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IB Diploma Chemistry HL Textbook.pdf

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CHEMISTRY IB Diploma Programme HIGHER LEVEL

for use with the

Lanna Derry Maria Connor Janette Ellis Faye Jeffery Carol Jordan Brian Ellett Pat O’Shea Sydney, Melbourne, Brisbane, Perth, Adelaide and associated companies around the world.

Pearson Heinemann An imprint of Pearson Education Australia A division of Pearson Australia Group Pty Ltd 20 Thackray Road, Port Melbourne, Victoria 3207 PO Box 460, Port Melbourne, Victoria 3207 www.pearson.com.au/schools Copyright © Pearson Education Australia 2009 (a division of Pearson Australia Group Pty Ltd) First published 2009 by Pearson Education Australia 2011 2010 2009 2008 10 9 8 7 6 5 4 3 2 1 Reproduction and communication for educational purposes The Australian Copyright Act 1968 (the Act) allows a maximum of one chapter or 10% of the pages of this work, whichever is the greater, to be reproduced and/or communicated by any educational institution for its educational purposes provided that that educational institution (or the body that administers it) has given a remuneration notice to Copyright Agency Limited (CAL) under the Act. For details of the CAL licence for educational institutions contact Copyright Agency Limited (www.copyright.com.au). Reproduction and communication for other purposes Except as permitted under the Act (for example any fair dealing for the purposes of study, research, criticism or review), no part of this book may be reproduced, stored in a retrieval system, communicated or transmitted in any form or by any means without prior written permission. All enquiries should be made to the publisher at the address above. This book is not to be treated as a blackline master; that is, any photocopying beyond fair dealing requires prior written permission.

Publisher: Editor: Text Design: Copyright & Pictures Editor: Project Editor: Typesetter: Production Controller: Cover Design: Cover image description: Cover image: Illustrator/s: Printed in China

Catriona McKenzie Marta Veroni Meaghan Barbuto Lauren Smith and Michelle Jellett Helen Withycombe Nikki M Group Aisling Coughlin Glen McClay Coloured scanning electron micrograph (SEM) of crystals of silver (symbol Ag, atomic number 47). Silver is a lustrous, white precious metal that has been known since pre-history. Its softness and malleability makes it useful in the production of jewellery and coinage. Silver has a high thermal and electrical conductivity, and so has been used in the electronics industry. Andrew Syred / Science Photo Library Nikki M Group

National Library of Australia Cataloguing-in-Publication entry Author: Derry, Lanna. Title: Chemistry: for use with the IB diploma programme higher level / Lanna Derry, Maria Connor and Carol Jordan. Edition: 1st ed. ISBN: 9780733993800 (pbk.) Target Audience: For secondary school age. Subjects: Chemistry--Textbooks. Chemistry--Examinations--Study guides. Other Authors/Contributors: Connor, Maria. Jordan, Carol. Dewey Number: 540 ISBN 978 0 7339 9380 0 Pearson Australia Group Pty Ltd ABN 40 004 245 943 Every effort has been made to trace and acknowledge copyright. The publisher would welcome any information from people who believe they own copyright to material in this book. Disclaimer/s The selection of Internet addresses (URLs) provided for this book/resource were valid at the time of publication and chosen as being appropriate for use as a secondary education research tool. However, due to the dynamic nature of the Internet, some addresses may have changed, may have ceased to exist since publication, or may inadvertently link to sites with content that could be considered offensive or inappropriate. While the authors and publisher regret any inconvenience this may cause readers, no responsibility for any such changes or unforeseeable errors can be accepted by either the authors or the publisher. Some of the images used in CHEMISTRY: For use with the IB Diploma Programme might have associations with deceased Indigenous Australians. Please be aware that these images might cause sadness or distress in Aboriginal or Torres Strait Islander communities. The publisher’s policy is to use paper manufactured from sustainable forests

CONTENTS 1 Atomic Structure

1

1.1 Ionization energies and electron arrangements 1.2 Orbitals, subshells and shells 1.3 Writing electron configurations Chapter 1 Summary Chapter 1 Review questions Chapter 1 Test

2 7 13 21 23 24

2 Bonding 2.1 Shapes of molecules and ions 2.2 Hybridization 2.3 Delocalization of electrons Chapter 2 Summary Chapter 2 Review questions Chapter 2 Test

3 Periodicity 3.1 Trends across period 3 3.2 First row d-block elements 3.3 Transition metal complexes Chapter 3 Summary Chapter 3 Review questions Chapter 3 Test

27 29 34 48 55 57 59

61 62 69 77 86 87 89

4 Energetics

91

4.1 Standard enthalpy changes of reaction 4.2 Lattice enthalpy and Born–Haber cycles 4.3 Entropy 4.4 Spontaneity Chapter 4 Summary Chapter 4 Review questions Chapter 4 Test

92 98 106 113 121 122 124

5 Kinetics 5.1 Rate expression 5.2 Reaction mechanism 5.3 Activation energy Chapter 5 Summary Chapter 5 Review questions Chapter 5 Test

6 Equilibrium 6.1 Liquid–vapour equilibrium 6.2 The equilibrium law Chapter 6 Summary Chapter 6 Review questions Chapter 6 Test

7 Acids and Bases 7.1 Calculations involving acids and bases 7.2 Acid and base dissociation constants 7.3 Buffer solutions 7.4 Acid–base titrations 7.5 Salt hydrolysis 7.6 Indicators Chapter 7 Summary Chapter 7 Review questions Chapter 7 Test

8 Oxidation and Reduction 8.1 Standard electrode potentials 8.2 Electrolysis Chapter 8 Summary Chapter 8 Review questions Chapter 8 Test

iv

127 128 145 152 160 161 164

167 168 174 185 186 188

191 192 201 208 216 227 229 236 238 240

243 245 255 272 273 276

9 Organic Chemistry

279

9.1 More functional groups 9.2 Reactions involving amines and nitriles 9.3 Nucleophilic substitution reactions 9.4 Elimination reactions 9.5 Reactions involving esters and amides 9.6 Reaction pathways 9.7 Isomers—the same but different Chapter 9 Summary Chapter 9 Review questions Chapter 9 Test

280 291 295 304 308 317 320 334 339 342

345

Solutions

351

Glossary

361

Index

368

CONTENTS

Appendix 1–6

CHEMISTRY: FOR USE WITH THE IB DIPLOMA PROGRAMME HIGHER LEVEL

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CHEMISTRY IB Diploma Programme HIGHER LEVEL

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CHEMISTRY: For use with the IB Diploma Programme Higher Level is the most comprehensive chemistry text specifically written for the IB Diploma Programme Chemistry course, Higher Level. The content is easy to follow and provides regular opportunities for revision and consolidation. All assessment statements of the IB Diploma Programme Chemistry syllabus are covered in highly structured and meaningful ways.

Coursebook includes Student CD

• Syllabus Assessment Statements given beside the relevant theory

• Comprehensive topic test of examination questions.

• stimulating photos and full colour illustrations to support learning of chemical concepts

Student CD contains:

• theory broken into manageable chunks for ease of learning • comprehensive exercises for ongoing review and consolidation

CHEMIST

• Review questions to revise all chapter content

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IB Diplom a Program me HIGHER LE VEL

CHEMIST for use with

• an electronic version of the coursebook

Maria Conn

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Carol Jorda

CD should launch automatical double click ly. If not, on ‘Setup ’. Consult the ‘ReadMe’ file on this CD for furthe r information.

• fully worked solutions to all coursebook questions

n

STUDENT

Orbitals

not fixed orbits. This is the electrons in orbitals, g Quantum mechanics places accurately (the Heisenber electron cannot be tracked within because the path of an cloud of negative charge The electrons are like a uncertainty principle). cannot be stated with certainty. position of the electron the orbital and the exact or electrons space in which an electron of region a is An orbital may be found.

CHEM COMPLEMENT Why s, p, d and f?

periodic orbitals of elements in the are used to name atomic g, h …; The letters s, p, d and f named alphabetically as are further atomic orbitals h orbital table. Theoretically there with electrons in a g or that is currently known however, there is no element

in the ground state. for their with no obvious reason appear to be rather random The letters s, p, d and f p orbitals, and d and f orbitals s orbital, dumbbell-shaped relationship to the spherical names. gives us the link to these of various shapes. History of lines in understand the patterns scientists were trying to on photographic film In the early 20th century, These lines were recorded metals. alkali of to the emission spectra these lines, scientists referred differing quality. In classifying the emission and they appeared with lines were also found on (meaning ‘fuzzy’). Other them as sharp or diffuse l lines were lines that only principal lines. The fundamenta spectrum of hydrogen— of the alkali metal. spectrum the on appeared to atomic orbitals: s, sharp; to give the letter names Scientists used these terms fundamental. p, principal; d, diffuse; f,

• Theory of Knowledge boxes, which allow easy integration of this requirement of the syllabus • ICT activities, which address Aim 7 for Experimental Sciences and are available on the Companion Website

Representations of methane

 SORBITAL $RAWTHESHAPEOFAN P ANDTHESHAPESOFTHEPX Y ANDP ORBITALS)"/ Z

vi 8

probability of finding gives us knowledge of the Schrödinger’s wave equation probability density or at a given instant. This an electron in a given position nsional axes. In any plotted on a set of three-dime of electron density can be in an s orbital. A graph energy electrons are found electron shell, the lowest for the lowest-energy distance from the nucleus of finding the probability against radial probability The shown in figure 1.2.2. –10 m) from the nucleus, electrons in an atom is × 10 distance of 0.529 Å (0.529 of 3 Å. electron is greatest at a lly until it is zero at a distance and then decreases dramatica spherical shape of the to three dimensions, the When this graph is extended ensity plot for an . This is seen in the electron-d of orbital can be recognized represents the probability The density of the dots s orbital (figure 1.2.3). more likely it is that an higher the density, the finding the electron. The that region of space. electron will be found in electrons also describes for the next lowest energy Bohr’s to The electron-density graph According orbital. orbital, this is also an s a sphere. Like the first energy level, or shell, so electrons are in the second model of the atom, these indicates the second energy the 2s orbital, where 2 this orbital can be named energy electrons in the type of orbital. The lowest level and s indicates the density graph. When the produce a spherical electron third energy level also energy levels have a compared, those at higher are orbitals s three spherical of finding an electron occurs. the greatest probability greater radius at which

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• a link to the live Companion Website.

• Chem Complement boxes, which engage students with interesting extension material and applications to Aims 8 and 9 for Experimental Sciences

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This leads to the 1s, 2s increasing radii.

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d as spheres of

and 3s orbitals being represente

y

Radial electron distribution z

x

7 8 0 1 2 3 4 5 6 (nm) distance from the nucleus

plot shows the a particular part probability of finding an electron in of the space surrounding the nucleus.

Figure 1.2.3 An electron-density Figure 1.2.2 The radial probability

graph for the

lowest energy s orbital.

y y z

y

z z

1s

Figure 1.2.4 1s, 2s and 3s orbitals.

x

x

x

2s

The radii of the spheres correspond

3s

to a 90% probability of finding an

electron within each sphere.

in the second energy of the remaining electrons The electron density plots The electron-density plots from the spherical s orbitals. The level are very different described as a dumbbell. on a shape that is best for these orbitals take there is a node at which and between the two lobes dumbbell has two lobes not surprising, as the is This zero. is an electron the probability of finding The dumbbell-shaped lobes contains the nucleus. region between the two the three p orbitals relates The difference between in figure 1.2.5 seen orbitals are the p orbitals. be can This the orbital is aligned. is aligned with the to the axis along which with the x-axis, a py orbital where a px orbital is aligned is aligned with the z-axis. y-axis and a pz orbital

WITH THE CHEMISTRY: FOR USE

PRAC 1.1 Modelling orbitals

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IB DIPLOMA PROGRAM

STRUCTURE CHAPTER 1 ATOMIC

• focus on the IB Additional Higher Level (AHL) Diploma Programme Chemistry syllabus, topics 12 to 20

• Chapter summary, which includes chapter glossary and key points

probability

Each chapter in the coursebook includes:

9

Teacher’s Resource CD The Teacher’s Resource CD provides a wealth of teacher support material, including:

Workshee

t 9.2

Nucleophil

ic substitu tio

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NAME:

CLASS: INTRODUCTIO N Nucleophilic subst two reaction mech itution reactions may be used to produ anisms possible: ce alcohols, amin 9.3. SN1 and S 2. es and nitriles. N These reactions There are are discussed in sections 9.2 and

• fully worked solutions to coursebook questions

No

Question

1

• worksheets for practising skills and consolidating theory; answers are also included • teacher demonstrations to engage students and enhance understanding of concepts

2

3

Answer

Explain what is meant by the term nucleophile.

Demons

tration

8.2

lating

Electrop

When two nucle ophiles of equa charge, such l as NH3 and H 2O are compared, one factor determine relative strength. s their and hence deter Outline this factor mine which of two is the stron these ger nucleophil e. Consider the relati strength of each ve nucleophilic of the following and state, with pairs an the stronger nucle explanation, which is a OH - and CN - ophile in each case: b CH3O - and OH c H2O and OH -

and nickel. AIM with copper late objects To electrop S II) sulfate MATERIAL -3 ified copper( acid dm 0.50 mol solution el nitrate solution cm) -3 ified nick . 3 cm ! 6 dm acid e (approx 0.50 mol cm) al electrod . 3 cm ! 6 Copper met electrode (approx al Nickel met

3 beakers ed objects washing plat key ring) spirits for spoon, Methylated to be plated (e.g. ct 6 V) obje tal Me ply (set to power sup 12 V DC wires Connecting

3 ! 250 cm

8.2 onstration ker. 250 mL bea beaker other in a the osite each the rim of plated opp tops over tion to the object to be bending the per(II) sulfate solu METHOD strip and the s of the beaker3 by per ified cop cop ct to be cm of acid to the side Place the 1 that the obje electrodes approximately 200 the ure Ens mp Cla er. ting wires. -Tack. Add with connec ply. Turn on the pow time to obtain or with Blu nt electrodes 5 ply to the pole of the power sup should be sufficie beaker. es. power sup e which the electrod e away Connect the nected to the negativ tely 10 minutes, supply to wip er 2 tly pow gen ima the to con need connect for approx plated is its. You may. object. Dis cell to run anode. per on the the methylated spir Allow the 3 ng as the per coating ting of cop a visile coa ct in water and with ct to reveal the copwith nickel metal acti obje from the obje of nickel nitrate, Rinse the 4 tion tals/sludge excess crys e process with a solu g. In latin trop sam in elec h Use the N 5 electrolysis a suitable object suc PLANATIO lication of r onto LTS & EX practical app be electroplated ation to thei ED RESU will see a ergo oxid will EXPECT ced at the and nickel ion students nickel) und demonstrat , the metals copper (copper or solution and are redu the object rather de this ano In Page 1 © Pearson Educa cedures This page from als at the trolyte form on al at the the Chemistry: tion Australia met separate pro these cells, the met pass through the elec k ‘sludge’ tends to to (a division of For use with ced In Pearson Austra the IB Diplom a Programme why a blac lia Group Pty se cations instantly redu as a spoon. HL Teacher’s Ltd) 2009. The reason al ions are crystals. cations. The Resource may be reproduced respective e object to be plated. ting is that the met students will allic for classroom coa uniform met mercial results that use. cathode—th ully uniform metal e to form lity com utif not have tim e the high quality duce such high qua of proprietary that a bea ct and do duc to pro ition of the obje will not pro n 1.1 s. In order , including the add ace io cess part restingly, surf pro at car g latin med ired stig sities. Inte This electrop silverware or on chro procedures are requ varying current den al inve tic ac in plex Pr the use of bitals r toxicity have seen g much more com CLASS: mixture and despite thei ling or electroplatin to the electrolyte used in industry, Model nts ely age wid anic still org are complexes cyanide s. mechanic TIONS NAME: quantum S QUES ined by STIMULU as determ orbitals n sizes of Questio . room use. relative No. Ltd) 2009 AIM and for class Page 1 rs Group Pty be reproduced the shapes Australia n skewe may of Pearson her’s Resource rt woode To model sho (a division Teac Australia ramme HL y of Education IB Diploma Prog LS delling cla © Pearson the For use with oured mo MATERIA Chemistry: other col s or from the ine page This our Plastic ferent col . at the several difcutting plasticine 1s orbital sphere is resents a Knife for This rep other so that the where the h diameter. an atom tely 1 cm right angles to eac the 1s orbital in the at approxima g of METHOD ere until sphere of rs into the sphere s the positionin sph a al m For origin 1 ee skewe rs. This represent nsional axes. of your sizes. Insert thr outside 2 the skewe the 3-dime clay, add to the in relative centre of at the centre of ference of e the dif modelling is ital. 2 cm) out nucleus colour of resents a 2s orb orbital and observ greater than ond sec gth This rep eal the inner 1s ped’ lobes (len Using a of the 3 dle is 6 cm. sha rev ter mid to arme f the ‘pe dia hal ether in sphere in orbital: Form two if possible.) est end tog Cut the 4 ent a 2p ir narrow d colour h To repres clay. (Use a thir skewer with the 5 les to eac a orbitals. to 6 at right ang the 2p modelling two lobes onto r two 2p 4 ement of in steps e anothe Thread the 6 to produc t you produced shows the arrang ce r. twi we ske 4 and 5 together lobes tha axes. This now orbitals, of peat steps ee skewers with z Re 2p set l 7 thr nsiona , 2py and 2p orbitals. Set up the resent a 3-dime of 1s, 2s, 2px 8 rep dels of the the combined set other to space. w your mo s and in results dra 1s and 2s orbital orbitals ent your To repres ss section of the pe 9 cro h of sha with the

4

• practical investigations to enhance the learning of chemical concepts and for use in meeting the mandated time allocation for practical work • practical notes for the teacher/lab technician • risk assessments for practical activities.

SAFETY sheet for Dem Assessment See Risk

Primary amin es may be produ ced by the reaction of a with a halogenoacertain nucleophile lkane. Draw the structure of the nucleophile, show its non-bondin ing g electrons. Using curly arrow movement, draw s to show electron for the synthesis the SN2 mechanism of ethanamin e from chloroethane.

This time-saving resource contains documents available as:

• M  icrosoft Word documents that can be edited, allowing you to modify and adapt any resources to meet your needs • PDFs to make printing easy.

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Companion Website www.pearsoned.com.au/schools The Companion Website addresses Aim 7 for Experimental Sciences by providing easy integration of technology into the classroom. It contains a wealth of support material for students and teachers to enhance teaching and learning in chemistry. The interactive material on the Companion Website allows students to review their work and revise fundamental concepts, as well as providing an opportunity for accelerated learning. The Companion Website contains: • Review Questions—auto-correcting multiplechoice questions for exam revision

O

H

O H

H

H H

• Interactive Animations—to engage students in exploring concepts • QuickTime Videos—to explore chemical concepts in a visually stimulating way • 3D Molecules Gallery—for interactive viewing and manipulating of molecular structures • Web Destinations—a list of reviewed websites that support further investigation and revision.

For more information on CHEMISTRY: For use with the IB Diploma Programme visit www.pearsoned.com.au/schools viii

MEET THE AUTHORS Lanna Derry Lanna Derry, the lead author of the CHEMISTRY: For use with the IB Diploma Programme series, is a highly experienced teacher of IB Chemistry. She has taught senior Chemistry in independent schools for more than twenty years and has authored revision guides for Year 12 Chemistry. Lanna is currently teaching IB Chemistry at Tintern Girls Grammar School, Ringwood East, Victoria, Australia.

Maria Connor Maria Connor is an experienced IB Chemistry examiner. She has taught IB and senior Chemistry for many years and is currently teaching IB Chemistry at Tintern Girls Grammar School, Ringwood East, Victoria, Australia.

Janette Ellis Janette Ellis has experience teaching both IB Chemistry and senior Chemistry. After teaching in Victoria for many years, she is now at Kambala, Rose Bay, New South Wales, Australia.

Faye Jeffery Faye Jeffery is currently teaching at Melbourne Centre for Adult Education. She has taught Chemistry and Biology for more than twenty years. Faye has written a number of texts for Chemistry and Science.

Carol Jordan

Brian Ellett has taught senior Chemistry for more than twenty years and has written a number of texts for Chemistry. He is currently Head of Science at Salesian College, Chadstone, Victoria, Australia. Pat O’Shea is a highly experienced teacher of Chemistry. He is currently Deputy Principal at Loreto College, Ballarat, Victoria, Australia. Pat has presented at many workshops for senior Chemistry teachers.

CHEMISTRY: FOR USE WITH THE IB DIPLOMA PROGRAMME HIGHER LEVEL

MEET THE AUTHORS

Carol Jordan is currently teaching at the Shanghai American School, Shanghai, China. She is an experienced teacher of IB Chemistry, IB Environmental Systems and Theory of Knowledge. She has been an assistant examiner and senior moderator for internal assessment for IB Chemistry. Carol is a workshop leader and was part of the team responsible for developing the new IB Diploma Programme Chemistry Guide.

ix

HOW TO USE THIS BOOK Our aim has been to present chemistry as exciting, accessible and relevant. The content is carefully structured with regular opportunities for revision and consolidation to prepare students for the IB Diploma Programme Higher Level Chemistry examinations.

Major features • Chapter opening pages that include a stimulating photo and a simple, student-friendly syllabus-related list of what students should be able to do by the end of the chapter • Chem Complement boxes that engage students with interesting extensions of the Chemistry theory and applications to Aims 8 and 9 for Experimental Sciences

Section 1.2 Exercises

)"/

Main energy Subshells in order of increasing energy level 1s 1 2 3

4

Orbital energy diagrams

• Theory of Knowledge boxes that address the links between the syllabus and aspects of the scientific way of knowing as required by the syllabus

1

2s

1

2p

3

3s

1

3p

3

3d

5

4s

1

4p

3

4d

5

4f

7

Maximum number of orbitals in energy level 1 4

1

Draw a 1s orbital.

2

Compare a 1s and a 2s orbital.

3

Draw three sets of 3D axes. p orbital. a On the first set of axes draw a x a py orbital. b On the second set of axes draw a pz orbital. c On the third set of axes draw between: Describe the spatial relationship a a p orbital and a py orbital

9

4 16

x

b a px orbital and a pz orbital compare with that of the: How does the energy of the 3p subshell

5

a 3s subshell? b 2p subshell? c 3d subshell? Order the following subshells in 4s, 3d, 2p, 1s, 3p

6

THEORY OF KNOWLEDGE

order of increasing energy:

orbitals in the 3rd energy level. a State the maximum number of in the 3rd energy level. b State the names of all the orbitals orbitals in the 4th energy level. a State the maximum number of with the highest energy in the b State the name of the subshell 4th energy level.

7

in the 1870s, through to Schrödinger’s From JJ Thomson’s discharge tube in model of today, advances made mathematical quantum mechanics affected by the atom have been significantly understanding the structure of modelling. advances in technology and mathematical (1908–1974) once said: The science philospher Jacob Bronowski of has been to give an exact picture sciences One of the aims of physical in the twentieth century has been the material world. One achievement to prove that this aim is unattainable. to death, science came one step closer In the 1980s, long after Bronowski’s Rohrer from atoms look like when Heinrich getting an exact picture of what the scanning invented Germany from Switzerland and Gerd Binnig and STM scans the surface of an element tunnelling microscope (STM). The l current over its surface. A three-dimensiona analyses the change in electric picture of computer modelling, enabling a image is then constructed using Nobel Prize two scientists shared the 1986 individual atoms to be seen. The in Chemistry for their invention.

STRUCTUREOFTHEATOMTHATWEDONT s #OULDTHEREBEKNOWLEDGEABOUTTHE models mathematical the or technology know about yet because the required do not exist?

OFSCIENTIlCKNOWLEDGEISULTIMATELY s $OYOUTHINKTHATTHEDEVELOPMENT What are the implications of this? limited by advances in technology? HAVEMEANTBYHISCLAIM s 7HATDOYOUTHINK"RONOWSKIMIGHT ARIN#HEMISTRYTOINDIVIDUALSOR s 4HE.OBEL0RIZEISAWARDEDEVERYYE to the acquisition of contribution major a made have groups who benefit to mankind’. Carry out some knowledge that is of the ‘greatest microscope and find out why Rohrer research on the scanning tunnelling of the prize. and Binnig were such worthy recipients

8

ONS

1.3 WRITING ELECTRON CONFIGURATI

energy and their arrangement in the main Knowledge of orbitals, subshells Three configurations for atoms and ions. levels enables us to write electron electrons are placed in these orbitals, principles govern the way in which subshells and energy levels: are principle determines that electrons 1 The Aufbau (or building-up) subshell available. This means that the placed into the lowest energy level lowest before the subshell with the next with the lowest energy is filled energy can start to be filled. (proposed by the Austrian scientist 2 The Pauli exclusion principle no more. but electrons, 2 or 1 0, hold can Wolfgang Pauli) states that orbitals lowest energy orbitals of the same subshell, the 3 Hund’s rule states that for Chem electrons with the same spin (see is attained when the number of all will occupy orbitals singly until Complement) is maximized. Electrons then a second electron will be added orbitals of a subshell are half-full, to orbitals.

This information is summarized

in table 1.3.1.

)

energy (kJ mol

first ionization

ATOMIC STRUC

CHAPTER 1

s !MAINENERGY pz orbital energy. For each of the following elements: A dumbbell-shape 1 Define the term first ionization LEVELSHELL IS 16 is d orbital aligned the z-axis. MADEUPOFAN of subshe energies the highest energy subshell that i State (sub-levofels) of UMBER with ionizationlls how the first energy. 2 Explain approximately being filled. These give the same number subshe Quantum mecha atomic lls (sub-levels) the elements in orderaofnumber are made up the number of electrons in and of orbitals levels of quantum theory. nics A model of the atom of the existence of energy the same energy. ii State evidence basedfor on that subshell. sub-levels (subshells).

a Lithium energies can be 3 Explain how successive ionization of an atom. b Fluorine used to determine the group number CHEMISTRY: FOR USE WITHyou would expect for the c Paladium THE IB DIPLOM 4 Describe the pattern that sulfur. A PROGRAMME ()'(%2 d Barium successive ionization energies of ,%6%, 21 e Xenon you would expect to 5 State which period 2 element energy. Explain f Germanium have the highest 3rd ionization your answer.

orbital 6 Distinguish between an atomic an orbit.

and

7 Describe the: a shape l axes b orientation on a set of three-dimensiona i of an s orbital ii of a py orbital. axes, sketch a pz 8 On a set of three-dimensional orbital and a px orbital. order of decreasing 9 List the following subshells in energy:

for each of the 17 State the electron configuration following elements. a Beryllium b Silicon c Iodine d Sodium e Niobium f Aluminium of each of the 18 State the electron configuration following negative ions.

a Cl b S2 3 c N 2 d Se of each of the 19 State the electron configuration following positive ions. 2 a Ba 2 b Ni 2p orbital and describe 12 Compare a 2px orbital with a z them. c K the similarities and differences between Chapter B3 d of: number total the 1 Test 13 State of: 20 State the electron configuration a p electrons in calcium Part A: Mult 5 inWhic zinch equa a Cr b d electrons iple-cho tion repre ice question energy of sents the c s electrons in potassium. an element third ioniz b Cu 1 Wha s t is the total ation M? ways in A M (g) to show number c Ag m M 4 (g)the six in an atom of electrons 14 a Use orbital diagrams 3e in the of iodine? B M 2 could distributed be3 in p orbit which 2 electrons (g) mM d Mo A 5 als (g) e level. energy main 2nd C the of M(g) p orbitals m M 3 (g) B 7

3e drawn. have you D M six3 diagrams b Consider the (g) m M 4 www.pearsoned.com.au/schools C 17 (g) e Hund’s rule. Identify which of these agree with D 23 following © IBO of the 6 HL Paper research which of for atoms 15 Draw orbitalIndiagrams 1 Nov the Companion Website to support learning and the follow Weblinks 05 are Q5available on ing groun elements. configurations © IBO HL 2 A trans d-state electrelated to this chapter. are unpa Paper 1 ition meta Nov 06aQ5Oxygen ron ired elect I 1s 2 2s 2 2 l ion 2 rons prese configura 2p tion [Ar]3 9 X has the elect nt? ronic d . What II 1s 2 2s 2 3 b Argon of the elem is the atom 2p ent? IB DIPLOMA PROGRAMME ()'(%2,%6%, ic number III 1s 2 2s 2 4 CHEMISTRY: FOR USE WITH THE A 27 2p A II only B 28 B I and C 29 II only C II and D 30 III only D I, II and III © IBO HL 3 Whic Paper 1 h statement May 07 Q4 © IBO HL 7 How is correct and energ Paper 1 many elect about elect y levels? Nov 01 Q5 rons are ron orbit in an atom A Yttrium, there in als of xenon? all the d Y (Z  39) orbitals A is the first periodic 10 table with element in the an electron B The maxi B 18 in a f sub-l mum numb evel. d orbital er of elect C 20 is 10. rons in one C The maxi D 36 mum numb main energ er of elect rons in the y level is 18. 4th D In a main © IBO HL 8 Whic Paper 1 energ h atom May 05 Q5 highest energ y level, the sub-l ion has the 1s 2 2s 2 2p 6 or evel with electron y is label 3s 2 3p 6 3d 7 the configura ed f. ? tion A Co © IBO HL 4 Whic B Mn Paper 1 h is corre May 06 Q6 ct about the element tin C Co 2 (Sn) (Z  50)? D Fe 3 Number of main energy level Number s © of IBO 9 HL Paper electrons containing A solid elem 1 Nov 01 in ent, X, conta Q6 electrons main energ highest its atoms ins unpa A y level and form ired elect s an ionic electron rons in 4 chloride, configura B XCl tion is possi A [Ne]3s 2 ble for elem 2 . Which 4 4 C ent X? B [Ar]3d 2 2 14 5 4s D C [He]2s 2 2 4 5 2p D 2 [Ne]3 14 s 3p 4

24

Paper 1

May 06 Q7

© IBO HL

Paper 1

Nov 00 Q6

CHAPTER 1 ATOMIC STRUCTURE

3p, 2s, 4s, 4p, 3d orbitals does not exist: 10 State which of the following 1p, 2s, 2d, 3d, 4f. Explain your answer. in the 4th main 11 State the total number of orbitals energy level and name four of them.

© IBO HL

NOT 1s 2s

2p

IB DIPLOMA PROGRAMME ()'(%2,%6%,

Quantum numbe uniquely describ rs A set of numbers that Terms and defini together e the position tions of an electron . Quantum theory Aufbau princip Energy can be le When filling in ‘packets’ or absorbed or emitted are placed into quanta. the lowest energy orbitals, electrons level availab s orbital A le. Electron configu spherical orbital ration Notatio that has the energy in each energy levels, lowest n showing main main energy subshells and level. number of electron subshells that Stability The represents the s amount of energy electron arrange in the an atom. particle. The ment in greater the stabilit possessed by a of that particle y, the lower the Electron spin and the lower energy An the chance of to form another causes each electron intrinsic property of electron it reacting species. s that to behave as about an axis. if it were spinnin Subshell (sub-le The spinning g vel) A group charge generat magnetic field the same energie of orbitals all es a whose directio having s. n depends on of spin. the direction Successive ionization energi First ionizat energy require es The amount ion energy d to remove the s of The amount required to remove electrons from of an elemen one mole of electron of energy t in the gaseous one mole of atoms of an state, one mole s from one mole electrons at a element in the of time. gaseous state. Heisenberg Wave-like proper uncertainty principle ties Physica limitation on There is a resemble those l properties that how precisely of waves rather we can know a subatomic than particle the location of particle such s. as an electron Concepts . Hund’s rule Electrons will occupy orbitals until all orbitals singly s !SHARPINCREAS of a subshell are half-full, second electron EFROMONESUC then a will be added CESSIVEIONIZAT energy to the to the orbitals ION next indicates . that the next Main energy is being remove electron level A group d from a full electron shell. of subshells with similar energie way a graph In of successive s. ionization energie this relates to the s electron configu Nodes Position ration of an atom. s on the electron s 4HEPATTERNOF the chance of finding the electron density graph where lRSTIONIZATION ENERGIESACROSS period gives is zero. evidence for A Orbital A the existence region of space energy levels of main in and which an electron be found. sub-levels (subshe due to peaks may lls) in atoms that correspond to full main energy levels and full Orbital diagra sub-levels. m A diagram matic represe electron configu ntation of ration showin g electrons as arrows in boxes. K and l 2500 Pauli exclus 2000 ion princip le Orbitals or 2 electron can hold 0, 1 s, but no more. 1500 Probability 1000 The chance of an event occurri 500 ng. px orbital A dumbbell-shape d orbital aligned the x-axis. 0 with 0 1 2 c Scandium 3 py orbital 4 5 6 A dumbbell-shape questions 7 1 Review Chapter atomic number questions 1 Review d orbital the y-axis. Chapter (Z)d Gallium 8 aligned with

• End-of-chapter tests that allow students to test their knowledge of the topic thoroughly using questions from past IBO examinations

2p

1s 2s

illustrate Hund’s rule: ‘Electrons will occupy orbitals singly until all orbitals of a subshell are half-full, then a second electron will be added

ary

• Review questions that draw together all aspects of the topic

)"/

Figure 1.3.1 These orbital diagrams for nitrogen

Chapter 1 Summ

• Glossary of terms and a summary of concepts at the end of each chapter

 !PPLYTHE!UFBAUPRINCIPLE (UNDlSRULEANDTHE0AULI EXCLUSIONPRINCIPLETOWRITE ELECTRONCONFIGURATIONSFOR ATOMSANDIONSUPTO:

to orbitals.’

CHEMISTRY: FOR USE WITH THE

12

• Comprehensive exercises that encourage students to consolidate their learning in a systematic manner while familiarising students with the IB command terms

x

Orbitals in the subshell

TURE

• ICT activities that address Aim 7 for Experimental Sciences and are available on the Companion Website

AND ORBITALS TABLE 1.2.2 ENERGY LEVELS

23

CHAPTER 1 ATOMIC STRUCTURE

 3TATETHEMAXIMUM NUMBEROFORBITALSIN AGIVENENERGYLEVEL

13

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ACKNOWLEDGEMENTS We would like to thank the following for permission to reproduce photographs, texts and illustrations. The following abbreviations are used in this list: t = top, b = bottom, c = centre, l = left, r = right. Argonne National Laboratory: p. 33. Corbis Australia Pty Ltd: pp. 74t, 76, 197, 267. Dorling Kindersley: p. 93b. Getty Images Australia Pty Ltd: p. 50. iStockphoto: p. 70. Jupiterimages Corporation © 2008: pp. 312(hose), 328. NASA: p. 39. Pearson Education Australia: Katherine Wynne: p. 206; Lauren Smith: pp. 114, 210, 261; Peter Saffin: p. 30. Photolibrary Pty Ltd: Cover, 1, 7, 27, 48, 51, 61, 63, 64, 67, 68, 69, 72 (mercury), 72b, 72tc, 72tl, 72tr, 74b, 82, 84, 93t, 108, 109, 115, 127, 128, 138, 145, 152, 154, 167, 168, 173, 177, 191, 205, 208, 230, 232b, 232t, 234, 243, 244b, 244t, 247, 256, 260b, 260t, 264, 268, 279 Shutterstock: pp. 91, 285, 286, 312 (fibres), 312 (styrofoam), 313. Thanks to the International Baccalaureate Organization (IB Organization) for permission to reproduce IB intellectual property. This material has been developed independently of the International Baccalaureate Organization (IB Organization) which in no way endorses it. Every effort has been made to trace and acknowledge copyright. However, should any infringement have occurred, the publishers tender their apologies and invite copyright owners to contact them. The Publishers wish to thank Maria Conner, Carol Jordan and Michael McCann for reviewing the text.

xii

1

Atomic structure

Chapter overview This chapter covers the IB Chemistry syllabus Topic 12: Atomic Structure.

By the end of this chapter, students should be able to: • explain how successive ionization energies relate to the electron configuration of an atom

• draw the shape of an s orbital and the shapes of the px, py and pz orbitals

• explain how the pattern in first ionization energies across a period gives evidence for the existence of main energy levels and sub-levels in atoms

• use the periodic table to determine electron configurations for atoms up to Z = 54

• define the terms orbital, subshell (sub-level) and shell (main energy level) • state the relative energies of the s, p, d and f subshells in a main energy level

• state the Aufbau principle, Hund’s rule and the Pauli exclusion principle • write electron configurations for atoms and ions up to Z = 54.

IBO Assessment statements 12.1.1 to 12.1.6

the story so far …

I

n 1900 the German physicist Max Planck was the first to apply quantum theory to a physical problem. His theory related to the radiation released by a body at high temperature. Planck discovered that the theoretical formula he had devised to explain this phenomenon required acceptance of what was then a radical assumption that energy was only released in distinct ‘chunks’ or quanta. In 1913, Niels Bohr applied Planck’s theory in his explanation of atomic structure. The problem with atomic theory up until that time was that classical physics could not explain the then current (Rutherford’s) model, which suggested that electrons spiralled around the nucleus in a series of orbits. If atoms obeyed the laws of classical physics, the energy they radiated would ultimately run out and they would collapse into the nucleus. Bohr used quantum theory to explain the behaviour of electrons around a nucleus. Electrons only existed in ‘fixed’ orbits or energy levels where they did not radiate energy. The energy levels could be assigned quantum numbers 1, 2, 3 etc. Energy would only be radiated if the electrons moved between a higher energy orbit and a lower energy orbit. Bohr’s application of his theory to hydrogen atoms matched physical observations. His predictions of the wavelengths of light that hydrogen atoms should release when electrons moved between orbits also matched physical observations—information that further validated quantum theory.

1.1 Ionization energies and electron arrangements You will recall from your study of standard level topic 3 that the first ionization energy is the amount of energy required to remove one mole of electrons from one mole of atoms of an element in the gaseous state (Chemistry: For use with the IB Diploma Programme Standard Level, p. 82). Similarly the successive ionization energies of an element are the amounts of energy required to remove all the electrons from one mole of an element in the gaseous state, one mole of electrons at a time. E

∞ E 3 3 2 1

E2

E1

energy required to leave the atom E3 < E2< E1

Figure 1.1.1 Electrons in higher energy levels need to gain less extra energy in order to leave the atom.

2

Successive ionization energies If an electron has energy equal to or greater than that of the highest possible energy level of the atom, it will leave the atom and a positive ion will be formed. Electrons in energy levels that are further from the nucleus have higher energy than those that are closer to the nucleus; therefore, it is these electrons that can be removed more easily by the addition of energy. The first electrons to be removed by the addition of energy will be those that already have the highest energy—those in the valence electron shell (see figure 1.1.1). This can also be explained in terms of the attraction of the electrons to the nucleus. As electrons in the valence shell are furthest from the nucleus, they are not attracted to it as strongly as electrons in other electron shells, and so they will be the ones that are removed first.

The first electron to be removed is one that already has a high energy and is least strongly attracted to the nucleus; that is, one in the valence shell. This will require the lowest ionization energy. If there are more electrons in this outer shell, then they will be removed next, with the ionization energy gradually increasing. When the outermost electron shell is empty, the next electron to be lost will come from the next closest shell to the nucleus. But since this shell is full, and therefore stable, a great deal more energy will be required to remove one electron from this shell than was needed to remove the previous electron.

12.1.2 Explain how successive ionization energy data is related to the electron configuration of an atom. © IBO 2007

The pattern formed by the successive ionization energies of an atom provides evidence for the existence of energy levels around the nucleus and allows us to work out the electron configuration of an atom. Consider the successive ionization energies for nitrogen and magnesium below: TABLE 1.1.1 Successive ionization energies for nitrogen

Ionization energy (kJ mol–1)

Ionization N(g) → N+(g) + eN+(g) → N2+(g) + e-

I1 =

1400

I2 =

2856

2+

3+

-

I3 =

4578

3+

4+

-

I4 =

7475

4+

5+

N (g) → N (g) + e

-

I5 =

9440

N5+(g) → N6+(g) + e-

I6 = 53 266

N6+(g) → N7+(g) + e-

I7 = 64 358

N (g) → N (g) + e N (g) → N (g) + e

Figure 1.1.2 An atom of nitrogen

has 5 electrons in its 2nd electron shell and 2 electrons in the 1st electron shell.

You will notice in table 1.1.1 that the first five ionization energies gradually increase from 1400 kJ mol−1 to 9440 kJ mol−1, then the 6th ionization energy is equal to 53 266 kJ mol−1. This large increase indicates that the 6th electron to be removed is coming from the next energy level, which is full and closer to the nucleus. A much greater amount of energy is needed to remove one electron from this new energy level than from the previous energy level because the electrons are experiencing a significantly larger attraction to the nucleus. This leads us to the electron configuration of 2,5 for nitrogen. A similar trend can be seen for the successive ionization energies of magnesium.

Ionization

Ionization energy (kJ mol–1)

Ionization

Ionization energy (kJ mol–1)

Mg(g) → Mg+(g) + e−

I1 =

736

Mg6+(g) → Mg7+(g) + e−

I7 = 21 703

Mg+(g) → Mg2+(g) + e−

I2 =

1451

Mg7+(g) → Mg8+(g) + e−

I8 = 25 656

Mg2+(g) → Mg3+(g) + e−

I3 =

7733

Mg8+(g) → Mg9+(g) + e−

I9 = 31 642

Mg3+(g) → Mg4+(g) + e−

I4 = 10 540

Mg9+(g) → Mg10+(g) + e−

I10 = 35 461

Mg4+(g) → Mg5+(g) + e−

I5 = 13 630

Mg10+(g) → Mg11+(g) + e−

I11 = 169 987

Mg5+(g) → Mg6+(g) + e−

I6 = 17 995

Mg11+(g) → Mg12+(g) + e−

I12 = 189 363

CHEMISTRY: FOR USE WITH THE IB DIPLOMA PROGRAMME HIGHER LEVEL

CHAPTER 1  ATOMIC STRUCTURE

TABLE 1.1.2 Successive ionization energies for magnesium

3

The first two successive ionization energies of magnesium increase from 736 kJ mol−1 to 1451 kJ mol−1. The next ionization energy increases greatly to 7732.6 kJ mol−1, indicating that the 3rd electron is being removed from the next energy level. This energy level is full and closer to the nucleus, so a much greater amount of energy is required for the 3rd electron to be removed than for the 2nd electron. The successive ionization energies increase gradually as electrons are removed from the second electron shell. There is another large jump from 35 461 kJ mol−1 for the 10th ionization energy to 169 987 kJ mol−1 for the 11th ionization energy, indicating that this electron is being removed from the next energy level, the inner electron shell. This leads us to the electron configuration of 2,8,2 for magnesium.

60 000 50 000 40 000

2nd energy level

30 000

1st energy level

10

20 000 10 000 0

1st energy level

6.00

70 000 log (ionization energy)

–1

ionization energy (kJ mol )

The pattern formed by successive ionization energies can be seen clearly in a graph of electron being removed against ionization energy (kJ mol−1). Figure 1.1.3 shows the successive ionization energies of nitrogen (Z = 7). In figure 1.1.3, the ionization energies for the first 5 electrons are low, suggesting that there are five electrons in the outer shell, then there is a sharp increase in the ionization energies for the 6th and 7th electrons, which are in the first electron shell.

0

1 2 3 4 5 6 number of the electron being removed

Figure 1.1.3 The successive ionization energies of nitrogen.

5.00

2nd energy level

4.00 3.00 2.00 1.00 0.00

7

3rd energy level

4th energy level 0

2

4 6 8 10 12 14 16 18 20 number of the electron being removed

Figure 1.1.4 The successive ionization energies of potassium. This graph is plotted as

log10(ionization energy), to make the increases more obvious.

Evidence for the four electron shells of potassium can be seen in figure 1.1.4. In this graph there are three sharp increases in ionization energy. The first occurs after 1 electron has been removed and the second and third each after another 8 electrons have been removed. This leads to the electron configuration of 2,8,8,1 for potassium. Table 1.1.3 shows that the successive ionization energies for the elements of period 3, Na to Ar, increase gradually until the next electron to be removed is from an electron shell that is full. This is shown by the shaded region of the table. When an electron is removed from a full electron shell, the ionization energy increases dramatically.

4

Table 1.1.3 Values of successive ionization energies, In, (kJ mol-1) for the elements of period 3

Element

I1

I2

I3

I4

I5

I6

I7

I8

I9

Na

  494

4562

6912

  9543

13 353

16 610

20 114

25 490

28 933

Mg

  736

1451

7733

10 540

13 630

17 995

21 703

25 656

31 642

Al

  577

1817

2745

11 575

14 839

18 376

23 293

27 457

31 857

Si

  786

1577

3231

  4356

16 091

19 784

23 786

29 252

33 876

P

1060

1903

2912

  4956

  6273

21 268

25 397

29 854

35 867

S

1000

2251

3361

  4564

  7013

  8495

27 106

31 669

36 578

Cl

1260

2297

3826

  5158

  6540

  9362

11 020

33 610

38 600

Ar

1520

2666

3928

  5770

  7238

  8811

12 021

13 844

40 759

First ionization energies

first ionization energy (kJ mol–1)

Recall that the first ionization energy of an element is the amount of energy required to remove one mole of electrons from one mole of atoms of an element in the gaseous state. The state of the element is important because if the element was solid or liquid, the input of energy would change its state before electrons could be removed. A graph of first ionization energies of the first 54 elements shows distinct patterns that lead us to a greater understanding of the electron structure of an atom.

2500

He

Ne

2000 F

1500 1000

O Be

C

Mg

Li

0

P Si

B

500 0

Ar

N H

WORKSHEET 1.1 Ionization energies

Na

10

Al

Kr

Cl S

Zn

As

Br Cd Sb Pd

I

Xe

Mn Co Se Te Y Nb Tc Ca Ti Sr Fe Ni Cu Ge Sn Mo Ru Rh Ag Zr Sc V Cr Ga In K Rb

20

30 40 atomic number (Z)

50

60

The most obvious feature of this graph is the periodic series of peaks corresponding to the first ionization energy of the noble gases (He, Ne, Ar, Kr, Xe). These elements have high first ionization energies because they have a full electron shell and an associated high degree of energetic stability. The next most obvious feature of the graph is the lowest point of each periodic series. These troughs correspond to the group 1 elements (Li, Na, K, Rb). These elements have only one electron in the outer shell and so the first ionization energy is small, as little energy is required to remove this electron from the atom. Recall that the attraction between the valence electrons and the nucleus is not great due to the low core charge (Chemistry: For use with the IB Diploma Programme Standard Level, p. 85).

12.1.1 Explain how evidence from first ionization energies across periods accounts for the existence of main energy levels and sub-levels in atoms. © IBO 2007

CHEMISTRY: FOR USE WITH THE IB DIPLOMA PROGRAMME HIGHER LEVEL

CHAPTER 1  ATOMIC STRUCTURE

Figure 1.1.5 The first ionization energies of hydrogen to xenon.

5

Both direct and indirect evidence are used to explain the nature of matter. Direct evidence comes from observations and experimentation, while indirect evidence comes from inductive reasoning; that is, by looking at the evidence provided by established laws and theories and using them to draw conclusions and offer explanations. As an IB Chemistry student you can’t observe electrons in energy levels or experimentally determine successive ionization energies, but you can interpret the work of scientists such as Bohr, Heisenberg and Schrödinger, and use ionization energy data to provide evidence for the existence of electrons in energy levels. • Describe examples of knowledge you acquired in Chemistry using direct and indirect evidence. • Compare the strengths and weaknesses of direct and indirect evidence. • Is it possible to have stronger and weaker types of direct evidence? Give some examples.

first ionization energy (kJ mol–1)

Theory of knowledge

If we examine the first ionization energies of the period 2 elements, we observe a gradual increase, with some notable exceptions (see figure 1.1.6). Beryllium has a higher first ionization energy than boron. This suggests that Be has a greater stability than B. Similarly, nitrogen has a higher first ionization energy than oxygen. This pattern is repeated in periods 3, 4 and 5, although it is more difficult to observe in periods 4 and 5 due to the transition metals between groups 2 and 3. Generally speaking, it can be observed that group 2 elements have a greater stability than group 3 elements, and group 5 elements have a greater stability than group 6 elements.

2500

Ne

2000 1500

C

Be

1000

Li

500 0

F

N

0

1

O

B

2

3

4 5 6 atomic number (Z)

7

8

Figure 1.1.6 The first ionization energies of the period 2 elements.

This apparent stability could be expected to correspond, as with the noble gases, to a full electron shell; however, these atoms are still filling the second main energy level. This suggests the presence of sub-levels (subshells) within the main energy levels. The first sub-level in the 2nd energy level is filled with 2 electrons. The second sub-level appears to be filled by 3 electrons, with a third sub-level filled with a further 3 electrons. The other possibility, which we will discuss in section 1.2, is that the second sub-level gains some stability in becoming half-full, and its greatest stability is when it is completely full with 6 electrons in it.

Section 1.1 Exercises 1

Define the term successive ionization energies.

2

The list below gives the successive ionization energies of lithium. Li(g) → Li+(g) + e−; I1 = 513.3 kJ mol−1

Li+(g) → Li2+(g) + e−; I2 = 7298 kJ mol−1

Li2+(g) → Li3+(g) + e−; I3 = 11 814.8 kJ mol−1 Explain how these ionization energies can be used to determine the number of electrons in the valence shell of lithium.

3

The first six ionization energies (I1 to I6) of an element (in kJ mol−1) are given below: 1060, 1890, 2905, 4950, 6270, 21 200

a State which group this element is in. b Discuss whether this information is sufficient to state which period this element is in. 6

4

Locate the elements sodium, aluminium and sulfur on the periodic table.

a State which group each of these elements is in. b State which of these three elements will have the highest second ionization energy and explain your answer. 5

Locate the elements magnesium and chlorine on the periodic table. State which of these two elements will have the higher third ionization energy and explain your answer.

6

Explain why the first ionization energy of magnesium is 900 kJ mol−1 while that of sodium is only 519 kJ mol−1.

7

Explain why the first ionization energy of helium is greater than that of hydrogen, yet the first ionization energy of lithium is much less than that of helium.

8

The first ionization energies of the period 3 elements are as follows: Element

Na

Mg

Al

Si

P

S

Cl

Ar

First ionization energy (kJ mol-1)

494

736

577

786

1060

1000

1260

1520

Describe how the first ionization energies of these elements give evidence for the existence of energy levels and sub-levels in atoms.

1.2 Orbitals, subshells and shells

The modern model of the atom is a consequence of the work of several scientists of the 1920s. The work of Louis de Broglie, a French physicist, followed on from that started by Albert Einstein and Max Planck. In 1924 de Broglie proposed that ‘any moving particle or object had an associated wave’. Electrons had been found to exhibit wave-like properties, so an acceptable model of atomic structure needed to take this into account. Werner Heisenberg proposed that the dual nature of matter places a limitation on how precisely we can know the location of a subatomic particle such as an electron. This is known as the Heisenberg uncertainty principle. In 1925 Wolfgang Pauli formulated a quantum mechanical principle that led to the understanding that no two electrons could have identical quantum numbers. The quantum numbers for an electron uniquely describe its position. This is known as the Pauli exclusion principle. Erwin Schrödinger used these ideas in 1926 to generate a mathematical model to describe the behaviour of electrons in atoms. His model correctly predicted the energy levels of the electrons in all atoms. Schrödinger’s model of the atom is known as quantum mechanics or wave mechanics.

Figure 1.2.1 Erwin Schrödinger, the Austrian

physicist whose series of papers published in 1926 founded the science of quantum mechanics.

CHEMISTRY: FOR USE WITH THE IB DIPLOMA PROGRAMME HIGHER LEVEL

CHAPTER 1  ATOMIC STRUCTURE

Although Bohr’s model of the atom offered an explanation for the line spectrum of hydrogen (Chemistry: For use with the IB Diploma Programme Standard Level, chapter 1) it could only explain the line spectra of other atoms in a rather crude manner. In addition, Bohr’s model of the atom could not explain the apparent existence of energetic sub-levels, as identified by a study of first ionization energies (see previous section).

7

Orbitals Quantum mechanics places the electrons in orbitals, not fixed orbits. This is because the path of an electron cannot be tracked accurately (the Heisenberg uncertainty principle). The electrons are like a cloud of negative charge within the orbital and the exact position of the electron cannot be stated with certainty. An orbital is a region of space in which an electron or electrons may be found.

Chem complement Why s, p, d and f? The letters s, p, d and f are used to name atomic orbitals of elements in the periodic table. Theoretically there are further atomic orbitals named alphabetically as g, h …; however, there is no element that is currently known with electrons in a g or h orbital in the ground state. The letters s, p, d and f appear to be rather random with no obvious reason for their relationship to the spherical s orbital, dumbbell-shaped p orbitals, and d and f orbitals of various shapes. History gives us the link to these names. In the early 20th century, scientists were trying to understand the patterns of lines in the emission spectra of alkali metals. These lines were recorded on photographic film and they appeared with differing quality. In classifying these lines, scientists referred to them as sharp or diffuse (meaning ‘fuzzy’). Other lines were also found on the emission spectrum of hydrogen—principal lines. The fundamental lines were lines that only appeared on the spectrum of the alkali metal. Scientists used these terms to give the letter names to atomic orbitals: s, sharp; p, principal; d, diffuse; f, fundamental.

Representations of methane

12.1.5 Draw the shape of an s orbital and the shapes of the px, py and pz orbitals. © IBO 2007

Schrödinger’s wave equation gives us knowledge of the probability of finding an electron in a given position at a given instant. This probability density or electron density can be plotted on a set of three-dimensional axes. In any electron shell, the lowest energy electrons are found in an s orbital. A graph of probability against radial distance from the nucleus for the lowest-energy electrons in an atom is shown in figure 1.2.2. The probability of finding the electron is greatest at a distance of 0.529 Å (0.529 × 10–10 m) from the nucleus, and then decreases dramatically until it is zero at a distance of 3 Å. When this graph is extended to three dimensions, the spherical shape of the orbital can be recognized. This is seen in the electron-density plot for an s orbital (figure 1.2.3). The density of the dots represents the probability of finding the electron. The higher the density, the more likely it is that an electron will be found in that region of space. The electron-density graph for the next lowest energy electrons also describes a sphere. Like the first orbital, this is also an s orbital. According to Bohr’s model of the atom, these electrons are in the second energy level, or shell, so this orbital can be named the 2s orbital, where 2 indicates the second energy level and s indicates the type of orbital. The lowest energy electrons in the third energy level also produce a spherical electron density graph. When the three spherical s orbitals are compared, those at higher energy levels have a greater radius at which the greatest probability of finding an electron occurs.

8

This leads to the 1s, 2s and 3s orbitals being represented as spheres of increasing radii. y

Radial electron distribution

probability

z

x

0 1 2 3 4 5 6 7 8 distance from the nucleus (nm)

Figure 1.2.3 An electron-density plot shows the

Figure 1.2.2 The radial probability graph for the

probability of finding an electron in a particular part of the space surrounding the nucleus.

lowest energy s orbital.

y y y

z

z

z

x

x

x

1s 2s 3s

The electron density plots of the remaining electrons in the second energy level are very different from the spherical s orbitals. The electron-density plots for these orbitals take on a shape that is best described as a dumbbell. The dumbbell has two lobes and between the two lobes there is a node at which the probability of finding an electron is zero. This is not surprising, as the region between the two lobes contains the nucleus. The dumbbell-shaped orbitals are the p orbitals. The difference between the three p orbitals relates to the axis along which the orbital is aligned. This can be seen in figure 1.2.5 where a px orbital is aligned with the x-axis, a py orbital is aligned with the y-axis and a pz orbital is aligned with the z-axis.

PRAC 1.1 Modelling orbitals

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Figure 1.2.4 1s, 2s and 3s orbitals. The radii of the spheres correspond to a 90% probability of finding an electron within each sphere.

9

z

z

x

y

z

x

x

y

pz

y

px

py

Figure 1.2.5 There are three different p orbitals, each of which is aligned with a different axis. z

z

z

y

y

x

y

x dyz

z

x dxz

dxy

z

12.1.3 State the relative energies of s, p, d and f orbitals in a single energy level. © IBO 2007 y x

y x

dx2 – y2

d z2

Figure 1.2.6 There are five different d orbitals, four of which are aligned in a plane defined by a pair of axes; the fifth is

aligned along the z-axis.

5f 5d 5p

4f energy

4d

1

1s

3d 3p 3s

2p 2s 2

3 shell

5s

4p 4s

4

A group of orbitals with the same energy is called a subshell. The number of orbitals that make up a subshell increases as the energy of the orbitals increases.

5

Figure 1.2.7 The energy of the subshells increases from s to p to d to f.

10

The Bohr model of the atom described the increasing energy of electron shells as their distance from a nucleus increases. In quantum mechanics, an electron shell or energy level is a collection of orbitals with the same approximate energy. Electron shells of increasing energy are made up of increasingly larger numbers of orbitals. Within the energy level, the orbitals have increasing energy. The s orbitals have the lowest energy in the energy level, the p orbitals have the next highest energy, the d orbitals have the next highest and the f orbitals have the highest energy.

There is only one s orbital in an s subshell, there are three p orbitals in a p subshell, five d orbitals in a d subshell and seven f orbitals in an f subshell. Figure 1.2.7 shows the increasing energy of the subshells.

Chem complement Quantum numbers In quantum mechanics, an electron is assigned a set of quantum numbers that describe its exact location. The principal quantum number, n, gives its main energy level or shell. Principal quantum numbers can have positive integral values such as 1, 2, 3, 4 etc. The second quantum number is called the angular momentum quantum number, l. It can have integral values from 0 to n − 1 for each value of n. This quantum number defines the shape of the orbital. The letters s, p, d and f correspond to values of the second quantum number, namely 0, 1, 2, and 3 respectively. The magnetic quantum number, ml, can have integral values between −1 and l, including 0. This quantum number describes the orientation of the orbital in space. Table 1.2.1 The relationship between the quantum numbers n, l and ml

n

Possible values of l

Subshell

1

0

1s

0

2

0

2s

0

1

2p

−1, 0, 1

0

3s

0

1

3p

−1, 0, 1

2

3d

−2, −1, 0, 1, 2

0

4s

0

1

4p

−1, 0, 1

2

4d

−2, −1, 0, 1, 2

3

4f

−3, −2, −1, 0, 1, 2, 3

3

4

Possible values of ml

In summary

• Within an atom there are major energy levels or shells. Each energy level is assigned a number, starting at the nucleus and increasing outwards (1, 2, 3 …). • The energy levels are further divided into subshells or sub-levels. These are assigned the letters s, p, d, f. Within an energy level, the energies of the subshells increase in the order s, p, d, f. • Within subshells, electrons move in regions of space called orbitals. • Subshells contain different numbers of orbitals. The s subshell contains 1 orbital, the p subshell 3 orbitals, the d subshell 5 orbitals and the f subshell 7 orbitals.

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The key descriptive features of the quantum mechanical model can be summarized as follows.

11

12.1.4 State the maximum number of orbitals in a given energy level. © IBO 2007

Table 1.2.2 Energy levels and orbitals

Main energy Subshells in order level of increasing energy

Maximum number of orbitals in energy level

1

1s

1

 1

2

2s

1

 4

2p

3

3s

1

3p

3

3d

5

3

Orbital energy diagrams

Orbitals in the subshell

4

4s

1

4p

3

4d

5

4f

7

 9

16

THEORY OF KNOWLEDGE From JJ Thomson’s discharge tube in the 1870s, through to Schrödinger’s mathematical quantum mechanics model of today, advances made in understanding the structure of the atom have been significantly affected by advances in technology and mathematical modelling. The science philospher Jacob Bronowski (1908–1974) once said: One of the aims of physical sciences has been to give an exact picture of the material world. One achievement in the twentieth century has been to prove that this aim is unattainable. In the 1980s, long after Bronowski’s death, science came one step closer to getting an exact picture of what atoms look like when Heinrich Rohrer from Switzerland and Gerd Binnig from Germany invented the scanning tunnelling microscope (STM). The STM scans the surface of an element and analyses the change in electric current over its surface. A three-dimensional image is then constructed using computer modelling, enabling a picture of individual atoms to be seen. The two scientists shared the 1986 Nobel Prize in Chemistry for their invention. • Could there be knowledge about the structure of the atom that we don’t know about yet because the technology or the mathematical models required do not exist? • Do you think that the development of scientific knowledge is ultimately limited by advances in technology? What are the implications of this? • What do you think Bronowski might have meant by his claim? • The Nobel Prize is awarded every year in Chemistry to individuals or groups who have made a major contribution to the acquisition of knowledge that is of the ‘greatest benefit to mankind’. Carry out some research on the scanning tunnelling microscope and find out why Rohrer and Binnig were such worthy recipients of the prize.

12

Section 1.2 Exercises 1

Draw a 1s orbital.

2

Compare a 1s and a 2s orbital.

3

Draw three sets of 3D axes.

a On the first set of axes draw a px orbital. b On the second set of axes draw a py orbital. c On the third set of axes draw a pz orbital. 4

Describe the spatial relationship between:

a a px orbital and a py orbital b a px orbital and a pz orbital 5

How does the energy of the 3p subshell compare with that of the: a 3s subshell? b 2p subshell? c 3d subshell?

6

Order the following subshells in order of increasing energy: 4s, 3d, 2p, 1s, 3p

7

a State the maximum number of orbitals in the 3rd energy level. b State the names of all the orbitals in the 3rd energy level.

8

a State the maximum number of orbitals in the 4th energy level. b State the name of the subshell with the highest energy in the 4th energy level.

1.3 Writing electron configurations

1 The Aufbau (or building-up) principle determines that electrons are placed into the lowest energy level available. This means that the subshell with the lowest energy is filled before the subshell with the next lowest energy can start to be filled. 2 The Pauli exclusion principle (proposed by the Austrian scientist Wolfgang Pauli) states that orbitals can hold 0, 1 or 2 electrons, but no more. 3 Hund’s rule states that for orbitals of the same subshell, the lowest energy is attained when the number of electrons with the same spin (see Chem Complement) is maximized. Electrons will occupy orbitals singly until all orbitals of a subshell are half-full, then a second electron will be added to orbitals. This information is summarized in table 1.3.1.

12.1.6 Apply the Aufbau principle, Hund’s rule and the Pauli exclusion principle to write electron configurations for atoms and ions up to Z = 54. © IBO 2007

1s 2s

2p NOT

1s 2s

2p

Figure 1.3.1 These orbital diagrams for nitrogen

illustrate Hund’s rule: ‘Electrons will occupy orbitals singly until all orbitals of a subshell are half-full, then a second electron will be added to orbitals.’

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Knowledge of orbitals, subshells and their arrangement in the main energy levels enables us to write electron configurations for atoms and ions. Three principles govern the way in which electrons are placed in these orbitals, subshells and energy levels:

13

Table 1.3.1 Maximum numbers of electrons in orbitals

Main energy level

Subshells

Orbitals in the subshell

Maximum number of electrons in the subshell

Maximum number of electrons in energy level

1

1s

1

 2

 2

2

2s

1

 2

 8

2p

3

 6

3s

1

 2

3p

3

 6

3d

5

10

4s

1

 2

4p

3

 6

4d

5

10

4f

7

14

Electron configurations

3

4

18

32

2n2

n

Chem complement Electron spin Detailed study of the line spectra of many-electron atoms (i.e. atoms other than hydrogen) showed that each line was actually a pair of very closely spaced lines. This was explained by two Dutch physicists George Uhlenbeck and Samuel Goudsmit after an experiment by Otto Stern and Walther Gerlach in 1922 in which electrons were deflected in a magnetic field (see figure 1.3.2). Electrons were postulated to have an intrinsic property, called electron spin, which causes each electron to behave as if it were spinning about an axis. The spinning charge generates a magnetic field whose direction depends on the direction of

spin. This gives two possible values for a fourth quantum number,the spin magnetic quantum number, ms. This quantum number completes the description of an electron and allows the Pauli exclusion principle to be completely satisfied. No two electrons can have the same set of four quantum numbers. This results in the need for two electrons occupying the same orbital to have opposite spins. These are represented in orbital diagrams (see figure 1.3.1) by arrows pointing in opposite directions.

S

slit



beam of atoms magnet

beam collector plate

Figure 1.3.2 Atoms in which the electron spin quantum number of the unpaired

electron is + 21  are deflected in one direction and those in which it is − 21  are deflected in the other.

14

N Figure 1.3.3 The direction in which an electron spins

determines the direction of its magnetic field.

Note that an orbital containing electrons is said to be ‘occupied’. Because they are regions of space, atomic orbitals exist whether they are occupied or not, but they are not included in an electron configuration unless they are occupied by one or more electrons. Electron configurations can be represented in two ways: as an orbital diagram (electrons in boxes) or in a form that uses superscripts. An orbital diagram represents each orbital as a box and each electron as an K or l arrow, so the positioning of the electrons in the individual orbitals can be shown clearly. Although orbital diagrams are a useful visual aid, they are not used routinely when a full electron configuration is required. The form of electron configuration with the superscripts is more commonly used to represent an atom or ions quickly. In this form the occupied subshells are written in order of increasing energy with the number of electrons in the subshell shown as a superscript.

shell number

3p 4 orbital type

number of electrons in this type of orbital in this shell

Figure 1.3.4 Electron configuration notation.

Table 1.3.2 Electron configurations of several lighter elements

Element

Total number of electrons

Orbital diagram 1s 2s 2p 3s

Full Electron configuration

Helium

2

1s2

Lithium

3

1s2 2s1

Boron

5

1s2 2s2 2p1

Nitrogen

7

1s2 2s2 2p3

Fluorine

9

1s2 2s2 2p5

Sodium

11

1s2 2s2 2p6 3s1

structure of some complexes that involve the lanthanides and actinides.

Chem complement While the theoretical existence of the f subshell is of interest to those of us who wonder about what comes next, it is only the lanthanides and actinides that have electrons in the f subshell. (The lanthanides are named after element 57, lanthanum because they were thought to all have similar chemical properties to lanthanum.) However, the f orbitals are not occupied in the ground state until element 58 (cerium), so lanthanum is not officially part of the ‘f-block’; however, on the basis of its chemical properties it is the first member of the lanthanides. Similarly, while actinium is the first member of the actinides, thorium is the first element with an electron in the 5f subshell. The f orbitals are buried deep beneath the valence shell and they rarely play an important role in chemical change or bonding. However, the orbital shapes can be useful in interpreting spectra and in understanding the

Figure 1.3.5 The shapes of the f orbitals are

varied and unusual.

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The f subshell

15

Electron configurations of elements with more than 18 electrons In figure 1.2.7, the energies of all the subshells up to 5f are shown. Notice that the energy of the 4s subshell is less than that of the 3d subshell. This means that the 4s subshell fills before the 3d subshell. As a consequence of this filling order, the majority of the elements of the first transition series (V to Zn) have a full 4s subshell. When the number of electrons becomes great (>20), even the shorter (subscript) form of writing the electron configuration becomes tedious. For note-taking purposes, a further shorthand form can be used in which the electron configuration of the previous noble gas is represented by the symbol for that element in square brackets (e.g. [Ar]) and is followed by the rest of the electron configuration for the element required. Using this shorthand the electron configuration of calcium would be [Ar]4s2. Note that this shorthand form should not be used when answering examination questions. The periodic table can be used to great advantage in writing electron configurations. The periodic table can be divided up into blocks that reflect the subshell that is being filled. s-block p-block

period

1 1s 2

2s

3

3s

d-block

3p

4

4s

3d

4p

5

5s

4d

5p

6

6s

5d

6p

7

7s

6d

Lanthanides Actinides

1s

2p

f-block 4f 5f

Figure 1.3.6 The periodic table can be divided up into blocks reflecting the subshell that is being filled.

This view of the periodic table can be most useful in determining the electron configuration of any element. To determine an electron configuration: Step 1: Locate the element on the periodic table. Periodic table and order of filling

Step 2: Determine the name of the block that the element is in (s, p, d, f) Step 3: Count how many groups (vertical columns) from the left of that block the element is in. Step 4: Determine which period the element is in by counting down from the first period, which consists of H and He. Step 5: Fill all shells and subshells that have a lower energy than this subshell (see figure 1.2.7).

16

This information is used as follows: 1 The period in which the element is found gives you the last main energy level that is being filled. For example, if the element is in period 3, the last main energy level being filled is the 3rd energy level. Note the exceptions to this rule occur in the transition metals, where the 10 transition metals in period 4 are filling the 3d subshell and the 10 transition metals in period 5 are filling the 4d subshell. 2 The block in which the element is found gives you the subshell (s, p, d, or f) that is being filled. For example, if the element is in group 2 of the periodic table, it is in the s-block, so the s subshell is being filled. 3 The number of groups from the left gives you the number of electrons in that subshell e.g. an element in group 4 is 2 groups from the left-hand side of the p-block, so there are 2 electrons in the p subshell.

Worked example 1 State the full electron configuration of calcium. Solution Calcium (Z = 20) is in period 4, group 2, which is in the fourth column of the s-block of the periodic table. This means that the last subshell being filled is the 4s subshell and this has 2 electrons in it. The electron configuration of calcium is therefore 1s2 2s2 2p6 3s2 3p6 4s2.

Worked example 2 State the electron configuration of tellurium. Solution Tellurium (Z = 52) is in period 5, group 6, which is in the fourth column of the p-block of the periodic table. This means that the last subshell being filled is the 5p subshell and this has 4 electrons in it.

The periodic table can be used as a means by which to recall the relative energies of the subshells (see figure 1.2.7) and hence their order of filling. If you follow through the elements of the periodic table in atomic number order (hydrogen, helium, lithium, beryllium etc.) the order in which the blocks of the periodic table (see figure 1.3.6) are encountered gives the filling order of the subshells. Referring to figure 1.3.6, for elements 1 to 54, this would be:

1s 2s

2p

3s

3p

3d

4s

4p

4d

4f

5s

5p

5d

5f

6s

6p

6d

1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p where the 1s subshell has the lowest energy and the 5p subshell has the highest energy.

Figure 1.3.7 Another popular way of

remembering the filling order of the subshells is to follow the arrows from top to bottom and right to left.

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The electron configuration of tellurium is therefore 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p4. In shorthand form this would be [Kr]5s2 4d10 5p4. Remember that this shorthand form is not acceptable in examination answers, for which the full electron configuration must be written.

17

1s 2s

2p

3s

3p

Figure 1.3.8 Orbital diagram for phosphorus.

When we were studying the first ionization energy of elements (see figure 1.1.5), it was clear that certain electron configurations possessed greater stability than others. In addition to the great stability offered by a full subshell, the peaks in figure 1.1.5 for N (1s2 2s2 2p3) and P (1s2 2s2 2p6 3s2 3p3) demonstrate a certain degree of stability in having a half-full subshell. The orbital diagrams for nitrogen (see table 1.3.2) and for phosphorus (see figure 1.3.8) show that these elements have one electron in each orbital of the p subshell. When we consider the transition metals, the stability of a half-full d subshell is highly sought after. So much so, that when there are 4 electrons in a d subshell, 1 electron is ‘borrowed’ from the full s subshell of closest energy, so that the d subshell can be half-full. A similar situation arises when the d subshell is nearly full. When there are 9 electrons in a d subshell, 1 electron is ‘borrowed’ from the full s subshell to fill the d subshell. The resulting configurations for a small number of transition metals are shown in table 1.3.3.

Table 1.3.3 Electron configurations of exceptional transition metals

Element and atomic number (Z) Chromium

Orbital diagram and electron configuration 1s 2s 2p 3s 3p 3d

4s

4p

4d

5s

24 1s2 2s2 2p6 3s2 3p6 3d 5 4s1

Manganese

25 1s2 2s2 2p6 3s2 3p6 3d 5 4s2

Copper

29 1s2 2s2 2p6 3s2 3p6 3d10 4s1

Zinc

30 1s2 2s2 2p6 3s2 3p6 3d10 4s2

Molybdenum

42 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d5 5s1

Technetium

43 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d 5 5s2

Silver

47 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s1

Cadmium

48 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2

While this ‘borrowing’ of electrons from the full s subshell appears to break the Aufbau principle, it does result in an electron arrangement with a greater stability (lower energy), and so supports the Aufbau principle in this way.

18

Electron configurations of ions An ion is formed when an atom loses or gains electrons. While an ion will have the same atomic number as an atom of the same element, it will have a different number of electrons and so will have a different electron configuration. The driving force behind formation of ions is the gaining of the same stability as the nearest noble gas, so ions of main group elements have the same electron configuration as the nearest noble gas.

WORKSHEET 1.2 Electron configurations

Table 1.3.4 Electron configurations of ions of some main group elements

compared to that of the nearest noble gas Element

Atomic number (Z)

Group

Symbol

Electron configuration

Neon

10

0

Ne

1s2 2s2 2p6

Magnesium

12

2

Mg2+

1s2 2s2 2p6

Nitrogen

 7

5

N3-

1s2 2s2 2p6

Argon

18

0

Ar

1s2 2s2 2p6 3s2 3p6

Potassium

19

1

K+

1s2 2s2 2p6 3s2 3p6

Sulfur

16

6

S2-

1s2 2s2 2p6 3s2 3p6

Note in the table above that Ne, Mg2+ and N3− ions have the same electron configuration, and Ar, K+ and S2− have the same electron configuration. These groups of ions are referred to as isoelectronic—having the same number of electrons.

Electron configurations of metal ions

Transition metal ions The majority of the 3d transition metals have a full 4s subshell, although as discussed above, chromium and copper have only one electron in a 4s subshell. A full 4s subshell has a slightly higher energy than a partially filled 3d subshell (see chapter 3), so when a 3d transition metal loses electrons, the electrons are initially lost from the 4s subshell. If more than 2 electrons are lost, then the subsequent electrons are taken from the 3d subshell. Similarly, electrons are taken from the 5s subshell before the 4d subshell. The electron configurations of some transition metal ions are shown below. These will be discussed in more detail in chapter 3.

Element

Atomic number (Z)

Ion symbol

Number of electrons in the ion

Electron configuration

Iron

26

Fe2+

24

1s2 2s2 2p6 3s2 3p6 3d 6

Iron

26

Fe3+

23

1s2 2s2 2p6 3s2 3p6 3d 5

Vanadium

22

V2+

20

1s2 2s2 2p6 3s2 3p6 3d 2

Chromium

24

Cr3+

21

1s2 2s2 2p6 3s2 3p6 3d 3

Copper

29

Cu+

28

1s2 2s2 2p6 3s2 3p6 3d 10

Copper

29

Cu2+

27

1s2 2s2 2p6 3s2 3p6 3d 9

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Table 1.3.5 Electron configurations of some transition metal ions

19

Section 1.3 Exercises 1

State the maximum number of electrons that can be found in: a any px orbital b the first energy level c any d subshell d the second energy level e any s orbital

2

Draw an orbital diagram for each of the following elements. a Magnesium b Silicon c Chlorine

3

Describe how three electrons would be distributed in a p subshell that was previously empty.

4

Name the block of the periodic table in which each of the following elements is found. a Sodium b Carbon c Iron d Chlorine e Uranium f Silver

5

For each of the following elements, state the highest energy subshell that is being filled. a Oxygen b Chromium c Strontium d Aluminium e Arsenic f Rhodium

6

State the electron configuration of each of the following elements. a Nitrogen b Chlorine c Calcium d Nickel e Selenium f Tin

7

State the electron configuration of each of the following negative ions. a F− b Te2− c P3− d Br−

8

State the electron configuration of each of the following positive ions.

a Mg2+

20

b Al3+

c Rb+

d Zn2+

Chapter 1 Summary

Quantum numbers  A set of numbers that together uniquely describe the position of an electron.

Terms and definitions

Quantum theory  Energy can be absorbed or emitted in ‘packets’ or quanta.

Aufbau principle  When filling orbitals, electrons are placed into the lowest energy level available.

s orbital  A spherical orbital that has the lowest energy in each main energy level.

Electron configuration  Notation showing main energy levels, subshells and number of electrons in the subshells that represents the electron arrangement in an atom.

Stability  The amount of energy possessed by a particle. The greater the stability, the lower the energy of that particle and the lower the chance of it reacting to form another species.

Electron spin  An intrinsic property of electrons that causes each electron to behave as if it were spinning about an axis. The spinning charge generates a magnetic field whose direction depends on the direction of spin.

Subshell (sub-level)  A group of orbitals all having the same energies.

Heisenberg uncertainty principle  There is a limitation on how precisely we can know the location of a subatomic particle such as an electron. Hund’s rule  Electrons will occupy orbitals singly until all orbitals of a subshell are half-full, then a second electron will be added to the orbitals. Main energy level  A group of subshells with similar energies. Nodes  Positions on the electron density graph where the chance of finding the electron is zero.

Wave-like properties  Physical properties that resemble those of waves rather than particles.

Concepts •

A sharp increase from one successive ionization energy to the next indicates that the next electron is being removed from a full electron shell. In this way a graph of successive ionization energies relates to the electron configuration of an atom.



The pattern of first ionization energies across a period gives evidence for the existence of main energy levels and sub-levels (subshells) in atoms due to peaks that correspond to full main energy levels and full sub-levels.

Orbital  A region of space in which an electron may be found.

first ionization energy (kJ mol–1)

Orbital diagram  A diagrammatic representation of electron configuration showing electrons as K and l arrows in boxes. Pauli exclusion principle  Orbitals can hold 0, 1 or 2 electrons, but no more. Probability  The chance of an event occurring. px orbital  A dumbbell-shaped orbital aligned with the x-axis. py orbital  A dumbbell-shaped orbital aligned with the y-axis. pz orbital  A dumbbell-shaped orbital aligned with the z-axis.



2500 2000 1500 1000 500 0

0

1

2

3

4 5 6 atomic number (Z)

7

8

A main energy level (shell) is made up of a number of subshells (sub-levels) of approximately the same energy. These subshells (sub-levels) are made up of a number of orbitals of the same energy.

Quantum mechanics  A model of the atom based on quantum theory.

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First ionization energy  The amount of energy required to remove one mole of electrons from one mole of atoms of an element in the gaseous state.

Successive ionization energies  The amounts of energy required to remove the electrons from one mole of an element in the gaseous state, one mole of electrons at a time.

21



An s orbital is spherical and p orbitals are dumbbell shaped.



y

Each subshell contains a fixed number of orbitals, and hence there is a maximum number of electrons that can be in the subshell. Subshell (sub-level)

Number of orbitals

Maximum number of electrons

s

1

2

p

3

6

d

5

10

f

7

14

z

x

z

z

x

y

x

y

y

px

Hund’s rule determines that when electrons fill a subshell, each orbital is half-filled before a second electron is added to any of the orbitals.



The Pauli exclusion principle states that an orbital can contain 0, 1 or 2 electrons but no more.



The above three rules are applied when an electron configuration is written.



The position of an element in the periodic table can be used to determine electron configurations by determining the period (energy level), block (subshell) and position in the block (number of electrons in the subshell).

py

In a main energy level the s subshell has the lowest energy, followed by the p subshell, the d subshell and finally the f subshell.

5f 5d 5p

4f energy

4d

1

22



1s

3d 3p 3s

2p 2s 2

3 shell

5s

4p 4s

4

s-block 2

2s

3

3s

d-block

3p

4

4s

3d

4p

5

5s

4d

5p

6

6s

5d

6p

7

7s

6d

Lanthanides Actinides

5

p-block

1 1s

period



The Aufbau principle determines that the electrons of an atom fill the energy levels in order of increasing energy.

z

x

pz



2p

f-block 4f 5f

1s

c Scandium d Gallium

  1 Define the term first ionization energy.   2 Explain how the first ionization energies of the elements in order of atomic number give evidence for the existence of energy levels and sub-levels (subshells).   3 Explain how successive ionization energies can be used to determine the group number of an atom.   4 Describe the pattern that you would expect for the successive ionization energies of sulfur.   5 State which period 2 element you would expect to have the highest 3rd ionization energy. Explain your answer.   6 Distinguish between an atomic orbital and an orbit.   7 Describe the: a shape b orientation on a set of three-dimensional axes i of an s orbital ii of a py orbital.   8 On a set of three-dimensional axes, sketch a pz orbital and a px orbital.   9 List the following subshells in order of decreasing energy: 3p, 2s, 4s, 4p, 3d

10 State which of the following orbitals does not exist: 1p, 2s, 2d, 3d, 4f. Explain your answer. 11 State the total number of orbitals in the 4th main energy level and name four of them. 12 Compare a 2px orbital with a 2pz orbital and describe the similarities and differences between them. 13 State the total number of: a p electrons in calcium b d electrons in zinc c s electrons in potassium. 14 a Use orbital diagrams to show the six ways in which 2 electrons could be distributed in the p orbitals of the 2nd main energy level. b Consider the six diagrams you have drawn. Identify which of these agree with Hund’s rule. 15 Draw orbital diagrams for atoms of the following elements. a Oxygen b Argon

16 For each of the following elements: i State the highest energy subshell that is being filled. ii State the number of electrons in that subshell. a Lithium b Fluorine c Paladium d Barium e Xenon f Germanium 17 State the electron configuration for each of the following elements. a Beryllium b Silicon c Iodine d Sodium e Niobium f Aluminium 18 State the electron configuration of each of the following negative ions. a Cl− b S2− c N3− d Se2− 19 State the electron configuration of each of the following positive ions. a Ba2+ b Ni2+ c K+ d B3+ 20 State the electron configuration of: a Cr b Cu c Ag d Mo www.pearsoned.com.au/schools Weblinks are available on the Companion Website to support learning and research related to this chapter.

CHEMISTRY: FOR USE WITH THE IB DIPLOMA PROGRAMME HIGHER LEVEL

CHAPTER 1  ATOMIC STRUCTURE

Chapter 1 Review questions Chapter 1 Review questions

23

Chapter 1 Test Part A: Multiple-choice questions   1 What is the total number of electrons in p orbitals in an atom of iodine? A 5 B 7 C 17 D 23

© IBO HL Paper 1 Nov 06 Q5

  2 A transition metal ion X2+ has the electronic configuration [Ar]3d9. What is the atomic number of the element? A 27 B 28 C 29 D 30

© IBO HL Paper 1 May 07 Q4

  3 Which statement is correct about electron orbitals and energy levels? A Yttrium, Y (Z = 39) is the first element in the periodic table with an electron in a f sub-level. B The maximum number of electrons in one d orbital is 10. C The maximum number of electrons in the 4th main energy level is 18. D In a main energy level, the sub-level with the highest energy is labeled f. © IBO HL Paper 1 May 06 Q6

  4 Which is correct about the element tin (Sn) (Z = 50)?

A B C D

Number of main energy levels containing electrons

Number of electrons in highest main energy level

4

 4

4

14

5

 4

5

14 © IBO HL Paper 1 May 06 Q7

24

  5 Which equation represents the third ionization energy of an element M? A M+(g) → M4+(g) + 3e− B M2+(g) → M3+(g) + e− C M(g) → M3+(g) + 3e− D M3+(g) → M4+(g) + e− © IBO HL Paper 1 Nov 05 Q5

  6 In which of the following ground-state electron configurations are unpaired electrons present? I 1s2 2s2 2p2 II 1s2 2s2 2p3 III 1s2 2s2 2p4 A II only B I and II only C II and III only D I, II and III

© IBO HL Paper 1 Nov 01 Q5

  7 How many electrons are there in all the d orbitals in an atom of xenon? A 10 B 18 C 20 D 36

© IBO HL Paper 1 May 05 Q5

  8 Which atom or ion has the electron configuration 1s2 2s2 2p6 3s2 3p6 3d7? A Co B Mn C Co2+ D Fe3+

© IBO HL Paper 1 Nov 01 Q6

  9 A solid element, X, contains unpaired electrons in its atoms and forms an ionic chloride, XCl2. Which electron configuration is possible for element X? A [Ne]3s2 B [Ar]3d2 4s2 C [He]2s2 2p2 D [Ne]3s2 3p4 © IBO HL Paper 1 Nov 00 Q6

10 Values for the first ionization energies (IE) for five successive elements in the periodic table are given below. Based on these values, which statement is correct? Element

K

L

M

N

O

P

IE / kJ mol−1

1060

1000

1260

1520

418

?



ii Explain how successive ionization energies account for the existence of three main energy levels in the sodium atom. (3 marks)

b State the formula of a stable ion formed from elemental vanadium. Identify which electrons are lost when the ion forms. (2 marks)

A The outermost electron in element K is in a higher energy level than that in element L. B Element M is probably a metal. C Element N is probably a non-metal. D Element P has a lower first ionization than element O.

© IBO HL Paper 2 Nov 07 Q2b

4

a Explain why the first ionization energy of magnesium is lower than that of fluorine. (2 marks)

b Write an equation to represent the third ionization energy of magnesium. Explain why the third ionization energy of magnesium is higher than that of fluorine.

© IBO HL Paper 1 Nov 99 Q6 (10 marks)

(3 marks) © IBO HL Paper 2 May 06 Q7e

www.pearsoned.com.au/schools

5

For more multiple-choice test questions, connect to the Companion Website and select Review Questions for this chapter.

Outline why the first ionization energy of Al is lower than that of Mg in terms of electronic configuration. (2 marks) © IBO HL Paper 2 Nov 03 Q6 b(i)

Part B: Short-answer questions

Part C: Data-based question

1

Using the periodic table inside the cover of this coursebook, give the symbol(s) of:

Explain the difference in the two values of ionization energy for each of the following pairs:

a an ion with a double positive charge (2+) with an electronic configuration of [Ar]3d5.

a

(1 mark)

The 1st ionization energy of beryllium is 900 kJ mol−1 whereas the 2nd ionization energy of beryllium is 1757 kJ mol–1. (2 marks)

b two elements with a ground state configuration of ns2np3. (1 mark) © IBO HL Paper 2 May 01 Q1a ii, iii

a State the full electron configuration for a bromide ion.

b The 1st ionization energy of aluminium is 577 kJ mol–1 whereas the 1st ionization energy of magnesium is 736 kJ mol−1. (2 marks) © IBO HL Paper 2 Nov 05 Q2d

(1 mark)

CHAPTER 1  ATOMIC STRUCTURE

2

© IBO HL Paper 2 May 99 Q6f

CHEMISTRY: FOR USE WITH THE IB DIPLOMA PROGRAMME STANDARD LEVEL

25

(1 mark)

b Write the symbol for the ion with a 2+ charge which has the electron configuration of 1s22s22p6.

c

The successive ionization energies of germanium are shown in the following table:

(1 mark)

c Write the symbols for three other species, which also have the electron configuration of 1s22s22p6. (2 marks) © IBO HL Paper 2 May 06 Q2c–e

3

a



i Explain why successive ionization energies of an element increase. (1 mark)

Ionization energy / kJ mol−1



1st

2nd

3rd

4th

5th

760

1540

3300

4390

8950

i Identify the sub-level from which the electron is removed when the first ionization energy of germanium is measured.



ii Write an equation, including state symbols, for the process occurring when measuring the second ionization energy of germanium. (1 mark)



iii Explain why the difference between the 4th and 5th ionization energies is much greater than the difference between any two other successive values. (2 marks)

Part D: Extended-response questions 1

a Describe how the first four ionization energies of aluminium vary. (You may wish to sketch a graph to illustrate your answer.) (2 marks)

b State the electronic configurations of aluminium, boron and magnesium. Explain how the first ionization energy of aluminium compares with the first ionization energies of boron and magnesium. (5 marks) © IBO HL Paper 2 Nov 01 Q5a, b

26

2

The successive ionization energies for boron are given below in kJ mol−1. 1st

2nd

3rd

4th

5th

799

2420

3660

25 000

32 800

a Explain the reason why there is a large increase between the third and fourth values. (2 marks)

b Explain the reason why the increase between the first and second ionization energies is more than the increase between the second and third ionization energies. (2 marks)

c State, with reasons, how the value for the second ionization energy of carbon would compare with that of the second ionization energy of boron. (2 marks) © IBO HL Paper 2 Nov 98 Q5b Total marks: 50

2

Bonding

Chapter overview This chapter covers the IB Chemistry syllabus Topic 14: Bonding.

By the end of this chapter, you should be able to: • use the VSEPR theory to predict and describe the shape and bond angles of molecules and ions with five and six pairs of bonding or non-bonding electrons • describe sigma (σ) bonds in terms of the end to end overlapping of hybrid or atomic orbitals • describe pi (π) bonds in terms of the side by side overlapping of p orbitals

• explain hybridization of atomic orbitals as the mixing of atomic orbitals to form new orbitals that are suitable for bonding • compare and explain the relationships between Lewis structures, molecular shapes and types of hybridization • describe the structures of selected molecules and ions in terms of delocalization of π electrons.

IBO Assessment statements 14.1.1 to 14.3.1

T

he shape of a molecule is important in determining how it interacts with other molecules around it. The shape may govern whether the molecule is polar or non-polar and hence how strongly it is attracted to other molecules by intermolecular forces. H

H H

H C

H

H

C H

C H pentane

H C H

H C

H

H

H

H

H C H

H C C C

H H C H

H H

H 2,2-dimethylpropane

Figure 2.0.1 The higher boiling point of pentane in comparison to its isomer 2,2-dimethylpropane can be attributed to the difference in their shapes.

When two non-polar hydrocarbons of equal molecular mass are compared, the ease with which the molecules can pack together in the liquid state will be governed by their shapes, and will cause a difference in their boiling points—the more linear the shape of the molecule, the more easily the molecules will pack together, the stronger the van der Waals’ forces between them and the higher the boiling point of the hydrocarbon. Two isomers of pentane, C5H12, illustrate this point clearly. The boiling point of pentane, the straight-chain isomer, is 36.1°C while the boiling point of its isomer 2,2-dimethylpropane, a much more compact molecule, is only 9.5°C.

VSEPR theory You will recall from your study of the shapes of molecules (Chemistry: For use with the IB Diploma Programme Standard Level, chapter 2,) that the repulsion between pairs of bonding electrons and non-bonding electrons determines the shape of a molecule. These pairs of electrons can be referred to as negative charge centres. A bonding electron pair is attracted by the nuclei of both bonding atoms. In comparison, a non-bonding electron pair is only attracted to one nucleus, so it experiences less nuclear attraction. Its negative charge centre spreads out more than does that of a bonding pair of electrons. As a result, the non-bonding pairs of electrons exert greater repulsive forces on adjacent pairs of electrons, so the repulsion between a non-bonding pair of electrons and a bonding pair of electrons is greater than that between two pairs of bonding electrons. An adjacent non-bonding pair of electrons will have the effect of compressing the bond angle. Order of repulsion: non-bonding–non-bonding > non-bonding–bonding > bonding–bonding

+

+

Bonding electron pair

+

Non-bonding electron pair

Figure 2.0.2 A bonding electron pair is attracted by the nuclei of both bonding atoms, while a non-bonding electron pair is only attracted to one nucleus.

28

Bonding electron pair

Nuclei

Non-bonding electron pair

Nucleus

Figure 2.0.3 A non-bonding electron pair spreads out more in space than does a bonding electron pair.

Double and triple bonds (multiple bonds) each constitute one negative charge centre. Like a non-bonding electron pair, the negative charge centre of a multiple bond is larger than that of a single bond and exerts greater repulsive force on adjacent negative charge centres. The basis of VSEPR theory is that the best arrangement of a given number of negative charge centres is the one that minimizes the repulsions between them. The shapes of different molecules or ions depend on the number of negative charge centres (electron domains) around the central atom.

2.1 Shapes of molecules and ions The shapes and bond angles of molecules and ions with 2, 3 and 4 negative charge centres were discussed as part of the Standard Level course. In this chapter we will discuss the shapes and bond angles of molecules and ions with 5 and 6 negative charge centres.

124.3°

Cl 111.4°

Cl

C

O 124.3°

Figure 2.0.4 In this Lewis structure of Cl2CO, the electrons of the C=O double bond exert a greater repulsive force than do the electrons in the C–Cl single bonds, so the O–C–Cl bond angle is 124.3° rather than 120° as found in trigonal planar molecules such as BF3.

In order to accommodate more than 4 pairs of electrons in the valence shell of an atom, the octet rule is broken. This ‘expansion’ of the valence shell is found to only occur for atoms in the 3rd period and beyond. It seems that the presence of an empty d orbital in the valence shell allows the extra electrons to be accommodated. In addition, the size of the central atom in the molecule is a factor. The larger the central atom, the larger the number of atoms that can surround it. The size of the surrounding atoms is also important. An expanded valence shell occurs most frequently when the surrounding atoms are small and highly electronegative, such as fluorine, oxygen and chlorine.

Five negative charge centres You will recall that in a Lewis structure, valence electrons are represented by dots or crosses. As a covalent bond consists of a pair of bonding electrons shared between two atoms, a dot and a cross between the atoms represent a bond, while pairs of dots or crosses represent non-bonding pairs of electrons. As with simpler molecules, the first step in determining the shape of a molecule with five or more negative charge centres is to find out how many electrons are in the valence shell of the central atom. Next you should use the formula of the molecule to determine how many bonds the central atom forms with other atoms. The difference between these two numbers will give you the number of non-bonding electrons around the central atom. These non-bonding electrons can be paired up as non-bonding electron pairs or negative charge centres. The combination of bonding electron pairs and non-bonding electron pairs leads to the shape of the molecule (see table 2.1.1). b a When five atoms are bonded to a central

DEMO 2.1 Modelling five and six negative charge centres

90° 120°

Figure 2.1.1 (a) A trigonal bipyramid has six faces

and five vertices. (b) When five atoms are bonded to a central atom, the shape of the molecule is called a trigonal bipyramid.

CHEMISTRY: FOR USE WITH THE IB DIPLOMA PROGRAMME HIGHER LEVEL

CHAPTER 2  bonding

atom, according to VSEPR theory, the pairs of bonding electrons (negative charge centres) will move as far apart as possible, in order to minimize the repulsions between them. The shape that the molecule takes on is called a trigonal bipyramid. This shape can be thought of as two trigonal pyramids fused together at their bases.

14.1.1 Predict the shape and bond angles for species with five and six negative charge centres using the VSEPR theory. ©IBO 2007

29

There are three atoms in a plane forming an equilateral triangle around the middle of the molecule (the equatorial positions), one atom at the top and one atom at the bottom of the molecule (axial positions). The plane of the equatorial atoms is at right angles to that of the axial atoms.

Example 1: PCl5 Phosphorus pentachloride (phosphorus(V) chloride) is a molecule in which a phosphorus atom has formed covalent bonds with five chlorine atoms. This results in the phosphorus atom having 10 electrons in its valence shell (an expanded valence shell). The shape of the molecule is described as trigonal bipyramidal and the molecule is non-polar. The bond angle between the equatorial atoms and the central phosphorus atom is 120°. This is consistent with the equilateral triangle formed by the three chlorine atoms in that position. The bond angle between an equatorial chlorine atom and an axial chlorine atom will be 90°. Cl Cl P Cl

Cl 120°

Cl Figure 2.1.2 Phosphorus pentachloride, PCl5, has a trigonal bipyramidal shape.

There are other molecules with five negative charge centres around them; however, phosphorus, being in group 5, is the only member of period 3 that bonds with five atoms and has no pairs of non-bonding electrons. The other molecules have combinations of bonding and non-bonding pairs of electrons, so with the greater electrostatic repulsion of the non-bonding charge centres and the smaller numbers of atoms bonded to the central atom, a range of shapes is assumed.

Example 2: ClF3 87.5° F

F Cl

185°

F Figure 2.1.3 Chlorine trifluoride is a ‘T-shaped’ molecule.

30

Chlorine trifluoride is a very reactive, poisonous and corrosive gas. Chlorine has seven electrons in its valence shell and the formula tells us that three bonds are formed with fluorine. This leaves four non-bonding electrons (two pairs) in the valence shell. Consequently there are five negative charge centres around this chlorine atom—-three bonding and two non-bonding. The non-bonding pairs of electrons take up equatorial positions around the chlorine to minimize the electrostatic repulsions in the molecule. The angle between the equatorial fluorine and the axial fluorine atoms is slightly less than 90° (87.5°) due to the repulsive forces exerted by the non-bonding electrons. The shape of this molecule is commonly referred to as ‘T-shaped’.

In table 2.1.1 a range of shapes for molecules with five negative charge centres is shown. Notice the effect of the non-bonding electrons on the shapes (in particular the angle between axial atoms) of these molecules. Table 2.1.1 Summary of molecules with five negative charge centres around the

central atom

Number of bonding pairs on the central atom

Number of nonbonding pairs on the central atom

Example

Shape of molecule

5

0

PCl5

Trigonal bipyramid

Representation of the shape

120°

Cl

Cl

90°

Cl

4

1

SF4

Seesaw or ‘sawhorse’

186.9°

Cl F

90°

F

S

116°

F

F

3

2

ICl3

T-shaped

Cl

P

Cl 86.2°

I

187.2°

Cl

Cl

2

3

XeF2

Linear

F Xe F

Six negative charge centres

In an octahedral molecule four atoms in a plane form a square around the middle of the molecule (the equatorial positions) and there is one atom at the top and one atom at the bottom of the molecule (axial positions). The plane of the equatorial atoms is at right angles to that of the axial atoms.

a

b

Figure 2.1.4 (a) An octahedron has eight faces and six vertices. (b) When six atoms are bonded to a central atom the molecule is octahedral.

CHEMISTRY: FOR USE WITH THE IB DIPLOMA PROGRAMME HIGHER LEVEL

CHAPTER 2  bonding

When there are six negative charge centres around a central atom they will move as far apart as possible, according to VSEPR theory, in order to minimize the repulsions between them. If 6 atoms are bonded to the central atom, all six negative charge centres are bonding electron pairs and the shape that the molecule takes on is called an octahedron. This name comes from the solid shape formed when the vertices are joined. An octahedron has eight faces and six vertices. This shape can be thought of as two square pyramids fused together at their bases, and so can be described as square bipyramidal.

31

Example 3: SF6 Sulfur hexafluoride (sulfur(VI) fluoride) is a molecule in which a sulfur atom has formed covalent bonds with six chlorine atoms. This results in the sulfur atom having 12 electrons in its valence shell (another example of an expanded valence shell). The shape of the molecule is described as octahedral or square bipyramidal and the molecule is non-polar. The bond angle between the equatorial atoms and the central sulfur atom (F–S–F) is 90°. This is consistent with the square formed by the four fluorine atoms in that position. The bond angle between an equatorial fluorine atom and an axial fluorine atom will be 90°. F

F

F S

F

F F

Figure 2.1.5 Sulfur hexafluoride, SF6, has an octahedral shape.

F

F

P

F

F

Example 4: PF6− F F

Figure 2.1.6 The phosphorus hexafluoride ion, PF6−, has an octahedral shape.

The phosphorus hexafluoride ion, PF6−, is a Lewis base that is formed when PF5 reacts with XeF2. While its chemistry is unusual, its shape is the same as that of sulfur hexafluoride—octahedral, or square bipyramidal. This molecule is an anion (negative ion) because in addition to the five fluorine atoms that would normally covalently bond with a phosphorus atom, an extra fluoride ion, F−, has bonded in a dative covalent (coordinate) bond with the phosphorus atom. In PF6−, the phosphorus atom has 12 electrons in its valence shell. There are other molecules with six negative charge centres around them; however, sulfur, being in group 6, is the only member of period 3 that bonds with six atoms to form a neutral molecule and has no non-bonding pairs of electrons. The other molecules have combinations of bonding and non-bonding pairs of electrons, so with the greater electrostatic repulsion of the nonbonding charge centres and the smaller numbers of atoms bonded to the central atom, a number of different shapes is assumed.

Example 5: XeF4 F F

Xe

F F

Figure 2.1.7 Xenon tetrafluoride, XeF4, is a square planar molecule.

32

Under conditions of high voltage and with the very electronegative element fluorine, xenon will react to form xenon tetrafluoride, XeF4. Xenon has 8 electrons in its valence shell. According to the formula, four bonds are formed, so there are four non-bonding electrons (two non-bonding pairs). These nonbonding pairs take up the axial positions in the octahedral arrangement of negative charge centres to minimize the electrostatic repulsions in the molecule. As a result, the XeF4 molecule takes on a square planar shape in which the F–Xe–F bond angle is 90°. The XeF4 molecule would be non-polar. Notice that there are 12 electrons in the valence shell of the xenon atom. The octet rule is certainly broken in this case!

Figure 2.1.8 When xenon tetrafluoride is isolated, it forms colourless, roughly cubic crystals.

Table 2.1.2 summarizes the shapes that can be taken on by molecules with six negative charge centres. Notice that in the case of square pyramidal molecules, the one non-bonding pair of electrons repels the square plane upwards to give an angle that is less than 90° with the atom at the ‘top’ of the pyramid. WORKSHEET 2.1 VSEPR for five and six negative charge centres

Table 2.1.2 Summary of molecules with six negative charge

centres around the central atom Number of Example non-bonding pairs on the central atom

6

0

SF6−, PF6−

Shape of molecule

Octahedral or square bipyramidal

Representation of the shape

F

F

90°

S

F

F

90°

F

F

5

1

BrF5

Square pyramidal

F 90°

F

4

2

XeF4

Square planar

F F

F Br

Xe

VSEPR: Basic molecular configurations table

85°

F F

F 90°

F

CHEMISTRY: FOR USE WITH THE IB DIPLOMA PROGRAMME HIGHER LEVEL

CHAPTER 2  bonding

Number of bonding pairs on the central atom

33

Section 2.1 Exercises 1

The valence-shell electron pair repulsion (VSEPR) theory is used to predict the shape of molecules. Explain how these predictions are made.

2

Outline the meaning of the term negative charge centre.

3

Draw and name the four shapes that can be taken on by molecules with five negative charge centres around the central atom.

4

Explain how the shape in which six atoms are arranged around a central atom can be called ‘octahedral’.

5

Describe how the repulsion between a non-bonding pair of electrons and a bonding pair of electrons compares with that between two bonding pairs of electrons.

6

Describe how the repulsion between a non-bonding pair of electrons and a bonding pair of electrons compares with that between two non-bonding pairs of electrons.

7

Draw and name the shape of a molecule with four bonding pairs of electrons and two non-bonding pairs of electrons around the central atom.

8

Draw each of the following molecules and state their shape according to VSEPR theory.

a b c d

PF5 IF5 XeF4 SF4

2.2 Hybridization In Lewis theory, a covalent bond is a shared pair of electrons. This suggests that there is a region of electron density between two atoms as a result of the electron pair being shared. Atomic orbital theory (see chapter 1, section 1.2) suggests that electrons exist in atomic orbitals that are regions of space surrounding the nucleus of an atom. Each atomic orbital can hold no more than two electrons. If covalent bonding is approached from the perspective of quantum mechanics, the question ‘How can we use atomic orbitals to explain bonding and to account for the shapes of molecules?’ can be asked. This leads to a model of chemical bonding called valence-bond theory.

Overlapping orbitals In valence-bond theory, a region of electron density between two atoms is explained in terms of the overlap of two atomic orbitals. Two orbitals from different atoms, each containing one unpaired electron, can merge in the region of space between the two atoms. This overlapping of two orbitals creates a bonding orbital between the two atoms.

34

When two p orbitals overlap in the same axis, or an s orbital and a p orbital or even two s orbitals overlap, the bond formed is called a σ (sigma) bond.

1s

1s

1s

3p

3p

14.2.1 Describe σ and π bonds. © IBO 2007

3p

Figure 2.2.1 (a) Overlap of two 1s orbitals to form a σ bond between two H atoms, (b) overlap of a 1s orbital and a 3p orbital to form a σ bond between an H atom and a Cl atom and (c) overlap of two 3p orbitals to form a σ bond between two Cl atoms.

This overlapping of orbitals explains the difference in bond length between two single bonds, such as an F–F bond and a Cl–Cl bond. The two orbitals overlapping in the case of the F–F bond are both 2p orbitals, whereas in the case of the Cl–Cl bond, the overlapping orbitals are both 3p orbitals. The 3p orbitals are larger than the 2p orbitals, so when two 3p orbitals overlap, they do so at a greater distance from the nucleus than do two 2p orbitals. This causes the bond length in Cl2 to be greater than that in F2. You will recall from chapter 1 that the three p orbitals are at right angles to each other. When two px orbitals are overlapping in a σ bond, the other two p orbitals, py and pz are not overlapping. When two atoms come close enough together, the py and/or the pz orbitals are able to overlap in a sideways fashion. This sideways overlapping of two orbitals is called a π (pi) bond. In a π bond the overlapping is not as great as in a σ bond, so a π bond is not as strong as a σ bond.

z

z

x

Figure 2.2.2 When the px orbitals are overlapping end to end in a σ bond, the py and pz orbitals are not. y py

π bonds occur most commonly when a multiple bond (a double or a triple bond) is formed. A double bond is formed by one σ bond and one π bond. Since a π bond is not as strong as a σ bond, a double bond is less than twice as strong as a single covalent bond between the same two elements. A triple bond is formed by one σ bond and two π bonds and is less than three times as strong as a single bond.

π bond py x

Figure 2.2.3 When py and pz orbitals are brought close enough

together they overlap sideways to form a π bond.

Bond lengths can tell us much about the equality (or otherwise) of covalent bonds between atoms. When the bond lengths of the four C–H bonds in methane, CH4, are measured, they are all found to be equal. This suggests that they are identical in nature. To explain these bonds using the concept of overlapping atomic orbitals, we need to consider the electronic configuration of the atoms involved. The electronic configuration of carbon is 1s2 2s2 2p2 and that of hydrogen is 1s1. An orbital that overlaps with another must have only one electron in it, so that the new overlapping (bonding) orbital has no more than 2 electrons in it. With this in mind, the valence shell configuration of carbon, 2s2 2p2, causes problems. There are only 2 orbitals with just one electron in them. The 2s orbital is full and the third 2p orbital has no electrons in it at all.

Hybridization

CHEMISTRY: FOR USE WITH THE IB DIPLOMA PROGRAMME HIGHER LEVEL

CHAPTER 2  bonding

Hybrid orbitals

y

y

35

This can be shown using an orbital diagram (electrons in boxes).

1s 2s

1s

2p

Figure 2.2.4 The orbital diagram for carbon suggests that there are only two orbitals that could be involved in bonding.



2sp 3

Figure 2.2.5 In the orbital diagram for a hybridized carbon atom, four orbitals could be involved in bonding.

If the four bonds that carbon is commonly observed to make are to exist, there must be four unpaired electrons in the valence shell. This is made possible if the atomic orbitals in the valence shell are assumed to mix together to make hybrid orbitals (figure 2.2.5). The shape of a hybrid orbital is different from that of the two orbitals from which it is formed, but the total number of orbitals in the valence shell remains the same. Note that there is no energy change involved in forming hybrid orbitals; that is, the total energy is the same before and after hybridization has occurred.

14.2.2 Explain hybridization in terms of the mixing of atomic orbitals to form new orbitals for bonding. © IBO 2007

The process of mixing atomic orbitals as atoms approach each other is called hybridization.

sp3 hybridization The case of carbon’s four equal bonds can be explained by hybridization. The four orbitals in the valence shell of carbon, 2s, 2px, 2py and 2pz, are mixed together. Whenever we mix a certain number of atomic orbitals, the same number of hybrid orbitals is formed. Hence, when the four valence shell orbitals of carbon are mixed together, we get four sp3 hybrid orbitals.

DEMO 2.2 Modelling hybridization

y z

+

x

s orbital p x orbital y

py

y z

hybridization z

+

x four sp3 hybrid orbitals

p y orbital

p z orbital

Figure 2.2.6 One s orbital and three p orbitals mix to form four sp3 hybrid orbitals.

36

The shape of an sp3 hybrid orbital is not the same as either the s orbital or the p orbital. Like p orbitals it has two lobes, but, unlike p orbitals, one lobe is much bigger than the other. When put together, the four lobes of the sp3 orbitals take up a tetrahedral shape in space. This shape allows the electron pairs to be as far apart as possible, so as to minimize electrostatic repulsion. The H–C–H bond angle is 109.5°. In this manner the hybridization of atomic orbitals agrees with the requirements of VSEPR theory. Each orbital is equal in size, hence allowing the four bonds to be equal in length and strength when they overlap with the orbital of another atom. Whenever there are four negative charge centres around a central atom, the hybridization of that central atom must be sp3. This is illustrated by the following examples.

14.2.3 Identify and explain the relationships between Lewis structures, molecular shapes and types of hybridization (sp, sp2 and sp3). © IBO 2007

sp hybridization

Example 1: Methane, CH4

A molecule of methane, CH4, is formed when each of the four sp3 hybrid orbitals of carbon overlap in σ bonds with the s orbitals of four hydrogen atoms. The shape of the molecule is tetrahedral. a sp3 hybrid orbitals

b

σ bond

H

×

H H C×H ×

×

C

H

H

H H

Figure 2.2.7 (a) The σ bonding between sp3 hybrid orbitals of carbon and s orbitals of hydrogen in methane. (b) The Lewis structure of methane.

Example 2: Ethane, C2H6

Each of the two carbon atoms in ethane exhibits sp3 hybridization. Each carbon forms four bonds—three to hydrogen atoms and one to the other carbon atom. There is a tetrahedral arrangement of bonding electron pairs around each carbon.

a

σ bonds

H sp3 hybrid orbitals

b

H

×

×

H H

H

Figure 2.2.8 (a) The σ bonding between sp3 hybrid orbitals of carbon and s orbitals of hydrogen in ethane.

(b) The Lewis structure of ethane.

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CHAPTER 2  bonding

H H

×

H×C C×H

C

H

×

H H C

37

Example 3: Ammonia, NH3 In ammonia, nitrogen has 5 electrons in its valence shell and forms three equal bonds with the hydrogen atoms. In addition there is a non-bonding pair of electrons on the nitrogen atom. This must be included in consideration of hybridized orbitals. The s orbital and three p orbitals of nitrogen mix to form four sp3 hybrid orbitals; however, one of these four hybrid orbitals is already filled with the non-bonding pair of electrons. The other three hybrid orbitals are able to form σ bonds with three hydrogen atoms. As in VSEPR theory, the 107° bond angles in ammonia can still be explained in terms of the greater repulsion between the non-bonding electron pair and the bonding electron pairs. The ammonia molecule is trigonal pyramidal in shape. a

b sp3 hybrid orbitals

H×N×H ×

N

H

H

H H

Figure 2.2.9 (a) The σ bonding between sp3 hybrid orbitals of nitrogen and s orbitals of

hydrogen in ammonia. (b) The Lewis structure of ammonia.

Example 4: Hydrazine, N2H4 In hydrazine, the two nitrogen atoms share a single covalent bond and each is bonded to two hydrogen atoms. There is a non-bonding pair of electrons on each nitrogen atom. Each of the two nitrogen atoms in hydrazine exhibits sp3 hybridization. a

b H

H ×

×

×

sp3 hybrid orbitals

N

N

H

H

H N

×

H

N

H

H

Figure 2.2.10 (a) Hydrazine exhibits σ bonding between sp3 hybrid orbitals and both s orbitals and other sp3 orbitals.

(b) The Lewis structure of hydrazine.

38

The s orbital and three p orbitals of oxygen mix to form four sp3 hybrid orbitals; however, two of these four hybrid orbitals are already filled with non-bonding pairs of electrons. The other two hybrid orbitals are able to form σ bonds with two hydrogen atoms. As in VSEPR theory, the 104.5° bond angles in water can be explained in terms of the greater repulsion between non-bonding electron pairs and bonding electron pairs. The water molecule is V-shaped.

b

a sp3 hybrid orbitals

O

H×O H

H H Figure 2.2.11 (a) The σ bonding between sp3 hybrid orbitals of oxygen and

s orbitals of hydrogen in water. (b) The Lewis structure of water.

Chem complement

the NASA space shuttles, and the Phoenix lander, which was launched on 4 August 2007.

Hydrazine and life on Mars

The development and launch of the Phoenix lander was a collaborative project between universities from USA, Canada, Switzerland, Denmark and Germany as well as NASA and the Canadian Space Agencies. The lander arrived on Mars on 26 May 2008 and searched for environments suitable for microbial life, including evidence of water—essential to life as we know it.

Hydrazine, N2H4, is a colourless liquid at room temperature with some similarity to water. Its melting point is 1°C and its boiling point is 114°C. Liquid hydrazine has a density of 1.01 g cm−3. While it is commonly used in the synthesis of other chemicals, a more dramatic use of hydrazine is as a rocket fuel. In World War II, hydrazine was used in mixtures with water, methanol or hydrogen peroxide as a rocket fuel for the Messerschmitt Me 163B, the first and only rocketpowered fighter aircraft used in that war. In more recent times it has been used as the propellent for rocket manoeuvring thrusters, because of its reactions that generate large quantities of hot gas from a small volume of hydrazine.

3N2H4(l) → 4NH3(g) + N2(g)



N2H4(l) → N2(g) + 2H2(g)



4NH3(g) + N2H4(l) → 3N2(g) + 8H2(g)

Hydrazine has been used in this way in the Viking landers (two spacecraft launched in the 1970s whose mission objectives were to obtain high resolution images of Mars),

Figure 2.2.12 Hydrazine was used in the

thrusters that helped land the Phoenix lander on Mars.

sp2 hybridization Boron is an element that only forms three bonds. It has three electrons in its valence shell and will form three covalent bonds with elements such as hydrogen (BH3) and fluorine (BF3). These three equal covalent bonds can be explained by the existence of hybrid orbitals. The electron configuration of boron is 1s2 2s2 2p1. If the 2s orbital and two 2p orbitals are mixed, the hybrid CHEMISTRY: FOR USE WITH THE IB DIPLOMA PROGRAMME HIGHER LEVEL

CHAPTER 2  bonding

In water, oxygen has 6 electrons in its valence shell and forms two equal bonds with the hydrogen atoms. In addition there are two non-bonding pairs of electrons on the oxygen atom. This must be included in consideration of hybridized orbitals.

×

Example 5: Water, H2O

39

orbitals will be called sp2 hybrid orbitals. The sp2 hybrid orbitals will be arranged around the central atom (boron) in such as way as to minimize electrostatic repulsion. This results in a trigonal planar arrangement. y z

x

+

p x orbital

hybridization

y

s orbital

z

x

three sp2 hybrid orbitals

p y orbital

Figure 2.2.13 One s orbital and two p orbitals mix to form three sp2 hybrid orbitals.

In BH3 and BF3 the sp2 hybrid orbitals can form σ bonds with the 1s orbital of hydrogen and a 2p orbital of fluorine. The H–B–H or F–B–F bond angle will be 120° and the molecule will have a trigonal planar shape. We can expect sp2 hybridization to occur whenever there is a trigonal planar arrangement of atoms around a central atom. p orbital F

a

F

b

H ××

F

Figure 2.2.14 (a) The sp2 hybrid orbitals of boron form σ bonds with hydrogen and fluorine. (b) The Lewis structures of BH3 and BF3.

×

F

××

×

B

F

××

××

×× ×

H

H

40

F

××

××

hybrid orbitals

B

××

××

σ bonds

sp2

H

×

B

H

×

B

×

H

Example 6: Ethene Ethene, C2H4, has a C–C double bond between the two carbon atoms and there is a trigonal planar arrangement of atoms around each carbon. Although carbon has four orbitals in the valence shell, only three of them (2s, 2px and 2py) are involved in the hybridization. The fourth orbital, the 2pz orbital, exists at right angles to the plane of the hybrid orbitals.

Multiple bonds

The sp2 hybridized orbitals of the carbon atom in ethene form σ bonds with hydrogen 1s orbitals and with the sp2 hybridized orbital of the other carbon atom. The atoms are sufficiently close that the 2pz orbitals of the two carbon atoms can overlap sideways forming a π bond. The combination of the σ bond between the two sp2 hybridized orbitals and the π bond between the two 2pz orbitals creates a double bond between the two carbon atoms.

C

H

σ bond

C

H

H

H

H

C

σ bond

C

sp2 hybridization orbitals

π bond

H

2pz

H C

Figure 2.2.15 Each carbon in ethene has three sp2 hybridized orbitals and a 2pz orbital at right angles to the plane of the sp2 hybridized orbitals.

H

Figure 2.2.16 Formation of π bonding in ethene.

π bond

a

H

C

C

H

× ×

σ bond

×

H

C

××

C H

PRAC 2.1 Modelling ethene

H

×

H

b

H

H

CHAPTER 2  bonding

Figure 2.2.17 (a) Formation of σ and π bonds in ethene. (b) Lewis diagram of ethene.

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41

Example 7: Diazene, N2H2 This compound has a double bond between the two nitrogen atoms and one hydrogen atom bonded to each nitrogen atom. The two nitrogen atoms exhibit sp2 hybridization and there is a π bond between the two nitrogen atoms to complete the double bond. The third sp2 hybrid orbital on each nitrogen atom is occupied by a non-bonding pair of electrons. This results in the molecule having a bent linear shape.

π bond

a

b H

H

N

××

σ bond

N

H

×× ×

N

×

N H

Figure 2.2.18 (a) Formation of σ and π bonds in diazene. (b) Lewis diagram of diazene.

sp hybridization Beryllium has only two electrons in its valence shell. Its electron configuration is 1s2 2s2. It makes two equal-sized bonds with elements such as hydrogen and the halogens. In order for two equal bonds to be made, the 2s and one 2p orbital must be mixed to form two sp hybrid orbitals. These orbitals will be arranged in such as way as to minimize electrostatic repulsion. This results in a linear arrangement.

s orbital

p orbital

two sp hybrid orbitals

two sp hybrid orbitals shown together (large lobe only)

Figure 2.2.19 One s orbital and one p orbital mix to form two sp hybrid orbitals.

In beryllium hydride, BeH2, the sp hybrid orbitals of beryllium form σ bonds with the 2s orbitals of hydrogen, and in beryllium fluoride, BeF2, the sp hybrid orbitals of beryllium form σ bonds with the 2p orbitals of fluorine. The H–Be–H and F–Be–F bond angles are both 180°. The shape of these molecules is linear.

42

a

b Be

H

H × Be × H

H σ bonds

××

××

F × Be × F

F

××

××

Be

××

F

××

Figure 2.2.20 (a) The sp hybrid orbitals of beryllium form σ bonds with hydrogen and fluorine. (b) The Lewis structures of BeH2 and BeF2.

Example 8: Ethyne, C2H2 Ethyne, C2H2, has a triple bond between the two carbon atoms (CIC) and there is a linear arrangement of atoms around each carbon. Only two of the valence shell orbitals of carbon (2s, 2px) are involved in the hybridization. The third orbital (2py) and the fourth orbital (2pz) exist in two different planes to the hybrid orbitals. y p

z

y

pz x

C

sp hybrid orbital Figure 2.2.21 Each carbon in ethyne has two sp hybridized orbitals, with the 2py and 2pz

orbitals in two different planes to the sp hybridized orbitals.

b

a H

C

C

H

×

H × C ×× C × H

Figure 2.2.22 (a) Formation of σ and π bonds in ethyne. (b) Lewis diagram of ethyne. CHEMISTRY: FOR USE WITH THE IB DIPLOMA PROGRAMME HIGHER LEVEL

CHAPTER 2  bonding

The sp hybridized orbitals of the carbon atom in ethyne form σ bonds with hydrogen 1s orbitals and with the sp hybridized orbital of the other carbon atom. The atoms are sufficiently close that the 2py and 2pz orbitals of the two carbon atoms can overlap sideways in planes that are at right angles to each other forming two π bonds. The combination of the σ bond between the two sp hybridized orbitals and the two π bonds between the two 2py and the two 2pz orbitals creates a triple bond between the two carbon atoms.

43

Example 9: Nitrogen, N2

WORKSHEET 2.2 Relating bonding to shapes

In molecules of nitrogen, N2, the 2s and one 2p orbital of the nitrogen atoms are mixed to form sp hybridized orbitals. One sp hybrid orbital from each atom forms a σ bond between the two atoms and the second sp hybrid orbital on each nitrogen atom is filled by a non-bonding pair of electrons. The remaining p orbitals, py and pz overlap sideways to form π bonds. π bond

b

a N π bond

×

N ×× N ××

N σ bond

Figure 2.2.23 (a) Formation of σ and π bonds in nitrogen. (b) Lewis diagram of nitrogen.

Theory of knowledge In ‘On Science and Uncertainty’, published in Discover, October 1980, Lewis Thomas (1913–1993) an American physician wrote: Science is founded on uncertainty. Each time we learn something new and surprising, the astonishment comes with the realization that we were wrong before. The body of science is not, as it is sometimes thought, a huge coherent mass of facts, neatly arranged in sequence, each one logically attached to the next. In truth, whenever we discover a new fact it involves the elimination of old ones. We are always, as it turns out, fundamentally in error. Scientists do not use the term ‘the theory of …’ except for those ideas that have been so thoroughly tested that when they are used to explain some phenomena they give the expected result every time. However, we have to be cautious about thinking that theories are absolutely true or have been proven right and assume this status will never change. No matter how many scientists confirm a theory through well-controlled experiments, there is always a possibility that someone

one day will be able to extend it, find it limiting, incorrect or even wrong, making each new theory in a sense closer to the truth than the previous one. • Is it likely that valence bond theory is closer to the truth about how atoms bond than VSEPR theory? Explain. • Is it possible for a theory to be proposed that has not been experimentally tested and verified? • If science can never prove anything to be true, why do we trust it so much? A good or ‘right’ theory is one that appears to work in all the cases that can be found. Can you recall any examples of theories in Chemistry that were once considered ‘right’ but we now know are ‘wrong’ or limiting, and ‘good’ theories that have been replaced by explanations that are even more right? • In what areas will scientific theories extend in the future? Do you think there is anything that scientific theories will never explain or will science one day give us the answers to everything; nothing will be beyond its scope?

Generally speaking, if there are four negative charge centres around the central atom, the hybridization will be sp3; if there are three negative charge centres, the hybridization will be sp2; and if there are two negative charge centres, the hybridization will be sp. 44

Table 2.2.1 Summary of hybridization

Number of negative charge centres around central atom

Number of non-bonding electron pairs on central atom

Type of hybridization observed

Example

Description of bonding between atoms

4

0

sp3

Methane, CH4

σ bonds between sp3 hybrid orbitals of carbon and s orbital of hydrogen

H

C H

H H

4

0

sp3

Ethane, C2H6 H

H

C

C

H

H

H

4

1

sp3

H

σ bonds between sp3 hybrid orbitals of carbon and s orbital of hydrogen AND sp3 hybrid orbitals of the two carbon atoms

σ bonds between sp3 hybrid orbitals of nitrogen and s orbital of hydrogen

Ammonia, NH3

N H

H H

4

1

sp3

Hydrazine, N2H4 H

H

N

N

H

4

2

sp3

σ bonds between sp3 hybrid orbitals of nitrogen and s orbital of hydrogen AND sp3 hybrid orbitals of the two nitrogen atoms

H

σ bonds between sp3 hybrid orbitals of oxygen and s orbital of hydrogen

Water, H2O O H

3

0

sp2

Borane, BH3 H

H

σ bonds between sp2 hybrid orbitals of boron and s orbital of hydrogen

B

H

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CHAPTER 2  bonding

H

45

Table 2.2.1 Summary of hybridization (continued)

Number of negative charge centres around central atom

Number of Type of non-bonding hybridization electron pairs observed on central atom

3

0

sp2

Example

Description of bonding between atoms

Ethene, C2H4

σ bonds between sp2 hybrid orbitals of carbon and s orbital of hydrogen, σ bond between sp2 hybrid orbitals of the two carbon atoms AND π bonds between two carbon atoms

H

H C

C

H

3

1

sp2

H

σ bonds between sp2 hybrid orbitals of nitrogen and s orbital of hydrogen, σ bond between sp2 hybrid orbitals of the two nitrogen atoms AND π bonds between two nitrogen atoms

Diazene, N2H2 H N

N

H

2

0

sp

Beryllium dihydride, BeH2 Be

H

2

0

sp

Ethyne, C2H2

H

2

1

sp

C

C

H

σ bonds between sp hybrid orbitals of carbon and s orbital of hydrogen, σ bond between sp hybrid orbitals of the two carbon atoms AND two π bonds between two carbon atoms.

σ bond between sp hybrid orbitals of the two nitrogen atoms AND two π bonds between two nitrogen atoms

Nitrogen, N2

N

46

H

σ bonds between sp hybrid orbitals of beryllium and s orbital of hydrogen

N

Section 2.2 Exercises   1 Both σ (sigma) and π (pi) bonds involve the overlapping of orbitals. Draw a diagram to show the difference between the overlapping that constitutes a σ bond and the overlapping to make a π bond.   2 Copy and complete the following table to give an example and to indicate what type of bond could occur between the orbitals or hybrid orbitals listed. Orbital 1

Orbital 2

Example of molecule in which the bond occurs

s orbital

s orbital

H2

s orbital

sp hybrid orbital

p orbital

p orbital

3

sp hybrid orbital

σ bond? (yes/no)

π bond? (yes/no)

sp2 hybrid orbital

  3 Consider the following diagrams and state whether they are illustrating a σ bond or a π bond.

a

b

c

i ii

  5 State the type of hybridization that would be occurring in order for the following geometry to occur around the central atom. a trigonal planar b linear c tetrahedral   6 Considering the bonding between two carbon atoms, outline how each of the following bonds are formed. a a single bond b a double bond c a triple bond   7 Describe the type of bonding and any hybridization involved in the bond between the carbon atom and the nitrogen atom in hydrogen cyanide, HCN. CHEMISTRY: FOR USE WITH THE IB DIPLOMA PROGRAMME HIGHER LEVEL

CHAPTER 2  bonding

  4 Use the concepts of σ and π bonding to explain why a carbon–carbon double bond is not twice as strong as a carbon–carbon single bond.

47

  8 Describe the hybridization of the nitrogen atom in ammonia.   9 The carbon atom in a molecule of methane, CH4, is sp3 hybridized. Explain how the geometry of the sp3 hybridization agrees with that suggested by the Lewis structure of methane. 10 The carbon atom in carbon dioxide, CO2, makes both sigma and pi bonds with the oxygen atoms. Describe the hybridization of orbitals around the carbon atom and around the oxygen atoms, and describe how the sigma and pi bonds are formed with oxygen. In your answer you should also account for any non-bonding pairs of electrons.

2.3 Delocalization of electrons In the previous section we saw that the Lewis structures of the molecules agreed with the structure explained by hybridization. However, not all molecular structures can be represented by just one Lewis structure.

Resonance structures In the case of some molecules and ions, the experimentally determined arrangement of atoms is not adequately described by a single Lewis structure. Usually in such cases, experimental measurements have determined the bond lengths of two or more bonds to be equal, yet it is not possible to construct one Lewis structure that represents this information. Ozone, O3, is one such molecule.

Figure 2.3.1 A molecular model of ozone with two equal length bonds.

Each oxygen atom in ozone has six valence electrons and needs to make two bonds to complete the valence octet, yet if two of the oxygen atoms each make two bonds, the third will need to make four bonds and this is not possible. A Lewis structure in which there is one O–O single bond and one O–O double bond meets the requirements of eight valence electrons on each oxygen atom; however, this does not match the experimental measurements. To explain the two equal O–O bond lengths of 0.128 nm we can use resonance structures.

A set of resonance structures shows a number of Lewis structures that are equivalent, having exactly the same placement of atoms, but differing placement of electrons; that is, double and single bonds are interchanged. Neither Lewis structure on its own truly represents the molecule; however, a blend of the two Lewis structures does the job well. The resonance structures of ozone are shown below. The doubleheaded arrow is used to indicate that the real molecule is described by an average of the two resonance structures. This average of the two structures is called a resonance hybrid.

O O

O

O

O O

Figure 2.3.2 There are two resonance structures of ozone.

48

Other molecules whose Lewis structures are best represented as a set of resonance structures include benzene (C6H6) and the nitrate (NO3−), nitrite (NO2−), carbonate (CO32−) and carboxylate (RCOO−) ions.

Example 1: Nitrate ion, NO3– There are three equivalent Lewis structures that can be drawn for the nitrate ion. In each case the arrangement of atoms is the same, but the placement of electrons differs. All three Lewis structures, when taken together, describe the nitrate ion in which all three N–O bond lengths are the same.

N

O

O

O

O

N

O

O

O

O

N

O

Figure 2.3.3 There are three resonance structures of the nitrate ion, NO3−.

Example 2: Nitrite ion, NO2− With one less oxygen atom than the nitrate ion, the nitrogen atom in nitrite has a non-bonding pair of electrons and a double and a single bond to the two oxygen atoms. There are two equivalent Lewis structures for the nitrite ion.

N

O

N O

O

O

Figure 2.3.4 There are two resonance structures of the nitrite ion, NO2−.

Example 3: Carbonate ion, CO32− The three resonance structures for the carbonate ion are very similar to those for the nitrate ion except that the ion has a 2– charge, rather than the 1– charge of the nitrate ion.

O

C

O

2

O

O

C

O

O

2

O C

O

Figure 2.3.5 There are three resonance structures of the carbonate ion, CO32−.

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CHAPTER 2  bonding

2

O

49

Example 4: Ethanoate ion, CH3COO−

Chem complement Dame Kathleen Lonsdale Kathleen Yardley was born in Ireland in 1903. She was educated in England, graduating in Physics from University College, London, in 1924. After graduating, she worked on molecular structure using X-ray crystallography in the research team of Sir William Bragg, also at University College. Kathleen married Thomas Lonsdale in 1927 and combined her work in science with motherhood, having three children. As a pacifist, she refused to become involved in civil defence duties during World War II, and was imprisoned. Her work in X-ray crystallography confirmed the structure of benzene as being that of a planar hexagonal ring of carbon atoms. Later work in X-ray diffraction measured all of the bond lengths in benzene to be the same. She also worked on the synthesis of diamond and lonsdaleite, a rare form of diamond found in meteorites, was named in her honour. In 1949, Kathleen Lonsdale was made professor of Chemistry at University College, London, and as such was the first female professor at that College.

The ethanoate ion is typical of a carboxylate ion, formed when a carboxylic acid donates a hydrogen ion in an acid–base reaction. While two resonance structures are possible for any carboxylate ion, note that when the acidic hydrogen atom is in place on the carboxylic acid, the C–O bond lengths are actually different, since they are locked in place as a C–O double bond and a C–O single bond.

O H3C

O

C

H3C

C

O

O

Figure 2.3.6 There are two resonance structures of the ethanoate ion, CH3COO−.

Example 5: Benzene, C6H6 In 1865, Friedrich August Kekulé von Stradonitz proposed that the structure of benzene was that of a single ring of six carbon atoms with alternating single and double C–C bonds. The length of the C–C bonds in benzene were found using X-ray crystallography to actually be equal in length (0.139 nm). To represent the true structure, benzene can be drawn as two resonance structures; once again remembering that resonance structures have the same arrangement of atoms but different placement of electrons. H H

H

C C

C

C

H

C C

H

H

H

H

H

C C

C

C H

also written as

Figure 2.3.7 There are two resonance structures of benzene, C6H6.

Figure 2.3.8 Dame Kathleen Lonsdale (1903–1971).

50

C C

H

H

Sometimes to make a scientific breakthrough, no matter how well tested the idea, you need to be able to think creatively. For German chemist Friedrich August Kekulé (1829–1896), the inspiration for his ‘out of the box’ thinking came from a dreamlike vision he had when he dozed off while thinking about the benzene molecule. Figure 2.3.9 Friedrich August Kekulé

von Stradonitz (1829–1896) discovered the ring structure of benzene.

Up until 1858 chemists did not really know what organic molecules looked like at the molecular level.

Kekulé, a theoretical chemist, was particularly curious as to why benzene, the liquid manufactured from whale oil and used for lighting, has such unusual chemical properties. Without a clear understanding of its structure, an explanation seemed impossible. One day while travelling home on the bus from his laboratory, he dozed off and in his dream saw benzene molecules twisting and turning in a snake-like motion. One of the snakes caught hold of its tail and made a ring-like formation. When he woke he had a flash of inspiration, and as soon as he got home he made sketches of his dream form of benzene. In 1865 Kekulé presented a paper to the Royal Academy of Belgium, proposing that the structure of benzene was a single hexagonal ring of six carbon atoms with alternating single and double carbon–carbon bonds. His theory met widespread approval from fellow chemists. However Kekulé’s structure did not stand the test of time. It wasn’t long before modern X-ray diffraction techniques gave chemists a deeper understanding of cyclic compounds, revealing that the carbon–carbon bonds in benzene were actually equal in length. Further experiments provided evidence that Kekulé’s alternating single and double bonds were not possible. First, benzene underwent substitution reactions instead of the addition reactions you would

expect of a molecule with three double bonds (Chemistry: For use with the IB Diploma Programme Standard Level, p. 376). Furthermore, Hess’s law data based on the hydrogenation of cyclohexene gave an expected enthalpy change of –360 kJ mol−1 for Kekulé’s structure, while that of benzene was −208 kJ mol−1, suggesting that benzene had a more stable structure.

3

+ 3H2 C6H10

∆H = −360 kJ mol−1

3 C6H12

+ 3H2

∆H = −208 kJ mol−1

C6H6 C6H12 Figure 2.3.10 A comparison between the hydrogenation of cyclohexene and that

of benzene.

The emergence of these problems resulted in Kekulé’s structure being replaced by a truer ‘resonance hybrid’ structure of benzene with the same arrangement of atoms but with the length of each C–C bond being somewhere between that of a single and a double bond. • Kekulé was more of a theoretical than experimental chemist. Distinguish between the two. • Kekulé was a talented linguist and artist, and his parents wanted him to become an architect. How do you think his experiences of language and art might have helped him as a chemist? • Kekulé’s story demonstrates that being able to make connections between seemingly unrelated phenomena can lead to significant advances in science. Since we can’t rely entirely on inductive or deductive reasoning as a way of knowing, scientists need to be able to tap into other creative ways of thinking. Edward de Bono is regarded by many as the leading authority in the field of creative thinking. What do you think de Bono meant when he said ‘Logic is the tool that is used to dig a hole deeper and bigger, but if a hole is in the wrong place, then no amount of digging will get you to your intended destination.’

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CHAPTER 2  bonding

Theory of knowledge

51

14.3.1 Describe the delocalization of π electrons and explain how this can account for the structures of some species. © IBO 2007

WORKSHEET 2.3 Delocalization of electrons

Delocalization of electrons (resonance) The molecules and ions that require sets of Lewis resonance structures to describe them can be accurately described in terms of hybridization of orbitals. Recall that π bonds were found to accommodate the second pair of electrons in a double bond such as the C–C double bond found in ethene, C2H4. In this case the bonding electrons in the π bond can be described as localized electrons, as they are restricted to the confines of region between the two carbon atoms. The equal length C–C bonds in benzene can also be described in terms of π bonds, but these electrons are delocalized. The distribution of electrons within a π bond can be called resonance, or the concept of resonance. This seems easy to confuse with the term ‘resonance structures’; however, it is important to note the context in which the term is being used.

Example 1: Benzene In a benzene molecule, the six carbon atoms are arranged in a ring in which the geometry around each carbon atom is trigonal planar. The C–C–H bond angle is 120°. The hybridization that matches this geometry is sp2, so we can deduce that each carbon atom has one 2s orbital and two 2p orbitals mixed together to make three sp2 hybrid orbitals. The third p orbital of each carbon is perpendicular to the plane of the sp2 hybrids and has one electron in it. These six p orbitals form a π bond right around the benzene ring, and the six unpaired electrons are delocalized within this π bond. This equality of bonds in benzene agrees with the experimentally determined bond length of 0.139 nm.

H C

H

C

C

C

C H

H

C

H

H

Figure 2.3.11 The six unpaired electrons can be represented in a delocalized π bond in the benzene ring.

– O O

N O

Figure 2.3.12 The unpaired electrons can

be represented in a delocalized π bond in the nitrate ion.

52

Example 2: Nitrate ion, NO3− In the nitrate ion, the nitrogen atom has a trigonal planar geometry, so we can assign sp2 hybridization to it. In addition, each oxygen atom will also be sp2 hybridized, with two pairs of non-bonding electrons occupying two of the three hybrid orbitals. A σ bond exists between the nitrogen and oxygen atoms and in addition to the extra electron which gives the ion a negative charge, the last electron in each atom occupies a p orbital perpendicular to the plane of the sp2 hybrid orbitals. These p orbitals create a delocalized π bond.

Example 3: Nitrite ion, NO2− The nitrite ion has a bent linear geometry, due to the presence of the nonbonding pair of electrons on the nitrogen atom. We can assign sp2 hybridization to it. In addition, each oxygen atom will also be sp2 hybridized, with two pairs of non-bonding electrons occupying two of the three hybrid orbitals. A σ bond exists between the nitrogen and oxygen atoms and, in addition to the extra electron that gives it the negative charge, the last electron in each atom occupies a p orbital perpendicular to the plane of the sp2 hybrid orbitals. These p orbitals create a delocalized π bond.

 – O  N O

Figure 2.3.13 The unpaired electrons can

be represented in a delocalized π bond in the nitrite ion.

Example 4: Carbonate ion, CO32− In the carbonate ion, the carbon atom has a trigonal planar geometry, so we can assign sp2 hybridization to it. In addition, each oxygen atom will also be sp2 hybridized, with two pairs of non-bonding electrons occupying two of the three hybrid orbitals. A σ bond exists between the carbon and oxygen atoms and, in addition to the extra two electrons that give it the 2– charge, the last electron in each atom occupies a p orbital perpendicular to the plane of the sp2 hybrid orbitals.These p orbitals create a delocalized π bond.

2– O O

 C O

Figure 2.3.14 The six unpaired electrons

can be represented in a delocalized π bond in the carbonate ion.

Example 5: Ethanoate ion, CH3COO−

 – O  H3C

 C O

Figure 2.3.15 The four unpaired electrons can be represented in a delocalized π bond in the ethanoate ion.

CHEMISTRY: FOR USE WITH THE IB DIPLOMA PROGRAMME HIGHER LEVEL

CHAPTER 2  bonding

In the ethanoate ion, the carbon atom in the carboxylate ion (COO−) has a trigonal planar geometry, so we can assign sp2 hybridization to it. In addition, each oxygen atom will also be sp2 hybridized, with two pairs of non-bonding electrons occupying two of the three hybrid orbitals. A σ bond exists between the carbon and oxygen atoms and, in addition to the extra electron that gives it the negative charge, the last electron in each atom occupies a p orbital perpendicular to the plane of the sp2 hybrid orbitals. These p orbitals create a delocalized π bond.

53

Section 2.3 Exercises 1

Explain why resonance structures were found to be necessary for molecules and ions such as C6H6 and CH3COO−.

2

Draw the resonance structures of:

a HCOO− b NO3− c SO42− 3

Describe how you would expect the actual sulfur–oxygen bond lengths in SO42− to compare with a sulfur–oxygen single bond and a sulfur–oxygen double bond.

4

a State the names of two molecules in which delocalized electrons may be found. b Name the type of obitals in which delocalized electrons are found.

5

a State which of the following molecules or ions will exhibit delocalized bonding: SO32−, H2CO, O3, NH4+. b Draw the resonance structures of the molecules or ions that exhibit delocalized bonding.

6

a Use the concept of resonance to explain why the six carbon–carbon bonds in benzene are of equal length. b Explain why the structure of benzene is drawn as a hexagon with a ring inside it, when the concept of resonance is being considered.

7

Naphthalene, C10H8, is a molecule made up of two six-membered rings with one shared side.

a Draw two Lewis resonance structures for naphthalene. b The observed carbon–carbon bond lengths in naphthalene are shorter than carbon–carbon single bonds and longer than carbon–carbon double bonds. Explain this observation using the concept of resonance. c Represent the resonance in naphthalene in a way that is analogous with that used to represent resonance in benzene. 8

54

a State and explain whether the π bond in CH3COO− is localized or delocalized. b State and explain whether the π bond in CH3COOH is localized or delocalized.

Chapter 2 Summary

Square planar   A molecular shape (geometry) in which four atoms surround a central atom in a plane.

Terms and definitions

Trigonal bipyramid  A molecular shape (geometry) in which five atoms surround a central atom, with three in an equatorial plane and two in axial positions.

Axial position  Atoms arranged above and below a central atom in a molecule. Delocalized electrons  Electrons that are free to move within a pi (π) bond. Equatorial position  Atoms arranged around the middle of a molecule at right angles to those in the axial position. Hybridization  The process of mixing atomic orbitals as atoms approach each other. Hybrid orbitals  Orbitals formed when atomic orbitals in the valence shell of an atom are mixed together. Localized electrons  Electrons in a pi (π) bond which cannot move beyond the two bonding atoms.

Valence-bond theory  A theory in which regions of electron density between two atoms is explained in terms of the overlap of two atomic orbitals.

Concepts •

Lewis structures (electron dot diagrams) and valence structures can be used to represent molecules with five or six negative charge centres.



The VSEPR (valence shell electron pair repulsion) model is used in determining the shapes of molecules and angles between the atoms in a molecule (see table 2.4.1).



In valence-bond theory, bonds are formed by the overlapping of orbitals. Such orbitals may form sigma (σ) bonds or pi (π) bonds.

Negative charge centres  Pairs of bonding or non-bonding electrons around a central atom. Octahedral (or square bipyramidal)  A molecular shape (geometry) in which six atoms surround a central atom, with four in an equatorial plane and two in axial positions.

1s

1s

1s

3p

Octahedron  A shape with eight faces, eight edges and six vertices.

Orbital diagram  A diagrammatic representation of electron configuration showing electrons as K and l in boxes. Pi (π) bond  Bond between two atoms formed by the sideways overlapping of two orbitals (usually p orbitals). Resonance  The concept of electrons being delocalized in a pi (π) bond rather than being bound in a double bond. Resonance hybrid  The average of two resonance structures, which represents the observed molecule accurately.

3p



Sigma bonds result in the formation of single covalent bonds.



Pi bonds, together with sigma bonds, result in the formation of double and triple covalent bonds.



Hybrid orbitals are formed by the mixing of two or more atomic orbitals, and support evidence of numbers of equal bonds around a central atom.



Hybrid orbitals are named according to the orbitals that are mixed together to form them.



Resonance structures are used to accurately represent a molecule for which one structure is not sufficient.

Resonance structures  Different Lewis structures of the same molecule in which double and single bonds alternate. Sigma (σ) bond  Bond between two atoms formed by the end to end overlapping of two orbitals.

3p

O

X

O

O

O

O

O

X

O

O

X

O

Square bipyramidal  see Octahedral.

CHEMISTRY: FOR USE WITH THE IB DIPLOMA PROGRAMME HIGHER LEVEL

CHAPTER 2  bonding

Octet rule  That the valence shell of an atom should contain no more than 8 electrons.

55

Number of negative charge centres

Number of bonding electron pairs

Number of non-bonding electron pairs

Shape of molecule

Example

5

5

0

Trigonal bipyramid

PCl5

Diagram of shape

Cl

Cl

120°

90°

Cl

5

4

1

Seesaw

SF4

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Table of contents

Stoichiometric Relationships
Atomic Structure
Periodicity
Chemical Bonding and Structure
Energistics/Thermochemistry
Chemical Kinetics
Equilibrium
Acids and Bases
Redox Processes
Organic Chemistry
Measurement and Data Processing
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Option C: Energy
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