removing the one electron electron in a hydrogen atom to the energies are usually given 1 energy and kJ mol the using in following kJ mol . You ionization in equation: needed electron 23 –1 mol i n 1000 3 1 U C kJ mol 1.31 × 10 one 1000 J × 6.02 × 10 = = remove from an atom) × N A o –18 2.18 × 10 to v = kJ mol p (energy energy convert the ionization , the number of atoms A 1 The c an Avogadro’s constant (N y 1mol) to r e in value f convergence from the atom. 1 Ionization Structure 1.4. × by using the O booklet. and = t i s are = n constant E y h × c constant E speed of light, equation Planck’ s equation 1 × 10 speed l The the m 15 J s these is ) 18 Alternatively, c Worked example 4 rst ionization booklet. the emission helium i the atom, the spectral –8 lines converge at f x C alculate the 5.05 × 10 m. The rst in ionization 590 kJ mol wavelength Å, for energy of –1 is the of . nd from kJ Na is 496 of kJ mol as given convergence for by the the IB data sodium atom the J energy and Then of ionization dividing it by 1 23 / 6.02 × 10 for one atom by converting the given value Avogadro’ s constant. 1 mol 19 = 8.24 × 10 J h × c c alculate the wavelength of light using E = λ –34 6.63 × 10 8 J s × 3.00 × 10 –1 m s 19 8.24 × 10 J = λ C alculate 7 λ = 2.41 × 10 m convergence, = 2410 Å. c alcium atomic spectrum. This corresponds to the UV region in the electromagnetic spectrum. E Successive ionization energies It requires atom while 58 to a the , of helium. v O c alcium kJ mol l 6. in u energy, First, 496 000 J mol rst ionization –1 of wavelength Solution spectrum of t a In r o 5. Å. the o d Practice questions in 1 energy C alculate spectrum n The more bec ause the energy the to remove number of electron–electron the protons repulsion second and exceeds the decreases. successive number of electrons remaining from an electrons y 6.63 × 10 λ, where × P = f 1 Hz 34 E = r Then, 3.29 × 10 c 1 m s 8 = 15 f × 10 e 8 3.00 equal Structure a the so result, electron increased that only remove the clouds electrostatic the stable next are noble electron pulled attraction. gas closer Once to all conguration increases sharply, as the the nucleus valence remains, shown the in and held electrons energy gure tighter are Electron congurations LHA As 1.3 by removed required to 29. Figure 29 ygrene from s s t Removing 10 electrons magnesium gives the noble-gas 2 conguration 1s or [He]. There is a 1– lom Jk( remove the 11th electron r 2 3 4 5 6 7 o 8 9 10 eletron 11 12 remoe The rst ve successive ionization energies for an unknown element X have 1 the group of values: the 403, periodic 2633, table 3860, in which 5080 and element 6850 kJ mol X is likely to be increase in energy occurs from the rst ionization second removed a stable energy to group level 1 of ). This noble of the the means gas that the second conguration element periodic contains table. of the one the periodic 5.5 v 3.5 E 3.0 2427, 3660, values: 25 026 and –1 32 827 kJ mol . Deduce the group of the periodic table in which this element is likely to befound. the two large jumps in the a gol l EI O 4.0 unknown element following u f x 5.0 4.5 explain sodium. i 6.0 table, for Therefore, the 801, the successive ionization an t a r o energies o and ) to of n U 30 ionization d gure successive ve electron, so the element Data-based question Using rst have 1 (403 kJ mol electron is likely to be atom. The energies C outermost belongs (2633 kJ mol from i n the 7 . o 1 Practice question found. v largest Deduce p Solution The . y following r e the O t i s Worked example 5 n Numer y 1 l 0 y P n o it a z i n o I e ) considerable increase in energy required to 2.5 1 2 3 4 number Figure 30 5 of 6 7 electrons 8 9 10 11 removed Successive ionization energies for sodium 59 LHA Structure 1 Models of the particulate nature of matter Ionization energy data Relevant Part 3: Graphing the logarithm of the ionization skills Tool 2: Extract • Tool 2: Use represent • data energies from databases 6. spreadsheets to manipulate data and data in graphic al Title the third (ionization form column energy)” in as your spreadsheet “log shown below: T ool 3: Construct and interpret graphs Part 1: Data collection C element name: 1 r 1. B Identify a database that contains successive 2 ionization energy data for the elements (for example, Choose one of the following elements: (ionization energy/ sulfur, l –1 chlorine, successive ionization labelling the energy data in a columns as 4 1 follows: 5 2 B 7 . Compute using a ionization energies 5 2 6 etc sure sc ales your and graph a 10. title. that outermost electrons. energy increases l the v E does properties trend in of metals large a increases change in in ionization main 12. each graph that electrons and “zoom will in in” allow energy IE values and across non-metals? C an you see any to of period is later the role of how and (Structure down LHA 60 do patterns variable of successive oxidation states of ionization these data is oen into of 2. Enlarge increases a group explain transition (Structure 3.1) graphic al of the elements relate to the organized into tables graphic al scientic knowledge? employed in other trends help to forms. What representations in the ] and ionization energies? (Tool 3, Reactivity 3.1) energies = increases in 3.1) elements? n the they + the energy level. you to closely inspect large Explain transformed advancement Why are log scales useful when discussing [H How energy) sublevels. Experimental and unusually energy? representations a energy level a O in graph existence with a the ionization existence of main energy levels in the atom. How successive closely. 11. Explain how the graph provides evidence for the Linking questions Construct the u f x electron. of (ionization explain Are subject graphic al areas? KOT ionization correspond to the i values logs log following questions: the indic ate examine suitably, with descriptive the plotting energy energies? vs ionization. t a why r o IE and successive c. present energy following questions: Identify the Explain you suitable the innermost b. ionization number. ionization LOG10) function. Part 4: Evidence for the existence of sublevels o labels, Answer a. that of d axis graph by (or Why is it useful to plot the logs of the ionization n M ake line the Identify b. U a a. that Part 2: Graphing successive ionization energies Plot Answer each showing o 1 ionization C i n 4 v 9. –1 kJ mol graph of LOG p energy/ ionization logarithm Construct vs 5. the spreadsheet y 8. ionization 3 the r e 2 4. etc element name: 1 O 6 A energy) kJ mol ionization 3 c alcium. t i s spreadsheet, or n Collect potassium y 3. argon, y P log ionization WebElements). 2. e A Instructions s s • Structure 1.3 Electron congurations End-of-topic questions 5. What is the maximum possible number of electrons in Topic review 1. Using your answer knowledge the guiding from the question as Structure 1.3 fully as third energy s s the level? topic, A. 3 B. 6 C. 9 D. 18 possible: e How can we model the energy states of electrons inatoms? the is the electron ground conguration state? Multiple-choice questions 7 A. Which row is correct for the following [Ar] 3d regions of the 2 [Ar] 4s D. [Ar] 4s 4 3d 1 (UV) Infrared low energy wavelength energy 1 (IR) low 7 . A. Which of frequency A. IE 5 4p the > 3 frequency low energy long short high long C. energy wavelength long low low frequency wavelength frequency energy The the following an III. gas an alkali metal salt to Which produce electron a line in n = 4 to n = 1 B. from n = 4 to n = 3 C. from n = 3 to n = 2 D. from n = 5 to v from E A. n = 3 Bunsen burner energy levels n = 4 and produces a line in the visible transition the a O likely between atom a 8. UV Ionization Which are energy 3+ (g) → X energies statement nitrogen o spectrum. transition isolated u an l f x in in i II and III only I, II and III placed 2+ produce a A. IE and (N) < IE 1 B. IE of measured in kJ. the atom X represents – (g) + e decrease across a period going (N) IE > in IE (N) its a < (O) IE (O) (N) higher > IE 1 The bec ause is ionization energies of correct? partly oxygen lled has two paired bec ause two paired sublevel lled oxygen has sublevel bec ause oxygen loses an electron an electron 1 rst sublevel (O) than bec ause nitrogen oxygen loses 1 from 9. its partly 1 IE rst 1 from D. (O) the atoms 1 in 1 C. about oxygen electrons electrons t a C. D. will prism? d I and II only I and III only r o A. electron a discharge tube B. n = 2 light inc andescent lamp ame An of behind process: X D. energies ionization from le to right. n a II. sources placed U I. when third C of spectrum i n Which line 4. C. ionization o wavelength high v frequency D. 3. Molar the correct? 4 B. p high wavelength is y low energy r e high B. following IE LHA short t i s Ultraviolet high 5 3d n [Ar] 4s C. spectrum? O B. y electromagnetic chromium (Z = 24) l 2. of y in P Exam-style questions What r 6. a higher ve unknown sublevel successive element are than nitrogen ionization 578, 1817, energies 2745, for an 11 577 and –1 in the same atom is spectrum? 14 842 kJ mol this element A. 1 B. 2 C. 13 D. 14 is . In which likely to group of the periodic table is be found? 61 Structure 1 Models of the particulate nature of matter 16. Extended-response questions Sketch an number 10. Explain, in your own words, why orbital of lling unpaired diagram for Al and deduce the electrons. [2] gaseous atoms 3+ produce line spectra instead of continuous spectra. [3] 17 . A transition element ion, X , has the electron 5 conguration [Ar] 3d 11. State the full and condensed electron congurations for following species in their atom silicon d. Ti the condensed germanium and orbital lling containing atom one or more electrons. [1] 19. 3+ c ation Describe, in your own words, how of an atom c an be determined anion [1] spectrum. 12. Determine which of the congurations below is why it c annot exist. Using the 2 2 2s 7 2p 2 3s Deduce and 2 1s The the 3p shape diagram of below energy 4s an s (not levels 10 3d 2 in state. 5 4p 10 4p to sc ale) the to [1] represents some of the = 6 X the the n 1 3 = 5 o represent of successive element are ionization given found. 2 = 4 n = 3 n = 2 n = 1 lowest emission [1] n n i to region be 4 t a hydrogen. visible four Table 3 C U diagram the = 7 n u of the in n n d spectrum 62 on trend and discontinuities in in energies table3. [3] for an Deduce the group of the periodic table in which element X is likely 5s a arrow transition rst unknown 1 hydrogen atom. l v E an energy The 1 orbital. r o f x O Draw 21. [2] 5s 6 3d electron i n electron 3s ground 2 4s 6 following p 15. 2p 6 3p 2 the general rst ionization from sodium to energies. o Sketch 2 3s 6 a of v 14. 2s represents 6 2p 2 which the the elements, emission [2] y 2s 2 these r e 2 1s Explain 6 3p explain, explore 3 5 2 3s congurations period its O 13. 6 2p booklet, the t i s 2 2s of 3p argon. 2 1s rst ionization from n energies 1s data [2] y Explain [2] l 20. impossible. for [1] –1 IE / kJ mol n 738 1451 7733 10543 Successive ionization energies for element X y P energy S the [1] 2– e. diagram deduce the total number of p orbitals [1] LHA c. Sketch [1] atom e selenium r titanium b. [1] ground states: 18. a. X. s s element the . Determine the atomic number of Counting particles Structure 1.4 by mass: the mole s s How do we quantify matter on the atomic scale? atoms of the in a glass oce ans substance, the of of so any these water than combined. mole, physic al particles. The enables glasses unit of chemists object comfortably There the of the to are water same relative amount for with time, large the molecular expressing numbers concepts masses masses of of of very molar, allow the atomic small particles. relative use of atomic small substance. mole entities contains given by the Structure 1.4.4 exactly the number of gives Avogadro constant. in the that M asses mass (A ) and C and of relative are atoms are expressed compared on a formula as relative atomic mass (M ). by r –1 Structure 1.4.3 — Molar mass (M) has the unitsg mol i n so small lead, Pb, –16 3.4 × 10 balance. This is too At the small same to time, be the about chemists and a unit very century , that large and which is allows hard to them imagine, to of work even Pb let bec ame one of the on of all and the present present gives the actual in a is molecule. determined volume of solution. law measured pressure element concentration Avogadro’s and each formula element molar solute — gases temperature states under contain the that equal same conditions equal numbers with mole, useful in 1 g directly. have most count. comfortably most measured would the atoms alone numbers of atoms. This unit, the quickly be element, weighed u 19th need f x masses 2.9 × 10 c annot stable number 21 huge, masses t a analytic al g. their heaviest i only r o of that the volumes C are of of molecular each The formula of a compound of n atoms o molecules amount of — atoms ofmolecules. U million a d and Even the Structure 1.4.6 of The mole (Structure 1.4.1) Atoms atoms The o v r of Structure 1.4.5 of p relative to empiric al y — 12 sc ale The ratio compound. number Structure 1.4.2 — simplest r e elementary — The mole (mol) is the SI unit of amount One O of n y t i s Structure 1.4.1 numbers species. de al Understandings At and l of all small, number y in huge P more extremely a e are contains r Atoms of a mass sensitive lead is Therefore, both was concepts very small devised in the in chemistry. l The mole (with the unit “mol”) is the SI unit of amount of substance that contains an will use atom, the a a be elementary molecule, v O 23 6.02214076 × 10 c an rounded value of entities an of electron the that or substance. An elementary entity any other species. In this book, we 23 mole: 1mol = 6.02 × 10 Figure 1 One mole quantities of dierent substances (le to right): E aluminium, water, copper, sucrose and sodium chloride 63 Structure 1 Models of the particulate nature of matter Avogadro’s constant Prex Symbol (N ) is the conversion factor linking the number of particles A F actor –1 and amount of substance in moles. It has the unit of mol : –12 pico p 10 23 N –9 nano n 10 micro µ 10 milli m 10 centi c 10 = –1 6.02 × 10 mol A –6 In chemic al other we –2 c alculations, conversion need to factor multiply substance (n) into the the Avogadro’s (table1). mass in number For kg of constant is example, by 1,000. atoms or used to in Similarly, any the convert other same way kilograms to 10 to multiply that amount by N : A 3 k 10 mega M 10 N = n×N 6 A In 10 chemistry texts, the term “amount of substance” is oen abbreviated to just “amount”. Table 1 Decimal prexes × 10 amount atoms of of this lead (Pb), element. Solution we c an rearrange n = N A n(Pb) = i n According is signic ant discussed in the 3 ≈ Tools for containing as follows: mol 3 mol, so 4.8 10 sample × 10 mol = 4.8 mmol. 3 mol and 4.8 mmol) have been n o i mole is t a are ). so a small. convey just Choose huge how one approximate number, Measuring of and it amounts is of useful for everyday counting objects particles in moles bec ause they c an help use to large this number is. the following and conduct the necessary research to reach an answer. • How many moles of grains • How many moles of water of sand are molecules in are a desert in a of large your sea or choice? ocean of choice? molecules in What the number of atoms in • How tall 0.25 mol • How many water v water E 64 The One of N a Research skills ATL • of = 21 • 0.25 mol c. of a number l O the 1 mmol × in rounded to two signicant gures, the same as in the least precise value used in u the number of atoms in 2.5 mol of copper metal b. d r o f x a. table1, the division (2.9 × 10 Activity 4.8 × 10 23 10 In this example, both answers (4.8 × 10 chapter. C alculate: to n 10 × 3 U chemistry correct = A C gures of N 21 × 6.02 use equation o v 2.9 Therefore, the mmol, p N and y n, mol r e To nd in O the 21 2.9 t i s C alculate n y Worked example 1 The l mole is of the is human age a of cells the represents universe, stack of one moles of air mole are in in of roughly moles of how many seconds? sheets of paper? your school building? people? your y G need P 9 giga r kilo we e d any grams, convert the amount of structural units (N), –1 deci as into s s –3 Structure Relative molecular mass and molar 1.4 Counting particles by mass: the mole mass (Structure 1.4.2 and 1.4.3) In Structure 1.2, we introduced the concept of relative atomic mass, A , which r is the ratio the Similarly, mass of a certain atom relative molecular to mass, one-twelh M , is the of ratio the of mass the of mass a c arbon-12 of a s s atom. of molecule r or other A and multiatomic M are ratios, so they to one-twelh of the mass of a c arbon-12 atom. Both have no units. r To nd the M of a molecule, we need to add together the A r that values for all atoms r r in molecule. e r species l M for a molecule of water. r Water, H O, is composed of two hydrogen atoms (A 2 atom (A 16.00). Therefore M (H r should always use the O) = 2 × 1.01 + 1.01) 16.00 = and actual (not rounded) values of A , which the keep data all booklet signic ant and the gures periodic in table at c alculated M the end values of and this book. never of ions (C aCl ) is an ionic compound instead of formula unit. that consists of as many bec ause ions. the masses r hydrates is compounds sulfate large, ve as of follows: means water. a O) = 2 A (Cu) + r E v O •5H 4 O” = 63.55 + = 249.72 + contains masses with the in ions. that one Ca which neutral atoms The in = composition ionic and compounds will be discussed Structure 2.1. 110.98. water molecules O. Copper(II) sulfate 2 Activity (gure2). The stoichiometric C alculate the M values for the r M value for this hydrate c an following r species: a. ammonia, NH b. sulfuric acid, H 3 4×A (O) + 5×M (H r r r O) 2 SO 2 32.07 structure of formula unit of copper(II) sulfate is Therefore, the A (S) + ) and one One of the most common •5H crystals as (2×35.45) 4 l c alculated 40.08 compounds 2.2) deep-blue 2 molecules same pentahydrate, CuSO clear, before “H = (C a 2+ the r hydrates: (Structure c ations unit negligible. u “5” with M (CuSO r forms f x coecient be form bonds copper(II) pentahydrate bound r 2+ c alcium formula i ionic coordination are A (C a) + 2×A (Cl) 2 many smallest t a form ) = Its approximately electrons r o M any of have ). o M (C aCl ions (Cl d Therefore, The anions for that substance c alcium chloride n and two Cl chloride M r example, U 2 twice molecules, the For p composed o is C substance c alculated using the smallest given v a number. i n If is integer are Similarly, round them to the r nearest oxygen y r in one 18.02. 2 r e You = r = r O t i s Solution n the y C alculate y P Worked example 2 + (4 × 16.00) + (5×18.02) c. sodium sulfate Na •10H SO 2 4 4 dec ahydrate, O 2 65 Structure 1 Models of the particulate nature of matter s s e r l M, is of a chemic al numeric ally substance equal to is relative the mass of molecular O 2 1 mol mass of n mass, mass •5H 4 y Molar p r e Molar Crystals of copper(II) sulfate pentahydrate, CuSO O Figure 2 that substance. (for substances with molecular and ionic structures) or relative atomic mass (for substances with atomic For example, M(Na) = –1 22.99 g mol and –1 M(H O) = 18.02 g mol 2 o i n v structure). Science as a shared endeavour shared understanding of common terminology helps scientists to Hi s to r i c a l l y, c o n ta i n e d 16 to This was as t he re as entities w e re is th e constantly amount (a to ms , a to ms in u l a E v O f x here m e a s u re physic al that v alue as m a ss 2018, all constants being updated. substance mo l e c u l e s , 0.0 12 k g of th e th e mo l e (o r t ha t ions, 12 g ) ( a p prox i m a t e l y i m prove me n ts with gre a te r scientists from of e l e c tro n s c a r b o n -1 2 . SI base instead units, of 6 .0 2 × 10 i n s t r u me n t a t i o n had the objects. in Versailles, mole, were France. It was dened in terms of Following these changes, one 23 mole of entities The no a substance of 2018 that through match is equals two the now dened exactly as 6.02214076 × 10 12 g exact SI numeric al of the mole exactly. As quantities, values of me ans a that result, the their the kilogram the mass experimentally dierences determined and respective between these mass of numeric al a A the or are 1 mol values mole) M no of of c arbon-12 M (dened longer (dened through r c arbon-12 v alues of numeric al r the elementary substance. redenition longer so atom). small However, the (approximately –8 4 × 10 %) Why are How do existing 66 ) a l l ow e d pre c i s i o n . including physic al in more than 60 countries met at the General Conference on Weights and Measures agreed of 23 n u me r i c a l f re qu e n tl y, November terminology defined e l e me n ta r y o the re v i s e d s c i e n ti s t s On ma ny i be mo l e pa r t i c l e s) How e ve r, to as t a r o d other eectively. th e n U communic ate or C A that they constants scientists c an and be achieve denitions? ignored values a for all continuously shared practic al being purposes. revised and updated? understanding of changes made to y P y t i s Structure The amount (n), mass (m) and molar mass (M) of any substance are 1.4 Counting particles by mass: the mole related as follows: m n = M all is probably the stoichiometric the masses of most common c alculations. chemic al expression Although substances are the in chemistry, base SI traditionally unit of as it is mass expressed in s s This used in almost is the kilogram, grams, and e –1 molar masses in g mol r sugar sucrose. H O 22 . sold is an in the form organic of cubes compound that with are made the almost molecular C alculate: 11 b. the amount c. the number of oxygen atoms in one cube of sugar of sucrose in one cube (2.80 g) of sugar r H 12 O 22 ) = 12 × 12.01 + 22 × 1.01 + 11 × 16.00 H 12 O 22 ) = 342.34 g mol 11 = M 2.80 g H n(C 12 O 22 ) = ≈ 11 0.00818 mol 1 342.34 g mol One mole = 11 of × sucrose n(C H 12 11 × n(O) 11 0.00818 mol × N = ≈ 23 × C alculate: a. the H molar mass of sulfuric acid, SO 4 oxygen atoms, so 0.0900 mol 0.0900mol A of 6.02 × 10 o = 22 11 mol ) d = N(O) contains O Activity 2 n n(O) U c. o n C i n m b. 342.34 v 1 M(C = 11 y M (C p r e Solution O the molar mass of sucrose t i s a. a. entirely formula n 12 oen y C is Sucrose l of y Table P Worked example 3 –1 mol b. the amount 1.00 g c. the of of substance sulfuric number of in acid hydrogen 22 ≈ 5.42 × 10 atoms in 1.00 g of sulfuric acid i t a r o u l f x a E v O 19 Figure 3 There are more oxygen atoms in one sugar cube than the estimated total insect population on E arth (10 ) and total grains of 21 sand on E arth’s beaches (10 ) 67 Structure 1 Models of the particulate nature of matter Empiric al formula, molecular formula and chemic al analysis (Structure 1.4.4) The composition represented each element formula present shows in the empiric al of in the molecule simplest substance. c an be formula ions the in the the of ratio The identic al is substance formula, dierent as compound the a molecular of the and O 2 O H 2 2 r e H 6 12 H 2 5 CH 6 O 22 11 C O 2 H 12 O 22 11 o v empiric al formulas of selected substances n U C i n Molecular and C p H C Table 2 HO O 12 O y C O 2 10 formula O H H C 4 Empiric al t i s 3 sucrose represents the simplest formula 2 glucose which n H are same l water that the O O of For ionic compounds, the unit, y ozone peroxide formulas o i d t a r o u l f x a E v O Figure 4 is used The Sodium uoride is an ionic compound in some countries as a food number of atoms element in mol (N = of a certain n×N ). with the empiric al formula NaF. It supplement to prevent tooth dec ay element Therefore, is the proportional to the amount of that empiric al formula also shows the A mole ratio water, H of O, elements contains in two a chemic al atoms of compound. hydrogen and For example, one atom of one molecule of oxygen, so the 2 atomic 68 ratio of hydrogen to oxygen in water is 2:1. Similarly, one mole of water y Molecular O c an be empiric al elements P Substance hydrogen contrast, the (gure4). oxygen butane In dierent empiric al (table2). formula structure shows the actual number of atoms substance. atoms molecular or same that of with which r ratio chemic al e substance a molecular s s of of by a Structure contains mole The two ratio of moles elemental mass, which ratio hydrogen to composition is c an atoms oxygen of a in and compound referred to as the be c alculate to mole of particles the is oen expressed in percent percentage composition, by ω. The percentage composition of a compound. percentage composition of water. r Solution O) = 1 mol, then n(H) = 2 mol and n(O) = 1 mol. Using m = n × M: 2 1 m(H) 2 mol = 1 mol × 1.01 g mol = 2.02 g l = 1 m(O) × 16.00 g mol = 16.00 g O) = 1 mol × 18.02 g mol × 100% = 18.02 g 2 2.02 g ≈ 11.2% 18.02 g ω (O) = 100% chemists empiric al experimental masses In a techniques, typic al products mass or the are as percentages ω (Fe) = 72.36%, then Let m(Fe O x compounds contains 72.36% ω (O) = 100% 100 g, then m(Fe) = y to determine 72.36 g ≈ = the be or Practice question in by elemental and weights elements C alculate measured. oxygen, These by decomposed, and determined excess (iron determined the the percentage composition of sulfuric acid, H SO 2 4 various analysis. the volatile are then original sample. of oxides). Deduce the empiric al iron. 72.36% 72.36 g and amount of = 27.64%. m(O) = 27.64 g each element: a M l m n = n(Fe) = v O Use ) c an in chemic al be are combustion weighed. u f x If of deducing o Solution several that burned products i form oxide is and of t a an sample c an combusted d oxygen of a automated sample trapped r o and formula is decomposition in problem percentage composition or composition compound fully Worked example 5 Iron its n into the elements such opposite from U converted of the percentage which experiment, combustion The combustion percentages analytic al face C mass the in oen compound i n The data. analysis, of a o the for v destruction more formula p other 88.8% y practice, the = r e In 11.2% O t i s ω (H) = n y 1 m(H y n(H P Let e the mass: the mole oxygen atoms, so the Worked example 4 C alculate by water is also 2:1. commonly used one Counting s s mole of hydrogen 1.4 1.296 mol –1 55.85 g mol E 27.64 g n(O) = ≈ 1.728 mol –1 16.00 g mol The mole ratio Therefore, the x : y = 1.296 : 1.728 empiric al formula of ≈ 1 : 1.333 the oxide ≈ is 3 : 4 Fe O 3 . 4 Figure 5 Fe O 3 magnetite, is the main component of the mineral 4 a common iron ore 69 Structure 1 Models of the particulate nature of matter Worked example 6 Hydroc arbons unknown to are organic hydroc arbon produce 26.41 g of compounds has undergone c arbon dioxide, of c arbon and combustion CO , and in hydrogen. excess 13.52 g of An oxygen water, H 2 the empiric al formula of the O. 2 s s Deduce hydroc arbon. Solution e 1 M(CO ) = 12.01 + 2×16.00 = 44.01 g mol 2 n(CO ) = ≈ r 26.41 g 0.6001 mol 2 –1 44.01 g mol n(CO ) = 0.6001 mol 2 1 M(H O) = 2×1.01 + 16.00 = 18.02 g mol 2 l 13.52 g O) = ≈ 0.7503 mol 2 –1 18.02 g mol n(H) = 2 O) = 2×0.7503 mol≈1.501 mol the hydrogen hydroc arbon, C H x The mole ratio Therefore, Multiplying this U whole number whole the ratios. ratio we divide term the you ratio ratio each of 1 : 1.333. should by of 3, in Then, multiply and then the ratio you Whole To numbers are also c alculated was convert it to a whole by the smallest number in the c an ratio 5 initially 1.296 and 1.728. from H 2 number 5, originate 2 : 5 hydroc arbon is C values: ratio which as example = products y by a worked 1 : 2.5 o factor you gives ≈ the C a ratio, This of combustion non-integer i n ratio. two formula formulas In v number empiric al of 0.6001 : 1.501 empiric al integers. comprised = the p express known as x:y in r e We the atoms , so: y O n(H and t i s × c arbon 2 All n y n(H to use trial obtain subsequently and the rounding error to determine whole the number result, ratio. gives a 3 : 4. n The molecular formula of a compound can be deduced from the empirical formula if we know the molar mass of the compound. For example, you might determine The masses of discussed in + of twice –1 (1.01 5) 29.07 is the = 29.07 g mol roughly number of half atoms of as 58.12, the therefore empiric al the molecular formula: C suggests be sure that about this l a E v O be with distinguished comparing their it hydroc arbon without the by same further could be analysis, butane, C molecular as formula. there is Butane formula must H 4 methylpropane, c an × 10 H 4 c annot Structure 1.5. 2) value have Table2 is o × . The molar mass of the empirical formula can be calculated: i molar (12.01 t a the substances –1 58.12 g mol u gaseous d r o f x Determining experimentally that the molar mass of the hydrocarbon in worked example 6 is . However, we 10 another and hydroc arbon, methylpropane measuring their boiling points ( Structure 1.1) or infrared spectra (Structure 3.2). Practice questions 1. Deduce a. an b. a the empiric al oxide of formulas manganese hydroc arbon that of that the following compounds: contains produces 36.81% 5.501 g of of c arbon oxygen dioxide and 2.253 g of water upon complete combustion 2. Deduce the molecular –1 is 70 42.09 g mol formula of the hydroc arbon from 1b if its molar mass y = P n(C) Structure 1.4 Counting particles by mass: the mole Experimental determination of empirical formula Relevant skills Tool 1: • Tool 3: Instructions Measure C arry mass out c alculations involving decimals 1. Weigh 2. Obtain andratios and • Tool 3: Use • Tool 3: Construct and • Inquiry 3: realistic piece from dry of crucible. magnesium your teacher. Explain and relevant 4. Heat the crucible. improvements the crucible, with its lid enter Periodic ally on, the li over crucible. equipment prec autions will get around open ames. very hot. around it and do not touch it while it burns with a heating until the magnesium no longer Then, up. 6. When 7 . Heat to the the directly at it. remove for Repeat a the heat source and minute. this its contents Allow to cool strongly and Q uestions balance 1. • pipeclay triangle • tripod • heat-proof mat • tongs • magnesium ribbon n i o d mass graph the empiric al formula oxide. experimental data of determine cycle until the from mass of other empiric al members magnesium formula to the of oxide your vs class. mass of 6. Identify t line Explain of any on what Explain why the Identify and applic able) and draw a best a graph shows about the composition oxide. you until (if graph. magnesium crucible 7 . anomalies the repeatedly constant explain two heated mass was major and weighed the achieved. sources of error in this procedure. 8. Suggest that realistic could improvements to the methodology minimize the sources of error you have l a Figure 6 u Bunsen burner The experimental set-up E v O f x t a r o ribbon to for an magnesium. 5. coiled magnesium a your one. Obtain Plot 4. crucible Compare actual 3. data C U lid 2. the magnesium o v Bunsen burner Process of i n • (±0.01 g) y crucible and lid • p r e • allow the re-weigh. heating-cooling-weighing mass is constant. M aterials and few minutes. crucible is cool, weigh it. crucible additional very bright light. Do not look cool O M agnesium allow lights t i s ishot. to Continue crucible Take suitable lid y prec autions roaring n The 5. protection. a crucible l • eye suitable the P to ame. y Take 0.3 g mass. Twist the magnesium into a loose coil and place it air Wear exact 3. graphs S afety • its e interpret investigation • (between approximation and estimation Bunsen • ribbon Measure r an clean, a 1.0 g) inside to a s s • identied. 9. Reect on empiric al round C an to the role formula the of approximation c alculations. nearest whole and rounding in When is it suitable to number? When is it not? you come up with a rule of thumb of when to round and when not to round? 71 Structure 1 Models of the particulate nature of matter Measurement Atoms, molecules impossible. As with to all Consider and ions concept mass, which mass of a so the c an measurements, the are of be is easily mass sample small mole has of that counting powerful them bec ause directly is virtually it relates number of s s particles The measured. an uncertainty c alcium associated with it. c arbonate, C aCO , is found to be 3 up to g ± 0.001 0.001 does c alculation a in it represent? nd particles, measurement you out. is How You quite that is clearly particles see that in a c an minuscule does moles it ever negligible? involve the DP making If so, inaccurate mass. uncertainty O properties and solutes. heterogeneous colourless mixtures are discussed in sugar (from the For the so sugar a solutions. substances more the solvent. example, than the of or two properties The other solution (white of Solutions solvent more The of the crystalline in meaning “water ”), in in easier to bec ause chemic al components. whole water powder), which are used it c an reactions. E ach solution solvent is usually the major components sugar is participate or solutes. a of is solution the so the are similar solution more like water solute. In this topic, we will consider only aqua is are water the (clear solvent aqueous solutions solvent is water. o i d solute t a r o u l f x a E v O 72 is L atin solution, of liquid) while Structure 1.1. the or in Sometimes mixtures one n U c alled Homogeneous of properties and out C the solvent gases. dissolved homogeneous i n component of c arried or o are consists of a are solids v Solutions to than tiny, but in course, particularly when y the reactions mix p aect and is measurements. t i s chemic al handle r e M any by many when? Think about these chemistry Solutions and concentration (Structure 1.4.5) How represent? Do a quick the large. through be n experiments many will uncertainty proceed This measurement l doing as direction. mass y questions it the In some c ases, the ethanol and water present it is is not water ” the identity water, in major rather Figure 7 each the of of the these mixture, component. than How a solution is formed “4% it solvent liquids is For solution of is unclear: c an be traditionally example, for c alled example, if we mix a solvent. regarded we as the However, if solvent, even if say “96% solution of ethanol in water in ethanol”. y Is of g and either means P terms This r moles g. e 3.500 Structure Solutions solute solute, small the are and oen and so has proportion term much a of high ratio solute, “concentrated” per 100 g less than of the 10 g according to the concentrated solution of of and refers solvent, the solute so has to the per or mole a ratio with term 100 g of much of the to more particles solvent. than has a Generally, 10 g of the refers to solutions with solvent. precisely would dened c all a precise denitions and their choice and dependent. and The should solution of be terms used “concentrated” with c are. 5 g of sulfuric acid (H For SO in much higher laboratories. At proportions the in same 100 g time, of a acid solution water would of in water 5 g be ) 100 g of water 4 to of are commonly potassium considered very 4 concentrated by permanganate The in antiseptic concentrations extent does above than could quantity composition to of solutions concentration, c, the is also volume of the solution: common units for molar be very dilute smaller 1 × 10 of 1 × 10 molar i –6 = –3 mol dm concentrations are sometimes –3 (for of Note the that whole “the solution For of is of is to to say is the same –3 µmol dm ) c an –3 a as M (for “2.5 MNaOH” sodium refers incorrect ratio of the mol dm ) means that hydroxide. specic that “the substance, not the concentration of a –3 1.0 mol dm ”, as chloride it is or not clear water. The whether we are talking correct statement –3 oen of sodium chloride represented by in a square solution brackets is 1.0 mol dm ”. around the solute –3 example, ammonia. is sodium concentration concentration formula. it of the (which or abbreviated expression 2.5 mol a be solution the concentration” example, concentration E Molar chloride the “molar For example, contains v O would solution term solution. sodium about the For u each dm ). l f x 3 mmol dm t a units r o –3 –3 mol dm is o –3 = 1 µmoldm mM mol dm (mmol dm d –3 or are –3 units used: 1 mmoldm The solutions, molarity, n also For water. –3 concentration –1 ). U most mol L of numeric ally. C solute V solution potassium expressed in terms of = as 100 g expressed known as solute The be per numeric ally help or hinder the n c 0.1 g knowledge? Molar solute a less of p a expressing are concentrations i n of examples typic al o the concentration. amount of the solutions as v Quantitatively, in worker, y what communic ation medic al r e To any O ) sulfuric t i s permanganate (KMnO of n used as “dilute” y 2 “dilute”, and example, most l chemists have y not not context P are is e do r words interpretation mass: the mole between the TOK Some by proportion of dilute solution solute “dilute” ratio large solvent, while a low solutions and solute to a mass contains Counting s s solute classied solvent. A 1.4 the expression [NH ] = 0.5 M refers to a 0.5 mol dm 3 – Similarly, the expression [Cl ] refers to the molar concentration of chloride ions in a solution. 73 Structure 1 Models of the particulate nature of matter Worked example 7 –3 C alculate a the solution molar concentration prepared by dissolving of sodium 3.60 g of chloride, N aCl(s) in in mol dm water to , in make 3 25.0 cm of the nal solution. s s Solution First, c alculate the molar mass of sodium chloride: e 1 M(NaCl) = 22.99 + 35.45 = 58.44g mol m n = to r Then use c alculate the amount of solute: M 3.60 g 0.0616 mol –1 58.44 g mol 3 Convert the volume to dm by dividing by 1,000: 25.0 cm 3 =0.0250 dm n C alculate the mass of sulfuric acid, Use c = to c alculate the concentration: SO 2 , in 50.0 cm of a solution 0.0616 mol 4 c(NaCl) –3 where [H SO 2 ] = ≈ = 1.50 mol dm of a ρ solution , of the solute sometimes solute V solution C i n Worked example 8 C alculate the mass example concentration acid, in mol dm ρ(H SO 2 f x mass mass of the solute to sodium chloride in the solution of the solute concentration and as the volume of the solution, we 3 = = 144 g dm 3 the mass and concentration molar mass, using of NaCl the c an be found relationship ρ from its molar = solute c × most common a mass, as follows: ρ solute c and = solute M solute units molar for mass concentration concentration of the are same M solute 3 ρ(NaCl) = c(NaCl)× M(NaCl) = 2.46 mol dm concentration c an follows: : solute 1 ×58.44 g mol –3 The from 3.60 g concentration u l E v O 74 of the 0.0250 dm Alternatively, ) = 4 –3 0.150 g cm mass mass i t a sulfuric the the ρ(NaCl) concentration –3 of in a solution with know o d molar r o the we of 7. Solution c alculate ratio n U worked C alculate the o v m = solute If expressed as the is volume of the solution: ρ Activity which p the is solute, y composition concentration, r e The 3 2.46 mol dm 3 0.0250 dm 4 O t i s V 3 H n = l 3 V(solution) y Activity g dm solute 3 ≈ 144 g dm –3 and are g cm related . M ass by molar y ≈ = P n(NaCl) Structure 1.4 Counting particles by mass: the mole Worked example 9 A standard solution was prepared by dissolving 6.624g of sodium c arbonate, Na CO 2 a , in deionized water 3 3 using 3 250 cm volumetric ask. An analytic al pipette was used to transfer 10.0 cm sample of this solution to a 3 100cm volumetric ask, and the ask was topped up to the graduation mark with deionized water. C alculate in moldm , of sodium c arbonate in the new solution. Solution we need to nd the concentration of sodium c arbonate in the e First, standard solution: 1 M(Na CO 2 ) = 2×22.99 + 12.01 + 3×16.00 = 105.99 g mol 3 r 6.624 g n(Na CO ) = ≈ 3 0.06250 mol –1 105.99 g mol 3 V = 3 250 cm = 0.250 dm standard that the accuracy of a typic al volumetric ask is three signic ant gures. 3 (Na standard CO 2 ) = = 3 0.250 mol dm 3 we First, need c alculate to c alculate the amount the of concentration Na CO 2 in = 10.0 cm of sodium sample. c arbonate Remember to in the new solution. convert 3 3 V the 3 = 0.0100 dm sample 3 (Na CO ) 2 3 CO ) = c (Na sample CO 2 ) = 0.250 mol dm 3 3 n (Na sample the 2 sample 0.250 mol dm 3 ×0.0100 dm = 3 is diluted with deionized water to produce (Na sample CO 2 ) = n 3 work out (Na new the CO 2 ) = 0.00250 mol 3 concentration of Na CO 2 of the new solution: 3 V = 3 100 cm = 0.100 dm new 0.00250 mol (Na new CO 2 ) = = 3 common solutions following = sample c × sample two of sample n = new new know that gives n = sample the following c E sample = new V n , so the of solute amount Na not CO 2 form the of change. by the 3 standard concentrated solutions required of Practice question concentration when the solute are c alled 3. A standard by copper(II) solution in worked example solution dissolving was prepared 2.497 g of sulfate pentahydrate, 9, CuSO • 5H O, 4 2 using a in deionized 3 water 100 cm volumetric 3 ask. A 5.00 cm sample of this 3 solution was C alculate diluted the to 250.0 cm . concentration, in –3 mol dm nal you of does c an substitute equation 1 into , of copper(II) sulfate in the solution. equation 2. This new expression: × V sample c a You V v O c l new 2. the the to concentration c alculations: V in them known dividing u n a concentration f x 1. the with chemic als dilute t a did the and r o determine we store i Stock standard solutions. To to d needed. practice stock solutions) by amount n a solution 3 o is (so-c alled new the 0.0250 mol dm 3 0.100 dm It the 3 U c in solution, C volume i n c an new o n you the v Therefore Now 0.00250 mol p When = 3 volumes to dm y standard all r e c O Then t i s 0.250 dm n y 0.06250 mol c l Note y P 2 the s s –3 concentration, The process solutions is for preparing standard discussed in the Tools new Therefore, need to to c alculate know the solution, and the c c V × 1 V = 1 × 2 for chemistry the original volume concentration concentration of the of of a solute the in solute, a new the solution, chapter. you just volume of the original new solution. In summary: 2 75 Structure 1 Models of the particulate nature of matter C ase study: spectrophotometry and c alibration curves Spectrophotometry the is intensity of commonly is visible, used for an analytic al ultraviolet technique and determining based near-infrared concentrations on the radiation. of measurement of This technique coloured substances in s s solutions. A spectrophotometer through a of value standard solutions substance are producing a determining or a c alibration electric al unknown the absorbed are result of measured. the unknown curve relates conductivity) of c an measurement be on 0.50 i r o 3 A typic al c alibration curve O a was obtained using a series of l The calibration curve in gure8 u f x Data-based question coloured (such as concentration of measuring that property curve. t a Figure 8 the plotted c alibration y 0.40 by the property c alibration p 0.30 found to of are The o 0.20 concentration/mmol dm C 0.10 (gure8). solution several o 0 d 0 absorbances measurable the Initially, by serial dilution n 0.10 a sample. concentration the r e i n U ecnabrosba 0.20 v 0.40 0.30 These passes measures the absorbance. Absorbance the c alibration curve concentration the by prepared studied solution. c ase, pH The plotting for converts it into the light which photodetector O solute. and used the general absorbance, the absorbances wavelength, The n in studied of certain Figure 9 A series of standard solutions of potassium permanganate Ideally, the calibration curve should be linear, pass through the origin and have a tilt of approximately 45°. If the curve does not meet any of these requirements, it standard solutions of potassium should be constructed again using a slightly dierent wavelength of light and/or . A solution v permanganate, KMnO 4 dierent set of standard solutions. Sometimes linearity can only be achieved within with unknown concentration of E KMnO a narrow range of concentrations. In this case, the studied solution can be diluted, has an absorbance of 4 so the concentration of the studied substance falls within the range of calibration 0.285. Determine the concentration of KMnO curve. In the last case, some additional calculations will be required to relate the in that solution. 4 concentrations of the studied substance in the diluted and original solutions. 76 y then and a solution. y the their light amount the of t i s In of light studied l is the concentrations, substance the P curve and of transmitted describing (gure9), against the produces sample r is small e intensity a Structure Another technique, spectrophotometry colorimetry, but “spectrophotometry” correct but is based on the 1.4 Counting particles by mass: the mole same principles as limited to visible light. The terms “colorimetry” and are oen used interchangeably, which is not entirely very common. s s Concentration uncertainty of a standard solution standard solution this activity, of copper(II) you is a will sulfate, solution prepare each by of known two using concentration. M aterials standard solutions dierent equipment. Wash • Weighing boats (2) • 100 cm • Stirring • Funnels (2) • Pipettes • Spatula • Reagent • Blank labels • Colorimeter • Cuvettes • C alibration bottle containing distilled water 3 By propagating will the determine a measurement the of allow You your solutions you to assess the 1: Measuring Tool 1: Standard 3: volume solution C alculate and and mass preparation interpret percentage error and curve relating sulfate and absorbance • Tool 3: Express quantities and uncertainties to an • appropriate number of signic ant 3: Record • Tool 3: Propagate uncertainties • Inquiry Assess accuracy and precision Wear • Solid copper(II) sulfate is an irritant and environment Dispose of all solutions pentahydrate, CuSO toxic to 3 • 100 cm • Milligram •5H 4 O 2 volumetric ask balance (three decimal places) Additional equipment for solution 2: 3 • 100 • Centigram n • protection. U the eye sulfate C i n S afety • Copper(II) Additional equipment for solution 1: v 2: gures measurement uncertainties p Tool o • concentration of copper(II) y r e percentage uncertainty bottles (2) O Tool beakers (2) rods (2) n skills • Tool then you will concentrations. • • concentration will values. t i s Relevant the This uncertainties, concentration y of the actual colorimeter. accuracy of l using the precision P assess y • r In e A cm measuring balance cylinder (two decimal places) Instructions appropriately. 1. Use the equipment sulfate provided to prepare two standard solutions, both with –3 concentration i solution 1, milligram 0.020 mol dm you should balance. use For . the When preparing volumetric ask and solution 2, use the measuring meniscus of the solution etched line cylinder 2. indic ating u volume, l a E v O f x t a r o d o copper(II) balance. measurements you make along the way, 3 e.g. 250 cm 3. Following the your teacher ’s colorimeter, instructions measure the on how to use absorbance of your solutions. 4. fixed volume of solution when the meniscus is on the the centigram including their uncertainties. etched Refer to the c alibration concentration volumetric flask contains a Record and of curve to determine the actual your solutions. Q uestions 1. line, Determine the uncertainty of the concentrations of solutions 1 and 2. 3 e.g. 250 cm 2. C alculate the percentage error of the concentrations of solutions 1 and 2. 3. Assess the precision and accuracy of the concentrations of solutions 1 and 2. 77 Structure 1 Models of the particulate nature of matter 5. 4. Consider the way you have presented The construction of c alibration curves involves your ATL preparing c alculations for the questions above. Do samples concentrations. think they convey your thinking? Do reader would be able to easily solutions that cover Instead of measuring a and range of dissolving you think a a of you follow certain mass of solute the way you have done your s s here, chemists oen start with a stock solution and thought process? How could you improve perform a the presentation of your c alculations? want to look serial dilution. the advantages disadvantages of using a serial dilution in the through some of the examples in this textbook for of samples for a c alibration ideas. r Avogadro’s law (Structure 1.4.6) Amedeo same Avogadro temperature has been pressure conrmed that equal contain in many of two a substance amount reacting proportional to the of gaseous V turn, 2 the amounts other volume of any gaseous gas a and consumed substances c an be S(g) + O C alculate the combusted this → of 2H O(l) 2 volume of + under the proportional Therefore, the same conditions are chemic al proportional to their equation. produced found without H S, in the As a result, if we reaction, the volumes c alculating their amounts. O(l) + 2SO proceeds as follows: 2 (g) 2 oxygen, O (g), consumed and sulfur dioxide, 2 the volume volumes are of hydrogen measured under sulde the combusted same was conditions. The ratio of the stoichiometric coecients of H you c an S and O 2 is 2 : 3. Therefore, 2 3 multiply the volume of combusted H S by to nd the volume of 2 2 combusted O : 2 3 V(O ) = 3 V(H 2 S) = The 3 × 0.908 dm 3 ≈ 1.36 dm 2 2 2 ratio of the stoichiometric coecients of H S and SO 2 is 1 : 1. Therefore, 2 2S(s) the volume of combusted H S is the same as the volume of produced SO 2 = V(H 2 S) = 0.908 dm 2 reaction was Note that the volume of liquid water c annot be found in the same manner, as consumed in Avogadro’s law applies to gases only. 3 1.25 dm Linking question Avogadro’s behaviour 78 : 2 3 ) V(SO hydrogen sulde if the oxygen E volume (g) 2 a S(g) 2 v O 2H produces instead of sulfur l dioxide: o i sulfur if are volume. 2 2H of its measured or sulde, particles to known as Solution u f x sulde elemental All t a r o d 3 . → volumes produced 0.908 dm Incomplete combustion of hydrogen (g), hydrogen (g) 2 the 2 Practice question 4. of 3O 2 C alculate SO + n U combustion 2H of molecules. now species: products balanced Worked example 10 The these C i n of the species of reactants coecients in number o know of v stoichiometric the proportional p In is is y 1 = 2 and gas amounts V 1 n a r e n of the of and n volumes are amount other, any two gases O the each t i s to of numbers experiments Avogadro’s law Since volumes equal y hypothesis suggested and law of a applies real gas to ideal gases. Under what conditions might the deviate most from an ideal gas? (Structure 1.5) y the l 1811, at P In This curve. e preparation worked Discuss You and may Structure 1.4 Counting particles by mass: the mole End-of-topic questions C alculate: Topic review Using your knowledge from the Structure 1.4 The molar ) 4 the guiding question as fully as of potassium alum, topic, KAl(SO answer mass •12H 2 O. [1] 2 possible: c. The d. The amount potassium total of substance, in mol, of atoms in alum. the number of oxygen atoms pentahydrate, in CuSO 0.400 mol of by •5H potassiumalum. 4 O? complete water, B. 9 D. 5.40 × e. 10 The × 10 of c alcium To visualize nitrate, , is dissolved in water to a volume of 200 cm composition, alum. the mole, is the concentration of NO 6.02 × 10 needed – What up . 2 ions in this solution? 3 a chemistry by B. 7.0 g dm –3 C. 0.021 mol dm D. 0.043 mol dm 50 kg –3 –3 9. 4. which the molecular CH 3 CH empiric al formula the CH 2 OH C. CH 2 CH 3 COOCH 3 CH 2 D. CH 3 CH 2 student Which volume of a 5.0 mol dm sulfuric acid (H ? 3 C. 3 B. student obtained an M ass of tin M ass of oxide mass= B. tin aer of these data, tin? SnO SnO 2 = empiric al E Alums are XAl(SO ) 4 salt the contains following an oxygen-containing organic compound, 5.00 g of A standard 59.95% of oxygen [1] [2] solution of potassium sulfate, K SO from volumetric 8.714 g ask. of the C alculate solid the salt mass using a was 250 cm concentration, in –3 –3 , and potassium 11. , 4 3 prepared g dm The ve formula of the molar sulfate c alibration standard potassium in concentration, the curve in solutions, nal mol dm , of solution. gure7 in in was which the permanganate, KMnO , [2] constructed using concentration of varied from 0.100 4 –3 to 0.500 mmol dm . Describe how you would prepare 1.78 g these 12. solutions C arbon using monoxide, produces serial c arbon CO, is dilution. dioxide, a [3] toxic gas. Its combustion CO (g). 2 what is the correct formula of a. Deduce of C. the c arbon balanced equation [1] 3 3 b. D. for the combustion monoxide. SnO C alculate the volumes, in dm , of consumed c arbon SnO 5 monoxide and oxygen if the combustion produced 3 2.00 dm Extended-response questions 7 . that for heating to a constant 2.26 g to the a A. oxide heating l the of v O According before of u f x 10 cm following data during an determination oxide of tin: cm t a experimental the sulfur [3] formulas 2 3 D. grain shovel of water upon complete combustion. 10. i A cm r o 6. 0.0050 5.0 c an of a 3 cm average which produces 9.55 g of carbon dioxide and 5.87 g o 0.010 d A. an an ) n U 0.10 mol dm of if student b. 4 solution whose concentration of hydrogen ions is –3 oxide the minute. compounds: 3 stock solution is required to prepare 0.50 dm per decided to sand. Estimate the time project and empiric al 3 SO 2 this 5 mg, [2] a. COOH 3 sand the CH 2 –3 5. same as of Deduce C B. the complete o CH is formula? i n A. molecule v For to weighs of [1] mass, of y 3.5 g dm grains p r e ofsand –3 A. produced 1.00 g of 23 pile 3 ) 3 be O 8. 0.70 g percentage potassium 24 C a(NO [1] c an n 2.16 containing of y C. t i s 3.60 sample that 2 A. A mol, decomposition 24 3. in l sulfate of P is amount [1] 1.00 g of r potassium Multiple-choice questions copper(II) 1.00 g of alum. number Exam-style questions What in y The e b. How do we quantify matter on the atomic scale? 2. s s a. 1. hydrates •12H 2 O, of the where X measured general is an of c arbon under dioxide. the same All volumes conditions. are [2] formula alkali metal or other 2 singly-charged c ation. decompose follows: XAl(SO as ) 4 •12H 2 When O(s) 2 → heated, most alums XAl(SO ) 4 (s) 2 + 12H O(l) 2 79 Structure 1.5 Ideal gases s s How does the model of ideal gas behaviour help us to predict the behaviour of real gases? As with a any theoretic al simplic ation many c ases, it predicts sucient model, has for its the the concept properties most of an ideal gas at advantages and limitations. In practic al of real gases with a purposes. low real temperatures gases ideal gas deviates model and high pressures signic antly c annot be from used under these conditions. gas consists are moving particles with — Real Structure 1.5.3 — The molar Structure 1.5.4 — The relationship deviate volume from of an the ideal ideal between gas gas the is a model, pressure, = no V nRT and the combined gas a law V 2 = T low temperature temperature temperature p 1 at specic intermolecular and and amount and pressure. pressure. of an ideal gas is shown in 2 T 1 high forces. All y pV and r e equation at volume, 1 gas particularly constant p ideal volume O Structure 1.5.2 the negligible t i s gases of considered elastic. n ideal particles l An 2 p o v Assumptions of the ideal gas i n The ideal gas model states that C model (Structure 1.5.1) an ideal gas collide inelastic sound. perfectly molecules with collisions However, elastic to the following ve another are gas not stationary. molecule or They the move side of a in straight lines container. and of larger the no objects, collisions energy is energy between lost from c an be transferred molecules the in an ideal as heat gas are system. The volume occupied by gas molecules is negligible compared to the volume of the container l Vaporized E v a (0 °C) and water 100 kPa gas conditions. gaseous space in occupies phase but is the which the about In both same volume the 1600 times the volume of liquid water at 273.15K pressure (standard temperature and pressure, STP). the same they occupy occupies Nitrogen changed, are that o they i or u f x O forces conforms Collisions between molecules are perfectly elastic In 4. Intermolecular means t a r o d until 3. gas Molecules of a gas are in constant random motion This 2. n 1. U assumptions: gas 650 and of times c ases, the the There are no intermolecular size gas molecules the is are the volume number of to of liquid nitrogen under molecules in liquid and individual >99.9% free of molecules has not empty space. This is the move. forces between gas particles studied in For an ideal gas, the intermolecular forces are negligible compared to the kinetic Structure 2.2. energy of the molecules. As such, an ideal gas will not condense into a liquid. 80 y P — between y collisions behaviour of prediction, so the However, Understandings Structure 1.5.1 the the r precision that e is Structure 5. 1.5 Ideal gases The kinetic energy of the molecules is directly proportional to Kelvin temperature This relationship is studied in Reactivity 2.2 Pressure–volume relationships Boyle pressure of (1627–1691) a given now of known as a gas is that, at constant inversely Boyle’s law, c an temperature, the proportional be expressed to as its volume. This e relationship, established amount s s Robert follows: Figure 1 An ideal gas consists of 1 p or ∝ pV = k (a constant) or p V 1 particles that V 2 collide elastic ally, r 1 =p 2 V intermolecular forces and gure 1, walls pressure. so the the volume every is is gas are constantly striking and bouncing o force halved, second pressure a The there these there are doubled of are twice impacts twice as as produces many a volume when compared molecules in each unit many impacts with the container (gure 2). Standard databases temperature and of temperature for to the behaviour of ideal gases. in pasc al (Pa), comparative 100.0 kPa where used mercury and y built on certain assumptions related between pressure and What commonly millimetres o C (atm), are o in phenomena. The ideal gas model is volume, 1/V a found E (psi). pressure representations of natural –2 pressure is the of atmosphere inch i t a l of v unit units u f x O SI other TOK Models are simplied reciproc al Graphs showing the inverse relationship volume of an ideal gas The ,erusserp d r o ,erusserp Figure 3 doubled n U p p volume, V pressure Halving the volume of a container doubles the pressure halved p r e v i n Figure 2 O t i s volume to the volume of the gas (the container) measurable n walls, If so of container. l space, molecules the y of the of y the P In have no occupy negligible in 1 Pa = 1 N m –3 = 1 J m . M any dierent countries, including the (mm Hg), pressure purposes. bar, and conditions STP for pounds (STP) gases is are 0 °C per square frequently or the is the role of assumptions development of scientic models? What not are the implic ations of acknowledging a model’s limitations? 273.15 K pressure. 81 Structure 1 Models of the particulate nature of matter Worked example 1 3 A weather is released balloon at sea lled level. with The 32.0 dm balloon of helium reaches an at a pressure altitude of of 100.0 kPa 4500 m, where 3 the atmospheric helium at in that the altitude. balloon is 57.7 kPa. Assume remain that C alculate the the volume, temperature and in the dm e × p 1 it follows that V 1 = , so: 2 r law, p 2 3 32.0 dm × 100 kPa ≈ 55.5 dm 2 57.7 kPa certain volume contains of altitude, a 0.250 m weather 16.0 g of . balloon C alculate helium, the has a temperature pressure, in of n a 3 a O At t i s 1. l y Practice question –35.0 °C and kPa, inside the balloon if He(g). y r e Real gases vs ideal gases (Structure 1.5.2) of between the the against no of become pressure. pressure pressure decreases and volume longer signic antly, the container. signic ant. This means volume for a real halves n U doubling gas proportion forces reducing pressure real C relationship no and molecules so a real applies. an ideal gas, gas. For volume. erusserp t a l u 0.5p a E v V 2V 0 volume 0 Figure an ideal 4 gas Doubling but not for the a pressure real gas halves the the Figure ideal gas p may little space to decreases the number for longer gas the This that, the With real gas o i d r o f x O 82 a large intermolecular collisions, graph of a i n of volume occupy o move, the to v begin p When volume for 4 inverse shows a the real gas, y P 3 = V the of constant. V Boyle’s of amount Solution From , s s balloon pressure Structure For a gas to deviate intermolecular the molecules from forces ideal and/or themselves. gas a behaviour, signic ant This there volume commonly must of occurs the at a be Ideal gases detectable gas low 1.5 must be occupied by temperature and high pressure. is reduced. form and As molecules At the may high not pressure, considered to be an c annot low negligible. At so high intermolecular molecules an ideal pressure, the space of gas there far are the attraction to apart only no and behaviour occupied temperature, forces part of in the the a reduced space longer very by few the prevent are low molecules molecules molecules are interaction per unit moving too fast assumptions from how ideal each of behind pressure the and behaviour. of the following might behaviour low high gas at low pressure temperature or at high E or which is more likely to exhibit HBr(g) (g) 4 dec ane, C e. ideal HF(g) bromide, methane, CH the a or hydrogen d. temperature fluoride, v O hydrogen l gas c. lead to u f x b. at predict reason: pressure or gas pairs, a i at give of to t a gas and likely validity volume following r o a. the the o ideal molecular of forces affect are n L arge each intermolecular d For Strong ideal gasmodel. temperature o conditions U b. 4. for y main gasmodel: a. allow C Consider to form. i n 3. volume p the what deviations of themselves is r e Outline Discuss between pressure and high v 2. between them, inverse, so the gas is Activity 1. space. volume of the gas. n At container, is attraction O the molecules signic ant volume molecules of elastic ally. compressed, and gas t i s in more the forces l for temperature. are of y keep them. conditions be rebound a energy intermolecular ideal gas. Ideal gas conditions The there becomes pressure kinetic y not between the another, P relationship one necessarily molecules themselves the with e of temperature, r As volume low collide molecules High pressure: The At they s s Low temperature: H 10 (g) 22 propanone, CH COCH 3 (g) 3 or butane, C H 4 (g) 10 83 Structure 1 Models of the particulate nature of matter Real gases Gases that deviate from the ideal gas model are known as n real gases. ( 2 ) ( p + a V V – nb ) nRT = s s Relevant skills measured Tool 2: • Inquiry Use spreadsheets to manipulate data. pressure 1: Select sufficient and relevant correction sources of correction information. forces for for of between volume molecules r molecules • Inquiry 1: Demonstrate creativity in the designing, measured presentation of the investigation. volume Instructions real gases is volume, amount and modelled by the parameter van der and b for b corrects various equation: a / 6 × 10 ammonia, NH Pa m OH 20.94 3 2 6 H ethanol, C OH 5 hydrogen uoride, 4 3 Ne H 5 3 E H 3 O 11 8 propan-1-ol, C 2 OH a 12 H propane, C xenon, l 5 v O H pentane, C pentan-1-ol, C u f x OH methanol, CH t a methane, CH HF r o Kr HBr HCl o chloride, i bromide, hydrogen d hydrogen OH 12.56 Van der Waals parameters, a and b, for a selection of gases strength and Values of –3 b / × 10 3 m 0.0371 0.0320 0.1164 0.1326 0.0648 0.0651 0.0871 0.0346 0.0238 4.500 0.0442 3.700 0.0406 9.565 0.0739 5.193 0.0106 2.303 0.0431 9.476 0.0659 0.208 0.0167 19.09 0.1449 25.88 0.1568 9.39 0.0905 16.26 0.1079 5.537 0.0305 4.192 0.0516 7 Xe Table 1 5.580 n U 2 He 7.566 C i n Cl o 9 H ethane, C v H 4 force volume. shown in table 1. p 10 butan-1-ol, C water, H are 13.89 chloromethane, CH neon, molecular 1.355 H 4 krypton, intermolecular y Ar butane, C helium, gases r e argon, for –2 mol 4.225 3 for O –1 Substance 84 a corrects t i s Waals of pressure, a n temperature between y relationship l Parameter The –1 mol y or P implementation e • Structure 1. Use of a the selection factors instance, of the data affecting the in table 1 values of to a explore some and You b. For will how you could look at: to need to analyse choose to decide it. how explore, you may and/or look up additional intermolecular force strength and the value of a • molar • the mass and the value of b 2. Consider on data which need to to select, and option perform you c alculations data. how you could present your data ATL graphic ally. effect of volume on the deviation from Prepare a one-page summary of ideal exploration to share with your behaviour. r conditions, do some gases deviate more from ideal volume of an ideal gas Avogadro’s temperature and pressure contain equal numbers of particles. The molar volume p of an ideal gas is a constant at specied temperature and pressure. For example, 3 at STP , the molar volume of an ideal gas, V = 0.0881 mol has a Molar volume of an ideal gas 1 70.90 g mol compared with a soccer ball pressure mass of 2.47 g. gas. u a 2.47 g = STP the = 22.7 dm 0.0881 mol l n 1 mol v O = = 3 22.7dm m M 3 2.00 dm = m at of o f x V V , C g mol Solution = gas 1 in n unknown mass, 3 V o an i of molar t a sample the covered in 28.3 cm Figure 5 1 a given temperature and d r o 3 2.00 dm . 2 32.00 g mol Molar volume of any gas is identic al at Determine n U 1 16.05 g mol –1 mol Cl 2 1 4.00 g mol Worked example 2 A O 4 1 2.02 g mol Figure 6 i n CH He v , is equal to 22.7dm m 2 is y r e Avogadro’ s law states that equal volumes of any two gases at the same H law Structure 1.4. (Structure 1.5.3) O t i s The molar why (Structure 2.2) n others? y than l comparable behaviour y P Linking question Under class. e your gas Ideal gases s s • much Depending 1.5 1 28.0 g mol Practice question E 2. Determine the –1 molar mass, in g mol that has a density of 3.12 g dm , of an elemental gaseous substance –3 at STP. Identify the substance if its molecules are diatomic. 85 Structure 1 Models of the particulate nature of matter Hypotheses Amedeo contain pressure. postulated numbers This of that particles bec ame equal under known as volumes the of same Avogadro’ s dierent gases would conditions of temperature hypothesis. A hypothesis is a tentative and falsiable explanation or description of a phenomenon, used What to test the predictions which predictions c an be deduced. Predictions hypothesis. might be derived from Avogadro’s hypothesis? r gas equation a gas by c an be used to determine the collecting a known Instructions volume of it 1. under known conditions of temperature and Measure ambient Alternatively, practic al you will experimentally determine the molar of butane found in disposable plastic lighters. the day you Relevant skills 2. 3: range Record to an uncertainties appropriate in Half-fill measurements as a precision and propagate Fill processed data. the Inquiry C alculate 2: 3: Assess Identify Inquiry 3: and and the and discuss random Evaluate percentage sources and impacts of error. implic ations limitations and of 3: Explain realistic assumptions on and clamp 4. an relevant and 5. improvements protection. o i d r o eye Wear • Butane gas is flammable. Keep away from open flames • Disposable plastic lighter for 100 cm • Balance (±0.01 g) • Clamp and stand • Thermometer example 3 measuring (if large plastic cylinder water. to the Measure the brim with water trough so that its mouth is under correctly, of the measuring cylinder water. Hold it in this position with a (figure 7). dry Hold it the the the lighter thoroughly in water, then take it out again with a paper towel. Weigh the lighter lighter under to water and press the button release the gas so that it bubbles up inside the measuring cylinder (figure 7). Continue until 3 you have exact collected around 100 cm of gas. Record the volume. 6. Release 7 . Dry 8. If the the gas lighter in as a well-ventilated thoroughly as area. possible and reweighit. you have time, repeat to get three sets of results. trough 3 100 cm cylinder available) water Figure 7 86 with for cylinder v E Barometer a experiment. water. data loc ation on a • • container, l L arge O • u f x Materials t a • and sparks. full weather geographic lighter. investigation. S afety done be Submerge on to If should n Inquiry water. methodologic al conclusions. • error. precision. U weaknesses, interpret the C systematic • and accuracy i n • 3: in barometer. o Inquiry v Tool • it a loc al your trough the measuring andinvert • of in p in the plastic temperature 3. uncertainties the pressure do with search y Tool r e • c an O atmospheric mass you t i s this pressure pressure. In n of y mass l ideal molar Experiment apparatus measuring y P Experimental determination of the molar mass of a gas The natural c an then e be from s s and Avogadro equal Structure 1.5 Ideal gases Questions 1. Design a suitable results table for 8. your data. Suggest realistic determining 2. Process your molar data mass of to obtain an experimental value 9. The the Propagate the uncertainties. 4. Compare pressure sum partial by the 6. Discuss the accuracy to percentage and precision the theoretic al the relative impacts of systematic and additional need to on butane. processing data and Consider alternative least two major cylinder is in fact water and the How to could account information methods random for do If you have time, show your ideas sources of presented the (Tools 2 as sketches advantages and 3, or as and accurately limitations Reactivity 2.2) plotted of each o C i n v representation? are p What your y be O t i s c an points. r e data to n y error. Linking question Graphs this? you for determining the teacher and try them out. experimental you mass of a gas that could be done in a school results. at of l Comment the measuring pressure research? laboratory. 7 . for y your of data What molar on your the vapour your data. 10. errors inside the pressure adjust error. of method r Assess c alculating value of P 5. experimental this e value your to mass of a gas. for butane. 3. improvements molar s s the the Pressure, volume, temperature and amount U There are four variables of ideal volume the gas occupies, 3. The absolute 4. The amount of the gas, He law: a he gas these This performed observed kept that E v proportional. variables constant. were u of of two a changed. two other Boyle’s temperature V T n l with any p temperature of the gas, f x O up of the the each other: an is on each what other Robert experiment constant, pressure o The aect i 2. eect by the gas, that t a pressure keeping gas d The r o 1. The exerted an n of an ideal gas (Structure 1.5.4) but and the c an Boyle be investigated did when he by c ame where the amount and the volume volume of of the the gas container were was inversely 87 Structure 1 Models of the particulate nature of matter Graphing the gas laws Online simulations relationships ideal of gas gas. In pressure this simulation, direct data and to easily task, and you which analysis inverse and temperature will will collect allow skills, as 3. explore the volume, you well as for data to Using a for fixed a five from an simulation, pressure certain amount different vary and the record of gas. temperatures temperature at a the resulting Collect in a data suitable ideas Compute the temperature values in both °C and K. Construct two graphs of V vs T; one with Use data from simulations. 6. Using the simulation, vary Understand direct and inverse Inquiry 1: Identify dependent, independent and certain five amount different the of Collect gas. volumes in variables. Construct Simulation and allows you temperature to for change 8. Use an ideal gas. It must have 9. option to hold one variable constant and p spreadsheet Construct vary the graph of a graph of p a vs to suitable Spreadsheet 1. software the for a certain and amount different vary the record of gas. temperatures temperature at a the resulting Collect in a data suitable table Construct a graph of p vs 2. Describe as direct least in orother. 3. When 1 for . 1 the c ase? Which relationship proportionality, studying temperature T. this is the variables shown inverse in were each graph proportionality, i o units for c an it into is SI important units temperature, vary to convert all (kelvin). Discuss whereas depending on the why pressure and source. The combined gas law We have seen u proportional l a E v O f x t a r o d volume gases, values c ase that to pressure absolute is inversely proportional to volume temperature. 1 p ∝ ; p ∝ T V Combining the two relationships gives: pV pV ∝ T = k (a or constant) or T p V p 1 V 2 1 2 = T T 1 This 88 values your n spreadsheet. 2. at each controlled? pressure for U five simulation, volume your your dependent and independent in C constant i n Using in for least vs V were variables at V o v Instructions 1. What table compute p • two. pressure for V Questions other data y an your pressure, r e volume that a resulting O 7 . Materials t i s spreadsheet. • volume at a constant record n • a proportionality. in °C and l Tool and y • controlled the spreadsheets to manipulate data. temperature 3: in K. equation is known as the 2 combined gas law and directly y Generate 2: T T r 2: Tool your 5. P Tool least in 4. the other with • at table proportionality. Relevant skills • volume for spreadsheet. practice reinforce the constant volume and e spreadsheet about and you pressure s s temperature, amount allow between Structure 1.5 Ideal gases Experiments The gas laws variables arose were experiments in which certain Inspect the while others apparatus were shown c arefully in gure8. pressure gauge What might is independent the be explored with variable? this set-up? What thermometer What variables must 250 controlled? What is the purpose of mL round-bottomed flask e be s s manipulated. from controlled, each of the containing air itemsdepicted? r water bath lled released 35,000 m, –50 °C. C alculate where level. the of The helium balloon pressure is at 25 °C 475 Pa and the 3 the volume, in m , of the gas conditions of the gas in the 100.0 kPa 3 = 32.0 dm 1 to convert temperature to kelvin: = 25 1 the conditions to make the balloon. sure f x the units = –50 into the = 298.15 K weather balloon at 35,000 m. 0.475kPa = + 2 unknown 273.15 combined p a v O E the = 2 numbers the 2 V T 273.15 in are consistent with the initial conditions of p p V 1 = 223.15 K gas law: V 2 1 2 = T T 3 × Practice question 2 1 100.0 kPa Rearranging + gas u the the l Substitute of t a Remember r o list i d T Then, is these o Remember under weather balloon. = 1 V an n p pressure temperature balloon U the the a reaches C Solution List in i n conditions. and eventually o of 32.0 dm sea v altitude with at p is n balloon 100.0 kPa O weather y 3 of r e Worked example 3 A l into the behaviour of a gas y P Apparatus for conducting an t i s experiment y u Figure 8 0.475kPa 32.0 dm × V 2 3. = A sample of an ideal gas has 223.15 K 298.15 K 3 a expression in terms of V volume of 1.00 dm at STP. 3 gives: C alculate the volume, in dm , 2 3 V ≈ 5.04 × 10 3 dm 3 = of that sample at 50.0 °C and 5.04 m 2 50.0 kPa. 89 Structure 1 Models of the particulate nature of matter TOK Throughout were chapter alpha particles you through with have explored observations gold atoms, to of the models the related natural world manipulation of to or the particulate obtained gases in the nature through gas laws, of matter. M any of these concepts experimentation: to from the interaction s s of this developed explorations of subatomic particles atCERN. e r l n O y used by Joseph Louis Gay-Lussac to investigate the thermal expansion of gases (le) CERN used to investigate elementary particles (right) p v How do scientists investigate the behaviour of particles that are too small to be observed directly? How have advances in o technology inuenced scientic research into what matter is made up of ? C i n Ideal gas equation combined three gas law parameters, suggests p, pV expression remains T n the U The V or T, that for aects constant. The o i d or in such a way that the value of that constant must be n: R is the nR universal gas constant, or simply known as the u = = T ideal gas equation, which is gas constant. traditionally The written as last expression follows: nRT The value and units of l a E v O f x t a r o where pV two pV ∝ n T is given gas, the change in one of other exact proportional to the amount of the gas, pV any the R depend on the units of p, V, T and n. If all four parameters 3 are expressed in –1 R ≈ The 8.31 J K same standard SI units (p in Pa, V in m , T in K and n in mol), then –1 mol . value and units of R c an be used if pressure 3 volume in dm is expressed in kPa and 3 , as the two conversion factors (10 –3 for kPa to Pa and 10 3 for dm 3 m ) c ancel each other out. Linking question How c an the experimental 90 ideal gas data? law be used to (Tool 1, Inquiry 2) c alculate the molar mass of a gas from to y P y t i s The set-up The ATLAS detector at r e Figure 9 Structure 1.5 Ideal gases Worked example 4 A 3.30 g sample of an unknown organic compound was vaporized at T =150 °C and p =101.3 kPa to produce 3 1.91 dm 3 of a gas. The gas was combusted in excess oxygen to produce 3.96 g of water, 2.49 dm of c arbon dioxide 3 and 1.25 dm of nitrogen at STP. s s Determine the following for the compound: molar mass b. empiric al formula c. molecular formula r Solution determine the molar mass, we need to nd out the amount of the compound using the ideal pV = T = n = gas equation: l n RT 273.15 = 423.15 K 3 101.3 kPa × 1.91 dm 0.0550 mol –1 mol × 423.15 K 3.30 g –1 Therefore, M = = 60.0 g mol 0.0550 mol c arbon, amounts hydrogen of these in atoms c arbon in the dioxide, combustion water and = = ≈ 0.220 mol –1 M = 2 × 18.02 g mol O) n(H = 2 × 0.220 mol = 0.440 mol 3 ) = = ≈ 2 3 V 22.7 dm 0.110 mol –1 mol M = ) n(CO = 0.110 mol 2 3 V n(N ) 1.25 dm = = ≈ 0.0551 mol 2 3 V × n(N ) = 2 × 0.0551 2 The m(C) = = 0.440 mol 0.110 mol m(total) + 1.32 g organic + = original sample: oxygen. and To check nitrogen) with this, the we mass need of to the compare original the total mass sample: 1.32 g ≈ 1.54 g 1.54 g compound a the the 0.444 g ≈ 14.01 g mol l N) 4 so the E M(CH ≈ –1 0.110 : 0.440 : 0.110 empiric al contain c arbon –1 1.01 g mol 12.01 g mol v O The = × 0.444 g Therefore, x : y : z (hydrogen, organic compound, so the in –1 0.110 mol = 0.110 mol u = × × f x m(N) c. elements = also the those t a m(H) three could i the compound r o of original o = 2 mol d n(N) –1 22.7 dm M from as n U n(C) same C 2.49 dm V n(CO i n 2 originate the o n(H) are v 2 products nitrogen p O) nitrogen 3.96 g m n(H and elements y All r e b. O t i s ≈ –1 8.31 J K n + y 150 y To P a. e a. ≈ did 3.30 g not contain 12.01 + so its formula c an be represented as C H N y . z 1 : 4 : 1 formula of the compound is CH molecular oxygen, x = N. 4 –1 4 + 1.01 formula of + the 14.01 = 30.06 g mol compound will have –1 . This twice value the is half number the of experimental atoms of value (60.0 g mol each element: C H 2 N 8 ), . 2 91 Structure 1 Models of the particulate nature of matter End-of-topic questions 0.58 Topic review × 8.31 × 373 B. 1. Using your knowledge from the Structure 1.5 6 × 10 250 × 10 topic, 3 answer the guiding question as fully as s s 3 100 × 0.58 × 100 0.58 × 100 × 10 6 × 250 × 10 × 10 C. possible: 8.31 × 100 × 10 e 3 How does the model of ideal gas behaviour help us to 6 × 250 D. predict the behaviour of real gases? 6 Which Multiple-choice questions of the following are shows volume assumptions of the ideal gas B. P by the gas particles is negligible II. There are no intermolecular forces V between r e gas particles Zero particle movement II and III only I, II and III The temperature is A gas syringe will at 27 °C. contains volume constant 40 of of gas an be pressure? 3 cm 3 cm 3 44 cm 84 cm 3 D. A 0.58 has a g E 5. sample volume expressions 0.58 of is × of an equal 8.31 ideal × cm to . the 100 3 92 10 6 × gas gas it is 100 250 × 10 molar of at mass and balloons atoms, at contains constant the largest temperature pressure? H NH (g) 2 (g) CH 3 (g) HF(g) 4 the 3 3 What 3 2 dm 2 dm are the 3 1 dm conditions for the ideal 4 dm gas behaviour of gases? A. Low temperature and low B. Low temperature and high pressure. C. High temperature and low D. High temperature and high pressure. pressure. 100 °C following of following hydrogen 27 °C. 8. kPa the the of warmed to real at Which of number and 3 250 A. 100 × ideal aer Which doubled and a 3.6 v O C. 1.9 l A. B. is 3 cm the 7 . What is the u 57 °C the pressure tripled? f x What is the i 1527 °C 1800 °C gas t a C. D. aer 1 V o 162 °C 450 °C ideal d A. B. an gas r o 4. volume of the n the of U temperature 1 V C D. P o I and III only C. v B. i n 3. I and II only p P A. V D. between y C. III. O occupied t i s volume relationship ideal gas, at constant n A. P The the an temperature? model? I. of l Which correctly and y 2. graph pressure gas? pressure. y the 373 P Exam-style questions × r 8.31 Structure 9. Which are of the following statements about an 14. ideal gas At forms constant temperature, the several gaseous empiric al compounds p I. C arbon Deduce correct? using and the compounds molecular data from the 1.5 with formulas table Ideal gases uorine. for these below. [3] = constant V 3 C arbon / M ass of 1.00 dm p At constant volume, s s Compound II. = constant mass % at STP / g T X 13.65 Y 24.02 Z 17 .40 3.88 V III. At constant pressure, = constant A. I and II only D. I, II and III 15. An organic compound 9.1% of hydrogen vaporized and sample of A contains 36.4% A with of 54.5% a oxygen mass of of by c arbon, mass. A 0.230g occupies l 3 a volume of own words, behaviour at low why real gases deviate a. Determine b. Determine c. Using your answers to parts a and b, determine the temperatures and high pressures. the empiric al [2] the relative formula of molecular molecular formula of A. 11. A c ar tyre inated to 2.50 bar (250 kPa) at 10 °C contains of compressed air. Aer a long journey, the A closed gas tyre temperature increases to 25 °C and the pressure and Determine the tyre volume under these mixture Assume that there was no air readily when ) 4 (s), heated: CO 2 (s) → 2NH 3 (g) + CO 3 (g) + H 2 gases the produced vessel heated vessel that with question12 to a each were of 1.50 dm C alculate the , initial The [1] [1] of hydrogen volume of the temperature of the gas pressure, in kPa, of the gas cylinder. gas mixture rises to [1] is completely cylinder inside ignited, and 800 the °C. both reactants are temperature inside C alculate the cylinder at that the pressure moment. [2] –3 An unknown gas 4.00 g sample of X has X was a density of 2.82 g dm combusted in excess at STP. A oxygen to 3 produce and the c arbon 2.50 g dioxide of at hydrogen uoride and 2.84 dm of STP. vessel Determine the following for X: pressure in the Assume that the gases do not other. the the initial the 0.32 mol gas. [2] A. transferred to a 3 volume 200 °C. temperature. u l a. molar b. empiric al mass c. molecular [1] [2] formula formula [2] [1] a E v O f x react at with up o was in 17 . [2] t a sealed produced the 2.25 g of i gases of When the in consumed O(l) contains oxygen and A. mass of 2 STP, of the individual decomposition c arbonate. r o The on at d ammonium 13. volumes, in dm b. n 3 Determine C alculate mixture decomposes 3 U (NH CO 2 a. C ) 4 i n c arbonate, (NH 25 dm 25 °C. o [2] Ammonium is is loss during the journey. 12. v conditions. of p 261 kPa. cylinder 3 cylinder to steel 0.16 mol y r e 16. 3 12.0 dm n ideal your p=102 kPa. O from in t i s Explain, T=95 °C and y Extended-response questions 10. at 0.0785 dm y II and III only r I and III only 6.08 P B. C. 4.41 e T 93 s s e r O n l y P y t i s y p r e n U C o v i n structure and bonding of Models 2 Structure o i d t a r o u l f x a E v O Structure 2.1 The ionic model s s What determines the ionic nature and properties of a compound? the strong electrostatic attractions between oppositely of positive and negative ions, which attract each other charged ions. Once liquid, however, ionic compounds are electrostatically. In solid ionic compounds, these ions electrical conductors due to the presence of mobile ions. are arranged in rigid crystalline lattices. Melting these Due to their charge, ions interact strongly with polar water solids requires a large amount of thermal energy due to molecules, so ionic compounds are oen water-soluble. r e Ionic compounds are characterized by the presence l c ations. electrons, they Structure 2.1.2 — form The atoms negative ionic lose electrons, they Structure 2.1.3 — When non-metal atoms bond ions is dimensional c alled anions. formed formulas. by electrostatic attractions between oppositely charged ions. by compounds the anion. are The named anion with the c ation adopts the sux rst, “ide”. exist c an The them rise is bonded varying to . are atoms arrangements certain nitrogen, N They to dierent However, in connected of of help properties. agriculture, crops grow. to that This of is the bec ause the nitrogenous and For together type, several to atoms features of the example, nitrogen structure in or fertilizers and compounds 78% dierent bonds of are bonding dierent of the empiric al between air around of added to soils nitrogen in air o dierent d are same atoms 2 to the three- by n us give isolation. U elements. in be p rarely Atoms C ways. i n Atoms as o v Introduction to bonds and structure exist represented y ionic followed compounds structures, r e Binary Ionic lattice O gain ions n metal c alled positive y When form t i s Structure 2.1.1 — y P Understandings found in fertilizers. Atoms are held together by chemical bonds. This chapter discusses three dierent i r o bonding models: ionic, covalent and metallic. These lead to four types of structure: t a ionic, molecular covalent, covalent network and metallic. Y ou may be wondering why there are four types of structure, given that there are only three types of bonds. u f x This is because covalent substances can be found in two arrangements: a continuous 3D network, or discrete groups of atoms known as molecules. ionic bond ionic E v types of structure covalent metallic a O l types of molecular covalent covalent network metallic metal ion deloc alised electron t Figure 1 bonds and There are three types of four types of structure 95 Structure 2 Models of bonding and structure Chemic al bonds Models Chemic al Structure models 2 .1 , of bonding models represent so are All charged on strong species which help we This bonding does are attraction that hold atoms or occur due negatively to electrostatic charged us to The electrostatic attraction between… c annot is one of models Type of bonding Positively charged are inadequate, but it Negatively charged species ionic c ations covalent atomic metallic c ations species As All bonding types involve a positively charged you limitations of these of the the are electrostatic ally attracted sections, strengths and various bonding a negatively charged to each other compounds compounds instance, However, also sodium elemental chlorine is a poor when are are electric al molten very chloride, sodium is are or ionic compounds. They characteristic of ionic compounds. conductors when solid, but good dissolved. The reactions and properties of dierent to those of their constituent elements. the a examples of properties main so poisonous gas. ingredient metal that in reacts table salt, violently is water-soluble. with water, and n U C For ionic i n these are conductors sulfate which o electric al copper(II) brittle, v Ionic and and y chloride crystalline p are r e Sodium o i d t a r o u l f x a v O Figure 3 Sodium Sodium chloride and chloride crystals on a tree branch and copper(II) sulfate crystals. copper(II) sulfate are ionic compounds Oshore oil platform in USA. E C alifornia, species and electrons electrons Ions (Structure 2.1.1) models. Figure 2 of n model. pair O some shared deloc alized y through identify nuclei t i s a species that work anions understand the Table 1 of between involved (table 1). necessarily make to together l models important ions attractions species. The type of bonding have limitations. not weaknesses 96 and species of bonds What examples of structure bonding are present in the photo? Before rst discussing look into ionic what ions bonds are. and the characteristics of ionic structures, we will y models the and forces chemic al useful. This is positively depends and bonds substance. P All re asons a r the s t r u c t u r e. in phenomena. that d i r e c t l y. d i s c u ss e observe and models things 2.3 simplify c o m p l ex Sometimes visualize and s s Scientific 2.2 Structure 2.1 The ionic model a C ations and anions b + Sodium chloride contains sodium ions, not sodium atoms. Sodium atoms and sodium ions have dierent numbers of electrons, and therefore behave dierently. Na Na + notice 1. number 2. electron 3. charge. of three dierences between Na and Na : electrons Figure 4 are neutral. Sodium ions have a 1+ charge, indic ated r atoms by a + sign next to the symbol: Na . ions have a 1+ charge. In a protons charge is 11 a sodium protons 10 ion = there (charge electrons = (charge 11 0 are: 11+) = charge is 11 10 10 2 a the slightly atom. Figure denoted dierent This is protons electrons You c an sodium atom of ignore now, we and a sodium a atoms (c alled these are b of means that sodium protons ions is c ations S are have (a) sulfur atom (b) sulde ion greater than 11 protons and Activity ions. They contain a greater number of electrons Show that the sulfur atom is neutral 2 the name: superscript the rst part by the sux in the symbol S corresponds to . Note that anions adopt the name of their parent ide. and of the 2– by sulde only consider monatomic ions. You will look at charged polyatomic ions) in a later ion counting particles. has their Determine of the a charge subatomic the electron sulfur atom groups and of as ion. shows a sulfur atom and a sulde ion. The sulde ion has a by followed will charge. neutrons in ionic c alculations conguration For charge 1– 2– This charge As a uncharged. Figure 5 electrons. have learned 1+ S electrons. positive a charge is 1+ charged 5 than combined charge overall negatively protons. charge, negative E than are the a electrons, more as l combined Anions with v O 10 ions charged, u are positively the 1 3s 6 2p f x C ations 6 2p 2 2s and charge you have n 2 2s 2 : 1s a i 2 + Na of protons t a r o Solution Na: 1s conguration o electron 1+ d the ) = Worked example 2 Deduce ) U O verall 11 = Structure 1.2, that C 11 (charge 11+) In o In = are: i n b. (charge electrons O verall there v 11 atom p 11 sodium y a. r e Solution O atoms are neutral sodium t i s sodium b. n Determine the number of subatomic particles to show that a. l y Worked example 1 y + superscript atom (b) sodium ion P Sodium (a) sodium arrangement e will s s You sulde ion. section. 97 Structure 2 Models of bonding and structure Predicting the charge of an ion The 18. main The group elements electron corresponding are in congurations c ations are periodic for shown table some main groups 1, 2, 13, 14, 15, 16, 17 and group element atoms and their below: s s 1e + Na 2 Na 2 1s 6 2s 1 2p 2 3s 2 1s 6 2s 2p e + Na has the same electron conguration as neon, Ne. Two dierent species with electron conguration are c alled r + same isoelectronic. Therefore, isoelectronic. 1 2 2 1s 1s 2 2s 6 2 2p 3s 6 3p 2 is isoelectronic with helium, Ca He. this all formation formed parent the resulting atoms gain gain they i n 1s 2 2s 6 2p 2 3s U 2 O Cl 5 2 3p 1s in is isoelectronic with argon, r o the seven outermost electrons. l O requires a large amount of energy a while the addition of a single 2Na(s) the + Cl u f x of the positively charged nucleus 1s 3p electrons of formation Consider noble gas conguration if it lost electrons from the attractive pull gain t a The Chlorine would also have a 3s o that formation a chlorine atom gains an electron. However, the removal of so many 2p 2 6 Ar. Atoms the 2s 2 i d To obtain a noble gas conguration, 6 2 anions of an is ionic formation (g) Noble ions, 2s 6 3p full (or achieve the the achieve and electrons charged. loss of examples a noble C ation electrons. below gas electron +2e 2 O 2 1s 2 2s 6 2p 2– O sodium is anions. reduction compound of have oen 2p become a involves at gases 2 3s have all done so by losing positively to Ar. they electrons, Look order argon, 6 2p 4 n – Cl 2 it electrons. electrons lost are bec ause C +1e have c ations of oxidation when atoms As conguration: Cl form p are the example electrons. o where charged, an congurations. elements conguration. The atoms above valence v Anions is gas group with y negatively noble main r e outermost have When noble gas electron their are c ations sublevels. isoelectronic 2 2s O resulting “closed”) is t i s The 1s n 2+ + Li 2+ Ca 2 4s y 2s are l 2 1s Ca Ne y 2e + Li and P 1e Li Na isoelectronic As reduction is with the neon, gain of Ne. electrons, process. from its elements is a chloride redox reaction. from its elements: 2NaCl(s) 2 + Sodium chloride, anions, Cl other is a . The NaCl, half reduction is made equations and up are therefore of sodium shown the c ations, below. formation The of Na rst NaCl is , and chloride an oxidation and the from its elements is a redoxreaction. electron releases energy. This is v + 2Na + 2Na 2e Electron loss = oxidation why chlorine instead will gain an E electron to become a chloride ion. + Cl 2e 2Cl Electron gain = reduction 2 The energetics of these processes, Once you have learned about oxidation states (Structure 3.1), you should also be called ionization energy and able to see that the sodium is undergoing oxidation bec ause its oxidation state bec ause its oxidation state electron anity, are discussed in increases (from 0 to +1) and Structure 3.1 and relevant in the decreases construction of Born–Haber cycles (Reactivity 1.2). 98 (from 0 to 1). the chlorine is reduced Structure Atoms tend to achieve a noble gas electron conguration 2.1 The ionic model through gaining, TOK losing, or, to as the as we will octet rule. see in It is Structure 2.2, c alled the octet sharing rule electrons. bec ause This most is noble oen gases referred have General eight rules in chemistry (such electrons in their outer shell. as exists element a and relationship its periodic between table the group. charge In of the ion formed by a main group general: octet rule) exceptions. have to oen How exist for a have many rule s s There the exceptions to cease to be useful? in groups 1, 2 • Elements in groups 15, • Elements in group and 16 13 and form 1+, 2+ 17 form 3–, and 3+ 2– and ions, 1– respectively ions, respectively The 18 (noble gases) do not e Elements electron conguration of 2 form ions c arbon, 1s r • 2 2s 2 2p , suggests that The relationship between periodic table group and ionic charge is illustrated in four electrons in order to achieve a noble 18 result 4+ of C 13 14 15 16 more 17 3 6 Cs 2+ 2 Te 2+ Ba is simply a electron gaining hydrogen in one. nucleus: a The charge density of a H combine with other species. ion is One O , formed when hydrogen f x 3 thus atoms forming c an also gain an – hydride anions, H . l a E v O + u Hydrogen very example + hydronium ions, H sublevel. proton therefore such 1s loss with t a + it. the Electron i which one or r o , only electron to process discussed in c arbon forms compounds c alled does covalent not involve Covalent bonding is Structure 2.2 n have electron o that d atoms I U The charges of some common ions losing Br i n 2+ + H Se Sr + either 2 Ca + Hydrogen Cl possible, which y Rb 2 S is formation. p 5 + 3 P ion formation respectively. o K 3+ Al C 4 2+ Mg a bonding, F v Na O r e + 3 Figure 6 2 N this commonly through + Li 2 the ions, O t i s 2 in 4– or C Although 1 conguration. This n y would 1 gas l gure 6. y P c arbon atoms could lose or gain no high, is the to form the electrons so these ions by formation of surrounding c ations readily formation of acidic c ations achieve They leads a bond noble with gas water. conguration, Hydride bases. anions You bases in will are very learn strong more about Reactivity 3.1 hydrogen – + + e –1 ion, H + 1 e hydride hydrogen + ion, H atom + Figure 7 The formation of H and H 99 Structure 2 Models of bonding and structure Practice questions 1. Determine the charge magnesium, Mg c. aluminium, Al d. uorine, F e. nitrogen, N f. selenium, Se g. barium, ion formed each 3. Complete the table: name of ions d, e and f above. of 15 2p 6 conguration 2 3s 5. noble group dierent charges. i (gure 8). t a u a E v O l 3d is an elements, o main gases do not n transition element to d r o f x Fe: [Ar] 100 why U A Explain given, 6 3p 2 a element o 1s 2 2s C i n b. v electron 2 18 p + H three 2 isoelectronic a partially element c an lled d sublevel. example, iron commonly In contrast form multiple ions with 2+ For species: form ions. with transition identify 0 y r e 8 1s Charge electrons O + K each Electron conguration t i s beryllium n Number of protons a. following elements. l Symbol y Number For the forms Fe 3+ and Fe ions + 2 – – Fe : [Ar] e 2 3d 4s iron(II) ion 4s – 3 e – + 3 iron atom Fe : [Ar] 3d 4s iron(III) ion Figure 8 Iron atoms c an form ions with a 2+ charge and ions with a 3+ charge y State 4. of Ba 2. Name by P the the e b. of Li s s lithium, r a. Structure Consider the electron congurations of the rst-row transition elements. As 2+ Most of them areformed. contain This two helps to 4s electrons, explain why which most of are lost when the M these 2.1 The ionic model elements seen in chapter the ions commonly 3d form 4s sublevel lls Structure 1.3, up before the sublevel. 2+c ations. s s t Table 2 Electron congurations of the Electron conguration Symbol rst-row transition elements Element Atom [Ar] 4s Ti titanium [Ar] 4s 2 2 iron [Ar] 4s Co cobalt [Ar] 4s Ni nickel [Ar] 4s 5 [Ar] 3d 5 6 7 8 the bec ause successive ionization to to a oset certain the previous subsequent in sublevels data by other then are lost before It it have energy data. variable in energy, important to Deduce the abbreviated of electron each of the following: 2+ a. Mn b. V c. Cu d. Cu oxidation as 3+ shown + realize that energy, release small could energy, amount be but of this such energy as additional lattice 7 . energy Zinc a. only forms 2+ ions. Deduce the full electron 2+ energetic ally favourable if it conguration of Zn b. Explain . why zinc is not a i transition element. t a 3 30 6. 2+ absorbs that a ionization together is Practice questions conguration 8. The has ion of mass a transition metal number 55, electron 5 conguration l [Ar] 3d and a charge of 2+. a. Write its symbol using nuclear notation. v E n o it a z i n o i a 01 / ygrene u lom Jk r o 3 f x O 3d 9). They process. 40 10 close requires ionization, 4. Ionization b. Identify a 1+ ion that has the same as electron the conguration above. 0 0 2 4 number Figure 9 are processes only exothermic 20 sublevel. successive period (gure isolation. ionization 4s o compared 3d in examining elements energy happens usually a and the electrons many of these elements oxidation states (Structure 3.1). This d If 4s to 4s in n is formation. leads the ionization rarely investment transition by energy the occur C the variable explored in U by on have be similar ionized, i n focus c an is are ionizations o elements characteristic states sublevel elements successive [Ar] 3d v Transition 3d transition Further 9 3d p bec ause Let’ s row [Ar] 4s 10 y rst electrons. 8 [Ar] 3d r e the 3d 7 [Ar] 3d 3d 1 the 6 [Ar] 3d 3d 2 When [Ar] 3d 3d 2 copper 5 3d 2 Cu 4 3d 2 y Fe 1 n [Ar] 4s l [Ar] 4s manganese O chromium Mn [Ar] 3d P Cr 3 3d y [Ar] 4s 2 [Ar] 3d 3 t i s vanadium 1 [Ar] 3d 3d 2 V 2+ ion 1 3d r sc andium e 2 Sc D ata for the rst of 6 8 electrons 12 ionization energies of iron. electrons are very close together in energy. 10 As you c an see, the 4s and The large jump electrons occurs bec ause the 9th electron is removed 12 removed between the 8th and 9th from the 3p energy level, which is closer to the nucleus 101 Structure 2 Models of bonding and structure Communic ation skills ATL You may have noticed that we c an refer to charge using dierent formats depending on context. When using chemical symbols, charges appear as 3+ writing, we say “the ion has a . In speech or s s a superscript number followed by + or –, for example, Fe 3+ charge”. Charge is related to oxidation state (Structure 3.1), where the + or – sign is given rst followed by the magnitude. Roman numerals are also used to and indic ate oxidation states in the names of Reactivity 3.2). For example, r compounds (Structure 3.1 the 2+ copper(II) ion is Cu own to charge example ionic to is help 2+, you and its symbol remember these distinct charge. ways of l the its the an element trend in variable 3.1) successive oxidation in the periodic ionization states? table relate to the energies of transition elements (Structure 1.3) o v p their of (Structure y does explain position ion(s)? r e How of O does charge n y How t i s Linking questions for a oxidation state is +2. their and bond bonding to way look at to a attraction element estimate measure electrons. periods u of has an so the in of Within and element, Values forms results c ations, in to the and each other bec ause of formation of ionic bonds. another forms anions, they c an whether up it the the the has in a ability groups. high Pauling of periodic sc ale electronegativity bond an between atom table, This to sc ales are value to attract that attract used 4.0. a given elements is ionic is to the by The of is of Electronegativity covalently increases the most covalently chemists and two. pair uorine pairs dimensionless of two between electronegativity means tendency electronegativity the a electronegativity is electronegative bonded c alled the range from electronegativity 0.8 of bonded across the electrons. Pauling sc ale. to 4.0. Fluorine c aesium, one of the If two elements have an a electronegativity dierence greater than 1.8, the bonding between them will have a high E v ionic character Electronegativity and other periodic is One l O Figure 10 This form an ionic compound. dierence i 1.0 ( χ) the t a r o 2.0 f x ytivitagenortcele One character 0 trends greater detail in are discussed in Structure 3.1 least electronegative electronegativity The larger the compound, bonding than the of 102 given ionic ally o ionic a electrostatic ally attracted Electronegativity (χ) greater d ecnereffid ionic if are charges. n U Therefore, 3.0 anions opposite C C ations i n Ionic bonds (Structure 2.1.2) 1.8 is dierence the main greater assumed (gure type bonding elements, 10). of is 0.8. Noble gases are generally not assigned values. to In in ionic occur reality, bonding present. electronegativity the in character when the bonding the between two elements in a of the bond dierence occurs compound is in across ionic, between them. Ionic electronegativity a continuum, but there so is greater above 1.8 may be other types y your referring its P Write , e For example, “oxygen has an oxidation state of –2”. Structure 2.1 The ionic model Data-based question Predict which of the compounds sodium Dierence χ uoride, (Na) ∆ NaF χ chloride, χ ionic structure. electronegativity 0.9 and ∆ χ (F) = (∆ χ ) 4.0 3.1 = χ 0.9 and (Cl) = 3.2 = 2.3 (Al) = 1.6 and χ (Cl) = 3.2 = 1.6 3 Table 3 Electronegativity dierences for selected metal chlorides l at c an the of dierences from in approximate constituent how electronegativity each ionic elements are in a compound the periodic generally found will be by looking table. Elements with at a greater horizontal other. p o i n v Worked example 3 y distance its r e large qualitatively positions O You n y t i s Periodic table position y ∆ have P chloride, AlCl χ will r χ aluminium = (Na) ∆ NaCl χ = in 3 e sodium table s s Compound in Compare the ionic character of bonding in the following pairs of compounds: c aesium b. magnesium uoride, C sF, oxide, and MgO, c aesium iodide, and I are dierence F, in in distance the periodic the bond 0.8 ∆χ(C sF) = in ∆χ both higher 4.0 and χ(I) 1.9 v alues are = are between them is more Qualitative comparison: In the from periodic e ach than ionic. must be bond ionic so the character than bonds C sF C sI. larger Mg C and and dierence than between χ(Mg) = has a Mg 1.3 (MgO) χ(C) 1.8, than Mg that and O O are are. between between O further must C be away Therefore, Mg and more and O, the O and the ionic. Q uantitative comparison: ∆χ However, table, other electronegativity = 2.6 ∆χ(CO) gre ater compounds percentage 2.7 a Both 0.8 = = 3.2 v O ∆χ(C sI) χ(F) l χ(C s) = and Therefore, u f x = other larger between Q uantitative comparison: χ(C s) is b. e ach table. electronegativity me aning ionic. from t a and gre ater r o Cs and a i the Cs are o than F d and n Qualitative comparison: Cs CO U a. C sI c arbon monoxide, Solution C a. = and than 1.8 ionic ally O and = χ(O) = 3.4 2.1 and χ(O) = 3.4 0.8 bond for this ionic ally bec ause compound. bec ause ∆χ is C lower and than ∆χ O is do gre ater not bond 1.8. E 103 Structure 2 Models of bonding and structure Activity Determine using the the following pairs of elements are likely to bond ionic ally s s i. whether following two methods: look at their positions in the periodic table ii. refer to their electronegativity values in the data booklet. d. As and S Rb e. P and Cl c. C a and I f. Ag and Ga and e Li and F b. r a. Br oen incorrectly said element , that do not only ionic together. t this bonds There form when a metallic element and are substances, such as aluminium description. Aluminium is a metal and chlorine 3 a non-metal, Polarized light properties that would are expect point and high volatility. + two elements (1.6) suggests they propellants ionic compounds contain + which is made up of NH polyatomic ions. As 4 several atoms. NO You are expected 3 in table 4. HCO c arbonate CO 3 2– 2– sulfate ATL SO 4 4 Table 4 Common polyatomic ions silver(I) sulde Cl, 4 + , are 4 polyatomic ions are ions that contain and formulas of the polyatomic ions shown engage spend in some table What with 4. time Some strategies memorizing students will you like use? the to names use How and formulas of the ashc ards, will you others make make sure you them? patterns the in list their of ionic compounds shown in table 5. C an you notice any names? You should notice that, in the names of ionic compounds: 3 hydroxide oxide suggests, names instance, NH anions. Ammonium ions, NH C aCO Ba(OH) E iron(III) 2 a barium Consider KF MgF c arbonate the For low Naming ionic compounds l c alcium uoride to ions mnemonics. actively Formula v O magnesium need t a uoride up u potassium will polyatomic i r o f x Name You o d PO elements. a Self-management skills 3– phosphate two as between these n 3 know U hydrogenc arbonate to name C nitrate i n OH their such dierence o NH hydroxide than But the compound ionic ally. 4 + ammonium more bond c ations and Cl v Formula ionic ally. compounds, p Some Name bond y Polyatomic ions rocket not r e Its uses include fertilizers and 3 to covalent electronegativity do 4 . of The nitrate crystals. Ammonium nitrate contains two polyatomic ions: NH and NO them characteristic micrograph melting of ammonium you O has Figure 11 so t i s is n chloride, AlCl that bond l non-metallic • the c ation name • c ations • monatomic is given rst and is followed by the anion 2 Fe O 2 Ag adopt the name of the parent atom and the name remains unchanged 3 S anions adopt the rst part of the name of the parent atom, 2 followed Table 5 Names and ionic compounds 104 by the sux -ide. If the anion is polyatomic, refer to table 4 formulas of some • the name of the compound does not reect the number of ions in the formula. y P is a y It Structure Practice questions Anions are common 9. State the name of each of the 2.1 The ionic model conjugate bases acids. The of strength of following compounds: acids and stability of their anions a. RbF d. Sr(OH) 2 Al S 2 c. e. c an using 3 AlN be compared quantitatively BaCO 3 f. NH HCO 4 K 3 , their which s s b. dissociation constants, will be introduced in a Reactivity 3.1 e The formulas of ionic compounds name ratio of the an ionic ions in charges and negative the c ation, of net charge charges then of must work you what elements it contains, but not the working the out c ancel out how the formula of an ionic compound compound out. many First, of is zero, so determine each ion you the the positive charge of the need to reach a zero. a. c alcium oxide b. c alcium nitride c. the formula of c alcium oxide, work through The the second method charges subscript, and ignoring i n Step 1: Determine the charges of the c ation and C alcium 2 1s has 2 2s 2p 3s 6 3p electrons, has Oxygen atoms . C alcium they form electron have a of 2 six atoms ions with have a conguration outer shell 2+ of two outer 2 1s 2 2s electrons, 4 2p so 2+ . they charge. Therefore, calcium ions = Ca criss-cross the into the rule. other Swap ion’s sign: Ca charge. o with 4s so d ions an conguration 2 the them n Oxygen form electron 2 U shell an 6 C the anion is turn o v following steps. aluminium nitrate y deduce the d. p To c arbonate r e Solution a. sodium O Deduce the formulas of the following ionic compounds: t i s Worked example 4 n charge the tells for y total that basis l anion and compound The y remembering it. P is of r The Ca Then, simplify the ratio: 2 and oxide ions = O O Ca r o i 1 1 t a Step 2: Determine how many of each ion are needed Step 3: Check that the net charge is zero in order to achieve a net charge of zero You check your working by adding up the charges of There are two methods you can use for this step. The rst e ach individual ion. If you did Step 2 correctly, the f x u is the bar diagram method. Write out the ions as blocks charges will add to zero: equal to the number of charges on each individual ion: l ion, so compound is a oxide diagram E bar v O The 2+ Ca + 2 Ca Total 2+ 2– O Total of = 2 contains ratio charge O one c alcium ion and c alcium 1:1. to oxide in negative charge = 2 one Net the positive charge = 2 2 = 0 the Step 4: Write the formula Ca This O 1 is a straightforward example where the 1 magnitude anion, and of charge hence the is equal formula for is the c ation and C aO. 105 Structure b. 2 Models C alcium ions as of bonding nitride have is a dierent and structure more complex charges. Work example as the c. Sodium through the steps must before. c arbonate not split polyatomic and draw up contains or cluster. brackets a change polyatomic the ion. You ratio of atoms in the Treat it like an indivisible entity around it if the formula contains Step 1: Determine the charges of the cation and s s more than one such ion. the anion 2 C a: 2 1s 2s 6 2 2p 3s 6 3p Step 1: Determine the charges of the cation and 2 4s the anion atoms have two outer shell electrons, so 2 N a: they form ions with a 2+ 1s 2 2s Sodium 2 2s have ve atoms have one outer shell outer shell electrons, form ions with a so 1+ charge. 2 ions, CO , have a 2 3 form ions with a 3 charge. + Sodium ions = Ca = Na and charge. c arbonate 2 ions = CO 3 3 and nitride ions = N needed in order to achieve a net charge of zero Step 2: Determine how many of each ion are t i s needed in order to achieve a net charge of zero Bar diagram method + + Na + 2 + 2 + 2 Ca Ca The bar diagram c arbonate in the ions, contains so the compound three ratio is of its simplest the ratio here form. o in i simplify t a alre ady N in the contains so compound Criss-cross the is a = 6 poly atomic not ratio 2:1 Na 3 CO Na you ion c an to change. draw remind Again, 1 brackets yourself there is around that no its the formula need to simplify here. + 2 CO + Na 3 3 N T otal negative charge = 6 3 T otal positive charge = 2+ Net charge = 2 2 = T otal negative charge = 2 0 Step 4: Write the formula fo r mu l a is Na CO 2 th e re a re no . Note th a t in th e fi n a l 3 bra c ke ts a ro u n d th e 0 po l y a t o m i c one Step 4: Write the formula Ca N 3 106 CO 2 Step 3: Check that the net charge is zero Th e = one it a n sw e r 6 and c arbonate rule Remember, the u l E charge ions to N Ca Net sodium sodium Na 2+ T otal positive charge = 6+ of 3 does bec ause 2+ Ca v O 2+ two ratio Step 3: Check that the net charge is zero Ca 3 n U to d r o need f x is no 2 rule Ca is and nitride N 3 There ions to 3:2 Ca Criss-cross c alcium c alcium C ions diagram ion, o nitride v bar two i n The 2– CO y 3 N Na p r e 3 N O Bar diagram method Ca n y Step 2: Determine how many of each ion are l 2+ C alcium ions so 2 ion c arbonate be c a u s e ion. the fo r mu l a c o n ta i n s only y C arbonate they electron, 3 2p P atoms 1 3s r 1s they Nitrogen 6 2p charge. 2 N: e C alcium Structure d. The nal example, polyatomic ion. aluminium Follow the nitrate, also contains a same steps as 2.1 The ionic model Criss-cross rule before. 1 Al Step 1: Determine the charges of the cation and the anion 1s 2 2s 6 2p 2 3s 1 3p 1 Aluminium so they atoms form ions have with three a 3+ outer shell , NO have a 1 there is no need to simplify r = Al and nitrate ions = NO 3 do not remember the charge on a 3+ polyatomic formulas table and 4. M ake charges o sure by that Al you heart. T otal positive charge = 3+ charge 3 3 = 0 The method to formula indic ate so compound is the one ratio aluminium of ) 3 3 each oxide of the n of e. lithium f. barium c. sodium sulde g. d. lithium nitride a of an ionic contain the written as are used nitrate AlN O 3 9 transition elements have oxidation number of the transition hydrogenc arbonate t a l formation be one ammonium is metal covered in ion in brackets. This Structure 3.1 phosphate. compound from its elements a redox (Reactivity 3.2) is formal E LHA How v O the not brackets than u f x Linking questions is that more nitrate i magnesium strontium chloride r o a. reaction? should of following compounds: b. Why Note Names of ionic compounds that o formula d the nitrate U Practice questions formula C (NO 1 and to 1:3. Al Deduce ion aluminium . 3 o the contains ions, 3 i n in diagram nitrate NO 3 3 presence y bar three NO 3 The v The ion. ) Al(NO p NO r e Al is the O t i s Step 4: Write the formula Bar diagram 3 T otal negative charge = 3 needed in order to achieve a net charge of zero + 3 10. = 3 NO n Net Step 2: Determine how many of each ion are 3 NO l these revise y learn ion, NO y you here. P If ratio Step 3: Check that the net charge is zero 3+ ions the charge. 3 Al e ions, 3 electrons, charge. Again, Nitrate s s 2 Al: sulfate? LHA Polyatomic the charge used to predict the preferred structure of (Structure 2.2) anions relationship are conjugate between dissociation constant, K their ? bases stability (Reactivity of common and the acids. conjugate What is acid’ s 3.1) a 107 Structure 2 Models of bonding and structure Ionic lattices and properties of ionic compounds (Structure 2.1.3) L attices ionic crystals, continuous, negative ions. The exact arranged in a networks arrangement of of lattice structure. repeating ions in a units of L attices are positive and lattice depends on the size and ratio of the ions. Ionic compounds are made of in a lattice structure l e asily c o mp o u n d s of ion in contain a the is an e mp i r i c a l s tr u c tu re. qu a dr i l l i o n A th a t the Na ions ra ti o. a r ra n ge d involves reliability. nding n and ions a re in it of a i n d i c a te s sodium c o n ti n u o u s p re s e n t in th e c h l o r i d e, t he l a tti c e. l a tt i c e in a o oen U Research Cl C i n Thinking skills ATL and g ra i n p i n d i c a te s v 1:1 fo r mu l a fo r m u l a : single + Th e O c an ionic ty pe y N aCl, of e ach r e of n y fo r mu l a ra ti o t i s Th e information but also evaluating its usefulness Consider the statement “each grain of NaCl can easily contain a quadrillion ions”. o i 108 up Ionic you with compare dierent? bonds Is are reasonable would reliably your it the to nd own this to you this dierence species Bec ause of surrounding to How of do between it, fact-check the statement? information? estimate one. non-directional. charged you? need This with the number they the two means the of ions compare? values that an attraction in a grain of salt Why might they be signic ant? ion will being attract equal in all all oppositely directions. – E ach ion attracts all oppositely ions around it sound could Come and Ionic bonding is non- E charged + How u directional. – • • + a Figure 13 – information – l + + v O – + f x – + – this What t a + Does • + r o – – + d – + • the in all an the ionic forces in this non-directional surrounding lattice an ionic are anions, very lattice, quality, and strong. but each vice Figure remember, c ation versa. 13 This shows actual in the means a 2D lattices ionic the lattice forces attracts of representation are 3D. attraction of the y P ions arranged are r u Figure 12 ions e charge the three-dimensional s s Within Structure 2.1 The ionic model L attice enthalpy L attice enthalpy values tell us how strong the ionic bonds are in particular ionic lattice. L attice enthalpy, ∆ , is the standard enthalpy change that occurs on lattice formation the strength all the need of of gaseous an ionic electrostatic to bond forces overcome. A of from one bec ause, attraction general mole in of order the for between equation for solid the ions c ations the lattice. to and lattice It is a measure of become anions in dissociation gaseous, the lattice process is below: + M (g) + X (g) ∆H > 0 r MX(s) lattice process is quoted forces are shall is enthalpy of and energy The gure formation of Two strength to overcome factors of the the aecting increasing lattice ions radius. enthalpy attraction Figure 14 required C –12 Ionic radius/10 U –1 /kJ mol lattice compound m C ation Anion +1 –1 n Anion 133 NaF 930 102 133 +1 –1 C aF 2651 100 133 +2 –1 greater The lattice enthalpy smaller C aF is KF bec ause between considerably 2 larger the radius of Ca in NaF ions and F than 2+ ionic c ations are smaller, + Na that of ions is greater. KF. This is in part + compared to K . However, it is mainly due 2+ greater charge between on the the Ca a attraction of than attraction l the enthalpy electrostatic c ations E v O to the the u therefore to lattice f x and due i a Lattice enthalpies of selected compounds t a has r o NaF o 138 d 829 Table 6 the lattice Ionic charge C ation KF 2 Lattice enthalpy is the energy to overcome the electrostatic forces of attraction holding ions together in enthalpy of the compounds in table 6. Ionic ∆H from o in lattice electrostatic electrostatic i n variations the lattice charge ionic for However, in gaseous charged ions: ionic of 14. 298 K are oen given in the data booklet. required increases. ionic charge. increasing with denition formation in at enthalpies y decreases the shown endothermic the ions exothermic that enthalpy p • the with as to lattice L attice v increases the with between oppositely • only the process of booklet. r e between at increases attraction ionic radius Look consider values data represent opposite consistent the O L attice we which that in n book lattice, values ions — the Experimental found t i s a negative gaseous be l this as c an y from endothermic. compounds y some P The e shown be ions s s the c ation, and which anions in results in greater electrostatic C aF 2 109 Structure 2 Models of bonding and structure F actors aecting the lattice enthalpy of the group 1 chlorides Charge density volume. the this group enthalpies 1 of is a term task, you c ations the used will and to describe explore relate the this to Part 3: Analysis charge per unit charge density the trend in 5. C alculate the volume 6. C alculate the charge 7. Relevant skills Plot two graphs: Tool 3: General mathematics • Inquiry 1: • Inquiry 2: Identify, explain describe and between Describe explain 9. patterns, Discuss the relationship enthalpy, and the the trends differences lattice enthalpy. shown in the between the two your graphs the relationship 11. prediction, including a comparison you obtained in 7 and the sketched O predict you n the t i s c ations, the and l Evaluate graphs 1 explain density y 10. group and charge graphs. relationships Part 1: Prediction the lattice graphs. predictions of For c ation. P and Instructions 1. and each showing r 8. State and radius graph spreadsheets to manipulate data other, trends c ation. of obtained in 2. Consider possible extensions to this investigation: between: what other aspects of ionic radius, charge density and charge y ionic radius and lattice enthalpy of their chlorides b. r e a. lattice enthalpy could you explore? density and lattice enthalpy of their chlorides. the graphs you expect to obtain for the above. 3. C i n Part 2: Data collection Collect the following data for the group 1 chlorides: ionic charge, ionic radius, lattice enthalpy. Possible U sources of information include the data booklet and Input your data into a spreadsheet table. and organize it into Practice questions 1. Write equations, u enthalpies of including KBr, state symbols, that represent the O a lattice C aO and MgCl 2 2. l a E v O f x t a r o i o suitable d a n online databases. Cite each source appropriately. 4. o relationships p Sketch v 2. State and explain whether you expect KF or K to have lower lattice 2 enthalpy 3. State have value. and the explain which of greatest lattice enthalpy the following value: ionic compounds NaCl, MgCl , Na 2 4. you O or expect to MgO. 2 Describe and explain the trend in lattice enthalpy of the group 1 chlorides down the group from LiCl to CsCl. Properties of ionic compounds The properties contain in 110 a c ations lattice. of ionic and compounds anions held are due together by to their strong structural features: they electrostatic attractive forces y • ionic one density e between Tool 2: Use each lattice group 1 chlorides. • of s s of In Structure 2.1 The ionic model Volatility Global impact Volatility (from the L atin volare, to y) refers to the tendency of a substance to of science vaporize (turn electrostatic into a forces gas). of For an attraction ionic compound holding the ions to turn into together a gas, must be the strong overcome. Some volatility of ionic compounds is therefore very low: they are ionic compounds said to be “nonuncharacteristic ally volatile”. This also means they points and sodium chloride is approximately 1 075 K. Magnesium oxide, frequently used in be furnaces due to its ability to withstand high temperatures, melts at around 3 098 K. bec ause c an Such described they they as and, are does transportation t i s + i n – – – + – – + move. They + + are fixed in the – + U – + regular + – – + – d i + + + f x conduct Electrolytes in batteries electricity bec ause they contain mobile ions. separated Used batteries are frequently from other types of waste to prevent them from ending up in landll. water metals and other substances, be recycled. which c an Do you separate your used + batteries from + other household waste? + + + + l + a v + + + + water + + + molecule + + + + E + + + O + (red) is for batteries. This is bec ause batteries contain valuable u + These are waste separation dissolve in t a r o + o + heat Figure 15 bins in Jakarta, Indonesia. The lemost bin n – – arrangement. C + ions cannot o + signic ant O + – – v + electricity have y compound p r e A solid ionic necessarily disposal and c an anions are free to move past one another , allowing them to conduct electricity when a potential dierence is applied. not environmental impacts. ions in a solid lattice are not mobile. When molten or aqueous, both cations and easily recycled. n y cannot change position. Solid ionic compounds do not conduct electricity because more oen, harmless. Their manufacturing, a solid ionic lattice, cations and anions can vibrate around a xed point, but they non-volatile. This be l they c an solvents” y however, mean able to move. Ionic compounds contain charged particles, cations and anions. In P This, In order to conduct electricity, substances must contain charged particles that are “green are c an Electrical conductivity contained used as r means be ionic liquids e solvents. because the low melting have high boiling points. Ionic compounds typically have high melting points too. The melting point of cannot conduct have s s The + + When an ionic compound is heated When an ionic compound is strongly and melts, the ions can dissolved in water, it can conduct move around and the molten electricity because its ions can compound conducts electricity. move among the water molecules. t Figure 16 Molten and aqueous ionic compounds are electric al conductors 111 Structure 2 Models of bonding and structure Solubility Ionic compounds insoluble in Water is discussed in in a polar solvents solvent. The such as dierence polar solvents hydrogen and in the partial atoms, water combined molecule positive an ionic charges with in the electronegativity bent geometry their positive the the water oxygen molecule, a lattice and solvent, the solvent added negative point to towards is no molecules, water. charges become there negative the surrounded attraction so the of the remain In ionic within δ‒ H δ+ ions the n + Na + Na Cl Cl + Na U + Na + H O + + Na Cl Na δ+ H H δ+ δ‒ Cl i u l a E v O f x ionic competing ionic • the Ionic O ions bonds + Na δ‒ δ‒ δ+ O H and and of O H H δ+ δ+ are water in the insoluble the lattice molecules. c alcium in water. This is bec ause H water molecules there are two present: c ations between anions include dissolve attraction between compounds water δ+ H δ+ compounds forces association c ations δ‒ O The dissolution of ionic compounds in water involves interactions between ions and • O δ‒ t a r o Cl all H δ+ + Na + + Not δ+ δ+ o d Na Na H O δ‒ H Cl Cl + H + Na Cl δ+ δ+ δ+ H Na Na Cl H H Cl δ+ δ‒ O Cl H n Na + Figure 17 ions anions and individual O H Cl the position c ations, molecules. δ+ δ+ C i n H + and the O p δ+ Na molecules result, water between o δ+ O v δ‒ Cl a y r e dissolved, but not when solid. Cl oxygen atom δ‒ electricity when molten or Cl the y diagram δ+ water As by c ations lattice. why ionic compounds H on towards anions. t i s labelled The point Activity explaining charge l and being partial charges the non-polar compound the of their partial hydrogen atoms. and ions anions and when are the the of and the lattice partial charges electrostatic stronger Examples c arbonate in ionic silver than the water attractions association compounds chloride. of that molecules between the between the are insoluble in y of out that a the P pulled c ase 112 between of r so partial having on compound themselves are conduct water, and e Imagine a as Structure 2.2. result Draw such hexane. greater and detail is typic ally soluble in s s Polarity are non-polar Structure 2.1 The ionic model Research skills ATL H e av y Some metal h e av y Use the and their as le ad t re a t m e n t metals internet to out and nickel, p ro c e ss e s of industrial re s e a rc h other often take form insoluble adv antage effluents ex a m p l e s of t h ro u g h of this salts. p ro p e r t y, p re c i p i t a t i o n . p re c i p i t a t i o n re a c t i o n s uses. Describe and p re c i p i t a t e is ex p l a i n the changes that a re o b s e r ve d when a forme d. r data demonstrate the physic al properties of ionic y (Tool 1, Inquiry 2) Close-up photograph of the O t i s Figure 18 How can lattice enthalpies and the bonding continuum explain the trend in n l experimental compounds? y P Linking questions What e • such s s re m o v i n g • ions, w a s t ew a t e r formation of a lead(II) chromate precipitate melting points of metal chlorides across period 3? (Structure 3.1) in the reaction between aqueous solutions of lead(II) nitrate and y o C n o i t a S alar de Uyuni in Bolivia, chloride. particularly lithium. p r e v i n U a ats at of sodium u l ions, S alt v E mineral form d r o f x O Figure 19 potassium chromate which are mainly made of halite, the The brine below the rock salt Global demand Lithium-ion batteries c an be used for lithium crust is rich in dissolved metal is increasing due to its use in batteries. to power mobile phones, laptops and electric vehicles 113 Structure 2 Models of bonding and structure Solubility of ionic salts The patterns ionic in aqueous compounds are solubility often of several common referred to as Instructions solubility Part 1: Solubility rules us We c an deduce use the these differences in solubility to help identity of will different solutions of ionic an unknown ionic compound. General In this task you mix table (known as a and observe precipitate) whether is an insoluble 7. solubility For rules c an be inferred from the data in example: product • All • Sulfates nitrates are soluble. produced. generally insoluble, except sulfates issues address in an relevant sulfate. safety and table 7 and infer at least three more investigation solubility rules. Inquiry 2: Interpret l • general qualitative data y • Dilute eye protection. A, will B, C, sodium c alcium nitrate and silver nitrate solutions be provided and D. chloride, are do not know solution, take c are when handling all solutions You bec ause you do not know exactly which is which. of are all potential irritants. and retain any precipitates formed. you tubes. i n • materials from by do You mixing not c an a of Dispose of waste solutions and precipitates according your school’s guidelines. Materials Dilute acid solution • S amples piece of on a copper black wire solutions – COO 3 + , Na knowledge of other areas solution observable, small-sc ale each on will a mixtures of solutions plastic become particularly if sheet. If a precipitate opaque. This will be you lay the sheet on a teacher . If they approve it and if you have time, try it out! C ations ammonium, barium, lead, 2+ c alcium, C a + , K ) + NH + silver, Ag 2+ 2+ Ba Pb 4 soluble soluble soluble soluble soluble soluble soluble soluble soluble soluble soluble soluble soluble soluble soluble soluble insoluble insoluble soluble soluble soluble slightly soluble insoluble insoluble soluble soluble insoluble slightly soluble soluble soluble insoluble insoluble a ethanoate, you each background. t a l – CH u + (Li v O f x group 1 c ations 3 but Devise a method, present it clearly, and show it to your cm long) labelled A, B, C and D nitrate, NO black nitrate, identify background (~0.5 r o of easily the to n Small • sheet formed, is o • plastic i Pipettes U Clear • d • is C to c alcium job labelled c arbonate, have to mix the solutions inside test prepare drop solutions potassium listed and the solubility rules. your o v Collect the draw and Your chemistry. Note • using also nitrate which. p They may is of are y should r e You samples solutions silver which irritants. • with These O Wear t i s You • n Part 2: Identification of ionic compounds S afety – chloride, Cl E Anions – hydroxide, OH 2– sulfate, SO slightly soluble insoluble 4 2– c arbonate, CO 3 Table 7 114 Aqueous solubility of common ionic compounds insoluble insoluble y and Study environmental ammonium P Tool 1: Recognize and group 1 r are Relevant skills • e compounds s s rules. Structure 2.1 The ionic model End-of-topic questions 7 . What is the formula of sodium nitrate? A. NaNO B. NaNO C. Na D. S s s Topic review 2 1. Using your answer the knowledge guiding from the question as Structure 2.1 topic, fully as possible: 3 e What determines the ionic nature and properties of a N 3 compound? 2. Explain why ionic substances are r N 3 2 always compounds. Which compound has the largest enthalpy? Exam-style questions The elements group 17 generally C aO C. K form ions with D. A. K 7+ 1– statement A. L attice 7– B. 2 2s 1s D. 1s 2 2 2s 2 Which 2 3s 6 2p 2 are made up correct? electrostatic of C aF L attice the Which of equation K a increases ions the energy needed to c ation to an anion. when the charge of the correctly when the charge density increases. represents the lattice enthalpy oxide? + A. radii of the increases. component potassium from decreases ions enthalpy the 2– O(s) → 2K (g) O(s) → 2K(s) O(s) → K + O (g) 2 B. K C. K + ½O 2 molecules. (g) 2 2+ 2 2– (g) + O (g) 2 2 contain held to together c ations. name of c arbon sulte c alcium sulte C. c arbon sulfate D. c alcium sulfate E B. a the is c ations and anions. deloc alized bec ause l A. is the D. represents electron enthalpy when increase. D. electrons when K O(s) → 2K(g) + ½O 2 (g) 2 dissolved, but not when solid. v O What from u Ions or is compounds an component 10. i uoride electrons 6. between f x D. results structures molten ionic L attice of t a Ionic 2 3d r o C. 2 4s about bonding C alcium 2 3d 6 3p statement Ionic 2 4s 6 3p 3s attraction B. 6 3p d A. 6 2p 2 2s 2 3s C. o 5. 6 2p conguration of this ion. n C. 2 2s electron 6 3p U 1s full 2 3s C 2 B. the 6 2p i n 1s enthalpy transfer o Give 2 A. L attice v protons. charge and contains 20 increases ions p D. The ion of element X has a 2 + correct? enthalpy component 4. is y C. Which r e 1+ O 2 9. B. S 2 O charge? t i s which in n B. Multiple-choice questions 3. lattice l C aS of y A. value y P 8. C aSO anions transfer 11. List the lithium halides in order of increasing lattice enthalpy. A. LiF, LiCl, LiBr, LiBr, LiCl, LiI B. LiF, C. LiI, LiBr, LiCl, LiF D. LiI, LiCl, LiBr, LiF ? 4 LiI 115 Structure 12. 2 Models Which of bonding substance has Melting and an structure ionic structure? Electric al Electric al conductivity conductivity Solubility point / °C in water when molten when solid high none none B 186 low none none C 1083 high good none D 1710 low good good r Describe applic ations in the eld of optics. for c alcium uoride. [1] c. Deduce d. Potassium the Chromium the structure and bonding in solid charge of the dichromate(VI) dichromate(VI) contains is a uoride. [3] i. c. Explain why solid form Cr Write c alcium uoride is a poor c an conduct The lattice but molten c alcium uoride electricity. [2] ii. a the Copy the arrows d. enthalpy of c alcium uoride is shows lattice the process enthalpy Explain why including of the state uoride. The lattice enthalpy of c alcium oxide, iii. [1] C aO, 15. 1 the potassium full the atom v E 116 Cr 2 [2] orange draw The [1] 4s Write the 3d full electron conguration of a 3+ Cr ion. equation uoride Li(s) ion. + from F (g) [1] below its → shows elements the formation of lithium under standard conditions. LiF(s) 2 a. Balance b. Identify the the charge equation. c. Identify the oxidized [1] O 2 7 of the lithium ion. [1] in mass species and the reduced [1] species in this reaction. [1] between a d. Sketch a diagram showing the structure of potassium ion is a negligible. a bright and represent the conguration of a dierence and the dichromate(VI), K l O potassium electron contain ion. why lattice greater than u Explain the o Write is uoride. breathalyser compound potassium b. c alcium f x a. of why oxide t a ionic of types Explain r o Certain . c alcium i that 14. of d is3 401 kJ mol enthalpy n iii. [2] process in part (i) is endothermic. to 2+ associated with the c alcium conguration [1] below boxes orbitals of a Cr symbols, U ii. equation, diagram the electron others. electronconguration in the 3d and 4s C that an in among atom. o Write i n i. . v 1 2 651 kJ mol abbreviated chromium ions y conductor, 3+ ions and Cr p r e of electric al [1] transition element that commonly 2+ forms c alcium ion. chromium. O b. has formula n the t i s Deduce l uoride a. y C alcium lithiumuoride. [1] [2] y P Extended-response questions 13. e 36 s s A Structure 2.2 The covalent model s s What determines the covalent nature and properties of a substance? Substances From water to diamond to nitrogen gas, from oils to characterized plastics to polyatomic ions, these species contain atoms low held together by strong covalent bonds. Covalent bonds molecular lead to the formation of two dierent types of structure: low melting points and boiling points. Their solubility and covalent network structures (also known as giant covalent volatility structures) and molecular covalent structures forces. poor electric al Structure 2.2, how positively shell pair with a two total of 8 electrons. three shared pairs of the same atom. Structure 2.2.4 — electron domains a molecular C arbon v E molecules. bond Electron from the repulsion of central atom. polarity results of from the the bonded atoms. < You structures. dipole–dipole forces < is more in a than Resonance one structures occur when possible position Structure 2.2.12 — Benzene, C H 6 example of a molecule Structure 2.2.13 — which for a double molecule. they have an Some for determine which has each Formal atom of in several , is an important 6 resonance. atoms expanded Structure 2.2.14 — c alculated that c an octet charge a form of values species possible molecules in electrons. and c an be used to Lewis formulas is preferred. The is Structure 2.2.15 — geometry. a O the covalent network describe Molecular polarity depends on both molecules (dispersion), predicted l and Structure 2.2.8 — of Bond structures. between Shell a u f x polarity Structure 2.2.7 — network be electronegativities Structure 2.2.6 — bond to around Structure 2.2.5 — in Valence is shared pair (VSEPR) model enables the molecules dierence the t a of bond of we forces. mobile and stationary phases. there i Repulsion shapes The r o Pair electrons how intermolecular their relative attractions involving intermolecular forces to Structure 2.2.11 — o from coordination the d originate A both forces covalent bond is, as LHA which covalent a well intermolecular used to separate the components of a mixture based on n in and their have Structure 2.2.10 — Chromatography is a technique electrons U Structure 2.2.3 — about (dispersion) Single, double and triple bonds and shapes hydrogen bonding. electrons charged nuclei. respectively. bond London by the of learn C one, formed refers to the tendency of atoms to gain a Structure 2.2.2 — involve is shared i n valence rule a their on what as Substances with generally y octet bond between learn water. hand, o the The covalent v and A attraction also will molecules p electrostatic explain depending r e Structure 2.2.1 — you represent in other O will we the t i s and Understandings greatly solubility on n In poor y are are also l substances structures r covalent and structures, vary network P general, conductors. volatility covalent by high melting points and boiling points, y In with e Covalent bonds lead to a vast range of dierent substances. and Sigma bonds (σ) form by the head-on combination of atomic orbitals where the electron density silicon form covalent is concentrated along the bond axis. Pi nature of the force that bonds (π) form by the lateral combination of p orbitals exists where determined the electron density is concentrated on opposite by the size and polarity sides of the bond axis. Intermolecular dipole–induced forces include London dipole, dipole–dipole and Structure 2.2.16 — mixing hydrogen bonding. atomic Hybridization is the concept of orbitals to form new hybrid orbitals for bonding. Structure 2.2.9 — relative strengths Given of comparable intermolecular molar forces mass, the are generally: 117 Structure 2 Models of bonding and structure Covalent bonds and molecules (Structure 2.2.1) Covalent covalent bonds bond Diatomic and formed the from positively hydrogen, H , is when the atoms charged the share electrostatic nuclei simplest pairs of attraction of the covalent valence electrons. A between atoms a shared pair of s s electrons are results involved in the bond. molecule. It consists of two 2 from hydrogen electrons, the atom forming formation together (gure single, of by the a covalent 1). double double Atoms or bond c an triple covalent made share bonds, bond in diatomic electrons gives both arrangement and forms a single covalent bond This is a double covalent bond (two shared pairs of electrons). Only the electrons in the highest energy level (outer shell) are shown here , and diatomic oxygen, O 2 2 o i t a u l a v Examples of substances that contain covalent bonds: plastics, graphite (in pencil leads), zzy water, and the c arbon dioxide y n A shared pair of y o formation in diatomic hydrogen, H d r o 118 E bubbles in it f x O Figure 2 bond O . 2 Oxygen molecule n Covalent U Figure 1 O C i n Oxygen atoms oxygen, O atoms a stable p r e O pairs Figure 1 also H Hydrogen molecule v O electron three O Hydrogen atoms or l H H one two respectively. y t i s H from one, P shows held r of each atoms e hydrogen Structure 2.2 The covalent model Activity Draw a diagram to show the triple covalent bond in diatomic nitrogen, N 2 s s Electronegativity bonds generally electronegativity, 1.8. when When the dierence bond in dierence them is relatively high between electronegativity between of Structure 2.1, between predominantly we two saw that ionic bonds atoms the is atoms character bond together, bond bonds they form are covalent bonds periods are found likely with to be each the referring O and following to c arbon b. sodium c. c arbon d. chlorine e. iodine and iodine f. aluminium and uorine to so the they same form a no between ionic bonding. a covalent borderline and Bonding continuum, character Structure 2.1, sharp at with one covalent occurs along greater ionic end character at and the greater other. groups. If two atoms of Structure 2.4 addresses this each other in the periodic covalent. For continuum example, in more detail with the oxygen and bonding triangle pairs of elements electronegativity bond covalently or values: oxygen hydrogen v and up zero, of is a a. and oxygen their and close is atoms discussed in there t a whether by u ionic ally, l f x Determine dierence other. Practice questions 1. non-metal i form across electronegativity r o uorine increases two o the electronegativity d table, high the When between them. Electronegativity relatively compounds. covalent substances n covalent or U element elements compounds, As y either always C be are i n c an which covalent o that p Electronegativity dierences are lower than 1.8 suggest bonding is present substances, v that n greater covalent O covalent bonding t i s character 2.0 1.0 l y greater ionic r e ecnereffid ytivitagenortcele ionic bonding 0 ionic less than 1.8 covalent. 3.0 Figure 3 Unlike greater than is y 3), the electronegativity atoms P (gure the between r form form typic ally non-metals. In e Covalent and oxygen E 119 Structure 2 Models of bonding and structure Lewis formulas Groups way of atoms valence formulas. of are Lewis dots or covalently are formulas, pairs of bonded arranged pairs crosses, in of or a valence a together molecule c an are be electrons combination of c alled are all molecules. The shown using Lewis represented as dashes, three. Consider the F various Lewis formulas of uoroamine, NH F, in gure 4. 2 H H H Here are the dashes rules are for drawing Lewis formulas regardless of whether dots, used: F Only electrons are shown. 2. Electrons 3. Each pair of electrons shared between two atoms represents a covalent bond. 4. Electrons are arranged in pairs. These are all acceptable Lewis formulas of uoroamine, NH l F 2 bond two (termed atoms region positioned in the (also between the referred to as two atoms lone pairs) are region positioned O the are involved in the bond. Non-bonding electrons from bonding electrons) n a t i s 5. in the y between away involved in the bond. Atoms in Lewis formulas generally have noble gas electron congurations. This is r e known as the octet rule because noble gases oen have eight valence electrons. y Noble gases already have full octets and do not readily gain, lose or share octet rule some stable of 1. Work c an these out the of t a 3. least always O u l a 5. Bond the Assign central them. help there by of the of than and limitations. situations. an (this draw the has octet is seen exceptions Lewis valence valence drawing is central to does of electrons. at AHL). the electrons by It not Sometimes in two to c an Conversely, rule. most for explain atoms Species with odd octet formulas electrons their usually c ases: each atom in the work out how many single pairs symbols on found the in the page. The element with centre. Hydrogen atoms atom(s). peripheral E ach achieve of also you it are. atoms non-bonding they but certain fewer are number atoms surround the will number number between until total in expanded octets total electrons Arrange the 4. N Divide pairs N v O Figure 5 f x + O 2. i r o + N F thumb, with o d F of bonds electrons steps the molecule. even form valence rule form n U Following + molecules atoms numbers useful gases C larger a i n form is noble o why v The p electrons. Hence, they are usually found in an unbonded, monatomic form. atoms bond of noble-gas together represents electrons to the an by drawing single bonds electron peripheral pair. atoms. Keep going congurations. N 6. Assign any 7 . Check that remaining electron pairs to the central atom(s). By sharing a pair of electrons, two uorine atoms c an obtain a full octet of E valence electrons. the central atom has a full octet. If it does not, try the following Oxygen atoms c an do two methods: this by sharing two pairs of electrons, and nitrogen atoms share three pairs • Reassign non-bonding additional • Check octet 120 bonds that rule the (see to pairs the molecule page on the peripheral atoms to become central atom 123 you for are looking examples) at is not an exception to the y 1. H P Figure 4 valence r H crosses e or s s pairs In that electrons Structure 2.2 The covalent model Worked example 1 Draw the Lewis formulas of each of the following molecules: a. water, H O b. nitrogen trichloride, NCl 2 c. c arbon dioxide, CO 3 2 s s Solution Follow the steps above for e ach molecule: O b. nitrogen trichloride, NCl 2 Step 1 Count hydrogen: 1 2 nitrogen: 5 + chlorine: 7 6 = 8 Total: 5 + c arbon: 4 × 21 3 = = 21 Total: 4 16 = 13 pairs = 8 Cl H is is likely only the to one be in periphery. centre. There is likely only to so far... already energy need more full electrons. H remaining central atom Check the central full remaining assigned Oxygen the full Lewis octet. E v a O octet has to 4 electron the c arbon, so it the centre. pairs O used so far... O The oxygen atoms now have Cl full octets. chlorine atoms now pairs used so far... have octets. pairs used so far... We have used all 8 available Cl electron pairs available Cl N pairs. for None the are central c arbon. Cl oxygen. electron formula, l atom has a are 2 u that f x Step 7 The t a the pairs on r o electron i any N o d Put far... 8 The 12 Step 6 2 Cl Cl Cl in level only holds up electrons. They do not any so one be O centre. N used n two bec ause the U to the have noble gas congurations rst pairs only to o peripheral atoms atoms hydrogen Cl C nitrogen, so it C pairs on peripheral v bonding The i n Put non- in is likely y used 3 Step 5 be is p single bonds pairs one O There Cl H 2 pairs Cl r e is Draw the N oxygen, so it the 12 16 O on Cl There t i s Hydrogens = = n O 2 2 pairs H × 12 l 2 Arrange + 2 y 26 pairs 2 the atoms Step 4 oxygen: 6 26 y Step 3 = 8 = 4 electron 2 P C alculate the number of × oxygen: 6 Total: 2 Step 2 3 r valence electrons c. c arbon dioxide, CO e a. water, H pairs in The remaining assigned to Nitrogen has electron pair is the 4 nitrogen. electron ! pairs in O therefore a the full Lewis octet. formula, therefore a The 2 c arbon atom has only electron pairs in this Lewis formula. We need to reassign two of the non-bonding pairs on the oxygen atoms to form double bonds: O Now the electron full c arbon atom has 4 pairs and therefore a octet. 121 Structure 2 Models of bonding and structure Polyatomic magnitude gained. charge the valence charged charge formulas indic ated subtract an of you a you add electron. of how poly atomic with electrons, charge, groups indic ates ions superscript also an need to electron, This is covalently many are factor illustrated in for in the in been square brackets. the charge: every the atoms. have enclosed outside and bonded electrons for set of every 2– b. c arbonate ion, + CO c. hydronium ion, H O 3 Solution steps above for e ach ion: b c arbonate ion, CO Count c arbon: 4 hydrogen: 1 oxygen: 6 electrons This This polyatomic ion has a has one C alculate the 2 1 + 1 = 8 Total: 4 4 pairs = O n i this Total: 3 8 2 4 meaning that it fewer + 6 electron. 1 = 8 pairs H O H do not between atoms H O O H C H O O 3 3 pairs used so pairs used so far... far... The peripheral hydrogen atoms O need have central and noble gas therefore congurations do not need any C here. more O electrons. O step. The full 12 oxygen atoms now have octets. pairs We used have so used far... all 12 available O H electron The electron you a remaining Skip E any Put so peripheral v O Step 6 it, distinguish and far... l atoms to in 24 3 C hydroxide ion has only two atoms = O O t a peripheral so u pairs used r o on The f x Put non- bonding pair 2 one = o single bonds + = H d Step 4 Draw the 18 3 polyatomic ion has a charge, has pairs U the atoms Step 5 + 1+ electrons. C i n pairs Arrange 1 additional 12 2 number of electron two 24 = Step 3 has This meaning that it o 8 + electron. polyatomic ion has a charge, × oxygen: 6 v Step 2 additional 2– 18 p Total: 6 meaning that it = y charge, 3 r e 1 hydrogen: 1 × O 3 O oxygen: 6 valence t i s Step 1 + c hydronium ion, H 3 n 2– a hydroxide ion, OH l the y Follow remaining 3 electron pairs. None H are pairs pairs available for the central c arbon. H are on the assigned to the oxygen central atom. The hydrogen already The fourth electron pair is atom has 122 a noble gas conguration. assigned to the oxygen atom. y P 3 charge, e r – hydroxide ion, OH additional examples. Worked example 2 a. with counting positive worked Draw the Lewis formulas of the following ions: or brackets, When additional next The lost s s negative are the Lewis the you ions of Structure 2.2 The covalent model + 2– a hydroxide ion, OH b c arbonate ion, CO c hydronium ion, H 3 Step 7 O 3 + Check O H O that the H central ! full All atoms C have noble-gas H octet O congurations. O All Note the square atoms have noble gas brackets and The c arbon atom has only which are 3pairs in this Lewis formula. charge Lewis formulas of Brackets and used in the are added to complete polyatomic We need to reassign one of the Lewis r the e congurations. charge, formula. ions. non-bonding pairs on the oxygen 2 l O triuoride, BF in Lewis Consider the Lewis atoms that formulas the do that arrangement formula shown three pairs of electrons around it. The follow in of the gure the atoms 12 6. octet with Here, boron atom is rule. At SL, F fewer than eight electron U 3 only not contain p electrons. contain of formula. o aware charge C valence molecules be therefore a and v should i n you and added to complete the Lewis Sometimes pairs Brackets y octet. are O c arbon atom has r e full t i s the 4electron n y C y P atoms to form a double bond: Now s s atom has a the pairs of boron boron atom has B electron decient. n F F Similar to boron, many other elements of groups 2 and 13 form stable electron- o d decient molecules. These include beryllium, magnesium and aluminium. In Lewis formulas of such molecules, group 2 elements (Be and Mg) have only two bonding Figure 6 Lewis formula of boron triuoride t a r o i electron pairs while group 13 elements (B and Al) have three bonding electron pairs. Practice questions Draw the Lewis formula of each BH 3 c. BeCl 2 l + 3 Draw the a. HBr c. O Lewis 2 a 3. CH formula E v O e. of of the u a. f x 2. each of the following: b. BCl d. AlCl 3 3 following: b. OF d. O 2 3 e. HCN f. CH g. CH h. N 2 4. Draw the Lewis Cl 3 O H 2 formula of each of the 4 following: + a. NH b. NO d. NO 4 3 + c. NO 2 e. 2 OCl 123 Structure 2 Models of bonding and structure Linking questions What do noble some gases of the form limitations covalent of bonds the octet less rule? readily than other s s Why are elements? (Structure 1.3) do bonds ionic c an bonds form only form between between atoms of the dierent same elements element? while (Structure e Why covalent 2.1) r hydrogen molecule, H , shown in gure 7, contains a molecule, O , contains a double bond. Diatomic 2 triple bond closer this order, bonded bonds order than double shows p 8 o C n than o u a E 124 (bond stronger atoms of Single strength dier in than length and pairs order bonds, together bond electron (bond 3) single triple increasing length between bond number bonds l are The order. v enthalpies Reactivity 1.2 increasing t a f x O Bond The relationship order double i d r o Figure 8 nitrogen molecules with diering bond v i n U single O oxygen and , contains 2 y Hydrogen, r e H Figure 7 nitrogen, N t i s a single bond. The n y 2 oxygen l The 1), their and Double pattern for bonds between double strength triple double bonds. bond are, in two bonds and bonds bonds, strength length. are and turn, atoms (bond still. triple shorter referred to as Double stronger hence is bond order 2) and triple than bonds are Triple bonds hold bonds single are shorter bonds. Figure c arbon–c arbon bonds. discussed in Bond strength bond enthalpy. and length c an be quantied. The stronger a bond is, the larger its y P Bond order (Structure 2.2.2) Structure 2.2 The covalent model Bond length is dened as the average distance between two bonded nuclei. Table 1 shows bond enthalpy and bond length data for carbon–carbon single, double and triple bonds. The carbon–carbon triple bond is the shortest but also the strongest. –1 t Table 1 –12 Bond enthalpy (at 298 K) / kJ mol Bond length / 10 Bond enthalpies and bond s s Bond m lengths of c arbon–c arbon bonds 346 154 614 134 C≡C 839 120 r e C–C C=C and its bond m. Use these data and the information in table 1 to predict the bond enthalpies and bond lengths of single and double carbon–nitrogen why two hydrogen between atoms. sketching a between potential Read graph of it and try potential The two paragraph hydrogen by energy and the distance to represent what it is energy (y-axis) hydrogen nuclei (x-axis). below atoms vs distance C between by relationship explanations. forms i n describing written bond Two hydrogen atoms have no eect on each other if they are o between the covalent v describing complement a O oen explains p Graphs y Communic ation skills r e ATL t i s bonds. Then compare your predictions with the values in the data booklet. n y 12 length is 116 × 10 l 1 The bond enthalpy of a carbon–nitrogen triple bond is 890 kJ mol y P Data-based question U separated by a suciently large distance. As the two atoms n approach each other, the electrostatic attraction between the hydrogen nuclei and each other ’s electrons increases. This process o d leads to a decrease in potential energy. As the atoms get closer, they reach a point where the two nuclei attract the two electrons i r o in the pair, eectively sharing them. This arrangement occurs at a t a potential energy minimum, meaning that the molecule is energetically stable. If the atoms were to get any closer together, the resulting repulsion between the two positively charged nuclei would outweigh f x the attraction for the shared pair of electrons, leading to a rise in What c an into c an search graphic al To one what typing your sketch “hydrogen graph to molecule a potential potential energy is a graphic al Linking question energy engine. extent description? compare by representations v above? a nished, nd a curve” have l You O you curve. u potential energy. When How add to explanations such as the one representation better than a textual and does triple inuence the presence of double bonds their in molecules reactivity? (Reactivity 2.2) E 125 Structure The 2 Models formation bonds is a of of bonding and structure Coordination bonds (Structure 2.2.3) coordination feature of electron pair We sharing reactions, as have seen that a covalent bond is formed when each of the two atoms in discussed in the bond contributes an electron to the bond. Sometimes both the electrons Reactivity 3.4. in the covalent bond come water molecule, hydronium the electron Dierent is direction bond, of and representations the which hydrogen formed, bonds arrow atom resulting c ation leading are shows oen the the atom + O with H O H H H an pair. + help to coordination explain bonds how are bonding Draw the Lewis i n the following + c arbon 4 3 monoxide, ammonia boron CO C ozone, O c. triuoride, NH molecules. any other molecules, indic ating the BF 3 n U LHA of clearly. ammonium ion, NH b. d. formulas bonds certain o coordination a. v 5. in from y Practice questions p r e covalentbond. occurs indistinguishable n bonds formed, O Once t i s Coordination H y coordination bond l H H arrow donated the electron coordination bond c alled formation of a + showing the H are encounters a indic ated which accepted to bonds 3 Coordination bonds in transition metal complexes their of complex ions reactions are discussed in cation and the surrounding atoms or groups of atoms, called A common form the feature of all ligands is a lone pair of electrons, ligands. which c an be used to coordination bond to the metal ion. l u Why do Lewis acid–base complex ions contain coordination bonds, which hold together the central metal t a r o f x Linking question Transition metals can form complex ions such as the one shown in Figure 10. These i Reactivity 3.4. o properties and d The reactions lead to the formation of O H E v a coordination bonds? (Reactivity 3.4) H + 2 O 2 OH O 2 2 Cu H O OH 2 H 2 O 2 + 2 [Cu(H Figure 10 O) 2 ] 6 Water molecules form coordination bonds with copper(II) c ations, the formation of a blue hexaaquacopper(II) complex ion 126 leading to y ion, the bond Coordination a The P of the hydronium to The 9). when atom. r u Figure 9 coordination (gure bond. pair a example, same e along ion For the s s coordination bonds from Structure The valence shell electron pair 2.2 The covalent model repulsion model (VSEPR) (Structure 2.2.4) You might have hydrogen come atoms This an c an across a angle. Such also be Lewis shown formula formulas for water better that shows the two illustrate the geometry of the s s molecule. at by ball-and-stick and space-lling models (gure 11). reect the feature antibiotic of a molecules geometry properties. be shown 1. Electron pairs is based repel on each the their properties. For example, Figure 11 gure 12. Electron following other three- and Space-lling model of a water molecule VSEPR model: Pair side chain Repulsion R premises: therefore arrange β-lactam ring H O model Shell The geometry, and it is an t i s VSEPR in to formulas do not molecule. known as a beta-lactam ring is key to its explored using the Valence The molecular contributes structure Lewis a N C n c an a as in l geometry is termed which within This is such atoms y Molecular molecule of y molecular of molecules P the shape of arrangement r dimensional essential representations three-dimensional e Two-dimensional S CH 3 themselves as far apart O Non-bonding triple region of domain space than a bonding • a double pair or pair of in of useful in density electrons electrons bond gure these due discussions. to 13 illustrates which one is a CH electron of electron 3 O C O HO Figure 12 The beta-lactam ring (in red pairs. above) is a common feature of antibiotic molecules such as penicillins. The geometry lone pair) multiple this N of this structure is vital to its reactivity. The (known as a involve C amide bond An presence point: pairs the amide bond of hydrolyses readily due to the strain c aused by the 90° bond angles. This means that it c an attach to enzymes electrons). responsible for building bacterial cell walls. central sulfur atom has i of the (single bond) (which r o domains, non-bonding t a domains. be pairs more space than single bonds. c an be: triple formula electron will electron bonding n • Lewis more o non-bonding high occupy U a bonding bonds d • three occupy C a electron The pairs o and electron domain is electron i n domain lone) bonds). Double The term (or possible. v 3. as p (single An other y 2. each r e from domain, and two are One of the bonding domains is a double bond, the other is a single bond. u l f x one non-bonding domain composed of a lone pair of electrons S one bonding domain number domain Determine how one bonding domain t Figure 13 composed of a The sulfur atom in sulfur single bond dioxide has three electron domains geometries Predicting the electron 2. explore VSEPR. Count a 1. now E will using double bond v O We composed of a the of involving VSEPR electron shape two, of domains a three and molecule around a four electron domains involves two steps: central atom to deduce the geometry. many of these are bonding domains and how many are non- bonding domains. 127 Structure 2 Models of bonding and structure Two domains: Linear If there other. are two They bonding a is straight Molecules in adopt the positions molecule is linear to electron at 180° known as the illustrate that pairs in (gure those 14). The domains angle bond angle. This the central atom this electron both domains domain include geometry beryllium also have chloride, BeCl , a linear c arbon dioxide, H 2 Figure 14 . These are shown in gure 15. r ethyne, C 2 Linear geometry has two Cl 180° C H in which the central atom has two electron domains. Note that double and triple bonds count as one domain there each Trigonal, A bec ause trigonal planar all domain on the domains only triangle, geometry are bonding of the three the molecule has type i o each are pairs adopt positions at 120° is and c alled c an trigonal planar. planar, have two domains shown in domains, the bec ause the atoms lie possible molecular molecule has trigonal are bonding domains, and one is a non- bent gure (or V-shaped) geometry. 17. (a) (b) Cl 120° N B Cl O Cl <120° O u A trigonal planar electron domain geometry c an lead to (a) trigonal planar molecular geometry if all three domains are bonding domains, l a E v O f x Figure 17 geometry domain, t a r o d of two a geometry electron presence of non-bonding domains: geometry. bonding Examples form n U When electron three domain domains depending the C When planar 128 the i n each other • electron plane. geometries, • domains, o Trigonal planar geometry has three electron domains at 120° angles to a This v Figure 16 on other. bonding p at three y from are r e If O t i s Three domains: Trigonal planar n Examples of linear molecules, O l O y Figure 15 Be , 2 or (b) bent (V-shaped) molecular geometry if one of the domains is a lone pair of electrons The bond angle in NO is smaller than the predicted 120°. This is bec ause the 2 non-bonding domains, and pair (or lone therefore pair) takes exerts up a more stronger space. repulsion than the bonding y 180° other Cl CO P 180° electron domains at a 180° angle to each are molecular e Examples each electron domain and 2 and repel between line. with geometry. a c alled geometry domains, s s on therefore pairs geometry electron Structure Four If the are each ends four bonding other. of might the This domains domains, electron domains expect conguration. The four form arrange the electron However, this the domain electron corners domains would themselves pairs geometry of to result a is positions at tetrahedral 109.5° bec ause tetrahedron. arrange in adopt c alled angles themselves of 90° in a square between the domains. tetrahedrally to maximize the bond angles and four electron Tetrahedral geometry depending all four on domains geometry the are gives rise to three possible other molecular presence of non-bonding domains: bonding domains, the molecule has 109.5° tetrahedral l geometry. three When only bonding of two of four are bonding domains and one is a non- molecule has trigonal pyramidal the four are domains, each domains the type the are domains molecule has shown in bonding bent gure (or geometry. domains V-shaped) and two are non- geometry. 20. A tetrahedral electron P trigonal pairs bond occupy angles. more space Methane, CH than , has 4 the bond angle corresponds to the pairs and therefore or bent (V-shaped) molecular geometries domains, non-bonding which pairs, 109.5°. Ammonia, NH an stacked and revolutionized packages bec ause they c an be easily packed. the food Their invention packaging industry in the mid-20th century therefore , has one 3 smaller bond angle (107°). Water, H has lled, Tetrahedral food leads to and o d non-bonding trigonal pyramidal V-shaped bonding no predicted non-bonding pair, and therefore has a two or are convenient n decreased bent U Non-bonding pyramidal Figure 19 leads to tetrahedral, C tetrahedral domain geometry o F i n F S v Si y t Figure 20 F p r e Examples the O • of domain, t i s bonding to n When y • has angles y When domain at P geometries, electron domains r e ach tetrahedral • 18 between them. e Figure distances A covalent model s s You The domains: Tetrahedral geometry there from 2.2 O, has 2 even smaller bond angle of 104.5°. i r o The angles and geometries are shown in gure 21. Note that some bonds are wedges ( ) and t a represented using dashes ( ) to better convey their 3D shape. Wedges represent bonds that are coming out of the plane of thepage at u f x an angle and dashes represent bonds that are going into the plane ofthe page. H l a H v O C H N O H H H H 109.5° H 107° H 104.5° E t Figure 21 In methane, CH , ammonia, 4 NH , and water, H 3 all slightly dierent repulsion exerted O, the bond angles are 2 due to the stronger by the lone pairs 129 Structure 2 Models of bonding and structure Multiple bonds Multiple bonds (double and triple bonds) count as one domain. A double bond represents one domain, but it contains two electron pairs. Similarly, a triple bond is one domain composed of three electron pairs. Since multiple bonds contain s s more than one pair of electrons, they exert a greater repulsion than single bonds. The increased repulsion causes the bond angles in the molecule to deviate from predicted values. 121.5° H H For example, hence each its c arbon molecular is ethene is has trigonal three bonding However, the H–C–H bond angle is 117° and the H–C =C bond Multiple bonds generally 121.5°, as the greater repulsion exerted by the double C=C bond a greater repulsion than single bonds pushes shows a C–H bonding summary of the domains. geometry This of is molecules geometry domains of non- Molecular Bond geometry angle 0 U 4 1 u domains using balloons. If l you hold several balloons a at a central point by their knots, they will arrange E v O f x You can model electron themselves as far away from each other as possible. The central point where the balloons meet represents the central atom in the molecule. How else might you model electron 130 o i Thinking skills domains? planar V-shaped) tetrahedral 2 four electron domains t a r o ATL three and (or and four BeCl 2 180° CO 2 HCN 120° BF , NO 3 <120° O , SO 3 , SO 3 3 , NO 2 2 + 109.5° CH , NH 4 2 Summary of the geometries for two, bent 0 3 d Table 2 4 1 n tetrahedral trigonal C 2 0 o 3 v planar i n 3 trigonal linear p 2 three Examples domains 2 two, y r e domains linear with bonding bonding gure 22. O t i s Number Number of Number of in n electron domains. Electron domain shown l 2 both y Table away electron domains planar. This would suggest 120° , ClO 4 4 + trigonal pyramidal <109.5° NH , H 3 bent (or V-shaped) <109.5° H O PBr 3 O, NH 2 , 3 , SF 2 2 y angles. angle in P bond atom geometry r and H Figure 22 exert 117° C e C H Structure 2.2 The covalent model VSEPR Unlike 3D a formulas, VSEPR molecular of molecular molecules. database to You develop models will represent the 3. your Relevant skills • Tool 2: Digital molecular models Identify and extract data • molecular geometry • molecular formula • bond • 3D domain geometry from a database angle sketch data of the (if available) molecule, including bonding and Write list of the molecular geometries that you need 4. know. Draw the Lewis of whether in the Tools for chemistry chapter) and extract two 5. Organize 6. If geometry. If possible, rotate the images to get a sense bond of their three-dimensional shape. that your angle data data the into is a and molecular numbers of bonding and non-bonding domains. the geometry a. NF c. BF multiple bond bonds (e.g. double or triple angle. electron of each of domain the geometry and molecular following: b. CH d. BeH 3 Cl 2 3 2 2 i e. HCN f. g. SF h. SO 3 NO t a 2 9. Predict 2 the bond angles in each of the species in question 8. u f x 10. Methanal is an organic compound with the molecular three bonding domains and one non-bonding domain. l O 7 . Deduce of on two bonding domains and two non-bonding domains e. r o d. two bonding domains only 8. o two bonding domains and one non-bonding domain c. d b. four bonding domains effect p geometry examples o domain following the for effect of non-bonding domains on the bond bonds) n the table. search y the C electron given U the geometry v i n Practice questions suitable available, angle • molecule is polar or non- illustrate: r e • a. deduce the O t i s examples of species belonging to each molecular Identify Then n y and Access a VSEPR database (such as those suggested polar. 6. each. number of bonding domains and non-bonding domains, 2. formula l to a y P non-bonding domains. Instructions 1. e 1: electron r Tool • understanding of geometry. • For each, write down the following: explore the models s s in Lewis geometry formula CH Identify the number of bonding and non-bonding around atoms v geometries. a domains linear b. trigonal c. tetrahedral d. trigonal planar e. bent E a. with the following O. 2 a. Draw b. Deduce the c. Suggest Lewis the formula molecular for methanal. geometry of methanal. molecular Explain values your for the bond angles in methanal. reasoning. pyramidal (there are two possible answers here). 131 Structure 2 Models of bonding and structure Linking question How useful is the VSEPR model at predicting molecular geometry? s s Bond polarity (Structure 2.2.5) is related molecules. the dierence c ase in of pulls the the atom the we electrons has shared a in much partial negative positive separation charge, of This polar charge, vector the less is δ+. charge The shown between dipole as an atom δ–, two and the strongly the other non-identic al be the bond. bond, v and shared two Figure 23 in HF . Note that the bond dipole in HF c an be represented as a vector the the shied the involved bond, in the towards dierence formation of the pure covalent bond. The however, hydrogen hydrogen does. This atom (uorine) (hydrogen) adopting by atoms a is vector more towards the c alled a (gure23). covalent is more electronegative atom in electronegativity charge the arrow starts as a plus sign points the in c ase, than bonded The there is an 2 the one atom from bond between the are identic al, the distributed evenly about meaning that it is non-polar and has no dipole moment. R ank the Explain following bonds in order C–H, why Br is non-polar, of C–C, increasing polarity: C–F, C–O whereas HBr is polar. 2 u a v E 132 l O shared. is greater than with , results o f x 12. bonding pair of electrons is shied electrons are transferred, not The greater this shi. atoms t a large (greater than 1.8), then the are considered ionic because electrons said to be a covalent 11. between the two atoms is very electronegative atom. Such bonds is necessarily Practice questions dierence of electronegativity almost entirely towards the more two bond the i r o As seen in Structure 2.1, if the d (arrow) If of bond. n bond U covalent atoms, Representation of the molecular electrostatic potential of the polar pair covalent C the o The δ i n F + δ not represented electronegative atom in the bond. H is electronegativity more c an along in This covalent bond, moment arrow electronegative pair pair. p at not bonded atoms. y dipole moment. greater r e The the bonding electron describe as a partial the H a of is O a electronegativities bond Bond polarity t i s what adopting covalent between the two atoms in the bond. uorine it a n so distributed within bonds and in l The to are electrons y in leads electrons of identic al atoms, such as the two uorine atoms in F sharing atom, F of equal HF. way pair Linking question What properties of ionic compounds might be expected in compounds with polar covalent bonding? (Structure 2.1) y the the P In equally to shared r shared The e Polarity Structure 2.2 The covalent model Molecular polarity (Structure 2.2.6) Molecular polarity distribution electron and a μ. The D. value is an to of the partial dipole such as on but A negative charge important depends are a polarity, molecule. on the moment trait of it describes the electron molecule is polar when the charge on one end of the other. is oen molecules This results in a dipole reported in the unit bec ause it gives rise to many volatility, solubility and boiling point. both bond polarity and molecular geometry. given in table 3. e polarity bond whole positive properties examples to r Molecular leads partial Polarity characteristic the s s moment, Some similar distribution molecule, debye, is throughout moment / D bond dipole O bond dipole polar 1.85 polar 1.01 O 2 net dipole trichloromethane, H bond dipole C Cl Cl triuoride, i n boron non- F polar bond dipole 3 B dioxide, non- bond dipole CO polar H H C H H 0 polar H u C non- non-polar molecules l Examples of polar and a E v O f x Table 3 H 0 o 6 i H 2 t a r o ethane, d 2 n c arbon F U F C 0 C BF o net dipole v Cl y 3 p r e CHCl O t i s water, H n l Polarity y Lewis formula y P Dipole Molecule 133 Structure 2 Models of bonding and structure Molecules are polar This may dierent they the occur due polarities. contain polar when their to the Both water bonds that do of the molecule, happens not c ancel out. net This the Examples contain polar other include bonds These out. bonds means the that of the geometry bonds dierence around each and triuoride are atoms virtually between c arbon atom hexane, C hydroc arbons 3 cyanide, HCN So State why we have CO is non-polar. and a only emulsiers, soaps and 2 and explain whether PH membranes. is or polar non-polar. and shows regions H H C C H C S H hydrophilic head A diagram hydrophilic regions l a the way that they a net to well bonds. The In addition, dipole of the zero, Long molecules interesting features of a tetrahedral or very close the phospholipid detergent c an have a polar applic ations including molecule. bilayer in cell C an you identify the H H H H H C C C C H H H H H H H showing the hydrophobic and molecule make it “infrared (IR) H H C hydrophobic tail (Structure 3.2) c arbon–c arbon electronegativity. non-polar. many as are Hydroc arbons Linking question What dioxide molecule? H of a synthetic detergent molecule, LHA E v O 134 H u f x Figure 24 i H H t a r o d O C o O in as typic al H + – Na a in molecules. leading detergents, 24 a c arbon bec ause of the small atoms. generally C C non-polar small region, n U H O Figure i n 3 polar discussed non-polar and same non-polar two results are the o Explain 15. region 14 v 14. far, H 6 electronegativity of y d Therefore, the have p hydrogen zero. r e c to in such bonds are non-polar. c arbon–hydrogen both 4 ammonia, NH in results dioxide. Both c arbon) such as ethane contain two types of bonds bec ause c arbon–hydrogen arrangement n non-polar electronegativity b across l The methane, CH dipole O a net c arbon dierence y molecules is polar or c arbon–c arbon and presence of polar bonds. when all their are their positioned boron and hydrogen and triuoride are (compounds of equal boron Molecules are non-polar bonds non-polar: a have bec ause bond dipoles c ancel each other out. are bec ause polar This despite dipoles t i s following in bonds polar active”? H H y non-polar, bonds: each of the results the are r involved. each when their bond Practice questions whether 3) P both when dipole. atoms c ancel Determine bec ause (table e no molecules 13. or trichloromethane s s This the and molecule. Molecules are non-polar in bond dipoles do not c ancel each other out. geometry Structure Covalent network structures 2.2 The covalent model Chemists talk about two types of stability: kinetic (Reactivity 2.2) (Structure 2.2.7) and thermodynamic (Reactivity Covalent bonds give rise to two dierent types of structure: molecular and 1), and diamond illustrates the covalent network (also known as giant covalent). The atoms in both cases are s s dierence well. At high pressures, joined by covalent bonds, but the two structures are very dierent in terms of their such as those occurring deep properties. Molecular substances consist of discrete groups of covalently bonded inside the Earth’ s crust, diamond atoms called molecules. By contrast, covalent network structures contain atoms e is the most stable allotrope. that are held together by covalent bonds in a continuous three-dimensional At standard temperature and lattice. Examples of covalent network structures include silicon, silicon dioxide, r pressure, graphite is the most and most of the allotropes of carbon, such as diamond and graphite. thermodynamically stable allotrope y P of carbon. Under these conditions, Allotropes of c arbon the transformation of diamond group of but substances have its allotropes known dierent as forms include fullerenes. chemic al and known as diamond, They physic al are all allotropes. graphite, due into graphite is thermodynamically graphene, and composed properties C arbon is to of spontaneous, but is so slow that c arbon their diamonds eectively last forever. dierent arrangements. Diamond is kinetically stable but O structural and structural t i s atoms element, dierent n a such have y one elements l Some thermodynamically unstable (we describe it as being metastable). (a) (b) You can read more about p o C n i d t a r o u l f x a v O Figure 25 (e) o (d) y r e v i n U (c) spontaneity in Reactivity 1.4 The allotropes of c arbon: (a) diamond, (b) graphite, (c) graphene, (d) c arbon nanotubes and (e) buckminsterfullerene, C 60 E 135 Structure 2 Models of bonding and structure In diamond, each arrangement. oen used in It is Diamond’ s make jewelry. high Diamond through the is cutting is a tools such and are all lattice as atoms For in this a tetrahedral reason, it is saws, polishing tools and dental mean conductor energy and other known. loc alized vibrational regular four durability electric al electrons to substances index poor conductor: highly bonded hardest in the c arriers strong that it is also used to bec ause it has no mobile bonds. c alled It is s s thermal atom the refractive particles — the excellent well of heavy-duty drills. charged c arbon one however an phonons travel covalent bonds. e r l n y South Afric a p r e Kimberley, O The Big Hole: a former diamond mine in Graphite is composed of layers of sheets made of carbon atoms. Each carbon v atom is bonded to three other carbon atoms in a hexagonal arrangement where o i n the geometry around each carbon atom is trigonal planar. The carbon atoms in graphite are bonded such that one electron per carbon atom is delocalized. These C delocalized electrons are free to move in the planes above and below each sheet and therefore graphite is a good electrical conductor. While the covalent bonds U between the carbon atoms within the sheets are strong, the forces of attraction n (called London (dispersion) forces) between the sheets are weak. This means that the sheets can be separated easily, making graphite a good lubricant, as well as an Graphene and time, to conductor. with bendable l a E v O Fullerenes hexagonal these are are vast are and a a single said It is to range group be also strong. electronic utilized in less a sheet exible, isolated Graphene is thus one-atom Like in 2004, graphite, transparent graphene applic ations, from it is and, is a an at excellent the same promising desalination new technology displays. of c arbon rings. allotropes Some with with atoms fullerenes c arbon nanotubes. objects graphite. lightweight, potential nanotechnology. and of two-dimensional. First of pentagonal known as molecules Due to their Nanotechnology dimensions of less form arranged in interlinking long size, hollow cylinders: c arbon nanotubes involves the use of atoms, than 100 nm (about 1000 atoms or across). nanotubes strength C arbon is due to and good are the nanotubes graphite are essentially therefore extraordinarily C arbon 136 is electric al u A microscope image of a is t a thick material, dentist’s drill showing the diamond chips o causes parts of the graphite sheets to come o, leaving a mark on the paper. i d r o f x Figure 27 ideal material for pencil leads. When you use a pencil to write, the force applied used have graphene, electric al to strong reinforce covalent potential the of materials. holding applic ations presence conductors. composite bonds in the electronics deloc alized Their c arbon exceptional atoms together. bec ause, like electrons means that they y P y t i s u Figure 26 Structure Buckminsterfullerenes, or buckyballs, have a covalent molecular formula C , the atoms in buckminsterfullerene are arranged in The covalent model structure. With Silicon the 2.2 and c arbon are found in hexagons and 60 group pentagons to suggest a very familiar shape: a football. Along with other 14 similar fullerenes, buckyballs could have exciting new roles in medicine as drug of its molecular structure, C has a low boiling point: they have some properties. Other patterns in c arriers. the Bec ause hence spheric al periodic table are explored in overcoming the 60 s s Structure 3.1. weak intermolecular forces of attraction does not require much thermal energy. Silicon and silicon dioxide four that the and where each silicon arrangement. This atom bonds in boiling The Si–Si the lattice point. bond is 4 we aker Therefore, in result Table the strength than Si–Si also in high contrasts its bond helps to strength, some C–C is of more explain re active the 114 346 226 3265 between E arth’s O in 1:2 to insoluble in atoms. of silic a of poor a and bond of it is table group, anticipate. Diamond Double unusual found the and is a triple For the properties example, poor electric al bonds are between silicon atoms. and sand (table 4) shows that single bonds a species containing Si–O bonds. forms . a covalent network containing E ach silicon atom is bonded 2 Figure 28 each oxygen atom is bonded Structure (le) and silicon dioxide, SiO form of silicon dioxide is crystals of covalently quartz , known as quartz. The 2 (gure 28). photograph also contains crystals of iron and glass. disulde, FeS , commonly known as 2 “fool’s gold” dioxide are high conductor in that formula SiO crystalline has metalloid. strengths mineral atoms include silicon water a is hence The but might strong. Silicon is one of the most abundant elements much oxygen E is ratio, four properties it C–O a forms quartz, and v The a atoms very a semiconductor. l Other crust silicon O two and are silicon dioxide, covalently to O is periodic one u and Si–O f x Silic a, or Si the and c arbon silicon same as i the Si a the t a between is r o Comparing while silicon to similar where X = Si or C o while as C non-metal, not d common belonging are silicon, n a conductor, i n is silicon 466 U c arbon despite and 787 358 A selection of features of c arbon and Interestingly, in 1086 o –1 p 1414 4827 enthalpy the melting r e 3500 Boiling point / °C c arbon than dierent of silicon v Melting point / °C of as O Si 75 ionization energy / kJ mol as y C (graphite) m –1 Table 4 well fe atures points. enthalpy / kJ mol X–O bond as the counterpart –12 First bonded to n radius. dierence Atomic radius / 10 X–X bond is arrangement is similar l atomic This boiling lattice tetrahedral t i s and cov alent point silicon. larger bond. a y a C–C and in y melting c arbon atoms c arbon atoms in diamond. extensive high has three-dimensional P The of a silicon r to forms other e Silicon of those melting heat and we associate point. In its with solid sand: it crystalline is hard, form, electricity. Linking question Practice questions Why 16. Explain why graphite and graphene conduct electricity, but diamond does not. 17 . Explain why are silicon–silicon bonds generally c arbon diamond and silicon dioxide weaker bonds? than c arbon– (Structure 3.1) have high melting and boiling points. 137 Structure 2 Models of bonding and structure Activity Summarize the properties of some of the covalent using Graphite a table like the Graphene one C below. fullerene Silicon dioxide s s Diamond substances 60 Element(s) Arrangement of atoms e Electric al conductivity r which to their forces name substances. The and of polarity collectively type the c ause water some that of are weak Figure 30 water water water forces such as δ polar l covalent bond a intermolecular of through to ice are and termed than strong stay close to What break answer of in exactly away each from c ase is chemic al bonds, attraction that occur dierence + force forces cubes? in weaker forces the + δ ice The forces. δ London volatility, solubility and boiling point of electrostatic highlights atom inside are intermolecular molecular molecules contain two molecules it? now turn our forces include molecules surrounding Intermolecular properties the to intermolecular oxygen molecules the water the central apply y intramolecular a three will dipole–dipole forces p o i and to rst We known as forces depend on the size intermolecular o forces. forces the the know forces liquid physic al molecules. t a occur within molecules. E v O 138 aect between u bonds, Intermolecular forces occur between molecules they Intermolecular Intramolecular forces, such f x as covalent d Figure 30 but about when molecular substances. r o ▸ intermolecular what the You bonded n the ice cubes join forces intermolecular of C Intermolecular forces hold the molecules together in water and inside But How of Types r e Figure 29 and water. each, overcome cube U the bonds. another? being of i n covalent glass atoms bonding. molecules, intermolecular Collectively v a hydrogen attraction that hold atoms and ions between strength Waals forces Consider is suggests, and of dipole–induced dipole forces, hydrogen bonding. one attraction molecules. (dispersion) forces, van der of forces referred to as O forces. As the are forces (Structure 2.2.8) discussing t i s attention been n together, have attraction δ between intermolecular y we l far, y So P Intermolecular Structure When molecular between oen the used as molecule indic ators do Structure 1.5, that no we high When of break during discussed forces in deviates account the the or sublime, Therefore, phase the are ideal for from gas in ideal volume the which gas. the gas The forces boiling points weak intermolecular interactions are between I 2 molecules break on changing state covalent bonds in c arries This ideal real gas and (gure31). behaviour, of intermolecular melting strength. law, the which forces using a actual force changes present temperatures, gas boil intermolecular intermolecular into negligible a of melt, overcome. is gas we assumption equation must model. particles, the generally true c an correct at low frequently for the Real gas models which is r take and are covalent model e used. presence also not intermolecular pressures be substances molecules The s s In the 2.2 considered ideal gas model. strong covalent bonds 500 K 2 molecule between do not break on changing state Intermolecular forces of O t i s Figure 31 2 in I n y atoms l 200 K y P 3 attraction are overcome when a molecular substance changes state, but the covalent bonds remain intact pV 1000 K ideal gas ideal gas behaviour at lower temperatures. The o dierent d behaviour at n Gases deviate more from horizontal black line represents an ideal gas. o 900 p (atm) Figure 32 C 600 U 300 v i n 0 p 1 y r e RT The curves show the deviations from ideal gas temperatures for a real gas i t a r o London (dispersion) forces (LDFs) All molecules experience other a which another dipole means molecule. electrons when negative around that one one These the (dispersion) from charge. a of a forces temporary side c arries London induced a partial c auses dipoles (or LDFs), which are instantaneous dipoles. A (dispersion) molecule molecule. E v O dipoles, in has partial London resulting u molecule forces l f x intermolecular (or occur positive forces induces) due to charge, and the involve a the induced temporary dipole random movement Figure 33 Intermolecular forces of attraction are overcome when iodine sublimes, and are formed when iodine is deposited on a surface 139 Structure 2 Models of bonding and structure In a simple non-polar molecule such as hydrogen, H , the electron distribution 2 is on average around of within electrons perfectly the in a molecule. molecule, symmetric al. with one in the If the of the There molecule 34). we However, could electron Instead, region negative (δ –). elsewhere (gure we somehow would a also see a be a would somewhat having positive are more region partial constantly very unlikely be unequal electrons, of lower electron rendering it electron density charge (δ+). e r H H is symmetric al 2 t i s n l y the electron distribution in H This temporary dipole then induces further temporary dipoles in the molecules O around it. Due to electrostatic repulsion, a region of partial negative charge (δ–) on one molecule will repel the electrons in neighbouring molecules. This temporarily creates regions of partial positive charge (δ+). The resulting y r e electrostatic attraction between the δ– region on one molecule and the δ+ region on the next is termed a London dispersion force. However, this arrangement p is temporary. In the next moment of time a dierent pattern of induced dipoles + δ H H n a – u l + H force a temporary dipole is present in a molecule, A weak London dispersion force forms between the opposite + + δ δ δ δ + δ δ δ + δ + + δ – δ δ δ δ + δ + E v O f x δ δ o – δ + δ – δ dispersion any given moment in time, t a – – δ partial charges δ δ At inducing a dipole in another. i d r o + δ + δ H London Figure 35 o U C i n δ v willemerge. δ + δ δ – δ – + δ δ δ + + δ δ + δ – δ δ δ + δ + δ + δ + δ + δ Figure 36 Dierent arrangements of LDFs between molecules, instantaneous dipole on one molecule and 140 an induced which result dipole on an adjacent from interactions between an molecule y H On average, P H Figure 34 moving freeze time and take a photo distribution would molecule with electrons s s density, slightly symmetric al Structure There are two main electrons and the how the easily electron polarizability, The the number electron elements, higher larger more boiling is all distorted of less you an form down LDFs: the number of electric of eld. non-polar the attracted polarized as by of polarizability the The molecule: greater forces. which going and easily points strength aect the dispersion increases become become the the They and descend group. to the the the diatomic As a nuclei. strength molecules. result, the This of means that LDFs increases, group (table 5). r Boiling point / °C –188.1 uorine, F 2 Br 58.78 Increasing iodine, I 184.4 2 LDF Table 5 Boiling point increases down group 17 due to an increasing number of resulting in stronger LDFs greater electrons polarizability Figure 37 boiling organic point also increases compounds: each for successive molecule has members of a one more CH n of be molecular quantied in terms in of and mass, terms of a larger so we their oen contradictory not c ausal. each is the responsible requires Therefore, proportional for the in the number of The have Homologous one series and isomers of before organic compounds are discussed Molecular size in molecular to larger in in the of electrons. the Structure 3.2. mass when molecules is number the number of of This of increasing may But here electrons is accompanied protons and mass. overall between with mass. number molecular maintain number of increases negligible and greater increase increase the to electrons. relationships strength mass mass electron any increase understand LDF electrons molecular electrons. an to say increasing greater of try oen bec ause number proton molecule. E neutrons an between R ather, greater to v relationship a due and we u mass, the by patterns a seem for example, l O molecular For of t a f x look variables. refer LDFs. Patterns and trends Scientists number i molecules r o comparing size o c an greater d hence point IMFs homologous series than 2 it, higher boiling overcome LDFs of molecular size on LDF strength and boiling point U The The eect energy needed to C i n size more stronger o more y v larger molecular p r e electrons, strength O t i s 2 n 2 bromine, l –34.04 y chlorine, Cl y P Substance covalent model e to stronger 17 increase shape. distribution electrons clouds electrons leading the group of that The s s Consider the the factors molecular 2.2 However, neutrality of the protons always leads to a electrons. 141 Structure 2 Models of bonding and structure Compare the boiling points of the two isomers of C H 5 CH CH 3 CH 2 and 2,2-dimethylpropane : pentane, 12 (CH 3 ) 3 C (table 6 and Space-lling models show pentane (top) has a greater area of contact gure 38). 4 e that CH 2 s s u Figure 38 CH 2 for LDFs than 2,2-dimethylpropane r (bottom) contain dierence is in their the same to with shapes. number Pentane of electrons molecules are and are long non-polar. The and are therefore able other across the length the better of 9.5 area pentane. As a interaction result, temperature. bec ause pentane has Molecules of of a so they pack together room is limited, the LDFs less in eciently. As the rst point H 6 H propane, C 3 will weaker, so it is a gas at shown below. Describe observe. 161 o forces. i dipole LDFs more 42.1 of 2,2-dimethylpropane respectively. between we are and between the 88 CH 3 If rounder 0.5 points 36.1 °C, occur electrons. are are interaction 10 t a contrast London forces and dipole- 8 2-methylbutane, CH u a induced l and (dispersion) boiling and LDFs v E Compare d r o f x O Activity 4 The you H butane, C 2. alkanes n U 4 2 there is Boiling point / °C methane, CH ethane, C four trend C Alkane is, molecules of y for boiling o data the p r e point explain i n and v Boiling the 2,2-dimethylpropane temperature. Data-based questions 1. that between relatively high boiling point and is a liquid at isomers of pentane molecules contact 2,2-dimethylpropane Boiling points of the two compact, molecule; O room of t i s 2,2-dimethylpropane large the n full 36.1 a Table 6 each y pentane interact l point / °C have all a Predict CH(CH 2 types a of value ) 3 . made pentane are 9.5 °C Explain your answer. 2 molecules substance and for the boiling point of up bec ause of only all molecules contain non-polar molecules, only occur. Dipole–induced dipole forces LDFs are forces of attraction between Dipole–induced dipole forces occurring presence between of temporary of a a polar permanent dipole intermolecular in the force are a molecule dipole in temporary, type and the neighbouring attracts of a nearby polar instantaneous, dipoles. intermolecular non-polar molecule non-polar non-polar or related oxygen induces molecule. force molecule. The For the formation of a example, this type molecules, O , to polar water 2 molecules. aqueous 142 Dipole–induced solubility of dipole oxygen is forces relatively are low. weak, which explains why the y isomers P Both Boiling Isomer Structure 2.2 The covalent model Dipole–dipole forces Dipole–dipole temporary therefore forces dipoles. involve When experiences a permanent molecule dipole–dipole is dipoles, whereas LDFs result + δ from forces of δ Cl H attraction with neighbouring polar Figure 39 Hydrogen chloride, HCl, (gure 39). molecules are polar and attract each other due to dipole–dipole forces Hydrogen chloride, HCl, and diatomic uorine, F , have similar sizes and 2 comparable molecular masses (table 7). They both therefore experience LDFs of a e similar strength. However , HCl molecules are polar and therefore experience dipole– are stronger than those between F r dipole forces in addition to LDFs. The intermolecular forces between HCl molecules molecules, so HCl has a higher boiling point. 2 38.00 Intermolecular forces –188.1 –85.05 (dispersion) (dispersion) Table 7 Diatomic uorine and hydrogen chloride have similar molecular masses but dierent boiling points of the following species B. c arbon dioxide, C. c arbon monoxide, experience dipole–dipole 2 CO 2 c arbonate ion, CO 3 The molecular masses of ICl and Br are very 2 Draw a diagram arrange and to 58.8 °C, show themselves how when propanone near one another. a are strong dipole bonded to a intermolecular involving highly forces hydrogen. that When electronegative u f x covalently bonds strong Label t a contains this dierence. COCH 3 i r o Hydrogen bonding Hydrogen Explain molecules, CH a atom any , might 3 dipole–dipole forces o occur. similar, while their boiling respectively. d that 97.4 °C n 20. are U points C 2– D. 19. i n CO forces? o oxygen, O v A. y Which p r e Practice questions 18. O t i s and dipole–dipole y 36.46 n London London 2 HCl l F Boiling point / °C y Molecular mass P Molecule s s molecules + δ δ Cl H polar, it has a permanent dipole and form when a molecule hydrogen atom is such as oxygen, nitrogen or uorine, the bond between them is very polar. The electrons in the covalent bond are drawn strong another more electrostatic electronegative intramolecular electronegative on a a the charge (δ+) v O form towards positive l partial the interaction — a atom. hydrogen This bonds E covalent atom, hydrogen. is c an resulting hydrogen hydrogen usually also This found in a bond — with on a considerable atom c an then the dierent electrons of molecule, but exist in certain situations. bond t Figure 40 hydrogen bond form electronegative electronegative Hydrogen bonds between the hydrogen in a very polar bond and electrons on hydrogen atom atom a highly electronegative atom atom (e.g. F, O, N) (e.g. F, O, N) δ+ lone electron pair 143 Structure 2 Models of bonding and structure Hydrogen 1. water bonds occur, for example, between: molecules 2. ammonia (NH 3. hydrogen 4. water ) molecules 3 and molecules dimethyl ether molecules, (CH ) 3 These hydrogen bonds are depicted with blue dashes in O 2 gure 41. F H H H H types the name they of are suggesting otherwise, intermolecular intermolecular force, forces. but much p Enthalpies of hydrogen bonds Single single covalent covalent bonds compared to way is sucient look at to remember what to overcome 1 (~20 kJ mol the ), but i in water happens o good is n U not that the the hydrogen when water hydrogen covalent chemic al covalent bonds. –1 force enthalpy / kJ mol 20–40 150–600 bonds boils. At bonds bonds are weaker 100 °C, between between the than water oxygen chemic al bonds energy available molecules and hydrogen atoms 1 molecules themselves (463 kJ mol ). Practice questions u l a E v O f x 21. Which their two of the following substances c an form hydrogen bonds between molecules? A. hydrogen B. uorine, F C. methanol, CH uoride, HF D. uoromethane, CH 2 OH 3 F 3 22. Draw a diagram molecules, CH to show a hydrogen bond between two methanol a hydrogen bond between a water molecule and a bond between a water molecule and a OH. 3 23. Draw a diagram methanamine to show molecule, CH NH 3 24. Draw a diagram methanal to show molecule, CH a 2 hydrogen O. 2 144 not stronger than other t a r o d A bonds than are 3 Typic al bond or intermolecular C Table 8 and bonds generally o v i n u bond are weaker Bond or intermolecular force type Hydrogen hydrogen They C 3 CH y bonds: r e Despite H H n H O O t i s H Hydrogen bonds c an form l F H between a variety of dierent molecules y N H y P H u Figure 41 H O N O e r H H O H molecules (HF) s s uoride Structure 2.2 The covalent model 150 H O 2 100 s s 50 Te r 2 period 2 3 4 nber 6 3 Se 2 H S 2 3 y SnH H 4 3 Hr 3 HCl GeH SiH 4 4 4 v boiling points of the group 14 hydrides (CH , SiH 4 cloud. you increasing down the Descending bottom trend up. predicts group the due This to 15 in is boiling 30 °C. hydrogen the group hydrides, expected 16 Again, this the molecule boiling E therefore precisely and what of and , PH rise of in the the around 130 °C, electron data show in the H O AsH point mass. is and SbH 3 c an O, point to form of O , in 3 H from when in H phosphine, Extrapolating fact it the is trend H O O 100 °C H H H H higher-than-expected boiling point is due to between group due ) 4 therefore an 3 boiling molecular , and SnH 4 down the 17 and ammonia molecules. O H hydrides hydrogen are similar uoride, to HF, that have of group 15: very high 2 v water The water, H molecules. single point bonds a O points. LDFs, 3 increasing l 2 of polarizability hydrides, NH u trends period their a approximately presence boiling A the see , f x the expected. and stronger 3 higher, The of We stibane, SbH ammonia the are , GeH 4 increases gure 42. 3 to is electrons there mass o curve point molecular i , of group, The t a next PH the number the r o the at of group? d look to boiling curve the The boiling points of the 14 to 17 hydrides show the eect of hydrogen bonding n increasing Now go U group, as C the i n do Figure 42 group o –200 change t y CH p r e –150 O t i s PH –100 How l H NH –50 n g n i l i ob H SbH y 1 e H 0 P C° / tniop HF the presence of hydrogen bonds H O water is to four very hydrogen high: bonds 100 °C. O between H up H H (gure 43) and Figure 43 H Water molecules c an form up to four hydrogen bonds each 145 Structure 2 Models of bonding and structure Hydrogen higher water the bonds densities bec ause formation this ice layer The to is that between dense of ice lakes notable when than on properties. liquid. liquid lake water. surfaces freezing into Most Solid This the have (ice) oats on liquid unusual during solid substances water property winter. blocks of ice, formation of a leads to Insulation from allowing aquatic survive. bonding network more O–H water of in ice results molecules random, covalent which bonds in the (gure results in water 44). in a are In the higher shorter regular, liquid density than H phase, very the ordered hydrogen compared - - - r Note its than O molecules. e bonding solid less layers prevents hydrogen open-c avity is water s s ecosystems it of give when to ice. hydrogen bonds y C i n n U o i t a u l a v The hexagonal arrangement of water molecules in the ice lattice leads to the characteristic shape of snowakes Figure 46 Extensive hydrogen bonding is responsible for the high surface tension of water, walk on the surface without submerged 146 y p to be less dense than water (b) Ice oats on water bec ause of this d r o E Figure 45 f x O o The open-c avity structure of ice c auses it O t i s r e (a) v Figure 44 n l (b) which allows small insects to sinking or becoming even partly y P (a) Structure 2.2 The covalent model TOK DNA (deoxyribonucleic accurate along composed bonds. two DNA two is generations. strands replic ation strands, molecules of allowing DNA organic involves copies has for a molecules breaking to genetic be this information. A highly information to be double-helix held these structure together hydrogen by hydrogen bonds to “unzip” made. r two store essential new DNA A A G G “unzips” into two T n forces have storage systems knowledge? are exist electrostatic discussed the four o of t a table which are in the attractions types of natural world, in that keep intermolecular 9. Type of intermolecular force a London Waals (dispersion) v dipole–induced forces dipole–dipole forces, u in We together by hydrogen bonds, i areas l der information intermolecular f x O van of other together. summarised U summary, molecules in r o In types and pairs Hydrogen bonds are shown by black dashes d other science T The DNA double helix is held broken during replic ation. What base G A Figure 47 C C C strands i n different G o v double helix A y T p A O t i s r e G C n y C G C T T A l T T G A T C A y form C A P C G molecules e the between of acid) mechanism s s passed DNA-copying forces dipole (LDFs) forces Where does it occur? between all molecules E between a between polar polar between a molecule and a non-polar molecule increasing forces hydrogen bonding molecules strength and a highly hydrogen electronegative covalently atom (F, O or N) bonded to another highly electronegative atom Table 9 Summary of types of intermolecular forces 147 Structure 2 Models of bonding and structure Figure force 48 shows present in the a overall method for determining the strongest intermolecular substance. Do the molecules contain a hydrogen directly bonded to s s a very electronegative atom (F , O, N)? e yes no r Are the molecules hydrogen bonding some molecules l are polar and some no t i s London (dispersion) forces forces Figure 48 Figuring out the strongest intermolecular force present dipole–induced dipole y r e ethoxyethane, CH CH 3 4 ethanamine, CH CH 3 NH 2 uoroethane, CH F n Cl, and methane, CH 3 water, H O, and o i substances u l a E v O f x t a r o d Solution in this Ethoxyethane is polar , so it forms dipole–dipole forces. Carbon tetrauoride is non-polar , so it only forms LDFs. d. Ethanamine is also polar, form it forms hydrogen bonded so only to forms London London c. c an it form b. directly so example Butane e. non-polar, worked a. atom is 4 hydrogen uoride, HF 2 the 3 2 chloromethane, CH All 2 CH 3 CH 2 c arbon tetrauoride, CF g. bonds nitrogen, bec ause which dipole–dipole is (dispersion) (dispersion) a it very forces. forces. contains a hydrogen electronegative atom. It forces. Fluoroethane is polar, so it forms dipole–dipole forces. Note that it does not form hydrogen bonds. The structural formula might suggest that uorine is directly bonded to hydrogen, but this is not the case. If you draw the Lewis formula, you will see that uorine is bonded to a carbon atom. f. Chloromethane form g. is polar dipole–induced Hydrogen bonds molecules bec ause that is directly experience 148 3 OCH 2 C i n f. CH 2 CH 3 U e. CH 2 o butane, CH b. d. determine the intermolecular forces the molecules to each other. a. c. p attract v that forces between a pair of interacting molecules Worked example 3 For each of the following substances, are non-polar O dipole–dipole n y yes are while dipole formed they bonded between both to a dipole–dipole methane is non-polar. They would therefore forces. contain hydrogen. forces. water a and highly They hydrogen uoride electronegative atom are both polar, so they also y P polar? Structure 2.2 The covalent model Practice questions 25. For each that of attract the following molecules to CH substances, a. propane, CH b. hydrogen bromide, c. hydrogen uoride, HF d. buckminsterfullerene, C e. ammonia, NH the intermolecular forces CH 2 3 s s 3 determine each other: HBr and diatomic ethanal, CH oxygen, O 2 r f. CHO 3 State and explain water which of the following species c an form hydrogen bonds molecules: propane, CH 3 c. y CH 3 CH 2 ethanoic acid, CH 3 COOH 3 extent c an intermolecular behaviour? do the c an of “bonds” Structure advances e.g. in the Chemistry 2.1, and “forces” technology (IUPAC) compare? Structure 2.3) updated real gases lead to changes in scientic International denition of Union the of Pure and hydrogen bond? n U C Applied terms 1.1, denitions, deviation i n How the o (Structure explain v How forces (Structure 1.5) p ideal y what from r e To O t i s Linking questions n ammonia, NH b. l a. y with P 26. e 60 , 3 The properties of covalent substances generally: (dispersion) 10 these forces. shows typic al The in forces forces. Intermolecular force type E London (dispersion) sipole–dipole hydrogen Table 10 forces forces bonds relative ranges lies forces strengths dipole–dipole enthalpy a dipole dipole–dipole < enthalpy trend v O dipole–induced forces the u Table mass, l f x London and molar the somewhere of forces associated reects o comparable t a Given r o Relative strength of intermolecular i d (Structure 2.2.9) < with trend in intermolecular forces are hydrogen bonding. overcoming strength. in-between The London each of strength of (dispersion) –1 Typic al enthalpy / kJ mol 1 to 5 3 to 25 20 to 40 Comparison of the strength of various intermolecular forces 149 Structure 2 Models of bonding and structure Data-based questions The relative mole cular of Table intermole cular 11 shows forces mole cules are best with illustrate d mole cular by comparing masses in the mole cules range of 44.0 that to have 46.1 similar and wildly points. Name and point moment / D / °C 44.11 0.084 –42.1 46.08 1.30 –24.8 44.06 2.69 Intermolecular forces London Dipole– Hydrogen (dispersion) dipole bond yes no mass formula H C C C H H H H 3 8 H H 2 O 6 C H O O 4 H H H C H H down are four shown more by the and data that are in important London then statements data 78.2 yes yes yes 1.41 100.7 yes yes yes intermolecular force data for organic compounds of similar molecular masses experience dipoles 1.69 u temporary E Write explaining molecules no H a c ausing and all C O boiling point v O that 46.03 O l Polarity, Interpreting show f x Table 11 H i 2 H t a O 2 O yes o C 46.08 d CH 150 H r o methanoic acid that H O 6 n H 2 U C H C H ethanol i n C yes no o C 20.2 v 2 H p C H yes y H ethanal r e H C yes O H O C l y H H t i s methoxymethane no n H P H r propane induce further describing table 11. skills in science. (dispersion) the Attempt to forces temporary relationship explain What do bec ause dipoles in between each the data they the all in 11 suggest? electrons that For example, they move around, surroundingmolecules. molecular statement table contain using structure your and intermolecular knowledge forces from this topic. y Boiling dipole Structural formula e Molecular Molecular molecular C dierent s s boiling strengths masses. Structure 2.2 The covalent model Physic al properties of covalent substances There are two covalent. properties of We will look covalent structures molecular with properties at a substances covalent are physic al by are covalent dierent governed structure their properties and very network dictated conductivity structures: have lattice of are network and molecular properties. Most of the by intermolecular not comprised of forces. molecules, so features. covalent substances including volatility, e electric al of two s s Substances their types These solubility. r Volatility Figure 49 Molecular covalent with a covalent network structure are solids at room temperature and intermolecular forces are overcome pressure. Vaporizing holding the them requires structure a lot together. of energy They are bec ause therefore of the strong non-volatile covalent and have to vaporize to molecular be substances, overcome. Since the intermolecular intermolecular forces forces are between the relatively 27. generally a. variety variation substances in we , the have CO at ), this chapter that and at which consist boiling each at much points. other. 500–600 °C, boils of This and a are free as to their move. and a Covalent electrons are f x some notable exceptions, due to O meaning conductors insulators. panels consist The cells E electricity. v Solar and of the that as of up” cells, larger of lower of CCl c arbon , is 4 uoromethane, CH dierence of F, is 3 –78.2 °C. volatility molecules 3 point 76.7 °C. The boiling point for are gases or Comment on this using your intermolecular knowledge forces. have buckminsterfullerene are in wax, usually to (C ), tetracosane not contain electric al charged generally do not contain loc alized covalent bonds, and graphite, which is a good deloc alized intermediate photovoltaic contain such presence its a semiconductor, l conductor have or NH boiling tetrachloride, molecular molecules, these c andle needs substances “locked the each 60 of molecular, u are electric al include substance they do not contain ions. There small of The however, a of in o that particles, network electricity, bec ause t a such both conduct is i d particles To r o substances, conductors. of M any molecules component 391 °C. Electrical conductivity larger Examples 50 Covalent consist hydroc arbons. M any 2 between sublimes H 24 in some is, species n (C and substances. U which O there temperature. melting LDFs molecular considered , H substances higher stronger of forces, C and intermolecular i n However, volatility 2 room and o 2 liquids sizes v example, N of CH 4 28. p large the He or Kr b. y to r e Due volatile. which pair has the highest boiling point. weak, the energy required to overcome them is low and therefore molecular substances are Identify O need t i s order molecules n y Practice questions very high melting and boiling points. In l bonds y P substances vaporize when the Substances electrons. Silicon is a conductivity which convert places solar it between energy into semiconductors, such as silicon. Figure 50 Solar panels containing silicon provide this satellite with electric al power 151 Structure 2 Models of bonding and structure Solubility When a substance substance network A structures holding molecular between is oen insoluble atoms substance the solute said is and in attraction the most are solvent. formed between the Substances solvents bec ause of with the covalent strong likely to solvent dissolve are in a stronger solvent than the if the intermolecular attraction chemistry in to polar solvent is that dissolve in solvents. non-polar, “like dissolves non-polar like”. solvents. In other Similarly, forces between the words, non-polar polar solutes are likely Dissolving is unlikely to occur if the solute is polar or vice versa. l O t i s n y TOK “Like dissolves like” is a useful rule of thumb, which is helpful in many cases but not all, and it has no explanatory power. In a small group, discuss the advantages and disadvantages of having rules like this. You may want to y r e consider: what are the alternatives? Do such rules have an “expiry date”? What rules of thumb do you use in other areas of knowledge? is non-polar. 2 almost molecules. they to form Methanol, the a in that water solution ethanol primary and alcohols, CH 3 2 − O − l H readily decreases and water such are as hexane. as with ethanol, said to be are water more miscible, as proportion. dissolve with in water. The solubility of increasing c arbon chain length Solubility g / 100 g of H O 2 3 OH CH miscible 2 (CH 3 ) 2 (CH 3 OH 6.8 OH 0.014 3 ) 2 8 The aqueous solubility of primary alcohols decreases with increasing Hydroc arbon a H v Ethanol and F ats and easily with oils chains are are largely non-polar observable oil to make immiscible, when salad which non-polar vinegar dressing. and means they and do not form (an aqueous The do therefore two not hydrogen not solution liquids mix, do bonds due form to dissolve of ethanoic two their in with water. layers water. This is acid) and is are mixed said to be diering polarities. water c an form Therefore, ethanol is Long hydroc arbon chains, such hydrophobic (water-hating). hydrophobic tail and a they associate greasy are 152 any bonds, such strongly O − H − hydrogen bonds. water soluble in solvents associate hydroc arbon chain length between ethanol and water Figure 51 Table 12 c annot hydrogen Ethanol mixed CH CH nonan-1-ol 51). non-polar it Structural formula CH butan-1-ol u 2 E f x CH 3 O (c) 2 between ethanol molecules CH o (b) CH 3 i CH form (gure in bec ause propan-1-ol Primary alcohol ethanol t a r o d O − H − c an however, − O − H readily water when (table12). CH in n U (a) between water molecules Substances dissolve c an dissolves C likely O − H − H It insoluble i n H is o Iodine p , v Iodine, I H − O − stains with those and in fats and detergent hydrophilic (water-loving) both (gure as Soap 52). water and the Molecules non-polar with c alled surface-active agents, or a oils, head. fats said to be oen They and hydrophobic surfactants are molecules oils tail work have a bec ause frequently and found in hydrophilic head y the in likely covalent together. P are dissolve and their of and molecules. solutes to are forces solute) r It the e solute dissolves, as s s bonds (known Structure 2.2 The covalent model Water Water s s Water e Grease (a) Surface of material Detergent 'heads' attracted by water (c) y The eect a high they known as potential opiates, In are diamorphine, groups for misuse and are reduces have on the body depends on 2. their ability to bind to opioid receptors in the brain. ability to cross the barrier. blood-brain barrier lipids the morphine, This mechanism i n times more same time, c an heroin, easily action an tolerance, to cross makes has is the analgesic hydroxyl which the blood-brain opioid metabolized receptor. diamorphine and greatly Diamorphine is quickly than more addiction both groups, molecule. binds diamorphine about morphine. severe side central ve At the eects, nervous system depression. In most countries the use of diamorphine they are permeable to lipid molecules. The presence U C including of potent the as ester diamorphine which prevent polar molecules from entering the central nervous system. These membranes are lipophilic, which means and brain, into of o membranes that coat the blood vessels in the brain and v The blood-brain barrier consists of a series of cell in In polarity with y their known p factors: 1. the r e soluble two also substituted n addiction. derivatives, with O its l y and painkillers t i s strong water In a surfactant molecule, the hydrophilic (water-loving) head and a hydrophobic (water-hating) tail li greasy stains o materials Global impact of science Morphine Grease washed aw ay by P (b) Surface of material Figure 52 r Grease is either banned or restricted to terminally ill patients of one amino and two hydroxyl groups in the morphine CH OH 2 u f x l c ancer or central nervous hydroxyl (b) CH system 2 CH 3 C H C 3 N CH 2 O groups O morphine derivatives) E v a O amino group of O (can be substituted in O tertiary Figure 53 i 2 t a N C 3 d r o H CH forms o in lipids and therefore limits its ability to reach the opioid (a) certain disorders. soluble in water but at the same time reduces its solubility receptors in the brain. n with molecule (gure 53) makes it suciently polar to be O C O CH 3 OH The structures of (a) morphine and (b) diamorphine (heroin) 153 Structure 2 Models of bonding and structure Physic al properties of ionic and covalent substances You its c an figure physic al out the structure properties such as of a substance electric al by testing conductivity, you do that could not know be which is which. Design a safe method used to identify A, B and C. s s aqueous solubility and melting point. Possible Tool 1: Recognize or Inquiry 1: and address environmental the issues in relevant an spatulas, safety, leads, investigation. beakers, ammeter, crucible, • substances you could use distilled lamp, Bunsen water, pipettes, conductivity burner, pipeclay probe, triangle, Demonstrate independent thinking, mat, mortar and pestle, melting point Inquiry 2: Identify and record relevant qualitative a proposed method risk assessment and show this and your presented bottle with three contains a bottles crystalline labelled A, B and solid. Extension One is sodium What chloride, another is sugar, and the third is white if would bec ause electric al • insoluble strong experimental data the physic al of covalent substances? • (Tool 1, Inquiry 2) group determine the • and compounds for explain graphite the of their molecules conductors form solubility, with Explain why and although 32. Explain E volatile, covalent are and not electric al 33. graphene. relationship they both and are non- are would require breaking the typic ally: together by intermolecular forces, they do not contain mobile charged on the strength of the intermolecular forces Which two in of the following substances are readily water? A. ethanoic acid, CH B. octane, CH C. octan-1-ol, CH (CH COOH ) 2 CH 6 3 (CH ) 2 CH 6 OH 2 volatile, D. propanone, CH COCH 3 3 covalent bonds. 34. CH Explain why ionic substances, such as sodium chloride, OH, is miscible do 2 not dissolve in oil. water. 35. Explain why hexane than diatomic bromine, Br , 2 154 graphene) molecules. soluble between substances substances 3 with held are 3 contain typic ally: covalent bonds graphite dissolution 3 why ethanol, CH strong are 3 network molecular bec ause depending solvent volatility and boiling point. 31. by include structures are structures weak electric al varying they a Describe except v O 30. covalent bec ause are soluble? How charged particles their covalent water you particles l why conductors, poor bec ause molecular together mobile not If u f x Explain • forces? Practice questions 29. which t a (Structure 3.2) nature volatile with i theintermolecular r o of extent does a functional solvents network (exceptions contain o what d To all not n Substances properties in do covalent bonds. U demonstrate they held C Linking questions What conductors i n bec ause covalent are were your method? o poor they v between atoms with substance y substances non-volatile • ionic modify p • summarize, you r e To the sand, but approval. n are E ach time, try it out. for O C. teacher t i s You have your y Instructions to l Complete observations. in water. is more soluble in y hood, combustion spoon. • heat-proof apparatus, fume P initiative or insight. power pack, electrodes, r ethic al and e • equipment mightinclude: Relevant skills Structure 2.2 The covalent model Chromatography (Structure 2.2.10) The components of chromatography. underlying most based involve of their forms the phase oen be separated chromatography existence of a chromatography, dierent intermolecular mobile c an of anities the for and (gure mobile phase components of identied using 54) have and a the the same stationary phase. mixture are separated each of the two phases. These anities forces. moves through the stationary phase, c arrying the components Methods the mixture. The components all start at the same point but are are through the the stationary two phase phases. The at dierent rates components due to their eventually diering anities for separated. (b) liquid column chromatography y sample O t i s solvent front move down the column at different rates, forming bands n o i t a 1 2 3 Sample is placed Solvent is Different components in the sample on the top of the poured down become separated into bands and are packed column the column collected at the bottom of the column a E v O l u chromatography o 2 C i n SiO U d r o f x gas–liquid p of the sample with adsorbent (c) components packed e lu b v oran ge y r e column water (the solvent) solvent n solvent paper clip l (a) paper chromatography are mixtures Structure 1.1. y of separating discussed in P each for transported r of e The principle: forms on mixture s s In a All detector heated chamber recorder sample injected t Figure 54 All types of here column – c apillary tube chromatography involve a mobile c arrier gas with liquid-coated walls phase and a stationary phase 155 Structure 2 Models of bonding and structure Research skills ATL Chromatographic be grouped methods according adsorption, we will the focus former latter is on the two on ion of c ategories. separation planar (for based gas on principle at notes in example, partition, a paper of In this and chapter, thin-layer. partition, whereas least one other method of chromatography) your c an column chromatography). the Research They vs and Venn compare r as of into example, planar chromatography: mixtures Organize (for exchange adsorption. (such methods. or types separates based classied mechanism exclusion chromatography dierent or be format e The size c an their s s chromatography), on and diagram. contrast the solvent. paper The 55). (made mixture is The of to a hydrated be of the paper. The paper is then chromatography experiment using dierent containing a small amount You c an see that the purple ink at the right-hand the vapour solvent atmosphere side is a mixture containing red, solvent yellow and allows and prevents blue dyes in the solvent at is placed the i n coloured ink are dropped onto piece of to onto chamber start line as a near the beaker, Placing a lid on top of the become through a such saturated with the evaporation. After Different components of the inks rise up the paper to different extents and are separated o i d u l a E v O f x t a r o solvent paper rectangular n U paper C chromatography paper a p different a dotted in o v of is bottom. chamber loss demonstrate cellulose) and the mobile phase is a Before Spots to phase y beaker of r e inks. way separated Simple paper bottom simple stationary O suitable Figure 55 pigments t i s chromatography of (gure n mixtures chromatography y Separating l chromatography The solvent The components to a their is allowed relative greater to the anities anities transported known as of for further for the up move up mixture the paper will paper. until dierent stationary solvent the the move and dissolve The results it reaches distances mobile more of a up a phases. readily point the in near paper the top. according Components with it and chromatography are therefore experiment are chromatogram. Chromatography paper is composed of hydrated cellulose, which contains many –OH groups. These groups are very polar and attract water molecules tightly, saturating the surface of cellulose with water . When coupled with a less polar organic solvent, this technique can be used to investigate mixtures such as leaf pigments or amino acid mixtures. The mixture components are partitioned between the water layer and the solvent: less polar components dissolve in the solvent and get carried further up the paper , whereas more polar components stay predominantly in the water layer . The separation can be optimized by using dierent solvents, or mixtures of solvents. 156 y P Paper Structure Thin layer 2.2 The covalent model chromatography (TLC) Thin layer chromatography (TLC) is more expensive than its paper counterpart, but it oers greater sensitivity. In TLC, the stationary phase is a rectangular plate made of glass or metal coated with silica (silicon dioxide, SiO ) or alumina (aluminium oxide, 2 O 2 s s Al ). The silica or alumina surfaces are very polar and contain many hydroxyl 3 groups. The mixture is spotted onto the plate and placed into an eluting chamber containing a non-polar organic solvent. The polar substances in the mixture travel surface is greater than their tendency to dissolve in the solvent. Conversely, non- r polar components dissolve in the solvent and travel further up the plate. e up the plate slowly because their tendency to adsorb onto the silica or alumina n l y reacts with the spots to form coloured products. Other purposes. Coloured substances are easily located on colourless compounds can be located in dierent ways. a chromatogram. When a substance is colourless, an For example, some might show up under UV light. C an additional step is needed. For example, when a mixture you think of any other examples of how existing scientic contains amino acids, the chromatogram can be sprayed methods were adapted to solve a problem? chromatography R distance travelled for a c an spot be on by the spot (b) to the = F a quantied by travelled by of R , are 56 Retardation c alculated factors, from the distances F measured on a chromatogram o paper type i on and front t a depending travel ratio of t Figure solvent u in the solvent (a): mixture on baseline a c annot always a l spots are v E the is the n U d r o f x O Since values c alculating the chromatogram distance b R a C the experiment value F i n ). The F y a o of v results retardation factor ( R p ) F The r e Retardation factor (R O t i s Experimental techniques are adapted to suit dierent with a locating agent. A locating agent is a substance that y P Experiments pencil baseline b further than the solvent that transports them, R F the factors range such used (in of as 0 the paper to 1. The distance composition of travelled the chromatography). solvent, R values by a spot c an vary temperature, pH are reproducible F and therefore values to c an be accepted used values, to identify provided substances that the by comparing conditions are the experimental same. 157 Structure 2 Models of bonding and structure Worked example 4 Consider the thin layer chromatogram shown in gure 57 . Given that the stationary phase is silica and the solvent is a Calculate the retardation factor for the red and blue spots. hydrocarbon, determine which component is more polar . s s cm 6.0 cm r 8.0 maximum solvent distance the front distance cm level of the solvent baseline: 2.0 cm and the (blue). solvent c alculate the at The front contains distinct 8.0 cm (red), distance is three spots factors (R ) F blue spots (top bottom) = by the red R (2 signic ant gures) l a v E This experimental between substances than the previous 158 do a each substances blue phase is for is blends signic ant than R spot solvent gures) (blue). of into the clear the red spot (non-polar This so the than me ans dots you next. you are will If c an phase. spot The that the frequently you obtain want using a soluble solvent) that in in the and a the gre ater phase substances the shown a have stationary contains substances oen try more substances blue the are hydroc arbon of stationary polar phase, the composed experiment, mobile (2 of polar, more composed the 0.17 larger spot anity colour blue the F mobile are in is component(s) chromatography which = (red) (silic a) Chromatograms chromatogram has less clear separation examples we looked at spot solvent u f x O u Figure 58 0.67 red the i = follows: t a 12.0 cm by r o 8.0 cm = travelled the by 12.0 cm the the by travelled 2.0 cm 6.0 cm for travelled distance o travelled (red) = F distance as d distance R and = F F and distance (blue) R dierent between 12.0 cm. retardation at o c an the as n baseline You from and mixture U (green) the appe ar C distances that which the resulting chromatogram (right) v shows i n 58 components, p The apparatus for thin layer chromatography (le) and Solution Figure solvent y the r e Figure 57 above O t i s solvent mixture n cm 0.0 travelled y 2.0 l the spot 12.0 cm in red that spot. books. If you chromatogram in clearer dierent separation solvents. between y P solvent travelled baseline e lid Structure 2.2 The covalent model Thin-layer chromatography of plant pigments The pigments found chromatography, in plants c an depending on be separated their by Part 2: Chromatography relative affinities to 3 4. mobile and stationary phases. The separation Place by changing the phases. In this few cm tank. Place will compare the chromatograms stationary phases are on top, the chamber to the in solvent vapours. becomes You saturated c an place a paper changed. towel soaked in the solvent vertic ally inside the tank Relevant skills Measuring length 1: Paper 5. or thin layer Then, Record 3: Propagate uncertainties in uncertainties in the starting sides of you must make attaching top of the Wear • Flammable eye Keep away from flames and solvents a are the toxic well-ventilated to the space environment. and dispense Chromatography paper • TLC plate • Pencil • green beaker the solvents ether:propanone mixture) Method using splint the before pigment it a the once line that paper, the placed in the tank by across the solvent pencil evaporates. spots Allow tank draw in Draw over circles are visible in the plate to dry. the distance travelled by the Measure the distance travelled by each pigment the base Determine a line to the reasonable you have solvent. centre of the spot. uncertainty for the measured. value C alculate the R of each pigment, and the F uncertainty Compare of your (9:1 values of each. your class. results to Compare those your of other results to members literature for the R values, if available. Search online F for information that will help you identify each of thepigments. petroleum O l u Chromatography 10. a are upright chromatography tank leaves f x • Fresh or for o Propanone tank foil i • aluminium t a Lid U Chromatography • d Ruler • r o • from Q uickly front the distances 9. to is Measure from 8. or o • spotter solvent C Chromatography n Pestle and mortar • v i n Materials plate top. chromatogram. 7 . lid it tank. the the around solvents cupboard. • near the Ifyou that y in fume or is sure p Work Remove r e Organic • a solvents. 6. sources of ignition. • in protection. to solvent, taking O S afety it the Ensure the plate does not touch beaker. t i s conclusions other line. the Discuss the impact of uncertainties on the • into below the pigment spot n 3: plate solvent processed data the Inquiry TLC the measurements on • the keep l 3: Tool to y Tool • process. chromatography c are • lower the y 1: Tool up P Tool • speed r to • solvent into ensure that the e and chromatography lid obtained when the with mobile the the investigation, atmosphere you of c an be the optimized a s s the 11. Comment accuracy, of on of the your measurement precision class’ s and, results. uncertainty on if possible, Discuss the impact the validity of your Part 1: Preparation of the chromatography plate a 1. conclusions. 3 Grind the leaves with a few cm of propanone until v Part 3: Changing the phases you have obtained a concentrated pigment extract. Handle the a pencil, line in bottom. TLC E 2. 12. plate c arefully by the edges. the plate, 1 to 2 cm Spot the plant spot taking of the solvent results would system, or a differ if you used different stationary phase. Then, try it out. pigment Clear c are to pigment. create Allow up as instructed by your teacher, taking c are extract onto the starting not line, how different from the 13. 3. Consider Draw a across it a to small, to dispose of the solvents down the sink. concentrated dry. 159 Structure 2 Models of bonding and structure Practice questions 36. C alculate the retardation factor, R , for the green spot F at 37 . 6.0 cm Suggest in gure 58. why the baseline in paper chromatograms is s s drawn in pencil. 38. Chromatographic and analysis individual of a pigments mixture (B, C of and dyes (A) D) produced chromatogram on the right. and explain which pigments are present in A. A B C r Identify D e the three l LHA a two molecule or molecule. double be drawn Resonance or for triple oen bond ozone, O alternative in a described happens resonance by when molecule. (gure 3 show be 59). Note For that a bonds identic al • intermediate of bond have a Lewis formulas arrow is c an used to O that ozone the double. In Bond two oxygen–oxygen reality, length the and ozone bonds molecule strength data in is (table ozone a 13) show are: each other bond distributed between a single O–O bond and a double O =O bond in terms order of 1.5, bec ause there are 3 bonding electron pairs across two domains. l u Average bond –12 Bond Length / 10 m Bond order –1 enthalpy / kJ mol E v a O–O O are hybrid of strength and length t a • to two o • one structures. in Instead, represent more than one position double-headed p resonance two i d r o f x O 160 two the suggest and is formula. collectively o that might single n U these one Lewis that Ozone has two possible resonance structures C structures dierent: there structures. v i n These single example, O Figure 59 a resonance structures y a c annot possible r e for more n the are O there t i s Sometimes, single O O=O bond bond 148 144 1 in 128 362 1.5 121 498 2 double Table 13 ozone bond Bond data for oxygen–oxygen bonds y P y Resonance (Structure 2.2.11) Structure 2.2 The covalent model LHA Global impact of science Diatomic oxygen, O , and ozone, O 2 allotropes that ultraviolet in O , (UV) with dierent radiation section bond are oxygen of order corresponding the 2, wavelengths UV-C is spectrum. stronger to The and of and resulting UV-B the Ozone double therefore bond has a (CFC s) c an in radiation skin layer in depletion the reach c ancer c an c ause DNA mutations, (melanomas) c aused atmosphere the E arth’ s by means surface. and c ataracts. s s electromagnetic , 3 absorb chlorouoroc arbons that more International UV radiation collaboration 2 O , bond with enthalpy bond order than 1.5. the oxygen–oxygen Breaking the bond in involving of oxygen–oxygen the 1987 Montreal ozone-depleting e higher Protocol and the banning substances has slowed down the 3 bond requires be of the double of depletion. concentration oxygen–oxygen bond in O of Evidence ozone in in the recent stratosphere 2 brought wavelengths be correspond electromagnetic in the The bonds 330 nm to the and UV-C spectrum, stratosphere in UV-C surface and UV-B radiation from and UV-B prevents the nearly all E arth’s (gure 60). the ozone layer has increased, (gure 61). y p r e C o v i n n U o i d Figure 61 S atellite images of a hole in the ozone layer over Antarctic a from 1979 to 2011. The largest ozone hole to date was recorded in 2006. Since then, data suggests that the hole has The ozone layer in the stratosphere absorbs radiation and most UV-B radiation been getting smaller again u l a E v O f x t a UV-C r o Figure 60 of ozone respectively. reaching recovery below. 3 solar the O presence of O of in n the below. resulting radiation with t i s of and wavelengths wavelengths regions The 240 nm by electromagnetic y These of broken by l c an about years suggests the y dissociation c an rate energy. P The higher r double t Figure 62 Former US president Barack Obama presents the Presidential Medal of Freedom to Mexic an chemist and environmental scientist helped and M ario Molina. Molina to uncover the link between CFC s ozone layer depletion 161 2 Models of bonding and structure LHA Structure Deloc alization Resonance of structures deloc alization. loc alized are shared loc alized c an between by more be described covalent the than bonds, two two atoms. atoms as two in a single atoms structure using the concept share Deloc alization a molecule or a pair occurs ion, as of electrons, which when electrons s s are In opposed to being between a pair of atoms. Consider the ion NO . It is represented by two resonance structures (gure63a). e 2 The nitrogen–oxygen bond has bond order 1.5, and its length electrons is deloc alized c an be the deloc alization across represented the with a two N–O structure bonding domains. containing a Practice questions Draw the structure of electron (a) deloc alized structure N the c arbonate ion, CO why the C–O bond O order O in this ion is 1.33. – Figure 63 (a) The ion NO c an be represented by two Lewis formulas (b) An alternative 2 and to those in in the c arbonate c arbon dioxide OH. 3 Note that the does oxygen dissociated by and Lewis ozone different wavelengths of light? actual molecule which is best is to a the or single the 63b and a Lewis of that “in-between” the single between be bec ause it does not pairs. Instead, c annot electron formula electron movement. molecule somewhere by not interconversion bond presence represented is non-bonding the represented deloc alization. the various deloc alized various resonance resonance structures by a single Therefore, the resonance structures, structure. o i Benzene and resonance (Structure 2.2.12) Benzene, C H 6 , is an important example of a molecule that demonstrates 6 resonance. u l a E v O f x t a r o d (Structure1.3) due involve rotation describe formula Figure n are U collectively Why not through in bonding C structures of i n Resonance Linking question structure loc ations electron pair o specify v and methanol, CH representation of the ion shows the deloc alized p ion strengths y Compare the C–O bond lengths r e 41. N N , and 3 explain O the 2– of (b) t i s Draw shows ozone that deloc alized pair. 40. reason, that n the y shows 63b). this line l 39. (gure For dashed Figure 64 The ball-and-stick structure and electron cloud of benzene, C H 6 162 are One pair of 6 y ion P the bond. strength r between those of a single N–O bond and a double N =O and Structure Lewis formula connected structures for through are benzene contains alternating possible single a six-membered and double ring bonds. of Two c arbon atoms resonance H H H H H H H electrons in the c arbon–c arbon bonds in of benzene benzene are deloc alized this reason, these deloc alized electrons are oen skeletal formulas, straight lines n y For l In ring. y the the Resonance structures P of Figure 65 r t H e H s s H H around covalent model (gure 65). H Some The LHA The 2.2 represent bonds. The ends and represented with a circle: vertices (corners) of the bonds hydrogen Detailed skeletal be found are is in a planar the and simplest large range stabilized aromatic of due substances (gure 66). structures of 1,2-dichlorobenzene structure. Cl benzene the 1,2-dichlorobenzene structures rings in of the l E v a O curcumin Figure 66 (a) Curcumin, that contains following below. drawn. drawing molecules are Structure 3.2. TOK Skeletal of formulas “chemistry extensively what extent using are and language? are an example shorthand” used formulas a of for by that chemists. chemic al are To symbols, equations like a What chemic al is gained or lost notation? circle to molecules and identify the (b) (c) Kevlar found shown formulas not electrons. each: (a) is o Research of deloc alized u f x 3. the i structure represent Cl t a a d Draw r o 2. deloc alized structures similar n resonance of given in describe are o resonance other and to C the the presence i n of Draw used U One the hydroc arbon, Activity 1. to is p c an that Benzene hydroc arbon. The term “aromatic” v it aromatic molecules electrons. to an atoms instructions y cyclic is r e Benzene c arbon atoms, and O t i s represent in the yellow food estradiol spice turmeric (b) Kevlar is used for making personal armour (c) Synthetic estradiol is used as medic ation in menopausal hormone therapy 163 2 Models of bonding and structure LHA Structure Communic ation skills ATL In ordinary conversation, something a to having indic ate compound. a a the term pleasant benzene However, “aromatic” smell. derivative some In or aromatic c an be chemistry, the used For presence of a benzene ring in compounds example, azulene — a component of some blue are non-benzenoids. pigments two planar rings that are joined r together: C an you think that of have other examples dierent of chemic al meanings in C C everydaylife? C H benzene, The deloc alized electrons in the p orbitals of benzene six plane p of one electron orbitals the in overlap, molecule. each of forming The a of electron the plane density of the above c arbon atoms are sp hybridized step H to c arbon (gure have hexagon of 69). come bec ause to of August him the (b) structure of 68 above deloc alized and (gure below the 67). The molecular with (gure 68). multiple alternating bonds. single Kekulé was credited dream. The structure diering formula The and with 1:1 for benzene, ratio of Eventually double the c arbon a cyclic bonds was structure, which is proposed was a lop-sided c arbon-c arbon bond lengths. CH 2 o i stamp the t a Kekulé on u for a August postage suggestion benzene l (a) E Kekulé’ s 164 69 German (a) v E ast d r o an f x O u Figure Figure a the of n CH 2 in was 6 structure presence atoms U H the its C i n said deduce suggested arrangement proposed to o hydrogen was v to next are the structure of benzene established that C 6 the ring y was electrondensity this p it in r e Once c arbon atoms occupies a p orbital. of (seeStructure 2.2.16). molecule Physic al evidence for six ring electrons 2 ring below the O The t i s 67 n H In Figure y C C H l C H Some of the 19th century suggestions for the (b) structure of benzene y terms P H H in e bonds H H a found in nature — contains alternating single and double H and describe H H form to however, this term is s s used as Structure diraction c arbon atoms suggesting The patterns in they were c arbon–c arbon that and they a double are also of all showed where all equivalent bond length regular in (gure hexagonal bond bonds arrangement of had the same length, 70). benzene c arbon–c arbon intermediate a c arbon–c arbon covalent model is intermediate (table 14). Bond between that of a enthalpy data s s single later benzene The LHA X-ray 2.2 shows strength. 0.140 nm e Average bond –12 Bond Bond length / 10 m –1 enthalpy / kJ mol single C=C in bond benzene double bond 154 346 140 507 134 614 length and strength data for the c arbon–c arbon bond in benzene diraction compared to single and double c arbon–c arbon bonds c arbon The chemic al does not end form readily where bonds the of addition double new the Instead, diers alkenes, bond species. benzene is reactions. breaks In deloc alized from what would normally be benzene and the tends to ring, which undergo value increases c alled cyclic comparing the thermochemistry bond, contains change in is This with hydrogenation two double is data 152 kJ mol for c arbon– 0.140 nm are and substitution discussed in reactions Reactivity 3.4 quantied bond. . benzene When 120 kJ mol This we and add value is those for hydrogen . 1 ΔH hyd = –120 kJ mol cyclohexane reaction, enthalpy of hydrogenation, molecule is 1 u a a c alled v similar nearly double: l is double energy is 2 f x O reaction referred to as the one E ach length 1 it t a r o H the cyclohexene a molecule. benzene o cyclohexene double + This benzene For reactions. six-membered alkenes. example, this the i to of d For by stability molecule. bond Addition energetic ally n established the resonance energy. be C i n a U by would substitution Resonance energy of benzene Resonance undergo addition c arbon atoms on either benzene, addition would disrupt the electron X-ray unsaturated, but unlike alkenes, Alkenes o unfavourable. undergo with eect benzene v stabilizing of Like p reactions, alkenes. c alled y it behaviour from r e expected technique electron density in an individual benzene the structure of benzene A shows a contour map of O t i s the Chemic al evidence for 70 n Bond l y Figure Table 14 y C P C r C–C and ∆H the . change in energy is 1,3-cyclohexadiene is hyd bonds, so its enthalpy of hydrogenation is 1 232 kJ mol E 1 ΔH + 1,3-cyclohexadiene 2H hyd = –232 kJ mol 2 cyclohexane 165 LHA Structure 2 Models of bonding and structure The Changes in energy next molecule theoretic al reaction are known as Enthalpy the molecule, series matches Kekulé’ s 1,3,5-cyclohexatriene. suggested structure and is the 1,3,5-cyclohexatriene contains three enthalpy double changes. in across a changes bonds and its enthalpy of hydrogenation is predicted to be 120 × 3 = are 1 1 360 kJ mol discussed in Reactivity 1.1. is However, the actual value for benzene is 208 kJ mol is due to the added stability provided by . resonance in the benzene ring. The s s compounds 71). Naming This cyclic (gure covered in 1 dierence between these enthalpy values is 152 kJ mol , which is the resonance Structure 3.2. energy. e r lom Jk / yplahtne (152 kJ mol theoretic al 1,3,5-cyclohexatriene of p an benzene is less exothermic C has additional o v i n Hydrogenation reaction ) energetic cost to than break n compound the 208 y to l benzene r e of compared O resonance –1 energy y The t i s 71 360 expected. the This deloc alized is bec ause the electron rings. 1,2-disubstituted benzene compounds and the single Kekulé bonds. structure When i whereas c arbon–c arbon evenly the in the you two distributed, , reacts with benzene to form 2 expect chlorine atoms that at two possible either end of a isomers c an be c arbon–c arbon single isomer, the chlorines would be at either end of a bond these benzene, containing alternating double would other double of chlorine, Cl two (gure 72a). isomers However, represent the bec ause same u l all electrons molecule (gure are 72b). (b) Cl Cl Cl Cl Cl a E v O f x (a) Cl Figure Kekulé only 166 with t a r o d bond, one o 1,2-dichlorobenzene, formed: n U Consider 72 (a) structure one isomer The of two theoretic al benzene bec ause (b) the The isomers actual c arbon of 1,2-dichlorobenzene structure atoms are all of for the 1,2-dichlorobenzene has chemic ally equivalent y P 1 u Figure 152 Structure 2.2 The covalent model LHA Practice questions 42. Describe structure 43. How many pieces ring State bond the the evidence — one deloc alized benzene of of physic al and one chemic al — for the benzene. electrons contribute angle in to the does each deloc alized benzene and the of the c arbon atoms in a s s 44. two of electron cloud? molecular geometry around each c arbon atoms. are resonance energy in benzene explain its relative unreactivity? the structural features of benzene reactions? that favour it undergoing (Reactivity 3.4) n substitution y electrophilic l What the 2.1, 2.2) (a) eight or in 12 these electrons. molecules electrons, , This c an the in the molecules is have respectively. and discussed with SL known as an ve or Examples six of pairs such of same containing do not (b) central electrons, amounting to , shown in gure 73. steps used expanded check that the expanded octets for drawing octets. Lewis However, this at SL for Cl Cl I Cl Figure 73 Cl Two species with expanded octets: PF and ICl 5 c ase, we omit the nal central atom has a full octet. ve the and six central pairs of atom, electrons , with 4 around respectively o d Worked example 5 in formulas n step — we The F F At more 4 U molecules the chapter. with C follow this species include phosphorus tetrachloroiodide ion, ICl Drawing Lewis formulas for c an of atoms expanded octet. 5 You section containing i n pentauoride, PF were familiar o 10 valence VSEPR be v atoms and also p than should P F y formulas you r e HL O t i s Expanded octets (Structure 2.2.13) Lewis F F y does (Reactivity P How e r Linking questions Draw the Lewis formulas of each of the following species with expanded octets: sulfur hexauoride, SF b. Count above for e ach Fluorine: 7 electrons Total: 6 E C alculate the v valence Step 2 c. + × 42 molecule: b triiodide ion, I 6 6 = = 42 48 48 This × 3 pairs 21 meaning that it has an 3 Xenon: 8 Oxygen: 6 Total: 8 + × 18 3 = = 18 26 electron. + 1 = 22 26 = 2 = polyatomic ion has a –1 charge 22 24 c xenon trioxide, XeO 3 Iodine: 7 additional = XeO 3 Total: 21 2 xenon trioxide, 3 a sulfur hexauoride, SF Sulfur: 6 a O Step 1 steps u the l f x Follow triiodide ion, I t a 6 Solution i r o a. 11 pairs = 13 pairs 2 number of electron pairs 167 2 Models of bonding LHA Structure and structure a sulfur hexauoride, SF b triiodide ion, I 6 Step 3 Arrange c xenon trioxide, XeO 3 3 O F the atoms F F I I Xe O F F s s I S O F Draw e Step 4 F O the single F I bonds I r F I S P Xe O F used so far... 2 pairs used 3 t i s F peripheral S now have noble gas congurations do F not need any atoms now have 8 full We remaining have pairs used pairs, above nal. 24 is so the There structure are 6 pairs electrons on the central of electrons around the central are three pairs of C atom. It has an expanded octet. U use is for n we far... O Xe All assign to 13 have the O the nal available been electron xenon atom. electron pairs used. iodine atom has an expanded XeO have , each 3 an atom expanded incorrect, deducing and an Lewis around it. has L ater octet. in formal exception this chapter, charge: structure of this XeO we will and will discuss help explain 3 formulas. to have an expanded octet. a E v O Practice questions 45. Draw the a. PCl b. BrF Lewis formula of each d. of the following: ICl 5 2 + e. 3 c. IF ICl 2 f. 5 168 so o structure method for used needed. i this none pairs We species is an ion, so have We can see that the central u the and are now octet, with 5 pairs of electrons l f x However, to deduced I atoms O square brackets and the charge t a structure electrons This oxygen octets. pair d the r o In eight I I sulfur 12 assigned to the central iodine atom. atom full far... I remaining i n pairs so I The electron used available The O o electron all pairs used. v any Put 24 more O p Step 6 octets. O y uorine far... Xe therefore r e electrons The and I F so peripheral iodine atoms atoms F used O The pairs n I pairs F on far... y F bonding so I I pairs Put non- l 6 Step 5 O SOF 4 us the concept deduce why it is the of true considered y F F Structure In the SL atoms occur for covalent model geometry of expanded octets section have The LHA Molecular 2.2 of two, this three expanded chapter, or four octets we discussed electron containing the domains. ve or six geometries that arise when Several additional electron geometries pairs. s s Five domains: trigonal bipyramidal geometry If there are and electron shape electron domain domains, adopt geometry consisting of two c alled a trigonal 120° in and those 90° domains from each trigonal bipyramidal. that share the bipyramidal same The repel ve triangular molecule fall each other. This into domains form base. The electron two c ategories: 1. Equatorial: the bonds forming the triangular base of the pyramids, at 120° from the bonds triangular that form the axis of the molecule, at 90° Figure from the plane of trigonal base. 74 A balloon model of the bipyramidal geometry A trigonal molecular When geometries, all ve four When of only of domains the the two the molecule has ve of the the domains are domains the two are non- and three geometry. Figure 75 75. The four types of are trigonal bipyramidal geometry: seesaw (c) (a) electron domain trigonal T-shaped (c) bipyramidal (d) (b) linear (d) F F I F Cl F I S F F a E e domains, i l B linear (b) u F v position bonding gure and geometry. t a r o f x O equatorial are F P domains, T-shaped o F F in d (a) F bonding molecular has shown geometry. n type are seesaw molecule has ve trigonal are bonding domains, and one is a non- molecule has domains, each B presence of non-bonding domains: domains, U of ve domains, non-bonding Examples the possible C • four geometry. domain, three bonding the to o When on bonding rise i n • are gives v bonding depending domains geometry p When domain y bipyramidal • electron r e • bipyramidal electron domain O t i s the n Axial: l 2. y each other . y in is pyramids electrons at P domains the positions r a ve therefore e other I F B a B a a B B B e a e 120° A A B B e 90° B less than 120° B e A A e B a B a less than 90° 180° a less than B a 90° axial position 169 LHA Structure 2 Models of bonding and structure Like other geometries, the 120° distorted the presence and 90° bond angles in a trigonal bipyramidal Practice question shape 46. Consider the T-shaped molecular Why do seesaw and of non-bonding pairs, due to their greater geometries. are Six domains: octahedral geometry s s domains by repulsion. you think the non- bonding are loc ated in If there are six domains, the electron pairs in those domains adopt positions at equatorial positions, not in axial 90° from each other. This electron domain geometry is c alled octahedral. This is positions? bec ause the electron is no electron six octahedral domains the are geometry bonding four domains, the each type 90° bond geometry presence F angles in pairs, o 170 77 i E Figure t a a B each other, so to three common molecular the molecule has octahedral in gure to domains, their and two are non-bonding geometry. 77. molecular due geometry. geometries greater are distorted by the repulsion. (c) F F F Br F Xe F F F u l v O f x B A (c) F F B F d r o F to F (b) F S shown octahedral non-bonding bonding C (a) are square planar n U F of domains are 90° are bonding domains, and one is a non-bonding o domain i n electron six domains, square pyramidal molecule has A balloon model of the The octahedral of domains v 76 of the rise p Examples Figure six at y When the molecule has r e • of the gives all equatorial domains. O ve domain, are and t i s When axial presence of non-bonding domains: geometry. • molecule between domain on polyhedron equilateral triangles. distinguish depending eight-sided n all an an l When in is B B B B B B A 90° A B B 90° <90° B B B B The square planar three types of y octahedral to which y • domains need geometries, identic al octahedron, P there An eight an r The of form e consisting domains octahedral electron domain geometry: (a) octahedral (b) square pyramidal Structure The covalent model LHA Electron 2.2 Number of Number of Number of bonding domain Molecular Bond geometry angle non-bonding domains Example domains geometry domains trigonal 5 5 0 trigonal 90°, 120° PF 5 s s bipyramidal bipyramidal 4 1 seesaw <90°, SF 4 <120° T-shaped 3 linear 1 octahedral 90° square <90° pyramidal six electron domains and molecular geometry of a two e. four bonding domains and two non-bonding domains the number following seesaw f. square g. linear the are l PCl 3 IF 5 d. ICl e. ICl 2 + 2 E Predict the bond a BrF v O c. possible domain following: 5 b. two electron answers geometry u the a. pyramidal f x Deduce bipyramidal (there around atoms n square planar e. domains o octahedral d. non-bonding i c. and geometries: t a T-shaped three non-bonding domains d trigonal b. bonding molecular r o a. of and U the domains C Identify bonding p ve bonding domains d. 4 o four bonding domains and one non-bonding domain c. v six bonding domains XeF O geometry b. of 4. domain following bonding and non-bonding domains: a. with 3. the i n 2. electron with 6 y the species r e Identify SF BrF 5 90° t i s Activity 1. planar 3 n Summary of molecular geometries with ve and square 3 I l Table 15 2 y 4 ClF y 0 5 180° P 6 <90° e 6 2 2 r octahedral 3 f. SOF g. SiF here) and molecular geometry of each 4 2 6 + h. IF 4 j. SCl 4 angles in molecules a to j above. Linking question How does position in the ability the periodic of some atoms table? to expand (Structure their octet relate to their 3.1) 171 2 Models of bonding and structure LHA Structure Formal charge (Structure 2.2.14) Formal charge in the at all. Lewis looking at the charge an were shared equally, We c an formula the assign when electrons charge formal several an = have its to of if all the bonding non-bonding atoms possible. has number if charge are atom would and in Formal a electrons electrons molecule charge is to were not select the c alculated electrons before bonding – number of electrons aer bonding = VE – (NBE + r FC ½BE) VE = number of number valence of electrons non-bonding of sum of overall the bonding formal charge of , all atoms Cl = 5 of the follows: (bec ause nitrogen nitrogen is in atom c an be f x t a VE charge as The u 6 (bec ause FC(N) Therefore, the Lewis which electrons) 5 the formula (2 + 0.5 formal E Lewis = in bonds, is zero. × formula are the 6) charge = of BE nitrogen = atom equivalent to six 7 = atom or ion Lewis Lewis formula should be equal to c an e ach (bec ause 6 has 2 of in the one in the bond, to determine the atom: is in group 17) Lewis non-bonding (bec ause has repe ated chlorine (bec ause six be chlorine formula each chlorine electrons) Lewis formula each chlorine therefore two bonding electrons) FC(Cl) 0 the nitrogen atom in this Therefore, this The Lewis sum overall 172 = NBE a bonding three v O has l = process charge VE group 15) atom has two non-bonding electrons) atom same formal NBE = 2 (because in the Lewis formula the nitrogen BE molecule the the o formal c alculated r o The i d Solution C N in in n U Cl atom is zero by c alculating the formal charges of the atoms in the 3 Cl a atom molecule or ion. is shown below. 3 in the p the overall charge of nitrogen trichloride, NCl The Lewis formula of NCl of to the o i n Show that molecule. charges that assigned v Worked example 6 electrons to y the of r e The number before bonding assigned O = atom electrons formula BE the t i s = charge n formal l = y FC = 7 the (6 formula of the charge + 0.5 formal is 2) = of 0 e ach chlorine atom in zero. formal of × charge the charges is molecule. zero, which is also the y P Where: NBE by before and aer bonding: e Formal atom s s shared best is molecule Structure 1. formal 2. a 3. negative charges dierence When these conditions in close to formal formal presented are generally favourable: zero charges charge(s) with covalent model LHA following The in the assigned alternative Lewis molecule to the close more formulas, to zero electronegative we look for the s s The 2.2 atom(s). one that satises requirements most. 2– Assign formal charges to the atoms in the sulfate ion, SO , e r Worked example 7 to deduce its Lewis formula. Solution l 2– We start by drawing the Lewis formula of SO in a We way know that sulfur is prone the octet octets, rule: so the FC(O) = 6 + 0.5 × 8) Lewis the sulfur atoms in the this FC(S) = FC(O (6 + 0.5 × 2) = –1 (all the oxygen atoms formal Lewis formula: = +2 S O y (0 this of 2 6 (0 singly p 6 on charges charges formula o = based formal C FC(S) atoms the v oxygen c alculate i n and c an proposed: O C alculating we expanded is O O r e O forming formula O O to Lewis t i s 2 O Now following n follows y 4 that y P 4 + 0.5 × bonded) U doubly more electronegative the sulfur and oxygen 12) = 0 = FC(O of gives: bonded) 6 = (6 6 + (4 0.5 + × 0.5 2) × = –1 4) = 0 are equivalent in this Lewis formula and therefore each The and conditions the are in this Lewis dierence not formula between O 2 2 OH E c. these draw charge the i CO 3: same structure. second Lewis a of range Therefore, In both which as 1: in these the equal to Lewis sums of to are v alues more formal of the negative suggested zero formula the charge the charge closer conditions the has previously formal are second the the the formula c ases, is that However, oxygen is and in this have favourable. preferred. charges are poly atomic 2, ion. u H b. formal to v a. the need a will l You is close t a r o f x O Determine them charge, not favourable. Practice questions 47 . are o zero, charges d to formal 1) n have the same formal charge of The of Lewis every atom formulas in each of the molecules below. rst. d. NO e. SF 3 6 f. BH 3 173 2 Models of bonding and structure LHA Structure Practice questions 48. Based on formal charge considerations, determine which Lewis formula is preferred: a. are in the the dierent assumptions c alculation of r made b. formal 49. in a of oxidation species? states the below suggest formal charge of each atom in the XeO (Structure for and an alternative Lewis formula in molecule P atoms and which 3.1, charges are all zero. Reactivity 3.2) the shown formal l t i s n y O Xe Draw two alternative of the Lewis two formulas is for ClO . Determine, using formal 3 y which r e charge, O 50. favourable. p o v Sigma bonds () and pi bonds (π) (Structure 2.2.15 ) this attraction will on delve that overlap with that i with that of a the double This is nuclei into are other, bond covalent and how electrons each bond dened atoms o know average l u than double a In a the the triple other bonding shared of pair electrons shared using allowing the the electrostatic electrons. In this section, occupying atomic orbitals valence bond theory. This is the electrons to pair their increases with greater bond spins order. Consider the 1 for a single C C bond (346 kJ mol ) and contrast it 1 double C=C bond is the bond (614 kJ mol signic antly higher but two bonds. This bonds in the is why the ). The double bond pi bond (π). strength average not double of Pi the are bond enthalpy for that of the single bond. not bonds double the are same: one is a generally bond is not weaker exactly strength of the single bond. c arbon–c arbon two are pi σ bond, bonds one (gure of the three bonds is a sigma bond and 78). σ E π bond. as of form bonds when orbitals in neighbouring atoms strength enthalpy bec ause sigma a pairs sigma bond () and the other is a v O f x t a r o d We valence Figure 78 174 deeper neighbouring idea we two n U we topic, between C in i n E arlier σ π π Single bonds are sigma bonds. Double bonds contain one sigma and one pi Triple bonds contain one sigma and two pi bonds y Determine 3 charge e What s s Linking question Structure 2.2 The covalent model LHA Sigma bonds () Imagine two unpaired they do so, is an the this axis to form a along of pair each Eventually up forming other. their a 1s They each orbitals will have one overlap. When covalent bond. s between the nuclei of the two bonding atoms. This line or internuclear axis. The sigma the orbitals bond. bond of This axis. means Figure 79 two that 1s orbitals there shows that is a will s s p region of high sigma bonds c an form dierent types. Figure 79 bonds are always sigma bonds. p Sigma bonds are formed y P Single p overlap r pairs line approaching orbital. will bond axis density between 1s e electron atoms the electrons invisible known as the along in s s Imagine hydrogen electron, when orbitals overlap along the bond axis The shapes of discussed in Pi bonds get lobes above other name from the Greek letter π, and things in are named aer things bonds, the the in two bond electron neighbouring axis. density This is type pi bond two lobes contains two at opposite electrons, sigma Therefore, count double the bond bonds there pi and (gure80). of the bond overlap above and below bond axis axis, as shown in Figure 80 Pi bonds are formed when occupy both lobes. porbitals overlap above and below the bond axis C=CH 2 . 2 in the H structural formula. Remember that single each of the multiple bonds contains one sigma are 7 sigma bonds. bonds. contains sideways, C H bonds bond u the sigma π p C a v Then, E bond. are l f x O count bond pi bonds in the molecule (NC) The structural formula is given below. bonds sides which i Determine the number of sigma and First, overlap pi t a r o Worked example 8 Solution a o A c an forms n bond consists of 80. d gure they overlap p concentrated at opposite sides of the bondaxis. One pi atoms, of o present C pi are below U In and resemble? v orbitals they i n p above corresponds to the below the axis. science Pi bonds (π) If which are you look along the bond axis, the pi bond looks like a p orbital, p What their When orbitals y with p. r e letter p O came from in the Latin alphabet. When you look directly along the bond axis, the sigma bond looks spherical, like an s orbital. and Structure 1.3. t i s Sigma bonds get their name from the Greek letter , which is where the letter s s n l y Thinking skills ATL E ach one pi triple bond. bond There contains two pi bonds. E ach are 5 pi bonds in total. 175 2 Models of bonding and structure LHA Structure Worked example 9 Determine the number of sigma and pi bonds in the phosphate anion, 3– PO . The Lewis formula is given below. 4 s s 3 O r O Practice questions and contrast sigma bonds () and pi Solution bonds (π). why two orbitals c annot overlapping s y Explain l The three single bonds and one of the bonds in the double bond are all sigma 52. the double bond. O t i s Activity n bonds, giving a total of 4 sigma bonds. There is one pi bond: the second bond in form a pi bond. r e number of sigma and pi ethanoic acid, CH bonds. p a. COOH 3 v propyne, CH c. ethanenitrile, CH CCH d. c arbonate e. ammonium ion, NH o b. 3 CN 3 i n phosphorus such be being phenomenon example, explained the explain on M any ion, HCOO • Valence sigma energies bond bond theory describes framework the strong, hexagonal benzene. on a • Molecular orbital theory shows that deloc alization more gives benzene a distinct energetic advantage. orbital theory. This are spread best in across the Molecular entire orbital mechanics. specic bonds. we use the C an theory you other draws think sciences of to heavily any other enhance from quantum examples our where understanding described using both of chemistry? aromatic stability of benzene by two models: E Hybridization (Structure 2.2.16) By now you covalent from should bonds. tetrahedral one If know these structure another. that around For c arbon is bonds are the example, tetravalent, single c arbon bonds, atom at methane, CH , 4 176 of discussions based loc alized is 3 molecular electron u c an For molecular electrons than are a theories. a that to l Sometimes magnetism. v O rather helps information chemists theory, assumes molecule as amongst sophistic ated theory theory give PCl o properties bonding to f x of bond fails methanoate i or it trichloride, 3 t a valence geometry, r o While + 4 azide ion, N d Theories 3 n h. U g. CO C f. 2– ion, y Look up the Lewis formula of each of the following molecules. Determine their meaning they are it approximately has a c an usually form four arranged in a 109.5° angles tetrahedral structure. y Compare P 51. e O Structure 2 the ground state electron conguration of c arbon, 1s 2 these (gure and observations. C arbon contains two unpaired 2p 2p , c an deduce therefore should form two bonds (one involving each not four. Furthermore, the two occupied 2p orbitals are at of an electron element from its unpaired atomic electron), the electrons conguration 81a) covalent model 2 2s You contradicts The LHA However, 2.2 90° number. This is covered in from Structure 1.3. one another, changes Hybridization c arbon for when is the There An means that the atomic orbitals must undergo form bonds. of mixing example of atomic this orbitals process is to form shown in new hybrid gure 81 for the are two steps: 2s electron is r A they concept bonding. atom. This promoted to one of the 2p orbitals. 2. Hybridization: The singly occupied 2s and 2p atomic orbitals are hybridized, meaning they combine and give rise to orbitals of new shapes. The resulting l orbitals are called hybrid orbitals and they all have the same energy. (c) 2s 2p state electron conguration of c arbon (b) One of the 2s electrons is promoted to a 2p orbital The four atomic orbitals are hybridized to form four sp Promotion the repulsion is energy does it energetic ally released not by require experiences favourable. the much when energy of combined in orbitals therefore has s character c an 2s electron hybrid 2s and of orbitals one 75% p part character are 2s three 2p) are produced. and three E ach of parts 2p, (gure 82). H central 109.5° + + + carbon atom 3 sp 3 sp H 3 atomic orbitals combine to form four equivalent sp H H 1s atomic orbitals orbitals angles. therefore three 2p v E hybrid a One 2s and bond and + (one 3 sp the equal to the number of atomic orbitals orbitals u + is composed l O four sp atomic mixture 2p 3 and a 25% hybrid orbitals 109.5° is 2s Figure 82 orbitals four equivalent f x + The four If t a and c arbon, hybrid hybrid them. r o these resulting make i to formation outweighs relieves o number combined it n hybrid orbitals d The promotion step absorbs bond bec ause paired. 3 sp The subsequent U of the o this. hybridization but C energy, i n O verall, hybrid orbitals v of identic al energy p 3 (c) y (a) The ground hybrid orbitals r e Figure 81 sp 2p O 3 2s n y t i s (b) (a) y Promotion: P 1. e orbitals 109.5°. s s certain not In arrange c arbon, themselves these orbitals tetrahedrally, are each leading to the occupied by one electron 3 Figure 83 form four sigma bonds. For example, in methane, CH , each 4 In methane, each c arbon sp orbital overlaps with a hydrogen 1s orbital. 3 of the sp forming hybrid four orbitals covalent overlaps bonds with a (gure 83). 1s atomic orbital on a hydrogen atom, The geometry around the c arbon atom is tetrahedral 177 Structure 2 Models of bonding and structure LHA 2 sp hybrid orbitals 2 p Carbon can also become sp z hybridized. The combination of one 2s and two 2p 2 atomic orbitals produces three sp 2 hybrid orbitals. These sp hybrid orbitals arrange themselves in a trigonal planar fashion, at 120° from one another (gure 84). 2 sp The hybridized remaining orbital contains unhybridized p C one electron, orbital c an so then they go on c an to s s E ach 2 form sigma bonds. form a pi bond with a z sp parallel p orbital on a dierent atom (gure 85). e z r 2 sp 2 2 sp sp pi 2 2 Figure 84 E ach of the three sp hybrid 2 sp sp H l orbitals (blue) contains one electron. The y C atomic orbital (white) also C z contains one electron t i s n unhybridized p sigma 2 sp p H 2 orbitals (white) overlap z side-by-side and v sp hybrid of one orbitals. 2s U them and They C combination one adopt 2p a hybrid orbitals form a pi bond, The remaining resulting in a o i n sp hybrid orbitals two 2 two hydrogen atoms. c arbon–c arbon double bond The z the two c arbon atoms each have three sp p unhybridized p , p 4 sigma bonds with each other and H 2 sp y In ethene, C (blue) which form r e Figure 85 z O H atomic linear orbital leads arrangement, to with the formation of 180° between (gure 86). n The hybrid orbitals can form sigma bonds. The remaining unhybridized p sp C sp E ach of the two sp hybrid orbitals (blue) contains one electron. The two unhybridized p and p y atomic orbitals z l u f x Figure 86 z pi t a p (gure 87). i y d r o p sp sp H H sigma (white) also contain one electron each a E v O 178 and y orbitals can form two pi bonds with parallel p orbitals on a neighbouring atom z o p p pi p Figure 87 In ethyne, C p , the one sp–sp overlap and z the two 1s–sp overlaps form three 2 The four unhybridized c arbon triple bond y z H 2 sigma bonds. p y p orbitals form two pi bonds, resulting in a c arbon– y P H Structure 2.2 The covalent model LHA Hybridization in other atoms Hybridization in the p also orbitals occurs will in dier, atoms but the other than general c arbon. principles The are number the same. of electrons For example, 3 consider the oxygen atom in water, H O, which is sp hybridized: 2s 2 • The • We oxygen c an distinguish contains 2 1s 2 2s ground 2 a of the electrons electron three and conguration is 1s 2p the orbitals other to two 2s 2p . show that one of are singly occupied: 1 2p 2p y z 3 The 2s and 2p orbitals hybridize to form four equivalent sp 3 hybrid orbitals sp six electrons in total (gure 88). Two of the sp hybrid orbitals Figure 88 contain a pair of electrons each and therefore do not form bonds: they hybrid orbitals The 2s and 2p atomic orbitals 3 the two lone pairs on oxygen that you to form four sp are familiar with. The y hybrid orbitals close the to oxygen atom orbitals However, has the on two lone bond are are two is occupied tetrahedrally of pairs. angle singly and go on the This not hybrid arranged, orbitals corresponds exactly 109.5° are hybridization is close to to the bent but bond but not molecular exactly sp diagrams described to for show the the above. and number its shape. remember This have a as is from single large the eect pi E v a O 2 sp by to an the of changes nitrogen involve sp by the atomic to hydrogen to 3 electron domain atom of a when is equal to the determining molecular molecule. Number Number of Electron domain of electron non-bonding Molecular geometry geometry domains 4 4 domains tetrahedral 0 1 2 Hybridization, bonds form ammonia, NH 3 trigonal planar 2 electron domain geometry and linear tetrahedral trigonal pyramidal 2 bent 0 trigonal planar 1 Table 16 it hybridized orbitals, so they do orbitals 3 undergone when VSEPR that double and triple bonds domains u l of hybrid 3 not geometry Number Hybridization sp the do linked formed of electron bonds on closely discussion additional bec ause f x not electron domains. are orbitals i will treated hybrid t a are bonding of r o You of sigma number Describe orbitals o The d geometry. 54. n U Hybridization and geometry Hybridization promotion oxygen atom in C water, process geometry. p box are o orbital hybridization i n and v Draw angles rather 104.5° suggests that Practice questions 53. form sigma already full, so the 3 the to hydrogen atoms. y that orbitals orbitals hybrid 109.5°. fact s r e The four hybrid the O Since two with t i s bonds n remaining l in oxygen are hybridized constitute y 3 containing P • r x between pair 1 2p state 4 e them atom’ s 2 s s 2p 2 0 bent linear molecular geometry 179 LHA Structure 2 Models of bonding and structure Practice questions 55. State the 56. hybridization of: Consider the following species: CF , HCN, N 4 an atom with b. the c arbon c. the oxygen d. the nitrogen e. an tetrahedral For electron domain each species, geometry atom in ethene, C H 2 atom in oxygen of the c arbon deduce of the a. number b. electron c. hybridization. and H 2 following: electron domains 4 diuoride, OF domain geometry 2 in molecular nitrogen, N 2 atom with trigonal pyramidal 57. molecular The Lewis domains, f. an atom with bent molecular formula Deduce electron domain for atoms A, B and C. H H O H O H H B C r e Hybridization and deloc alization Consider the ethanoate ion, CH COO . It forms when 3 hydrogen H bond orders bond become change in gure from the 2 c arbon–oxygen and 1 across to 1.5 the 89. In the C=O bond of ethanoic acid, CH O 2 0.128 nm sp hybridized. U bond H C 3 C unhybridized the them. –OH The When ethanoic acid loses a hydrogen ion, atom orbitals shown in the c arbon loses in the double domains. and a the This oxygen is orbital electron in each atom are both forms the pi figure is oxygen atom, in –OH, has sp lost to have thre e form As a ele ctron ethanoate, result, the domains the remaining oxygen around hybridization. atoms them. oxygen and Thre e the ele ctron 2 to sp overlap hy bridization allowing the p and one unhy bridize d ele ctrons in them to 2p be orbital deloc alize d 90. ethanoate ion shown as O E v H C CH C 3 3 — C — O O Figure 90 180 COOH, acid change: O a O of the ethanoate ion. 2p hy bridization. correspond The i structures as electrons c arbon–oxygen u resonance l two e ach. t a r o f x Activity domains sp e ach the C–O bonds become equivalent the c arbon an o d Figure 89 adopts other hydrogen 2 atom The ethanoic lengths 3 between When O bond 3 The n 0.128 nm e ach. two C (b) so, deloc alized i n shown doing o O By v 0.133 nm ion. p C C 3 y O 0.123 nm Draw H O t i s H electron molecular n C A (a) and shown of l y H H geometry here). is number The deloc alization of electrons due to resonance in the ethanoate anion y possible propanoate P are methyl hybridization, geometry (two geometry answers of the r below. geometry e atom . 2 nitrogen atoms in these s s a. Structure 2.2 The covalent model End-of-topic questions 7 . The hydronium ion is formed when a water molecule, s s Topic review + H O, reacts with a hydrogen ion, H . Which type of 2 1. Using your answer knowledge the guiding from the question as Structure 2.2 fully as topic, bond is formed in this reaction? possible: coordination bond B. ionic bond C. hydrogen bond e A. What determines the covalent nature and properties of a Compare and contrast ionic bonding, covalent bonding D. intermolecularforces. and contrast molecular and 8. covalent network intermolecular The force electronegativities, Multiple-choice questions Which of the following species is molecular? A. 2 NO 4 N 4 CH 3 ) ) CO What is the electron domain Cl–P–Cl trichloride, Electron domain A. 3.4 3.2 C–O Which of the 10. following allotropes A. graphite B. graphene C. buckminsterfullerene D. diamond What are the intermolecular of forces c arbon is molecules of CH present between Cl–P–Cl bond angle / ° F? 3 molecule A. London B. London (dispersion) forces and dipole–dipole forces C. London and 109.5 109.5 D. (dispersion) (dispersion) forces forces, dipole–dipole forces hydrogen bonding hydrogen bonding pyramidal 11. trigonal What are the formal charges on P and O in the Lewis 100.3 formula of the phosphate oxoanion? LHA trigonal pyramidal 2.6 molecular the geometry trigonal a D. tetrahedral the for Molecular tetrahedral tetrahedral v O C. ? 3 l B. tetrahedral PCl angle u f x geometry geometry, bond H–Cl D. molecular? t a phosphorus the i and r o geometry of CO 2 o 6. O d D. OH 2 3 9. C. c arbon–oxygen bond? n (CH shortest 2 3 C. the U (CH has O–H C molecule CH B. i n A. 3 O 2 Which Cl C–H B. o D. 5. O p NH v KBr C. are: C Which bond is the most polar? O B. elements y Na 2.2 r e A. four O t i s Exam-style questions 4. H χ of n Element χ, y substances. y 3. and l Compare P 2. r substance? pyramidal trigonal 100.3 A. P is 3 1 and O is 0 E O pyramidal B. P is +5 and O is C. P is 0 and O is 0 and D. Both are 2 1 O 3 O 181 Structure LHA 12. 2 What Models is the of bonding electron and structure domain geometry of the sulte 17 . What is the hybridization of the oxygen atom in ethanol, of the c arbon atom in 2 oxoanion, [SO ] ? CH 3 CH 3 OH? 2 A. trigonal planar A. sp B. trigonal B. sp C. tetrahedral C. sp D. bent D. It 2 pyramidal s s 3 What is the molecular geometry of BrF 18. ? What is not is the hybridized. hybridization hydrogen C. T-shaped D. square A. sp B. sp C. sp D. It 2 is the molecular geometry of [PF ] is not ? 6 19. A. How many trigonal planar molecule octahedral following molecules is non-polar? 4 B. ClF C. BrCl 3 U D. SeF 20. 21. atomic orbitals results in a sigma bond? I and II only I and III only I, II and III pi 3 1 2 2 3 1 Describe the meaning of the term covalent bond. [2] + The phosphonium ion, PH , is formed when a 4 + hydrogen ion, H PH . For each , reacts with a molecule of phosphine, species, phosphine and phosphonium: a. Draw the Lewis formula. Identify any coordination bonds [3] Deduce c. Suggest d. Deduce the molecular a v O b. electron domain geometry and geometry. the bond [2] angle. [2] + I. Explain whether your it is polar or non-polar. reasoning. [2] + II. E III. 22. Methane, CH , ammonia, NH 4 have tetrahedral angles of , and water, H 3 electron 109.5°, 107° domain and O, all 2 geometry with bond 104.5°, respectively. + Explain 23. these Diamond Describe these 182 present in a 3 u II and III only D. l f x C. shown o of are Extended-response questions i B. le combinations 2 D. t a A. the r o below of d Which B. C. 6 16. 1 n 5 sigma A. C SF bonds cyanide, HCN? o the v A. of pyramidal pi y square D. Which bipyramidal and hydrogen p C. sigma of r e trigonal i n 15. B. hybridized. O What t i s 14. l y 3 pyramidal and and two dierences graphite explain materials. in are the bond angle allotropes of values. [2] c arbon. electric al conductivity of [3] y square planar n octahedral B. cyanide, HCN? P A. r 5 e 13. (V-shaped) Structure 24. Household acid, CH vinegar is made from aqueous ethanoic 25. The following substances all have 2.2 The similar covalent model molar masses. COOH. 3 i ethanoic acid, CH ii propan-1-ol, CH COOH iii methoxyethane, CH iv butane, CH 3 a. Draw the Lewis formula of ethanoic acid. [2] CH 3 b. Draw a diagram to represent the intermolecular CH 2 3 between a and a water molecule. CH acid molecules when c an gaseous form or Use your and structure solvent. acid Dimerization molecules c an as of your Explain why, molar 26. Explain 27. Water O why increases is mass of an ethanoic dimer. [1] a. the down an Oxygen ethanoic acid is more likely to of of the solvent. in aquatic [1] group 17 elements [2] water bodies supports the y why point strength of have similar molar group. dissolved r e Suggest the excellent presence ii. boiling that O the H 3 the forces, it is helpful to n CH C [4] l y acid comparing substances t i s C alculate when intermolecular masses. i. reasoning. O C forces four substances in y H O intermolecular increasing boiling point. Explain compare C of these P b. 3 list shown dierent H to means that two associate below. O knowledge dissolved in a r ethanoic 3 dimers, order non-polar 3 CH 2 e particularly CH 2 [2] a. Ethanoic 2 molecule of ethanoic acid 3 c. OCH s s interaction OH 2 CH organisms. Describe the type of intermolecular forces that occur between oxygen form dimers when dissolved in non-polar hexane) than in polar as water). COOH, [2] dissolves readily in 3 water. The CH 3 Explain CH 2 this solubility CH 2 dierence in low. in ethanoic c arbon and acid. a. [2] b. as a of and higher dissolve Explain dissolve group shown Identify has in cross from brain barrier two of in water but not strong substances morphine solvents. [2] painkillers known as opiates. produces diamorphine below: explain which of these two opiates aqueous solubility. [3] the is blood Identify opiates into the composed and dissolves of brain. The blood– lipids which are explain which of these more readily in non-polar environments. CH [1] 2 O CH 3 a CH [1] why ionic compounds hydroc arbon are OH N O. The potency of opiates depends on their ability to non-polar. u l 2 the a molecules, H 2 diamorphine esteric ation (heroin), t a r o CH of [1] n hybridization atoms f x E v O C acid. to water oen solvents. readily and belonging The i oxygen the ethanoic not Morphine [2] and compounds organic do 28. o Deduce of in hexanoic acid, very solubility. d f. 3 is State the number of pi and sigma bonds in a molecule H of COOH, 2 U LHA e. aqueous CH 2 Ionic C CH b. , 2 o Ethanoic acid, CH v as (such i n d. (such solvents p molecules, O solvents C H 2 3 CH 2 O O 2H O C CH 3 C C O O 3 O 2CH 3 COOH O C OH morphine O CH 3 diamorphine (heroin) 183 Structure A Models paper of bonding and chromatogram obtained using a structure for a non-polar mixture of amino acids is 31. The c arbonate ion contains deloc alized LHA 29. 2 electrons. solvent. a. Draw the three resonance structures of the 2 a. C alculate the retardation factor, R , for the spot c arbonate ion, CO F X in the Deduce which chromatogram. b. b. to have the of the lowest [2] [2] three components is likely polarity. State the bond in bond the order of c arbonate the c arbon–oxygen s s labelled . 3 ion. [1] [1] 32. Benzene, C H , contains deloc alized electrons and is 6 represented as e 6 oen follows. r X l a. Explain, with reference Y deloc alized Z b. and Outline The diagram a the a chromatogram of sauce food chromatogram. of (labelled colourings Outline an S). have extract further Four also three conclusions 33. c an be derived from the information in the chromatogram. Sulfur leading a. n U a S E122 E124 o i t a u l v E 184 d r o f x O E110 to Draw the front the molecule. [2] of the , is an [1] chemic al evidence that structure of benzene shown [2] acidic gas. If present in the 3 b. E102 trioxide, SO atmosphere, [3] solvent supports an acid two octet Lewis Deduce, which of Explain c an dissolve in rainwater formulas for SO , one that follows 3 rule, Determine in trioxide rain. expanded atoms c. sulfur C i n that piece above. been of why the form a ring benzene. one o on samples is curry v reference run below supermarket of Describe p from plane y r e 30. the hybridization, benzene how bond length data conrms the structure c. below to in O t i s above 1.70 cm 0.75 cm electrons n y 4.10 cm and your formal Lewis using the where the sulfur atom has octet. the the one the two [3] charges formulas concept Lewis of you of formal formulas reasoning. each of the drew is in (a). [2] charge, preferred. [1] y P 4.90 cm 5.40 cm Structure LHA 34. Ethene, C H 2 , belongs to a group of substances known 36. The condensed structural formula 2.2 of The covalent model phenylamine is 4 as the alkenes. C H 6 NH 5 . A molecular model of phenylamine is shown 2 below. a. Draw the b. Deduce Lewis formula for ethene. [1] H c. the molecular atoms Suggest in values geometry of each of the ethene. for the [1] H following bond angles in H ii. HCC r HCH N C Deduce f. Explain why around the of ethene. [1] H C of the H c arbon atoms in ethene. main type between your of in ethene. intermolecular ethene [2] force molecules. reasoning. [2] the hybridization in graphite. electric al conductivity of c arbon atoms in Using each these of the data, two explain the materials. Deduce the molecular atoms in [4] b. State the U C A domain of the geometry and c arbon and nitrogen phenylamine. nitrogen c. electron geometry o and the i n diamond of v State H H p a. 35. C C y Explain the bond restricted r e present c arbon is O Identify c arbon rotation t i s g. [1] molecular n hybridization y C the hybridization atoms theoretic al y e. l State the number of sigma and pi bonds in a P d. molecule C [2] e ethene: i. s s c arbon in [2] of the c arbon and phenylamine. study of the [2] electron structure of phenylamine suggests that the H–N–H bond n angle is in very o i u l you to may geometry the should conclude around be experimental the about 112.79°, which value. the Discuss molecular nitrogen in the –NH group 2 in the structure hybridization 37 . C arbon bonds c arbon with atoms with atoms each are other. phenylamine, state c an other of of form nitrogen single, c arbon to on deduce the this double atoms. unlikely and or Suggest basis. triple [3] covalent why two form quadruple bonds [2] a E v O f x t a r o d what phenylamine close 185 The metallic model Structure 2.3 s s What determines the metallic nature and properties of an element? A large proportion elements, millennia. into very ability wires, useful of and and E arth’ s humans metals to conduct versatile. crust have be is The composed of been using metals metals moulded into shapes, electricity These and are a are Metal result c an many found as elementary substances in are found in ores, in their oxidized reduced be and state. posing treated is the c ations electrostatic and provided safely Structure 2.3.3 deloc alized require and — that waste many metals is collected Transition elements have deloc alized o C i n v on the charge of the ions and the radius of the metal ion. However, metallic correctly. p Structure 2.3.2 — The strength of a metallic bond depends large amounts of risks. d-electrons. electrons. reactivity of their molten Metallic structures (Structure 2.3.1) 100 described electrons. related to 6 ,0 0 0 metal y e a rs artefacts a v E 186 be oldest a Jordan o the example, known elements of electron groups. the they Therefore, of metals. between deloc alization which increase properties are attraction electronegativity, trend: i of the For 118 electrostatic degree down opposite t a estimated O k n ow n is one aw l , The the gradually the the their decreases u It and in l old, copper discovere d d Va l l e y. Ancient tool, r o piercing f x Figure 1 to of as n U Nearly period 3 in generally metallic down Bonding metal pure and elements, from metallic to non-metallic (table 1). and of is c an be deloc alized inversely across periods and elements decrease from metals elements increases properties groups in c ations demonstrate across periods. sodium to argon, change LHA bond of achieved with the Higher n lattice is O metallic ore y a processes t i s A between its furnace. electrolysis r e attraction by from blast environmental recycled, Understandings Structure 2.3.1 — a y but in y Some nature, iron(II) l energy, are of c arbon extraction of metallic bonding. metals of compounds. heat makes properties reduction addition for P them the r drawn The of and e metallic Structure 2.3 The metallic model Na Mg Al Si P S Cl Ar metal metal metal metalloid non-metal non-metal non-metal non-metal metallic metallic metallic covalent molecular molecular molecular monatomic network covalent covalent covalent semi- low low low 2.2 2.6 3.2 Metal, metalloid, or non-metal? Structure high high high conductivity 3 1.6 to the notes to way a properties eort their peer ’s to particles and is a connect see if central each of behave. there any are non-metallic highly are shown values electronegative somewhere in the properties. deloc alized. rather move negatively of results electrons lattice electrons from the of the greater greater metallic covalent character character metallic covalent bonding bonding electrons c ations. This creates a 1.0 2.0 3.0 4.0 mean electronegativity between metal c ations (gure 3). + p Figure 2 bonding The metallic–cov alent continuum + + + - + + + + + + + + + + + an form c ations. + s h ow i n g that have both attached to individual ions, attraction them On the le, form metallic valence packed t a metal not their surrounds around u a are closely that + a of deloc alized atoms, electrostatic electrons l Structure of the + v “s e a ” of deloc alized E a “sea” f x O p Figure 3 by within metallic 2. to they o bonding sea Deloc alized tend non-metals i the charged other r o and freely to d Metallic bond gure compare dierences. n become atoms U metallic nished, and/or middle: read, C The metallic bonding model When in that you o and there Metalloids have As characteristics of y right, is electronegativity you similarities i n metallic the bonds. low chapter. p On covalent with this v bonds. elements in observable aects between r e are behaviour connection idea the When are metallic–covalent bonding continuum there microscopic The y your physic al how materials. n metals and conscious about and l a thinking matter O make of t i s structure involves features N/A P oen observable but 1.9 elements e 1.3 period r of y Chemistry The 0.9 Properties Thinking skills ATL low conductor Electronegativity p Table 1 s s Electric al a r ray of positive ions (c a t i o n s) surrounde d ele ctrons 187 Structure 2 Models of bonding and structure + – The properties of metallic structures + − − The presence of deloc alized electrons gives rise to many characteristic properties − of + + metals. Metals are good electric al conductors, thermal conductors, and they + are malleable and ductile. − + + Substances + − Consider − a c an conduct sample of electricity metal. It when contains a mobile lattice charged of c ations particles are surrounded s s − − present. by − free-moving, applied, donated deloc alized the negative the difference ele ctrons terminal positive are randomly applied, and and negatively throughout there towards is the a net charged the electrons. metallic movement positive structure. of terminal The deloc alized When electrons away electrons potential from the move dierence is negative terminal (gure 4). terminal objects. In some They Mirrors have have metal types of and to knowledge some appeared a most glass might in why they are used to make believed as to have magic al valuable, status-signalling literature, both as instruments of clarity and we know the household surface. approach knowledge were perceived is sparked human fascination more These the deloc alized mirrors are are electrons in metals made examples by applying a of how dierent same topic. open to interpretation than others? n U C i n o Are of light, mirrors been Nowadays, v areas silver also which y of cultures, well, historic ally p layer incident light have r e misrepresentation. reect reect n curiosity. properties. and surfaces O and lustrous Reective l are mirrors. t i s Metals y TOK o i d t a r o u l f x a E v O p Figure 5 p e o p l e, 188 a Anish buildings K a p o o r ’s and sky Cloud around G ate it on in its Millennium curved, Park, s t a i n l e ss C h i c a g o, steel refle cts surface the y t ow a r d s potential P move the by a r is When e p Figure 4 Structure 2.3 The metallic model One of the reasons conductivity, substance is heated, become is its particles more a vibrations vibrations passed vibrations and movement this metals collisions each which collision. increases reduces path. In this in electrons the other lattice ions. collisions the in energy to the metals parts of vibrate Some are temperature frequency c ase, transfer electrons to magnitude. of are the lattice. heat. the and These with of in bec ause of their the energy is c ause of (gure collisions, resistance more, so kinetic 6). electric al Conversely, so the electrons decreases. Figure 6 400 Re s i s t a n c e line arly t e m p e ra t u r e 300 Ω / e c n at s i s e r 300 600 900 temperature / K without wide to array this ability This of of oer very no low resistance a solid useful shapes, to be property ranging anchors. to temperatures property. pressed is the l u a zippers the breaking. f x such is exhibit that At t a M alleability materials i materials are temperature. r o other “critic al” from o certain d Superconductors or – + – – + + + + + + + + + + + + + + – push – current – – below many metals and some + – pounded reason sewing electric – – 1200 n 0 C U –100 o v i n 100 with most p metals 200 0 a for y r e inc re ases O direct ions also the energy conductors increase c ations c ations Since kinetic increases, between temperature more vibrating good c ations other n with its to collisions. increased are of along y in The through metals metals, in their energy of a When a substance y a thermal t i s in pass heat in the easily arrangement. why to decrease move are energy resistance a heated, c an more converted so The particles. l are energy, well. its e thermal there of P As explains kinetic cookware is their thermal heat kinetic energy gain electrons they make r This is packed surrounding to conduct s s metal used they vigorous. These mobile, are that related to the When closely metals meaning into – dierent shapes electron ‘sea’ why metallic needles to objects have aeroplanes to – – + + + – + – + a O Metallic bonding is non-directional. When a force is applied to a metallic structure, + + + + the layers of cations can slide past each other without breaking the electrostatic – – – attraction to the surrounding delocalized electrons. The cations can therefore v + + + + + + be rearranged, allowing the metal to take on a new shape (gure 7). Aluminium foil, oen used to wrap food, demonstrates that the metal aluminium is readily E malleable. A related Ductility that property, and allows + – ductility, electric al electric al is the ability conductivity wires to be are a of a solid very produced at to useful the be stretched combination industrial sc ale. into of wires. properties p Figure 7 Metallic intact even into sheet a after or a bonding metal other – is remains hammere d shape 189 Structure 2 Models of bonding and structure To summarize, metals are typic ally: • good electrical conductors because they contain mobile delocalized electrons • good and malleable without conductors packed and bec ause they contain mobile deloc alized electrons c ations ductile, bec ause the layers of c ations c an slide past s s • thermal closely each other breaking the metallic bonds. e r l blast furnace p Figure 9 Aluminium aluminium c ans retains C i n o metals, n a O f rom y metal p iron its in a re cycling properties plant. after Like being other re cycle d Database investigation of the properties of metals this skills task, you will Relevant skills Inquiry 2: Identify • Inquiry 1: and Formulate 2: a Collect database extract o Tool • i d data to explore the properties of metals. from databases research questions and record sufficient relevant quantitative data Instructions Using that u l a E v O f x t a r o • use n U In This at will least may an you involve conductivity are one allow AHL of database to looking a at number student, of your investigate you choice, the trend electric al of metals may wish extract to the and analyse data properties of metals. conductivity, melting point or thermal across to identify, relating a period explore a or down property of a group. the (If you transition elements.) State a research question that is addressed by your data collection and processing. Before starting this task, create a plan of the steps you think it will ATL require do the and task estimate and keep the a time log of needed the to actual complete time each step. Then, you spent completing the various steps. When you have finished, compare the two sets of steps and times. How do they compare? How can these observations help you plan extended tasks in the future? 190 y P y t i s r e Liquid v p Figure 8 Structure 2.3 The metallic model Practice questions 1. State the property (or properties) of metals that makes them suitable for manufacturing: kitchen pans b. electric c. radiators d. c ars e. articial hip joints f. guitar strings. power labelled labelled diagram diagrams electric al good thermal bonding metals occurs in metals. are: conductors conductors. O t i s Linking questions n good c. how why l malleable b. explain explain y a. to to y Draw e a 3. r Draw c ables P 2. s s a. r e What experimental data demonstrate the physical properties of metals, and trends in reactivity of metals c an be predicted from the periodic table? C i n The strength of metallic bonds (Structure 2.3.2) stronger the attractions between the U The c ations strength decreases greater ionic charge factor, point the group and boiling bonding 1 metals data a melting and A the trends c an be density used to of the with “sea” of compare the strength Consider the melting and boiling points of and potassium (table 2). Table 2 Melting data 10 point / °C point / °C 181 1342 1+ 102 98 883 1+ 138 63 759 points ionic m decrease radius deloc alized Melting and boiling point Boiling for group 1 metals –12 76 boiling greater electron increases 1+ v E potassium and the Ionic radius / the c ation increases. is metals. sodium Charge of sodium attraction point dierent lithium, lithium c ations in l O Element The charge, electrons, the u metallic ionic electrons. f x of to t a Melting related greater ionic radius, and i third deloc alized r o A with deloc alized o bond d Metallic and n stronger the metallic bond. o v (Reactivity 3.2) p What y trends in these properties, in the periodic table? (Tool 1, Inquiry 2, Structure 3.1) means electrons. a down the greater This group average weakens the as the ionic distance radius between the electrostatic forces of between them. 191 Structure 2 Models of bonding and structure Now compare sodium against two other period 3 metals, magnesium and aluminium. Table 3 Melting and boiling point Element data for period 3 Charge of Ionic radius / Melting Boiling point / °C point / °C metals –12 the c ation m s s sodium 10 102 98 883 2+ 72 650 1090 aluminium 3+ 54 660 melting and boiling The points Decrease in ionic radius: Greater number contributes one the distance attraction increases and valence the valence electron that to the between strength deloc alized electrons increase sea becomes the c ations and between them of the electrostatic electrons of deloc alized electrons per ion: three aluminium of E ach aluminium atom electrons. Sodium atoms y only this c ations r e have the and O 3. between reduces electrostatic Increase in ionic charge: attraction this magnesium are: n the this t i s 2. increasing sodium, for y electrons, of reasons deloc alized. p v Using trends in data to predict properties skills the task, you will properties examine of U • that result, e asily in cut 1. Examine of is with 2: C arry Instructions sodium sodium these trends a 2. u l a E v O 192 be f x knife c an bonding a Inquiry 3. the Look your down Briefly 5. Draw 6. Reflect of that in your second on relevant point will help how second graph group values evaluate a interpret charts and graphs and accurate data processing and you boiling point data in table 2. Plot a graph analyse the melting and boiling point group 1. predicted 4. group 1 melting points to graphs out your metals the and melting data Extrapolate other in o metal As 3: Extrapolate i soft Metallic we ak. 3: Construct Tool t a a relatively d is r o p Figure 10 Tool n • trends unknown elements. Relevant skills • the C predict o this i n In up in predict the melting and boiling points of the (rubidium the values. the to 1 chemistry C alculate validity graph of plotting graph. and of the these data data the booklet and percentage compare them to errors. extrapolation. data, that c aesium). including covers a Rb larger and Cs range this time. affects the quality y period. l 1. the P across 2519 r The e 1+ magnesium Structure 2.3 The metallic model Practice questions 4. The melting points respectively. than List of A. Be, C. C a, Which C a, of and c alcium why the are 63 °C melting and point of 842 °C, c alcium is higher Sr, in order of increasing melting point. Ba Ba, Be the following explains B. Be, Ba, D. Ba, Sr, C a, Sr C a, Be why aluminium has a higher melting point r than reasons potassium. metals Sr, potassium magnesium? The contains ionic more III. The charge deloc alized electrons per ion than magnesium 3+ radius of Al of an 2+ is smaller than that of Mg aluminium ion is greater than that of a magnesium B. II only II and III only D. I, II and III features of metallic bonding that make it possible for metals to (Structure 2.4) p The position transition table are of the elements discussed d-block in the in LHA o C (Structure 2.3.3) v i n Bonding in transition elements O the y are alloys? r e form n I only C. t i s A. Linking question What l y ion. y Al II. P I. e 6. that these of three s s 5. State and periodic Structure 3.1. Transition elements are found in the d-block in the periodic table, and they have U many properties typically associated with metals: hardness, strength and high n density. They exhibit variable oxidation states and, frequently, catalytic properties. Transition elements are dened as those that have atoms, or give rise to ions, with o d incomplete d sublevels. Most of the d-block elements are transition elements, but there are some exceptions. For example, zinc is not a transition element because its i r o 2+ 3d sublevel is full, both when it is an atom and when it forms the only stable ion Zn t a Observations obtain are alludes by analysed precipitating new to the French a lead out element colourful the E arth and all The other name compounds Nicolas-Louis sample lead, aer at phenomena. chemist ore the looking natural a discovered this about v O chromium and observers, data u to l f x Scientists it parts of the forms. Vauquelin, of the universe, transition element Chromium who was observed p Figure 11 made containing chromium from Siberia. f rom palladium. Vauquelin noted the dierent colours and word for colour, c atalytic c o nv e r t e r rhodium Tra n s i t i o n metals and often h av e named c atalytic Greek C ar platinum, Aer properties due to their v ariable chromos. E ox i d a t i o n states 193 2 Models of bonding and structure LHA Structure In transition well the as the electrostatic that surrounds compared chemic al properties to the of forces them. group of outer results and in the level. attraction This 1 electrons the in d The sublevel greater between higher the become electron c ations melting and points deloc alized, as density for the strengthens electron sea transition elements group 2 metals. s s The elements, electrons of The large number of delocalized electrons in transition elements also allows for transition elements are discussed good electrical conductivity. This is because there are more delocalized electrons in Reactivity 3.4. e that can move along the metal sample when a potential dierence is applied. The inc andescent Tu n g s t e n ’s melting point high t e m p e ra t u r e stone: suitable and f rom “ tung It d e n s e. the is In Swedish for also fact, for sten” o of the presence following properties deloc alized d coloured compounds B. high density Which the Transition M any i B. oen C. elements due contain Transition u D. the element Transition of are elements C. good D. c atalytic is a result of the d d oen electric al conductivity properties incorrect? have high melting points and boiling unpaired d electrons. are coloured bec ause their ions sublevels. good deloc alized elements transition compounds incomplete elements of generally presence transition presence electric al conductors due to the electrons. have multiple oxidation states. 2+ 9. l a v O f x t a r o d points of electrons? following statements is o A. of n A. U 8. of C i n Which y p r e v Practice questions 7 . O t i s h e av y it applic ations. strong derives ex t ra o r d i n a r i l y makes n name in contains y exc e p t i o n a l l y filament bulb l high its light Explain, zinc is with not a reference to the electron congurations of Zn and Zn , why transition element. Thinking skills ATL As you know, there are various linking questions in this book that aim to Why is the E Linking question interconnect networked of metals evident that 194 explores least one other 3.1) the part across the d-block? compare (Structure content understanding of and encourage chemistry. Write a the development of a linking connection between the question content in this of your section and across a period at less course trend in melting own, points the answers. of the DP chemistry course. Share it with a partner and y tungsten. g l ow i n g P this r Figure 12 Structure 2.3 The metallic model End-of-topic questions 1. Using your question knowledge as fully as from the Structure 2.3 topic, s s Topic review answer the guiding possible: and contrast ionic, metallic Compare and contrast ionic, metallic, and r Compare 3. covalent bonding. covalent network, and molecular structures. Electrostatic deloc alized attractions a lattice of positive ions and Electrostatic attractions between a lattice of negative ions and attractions between protons Electrostatic attractions between oppositely has a metallic structure? Solubility in Electric al water conductivity when A 36 high B 186 insoluble low C 1083 high high D 1710 molten when solid low t a high o i d r o insoluble Electric al conductivity n U Melting point / °C C substance electrons. charged ions. i n Which and o Electrostatic D. v C. p protons. y between electrons. deloc alized 5. describes metallic bonding? r e B. best n A. statement O Which t i s Multiple-choice questions 4. l y Exam-style questions y P 2. e What determines the metallic nature and properties of an element? low low low high u l f x a E v O 195 Structure 2 Models of bonding and structure Extended-response questions 6. Metals have many useful properties including malleability and electric al conductivity. s s sodium, magnesium, aluminium, copper, Na Mg Al Cu 2.1 2.3 3.8 5.9 Element Electric al 7 × 10 e conductivity / –1 S m r why metals are malleable. Include a labelled answer. conductivity Describe Explain why increasing of metals decreases with [2] y data for the period 3 elements Si Identify b. Plot c. Explain d. Describe a of trend and of boiling o the 3 structure graph i t a the the 519 n U d a. 2 in C 090 each point the of vs shown P the S atomic Cl in period 2 –34 3. number. across in Chlorine, 8 445 elements [1] below. Sulfur, 4 281 point trend is Phosphorus, 265 boiling explain o v Silicon, Al i n Aluminium, 1 p point Mg the electric al Argon, Ar –186 [1] [1] period. conductivity [3] across period. [3] u elements 8. Explain of 196 conductivity [2] a 3 electric al M agnesium, l v E period electric al conductivity of temperature. Na 883 r o f x O The the below: Suggest why the electrical conductivity of copper is signicantly higher Boiling point / °C Figure 13 in [3] shown aluminium. Sodium, Boiling ▴ trend is than that of the other three elements shown in the table. 7 . Element the the and r e LHA iii. explain elements O ii. and magnesium several t i s sodium, of your n i. electric al in l The y b. diagram why metallic melting bond point strength. and boiling point are good indic ators [1] y Explain P a. From models to materials Structure 2.4 s s What role do bonding and structure have in the design of materials? So far in Structure 2, classied and metallic. use, were chemistry, civilizations Bronze developed before development according to Age, based explanations on for Another by the materials and Iron Age. Uses into for these observations of their those From metals materials to nanotechnology, sometimes are discovered and by the uses sometimes into their structures, of Polymers while molecules chance. and polymers. will help Alloys polymers very are p macromolecules, represented by a bonding triangle. monomers. C Structure 2.4.4 — between the ionic, covalent and metallic models, and can be Polymers made grouping them ceramics. Two of Structure 2.4: metals Your you are the from knowledge of understand the nature predominantly metallic covalent compounds. structure, although their large. o i n Structure 2.4.1 — Bonding is best described as a continuum and and and heavily involves polymers molecular v Understandings are draws considered in polymers. have It materials metals, bonding structure y properties research deliberate, and r e unique is alloys the applic ation, engineering. classify alloys) and of their materials. and are study and O of to c ategories structure the new types: (specic ally, properties had proposed. of physics way three is materials t i s been Age, of n properties, classied science properties y materials be M aterials c an covalent properties. characterized Stone also materials ionic, y has use: and c an how bonding: l they origin, discussed their P History M aterials have to r their we according e be are from large molecules, or repeating subunits c alled Structure 2.4.5 — Addition polymers form by the breaking bonding triangle is determined by the relative contributions of Structure 2.4.3 — non-metals. They mixtures of a metal and other have in each monomer. Condensation reaction functional with the between release of a small polymers groups in form by the each monomer molecule. i u printer bond a 3D l A v E 1 properties. double t a r o f x O p Figure enhanced o or are d metals Alloys a Structure 2.4.6 — LHA of the three bonding types to the overall bond. n U Structure 2.4.2 — The position of a compound in the cre ating a model from a polymer material, and t ra d i t i o n a l J av a n e s e gamelan metallophones 197 Structure 2 Models of bonding and structure The bonding continuum (Structure 2.4.1) strongly ionic, Some materials cannot be categorized easily into substances with metallic, ionic and e.g. CsF covalent bonding. We saw this in Structure 2.1, with compounds such as aluminium chloride, AlCl , having a hybrid of covalent and ionic bonding. We also discussed 3 s s metalloids in Structure 2.3, which exhibit covalent and metallic properties. Therefore, bonding is best described as a continuum between these models. chemists diagram to Anton represent Eduard the van Arkel continuum of and the Jan Ketelaar three constructed a triangular bonding types. The three strongly corners represent sides the “pure” metallic, ionic and covalent bonding r strongly (gure 2). The covalent, metallic of bonding example, an the Al–F intermediates bond c an be between one bonding type and positioned roughly ionic–covalent continuum. Arkel-Ketelaar halfway along the d i a g ra m classic ation This anomalies. M aterials of us c an properties, disadvantages science each to be to organize recognize classied structure, and objects, trends and according bonding. to information and patterns as well as dierent What system? are v p y origin, in help r e use, c an O use concepts. t i s We n y Patterns and trends criteria such as the advantages and solubility bond type, conductivity U particles. and volatility structure of Table a 1 and, if of a substance C i n The applic able, substance depends summarizes o The properties of materials with dierent bonding types these on and c an be explained in terms of intermolecular the presence other forces. The of mobile electric al charged properties of substances with Metallic soluble conductor corrosion Ty p i c a l where properties substances Molecular covalent Covalent network low high low high low varies varies no no when solid no no yes when liquid/ (except and for graphite graphene) aqueous a v exc e p t i o n s to E p Table 1 198 yes conductor susceptible no l O thermal low u electric al brittle f x water high i volatility varies Bonding type t a boiling point r o melting point and Ionic o d Property n dierent bonding types. yes no no no yes no, varies except some yes polymers yes of varies metallic, do not cov alent ex h i b i t and typic al varies ionic substances. b e h av i o u r This table no is me ant as a g e n e ra l guide and there are y v an For l triangular The represent 2 another. p Figure 2 triangle P e.g. F e.g. Cs e Dutch Structure Brittleness malleable. they c annot unable to is and coil up again. property is even aer with describe the force the oxidation of a metal, such as more surrounding a manner Graphite molecular more than This size and one to its is generally refers to a environment, which bonding some bonding explain are vary have widely. molecular type the For covalent network substances polarity type. volatile, despite the fact graphene covalent and of mercury whereas helps bonding typic al while variable Waxes are covalent plastics c an contribute to the variations c ategory. of properties C i n n U o i d t a u a means that removed. a is is O in density l v Rusting observe this n its external c an which plasticity. Corrosion and You plasticity, l water. Molecular r o E and return to y to force, contrast, rubber is elastic due to its and material and a e shape a to way. low f x O p Figure 3 are y given substance are bec ause p a metals force o within In opposite metals. substance. subjected uncoil substances, c ases, the while a y their within to v observed a a are ions brittle removed. In a metal spring, the metallic material behave covalent some of a when is to used electricity. bec ause In properties of some has or are subjected t i s hard. not crystals when r e are do The oxygen between conduct molecular oen in atoms elasticity. able deformed of substances that properties so is aluminium both the band. material materials for shape force example presence the example, solids an reaction damages that is corrosion the chemic al M any clay the being its The Ionic break other. change aer elastic retains easily. and r in will malleability. snap P The term each chains an material iron, deformed responsible in of substances shape polymer Modelling opposite past materials property the be slide original bonding long the From models to materials s s Elastic their is Brittle 2.4 type of corrosion. Iron and steel rust in the presence of water ox y g e n 199 Structure 2 Models of bonding and structure Worked example 1 Explain the given that those in trends the in bonds potassium melting in points silver halides. for halides Use the silver have data halides greater given in and potassium covalent table 2 halides, character and table than 3. s s Melting point (˚C) e point AgBr 432 AgI 558 generally increases points of the silver halides KBr 734 O t i s point 773 n y Melting point (˚C) KCl l Melting P p Table 2 681 points of the p o t a ss i u m of the electrons the going London ions u l a E v O f x You ionic and try 2. to The region of the loc ate silver section. so In halides the these halides contrast, from is the the all in to the The AgI that their melting increasing increases melting point the to strength increase one of the another. attraction melting the in also halides depends on how well region are oppositely charged decreases (table 3). triangular covalent potassium between point the bonds From KF to KI, the size of the all bonding of the rmly diagram in diagram, near the loc ated in the ionic triangle. Data-based questions Plot a graph of the data given in tables 2 and 3. Compare your graph with the tables and suggest which is a better method for illustrating the data. 200 points number predominantly ionic, with minimal compounds are AgF strength with means forces. c ausing electrostatic therefore halides Waals group interact the and silver der forces, Therefore, anions of van the potassium increases, decreases c an gure o c ations anion in i the down character. t a r o d bonding covalent of (dispersion) (table2). The character strength n U of covalent on halides o partial depend C i n Solution The p Melting v p Table 3 y KI r e decreases y 455 r AgCl Structure Use of triangular 2.4 From models to materials bonding diagrams (Structure 2.4.2) O ver time, chemists have further developed van Arkel and Ketelaar ’ s original You work. Eventually, electronegativity was incorporated in a way that gures 4a and encountered Structure resulting triangular and bonding diagram (or “bonding triangle”) is 2.3. shown in greater covalent character character greater ionic covalent greater covalent bonding character 2.0 2.0 3.0 O 1.0 l covalent bonding y 1.0 metallic bonding 4.0 mean electronegativity 0 Ionic to bond. of a of know We E position • dierence • mean and three the to out χ, 50 75 25 OF 3.0 which triangle to values two (b) diagram The 3.5 the of is two dierent Examples of determined overall the metallic–covalent incorporates electronegativity. types electronegativities, electronegativity, in This bonding bonding work 50 NO 2.5 continuum diagram. dierence in 75 0 .0 Covalent electronegativity need in the 25 Mean electronegativity (휒 bonding bonding substance the the then a ionic–covalent triangular electronegativity contributions need The v O The (a) resulting 92 100 2.0 l Metallic p Figure 4 u 1.5 8 o i Mg 1.0 mean n AIN t a f x 0.0 o d 1.0 r o ytivitagenortcelE 1.5 AILi The C 2.0 0.5 (c) v MgO U ecnereffid i n )휒∆ ( 2.5 %Ionic p CsF 3.0 y 3.5 %Covalent r e (c) n character P bonding in y greater metallic r ionic t i s ecnereffid ytivitagenortcele (b) 3.0 4a 4b e gure 4c. (a) gure gure 4b. Structure The 2.2, s s in rst merges the two in continua bond. To elements bonding continuum bonding c ategories, several by the substances are the included relative loc ate it, we involved in the parameters: ∆χ, is which the is the ionic–covalent metallic–covalent parameter parameter. 201 Structure 2 Models of bonding and structure We do more not need dierent c arbon–oxygen diagram as to consider bonds the may bond triple in molecular have formula identic al c arbon dioxide c arbon–oxygen or bond parameters. is in bond the in order. For As a result, two or example, the double same place in the bonding c arbon monoxide. s s Worked example 2 Determine the position of the following substances in the triangular bonding diagram: pure silicon, Si c. barium iodide, BaI 4 2 example, between C and F in CF = c alculate 4.0 of 2.6 the and 4 . In C arbon has an uorine is electronegativity 4.0. From this, dierence, ∆χ: this we atoms are all the same electronegativity. dierence 2.6 consider atoms. In elemental mean = example, two Si type 1.4 and Therefore, between two electronegativity Si is the bond between substances like this, the thus the the same electronegative atoms equal have O ∆χ c an pure silicon, Si t i s electronegativity we b. 4 need to look at the n bond we l tetrauoride, CF this y c arbon For is to zero, and the the Pauling r e electronegativity of one Si atom (1.9). c alculate the mean electronegativity, χ: This 2.6 + 4.0 χ = places 3.3 these the triangle between C in the (gure and F is we upper 5). c an part This polar loc ate of the the c arbon– covalent ∆χ = atoms 2.7 2.7 χ U = = + in BaI 2 in region the bond 0.9 = these ionic region n With parameters, of the we c an diagram %Ionic Ionic l a v E ytivitagenortcelE Bal 8 92 25 75 50 50 75 25 CF 1.0 0.5 Si 100 0 1.5 p Figure 5 2.0 2.5 3.0 ean electronegativity (휒 ) Metallic The loc ation of 0.9), and certain bonds in the triangular 3.5 4.0 Covalent bonding d i a g ra m loc ate the bond in the (gure 5). %Covalent 1.5 1.0 = = 1.8 o 2.0 are barium ( χ 1.8 u ecnereffid O f x 2.5 (gure 5). 0.9 t a 3.0 between metallic and triangle 2.7). 0.0 202 the bonding 2 i d )휒∆ ( r o 3.5 two iodine ( χ region means that the bond covalent. The iodide, C of parameters, bond i n uorine barium the o With c. v = silicon in p covalent 2 y Then, y P Solution a. e c arbon tetrauoride, CF b. r a. Structure 2.4 From models to materials Practice questions 1. Using electronegativity of following the values, substances deduce the main type of bonding in each by determining their position in the bonding triangle. c. lithium d. brass, e. aluminium f. silicon be discussed NO, and bonding typic al the three nitrogen dioxide, NO of pure ionic look at a selection of substances with signic ant contributions covalent Silicon network substances, is brittle conductivity, (like most Semiconductors is and an covalent are forms atoms a poor and in the is regions in the bonding both the covalent and metallic other electrons. acidic between electric al or On share (weakly) substances) illuminated, covalent exhibits metal. intermediate generally heated, which a it hand, Like it forms covalent oxide. In terms of conductors classied as a (like metals) and semiconductor. conductors, but their conductivity o when it in like presence of certain impurities. i Figure 6 hundre ds of A silicon t i ny wafer containing mic rochips for use in ele ctronics. u l a E v O f x t a r o d increases structure (shiny) and that n insulators silicon lustrous metallic means U electric al is the which C a between metalloid, p properties. a i n is o positioned It from v is , are covalent (Structure 2.2), and metallic (Structure 2.3) substances. Silicon: metallic and covalent character triangle. 2 by its position in the bonding properties more than one type of bonding. Silicon dierent types diagram. predicted the of previous question. n c an the the l in in y contributions e will monoxide, point substance already relative substances y we a same the the r e Now of of nitrogen the have (Structure 2.1), each 3 2 O We describe in r properties triangle. chloride, AlCl t i s The at C sF O dioxide, SiO why loc ated 2 composed of Cu and Zn bonding Explain Li y 3. 2 uoride, oxide, Qualitatively of s s uorine, F c aesium P 2. a. b. 203 Structure 2 Models of bonding and structure M agnesium iodide: ionic and covalent character L attice enthalpy Structure at 2.1, additional is and discussed in higher in Reactivity 1.2 In level. magnesium character are unusually high iodide, both ∆χ = 1.4 and about lattice 50%. As χ a = 2.0. The percentage ionic and covalent result of this, magnesium iodide has an enthalpy. s s Data-based questions c an test the compound ions are signic antly larger ions, lattice electron the enthalpy sodium Compound are with lower between chloride silver Theoretic al lattice in 2 the Cl 6 of between and 734 the oretic al theoretic al compare the and ionic and l a tt i c e e n t h a l py experimental covalent lattice character of Aluminium chloride: ionic and covalent character compound the aluminium chloride unusually c annot true In if which a at the is ionic melting as used in classied which compound, to as ∆χ predict ionic but solid. very the low aluminium ionic χ = for high an a has a signic ant covalent appears in the of is region of a substance. This is particularly character. covalent covalent in and the bonding ionic pressure, it melts into Al ionic same Therefore, aluminium chloride is polar both it shows that the bonding triangle mixed 2.4. and chloride This properties has exhibits At though iodide. = 1.6 and chloride when is even substance between aluminium lattices point, magnesium that chloride, boundary 7) at 190 °C, ionic be bonding result, forms low triangle always the aluminium right As representing an bonding trichloromethane. 204 y p o Al w ays o d i m e r, Cl Cl Tw o ex p e r i m e n t a l and –1 each compound. i bonding Al E p Figure 7 a Cl v O Cl in has Cl Cl Al bonding assess values of contribution is aluminium chloride. Unlike magnesium iodide, aluminium chloride Cl Cl dierences values, Another Al l Cl the enthalpy u Al Cl of compounds t a Cl d r o f x Cl on Comparison two theoretic al enthalpy / kJ mol 918 n U Based for and 750 C AgCl values, allows a certain chloride. –1 chloride, p Table 4 This experimental and r e v i n silver enthalpy tends to electronegativity 790 NaCl v alues lattice polarisable. enthalpy / kJ mol chloride, for an ionic prediction. This oen happens Experimental lattice sodium enthalpy However, if the bonds experimental more predicted and lattice (covalent bonding) in addition to ionic bonding. dierence for behaviour. theoretic al those clouds sharing the comparing O shows the by theoretic al t i s of Table ionic covalent, than particularly electron degree 4 partly the n their completely greater model y bec ause a l with of example, compound. soluble in Also, non-polar triangle. properties. AlCl 2 Cl 6 unlike dimers a 3 (gure typic al solvents such as y be assumes For P between assumptions values. r experimental e You Structure 2.4 From models to materials Practice questions Activity 4. Use your knowledge of the behaviour of compounds containing a Mean electronegativity and hydrogen– dierence oxygen bond to predict where in the bonding triangle you c alculate the ∆χ and χ values to see if your predictions for electronegativity dierent bonds were shown below. O utline the close. properties are s s Then in expect to nd this values bond. you would expect a compound with these bonds to e have. Linking question composites bonded reinforced concretes, components and steel which bars, are made have c. χ = 2.3 from ionic unique and ∆χ properties? 2.1, 2.2, 2.3) = 1.4 ∆χ and = ∆χ l metallic elements c an be mixed with other elements, metallic or non- Structure a metallic, therefore maintaining the retain metallic by of The identity from their are held electrons such as of traces original ratio the to of the components substance. about properties. 2%. All In For to and those the of melting an example, addition, these in alloy the c an characteristics are the c ations layers of in the metallic c ations slide lattice past are the i the c an be of of typic al discussed c ations in a contain sea of bonding between and electrons This accounts is and the the fact c ations non-directional. for the malleability ofmetals. c arbon in alloys of electrons, retain mixtures. n all applied, that vary without components of deloc alized pure metals. proportion each o metals, same other size. When a easily. t a r o is d pure force lattice we structures structure, Alloys point, Properties of alloys In the electric al conductivity and dierent resistance within throughout. U many alloys be C ranges of corrosion c an alloying a metal with other elements. mixtures. the of characteristics properties hardness, sea ions i n are changing deloc alized or o enhanced the as atoms v such steel many However, Properties Alloys Dierent p lustre. alloys. 2.3, metallic y still form r e while to O In how Pure n y t i s Alloys (Structure 2.4.3) 0.3 = 1.9 y covalently (Structure like χ = 1.9 and r and do χ = 2.9 b. P Why a. While malleability is a useful property of metals, sometimes we require a metallic structure to be stronger. Alloying involves the addition of atoms or ions with a u f x dierent radius to the cations of the pure metal, which disrupts the regular structure of the lattice. When an alloy is struck with a force, the layers of cations do not slide past each other as easily. Therefore, alloys are usually stronger than pure metals. l a E v O Figure 8 iron alloy are packed where as element In in the to pure a metals, re gular addition form a r ra n g e m e n t , an of ions another a l l oy re ducing the p a tt e r n , disrupts the this malle ability 205 Structure 2 Models of bonding and structure Examples of alloys NaK is a sodium–potassium alloy that has a lower melting point than its constituent elements, so it is a liquid at room temperature. Because of this, NaK is used as a nuclear reactor coolant, which is non-volatile and can be pumped as any other liquid. metals are to used spectacle is an abundant are harder from original being shape upon deformed than heating. They through use, such as tools. presence not the objects from rusting have methods (such as to barrier be deformed c arbon them to ideal many varieties relatively create for of a steel. variety of steel, each and into from (for preserve methods the of valuable metal iron and and instance, their (such galvanising). elemental hydrated properties exposing steel ships physic al as oxygen, O iron(III) has and and of metals. the iron economic bridges) integrity. painting Rust oiling) 2 . Rusting oxide. This is an Rust also underneath implic ations. need to be protection and sacricial i n o include made structures water iron v methods are p protected making There c an with y of of volume corrosion. steel-containing iron, it c arbon and other elements. transforms does iron n it pure Pure alloying O so it to by l the of crust. y in lowering degradation L arge amounts rusts E arth’ s overcome r e The the stronger bec ause easily, further to t i s to o in be construction dierent steel c an and compound, akes their Stainless steels are iron alloys that contain at least 11% chromium. The chromium C reacts with oxygen in the air or water to form a thin layer of chromium oxide on the surface of the steel. This chromium oxide layer prevents rusting, and therefore U stainless steels have useful applications in cooking and medical-grade equipment. o t a u Tow e r a Eiffel several l The lasts v (a) i d r o 206 itself E process f x O p Figure 9 (b) n (a) in months Paris, (b) F ra n c e, Adding is made chromium to of iron. steel It is painted cre ates every stainless f ew steel, a y e a rs to p ro t e c t rust-resistant alloy it f ro m rusting. suitable for The kitchen painting equipment y ranging problematic ionic metal limitation containing iron, to prone r Steels is return are P This Like that that frames. easily. uses alloys objects e Iron are make s s Memory Structure 2.4 From models to materials How does the c arbon content aect the hardness of steel? Steels are that are known as made up medium of 0.3% steels. to The 0.6% c arbon hardness of by mass Relevant skills medium• samples with varying c arbon contents were Inquiry 1: Demonstrate independent thinking, s s steel tested, initiative or insight giving the results shown in figure 10. • Hardness known materials have (units: as the HV) were Vickers greater hardness Vickers Inquiry determined using a hardness test. 1: Formulate research questions and hypotheses Harder • Inquiry 2: Interpret graphs • Inquiry 3: Interpret processed values. data research information 0.45 0.50 0.55 0.60 Medium of of data: C alik, C arbon C arbon A., Akin, Content Steels”, Z. on D. , the of S ahin, Natur forsch. O. and 65a, U c a r, of copper and in tin. It is harder common, 6. than bronze pure i bec ame What 4 6 8 – 472 was t a r o steel 5. to system copper do steels of you your O utline an and least the performed. compose a answered possible by the you include the: variable studied trend one might c arbon is (in this c ase, steels). explain at think with Outline your being c arbon and think N. n alloy Before Try o an Describe 4. Properties U is corrosion. d Bronze specific 3. sure independent – c arbon-containing Mechanic al the o “Effect h a r d n e ss of methodology C Source (2010) the i n steels. s h ow i n g v G ra p h range – medium Carbon content (wt. %) p Figure 10 – M ake be y 0.40 above. variables given, test O shown – hardness would p 0.25 that t i s 0.30 question r e 600 and the 1200 800 of on Vickers y Based the n 2. how l Research y VH ,ssendraH 1. data analysis to P Instructions 1400 1000 and r draw and justify conclusions 1600 0.25 e method values shown alternative happen content to in the graph. interpretation. the greater hardness than 0.55%? ideas. experiment answers to that questions could 4 and be 5 done to explore above. resistant to used in shipbuilding, tools various household artefacts including coins. Brass is a highly malleable properties, due to which its is alloy acoustic why it is of copper properties. oen such as hospitals. and The zinc. It is copper used in to make brass has music al antimicrobial a E v O l u f x instruments used to make door handles in public buildings Figure 11 coins f rom Old bronze China p Figure 12 made for of An b ra ss . me asuring 19th This century theodolite instrument was used angles 207 Structure 2 Models of bonding and structure Research skills and social skills ATL The life cycle of a product present properties, determines involves undergo the various many of the process and materials transformations processes involved throughout involved in a in (figure product life its production, 13). Structure, usage as well and as disposal. The chemic al and physic al s s substances cycle. crushing the ore e r mineral pure mineral recycle end of waste tip waste metal useful life v U products from task, you will work particular skills involved. people. Collaboratively, plastic the life 4. Plan to research E of the a. extraction b. manufacturing c. use d. disposal. Consider reflect on the – if and how you will divide up the work within your to of the team – how you with each and your team will share your findings other. following investigate: 5. C arry 6. Prepare out your a research. visual depicting the representation life cycle of your of your research, product. Before bottle. you various to stages raw materials of the start, take everyone’ s Briefly reflect on how you collectively reached a choice. lifecycle: 208 a research a 3. in one aircraft v 2. c arbon to also ingot Work with one or two other products an will l – in iron in a ship O – and aluminium group u substances – choose f x 1. a You t a collaboration with product. i a r o of o this cycle the metal metal The life cycle of a metal d In extracting n the metal p Figure 13 scrap metal C manufacturing i n the metal o recycle shaping y scrap metal in use p r e recycle metal O t i s rubbish to n y l crushed ore a moment opinions represent into to reflect account on how to take when deciding how your work. product’ s 7 . Share 8. Individually, and your work identify with reflect one IB your on class. your learner role in profile developed during this task. your group, attribute that you y P purifying the mining the ore Structure 2.4 From models to materials Practice questions 5. Explain 6. Suggest of a why alloys why the such as steel electric al are harder conductivity than of an pure metals. alloy is oen lower than that pure metal. s s are alloys compounds? more correctly (Structure described as mixtures rather than as r Why 1.1) e Linking question and as Polymers synthetic N atural many of covalently macromolecules, such as nylon include biologic al which are brous molecules characterized density compared are very examples materials involved in of like life bonded large polymer chains are end-to-end (gure 15), This process and is c alled like when small polymerization. There m o n o m e rs be ads represent repeating brackets, subscript the n polymer. n aer A atoms that appears polymers c an vary, with unit is the the represents a structural shown indic ating typic al E units, so is of that closing large so p Figure 14 polyethene A digital model terephthalate of (PET), a polymer to make and a chain the c alled resulting repeatedly along a polymer the formula enclosed the in polymer bracket polymer very up i polymer of are two types of t a the the group a The a a l across of is lengths v O formula, exact u The structure in f x repeating unit chain. join m o n o m e rs polymer Repeating units A as as n small silk, such sy n t h e t i c o the and molecules U the polymerization, r o chain, In Here molecules. synthetic lignin low types beads strung together to condensation d p o l y m e r. their C addition made by other o join necklace. p Figure 15 a to processes, i n a polymerization: materials low proteins. monomers create and p Long, c alled and bonded v starch are textiles polymers the DNA, cov alently conductivity, y well of r e polymers. class n materials. Plastics a electric al l of are and O thermal t i s Polymers y P y Polymers (Structure 2.4.4) the large chain represents contains best of to brackets or describe the with continues the hundreds way repeating unit. In the number on of thousands bonds either drawn side. repeating units of repeating number. 209 Structure 2 Models of bonding and structure Worked example 3 A section Draw of the the polypropene repeating unit of CH H CH H CH H H H H H 3 3 molecule is shown below. polypropene. H 3 CH H H H 3 s s r Solution the section, pattern you every c an two see that When drawing a repeating unit, make sure that the there is a connecting bonds on both sides cross the brackets. If you c arbon atoms along the l need to represent the whole polymer chain, do not forget polymer chain. n y to include the subscript n to signify that the polymer is Therefore the repeating unit is composed of many repeating units: 3 H H 3 H H n the repeating unit for each of the polymer H H H H H H — — — — — — C — C — C — C — C — C — C — H Cl H H C C C H 2 H H CH 5 H O CH structure n 9. The a E Draw section structure of of the also (CH polymer nylon-6,6 ) 2 6 H 2 OH H O H CH 5 3 OH OH H H H H as C O H H O CH OH polystyrene, is showing shown three is shown 2 OH below. below. repeating units. Draw a section O (CH ) 2 C 4 n 210 H 3 O OH known O NH 2 H 2 H O CH H a — H H poly(phenylethene), v 2 OH OH l O — CH — CH H H — of O u The OH CH OH H H f x 8. 2 H t a H OH i OH r o O 2 O H o CH H d OH H C c. H H n H Cl U — H — Cl — — — — H shown. C H sections i n b. p Draw a. o 7 . v Practice questions y r e H O CH H t i s CH of the polymer showing two repeating units. y at repeating P Looking e H Structure 2.4 From models to materials The structure of the repeating unit originates from the monomer used to make the polymer. the As prex a result, poly-. the some However, refer to their commercial monomer. polymer many Similarly, brand names. names natural many C an are monomer polymers synthetic you think polymers of name in brackets, with have unique names that do not any names are referred to by examples? s s Polymers can be classied according to their source (natural or synthetic), or type of polymerization reaction undergone by the monomers (addition or condensation). examples are e Some shown in table 5. Repeating unit synthetic? Type of balloons, elastic polymer natural addition 2 (natural rubber) c ar tyres (vulc anized t i s n bands, y poly(isoprene) rubber) polyethene C H 3 H c arrier bags, H packaging, H OH walls, toys paper, U natural condensation synthetic condensation o HO cell cellophane C i n O O v cellulose addition y H insulation, synthetic c able p C r e C O H y CH 2 Natural or uses l CH Examples of P cis-1,4- r Name OH H equipment, C personal armour t a of sports O O H Examples C o N i r o d N p Table 5 n Kevlar p o l y m e rs u l a E v O f x Figure 16 chemist dire ctly Kev lar Stephanie linked to the anti-stab clothing. a ny t h i n g like and was f i rs t Kw o l e k . use of Kw o l e k s av i n g sy n t h e s i z e d O ver 3 ,0 0 0 Kev lar in life Po l i s h - A m e r i c a n lives bulletproof remarke d, s o m e o n e’s by s av e d to “I d o n ’t bring think you h av e vests been and t h e r e’s satisfaction h a p p i n e s s .” 211 Structure 2 Models of bonding and structure Properties of polymers In general, most thermal and points. length is of due the each strong, they the and have number polymer other not are their macromolecular moving also intermolecular The long entangled, forces that occur along polymer requiring strands c an also wind energy to disentangle and the which houses a Starch a collagen, an natural which organism’s c arbohydrate through is and a photosynthesis. structures such as proteins, structure gives genetic natural as well our linear as is n a t u ra l of plants. polymer way t a tightly as u l a unit of starch in polymers stalks polymers by and oil-based other are important store of as by therefore humanmade. coating These and M any crude on some from have oil. pans of for a an The them as and important v Explain why polymers have 11. Explain why polymers are The see the resulting structural bres solvents. The integrity of in starch starch are polymer, like so the chains c an dierence lies in the have an irregular c annot pack as bonds with one another. petroleum-based, which include PET rubber biomedic al bottles, produced alternative applic ations. insulators. nylon tyres. Semi- materials sustainable electric al c an by using cellulose to relatively high melting points. generally Plants make You common Practice questions 10. store in elasticity, and means that the strong, biologic al a trunks. main synthetic provide others cellophane. chains in while energy glucose. glucose not Examples and the hydrogen renewable polymers is chains fewer many like also it polymer form tree properties humans. the the from is way these walls. It supports another. as and properties: of cell cellulose material polymer derived well these a result, in in digested in one materials plants as structure and polymer dierent derived bacteria. polymers, an water with connected: Teon synthetic it in skin acts polymer groups) bonds advantage and and are are non-stick plants As cellulose they has are structure. that textiles, OH it easily monomers branched means repe ating E 212 the down and energy However broken taken o be main insoluble makes have textiles, i d r o f x O The the Synthetic HO p Figure 18 is cellulose. O O cellulose paper, Starch and (hydroxyl hydrogen n make OH of strong Humans groups form organisms, that found o a up c an C is made U c e l l u l o s e, Co tt o n , chains uniquely strength p Figure 17 –OH i n are of v presence polymer the information. Cellulose cotton, living polymer repeating unit of cellulose in table 5. The and c arbohydrates and nucleic acids, to p brous is monomers y glucose include including provide r e Cellulose of n DNA, Some processes. Others type O plants. molecules, polymers. life the t i s support together, l biologic al natural chains arranged. to crude y are y are r monomers M any properties of a polymer depend on its chain length, the type of holding P specic forces Like low. separatethem. The nature. charged particles, so their relatively high melting points and boiling of molecules. become of contain conductivities polymers to bec ause do e around electric al molecular, This are materials, s s Though the polymers covalent Structure 2.4 Figure 19 This c re ate d a f rom From models to materials artificial mixture of s e m i - sy n t h e t i c p o l y m e r. c an organ be used in t i ss u e living T i ss u e s was cells like and this t ra n s p l a n t a t i o n s s e r mainly although of they c arbon c an and c atch hydrogen. re and sustain in packing chains be it is it chains widely overcome On It is a forces by non-polar new hydroc arbon (LDFs) hold the polymer heating, cooling, used plastic so LDFs polystyrene c an form CH — CH 2 n c an between p Figure 20 used plastics inert biodegrade are covalent network The structure of p o l ys t y r e n e is as that, and by phenyl insulating due to As groups (C 6 H 5 ), Polymers crosslinking covalent many contain sites. which London dispersion and other is In provides very of the (LDFs), discussed intermolecular in Structure forces 2.2. bonds, plastics do materials, reactive some known forces for the breakdown rarely binding together, strong result, the panels are their a involves bacterial bonded formed bulky insulator, particularly in its durable. molecules. act the plastics, as the crosslinking. few entry points a v E Activity small covalently microorganisms. useful Biodegradation could to coolers therefore l chains into that make due thermal u groups are and easily good o is a industry. f x O dense for to it microorganisms, polymer A so reasons typic ally rotate is chemic ally functional 20. i the be gure dispersion t a are c annot Polystyrene r o of they in easily. d form, construction One c an a molecules, so the polymer solidies again. brittle. expanded LDFs shown London reshaped is n making and is water. U Polystyrene in These down polystyrene by polystyrene together. melted not of insoluble polystyrene, containers. o so known as food C structure and and i n The commonly peanuts v found p Poly(phenylethene), y combustion. — composed unreactive, n l polymers chemic ally O synthetic oen y P y are are r e They t i s Plastics Compare and contrast two of the polymers discussed in this topic. 213 Structure 2 Models of bonding and structure Investigating hydrogels Smart materials response to undergo a surrounding change in conditions Instructions properties in (such as pH, Part 1: The effect of pH variations on the hydrogel materials or used applic ations. to swell. 21). basic water, investigate surrounding conditions. numerous groups the environments, ions, –COO c arboxylate . dishes, c ausing the gel the c ausing Prepare the response of a c arboxyl form groups. hydrogen hydrogel to become ionized, The negative charges groups repel each – weak acid solution – weak base solution Place the 3. Place a hydrogel 1: on a lens is the following: white in the dicult to surface. rst dish and measure its see, it might be inside out. Reverse it and try again.) 5. Swap the contact lenses in the acidic and basic other, changes swell. the it of Place contact lenses in the other two dishes and note solutions Justify (If one buer solution 4. changes Inquiry dishes contact diameter. Relevant skills • lens in and diameter. observe what happens. O the shallow dishes, such as petri contact any making clear containing – 2. contract they three each n neighbouring hold –COOH 1. biomedic al will in another, c arboxylate and crosslinked polymer t i s on In of Are the reversible? range and quantity of y r e Part 2: The effect of salinity variations on the hydrogel measurements • Inquiry 2: Interpret qualitative and quantitative data 6. Repeat the Part 1 experiment, using these solutions p instead: S afety protection. Commercially available soft contact lenses (manufactured from poly(2-hydroxyethyl • Contact lens buffer solution Sodium chloride base solution (for –3 ) equipment including petri pipettes and balances. u l a E v O f x Ruler laboratory beakers, 0.06 mol dm i Standard dishes, • water t a r o • Distilled example, sodium o Weak hydrogenc arbonate 214 d Weak acid solution (for example, household vinegar) reuse Describe Propose of 9. lens buer solution water saturated sodium chloride solution c an damaged. 8. • • – You 7 . • distilled n U methacrylate-co-methacrylic acid)) • contact – C • i n Materials – o eye v Wear the the your a Part Devise Part an 1 lenses, they are not results. hypothesis 2 provided to explain the observed results experiment. experiment that you could do to test your hypothesis. Identify the independent, dependent and control you make? variables. Justify measurements. the What measurements would range and quantity of these y (figure one and contain these up l with you chains industrial y bonds forming task, media, of made absorb variations hydrogel acidic range are are smart r In c an this to a Hydrogels P The that In hydrogel in They fields). e networks magnetic s s temperature Structure acid form CO 2.4 From models to materials base form COO H 2 O O O OH s s OH O + H O O e + H HO O O O O O r OH O COO H electrostatic repulsion groups causes hydrogel to contract to expand between – CO O H groups and a result, disposed challenge, durable the a and longer term. petroleum-based risk an enough Much to area be Work light-sensitive. must be taken to For action. c an down slow or gas also are quality is bacterial functional oxygen in action up time ways by groups largely unaected by end for to in the but in the D isc arde d oce an. plastics The cold c an and oce an oceans, s l ows d ow n dark plastic d e g ra d a t i o n biodegradable make e nv i r o n m e n t where it Greenhouse at AHL in gases Structure are discussed 3.2. existing microorganisms. having that up incinerated or are increased surface attractive to bacteria hydro-biodegradable plastics down in conditions environments produce nd of feedstock plant-based them is instead ends period to adjustments recycled. interest, is designing plastic materials that done broken cleaned prevents some p Figure 22 and plastic certain appetizing or are of a being more that even E greenhouse low for facilitate dispose In growing example, v bacterial of chains, c arbohydrates disc arded pH release atmospheric pollutants the wildlife. a O contain recycled u polymer not l f x or shorter of used is plastics Biodegradable plastics area, to is they and which o poses waste when sorted melted, they where be for mitigating the several challenges. i in and landlls. deteriorate must and t a are in plastic up bec ause aluminium, r o The of oen and d persists most recycled glass waste poses n As plastics be U recycling, being Unlike to approach C from needs third recycling i n plastic broken plastic types the then It heating. is Plastic energy-intensive: beforehand. on recycling plastics. the y is of with o process reuse, v The and impact a ss o c i a t e d p reduction groups r e Polymers and the environment environmental – CO O O t i s – I n t e r c o nv e rs i o n n hydrogel Aer l between carboxylate groups cause y between carboxyl y 2 hydrogen bonds p Figure 21 P CO a process that found environmentally c alled provide in landlls, harmful hydrolysis. enough C are oxygen for decomposition products, such as the methane. p Figure 23 m i c r o g ra p h embedded c arrier and are a bag. sw e l l s for A of in sc anning starch a b i o d e g ra d a b l e The u p, ele ctron g ra n u l e s starch absorbs providing bacterial plastic water gre ater surface action 215 Structure 2 Models of bonding and structure Global impact of science The 12 the green chemistry benets ecient reduction use of of waste, of The adopted choice by many of and reactants, renewable materials, materials philosophy and of and prevention of green chemistry has educ ational eventually and commercial r organisations chemic als and energy passed into national and internationallaws. of renewable their Even c arbon c arbon-neutral biosynthesis is of renewable, such as polymers bec ause removed from the the releases are sometimes same quantity atmosphere by monomers in plants. evaluate have for require factors. For a product, household example, used fertilizers or to grow other pesticides, crops, which which have their c an 12. Outline two challenges remain recycling plastic waste. two structural features of and Their impact be auty banned in m i c r o g ra p h on are a single sometimes products, m a ny particles, but countries re d of this s h ow i n g k n ow n blood added use to is is into polyethene t a Ele ctron polyethene mic robe ads, Mic robe ads 216 health small way a v E mic roplastic such as cell. he alth n ow food samples banana shortages. environmental fragments not unchanged. their u l f x O Figure 25 and nd polystyrene biodegradable plastics. as very process chemic ally and i Outline r o 13. chain are This o with d associated 5 mm. c ause associated n U under potentially own Microplastics Practice questions bioplastic waste p i n e a p p l e c r ow n s plant-based plastics uses land that be These food C may maize otherwise many of i n could consider impact o growing to environmental v you the p p Figure 24 To the of same Bec ause were as is still made onion f rom skins and even Additionally, farming maize broken down biodegradation their size, places. discovered microplastics were peel, implic ations. plastic, of unexpected y the are biodegradation plant-based dioxide petroleum- r e c arbon plastic are non-renewable. The bioplastics though dioxide, considered of for plastics is O materials maize. Most feedstock n so t i s based, raw feedstocks and designing materials biodegradable. y be bioplastics, green chemistry: l use to biodegradable plastics, or aligned with two principles of in In they 2022, human unknown. over time to sizes bec ause the plastics c an enter the microplastics blood for the food of rst PET, time. y are P Plant-based e been use biodegradable accidents. principles emphasize non-hazardous s s solvents, of Structure 2.4 From models to materials Addition polymers (Structure 2.4.5) During way polymerization, this contain c an happen is a double bond due to and chemic al their 2 H 4 properties , is the polymerizes Consider c arbon-3 that involves very large number of ethene o i d H H H H H H H H H H H H H H H H a H H H H E v O l u f x t a r o H H H a n U process form a o 4 H to C i n H similar and O c arbon-2 y between p shared n up… monomers together: H H physic al r e are the 3 a open H H polyethene. H H H H form 4 H imagine bonds monomers: H to v joining 2 H dierent monomers. H electrons H H form a order, and y double H H 1 quite bond t i s their 3 these H have in H H bond, H Now in 2 and electrons y 4 H H single molecules two change One monomers that l C 3 H H It the the polymer. H C 1 and of corresponding alkene. 2 electrons H simplest breaks Bec ause polymer their H H The length, than C H bond a P 1 double monomer. form involves monomers: H C … extraordinary ethene H The another to which r two together e Ethene, C bond. with combine s s covalent monomers addition polymerization, H H H H H H H H H H H H H H H H H H C C C C C C C C C C C C C C H H H H H H H H H H H H H H 217 Structure 2 Models of bonding and structure We c an convey simplify that the there H equation are a very for the large addition number H H H H of polymerization reaction using n to monomers: H s s n H H n n H also use structural CH 2 formulas: CH 2 2 n Any monomer that contains a c arbon–c arbon double bond c an undergo addition non-polar therefore in chlorine where stronger of the with in past, People u l f x a E v O the and and of why required is CH is polar, are present. PVC is H CH H H 3 3 H p o l y (c h l o r o e t h e n e) the and the and they worked. materials of practic al materials the ( b) polypropene are a the centuries the particular to in scientic scientic time. Science is and scientic natural world. knowledge. knowledge. understanding the properties. scientic Empiric al understanding of research are scientic science technology, and over before applic ation using observation of observable contributed and physic al preceded materials’ engineered empiric al science disciplines applic ation for Nowadays, for two understanding technologies properties between these developments behind early role dierence developed reasons Desirable What is technologic al observation rst. H between observing technology used underlying how forces y focus the relationship i concerned In dispersion p o illustrates changing t a r o d Structure 2.4 The (a) bond n U TOK the London n form chloroethene c arbon in the polymers. n to from other c arbon–chlorine H H re actions The made Every o polymerization only two C Ad d i t i o n polymer between the polymer chains, unlike in r e H a substituent. forces v i n H H is (b) Cl n 218 the Cl H p Figure 26 a PVC, poly(chloroethene). dipole–dipole polyethene the (a) H has is n polymer results chloride, name O which Polyvinyl IUPAC t i s resulting Its l molecules. y polymerization. oen occurs identied, and then knowledge. and technology? n y P 2 condensed e c an r We Structure 2.4 From models to materials Practice questions 14. For each of draw ii. deduce following section the of monomers: the polymer structure of the F F F polymer. H H H — — — — structure of of the the shown: repeating unit monomer. CH − 2 CH − 2 CH− CH 3 H H and C CH − … 3 C C Linking questions What C functional molecules act H CH as groups in c an enable them to monomers for addition 3 reactions? (Structure 3.2) Why atom C 3 is the economy 100% for an addition polymerization ethene point H i contrast 3 u melting a and bonding, C l Compare CH 3 t a r o f x O 16. H CH o H d C − 2 3 n b. CH− CH 3 − − CH CH − − CH 2 3 U − CH CH 3 − − … C a. o the sections structure i n draw polymer the p the v of deduce H y — — H each r e — C — C = C — C — H O t i s but-2-ene y H n H l C r C (prop-2-enenitrile) P CN H ii. repeating units the y H H i. of e F acrylonitrile c. For three unit tetrauoroethene b. 15. containing repeating s s a. a the i. and polyethene in terms of their structure, reaction? electric al (Reactivity 2.1) conductivity. v E 219 2 Models of bonding and structure LHA Structure Condensation polymers (Structure 2.4.6) Condensation reaction For every balance webbing typic ally tensioned Nylon is a is a between have two nylon and reactive cellulose, functional monomers, a small are produced in a groups on either end. molecule is released water or HCl). p ra c t i c e uses ny l o n between two condensation anchor r points. Slacklining that formed include that (for polymer e in bond which monomers s s instance, Figure 27 polymers, between l and between an alcohol and the a hydroxyl water H O H functional used ethyl formed to make in the condensation condensation H H ethanoate re action polymers H−O + − H − reactions two H − The released. This − heat H is is H−C−C−O−C−C−H − e s t e r, c arboxyl − − 4 H = SO 2 ethanol an and molecule esteric ation conc. H − H n reaction reaction, linkage, known as O ester y is this H − t a 3.2. further e t h a n o a t e, u properties E 220 a Structure and discussed v are l structure esters ethanol p i − f x O The and O lost 2 E t hy l an In H−O−C−C−H + acid 28). − − − = r o ethanoic p Figure 28 form reaction H H H condensation o O d H H−C−C−O−H a o groups condensation in (gure n U functional acid C formed c arboxylic y P y t i s r e are v a i n Esters are between similar to water ethanoic acid esteric ation: of groups react to form a linkage, and a small molecule is released. in However, groups, We will in polymerization, one at each look at two end of the the examples reacting monomers molecule, of how this which have allows c an happen. a two reactive functional polymer chain to form. Structure 2.4 From models to materials Condensation between two dierent monomers alternating monomer a between to form = = HO − C an ester pattern acid linkage is for ABABABAB in and gure H O− − form − + OH H 29 based on two each ends structure, polymer product OH + H O n − − O + − OH 2n H number the length condensation = 31). formed 4 during polymer the is and chain is known as a condensation H − N − (CH ) 2 6 O = = − N − H − H HO − C − (CH ) 2 4 − C HO − − H − − N (CH − ) 2 N − (CH 4 − − 6 − H − … C (CH − O + O 2 repe ating unit is H ) 2 … 4 − − C − C = The − N = ny l o n - 6 , 6 . 6 = − O H ) 2 C = E ) 2 H − of resulting n i nv o l v e d between product − C − HO N − (CH formation where p a i rs − ) 2 + The the condensation p o l y e s t e r, = C − (CH H p Figure 31 a monomer re action The O O linkages g e n e ra l i z e d − − (gure u − N amide polyester. o multiple of of O i the A make the t a in to 2 d the − produced a 6 is l ) 2 each monomer c alled a n − hexane-1,6-diamine v O N − (CH H and to is U C− O− subsequently o − f x − H due with re action nylon-6,6 acid c an Figure 30 r o polymer hexanedioic that forms a long polymer chain with monomer resulting = = − reactive This p The has 30). y (gure LHA gure O in O − O 2 n − y (diol) n process. m o n o m e rs ester C C poly amide an t i s C O The two OH v O C condensation linkage linkage. O − A between i n − pattern ester = HO an O monomers = n by (diol) l − molecule more alternating joined general alcohol r e an reacts = = C product with A The O C Ester react formed. y O Alcohol creates P acid This 29. re action Dic arboxylic The is react with monomer Figure 29 C − OH − if c an end. difunctional to HO end either example, monomer shown at e O dic arboxylic structure: the either r O a with at group H monomer monomer polymer group functional H reaction B, functional dierent s s with of same a H polymer the with − a with monomer HO monomer another O One O + O 2 s h ow n in H O O 2 + H O 2 gre en 221 2 Models of bonding and structure LHA Structure Condensation of the same monomer If there is a dierent polymerize OH O acid with (gure 32) of by contains group on condensation. both a each For hydroxyl 3-hydroxypentanoic acid end group c an of a example, and react a monomer, then it c an 3-hydroxypentanoic c arboxyl with one group. another to Multiple form apolymer. OH CH Proteins The structure of biopolymers that play structural a central integrity, role and in life. Proteins are 2-amino processes, polypeptides, which are polymers a as acids. building –COOH group, There are blocks so they of 20 c an polymerization. types proteins. In form a of 2-amino 2-amino amide polypeptide, H v COOH R is group also through known as a COOH O NH 2 R C H 1 H N C R H 2 2 NH o O n 3 2 H 2 C R i H H 2 H N R C H C H 3 COOH NH 2 C R H R 4 4 COOH O 2 NH u l 2 C R H H N C R 5 H 5 COOH H O 2 NH a E v O f x t a r o d H 2 C C i n U H NH o O link 2 other n 2 p C 1 2 amide each O NH H by living y r e a R with aspects monomers have an –NH t i s H of the used of y O formation acids acids linkages peptide bond: The up l condensation chain made transport molecular 2 C R H 6 H N C R H 6 COOH COOH amino acid molecules (where R 1 , R 2 , R 3 etc. represent any of the 20 or so groups found in naturally occurring amino acids) polypeptide (part of) y and polypeptide other P organisms Figure 33 all monomer c alled 222 metabolic virtually r acid functions, the of 3 - hy d r ox y p e n t a n o i c sensory e and p Figure 32 are 3 s s monomers functional itself Structure 2.4 From models to materials LHA Activity Figure 34 shows the 3D structure of lysozyme, the natural antibacterial enzyme contained in saliva, milk, mucus, tears and egg whites. The carbon atoms are s s shown as white, the oxygen atoms are red and the nitrogen atoms are blue. C an you nd any amide linkages? e r Identify the type of v monomers a between polymer containing condensation and functional hydroxyl group, –OH, A c arboxyl group, –COOH, c. Two d. An E A b. acyl addition made these a. hydroxyl polymer made from (gure 32). a 19. n l dierences O two polymerization. of acid l Outline O 18. y unit p repeating u the 3-hydroxypentanoic o t a r o f x Draw C l ys oz y m e Practice questions 17 . n of o structure i 3D y P y t i s r e v i n U The d p Figure 34 and a the and small groups c arboxyl and an condensation molecule form group, amino released when polymers: –COOH group, –NH 2 groups, –OH chloride group, –COCl, and an amino group, NH 2 223 2 Models of bonding and structure LHA Structure Hydrolysis The reverse formed by of condensation condensation macromolecules 35 formed shows again by form how the monosaccharides energy hydrolysis, of condensation reactions and hydrolysis hydroxyl reserves up in Note form c arbohydrates, living CH down via the two are by hydrolysis. reactive ends of with used a glycosidic as energy organisms. OH linkage condensation and polymers which the biologic al r OH 2 which where All break polymerized reaction. groups, make are water. 2 H monosaccharides l y (water added) OH CH 2 H O HO OH i n glycosidic C bond condensation OH t a Some —OH lactose CH OH 2 O O H polysaccharide e.g., starch, glycogen O O OH Condens ation c a r b o hy d ra t e s . OH 2 o i r o O p Figure 35 CH 2 O HO e.g., maltose, sucrose, hydrolysis n U CH O d H OH 2 disaccharide o v p H OH 2 O O y r e CH n hydrolysis O 2 (water removed) CH O H galactose t i s condensation e.g., glucose, fructose, and hy d r o l ys i s groups are of monosaccharides, o m i tt e d for the p o l y m e rs that make up clarity u l a E v O f x Evidence Water is oen biologic al for water forms to referred reactions. on M ars, as to For as the this the ‘molecule reason, presence of NASA of water life’ due scientists is to its have thought to vital role in been be searching required for life exist. Linking question What functional condensation 224 groups in reactions? molecules c an (Structure 3.2) enable them to act as monomers for y P CH and c alled molecule monosaccharides Polysaccharides sources by is a reverse are up e linkage. by reactions split s s Figure is Structure 2.4 From models to materials End-of-topic questions Which of 1. Using your knowledge from the Structure 2.4 to the following pairs of molecules will form a polymer? topic, A. answer the guiding question as fully as OH O OH possible: OH H H (CH mean electronegativity χ PbBr (Pb) = 1.8 and χ 2 ∆χ, for 6 the bonds H found in . (Br) = O O 3.0. ∆χ ) 2 4 HO 4.2 C 2.4 1.2 D 1.2 2.4 Which of the ionic values diagram and the properties of electronegativity Cs F Zn S 0.8 4.0 1.6 2.6 a sodium-potassium C. Zinc D. Bromine, uoride, sulde, ZnS the iron and mixtures and is an 2 Br and H H C N COOH 2 CH 2 alloy? SH a B. following chloride water D. H H C C H H O H vanadium v sodium 2 CH COOH 3.0 u of A. Br l Which alloy C sF 1.9 t a NaK, C aesium Ag i K 0.8 A. C N 2 o d Na 0.9 r o Element B. H Refer to the below. f x C. c arbon D. magnesium and chlorine bre and a polymer E 5. bonding has compounds? Electronegativity O 4. substances covalent H OH n triangular following and U both C. p 2.6 o B C 2.6 v 4.2 i n A y Electronegativity dierence, (CH r e Mean electronegativity, χ 3. H and O dierence, bromide, ) 2 χ, and t i s lead(II) electronegativity, n the y C alculate l Multiple-choice questions 2. OH 3 y B. CH 3 O P Exam-style questions r ofmaterials? e and What role do bonding and structure have in the design CH not s s react LHA 6. Topic review Which of these substances does H not H and H C H have the ability to polymerize? A. ethane, CH B. tetrauoroethene, CF C. propene, CH D. propenenitrile, CH 3 CH 2 3 CHCH 2 2 CF 2 3 CHCN 225 Structure 2 Models of bonding and structure Extended-response questions 7 . Tin(II) chloride lead(II) used in the a melting has a production glassware. point melting One of of of point aurene these 10. 247 °C while of 500 °C. Both Chloroethene shown glass, an iridescent two substances in the vapour substance and explain C= C — H Cl [3] chloride, AlCl 3 , forms white Identify the structural feature of it to form a polymer. the mean State values type of [2] The the a mean repeating molecular sample reference to the triangular of unit of the polymer. poly(chloroethene) is O diagram, why aluminium chloride characteristics of both covalent and d. ionic substances. Deduce [1] per Using your 2 Cl 6 VSEPR chloride. theory, deduce [2] is oen 11. U Al The dimer contains coordination bonds. Outline the meaning of the term the is broken point Explain of –154 °C, poly(chloroethene) this dierence in (PLA) [2] is a feedstocks down by bioplastic such as made from sugarc ane and maize. microorganisms under high O nHO C CH OH CH 3 2-hydroxypropanoic (lactic) acid [1] formal charges l O of the atoms inthe aluminium Explain why metals. Include dimer alloys a are a chloride above. generally v O LHA 9. Determine It [2] is points. acid renewable answer to two chloroethene melting 210 °C. repeating units u f x coordination bond. ii. Polylactic t a two melting o i d r o i. the around of of your temperature and high humidity conditions. Al Cl is point n Cl Cl melting found as the dimer Cl Cl The whereas chloride. . Cl e. geometry and bond angle of chloride signic antgures. [2] C Al of aluminium State o Aluminium for i n d. knowledge molecular aluminium formula v the Lewis number chain. y the mean p c. Draw r e b. the polymer [2] found to be –1 displays [1] mass of the polymer chains 69,000 g mol bonding [1] polymerization. n with Draw for t i s Explain, this l dierence chloride. in ii. of y aluminium name electronegativity and c. electronegativity the diagram [2] O stronger than CH + C nH O 2 pure CH 3 in your answer. n [3] E polylactic acid (PLA) a. Deduce action why than PLA is more susceptible to bacterial petroleum-based plastics such as poly(propene). b. Outline of 226 two [2] advantages bioplastics such as and PLA. two disadvantages [4] y b. P low melting point and boiling point. C alculate chloroethene crystals with a thatallows i. r a. Aluminium a. known as chloroethene is your reasoning. 8. of e that polymer — identify a H phase. Using electronegativity tables and the triangular bonding diagram, form structure below. — molecules to The — discrete react exists H as c an poly(chloroethene). are s s artisan has chloride Structure Kevlar is a high synthesized The structure … tensile-strength in 1964 of by Kevlar polymer chemist is shown 13. rst Stephanie The repeating units of polymers A From models to materials and B LHA LHA 12. 2.4 are shownbelow. Kwolek. below. − = O s s C − CH — CH 2 n − H r N − H C C C H H H O O C − O C C C H H N = O C − polymer B O − For H or − For each − N 14. − … linkage State the formula of the other compound produced is the monomers made. Cl [2] name of this type of polymerization. chains hydrogen Draw a bonding diagram two showing adjacent to one another which occur. Draw b. Deduce the monomers formed in structural [3] are shown below: O C H N NH 2 2 Cl repeating the [1] the unit formula the of of the the resulting inorganic polymerization polymer. [2] product process. [1] through features of i hydrogen to Identify Kevlar [1] hydrogen bonds Kevlar polymer chains. [2] u l a E v O f x t a allow between attracted r o e. are bonding. a. draw o Kevlar d d. [2] n U when Kevlar monomers polymerize and state the C c. Kevlar of C and monomer(s). condensation O [1] Draw which formulas Kevlar. i n structural in b. from the amide its o the v Identify Two of deduce p − polymer. polymer, y r e formulas H polymer, state whether it is an addition condensation N b. each O a. − t i s − = C n n y H a. O − l − H H y H P H e − polymer A − = C O 227 s s e r n l O of y p o v i n C l a ss i f i c a t i o n m a tt e r y P y t i s 3 r e Structure C n U o i d t a r o u l f x a E v O Structure 3.1 The periodic table: Classication of elements s s How does the periodic table help us to predict patterns and trends in the properties of elements? of use the it according to we their we classify the oils bec ause depending classify matter motor viscosity, purposes c an c an example, on is based on what c an be they how are thick The classied used they for are. increase correlates or So, down and elements? the trends with and with decrease the groups in the of position of the periodic certain of periodic table number shows the outer energy r e Structure 3.1.8 — Transition occupied by electrons. Elements in a d-sublevels have a common number of valence electrons. that give them Structure 3.1.3 — Periodicity elements across a period include and in properties increasing decreasing metallic a in properties group. of elements character non-metallic of character metal a continuum. oxides This includes if forming the a show the bond. compound It number is the were of a v E nd a detailed to the between be ionization the of orbitals oxidation explained energies element absorption properties. variable complexes light in by the are close when the split an are electron colour d-sublevels. absorbed is complementary to the colour observed. electrons charge that atom composed of ions. u l f x O c an promoted The from basic Periods, groups and blocks in the periodic table (Structure 3.1.1 and Structure 3.1.2) The periodic table consists of horizontal periods and vertical groups. The periodic table is also divided into four blocks: s, p, d and f. Elements are organized into these blocks according to the arrangement of their outermost valence electrons. For example, elements in groups 1 and 2 have their outermost valence electrons in the s sublevel, so these are in the s-block. The blocks are shown in gure 1 on the next page. We You is due of o have coloured have incomplete properties oxidation state is a number to value. group i in atom successive c an trend of t a would an r o transferred trend d Structure 3.1.6 — The to the in elements Structure 3.1.10 — Transition through amphoteric to acidic non-metal oxides. assigned non-metallic their formation the n show and that transition down a U Structure 3.1.5 — Metallic fact in group 1 of 17 elements. states C elements the trends i n group to down o Structure 3.1.4 — Trends refers and v of elements characteristic p Structure 3.1.9 — The in y is occur across a period. LHA period ionization n Structure 3.1.2 — The rst patterns l increasing group These described visually O Structure 3.1.7 — Discontinuities groups and blocks. that be y t i s Structure 3.1.1 — The periodic table consists of periods, level across the periods table. c an graphs. Understandings energy properties closely elements y how For P dierent ways for. r we e One c an also classify elements in the periodic table as metals, non-metals and version of metalloids. Metals generally have three or fewer valence electrons, which are the periodic table at the back of deloc alized and thus contribute to metallic bonding ( Structure 2.3). Non-metals this book. usually have four or more valence electrons which are not deloc alized. 229 Structure 3 Classic ation of matter s-block 1 18 transition elements 1s 1s 2s 14 15 16 2p d-block e 3s 3p 3 4 5 6 7 8 4f 7p y r e 5f p, d and f blocks of the periodic table ionic and show number(s) directly 1 either alkali metals transition Groups elements groups 18 noble gases Periods periodic given are the a red zig-zag in table are which to 7 . The n, of the this period. are Elements in u groups For l a E v O f x t a electrons 1 to the ve You This 3 it the and is use group valence this all 2p They metalloids Metals c an form both are usually are to the le of the have to period summarized in table 1. number outermost in the is period 18, electrons. same equal group 1, the 13 from 1 to 18. Elements in some are 2 corresponds to the electron have n = sublevels 2, and for elements their valence sublevels. number information n = 3 electrons gas in or groups straightforward ve noble of For elements 2s group 1. be to number the alkali metals, the For last digit example, of valence number all of of have the one group elements electrons. valence in For electrons. electron in their number group 15 is all equal have electrons. means that has the the elements number c an in same 2, sublevel. valence example, found example, valence For from numbered name, 1 in gure behaviour. to are on the right. collective numbered non-metallic considered principal quantum number, Names of groups in the periodic table the i r o Table 1 in are of and Elements non-metals o halogens d 17 side and metallic bonds. n U metalloids, 3 – 11 hybrid covalent o Group name C i n Metalloids Group p The s, v Figure 1 6p to for in the 2 [Ne] 3s 3 3p deduce valence these represent conguration: to the electron congurations of elements. for periods 2 and 3. Consider phosphorus: being in period the sublevels, sublevels. inner You electrons and c an to being use give the the in group symbol of 15 means that the condensed previous electron y 6d n 7s l 5d 4p O 6s r 4d 12 P 5s 11 y 3d 10 t i s 4s 9 f-block 230 17 s s 13 2 Structure With period 4, we need to take the 3d 3.1 The periodic table: sublevel into account, which lls aer the Remember 4s sublevel and before the 4p sublevel. The group number is of electrons occupying the 4s, 3d and 4p sublevels. Consider all in group 17 means that 17 electrons ll the outer sublevels. go into 3d and the remaining ve go into 4p, so the electron dierent of electrons. on how to ll conguration of with electrons is 5 3d 4p Structure1.3 of a properties By of the also found certain table as the we know elements time, but table also depicted their very chemist Dmitri periodic which properties. mass and Mendeleev system eventually today. like be this not only based helped presence of gaps in signalled elements. the possible G allium, for existence of example, was not known to Mendeleev, but he had assigned an element close to was of For you the have periodic of this table. positions book E a. element to not name of the element name the Sc, you think Y , La table and do it Ac, you matters? or Sc, think Y , Lu and hydrogen What is the purpose of properties. properties clear-cut. the loc ation of belong to group Figure 2 1, Some of Mendeleev’s notes leading up to his 1869 periodic system a, b and following c, use the periodic table at information: How has the organization of elements in the periodic table facilitated the elements? the discovery of new (Structure 1.2) group principal quantum number iv. full v. condensed electron 2, it periodic Linking question iii. period it described in ii. its these always Does deduce i. of is a back v the the of questioned l O Activity For many is second scribbles at the of predictions. elements may the some u in (near discovered, the f x hydrogen 68 predicted Mendeleev’ s Classic ation example, around and i gallium of 2) t a were mass gure r o When of 3: A o atomic centre d with Do led to the properties of the elements the the altogether? discussion about which elements n the Mendeleev’ s to atomic U known undiscovered the it of the elements, understand other order of be? understanding would in classic ation? century, chemists had long according in Russian scientists at should understand patterns and group C Organizing and elements 19th version of better i n periodic masses to order known late them early us to Where loc ation ongoing o atomic the patterns. an Lr? dierent v developed the elements put belong refer a the p had helps remembering to you lesson? or is y classifying Some all mid do 17 r e been on table, oen universal exploring their similarities, trends knowing periodic dicult. and How this O the Without elements? chemistry to t i s the and connected n things dierences the typic al are y Classifying in example organizing during them. group concepts y it learn chemistry without the periodic l to to many r way trying How P table. given in e Patterns and trends Imagine s s 10 number guidance sublevels 2 bromine is [Ar] 4s p, d and f a Two go into 4s, Detailed ten s, have bromine: maximum being that equal to the total sublevels number Classic ation of elements conguration electron group 16 b. conguration period 3, group 2 c. period 4, group 8 231 Structure 3 Classic ation of matter Periodicity: Trends in the periodic table (Structure 3.1.3) Elements are structure. As of to make trends for in various You predictions the properties graph periodic observe table going the is table chemic al known as across trends using in the to their periodic and down Trends electron anity values for each and property sections 9 of atoms with increasing atomic number. 10 atomic number. o start decreases. and so a period u the As l a E v O f x t a r o the i d blue, and the halogens in From there The This higher means no end nuclear that are radius in are electrons being is the in 60 radius and to for period The inner become same by the alkali higher from levels the the to nuclear smaller down (red nuclear attractive atomic additional going valence each going radius proton, across a electrons and additional proton. across the period, a group in the periodic table. dots). Although the number of charge level. electrons. the charge the with level an electrons. going metals blue), has between energy inner energy of red element increased increases the (from attraction to the therefore a shielded of to shielding 3 The alkali metals have data points highlighted in successive greater generally added number each each added gure increases, electrons of charge. there c ausing shown greater 50 green bec ause increase atomic protons the is electrons is is to This nucleus, the 40 atomic number of the atomic radius against red, the noble gases in 30 n Plot 20 y U 0 Figure 3 C 0 p 50 o v i n c im ot a r e 01 / suidar 100 O 21 150 n y t i s m 200 This also increases, the valence results in the outermost electrostatic force of the nucleus by a y radii l atomic 250 232 from booklet. the atomic electronegativity. r shows observed P 3 are groups, such as Atomic radius Figure electron table, which helps reactivity of elements. The study periodicity. periods the according trends ionization energy, these and 10 of the data the c an about periodic ionic radius, c an in you e radius, result, s s you organized a Structure The diminished known as the electrostatic force eective nuclear experienced charge, Z e by a shielded valence 3.1 The periodic table: Classic ation of elements electron is . positively charged nucleus s s O t i s nucleus shield some of the positive nuclear charge The concept of eective nuclear charge based on electron shielding with or a larger eective than the positive electrons. from 13 atomic group radius, 2 2 a negative its electrons to the 17 tend are nuclear is the to the a the gas going and the ionic charge radius magnitude, equal to the number of electron from congurations. group increases. + 1 to For group 2, + example, Na form an anion, an atom gains more these electrons additional radius. The than protons. The electrons is smaller, greater the magnitude with radius the charge decreases equal to the number of from group 3 For electrons. the radius. anions Ionic To in atomic ionic form eight. charge charge anions. resulting nucleus than with larger than Al c alled increases. down This turn sublevel, larger . increasing nucleus. in greater minus E the is charge going the the noble than and example, N 15 to group 17, 2 is larger than O , which in turn a due charge v However, of and larger than F with have decreases the l nuclear O the as radii. u f x 15, valence is between charge, Groups as which valence radius 16 ions c ation, an atom loses protons c ations, increases 3 , attraction ionic c ations, for a i of the eect smaller radius 13, greater t a and of to these r o force ll group This is + negative to charge form ionic more d with electrons is to larger than Mg Ions to form in o is tend Therefore, To n and to and lead c ations. resulting U 1, c alled radius. charges valence are electrons, nuclear atomic Groups Similar 2 charge valence C so the its i n smaller of o Therefore, is positive more v one y Ions p r e Ionic radius y valence electron and n l y electrons between the Figure 4 e r P valence electron group, number similar of to ionic radius electron the trend increases energy for for levels atomic both and c ations and anions increased shielding radius. 233 Structure 3 Classic ation of matter Figure 5 shows atomic and ionic radii for c ations and anions. + Li Be Li + 2 Be F F 45 76 60 s s 99 133 130 + Mg Na + 2 Al Na + 3 Al Mg Cl Cl e 54 72 100 102 124 140 181 160 Ca + + 2 r K Data-based question Ca Br 117 Use the values from the data booklet 138 174 12 m t i s explain the trends in your graph. R adii of c ations are smaller than those of the parent larger than those of the parent atoms Ne, F from a dierent protons F to 2 2s For example, Mg + , 6 2p number electrons . Bec ause of they are protons, their size (gure 6). o to + Ne 2+ Na Mg 2 rst Structure 1.3 (AHL ), eject 2 2s 6 2p , an + IE → ionization table. shielding electron out of ionization a neutral energy (IE) is atom or the molecule minimum in energy itsground state. + X(g) (g) + e energy (IE This eect X is 1 ) bec ause becomes generally the greater, decreases number so less of down energy energy is the levels groups of the increases and the required to ionize atoms. a Going across a period, the number of protons in the nucleus increases, so the outermost electrons are held closer to the nucleus by the increased nuclear charge. At the same time, the shielding eect remains nearly constant because the number of inner electrons does not change. Therefore, more energy is required to remove outermost electrons, so ionization energy increases across a period. are general trend of decreasing ionization Structure 3.1.7 (AHL ). across 234 with C required The discussed in 2 o discussed in t a As u l v E which of 2 conguration. All of these species have the same electron conguration, 1s i d r o f x O period, ratio electron conguration 1s Ionization energy periodic increasing ionization a same the elements, the dierent sizes The are some discontinuities in across the have 220 R adii of anions are n U Figure 6 energy on i n 2 of dierent depending O trend all v varies and O p formed but have 2 , atoms. y species + Na r e Isoelectronic I 136 O Figure 5 I n values in 10 against their atomic number . Try to the l y the period 2 and period 3 elements There 196 200 a period is shown in gure 7 . energy down a group and increasing y 100 to plot a graph of the ionic radii for Br P K Structure 3.1 The periodic table: Classic ation of elements 2500 He Ne 1 Ar 1500 N Kr Xe O r 1000 Be B 500 Al Na 36 atoi nber Plot of rst ionization energy against atomic number for the elements from a more is trend for similar energy increasing is to (g) this of anity when across an a given an additional element, in energy: electron period is the the is the added. in table charge The the electron going across increases and general gure 8. trend of i o Cl electron is higher periodic nuclear shown F Br t a u l a 150 S O C 100 H E 50 v nortcele O k / ytiniffa f x lo 200 a d r o 1– 250 for ionization 350 300 is electron anity (EA) released electron EA reaction rst that + when n so favourable The period X released molecule. U a → or o more anity. e energy C The + the atom i n X(g) is neutral y to v attached p Electron anity (EA) 54 hydrogen to xenon r e Electron anity n 18 O 10 2 t i s 0 y Rb K Figure 7 l Li y tsrfi P n o it a i n o i H e grene s s lo 2000 Si Li Na P K Al B 0 0 5 10 15 20 25 30 35 40 atoic ner Figure 8 Plot of rst electron anity against atomic number for the elements from hydrogen to krypton 235 Structure 3 Classic ation of matter Going TOK US atom chemist Linus down gains the data in but not For some a an group, you electron gure 8 would due does to not predict the show that increased a clear less energy shielding trend: this is of is released when an the true nucleus. for However, group 1 elements, Pauling for the other groups. introduced the denition of Pauling described was also the electron dierent 3p was In 1954, he anity will between an electron reduce silicon the and to an energy atom results released. phosphorus, as in This a less stable explains phosphorus the has a sublevel. awarded the Nobel Prize in bonding for and his work on structure. do not always usually the c ase the Nobel data with full for or an element's half-lled electron sublevels, magnesium, manganese, zinc and the noble adding an electron is not favourable. weapons of half-lled anity. This is such gases, Peace Prize of to elements electron drop in as beryllium, where the process l mass destruction. scientists play in the Electronegativity peace in the world As dened in Structure 2.2, electronegativity today? attract What of are the ethic al responsibilities scientists? a electrons group, and complete increases valence 9 chemic al across shells will electrons. shows a a bond to period. attract summary of the periodic is of an atom’ s ability to bec ause more smaller easily than decreases atoms with down nearly larger atoms with discussed. p C n U o i d u l a E v O f x cimota t a r o suidar 236 measure Electronegativity trends atomic radius electron affinity Figure 9 a o v i n electronegativity This electrons ionization energy is itself. y valence Figure a r e fewer from O t i s of The direction of the arrow indic ates the direction of increasing value for each property ytivitagenortcele c an n role promotion y What y won opposition in years P he his accurate chemic al Eight nitrogen, later, have r We Chemistry for adding which e person to win Nobel Prizes in two c ategories. elements, conguration, rst ygrene noitazinoi here. s s electronegativity Structure 3.1 The periodic table: Classic ation of elements Activity Identify the element with the Property ionic radius each property. Ca C or The rst Si column Li or F has been Ba or done Cl for K or you. Br C or Pb C s or F I e anity ionization for Mg or Sr radius electron value iodine s s atomic greatest Strontium or I energy I r electronegativity I chemic al will who follow some explored elements in the of the 18th the work periodicity done of 3. by Julius the • Tool 2: Represent spreadsheets to manipulate data C alculate the atomic volume the volume is in graphic al form formula Use • Tool 3: Apply Inquiry trends 2: interpret scientific notation and Identify, and use SI prefixes and units describe and explain In a spreadsheet, radius for the the volume equation table a single of for from of the hydrogen to format hydrogen atom, volume A of B u l 4 H Li a 2 the a a graph 7 . At the 8. the time, properties To what the law Are there Again, column of atomic features of of every to Atomic by one mole of an copy and paste the obtain volume the hydrogen. down vs graph data atomic that for the other mass. illustrate the periodicity. chemists extent of . of up formula described eighth does the element the pattern as in pattern of similar the law your of octaves. graph support octaves? any elements that do not fit the general trend? sphere. number E v O f x Atomic below). t a r o using elements o C alculate the example mass and atomic i (see atomic d 2. data strontium collect Discuss n 1. Plot 6. phenomenon U Method 5. patterns, relationships the elements. C • and mol down taken o 3: v Tool i n • pasting volume 1 3 element, in m data and y Use volume of a single atom of the other copying p 2: r e 4. Tool the by the column. century. Relevant skills • C alculate elements known n you Meyer, l task, O this Lothar t i s In y P y Periodic trends in atomic volume E Volume of a Atomic Atomic radius mass /×10 Atomic single atom radius/m /m 3 12 m Atomic volume 3 /m 1 mol 1 2 3 4 237 Structure 3 Classic ation of matter TOK You have concepts now in keep It about chemistry c an asking be ‘why’. various that c an interesting For Because uorine atoms have a higher nuclear charge than interconnected be to carbon atoms, and both types of atoms have the same used to construct see how far back s s explanations. c an learnt DP number of energy levels. you example: Why? The C—F bond is highly polar. Because uorine atoms contain nine protons each, e whereas carbon atoms contain six. Fluorine and carbon Why? atoms have two electrons in the rst energy level, and their Because the electrons in the bond are not distributed r valence electrons are in the second energy level. equally between the two atoms. Because the rst energy level holds up to two electrons. Because the dierence in electronegativity is large. l Why? Why? And so on. use to How build Because uorine has a stronger tendency to attract a good about the Why? explanations Why? acquire the such knowledge that as this? What makes a “explain” How c an should questions you know in your IB how detailed your y r e examinations. Because uorine has a smaller atomic radius than carbon. you explanation? bonded pair of electrons towards itself than carbon. Think did explanations O you t i s Why? n y Because the rst energy level only has one orbital. Because uorine is more electronegative than carbon. be? p o v i n Metallic character and periodicity C (Structure 3.1.4) character value it. to ionization electrons not in energy. their a the quantitative metallic Elements structure. For example, the that greater u The in they group of l character 17 is a E v to accept We c an some the an more donate linked of are to the group high that an elements is, we element metallic in these chemic al known metallic reactive their do is character with known as as lower the not assign a closely have linked to deloc alized ionization decrease chemic al energies trends in the reactions. alkali the We have down group, in the as decreasing group. This they chemic al have a reactions. describe the halogens property is the opposite to metallic The non-metallic going reactive, reactions metals, going electron halogens. electronegativity. more alkali down valence character — this therefore the character going single halogens are electron illustrate typic al also to elements, non-metallic electronegativity up 1 increasing elements their and and are tendency terms higher group energy t a means with of o more likely to be metallic. ionization property, character Therefore, are i d r o f x O 238 is However, n its U Metallic and down as ll metals the they their and character and group. have a valence the Halogens greater tendency sublevel. halogens by looking at y P Why? Why? Structure 3.1 The periodic table: Classic ation of elements Reactions of alkali metals with water + A Brønsted–Lowry acid is a substance that donates protons, H . A Brønsted– Brønsted–Lowry acids and bases, + Lowry base is a substance that accepts protons, H . The alkali metals all form and aqueous Brønsted–Lowry bases when reacted with water. The general the alkaline the reaction of a group 1 metal with water is as and basic between substances follows: covered in Reactivity 3.1. 1 M(s) + H 2 O(l) H → 2 (g) + MOH(aq) 2 example, lithium reacts with water as e For follows: 1 + H 2 O(l) → H 2 (g) + r Li(s) LiOH(aq) 2 going down the group. As a result, reactivity increases and the reactions of alkali character octet, of the react with becoming halogens other species anions. decrease The to gain electrons, and to O halogens complete achieve electronegativity and non-metallic going down the group, and hence their Figure 10 c an illustrate elemental and halogens a elemental as other uorine + Cl → F Cl + 2 is bec ause not not happen, uorine is more however: chlorine electronegative favourable. an c annot than When elemental electron, This eect increases going down group 1 reactions forming a In this reaction, chloride loses electrons and (undergoes uorine gains oxidation) electrons oxidize uoride (undergoes reduction). Oxidation chlorine, so the and n is will the halogens. follows: 2 reaction at gains U This chlorine of looking Elemental sodium metal reacts vigorously with water. C 2 reverse reaction anions elemental by 1 F ions. halogens the ion, group. i n 1 the and chloride 2 The of the o ion with reactivity down v reacts uoride the going p between uorine decreases y We also r e reactivity n y a t i s Elemental l metals with water occur faster and more vigorously going down the group. y P Ionization energy decreases and the metallic character of the alkali metals increases Reactions of halide ions with halogens are s s for distinction equation reduction are discussed in Reactivity 3.2. The greater the electronegativity dierence between the two reacting species, the 2 o d more readily the reaction occurs. Hence, F will react faster with I than with Cl This is also the case for the reactions between halogens and alkali metals: F 2 . reacts i r o faster with K than with Li. The equation for the former reaction is shown below: t a 1 + F 2 + K → K 2 + F u f x Activity the row table has to been show a rst l The done v O Complete which for reactions F F Cl Br I 2 2 2 No reaction E 2 c an occur between halogen Cl F species. you. 2 + No 2Cl → 2F Br + Cl 2 F 2 + 2Br → 2F I + Br 2 F 2 + 2I → 2F + I 2 reaction No reaction No reaction 239 Structure 3 Classic ation of matter Practice questions 1. Write the between 2. List the word alkali symbol in metals terms of and equations from the their reactions most with 1 17 least b. Ba c. K d. F I + most or 2 I 2 Li or reactive + Ba F pair of reactants in to + Cl 2 water. ATL + 2 Cl + 2 or Li + KCl or F 2 d Cl + 2 KI Thinking skills The modern periodic table is a very useful way of organizing the elements. Conduct an online search for alternative representations of the periodic table. exploring An example is shown in gure 11. reactivity of elements? Tool 2) O t i s y p r e C o v i n n U t a Figure 11 o i d r o This periodic table has a “bicycle wheel” structure, with each group being a spoke of the wheel u l f x a E v O 240 a 2 n 2, group the the + y (Inquiry and chemic al to Li l group in in Identify a. Select two or three alternative periodic tables and consider: • What • To features • What are the advantages and disadvantages of the dierent periodic tables? what of extent the are elements they are essentially being the highlighted same as the in each standard c ase? periodic table? y trends used 3. r often reaction P the reactive are simulations and online reactions the e Linking question Why for water. s s reactive, and potassium Structure 3.1 The periodic table: Classic ation of elements Metal oxides and non-metal oxides (Structure 3.1.5) Another way to describe acids and bases is through their ability to accept or donate a pair of electrons: a Lewis acid can accept an electron pair, and a Lewis base can s s donate an electron pair. Many metal oxides are Lewis bases. They react with water to form hydroxides, also bases, by donating an electron pair to hydrogen in water. Reactions of alkali metal oxides with water have the general equation: are O(s) some 2 O(l) → forms that H a 2 2 6H O(l) O(l) or donate an acids by CO 3 (aq) H 2 2 SO 3 (aq) sulfurous acid H 2 SO 4 (aq) sulfuric acid O(l) 4H oxides a of the elements stronger base stronger acid than P as both a Lewis For example, aluminium oxide, Al pair other examples: 3 PO 4 (aq) phosphoric acid forms behaves 2 p species (l) form c arbonic acid u 3 the O(s) to some O(l) a v period, 2 are (aq) hydroxide o H water Here i H l + oxide Na with water. 2 o 2 react in (s) n H + (s) They oxygen Ba(OH) barium 2 hydroxide C O(l) t a (l) any form v acids. from + water and SO electron 2 trioxide 10 to Mg(OH) i n (g) 3 water O(l) d 2 example, chemic al H + dioxide amphoteric. an Lewis pair r o A are (g) E with For 2 U 2 with (aq) H + dioxide across react magnesium f x O acidic. 2 also equation: oxide phosphorus(V) Going M(OH) + electron SO They general oxide oxides an bases. hydroxide examples: accepting O 2LiOH(aq) n H Lewis following Non-metallic 4 O(l) l also the BaO(s) barium P 2 hydroxide O are + magnesium sulfur H lithium MgO(s) SO 2NaOH(aq) y some sulfur O(l) sodium + with CO 2 oxide oxides c arbon H r e are + O(s) MO(s) Here examples: t i s 2 hydroxides, 2MOH(aq) oxide lithium Group → y 2 O(l) O(s) sodium Li 2 y 2 H P Na + r Here 2 e M electron pair 2 4 become than O 10 acid O 3 less MgO(s) basic in its and more reaction (s). and a Lewis base is termed (s), is amphoteric: it c an accept depending on what it is reacting with. 241 Structure 3 Classic ation of matter For example, oxide Al 2 acts O 3 as when a reacting (s) + with (s) + 3H 2 → O(l) 2Na[Al(OH) 4 acid, aluminium → 6HCl(aq) 2AlCl (aq) as a Lewis base: + 3H 2 r chloride neutralization reactions: an acid each other to form a salt. and a O(l) base react standard Na level of DP 2 O(s) MgO(s) Al know and do not denition bases. of need to Lewis acids Acid or Lewis acids and bases in basic 3 (s) Trend O(l) o dioxide and atmosphere by 2 is of nitrogen to P 4 O 10 (s) SO acidic SO 3 2 (l) and (g) acidic 3 corresponds naturally weak acidic to a due neutral solution: it to the presence of c arbonic acid: (aq) 5.6. We know the right rain is in 2 , the more that are other more periodic acidic oxides, such as acidic than than CO 2 bec ause table. If these gases normal, and is known as less than 5.6. oxides vegetation. industrial is dioxide, SO ensuing pH which forms CO sulfur further the a H rainwater are has 7 .0, which → and rainwater, decomposing are These processes, produced pollutants such as naturally c an the also by be volc anic eruptions released into the combustion of fossil fuels with levels of sulfur impurities. u You have seen the reaction of SO l 2 with water that produces sulfurous acid. The reaction between water and nitrogen dioxide produces nitric acid and nitrous acid: a v O f x high 2 for sulfur i and H value which t a r o d in acid rain, Sulfur + of R ainwater dioxide, oxides, NO and pH C pH (g) a basic. x nitrogen dissolve 2 has n U typic al nitrogen nor c arbon CO A 298 K i n dissolved (s) acidic p at acidic o water neither v Acid rain and ocean acidic ation is 2 in the acid–base properties of the oxides of some period 3 elements Reactivity 3.4 (AHL ) Pure SiO amphoteric base? more detail in Table 2 O y discussed basic 2 r e are the you O oxide chemistry, n the t i s Formula of At l y The acid–base properties of some period 3 elements are shown in table 2. with 2NO nitrogen 2 (g) + H 2 → O(l) dioxide HNO nitric 3 (aq) + acid HNO 2 (aq) nitrous acid Linking question do explain E How Ultimately, to nitric nitrous acid and acid and sulfuric sulfurous acid, acid are oxidized by atmospheric oxygen respectively: differences in bonding the differences in the 2HNO 2 (aq) + O 2 (g) → 2HNO 3 (aq) properties of metal and non2H metal oxides? Structure 2.2) (Structure 2 SO (aq) + O 2 (g) → 2H 2 SO 4 (aq) 2.1 and Oceans absorb atmosphere. acidity 242 3 c an As a large a aect proportion result, the c arbonic ability of of the acid coral is reefs c arbon dioxide formed in and the shellsh released into the ocean. to Increased ocean c alcify their skeletons. y are 3 acts P reactions oxide aluminium acid these ](aq) sodium aluminate hydrochloric oxide of + hydrochloric aluminium Both base, aluminium e 3 a s s O hydroxide, hydroxide reacting 2 sodium sodium oxide Al aqueous 2NaOH(aq) aluminium When with Lewis acid: Structure 3.1 The periodic table: Figure 12 reacts water and Classic ation of elements Sulfur dioxide reacts with oxygen in the atmosphere to in the air form acid sulfur dioxide sulfur sulfuric acid, c ausing acid rain rain trioxide H SO 2 4 s s e r corrodes metals burning kills erodes trees wet acidifies limestone other surfaces lakes (fish t i s buildings fuels die) O Oxidation states (Structure 3.1.6) purely ionic or covalent, but rather occurs on a continuum with unequal sharing v represent the charge on an atom in a compound if it were composed of ions. It also C i n Water , for example, is a predominantly covalent molecule: H o describes the number of electrons shared or transferred when forming a bond. p of electrons. The concept of oxidation state, or oxidation number, is used to y r e The bonding triangle introduced in Structure 2.4 shows that bonding is oen not n and y erodes coal l power stations y P volc anoes 2 O O. If it were an ionic 2 H +1 H +1 compound, then oxygen would gain two electrons and each hydrogen atom would Figure 13 assigned an oxidation state of –2 and each hydrogen has an oxidation state of +1. elements in water a more electronegative electronegative number of oxidation useful deducing oxidation states of atoms: u of all zero. atoms in a polyatomic Rule 4 The oxidation state of uorine is –1 in all compounds. Rule 5 The oxidation the E The of group oxidation more state v state Rule 6 of 2 state The oxidation state electronegative with state the less is of 2 so O of a is is equal to 2 , 1 metals is always +1 and the oxidation always +2. oxygen is –2 oxygen except in OF has an 2 , where uorine is oxidation state of +2, and in where it is –1. hydrogen elements electronegative c alled group metals electronegative, peroxides such as H Rule 7 ion charge of the ion. a O l f x The states is oxidations oxidation states. The sum of the oxidation states of all atoms in a neutral compound is zero. oxidation element the negative positive Rule 3 the free assigned Rule 2 of any are assigned The sum of for atoms are Rule 1 the state rules atoms i are the less t a There the r o and o compounds, states d Deducing oxidation states in compounds In The oxidation states of n U lose one, as oxygen is more electronegative than hydrogen. Therefore, oxygen is (most is +1 when it non-metals), elements (metals). is combined and –1 when Hydrogen in with it a is –1 more combined oxidation hydride ion. 243 Structure 3 Classic ation of matter The is rst rule bec ause states atoms that of electrons are example, consider shared electronegativity is oxidation same evenly across elemental of the state element bonds oxygen, O atoms is of have 2 in : identic al. an element equal elemental the is always zero. This electronegativity, so the compounds. molecule Therefore, is the For symmetric al, and the oxidation state zero. Deduce the oxidation state of the atoms in H c. P in the phosphate ion, d. all the elements in LiAlF 4 Cu, Zn, Xe, and Au P , 3– PO 4 states of its (oxidation state of K) 0 multiply the 4 is a neutral compound, the sum of the equal to zero: (oxidation state of Mn) oxidation state value for rule the + the above to rule 5, oxidation equation the b. the a. 2 Cr 2 O CO d. HClO 2 4 2 2 3 Fe j. CH k. CH l. Na 2 O 3 2 2 3 OH O O 2 E h. v NaH a SO 4 of its to 3, is as + rule 6, above oxidation state of P Using rule state of F 2: According 4, the above 1 + to is of –2. and then states nd there the are value four for Mn zero. potassium is Substituting (–2 × 4) = +1. According to these values into 0 so each atom is 4 to has an oxidation state of 0 polyatomic (–2 × state above 4) ion, the sum of the oxidation charge on the ion: oxidation equation + a the (oxidation state of O = of × 4) = oxygen –3 is –2. Substituting this gives: –3 = +5 (oxidation × 4) K, as gives: + PO equal the the state oxygen species, According into and four, 3 rule atoms to O oxidation by = +7 elemental (oxidation state of P) d. of above (oxidation state of P) value l O OF f. According states u CH c. e. 7 c. following f x b. g. 244 K all o in are (rule 1). i atom compounds: These to the oxidation (oxidation state of Mn) t a each oxidation states to d Assign r o 4. of state oxidation state of Mn Practice questions states sum n U 1 6, the C i n According oxidation make p assign o would v c an that (oxidation state of O oxygen oxygen atoms in the compound. You + y We = + is r e 4) atoms O oxidation t i s According to rule 2, as KMnO n y Solution × l 4 state of Li) + (oxidation state of Al) + (oxidation = 0 rule oxidation equation 5, the state of above oxidation state of lithium is uorine –1. Substituting gives: (oxidation state of Al) oxidation state of Al is + = +3 (–1 × 4) = 0 +1. According to rule these values into the y Mn in KMnO b. r a. 2 each e Worked example 1 a. of s s atom the the Structure 3.1 The periodic table: Classic ation of elements Naming of oxyanions Oxyanions are oxyanions, you a Roman with no at anions include the space end of that include the oxidation the name. between the ion example, consider the anion MnO Mn has an potassium This . 4 The Roman name The and atoms. the numerals the anion is When naming non-oxygen atom as are named manganate(VII), neutral compound KMnO compound is also enclosed in bracket. known under 4 its is c alled traditional Practice questions permanganate. 5. the KClO , metal, for non-oxygen is c alled Roman used atom in potassium numeral is to an refer to the oxyanion c an also be a non-metal. omitted for some common oxyanions. These nitrite nitrate(III) nitrate nitrate(V) 2 3 2 SO sulte 3 systematic names of some oxyanions. have nomenclature the to Your data IUPAC and study booklet, brackets by of in the periodic Linking question How c an used to will chemic al oxidation states be analyse redox reactions? for a long time table, and the International chemistry writing where elements the i subscripts dierent symbols bromate(III) Union involve formulas of Pure and using symbols correctly necessary) and naming substances t a according in identify the encouraged (IUPAC). r o (including stated to use conventions Chemistry are we o they symbols d Applied used Nowadays, n 14). U (gure Both are C Communic ation skills Scientists as i n acceptable in examinations ATL magnesium potassium sulte o The common names and c. (Reactivity 3.2) sulfate(VI) v Table 3 sulfate 4 for nitrate d. p SO hydrogen sodium phosphate(V) sulfate(IV) 2 formulas a. O Systematic name chemic al b. y Common name r e NO t i s NO the following compounds: chlorate(VII). shown in table 3. Ionic formula Write the oxidation state of a n the oen l practice, are the example, are y In 4 numerals y transition Roman P While r name, manganate(VII). +7 . The of e potassium oxidation state of oxygen state s s as should numeral brackets For polyatomic guidance. T est your chemical communication skills by nding the error in each of the following and writing the correct answer: 3. O3 iron, 4. MgOH2 5. cuprum(II) oxide a E v O Figure 14 u NE l HCL 2. f x 1. The symbols used by alchemists for copper and respectively 245 3 Classic ation of matter LHA Structure Discontinuities in ionization energy trends (Structure 3.1.7) E arlier going in this topic, across 13, you period. and from saw You group that may 15 there have to was a trend noticed group 16 two of increasing ionization discontinuities from (gure 15). e 2 Ne l N Ar n o it a i n o i Be t i s B t s r if O Mg Al Na the for o u between l a E v O f x t a come is 2s the in therefore the between two of to higher This of c an be s remove energy drop and nitrogen in p explained conguration 2 beryllium is 1s easier a boron, electron an 2s by boron 1 2p . Despite the electron sublevel, ionization of from this which energy is shielded therefore sublevels. and oxygen, c an be explained by 2 close space in existence 2 into of is is sublevel. orbitals. oxygen it is 1s region electron electron it and The third period elements 2 conguration boron, discontinuity, repulsion for not same by evidence i d r o while in valence nucleus second electron p the the electron charge n U as provides The the nuclear beryllium congurations. Z and o and between electron C 2 element from their y r e discontinuity, at 2s greater v 2 is 1s The ionization energies of the second i n rst looking 3rd period Si Li The Cl S P n y O C The 2 2s . have easier conguration of nitrogen is 1s In to nitrogen, The paired increased remove. the three 3 2p electrons repulsion. This electrons in the 2p orbitals explains in The the oxygen occupy the higher-energy drop in + N N 3 2 2p 2p 2 2 2s 2s + O O 4 3 2p 2p 2 2 2s 2s A half-lled electron repulsion, p sublevel is stable due to decreased as none of the electrons are paired paired ionization groups. Figure 16 2 2s 4 2p proximity. and electron energy y P grene F Figure 15 246 r lom 2nd period 2 atomic number, do energy group 2 to s s group a Structure 3.1 The periodic table: Classic ation of elements LHA Properties of the transition elements (Structure 3.1.8) A transition or c an element form gives example, the metallic ions elements is elements have This not which considered a not also c atalyst occur, a type solid oen of that the 2 incomplete exist as up in a rhodium in the 2CO of d-sublevel. electrons the phase to the required the reactants. Gas are transition exhaust of elements converters vehicle, 2 used adsorb a reaction and Transition chemic al as C atalysts desorb activation energy are Reactivity 2.2 monoxide c arbon dioxide: Linking question n What are the arguments for and against including scandium as a transition element? (Structure 2.3) adsorbs onto surface of solid catalyst at active site u E v a tsylatac l tsylatac tcudorp O f x t a r o i o d c u d A and discussed further in heterogeneous c arbon reactant A catalyst for undergo t o r p from catalyst electron molecules adsorb onto molecules heterogeneous catalyst product desorbs Its form 2+ ions, electrons in elemental (gure 17). C atalytic to Heterogeneous c atalysts c atalysts. and d energy reaction. separate element, 4s means that zinc is not magnetic. activation that heterogeneous desorb its All which the in loses intact. state, reduces converters. → an o O with readily remains used then and it U + so transition molecules 2CO , C c atalytic oxygen that used palladium in c ations neutral being products creates a v c atalysts The result they external magnetic eld. i n Platinum, not c atalyst surface changes. and of are its species a p elements a a as electron y are is while d r e to in unpaired have high melting and 10 3d d-sublevel paired elements electrons, throughout electrons and the O A complete are an d an deloc alized t i s the [Ar] 4s with of properties. deloc alized these d-sublevel transition element despite being in the d-block. This form 2 is align spin are lled Having an incomplete n does c an and the unpaired the partially y it bec ause so a characteristic energy strong, has d-sublevel. y moment, is some between transition conguration zinc high attraction transition bec ause but elements have that l is The element incomplete in paramagnetic. Zinc electrons an an Most magnetic is d transition structure. as with P are the the dened r points. is c ation e For metal stable s s d-sublevel a A B reaction occurs reactant B adsorbs onto surface on catalyst catalyst of solid catalyst at active site B Figure 17 c atalysts. The action of heterogeneous Transition elements are oen A good heterogeneous c atalysts 247 3 Classic ation of matter LHA Structure This reduces when fuel H 2 cells. In emission fuel. this of the Platinum process, H 2 is products also used (g) and O 2 of as (g) incomplete a combustion heterogeneous adsorb onto the c atalyst created in hydrogen c atalyst surface and O(l) desorbs. some of elements which are have also able distinctive to form compounds known as s s Transition is the burning complex ions, colours. The chemistry behind this phenomenon detailed later in this topic. e r Variable oxidation states in transition elements (Structure 3.1.9) other table, element, you c an if you know determine its 3d. c ase sublevels. into 4s 2 t a argon: 1s The two u sublevel 2 and to in achieve to a Iron 3d valence 8, electron the is in period comes aer electrons. so it six has previous noble gas in go into 4, 4s This eight 6 2 3p 4s the In so its c an be valence the valence bec ause it is higher 3d deduced electrons. sublevel. the of iron is electron [Ar] 4s 6 3d . If you conguration of 6 method described c ases, stable l [Ar] 4s a 4s electron 1 is expand from Two go 3d both more copper conguration conguration, 2 3s copper. and with argon: [Ar] remaining 5 3d starts Fe. 2 6 2p of group the condensed exceptions 1 [Ar] 4s and electron 2s is o the full number Iron sublevel, i the the number. Therefore, need in the electron worked is example are promoted into the 3d conguration. Therefore, chromium is 10 3d . a Variable oxidation states Another property of the transition elements is their ability to form a variety of oxidation states of elements stable are ions in dierent oxidation states. These are known as variable oxidation given in states. Some of the booklet. given 248 the chromium v transition the data d r o common E the f x O The determine group n U the is Remember, inenergy. Then, O valence and transition element on the y the 4s this C are conguration in o identify sublevels which a electron conguration of iron, p electron brackets, i n Then, v condensed square of conguration. r e Solution The position t i s Worked example 2 Determine the condensed the electron n any l with periodic y As in gure 18. most common oxidation states of transition elements are y P Deducing electron congurations of transition elements Structure Ti V Cr Mn Fe Co Ni The periodic table: Classic ation of elements LHA Sc 3.1 Cu 7 6 6 6 5 5 5 5 5 4 4 4 4 4 4 3 3 3 3 3 3 3 3 3 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 oxidation that occur in s s states 4 compounds +2, c an +4 of formation most manganese(IV) variable oxidation common means that oxide, states in and it oxidation states of forms the following manganese(VII) transition elements oxides: oxide. c an be explained Methods by the fact that their successive ionization energies are close in value. The lling successive ionization energies of chromium and copper are shown in gure for drawing orbital diagrams are 19, Structure 1.3 along with orbital lling diagrams. 0 (b) copper om 30000 l 10000 5000 v E n o it a z i n o i 15000 10 12 14 10 12 14 [Ar] 4s 3d [Ar] 4s 3d a O k / ygrene 20000 8 ionization 35000 25000 6 u 1– f x 40000 4 i 2 t a r o 0 o 5000 n 10000 d n o it a z i n o i 15000 o 20000 C 25000 U k / y g r e n e i n om 30000 v 1– 35000 described in y chromium 40000 p r e (a) their n The the This O oxide, that +7. t i s manganese(II) see and y you are y 18, 4 transition elements, common oxidation states are shown in a larger font l gure manganese The most e From Common oxidation states of the period to zinc. P sc andium r Figure 18 1 1 0 0 2 4 6 8 ionization Figure 19 The rst 13 successive ionization energies and orbital lling diagrams for (a) chromium (b) copper 249 Structure 3 Classic ation of matter LHA 1 Chromium forms ions with oxidation states ranging from +1 to +6, losing the 4s 3 electron rst and then the d electrons. The most common an electron chromium energy has conguration of [Ar] 3d an between electron losing , and CrO conguration the rst of electron [Ar]. to , which 2 and Cr 4 There through + are Cr 2 3 has ions is the a 2 O very sixth , in which 7 small increase in electron, so all six s s Activity oxidation Find data for the rst 13 ionization The states increments are possible. between successive ionization energies for copper are larger + 4 graph for your which has an electron forms fewer conguration conguration of [Ar] 3d for the it stable ions. of [Ar] 2 3d the . Cu(III) superconductors, but higher Sc andium electron and Cu(IV) oxidation the up until the 1920s, are conguration [Ar] 4s and it was not has an are Cu not stable. 1 3d 3 + . Sc considered a was the only known transition element, as a energies, for element has to be able to form ions with each However, sc andium c an also exist in the +2 t i s 1 conguration of [Ar] 3d . An example both lling here: c an exist in diagrams v atom/ion of d-sublevels. having compound an is electron C sScCl +2 and these +3 3 oxidation states. The elements and their ions are electron structure o 4s 3d C i n Mn iron sc andium(II) p given and orbital a y condensed r e M anganese of lled state, O Practice questions 6. partially oxidation transition element. n ions [Ar] U 2+ Mn n [Ar] 3+ Mn Fe o i d t a r o [Ar] 2+ Fe u l f x a E v O 250 [Ar] [Ar] 3+ Fe [Ar] 2 Explain why Mn 3 oxidized 7 . to Vanadium + 3 is + more common than Mn 2 and why Fe + c an be easily + Fe commonly a. Determine b. Explain the forms compounds condensed electron where it has a +5 oxidation state. conguration of elemental vanadium. why the most common , electron l ionization common has ions exist in some compounds, such as in states y the which dierence in transition and , electron ion successive common + and Cu 2 congurations, The oxidation state of vanadium is +5. y between so P connections look chromium, 9 each element. graphs, for 10 transition elements and plot Using those r a than e energies of the other period Structure 3.1 The periodic table: Classic ation of elements LHA Transition element complexes (Structure 3.1.10) Transition ions elements when form bonded with coloured ligands. compounds Ligands are known as complexes or complex electrons that coordination c an bond be donated formed to a between transition the element ligands and c ation. the This central s s OH of 3+ molecules or ions with a lone pair 2 results in a c ation H O OH 2 (gure 20). When ligands the form Normally, visible gap light all ve d between region of are bonds split orbitals into have with two the orbitals transition sets same the split the electromagnetic with energy: corresponds element dierent to they a c ations, the energies are H (gure O 2 degenerate. The OH wavelength within the spectrum. 2 2 3 The complex ion [Cr(H O) y l Figure 20 OH y energy coordination d-sublevel P 21). in r orbitals 2 e Cr 2 6 + ] has six coordination bonds from the lone + 2 t i s ygrene The formation detailed in (b) gaseous ion with non-degenerate in wavelength orbitals. absorbed. gure 22. the The the split higher corresponding colour colour wheel is colours also electrons happens the energy in when gap the the lower complex between the are opposites on the colour wheel shown in the data booklet. i 585 nm orange u l yellow 575 nm a 400 nm E v O f x red to t a r o 647 nm 700 nm allows This observed is the complementary colour to Complementary The orbitals orbitals. o colour given a between into d the energy complex ions is o light in promoted C absorbs be orbitals of the transition element ion n to of ligands on the d i n dierence orbitals degenerate orbitals U The The eect ion — non- v — degenerate orbitals Figure 21 five d orbitals in a complex of Reactivity 3.4 y isolated p five d orbitals in an r e (a) O molecules to the central ion. It is violet in colour O level 3 Cr n pairs on the H green violet blue 424 nm Figure 22 491 nm The colour wheel 251 Structure 3 Classic ation of matter LHA 3 For example, [Cr(H between the opposite colour The larger shorter the an d on O) + ] absorbs orbitals. the splitting electron. wavelength wavelength, 6 As colour of the and frequency light (3.00 × 10 light when electrons the complex the greater orbitals, for a larger frequency. the splitting, energy the Remember light the 1 m s frequency (in s ) f have a between × wavelength (in m) × λ r = constant × J s) (6.63 × 10 1 frequency (in s × ) f 2 The colour of [CuCl ] ions is equivalent to visible light with a wavelength of 4 2 647 nm. Calculate the energy gap of the split d orbitals in [CuCl of the of light complex the wavelength metres. U splitting l stronger a gap × 10 to determine of d the of 491 × 6.11 × 10 f × λ. wheel 647 nm, the (gure 22): Remember to convert 10 m × f 1 s energy gap between the orbitals: 34 E = 6.63 × 10 ligand the the split and the complex orbitals coordination 1 14 J s × 6.11 × 10 depending bonds d transition observed. with orbitals higher on the s the J element The same central central ion c ation ligand metal will are will c ation. produce a two factors that c ause dierent Ligands greater that form energy (gure 23). greater splitting of of light bonds orbitals absorbed strength and wavelength colour 9 = the stronger between ligand = 14 f = coordination The relationship the 19 the between c E = 4.05 × 10 colour in × f to the 1 m s n u the o i t a identities aect v Figure 23 E strength d r o f x O greater ligand 8 E = h on frequency of the light, use to 3 Then use The opposite C i n determine the corresponds is o 491 nm. To ion absorbed y the v colour wavelength ] 4 p r e Solution If O t i s Worked example 3 n y l h energy smaller greater wavelength of wavelength of light absorbed complex colour the colour of complexes The third factor aecting the colour observed is the charge on the central transitional + 3 element cation. A Cr + 2 complex will have a dierent splitting of d orbitals than a Cr complex with the same ligands, and they will therefore be of dierent colours. 252 y Planck to 1 = ) P = required to needs relationships 34 = E promoted energy: c energy (in J) are appears violet, as this is the e of d higher and yellow result, wheel. Therefore, 8 speed a s s promote split 2 Structure 3.1 The periodic table: Classic ation of elements LHA Practice question 2 8. A complex Chloride colour [Cu(H ions 2 are change O) 6 + ] (aq) changes considered to be colour weaker when excess Cl ligands than (aq) is added. water. Explain the observed. s s Colorimetric analysis of a solution of unknown concentration this of a task, you transition curve. app that in You c an the solution will need measure unknown Method concentration by interpolating a a 1. Review 2. Read the colorimetry section of the colorimeter, or a mobile RGB through the procedure values. equipment you will chapter. need. Relevant skills • Tool • Tool 1: Prepare • Tool 1: C arry out dilutions • Tool 3: Construct • Tool 3: Interpolate 1: Use and apply colorimetry standard solution Prepare a interpret C alibrate uncertainty measuring bars apparatus, including • Inquiry 1: research questions transition metal solutions include U Possible Formulate those containing manganate(VII) ions, MnO and complex ions, pentaaquathiocyanatoiron(III) O) 5 (SCN)] is also of a of concentration task of a You in , complex ions is not you will be to obtain concentrations. five colorimeter. the absorbances a c alibration curve that of the shows five standard absorbance vs Check 9. for errors Propagate and by verifying that your line of best fit through the origin. uncertainties graph. and include error bars on Draw lines to determine the maximum minimum gradients. Part 2: Analysis of solution of unknown concentration 10. Obtain a small sample of the solution of unknown technique, so for SL students. this complex technician. of analytic al Plot goes concentration. Determine the absorbance of the analysed solution. involves determining the solution prepared by 11. Use your c alibration curve to determine the your concentration of the analysed ion in the solution. c an also use it to determine particular u f x the or important suitable version concentration teacher Knowledge ] i task simplied . this 6 t a A use r o this to O) ion dilution concentration. 8. o 2 necessary 2 7 . your complex ions, d 2+ [Fe(H [Cu(H Determine , 2+ hexaaquacopper(II) 6. n 4 the C i n conclusions C alibrate o Discuss the impact of uncertainties on the v Inquiry 3: the wavelength of incident light. 5. solutions. sensors of serial y Inquiry 1: and different p • Draw graphs r e Tool 3: of a Depending on the colour of the solution, determine a graphs suitable • solution perform O 4. interpret and t i s solutions and standard analysing, n a y 3. • l Part 1: Calibration curve preparation y the Skills below and make a list of P phone determine r c alibration will metal e In coloured component in everyday substances. 12. Determine uncertainty the of absolute, the relative and percentage concentration. l a E v O Part 3: Developing a research question 13. Brainstorm involve possible using this research questions that could analytic al technique. Linking question What is ligands How the in c an nature forming of colorimetry concentration of a the reaction complex or ions? between transition element ions and (Reactivity 3.4) spectrophotometry solution of coloured be ions? used to c alculate the (Tool 1, Inquiry 2) 253 Structure 3 Classic ation of matter End of topic questions A. –141 B. –141 C. (–141 D. –141 Topic review 23 Using your knowledge from the Structure 3.1 × 2 × s s 1. 6.02 × 10 topic, 23 answer the guiding question as fully as possible: + 753) × 6.02 × 10 753 e How does the periodic table help us to predict patterns + and trends in the properties of elements? 7 . Which of Li and I of the following does C. C s and I D. Li and F transition element D. element Which the following added SiO forms an acidic solution water? 2 9. SO II and III only Which of the following charge constant shielding following C. X D. X (g) + e → + 3 (g) → 2 of energy, from → 2e the in one kJ, (g) (g) + X (g) E (g) X + X + Which + 2 (g) 2 + X + + EA of oxygen 254 N A = a tendency lower reduce I the +1 oxidation B. of the state +2 C. following is to be of reduced D. a form variable low oxidation states D. complex electric al conductivity of the ion formation following metal Electrons statements compounds absorb Electrons release lower-energy one mole of = oxide ions atoms? –141 kJ mol 6.02 x 10 ; 2nd 1 mol ) EA of oxygen 2 property of are energy explains why coloured? when they move to a C. Transition D. Electrons What is the energy when they drop elements are 4 ] = A. 2 have variable deloc alized oxidation state of in oxidation states transition elements cobalt in the ? B. 0 down to a level represents the total amount of oxygen Br +6 typic al 2 to than nitrogen in sodium +5 not 2 properties C. Which correct? elements? c atalytic [CoCl 23 ; Br will is reducing agent than F oxidize B. B. 12. 1 753 kJ mol 2 weaker halogens higher-energy d-orbital 1 (1st has 2 Which A. e following of a nitrate. second ionization e required mole a Cl transition a X → v O B. X(g) shows l 2 A. increases 11. u the electrons o D. of valence is transition why t a nuclear Which of explain periods? decreases C. energy? 6. number radius not D. A. i atomic across r o increasing B. f x 5. A. trends does increases d electronegativity I n I and II only D. U C. 2 C. A. 10. 2 will the C. +2 D. +4 complex ion LHA III only I C i n B. 4. SrO II only Cl B. about o III. A. 2 A. Deduce v II. statement p I. to oxides Which y when of 8. 2 2 n d-block element C. zinc? O B. to vigorously? y metal apply most 2 2 t i s A. not r e 3. Which react l 2. would y B. Multiple-choice questions elements P C s and F r Exam-style questions pair A. Structure LHA 13. What [NH 4 ] is 2 the ligand [Fe(H 2 O) 6 in the ][SO 4 ] 2 17 . complex The periodic table: Classic ation of elements This question is about acid deposition. ? a. 2 A. 3.1 Explain, using relevant chemic al equation(s), + why Fe rain is naturally acidic. [2] 2 [SO C. H D. [NH 4 b. ] Explain why production 2 burning of acid coal leads to the s s B. rain. [2] O c. + 4 Figure 12 on page 243 shows some of the ] consequences of acid deposition. e x 14. [Ni(NH 3 ) 6 ] is a complex ion with nickel in the +2 i. oxidation state. What is the overall charge, x, of List the ve consequences of acid deposition the in gure r shown 12. complex? 0 B. 1+ rain. Suggest graph energies how State of c. State the c an the be copper. abbreviated electron 15. This question is Explain why the has a greater the chloride chlorine ion has atom. gas is bubbled an through balanced symbols, metals State Explain alkali why metals the Describe E of Suggest the Write a a added indic ator water. to and the a for the pink with a split b. [2] d colourless. to reaction solution to coloured [2] aqueous takes change place, from pink 2 + 4Cl ⇌ [CoCl 4 ] + 6H 2 O blue [Co(H the 2 O) how that 6 + ] complex ion absorbs light of approximately energy orbitals Explain is added + ] wavelength C alculate nitrate is a of the dierence cobalt colour chloride in this complex change ions are 540 nm. between the in this ion. [2] reaction weaker ligands than to small a molecules. [2] [1] 21. Explain, why [2] by referring titanium c alcium only c an to exist successive in variable occurs in the +2 ionization energies, oxidation states, but oxidation state. [2] large piece of water. explain the change in the value balanced The water metals. group. added a 6 forms exchange the 2 a. [2] energy of the is O) of nitrate water. alkali the ligand 2 shows indic ator. container symbols, with the Then, the above. occurs. down l full ii. of of ionization sodium colour vigorously decreases container is reaction number Phenolphthalein i. d. react group v O c. the this described u b. alkali f x a. reaction i The why aqueous t a 16. Explain the r o ii. for a zinc acid pink orange-brown liquid is equation, including state aqueous colour 2 copper(II) hydrochloric ions, the [Co(H o a whereas to blue: radius n bromide, c ausing [1] aqueous excess cobalt(II) [2] d Write greater why solution, electronegativity a produced. i. When [2] [2] chlorine potassium 20. radius of the halogens U When Explain p Explain 19. o uorine [1] conguration of a copper(II) ion. are in group. oxygen. than d. why the halogens C c. atomic down The i n than the halogens. table. v Explain the periodic why increases b. about the electron [2] LHA of full y a. 17 r e group the [2] conguration of copper. Extended-response questions oset. rst 10 ionization O b. it showing n 3+ a [1] consequences of acid y 2+ D. Sketch the t i s C. a. of l 18. one y A. Choose P ii. [1] for the pH of the solution in once the reaction has ended. [1] equation, including state reaction of rubidium with water. [2] 255 Functional groups: Structure 3.2 Classic ation of organic compounds s s e How does the classication of organic molecules help us to predict their properties? r a common on compounds the that human basis share of approach shared important utilized in many characteristics, features of elds chemists structure and of science. utilize a Just unique as biology system of uses scientic nomenclature group and reactivity. (atom but dierent Structure 3.2.8 — M ass c an c ause spectrometry o v type of U C i n the bond (IR) spectra present Structure 3.2.10 — Proton the same arrangements of atoms. fragmentation Structure 3.2.9 — Infrared identify spatial p compounds have connectivities and bond y r e multiplicities) identities, in nuclear a (MS) of of organic molecules. c an be used to molecule. magnetic resonance 1 spectroscopy ( n chemic al H NMR) environments gives of hydrogen Structure 3.2.11 — Individual o i u l a E v O f x t a r o d clusters of signals on atoms the in a dierent molecule. c an be split into peaks. Structure 3.2.12 — D ata oen information combined in from structural dierent analysis. techniques are LHA constitution O Structure 3.2.7 — Stereoisomers n y t i s Understandings 256 taxonomy to to l name is organisms y classify P Classic ation Structure 3.2 Functional groups: Classic ation of organic compounds Formulas of organic compounds (Structure 3.2.1) Introduction to organic chemistry are chemistry studie d. four to process other by This spe cial fe ature alcohol, plastics, fertilizers. are why drugs mole cules, considere d to be or together and fuels, me dicines, such as DNA, by which is This such paints, foods, contain a wide dyes, pesticides c arbon organic. represent empiric al of present the 6 H 12 the 6 and glucose an the formula Both of the be types structures of of formula ratio the of several formula of deduced. by using oer 2 O. you or no Empiric al molecules. has the molecular From this information, the were and molecular introduced in formulas Structure 1.4. C n o i 12 a t a H little complex U 6 of in atoms u C of a structure l G l u c o s e, v The E groups. d r o f x O p Figure 1 ways present number more glucose CH atoms actual larger, c arbohydrate empiric al c annot simplest describes these possible molecule O in i n of compound p C structure molecule. a organic o formula represents molecular about example, an v For of y The in structure formula molecule. information the formulas. r e An types O c an dierent t i s Types of formulas You is cov alent structures. chemistry include form c atenation , cyclic organic compounds solvents, biologic al joine d branche d, re ason O rganic industrial also the are compounds c an n and study. Most atoms undergo they y atoms of is c an so l and field identic al c arbon-base d ele ctrons, y v arie d atoms straight-chain, c arbon where v alence P and chemistry four C arbon many produces of have atoms. which of field r bonds. the atoms e the bonds is C arbon s s O rganic O the 6 , is ring a is simple not s i x- c a r b o n refle cte d by sugar the with hy d r ox y l molecular functional formula 257 Structure 3 Classic ation of matter Structural formulas, unlike molecular formulas, describe the structure of a compound. There are three types of structural formula: full, condensed and skeletal. • Full structural formulas to one another in a • In a A the but some skeletal formula and functional is all the the most basic included c arbon H C C CH H o CH H C H H C C structural The atoms Condensed H H the formula, present in 3 3 Skeletal formula CH 2 CH formula 3 y H H O atom. p H H of shown but the end of shown in table 1. r e H OH CH(OH)CH 3 O n o i H C H O H C C C t a u l a E v structural line CH 2 CHO O H H CH 3 C(O)CH 3 H H H C CH C 3 CH=CH 2 or H Full, of 3 H H condensed organic and a a and skeletal covalent describes for 3 CHCH 2 formulas c an all be used to represent compounds formulas, that CH H H p Table 1 CH H C used, not n C relative positions l C their O C H propene are a as are t i s H H i n U d r o f x O 258 H and relative O propanone single atoms structural H C H propanal structures positions omitted. representation structural v H propan-2-ol the are atoms represents also formula H all bonds hydrogen vertex are Full propane their y Name or and each groups and triple two bond bonding bond, three between two electrons. lines are atoms is represented by a For a double bond, two lines used. y line c arbon are two-dimensional bonds, P each the and compound. condensed structural formula, shown, where In atoms r • displayed formulas all e are or showing s s representations Structure Constructing technique, of individual from the optic al 3D models enhancing atoms. naming isomers) of to of your organic ability Models organic c an compounds to visualize enhance molecules complex reaction of is an excellent Functional groups: Classic ation of organic compounds interactive molecules and mutual orientation understanding and 3.2 visualizing of a variety of concepts, stereoisomers (including mechanisms. s s Practice questions Draw the full structural formula and skeletal formula for each of the following organic compounds. C III. C 2 2 5 H H H 6 4 O formulas are compounds C 4 3 H H H 4 8 62 O O 88 72 t a r o Linking question the skeletal advantages compound formula, 3D u are organic that elements’ a an c arbon other l of about the E v O f x unique 90 O o D 2 –1 i C formula and Molar mass / g mol d C 2 empiric al C CH What same n CO B all the U A of has y following formula? p I, II and III is formulas? o D. sum empiric al r e II and III only Empiric al formula the also v I and III only What 3 2 C. the CH 12 B. of 2 2 OH O I and II only molecular CH C=CHCl molecular A. Which 2 2 CH O C II. 2 CH t i s I. 2 2 CH n these CH i n 3. of 3 2 CH y chloroethene, H Which CH l c. 3 r butan-1-ol, CH P hexane, CH b. y 2. a. e 1. enables it to form compounds? and disadvantages (structural models, more etc.)? compounds (Structure formula, of dierent depictions stereochemic al (Structure than 2.2) formula, 2.2) 259 Structure 3 Classic ation of matter Models Models are useful representations of the physic al more and complex computer 3D models in the direct molecules Skills observation (such as features models. that You is dicult. structural Two-dimensional formulas) become clear explore some c an in of do not three-dimensional the dierent types of chapter. laboratories animals and are group of atoms, synthesized research synthesized in by living found fe atures n a t u ra l l y hy d r ox y l , O O ester o C n It y p r e steroid fibrillation. found in OH H OH O O H o i c ardioactive O OH hydroxyl in the a l k ox y f oxg l o v e and plant. ester It is functional used to tre at congestive he art failure and groups u f x All organic compounds are divided into classes of organic compounds l a E v O 260 a atrial D-glucose- t a is or Natural compounds organisms. H alkoxy organic pharmaceutic al companies, universities. v i n U D igitalin he artbe ats d irre gular r o p Figure 2 gives organic compounds, and this number is constantly are OMe β that characteristics. O and of a dependent on the specic functional group found in their molecules. Compounds of the same class have similar chemical properties. Some functional groups and their corresponding classes are given in table 2. R symbolize formulas other any for c arbon-containing classes functionalities of organic are groups. They compounds. c alled alkyl groups groups are useful are for used in table 2 to writing C arbon-containing general groups with no y millions compounds or chemic al n of and t i s industrial plants tens new atom, l are as an e rising is physic al y There their P functional group compounds r Functional groups (Structure 3.2.2) A represent s s some when of Structure Functional 3.2 Functional Name chloro, of organic compounds Sux Formula Class Example chloromethane bromo R X — halogenoalkanes H C Cl 3 s s or Classic ation group halogeno (uoro, groups: iodo) ethanol R OH -ol H alcohols C e hydroxyl H OH C 3 r H H aldehydes H c arbonyl -one ketones C -oic O R′ — U methoxymethane H C O 3 CH 3 H H C C i o 3 NH 2 H t a ethanamide O H amides C C 3 u f x l a E v O R C OH primary amines -amide NH NH 2 2 methyl ethanoate O O C -oate esters H C C 3 O O R’ CH 3 methylbenzene R phenyl S ummary C C 3 ethanamine -amine 2 O R ester p Table 2 NH d r o amido R H n amino ethers 3 O o R ethanoic acid c arboxylic acids C i n OH alkoxy acid v R c arboxyl CH O y O C 3 C p r e O H O propanone R’ C C H t i s R C 3 n y -al y C O l R c arbonyl P ethanal O of c l a ss e s — of organic compounds and aromatics their functional H C 3 groups 261 Structure 3 Classic ation of matter Self-management skills ATL How c an names, build the of and various develop familiarity structures? maintain functional You your may with uency. groups the wish will to As various create you help will you to a functional ashc ard see in groups, their deck Reactivity predict the help you recognizing chemic al behaviour various substances. or c an unsaturated. , a single. hydroc arbon must C=CH saturated double 2 . Alkanes no or and are c arbon methane, unsaturated c arbon bonds. alkenes. The are both aliphatic. H p c arbon is H o a an y H H with in c arbon c alled alkenes C H (left) chain are c arbon hydroc arbon triple H they c arbon aromatic rings. C C s a t u ra t e d whether the c arbon bonds r e is 2 a primary double v i n Ethene (right) more contain of The H C ethane or c arbon ethene, H molecules H p Figure 3 one c arbon H example family. to all O is their alkane contain with alkene means simplest according compound, t i s Hydroc arbons simplest The classied n member of the 4 be saturated l are CH a y bonds also In double bond, is u n s a t u ra t e d , where as U TOK d the organic word i context: What is c an organic. the role dier This food, organic t a r o u to of and from word takes organic signify context a in chemical their have meanings on a variety chemistry and in of precise meanings in everyday language. meanings depending on its organic household waste all use dierent thing. the choice and interpretation of language? Linking questions E v a LHA l f x O 262 Consider natural, organic which o chemistry, n Terms such as What form is a How a the nature dipeptide? c an the functional reaction ethene of pathway into reaction (Structure group reactivity between ethanoic that occurs when two amino 2.4) acid? be used compounds, (Reactivity 3.2, to e.g., 3.4) determine converting acids y compounds saturated P Organic e r S aturated and unsaturated hydroc arbons This to 3, s s to you and Structure 3.2 Functional groups: Classic ation of organic compounds Homologous series (Structure 3.2.3) A homologous series based the on same similarities general you its c an family formula, group. identify a their If you which of compounds structure which know group a and varies the from general particular that c an be grouped together reactivity. A homologous series has one member formula for compound to another by one CH 2 s s (methylene) is in each homologous series, belongs to without drawing structure. contain series homologous series. general formula C hydrogen You should member diers be able from to the n H 2n+2 only. . Alkanes Table identify previous are shows how the hydroc arbons the alkane structural by a single CH 2 formula of group. structural formula structural formula formula t i s H CH CH 4 O methane C 4 H 3 CH 2 H 3 CH 2 CH 2 of 3 CH CH 2 2 CH CH 2 2 CH CH 2 2 CH CH 2 H o series 3 CH i a homologous CH t a CH 14 3 u H CH 12 l The 6 H 3 d C 5 CH 10 E v O p Table 3 f x hexane C H r o pentane 4 H n C C H H 3 U butane C H H H C C C H H H H H H H H C C C C H H H y CH 8 CH o H 3 C 3 CH 6 v C H i n propane 2 H p C r e ethane H n Full Name l Condensed 3 y Molecular atoms y successive the and P each has c arbon e alkane that r The H H H H H H H H H C C C C C H H H 3 CH H H H H H H H H H H C C C C C C H H H H H H H 3 alkanes 263 Structure 3 Classic ation of matter The You c an alkenes are c arbon–c arbon halogenoalkanes from a homologous process known as This form alkyne series The of unsaturated alkene series has hydroc arbons that contain a the family have c arbon–c arbon triple bonds, general with the formula of C general what c arbon–halogen was a detailed R adic al in a n H 2n–2 . Halogenoalkanes substituted with a are alkanes halogen atom. where They one have of the the formula C n H 2n+1 bond into alkane substitution is 3.3. r formula C n H alkynes C 2n H C 2n–2 C C 2 H H ethyne ethene r e 3 C H CH 3 U H C C C 3 CH 3 3 i of s t ra i g h t - c h a i n alkenes, Members u f x group ( of H 3 H H H C C C H H H Cl 1-chloropropane H H H H l Another a contain the as aldehydes, c arbon the and H 3 C C C C H H H Cl halogenoalkanes have a series general of chain. but ends. series, the Ketones the the c arbonyl These and aldehydes, the group homologous H contain 2n+1 have group same is all n oxidation functional have c arbonyl alcohols formula of C reactions (Reactivity 1.3) of except CH 1-chlorobutane homologous They homologous Aldehydes C H alkynes the OH). combustion E v O 264 series t a homologous CH o d r o The H but-2-yne but-2-ene p Table 4 n C H Cl H C 4 C propyne C i n propene C H o v C C H chloroethane y H X 2n+1 C H p H H H O C n H n y H C t i s H n halogenoalkanes hydroxyl functional c an undergo general formula of C n H 2n O. (C=O) in the terminal position formula anywhere are the Alcohols reactions (Reactivity 3.2). general found series a OH. on and the functional group c arbon chain summarized in table 5. y alkenes l General series X, P Homologous . hydrogen atoms has general highly inert Reactivity 2n e molecule. relatively H s s introduces where X is the halogen atom. polar n formula of been substitution bond. radic al C substitution. double alkanes The by a synthesize Structure 3.2 Functional groups: Classic ation of organic compounds Homologous alcohols aldehydes ketones series General C n H 2n+1 OH C n H 2n O C n H 2n O formula H s s H H O C C H OH C C no C 2 e H C ketones 2 H H H ethanol ethanal H H H C H H H H OH C C H H C 3 H t i s C C C H O H n C C H H H H H H H H H C C C O H H H O C C OH H H H H H C C C H H H H H C C C C C H H H H OH C have OR c arbon atom: of is formula c alled OCH 3 . where an This is ethers: methoxymethane 3 = 1, ROR’, is n a l d e hy d e s H the where group. known a H 3 formula C n alkoxy l 3 general E v O examples H the group general u The family. f x Ethers the as a 2n O 2 R and CH and and R’ C C C H H H H butanone H H C C H H H C C C O H H O C H H pentanal H pentan-3-one ketones have a c arboxyl group. c arboxylic acid of this represent simplest methoxy 2 , simplest The H C n homologous have HCOOH, alcohols, i acid, of H H H H t a acids r o C arboxylic Methanoic series H C o homologous d The H H pentan-1-ol p Table 5 H U 5 butanal o H H H C H i n H H v butan-1-ol C y C 4 p C r e H C H propanone propanal H O propan-1-ol l C y H H H O y H P H r H alkoxy group. alkyl group Here are groups. has one some 3 methoxyethane 265 Structure 3 Classic ation of matter Members There are amines. atom. and in of In In the three amine types primary amines: amines, secondary tertiary homologous of the amines, amines, it is series primary nitrogen the atom nitrogen bonded to all have amines, is three is the general formula C n H bonded bonded c arbon to to two atoms. only one c arbon This is summarized have the have medic ations of c arbon also derived They atoms and bonded the to are c arboxylic amido acids and functional nitrogen H H H C C C H H H p C o H n H o i tertiary l a E v O 266 and formula of . Example H NH 2 H C N C H H H H 3 2 H dimethylamine H C N C H H H H t a secondary general C H H trimethylamine u P r i m a r y, a –C(O)NH y r e v i n U d r o f x p Table 6 have propanamine 2 tertiary c arboxylic O t i s 1 secondary from avouring agents and group, H primary derived from explosives. from contain They ranging amines Linking question How useful are 3D models (real or virtual) to visualize the invisible? (Tool 2) y Number are NO. R’. n amine 2n+1 2 l of H RCO applic ations y Type n solvents of P C formula variety r Amides to general a e and s s acids N. c arbon atoms, in table 6. Esters 2n+3 secondary amines and tertiary Structure 3.2 Functional groups: Classic ation of organic compounds Practice questions 4. Which functional group is circled in this molecule? 7 . Which of the following functional groups is present in paracetamol? A. hydroxyl B. c arboxyl s s O OH C C. c arbonyl D. ether H Which functional groups are present in this molecule? r O CO 3 N following II. c arboxyl are present i H O H drug used Its to treat structure is symptoms of shown below. OH O O 3 N CH 3 H Which of the following functional groups are present C C C in this molecule? A. amido, hydroxyl, ester B. amino, hydroxide, ether C. amino, hydroxyl, ether D. amido, H O a E C l C v C ester t a C III. u f x O H r o O H O groups a o ether functional molecule? O the d I. of aspirin CH is disease. y the G alantamine n in hydroxyl p ester, amino Which phenyl phenyl n y ether, amino c arboxyl, amino D. o c arboxyl, D. B. alkene C C. c arboxyl v ester, i n ether, c arbonyl, U 6. c arbonyl, 3 C. Alzheimer ’s CN CH A. r e 8. CN C t i s O B. O 3 H A. l CH y P H e 5. C hydroxide, ester H C H A. I and II only B. I and III only C. II and III only D. I, II and III 267 Structure 3 Classic ation of matter Research skills ATL Our atmosphere oxygen, O . precursors formed is mostly Organic of billions made biomolecules, of up compounds, of nitrogen, N which are may thought 2 , and have to ammonia water vapour been (NH was these as the Miller very dierent. conditions Urey in An the attempt mid-20th was made to century, 4 electrode ) 2 ) known e recreate ) hydrogen years ago when the composition of the (H atmosphere 3 methane (CH have experiment. condenser (or two another many in of your own possible document the Urey experiment biomolecules), and to words, your drawing knowledge of information sources, and cold water in reliability. cooled water containing organic compounds y sample taken for chemical analysis a p p a ra t u s for the M i l l e r– U r e y ex p e r i m e n t o i n v p The O t i s r e p Figure 4 n y their as origin l Fully the ndings connections chemistry. evaluate your of Physic al trends in homologous series C (Structure 3.2.4) of they the have types given the forces The c arbon and u a 7. points These the and melting as point chemic al group is properties, as responsible for the undergoes. member. This time, altering is the bec ause the length intermolecular properties. points increase boiling each 2 similar properties such as melting and boiling successive by CH physic al have functional compound physic al with each increases values consequence, series, the The for the and the straight-chain number melting of c arbon point alkanes atoms increase are shown increases. with increasing molarmass. l a Name Formula methane CH E ethane C propane C butane C pentane C hexane C p Table 7 268 chain therefore boiling table As v O f x in gradually series group. reactions homologous change i of homologous functional o a points same same characteristic t a r o d For of the n U Members The 2 3 4 5 6 H H H H 6 8 10 12 14 and point / °C Melting 161 4 H melting Boiling boiling points point / °C 183 89 172 42 188 0.5 135 36 130 69 95 of the s t ra i g h t - c h a i n alkanes y as theory P summarize sources on the Miller r Research s s the 2 Structure 3.2 Functional groups: Classic ation of organic compounds The close correlation between c arbon chain length and boiling point enables you to predict The the properties of neighbouring members of the homologous series. trend in melting points is not as smooth as that of boiling points and would have less certainty in terms of the prediction of properties. You c an boiling use simple points for distillation successive apparatus and a temperature probe to measure the learned techniques exist between London non-polar polar results in forces, (for is HCl). or H London are 2 ). molecule polar. (dispersion) forces intermolecular covered in Structure and forces 2.2. were Which of will the so homologous far series experience discussed hydrogen bonding? n number of carbon atoms 5 6 7 8 9 10 alkane alkene 150 other o E v 50 3 Every non-polar i 2 a 1 forces hydrogen u l f x O C° / erutarepmet 0 (dispersion) t a r o 150 d 200 100 example, example, it London (dispersion) U 250 50 (for whether 300 100 stronger London molecules molecules (dispersion) from length. C experience between point increases y forces boiling chain i n also c arbon p increasing probe o attractive They in the t e m p e ra t u r e v as a r e trend forces i n c o r p o ra t i n g n distillate from faucet a p p a ra t u s l water O t i s to sink y P y water The 1.1. e r probe condenser D istillation distillation Structure members of a homologous series. temperature p Figure 5 about in s s You alcohol carboxlc ac Figure 6 inc re ases The with member of series the g r ows as in a boiling e ach point s u c c e ss i v e homologous c arbon chain length 200 269 Structure 3 Classic ation of matter Constructing a simple graphic al model In this task model will construct and test a simple Part 3: Graph graphic al homologous series of 7 . Create 8. Using a sc atter graph of the data. choice. software, draw a line or curve of best fit that describes the data well. Relevant skills 2 9. Inquiry 1: State and explain Determine the R value and the equation of predictions using scientific or curve. understanding Use • Tool 2: Represent • Tool 3: Sketch and extract data from databases Part 4: Test the model spreadsheets to manipulate data data graphs in to graphic al You have now created a simple graphical model of the relationship between boiling point and carbon chain length. form qualitatively describe You can test your model by using it to make predictions and trends Tool 3: Apply the coefficient of determination ( R • Tool 3: Construct graphs and draw lines or ) 10. curves of 3: Tool • Inquiry 3: C alculate 2: possible, the 20th members of the homologous series. graphs and Describe interpret and percentage explain patterns, error trends and 11. Look 12. C alculate up their actual the boiling percentage point errors predictions. relationships p Instructions values. of each of the y r e • Extrapolate O Tool t i s • Use the equation of the line or curve of best fit to predict the boiling point of the 7th, 10th, and if best fit n • y then checking the accuracy of these predictions. 2 Part 5: Interpret the results 2. Predict your first the chain length. knowledge a your that qualitatively have i t a r o for a variety Chemistry data on the you boiling have do This their the the accuracies is accuracy c arbon an data. a value How errors for the following questions: compare? Why might and of the extrapolation that the explain predictions change for lengths? might interpolated Describe of chain example predicted had 14. percentage be? Does the the Consider was the accuracy data the bec ause outside the you range of change if you instead? number data: of point enter c arbon the a the v O and instance, for the six you trend shown in the graph plotted. NIST 15. If uncertainties were reported along with the boiling 16. Briefly point data, what impact did these have on your analysis? organic evaluate your model, giving examples of its selected and enter the data into spreadsheet. Organize (for u a substances l f x Collect of Web Book). compounds 6. • describes the contrast Identify an online database that contains boiling point data 5. How longer predicted. Part 2: Data collection • referring and predictions. this o you • forces and d relationship 4. prediction, intermolecular groups. graph three between boiling point and Justify of Compare n Sketch relationship U your functional 3. members of a homologous series of choice. c arbon to six o the C List i n 1. v 13. Part 1: Prediction 17 . name, atoms, specific How strengths might the and weaknesses. accuracy of this simple model be formula, improved? and convert all temperatures into a consistent unit. E 270 Extension Choose task. a different Compare homologous and contrast the series two and sets of repeat the results. y Identify 2: l 2: Tool r Tool • line P • the e • s s your you of the boiling points of a Structure The dierent members this principle mixture of is passed Therefore, the as a through length of a their similar will column. of mean that petrochemic al example, oil dierent industry, crude oil is a c arbon chain. crude organic compounds is Fractional vaporized and through the column, it cools. levels in the column depending Compounds with a similar boiling point, condense down the where members of at the column, have a same as they level. Long-chain have a higher lower boiling point, so will it is cooler. a homologous series also increase with y What the is the inuence functional of groups the on c arbon chain intermolecular length, forces? branching (Structure and the nature 2.2) O t i s Linking question of existence of an example, is substituents organic any class alkyl group a in of and compound considered dierent an the is compound compound, functional that an chemic al substances, names help us to organic not the groups part of the needs length present. longest to of A provide enough its longest substituent c arbon chain. c arbon is any For alkane compound that is not part of the main substituent. An U chain any the of of o part and identify name C chain, to millions The i n information of them. p the dierentiate v With y r e Naming organic compounds (Structure 3.2.5) n l c arbon chain length. y viscosity series the their rises dierent compounds column, For of where vapour at will further Shorter-chain and the chain. length, length process condense c arbon the As the In of P density will homologous technique. in separation chain up vary of temperatures. separation that condense further members Classic ation r point. condense physic al compounds compounds The a therefore boiling used of dierent groups: e and is at Functional s s then points condense hydroc arbons distillation on boiling will 3.2 alkane with alkyl substituents is known as a n branched-chain alkane d o The International Union of Pure and Applied Chemistry (IUPAC) has developed a universal chemical naming system for both organic and inorganic compounds: r o i IUPAC nomenclature. IUPAC nomenclature rules helps chemists to avoid confusion, eliminate language barriers, and freely communicate knowledge. An t a overview of the IUPAC naming system for organic compounds is shown in gure 7: l longest includes or the or a E v b u f x O a the c arbon chain that the principal most complex heterocyclic alphabetic al cyclic suffix system the principal (parent) chain O utline of the group -ene -yne suffixes order saturation indic ating or unsaturation of the principal p Figure 7 indic ating principal -ane c prefixes in group, nomenclature of organic chain compounds 271 Structure 3 Classic ation of matter Naming alkanes Alkanes are saturated organic molecules that only contain In the example below, numbering from le to right results in the carbon methyl substituent being on carbon2. Numbering from right to carbon single bonds. You can use the following steps to deduce the IUPAC name of an alkane: le would incorrectly have the substituent on carbon 5. s s H 1 Examine the structure of the compound and determine H the longest root name atoms the c arbon chain. This (table8). If an alkane is a C H C C 1 name is the same as its root have the same sux: H C C H 2 3 4 H C 5 6 name. All H alkanes H C H H H H r its H straight-chain H molecule, H provides the alkane depending on the number of H -ane. name name as of the alkyl prex, number and and substituent a the hyphen prex. goes is used before to Therefore, the root separate the the l loc ant 1 a above methane 3 propane 4 butane 5 pentane 6 hexane 5. is c alled 2-methylhexane. n ethane y compound 2 If there are multiple dierent substituents, they are arranged t i s in alphabetical order prior to the root name. For example, O the below compound is called 3-ethyl-2-methylhexane. H ethyl alkyl substituents of a are substituent for the alkane series determined by the number of H Some with the examples sux are changing shown in from table “-ane” to 9. H methyl H H C C H H H H specify the numbers, E c arbon the chain. The substituent is the c arbon given are substituent multiple is identic al substituents, the number indic ated name, as by a in before the table10. Multiplier name 2 di 3 tri 4 tetra 5 penta CH 3 2 2 2 CH CH CH 3 2 2 CH CH p Table 10 3 2 CH 3 The Numeric al loc ants of For m u l t i p l i e rs each in the substituent example, the I U PAC are below nomenclature separated compound substituents a substituent H by loc ants, to the main H chain is C H numbered such that H the multiplier summarized Number of identic al substituents lowest value H H loc ant. H C C H H H C C H 272 3-ethyl-2-methylhexane 2,3-dimethylhexane. of known as H 6 H methyl substituents comma. alkyl position there C 5 H H t a c an assigning for u You names CH a 3. I U PAC CH propyl v O p Table 9 CH ethyl butyl If 4 H formula l 4 3 H C H C Condensed name methyl f x 3 substituent H 2 H of i Substituent 1 C substituent r o Number of c arbon C 1 H o C C H 6. d C atoms in alkyl group 2 substituent H C n H C U H H C i n “-yl”. atoms, C o c arbon H H present in the compound, the is v name names H y If root C p 2. I U PAC r e p Table 8 H C C C H H H H H by a is c alled y The Name P 4. Number of c arbon atoms e c arbon continuous for Structure 3.2 Functional groups: Classic ation of organic compounds Structural isomers are compounds that have the same molecular formula but dierent structural formulas. Dierent isomers have unique physical and chemical properties. In the following worked example, we will look at the structural isomers of the hydrocarbon hexane, and practise using the rules above to name these isomers. s s Worked example 1 Three dierent structural isomers of hexane, C 6 H 12 , are shown below. Using the IUPAC nomenclature, e name these isomers. a. H H H H C C C C C b. c. H H H C H H H H H C C C C H H H H C H H H H C C H H H H H H Solution a. First, count straight so the c arbon of There c arbon atoms in the chain. parent chain is There are ve c arbon bonded pentane Since number has CH ) the lowest possible substituent name is loc ant. The bonded to c arbon 2. for this compound is 2-methylpentane. There are In this chain, c ase, the is so the irrelevant for the longest parent chain is direction both the of numbering as the would 3 CH 2 CH 2 a CH 3 CH E CH 2 C CH CH CH 3 CH CH 2 3 ) are substituents, to c arbon identic al, one 3. a numeric al indic ate the number of this c ase, di indic ates two name for this compound is between Like ( b, CH The 3 ) a the Remember numbers, number and and a parent chain is substituents IUPAC name are for commas are are used letter. butane. bonded this that hyphens to Two methyl c arbon 2. compound is therefore 2,2-dimethylbutane. 10. Draw the structural formulas of the following molecules: 2 CH 3 a. 2,2,3-trimethylpentane b. 5-ethyl-4,4-dimethyloctane 3 3 CH 2 3 CH 3 CH c. the nomenclature rules, state the name of following compounds: v O b. CH IUPAC the l a. of u each is give to In 3 one t a r o f x Applying the substituents. Practice questions 9. IUPAC used c. of compound approaches substituents. between butane CH and substituents required 2,3-dimethylbutane. straight i loc ants so in The o chain symmetric al, same and atoms d c arbon c arbon is 2 ( substituents. n c arbon four U b. these c arbon two C IUPAC 3 i n The ( the H H methyl o methyl to multiplier c arbon chain so that the v substituent the two p Then, are C y atoms, number r e longest the H H O t i s H C n C y H C l H y C P H H H r H H 2 CH 3 CH 3 3 3 273 Structure 3 Classic ation of matter Naming alkenes Alkenes bond. same as unsaturated IUPAC those of for -ane following organic nomenclature the used alkane for molecules rules series. for All that the contain naming alkene names of a c arbon alkene end with c arbon double molecules the sux are the -ene, s s instead The are The alkanes. compound has both a c arbon c arbon double bond, and an side chain. CH 3 1 2 3 methyl ( on from the position taking right c arbon-4. to 3 ) substituent le at follows: loc ant 4, the of the over would Therefore, example. CH priority c arbon incorrectly place the numbering name of this occurs from molecule is double substituent c arbon any substituents bond y r e loc ant name p 4-methylhex-2-ene C C i C r o H C H H molecule so the Numbering the atoms the possible in the a c arbon straight from C C H right le There The so -ene. to b. are gives sux double c arbon loc ant. bonded double H C C H H C bond follows c arbon molecule the The six chain, c arbon–c arbon double bond the The H H H c arbon butane. from -ane to chain H C H H H H c arbon–c arbon double changes H H longest parent chain is contains unsaturated lowest the sux E bond, c arbon so indic ates unsaturation H H u The four chain, a are c arbon l There v O a. H f x Solution b. t a H o H o H H suffix n name the following alkenes. d a. C chain U Worked example 2 Using the IUPAC rules, v parent loc ant i n methyl substituent atoms the contains bond the longest a -ene. c arbon-2 in the longest methyl (–CH 3 ) substituents c arbon atoms number 4 and 5. c arbon-1. The The 274 IUPAC name for this compound is but-1-ene. IUPAC straight hexane. c arbon-c arbon double bond, from -ane to follows Two in parent chain is changes chain. to so name for this 4,5-dimethylhex-2-ene. molecule is are y the this Numbering bond chain, loc ant, O as in double c arbon lowest t i s With molecule. c arbon right the n to longest y le the the have 3 l in c arbon must r present number bond 3 6 P you double CH 2 5 CH When CH 4 e H alkyl Structure 3.2 Functional groups: Classic ation of organic compounds Naming halogenoalkanes When naming identied each halogenoalkanes, the position of the halogen substituent is the halogeno derivatives loc ants for the loc ant on the parent chain. Table 11 lists the prexes used for H alogeno Substituent group name group. of methane substituent and are ethane featuring equivalent, so only they are one halogeno omitted. s s For by F uoro Cl chloro group, all Consider the following compound: C H H I H Br N ames substituents this atom halogenoalkane alphabetic al order. to other molecule with For is positions will bromoethane, more than example, the one form not halogeno following the same molecule. The 1-bromoethane. group, list the halogens in compound is named 2-chloro- 1,1,1-triuoroethane. C loc ants. So compound takes the also priority following features over the molecule c arbon is C C C i H C C t a C E b. u H H a H H rules assigning o Cl nomenclature of 2-chlorobut-3-ene. to deduce the c. Cl following compounds. v O H IUPAC the l a. of not H 2 d r o f x the names terms n H Practice questions Apply in 3-chlorobut-1-ene, H H 11. c arbon double bond, the group U H a halogeno o bond C halogeno i n a double v Cl F If H y C p H r e F F n for O a bromine t i s For the name halogeno l IUPAC of y Moving P p Table 11 iodo y C bromo r H e Br H C Cl H C Cl H C Br H 12. C Deduce the structural formulas for the following H compounds. a. 2,3,4-tribromohexane b. 1,3-dibromo-2-chlorobutane c. 2-uorobut-2-ene H H H C C Cl H H H l 275 Structure 3 Classic ation of matter Naming alcohols All alcohols you is need contain to numbered The over group ethanol the in this any or functional sux group c arbon series position do results from has the c arbon chain loc ant, taking methanol. loc ants. The This molecule, as is rst two bec ause the seen in methanol more The H 2 C H H H 1 H H H propan-1-ol loc ants to specify H 2 C p C require before the -ol sux, H 3 C o v H atoms comes y r e 1 HO H c arbon loc ant n or group. example: H H ethanol 3 C C OH H H propan-2-ol C i n Naming compounds with c arbonyl groups c arbonyl aldehydes at the end of to the c arbon the chain. atoms aldehyde t a When the primary u is priority l have over the other the chain. that is group for c arbonyl known as the group in two classes of compounds: group is terminal aldehydes, as it always positioned position. always There is occurs at the aldehyde includes a substituent, the numbering of c arbon the the the so This the the sux -al, attached so group c arbonyl or propane to ethane is compound substituents -one, functional the functional chain, such sux at have is aldehydes, c arbonyl When c arbonyl c arbon numbered the begins names In c arbon number i d r o f x a E v O 276 need of (C=O) ketones. o no end group and n U The not is a the has the c arbon–c arbon becomes group. For example, loc ated at the terminal position in the ketone. group c arbonyl becomes ethanal. In ketones, lowest double propanone. the c arbon chain possible bonds. loc ant, taking Ketone names y H three l with hydroxyl H O the C y of C t i s compounds H r HO H P H C H for is same methanol position The possible below. HO the When naming alcohols, -anol. lowest require H Alcohol the series not in group. -ane to c arbon double bonds. homologous homologous in OH) chain hydroxyl substituents molecule this parent e hydroxyl of that other simplest members and such hydroxyl ( the s s priority the change Structure 3.2 Functional groups: Classic ation of organic compounds Worked example 3 Using the IUPAC rules, name the following aldehydes. There are no substituents, so the IUPAC name is methanal. a. b. O CH H H 3 H O b. parent and the chain H C C C C H H H H The c arbon sux comes from the C 5 alkane, pentane becomes -al. C H numbering starts from the c arbonyl H and attached c. H F therefore to the methyl substituent is c arbon-4. H r O The C C C is 4-methylpentanal. This is an example of an aldehyde featuring a halogeno substituent. The parent chain comes from the C H H 4 l H name alkane, butane, and the sux becomes -al. The The parent chain comes from the C alkane, group, molecule terminal contains position, so a the c arbonyl sux becomes -al. The O H molecule the terminal ketone bec ause that c an Therefore, parent and the so the becomes required, as c an exist. So The parent and the or sux propanone. propanone exist; if the c arbon-3, it is The the all the information in the name -one. b. chain sux H CH H H 3 C C C H H H O H the nal attached C comes is the is C becomes this c arbonyl from the C -one. only The 4 alkane, H butane, loc ant is not four-c arbon ketone that IUPAC name will be butanone. loc ant is not only c arbonyl would propane. be three-c arbon group an required were aldehyde. to deduce the c. numbered group chain sux has at Therefore, from the c arbon-2, comes becomes right lowest and the the to from the C -one. le such loc ant. methyl IUPAC name The The alkane, that the is at pentane, atoms are c arbonyl c arbonyl group is 5 c arbon group is c arbon-4. 4-methylpentan-2-one. a v O ‘propanone’. l is The alkane, 3 the substituent 3-uorobutanal. H group that is not in u f x c arbon-1 structure is H i required position, name O H c arbonyl from the C a H t a IUPAC comes contains r o The at chain C o parent The d The H C n Solution a. is c. H C from uoro y H C H H starts the p C H C C H U C name o b. i n H name the following ketones. H H v a. IUPAC r e Worked example 4 Using the IUPAC rules, numbering therefore c arbon-3. group in the -ane and O to The c arbon methane. t i s 1 n y Solution a. y c. H IUPAC C P H e group, s s The H Naming c arboxylic acids E C arboxylic acids are C arboxylic acids have the sux -oic acid. parent chain is a class of compound featuring So a a c arboxyl group (–COOH). c arboxylic acid with an ethane c alled ethanoic acid. 277 Structure 3 Classic ation of matter In c arboxylic group, aldehydes, always acids, including in there the is chains c arbon no terminal need from to are the assign position. Here H numbered starting with the functional c arboxyl loc ants are two H group to in examples CH the c arboxyl 3 of H H chain. groups, C H H O C H C C C C H H H H C OH OH acid 4-methylpentanoic acid the structures of the following molecules 2-chloropropan-1-ol d. 5-ethyl-2-methylhept-3-ene e. 2-methylpentan-2-ol CH O the following c. CH molecules: CH 3 CCH 3 CH e. OH 2 3 O H 3 i u l a E v O f x t a r o Science as a shared endeavour Scientists collaborate facilitates unambiguous in this chapter structures bottle to of are that still foll ow u se d aci d, four m i, or of c ases, name the of and the phenyl)propanoic of and you such stems ant: will names naming chemic al as and bu t to ac id for mic a . re co gnize you have compounds listed on you the will learnt using label now both of be a able suxes. enc ourages cer tain a llude terminology rules shampoo, n omenc lature these for agreed-upon The ingredients me thano ic hor miga, acid. the guideline s, L atin mild at product naming M any IUPAC for look example, the world, communic ate orga nic Span ish, for miga some IUPAC after you IUPAC Fo r to hygiene widel y. su bstan ces. formi c In some If the communic ation. you names. s tanda rdization names across enable personal recognize The of a and Por tugue se 278 CH 2 name 3 o in 3 3 CCHCH to CH 3 CH 3 2 C n a t u ra l l y rules CH n o c c u rs 2 o acid, CH v formic ants 3 i n as U some 3 CH d k n ow n nomenclature CH=CHCH p CH d. commonly 2 y 3 O IUPAC CH r e Apply n b. 3,3-dimethylpentane t i s hexan-3-one CH acid, IUPAC name c. b. Methanoic the a. a. p Figure 8 on y given: 14. based l Draw If the is us es an d commonly you this the u se non-systemati c as spe ak the cha ra cteri stics known Italian , roo t of names of as Fren ch, for mic a, respe ctively. are far painkiller too long ibuprofen to is be useful. For example, the (RS)-2-(4-(2-methylpropyl) y P Practice questions 13. r 3-bromopropanoic are e C they c arboxylic acids: O Br Like in as s s H c arbon the Structure 3.2 Functional groups: Classic ation of organic compounds Structural isomers (Structure 3.2.6) Structural molecular of or constitutional isomers formula structural but dierent are compounds connectivity of the that atoms. have the There same are several types isomers. s s Chain isomers Chain isomers Butane is a structural and straight-chain However, isomers with methylpropane molecule methylpropane is dierent both have bec ause a it lengths the has of c arbon molecular no branched-chain chains. formula C substituents molecule, on the 4 H 10 For . c arbon r chain. are butane bec ause it has an methylpropane C C C H H H CH 3 CH 2 CH m e t hy l p r o p a n e are chain i s o m e rs Positional isomers changes. isomers, all with the molecular isomers formula C 5 H 11 has C H H Skeletal E p Table 13 v formula Po s i t i o n a l H i s o m e rs H C H H CH(CH 3 ) 3 H H C Br H 2-bromopentane C H C dierent positional o i C t a C a O H u H H l f x structural H three Br. d r o formula H where the position of the functional bromopentane 1-bromopentane Molecular formula structural example, Isomer Full are For C H n group U Positional isomers C H o and 3 C Butane i n p Table 12 CH C H H v Skeletal formula 2 H p Condensed structural formula H H n C H O H y H 10 r e H H t i s H Full structural formula H 4 y C y butane Molecular formula l Isomer P alkyl substituent. e example, H 5 H 11 3-bromopentane Br H H H H H C C C C C H H H Br H H H H H H H H C C C C C H H Br H H H Br Br Br of bromopentane 279 Structure 3 Classic ation of matter Worked example 5 Draw the structural formulas of all possible structural isomers of a compound with the molecular formula C 4 H 9 F . Name the isomers using IUPAC nomenclature. First, draw the molecule, longest it is a possible straight c arbon chain. Having For you four-c arbon chain. identied need longest C C look chain by and at named possible one c arbon, and chain in place as the many uorine positions share add the hydrogen atoms. Remember numbering of the c arbon chain c an occur in to same add molecule, the either c arbon in the else H H 3 4 C C H H H H H 2 C 3 C 4 C H H in H are to to the ways 2-uorobutane, H as is methyl on C C uorine to the group For central anywhere H C substituent possible above, chain. create on the unique c arbon chain molecules. H C H 3 isomers H H H H 2 1 H C C C H H H C 3 F H H 2 1 H C C H F H 1-uorobutane, and favoured. u l f x a E v O 280 equivalent possible o loc ants is equivalent not i lower is are t a the 3-uorobutane 4-uorobutane d and 4-uorobutane r o isomers: and position many the propane H 2-fluorobutane 3-uorobutane as n F C C Then, U H H as the position the H H C 1 H i n H adding to four-c arbon butane chain. o 1-fluorobutane H as possible p C a v F create formula group y 2 C only r e 1 H H will methyl chain, direction. H molecular a C O this the need that t i s then the as you possible, C 1-fluoro-2-methylpropane 2-fluoro-2-methylpropane Activity Draw and name all the structural isomers of chloropentane, C H 5 Cl. 11 y chain propane chain. n c arbon a l c arbon the isomers, P the To on butane isomers. Shorten the creating y number substituent the C C Next, all chain r C to e this s s Solution Structure 3.2 Functional groups: Classic ation of organic compounds Primary, secondary and tertiary compounds Changing the the chemic al alcohols, the You two to halogeno the c arbon atoms other group is is to bonded this secondary to a c arbon secondary or tertiary (3°) H H H 2° H c arbon atoms class of H C C C H Cl H secondary in attached the H H H alcohol the to (2°) same the and way c arbon as H c arbon H C C C H Cl H H 2-chloro-2-methylpropane tertiary (3°) atom next to the functional group C C C H H H OH H of to the primary, and groups functional OH used secondary halogenoalkanes Reactivity to and are form tertiary discussed in 3.3. c arbon H Primary, secondary and tertiary H alcohols form dierent products 2-methylpropan-2-ol and tertiary secondary or in (3°) directly to the and is Amine slightly reactions. in Reactivity This is 3.2. dierent to the classic ation depends nitrogen alcohols oxidation discussed alcohols tertiary halogenoalkanes. bonded 3° C C H halogenoalkanes a atom alkyl unlike E v O group, as alcohols C H H (2°) H H i number secondary C H t a amines with functional next C u the of (1°), l on used H OH r o f x method C c arbon propan-2-ol Primary Classic ation C H propan-1-ol p Figure 10 H H primary, mechanisms n C H H o H U H 2° d H H c arbon conditions indic ates H 1° reaction and halogenoalkanes halogenoalkanes: the number of (gure 10). The p classied bonded to o the (1°), is H i n c arbon are H 3° C v Alcohols H c arbon 2-chloropropane Primary atom y p Figure 9 c arbon c arbon atom is r e 1-chloropropane primary (1°) (gure 9). c arbon Cl order to y C In n C atom. l C tertiary, O H and y H aects t i s H H chain tertiary, we count secondary (2°) H 1° c arbon c arbon atom. A c arbon atom, a c arbon atoms, and a atoms parent group. primary, bonded other primary, functional compound one c arbon the the classify compounds such as e to other a of on c an P three of bonded bonded group r is functional compounds. position whether number atom the of s s on halogenoalkanes, determine the of halogenoalkanes and amines as depending In position properties that atom of the consider the c arbon group. 281 Structure 3 Classic ation of matter When naming “amine”. amines, When the substituents are methyl are parent name loses the bonded to the nitrogen atom. 2° is replaced H C H C H gure 11, the substituents H C C C H H H H H H C tertiary (3°) amines Classify a useful skeletal organic formula propan-2-ol of the as a propan-2-ol. primary, following is a secondary H H H H H H H i t a H C H H a OH E v O l C H u H H a C H H C CH reason. H C 3 H H D. H C C H d H f x C H OH n C o C U C r o H C. 282 C H giving C i n H H alcohol, o B. OH H tertiary v A. H or secondary alcohol. p which solvent. of H CH 3 O b. Deduce is the H H C C C C H OH H H H y Draw n y a. r e 16. Propan-2-ol 3 3 H 1-amine and t i s Practice questions CH y (2°) CH N l secondary C N,N-dimethylpropan- 1-amine (1°), H H amine P N-methylpropan- Primary C H H p Figure 11 H 3 N r H 3° H CH N propan-1-amine 15. by used to signify that amine H H which N is e C In -e, s s H amine H H sux secondary and tertiary amines, groups. 1° H the naming Structure 3.2 Functional groups: Classic ation of organic compounds Functional group isomers Functional group isomers dierently such Compounds each group and with other. For isomers of they are have hydroxyl and example, this an alkoxy consider molecule methoxymethane, C H H H of propanal. C C H H group, H O C H H H be functional Propanone, isomers group the isomers simplest have the of each other, as they ketone, is a functional molecular H formula C 3 H 6 O. H O C H C C H O H H H simplest The ester and c arboxyl ester, groups c an also methyl methanoate C 2 4 be O 2 functional group , is an isomer of the C c arboxylic acid, ethanoic acid. H H U O H C C O O H O H i r o ethanoic acid t a Molecular models H 8 and O. obtained In u information 3 methoxyethane this from task, you different all will types have the 3. of Tool v • skills 1: Physic al and digital models molecular MolView). formulas of the three visualization Explore the isomers using software (for example, different visualization possibilities, including: ball and stick model, space- and models of these substances. Relevant Create online reflect on a O the propan-2-ol formula C l f x molecular H o d H methyl methanoate Propan-1-ol, C C H n H o with isomers. propanone i n Compounds v propanal p C r e H y H hydroxyl methoxymethane Both H a isomers Functional n isomer c an groups. O. l ketones with 6 O group and c arbonyl structural H y have alcohol 2 t i s both be group. C ethanol Aldehydes an c an formula C H OH arranged y C alkoxy atoms e H groups molecular ethanol, an the groups. P H with with functional the include ether isomers functional r H structural dierent s s of that filling model, molecular electrostatic potential (MEP)surface. digital molecular modelling E 4. Compare and contrast the strengths and uses of each Instructions 1. State the structural full of structural formula formula, skeletal various formula for e ach thre e and molecular models. How does the molecular structure of these three of compounds the formulas condense d 5. and the relate to their physic al properties and isomers. reactivity? 2. Using three a molecular model kit, construct models of the isomers. 283 LHA Structure 3 Classic ation of matter Linking question LHA How does there fact In the Molecular models task, you gathered information about three that three compounds, including: types of formulas, types of models, discussion of isomers strengths and properties. Reect on how this information could be organized. dibromobenzene support of the only the current benzene’s Would you opt for a table, Venn diagram, concept map, or anything dierent? model Do you have a preferred method? Why does eective organization of structure? information support learning? 2.2) r e (Structure s s of are Thinking skills ATL Another type of isomerism arrangements of the multiplicity. atoms. organic compounds is have However, Stereoisomers c an the they same be have and formula, congurational isomers (gure 12). O conformational isomers molecular dierent spatial subdivided into two major t i s classes: bond some n and by stereoisomers y connectivities exhibited pair of l stereoisomerism. A isomerism H H H H staggered of e t h a n e, eclipsed The two v i ew e d configurational isomerism along isomerism the bond p Figure 12 c an For about the f x This process u separate the groups the in bond isomers or does not a O l Congurational bond ethane, CH c arbon 3 CH breaking together to 3 resulting conformers involve individual bonded single form by a single bond, these dierent , the two CH in two 3 conformational groups possible c an rotate orientations, known as (gure 13). any bonds, and it is virtually impossible to conformers. isomers c an only be interconverted by breaking and reforming optic al cis–trans bonds. Congurational isomers c an be subdivided into cis trans isomers and isomerism of v Ty p e s optic al isomers (gure 14). c o n f i g u ra t i o n a l isomerism Cis–trans isomers E In aliphatic able to on the two alkenes, rotate dierent, two same “this about groups the isomers side identic al means 284 c arbon two about example, conformational isomerism isomerism contains rotate i isomers. configurational p Figure 14 molecule of o a groups t a r o d If Ty p e s n c arbon–c arbon conformational c o n f o r m e rs U p Figure 13 isomerism C i n H isomerism o H structural v stereo- H y H H H p r e H H of c an the of ” and are the cis unsaturated double isomer bond. where c arbon If two these opposite means “the sides other of the side atoms two identic al c arbon double bond, and a on trans to c arbon exist: a c arbon substituents side bonded c arbon are substituents trans double of ”. are not groups isomer bond. In are where L atin, cis y P Stereoisomers (Structure 3.2.7) Structure example, isomers bond is the trans, the about symmetric al the double reference plane: depending on molecule of but-2-ene has bond which it trans groups: Classic ation of organic compounds congurational (gure 15). The horizontal axis of the double is used to determine whether an isomer is groups are either H H side of the cis or plane. C s s H cis Functional LHA For 3.2 H 3 reference C C C C plane H C 3 CH H CH 3 3 trans-but-2-ene r e cis-but-2-ene an C=CHCH atoms has be 3 sides of cis–trans c arbon atom the , will not experience two identic al identic al groups reference isomerism, have two to be the two dierent. cis–trans (hydrogen groups For the atoms), H plane H C so the potential C C o also propene not ex h i b i t i does cis trans isomerism t a Cis Propene r o p Figure 16 d propene isomers will 3 H 3 propene, n C methyl attached to example, reference C two isomerism bec ause one of the c arbon U H has plane. (gure 16). H methyl substituents on the C also exhibit the trans-isomer o 2 to has i n H alkene unsaturated but-2-ene plane, while the v each opposite of p For on reference but-2-ene n l substituents cis-isomer the of y of i s o m e rs r e side trans O gure 15, the same cis y P y In The t i s p Figure 15 trans isomerism also occurs in disubstituted cycloalkane molecules. In this case, the reference plane is the at face of the ring structure. For example, the compound trans congurational isomers (gure 17). u f x 1,2-dimethylcyclobutane has cis The cis-isomer has both methyl substituents on the same side of the plane of the ring. The trans-isomer has methyl substituents on opposite sides of the plane of the ring. l O Disubstituted three-carbon rings also experience cis CH 2 H E 2 C 3 CH v C CH a H H H CH 2 H 2 H The cis–trans 3 CH 2 H C C i s o m e rs C C CH 3 trans 1 , 2 - d i m e t hy l c y c l o b u t a n e CH H 2 C CH 3 cis p Figure 17 3 trans isomerism (gure 18). of H H C 3 CH CH 3 cis p Figure 18 3 trans The cis–trans i s o m e rs of 1 , 2 - d i m e t hy l c y c l o p r o p a n e 285 3 Classic ation of matter LHA Structure Practice questions 17 . Deduce and draw the cis- and trans-isomers of hex-2-ene. For each of the following molecules, state whether they exhibit cis s s 18. trans isomerism. For the molecules that do, deduce whether they are in the or trans conguration. Name each molecule using IUPAC rules. b. CH H CH 2 3 CH H d. 2 H H structural c an form formula cis trans of e. 1-ethyl-2-methylcyclopropane f. 2,3-dimethylpent-2-ene is dened atoms. one c alled is or C 2 Chiral t a r o Pe n i c i l l a m i n e 2 - a m i n o - 3 - m e t hy l 3 - s u l f a ny l b u t a n o i c molecular the C arbon-1 - and acid l a E v O 286 has Penicillamine u name H f x p Figure 19 I U PAC C 3 c arbon have more the atom chiral c arbon ability to A bonded to stereocentre atoms pair rotate of four or exhibit optic al dierent atoms asymmetric a type isomers plane-polarized of are light; centre. congurational c alled this is known as optic al activity. OH H molecules 1 C a known as a optic al isomerism. o C as also n with It C of o v c arbon groups p 1,1-dibromobut-1-ene i HS y r e pent-2-ene d. i n O 3 d 2 deduce b. U NH then 1-bromo-2-chloroethene enantiomers. C and 2-bromoprop-1-ene isomerism 3 compounds, c. Molecules H following a. chiral or the H C 3 isomers. Optic al isomers A H C O the they t i s if C 3 C H Draw H 3 C C 19. CH 2 n 3 CH C l CH y H is a drug formula C is only c arbon-3 bonded achiral. C arbon-2 chiral. used H 11 NO bonded each groups therefore 5 to have to in 2 is Its three four c arbon-3 bonded The the S. treatment of rheumatoid arthritis. It has the structural atoms, bonded are to presence so four it atoms identic al of formula (the dierent this chiral is or not is shown chiral, or groups. two or gure achiral. However, methyl atoms c arbon in 19. C arbon-2 two groups), so of the four c arbon-3 is groups of atoms, and it is atom means that penicillamine is optic ally active and c an exist as a pair of two dierent enantiomers. Enantiomers are non-superimposable ofsymmetry. mirror images of each other, and they have no plane y H H c. 3 P H C CH 2 C C 3 CH 2 r C C H e a. cis Structure atoms formulas, and groups like the present one in a in gure 19, molecule are provide information connected. on However, Functional groups: Classic ation how the these of organic compounds LHA Structural 3.2 H formulas * do not allow centre. As a you to result, dierentiate enantiomers between are enantiomers drawn as in a molecule three-dimensional with a structures N chiral known CH N as stereochemic al formulas the 20), where specic symbols are 3 used to s s designate (gure direction of a bond (table 14). OH Name Symbol Directionality * bond C — aligned with the plane of paper c coming dash bond c going out of the plane (towards the viewer) r wedge bond HO behind the plane (away from the viewer) NH 2 e line P adopts tetrahedral and as line dash are central arrangement around bonds, bonds a a one chiral as a in stere ochemic al atom with with bond c arbon four angles atom, two regions of of formulas of 109.5°. the electron For the bonds are tapered, meaning that they start o narrow enantiomer are of a example, shown in stereochemic al the formula, stereochemic al you need gure 21. C 2 HOOC COOH H were to the to align same enantiomers, the in gure models the models direction. mirror 21, of are them such that However, image non-superimposable. using would if the a you show SH Nicotine the All c arbon ( t o p) is tobacco ( m i d d l e) is t hy r ox i n e t hy r o i d a of these atom, molecules designated n a t u ra l l y plant, is h av e *. sy n t h e s i z e d norepinephrine n e u r o t ra n s m i t t e r, ( b o tt o m ) by a and hormone f rom the gland 2 This means that if molecular model kit, it would be i in 3D penicillamine functional u l were the to groups hold structure a of of the same type mirror to one of the the other enantiomer. Figure 22 e n a n t i o m e rs HOCH the 2 Molecular of C H (O H ) C H O. c h i ra l models of the 2 , 3 - d i hy d r ox y p r o p a n a l , C an you identify c e n t r e s? a E v O f x t a point those r o impossible like create ) 3 c h i ra l by I H o you of d Enantiomers, E n a n t i o m e rs C(CH a 2 n p Figure 21 2 C U C) 3 I p Figure 20 o NH C i n N 2 C HS(H draw its v mirror plane H to formulas of the penicillamine p enantiomers For O HO NH y the image. I I from the r e draw mirror OH * wedge bond, and one as a dash bond. c arbon atom and widen. To O O drawn Wedge a arrangement that used t i s tetrahedral usually from Structure 2.2 bonds n density rec all dire ctional l will of y You S ummary y OH p Table 14 287 3 Classic ation of matter LHA Structure Worked example 6 Draw the stereochemic al 2-aminopropanoic chiral c arbon acid formulas showing of a the two enantiomers tetrahedral of arrangement around the atom. s s Solution draw the structural formula to deduce C C H NH atom O chiral. P * H is C the to mirror and the draw of is the attached and atoms to four methyl, amino and and groups around projected out of the plane of the behind atoms the plane of the paper (dash are in the plane of the paper (line enantiomer. H which bond bonds dashes draw t a a is It atom C o Finally, and rst H the i d r o are in just other the C 3 is the assigned 2 to tetrahedral used to give enantiomer, reected COOH NH you draw image of which group structure a a sense dashed your are of for all the 3D vertic al existing this rst equivalent. The structure. line to represent enantiomer. H H u l f x a E v O 288 wedges as your arrange projected groups asterisk. o matter one C not is an hydrogen p you that n does enantiomer, or v gives formula, such one atoms a y bond), two This atom i n U It with atoms: r e and bonds). of stereochemic al c arbon (wedge bond), marked groups. chiral paper is groups n draw the atom or O c arboxyl To c arbon atoms y chiral dierent t i s The l OH 2 C C H C COOH 3 NH M ake are sure that correct. le-hand the For image to the correctly H 2 connectivities example, HOOC between methyl show a the group c arbon central has to CH 3 N 2 been atom and the reversed (H c arbon bond. 3 groups C) in the y H c arbon r H which e First, we discussed the cis trans 1,2-dimethylcyclobutane. The isomerism trans-isomer of of a 3.2 Functional exhibits optic al groups: cycloalkane, this molecule Dierent This is bec ause enantiomers taste c an of also the be mirror same images are compound aected by a non-superimposable have specic dierent odours. enantiomer CH C 3 dierent chemic al environments, Our present pair light of by molecules enantiomers the other plane under angle, of the in rotates the human same in foods. plane of other chiral compounds, and is will rotate (gure p Figure 23 polarization enantiomer. anti-clockwise and is a explain E Draw v 2-hydroxypropanoic acid, C b. 4 H the Construct chiral 3D 10 (or racemate), the tasks (a)-(c): O which stereochemic al around racemic mixture complete ii. and O molecules, butan-2-ol, C Identify y following i. a. c. c alled a viewer enantiomer u the l of is o enantiomers f x O each pure plane-polarized light. Activity For a of o two by i of rotate light isomerism polarizer t a not plane-polarized r o mixture does of d Ro t a t i o n n U C i n pure enantiomer v polarizer p r e polarized light p Figure 24 3 n y t i s light it optic al enantiomer. unpolarized 50:50 The One enantiomer designated the (+) light and CH plane-polarized 24). source A C 3 t r a n s - 1 , 2 - d i m e t hy l c y c l o b u t a n e directions clockwise of non-chiral body. conditions opposite the presence in l ) the polarization enantiomer designated the ( but in in identic al y the same dierent are P rotates The biologic al enantiomers r A as become of e such but properties 3 (gure 23). H The organic compounds s s perceived its of mirror plane also H isomerism. Classic ation LHA E arlier, Structure c arbon formulas 3 H 6 O atom showing 3 is chiral. the tetrahedral arrangement c arbon atom. models for the pair of enantiomers. 289 LHA Structure 3 Classic ation of matter Rotation of plane-polarized light by optic ally active compounds Optic ally active Here will you rotate plane-polarized light. 5. Place 6. Dim the filters, and consider ways to the look filter the above light the source tube. through the polarizing filter. further. skills Inquiry at 1: polarizing filter e • polarizing and develop the second investigation Relevant second lights s s polarizing compounds explore this phenomenon using simple Demonstrate independent thinking and Identify the design • Inquiry 1: Design • Inquiry 2: of an control investigation variables sample and Identify explain and record a valid methodology relevant qualitative l observations S afety n y polarizing filter light source eye protection. teacher will provide you with further Rotate the safety changes 8. Repe at analysed. M aterials • Light source, filters for example, a mobile phone flashlight glass a measuring clear cover at one end, cylinder Clamp and stand S amples optic ally glucose or in on 2. to the Place a Pour of flat surface, the water in light the l of a details are a over pointing method in on of the in time the the variables observe the placing you of one Explore rotation that effect tube. of must of the plane- keep constant different compounds plane-polarized light. explain measure investigate • the of the order changes to do you would make one or more of the • of Show your on of effect rotation of concentration on the the effect of path length on the plane-polarized light investigate mixture angle the plane-polarized light investigate rotation 3. the the ideas effect of the composition of a rotation to your of plane-polarized light. teacher, and if you have time, try them out. M ass spectrometry (MS) (Structure 3.2.8) (MS) compounds into is an analytic al fragments, technique some of that which will c an be be used ions. to This is break up known mass as fragmentation. The molecular masses of these fragments will correspond to in the masses MS c an of certain functional groups or parts of the c arbon chain. As a result, 1.2. be fragments The a 290 and rotation up. tube. organic discussed rotation • source. M ass spectrometry the the explore • plane-polarized light. v Structure E spectrometer of height operational on filter u the aside. you water into the tube and clamp it in place source f x O The the Record source polarizing distilled above 4. light them substance(s) i Place Set the t a 3. test. of r o 2. to solutions o wish aqueous d Prepare this of to and following: Instructions 1. the Identify active substances, such as fructose. some order n sucrose, of List for U • a with solution light. C example, • tube i n Clear this Q uestions torch 1. • of polarize d this solutions o or effe ct polarizing experiment, prepare d filter intensity. p Two v • polarizing light y your the the r e being second to prec autions, depending on the identity of the substances O 7 . Your t i s Wear used to present molecular obtain in an masses mass spectrum. The information about the functional groups and structural organic compound. of the dierent dierent fragments peaks within a are mass recorded spectrum in a are graph c alled known y 1: r in Inquiry P creativity • Structure charge is fragmentation pattern. ratio (m/z) known as the in a mass molecular spectrum The peak with corresponds to the the highest Functional groups: Classic ation of organic compounds LHA collectively as a 3.2 mass to parent compound. This ion peak. The mass spectrum s s Worked example 7 for propan-1-ol is shown in gure 25. r 80 15 20 25 30 35 40 45 50 55 60 m/z spectrum of p r o p a n -1 - o l to the structure of propan-1-ol. you expected structural formula the will have analysed help more molecules. this option, c ase, C C C H H and u Molecular fragment the only fragmentation structural one formulas possible OH H identify for drawing its o H is by possible i H begin the draw there H structures out t a spectrum propose the each peaks and their of the m/z values. Use fragments: Explanation l a Molecular ion (the original molecule that + [CH 3 CH 2 CH 2 v O 60 the to In gure one d m/z values r o examine m/z f x the than compound, to (propan-1-ol): H Then, you o the When of This n all name U patterns. the formula. C know v you structural i n Solution p to show that the fragmentation pattern corresponds y M a ss Use the mass spectrum r e p Figure 25 O 12 n y t i s 0 of l evitaler 40 y P ytisnetni 60 20 If e 100 OH] has lost one electron) + 31 [CH 29 [CH 2 OH] Loss of CH 3 CH 2 from the m/z 60 + CH E 3 2 ] Loss of CH H H H H C C C H H H M r = 29 2 M OH from the m/z 60 OH r = 31 291 3 Classic ation of matter LHA Structure D ata Here on specic are some MS fragments are provided in section 21 of the data booklet. examples: peak with m/z = 15 corresponds to a CH • A peak with m/z = 17 corresponds • A peak with m/z = 29 • A peak with m/z = 31 • A peak with m/z = 45 to an fragment 3 OH fragment corresponds to a CHO or CH corresponds to a CH corresponds to a 3 2 O or CH COOH CH 2 3 OH fragment fragment fragment r straight-chain molecular formula C the structural spectra Explain A which B formulas of spectrum booklet. the is of two the produced C 29 60 o v 40 n U evitaler i n ytisnetni 80 20 o i 0.0 20 40 60 80 100 m/z B 43 100 evitaler a ytisnetni u l E v O f x t a r o d 0.0 86 80 60 40 86 20 0.0 0.0 20 40 60 m/z 292 80 100 10 O has two isomers. given. each p 57 100 by H possible are compound y A two isomers r e 22 of the data and 5 O b. the n Deduce M ass with isomers. t i s a. ketone y structural l A using section y P Practice questions 20. e A s s • Structure 3.2 Functional groups: Classic ation of organic compounds LHA Infrared (IR) spectroscopy (Structure 3.2.9) Infrared (IR) spectroscopy the types of functional IR groups. radiation. present IR is an analytic al in an organic spectroscopy Molecules spectroscopy is a type of absorb technique exposes IR that compound, a and sample radiation, of c ausing c an be used to identify therefore determine its an organic compound certain bonds to s s to bonds vibrate. IR vibrational spectroscopy e There are two basic types of vibration that are possible in molecules: stretching/ compression and bending. The type of movement observed depends on whether r the molecule is diatomic or polyatomic. For example, the diatomic molecule of hydrogen chloride can only undergo stretching and compression (gure 26). l a molecule the absorbs bond IR radiation, enthalpy and the the HCl molecule frequency of the a shorter of the enthalpy, iodide so contains vibration halide lower a the large occurs at frequency frequency iodine a of atom lower of the vibration. the and molecules (table 15). has frequency a relatively when Frequency of Br I Bond Wavenumber the fourth column 1 = λ is of –1 / cm 2886 usually table of halogen The described 1.3, in frequency Structure (f ) is related to 2230 wavelength halide (λ) via the relationship c molecules f = , where c is the speed of λ described in terms of its 14. As 2559 wavenumber, ν, light. is wavenumber is the This related means to that frequency the wavenumber c f. as reciproc al of the follows: ν = × a ν E v O wavelength: absorption w av e n u m b e rs l in the and –1 s 6.69 u spectra, shown e n t h a l py f x IR 7.68 298 p Table 15 In 8.66 t a H 431 366 i Cl H r o H vibration / 10 o 11 d –1 low bond n Bond in turn result larger the compared to the other Molecule enthalpy / kJ mol The vibration. U hydrogen its the higher C Hydrogen atoms, and i n masses wavelength, o v bond. The higher the bond enthalpy, the stronger the bond. This will in resultant masses of the atoms in the p on in n in depends c o m p re ss i o n O bond and y a vibration Stretching y P y When Cl r e p Figure 26 t i s H 293 3 Classic ation of matter LHA Structure In polyatomic Polyatomic molecules, molecules bond angle changes. bend when they spectroscopy. In the c an Both absorb other stretching also IR water and radiation words, c an undergo such be symmetric bending, c arbon (gure a dioxide 27), molecules so are they c an be stretch and analysed O H H H H r IR All of water thre e C C and modes of active compound dipole, such hydrogen, H performing change in the molecule homonuclear heteronuclear of the vibration; 2 , dipole molecules oxygen, O molecules, the such stretching moment, must otherwise, 2 as and such , and no that experience a change in peak do will not hydrogen 2 , are observed in the are uoride, compression molecules be have a permanent chlorine, Cl p a as i n is Diatomic active, the o c apable IR v Diatomic be y spectrum. to during r e IR moment O a dipole t i s For C n are v i b ra t i o n bending l v i b ra t i o n of molecules. stretching IR HF, vibrations. IR inactive. are only Bec ause there active. C Plotting the wavenumbers of the molecular vibrations against their intensity gives a graph known as an IR spectrum. Certain wavenumbers are characteristic of specic U functional groups, which are listed in section 20 of the data booklet. We can n therefore analyse the IR spectrum to determine the types of functional group present in a molecule. For example, the IR spectrum of butanoic acid is shown in gure 28. o i u ecnattimsnart l 50 a E v O f x t a )% ( r o d 100 0 4000 3000 2000 1500 1000 500 1 wavenumber / cm p Figure 28 IR spectrum of butanoic acid –1 There is a strong, very broad peak in the range 2500 3000 cm characteristic of –1 the O is 294 H bond in c arboxylic acids. characteristic of the C=O bond. The strong peak in the range 1700 1750 cm y Modes d i ox i d e asymmetric stretching y c arbon symmetric P Figure 27 by IR IR-active. e H where the s s H asymmetric. molecules O O or vibration Structure spectroscopy in a of molecule, an must unknown be used techniques, only so in with about c annot be information information spectroscopy, the functional used groups determine gathered derived nuclear to from from IR other groups: Classic ation of organic compounds present the structure spectroscopy analytic al magnetic resonance (NMR) s s or mass alone Therefore, conjunction as information technique molecule. such spectroscopy provides this Functional LHA IR 3.2 combustion analysis. composed hydrogen and oxygen is 10.43% hydrogen by mass. showing your spectrum of the compound functional group present is shown below. Deduce the in the molecule. 3000 2500 2000 1500 1000 o 3500 U C 0.0 i n 0.2 v 0.4 y 0.6 p r e % / e c n att im s n a rt 0.8 O t i s 1.0 n The infrared y b. l working. y Determine the empiric al formula of the compound, P a. containing only c arbon, of 62.02% c arbon and r An organic compound e Worked example 8 500 1 wavenumber / cm The mass spectrum of the compound is shown below. n c. Deduce the o d relative molecular mass of the compound, and hence determine the molecular formula. 16 E 15 u 14 0 v 15 20 a 40 20 d. l O evitaler f x ytisnetni 60 i 80 43 t a r o 100 58 42 26 27 29 25 28 25 37 31 30 Through experimental analysis, did not 36 35 39 38 41 44 57 40 40 45 50 55 60 m/z it was determined that the compound contain a terminal c arbonyl group. Deduce the structural formula of the compound. 295 3 Classic ation of matter LHA Structure Solution a. The method for determining an empiric al formula C 1: O 62.02 10.43 = 5.16 = 12.01 Step Divide result = by the value (1.72) 5.16 10.33 = 3.00 1.72 = 6.01 to determine the rounding if ratio, necessary is therefore C 3 H 6 O. 1.00 n l formula y empiric al = 1.72 1 b. A strong peak c arbonyl to the c arboxylic deduce range the of 1700 Possible acids functional and molecule atom terminal middle of in equal not the an is 58, or c arbon of the so up the formula. so and the its is molecular acid, As the the masses molecule c arbonyl is a by the the 3 empiric al H 6 O. there is only one c arbonyl structural of formula is C as you answer is represented atomic c arboxylic molecule, chain, the the aldehydes, asked group is not in group must be in the ketone. Therefore, the formula is CH 3 COCH 3 . n U magnetic resonance o i t a 1 Proton nuclear the u a spe ctrometer ester compound so has and Structure 3.2.11) in magnetic the add presence of H NMR) spectroscopy (Structure 3.2.10 most useful determine l (NMR) 1 molecules. v nucle ar d r o 296 A E resonance f x O p Figure 29 to propanone, Proton nuclear ( of you molecular position the compound is If C the i n oxygen also mass 58. the include question compound, o The is v d. this peak: the the p formula, molecular ion of but shows molecules y relative molecular r e The esters, group c arbonyl/C=O. c. 1750 cm organic O ketones, in group, C=O. t i s a the magnetic resonance ( analytic al technique for H NMR) spectroscopy structural is perhaps determination of organic 1 The the spectra number molecule. An produced of dierent atom's by H NMR chemic al chemic al spectroscopy environments environment is help of dened you to hydrogen atoms by the other atoms attached to it. 1 H NMR spectroscopy is an example of spin resonance spectroscopy. The 1 nucleus of a nucleus has hydrogen atom two possible quantum number (m I ), H spin with contains a single orientations, possible as positively described values of m I charged by the = +½ or m I proton. This nuclear spin = –½. y number 1.72 P 1.72 The 1.72 16.00 r lowest 10.33 1.01 mass e atomic 2: 27.55 by the relative whole below: Divide the percentage each shown s s Step is H Structure a hydrogen small nucleus magnet. behaves Similar to the as a spinning electron spin, charged the two particle that Functional groups: Classic ation of organic compounds LHA The 3.2 c an act as possible spin states of a 1 H nucleus are degenerate: they is the exist at the same energy level. When a strong 1 magnetic become eld applied non-equivalent to and sample containing split into two H nuclei, the two spin states dierent energy levels (gure 30). s s the magnetic field generated by the higher energy spin state is e high energy opposite to the applied external 1 spin = magnetic field – r 2 1 spin + = l = – 2 n y 2 the magnetic field generated by low energy = + O t i s the lower energy spin state is 1 spin aligned with the applied external 2 magnetic field no magnetic field magnetic field applied magnetic magnetic equal required in for the to dierence process of Chemic al is radiation molecule with four 1 to and E spectrum small different a be two hydrogen by as be spin of dierent signals. when CH H 3 Si C CH 3 CH promoted to from radio the p Figure 31 3 states. The waves of exact environment in dierent dierent is 3 the accepted reference u n i v e rs a l l y standard used in 1 NMR s p e c t r o s c o py energy hydrogen chemic al A summary of chemic al shi frequencies. hydrogen These of T MS lower H atoms radiation energy increases along with the exposed chemic al relative is values are dierent environments environments signals for of given in the data is hydrogen given in section 21 booklet. sc ale, assigned a where chemic al methyl the signal shi groups of 0 produced ppm. by a TMS is a (gure 31). 1 resolution H NMR spectrum, the area under each signal 1 dicult indic ate the to in H NMR low a is must equivalent proportional to the number of signal are expressed in units of parts per million (ppm). using standard v a a δ, it the absorbed l O is examining So, states nucleus u (TMS) Low-resolution When molecule. on H NMR spectrum f x symmetric al between electromagnetic measured tetramethylsilane state, depends chemic al shi, shi the spin t a displayed in a terms of their nucleus i frequencies 1 are the d absorb For energy r o The the between strength. higher analysed environments ΔE, eld the this hy d r o g e n o atoms to the n energy dierence, state of U energy states o external spin applied C energy two is i n The The field v a y p Figure 30 p r e applied y P 훥 E 1 spin to H nuclei generating measure, so an this signal. integration trace The is area under added to the 1 the relative number of H nuclei represented by each signal. 297 Structure 3 Classic ation of matter LHA 1 a Cl b CH CH 2 Consider the H two the an signals in integration NMR spectrum spectrum trace of 2, of chloroethane, indic ating and one with two an shown hydrogen in gure 32. There integration trace of 3. These signals 3 a b therefore correspond to the CH and CH 2 groups 3 in chloroethane, respectively. Figure 33 groups shows the with spectrum. This identic al is and NMR spectrum atoms bec ause in the therefore the of molecule form a 2-bromopropane. There are is single symmetric al, hydrogen so the two methyl an integration hydrogen chemical shift trace of 6 corresponds to these methyl groups, as 1 atoms. The other signal corresponds to the CH they have six r 2 groups environment. The signal with 3 3 three molecule, but only two signals in the e are 2 H hydrogen s s 1 4 are environments, one with group. δ/ppm l a c h l o ro e t h a n e, C 2 H H 5 NMR spectrum of CH CH 3 CH 1 3 2 1 v area under U hydrogen i interactions t a The radiation These u l a R aman, interactions bending the c ow o r k e r, The his R aman Sir K ariamanikkam 298 many result When the 2 in the of 2 - b ro m o p ro p a n e, a For and the 4, actual or 3 H 7 Br instance, imagine a one signal correspond so C 1 ratio. under 3 in this sc attering to for sc attering to the is 6, or H NMR equal to 1, and the relative number of numbers and dierent spin radiation of R aman is of any hydrogen atoms other values in a (NMR) radiation known who of of a electromagnetic we or have bond certain looked at. stretching wavelength sc attered. incident is types methods changes electromagnetic was and as sc attered “R aman awarded radiation are sc attering” aer Sir the 1930 Nobel Physics area. eect formed determine was experiments, and spectroscopic nuclear the Venkata work particles of phenomenon used R aman as area These be wavelengths This nowadays the structural very R aman basis dicult to and team his of features obtain R aman of at spectroscopy, which is various rst. arrived at materials. Evidence Eventually, with conclusive evidence h e re, existence of the eect. his Sriniv asa How Krishnan (IR). same. for the 2. could molecule, Chandrasekhara the with a Occ asionally, not to given environment, between underpin encounters Prize picture d effect equal environments TOK and v E V. R aman is each are where spectrum o d r o f x O C. the other in traces signals ratio. ongoing discovere d the atoms dierent 1:2 two NMR n in integration with H o spectrum 1 The C i n Sometimes δ / ppm p chemic al shift p Figure 33 6 y r e 4 O t i s b p Figure 34 3 b Cl n y a The much evidence is required in order to support a robust conclusion? for y P Br a 1 p Figure 32 Structure 3.2 Functional groups: Classic ation of organic compounds LHA Worked example 9 1 Sketch the H NMR spectra for the following molecules: propanone, CH b. ethanoic acid, CH 3 C(O)CH 3 3 COOH s s a. Solution Propanone, CH The structural and C(O)CH two is a symmetric al shows a c arbonyl functional methyl 3 , groups. The ketone. b. Ethanoic acid, CH in two the hydrogen atoms is O H signal methy l C H refle ct the 21). Integration there is in propanone range of 2.2 only traces one will 2.7 ppm are not hydrogen produce (data booklet, useful here environment. A H NMR spectrum is shown below: formulas E v COOH the two organic CH number 1 two NMR environments, spe ctrum is 10 8 6 chemic al shift 4 2 0 –2 δ / ppm 0 compounds are COOH 3 H 3 H of H of should shown below. b. C OH Deduce of a a. H l structural the signals 1 the the re gion u f x O The 2 in and the δ / ppm Practice questions 21. 3 of the in n 4 under protons produce o 5 i 6 chemic al shift t a 7 C i n 12 U 8 d 9 r o 10 of sketch signal protons o 1 sketch of the A a will 13.0 ppm, below: v bec ause protons the ratio 1:3. are as p section in group 9.0 produce The four O signal will of of y equivalent c arboxy l re gion r e one the t i s chemic al environments The group is shown H identic al of the 2.5 ppm. which H total n C in y C a l 2.0 H proton identic al: a H has environments. y all COOH, P for 3 chemic al chemic al The environment dierent r group 3 formula e a. signals the ratio 3 C 3 H and CH of OH areas under the signals in the 1 H NMR spectra of these compounds. 299 Structure 3 Classic ation of matter LHA 1 High-resolution H NMR Advances in technology have led to improvements in the accuracy of analytical 1 instruments like H NMR. There are distinct dierences between low resolution and 1 high resolution H NMR spectra and the information that you can extract from them. s s 1 We have already information The demonstrated about number of signals while that chemic al corresponds the low resolution environment to integration the trace of H NMR hydrogen number shows of the provides us with atoms dierent number in a molecule. chemic al of e environments, the hydrogen atoms 1 signals are split each into atoms in a clusters about the In a of high resolution peaks. number The of H splitting hydrogen NMR spectrum, individual pattern atoms of a signal provides us attached to neighbouring molecule. H The signals a signal spectrum split for the H into a atom b is into a peaks and cluster of (gure 35). The is referred to as a three peaks and is a noitargetni noitargetni 7 two = = n o i u l a E v O f x t a 8 of 2 1 C i n U d r o 9 1,1,2-trichloroethane split triplet. b 6 5 4 chemic al shift 3 2 1 0 δ / ppm 1 p Figure 35 are dierent c aused discussed with or The patterns previously, against close other, splitting by the spin the proximity, resulting The number c an be of in external the high resolution of in proton cluster has has H N eld. elds for NMR spectrum for the hydrogen neighbouring If spin This that hydrogen generated an N + 1 of nuclear splitting. by using the x signals magnetic signal a the interactions magnetic NMR peaks c alculated spin each If the proton H of is by c an known as individual either atoms their in nuclei spin signal, environments hydrogen a nuclei. be molecule c an x aect N spin coupling known as + are each multiplicity, rule: will be split into As aligned protons attached to its nearest neighbours, the NMR signal of H 300 of cluster 1 ,1 , 2 - t r i c h l o r o e t h a n e The in the is o referred to as a in atoms v doublet. two 1 peaks. : b b y for the H H a Cl p are signal r e There C and H n C a O t i s H Cl a l Cl y Consider 1,1,2-trichloroethane, which has two hydrogen environments: H y information signal. P with with r associated Structure the are given peak ratio, is in table deduced 16. The using ratio of Pasc al’s the heights of the peaks in a Functional groups: Classic ation Number of Number of peaks in hydrogen atoms (N) cluster (N + 1) Splitting pattern cluster, singlet doublet (s) 2 3 triplet 3 4 quartet 1 (d) 1:1 (t) 1:2:1 (q) 1:3:3:1 1,1,2-trichloroethane, the spins of the nuclei in the H The H a b environment doublet a environment triplet (t), and couple the with has + two peak proton, 1 rule. protons, ratio each and The will be therefore the H heights and of the signal a two peaks therefore the H 1:2:1. The b signal magnetic ally In to new form the is the row row. of of the parts high-resolution pattern triangle, Evidence dierent role in of exists the mathematics in known adjacent that world this H as numbers pattern including scientic NMR spectra correspond Pasc al’ s triangle was China are added together known and by ancient Persia. n What in observed mathematic al U civilizations each nuclei p (gure36). a a C in equal. be split into o patterns numbers i n splitting the will v 1 to be split into be equivalent H other. TOK The will will y not one N r e do has (d) as per the O a a environment. t i s The H b n in environment will couple with the spins in the adjacent H l example, NMR spectra. y For H more peaks are oen c alled “multiplets” y P 1 Splitting patterns due to neighbouring hydrogen atoms in e 2 s s 1 1 Peak ratio r 0 Splitting patterns of ve and organic compounds triangle. neighbouring p Table 16 of LHA Examples 3.2 knowledge-building? o i d t a r o u l f x a E v O etc p Figure 36 etc P a s c a l ’s etc triangle 301 3 Classic ation of matter LHA Structure Worked example 10 1 Show that the splitting patterns in the signals of the gure 37 correspond H to the structure of butanone, H 3 NMR spectrum in CC(O)CH 2 CH 3 s s e r 9 8 7 6 5 4 3 draw the structure of butanone v three Hydrogen the to Environment B i in The A 0. is adjacent to not adjacent in to the of hydrogen 3 any N + 1 which other rule, correspond to the hydrogen this environment spectrum. B environment ratio The u l a E v O f x predicted in each cluster chemic al values from shis the C C, which has three hydrogen corresponds to the quartet. The signal is split into corresponds of data the signals booklet. Hydrogen a triplet by the two hydrogen to that predicted by Pasc al’ s for B and C also closely match the This is summarized in table 17: Splitting Peak Predicted Actual pattern ratio δ / ppm δ / ppm 1 2.2–2.7 2.1 1:3:3:1 2.2–2.7 2.4 1:2:1 0.9–1.0 1.1 N + 1 rule environment N = 0 A singlet 0 + 1 = N = (s) 1 3 B quartet (q) 3 + 1 = N = 4 2 C triplet 2 p Table 17 Yo u c an + 1 (t) = 3 d i s p l ay data like this when 1 compound 302 to environment environment number C environments, According to the singlet CH 2 butanone environment B. peak triangle. = o is corresponding atoms the Therefore, t a r o d atoms. hydrogen n U corresponds N CH B spectrum. environment environments, so C A dierent in the o are signals deduce O C three i n There C 3 and of y H spectrum p r e environments. NMR O Solution First, H 0 n resolution t i s High 1 δ / ppm 1 p Figure 37 2 y chemic al shift l 10 to signals in a H NMR spectrum matching the s t r u c t u ra l details of a y P 11 Structure 3.2 Functional groups: Classic ation of organic compounds LHA Combining analytic al techniques (Structure 3.2.12) To determine the in a variety types of the of data It is the of a molecule, example, group you dierent present combination and of the data may be methods required are arrangement that will to needed enable of the you to interpret a to deduce c arbon atoms deduce the structure. may include: analysis • the • the • information from the • description of spectroscopy formula) (MS) data spectrum 1 H NMR spectrum reactions that the compound c an undergo section of DP chemistry). (covered in the y i n propan-1-ol, propan-2-ol and methoxyethane. Now you will examine these C o v In the skills task on page 283, you were asked to create molecular models of p Predict spectra from molecular models O t i s data booklet r e Reactivity l (IR) empiric al y infrared c alculate y mass (to n • data r combustion P • e This For functional molecule. unknown structure data. s s wide molecular models and consider the spectroscopic data of these substances. Tool 2: Identify and extract data from databases n • U Skills Instructions using models molecular the spectrum of each patterns the done determine molecule. number of so the and methoxyethane already. number of different 1 Then, predict the signals, area H NMR under each signal, shifts). u For each compound, predict some of the peaks found in its mass spectrum. spectra a Spectral l the compare to in your D atabase E v O AIST for spectrum propan-2-ol not Predict spectra IR and each chemic al have 4. Search major you 3. 5. the in (consider and if models environments f x splitting molecular propan-1-ol, t a hydrogen of kit, i Examine model r o 2. molecular a o Create d 1. a signals suitable expected database for and each compound. see how the actual predictions. Examples of databases include the for Organic Compounds. 303 LHA Structure 3 Classic ation of matter Worked example 11 Ethane reacts with chlorine to form Combustion analysis was used product X. to determine that a. Determine the empiric al formula of product X. b. The mass and product 1 X has the following composition by mass: H NMR spectra of product X are shown c arbon s s below. Deduce, giving your reasons, its structural 24.27%, hydrogen 4.08%, chlorine 71.65%. formula, and hence the name of the compound. mass spectrum 1 NMR spectrum e H 100 r 10 20 30 40 50 60 70 80 90 100 110 10 percentages by the relative atomic mass for = 2.021 hydrogen: = 71.65 = 2.021 35.45 When analysing containing abundant chlorine-35 of isotopes and of dierent chlorine-37. product f x X will H 4 Cl and C 2 H 4 . the We c an to molecular c an elements Therefore, 3 35 also occur the 2 . [C 27 [C H 2 ] + Cl] 4 35 2 H chlorine-37 and deduce that the in We 304 formula of c an conrm 1 H NMR spectroscopy. The H NMR a mixture of dierent spectrum The hydrogen has two signals, indic ating two integration there are three environment environments. trace times than in shows as The a signal at a many the ratio of 3:1, indic ating hydrogen atoms in one other. 2.0 ppm is a doublet (H neighbouring hydrogen proton. a At 5.8 ppm neighbouring A ), indic ating environment with one there is a quartet (H hydrogen protons. We c an environment therefore conclude 1,1-dichloroethane: loss chlorine-35 molecular ion atom peak (m/z 3 A Cl from 100) H A C C + Cl] 4 H of loss of chlorine-35 ion atom peak (m/z from 98) H p Table 18 molecule: 1 molecular ion containing two molecular H the B ), indic ating chlorine-35 + 2 m/z mass of the structural CHCl formula peak molecular ion containing + Cl 4 E 63 2 0 with three isotopes 37 [C ion molecular chlorine-35 isotopes 65 1 deduce that the the molecular relative (99). structural 2 relative atomic a 2 + Cl] H the formula following 2 to we identic al of 3 δ / ppm Reason l [C 4 37 Cl v O 98 2 H empiric al is average equal this using Therefore, the contain is 4 product X is likely to be CH u 35 [C 2 need to consider that chlorine fragment 100 Cl chlorine-37 atoms (table 18). Molecular m/z 4 t a and H i ions chlorine-35 2 spectra of compounds we r o fragment therefore C values C o two mass: mass chlorine, is d has formula the n b. empiric al U The bec ause 5 spectrum, formula 4.04 1.01 chlorine: mass C 12.01 the empiric al 6 chemic al shift o 4.08 From v c arbon: i n 24.27 7 p the each element. 8 y Solution Divide 9 r e m/z n 0 O t i s 20 a. l 40 ] F ra g m e n t loss of HCl and Cl ions of product X A Cl H B that product X is y 60 y evitaler P ytisnetni 80 Structure 3.2 Functional groups: Classic ation of organic compounds End of topic questions Which of the following analytic al LHA 4. techniques would Topic review 1. Using your answer the knowledge guiding from the question as Structure 3.2 fully as topic, CH 3 the dierence CH(OH)CH 3 , between and propan-2-ol, propanal, CH 3 CH 2 mass II. infrared e I. to predict their properties? r 1 III. H NMR I and II only Multiple-choice questions name of the I and III only II and III only D. I, II and III rules? 5. H Which C 2 H H 3 2-ethylpent-3-ene D. 3-methylhex-4-ene Which of the 3 3 3 COOCH CH CH 2 2 CH CH 2 2 the same homologous 3 i CH COCH in OH OCH 3 H NMR spectrum? 4 3 2 1 0 PPM A. propane B. propanal C. ethyl ethanoate D. butanoic acid Extended-response questions 6. a. State, giving a reason, if but-1-ene exhibits cis–trans isomerism. b. The a the a [1] reaction hydrogen which iodide at occurs room halogenoalkane, reason, if LHA u l E v O f x t a CH D. 3 is ? r o C. 3 3 o CH CH OCH d A. B. following 3 n series as CH 5 following o C. C 4-ethylpent-2-ene v 4-methylhex-2-ene B. i n A. p H U 3. CH C 3 the y C C 1 has r e HC substance O 3 t i s CH n B. C. following compound, applying l IUPAC the y is y A. P Exam-style questions What CHO? possible: How does the classication of organic molecules help us 2. s s show the 2-iodobutane. product stereoisomerism. between of but-1-ene and temperature this produces State, giving reaction exhibits [1] 305 Structure 3 Classic ation of matter LHA 7 . Organic analysis chemistry of hydrogen a. an and Deduce c an be unknown used organic to synthesize compound a variety indic ated of products. that it Combustion contained only c arbon, oxygen. two features of this molecule m/z 58 and that m/z 43. c an Use be obtained section 22 of from the the data mass booklet. 120 r P 60 40 l 58 42 27 26 14 16 25 41 44 40 35 30 40 57 O 20 38 36 31 t i s 15 39 37 29 28 25 0 45 50 m/z b. Identify the bond 20 of responsible the data n 0.2 3500 3000 i using infrared [1] A 2500 2000 1500 1000 500 1 1 from the H NMR and then identify the unknown compound 1 previous information, the H NMR spectrum and section 21 of the u l [3] a E 10 spectrum. wavenumber / cm information the the databooklet. v O f x t a r o Deduce o d 0.0 in o 0.4 c. C i n 0.6 U ecnattimsnart 0.8 v 1.0 A booklet. p 1.2 for the absorption at 55 y section r e Use n 15 y 20 306 y ytisnetni evitaler 80 [2] e 43 100 s s spectrum, namely 9 8 7 6 5 chemic al 4 3 shi / ppm 2 1 0 Structure a. Draw the stereoisomers of butan-2-ol using Functional wedge-dash type representations. b. Compound how A is enantiomers c an be distinguished equilibrium with compound OH using a polarimeter. B and B are present isomers. in Draw each two molecule. other [2] structural isomers one [2] of the compounds is shown: 0.8 0.7 0.3 0.2 3500 3000 2500 1 wavenumber / cm mass chlorine shown. 1000 [1] o i 40 50 60 70 m/z reason for presence structural spectrum. 1500 t a the the 30 this 2000 u Explain Draw the is producing d 20 a Outline compound l 10 E b. of v O a. spectrum the r o evitaler b. reason, f x ecnadnuba 0 11. a n The giving U Deduce, C 4000 i n 4500 o 0.1 0.0 y 0.4 p r e 0.5 v ecnattimsnart 0.6 O t i s 0.9 LHA of n spectrum l formula. y IR A same groups y Compound the functional e b. The the r Identify P a. a. [2] B with 10. organic compounds O A c. of [1] two in Classic ation s s 9. Outline groups: LHA 8. 3.2 the and two peaks at relative formula of m/z = abundance 35 of and the 37. peak at [1] m/z = 74. ethoxyethane. [2] [1] Deduce the number of signals and chemical shis with splitting patterns in the 1 H NMR spectrum of ethoxyethane. Use section 21 of the data booklet. [3] 307 To o l s for Chemistry s s Introduction purpose that develop this uency with work, these is to provide inquiry skills lab-based in and DP you with a chemistry. practise questions, toolkit You applying the containing should them collaborative in refer a key to variety sciences experimental, this of section contexts. project, and technologic al frequently Such your and throughout mathematic al the course to contexts might include skills scientic investigation assessment (IA). for internal O y p o C n o i t a physician and alchemist n l y t i s r e v i n U u a 860–930) was an Islamic Persian scholar, v E 308 l Figure 1 Al-R āzī (c. d r o f x O y practic al section scientic P tasks, of support r skills e The Tool 1: Experimental s s techniques Being take is and one responsibility the work, for learner with your being a prole strong actions principled attributes. sense and of their involves This fairness means and that you act justice, and that consequences. In conducting lab safety and ac ademic integrity. the twelve prevention protective place, is to prevention: accident and using integral eliminating must the and conducting risks protect laboratory important risk the part of assessments, by opting for experimenter, labwork. It includes putting safer the the correct alternatives if safety possible. environment, as well as space. v p else an gear, considerations anybody relate Familiar personal protective equipment (PPE) includes safety goggles and lab coats i n schools and laboratories these days, but it has not always been the case. When C o for the people present during the experiment. The use of PPE is commonplace in y S afety in principles r e measures chemistry waste prevention O wearing green t i s Accident and n of prevention l Two y S afety y experimental of honesty, P you principled integrity r with e S afety and respect of self, others and environment looking at historical photographs or images of science experiments, see if you can identify any safety risks and discuss suitable control measures. risk assessment to do something hot the could the is not Chemists need should with people be be think it on be the if the our they mass done to c arrying out experiments. ways of minimizing risk, and what of a the risk assessment potential to are: c ause harm (for to example, touch of the the person doing the crucible minimize should the crucible. This is likely to the wait 5 and risk its and, minutes contents). if for possible, the crucible balance, and pick it up using tongs. The near the experimenter ’ s lab bench). Figure 2 A selection of hazard that during the about the products of their of chemic als loc al rules involved in for and the disposal policies disposal future bottles in order process. to If Do you know what each experiments and identify of them The symbols you are likely to see on reagent experiment, but also aer its completion. in a chemistry lab. procedures. recovered have experimenter loc ated relevant to disposal accordance as should only E suitable placing will a balance S afety measure (the v O before to what hazards parts with themselves l measures: remove cool (continuing burn need that u they main activities prior hazards, severity and likelihood of harm, including who could how Control to well and f x • if or The of i experiment happen wrong. crucible). evaluating harmed completed t a get goes substances a be r o Risks: to identic ation o Hazards: example, • needs involves d • if risk n Assessing U A means? needs to be done in protect possible, the environment, as le-over reagents use. 309 Tools ATL for Chemistry Research skills Research the 12 principles I’m green chemistry and try to complete the blanks gure 3. Well, interviewing It involves making merald, who we use renewable feedstocks chemic al How r in s s oday of do you do that? where leads possible. So our company processes as sustainable the way in green makes chemistry. as So, r is polyester from corn possible. starch merald, ust what instead of crude oil. green chemistry ? e r ..................................................... ..................................................... ..................................................... ..................................................... ..................................................... o C n you In lots of ways spews out wo of them are water ................................. ground? ............ ..................................................... o do i the f x es. Anything want else to you We always aim to replace toic feedstocks with ones that do not say? threaten ................................. or the human health environment. ................................. u l a E v O How that atmosphere, and 310 waste t a pollution? the the es, of course. We do this by energy v i n U into d r o about reduce you try to p ..................................................... What guess processes efficient? ..................................................... ..................................................... I Figure 3 This comic strip ................................. , that’ s ................................. great. hanks ................................. ................................. ................................. ................................. shows two people discussing green chemistry. Think of examples to ll in the blanks n l make ..................................................... O And We reduce waste by ........................ you make sustainable? y do y P y t i s else processes r e How Tool Experiments Doing so should further involve minimizes using the the smallest environmental possible and amounts of 1: Experimental techniques reagents. economic impact of practic alwork. Ac ademic integrity citing and those others citations titled also should bibliography, the numbers and name URL, if be of the marks when but cited it is author, in or giving by them between an paraphrase a credit your by own words appropriate citation. another person’ s ideas. section at the end of the document references). The IB does not mandate expected date and followed you included works style, ideas distinguish of that the minimum public ation, title of information source, page applic able. which hazard symbols are solid copper(II) sulfate d. solid sodium metal e. cyclohexane. on a bottle containing: 3 out what What are system is used at your school, then write a full some of the hazards associated with U 3. referencing for this book. gure4? the is bec ause the used type it of scientic Quantitative Measurement provides research. obtained data is a from are core quantitative instrument Figure 4 Bottles containing concentrated solutions of sulfuric acid and 55 °C 55 °C tr ue Qualitative data include all value measurement, such as colour numeric al information obtained aspect of science and is important to data. One 50 °C of the factors 50 °C aecting data u quality f x scientists was bubbling. measurement. of not n or skill i from vital t a changes a information o d is r o data non-numeric al reagents in hydrogen peroxide Measuring variables Handling two C reference v Find i n 2. sulfuric acid y c. nd p 0.10 mol dm to o concentrated nitric acid b. likely r e a. you O State n y 1. t i s Practice questions y are required their clearly l included quotation then referencing must P particular are using respecting You r references (usually a of includes sources. e In-text Full others information s s Respecting used to obtain these data, as well as whether it correctly or not. l a O Accuracy and precision precision and allows measurement us if a to to 5). assess their times, how c an Repeating be the close a precise instrument measured measured is under values are the will occur processed over close value is a to to small the not accurate) (but not precise) same each used to determine a particular values or accurate evaluated in terms of their measurements precision: high-precision several refers precise values) (but (gure 45 °C 45 °C processed accuracy Therefore, Accuracy (and E conditions other. values v Measured range. true Figure 5 Precise values are close to each other. The mean of accurate values is close to the true value value. 311 Tools for Chemistry low and precision high high accuracy and Figure 6 The dierence between accuracy and precision low precision. high accuracy and are group of How shown you A. high accuracy, ± B. low C. high D. low below. 1 2 3 Mean −705 −707 −701 −704 accuracy, standard enthalpy of combustion is reported in booklet as −1 367 kJ mol . assumption that the Validity specic heat is usually be a think this valid? validity concerned other valid of Measuring with reasons why research question assumption? any situations in assumption would not Reliability the same is under l a E v O school of trials repeatability. precision results, In but some to c ases of entire something improperly whether when c an you an c an identify pieces or measures c alibrated of research, is oen analysis is what pH valid (certain doing are clear trials is the is using same it is probe when it meant to will result in an addresses the investigation good of trials allows you usually tests a for produces methods, When dierent to more good identic al practice. results The again repeatability: the c arried experimenter. the outliers. statistic al same repeated by experiment several the assessed in terms of and compare any three be trials conditions during doing an methodology Conducting and indeed reliability. whether with Reliability same practic al, and and accuracy appropriately. results the Repeating A connected results. agreement out pH measurement. are mostly water (Reactivity 1.1). of invalid u which analysis solutions f x you water i is C an of think the of t a this you that r o C an to invalidate dilute composed of o not bec ause equal d does is measurements, terms c apacity of a dilute measure. solution of in accuracy o the quality assessed for n example, experiment, C c alorimetry U a high low v i n The In precision precision, Validity and reliability Thinking skills ATL high p 1 the data precision y Ethanol’ s results in terms of low precision, r e 1 5 kJ mol these precision? 1 ΔH / kJ mol describe and O Trial would accuracy rule example) c alculate repeats, of you do a trials to a the assess their mean, more assess reliable the thumb. greater numbers of replic ates are required. In others (such as titrations), measurements should be repeated until several agree concordant with each values other, are and in obtained. titrations Concordance this generally means means a that the range of values less than 3 0.1 cm writing 312 (although up lab this value is investigations). not set in stone, and it is best to specify it when y results a n Their experiment, enthalpy of combustion of t i s ethanol. the l c alorimetry accuracy P a determined y performing students high The bullseye represents the true value Data-based question Aer precision e accuracy r precision low s s low and Tool Reproducibility is the consistency of results obtained by following the 1: Experimental techniques same Practice questions method and reagent by sources. dierent results. conditions of the reporting Next, we reasons ndings consider time, is scientists scientists allow several temperature, dierent outcomes gives to by key types pH, an experiment insight include other length, of into detailed scientists of people, instruments and to the those obtained 4. Explain reproducibility of their taken experimental methods more reproduce these ndings. measurements electric to current in and chemistry: why with a C an probe than using but not r M ass (e.g. a more air. reaction. in mass all on of involve For over is how time containers, mass record the gas mixture was mass before produced trailing one similar to determine mass was they the and hydrated M ass of crucible and anhydrous C. 2.030 g D. 100.10 g aer the reaction provides seem, have the same mass. mass most shown in hydration Identify sulfate copper(II) sulfate measurements precise the of of the copper(II) mistake sulfate and Figure 7 A digital balance. nothing on top collected suggest Why is there a number on the display even though there is what the of the balance? following data. A mistake might have c aused it: M ass / g ± 0.01 g 13.55 11.96 11.60 u 2.0 g feather a 0.0007 g B. E A. the copper(II) l following v O Practice questions using may measured. crucible the produces a gas will absorbed. It is worth noting i crucible of that esc aped gas will be lost. t a r o empty of f x of obtained the and or enthalpy being M ass Which mass changes during a o the d when of was decimal mass of the empty container when conducting a M ass the only n wishing made of mixture U student Is to any 0.5000 g) increase that happens when magnesium burns however the reaction Data-based question 7 . out (e.g. C analysis. 6. recorded lab, the decimal places it leave places i n gravimetric was mass measurement the bec ause reaction much to the example, the the Remember A not of o not do decimal a chemistry v information that mass example The value, than a p in balance a four In number y Another to the r e decrease down recorded precise precision. 0.5 g). Gravimetric analyses chemic al writing mass of greater y with When a the O place to. levels balance, t i s measured mass example, dierent a n For of paper. l measure zeroes. have precision y will balances the are values P greater pH valid? dierence. Dierent pH experimental outcomes be reliable electric potential pH precise determined 5. mass, measurements e volume, will teams the out s s when c arried Comparing research One but photo was balance? really massless? 313 Tools for Chemistry Volume The its suitability intended designed to of a particular function and measure chemic al reaction. asks needed. the volume approximate involve For and are the and level volume for a asks holding equipment depends on precision. are analyses, conic al designed of measurement volumetric Beakers volume-measuring not of very Measuring cylinders precise. volume changes during a pipette, give an liquids, burette and/or approximate not volumetric measurement measuring e of are of desired s s Volumetric analyses type the volume. r – burette, including vessel, measurements such should a be solutions, burette read or from the the form neck of bottom a a U-shaped volumetric of the ask. meniscus o i u l a v E Figure 10 Water and form 314 aqueous solutions a U-shaped meniscus Figure 11 Mercury forms an upside-down meniscus, where the centre bulges upwards above the sides inside Volume (gure 10). tap B – volumetric meniscus t a r o f x O burette aqueous n U d read from the bottom of the meniscus as C liquids, narrow D – volumetric ask i n M any C D A – measuring cylinder, o pipette, a C Figure 9 Various volume-measurement devices. n l volumetric pipette into a conic al ask B O y from a p being released v Figure 8 A liquid y P y t i s r e A Tool 1: Experimental techniques Practice questions 8. Which of the following volumetric • volumetric ask appropriate for precise volume measurements? pipette s s • are cm 100 • burette • beaker • conic al ask • measuring 90 80 e 70 60 60 r cylinder 50 P What glass container would you use to: 30 determine b. prepare c. c arry the volume of liquid added during a titration l a. 50 20 you need for from a pH a solid reagent experiment from 10 the reagent bottle 10. approximately measure 25 cm at of liquid? Figure 12 The meniscus should eye-level be read What is the volume of liquid contained in the measuring cylinder in gure12? for time is the example, measurements measured. certain The volume uncertainty arising from of usually of the gas a to be depends the measuring stopwatch to the the rate record produced. on The of a chemic al how long it takes, in uncertainty associated context in which time is being stopwatch itself is oen far smaller than the experimenter ’ s reaction time. n U uncertainty a When use C for with (s). might i n seconds, time second you o for y unit v SI reaction, p r e Time The O your lab bench 3 d. n acid solution y the standard t i s to a y 40 9. For example, if you are recording how long it takes for a clock reaction to produce a change in colour, there will be a short delay between the instant the colour o d changes and your nger stopping the stopwatch (gure 13). The stopwatch display might indicate a time of 16.45 seconds, which might suggest an uncertainty of i r o ± 0.01 s indicated by the two-decimal-places display (gure 13). A more realistic u l f x ATL Figure 13 In the “iodine clock” Thinking skills Some a E v O t a uncertainty, given your reaction time, would be around 0.5 to 1second. over as hours, these to experiments take place long periods days, c ases, record is or it times of time, such even always to the weeks. In necessary nearest second? reaction, the clear and colourless reaction mixture suddenly turns blue-black aer a certain period of time 315 Tools for Chemistry Temperature You c an use required digital accuracy dependent on analogue the context monitoring temperature probe to thermometers precision a of of of the a measure temperature. The measurement is highly experiment. temperature data to temperature logger as c an be shown achieved in by connecting a gure 14 e interface r O dissolving in water n l y t i s ammonium chloride temperature sensor y p r e o v i n Figure 14 Temperature sensors and data loggers c an be used to C monitor the temperature changes over time U Length gures are measurements discussed in in Tool 3: Mathematics. involves loc ation the be usually used. estimating between on i impact u 2 l a v Figure 15 usually also two number As with the sc ale of involve last rulers although other devices, such as other digit divisions. signic ant analogue in the Sc ale gures devices, resolution in reading the sc ale measurement, depending on its the will therefore measurement have an (gure 15). t a r o f x O E 316 1 This ruler gives a d detail c an o c alipers, n Length Signic ant 3 4 1 1.7 cm reading to two significant figures The relationship between sc ale resolution and 2 3 4 1.75 cm This ruler gives a signic ant gures reading to three significant figures y P .pmet time s s Continuous or and Tool 1: Experimental techniques Practice questions 11. A student conducted concentrations results and obtained, in trials of minutes, an iodine were seconds the and clock same reaction. in all The reagent three trials. The milliseconds (min:s.ms) are s s shown three temperatures below. State the b. Suggest times a to the suitable nearest second. uncertainty for the time measurements, in seconds. C alculate the mean seconds. for the colour change the limitations and of each propose an of the following 3 A student heats temperature student from the wishing down ice to bath, to of the and a beaker in and the checks the water, with the bulb beaker. perform this in thermometer a reaction temperature adds the in a at 2 °C cools one of the beaker, second removes beaker reagent. U c. the C reagents bottom water a o A of i n b. the 5 cm placing v touching by p a. temperature alternative method. y scenarios r e measurement n 01:12.985 explain O 01:15.120 01:10.400 t i s 1 and l Mean / s 1 ms 2 3 Identify occur y / s ± 12. to / min:s.ms y taken in P Time Trial time, r c. e a. In a study of an enzyme action at 40 °C, a student places a test tube n containing a sample of the enzyme solution into a 40 °C water bath. The student then immediately adds the substrate to the enzyme solution. is 16 orange bar? a 14 device you 15 c alled a 16 c aliper. u might obtain l using a 17 What 18 19 20 chemistry-related c aliper? a E v O 13 shows measurements the i Figure of t a 12 f x 14. length r o 11 the o What d 13. Figure 16 A vernier c aliper 317 Tools for Chemistry pH of solution The pH of solution. c an be is related sc ale. be estimated pH of probe a The red colour concentration indic ators, c abbage. shown is of such These hydrogen ions in that as universal indic ators compared to a indic ator, or the vary key, in colour allowing a (gure 17). solution c an be more reliably determined with a properly (gure 18). paper turns dierent r colours depending on the pH of the substance being tested across pH value c alibrated e pH pH from the with s s the entire to estimated extracted The Figure 17 pH solution anthocyanins to a pH l t i s Figure 18 A pH use the indic ator juice c abbage as a anthocyanins (HA), which U amphiprotic Materials • red • two are 2 A . In basic accept with protons form anions, A high : + and form • c ations, H H vinegar or neutral visible producing Relevant skills • Tool • Inquiry 1: Designing Measuring at solid household products, such as baking soda or • five • cloth or filter paper • standard clear glasses laboratory glassware Method 1. Prepare the indic ator by blending several leaves anionic) of different of red c abbage with a glass of water and filter the wavelengths resulting mixture through cloth or paper. characteristic colour a c abbage 2. juice. v O red and light the two anionic form u (c ationic, absorb 1.3), 1: as (green to yellow) l f x forms anthocyanins of such + H form (blue) (Structure products, A + neutral t a + (red) changes r o + three + HA A 2 cationic form All household detergent antacid tablets anthocyanins lose i H + H pH, o and solutions d protons liquid organic compounds. In acidic solutions with anthocyanins + H c abbage liquid n pH, red products. c abbage juice is rich in low of for determining the pH of y will C household Red you acid–base p task, to the colour of the universal indic ator o this natural i n In probe reading compared v Using a natural acid–base indic ator r e O To test one of your of the liquid household products, pour 3 20–30 cm filtered c abbage juice into and add 2–3 cm of the E gloves and 3. To test one of the solid and the c abbage eye Compare your solid dissolve it household in water, products, 318 Do not use products colours of your solutions with protection with a corrosive safety warning in figure19 to determine their approximate pH. crush before mixing with juice. the thoseshown • clear analysed solution. variables 4. Wear a 3 glass S afety • n y y P Tool 1: Experimental techniques s s e r l Figure19 Red c abbage juice at pH 1 (le) to 10 (right) What 18? the pH might values a of suitable indic ators advantage b. pH probes. substances and one be in each the a conductive liquid made ensure on the of Electric al conductivity, is a related be used E measured property, directly to monitor produced bec ause identity mobile of such and charge the as not so be in A amperes connected in where all the components path, the with no ammeter branches. c an be solid to be tested placed circuit 2 switch samples in a series solid into known a circuit is usually touch as electrode materials the A as graphite the within one measured must leads to either end of the solid. Inserting such do is liquid a is electrolytes) or each surface platinum. other. area, involved. circuit, requires C are must be distance These known as beaker between the factors all aect the carbon electrodes resistance. oen solution a given in microsiemens per centimetre, µS cm to be , tested is the probe. ability to Electric al conduct charge conductivity and which measurements c an be c an the progress conductivity ions switch 1 which with and It ammeter single circuit, (collectively electrodes factors a Inserting materials, applied ow solid pack. a to the series connect solution v O opposition battery inert that voltage connect simply or of depends electrodes, a l Current a you to circuit following u to by f x electrodes taken how of charge. the t a powered straightforward: throughout electric al current, type another, circuit. shows of o the simplest aer same ow measure r o circuit 20 in The one of i Figure the To d anywhere rate n is the circuit. connected Current is ammeter. circuit 1 o in an U series are current using gures 17 and C i n (A), in disadvantage of Electric current Electric tested c ase? y pH the p one a. of uncertainty v O utline each r e 16. are What O 15. n y t i s Practice questions y P present. of of a a reaction solution in which depends ions on are the consumed or concentration and Figure 20 series. solid Circuit Ammeters are connected in 1 shows how to connect a sample while circuit 2 shows how to connect a solution 319 Tools for Chemistry Electric potential dierence Potential difference in a circuit. measure It the 6 8 potential in In chemistry, in the is the expressed parallel you volts dierence to will energy in the most dierence (V) and across circuit component likely per unit measured a measured potential between two points voltmeter. components being encounter charge using and are Voltmeters therefore s s connected 4 2 0 is (gure 21). dierence measurements 10 V context of the known to as the cell potential potential. two half-cells dierence Measuring and cell completing between the anode and potential the involves circuit with Figure 21 Voltmeters are connected shown in parallel to a component in a connecting salt bridge as r is voltmeter The e c athode voltaic cells. gure 22. to measure V potential dierence of the lamp is being voltmeter l measured) salt bridge how current passes through solid materials, such as a copper wire or graphite O utline how through current passes electrolytes, such as solution containing solution containing lead(II) v molten bromide or 2+ A sodium chloride. i n (aq) (aq) Figure 22 The potential dierence across the two half-cells is measured Another term you value This reason is the potential. Their why high the come potential across high-resistance resistance cell is dierence i o hence potential means electromotive across a cell voltmeters that measured are virtually is close force when used no to (emf ), which is the no to current current the is owing. measure cell ows through emf. Practice questions 19. Explain the u l a E v O f x t a r o d them, might of by connecting a B represent metals n U maximum A and C voltmeter to the electrodes. 320 2+ B o aqueous y 18. p r e electrode. O O utline t i s 17 . B n y A Practice questions a. In the AgNO this mass, For 3 + + you volume, each NaOH(aq) (aq) section involved following mixtures undergo a change in conductivity as proceeds. HCl(aq) b. 20. why reaction NaBr(aq) have time, variable, → → of that an + AgBr(s) considered temperature, think measuring NaCl(aq) + how pH, H and O(l) NaNO to you state 3 (aq) measure current experiment variable, 2 and key types potential of variables: dierence. have done in the past which the instrument used. y P its potential dierence (in this c ase, the Tool 1: Experimental techniques Applying experimental techniques In this and section, analysing we will discuss chemic al experimental techniques for preparing, isolating species and phenomena. s s Preparation Preparation of a standard solution volumes dilution high of standard precision. volumetric glassware, This such is as solutions requires achieved burettes, by measuring using pipettes an masses and analytic al balance and r and and with volumetric asks (gure23). The solid volume with a solution 4. The beaker, each rod water solution formula of the or is is added of and who the volumetric funnel into to are the the to ask beaker rinsed the beaker, and over at using three a to a reagent substance, the its bottle mixture is glass funnel. times ask ask least the completely. volumetric volumetric turned prepared to with deionized until ten with water, ensure that the solute its times a Figure23 level to label reaches the Volumetric ask and pipette ensure that the showing the concentration, the date and initials solution, as well as any hazard labels, if o i d f x t a r o stirring rod paper u wedge stirring rod l wash bottle 18/08/2023, P.M.D., 1.51 –3 mol a O 1 2 3 4 5 dm v Figure 24 Preparation of a standard solution E Measuring for added dissolves thoroughly. relevant. a is solid on the ask. transferred name person to water stoppered mixed dry n The and this water the U 7 . is is and C ask solution until i n The clean completely. graduation mark 6. a o Deionized in v 5. glass transferred rod transferred adding weighed p is time deionized glass is is y The solute balance. r e 3. of liquid) follows: n small stirred sometimes as O A (or analytic al prepared t i s 2. an usually y using is l 1. solution y standard P A e Preparation cylinders preparing volumes sample with with or and diluting sucient at least beakers standard precision. four with graduation solutions, Similarly, signic ant as the marks they do balance should not never allow must be used measurement of show the mass of a gures. 321 Tools for Chemistry Practice questions 3 22. 21. These steps outline how to prepare a A student solution solution. List them in the correct prepares 100 cm 3 of a 0.50 mol dm standard of copper(II) sulfate by c arrying out the order. following steps. Identify the mistakes. Add deionized s s A. water up to the mark on the A. A sample of 7.98 g of hydrated copper(II) sulfate volumetric ask. was B. Dissolve the solute in a small amount of weighed and transferred into a volumetric deionized ask using a funnel. in a beaker. B. the solution to a labelled reagent Tap it for was solution into a water volumetric ask. to mark the on volumetric ask until the desired mass of the was neck beaker, invert and the the mark. ask. stirring until the meniscus rod with C. The D. A ask was label several times. for 23. was stoppered placed on water was c arefully was back at the mark. and the inverted hydroxide air property sodium is hygroscopic, moisture. for the Suggest preparation hydroxide. the meaning that it implic ations of this of standard solutions of o v p y r e absorbs twice. volumetric ask storage. Sodium At added, so the meniscus Excess solute. volumetric ask. ask the Too much O and funnel over the t i s Stopper water accidentally of dropwise. n the deionized added y Rinse was l the water above removed G. added the Preparation of solutions by dilution dilute burette water is a standard or added U several solution, volumetric times to to the certain and graduation achieve a volume transferred mark, and of the to the a solution is volumetric stoppered measured using ask. ask is Deionized turned over uniform composition of the nal solution. n o i d t a r o u l f x a E v O bottle Figure 25 A The solution in bottle B c an be prepared stock solution in bottle A 322 a pipette C a i n To bottle B by performing a simple dilution of the y the Weigh F. to P Pour point went E. was close storage. this D. water bottle r Transfer e water C. Tool 1: Experimental techniques Worked example 1 3 3 A teacher wishes to prepare 0.500 dm of 0.400 mol dm HCl for a class 3 practic al. A 11.0 mol dm HCl stock solution is available. Determine the volume of stock solution required to prepare the desired solution. s s Solution We know that: × V 1 1 c indic ates V stock known solution values and 2 indic ates 3 (11.0 mol dm the desired solution. gives: 3 3 ) × V = (0.400 mol dm ) × (0.500 dm ) l 1 V gives: V = 1 3 3 0.400 mol dm × 0.500 dm 3 11.0 mol dm 3 V = 0.01818182… dm 1 3 to cm , which are more easily measured in a lab, and round 3 = 18.1818182… 3 cm ≈ 18.2 cm (3 sf ) 1 as extremely c arry times out more a accurately low serial dilute for prepare c alibration concentrations dilution than that where in the are the prepared water. This 1 cm of procedure is often previous the required. one. in each The Figure solution until solution the and desired 3 +9 cm in 1 C alibration in are introduced 1.4 shows final 9 cm of Practice questions concentration 24. A technician the wishes following to solutions prepare by 3 diluting HCl 3 a 5.00 mol dm solution. What stock volume of cm the stock each 3 +9 cm curves Structure each test tube solution is needed in c ase? 3 3 a. +9 cm 750 cm u HCl 3 of 1.00 mol dm (aq) 3 b. 3.00 dm of 3 0.150 mol dm a 25. Suppose of the that stock the HCl (aq) concentration solution in gure 3 26 is 1 0.1 0.01 0.001 10 10 10 10 of 1 ( 100 1 ( 1000 0.50 mol dm . Determine 3 the ( v E l f x O stock cm 3 +9 cm solution 3 1 26 adding o 3 cm t a r o 1 i d 3 cm increasingly successive tube is obtained. 1 of prepared using this 3 preceding repeated solutions often concentration, each of the in mol dm , diluted solutions. 1 ( is transferring of are n deionized by series solution 3 is a curves p to to Solutions U ten used C method, how are concentration. o dilutions i n lower v Serial three y gures: V to r e signic ant ( 10000 ( Convert O t i s 1 n for y Solving y the 2 the P Substituting × 2 r where = 1 e c Figure 26 A serial dilution 323 Tools for Chemistry Reux and distillation Reux and (gure is a A used piece condenser a we in will equipment essentially ow the both reaction of inner cold the known as a tube within As a they very whereas between similar is used The is in them to a of c ases. tube the condense. to Reux separation both outer temperature granules equipment purposes. distillation tube. the c ausing involve dierent condenser another anti-bumping dierences for result, decreases, use is cooled gaseous You will also ensure smooth boiling. reux and distillation. (b) l thermometer y C purified product heat i Figure 27 (a) Reux apparatus (b) simple distillation apparatus is heating u l a E v O f x t a Reflux that any into the used of a boiling ask still is of liquid 27(b) a to is The in that shown on volatile substances during continuous reaction the is placed mixture components until substance the the the with condenser, temperature above condense being the reaction ask, so and drip back down two solution of a mixture point according to their of the more volatile lower boiling point rises into and is collected at the bottom of the stays constant during state changes, so will distilled liquids remain constant at the boiling o. simple distillation, (or of boiling the thermometer is solid aqueous and of condenser heated substance non-volatile sulfate the separate mixture condenses an loss A esc aping. above depicts example, copper(II) of Remember whichever from from reached. temperature point For The the mixture. rising used is head, condenser. the minimize instead points. component the to reaction vapours Distillation Figure 324 anti-bumping granules o d r o water in impure product n granules o v i n anti-bumping condenser heat U reaction mixture water out p r e water in O t i s out n y water condenser which with copper(II) water using this method. separates very sulfate dierent c an be a volatile boiling points). separated into y P (a) bec ause used mixtures, water. tube techniques discuss a are r Next, that heat confused methods e substances of is continuous notice to easily these s s The are However, method method. by distillation 27). Tool Fractional distillation is a related technique, used to separate 1: Experimental techniques mixtures of two Practice questions or more as an miscible and ethanol–water mixture is column, heated, volatile mixture. the condense liquids A the similar fractionating substances on with glass in the beads column mixture and boiling drip is points tted continually back to (gure 28), such the ask. As the 26. vaporize, rise into the cycle repeats many times. The vapours the lower-boiling-point column and a greater Compare collect bath, each the component, condenser. particularly if the The which eventually receiving distillate is ask volatile. A is and fractional distillation. reaches the top of the sometimes fresh 28. placed in an receiver is Primary alcohols such as used to ethanol by distillate. are acidied distillation is used air. to separate o fractional liquid i and d 29) r o (gure y Figure 28 Fractional distillation apparatus Industrial-sc ale C if you obtain (b) a (a) technique, reux, you would were looking to an aldehyde and c arboxylic acid. n U heat use which or y receiver oxidized o i n v distillation flask outline distillation p r e cold water in further n condenser large surface area are O t i s column with a aldehydes. there is c arboxylic acids. Identify and water out to l y to if oxidizing agent, the aldehydes granules r enough thermometer anti-bumping rst and P Eventually, oxidized potassium dichromate, fractionating contrast simple proportion e ice-water enters reux down into the distillation contain and of contrast s s The and and simple distillation. 27. ask. Compare mixtures such as crude oil Figure 29 An industrial fractional distillation facility, used to separate the components of crude oil u l a E v O f x t a t 325 Tools for Chemistry Isolation Drying to a constant mass When a solid is obtained in the lab, it may contain trace amounts of water, even if it dried to a constant mass to ensure that they do s s looks dry. Solid samples are oen not contain any water (or other volatile impurities). Drying to a constant mass involves: allowing 3. weighing it 4. repeating cool for the until method drying two should Common be mass before the 2 when of aer hydration the salt to a hydrated heating: heat O(s) xH determining magnesium of remove p why, sulfate water be obtained. desired solid Bunsen burners, of 2 the O(g) salt, such as water and + MgSO 4 measuring (s) hydration of magnesium heated to a constant mass. n U Separation of mixtures the should C i n sulfate, •xH water heating o Explain 4 and v MgSO by the includes y the the sulfate, that are O determine masses ensure equipment t i s c an magnesium to equal ovens. r e You chosen heating Practice question 29. consecutive n and few minutes l heating process decompose. hotplates a y not to r suitable does it sample The method used to separate a mixture depends on the properties of the mixture components. For example, if one of the components is magnetic, it can be removed o crystallization or simple distillation. Mixtures of liquids can be separated with i fractional distillation or using a piece of equipment known as a separating funnel. t a Filtration medium, used to u usually the l a v Figure 30 Gravity ltration E 326 d r o f x O using a magnet. Mixtures of solids and liquids can be separated using ltration, In the as separate much lter and separates such paper. liquid paper due The that to paper, insoluble smaller than solid (gure gravity. according which solids solid le passes gravity filtration the particles lter to from behind in as its so the sizes tiny liquids. particles, through is the 30), their contains Liquid they lter by holes or easily paper passing or them pores. aqueous pass is through a Filtration is oen particles are through the holes in known as the residue filtrate. name suggests, the ltrate passes through y 2. A the e heating P 1. Tool This process involves c an folding be very the lter slow, so paper it is into sped an up by fluting accordion shape the lter (gure 1: Experimental techniques paper. Fluting 31) before placing it inside the funnel. s s fold to crease the this is the open filter quarter sections eighth sections “accordion” shape cone Figure 31 Figure showing the steps for making a uted funnel to usually a the three pump solid or four mixture. on the perforated draws (gure from the 32). slipping times and the with liquid C are it is oen mixture plate used when the is loaded onto a inside a Buchner funnel. through the lter paper and into must under the The the solvent be taken lter to not paper. to The Buchner residue is funnel remove soluble impurities. out of a solution by removing the solvent. Most of the solvent is evaporated o by heating it over boiling water until the solution p flask v drop of it onto a cold tile and watching for the formation of crystals as it cools. The i n evaporating r o boiling burner l u f x a O ltration apparatus solution from evaporating basin cold tile leave for a few days to crystallize Figure 33 Crystallization apparatus E v Bunsen t a water Figure 32 Vacuum o i d basin n U solution C the crystals are allowed to dry to a constant mass on a watchglass. o bulk of the remaining solution is le to cool slowly to allow crystals to form. Finally, any remaining solvent is removed from the crystals by ltration (if necessary), and vacuum pump conical becomes saturated. The solution c an be checked for saturation by spotting a copper(II) sulfate filter paper overll the y crystallized at ltration, of r e A solid can be laid underneath prevent rinsed paper vacuum ask gravity O receiving lter than component t i s a moist from quicker y Suction is desired n of the y sheet is l residue cone from circular lter paper P Vacuum filtration r e fold to crease the 327 Tools for Chemistry A solvent c an evaporation, the also be which is removed a type of from a solute distillation through performed simple at low distillation, pressures or to solvent’ s boiling point. s s Practice questions 30. Figure base. 34 shows Copper(II) the preparation sulfate crystals and c an isolation be made of salt this crystals way, from from the the reaction reaction between an acid and an insoluble between base and filter l crystals funnel form O t i s r e y filter to remove heat evaporate the filtrate until unreacted base the chapter solution is not mixture isolation of an inorganic salt ltration by and evaporating the evaporating dish disadvantages of doing dish over a (gure 33). evaporated by reacting an insoluble base with an acid step. heating advantages p the rst o heating a the the l why E v O Suggest in the evaporated Discuss with earlier are in C solvents shown. compared discussed removed o i as is cool crystals filter the cooled t a Sometimes so, what u b. directly 328 d Identify f x a. c. r o Figure 34 Summary of the preparation and leave to salt n constant mass crystals start to form v i n U dry crystals to to dryness. hot n y start to water bath as y base and acid oxide. P mixture of excess base copper(II) e unreacted acid r salt solution sulfuric warmed rotary reduce Tool Miscible liquids liquids (such as funnel (gure c an oil be and 35). separated water) Aer c an ensuring by be fractional distillation. separated that the is Experimental techniques Two immiscible by placing them in a stopcock 1: closed, the separating mixture is oil poured into denser, lower narrower layer slowing neck Separating at is then down the funnels Soon aer, the drained as the liquids into an interface form two underlying between distinct beaker the two liquid layers. The layers interface water by opening the bottom end of the funnel. are also used to selectively extract solutes from one solvent another. r Recrystallization used to the is a puric ation desired organic solid synthesis selectively process from a based mixture products. dissolves dierent of This on selective solids. solubility that c an be Recrystallization is oen isolation components technique of the requires a mixture depending temperature. mixture of solids typic ally contains three types of components, of funnel Figure 35 Separating an oil–water O t i s The water runs out n on purify that a y solvent isolate y to l used P Recrystallization oil approaches the e into funnel. s s stopcock, the each with mixture with a separating funnel dierent solubilities in the identied • impurities that are soluble • the The recrystallization desired solid that we in in the the wish solvent solvent to at at isolate, all all temperatures temperatures which must be when hot, but insoluble when cold. process is detailed in gure 36. filter t a r o Figure 36 Recrystallization process the solid to insoluble The ltrate the a more starts o dissolves desired usual, the solid are form but to hot solvent. liquids, molecular intermolecular discussed in forces Structure 2.2. pure crystals vacuum pump pure crystals forming This c auses the desired any soluble impurities. removed cool slowly. follows, then of aer remaining are in all of and by hot Crystals ltration. of the desired solid form while remain in solution. ltration which complex as to impurities vacuum crystals v Cold allowed dissolved with impurities E 4. is soluble is along a Any the solid l 3. impure dissolve, u First, O 2. f x 1. In i d water are cold o hot polarity solvent n U solvent the C i n solid heat in insoluble impurities impure soluble Miscibility y insoluble p are o that r e impurities v • solvent: removing the soluble impurities and isolating dried. recrystallization, the hot impurities ltration but not two solvents step, the a are second desired used. solvent solid. This The is process added that c auses the precipitate out. 329 Tools for Chemistry Analysis Practice questions Melting point determination 31. What use method to would separate the you following The mixtures? purity solids a solid is placed in c an melt a be at assessed low c apillary to by measuring moderate tube. The its melting temperatures. sample is heated, A point. small Organic sample of the observed s s solid a. of frequently closely, noting a suspension containing down the temperature(s) at which it starts and nishes melting. Pure substances solid barium sulfate and an have sharp melting points that agree with published values. If impurities present, the melting point is usually lowered and the solid melts over a solution temperatures. ethanol–water mixture a sodium chloride solution Figure The d. a mixture of 37 lowered water and as into the from a oil bath. expected rst, Thiele The melting when the tiny tube the oil c an bath point yet be sample is is is used in a melting point determination. attached to a thermometer and then heated, approached. visible droplets sample has just melted. down and O impurities heating are second, when the t i s of soluble and insoluble the temperatures appear, mixture entire slowing Two thermometer y r e opening to allow air in n aspirin, an a containing y recorded: e. how tube l rate cyclohexane shows c apillary range of and out of the apparatus p v o or water bath C i n rubber band oil substance being tested n U o i d heat Figure 37 Melting point determination with a Thiele tube melting point observation of apparatus the may contain a built-in magnifying glass that sample. u l O f x t a r o Electronic facilitates capillary tube Data-based question following melting about the point E v concluded purity a The data of were the collected aer performing certain organic syntheses. O utline what c an be products. Experimental melting point of Theoretic al melting Experiment 330 Synthesis of aspirin Synthesis of methyl 3-nitrobenzoate product / °C ± 1 °C product / °C 122–129 136 76–77 78 point of y c. r an P b. are e aqueous sodium chloride Tool 1: Experimental techniques Chromatography Chromatography analysed to the the mobile likely to (gure38) is a collective separation and stationary encounter and of as thin-layer a term its phases. DP for a group components The chemistry two of types student chromatography methods according (TLC) of are to in which their a chromatography paper mixture is relative anities you s s are by chromatography (gure 39). front B 0 t i s (stationary phase) L B A O A L C L start line sample A A B phase) v Figure 38 The set-up for paper chromatography, and the resulting chromatogram C the tank is saturated adsorbent coating – SiO with solvent vapour 2 U components separate out on plate solvent 3 amount of the The end of the TLC l sample in solution is plate is immersed in spotted onto the plate a pool of solvent a O Figure 39 Thin-layer chromatography (TLC) 38, the c alculate R lengths A, E components v gure by to n thin layer of o u A small adsorbent coating i f x 2 TLC plate is prepared with a front t a (plastic or metal) solvent d r o plate 1 o i n atmosphere within thin layer of In p (mobile y r e start line solvent n l L paper y C chromatographic y P with lid B e r solvent beaker B, C marked L , L A and the , L B solvent, and L C indic ate the distances travelled 0 respectively. These lengths c an be used values. F The c alculation of R values was discussed in Structure 2.2. Review the content of F Structure 2.2 and answer the practice questions on the next page. 331 Tools for Chemistry Practice questions 32. 40 shows a. O utline b. Explain A student a why a which chromatography lid has investigating chromatography been component the experiment placed in the on components experiment as the mixture of a involving leaf pigments. container. has the black greatest anity for the s s 33. Figure solvent. food colouring sets up a shownbelow: e beaker with lid r C D E F drawn in ink black food colouring explain errors in the separated chromatography. rotated 90°, and run set-up using The again in a a shown. technique second origin n i paper acid / 90° and which answer the amino acid questions has the run the chromatogram (phenol / water then add locating agent v Between E attached. glutamic these H The acid amino these identity (Glu) acids CH aspartic have COOH 2 acid, of and COOH 2 two Asp ) R group leucine (Leu) dierent anity and functional the H 2 is for is a dierent are shown for CH 2 two glutamic COOH acid, Glu to With which which amino a acids. reference solvent H groups, atom, dierent below. the 2 (–COOH) c arbon for COOH 2 solvent 1. c arboxyl groups anities CH mixture below: greatest Amino acids all contain amino (–NH Asp in solvent 2 u 41 a State bonds. 332 the l gure O b. turn Figure 41 Two-dimensional chromatography of an amino acid Study a. f x miture in t a 1 (butanol / ethanoic water d r o chromatogram Glu o Asp the Leu U Glu solvent (gure 41). C Leu solvent i n = run known as chromatogram is run in one o solvent, be v two-dimensional two c an p mixtures Figure 40 Paper chromatogram of leaf pigments y and Complex r e Describe 34. O t i s B n y solvent start line A l paper to CH(CH 2 leucine, Leu and their structural ) 3 2 form known structures COOH CH polar The systems. 2 are substituent of hydrogen as an aspartic features, R group is acid (Asp), suggest why y P chromatography Tool 1: Experimental techniques Practice questions 35. Remember that the R value is c alculated as follows: F distance R travelled by the spot = F distance to the travelled by the chromatogram solvent s s Refer B below and: A a. c alculate the b. estimate the R value of the top spot. Assume that length A = 3.95 cm and B = 5.02 cm. F value of the middle and bottom black spots. F r Calorimetry measure the amount of heat released (or absorbed) during a thermometer reaction measurable heat c an be is a no illustrates out in an lost of to the error in of a c alorimetry is that the known surroundings, c alorimetry a c alorimetry and dilute insulated loss to the experiment involving container, such as a polystyrene a excess is that rise explain Determine e. Identify v C alculate d. E c. the whether amount the two equation a O and of to was for the further heat in an that will minimize determine and reaction, reaction hole) insulated container magnesium powder hydrochloric acid Figure 42 C alorimetry of the reaction between magnesium and dilute hydrochloric acid the enthalpy of lower heat the heat loss. insulated container using to 11.0 °C this determine diagram acid (with hydrochloric acid. reaction. would performed temperature balanced State the the hydrochloric 200 g. b. cup, to o Write experiment in during l a. modic ation The was shown used dilute i solution measures is and t a a 42 u reaction. gure surroundings and f x of zinc in magnesium dilute C two the c alorimetry of set-up between Suggest 5.0 g loss answer these questions. r o A to practice Heat reaction o lost 37 . and experimental reaction Identify b. 1.1 the surroundings. d a. in lid n the although i n Reactivity The c auses water. It is hydrochloric acid. Note that it is U 36. of experiments. Practice questions Review directly) mass v heat temperature of measure by insulation. magnesium minimize c annot p c arried is source principle we y 42 the energy large minimized between in basic (which r e Figure change that loss The c alorimetry, which is O assumed 1.1. change known as t i s a Reactivity enthalpy technique n reaction’ s a y discussed in using l chemic al y c an P We e R is the the enthalpy mass of the acid including state symbols. exothermic or endothermic. released, in kJ. 1 enthalpy change assumptions made for in this the reaction, c alculation in kJ mol above. 333 Tools for Chemistry Enthalpies gure43. a copper of A combustion measured c alorimeter c an mass be of determined the containing a fuel is using the combusted known mass of apparatus in a water. spirit In shown in burner this below c ase, the thermometer c alorimeter is made of a thermally conductive material to facilitate the energy calorimeter transfer from is also combustion transferred reaction to the to the water. c alorimeter, and Heat this from the combustion should be considered water when processing this dicult is to the results. quantify Heat and is should also transferred therefore be to the surrounding e wick fuel r Data-based question Figure 43 Determining the combustion student the c arried enthalpy of out the experiment combustion of shown butan-1-ol, C 4 in H 9 gure 43 to determine OH. The following results l wereobtained: A student of copper c alorimeter / g ± 0.01 g M ass of copper c alorimeter performed the took shown the in gure 43. and water / g ± 0.01 g following steps: Temperature of water before poured 20 cm of water the butan-1-ol spirit c alorimeter g ± 0.01 g weighed and lit the spirit Temperature burner without the temperature for one minute Use these extinguished the spirit that the burner, c apacity keeping replenished the c ap o to of c alculate heat water in c alorimeter copper weighed the spirit burner mass fuel reliability the titration (gure 44). measures experiment. 0.1 °C 60.1 with a standard of a is enthalpy of combustion. Note 4.18 kJ kg 1 K and the specic heat 1 K substance solution 74.38 of in known use a to fill solution c an be concentration determined and by volume in a beaker and funnel burette u l when filling 0 a E v O f x leave air gap the burette reading is taken from the bottom of the meniscus use left swirl the flask hand to with right hand while control the the drops are being added flow rate Figure 44 Titration apparatus. le hand 334 concentration it improve the of unknown reacting i would consumed. practic al ± –1 water t a that of two r o Suggest The experimental of 0.385 kJ kg o second time to nd the d a the –1 is Titration • combustion / °C c apacity n the the data specic U • the water aer C • the i n of water 21.7 75.47 butan-1-ol spirit burner aer combustion / g ± 0.01 g o measured of v M ass • of c ap before combustion / p • burner y into of r e M ass 99.92 combustion / °C ± 0.1 °C 3 • 24.03 O experiment They M ass t i s 38. n y Practice question and Note: le-handed control the ow rate with their right chemists usually swirl the ask with their y P enthalpy of a fuel A air, but minimized. spirit burner s s reaction clamp the Tool The equivalence point, stoichiometric by observing an indic ator at or near a is the added the solution to the a the burette of In point. the acid–base This redox oxidation present in this and is oen many achieved redox titrations, indic ator changes colour titrations state are and titrations, mixture. Some in reagents identiable, of at are said to be self- least one of the reagents is change. holds unknown which reaction change colour at easily the standard concentration solution, (or and analyte). the The ask below contains titration is performed r asfollows: burette, into contact through from by mark the tip taking a With or by just a solution. and at the funnel, below. draining reading the standard to The process the The some the of rotated of the is that solution is burette funnel solution ensure is is lled titrant, and drained twice with the added to the entire inner surface then repeated removed, is its from the more. standard solution to bubbles burette are removed volume is noted bottom of the meniscus. Figure 45 A student redox titration. is potassium Rinse the volumetric pipette: analyte and white ask the conic al the you end (when a point. burette. The is end is approached until at will the conic al with not two is and allow is indic ator be read are may choose to place it the delivered solution noted, ask distilled be titration volume least 3 of of In this c ase the top of the meniscus should volume of colour from change. the burette are and is required the burette disc arded water and it is in a to is reach relled if suitable placed back waste below aected if it is slightly wet, as long as it repeated. the This addition addition an exact of titrant of (to is concordant the the is time, standard solution down within a few standard solution as the end nearest recorded, results the slowed are drop) and the volume of accurate obtained titrant titration (typic ally within a cm You , which swirled, until the end point is standard volume Dropwise The each other). E v O repeated 0.1 much swirling, point. established. is will The with to be of rinsed results point quickly known drops distinguish i the is continuously u of how l f x 3 cm swily, burette. t a added the better solution burette contents Accurate titrations: is of nal ask clean. is to few 4 obscures the meniscus. indic ator changes colour). The purpose of this step is to r o is under able d the The The which idea The be standard ask, the to A o necessary. container. The rough placed ask. a n give is tile conic al U reached or clean pipette, is dark purple and C Rough titration: into • the paper a volumetric permanganate, KMnO rinsed with a small p a into the i n on transferred Using performing a The solution in the burette o added, is v is pipette three times. Prepare the conic al flask: analyte • of volumetric y amount • The r e • O zero tap and t i s the amount tilted with the Fill the burette: small n • A then y burette is l comes which y Rinse the burette: P • techniques e the by be change. equivalence bec ause accompanied Normally colour point must Experimental s s indic ating the amounts, 1: 335 Tools for Chemistry Titrations changes do in not temperature, titrations always certain are need properties which c an known as be indic ators. of the The end solution, measured over conductometric the and point such as c an be determined course of the thermometric titration. These titrations, respectively. a. Explain why of the burette reagents, a rough c. Explain why it does titration not is matter need to be rinsed with the purewater. needed. if the conic al ask holding the analyte is wet. at group in their lab c arbonate, is a water The or is balanced and sodium students dioxide equation that by Determine solution from c arbonate in a the the . a They from for the the 0.300 g a titration. hydrochloric nd the exact a primary reaction of standard: form and does not atmosphere. between sodium hydrochloric c arbonate was 3 of the hydrochloric acid 3 concentration, students’ is high-purity 23.83 cm U C i n solution. Sodium available found of 3 0.2 mol dm c arbonate. exactly performing titrating it against a solution of sodium o neutralized the (aq). c arbon v ii. a 3 that by p acid CO when concentration y Write 2 avoid the approximately HCl(aq) solid to that r e absorb i. Na stable is of mistakes know O concentration it three students t i s acid least of n A l Identify e. y d. in mol dm , of HCl(aq) in results. In chemistry you need and o electrolytic complete circuit, electrode (reduction) i d t a r o Electrolytic cells Electrolysis energy u f x are is input, usually l electrodes a a site for a allowing and convert usually of such that to know voltaic. oxidation an as are in at the electric al the inert form to both electrons lost to be a electroplating, cell reduction to into and as or occur. by of electrochemic al electrolytes, and a chemic al chemic al therefore battery involve oxidized types species at one (oxidation). or such two electrodes, gained energy process of construct involve other material, themselves or how They non-spontaneous made applic ations, E v O 336 DP cells: n Construction of electrochemical cells energy (Reactivity3.2). it pack. requires a continuous The electrodes graphite or platinum. Some active electrodes. reduced, These are rather than simply acting as y why of r Explain pipette P b. and instead e solutions s s Practice questions 39. from electric al conductivity or Tool The silver anode in gure 46 is an example of an active 1: Experimental techniques electrode. + s s + Ag→ Ag + e silver nickel Ag + e spoon to be plated → Ag (cathode) r nitrate solution Figure 46 An electrolytic cell used to electroplate an object placed at the c athode electrodes. Gaseous products can be collected in inverted tubes. Figure47 O t i s showsa Hofmann voltameter, in which slightly acidied water is being electrolysed, producing hydrogen gas and oxygen gas. The electrodes are located at the bottom of the tube to allow the gas bubbles to rise for collection at the top. i the top readily requires collecting of a C an you explain the of the tubes? How could the identity of the solid the diagram product, it will usually plate the observable. measuring from The the quantic ation of solids dierence in mass of the any solid particles or akes that might fall to the electrolytic cell. electrolytic electrode The t a as E applied. well the Quantitative including electrolysis as formation be for electrolysing water. u of the a electrode, in to therefore ) the power supply has been omitted l produced bottom leads and v O electrolysis electrode at o d r o f x gases be conrmed? Note that If cathode ( Figure 47 A Hofmann voltameter is used relative volumes of the gases collected y foils p platinum anode (+) o n U C i n v hydrogen r e oxygen n l y Electrode placement depends on the nature of the products released at the y P silver e + anode investigations surface electrodes area, should require c areful temperature, be thoroughly time control and cleaned of many potential and dried variables, dierence before the experiment. 337 Tools for Chemistry In voltaic energy. arises, Electrolytic and voltaic cells Reactivity chemic al two c ausing energy half-cells current to are ow is spontaneously connected as oxidation in a converted circuit, occurs in a into electric al potential one half-cell, dierence and reduction are occurs discussed in cells, When in the other (see Reactivity 3.2). The potential dierence c an be measured 3.2. by connecting a high-resistance voltmeter to the electrodes (gure 48). s s e e V + Na 3 e + NO Zn anode r Cu cathode 3 NO t i s → Zn + 2 (aq) + 2e Cu (aq) + 2e → Cu(s) O movement of cations movement of anions bridge wool. A solution lling chloride simpler of the a or It a half-cells source U-tube sodium alternative inert two is is with of agar sulfate) using a to complete mobile inert mixed and C potassium by the charge. i n constructed connects of o build-up Electrons ow in the external circuit from anode to c athode. In anions ow towards the c athode and strip p salt the c ations and v A bridge, y Figure 48 A voltaic cell. the salt r e n + 2 Cu 3 + 2 Zn(s) 3 y NO + l + 2 Zn 3 with plugging of lter the ions. an it anode, circuit A salt inert at paper respectively and prevent bridge c an be electrolyte either end soaked in (e.g. with a cotton saturated electrolyte and dipping its ends in the two half-cells U (gure49). n o i d t a r o u l f x a E v O Figure 49 The zinc and copper half-cells in this voltaic cell are connected by a paper salt bridge Electrode factors identity, aecting electrolyte cell how concentration potentials. placement and long the electrolyte concentrations). Other cell As has with and factors been temperature include running electrolytic (bec ause cells, the surface this cleaned and dried before connecting them electrodes should be in LED. It is a series of alternating copper coins, paper soaked in vinegar and similar to the rst zinc washers, battery constructed by Practice question Alessandro Volta in the 18th century 40. Compare and contrast electrolytic and voltaic cells. area, aects the Figure 50 A voltaic pile powering an thoroughly 338 are among the electrode the circuit. y NO P NO Tool 1: Experimental techniques Colorimetry and spectrophotometry Spectrophotometry interaction of UV, with visible or light IR is of an a analytic al certain radiation. technique wavelength. Colorimetry is a that is based on a sample’ s Spectrophotometry similar technique uses that is a range limited to s s wavelengths of visible light. In both spectrophotometry and colorimetry, light is passed through the sample, then detected (gure 51). The incident and transmitted intensities of light are r and concentration is quantiable, and it is oen presented in the form of a calibration curve. The calibration curve can be used to determine the concentration e compared to determine the absorbance. The relationship between absorbance l 280 nm t i s n y wavelength absorbance O 0.347 detector with and sample amplifier source Figure 51 A single-beam UV-vis spectrophotometer c alculated = 0 log ( ) I t where I is the intensity of the incident light and I 0 the depends the used, solvent same the the vs product of low in cm, and 2.0 ε the temperature experimental and l c an also be measurement concentration non-linear a with and and concentration becomes length, extinction coecient) nature cuvette absorbance absorbance solutions container) molar u an relationship concentrations v precise the the l of O this the as εcl to becomes a determined a c alibration (gure 52). Note that at high should ecnabrosba are If Therefore, comparing curve so on f x constant. (sample (known solution. conditions by cuvette = o of is A follows: concentration of the solute (c): t a that l constant the i a to as n U is proportional r o ε also d where is intensity is the intensity of the t transmitted light. Absorbance light C A from i n I o is v M athematic ally, the absorbance (A) device p digital output y cuvette monochromator r e light y P of a solution by measuring its absorbance of light of a particular wavelength. absorbance of unknown sample 1.0 concentrations, be used for concentration of measurements. unknown sample E 0 10 20 30 40 concentration/µg cm 50 3 Figure 52 Use of a c alibration curve to determine the concentration of an unknown sample 339 Tools for Chemistry While spectrophotometers wavelengths, at a specic should light a simple wavelength correspond absorbed by (gure 53). For wavelengths a of The that coloured solution is example, between are a placed blue 585 therefore and use is determine over the colorimeter ’ s strongly solution each absorbs A other orange colorimetry wavelength that falls range of wavelength to in by the the light the colour colour this 585 nm involving this range. l yellow t i s 575 nm O 400 nm green violet n y red 700 nm r e blue y 424 nm 491 nm p o C i n v Figure 53 The colour wheel Data-based question Practice questions A student constructed a 0.285. the c alibration absorbance at recording 490 nm of containing varying 2+ [FeSCN] . 0.40 Suggest the colour of the 2+ [FeSCN] linear the A unknown below, concentration determine the was found to concentration of 0.20 is absorbance 490 nm and the c of curve a at l where equation: 14 000 × c is v O = solution 0.30 gave relationship with following A curve u a c alibration ecnabrosba The solution. f x b. protein c alibration t a r o a. of i concentrations a the protein in this solution. d solutions by of o the curve Using n be 41. absorbance U The 0.10 concentration of 2+ E [FeSCN] , in 3 mol dm . Determine the 0.00 concentration of a sample 0.00 0.10 0.20 0.30 0.40 2+ of [FeSCN] (aq) with an protein absorbance 340 of 0.225. concentration / mmol dm 3 0.50 y P orange observed. wheel corresponding to experiment within setting sample. The r 647 nm a sample’s absorbance absorbed complementary opposite 647 nm. a absorbance only light. wavelength colours should visible c an e substance a measure s s Complementary to oen colorimeter Tool 1: Experimental techniques Molecular models Certain from features Lewis structure of observe allowing Models c an be be might interact kits or are the not apparent three-dimensional rotated (manually or digitally) molecular model kits c an visualize geometry, as well as with one another. constructed using digital are usually ball-and-stick models, digital based (MEP) on van surfaces. der Waals Digital radii) models and c an molecular easily be saved as les communic ation. Figure 54 A molecular model of ice H H O O H O O H O O O H H H O H a. in The interactions biologic al of digital biomolecules, E other research. computer-aided are computationally (in laboratory (in technology and vast silico), of If they Drug computing numbers of very drug explored design. vitro). of digital molecular models following scenarios: large revolutionized potential now drug models has between targets the a scientic construction screening disadvantages school science labs and a and models l The physic al v O b. advantages to y the compared u f x Discuss o C n i t a r o Practice questions in o MEP surface Figure 55 Various representations of sucrose 42. p r e v i n U d space-filling model O t i s O O ball-and-stick model n l y O structural formula y P or and geometry, us between this and other types of visualizations, including (oen potential molecules model help r models molecular models features molecular While toggled reference addition, how from as Models e be electrostatic future built soware. c an In structural predict such formulas. molecule. to space-lling for molecules, dierent us modelling models a of structural s s to or a c andidate using then be development power c an dierent group drug This of help extremely to molecules proteins and is is known as identied investigated resource-intensive; streamline and as process molecules synthesized is such development. molecules and their soware. promising c an molecules, modern the targets. process by quickly Figure 56 Molecular model of the protein rhodopsin, which is involved in vision 341 Tool 2: Technology Sensors are digital transmit labs include probes. the devices results temperature Digital that c an be electronic ally. sensors probes, and other pH used to measure Common probes, devices, physic al light such as sensors and conductivity colorimeters, practic al and easy way to measure certain properties that would otherwise require bulky equipment or time-consuming analyses (for example, produce at be probes are a Jacobson c ardboard 3 cuvette (top) and a 0.50 mol dm copper(II) app and devised values were used as I Bec ause the phones laboratory, are c are and create a paper c ase is I , of along lab. chamber, a that US blue absorbed, use. was yet respectively, compared to you are going to use Colorimeters and frequent apps Thomas c alibration. they c an make Kuntzleman and eective colorimeter using red green blue (RGB) published in the Journal of sulfate incident to as large amounts of data rst. certain copper(II) the If require smartphone that save manual chemists simple a and with equipment or sensors and c alculate solution, where the transmitted absorbance R (red) (gure 57). t items must i cuvette and the pH that be people taken holder to could frequently keep be them designed touch well for and away c arry outside from hazards. sturdier support of the analysed u l a E v O f x t a r o A3D-printed sample. Activity 1. M ake 2. If a list of available, Investigate app. the sensors download how the Remember to an available your school. RGB app onto a smartphone or tablet. following get in aect approval the from RGB your values teacher registered before by the trying any of these out. a. Colour of dierent a solution food Concentration c. Colour d. Path of the length of a could dyes, coloured background (i.e. the (you colouring b. reaching 342 the red o d for the copper(II) In before devices sensors, to collect checking chemistry way 0 the analysed solution (53), an absorbance sulfate sample a colour Given the R values for the blank (190) and of 0.554 was obtained a construction Educ ation. complementary B values of the pixels within the circle. in cuvette-holding coloured Chemic al applic ation records the average R, G, and contain c alibration by digital n U sulfate solution (bottom). The phone of it C Erik also that o Figure 57 The view through the examples measurements smartphone RGB analyser on a blank require c alibrate v to i n useful oen how Smartphones loggers intervals. sensors out alternative p pH nd data the example, y one, to than (for r e However, results techniques indic ators) connected regular sensors) O c an precise laboratory acid–base • more oxygen t i s traditional dissolved n oen or y • sensors l light the length detector). experiment for solution placed of with distilled water and example). of your behind sample the choice. the light sample. travels through before y a P provide are r convenientbec ausethey: • properties sensors in school chemistry e and s s Sensors Tool 2: Technology s s e r t Figure 58 Modern-day scientists What might about they be investigating? Find some of the current chemistry y l out y P researching a frozen lake in Antarctic a. research being done in the Antarctic use for tell data processing the soware manipulation large to and amounts perform of certain analysis data. tools that Operators, operations. you c an formulas Even with a small selection of simple operators, spreadsheets will help you to process experimental quickly shows the Note that formulas each references are recognize formula (e.g. E7) used should shown be in and to in that Excel the order of into an been of The the values in the last c alculated second equal sign (=) automatic ally screenshot. followed by cell multiplic ation (*) and division (/). operations. most the Note cell as t a Change in water / g in all processed Excel do not letters. o i C Mass of that operators same d r o alcohol as bec ause entered spreadsheet. have with such B A Primary Excel columns begins numeric al are an same operators, (a) 1 of screenshot the change strictly numbers rst n values the U Brackets screenshots in C using two shown o 59 columns i n two eciently. v Figure and p data y functions powerful and r e and are entering n Spreadsheets O t i s Spreadsheets D E Change in spirit G Amount of Enthalpy change / kJ mol-1 alcohol / g mol-1 alcohol / mol 1.2 32.04 0.039325843 319. 11302 ethanol 80 38.8 1.22 4.07 0.02481441 489.9552052 4 butan-1-ol 79.9 38.4 1.09 74. 1 2 0.014705882 872.0937984 5 pentan-1-ol 80.3 41 1.12 88. 1 5 0.01270515 1083.12848 B alcohol E F G Change in spirit Molar mass of Amount of Enthalpy change water / g temperature / K burner alcohol / g mol-1 alcohol / mol / kJ mol-1 38.1 1.2 32.04 =D2 E2 =–(B2 4.18C2 ) (1000F2) 80 38.8 1.22 4.07 =D3 E3 =–(B34.18C3 ) (1000F3) butan-1-ol 79.9 38.4 1.09 74. 1 2 =D4 E4 =–(B44.18C4 ) (1000F4) pentan-1-ol 80.3 41 1.12 88. 1 5 =D5 E5 =–(B54.18C5 ) (1000F5) methanol 3 ethanol 4 5 Figure 59 An Excel spreadsheet and D Change in 78.8 2 C Mass of E 1 a A Primary v O (b) f x 3 u burner 38.1 methanol l temperature / K 78.8 2 mass / g F Molar mass of used formulas are shown in the upper and mass / g to c alculate the enthalpy of combustion of primary alcohols from lower images, experimental data. The values respectively 343 Tools for Chemistry While symbols c alculators), when function your and this are ^ are − subtraction * multiplic ation / division ^ power =EXP() exponent used notation c alculations functions as part in spreadsheets (and some outside of a lab the spreadsheet, report. Common for example, spreadsheet summarized in table 1. Example =A5+B5 the sum of values in cells A5 and B5. =A5−B5 the value in cell B5 r Subtracts from that in A5. P =A5*B5 Finds the product of e Finds values in cells A5 and B5 l =A5/B5 value in cell A5 square of the value in cell A5. =EXP(A5) R aises =MAX() maximum =MIN() minimum =SUM() sum =LOG() logarithm (base 10) natural This logarithm, function is 2.718282…) inverse of the mean (average) of the maximum value in cells A5 to A8. i n the minimum together =LOG(A5) values in cells A5 to A8. log the natural log (base o d i t a r o the (base 10) of the value in cell A5. =LN(A5) is u sorting l recording a You inverse of advanced involve E v O f x More all the n U Adds Computes Table 1 Common spreadsheet functions value in cells A5 to A8. =SUM(A5:A8) Computes natural logarithm C =MIN(A5:A8) Finds o the power of the values in cells A5 to A8. =MAX(A5:A8) Finds the =LN() =AVERAGE(A5:A8) Finds to y (average) of A5. p mean (base cell v =AVERAGE() e in r e value O the by that in B5. t i s Finds n the =A5^2 y Divides c an the spreadsheet data, e) of the value in cell A5. This function =EXP(). functions conditional that you formatting, might inserting want scroll to explore bars, next could vertic al lookup, macros, etc. easily features plot of various types good-quality of graphs graphs (see in spreadsheets. the Graphing Remember to include section in Tool 3 for details). Most spreadsheets sc atter some bars line plot, line of functionality on of graph 344 / use each best and may point, t. In draw and best graphing t be and limited, drawing such these c ases, on soware error of best might by hand. packages c an Depending particularly curves you bars. be t, when or able on be trying omitting to nd a used to plot a simple what to program plot certain you dierent points use, error from the workaround, or print the y addition and Action + =LN() * not s s or and as must explaining operators Operator such you Tool 2: Technology Practice questions 43. State the formulas you would enter into a spreadsheet to c alculate the following: 0.400 s s a. 5 × 4.18 The mean of 42, 32, 45, 46 and 48. c. −log(0.0034) d. 10 e. the 12.5 values in cells A5 to A8. operation performed by each of the following spreadsheet formulas: investigation gave the into following the eect of an impurity on 45 −5.3 into a spreadsheet spreadsheet facilitate gather and their at Spectral data (e.g. of c alculate these amounts (e.g. In of the mean results. information, chemistry WebElements, (e.g. Chemistry Institute SDBSWeb, and geometry College Chemspider, NIST National Science Molecular Olaf data the (e.g. v E Energetics St. properties properties Industrial • information on: large updates. a Physic al M allard • of continuous and all organized you might use created by M ark Winter at University of Sheeld, UK) Chemistry, UK) • repositories and l O • to Elements the digital retrieval graph −5.3 u databases • are f x to a −5.5 i D atabases plot −3.3 t a r o D atabases to −1.1 −3.1 o the −1.4 n values −1.7 −1.2 o −3.2 Mean C 30 0.5 °C 0.7 i n −1.0 0.7 U −1.6 15 d Use ± 3 water p 0.8 freezing points. b. 2 10 these of v 0 Enter point / °C impurity / g ± 1 g 1 a. point y of freezing r e Freezing M ass the results. O =−(A3*B3*4.18)/(C4/46.07). An t i s c. 45. n =SUM(D5,E7,G3) l =MAX(C6:C10) b. y a. y the the r Describe of P 44. range e b. and of created by WebBook, Standards created by Royal Society of edited and the the by Linstrom and Technology, National USA) Institute of Advanced Technology, Japan) data (e.g. CoolMolecules, Washington State created University, by Hanson et al. at USA). 345 Tools for Chemistry You may you will of data. use These extended some are up will help you two three when entries • M any a time note of the extracting data the the cite terms collect the reference and into original to the a search information database source, that you Come could your of your at peers’ independent database independent data and what types extracted of three information it contains and ideas three and dependent identify one variables. that you would like to o database of your choice, investigate the melting points of: i group 1 elements u l a E v O f x t a • group 17 elements • period 3 elements. Explain any physic al 2. Explore group the or database (Note: trends you observe. Refer to your knowledge of bonding and properties (Structure 2). electric al conductivity of the elements in a periodic table period of you of your might is your choice by identifying and extracting data from a choice. only electric al ow its electric al conductivity nd resistivity resistivity. The values. A relationship material’ s between resistance to resistivity and 1 is electric al conductivity ). = resistivity 346 and from the n U d r o • with easy and dicult when using the database three Activity a section. Spend 15 minutes three Tell them: further. Using explore C to at found i n look you list looked you should note this y what • you class. this p • database your in with o which with up v • ideas r e your so itself. O variables use. y mentioned database. c an n databases the you you go along. l the as y of database. Listen large amounts from databases as part of parameters spreadsheet t i s one around dependent Share time, but sometimes P as search you Communic ation skills looking a involving exploring the database to get a sense of the types of data data databases well Choose at investigations e Organize 1. or database r M ake • ATL look available. • as tips to complex investigations: Spend that more s s • databases conduct Tool 2: Technology Modelling You have already come across models in your study of chemistry, for instance the ideal gas model, or atomic models. Scatter graphs help us model the relationship between two variables. As described in the NOS section of the subject guide, s s “models are simplications of complex systems”. With the aid of technology, scientists can model the structure and reactivity of intricate chemical phenomena. The best way to develop a model is to start with a simple relationship or concept are facilitated by technology: spreadsheet modelling and molecular mechanics. r Spreadsheet modelling set disturbances M arzzacco to on transform such as equilibrium proposed an input variables equilibria. You concentrations, introductory into c an for output explore variables in the example. eect of US chemist spreadsheet model of a simple equilibrium: k f heads (H) ⇌ tails (T) k r O t i s coin-ipping up system, and k . These parameters then determine the H and T r v We can now explore the eects of changing the initial concentrations and the rate spreadsheet scrollbars c alculation of the to values the and formulas spreadsheet equilibrium constant, K, k k and f k , where K = r k r C D [H] an =C5/100 [T] 8 0 9 k =C11/1000 12 14 19 =J31 v 20 a 0.1 0.05 Here we and reverse rate G coinfliin H eqilirim I J K tem time / [H] [T] 0 =B4 =B = 3+1 =J3B10J3+B13K3 =K3B13K3+B10J3 = 4+1 =J4B10J4+B13K4 =K4B13K4+B10J4 = 5+1 =J5B10J5+B13K5 =K5B13K5+B10J5 = +1 =JB10J+B13K =KB13K+B10J = +1 =JB10J+B13K =KB13K+B10J = 8+1 =J8B10J8+B13K8 =K8B13K8+B10J8 = +1 =JB10J+B13K =KB13K+B10J = 10+1 =J10B10J10+B13K10 =K10B13K10+B10J10 = 11+1 =J11B10J11+B13K11 =K11B13K11+B10J11 = 12+1 =J12B10J12+B13K12 =K12B13K12+B10J12 = 13+1 =J13B10J13+B13K13 =K13B13K13+B10J13 = 14+1 =J14B10J14+B13K14 =K14B13K14+B10J14 = 15+1 =J15B10J15+B13K15 =K15B13K15+B10J15 = 1+1 =J1B10J1+B13K1 =K1B13K1+B10J1 = 1+1 =J1B10J1+B13K1 =K1B13K1+B10J1 0 Outputs [H]eq 0.15 a 18 O 17 220 0.2 l 15 16 145 =C14/1000 k for 60. u 11 13 f x 10 forward F time gure t a 7 =C8/100 lom / noitartnecnoC 6 m 23 the in M arzzacco, as well as i [H] 5 E 3 4 r o 0.25 3 [T] from by o B d A 1 2 shown n f constants are proposed U a added C resulting have i n The o constants on the equilibrium concentrations and the equilibrium constant, K p f concentrations over time, from which a concentration vs time graph is generated. y r e The input parameters are the initial concentrations of H and T , and the forward and reverse rate constants, k n Charles be given y external of 0 5 10 15 20 25 30 35 40 Time / = 18+1 =J18B10J18+B13K18 =K18B13K18+B10J18 = 1+1 =J1B10J1+B13K1 =K1B13K1+B10J1 = 20+1 =J20B10J20+B13K20 =K20B13K20+B10J20 = 21+1 =J21B10J21+B13K21 =K21B13K21+B10J21 = 22+1 =J22B10J22+B13K22 =K22B13K22+B10J22 =J23B10J23+B13K23 =K23B13K23+B10J23 = 24+1 =J24B10J24+B13K24 =K24B13K24+B10J24 = 25+1 =J25B10J25+B13K25 =K25B13K25+B10J25 27 = 2+1 =J2B10J2+B13K2 =K2B13K2+B10J2 28 = 2+1 =J2B10J2+B13K2 =K2B13K2+B10J2 29 = 28+1 =J28B10J28+B13K28 =K28B13K28+B10J28 30 = 2+1 =J2B10J2+B13K2 =K2B13K2+B10J2 31 = 30+1 =J30B10J30+B13K30 =K30B13K30+B10J30 21 [T]eq =K31 22 24 25 26 Keq E 23 y a context l c an the P Spreadsheets e and build up from there. Here we will look at two examples of modelling tasks that =B10/B13 Figure 60 Spreadsheet = 23+1 model of a simple equilibrium system. Source of data: C. M arzzacco, J. Chem. Ed. , 1993, 70(12), p.993 347 Tools for Chemistry Activity Replic ate bars, the leave Adjust out the following and input ii. rate constants. study. molecular Here beyond we modelling mechanics. the mechanic al by scope of principles by a A the to will more DP study common modelling greatly that three has H species, species, profoundly for or for T. example, example, impacted our extending the scope of what chemists introduce treats molecular atoms complex chemistry the of removing O classic al related molecular chemistry values). equilibrium y of composed C of of investigate: C. n with system eect p ⇌ to the o B o of C mechanics i substituents) the high currently in that cell. explore r e and eld could equilibria ⇌ development “see” ⇌ v i n 3 number spreadsheet as area balls of course, distribution of mechanics, which is a joined by springs, in line molecular modelling, which involves the use of quantum electron density in reacting species. Figure 61 shows c an be molecular analysed oxygen on the R–O–R′ model are to models explore bond of the angle. water, eect The methanol of bond and substituents angle data methoxymethane, bonded to the derived from each shown in table 2. a E v O l H or CH to equilibrium which R’ represent either H very by manually entering a number into cell J18 i. two you of and the an how the addition low Extend u methoxymethane (R and than sudden you input both (very e. t a angles in molecular models of water, methanol and a when extremes Determine some of the limitations of this model. U d Table 2 Bond f x 111.5° of seconds happens the n larger bond r o methoxymethane eect 15 what and l the at check values *1 *1 *1 *2 *3 *2 *3 *2 348 Figure 61 Molecular models of water, *3 methanol and methoxymethane y to scroll manually.) eect of changing the Suggest understanding is the d. Molecular 107.1° include equilibrium: c. A 104.0° to P is ii. methanol explore able parameters concentrations A + B water not t i s that c an and the are input y Explore mixture angle on you the e initial (If r LHA b. R–O–R′ 60. change parameters i. intermediate Molecule gure parameters Remember The in simply s s a. spreadsheet them Tool 2: Technology Activity a. Outline bond Using in eect of replacing the H atoms with methyl groups on the a molecular editor of your choice, replic ate the molecules s s b. the angle. shown gure 61. Compare the bond angle values in your models with those given in e i. table 2. 62 each shows a model distance, is two water this activity. molecules energy between approaching the two one water another. molecules was summarized in table 3. *1 *2 Potential −0.15 −0.45 5.07 −3.88 U 1.78 Table 3 Distances between two water molecules and the hydrogen atom 10 other, formed. the The the 1.78 × 10 by and and an model Explore c. Add d. Using a the these the e. Find out the two what is loc ated 15.5 Å bond to form (or molecule, and the potential –0.15 kJ mol . As the molecules TOK approach We decrease hydrogen of until a . observe will have a length of our validity of the model allows with naked at a which eye. are To extended a observe molecular molecular editor of our invisible to the what advancements using Molecular to experimental data. sc ale, molecule us c an interactions values natural world senses. 1.78 Å modelling The the hydrogen bond is through bond –27.7 kJ mol predicted water and water molecular the c ase, the in extent have technology senses? your Avogadro. tools second between of as E b. such other 1 a a choice, molecule the both the l Build v O a. is energy that Activity in In the last 1 energy comparing one u veried m) in atom molecules predicts f x be two distance model 10 (or oxygen i between the t a each from r o energy m) 1.78 Å apart. molecules are close enough for a hydrogen the corresponding potential o with, d begin 15.5 × 10 5.07 and −27.66 energy values To Figure 62 A molecular model of two water molecules at distances of 15.5, 11.2, n C 11.2 o i n 15.5 1 energy / kJ mol p molecules / Å *2 *2 v between *1 y r e *2 n *1 O t i s *1 Distance l This of potential to y c alculated. the extensions y For possible P Figure Suggest r ii. visualizations available in that editor. molecule. editor, water happens c alculate the energy of the interaction molecules. when you add more water molecules and run an energy optimization. f. Build a dierent molecule and explore its interactions with other molecules. 349 Tool 3: M athematics The purpose of this reliability of your communic ation. graphs is to M athematic al relevant opportunity to to help quantitative nearly skills, every reinforce you ensure, measurements units, topic these in understand and signic ant DP they will If you take International Bureau • Dierent numeric al should be ounce, • foot, oz cubic • degree • millimetre • pasc al, • kelvin, K • metre for each of the volume v a energy c. amount of substance e. pressure E d. following quantities? o unit l b. i second, m/s u SI t a per the mercury, mmHg Pa time O 350 is of d a. cm F ahrenheit, °F r o What centimetre, f x 47 . • a space. correct the following: a. time b. temperature c. enthalpy d. mass n U 3 and by for = mistake o • c al Identify reaction to happen C c alorie, i n metre, m • separated y 48. are value and its unit. O units a reaction e. melting point f. density sec = −77 .5 kJmol = 277 °F g. volume h. amount = .5 mol i. amount = 3.0 g j. R 3 = 16 = 304 °K 0.33g = = each of the 1 of = in 0.78 g / dm 500 cc 8.31 J/(Kmol) n space y between Decimal markers should be preceded by a number, even if this number is zero. l A units? • used, and these • p following be y the Measures (Bureau should t i s of and • v SI SI units followed: r e Which Weights conventions Practice questions 46. of BIPM), r the P to International des Poids et Mesures, advantage of become habitual. Units According evaluate gures, uncertainties and chemistry. concepts, and analyses as well as good e every are section s s the Tool 3: M athematics Uncertainties Measurements Repeating a instrument during produces known as readings uncertainty. randomly human uncertainty the and 0.1) °C varying range should unit. of For gure 36.5 °C them) values. time. R andom example, the This tells inexact. with the inuencing instrument be to always and F actors therefore 63. are conditions include reaction in from same same this random imprecision, uctuations error is described by the recorded along with their temperature us that the is given measured as 36.6 °C value is r in the random error, and ± derived under Measurements 0.1 °C or (36.6 somewhere values 36.7 °C. e associated ± the s s variation, (and measurement rise to measurements uncertainty of 3 around ± 0.5 cm Measured burette burette are with is therefore the correct ± more level may 3 typic ally 0.05 cm . The precise. of precision and the o involves one the the based following: on the manufacturer over i by range of display or sc ale on the instrument t a stated the a lower of the measuring instrument u estimating the recorded uncertainty f x uncertainty • be r o the are using • using of have cylinder stated. The determination of the uncertainty of a particular usually estimating you should be uncertainty d must measurement • obtained values uncertainty the that measuring n measurements , U be a n l give the O while y precision instance, p greater For o with uncertainties. C Instruments y P y t i s r e v Figure 63 The temperature shown here is 36.6 °C ± 0.1 °C or (36.6 ± 0.1) °C i n which a value uctuates. way to estimate analogue (has a least that decimal count recorded the as is 0.7 g v lowest The E the zero instrument’ s uncertainty depends on whether it is digital sc ale). The uncertainty of a above an a or O The l Estimating the uncertainty of an instrument digital instrument is its instrument place on c an the least count, register. display. 0.1 g. is Consider 0.1 g, so the uncertainty is ± This ± that usually the 0.1 g mass and is, the the the lowest value reading of “1” in shown mass in of gure 64. the powder is Figure 64 The uncertainty of the measurement is ± 0.1 g 351 Tools for Chemistry Analogue instruments as smallest half the divisions at Therefore, that the last The in that the the marked The should be temperature reading is on on them. thermometer uncertainty temperature digit means 1 °C. sc ales division. is half the recorded value level is The in smallest as the In 37 °C 65 has is ± 0.5 °C. this tick c ase, mark it on the uncertainty of the measuring cylinder shown below. is tick marks at every 1 cm zero, the The r 3 has 0.5 °C. sc ale. e State ± Remember Practice questions 49. estimated sc ale sc ale division, 37 .0 °C estimated. with uncertainty gure s s which every the have sc ale sc ale n l y o C n the measurement, i three o State p r e v i n U images including units and uncertainties, shown in the below. 3 2 4 1 5 0 u l V a E v O f x t a r o d 50. O Figure 65 The temperature is 37 .0 °C ± 0.5 °C 80 °C 90 °C Uncertainty given by the manufacturer Figure 66 The uncertainty of this burette is stated in the label: ± 0.05 ml Sometimes 3 (or ± 0.05 cm 352 ) (gure 66). the uncertainty is stated on a label somewhere on the instrument y P y t i s Tool 3: M athematics Values that uctuate Sometimes certain uncertainty 3 ) and dioxide gas volume the of the at acid using specic meniscus. smallest sc ale of on their to use a is the the gas chips by (c alcium measuring gure 67. dicult pause syringe at make dierent instead c arbonate, the rate uncertainty cylinder. uncertainties, could it c ase of of c arbon bec ause the bubbles measuring you the However, determining measurement associated Alternatively, subsequently is time c ase in of a To greater you could try video points the is better estimate recording to examine the measuring cylinder. time include with cross is human no to the by quantify. Examples appearance of the the cloudy experimenter. solution The it takes the making kinetics for of a this the picture precipitate reaction reaction cross c an be placed white paper below (gure 68). seconds, and the uncertainty of the cross invisible is count). However, the actual uncertainty probably much higher, as it must reaction experimenter ’ s becomes by 42 least measurement the the (the noted precipitate, long obscured E the how reading ± 1 s dicult in o be stopwatch instrument is of i The to a opaque. a mixture be l measuring be change clear solution between sodium thiosulfate and produces v O by or should reaction progressively studied the the acid c an u example, These time from above t a mixture. hydrochloric mixture uncertainty reaction f x For of human reaction d include increase the uncertainty sources of uncertainty sources time and the uncertainty associated perception of the moment when the longer y the bubbles interfere with the r o Some o for measuring the rate of a volume measurements and Further C gas-producing reaction, look down at cross n Figure 67 In this set-up p r e and marble chips v U water i n hydrochloric acid O t i s gas collecting in measuring cylinder y option this in instance c an try to estimate the observe. marble explored for you n Another and you apparatus points and reaction. reaction be division measurements the between c an the In values time, c ases, y sc ale. gas the down the reaction in over these l during of variation In P volume slowing rate uctuates displays. r than the the production with half on hydrochloric interfere value probe e the based example, C aCO measured s s For the conductivity visible. Figure 68 Studying the kinetics of the reaction between hydrochloric acid and sodium thiosulfate involves measuring the time taken for the reaction mixture to obscure a cross on a piece of paper under the ask 353 Tools for Chemistry In TOK the c ase (gure of 68), determining M athematic al reaction random the range between error over c an sodium be which thiosulfate estimated they vary. by and hydrochloric acid repeating Suppose that measurements and the measurement techniques is help the the scientists repeated ve times under the same conditions and the following data are communic ate obtained for the time taken for the cross to become invisible: s s measurements and their uncertainties, process and make 42 s predictions, analyse data The What is the in the mean this knowledge? compare to 42.6 s ways: ≈ 43 s. one The based uncertainty on the of range a of set of replic ate values other value. role Estimating the uncertainty by halving the range of readings areas of We knowledge? c an be readings, and the other on the c an estimate the uncertainty by halving the range of the readings. Here, the l 6 is 45 s − 39 s = 6 s. Half the range is = 3 s. invisible is 43 s ± 3 s. time for the Estimating the uncertainty by finding the furthest reading from the meanvalue The values range s). mean larger time = of s) to and two for 45 s. the 4 s. 4 is s The away used cross as to mean from an value the is 2 s bottom estimate be obscured of away of the the under from the top of the range (43 s uncertainty. these − 39 s Therefore, conditions c an be o C i n v ± 39 s the taken recorded as 43 s 2 p the The from 43 s y 4 − r e range (45 s = the O t i s 2. become n 2 to Therefore, y range cross Data-based question A student measured the magnesium following volume and results. i u l a E v O f x t a r o d o Volume 354 dilute n U between of H 2 of hydrogen sulfuric (g) produced aer produced cm 2 in the minutes, aer 3 / 47 .7 the acid 2 minutes 3 ± 0.5 cm 43.5 mean reaction producing the a. C alculate b. Estimate the uncertainty by halving c. Estimate the uncertainty by nding 44.0 volume. the the range of furthest readings. reading from the meanvalue. d. Which uncertainty value would you report with the data? Why? y in is two P mathematics in How the 1. of time r scientic does 45 s production mean of 44 s role of estimated mathematics 43 s e experiments. 39 s from Tool 3: M athematics Measurement Measurements science are in the Dening SI in have system. precision no These constant and uncertainty, constants Symbol Some instance, are shown Numeric al Δv transition accuracy. for the values in seven dening below: value Unit 9 192 631 770 Hz Cs frequency of C s–133 r 1 speed of light in c vacuum 299 792 458 m s 34 6.626 × 10 elementary e 1.602 176 634 × 10 J s 19 charge C Boltzmann constant 1 1.380 649 × 10 J K 1 6.022 140 76 × 10 mol A K ec acy 683 lm W cd we 4 The seven dening constants of the SI have seen, division) additional includes it noting (Was based it on this, you least the might estimated count? uctuations understand you communic ation how the do impact in of Was the it be 0.2 cm . the values?) 3 ± 0.2 cm is the the The value absolute of the value is The it is x, in this associated associated absolute with. In l a v O the example Doing of the with dierent manufacturer? so will uncertainty help on you your relative or measurement: c ase the uncertainty for a 3 15.0 cm given example , and the value. It has above, expressed as a associated with. absolute uncertainty of percentage uncertainty = Using the volume 3 uncertainty it by absolute, 0.2 cm u f x percentage value be ± uncertainty. Percentage uncertainty: formats: following i as the uncertainty measurement t a units measured r o same dierent u(x), which is Absolute uncertainty: the • Let uncertainty, in consider 3 ± associated • this, d 3 15.0 cm expressed illustrate account n c an To to collected during o percentage. need data given U Uncertainties of the conclusions. Expressing uncertainties measurement (least count or half the smallest C reader) to determining o estimate your way Eective of uncertainty v you (and common measured. ways i n Did you a several equipment uncertainties. experiments data is are the p sc ale there While y uncertainties. r e As Table O t i s 1 luminous n N y 23 Avogadro constant l 23 k y h P Planck constant e hyperne limited and s s constants are exact x × 100% x u(x) = × 100% x above: E 3 0.2 cm percentage uncertainty = 3 × 100% 15.0 cm = 1.3% (2 sf ) 355 Tools for Chemistry • Relative (or the fractional) uncertainty: absolute uncertainty to the The ratio comparing the magnitude of magnitude of the associated absolute uncertainty of value: x relative uncertainty = x u(x) s s = x Using the example above: 3 0.2 cm e relative uncertainty = 3 15.0 cm the oen estimate as given is 3 so only ± that to 1 the In as the (which 10.30 g ± the is decimal not ± be Sometimes two signic ant very small. Either precision other experimenter should is consistently. of words, a uncertainty gives place. 0.1 g, associated with it. in terms of to 2 precision decimal Perhaps the uncertainty have omitted value value itself should readings giving might measured the inconsistent balance down and it is in fact the it 0.01 g U C mass the use o aer you places 10.3 g suggests gure. uncertainty uncertainty. decimal Alternatively, i n correct is written as precision, its v 0.1 g. zero of long signic ant the y mass of of given uncertainty the as if p but mass one r e the that number a incorrectly trailing the same to to particularly expressed as a n an the example, ± uncertainty O is have bec ause stated acceptable, correspond 10.3 g percentage t i s is must was the l are acceptable, Uncertainty places, is y are convention For uncertainty fraction. Uncertainties gures relative sf ) a correct failed bec ause it to is mass write down a signic ant), so 0.01 g. Activity n Copy and complete the table: Absolute Percentage Relative uncertainty uncertainty uncertainty 3 3 19.96 cm ± 0.04 cm 1.08 V ± 2% ± 0.01 u l a E v O f x 78.5 °C Decimal places and signic ant gures Any has measurement a involves limited number of measured 0 356 o i value t a r o d Measured with 1 a ruler uncertainty, so the signific ant figures typic ally 2 has no more 3 result (sf ). The than of the length three 4 Figure 69 Measurement uncertainty when using a ruler sf measurement of as a small shown 5 in always object gure69. y that decimal (2 P Note 0.013 r = Tool In gure certain, 69, as uncertain, way of third as measured length the getting actual the measurements these of fourth 3.67 cm. 3.6 could gure be using but The less 3.66 this or ruler, rst than two 3.68 cm. as we gures, 3.7 cm. are The 3 and last 6, are gure, 7 , is There is absolutely no not even sure about the never being subtracting number of are the two results rules of of the c alculations thumb, which depend on c arried out: measured decimal dividing number are values: places measured signic ant the present values: gures in answer the the present should values answer in the be recorded to used. should values be recorded to used. the initial burette reading was 1.03 cm . At 3 point, of the titrant the nal burette reading was 24.13 cm . C alculate the volume used. used = nal volume − initial volume 24.13 cm = 23.10 cm 3 − 1.03 cm 3 values answer used have should two also be decimal stated to places. two 24.31 = M 0.500 g = −1 40.31 g mol least therefore rounded E M any involves number c alculations incorrect required answers number of to a three of to several decimal intermediate steps. Always intermediate c alculations. gures signic ant steps. rounding 16.00 in a 0.500 g 1 = 40.31 g mol or two mol (3 so the answer in the values Rounding Round signic ant or sf ) should used. be The rounded answer is gures. errors. places c arry + 0.0124 multiplic ation, signic ant require due ≈ a operation the l to 0.012404… mol v O This = o = therefore u f x n(MgO) in mol, i M(MgO) m n MgO, t a = and n U mass of MgO subtracted oxide. r o Molar d of magnesium oxide, sample of pure magnesium Solution are decimal places. Worked example 3 C alculate the amount They C the dp) i n Both (2 o = v 3 y Volume p r e Solution n the beginning of a titration, the end O t i s 3 At l y Worked example 2 y and of neither There P least exact, r and least Multiplying the are measurements. c alculation Adding the 2. length is than e type 1. length greater gure. involving the is M athematics s s As the the 3: three too the early nal gures, extra could lead to answer to the but not signic ant the answers to gures through 357 Tools for Chemistry Propagating uncertainties Processed also have data are result by the of c alculation c an by be with the raw The data 2. Multiplic ation and division 3. Exponents (AHL only) variables. be and as an to consider these are only involving addition of t a u l a E v O f x m = 0.621 g propagate the associated of speed of purposes you c an uncertainty. uncertainty propagation uncertainty we propagation will not propagate uncertainties that are being added or before and aer respectively. The uncertainty for C alculate the mass of ethanol combusted and – in mass: combusted = initial the uncertainties. = ± 0.001 g u(m) = ± 0.002 g the 0.621 g + overall ± mass – nal mass 70.350 g u(m) = of values 70.350 g, You must add the absolute uncertainties: write the C change ethanol 70.971 g Finally, no most n i o the = Then values or constants uncertainty. m m 358 of three types: o i n U d r o mass ±0.001 g. of include For methods burner is measured combustion giving 70.971 g and both mass values is have subtraction, uncertainties The mass of an ethanol spirit c alculate into experiments with a small number of methods or uncertainty Examples and simplied statistic al Worked example 4 First, c ategorized c apacities. exact v subtracted. heat laboratory course. absolute Solution be the uncertainty. specic discuss typic al advanced the associated c an p the will in this c alculation adding it y a by in such need with r e In More assessed rarely you values we used introduced n be to values section, c an uncertainty O this that and t i s In that the subtraction will mass propagating l molar assume you given therefore way this is done depends on the type of y light, not and (± 0.001 g) result: 0.002 g are subtracting the mass values, so you y level, are measurements overall uncertainty of a P this that raw The e Addition by them. r 1. At and with estimated measurements. done manipulating associated s s c alculated each obtained uncertainties Tool 3: M athematics Worked example 5 3 A student uses a volumetric pipette to transfer 25.00 cm 3 the ask, 3 ± 0.05 cm of dilute acid into a conic al ask and then adds a 3 further 10.0 cm ± 0.5 cm of the acid along with its associated solution using a measuring cylinder. C alculate the total volume of acid added to uncertainty. s s Solution First, c alculate the total volume added: You are adding total volume of acid = rst volume + second the the volume values uncertainties and therefore to e propagate you must add the absolute volume uncertainties: V = 3 25.00 cm + 10.0 cm 3 = ± 0.05 cm 3 + (± 0.5 cm = 35.0 cm (1 dp) 3 = ± 0.55 cm the result consistent rounded to one decimal place 3 ≈ be least number of the second decimal ± 0.6 cm (1 sf ) volume, which has places in the raw data. This gives the overall result: 3 V The = 35.0 cm large or divided. in Since decimal and c alculations form, percentage you expressing c an them involving uncertainties think f x least c alculation is given number of of the the relative measuring of the largely pipette. sucrose in water to produce –3 in g dm , and the associated sucrose: This gives the overall uncertainty. result: –3 c This = 103.5 g dm percentage ± 0.3% uncertainty c an be converted to an absolute uncertainty: 0.3 –3 u(c) = × 103.5 g dm 100 four signic ant gures, –3 = signic ant 0.3105 g dm gures in the –3 a division, therefore to 0.3 g dm (1 sf ) propagate –3 c you cylinder volumetric c alculation. ≈ involves uncertainties to of u the result a is the E the 3 103.5 g dm rawdata. The dm l this 3 3 × 10 v O as end n concentration 10.35 g 100.00 that adding the o mass i V Note as the t a = = at are being relative C alculate the concentration of sucrose, d the m = this that the by dissolving 10.35 g ± 0.02 g of solid of solution. r o c alculate c just 3 ± 0.10 cm Solution First values are U 3 100.00 cm of percentages Worked example 6 A sucrose solution is prepared the uncertainties also as multiplic ation or division of C uncertainties in percentage i n uncertainty the of precision o multiplied add high y to you the p uncertainties requires v Propagating uncertainty r e eclipses 3 ± 0.6 cm O t i s the with n to is l that y Note ) y V P u(V) 3 r 3 must compute the = 103.5 g dm –3 ± 0.3 g dm percentage uncertainties and then add them together: 3 ± 0.02 g u(c) = ( ± 0.10 cm + 10.35 g 3 100.00 cm = ± 0.29324…% ≈ ) ± 0.3% × 100% (1 sf ) 359 Tools for Chemistry Worked example 7 The temperature of a 50.00 g spirit burner until it –1 0.01 g sample of water is 23.0 °C absorbed ± ± 0.1 °C. The sample is then heated with an ethanol 0.1 °C. by the water, in J, and its associated uncertainty. The specic heat s s C alculate the heat ± reaches 33.0 °C c apacity of water is –1 4.18 J g K First, nd = temperature T – – = 2090 J × (3 4.18 J g 10.0 K is rounded the to three a subtraction, so to propagate we are ( ± 0.2 K + 50.00 g 10.0 K heat have no × 2090 J 100 ≈ Q 360 = E = v = 42.8 J 40 J a uncertainty 2 (1 value uncertainty. 2090 J ± 2% percentage u(Q) c apacity sf ) 2090 J ± 40 J i to therefore 100% c an is The l O This sf ) × u specic assumed = (1 f x the ± 2% and uncertainties in the we must add the least number temperature of signic ant gures in you add them together: percentage uncertainties: t a r o = ± 2.02% ≈ ) values with o = multiplying d ± 0.01 g u(Q) the n mc∆T, consistent C + (± 0.1 °C) U = be o involves = ± 0.2 K Q to i n = ± 0.1 °C gures v c alculation signic ant c alculation. p in × sf ) result used –1 K O 50.00 g y = the Q: t i s mc∆T that Q absorbed, 23.0 °C = ± 0.2 °C is heat be assumed overall to be result is exact as (bec ause it was follows: converted to an absolute uncertainty: not measured in this experiment) and hence y 10.0 K u(T) Note the n = Then, in determine l 10.0 °C rst to P = values The this r e the use y 33.0 °C –1 Note and initial = = ∆T, T nal Q dierence, r ∆T the e Solution Tool percentage multiplied (raised the to uncertainties in uncertainties. by the the value third percentage of c alculations The the power), exponent. then uncertainty involving percentage by exponents uncertainty For instance, propagating its of if the the requires raw data is raw uncertainty data is cubed involves multiplying three. s s Worked example 8 e The rate equation for the decomposition of hydrogen iodide is found to be: 2 = k[HI] k is the rate constant and [HI] is the concentration of hydrogen C alculate the rate of the decomposition of hydrogen iodide when –3 [HI] = –3 0.60 mol dm –1 ± 0.03 mol dm . The value of k at this temperature is –1 s 2 rate = k[HI] 3 –1 mol –1 s = 0.5688 mol dm –3 ) –3 result –1 s rounded signic ant to sf ) two signic ant gures involves an in the gures values exponent to therefore percentage uncertainty: = × 100% 3 = ± 5% 2 overall also –1 s ± 10% 100 = 0.057 mol dm 0.57 mol dm notice processed, –3 0.06 mol dm E will a –3 = that the 2 in this c ase: ATL –3 of is good mathematic al communic ation? important? Why Why is it are you –1 s encouraged –1 out for all to your show your working c alculations? –1 s (1 sf ) –1 s Communic ation skills What s –3 ± 0.06 mol dm propagation range –3 0.57 mol dm v O You × l = ≈ rate is express this as an absolute uncertainty: 10 u(rate) which u c an exponent, result: 0.57 mol dm f x We = the = ±10% –3 rate of least rst compute the o the × value i gives = ± 5% the must the t a This by r o u(rate) it d multiply we with n 0.60 mol dm Then, consistent U 3 ±0.03 mol dm u([HI]) be used. C c alculation (2 i n The is of –1 s o number 2 ) v The 0.57 mol dm (0.60 mol dm p –3 ≈ × y (1.58 dm r e = O t i s Solution n y mol l 3 1.58 dm y iodide. P where r rate M athematics LHA Propagating 3: values increases they c an –1 s the cover overall uncertainty. becomes As raw data are greater. 361 Tools for Chemistry Uncertainties and means Trials are oen Uncertainties of repeated do not to need check to be for repeatability propagated and when minimize c alculating consider the following set of results from a titration: trial 2 0.00 0.00 11.00 11.25 11.00 21.95 burette 3 3 P ± reading / cm 3 0.03 cm burette 0.03 cm 11.00 3 volume / cm 3 ± 0.06 cm nal volume of the titrations, = omitting same is as ≈ 10.97 cm 3 rounded to decimal = ± rough gives 0.06 the trial, is cm titrant volume, as . follows: 3 + 3 3 10.966... cm the the 10.95 cm = two 0.03 3 + 10.95 cm 3 ± 0.06 cm n the to volume 3 × ± C i n answer given initial 10.95 o volume U The the 3 11.00 cm mean from p mean the volume uncertainty is 2 v The the 10.95 y hence r e Subtracting O t i s 11.25 Titrant n 3 ± 11.05 y 3 reading / cm l Final 0.10 two decimal places. uncertainty of Note each places that of the bec ause the the uncertainty measured of the values mean are volume is three trials. o i d A student results u l are performed given Volume of an experiment to nd the density of butan-1-ol. The below. butan-1-ol / 3 cm M ass of measuring M ass of butan-1-ol and 3 cylinder / ± 0.5 cm measuring g ± 0.01 g a E v O f x t a r o Data-based question 8.1 Determine b. C alculate the density c. C alculate the uncertainty absolute g ± 0.01 g 42.82 a. the mass uncertainty of the of butan-1-ol the and of (ii) a sample. density. relative 49.38 sample. butan-1-ol the cylinder / Express your uncertainty. answer as (i) an y 1 e Rough r Initial s s example, Titration 362 error. mean of a set values. For all random the Tool 3: M athematics Graphs and tables Graphs show variable. how changes Graphic al relationship in an techniques between the independent c an variables be used and to to variable aect a dependent examine predict the nature of the unknown quantities. s s Sketching graphs Sketched graphs to each have trends, other. labelled such as Examples but variables of unsc aled that sketched are axes, and they proportional graphs are shown or in are used inversely gure to show e qualitative proportional 70. r (b) 3 2 mol dm acid (highest concentration) how pH n c. rebmun melting i n c r e i g t m p t u r e evah the ,ygrene b. a E (a) curves showing the eect of concentration on the rate of the reaction between an acid and a gas volume is directly proportional to temperature (c) two graphs showing that pressure is inversely mass of gas at three dierent temperatures show: concentration system r a to v graphs E a. E l O Sketch e energy, proportional to volume (d) M axwell–Boltzmann distributions of molecular energies for a xed Activity taht selucelom fo t a graphs: (K) 3 u f x Figure 70 Examples of sketched T 0 1/V / dm metal c arbonate (b) graph showing that o C n i o ,erusserp r o d ,erusserp V/dm temperature, (d) U aP/P aP/P 3 volume, v i n (c) y acid (lowest concentration) p r e emulov ,emulov OC fo 3 0.5 mol dm O V t i s 2 acid time l y decudorp 3 1 mol dm y P (a) of reactants point changes of the when and products elements a strong over time in an equilibrium across period 3 acid is added to a strong base. 363 Tools for Chemistry D ata, particularly processed into quantitative charts. There data, are c an be dierent organized into tables and then types of charts, including graphs. Tables following • data c an be presented as a table. It is customary to include the features in data tables: Descriptive title to aid communic ation. each should be If more than one table is being numbered as well. listed in order value. increasing Dependent variable(s) in the columns to the right of the independent variable. List the results from repeating trials next to each other, and the l mean in the rightmost column. Units and uncertainties in column headings. • Consistent precision in each column. • Clear identific ation of values of importance a titration, anomalous example, concordant p C i n o clear table title units and uncertainties (for results). v descriptive colun eadings ic include or y in r e values O • n y Descriptive column headings. t i s • Table 1. Boilling points of aqueous solutions containing different asses of NaCl. 0.01 g / °C ± Trial 1 96.9 97 .2 97 .0 97 .4 98.0 98.0 o i r o Mean 98.8 105.2* 99.7 99.3 3.00 99.9 100.0 100.2 100.0 101.6 100.5 101.8 101.3 4.00 t a 2.00 consistent decial tis value is an anoal leost colun E v a O l u f x 98.5 variable in te 364 97 .0 Trial 1.00 independent d 0.00 Trial 0.1°C n dissolved / g ± Boilling point of solution U M ass of NaCl Figure 71 Common features of a data table places for values obtained using te anoalies identified if applic able sae equipent y P • of r Independent variable in the leftmost column , e presented, • s s Quantitative Tool Control tables variables bec ause sometimes qualitative include data you information data in that a c an you for need to found in of a are not data, where above way to usually the or applic able. below eectively variables that need databases). You c an included in experiment. to the Tables c an Otherwise, table. record additional be nd monitored or citations examples of how to do articles. r Activity an experiment data looking dierent during the at how the combinations practic al. potential of dierence of a half-cells and wrote down Organize the data into a suitable table. magnesium – copper: 2.05 V, 2.00 V, 1.34 V each and and used graphs represent patterns to in present the all the data. information are c ategoric al compare used to variables, group such data as straight means • that a Sc atter data a they c an graphs trendline) line bec ause are is be are usually according (e.g. data, along to 0–1 cm, but a explore graphs, known as forces, or identity of a range in 1–2 cm, they useful a to and expressed as c ategories intermolecular particularly organized represent added what data length us interpretation. Examples whole, distinct charts the value of the 2–3 cm). contain points joined by for continuous data, which numbered line and have any value range. plotted graphs, as quantitative graphs continuous points Unlike show Line E within also lines. such a Line graphs grouping v O • involve variable l quantitative of a aids u f x Histograms of into type substituent atom. • parts that of o Bar charts to way allowing types t a • us a dierent i allow percentages. r o Pie charts in simultaneously, are d include: • data There n trends U Charts C i n Charts and graphs o v magnesium – zinc: 1.54 V, 1.43 V, 1.23 V y zinc – silver: 1.30 V, 1.25 V, 1.27 V p r e copper – silver: 0.50 V, 0.53 V, 0.51 V O t i s zinc – copper: 0.80 V, 0.78 V, 0.74 V n following with l changes y the did cell y student voltaic P A e this in scientic journal directly as they throughout qualitative think (such but change included table have recorded not space be will your be should M athematics s s Sometimes for should they 3: to the matters the and show points is relationship then the a the on line or between curve relationship sc atter overall of two best between graphs relationship are not they variables. The t (also the known as two variables. necessarily joined up show. 365 Tools for Chemistry (2) (1) carbon dioxide argon 40 )%( and methane (1%) (trace) 21% s s ecnadnuba oxygen 30 20 10 4.0–7.9 8.0–11.9 12.0–15.9 r 3.9 e 0 nitrogen 78% and 16.0 to 19.9 size of microplastic (mm) vmpp 0 2 / OC 6 0 220 8 0.2 1880 1900 1920 1940 1960 2000 C t a 6000 ni 5000 u 4000 l nobrac noitartnecnoc a E v O f x edixoid mpp r o i o d 7000 3000 0 100 200 300 400 500 600 time in seconds types of charts and graphs. (1) Pie chart showing the present composition of the E arth’s atmosphere (2) Histogram showing the distribution in the size of microplastics in a sample (3) Line graph showing global temperature variation and concentration obtained 1980 n U (5) Figure 72 Dierent 0.4 o v parts per million by volume p 0 10 i n years before present = 000 05 000 001 000 051 000 002 000 052 000 003 000 053 000 004 180 ppmv 0.2 y 2 4 260 r e C° 300 mean 5-year mean 0.4 340 n annual 2 l 0.6 4 380 O 2 concentration t i s CO temperature variation y (4) (3) atmospheric CO through isotopic analysis of ice cores (4) Line graph showing global mean temperatures in recent 2 years (5) Sc atter graph showing the linear increase of c arbon dioxide concentration over time. Note that 366 the line of best t does not go through all the points bec ause it represents the overall relationship between the two variables y P under Tool 3: M athematics Plotting graphs All graphs, regardless technology • should Descriptive title to each whether the aid they are drawn communic ation. should be If more Axis variable on the • Axis labels, including units and uncertainties • Accurately cover you your sc ales changes in c an be the which scientists identify results, c an compacts signic ant equally used when the data in (gure points by eye. underlying plotting the that advanced outliers possible information are use It is that and not t of an cover allows a us very to range. 4 6 18 20 22 24 16 18 20 22 24 o 16 i t a v 2 14 time / h a 10 12 u l 100 10 d r o f x 10000 O airetcab fo noitalupop 100 000 1000 8 y 2 p 0 n 2000 000 o 4000 000 C 6000 000 1 large easily identify r e 8000 000 1000 000 consider, v i n 10 000 000 U airetcab fo noitalupop 12 000 000 10 000 000 to 73). 14 000 000 100 000 000 trend of outlier. 16 000 000 1000 000 000 the important c auses data graph do statistic al tests to determine 18 000 000 E y-axis spaced marks O so variable on the contain t i s Doing outliers, Sometimes level, discussing Logarithmic and n this dependent range y curve. At x-axis, suitable l when graph is being y outliers. one r or than P line plotted using plotted points. D ata sometimes contains the a or e Independent • that hand numbered as well. • sc ales by following elements: s s presented, of have 4 6 8 12 14 time / h Figure 73 The sc ales on both axes on the rst logarithmic sc ale on the y-axis in second 10 graph are linear. graph gives a straight Plotting the same data with a line, conrming that the trend in the data is exponential 367 Tools for Chemistry Data-based questions 1. Identify the outlier in the graph below: 120 s s 100 r 1 2 3 4 5 6 7 a the logarithmic logarithm of sc ale on the the successive y-axis is in for this potassium. c ase. o 2 electrons n to nucleus, n = 1 shell o 8 electrons close to i nucleus, n = 2 shell u 8 electrons far from nucleus, n = 3 shell l a 1 electron very far from 2 nucleus, 1 E v O f x gol t a 01 r o noitazinoi d ygrene U very close 4 3 energies C i n 5 368 useful ionization 2 3 4 n = 4 shell 5 6 7 8 9 10 11 12 number of the ionization 13 14 15 16 17 18 19 LHA why shows v Explain below y graph p r e time (minutes) n 0 O t i s 0 The l y 40 y P erutarepmet 60 20 2. e )Cº ( 80 Tool Error Error (also uncertainties, error bars known as eectively bars uncertainty showing bars) that a are data show the uncertainty of the show the uncertainty of the x-axis graphic al point representations of represents y-axis variable, a range of values. whereas horizontal s s error M athematics bars bars Vertic al 3: variable. 3.0 e 1 s r 1.0 Figure 74 Error bars show the uncertainty of a point 0.1 0.2 0.3 0.4 on a graph. The x and for this point time/s −1 speed = 1.4 m s Graph-plotting soware oen has an option to include error bars when the absolute associated with each point, the soware might not accommodate this. In this case p you can either draw the error bars for each point by hand or have the program draw error bars corresponding to the largest uncertainty in the data set (and add a note to v this eect). You should determine which option is best in the context of your analysis. o curves of best t A line or curve of best fit C i n Lines or −1 ± 0.2 m s y r e or percentage uncertainty is specied. However , if there is a dierent uncertainty y values are, time = 0.2 s ± 0.05 s, and O t i s 0 n l y t 0 y P m /d e ep s 2.0 represents the general trend in a graph and can be used for further analysis of the relationship between the variables. The line or curve of U best t does not have to pass through every single point, but it must have roughly n the same number of points on either side of it. If there are any obvious outliers amongst your data, you might need to ignore them when drawing the line or o d curve of best t. This does not mean, however, that they should be excluded from the analysis. It is good practice to circle and label outliers on graphs, as well as i r o discussing their possible sources and signicance in the analysis. t a u l a v E elbairav tnedneped O f x line of best fit t Figure 75 A generalized graph showing a line of best t independent variable 369 Tools for Chemistry s s c an be whether useful c an be simple (a) or complex (b) when drawing computer-generated lines lines and or curves of best t, but you must l consider t curves of best t make sense in n y the context of the data. For example, soware may skew the line due to an outlier, it may If show of points best on through as a a straight t do graph many uncertainty of have the best is where error error t, a have bars, bars as curve to it may greater then investigation U of that If or the the line error mol dm point the is of reactant R ate A 0.0002 / mol of 0.015 0.065 0.0050 0.089 u l E v a c. Draw a line of best t. d. Identify on outlier results. your and graph. suggest possibly 1 s 0.0040 the or what might have c aused it. ± pass are reaction dm 0.031 Draw should point outlier, 0.027 Plot the t a 0.0030 b. bars an of –3 ± n o t a graph best 0.0020 a. error of Lines and following data: 0.0010 a suitable. bars 3 / i d r o f x O 370 produced C Concentration curve p rate more o v i n reaction the expected. Data-based questions A be through the origin. possible. indic ate than would pass y the that line always r e outside the line of not O curves t i s or 5% y Technology curves of best e Figure 76 Depending on the nature of the data, temperature r (b) P time noitca er emyzne fo etar yticolev (a) Tool 3: M athematics Interpretation of features of graphs The relationship features and of area. a The an main variables correlation, thing aspect of to the c an form, remember idea be further gradient, being is that the studied. analysed by interpreting the intercept, maxima and minima, features We will of also a graph c an oen s s represent between graph: discuss the use of a 2 coecient of determination (R The rst thing to consider when any). If the dependent (y) increases, describes then the variables variable correlation is where to positively increase decrease. weak or If the absent of the the points sc attered in the c auses absent correlation n negative and v graphs showing positive, l decreases. y No correlation. increases, p Y X r e When X O variable t i s elbairav y Y X Negative correlation. Figure 77 Sketched randomly (gure 77). variable also increases correlation are When Y negative shows variable elbairav elbairav X increases, A it variable independent Positive correlation. X correlation independent correlated. Y Y variable are an graph is the when y dependent graph, the relationship a increases extrapolation. P the a interpreting variable and r (x) interpolation e (if ), o i n The relationships shown in graphs c an be linear or non-linear. A linear relationship y = mx + c, to a straight where m is variables its the same in a rate. temperature, For example, provided directly special ratio. that type When the the of volume the equation y-intercept. increases, gas is is in straight-line directly kelvin graph the other increases proportional to (gure78). If two must go through i pressure of gas a Figure 78 Pressure and a the relationship, in which the two variable of by c is the u l E v 100 described t a r o f x O 200 linear one pressure of gas 300 is temperature proportional, theorigin. that (slope) of the line and o are a graph d variables is constant the n by are in gradient U Direct proportionality line the C corresponds 0 0 +100 temperature (K) +200 temperature (°C) temperature of an ideal gas are linearly related, but the relationship is only directly proportional when the temperature is in kelvin 371 Tools for Chemistry Variables by the are same inversely rate. proportionality are however, a volume gives of an proportional Unlike not direct linear. straight-line ideal gas is a when one increases proportionality, Plotting graph. good one The of graphs the variables relationship example of and inverse the showing in its between other decreases inverse reciproc al form, pressure and proportionality (gure 79). s s e be C rep g( ytilibulos f x 6 40 30 20 10 0 250 0 10 20 30 40 v E 10 15 20 25 30 35 40 45 50 volume of 0.1 mol dm 3 3 acid Figure 80 added / cm Graphs showing non-linear relationships: (b) a solubility curve and 372 (c) a pH 50 60 70 temperature (°C) 0 5 straight lines. 50 2 0 not by functions, such as 60 a 8 are 70 u l Hp O 10 that described time / s 12 y relationships c an quadratic functions. )retaw g 001 200 all o and n 150 o 100 14 4 i 50 t a r o N[ 0.200 d 2 lom / ])g(O 0.400 includes relationships (b) U md 0.600 0 (c) that logarithmic i n 3 0.800 0.000 term v exponential, (a) a non-linear p is Sometimes, 1.000 r e gives a straight-line graph V Non-linear 3 1 /V/ dm volume of an ideal gas are inversely proportional to each other: 1 plotting p vs O Figure 79 Pressure and n l y t i s V/ dm (a) the decomposition of nitrous oxide, curve for the addition of a strong acid to a strong base 80 90 100 y P ,erusserp ,erusserp r aP/p aP/p 3 volume, Tool Gradient (or The gradient physic al (or meaning: it is of the a line rate is of separated on from whether each the a measure change of of its one steepness, but it also has a variable with respect to the other line, you need to choose two points on that line, other. graph Gradients shows a c an positive or be a positive negative or s s well depending negative, correlation, respectively. y − y − x 2 = 1 = x Δx For example, a graph 1 of mass of a substance (along the y-axis, in g) vs its r 2 volume x-axis, in cm ) gives a straight line. The gradient of the line has units 3 g cm , which gives two for lines gradient of best maximum c an be and used to c an t be substance. determined possible minimum estimate within by the gradients. the drawing constraints Half uncertainty the of the of steepest the error dierence the bars, between gradient. a steepest gradient v gradient reaction curves vs time have changing will produce a curve, (gure 82). t a reaction u a v E the while volume no further l f x O volume of gas gradients product during o decreases for n rate constant data least steep lines of best t i the have instance, r o bec ause graphs For and o gradients. minimum gradients on a graph c an from the most d Straight-line and C i n Figure 81 M aximum be determined U b y p r e ytitnauq most gradual quantity O these values the lines that t i s giving for gradual of n most density l uncertainty and the y The us y P 3 (along the e Δy m M athematics slope) slope) variable. To nd the gradient of a ideally 3: reaction is slower where gradient is less steep y reaction is faster where gradient is steeper x time Figure 82 The rate decreases over the course of a reaction, so the gradient decreases over time 373 Tools for Chemistry We can determine the gradient at a specic point on the curve by drawing a tangent line and working out its gradient. In gure 82, two tangent lines are shown in green. The one at t = 0 is clearly steeper because the rate is higher at that point in time. Figure 83 shows how to calculate the gradient of a tangent line. s s A the gradient To hinder on a the global 0, for of a instance kinetics numeric al pursuit level? the c quantity use (or b) the In the some plotting In line the of equation the c ases, the or line represents the signic ance plotted. when also investigation. intercepts you would concentration other c ases, the line of a in the the the a knl E v a R 0 is lnA, T product intercept A T 374 meets depends (gure 84). ln crosses axes. axis form on expect the E of a graph showing lnk vs where A is the Arrhenius factor the u l f x O y-intercept of it (this is c alled y = mx + c y-intercept. 1 Figure 84 The curve until m = 1 n is y-intercept. shared codes and or knowledge m being t a of do help gradient, data i extent symbols where b is the r o what or where where extending l c with the c an by y and d y = mx + b, As You y = mx + c or points n oen y = mx + b) the found o given as is (or graph. 1 o be p time / s C are c an U It exist in the s O t i s r e v i n intercepts extrapolation). straight-line y noitartnecnoc C Intercepts variations P fo B The a 3 dm BC Intercepts for gradient of tangent AB mol t t AC = by c alculating the equation time Δt = t International at c an the context of y-intercept to be vs be time as a used to part derive a y lom/ ]X[ ,X Δ[X] gradient of the tangent line at that point TOK t gradient of graph at time = Figure 83 The rate of reaction at time t c an be determined r md time e 3 reaction rate at = Tool Similar to drawing the error which gradient, the uncertainty of the the most bars. you and This c an least gives nd the steep lines maximum uncertainty of and y-intercept best t c an be possible minimum values determined within for the the 3: M athematics by constraints of y-intercept, from (gure 85). a s s ytitnauq r of intercept y of intercept from and the most minimum values for the y-intercept and least c an be steep lines of best t variables visible at The involve curve. inspecting The represented solution 520 nm, colour of of an which the maximum signic ance in the graph. unknown of values, each of these substance corresponds unknown green. to solution the (gure 86) green must be region red, as it is n o i d u f x t a r o ecnabrosba 400 500 600 700 l wavelength / nm a v O a a U to of around spectrum. complementary being spectrum maximum may under C the a graphs area o absorbance shows of the of the i n The on or v depends values y interpretation minimum p Further r e M axima, minima and areas on graphs O Figure 85 M aximum determined b t i s quantity n l minimum value for intercept y P best value e maximum value Figure 86 The maximum absorbance shown in this graph helps us to determine the colour of the solution changes ClO•, 18 km gure87. E The in The concentrations in the minimum maximum in levels high and the high ClO• in ozone concentration, ClO• of atmosphere ozone, O across 3 , and dierent concentration suggesting chlorine latitudes occurs an at the association monoxide are radic als, shown in same point as a between low ozone levels. 375 Tools for Chemistry 0.8 3.0 O s s 2.0 72 70 O a curve x-axis. It under x-axis the area variable curve under a graph) the number of Boltzmann distributions is change. useful in particles is the for the same U sample. species at Figure two distributions, 88 shows dierent by the over the the where it M axwell– temperatures. 2 o i d t a u l a E v O f x noitcarf r o citenik htiw selcitrap fo T the enclosed variable n ygrene 1 area y-axis M axwell–Boltzmann in C i n represents T the o area (or represents an accumulation of the v The concentration corresponds to a maximum chlorine p course of a certain , concentration y under 3 r e area curve and ozone, O E a kinetic energy Figure 88 The area under the curve in a M axwell-Boltzmann distribution represents the number of particles, sucient which c an be used energy to react at to compare the proportion of particles that two temperatures T (red) and T 1 The particles regions. E . At that Their the have kinetic higher sucient energy is temperature, a , 1 T , the to where T to react or greater shaded are area represented than is the larger by + 1 the activation (red have > T 2 shaded energy, green) and 2 therefore T energy equal (green), 2 at the which number the of shaded particles area is with smaller sucient energy (red only). to react is higher than at y 68 latitude (°S) ClO•, O 66 n l y 64 Figure 87 The minimum The 1.0 0.0 t i s 62 monoxide radic al, 3 r 0.2 0.0 376 e 0.4 CIO mpp / n oitartnecnoc 0.6 P bpp / n oitartnecnoc OIC 3 Tool 3: M athematics Data-based questions 1. Students investigating the eect of temperature on air volume plotted the following graph of their results: 140 s s 120 40 0 5 10 15 20 25 30 35 40 temperature (°C) following aqueous gradient briey data direct the were copper(II) of between the physic al line using using a mol dm solutions c alled atomic in order emission to signic ant gures and include the units. determine the concentration of a sample spectroscopy. n this two 0.130 o equation whether to negative. t a the these results and include a line of best t. 3 concentration, a Discuss of l Deduce d. E v O c. graph the answer or of 0.460 u suitable your gradient. standard i r o f x a positive 0.810 Unknown Determine is why not? 0.685 0.6000 Plot temperature or 0.540 0.5082 b. State the 0.380 0.4000 a. of 0.270 d 0.2819 t. and Why Absorbance, A 0.1002 0.2008 six U / best technique 3 Concentration, c of signic ance recorded sulfate volume proportionality? C of the correlation show y Describe the graph p Determine d. the o c. The whether r e Does v State b. i n 2. a. O t i s 0 n y 20 y 60 l emulov P ria fo r mm( 80 e 3 ) 100 the graph in line mol dm of shows best a , t, direct of copper(II) stating the sulfate. gradient and proportionality between y-intercept both absorbance and to two signic ant gures. concentration. 377 Tools 3. for Chemistry The graph hydrogen (right) gas experiments. drawn at 25 shows the collected A tangent seconds volume of over line for time has 50 for two been experiment 1. 45 s s 3 a. C alculate reaction the for instantaneous experiment 1 at 15 cm rate of t = experiment 2 40 25 s. (high temperature) Describe and explain whether the than that of reaction or in experiment 25 s 30 experiment 1 25 (low temperature) less than that in 20 10 graph shows dinitrogen tangent C alculate your the been initial answer concentration of O 5 rate to , changes drawn at of two t = over 0. reaction. 4 signic ant md lom d gures and state the units. in 30 the rst seconds. your answers to (a) and (b) 3 concentration of N O 2 5 / mol dm 0 3.40 10 2.60 20 1.95 30 1.50 40 1.15 × × –8 10 –8 10 –8 × 10 × 10 –8 × 10 2 5 O Tangent 2 N[ l u / ])g( dierent. f x at t = 0 1 a E v O 378 reaction 35 × why are of 01 above rate t a Explain average 30 –3 /s 8– the r o c. C alculate i b. 25 time o 3– Give has 2 20 time / s n a. A the U time. how pentoxide, N 15 C The i n 4. 10 y 5 o v 0 p r e 5 O t i s 15 0 n y experiment 1. 10 20 30 time/s 40 –8 y greater experiment 1. explain whether the 35 l is for greater or em u l ov 2 rate is fo overall of and reaction 25 s P Describe of = r less c. t e rate experiment 2 at mc / negordyh instantaneous 3 b. Tool 3: M athematics 2 Coecient of determination, R We have se en associate d likely to of fall it. all me asurements, dire ctly One of on the As the tools a result line we of of and this, best c an t, use to data the but derive d data rather assess from points the be in them, a have graph sc attere d pre dictive are on an not either power s s side that uncertainty. of 2 the relationship betwe en the v ariables in a graph is R , the coefficient of determination . e 2 In DP chemistry, you will use technology to nd R values, and you will be 2 to interpret them. However, you will not be required to c alculate R r expected 2 yourself. data R points lies between match the line 0 and of 1. best The t closer (gure it 89) is to and 1, the the more more 2 successful for model is at predicting the changes in the y-axis variable. R is only linear models. 0.7 fo 0.4 0.2 0.1 0.3 0.4 0.6 0.5 of ethanol / cm U = 0.862 0.2 0.3 of R 0.4 0.5 0.6 0.7 0.8 3 of ethanol / cm 2 R = 0.250 0.6 0.6 0.5 0.2 0.1 0 0 0.7 0.1 0.3 0.4 0.5 0.6 0.7 0.8 3 3 volume value is 1 when the points are all on the line of best l O points in graphs 2 to 4 become increasingly more spread 0.2 0.8 ethanol / cm 2 Figure 89 The 0.5 u volume 0.4 0.3 volume 0.3 s s am 0.1 f x 0 0.2 0.4 fo 0 g / l o n a ht e n 0.1 t a r o s s am 0.2 i fo 0.3 o 0.4 0.1 0.7 0.5 d g / l o n a ht e 0.6 0 C 2 R 0.7 0 0.8 i n 3 volume 0.7 0.1 o 0.2 0.2 v 0.1 0.3 p s s am s s am fo 0.3 0 0.5 r e 0.4 0.6 O 0.5 = 0.989 y g / l o n a ht e g / l o n a ht e 0.6 2 R t i s = 1.000 n y 2 R 0.7 0 l suitable the y the value of P closely The t of ethanol / cm (graph 1). The 2 out, c ausing the R to decrease a v E 379 Tools for Chemistry Interpolation and extrapolation Lines and involves curves of predicting extrapolation best t values involves c an be within interpolated the predicting range values of or the outside extrapolated. Interpolation experimental the data, whereas range of these data s s (gure90). y e range of r interpolation l O t i s y intercept line known versus were y of the concentration obtained concentration. by graph shown in gure 91. measuring the absorbance of six plot of A versus c 0.9 n U C solutions absorbance on x extrapolation using a straight line o the points i n The p Figure 90 Interpolation and Consider 0.8 (x , 2 y ) 2 o 0.7 d 0.6 i 0.5 t a r o 0.4 0.3 u l f x 0.2 (x , 1 y ) 1 0.1 0 a E v O 380 v intercept r e x n y extrapolation 0 0.1 0.2 0.3 c/mol 0.4 dm 0.5 0.6 0.7 3 Figure 91 Interpolation involves nding values within the range of the experimental data y P extrapolation Tool The Its absorbance of concentration a sample c an then of be unknown found by concentration interpolation, is as found to be 3: M athematics 0.460. shown, giving 3 0.34 mol dm . concentration, a c alorimetry copper(II) are c alculate ∆T ∆H). lowered due 92). time) metal The the curve the is could solving of the of be over mixture containing time temperature heat, measured Q used to nd the x reaction monitored amount also for (and during the and value later, the is the results needed enthalpy experiment is surroundings. (the backwards solution, line maximum temperature to the and allows part to us of the to the time graph at predict a that which shows the corrected zinc a decreasing was value of added to the T . Doing so a constant gure For 91 is involves same example, linear, the but the beyond during assumption the range relationship the the linearity is of reaction. that between lost at the data. In relationship practice, absorbance higher and temperature of 0 highest temperature = temperature that would have actually reached 2 T been reached if there were no 2 1 rate of cooling i ΔT for reaction 0 l a O = T T 2 0 u reactants mixed o extrapolation at same t a r o f x T d erutarepmet ΔT n U T C i n T o v = 1 p reactants T always y initial not r e = is concentration concentrations. compensating for heat lost T between this n in c ase. the loss O the always stays heat t i s variables of y Extrapolation rate l max assumes y over sulfate loss cooling y temperature zinc subsequently highest heat the (gure of for P temperature and The to Extrapolating graph the and in r change, copper(II) a solution equation 0.460 e to on the entering experiment, sulfate plotted known, s s In If by time Figure 92 Extrapolation of temperature data during a zinc metal and copper(II) v sulfate reaction to determine a corrected value of T , the maximum temperature 2 E 381 Tools for Chemistry Data-based questions 1. A student metal performed and copper(II) a c alorimetry sulfate. A experiment graph of the to determine student’s results the is enthalpy shown change for the reaction between zinc below. s s 68 66 60 r 58 56 52 C 50 44.5 19.5 º C 42 38 34 28 26 24 20 0 1 2 3 4 5 6 time / min the Explain d. Outline why a the what temperature reason, whether student other collected hydrochloric NaOH(aq). D ata f x Plot c. The maximum cooling volume need data every minute to c alculate while the titrating t = 2.5 min. heat, Q, and enthalpy Volume of change, N aOH ∆H, for this volume 0.600 mol dm of sodium sodium hydroxide hydroxide, Temperature / °C 3 (aq) / cm added and 0.00 22.4 shown in the table on the right. graph of these data. the rate of cooling back to would temperature by extrapolation a and the with l Find for exothermic or endothermic. 3 HCl(aq), of the heating 5.00 23.1 10.00 23.9 15.00 24.6 20.00 25.4 curves. v O b. a are temperature the is u temperature a. acid, you t a student r o A reaction i reaction. measured. the extrapolated information 8 n c. minimum giving o State, U State b. d a. 7 C i n 18 o v 22 y 30 p r e 32 O 40 of hydrochloric 3 acid was 25.00 cm . Determine the 25.00 25.6 30.00 25.5 concentration of the acid. E d. C alculate reaction, 382 ⦵ the in enthalpy change of n = y ΔT = 64.0 44 t i s / e r u t a r ep m e t 46 36 2. l º 48 neutralization, ∆H , for this 35.00 25.4 40.00 25.3 45.00 25.2 50.00 25.1 1 kJ mol y P 54 e 62 Tool 3: M athematics Sources of experimental error Experimental always have systematic has the or in dierent everyday meaning. consideration outcome of the fact that with measurements them. are Experimental never errors exact: they c an be random. “error ” very from associated an of language One sources investigation, of of c an the mean features experimental and possible “mistake”. of a In thorough science, “error ” scientic research errors, their impact on the ways to minimize them. r R andom and systematic error be minimized, random an include associated variations in the be by bec ause provides errors uncertainty random eliminated, precision uncertainty to in too measured high as repeating the estimate values of too values, low. L arge are a random spread over readings that uctuate with reading a sc ale, measurements made with a in the same direction y results p results greater error. r e aect (either errors measurements. particular instrument. Systematic errors leading to R andom O and of time, reduce not variations likely t i s Sources over errors The natural n range. but to equally y random due are l c an arise that y measurements P R andom errors e is word a result s s The errors uncertainties meniscus correct reading of volume making meniscus (gure the and c an be values errors background c an then include radiation, during a be corrected misc alibrated leaks in c alorimetry random uncertainties, a basis where evaluating error how (not far to an available, the be that literature the = E Note error theoretic al and with value value collection uncertainty − propagation, values accuracy and of percentage is from the theoretic al value is and Figure 93 Examples of systematic error when taking burette readings heat a are useful in between percentage percentage processed uncertainty) theoretic al is errors c an result. a way of value: value × theoretic al probe, apparatus, Comparisons theoretic al confused experimental experimental Percentage by results. precision a quantifying for obtained experimental v O Percentage and, errors, in l give errors u of temperature experiments. t a of analysis gas for it. i surroundings f x the at bottom of meniscus by making changes to the Errors and processed results Estimates are too o systematic of the that methodology. If a particular systematic error r o to accuracy eliminated, d of presence loss reduce not quantied, Sources readings from n be in For reading U but experimental c an burette low). 93). errors minimized, result too C Systematic will a or i n high the taking high o above consistently too v example, the 100% value sometimes c alled the accepted value or value. 383 Tools for Chemistry For example, the theoretic al value for the activation energy of a reaction is 1 reported in activation the literature energy as 65 kJ mol experimentally ± 1%. Students A and B determine this and obtain the data shown in table 5. Student A 1 nds the activation energy to be 60 kJ mol ± 10%. The ± 10% part is an estimate 1 of the c an value, see that (10%) that theoretic al that the two indic ates the value are low result lies is in within aligned. precision, the the range range However, the particularly of of 54–66 kJ mol the high when percentage compared to that of value (1%). Uncertainty Absolute r Experimental E R ange Percentage a 1 1 value / kJ mol / % uncertainty / kJ mol 6 4 2 Table 5 Student results for the experimentally determined the result obtained by 1 ± 4%. not result lie in is this be (though the attributed the random the is errors. loss of to following the precise This reaction: obtained as the by student A, theoretic al time, the theoretic al value). value does inaccurate. In addition, since the This C use the same errors surroundings suggests in between as systematic an that student or systematic B’s errors result and the random: investigation involving c alorimetry o heat b. a misc alibrated colorimeter to determine the absorbance of i asample c. taking a burette reading by positioning the viewer slightly below the meniscus rather than positioning the eyes on a level with themeniscus d. u 52. l a E v O f x t a r o d a. of as the that error is 23%, only up to 4% of the 23% dierence n U each for o to . value. Classify A use of student and pH probe c alculated obtained C alculate 53. a that the reports enthalpy values change to the for nearest the whole combustion number. of butan-1- 1 ol the Two students with a strong −1 430 kJ mol percentage error investigated base. Their 1 . the The in theoretic al the student C: −35 kJ mol student D: −50 kJ mol student’s enthalpy results of value is −2 676 kJ mol neutralization of a strong acid 1 ± 4 kJ mol 1 1 ± 10 kJ mol 1 theoretic al a. C alculate b. Comment value the on theresults. for the enthalpy percentage the relative errors impact of of of neutralization the . result. were: 1 The 384 not B than 1 result Practice questions 51. still percentage to signic antly smaller 48–52 kJ mol therefore and i n theoretic al 4% of v c an range range, is contributed the student is p uncertainty error in precision error y The greater random r e indic ating The 23 O consider 50 kJ mol 7.7 48–52 activation energy of a reaction t i s Now 54–66 students’ random is −57 kJ mol results. errors on each of . y 10 50 n 60 B l A student y student error / % P 1 / ± kJ mol . experimental e theoretic al indic ating the suggesting uncertainty the error, s s We random Tool 3: M athematics Errors and graphs A ytitnauq Graphs c an also provide information regarding systematic and random error line (gure of best R andom Systematic (either up or errors errors down). c ause c ause On data all points the graphs, data this to to deviate shi in from the the same means that the points perfect results random error s s direction 94). t. systematic error are systematic ally especially loc ated higher noticeable in the (or lower) y-intercept: a than expected. This is y-intercept that is not where quantity B would expect it to be suggests the presence of a systematic error. Figure 94 The presence of systematic and errors c an be inferred sodium hydrogencarbonate on the rate of its reaction n experimentally (table 6). y A student investigated the eect of concentration of y determined l 55. The combustion of primary alcohols methanol to was graphic al data P Practice questions octan-1-ol from r random 54. e you with citric acid. The reaction produces carbon dioxide, atoms in primary of combustion alcohol / kJ mol - –400 ¹ –814 –1300 –1440 8 –1700 n Table 6 U Experimental enthalpy of combustion of the primary alcohols dioxide / s ± 1 s 1 2 3 210 277 166 0.50 195 135 118 1.00 66 93 53 1.50 21 27 35 2.00 15 25 15 0.25 C 7 ³ ± 5 % o –990 6 i n 5 - p 4 v –680 25 cm³ of c arbon hydrogenc arbonate / mol dm 3 Time taken to produce Concentration of sodium y 2 - r e –230 of this gas. The student’ s data collected is shown in table 7 . ¹ ± 40 kJ mol 1 3 so the student timed how long it took to collect 25 cm O Experimental enthalpy t i s Number of c arbon 3 Table 7 Time taken to produce 25 cm of c arbon dioxide in the reaction of sodium hydrogenc arbonate and citric acid at a line your or and published dierent concentrations of sodium hydrogenc arbonate error curve of best t. graph, o and including i Using graph of the data in table 6, a. enthalpy of t a b. bars a d Plot r o a. C alculate the mean time hydrogenc arbonate for each of the sodium concentrations. combustion data, comment on the impact of systematic Using your random on the results. u f x c. error b. Determine halving an the estimate range of of each the set time of uncertainty by three trials. graph, comment on the impact of error on the c. Plot a graph to show the relationship between results. concentration l a E v O time, of including sodium error hydrogenc arbonate and bars and a line or curve of best t. d. Using your random graph, comment on the impact of error on the analysis. 1 e. C alculate the rate by nding for each T concentration. f. Plot a graph to concentration and g. show of the sodium relationship between hydrogenc arbonate rate. Using your systematic graph, comment on the impact of and random error on the analysis. 385 s s e r n l O y p chemic al C n U r e a c t i o n s? o drives y P y t i s v i n What 1 r e Re a c t i v i t y o i d t a r o u l f x a E v O Reactivity 1.1 Measuring enthalpy changes s s What c an be deduced from the temperature change that accompanies chemic al or Chemistry involves the study of chemic al An reactions and understanding energy principle of Conservation science, which of is energy is a fundamental examined empiric al experimental data, the language of or scientic understanding of progress energy changes associated with reactions. is reactions system and the involve a Reactivity 1.1.4 — The transfer of chemic al surroundings, while conserved. pressure It determined c an be the system Reactivity 1.1.3 — The whether the of reactions In l a E v O surroundings a chemic al stored of the in the control enthalpy over the refers to the change heat for a transferred at standard conditions and states. from the change in temperature of a in chemic al reactions All reaction, energy mixture is a is conserved. reactants function of the and Chemic al products, kinetic potential while the energy is temperature energy of the atoms, ions and present. chemic al reactions surroundings system total chemic al bonds of reaction molecules from the heat, involve from the energy surroundings. form of but it In an open system, changes. reaction system may also Most be in or it Energy may commonly, the form of be the may be absorbed energy is released into the by the reaction transferred in the electricity, sound or light. (contents the transfer of matter and energy is possible across its of flask) boundary Figure 1 better (Reactivity 1.1.1) surroundings system used o i t a + Energy transfer u system substance. reactants and be relative stability of are endothermic d r o f x = pure energy surroundings. stability exothermic. universe of to system and turn n or determines and relative direction U products the , under o between on the in studied. standard ΔH constant C transfer depending i n exothermic, being c an ⦵ reaction, Reactivity 1.1.2 — Reactions are described as endothermic or of leading reaction p energy the v total between products, the chemic al y Reactivity 1.1.1 — Chemic al understanding r e Understandings energy and of a information between how O chemic al reactants terminology, all contribute to an exist n and develop between This that understanding t i s our to experimentation. The use of models, transferred surroundings. relationships involves l and is the y observation mathematics the through of energy y compounds. and r chemistry P physic al changes of state of the elements and their e physic alchange? The universe is the (for transferred combination of the system and its though surroundings matter example, through energy and may energy its c an sides). A be c an matter added to closed system transferred neither be across enter nor the exit a beaker, allows no and transfer boundary. In an the energy of c an be matter, isolated system, system. 387 Reactivity 1 What drives chemic al reactions? s s e r In each of the above scenarios, energy is transferred. In hot springs, energy is transferred and in reworks, as heat, industrial during reaction, an but processes industrial also contributes place in chemic al c an a be industries in not loss used as a open only of or useful to closed aects map, systems. eciency energy, model heat the p Thermography structures to heat an ow where The of loss of the chemic al increase in thermal and loss from red is hot and purple n U C i n o v pollution. iscold. take process y heat r e Most O t i s Models o i d t a r o u l f x a E v O 388 as heat, Both scenarios are opensystems n light. l and y sound Figure 3 What c an are Thermograph of industrial engineering system the chemic al industrial advantages engineers processes? of use How modelling the does data this heat distribution collected help our to and transfer? How improve the eciency of environment? y P Figure 2 Reactivity 1.1 Measuring enthalpy changes What is the dierence between heat and temperature? Temperature, T, change in value is is an example of a independent of state function. the pathway For a state function, any between the initial and nal measurements. in it example, the will you take (the initial not tell you occurred the the temperature value) complete throughout the and then story day. of of any The the again water in the in temperature c alculation of a swimming aernoon uctuations the pool (nal early value), that may temperature change = T nal − T initial of energy thermal energy. convection and It c an absolute a E v and temperature a (−273.15 entropy, S, (in When from of a phase °C), all system kelvin) is by heat kinetic energy potentially absolute zero, 0 K stops and the transferred work. average from liquid to gas. At be radiation. the temperature transferred l O in in is temperature gradient. Heat has the ability to do increase that volume, is of enthalpy a Heat the motion reaches for of its proportional is sometimes to an This referred object, c auses the an result is increase example, a change of state the particles minimum to pressure. processes of conduction, particles. change, of a day, throughout the warmer body to a cooler transferred its and O form f x to as a result of the include y a i is as functions t a q, body, state u Heat, of r o examples the end has occurred p only the overall temperature change Other an the beginning and cooling that o C the temperature of a pool at n day, If you record give you an indic ation of the heating and o will not d it the theoretic ally possible average value. The kinetic energy of Entropy the particles of matter. As the temperature n l y t i s r e v i n U Figure 4 y P ΔT r issimple: e have if morning s s For increases, the kinetic is dened and explored in energy of the Reactivity 1.4 particles also increases. 389 Reactivity 1 What drives reactions? Communic ation skills ATL Your communic ation throughout the terms of “heat” solution When you describing particles within → MgCl released surroundings, considering rises. the refer the that between When the (aq) from and transfer to 2 the the you of + H reaction think thermal kinetic (g) temperature of about heat, energy in the temperature average 2 of a system, energy of the system. y p r e M agnesium ribbon reacting with hydrochloric acid n U C o v i n Figure 5 Thermochemistry is the study of heat changes that occur during chemical reactions. Heat changes are oen described in terms of enthalpy. At constant o a closed system to the surroundings during a chemical reaction. The terms i “enthalpy change” and “heat of reaction” are commonly used when describing the thermodynamics of a reaction. The most common unit of enthalpy change is kJ. u l a E v O f x t a r o d pressure, the enthalpy change, ∆H, is dened as the heat transferred from Activity Imagine a undergo glass a the of • Identify the system • Explain the movement the What of containing ice cubes sitting in the summer sun. It will enthalpy. Is • glass in • and water an open, and the of closed or isolated system? surroundings. energy in the form of heat, between the surroundings. would observation 390 water change you in observe terms of a on the change outside of state of the and glass? Explain this transfer of energy. system y are is reaction n you 2HCl(aq) heat simple l system. the hydrochloric acid. O are and e aqueous explanations will t i s you the accurate y the into in and Your use of the r reaction, system consistent terminology. P this + incrementally programme. understanding of theseconcepts. consider metal Mg(s) develop “temperature” your example, magnesium include scientic and demonstrate For skills will chemistry s s applic ation skills entire Communic ation In chemic al Reactivity 1.1 Measuring enthalpy changes Exothermic and endothermic reactions (Reactivity 1.1.2) A chemic al reaction surroundings in dened as to a to it c an be an the chemic al surroundings are place, to form chemic al This energy each process. The part the bond. are the reactants Energy and is are rearranged broken, and absorbed bond by the breaking is bond dissociation energy Energy made, of energy are therefore termed the bonds your of reactants products. of transfer of atoms in bonds, is type chemic al important the bonds is released into the and therefore bond making between understanding of the the surroundings and energy changes in Exothermic reactions example, so determine is the reaction in is between experiments which most any the focus c ases the reaction, the change or to heat is exothermic, measure For system and so the the absorbed we by c an use a amount of heat surroundings. In the school in temperature, ΔT, of the reaction n collect used enthalpy change surroundings. o ofchemistry. to as by i technology such and water. observations instruments the the c an be made using human senses, or with data-logging data is one of equipment. the The applic ation of essential skills in the study u l f x t a digital of lost d laboratory, aid r o the the being endothermic the of not positive. apparatus on is is is endothermic is system Observations In an heat sign and U solvent, a For change system C exchanged laboratory, negative. c alorimeter the reaction, enthalpy whether A of change, and endothermic The i n c alorimeter. being exothermic change system, perspective enthalpy change. o To an negative v the in the a enthalpy p enthalpy from have positive y dened a r e is reactions have O areaction. n is from y exothermic system made for new system system to the contrast, t i s the takes the process. when the the In l surroundings is an are quantied into from reaction. reactions. Chemic al break transferred y endothermic and absorbed reaction bonds system is exothermic P an is products. chemic al reaction heat endothermic new heat r new which chemic al create which e When in dened as an s s reactions is a O Energy proles (Reactivity 1.1.3) Energy proles are a visual representation of the enthalpy change during a reaction. Activation energy, E , is the a From an energy prole, you can determine the enthalpy of the reactants and the v minimum energy required for the products, the activation energy (E from E M any chemic al the energy reactions system level reactions to and are at a the ), and the enthalpy change for the reaction. a reaction to take place. You will study are exothermic. surroundings. considered lower to energy be The lower level and In these reactions, reactants in of stability. this energy reaction Products considered to be for more is released activation energy in Reactivity 2.2. are at a higher exothermic energetic ally favourable. 391 Reactivity 1 What drives chemic al reactions? Consider the reaction displacement Zn(s) Measured + CuSO quantities The is using measured is (aq) of a → the Cu(s) copper(II) is the the heat + ZnSO and or and change data-logging As the copper(II) a transferred in are the the r NO i 4 This heat is apparatus example the 3 (s) 3 , to is are an form the less products stable important aqueous than of the the component ammonium reaction of and fertilizers. When nitrate ions, the decreases. NH by 4 the the (aq) + reaction reaction endothermic NO 3 (aq) system mixture reaction from will the surroundings. The feel cold to touch. This is an (gure 7). endothermic reaction a ygrene products E a l a i t n e t op ΔH reactants reaction Figure 7 coordinate Using a thermometer or a temperature probe, you would in the temperature of the reaction mixture in an endothermic reaction. the products is greater than that of the reactants. energetic ally less stable than the reactants 392 are at a reactants. + → absorbed an NO solution containing of 4 water o of in energy prole reaction, therefore u l E v O f x t a r o d NH and nitrate, NH dissolves temperature example of an endothermic level describe the products as being o solid an C Ammonium an is n U energy 6 n p You would observe an increase The enthalpy of the observe a decrease The enthalpy of The products are described as being y l you would O products y r e v consider higher the gure reaction is y t i s laitnetop i n you in The P ygrene of the reactants. energetic ally more stable than the reactants graph reaction, ΔH Using a thermometer or a temperature probe, products is lower than that this a reactants in the temperature of the reaction mixture in an exothermic reaction. If In temperature of the surroundings. reaction coordinate The mixed in a equipment. result, to zinc exothermic reaction Figure 6 by zinc: temperature of the solution exothermic. E ion (aq) solution the system. is 4 of e and reaction displacement sulfate stirred, thermometer by increases, therefore 4 mixture generated solution involving s s c alorimeter. heat between zinc and copper(II) sulfate solution. It is a single reaction Reactivity 1.1 Measuring enthalpy changes Global impact of science Developments in science may have ethical, environmental, political, social, cultural and economic consequences, which must be considered during s s decision making. The pursuit of science may have unintended consequences. German chemist Fritz Haber was awarded the Nobel Prize in Chemistry in 1918 for developing a method to chemically extract nitrogen from the air production of fertilizers that began during the green revolution and continues r today. However, his process also provided Germany with a source of ammonia that was used for the production of explosives during the First World oen and you would graphs. exothermic to in able writing sketch C an and be you read texts write writing reports graphs and and endothermic thinking communic ation of to are intended formal styles. and also for answering extract analyse data the reactions? involves form to of meaningful that of the 4 Cl(s) → BaCl following is 2 (aq) correct l a enthalpy decreases v D enthalpy products E C products decreases for 3 (g) + 2H this chloride, NH products enthalpy products have than have than reactants lower the reactants have higher than the Cl: O(l) ΔH = +164 kJ mol reaction? lower the 4 1 2 Enthalpy increases O B 2NH ammonium u f x Temperature A + with represent n 2NH reacts specic information o Which + , using communic ation proles i (s) 2 t a 2 Ba(OH) r o Ba(OH) hydroxide, your ability to reports U Barium d 1. written accurately sketch these diagrams, including all of the components. Practice question in examination questions, and energy Try and improving dierent audiences. In laboratory Another verbally about C from utilize ability to accepted you i n your need forms forms of o you However, dierent and when and v terminology skills. mind skills communic ate p science, write of y read to range eectively r e communic ation comes wide to n most a ability l form cover Your O skills communic ation. t i s Communic ation is y Communic ation skills y P War. The global impact of science is evident in Haber ’ s research. ATL e by reacting it with hydrogen. Haber ’ s discovery allowed for the large-sc ale reactants have higher Stability products than the products stable are than products than are the less stable reactants more the are reactants less stable reactants products are more increases enthalpy than the reactants stable than the reactants 393 Reactivity 1 What drives chemic al reactions? ⦵ Standard temperature and pressure (STP) conditions are denoted by the symbol ⦵. STP is a Standard enthalpy change, ΔH (Reactivity1.1.4) temperature of ⦵ The 273.15 K and a pressure of standard enthalpy change for transferred ambient at pressure (SATP) refer to from ⦵ , refers conditions to the and heat states. the change in temperature of a pure It c an be substance. The units of 1 are kJ mol reaction conditions of K and 100 kPa. STP and ⦵ T o calculate ∆H SATP ΔH standard conditions are e 298.15 under more ΔH practic al pressure temperature determined and constant s s Standard a reaction, 100kPa. for a reaction, you therefore need to determine the amount of heat given in the released or absorbed in the course of that reaction. This can be done by measuring r section 2 of the data booklet. the change in temperature of a pure substance to or from which this heat is such as water , you need to know the specic heat capacity, c, of thatsubstance. Specic heat l Substance 1 c apacity / kJ kg 1 The specic example, a pure kJ to the specic raise the heat of copper 0.385 of 1 c apacity same amount Specic heat with size the is heat up has a given heat is substance, transferred c apacity is an of the the pure system same dened as of the to heat the rise higher the the by 1 amount c apacity in 1 the • the amount supplied. K. rise of that For as a heat does ton For , so it takes lower the specic temperature when the not example, 1 K. 1 K The in 1 vary in magnitude 3 a block 10 of cm sample of copper. When you o mass of that substance 1 or temperature depends on: C the by °C sample. v • i n the identity of that substance heat ethanol described. • of substance intensive property being specic substance, that p copper of a of y c apacities of water, ethanol and copper of kg kg r e The specic heat substance 1 c apacity of ethanol is 2.44 kJ kg temperature 2.44 heat Table 1 of temperature O ethanol c apacity the t i s 2.44 raise n 4.18 to y needed water heat K n U Practice questions 2. Using table 1, o 1000 kg of copper When u l a E v O f x Which of masses energy, energy is the of two their following dierent substances, temperatures is rise by specic heat c apacity of X is twice b. The specic heat c apacity of X is half c. The specic heat c apacity of X is one d. The specic heat c apacity of X is the a. table More 50 b. If °C the c. If 1, state heat than same equal is If equal which of the needed to increase 50 g heat of masses together, d. to raise the the is of of ethanol supplied the nal masses temperature, of 5 X °C and and Y , 10 absorb °C, the same respectively. correct? The Using required by 1 K. a. temperature 394 much water equal of how following 1 amount 4. of the b. i 3. kg c alculate of a. t a r o d temperature water water h the of of Y . Y . that same following as of Y . that statements of is temperature Y . correct. of 50 g of water to equal masses increase 20 °C and of ethanol and water, the more. ethanol at 50 °C are mixed temperature will be 35 °C. water ethanol and gives by by 50 °C. will at that that ethanol out at more 50 °C heat. cool down to room y P transferred. When calculating the amount of heat lost or gained by a pure substance Reactivity Specic a pure heat Q where c apacity, substance m = is c, using is the used to c alculate the amount of heat, Q, absorbed 1.1 Measuring enthalpy changes by relationship: mcΔT mass of the pure substance in kg and ∆T is the change in temperature s s of that substance in K. Heat, Q, is related to enthalpy change, ∆H, by the following equation: e Q ΔH = − n n is the number of moles of the the reacting therefore substances how are much limiting reactant. with the product said to be in least c an In a reaction, the limiting stoichiometric be formed. In amount present, contrast, the other excess experimental that directionality the are and on heat be direction. With change the experimental in the coee-cup enthalpy be transferred surroundings in in solution A coee-cup c alorimeter n 2. to 25.0 g of water in a a temperature rise of 3.80K is transferred to a 100.0 g sample resulting in a temperature rise from 22.0°C to 26.0 °C. C alculate the specic heat c apacity ofiron. Assume that u f x chloride. 180.0 J of heat ofiron, C alculate the enthalpy change of dissolution for 1 mol of lithium containing reactants Figure 8 t a c alorimeter, was recorded. i was added o d coee-cup r o LiCl, nested together experiment. When a 1.15 g sample of anhydrous lithium chloride, cork stopper for a between the every glass stirrer two polystyrene cups lower in magnitude than the actual will Worked example 1 1. of empiric al data is constant measured the eect of U contents the C always some consequence same reaction will a eect the as introduces and assessed. c alorimeter, value, method analysed thermometer o in be coee cup convenient i n always Their This a p errors procedure. c an is v Systematic and procedure. errors polystyrene change y their a r e systematic in enthalpy O the measure n y reactions to t i s Measurement Performing l reacting limits substance y which is P reactant r where the heat capacity of lithium chloride itself is negligible. = mcΔT = 0.025 kg = 0.397 kJ 1 mol E Now you × need a Q 2. 1 4.18 kJ kg v O 1. l Solution to First, determine the change in temperature, ∆T: 1 K × 3.80 K ∆T = Substitute determine the energy gained for (299 the 0.180 kJ = − 295) K = values into 0.100 kg 4 K. Q × c = mcΔT: × 4 K ofLiCl. M ake c the subject of the equation and solve: 1.15 g n(LiCl) = = 0.180kJ 0.0271 mol 1 c 42.39gmol − = = 0.450 kJ kg − 1 1 K 0.100kg×4K −0.397kJ Q 1 = ΔH = − n = −14.6 kJ mol 0.0271mol 395 Reactivity 1 What drives chemic al reactions? Practice questions 5. C alculate the energy absorbed by water when the temperature of 30 g of 1 water 6. 0.675 is raised kJ of by 30 heat is °C. The specic transferred to 125 heat g copper of water metal. is 4.18 J g 1 specic heat c apacity of 385 J kg K . C alculate the change in temperature of the copper metal. task, the we will enthalpy at reaction the for method the 3. used to At of exothermic metal three zinc minutes, powder, introduce record the 4 (aq) → Cu(s) + ZnSO 4 Continue minutes (aq) to after 5. Tool 1: Measuring Produce a variables Applying techniques 6. Use Q, 2: Applying • Tool 3: Processing uncertainties • Tool 3: Graphing • Inquiry 2: Inquiry 3: Processing data Evaluating Materials • The • the • measuring • thermometer c alorimeter r o or temperature zinc copper(II) sulfate solution powder l O Method Using the a the 25 cm of solution E Using of temperature The the or the to threeminutes, or c apacity but The Loss of measure specific same heat 396 is to as graph to c alculate for the the heat released, reaction, ∆H are made when using this from the reaction is completely as an insulator However, and heat much is the heat the against heat loss to coffee cup also has transferred smaller temperature of five water. acts heat that from source of from a c apacity of the error quantify. c alculated CuSO coffee-cup 4 solution. c alorimeter. This a temperature probe, loss solution every 30 of recorded until a constant heat than that reached to of is it from the water. an accurate evolved during the system in of an aqueous solution is water. this The to the change graph surroundings is the experiment and one that is will in temperature, include a ∆T, systematic will means be c alculated lower value of that than Q the the lower maximum error. temperature theoretic al than the value, making actual value. The record seconds of errors on the result of subsequent c alculations is for in temperature is procedures. achieved. it maximum important up cup to been reaction. • effect of the representation the thermometer to time up has 3 1.0 mol dm to • the accurately coffee heat difficult balance, released surroundings. water, main 3 mass Transfer 2. electronic v the an a 1. probe u • f x 3 1.0 mol dm t a cylinder o coffee-cup i • • d electronic balance assumptions heat The a • of transferred n U • variables number method: for temperature. change Assumptions and errors A Controlling ∆T to enthalpy C 1: process data v Inquiry to i n • technology versus in p Tool value of the o • your and change readings temperature y 1: the r e Tool maximum temperature determine • temperature the O reached. Relevant skills • take t i s CuSO n y 4. + mass of zinc and commence stirring. between zinc and copper(II) sulfate: Zn(s) between 1.3 g and 1.4g exact l displacement look change considering improvements in experimental y skills P this c alculate e r Investigation to nd the enthalpy change for a reaction In 1 K Copper metal has a s s 1 of c apacity Reactivity An accepted temperature is to look is in figure cooling complete, when 9. zinc Amore is Measuring enthalpy changes c alculating the maximum for section and systematic of the errors in data curve after the extrapolate this back to the introduced accurate at value 3 for minutes, ΔT as shown c an then be determined. s s moment the of compensate C° / erutarepmet reaction at method to 1.1 ΔT e r 4 6 8 10 time / min to measure the temperature change for the a 1.0 mol dm solution of copper(II) sulfate. The following results were recorded: determine = 4.69 kJ the 4.18 kJ kg limiting heat released: 1 K × reactant for In 39.0 K the f x n(Zn) 12 These truth, u l E in and Scientists certainty, power, explanation, responsibility. perform experiments r and process the raw data to 1.37g = enable us to draw conclusions. 1 We = compare experimental 0.0210 mol and theoretic al values. What concepts do we utilize when n(CuSO 4 ) = c × v justifying 3 = 0.0288 dm How so it is the do our we conclusions? use evidence? judgments subjective 0.0288 mol or amount, our 3 × Are smaller evidence, interpretation, values = 1.00 mol dm a there focus. = M a v O present are: in objectivity, perspective, culture, objective? and is knowledge, 65.38gmol Number of moles of copper(II) sulfate, Zinc of concepts justic ation, reaction. m Number of moles of zinc, theory are t a determine × of o 0.0288 kg amount 1 i = the o to r o Then, mcΔT d Q = TOK n Q the enthalpy change for this U Solution First, use and 39.0 C i n released reaction. v 1.37 M ass of zinc / g ∆T / °C p 28.8 M ass of copper(II) sulfate solution / g y r e 3 reaction between zinc powder and of heat n c alorimeter was used O t i s Worked example 2 Determine the amount l Example of a temperature vs time graph for a c alorimetry experiment A coee-cup y Figure 9 y 2 P 0 limiting reactant. You c an When appraising analysing experimental c alculate limitations, how do Q the enthalpy change of reaction from ΔH = − : assumptions have an impact on n 4.69kJ 1 ∆H = – = our perceptions? −223 kJ mol 0.0210mol 397 Reactivity 1 What drives chemic al reactions? Combustion of primary alcohols You c an determine common experiment the in a enthalpy change of combustion of school laboratory. After 2. Determine an repeating alcohols, you c an subsequently mass of the spirit burners using 3 Accurately determine the analyse this data and mass of 30 beaker or metal cm in a 250 cm Using either determine a temperature and record the probe initial or a temperature of 1: Recognise ethic al or and address environmental the issues in relevant an safety, investigation Measuring temperature and Ignite a period 1: C alorimetry Inquiry 2: when and how to insulate a. loss or gain Identify observations and and record relevant relevant qualitative b. quantitative data which each allow of Determine the burners, each containing one of the c are alcohols: methanol, ethanol, propan-1-ol, and pentan-1-ol beaker • tripod • temperature or metal c alorimeter S afety they be are handled generally suitable your level of v • identify suitable the this a final to mass the temperature burn for a period of two of flame each spirit burner is extinguished. Take burner to calculate the amount of heat released, Q, and the enthalpy change of combustion, ∆H, for eachalcohol. thermometer beaker clamp water spirit burner wick methanol Figure 10 E aligned with safety policies. extra bec ause the burner will be hot. A typic al arrangement of experimental apparatus for an enthalpy of combustion determination disposal methods and until hazards and control measures health burn experiment. In risk relevant school’s hazardous and a the determine for you should: hazards • your 398 assessment, assess identify assessment l • sources, risk identify the O • risk a c are u complete with i Using f x 1. of t a Instructions disposed flammable, r o volatile. and to reached o bec ause should d Alcohols water. The c an be set in one of two Use your values of ∆T of the water and ∆m of the n U probe or thermometer c alorimeter o • alcohol after C electronic balance i n • v 7 . butan-1-ol is p following or the y immediately spirit burns alcohol 30°C each r e 6. five it heat minutes. Materials • beaker to ways: allow change sufficient the burn O • different Appreciate heat over t i s against 1: to n Inquiry under alcohol l Tool • burner the y • spirit allow mass y 1: P 5. and Tool r thewater. Tool thermometer, e 4. • water c alorimeter. patterns. Relevant skills • of 3 contained identify initial balance. several times with a homologous series 3. of the electronic s s the alcohols Reactivity 1.1 Measuring enthalpy changes Research skills ATL Cite your sources fully, according to your school’s citing and referencing system. s s Worked example 3 to measure the amount of heat by The following results were recorded: the enthalpy change of Solution = mcΔT to determine the amount 1 Q 3.91 kJ with for oxygen this in a reaction K × 30.0 K combustion bec ause reaction. Methanol is the oxygen Number of moles of methanol, n(CH 3 OH) = Mr is present in U 1 32.05gmol the = − change of 1 −359 kJ mol = 0.0109mol experimental identic ation to measure. environment discrepancies the give values. E the of typic ally theoretic al dicult to aspects This smaller is the a useful and result TOK experimental set from of for Q : ΔH = − n of helps and data, examination in the us results. loss from involve systematic their errors. directionality C alorimetry than the assumption to and and of temperature heat make raw random experimental change usually negligible. between of reaction evaluated errors analysis a Scientists is be systematic v from the provide c an a essential experiments of that l are O The experiments procedures u f x Thermochemistry air. t a r o 3.91kJ ΔH enthalpy o c alculate the 0.0109 mol i c an d You in n 0.348g = = excess C m o reacts reactant 4.18 kJ kg p = × released: v limiting 0.0312 kg heat i n Methanol = of 1 y Q r e First, use O combustion for methanol. n and t i s released l y 30.0 ∆T / °C y 0.348 Change in mass of methanol / g of heat P 31.2 M ass of water / g Determine the amount released r the combustion of methanol. e A metal c alorimeter was used is predicted system, which is that the heat lost understand our judgments of theoretic al values. 399 Reactivity 1 What drives chemic al reactions? Thermometric titration The neutralization base reaction exothermic. unknown In this 2. between an acid and a skills concentration of task, Review hydrochloric acid the change in is added to percentage sections in the error and Skills chapter. Rinse and fill the burette with sodium hydroxide temperature while sodium solution. hydroxide titration, by 3. measuring the uncertainties you will determine the acid. The Record its concentration. temperature will 3 a maximum when the acid and base are Add 25 cm of acid solution to the cup and place it e 4. reach s s the is mixed under the burette. Nest it inside a second cup, for together in stoichiometric amounts. r additional thermal insulation. For safety,these cups should be placed inside a beaker to avoid tipping over. Relevant skills • Tool 2: Use • Tool 3: C alculate C alorimetry and acid–base titration 5. Position the the initial temperature temperature probe of the in acid the in acid the and record cup. sensors percentage Add a small solution Tool 3: Understand and the signific ance of uncertainties in temperature processed data 7 . • Tool 3: Continue Propagate uncertainties and state them to an appropriate level of the (~5 solution precision acid, cm stirring and small the graphs decreases 8. Inquiry 3: when and Identify random and discuss sources of systematic error • hydroxide Hydrochloric acid solution is is thermometer • graduated • burette • ~50.0 cm or temperature probe pipette and filler l O over v this a through the information, risk your 400 E Read by showing hydroxide Extrapolate the two Determine the with absolute sure you consecutive teacher. temperature vs volume of added. sections of the reached graph to find the during the titration. concentration of the acid, along and state several your solution temperature hydroxide all percentage values to an uncertainties. appropriate M ake level of precision. 4. C alculate acid 5. the percentage error of your experimental concentration. Determine absolute values the and to an C alculate enthalpy of neutralization, along with percentage uncertainties. State all appropriate the level percentage of error precision. enthalpy 7 . of of your experimental Comment neutralization. on the relative impacts of systematic and hydrochloric acid of unknown concentration Method 1. graph maximum 6. a 3 30.0 cm instructed sodium sodium hydroxide solution of known concentration • u f x 3 t a • polystyrene cups o cm 2. 3. i 250 d r o 3 two corrosive. a sodium corrosive. Materials • as C Sodium protection. Plot n • eye 1. U Wear up Questions S afety • Clear how to insulate loss or gain i n and Appreciate heat o • 1: v against of y Inquiry Extrapolate hydroxide Record the highest temperature until the p • 3: sodium volumes recording temperature r e Tool of reached with this addition. adding readings. • ) gently. O raw to t i s • volume error n interpret y 3 6. and teacher. for 8. relevant this error on concentration safety, materials and method. Use and assessment random safety data, to complete practic al work and show it to Suggest and the and values obtained enthalpy explain two made to this method. of for the acid neutralization. improvements that could be y 1: l Tool P • Reactivity 1.1 Measuring enthalpy changes Measurement Experimental precision. errors systematic extent are c ause errors these values errors c an be an assessed measurement inaccuracy. in errors in What enthalpy of in lead are terms to of their accuracy and imprecision, some of neutralization the whereas sources experiment? of s s systematic and enthalpy R andom random To what quantiable? water is a bomb c alorimeter “bomb”), and laboratories described the in (gure this 11). resulting utilize the same chapter. The A sample is burned temperature change of measured. ignition O surrounding c alled a (c alled research p oxygen of o list all the in research laboratories features that are labelled. feature. measurements and properties i the and each a why accurate of l What c arefully purpose v O Consider highly diagram the t a the c alorimeter used u Deduce d r o Study 2. 4. of a bomb f x 1. 3. Diagram o n water bath bomb Figure 11 temperature probe U calorimeter supply C i n stirrer v jacket y r e device y is chamber in experiments n used conducted c alorimetry t i s the a the y instrument inside experiments as l principles e r C alorimetry P Thinking skills ATL obtained with a bomb c alorimeter are precise. water make it suitable for c alorimetry experiments? E 401 Reactivity 1 What drives chemic al reactions? End of topic questions Topic review Using your knowledge from the Reactivity 1.1 topic, answer the guiding question as fully as s s 1. possible: What can be deduced from the temperature change that accompanies chemical or physical change? r Multiple-choice questions is correct for + 6HCl(aq) the → following 2AlCl 3 (aq) reaction? + 3H 2 (g) Reactants are less B Reactants are more stable than products and the reaction is endothermic. C Reactants are more stable than products and the reaction D Reactants are less exothermic reaction, B In an exothermic reversible the reaction, endothermic reversible the products reaction, are the is correct → for 2Fe(s) + this more stable activation 2 (g) 1 26.6 kJ are absorbed C 53.2 kJ are released for every mole of Fe produced. D 26.6 kJ are released for every mole of Fe produced. 4 reaction (g) 2O → do 2 (g) H(g) + the → (g) + Cl NaOH(aq) (g) + → + have 2H 2 Fe of → forward produced. a lower O(g) NaCl(aq) reactants. the produced. Fe NaCl(s) HCl(aq) E 402 2 (g) of mole a Na v D mole every reactants CO Br(g) + C every for l HBr(g) + for + H 2 O(l) energy than the reaction the t a CH O B which released u A are i 13.3 B r o A In kJ of reactants. forward o −26.6 kJ mol f x 5. = d ΔH the the than energy reaction? 3CO than of n 3 energy energy U 2 more activation reaction. 3CO(g) exothermic. exothermic. products? is greater than that of the o endothermic an statement is is C an In + have the i n In (s) reaction reaction. D O products reaction, C Fe the reaction is endothermic. p an Which and the v In reverse products and correct? A reverse 4. is than products O statement stable than y Which stable t i s A r e 3. = −1049 kJ mol n y 1 ΔH l 2Al(s) reaction is greater than that of the y Which P 2. e Exam-style questions Reactivity 1.1 Measuring enthalpy changes Extended-response questions 6. Nitrogen NO 2 (g) dioxide + CO(g) and → c arbon NO(g) + monoxide CO 2 react according to the following equation: (g) 1 ΔH = −226 kJ mol C alculate b. State the enthalpy acid for change the for the reverse reaction of NO in 2 reaction. the [1] atmosphere to deposition. [1] 3 3 Powdered an zinc insulated was reacted beaker. with 25.00 Temperature was cm of plotted 1.000 mol dm against copper(II) sulfate solution in r 7 . time. 40 Y 30 25 2 4 6 8 10 time / min a point State the d. To in f. Predict, giving a experiment chip each made reason, would (crisp) experiment copper(II) c alculate were when how m the using nal with ignited, = and E the the was [1] c arried was used out ve times. but the mass of of [1] energy released. 1 values enthalpy of c = for 4.18 J g m and reaction the theoretic al the ame was 1 K . c. [1] c alculated from value. used to [1] heat a test 7.8 1.2 21.3 Final temperature / °C in assumption water. Initial temperature / °C data one was chosen line. sulfate amount these M ass of chip / g C alculate the reaction [1] [1] reason, if zinc or copper(II) sulfate 25.00 g and M ass of water / g a. a of beaker. used, and outline the with the change extrapolated trial. to compare was be in a containing used enthalpy the of Suggest, l potato tube for time. c alculation v O A the reaction, concentration was assumption f x this an in should on placed i State used that was t a e. excess of the line. u values and each q = mcΔT c alculate (red) temperature enthalpy volume in to zinc represents. temperature this r o The the dierent be formula used extrapolated maximum powdered o should the graph d was The 8. the choosing same zinc the temperature on determine The which on n made at Y C maximum time U c. the o State what point v Estimate b. i n a. y 0 p r e 20 O t i s ΔT 35 n y C° / erutarepmet 45 at l 50 y P 55 The e produce the equation s s a. heat table 22.6 required, above and in kJ, from to raise section the 2 of temperature the data of the water, using booklet. [1] 1 b. Determine the enthalpy of combustion of the potato chip, in kJ g . [1] 403 Reactivity 1.2 Energy cycles in reactions The utilization of energy energy to its consume lives. Thermodynamics interconversions. states another that and energy that the c an total rst nor of to law of converted amount law is oen According vast is the The be This Agricultural, energy another. of for the for most c alled the this destroyed; account from to it This law, c an only means and law of conservation of energy. energy be that quantify fundamental c an for all enthalpy change for a reaction is independent of the pathway in and thermodynamic uses enthalpy of a data to reaction U changes in of law, the used of or , f data are Hess’ s law enthalpy of enthalpy change Born–Haber Hess’ s the data c alculate Reactivity 1.2.5 — A applic ation applic ation formation C i n combustion of ⦵ ΔH cycle is an to show energy formation of an ionic compound. n o i (Reactivity 1.2.1) When to a chemic al create chemic al u chemic al when l a E v O f x t a r o d Bond-breaking and bond-forming reaction products. bonds of made to place, bonds form in the bond-breaking is an chemic al energy understanding takes Chemic al are bonds: new transfer bonds between the energy the atoms the reactants products. of the system and the involved in reactants are Energy endothermic form: bond-making is an changes the broken is required process. chemic al is an rearranged new to Energy exothermic surroundings a are and break the is released process. This essential part of reaction (Reactivity 1.1). Laws and theories The nor law of conservation of energy destroyed; predictions power. 404 science. c alculations. o v Reactivity 1.2.4 — An between the initial and nal states. c an enthalpy changes of formation, p used , form we LHA the c one reaction O that of y r e states ΔH created from energy changes. This is one ⦵ combustion, law given principles Reactivity 1.2.3 — Standard energy. Reactivity 1.2.2 — Hess’s neither t i s Reactivity 1.2.1 — Bond-breaking absorbs and bond- releases a the remains constant. Understandings forming be converted n system our l form given to all y a central activities daily. and thermodynamics one is to What it c an be only made other laws be but, unlike have states converted you that theories, come energy between laws across in c an dierent do not be neither forms. have created Laws allow explanatory your study of chemistry? y study of energy domestic P amounts of and r industrial, e during reactions? s s How does application of the law of conservation of energy help us to predict energy changes Reactivity 1.2 Energy cycles in reactions Bond enthalpy Imagine a molecule (BE) is The dened in process between the of energy of new required gaseous homolytic two to covalent ssion species. . The break results one the in of mole of under electrons the the hydrogen energy. The molecules distributes This breaking requires bond enthalpy bonds by homolytic standard conditions. from formation of the bond radic als, equally indic ated symbol. example, the mole of the the 2 atoms the following bond enthalpy of the H–H bond is equal to the enthalpy change r For • as molecule, H hydrogen reaction: 2 (g) → is an ∆H = endothermic +436 kJ mol process, and therefore it has a positive the breaking the C–H values booklet. of bond are provided Bond the enthalpy same enthalpy bond will in in vary table values a 1 (on are wide the variety through next derived the of page) from and experimental data on compounds. alkane series For bec ause molecule methane This is bec ause 3 series chemic al of steps enthalpy for environment •CH 3 (g) in which each → •CH (g) → •CH(g) → •C(g) + + + •H(g) BE •H(g) BE •H(g) BE dierences changes (C‒H) 2 (C in the changes 3 4 the removal of 1 1 H) = +462 kJ mol (C H) = +424 kJ mol (C H) = +338 kJ mol 1 products for upon = +439 kJ mol each of each equation 1 and explain successivestep? t a r o i environment 1 hydrogen o the (g) BE d recognize chemic al 2 •H(g) one removal would be n (g) + U 2 → •CH(g) the a bond C (g) •CH you the the i n 4 •CH how underwent time, hydrogen atoms. CH C an a o successive at v dierent. of removed p was chemic al y a atom example, the r e environment of the individual bonds changes. If section 12 of O data t i s BE the n y enthalpyvalue. Selected l Bond-breaking 2H•(g) y P 1 H e by one hydrogen individual s s ssion simple into Bond enthalpies are average values and are therefore only an approximation. The average bond enthalpy value of the C u l f x H H H H C H a C H C H v O H 1 H bond is +414 kJ mol C C H C E H H H H Figure 1 H When you examine the structure of this hexane molecule, chemic al environment of the C–H c an you see how the bonds diers throughout the molecule? 405 Reactivity 1 What drives chemic al reactions? Average bond Average bond Bond Bond 1 1 enthalpy / kJ mol C =O 804 144 C =N 615 O=O 498 C ≡N 890 O−H 463 C−H 414 N =N 346 N ≡N C−O 358 I natural What sciences? used in do further do play we areas of n U H data models How other chemic al bonds reactions formed. c an in be the c an These obtained whether be explained explanations to acquisition determine C models and role i n model. data in and o empiric al changes broken y energy bonds p of v model 151 298 K r e Measured I O enthalpies at TOK 193 t i s Average bond 242 a of modify by the are and based rene the knowledge in the model is useful? How are knowledge? You can use bond enthalpy data to calculate the enthalpy change of reaction, ΔH: C H ΔH = Σ(BE H Figure 2 H 2 and A molecule of bromoethane, Br, has one C 5 ve C C bond, one C Br bond When data value, do when you of as not formulas c an bonds a take bond of – Σ(BE c alculated enthalpies are intermolecular in a reaction enthalpy the determine broken) reaction bond substances performing structural type and and bonds from formed) bond average enthalpies values. will dier Additionally, bond forces into account, which is particularly are solid or liquid. c alculations, reactants the of it is products number of useful to involved each draw out the full in the chemic al reaction. bond From present. l a E v O Models The model of bond changes taking to this which theoretic al evidence. central 406 change actual important this, H bonds its enthalpy u C f x enthalpy from o H The i Br t a r o d C of breaking place model values Using within is in empiric al in formation reaction agreement generated methodology and a from data to science. is used system. with bond We empiric al energy replace or to represent c an assess data, data modify the the energy extent by comparing the with experimental proposed models is a y Br n Cl Br 945 l Cl 839 470 r 614 C≡C 158 y C=C Table 1 H N P C N e 436 s s H−H O−O C on enthalpy / kJ mol Reactivity 1.2 Energy cycles in reactions Worked example 1 Using bond C 2 enthalpy data, H 4 (g) + HBr(g) determine the enthalpy change of reaction for the following process: → C 2 H 5 Br(g) s s Solution Start by drawing the full structural formulas of all of the H H bond. values Br)] − [5BE(C H) + 2701 gas to data formed. changes (g) → this gas 2HCl(g) c alculate on c an CH be the 4 (g) + H enthalpy 2 O(g) change, −185 kJ mol bond number formed → O l Determine the the steam: there as are ve C follows: Br)] H is combined with 1 reaction is u with gas hydrogen chloride: depend Hydrogen f x gas 2 for is used to make chlorine chlorine 3H 2 + Using section Linking question 1 in kJ mol , for the How and type of bonds by the reaction of expect bond to relate to bond length and polarity? (Structure 2.2) How CO(g) this you data natural reaction, in kJ mol does the strength of a c arbon–halogen 1 for would broken and enthalpy industrially (g) ΔH, . enthalpy, t a Enthalpy H booklet, H–Clbond. 2. + process the o the (g) of i of 2 change r o 12 enthalpy produce industrial Much d Cl The an product, reaction n hydrogen is hydroxide. the o process sodium U chloralkali BE(C C i n Practice questions and + v 1 In the y 2636 gas C) for p = The BE(C bond. r e [(4 × 414) + 614 + 366] − [(5 × 414) + 346 + 285] = −65 kJ mol 1. Br change from the data booklet: = − enthalpy n BE the O Then, substitute in the Therefore, Σ(BE of bonds formed) H) + BE(C=C) + BE(H Br y = [4BE(C H t i s ΔH = Σ(BE of bonds broken) − Br C H H H bonds, one C=C bond and one H C bond, and one C H C y four C H l are Br e there H + H C r reactants, products: P the bonds, one C and H C In reactants bond aect the , using rate of a nucleophilic substitution 3. Methane and chlorine Using the c. Deduce E Write b. v chloride. a. reaction? (Reactivity 3.4) react balanced bond enthalpy a section 12 of the data booklet. change produce chemic al enthalpy and to for explain values this chloromethane equation for this and hydrogen reaction. from the data booklet, determine the reaction. whether this reaction is exothermic or endothermic. d. State which are more energetic ally stable, the reactants or theproducts. 407 Reactivity 1 What drives chemic al reactions? Hess’s law (Reactivity 1.2.2) If you have Delhi, and travelled have transport one the millions way fun is to New York experienced to of travel oen the people between working out City, Shanghai, subways every point the that day. A In and fastest Paris, Seoul, London, M adrid or criss-cross these enormous cities any B. If transport you are network an there adventurous is more s s than half you traveller, route. e r l in Seoul provides many alternative y r e same net added true c an together, example, the 3 (l), a is chemistry. summary result overall trioxide, SO for be in an reaction as for follows: 1: S(s) (l) usually dierent shows reactions, which, reaction. oxidation of elemental sulfur, S(s), to dicult to + O 2 (g) 2 c arry (g), → SO 2 out and in one then step. sulfur Typic ally, sulfur is dioxide is oxidized further (g) SO 2 (g) O + 2 (g) → SO 3 (l) 2 up of is 1 2: the would basis enthalpy be Hess’s the changes same as the that occur enthalpy in each of these individual steps, change of the overall reaction. This is law: Regardless of the route by which a chemic al reaction proceeds, the statesof the system are the same. z Hess’s law Hess’s law: is an the applic ation enthalpy of the change conservation for reaction A of energy → B (ΔH x ) law. is Figure 4 illustrates equal to the sum of Hess’s law the enthalpy Hess’s law reaction 408 3 dioxide, SO o E Figure 4 t a ∆H trioxide: add total SO equation of enthalpychange will always be the same, as long as the initial and nal a C B l x v O ∆H u the y sulfur i d r o f x ∆H you → process to Step the A this rst Step If (g) n U oxidized sulfur 2 the chemic al number 2 However, to O + A a overall 3 S(s) of C sulfur be it i n For c an o when idea reaction – v the p The n The extensive metro system O Figure 3 pathways for navigating the urban sprawl changes c an by be for the applied combining to other reactions A nd the → C and C → B (ΔH unknown reactions with enthalpy known y + ΔH change for z a ). given enthalpy changes. y P y t i s Reactivity 1.2 Energy cycles in reactions Worked example 2 The reaction for the formation of methanol is shown below: 1 C(s) + 2H 2 (g) O + 2 → CH (g) 3 OH(l) Use Hess’s law and s s 2 equations 1–3 to c alculate the enthalpy change for the reaction above. 3 1 3 OH(l) O + 2 (g) → CO 2 (g) + 2H 2 1 O(l) = −726 kJ mol ∆H = −394 kJ mol ∆H = −286 kJ mol 2 1 C(s) + O 2 (g) → O (g) CO 2 (g) r 2 P 1 3 1 H 2 (g) + 2 → H 2 O(l) 2 are two methods for using Hess’s law for a question like this: the summation of equations method, and the Summation of equations method inspect the the reactants. methanol that reaction involve C arbon is a c arbon, reactant two moles of hydrogen as coecients. a reactant. This 2H 2 → CO (g) means (g) + O 2 2 (g) the → 2H equation ΔH = equation enthalpy 2 you need reactants, c an be used −394 kJ mol 3 value O(l) ΔH c an be must = used also to and 1 (g) Therefore, that this as U C stoichiometric 2 so c alculation, oxygen o require doubled O your and equation2, i n You + In v C(s) in scheme. hydrogen be use in enthalpy y with of reactions methanol as p Start formation to r e corresponding O t i s enthalpy cycle diagram method First, n There l y Solution y ∆H e CH the as a values product. written: direction as written, but with doubled: 1 −572 kJ mol For the product a methanol, reactant. you When need 2 O 2 2 (g) (g) + together, O + + + 2H 2 2 2H O(l) → CH c ancelling (g) O 2 chemic al u CO 2 3 However, equation, OH(l) the the equation sign of must the v the common are reversed to make methanol a change must also be changed: 3 1 O + 2 (g) ΔH = +726 kJ mol 2 the species common to both sides of the reaction, and add the 1 → CO 2 (g) ΔH = −394 kJ mol 1 → (g) 2H 2 O(l) ΔH = −572 kJ mol ΔH = +726 kJ mol ΔH = −240 kJ mol 3 1 2 → O(l) CH 3 OH(l) + O 2 (g) 2 3 1 2H 2 + 2 + 2 O → 2 + 2 O + CH 3 OH + 2 2 1 species gives the overall equation C + 2H 2 O + 2 → CH 3 OH, which is identic al to the reaction 2 –1 Therefore, omitted be enthalpy 2 question. states 2H + a + E the C(s) 1 C C ancelling the equations (g) l f x O Total in three 2 equation1. the t a the values: r o add enthalpy use i CO Now to reversing o not d product, n We also need half a mole of oxygen. Oxygen is present asa reactant in two of the above equations, so ignore this for now. to the save total enthalpy change for the reaction is equal to –240 kJ mol . Note that here and below space. 409 Reactivity 1 What drives chemic al reactions? Enthalpy cycle diagram method You c an cycle represent the c alculations needed to determine the enthalpy change of reaction in the form of an enthalpy diagram. 1 write the equation you are trying to nd the enthalpy change for: C + 2H 2 O + 2 → CH 3 OH. Below this, write the s s First, 2 other species included draw equations enthalpy two cycle arrows 1, 2 and 3 not included in the original reaction: CO + 2 2H 2 O. Oxygen, O 2 , is usually not diagrams. from the reactants to the species at the bottom to represent 1 equations 2 and 3. Draw an arrow from the products to the species at the C bottom to + 2H 2 + O equation 1. the arrows method, the with the enthalpy enthalpy change change for values. equation 3 As with the summation of needs to be doubled to give 394 value for two moles of hydrogen. The resulting enthalpy cycle diagram is 286 726 shown in l the × 2 Then, calculate the sum of the enthalpy change values, following the alternative and 2H 2 t i s pathway from the reactants to the products, via the CO O intermediates. If the 2 2 + O CO 2H pathway you are following is in the opposite direction of an arrow (in this case, the last arrow), reverse the sign of the enthalpy change. This calculation is summarized below: (−286 × 2) + (+726) = −240 kJ mol your method reasoning calculations: the summation of equations method and the make enthalpy cycle diagram method. In what situations would reaction. arithmetic al you use an enthalpy cycle? In what situations would you rather It than U the strengths and limitations of each method? T o what extent are where errors advisable simply examiner the prefer, in make sure assessments. when to c alculating clearly recording the opportunity to show nal you Oen show c andidates the enthalpy of your full working answer. This gives assign partial marks n use the summation of equations method? What are the is you clearly C i n methods for representing and solving Hess’ s law o v Whichever In the previous worked example, you have seen two Enthalpy cycle diagram y Communic ation skills Figure 5 p r e ATL + O 2 for the formation of methanol 1 (−394) n y gure 5. OH 3 the methods dierent if they represent the same concept? applic able. o i 4. C alculate the summation enthalpy of change equations for method, the and following reaction, using the equations 1–3 below: 1 3H u l a E v O f x t a r o d Practice questions 2 (g) + 2C(s) O + 2 (g) → C 2 H 5 OH(l) 2 1 1 C 2 H 5 OH(l) + 3O 2 (g) → 2CO 2 (g) + 3H 2 O(l) ΔH = −1367 kJ mol ΔH = −394 kJ mol ΔH = −286 kJ mol 1 2 C(s) 3 H + O 2 (g) → CO 2 (g) 1 1 2 (g) O + 2 (g) → H 2 O(l) 2 5. Determine enthalpy the cycle 4NH 3 (g) + enthalpy diagram 5O 2 (g) change method for the and following equations → 4NO(g) + 6H 2 reaction, using the 1–3 below: O(l) 1 1 N 2 N 3 2H 2 (g) + O (g) + 3H 2 (g) → 2NO(g) ΔH = +66 kJ mol ΔH = −92 kJ mol ΔH = −572 kJ mol 1 2 2 (g) → 2NH (g) → 2H 3 (g) 1 410 2 (g) + O 2 2 O(l) y label equations P Finally, CH 2 r 2 represent e Then, in from Reactivity 1.2 Energy cycles in reactions Using Hess’s law to nd enthalpy change When heated, decomposes and water. decomposition determine the then the is The enthalpy dicult enthalpy apply Hess’s decomposition of 6. hydrogenc arbonate potassium to change determine change law c arbonate, to for nd potassium two the (Note for this directly. other Repeat the procedure, hydrogenc arbonate c arbon You will that reached this time, bec ause using instead record the potassium of potassium the lowest reaction is endothermic.) reactions, enthalpy change Analysis for Part 1: Known enthalpy changes of reaction hydrogenc arbonate. Write a balanced equation, r 1. including Relevant skills for Inquiry Record, propagate 2: out and express uncertainties a. potassium c arbonate b. potassium hydrogenc arbonate acid C arry relevant and accurate data processing 3: Evaluate weaknesses, the implic ations limitations and of methodologic al 2. Show 3. Process that enthalpy assumptions on the conclusions a. the your acid is in excess change potassium the Anhydrous • The effervesces. prevent loss between Cover of the i n reaction c arbonate is an irritant. c arbonates and acids reaction reagents and v potassium 3 • 2 mol dm • anhydrous • potassium hydrochloric acid hydrochloric c alorimeter. Add stir the 5. the the Dispose and potassium mixture, of of exact transfer the it to your reaction a hydrochloric acid and hydrochloric acid thermal decomposition of potassium Apply Hess’ s law, and your processed data from part1, 8. Look up change Record of the for the enthalpy theoretic al the determine 9. Identify the two 10. Identify error in and this online. percentage assumptions investigation values value for decomposition hydrogenc arbonate coffee-cup temperature. and Propagate uncertainties to determine the absolute of reached. mixture hydrogenc arbonate 7 . c arbonate to the acid solution and the c arbonate potassium decomposition of potassium hydrogencarbonate. it temperature of the acid. monitoring potassium • to determine the enthalpy change for the thermal c arbonate into a 30 cm into • hydrogencarbonate 6. mass. collect hydrochloric Draw an enthalpy cycle connecting the three reactions: • 3 to temperature the Show and Write a balanced equation, including state symbols, for uncertainty potassium the cylinder Record maximum 3 g Record acid E 4. 5. o boat. measuring practic al. between: hydrochloric acid to produce potassium carbonate and other products. before starting. a a v O Use i approximately weighing 3. approval the u Place for for l f x teacher required and the decomposition of potassium hydrogencarbonate through the method and write a list of equipment 2. (solid) (solid) t a r o Read c arbonate hydrogenc arbonate Method 1. d potassium 4. n U Materials c ases. Part 2: Applying Hess’s law vessel with a lid to spillage. reaction hydrogenc arbonate p Hydrochloric acid is an irritant. • protection. hydrochloric o • eye C Wear both y potassium for c arbonate acid • in and experimental data to determine the r e b. S afety hydrochloric acid O Inquiry t i s • and symbols, y • C alorimetry state n 3: between: l 1: Tool reaction P Tool • the y • c arbonate. temperature e and produce s s dioxide potassium to and discuss explain two of Use error made their major you the have obtained. enthalpy potassium this of value to your result. throughout this validity. sources of systematic investigation. appropriately. 411 1 What drives chemic al reactions? LHA Reactivity Standard enthalpy changes of combustion, ⦵ ΔHc ⦵ , and formation, ΔHf (Reactivity 1.2.3 and Reactivity 1.2.4) must describing be a standard enthalpy change, there are a number of steps that s s When completed. Write 3. Include The standard enthalpy change of combustion , that occurs an the type equation reaction. describe symbols in the the reaction. r state of to equation. P ⦵ when standard The a substance standard conditions: state in of 25.00 °C its a ΔHc pure temperature, gas hydroc arbon dioxide, and a CO 2 , undergoes and component as of liqueed follows: 13 C H 10 (g) O + 2 (g) 2 standard section the are data equation of 4 H 10 the (g) O + 2 2 (g) the (g) of with included exothermic is 2 H their of in 10 burned the form that it (LPG). for on to produce The equation for its 1 , is −2878 kJ mol other substances, . This are given standard states. c an also equation. reaction, included oxygen ⦵ ΔHc butane the with , is a highly ammable O(l) butane, combustion value gas + 5H for reacts 4 combustion along value indic ates an o d so the be The energy is written with negative released into the product side: 1 → 4CO 2 (g) + 5H 2 O(l) + 2878 kJ mol 2 i Activity t a r o f x Write equations to describe the standard enthalpy change of combustion for the following compounds, and state the enthalpy change values by referring u l a E v O 412 for 13 C booklet, Therefore, Reactivity 1.1. 4CO combustion combustion change surroundings. introduced in the petroleum enthalpies n U reactions of chemic al enthalpy enthalpy Exothermic and endothermic 14 of standard → it Butane, C C The enthalpy the i n in and O. is is Pa. o value, v The 2 substance p 4 1.00 × 10 combustion, water, H enthalpy change y is of r e combustion pressure the state O c arbon a t i s When a and is (298.15 K), which is taken as being 5 room , standard n under oxygen. of y takes in mole l completely one to section 14 of the data booklet. Ensure that you include state symbols and that the equations are balanced: a. octane, C b. cyclohexanol, C c. methanoic acid, CH d. glucose, C e. chloroethane, C 8 H 6 18 H 12 6 O H 12 O 2 O 2 6 H 2 Cl (hint: a strong, corrosive acid is one of the products) 5 y Determine 2. e 1. Reactivity 1.2 Energy cycles in reactions LHA Global impact of science Much of the world’ s hydroc arbons. will sources enable the fuel high are do and make with Industrial set in investigated costs factors decisions by international as a possible new processes the are alternative Glasgow. this associated measure researching are with c arbon- technology rarely totally every method of environmental impact of most about important what fuel for a fuel to be sources are used to produceenergy? change that occurs are there argon, Ar, and Hg, uorine, F and Two 2 (g) change of is → of elements butane 4 are are a solids standard liquid substance in their are H 10 by c arbon, the C, (g) formation of and following butane, energy for state, example standard conditions: hydrogen, H 2 . Therefore, equation: ⦵ ΔHf the standard conditions, under is from its constituent 2 described C under of formed , is C 5H Br elements gases 1 −126 kJ mol . This value, n U enthalpy butane + . Most are , is i n 4C(s) 2 bromine, elements of states. that substance o constituent formation The standard elements a v The the 11 of p mercury, their mole y in but one r e elements when O standard enthalpy change of formation, ΔHf t i s ⦵ The n should we which targets COP26 problem is greenhouse gases. This y Who being one emissions. x How sources, sustainable? is and of l fuel emission Paris environmental production. dierent in However, level of NO emissions reach ammonia source. there to by the combustion of community y energy of harmful COP21 produced P “green”: as is scientic r is such combustion neutral reduce the e The to governments agreements, supply s s energy energy Increasingly, and other values for the standard enthalpy change of formation for many common compounds, of these c an be compounds 8 notice of to that 6 O when combustion necessary correct H v will 2 to use Cl writing and balance fractions 3 the to the data the equation. the The enthalpy enthalpy change change of formation benzoic acid, C e. c arbon f. methylamine, CH equations fractional This nal is 6 H 3 of chemic al the few by for referring COOH CO NH standard stoichiometric one 5 monoxide, for standard states values d. chemic al formation, balance booklet. section. standard state a ethanol, C of o H chloromethane, CH changes be 3 13 same i propane, C E You the and l c. the section 13 of the data booklet: O b. describe in u a. to compounds, f x to equations following section t a the in given d Write are r o Activity found 2 enthalpy coecients may c ases when it is equation. 413 Reactivity 1 What drives chemic al reactions? LHA Under standard example, conditions, diamond, the element graphite and values. an c arbon exists as buckminsterfullerene. several allotropes, Allotropes of for c arbon ⦵ have dierent convention The allotropes of is ΔHf that the most If element stable has allotrope several is the allotropes, the normal standard state. The enthalpy c arbon and 1 of structures are formation is 0 kJ mol for all elemental substances in their standard states. discussed in Graphite is standard enthalpy taken as the standard for the c arbon allotropes, and so it has a Structure2.2 1 Would for you expect allotropes of an element, such as diamond ⦵ averaging the have dierent ∆H values? f (Structure 2.2) and graphite, strengths and this approach. n of to values. y limitations enthalpy l Evaluate bond Applying Hess’s law to enthalpy changes of combustion the of a enthalpy overall the of of the of i n the U is a direct reactants) applic ation of a reaction, reactants products ⦵ = Σ(ΔHc is equal so and the nal enthalpy values of change. ΔHr and to , c an be c alculated between the sum of the the the sum overall of the enthalpy enthalpy change ⦵ − Σ(ΔHc Hess’ s function, initial enthalpy dierence C ⦵ the ⦵ for The reaction: ΔHr This of state between this initial and nal state of a overall change combustion combustion a o of the data. is between pathway aect enthalpy v changes changes ofthe The not combustion enthalpy enthalpy dierence system. does standard enthalpy topic, the p using of reaction this is y The in r e chemic al earlier reaction O discussed change t i s As products) law. n Using Hess’ s i d formation law and enthalpy of combustion data, the enthalpy change of −1 of t a r o in the data pentane is determined to be −177 kJ mol , but the value given 1 booklet is −173 kJ mol . Suggest why the values aredierent. Applying Hess’s law to enthalpy changes of formation u f x The l a E v O 414 6. o Practice question standard enthalpy change enthalpy of formation data. changes of formation of formation of the ΔH r the reactants ⦵ for The is a reaction dierence products equal to and the the ∑(ΔH f also sum overall ⦵ = c an between be the of the enthalpy ⦵ products) − ∑(ΔH f c alculated using sum reactants) of the enthalpy enthalpy changes of change of the reaction: y rationale . e the 0 kJ mol P on of Linking question Thinking skills Reect formation r ATL of s s their Reactivity 1.2 Energy cycles in reactions LHA Worked example 3 C alculate the standard enthalpy change of formation of pentane, C 5 H 12 , using the enthalpy of combustion data from section 14of the data booklet. s s Solution Step 1: Write values Write from balanced equations chemic al for the equation combustion for of the formation c arbon, of 1 hydrogen mol of and pentane: pentane 5C(s) and nd + 6H their 2 (g) standard C(s) + O 2 (g) → CO 2 (g) 1 ∆H 12 (l) enthalpy change O + 2 (g) → H 2 −286 kJ mol O(l) equations 12 with + 8O either for the 2 the (g) → 5CO 2 (g) summation reaction. These of + 6H 2 O(l) equations methods ∆H method were formation required for pentane the that overall reaction involve equation. scheme. c arbon, Therefore + 5O 2 your you (g) c alculation, and should → 5CO 2 you pentane. need multiply (g) C arbon the ΔH enthalpy = to is cycle diagram method to earlier in this topic. a use enthalpy reactant of in values equation combustion p 5C(s) In hydrogen enthalpy y are of reactions r e combustion or introduced Summation of equations method the 1 = −3509 kJ mol O change (l) t i s enthalpy H n these the 5 y Use C l ⦵ 3 y (g) = 1 ∆H 2 −394 kJ mol 2 the H P 1 H = ⦵ 2 Inspect 5 r ⦵ determine C section 14 of the data booklet. 1 Step 3: → e Step 2: a of corresponding to 1, but c arbon by ve moles ve: 1 −1970 kJ mol v a the product reactant. As a pentane, result, you the should enthalpy the three equations use 3O + 6H together, (g) → equation change (g) 2 sign 2 O(l) 6H 1, but C c ancelling 5 H 12 the 2 + 6H common the reaction is for 2 species 2 (l) + 8O 2 (g) O(l) + 2 + gives the to be reversed to make common pentane 1 to both sides of the reaction, and add the ⦵ 5CO 2 (g) ΔH c 1 = −1970 kJ mol = −1716 kJ mol ⦵ 6H 2 O(l) ΔH c 1 ⦵ → C 5 H 12 (l) + 8O 2 (g) ΔH 1 = +3509 kJ mol ⦵ → 2 2 overall equation 5C(s) + + 2 + 6H 2 (g) C → 5 H C 5 12 H + 12 (l). 2 1 ΔH Therefore, = the −177 kJ mol total against the to the sum E reactants the diagram of the kJ mol ΔH is 5C(s) + is on the enthalpy right. change Oxygen, O values, 2 (g), is following via the CO 2 (g) and H 2 O(l) the included arrows intermediates. in 6H (g) C 2 H 5 (I) 12 enthalpy If from the you 3509 are going 5 × of an arrow, reverse the sign of the enthalpy ( 286) change. ( summarized 394) below: ⦵ ΔH not 6 × products, direction c alculation a diagrams. C alculate This cycle v cycle enthalpy 1 −177 l O enthalpy product, not = +3509 kJ mol Enthalpy cycle diagram method The a positive: ΔH → (g) u f x the change 2 6H + 2 C ancelling + must negative o + (g) 3O → (g) t a 5C 2 + 2 1 = −1716 kJ mol c equation from species i 2 5CO (g) 5O the ΔH ⦵ → d r o 6H Total + ⦵ O(l) changes values: 5C(s) 2 n add enthalpy 2 + U 5CO Now (g) C For 2 i n 6H o You also require six moles of hydrogen as a reactant. Equation 2 should be used, with the enthalpy change multiplied by six: 1 = 5 × (−394) + 6 × (−286) + 3509 = −177 kJ mol 5CO (g) + 2 6H O(I) 2 415 LHA Reactivity 1 What drives chemic al reactions? Worked example 4 C alculate the enthalpy of combustion of pentane, C 5 H 12 (l), using the enthalpy of formation data from section 13 of the data booklet. s s Solution Step 1: Write a balanced chemic al equation Write enthalpy for change the formation values from 5 H of 12 (l) + 8O pentane, 2 (g) → 5CO c arbon 2 (g) + dioxide, 6H and 2 O(l) water from their elements and nd their r standard equations section 12 of the data booklet. + 6H 2 (g) → C 5 H 12 (l) ∆H 1 f −173 kJ mol 3 H + O 2 (g) → CO 2 (g) ∆H = −394 kJ mol = −286 kJ mol 1 f 1 ⦵ 2 (g) O + 2 (g) → H 2 O(l) ∆H 1 f 2 these the equations enthalpy with change either for the the summation of equations method reaction. Summation of equations method c alculation, must be you and reversed. need water. As a to use Pentane result, the enthalpy is a product in an → equation formation of 2, c arbon 5C(s) + 5C(s) but + ve dioxide 5O 2 (g) sign → 6H changes 2 (g) moles by to the equation, from a negative 2 for ⦵ ∆H the diagram method to that equation involve 1. pentane, Therefore, equation positive: 1 required (g) to in = +173 kJ mol ve: 5CO reactions product ⦵ ∆H are formation but overall U C of (l) change overall o is enthalpy 12 corresponding the v the H in i n dioxide multiply 5 values reactant enthalpy C C arbon a cycle p 1 dioxide enthalpy y your c arbon r e In or O Use determine t i s Step 3: l C(s) y 2 n = ⦵ equation. Therefore, you should 1 = −1970 kJ mol f 6H the three equations together, values: 6H 2 C 5 H 12 + 5C + (g) + 12 + 2 3O 5O 2 → → (l) 5O (g) c ancelling 2 + 6H the 2 O(l) ∆H species 6H 2 + 3O 1 = −1716 kJ mol to both sides of the reaction, and add the ⦵ 5C(s) + 6H 2 (g) 1 ∆H = +173 kJ mol ⦵ 5CO 2 (g) ∆H f 1 = −1970 kJ mol = −1716 kJ mol ⦵ → (g) ⦵ f common → (g) 6H 2 O(l) ∆H 1 f ⦵ → l u f x Total H 2 t a 5C(s) 5 3O i r o C + o add enthalpy (g) d Now 2 n You require six moles of water as a product. Equation 3 should be used, with the enthalpy change multiplied by six: 5C 2 + 6H 5 12 2 + 5CO (l) + 8O + 6H 2 2 O (g) → 5CO 2 ∆H (g) + 6H 2 1 = −3513 kJ mol O(l). Therefore, the total 2 1 a O enthalpy change for the reaction is−3513 kJ mol ΔH Enthalpy cycle diagram method enthalpy from If the you are enthalpy the sum reactants going of to the enthalpy the change products, via values, the following C(s) and H 2 (g) + O 5CO (g) 2 the This the direction c alculation is of an arrow, summarized reverse the sign of the 5 ( below: (+173) + 5 × (−394) + 6 × (−286) = −3513 kJ mol 5C(s) + 416 × 394) 1 = (g) 2 + 6H 6H (g) 2 O(I) 2 arrows intermediates. ⦵ ΔH (I) 12 6 × ( 286) 173 against change. H 5 diagram is on the right. E C alculate cycle v The C y 5C(s) P ⦵ 1 e C Step 2: for the combustion of 1 mol of pentane: Reactivity 1.2 Energy cycles in reactions LHA Practice questions 7 . Calculate the enthalpy change of formation of propanone, CH COCH 3 9. Consider the following enthalpy cycle diagram: (l), using enthalpy change of combustion data 3 Step 1 from section 14 of the data booklet. the summation enthalpy cycle of CH CH (g) 2 + H CH 2 3 equations method and the diagram method to support your 7 answer. 7 + O (g) 2 gas c an be made using the 2 (g) 2 2 Hydrogen O e + 8. (g) 3 s s Use following 4 (g) + Step 2 H 2 O(g) → 3H of the enthalpies of formation, + CO(g) booklet lists the standard 2CO ⦵ ΔH , f for some of the (g) + 2 species 3H y data 2 13 (g) P Section r Step 3 reaction: CH O(I) 2 l ⦵ in the reaction above. b. Determine no a. value is listed for H 2 ⦵ the using formation value of the ΔH 1 , standard step in kJ mol , for the 2 the using standard b. Determine enthalpy change of enthalpy the standard enthalpy of step 1. values. c. Suggest one reason why the O utline the of c alculated enthalpy change of ΔH from bond enthalpies is accurate lessaccurate. with some you of will of propan-1-ol is the compare methods this by Inquiry the C alculate 3: Compare accepted 3: and the and of context discuss random percentage outcomes scientific Identify systematic interpret errors a E v O l u f x Inquiry and an each of the the a. Refer b. Conduct c alculations to i • 3: algebraic Determine o and 1. reason why c an it c an be , value of considered be considered enthalpy of combustion of propan-1-ol to following methods: average real (as you have a bond c alorimetry described in Record access the to values. experiment. This could be Reactivity 1.1), a relevant enthalpy suitable or simulated, if simulation software. measurement uncertainties error t a Tool • arithmetic r o • to Use problems d 3: solve one b ⦵ ΔH described in Relevant skills Tool Method value to n this topic. • and law in part p combustion task, , of o obtained this U those In C of 1 . v enthalpy i n standard −2021 kJ mol Hess’s approximate. Comparing ∆H values The using y reaction value r e this why ΔH change, c alculated ⦵ c. change, section 14 of the data booklet. O reaction C alculate (g). n why y O utline t i s a. and propagate them. investigation c. 2. Refer to Determine standard the enthalpies percentage error of formation. of each of the values sources and impacts of obtained in methods 3. Discuss and standard each a, of your b and explain enthalpy a, b the of and c differences combustion c alculated values of between the propan-1-ol and obtained by methods c 417 1 What drives chemic al reactions? LHA Reactivity Born–Haber Born–Haber cycles the changes an enthalpy ionic are compound. ions in of the gaseous and anity, IE, mole steps is of with in the lattice the state. sometimes cycle For further include and enthalpy from metal Hess’ s steps enthalpy, standard electrons of several one ions ionization These in ionization enthalpy change mole with law. involved of multiple energies occurs on the or positively dened: M(g) → M (g) + IE in of (g) → ionization M is (g) + e endothermic. endothermic. of For gaseous a solid , of is species, the an X 2 equal electrons + anity c an be – e electron anity one → of X anity found t a ionic the atoms EA of is helium value of to half ΔH that c an occurs process is is at equal to the of the bond enthalpy enthalpy change on the addition of one in the gaseous state: < 0 typic ally or change This energy > 0 at standard of (g) in negative, nitrogen. but Values of there rst are and exceptions, such second electron section 9 of the data booklet. ⦵ lattice enthalpy, occurs is mole non-metals o the to – X(g) i d r o when ∆H gaseous lattice ions , is are dened formed as the from standard one mole of enthalpy change that structural units of a solid lattice: u l f x a E v O 418 ΔH n U of Electron The ⦵ X(g) electron anity, EA, mole as → C The (g) enthalpy formed. > 0 is at ionization ⦵ the ⦵ ΔH rst o i n 1 is of ⦵ at (section 12 of the data booklet): 2 0 standard element species, ΔH diatomic > y gaseous IE p a M(g) v For → at r e M(s) ΔH atoms monatomic enthalpy of sublimation: 0 O mole > values t i s enthalpy of atomization, one The section 9 of the data booklet. ⦵ The when IE n process found M e l be : + + y The 2 2 rst, + MX(s) The process c an be L attice is found → M chemistry, gaseous ions ⦵ (g) ∆H lattice > Experimental 0 values of lattice enthalpy at 298.15 K section 16 of the data booklet. are oen formation the + X endothermic. in enthalpies exothermic – (g) lattice from of quoted the solid enthalpy the solid as negative lattice always lattice. from refers values gaseous to the that represent the ions. However, in DP endothermic formation of y : the P 1 charged electrons, + IE enthalpy of formation. valence are formation of energy, of that atoms cycles combine the r second, one applic ation e removal The electron Ionization energy, another associated s s atomization, cycles (Reactivity 1.2.5) Reactivity 1.2 Energy cycles in reactions ⦵ have already substance formed states. be as dened the under enthalpy standard Standard represented the a change of single enthalpy that conditions enthalpy by standard ∆H from formation its of constituent an ionic elements substance, 2 (g) → in such their as standard NaCl(s), c an equation: ⦵ Cl , of a f NaCl(s) ∆H − s s + formation, occurs when one mole of a substance is 1 Na(s) of LHA We 1 = −411 kJ mol f 2 correct use of subject-specic your knowledge terminology and is essential understanding of to your ability new concepts in encounter electron the Term key terms standard anity. in state, M ake a this section: lattice enthalpy, glossary that has an entry for Denition Equation ∆H + Standard lattice of gaseous add terms a to ions solid your from one lattice chemistry lattice atomization, and enthalpy of enthalpy formation be is (X) + (g) X(g) f x at 1 X (g) (g) 2 2 l a time. electron anity Born–Haber cycle. A e EA(X) e a v O IE(M) + u + gure 6. i ΔH in over energy, construct (g) t a r o + M shown ionization to X o cycle combined d Born–Haber of c an + C The (g) n cycle enthalpy, glossary U Using the Born–Haber M → M o to of i n continue generalized MX(s) v mole c an enthalpy change that occurs on the formation p enthalpy y r e ⦵ You each term, following: Symbol lattice standard enthalpy of atomization, ionization O and detailing several formation, t i s energy of n will enthalpy y You l chemistry. y demonstrate e to P The r Self-management skills ATL + M (g) + X (g) 1 + M(g) X E ΔH (g) 2 2 ΔH (M) at ΔH f 1 + X 2 Figure 6 (g) 2 A generalized Born–Haber cycle 419 Reactivity 1 What drives chemic al reactions? LHA The be unknown you know anities an value determined ionic values lattice each the enthalpy of The the before, pathway of you you goes of nd the step Much should like reverse the in in a cycle. ionization is to the sign of of the For the of sum enthalpy any arrow cycle c an example, if energies, enthalpy equal with the direction Born–Haber the standard formation steps. against any pathway atomization, c an enthalpy other for opposite enthalpy enthalpy, of change the electron formation of of the cycle enthalpy s s for constructed where of following compound. changes you the and by diagrams enthalpy changes in the cycle. e r Worked example 5 cycle in gure 7. mol Br (g) 2 2 p + 1 K +419 kJ mol 1 Br U C + K(g) (l) 2 2 (g) 1 x kJ mol ΔH lattice (K) +89 kJ mol n at ΔH f 1 o i d Br (l) 2 2 1 392 kJ mol Born–Haber cycle to c alculate the lattice enthalpy of potassium bromide Solution The u l a E v O f x t a r o Figure 7 + lattice enthalpy is the enthalpy change for the reaction KBr(s) → K (g) + − Br (g). In gure 7, this step is shown from the bottom right of the cycle, to ⦵ the middle right, labelled ∆H lattice . To nd the lattice enthalpy, follow the + alternative Add the pathway, values of going the lattice from enthalpy ⦵ ∆H ⦵ (KBr) = −∆H that arrow, your so you the answer bec ause it is KBr(s) clockwise change for around to K at + ΔH at −(−392) + (+89) + (+419) + (+112) + (−325) +687 kJ mol for an was the the reverse of the +392 enthalpy endothermic (g). 1 reversed: lattice Br (Br) + EA(Br) = followed + ⦵ (K) + IE(K) = sign − (g) each step: ⦵ + ΔH f – Note 420 Br 1 ΔH + + (g) o i n v IE (K) 1 325 kJ mol EA (Br) e (l) e y + K r e 1 n +112 kJ + O (Br) at Br(g) t i s 1 ΔH + (g) l y + K of an process. enthalpy instead of ionic of formation −392. reaction Remember compound as a to positive express value, y P Determine the lattice enthalpy of potassium bromide using the Born–Haber Reactivity F actors lattice 10. Interpret and utilize the values from the Born–Haber cycle in Energy aecting enthalpy the are cycles in reactions LHA Practice question 1.2 value of detailed in gure 8 to Structure2.1. c alculate oxide, the MgO. enthalpy MgO of has formation doubly for the ionic compound magnesium charged ions. s s 2+ Mg (g) + O(g) + 2e +249 kJ mol at EA 1 (O) 141 kJ mol (1) (g) + O (g) + 2e 2 2 (g) + O + (g) e y 1 IE (Mg) +1451 kJ mol (2) 1 (O) +753 kJ mol (2) (g) + O (g) + e 2 2 (g) + O (g) (1) O (g) 2 2 (MgO) +148 kJ mol ΔH (MgO) U f 1 + O 2 C (Mg) at i n 1 ΔH (g) 2 1 +3791 kJ mol lattice o ΔH v + p 1 Mg(g) y r e Mg +738 kJ mol (Mg) 2 2+ 1 IE O t i s EA 1 + Mg n l 2+ Mg y P 1 2+ Mg e (O) r 1 ΔH 1 n x kJ mol Figure 8 Born–Haber cycle to c alculate the enthalpy of formation i d t a r o u l f x a 2 The lattice structure of crystalline magnesium oxide: E v O Figure 9 o for magnesium oxide grey = Mg + 2– , red = O Linking question What are the factors that inuence the strength of lattice enthalpy in an ionic compound? (Structure 2.1) 421 Reactivity 1 What drives chemic al reactions? End of topic questions Using the Born–Haber 1. Using your answer knowledge the guiding from the question as Reactivity 1.2 fully as represents cycle the below, determine which lattice enthalpy of s s c alculation strontium 1 topic, chloride, SrCl 2 , in kJ mol possible: 2+ (g) + 2Cl(g) 698 of energy help us to predict energy changes during r +1064 reactions? 2+ (g) + 2Cl Sr l Multiple-choice questions Which 3 the N–H bond ? 3 (g) → N(g) 1 + 3H(g) 1 NH B 3 (g) Sr(g) + 2Cl(g) N(g) → + H(g) 3 D NH 3 (g) → (g) → 3 N 2 (g) H + 2 2 +242 Sr(g) + Cl (g) 2 Sr(s) + Cl enthalpies 2 (g) c an + be •H(g) used to c alculate the enthalpy of the following Referring of change of B (3 × 391) − (436 C −[(3 D −(6 + equation × 945) + (3 436) Na C NaCl(s) → Na → Na 391) 945] lattice enthalpy? + Na (g) + Cl (g) + (g) + Cl (g) + (aq) + Cl + (s) + Cl (aq) (s) SrCl (s) 2 A −(−829) B −829 deduce c alculation of the u → × + a NaCl(s) booklet, the 945] l B E v O 422 + represents NaCl(g) NaCl(s) + 945] [(3 A D → 436) (g) from bond enthalpies. r o Which 391) × data represents 3 o [(3 2NH i − × the ⇌ t a × 391) + (g) d (6 436) of 2 reaction A × 3H 12 following f x LHA 4. section the + n enthalpy to (g) kJ mol U which 2 in C N reaction, i n 1 change (g) + C −(−829) D −829 + + 164 164 + + + 829 + 242 242 164 164 lattice enthalpy +164 2 o Bond •NH v 3. 3 (g) 2 p NH y 1 C r e 3 +549 O NH (g) + 2Cl(g) enthalpy t i s A represents yplahtne inNH equation n y + Sr 2. + + + + 549 549 242 242 (g) + + 549 549 + + 1064 1064 + − 1064 1064 − − (−698) 698 − 698 (−698) y P Exam-style questions e Sr How does application of the law of conservation LHA 5. Topic review Reactivity LHA 6. The Born–Haber shown cycle for potassium oxide, K 2 O, is 1.2 Energy cycles Extended-response questions 7 . + The + O following production 2 (g) equation of describes the industrial ammonia using the Haber process. (g) N 2 (g) + 3H 2 (g) ⇌ 2NH 3 (g) s s 1 +612 kJ mol a. C alculate this + 2K the process enthalpy using change bond of reaction enthalpy data. (g) + 2e + O(g) and explain exothermic or whether the e State reaction is endothermic. 1 8. lattice Ammonia reacts with oxygen to steam. enthalpy a. Write a balanced form chemic al 1 nitrogen and equation for this showing + 2 O full structural (g) 2 bonds 1 c. Using data O(s) 2 this 9. represents the lattice enthalpy 428 −(−361) Octane, C −361 D −(−361) + 428 428 + 838 + + + 612 838 838 428 + − + 612 612 838 − 612 oxygen 8 H 18 U The , is a for all reactants and nature (l) for values the fuel, of from [1] the covalent [4] section 12 of the enthalpy change for [3] component as following of octane liquid is gasoline. When combusted in reaction: 25 O + bond change b. 18 the H Using the determine burned via C a. 8 is C C + + booklet, o + v −361 i n B ? enthalpy reaction. gasoline 1 bond p c alculation inkJ
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