chemistry new edition

Trishna’s Pearson IIT Foundation Series Chemistry 9C L A S S > Provides student-friendly content, application- based problems and hints and solutions to master the art of problem solving > Uses a graded approach to generate, build and retain interest in concepts and their applications

About Pearson Pearson is the world’s learning company, with presence across 70 countries worldwide. Our unique insights and world-class expertise comes from a long history of working closely with renowned teachers, authors and thought leaders, as a result of which, we have emerged as the preferred choice for millions of teachers and learners across the world. We believe learning opens up opportunities, creates fulfilling careers and hence better lives. We hence collaborate with the best of minds to deliver you class-leading products, spread across the Higher Education and K12 spectrum. Superior learning experience and improved outcomes are at the heart of everything we do. This product is the result of one such effort. Your feedback plays a critical role in the evolution of our products and you can contact us - [email protected]. We look forward to it. A01_GATE_978-93-325-7606-3_PRELIM.indd 1 6/16/2017 6:06:40 PM

This page is intentionally left blank.

CLASS 9 Pearson IIT Foundation Series Chemistry Seventh Edition

This page is intentionally left blank.

CLASS 9 Pearson IIT Foundation Series Chemistry Seventh Edition Trishna Knowledge Systems

Photo Credits Chapter 1 Opener: Martyn F Chillmaid. Pearson Education Ltd, Dotta 2. Shutterstock Chapter 2 Opener: Konstantin Faraktinov. Shutterstock Chapter 3 Opener: isak55. Shutterstock Chapter 4 Opener: Vasilyev. Shutterstock Chapter 5 Opener: Oxford Designers & Illustrators Ltd. Pearson Education Ltd Chapter 6 Opener: Oxford Designers & Illustrators Ltd. Pearson Education Ltd Chapter 7 Opener: Sozaijiten Chapter 8 Opener: Pearson Education Ltd Chapter 9 Opener: Eclipse Digital. Shutterstock Icons of Practice Questions: graphixmania. Shutterstock Icons of Answer Keys: Viktor88. Shutterstock Icons of Hints and Explanation: graphixmania. Shutterstock Senior Editor—Acquisitions: Nandini Basu Editor—Production: Sakshi Kansal The aim of this publication is to supply information taken from sources believed to be valid and reliable. This is not an attempt to render any type of professional advice or analysis, nor is it to be treated as such. While much care has been taken to ensure the veracity and currency of the information presented within, neither the publisher nor its authors bear any responsibility for any damage arising from inadvertent omissions, negligence or inaccuracies (typographical or factual) that may have found their way into this book. Copyright © 2018 Trishna Knowledge Systems Copyright © 2012, 2014, 2015, 2016, 2017 Trishna Knowledge Systems This book is sold subject to the condition that it shall not, by way of trade or otherwise, be lent, resold, hired out, or otherwise circulated without the publisher’s prior written consent in any form of binding or cover other than that in which it is published and without a similar condition including this condition being imposed on the subsequent purchaser and without limiting the rights under copyright reserved above, no part of this publication may be reproduced, stored in or introduced into a retrieval system, or transmitted in any form or by any means (electronic, mechanical, photocopying, recording or otherwise), without the prior written permission of both the copyright owner and the publisher of this book. ISBN 978-93-528-6675-5 First Impression Published by Pearson India Education Services Pvt. Ltd, CIN: U72200TN2005PTC057128 Head Office: 15th Floor, Tower-B, World Trade Tower, Plot No. 1, Block-C, Sector-16, Noida 201 301, Uttar Pradesh, India. Registered Office: 4th Floor, Software Block, Elnet Software City, TS-140, Block 2 & 9, Rajiv Gandhi Salai, Taramani, Chennai 600 113, Tamil Nadu, India. Fax: 080-30461003, Phone: 080-30461060 www.pearson.co.in, Email: [email protected] Compositor: Saksham Printographics, Delhi Printer in India at

Brief Contents Prefacexiii Chapter Insights xiv Series Chapter Flow xvi Chapter 1   Nature of Matter 1.1 Chapter 2  Atomic S­ tructure  2.1 Chapter 3   Periodic Classification of Elements 3.1 Chapter 4   Chemical Bonding 4.1 Chapter 5  Mole Concept, Stoichiometry and Behaviour of Gases 5.1 Chapter 6  Chemical Kinetics and Chemical Equilibrium 6.1 Chapter 7   Water, Solution, Solubility and Hydrogen 7.1 Chapter 8   Metals and Non-Metals 8.1 Chapter 9   Organic Chemistry 9.1

This page is intentionally left blank.

Contents Prefacexiii CHAPTER 3 PERIODIC CLASSIFICATION Chapter Insights xiv OF ELEMENTS 3.1 Series Chapter Flow xvi Introduction3.2 CHAPTER 1  NATURE OF MATTER 1.1 Newland’s Arrangement of 3.2 Elements in Octaves Introduction1.2 3.3 Mendeleev’s Periodic Table Postulates of Kinetic Molecular 1.2 3.5 Theory of Matter Modern Periodic Table—the Long 1.2 Form of the Periodic Table Different States of Matter and 1.3 Their Properties 1.7 Periodicity3.9 1.7 Interconversion of States of Matter Atomic Size 3.9 Latent Heat Ionisation Energy 3.10 Applications of Latent Heat Electron Affinity 3.10 Electronegativity3.10 Classification of Matter on the Electropositive Character 3.11 Basis of Chemical Composition 1.9 Oxidizing and Reducing Property 3.11 Separation of a Mixture 1.11 Metallic and Non-metallic Property 3.11 Conditions Required for Chemical Changes 1.15 Practice Questions 3.14 Practice Questions 1.18 Hints and Explanation 1.25 Hints and Explanation 3.21 CHAPTER 2  ATOMIC STRUCTURE 2.1 CHAPTER 4  CHEMICAL BONDING 4.1 Introduction2.2 Introduction4.2 Discovery of Fundamental Particles 2.2 Reasons for the Formation of a 4.2 Chemical Bond Millikan’s Oil Drop Experiment 2.4 4.2 Types of Chemical Bonds 4.3 Discovery of Protons 2.4 Lewis Dot Formula 4.4 Covalent Bond 4.5 Thomson’s Atomic Model 2.5 Polar Covalent Bond 4.8 Coordinate Bond 4.9 Rutherford’s α-ray Scattering Experiments2.7 Hydrogen Bonding 4.10 Metallic Bond 4.11 Rutherford’s Atomic Model 2.7 Redox Reactions 4.12 Quantum Theory of Radiation 2.9 Practice Questions 4.19 Hints and Explanation Bohr’s Model of an Atom 2.10 Discovery of Neutrons 2.11 Practice Questions 2.15 Hints and Explanation 2.22

x Contents CHAPTER 5 MOLE CONCEPT, Important Properties of Water 7.3 STOICHIOMETRY AND BEHAVIOUR OF GASES 5.1 Solvent Property 7.3 Molarity7.6 Introduction5.2 Action of Water on Some Metals 7.9 Symbols and Formulae 5.2 Electrolysis of Water 7.10 Radicals or Ions 5.2 Mechanism of Electrolysis 7.10 Derivation of Formulae of Compounds 5.4 Faraday’s Laws of Electrolysis 7.10 Naming of Compounds 5.4 Boyle’s Law 5.8 Practice Questions 7.15 Charles’ Law 5.10 Hints and Explanation 7.21 The Gas Equation 5.12 Standard Temperature and Pressure or CHAPTER 8 METALS AND 5.12 NON-METALS8.1 Normal Temperature and Pressure Measurement of the Mass of Atoms 5.13 Introduction8.2 5.14 and Molecules 5.16 Metallurgy8.2 Avogadro’s Number and Mole Concept Graham’s Law of Diffusion Metallurgy of Iron 8.5 Open-hearth Process for Making Stoichiometry5.18 Special Type of Steel 8.7 Balancing of Chemical Equations by Trial and Non-metals8.9 Error Method 5.18 Carbon8.9 Practice Questions 5.24 Nitrogen8.12 Hints and Explanation 5.31 Phosphorous8.17 Oxygen8.19 CHAPTER 6 CHEMICAL KINETICS Sulphur8.21 AND CHEMICAL EQUILIBRIUM6.1 Transition Between Rhombic Sulphur and Monoclinic Sulphur 8.23 Introduction6.2 Practice Questions 8.30 Hints and Explanation 8.36 Classification of Reactions 6.2 CHAPTER 9  ORGANIC CHEMISTRY 9.1 Rate of a Reaction 6.3 Chemical Equilibrium 6.7 Introduction9.2 Reversible and Irreversible Reactions 6.7 Unique Properties of Carbon Dynamic Equilibrium 6.8 Compounds9.2 Law of Chemical Equilibrium or 6.9 Classification of Hydrocarbons 9.3 Law of Mass Action 6.14 6.22 Nomenclature of Hydrocarbons 9.5 Practice Questions Hints and Explanation Isomerism9.8 CHAPTER 7 WATER, SOLUTION, Homologous Series 9.9 SOLUBILITY AND HYDROGEN7.1 Alkanes9.12 Methane9.12 Introduction7.2 Alkenes and Alkynes 9.14 Sources of Water 7.2 Preparation Method of Ethene 9.15 Removal of Hardness 7.2

Natural Sources of Energy 9.19 Use of Organic Compounds in xi Soaps and Detergents Formation9.19 9.24 Practice Questions 9.27 Coal9.20 Hints and Explanation 9.34 Fractional Distillation of Coal Tar 9.20 Some Important Applications Of 9.23 Organic Chemistry 9.23 Use of Organic Compounds as Ingredients of Food

This page is intentionally left blank.

Preface Pearson IIT Foundation Series has evolved into a trusted resource for students who aspire to be a part of the elite undergraduate institutions of India. As a result, it has become one of the best-selling series, providing authentic and class-tested content for effective preparation—strong foundation, and better scoring. The structure of the content is not only student-friendly but also designed in such a manner that it motivates students to go beyond the usual school curriculum, and acts as a source of higher learning to strengthen the fundamental concepts of Physics, Chemistry, and Mathematics. The core objective of the series is to be a one-stop solution for students preparing for various competitive examinations. Irrespective of the field of study that the student may choose to take up later, it is important to understand that Mathematics and Science form the basis for most modern-day activities. Hence, utmost effort has been made to develop student interest in these basic blocks through real-life examples and application-based problems. Ultimately, the aim is to ingrain the art of problem-solving in the mind of the reader. To ensure high level of accuracy and practicality, this series has been authored by a team of highly qualified teachers with a rich experience, and are actively involved in grooming young minds. That said, we believe that there is always scope for doing things better and hence invite you to provide us with your feedback and suggestions on how this series can be improved further.

1CChhaappMCttSNeelrraayustsmtsteeibfiCMrmSNelcaaysarCuststmhtstieaeoipbfirmntecersaroItfniosinghotsf Remember Remember be ablRhe eelpmtehmembertosemcteiomnowr ieilsl and review the previous Before beginning this chBaepfoterer, byeoguinsnhionugldthbise cahbalepter, you should learning on a particular to: to: • define substance, mixtu•re,dmefiansse, swuebisgtahnt,ce,tcm. ixture, mass, weight, etc. • know the states of matte•r ksonloidw, ltihqeuisdtataensdogf ams atter solid, liquid and gas topics • basic properties of matte•r basic properties of matter 4.2 Air and Oxygen KINeyTRpOoiDntUs CwKTiIlOlehyNelpIdeas Key Ideas tThhee sptlaundeteenatrtshAthoafstiednrednocwtoiefmdyups lweittihntghe bthasicsAnecfchteesasriptitecesoro,mf liypfeoleluikteinsahgir,owtuhaltdiesr, bfcoehodapettce. rA,myonogu should be tctowhhhfheeIagsinceapeh,stethasbesisereerwceinsohartmtihlciiheeearslmstcu•a•hpolraibosrnsdktoslnieeuvennesinf,rottitdawwnswolseefii:totnthhhrmheadleavipaifeseatltatarlanennrearcadrt,ennhemftgraeoretnohtmmchaleetteicihsannuiecrdtlaeiielsoasla••p,erfaebtdahnmkdhtlsaeoneaatonbfmmomidtllneweooo.tcegh:suettepmhnclhree.eeaosgratuiietrosne.rnmraAt,anhitbmxmgroteeuovoermeslepedthehocieneffurftegaleetaormis,seefoansam.sttpocTohlmohelueirsedceeuwtonclvhfe.eigscloahispneiessthree different states of matter. 4.2 Air anadbsOolxuytegleynfree of airstias tceaslleodfomutaetrtesrp.ace. nINatTuThraRel aOptrmoDcoesUspsheCser•TleikIOmuheonfaolNddtrstmeetrahrstetiaoa4annni.rdd1ow2fcectolhobreurrAeedailsrtia,hmatpeenr,pedmicotiOoprwtii•xstatayinuttumgirhcoeneenanddatneteateodyrcrf.sdttauaoinsnntdddtpeaacyrrtocthiaorcecrlneetsvilivaewmtirehtsipiiecioothsnrwatareinothacfsesdocaoiyaftetodinwdteiatyrhcaocntviveirtsiieosn of ctThoheAvesetmp,rlawaoinrshpeiishtcehthraeiersthmaful•sohrotsahltspeeverariontrdtadneliovcfwitoadsrebeudldoisfueiunfsrtaoownmddiftihhvfhefaetnrhdrmceieestfnbiuinantlscdirtkciasldi•pannyieeaendlctreisesoaas.bnsroilstne.if.eHsmaoebnfaoctluitefe,etriTltdihkliiisiesfktlfeaceeiforru,ennenwsnleiedtal mteerkree,idnnfodatosssdaoeptfcr.omAteamctDttoeienlvivregerlyitkubee elementTs ext: concepts are F(LdOonwaiibeaafg)hustnITugouyTinrscahrrlr(iehutsoaaeateyehlt)trp,bsemCaepstleotTwyremoaceosomhoshafpcpoprrefmipeetlshhciispeeloAheeesaerhersyrnoartesesettfecmur•uhrlleaiirarikosirneosracumheofsndeanoiosuanfesitsitolpnvdm,rdmdxcdridhswadwmteiosnplfuetlrefeiseahedototsprhtrdhrehteihnueeeoaiasodenna-nnvn.ripiuttmersedodlotthaawwslefneeenfreiiceathtcgvaralhtseioehbnnclpeeurstfhtdaseral.daortacbfyshtdtrme,emhehoai.grptemfstiiir,fhnsaxeeomietsccrtrhniauaeeopeitlasnilri•tetstrsehhatatutdushetemcaurihrfoanomeaenaobatnincadmmanmdxareedsnetdteiotnhptoosduthrcgh’fosos.sroeadoepttesnunflhuauhdrsse-Mnesnerv.oesotfmgaaddreauprriecabsoicstea.erlaeonroahDttttmaAthaiccyahiimolannhdiu.cidltbeirxsnstmdlpodeeeo.ptsscuseovphadxirelmswonarepitretrefihhchiofancxieeefhctdmrtreheesgauemtaeuanmrisstptreietehcaossata.lseoaslpmcTsahs2lcnooh0eoeucdirmtskiedhameptwaooenolddvhfhsseioiesgtwcliioahogiscpthtneioheosst, mseppaareratme etthalCmesestaxrrbapuonnlcandtiiouenxirrdeeedd in a well and lucid ptiFoInG U R E 1.1 Atmosips hcearlleedaltsoroproostpechtes rues. ipnfertorocmefnivhteaagrdemisotfiufnltchrtaeldatMyioaettreaisolt.hnmos.adsHsooeffncctohe,elleiatctmFtisiigocsunoprhnoeesfir4dgeLe..aa4rbse:odLrCaaatbOosor2yraabptpreoeirrpnyoagtrpaehrtceieoptaniavvorieeaf rctiaotrhnbaoonnf dciaorxbidoen dioxide upward displacement cove(rb)wIhtichholids sfuthrtehmeradxiivmiduemd air is collected by aNLdOdeaoudnyrst-i(eetaocy)tr,mnbstuTieoonopmihscxnppefrpeeehfedraAoesrserlaerteanatmymuisreniiretsreoyashdasstetoithoipvfgoedimthhdihntfteeefeedeamrortaeeipitfrnnhettareonahtrdeuafiitrtgveeeehmodtslfrpaof6yerpe°orsaCrmstfu/rookrtehnmmegtarishfNosuaboefdrcorofuaamuatbilacretlael.e:ls1dyi7Dosbld°faioeCeltpufichnsrtevegeeoaadeirrs5aneiaH1rastttow°he2iCloS..uitnO.bhTsl4ehhineieisnifgncawhhlolteat.itmnAeuristscefawdomlrefpmcoemorsrmaoatthvpuleeaoryestpeiotwrrieotoiphtnvhae,erratmioanrbolef CO2 because the calcium sulphate chips and thus marble chips do not 2.8 Atomriec lSattr(edu)cdtuMtroeopsatniocdfstThraenwsefoarthmeartpiohnenoofmMeantateorccur incothmiselainyecrodnuteacttowraitphiddivlaHria2StiOon4sfiunrttheemr.perature. (i) Tr((oae))poTTshhpeehlhaeeyrieegrhotfaantdmtohsipchkenreesswohficthhisbelagyienrs varies from the equatorCtoaCthOe3p+oleHs.2SO4(dil) → CaSO4↓ + H2O + CO2↑ at the earth’s surface and extends up to 20 km height (iiE)(sG(xaaSo(((i(())adbabvcctr)))))elFmaFutlleiTuiTATIIMtohosnottptapeuehhbrcrchhosimlpiroteeraoenpoosEhieevltplneldllraeh’lenesddeeaosecZldeyessrsatftnauehrryt1titsto=rehrenihiru9sftnreoe,ary%os9hiecpctfmwetoeeoohd(icamfaegbifesootxsathmpp)tenhtitchthmhesefeoaSeirameenlgsuaotalrpraueotimpdetuthrpdrthmei.arehuareaesptrnaeomilteottrnidrseunrapnaocartrcthteebmeearomenoemnoderstavudfpianrpsecocge6ahgehpseg°treeohsaCaworouc(tsefe(muce/fpcirPPb.tos)kuCOrthhhe.e)thrmmrtooyyodCerShgislspoinPooieaharicitcoushcaobiudalagthosdraylcotirhlphuaersuarluiphliitiemlsacnPgtlrepmledaloherlr1ardyulZoeptaey7pespseeld°osewa=rrrCeetofnptoochdript1aefsrtatuibeitee1scotteheroahiootsers5iuetefe(ano1oetadsatCw°xbfoe2t)rCTtmsOa.fC5ieoACtp.tnhoTra2hikbapddsreamhrhpCOtLsbbriveehfeeooooduoaifefnlgrornsnlpoar4olhihueoluaolm.o-t.rttwr6.lfoiiwledeonAihstsnnsnis5hPiotsggse0ehixwhmniyeinekdlesepTeaimtenrcmemaet(aerbmhCeaorall’tngesbvpOPueteoesr4srtuo2.veri.tac)6repowtf.euaUttcrirhhttVeehiee..esCohfaCraacrtbeorinstidcsiopEwcxoaxiidrnsaetecmicetpuopllteaassrplgecpiahlvyraenpntehtdeteoripnica- S((((Mbdaetr))))eaEsTIetAwrrTltehoaatoahehcddierhststrteeiihproophaaeotthplah’lhsiihdnaereooleyeaseits–innchunegirr1gssere’reehcs9.h–fsaota%osHtncnufafldeinefyrosiafrdgitaftiecshunmcttaucethrhlhtoaaelseihei1tle–cs9slipd0ipaekposntrcthdnneomilaeeamnetlsy–=rleosvteeespcrsedr2pteaoe.rasr,h–tofsbalo7ieiozttofhrsuveopinirecnseohoglnlteeraaha–siyrysstepeeeehCer.hssVNSFetr.hrevo.ofaoaerearEplroTtapuermmuolrmioebhetruoichsaieescrlprtt–.iarfithrdoor6oyeloee0nnzgPrmin°ioenicroCsooniftnwwcthpeyotdaoheeenrltixrayeefc–tityrqhig1eieeeucu5n–sreear°d.eatxCoeso–tTt–ife.rfohrnCTtnoio11dseamh12HAHAn=s–lpnrtioacthsbieuyzg2i2avediozpehv,5nelirlniepcre8cyety.–2ekoarr,oi/-sH–teebm2hlo1t–ee/afhsdi25elo–sgsou0an.ei0erht1nbromoc7–bloepxeakss,2.fri4im:redsc4hS,0tos2heoeasv:l2uamirlaka8dunrmpbmPebgpocoMseifeaulvuswirrarteblabybhdotlUoeuutiennhnvenrVceeedsilriitiettoymaixss=ieuadslsel2wot2wowitehrhedicdinht2.o/c2ir/see2x0ka1psnsIs7eaotolni2lwen:dul4pnv2ps:s2uer-at8edswrdPsddsMauertiirnyesinlvyeicetahem.reoloauxgngahnmiacesapmrllaeallsnd (ii) (iii) ((ac)) ATrTisaebhdcmoeiaavlptlleaeieoydrtenahmrtsiu.sa,reHbetsoeeeod–vmrseoeppettheshhreenearotseteutte–.rrmaaetloptienserpcrahrmeteua–ursreceehswrwiuhsepiiCctshhhtforeheomexamitighechn–eatil6dgp0sdh°ruttooCepoateftbooraotbti–ueho1tseu58oeta°5–fbCT29CsA5.o0OTrbkkpLm2hmteiieaos–rfan4eribno.osocm7hvrfoeehtawChtishgheneheeoeimneefaanritTcrteethaamrh’blgs’splePestesruiuor4crarpf.tfa7Ueuac.crrVetee.ies of Carbon dioxide (b) wTeimthphereaigtuhrteids eccarleleadsesinfrvoemrsiaobnoultay–e1r5.°CThteo r–e1g2i0oP°nrCope.xetretinesds from 25 km to 40 km aboveObservation and inference tMheeteeaorrtihte’sssbuurfranceupis icnaltlheids olazyeorn. osph(ie) rCeoomr bouzsotinbeilitlyayer. absIot rdboseshna’trmbufrunl aUndVit doesn’t support burning of other substances in (c) radiations and thus protects life on the earth. This layer its atmosphere. Hence it is neither combustible nor a supporter of combustion

Chapter Insights xv 1.14 Classification of Matter TeST YOUR CONCepTS Very Short Answer Type Questions 16. Which among the following is an element? Different levels (a) Calcium oxide of questions have Directions for questions 1 to 10: Fill in the blanks. (b) Common salt been included (c) Ozone in the Test Your 1. In a ______ properties of the constituents are retained. (d) Water Concept as well as on Concept 2. A mixture of alcohol and water is an example of 17. Evaporation is the process of conversion of Application which ______ mixture. (a) a liquid to its gaseous state below the boiling will help students point of the substance 3. A substance which is formed by the chemical com- (b) a liquid to its gaseous state at the boiling point bination of two or more elements is called a ______. of the substance 4. Boiling point is the temperature at which ______ is converted to ______ at one atmospheric pressure. 5. Nonmetals usually exist in ______ state. 6. ______ show the properties of metals and nonmetals. (c) a solid to its liquid state at the melting point of to develop the the substance problem-solving 7. With an increase in the surface area of a liquid, the skills rate of evaporation______. (d) a solid to its liquid state below the melting point of the substance 8. The conversion of a liquid to its solid state on cooling is called ______. 18. The most convenient way of separating saw dust from water is 9. During freezing, heat is ______. (a) distillation 10. Atomicity of ozone is ______. (b) evaporation ‘Test Your Classification of Matter 1.27 (c) filtration 1DtSh1eiCtte.lrehheGccqoteetueionntenechescsrtehnaiefolcoadlpynropsrmtq,otrsueee’ffetcosarattulistoranltcsehrfornoiacmteivs1e1h. atvoe3b0:eeFnCoLrepOrveoeNavclihd1Ceodef.pT (d) sedimentation and decantation a fo((abr)) caarrleeasgsooslirodods ocomnductors of heat True or false pRACTICe QUeSTIONS 1A9.pWphlicIhCamAoTngIOtheNfollowing statements is true? (a) Compounds are heterogeneous in nature. (b) The proportion of constituent elements in compound is fixed. pr(ce) phaavreahtiigohntesnsile strength 1. Gases have max(cim) uiTnmhaecicnootmnesprtmiotuuoenlndet.ceulelamresnptascreetsa.in their prope7rt.iesAs the density of sodium is less than that of water it (d) all the above 2. Metals are high(ldy) dTuhcetfiolermaantidonmoaflalecaobmlpeo. und is a physical process.floats on water. 12. Aridme icxatnurbeeosef pcahraaltkedpobwy d_e_r__an__d_a.mmo3n.iumEvcahploor-ation20t.akWeshipclhacoef tfhroe mfolltohweinsugrifsaacheeotefroagleiqneuoidus mixt8ur.e?Ammonium chloride undergoes sublimation and hence it is a sublimable substance. (a) distillation (b) evaporationand hence it is (aa)suArfmaciextpurheeonfowmaetenroand. sugar 1143s‘dpA..Ceri(((IeWpaccdvcloe)))oehtpicnbiidimftnrctilrololiihieatfcecncyrngiadaemtnattmyihepot?wseaosinitunossiomgtnfpthth’emereftoallloamwionn(((gdbdg)))isthapsaueloubbtfmolaaisdmlislnoicuaiwo465utminmi...ondngtPaTIu:ionceeudtdlsrolieuornhrfseoeiuufbnibsmocsetathaial2snurm1caes.etemrhsmW(((((tocdbacoaae)))ui))lhmxrtsseasimAtAAcoualonlhmnidromgmmloekadie?anaidiiwnxdxxmnmttteeaauouuotnoererrnnuetdueergampsooolh.feoitffenhtwwwfnaenailaacdstttefa.eeeoetrrriulnltaaaortnnnsiewhcddd.aio((nsgcldbwagol))wmumscslioisduopmqlugsrueaeooasictrndphuweesalaermtale-sttremorgoenni1ae90o..usspSpTouhrholybiedpsliiecmctrooatanlietismtistosietangutnahedsiosnedothtssuhe.onsefcspatearctoteohc.meesspes ocouafnntdnhodetocbnoeonsvteerperastiaroainntetdohfebiyar co(am) zipnclexity: (b) copper 1.28 ClassificLat(eicov) nealloumf1Mi;niaLutmteevr el (d) phosphFoirlulsin the blan22k.sWhich among the following is a pure substance? (a) dilute sulphuric acid iAWbDeifTnsonehiutranrttomewtrosieisuaneartcbteeagClihsimnshotqhatapitantucopwnhenrkiemg1eoadc_sre1tf1e.aoiefnits5tomdisgro2dun.arecari1rm;((Tmn4xagenmeaclhet)l)aadalaumeyaooetsnactridgulstnooieseodlxmsowsmnacmdme.tsmuaudhpseLlhol.eloiaileapcftxuresvithhttnneoavcuydpgoireasnamaeimrpdlcdtsiooeaiohpxc3plifllelotneeiyudsdsntiiroivadtfefniegfoenedatrnfnttheoshee((ndredbexccltu))nieceosgmrtcosnimeiaoennenvsanonesgectclthtseretitittecwinheuusr111111dghee-lc412356eiac......lhleAiTbGEIcSdsomhoovh334rsdaepliaaate089il/iymrpupdina...onohrcnoeisnegScnrhisCRsocritrsupatgaaaiaauemseitasahtrrtcnlusnemirefbatoaepodmiabbdceanosrilbconeosllsdenbeuteeai,bennoiastucsryniwiehniniftnttlasiscnroekoc(((nofrfthoodcbcdidtieefeedsnb)isluaa))pmuovartcssmnirssrhm.auochalscmeeapupsqeotnor,feptbhuoirobufnoignftoleaeaoaorlelaleaorcartesnceflomntaeeaisasmcceahmnintunet.acruinobioeesnnpeteomeeNGlldmrlkgceaNadnyantelioer.rtwaalhopipnaeuaanCnilqlssodaCn.hsaeteiceptulesstttcu.eloalosaehiidyuuhnncdncruedgllio;d.ebpeesgiodefm-hryshohiinTeoeaDusneraverceddrhtclernetnreineieeilciglsccnsoyssaostyoseoaeeitsnp.sl,cuiinfxplo,slpeiaareapddianldrtetturbouoiwts.daorbtcescptraahenaclteoeelecodneadlnsrtoaurrcitigtneotbinabsoorotaedpneyohsfnweap.wmasteaemliatnpoteottaoer1211rooe--rf;08792..../2/2TnDremTAw0a1veo7htheeaome2cetpis:du4lhre2oeime:s3xopbr1aatdaPiaiysuMnxtresmiartmonaetuineraxareio.tengtsufuoenorHfrvtleeefwuoat.hapbiooriucnlonfemssrkiatoearsinseoptoldiiyanylodiaitrdsnnyafnsqiutilaisdimfounnrignondgbeeE.mcendsarrxenetisarioadptaosiflellislllgnqsayaaanutnssihdioetdaeahsncttletie.aioshnrooehnbgnrosageeumtneistoeeaogpbeuaole--sf 29. HINTS AND explANATION 30. 31. 32. 33. Bromine is a liquid and hence it possesses stronger electriactimty.osphere. with highlights on the intermolecular forces of attractions than gases. Multi4p1l.e c(ih) oAicme iqxutuerestioofnssawdust and water is taken in a common mistakes that 34. Gases have maximum intermolecular spaces and beaker hence they diffuse in each other, forming a homo2-1. hSoavliedso(hniial)yvevAanimbdfraialnfttiyeottrrefyrdpemaeipnsoeuttroriofiaasncfefuaosnnlbddneeehcdlaebuniynsceemtthahoreeeismftfoeixornemliednc.goulfiet aswcitohn2ea5. A drop of watsetrucdonetnaintss. usually make geneous mixture Millions of moinlectuhlees oefxwaamterinations 35. In washing soda (Na2CO3), the constituents ar2e2. Sulphur hafeswthderhoipgshoefstwaatotemr icity of eight. 26. Germanium has the properties of metals as well as combined chemically in a fixed ratio of the2ir3. Sublim(iaiti)ioTn h→e Imodixintuere+ issanpdo,uiroeddingeeunntldyeringotosutbh-e filter nonmetals; hence, it is a metalloid weights and hence it is a compound. limation oncohneeataingd collected into another beaker whic2h7. The process is decantation as mud being heavier 36. Solids have strong intermolecular forces of attrac- Filtration →is caSlalewddausftilt+ratwe.ater, sawdust floats on than water settles down at the bottom due to grav- tion and hence carbon and potassium possess strong forces of attraction. water (iv) The solid retained on the filter paper is called a ity and separates from water Evaporatiornes→iduNe. aCl + water, NaCl is soluble in 28. Lime water is a saturated solution of calcium 37. Soft metals are sodium and potassium water and undergoes evaporation on heating hydroxide; hence, it is a mixture Match the following 24. A mixture has a variable composition. ION 42. A → b Sodium is an element as it is made up of 44. A → b In distillation process, the non-volatile same atoms component sodium chloride can be separated from

Series Chapter Flow Class 7 Atomic Structure and 3 Air and 5 Chemistry in Transformation of Oxygen Water Daily Life 1 Matter Acids, Bases Classification of and Salts 4 6 2 Matter Class 9 Atomic 3 Chemical Bonding 5 Structure 1 Periodic 4 Mole Concept, 2 Classification of Stoichiometry and Nature of Chemical Kinetics Behaviour of Gases Matter Metals and Elements and Chemical Non-Metals Equilibrium 9 7 8 6 Organic Water, Solution, Chemistry Solubility and Hydrogen

Series Chapter Flow xvii Class 8 Classification of Air and Matter Oxygen 13 24 Atomic Structure Language of Chemistry and Transformation of Carbon and its Substances Compounds 75 6 Some Important Water, Solution, Elements and their Solubility and Hydrogen Compounds Class 10 Chemical Kinetics and 1 Atomic 3 Chemical 5 Equilibrium Structure Periodic Table Bonding Mole Concept, Acids, Bases 6 Stoichiometry and 2 9 4 and Salts Behaviour of Gases Industrial Organic Chemistry Metallurgy 7 11 Chemistry–I 8 Electrochemistry Organic 10 Chemistry–II

This page is intentionally left blank.

1CChhaapptteerr NMSNaayuttsumtteerbremeosrf F I G U R E 1 . 1   Figure Caption Remember Before beginning this chapter, you should be able to: • understand basic concepts of kinetic molecular theory. • comprehend classification of matter and different states of matter. Key Ideas After completing this chapter, you should be able to: • understand the comparison between different states of matter in correlation with kinetic molecular theory. • study the interconversion of different states of matter and correlate it to day-to-day activities and natural phenomena. • know about the characteristics of elements, compounds, homogeneous and heterogeneous mixtures. • illustrate the different separation methods for separation of different types of mixtures.

1.2 Chapter 1 INTRODUCTION The subject of chemistry basically deals with the study of matter in different perceptions. One of the important aspects of study of matter pertains to the various physical states in which matter exists. Another equally important perspective of study is the structure and molecular composition of matter. In addition to these two areas, study of matter is also significant with reference to the type of transformations matter undergoes under various conditions. Matter basically exists in three states: solids, liquids and gases. Irrespective of the state of matter, the basic units of matter are only molecules. The three states of matter differ with respect to the pattern of molecular arrangement which brings about a change in physical behaviour of the substances in their respective states. The various characteristics of matter which determine the physical behaviour are envisaged in the kinetic molecular theory of matter. POSTULATES OF KINETIC MOLECULAR THEORY OF MATTER 1. Matter is composed of small, tiny particles called molecules. 2. The empty spaces existing between the molecules are called intermolecular spaces. 3. M olecules have forces of attraction between them known as intermolecular forces of attraction. The force of attraction between similar molecules is called cohesive force and that between dissimilar molecules is called adhesive force. 4. The molecules possess kinetic energy due to their ceaseless motion. Different States of Matter and Their Properties The matter around us can be classified into three different states: solids, liquids and gases. The following table provides a comparison of the three states of matter: TABLE 1.1  Different states of matter and their properties Properties Solids Liquids Gases Mass Definite mass Definite mass Definite mass Volume Definite volume Shape Definite shape Definite volume No definite volume No definite shape, take the No definite shape shape of the container Density High density Lesser density than solids Least density Compressibility Incompressible Rigidity Rigid (cannot flow) Slightly compressible Highly compressible Free surfaces Any number of free surfaces Fluid (can flow) Fluid (can flow) Thermal Very low expansion One free surface, i.e., only the No free surfaces Diffusion Do not diffuse upper surface Higher than solids Much greater than both solids and liquids Some liquids can diffuse Diffuse spontaneously and spontaneously into another (e.g., rapidly water and alcohol), but others do not diffuse. (e.g., oil and water)

Nature of Matter 1.3 A comparative study of molecular arrangements in solids, liquids and gases, based on the kinetic molecular theory, is provided in the following table: Parameters Solids Liquids Gases Closely packed Loosely packed Very loosely packed Packing of molecules Very low More than those in solids Highest Intermolecular Strong intermolecular Moderate force of attraction, Negligible intermolecular space force of attraction less than that in solids force of attraction Intermolecular Possess only vibratory Possess translatory and Possess translatory, rotatory force of motion, but the mean rotatory motion in addition and vibratory motion in all attraction position is fixed to vibratory motion in one directions motion Molecular High kinetic energy, more Highest kinetic energy movement than that of solids Kinetic energy Very low kinetic energy EXAMPLE Why do solids have any number of free surfaces and gases have no free surfaces at all? Also compare with liquids. SOLUTION In solids, molecules do not possess translatory motion. Therefore, they occupy fixed positions. As a result, they have any number of free surfaces. In gases, molecules possess translatory motion in all directions randomly. Therefore, they do not have fixed positions and do not possess definite shape. Hence, they do not have free surfaces. In liquids, molecules possess translatory motion which is not random like gaseous molecules and the upper surface is free. Hence, they have only one free surface. EXAMPLE A rrange different states of matter in the increasing order of their densities with appropriate reason. SOLUTION In solids, the molecules are tightly packed. Hence, the number of molecules per unit volume is more and density is the highest. In gases, the molecules are very loosely packed and the density is the least. Molecules of liquids are loosely packed, but intermolecular space is much less than that in gas. Hence, their denisities are generally less than solids, but more than gases. Interconversion of States of Matter Since the physical behaviour of matter in various states depends upon the molecular arrangement which can be changed by changing the conditions of temperature and pressure, therefore, matter can be converted from one state to another under suitable conditions. This is also termed as phase transition.

1.4 Chapter 1 Interconversion Between Solid and Liquid States The process of conversion of a solid state to a liquid state is called melting or fusion and the opposite process from the liquid to the solid states is called freezing. Melting is carried out by supplying heat energy to the solid substance and freezing is carried out by extracting heat energy from the liquid substance. Melting: When the substance in the solid state is subjected to heating, the molecules absorb heat energy which increases the kinetic energy of molecules. This results in an increase of the kinetic energy of the molecules which in turn leads to increase in temperature; however, further supply of heat energy does not increase the kinetic energy of molecules. Instead, the heat energy is stored in the form of potential energy. With increase in the potential energy, the intermolecular forces of attraction decrease which results in increase in intermolecular spaces. The molecular arrangement of the solid changes to that of the liquid. The temperature at which the solid gets converted to the liquid state at the atmospheric pressure is called melting point of that solid. Although each substance has a specific value of melting point under normal atmospheric conditions, certain factors affect the melting point of the solids. Factors Affecting the Melting Point 1. Effect of pressure:  The effect of pressure on the melting point of the solids depends upon the nature of the solid. For solids which expand on melting, increase in pressure increases the melting point. This is because increase in pressure opposes expansion. Examples: Paraffin wax, silver, gold, copper, etc. For solids which contract on melting, increase in pressure decreases the melting point. This is because increase in pressure favours contraction. Examples: Ice, cast iron, brass, etc. The principle of regelation is based on the above concept. Since ice is a solid which contracts on melting, its melting point decreases. When two ice cubes are pressed together, the ice at the interface melts due to the application of slight pressure. When pressure is released, it solidifies again thereby joining the two ice cubes. This principle is called regelation. 2. Effect of addition of impurities: Addition of impurities to a solid decreases the melting point of the solid thereby allowing the substance to melt at a lower temperature. Example: Rose’s metal, an alloy of tin, lead and bismuth. Melting point of Rose’s metal → 94.5°C Melting point of Pb → 327°C Melting point of Sn → 231.9°C Melting point of Bi → 271°C Freezing: The process of conversion of a liquid state to a solid state is called freezing. It is the opposite process of melting. This is carried out by extracting heat from the liquid. When a substance in the liquid state is subjected to cooling, heat is extracted from the liquid. Then kinetic energy of molecules decreases which results in decrease in temperature. On reaching a certain temperature, further extraction of heat from the liquid results in decrease in potential energy

Nature of Matter 1.5 instead of decrease in kinetic energy. Decrease in potential energy leads to increase in intermolecular forces of attraction, and hence, decrease in intermolecular spaces. Hence, the molecular arrangement of the liquid changes to that of solid. The temperature at which the liquid converts into the solid at the atmospheric pressure is called freezing point of the liquid. The freezing point of the liquid is equal to the melting point of the solid for the same substance. Therefore, the factors which affect the melting point of a substance obviously affect the freezing point. For example, addition of impurities to the solids decreases the melting point of the solid and at the same time decreases the freezing point of the liquid to the same extent. The above principle can be made use of in the preparation of freezing mixtures. A mixture of three parts of ice and one part of common salt is called freezing mixture which is used to produce a lower temperature of –21°C. Freezing mixtures are used for preservation of food stuffs especially perishables such as fish and meat. Interconversion Between the Liquid and Gaseous States Vaporisation:The process of conversion of the liquid state to the gaseous state is called vaporisation. When the liquid is subjected to heating, kinetic energy of molecules increases which results in increase in temperature. At a particular temperature, there is no further increase in the kinetic energy of molecules.The heat energy supplied increases the potential energy of the molecules. As a result, the intermolecular forces of attraction decrease.The molecules move apart and the substance passes from liquid to gaseous state. This process is called boiling and the temperature at which the conversion of the liquid to the vapour states takes place at normal atmospheric pressure is called boiling point. Factors Affecting the Boiling Point Although every liquid is characterised by a specific value for the boiling point, it depends on some factors as discussed hereunder. 1. Effect of pressure:  The boiling point of a liquid increases with increase in external pressure. This is the reason why water boils at a lower temperature than 100°C at higher altitudes. This principle is made use of in the working of a pressure cooker. In the pressure cooker, water is subjected to heating in a closed vessel in confined space. The steam generated in fixed volume increases the pressure beyond the normal atmospheric pressure. Since the external pressure is more, the boiling point of water rises beyond 100°C. The temperature of cooking medium being greater than the normal boiling point, therefore, food gets cooked at a faster rate, and thus, saving time and fuel. 2. Effect of impurities:  When solid substances are dissolved in a liquid, the boiling point of the liquid increases beyond the normal boiling point. For example, when common salt is dissolved in water, the solution boils at a temperature greater than 100°C. Conversion of liquid to gaseous states can also be brought about even without supplying heat energy to the liquid. In the liquid, the surface molecules possess higher kinetic energies than the molecules in the bulk of the liquid. Due to this reason, the molecules break away from the forces of attraction of the other molecules and go into a vapour state. This process of conversion of the liquid into the vapour takes place at a temperature below the boiling point and is called evaporation. In contrast to boiling, evaporation is considered as a surface phenomenon since it is confined to only surface molecules. It is also a slow process taking place over a period of time at any temperature, whereas boiling takes place rapidly at a specific temperature. Vaporisation is a common term applicable for both evaporation and boiling.

1.6 Chapter 1 Factors Affecting Evaporation 1. Surface area: Increase in surface area increases the rate of evaporation. 2. Temperature: Increase in temperature increases the rate of evaporation. 3. Humidity: The amount of water vapour which the atmospheric air holds is called humidity. Higher humidity decreases the rate of evaporation. 4. Wind speed: Increase in wind speed increases the rate of evaporation. Condensation: The conversion of a gaseous substance into a liquid state is called condensation. This can be carried out by cooling the gas below a particular temperature. When heat is extracted from the gas, the kinetic energy of molecules decreases, and hence, temperature falls. When sufficiently low temperature is reached, further extraction of heat from the gas does not reduce the kinetic energy. This in turn results in increase in intermolecular forces of attraction bringing the molecules closer. At this point, the gas passes into the liquid state. The conversion of the gaseous to the liquid state can also TABLE 1.2  Critical tempera- be brought about by the application of pressure. The gases tures of common gases have to be cooled below a certain temperature and then Gases Critical temperatures subjected to application of high pressure. This entire process CO2 31.1°C is called liquefaction. Every gas requires a certain minimum NH3 132°C temperature for passing into the liquid state. That means, O2 –118°C above that minimum temperature, application of any amount N2 –147°C of pressure cannot bring about the transformation from the H2 –165°C gaseous to the liquid state. This temperature above which a He –240°C gas cannot be liquefied, howsoever high pressure is applied, 20°C is called critical temperature. Every gas is associated with a SO2 specific critical temperature. 1. Effects of condensation on climate:  The formation of dew, fog and clouds is the application of condensation. When the temperature is high, water in water bodies evaporates and during day time, the process goes on continuously and the air does not become saturated with water vapour. When the temperature falls during night time, the air becomes saturated with water vapour. The temperature at which the atmospheric air becomes saturated with water vapour is called dew point. On further lowering of temperature, some of the water vapour condenses and the water droplets condense as dew. If the condensation of water vapour takes place on floating dust particles, it results in the formation of fog or mist. In the upper part of the atmosphere, the condensed water droplets appear as clouds. Apart from these interconversions, the conversion of the solid to the gaseous states is also possible in some cases. This process of conversion of the solid into the vapour states directly without passing through the intermittent liquid state is called sublimation. The vapours when condensed give back the solids. Such solids formed from the vapour are called sublimates. Only some solids undergo sublimation. Examples: Camphor, naphthalene, ammonium chloride, iodine, etc.

Nature of Matter 1.7 Latent Heat The change of state of a substance is invariably associated with absorption or liberation of heat. However, the change is not associated with a change in temperature of the substance in any process. It is considered that the heat energy gets hidden into the substances involved in the process and is called latent heat. For example, when ice melts, the temperature of ice remains constant only at the melting point until the entire process is completed. The entire heat absorbed during the process is utilised to carry out the change. Therefore, the amount of heat energy required to convert 1 kg of a solid into a liquid at atmospheric pressure at its melting point is known as the latent heat of fusion. For ice, it is equal to 80 cal/g/°C. Example: When water is subjected to boiling, the temperature of water remains constant at the boiling point until all the water is converted to steam. The heat energy supplied during the process is stored in the steam and is called latent heat of vaporisation. For steam, this equals 540 cal/g/°C. Applications of Latent Heat Cooling Produced Due to Evaporation The liquid molecules absorb energy from the surroundings and overcome the forces of attraction thereby going into a vapour state. Since surroundings lose energy, they becomes cold.This principle is made use of in various daily life activities as listed hereunder: 1. Cool sensation produced when alcohol is poured on palm, 2. Cotton clothes producing cooling effect during summer, 3. Formation of water droplets on the outer surface of glass containing ice cold water and 4. Cooling of water in earthen pots during summer. EXAMPLE (i) Pressure cooker reduces the cooking time. Explain the principle involved. (ii) F ish and meat can be preserved for a longer time in ice if common salt is added to it. Give reason. (iii) When a glass of ice cold water is kept open at room temperature, what observation is found? Give reason. SOLUTION (i) In a pressure cooker, water is subjected to heating in a closed vessel in a confined space. The steam generated in fixed volume increases the pressure beyond the normal atmospheric pressure. Since the external pressure is more, the boiling point of water rises beyond 100°C.The temperature of cooking medium being greater than the normal boiling point, therefore, food gets cooked at a faster rate, and thus, saving time and fuel. (ii) A ddition of common salt to ice reduces the freezing point of ice to below 0°C.A mixture of three parts of ice and one part of common salt called freezing mixture can produce a much lower temperature (–21°C).Thus, fish and meat can be prevented from spoilage for a longer time in the freezing mixture. (iii) When a glass of water is kept at room temperature, water droplets are observed on the outer walls of glass.This is because the water vapour present in air when comes in contact with glass containing ice cold water loses the heat energy, and thus, the water vapour present in air gets condensed.

1.8 Chapter 1 EXAMPLE When two pieces of ice are pressed together, they form a single lump. Explain SOLUTION When two pieces of ice are pressed together, they form a single lump, it is due to regelation. When two ice cubes are pressed together, the ice at the interface melts due to the application of pressure and when the pressure is released, it solidifies again thereby joining the two ice cubes. EXAMPLE W hy does temperature remain constant as heated liquid gets converted to its gaseous state at its boiling point? SOLUTION When the heated liquid is converted to its gaseous state at its boiling point, the temperature remains constant because the change in state of matter is associated with increase in potential energy, and the kinetic energy remains constant. EXAMPLE The critical temperature of gases A, B, C and D are –118°C, –240°C, 132°C and 20°C, respectively. Arrange them in the decreasing order of intermolecular force of attraction. SOLUTION The higher the critical temperature the more is the intermolecular force of attraction and they can be liquefied easily. Decreasing order of intermolecular force of attraction is C > D > A > B. EXAMPLE What happens when water is kept in a plastic bottle wrapped with a wet towel? SOLUTION During the evaporation of water from the wet towel, heat is absorbed from the bottle and its content. Due to this, the water present in the bottle becomes cool. EXAMPLE W hy are cotton clothes preferred in summer? SOLUTION People wear cotton clothes to keep themselves cool in summer. Cotton fabric absorbs sweat more due to greater adhesive forces between cotton and water and allows the sweat to be evaporated at a faster rate. The sweat absorbs heat energy from the body and gets evaporated, thus. keeping the body cool.

Nature of Matter 1.9 EXAMPLE D iscuss the change in energy and arrangement of molecules on increasing the temperature of ice from –5°C to 10°C at 1 atm pressure. SOLUTION When ice is heated at −5°C, its temperature increases upto 0°C , i.e., the kinetic energy of the molecules increases. At 0°C, the ice starts melting. During this process, the energy supplied is utilised to increase the potential energy of the molecules keeping kinetic energy constant, and the arrangement of molecules changes. Once the process ends, the heat supplied is again used to increase the temperature of water by increasing the kinetic energy. However, from 0°C to 4°C, the molecules of water come closer and above 4°C, the molecules move farther away. The basic units of matter in any state are considered as molecules as far as the physical behaviour of matter in different states and their interconversion under different conditions are concerned. This is because a molecule is considered as the smallest particle of matter which has an independent existence. However, when focus is laid on the chemical behaviour of matter, the classification of matter with respect to molecular composition becomes inevitable. This is because a molecule is broken down into smaller particles called atoms which take part in chemical reactions. CLASSIFICATION OF MATTER ON THE BASIS OF CHEMICAL COMPOSITION Matter Pure substances Mixtures Elements Compounds Homogeneous Heterogeneous mixtures mixtures Metals Non-metals Metalloids A pure substance is the one which is made up of molecules containing same kind of atoms. For example, in case of pure water, all the molecules are made up of two hydrogen atoms and one oxygen atom. In case of a pure substance, the molecule may contain similar atoms or dissimilar atoms. The first category of pure substances in which a molecule is made up of atoms of the same kind are called elements. For example, in hydrogen gas, a molecule of hydrogen is made up of two hydrogen atoms. The second category of substances is called compounds. A molecule of carbon dioxide is made up of one carbon atom and two oxygen atoms. A mixture is such a substance which contains two or more kinds of molecules. For example, common salt solution contains molecules of NaCl and molecules of water. A mixture may include constituents as elements, an element and a compound, or only compounds. Depending on the distribution of the different kinds of molecules within the mixture, the mixture is classified into two types: homogeneous and heterogeneous. A homogeneous mixture has uniform distribution of the different types of molecules in the mixture. A heterogeneous mixture has non- uniform distribution of the different types of molecules. For example, aqueous solution of glucose is a homogeneous mixture, whereas muddy water is a heterogeneous mixture.

1.10 Chapter 1 Characteristics of Elements 1. Based on their nature, elements are classified as metals, non-metals and metalloids. Examples: Na, Mg, Al, Cu, etc., are metals. H2, O2, He, S, etc., are non-metals. As, Sb, Se, Te, Ge are metalloids. 2. E lements can also be classified on the basis of their atomicity. The number of atoms present in the molecule of an element is called atomicity. (a) M onoatomic elements: The molecule of the element is made up of only one atom. All metals, noble gases and some non-metals are monoatomic. Examples: Ag, Al, Au He, Ne B, C (b) D iatomic elements: The molecule of the element is made up of two atoms. Only some gaseous non-metals are diatomic. Examples: H2, O2, N2 (c) Polyatomic elements: The molecule of the element is made up of more than two atoms. Very few elements are polyatomic. Examples: O3, P4, S8 Characteristics of Compounds 1. The constituent elements are present in a fixed proportion by weight. Example: In CO2, carbon and oxygen are in the ratio of 12 : 32 (3 : 8) 2. The elements do not retain their properties. Example: The properties of water are entirely different from those of hydrogen and oxygen. 3. The constituent elements can be separated only by chemical methods. Example: Water can be decomposed into hydrogen and oxygen by electrolysis method. 4. All compounds are invariably homogeneous. 5. T he formation of a compound is generally associated with significant energy changes. C + O2 → CO2 – 94 k cal Characteristics of a Mixture 1. The constituents may be present in any proportion. 2. The constituents retain their individual properties. Example: Aqueous solution of sugar. 3. Separation of constituents is carried out by physical methods. 4. A mixture can be either heterogeneous or homogeneous. 5. Formation of a mixture does not involve significant energy change.

Nature of Matter 1.11 Classification of a Mixture TABLE 1.3  Classification of mixtures Types of mixture Homogeneous Heterogeneous Solid–solid Stainless steel, brass and bronze Iron and sulphur Solid–liquid NaCl in water and iodine in CCl4 Sulphur in water Liquid–liquid Alcohol and water, and benzene and toluene Oil and water Liquid–gas Liquor ammonia and soda water – Gas–gas Air – Solid–gas – Hydrogen gas adsorbed on Pd EXAMPLE “All pure substances are homogeneous. But, all homogeneous substances are not pure”. Justify SOLUTION A pure substance is the one which is made up of same kind of molecules. So, it is homogeneous in nature. A mixture which has uniform distribution of different kinds of molecules is also homogeneous but it is not pure as it contain two or more kind of molecules. EXAMPLE Water is a compound. Explain. SOLUTION Water is made up of hydrogen and oxygen which are present in fixed proportion and the properties of water are entirely different form its constituents, hence, it is a compound. Separation of a Mixture Most of the substances available in nature are not pure substances and are actually a mixture. The useful component present in the mixture can be obtained only by separating the individual components of the mixture by following a suitable method. The method of separation employed depends upon the nature of components in the mixture. TABLE 1.4  Types of separation methods of a solid–solid mixture Methods Properties exploited Description with examples Solvent extraction Solubility of one component in a A mixture of sulphur and sand. Sulphur is soluble solvent in CS2 and sand is insoluble Magnetic Magnetic property of one Mixture of iron ore and sand. Iron ore is attracted separation component by magnet and sand is left behind Gravity method Difference in densities of Mixture of sand and chalk powder. Sand being components heavier than chalk powder sinks in water, whereas chalk powder floats on water Sublimation Ability of one component to sublime Mixture of iodine and sand. On heating, iodine sublimes leaving behind sand. The vapours on cooling give solid iodine

1.12 Chapter 1 Methods Properties exploited Description with examples Fractional Difference in solubility of the Mixture of KNO3 and NaCl. KNO3 being more crystallisation components in the same solvent soluble than NaCl, when the aqueous solution of this mixture is subjected to heating, the more soluble KNO3 escapes out along with water vapour and less soluble NaCl is left behind and crystallises TABLE 1.5  Types of separation method af a solid–liquid mixture Methods Properties exploited Description with examples Sedimentation and High density of an insoluble solid decantation component Mixture of sand and water. Sand being Filtration heavier, settles at the bottom and liquid Size of the particles of an insoluble solid is slowly transferred into another Evaporation component container Distillation The ability of a solid to remain Mixture of BaSO4 and H2O. On undecomposed when a solution is heated passing through filter paper, water passes Centrifugation up to the boiling point of the liquid through and BaSO4 remains on the filter component paper Heating the solution to the boiling point of the liquid component followed by Mixture of sugar and water. Water condensation of the vapours evaporates on heating leaving behind crystals of sugar Size of density of solid particles in comparison to the size of liquid particles Mixture of NaCl and water. Water evaporates and condenses back to water, and NaCl is left behind in the distillation flask Milk contains solid fat particles in water. Size of solid particles is less, and hence, they pass through the filter paper. When this is subjected to centrifugation, heavier fat particles settle down at the bottom leaving behind lighter water on the top TABLE 1.6  Types of separation method of a liquid–liquid mixture Methods Properties exploited Description with examples Separating Difference in densities of the two funnel liquid components Kerosene oil and water are immiscible liquids. When the mixture is taken in the separating funnel, the Fractional Difference in boiling points of the lighter liquid (kerosene) forms top layer and the distillation liquids heavier liquid (water) settles down Distillation carried out by including a fractionating column. E.g., ethyl alcohol + water Ethyl alcohol has a lower boiling point than water, and hence vaporises. On passing through fractionating column, the vapours condense to give alcohol in a receiver. Water is left behind in the distillation flask

Nature of Matter 1.13 TABLE 1.7  Types of separation method af a gas–gas mixture Methods Properties exploited Description with examples Diffusion Difference in densities of component gases The gas with lower molecular weight diffuses faster Dissolution in than the gas with higher molecular weight suitable solvents Difference in solubility of E.g., H2 and CH4, and He and SO2 component gases in a given A mixture of CO2 and CO. CO2 is soluble in KOH Preferential solvent leaving behind CO liquefaction Fractional Difference in liquefaction of In a mixture of NH3 and N2, NH3 is soluble in evaporation component gases under pressure water and N2 is insoluble Difference in boiling points of the A mixture of NH3 and H2. Ammonia gets liquefied component gases under high pressure and hydrogen gas is left behind. When air is liquefied, the major components of air, N2 and O2 can be separated by subjecting the liquid to evaporation. N2 has a lower boiling point, and hence boils off, whereas O2 has a higher boiling point and remains behind TABLE 1.8  Types of separation method of a liquid–gas mixture Methods Properties exploited Description with examples Heating Decrease in solubility of a gas with When a solution containing a gas is subjected to slight Lowering the increase in temperature heating below the boiling point of the liquid, the gas pressure escapes out leaving behind the liquid component Difference in solubility of a gas in E.g., Separation of dissolved O2 by heating water the liquid at different pressures When a soda water bottle is opened, the pressure inside the bottle decreases. CO2 gas fizzes out of the bottle Paper Chromatography Apart from all different methods of separation of types of a mixture, there is a special technique for separation and identification of the constituents in the mixture. Principle: Based on difference in adsorption of constituents by a surface of an appropriate adsorbent material or solid medium (stationary phase). The rate of adsorption of a particular constituent depends upon its solubility in the solvent (moving phase). Example: Separation of a coloured constituent in a mixture of ink by paper chromatography. Process: A filter paper is taken which generally absorbs water. It acts as a stationary phase. The mixture containing different constituents is taken on the filter paper which is then dipped in another solvent called moving phase. If the constituent has more affinity for the adsorbent material, it moves slowly on the filter paper. If the constituent has more affinity for the solvent acting as a moving phase, it moves rapidly on the filter paper. Therefore, depending on the relative affinities of the various constituents with the stationary and moving phases, spots or lines appear on the filter paper at different positions. Example: Separation of coloured constituents in a mixture of ink.

1.14 Chapter 1 EXAMPLE How is oxygen prepared from air? SOLUTION When air is liquefied, the major components of air, nitrogen and oxygen can be separated by subjecting the liquids to fractional evaporation. Nitrogen has a lower boiling point, and hence boils off and oxygen has a higher boiling point, and hence remains behind. EXAMPLE Suggest a method of separation for a mixture of sodium chloride and ammonium chloride. Explain the process. SOLUTION Mixture of NaCl and NH4Cl can be separated by sublimation. PROCESS Ammonium chloride and sodium chloride mixture is taken in a porcelain dish. It is covered by an inverted funnel. The opening of the stem of the funnel is closed by means of a cotton plug. The outside of the funnel is kept cool by wrapping with wet blotting paper. On heating the porcelain dish, ammonium chloride changes into vapour state, which on coming in contact with the inner wall of the cooled funnel condenses into the original solid and gets collected there. EXAMPLE Mention the separation method and property exploited in the separation of the following solid-solid mixtures. (a) S + sand (b) Ι2 + sand (c) KNO3 + NaCl SOLUTION (a) S + sand Solvent extraction Solubility of one component in a given solvent Sublimation (b) I2 + sand Fractional crystallisation Ability of one component to sublime. (c) KNO3 + NaCl Difference in solubility of the components in the same solvent. EXAMPLE How will you separate sulphur dioxide gas from the gaseous mixture of SO2 and O2? SOLUTION SO2 gas from O2 can be separated by preferential liquefaction. SO2 gas liquefied under high pressure and O2 is left behind. Since SO2 is acidic in nature, it can also be separated by passing the gaseous mixture through KOH solution.

Nature of Matter 1.15 EXAMPLE When soda water bottle is kept open for sometime, it loses the tangy taste. Give reason. SOLUTION At high pressure, more amount of CO2 is dissolved in water. When the bottle is opened, excess amount of CO2 comes out as solubility of gas decreases with decrease in pressure, and hence, the soda water loses its tangy taste. Matter has an inherent tendency to undergo transformation under suitable conditions. The transformations are basically of two types. Firstly, the changes which do not involve any change in molecular composition of the substance. They are termed as physical changes. Secondly, the changes which involve changes in molecular composition of the substance. These are termed as chemical changes. Conditions Required for Chemical Changes 1. P hysical contact between reactants: A chemical reaction takes place only when the two reactants come in contact with each other. For example, sodium on exposure to water or even moisture reacts to give sodium hydroxide and hydrogen, and hence, sodium catches fire. Due to this reason, only sodium is stored in kerosene. 2. H eat: Generally, chemical reaction involves absorption or release of heat energy. Based on the heat changes, the reactions are classified as exothermic and endothermic reactions. The change in energy is represented as ∆H. Many compounds decompose on absorption of heat. These are endothermic reactions and their ∆H is positive. Examples: KNO3 ∆→ KNO2 + O2 ∆H = 244.7 kJ CaCO3 ∆→ CaO + CO2 ∆H = 178 kJ Certain chemical reactions, such as combustion, are associated with release of heat energy. These are exothermic reactions and their ∆H is negative. Example: CH4 + 2O2 → CO2 + 2H2O ∆H = −890.4 kJ 3. Light: Certain chemical reactions take place with the help of light energy. These are called photochemical reactions. Example: H2 + Cl2 light→ 2HCl 6CO2 + 6H2O sunlight→ C6H12O6 (photosynthesis) Some chemical reactions are also associated with release of light energy: Example: Mg + O → 2MgO + Light 4. P ressure: There are some chemical reactions which require high pressure conditions for the reaction to take place at a reasonable rate: Example: 2SO + O2 ¾V¾2O2¾® 2SO2 (manufacture of H2SO4 by contact process) Pb + S → PbS

1.16 Chapter 1 5. Catalyst: A catalyst is a substance which alters the rate of a chemical reaction. Generally, a catalyst speeds up a chemical reaction. They are termed as positive catalysts in contrast to the negative catalysts which deteriorate the rate of a chemical reaction. Example: N2 + 3H2 Fe→ 2NH3 450°C 200 atm For the manufacture of ammonia by Haber process, iron acts as a positive catalyst. Example: H2O2 → H2O + O2 Acetanilide acts as a negative catalyst for the decomposition of H2O2, and hence, acts as a stabiliser for the solution of H2O2. Since there are vast numbers of chemical reactions, the study of these reactions becomes easier and convenient by classifying them into various types. TABLE 1.9  Chemical reactions and their examples Types of chemical Explanation Examples reactions Chemical Combination of two or more N2 + O2 → 2NO (synthesis) and 2CO + O2 → combination substances 2CO2 (compound–element) Chemical NH3 + H2O → NH4OH (compound–compound) decomposition Splitting of a compound into 2NaNO3 ∆→ 2NaNO2 + O2 (thermal two or more simpler substances decomposition) 2HgO ∆→ 2Hg + O2 2HOC light→ 2HCl + O2 (photolytic decomposition) Double The radicals of the reactants 2H2O electric→ 2H2 + O2 (electrolyte decomposition are interchanged. A double decomposition) decomposition reaction Na2SO4 + Zn(NO3)2 → 2NaNO3 + ZnSO4 Displacement between an acid and a base is 2KOH + H2SO4 → K2SO4 + 2H2O called neutralisation reaction CaBr2 + K2SO4 Precipitation reactions also ↓ are double decomposition CaSO4 + 2KBr reactions Zn + CuSO4 → ZnSO4 + Cu A less reactive element is 2KCl + F2 → 2KF + C12 displaced by a more reactive element Since the displacement reaction depends upon the relative reactivities of various metals, they are arranged in a decreasing order of reactivity. Displacement Reaction In a displacement reaction, the more reactive element displaces the less reactive element from its compound.

Nature of Matter 1.17 Representation: AB + C → CB + A XY + Z → XZ + Y The ability of an element to displace another element is known by its relative position in the reactivity series. Metal Reactivity Series Potassium K (Most reactive metal) Sodium Na Reactivity decreases Calcium Ca Magnesium Mg (Least reactive metal) Aluminium Al Zinc Zn Iron Fe Nickel Ni Tin Sn Lead Pb Hydrogen H Copper Cu Mercury Hg Silver Ag Gold Au Platinum Pt The more reactive metal displaces the less reactive metal from its compound: Example: Fe + CuSO4 → FeSO4 + Cu Reactivity (copper (ferrous decreases sulphate) sulphate) As iron is more reactive than copper, it displaces copper from copper sulphate solution, thereby forming ferrous sulphate. Similarly, halogens also can be arranged in the order of reactivities which represents their relative abilities for displacement from the respective compounds. Halogen Reactivity Series F2 (Most reactive halogen) Cl2 Br2 I2 (Least reactive halogen) The more reactive halogen displaces the less reactive halogen from its compound. Example: 2NaBr + Cl2 → 2NaCl + Br2

1.18 Chapter 1 PRACTICE QUESTIONS TEST YOUR CONCEPTS 17. Solubility of a gas in a liquid _____ with rise of temperature. Very Short Answer Type Questions 18. Give an example for a liquid–liquid heterogeneous 1. What is the effect of addition of impurities on mixture. Suggest a method of separation for it. freezing and boiling points of a substance? 19. The forces of attraction existing between similar 2. What is a freezing mixture? Give an example. molecules are _____. 3. Define melting point. What is the effect of pressure 20. How do you separate a mixture of on melting point? (a) CO2 and O2 and (b)  H2 and O2? 4. What are cohesive and adhesive forces? 21. The formation of phosphorus pentachloride from 5. Cotton clothes producing cooling effect during phosphorus trichloride and chlorine gas is a _____ summer is an application of ______. type of combination reaction. 6. Why do wet clothes dry faster on a windy day? 22. The addition of MnO2 to KClO3 decreases the temperature at which KClO3 decomposes because 7. Is it possible to liquefy CO2 gas at 50°C? Give a MnO2 added acts as a ______ catalyst. reason. 23. What is paper chromatography? 8. A mixture of sand and NH4Cl can be separated by _____ method. 24. A mixture of NH3 and H2 can be separated by applying pressure, because of their high difference 9. Define latent heat of fusion. Give its value for ice. in _____. 10. What is meant by atomicity of an element? Give 25. Why is sodium stored in kerosene? some examples of polyatomic elements. 26. A compound is always _____. 11. Distinguish between filtration and centrifugation. 27. A reaction of decomposition of a compound AB is 12. Silver is a substance which expands on melting, so, accompanied by absorption of some heat energy. its melting point _____ when pressure is raised. What is the sign of ΔH? 13. Give an example each for solid–liquid and solid– solid types of a homogeneous mixture. 28. The temperature above which no amount of p­ ressure can cause a gas to liquefy is _____. 14. The atomicity of the element phosphorus is _____. 29. What is sublimation? Give some examples. 15. _____ is the method of separating heavier fat ­particles of milk from lighter water. 16. A metal “A” reacts with a metallic chloride of “B” to give metal “B.” But metallic chloride of “A” cannot give metal “A” on reaction with metal “B.” What conclusion can you draw from this? Short Answer Type Questions 35. What is a double decomposition reaction? Give an example each for neutralisation and precipitation 30. When two ice cubes are pressed, they join together. reactions. Explain the principle involved. 36. Why is it not possible to displace fluorine from 31. Why are cotton clothes preferable in summer? metallic fluorides by any other halogens? 32. How is liquefaction different from condensation? 37. Explain the process of condensation with respect to kinetic molecular theory. 33. Distinguish between evaporation and boiling. 34. When a glass of ice cold water is taken in a glass, water drops are formed on the outer surface of the glass. Give a reason.

38. What is critical temperature? How does it affect the Nature of Matter 1.19 liquefaction of gas? 40. List out the differences between physical and 39. Why does steam cause more burns than boiling c­ hemical changes. water? 41. What are the different types of combination ­reactions? Give an example of each. Essay Type Questions 42. Suggest the possible methods of separation of 45. Explain the process of separation of different l­iquid–gas mixtures. Explain each method with an ­constituents in coloured ink. example. 46. Explain the following methods of separation with 43. Explain the factors affecting an example each. (a)  melting point (b)  boiling point (a)  Fractional crystallisation (c) evaporation (b)  Sedimentation and decantation (c)  Preferential liquefaction 44. Give the postulates of kinetic molecular theory of (d)  Fractional distillation matter. Explain the process of melting on the basis of this theory. For Answer key, Hints and Explanations, please visit: www.pearsoned.co.in/IITFoundationSeries CONCEPT APPLICATION Level 1 ­separation techniques involved are __________, PRACTICE QUESTIONS ___________ and _________. Direction for questions from 1 to 7: State whether the following statements are true 11. The reactions of a compound KX with fluorine, or false. chlorine, bromine in three different reactions are given below: 1. Boiling point of rainwater is less than that of sea water. 2. Liquid has only one free surface. 2KX + F2 → 2KF + X2 3. The rate of evaporation decreases with increase in 2KX + Cl2 → 2KCl + X2 2KX + Br2 → 2KBr + X2 humidity. then KX is ____________. 4. During melting of a solid, there is no change in the 12. In two closed containers, substances A and B are kinetic energy of molecules. present. After sometime, the lid of the container 5. The reaction of sodium chloride with bromine gas containing substance A alone got off with a lot of pressure. This is because substance A _______. is a chemical displacement reaction. 6. The strong intermolecular forces of attraction are 13. If cold water is poured on a flask containing very hot distilled water and water vapour, the water responsible for high rigidity of solids. inside the flask started to boil below 100°C. The 7. Burning of a piece of magnesium wire is a synthesis principle involved is __________. reaction. 14. Three solid substances x, y and z are heated, the number of free surfaces of substances x and y Direction for questions from 8 to 14: decreases to zero and one, respectively. The num- Fill in the blanks. ber of free surfaces of z does not change. The sub- stances associated with an increase in PE is/are 8. A mixture of SO2, H2 and Cl2 can be separated by __________. _________ followed by _________. 9. A liquid wets the given surface if __________ forces are predominant over ________forces. 10. A mixture contains nitre, common salt and s­ilver chloride as the components. The different

1.20 Chapter 1 Direction for question 15: (a)  melting of tungsten Match the entries in Column A with the (b)  sublimation of tungsten appropriate ones in Column B. (c)  oxidation of tungsten (d)  reduction of tungsten 15. Column B Column A (  ) a. Fe + S → FeS 20. Identify a physical change among the following: (  ) b. 2KI + Cl2 → 2KCl + I2 (a) respiration A. Photolysis (  ) c. CaO + H2O → (b)  digestion of food (c)  burning of wax B. Electrolysis Ca(OH)2 (d)  glowing of an electric bulb ( ) d. PCl3 + Cl2 → PCl5 C. Element–element 21. During a phase transition of a substance the combination ­temperature (T) versus heat energy (Q) graph is shown below. Identify the regions of the graph D. Compound– which show an increase in only PE. compound combination E. Chemical (  ) e. CH4 + 2O2 → CO2 Q F displacement + 2H2O E F. Double (  ) f. 2H2O → 2H2 + O2 D decomposition C G. Element– (  ) g. Pb(NO)2 + 2HCl + B compound PbCl2 + 2HNO3 combination T A0 H. Combustion (  ) h. CaO + CO2 → I. Hydrolysis CaCO3 (a)  AB, BC (b)  BC, DE (  ) i. AgBr2 → Ag + Br2 (c)  CD, EF (d)  all the regions in the given graph Direction for questions from 16 to 45: For each of the questions, four choices have been 22. A chromatogram of pure samples of food colours provided. Select the correct alternative. X, Y and Z is given in the following illustration 1. Three samples of same food materials A, B and C are PRACTICE QUESTIONS 16. The bulb of a thermometer when dipped in petrol analysed for purity, with the help of the chromato- and then taken out, the level of the mercury thread gram in illustration 2. Identify the impure sample. in the thermometer • • • (a)  starts falling • • • (b)  starts rising (c)  remains at the same level (d)  initially falls and then rises 17. The high diffusibility of gases is due to • Z • (a)  high intermolecular forces of attraction XY A BC (b)  high KE of molecules (c)  restricted translatory motion in upward direction (a) A (b) B (d)  all the above (c)  C (d)  A and C 23. Maximum intermolecular forces of attraction 18. Water kept in an earthen pitcher, during summer exists in days becomes very cold due to (a) bromine (b) air (a) condensation (b) evaporation (c) oxygen (d) copper (c) freezing (d) fusion 24. A gaseous mixture of A, B and C is passed through water. The gaseous mixture B and C remains. If this 19. The electric bulb on long use forms a black coat- gaseous mixture of B and C is subjected to sudden ing on its inner surface. The process associated with expansion followed by application of high pressure, this is

Nature of Matter 1.21 B liquefies leaving behind C. Identify the set of (2) potential energy decreases and kinetic energy gases. remains constant (a) SO3, NO2, O2 (b) Cl2, SO2, H2 (3) potential energy increases and kinetic energy (c) CO2, CO, N2 (d) NH3, N2, H2 increases 25. Identify the element from the following: (4) potential energy decreases and kinetic energy (a) air (b) iodine vapour decreases (c) water (d) amalgam (a)  4 2 3 1 (b)  1 3 2 4 (c)  1 3 4 2 (d)  3 4 1 2 26. In which of the following cases, cooking is very 32. For the separation of red ink from blue ink a slow? ­technique is used which is described below. Arrange the statements in a proper sequence. (a)  pressure cooker at sea level (b)  pressure cooker at higher altitude (1) T he blue and red ink form spots at certain (c)  open vessel at sea level ­distances on the paper. (d)  open vessel at higher altitude (2)  Paper chromatography is used for the separation. 27. Which of the following reactions is not a combina- (3) Paper and solvent are taken as stationary and tion reaction? mobile phases, respectively. (a)  reaction of iodine with white phosphorus (4) A narrow strip of paper with a line drawn is cut (b)  reaction of iron with sulphur and a mixture of red and blue ink with the help (c)  addition of water to lime of capillary is placed on the line marked on the (d)  addition of concentrated sulphuric acid to sugar paper. 28. A mixture of three liquids X, Y and Z when (5) The paper is suspended in the closed jar with ­subject to fractional distillation, the order in the help of hook. (a)  2 4 3 5 1 (b)  2 5 3 1 4 which the vapours condense back to liquid state in (c)  2 3 4 5 1 (d)  1 4 5 3 2 ­fractionating tower is Y, X and Z. Arrange them in 33. Among the following, identify the substance in which the correct order of vapour pressures. molecules possess vibratory, rotatory and translatory motions in all directions except in one direction. (a)  Z < X < Y (b)  Y < X < Z (c)  X < Z < Y (d)  X < Y < Z (a) bromine (b) iodine 29. Grease spots from garments can be separated by a (c) ammonia (d) silicon dioxide PRACTICE QUESTIONS method of 34. In Darjeeling, distilled water boils at a temperature (a) chromatography (b)  solvent extraction (a)  above 373 K (b)  above 473 K (c) sublimation (d)  dissolution in suitable solvents (c)  below 373 K (d)  at 373 K 30. Which of the following involves both neutralisation 35. At melting point, as well as precipitation? (a) kinetic energy remains constant and potential (a)  reaction between baking powder and H2SO4 energy increases (b)  reaction between BaCl2 and Na2SO4 (c)  reaction between AgNO3 and HCl (b) kinetic energy increases and potential energy (d)  reaction between slaked lime and H2SO4 remains constant 31. Arrange the following changes of energy during (c)  both potential energy and kinetic energy increase following phase transition in a proper order. Ice (0°C) → water (50°C) → ice (0°C) (d) potential energy increases with a decrease in kinetic energy (1) potential energy increases and kinetic energy remains constant 36. Which among the following statements is true? (a) The rate of evaporation in a coastal area is less when compared to a non-coastal area. (b) The rate of evaporation in a non-coastal area is less when compared to a coastal area.

1.22 Chapter 1 (c) In both the areas the rate of evaporation is the same. 42. Which of the following phenomena is based on the (d)  None of the above principle that cooling results due to evaporation? 37. Which among the following is not a homogeneous (a) formation of water drops on the surface of cold mixture? drink bottle (a) solder (b) formation of crystals of ice on the inner surface (b)  aqueous solution of NaCl of the lid of an ice cream box in freezer (c)  sulphur in carbon disulphide (d)  sulphur in water (c) white foggy appearance on the surface of large ice blocks (d)  stretching out of tongues by dogs during summer 38. Gunpowder is a _____ 43. Which among the following is the false statement? (a)  solid–liquid homogeneous mixture (a)  Water boils below 100°C on mountain peaks. (b)  solid–liquid heterogeneous mixture (b)  Ice undergoes sublimation on surface of moon. (c)  solid–solid homogeneous mixture (c)  Ice melts above 0°C on mountain peaks. (d)  solid–solid heterogeneous mixture (d)  Cooking of food is faster on mountain peaks. 39. Which of the following gases can be separated 44. The order of vapour pressures of four solids is ­completely from a mixture by using water as a solvent? P << R < Q < S. Which of the following has the ­maximum tendency to sublime? (a) CO2 and O2 (b) N2 and NH3 (c) CO2 and NH3 (d) H2 and N2 (a) P (b) Q 40. Identify the mixture which can be separated by (c) R (d) S magnetic separation method. 45. Identify the methods by which the i­ndividual (a)  chalk powder + sand (b)  iron + sand ­components of mixture containing water, ­potassium nitrate, sodium chloride, alcohol and carbon (c)  common salt + sand (d)  sulphur + sand ­tetrachloride can be separated. PRACTICE QUESTIONS 41. Which among the following is true? (a) separating funnel, fractional distillation, f­ractional crystallisation, distillation (a) A ir is a bad conductor of heat and thermal expansion of solids is more than that of gases. (b) fractional distillation, distillation, fractional crystallisation (b) Air is a good conductor of heat and thermal expansion of solids is less than that of gases. (c) separating funnel, fractional distillation, ­filtration, distillation (c) A ir is a bad conductor of heat and thermal expansion of solids is less than that of gases. (d) separating funnel, fractional distillation, ­sedimentation and decantation (d) Air is a good conductor of heat and thermal expansion of solids is more than that of gases. Level 2 5. In summer, Khus Khus mats are used for reducing the heating effect. Explain the principle involved in this. 1. Can water be made to boil in a paper cup without the paper being burnt? Explain. 6. A statue coated with chemical substance X on long exposure to polluted atmosphere becomes black. 2. “How is the principle of regelation applicable for This colour can be restored by treatment with welding?” Explain. H2O2. Identify element X and also the types of chemical changes involved. 3. Why molten silver cannot be used to make sharp castings? 4. A tarnished silver rod when kept in water contain- ing magnesium bars regains its lustre. Justify.

7. A mixture of X and Y on subjecting to paper chro- Nature of Matter 1.23 matography gave the chromatogram ‘A.’ When the same mixture is subjected to heating, chromato- 14. H2SO4 is always diluted by adding it to water but gram B was obtained. What do you infer from the not by adding water to it. Justify. chromatograms? 15. What type of reaction is involved in the usage of •• AgBr in photography? The positive print developed • is dipped in AuCl3 solution at the end to impart beautiful appearance to the photograph. Explain •• the reaction involved in this. Justify. AB Directions for questions from 16 to 25: 8. Why is solid CaCl2 spread on roads in cold c­ ountries Application- Based Questions during winter season? 16. In a chemistry lab, Rina took some mercury and 9. Explain the separation techniques involved in the water in two test tubes separately. Then she drained separations of constituents in gunpowder. What off both the liquids and on observing the empty test types of reactions are involved in the explosion of tubes, found some difference. Can you guess what gunpowder? the difference is? Explain with appropriate reasons. 10. Sodium cannot be preserved in water. However, 17. If we keep a box of ice cream in a freezer for too long, sodium amalgam can be kept in water. Justify. crystals of ice are formed inside the box. Give reasons. CH3I Acetone 18. Why are cotton clothes preferred in summer? CH3OH Vapour pressure 19. Small pieces of steel and some powdered rust are CCl4 taken in two test tubes separately. What will you observe when concentrated hydrochloric acid Temperature is poured into both the test tubes? Justify your PRACTICE QUESTIONS observation. 11. When a mixture of these four liquids is taken, how can they be separated? Justify. 20. Explain how individual gases can be separated from a gaseous mixture of O2, H2 and CO2. 12. Explain why N2O supports combustion more v­ igorously than air. 21. The critical temperatures of CO2 gas and N2 gas are 31°C and –147°C, respectively. Which gas is 13. A copper rod is placed in AgNO3 solution and FeSO4 l­iquefied easily and why? solution. What changes do you observe? What type of reactions takes place? Justify your observation. 22. Can water be boiled below 100°C temperature? If yes, give a reason. Level 3 4. In cold countries, ethylene glycol is used in car 1. When glass of water is freezed, formation of ice radiators for both winter as well as summer seasons. starts from the top layer but melting of ice starts Explain. from the bottom. Justify. 5. When a mixture of three miscible liquids is sub- 2. Solids generally undergo melting on heating. jected to fractional distillation, liquid B is obtained But only certain specific solids like naphthalene in the receiver flask. The remaining mixture on fur- ­camphor undergo sublimation. Give a reason. ther fractional distillation, A, is left behind in the distillation flask. On the basis of the results, com- 3. Fractional distillation of ethyl alcohol–water ment on the critical temperatures of A, B and C ­mixture gives a mixture of 95.6% ethyl alcohol when they are in gaseous state. and 4.4% water. Further separation can be brought about either by the addition of CaO or by the addi- tion of a water soluble salt such as potassium acetate. Justify the formation of pure ethyl alcohol in both the cases.

1.24 Chapter 1 7. Two ice blocks of 10 g each are placed in 2 L distilled water at 273 K. One of the ice blocks is made up of Directions for questions from 6 to 10: sea water and other one is made up of distilled water. Application-Based Questions What will you observe if the ambient temperature is also 273 K? Give reasons to support your observation. 6. Dicky, Micky and Vicky had three liquids, A, B and C, respectively. They mixed these liquids and observed 8. Can phase transition be used to test the purity of gold? that they form a homogeneous mixture. They were unable to separate the liquids and asked their teacher 9. The melting point of a non-sublimable solid is 100oC. to separate these for them. The teacher subjected the What do you observe when a small piece of this solid given mixture to fractional distillation. Liquid B was is taken in a test tube and placed in boiling water? obtained in the receiver flask. On further distillation, A was left behind in the distillation flask. On the basis 10. What is the shape of the meniscus observed when of the results, comment on the critical temperatures of water and mercury are taken in two different A, B and C in their respective gaseous states. c­ apillary tubes and why? PRACTICE QUESTIONS

Nature of Matter 1.25 CONCEPT APPLICATION Level 1 True or false 2. True 3. True 4. True 1. True 6. True 7. True 5. False Fill in the blanks 12. sublimes at normal atmospheric pressure 8. diffusion, fractional evaporation 13. decreasing pressure decreases the boiling point 9. adhesive forces, cohesive forces 14. x, y 10. solvent extraction, filtration, fractional crystallisation 11. KI Match the following B : f C : a D : h E:b 15. A : i G : d H : e I :c F : g Multiple choice questions 16. a 20. d 24. d 28. c 17. b 21. b 25. b 29. b 18. b 22. c 26. d 30. d HINTS AND EXPLANATION 19. b 23. d 27. d 31. (i) potential energy increases and kinetic energy 34. Pressure decreases with increase in altitude, and hence, remains constant boiling point decreases. Since Darjeeling is at higher attitude, water boils below 373 K, i.e., below 100°C. (ii) potential energy increases and kinetic energy increases 35. Kinetic energy does not change but potential energy increases at melting point. (iii) potential energy decreases and kinetic energy decreases 36. The more the humidity, the lesser is the rate of evaporation. In coastal areas humidity is more, and (iv) potential energy decreases and kinetic energy hence, the rate of evaporation is less. remains constant 32. (i) Paper chromatography is used for the separation 37. Sulphur is insoluble in water and forms a solid–­ (ii) paper and solvent are taken as stationary and liquid heterogeneous mixture. mobile phases, respectively. 38. Gunpowder is a heterogeneous mixture of KNO3, (iii) A narrow strip of paper with a line drawn is cut C and S. and a mixture of red and blue ink with the help of 39. Among the given gases, ammonia can dissolve capillary is placed on the line marked on the paper. ­readily in water. Therefore, it can be separated by (iv) T he paper is suspended in the closed jar with using water as a solvent. the help of hook. (v) T he blue and red inks are separated on the paper by adsorption technique. 33. Bromine is a liquid. The molecules of liquid possess 40. Since iron is magnetic in nature iron and sand vibratory, rotatory motion and translatory motions ­mixture can be separated by magnetic separation. in all directions except in upward direction.

1.26 Chapter 1 not sweat like human beings. All other phenomena are due to the condensation of water vapour due to 41. Conduction of heat in matter takes place via the very low temperature. molecules. One molecule absorbs heat and passes it to the other molecule. As in gaseous state the mol- 43. With increasing altitude, pressure decreases, and ecules are far away from each other, the heat cannot hence, boiling point of water also decreases. As a be transferred from molecule to molecule, and thus, result, cooking of food is delayed on mountain peaks. there is no conduction of heat through air which is a mixture of gases. 44. Rate of sublimation is directly proportional to vapour pressure. Hence, ‘S’ undergo sublimation In case of solids, the molecules are closely packed more easily at room temperature. and the intermolecular forces of attraction are max- imum. Therefore, on heating, the molecules can- 45. Both KNO3 and NaCl are soluble in water forming not move freely there by showing less expansion a homogeneous solution which is soluble in alcohol than gases in which the forces of attraction between and forming a mixture. This mixture is insoluble molecules is less, and hence, on heating KE of mol- in CCl4. Hence, CCl4 is separated by separating ecules increases, and therefore, the volume increases. f­unnel. Alcohol is separated by fractional distilla- tion. As KNO3 is more soluble than NaCl in water, 42. Dogs stretch out their tongues during summer therefore, it is separated by fractional crystallisation. because the surface area of tongue provides scope NaCl from water is separated by distillation. for evaporation which protects the body from over- heating during summer. This is because they can- HINTS AND EXPLANATION Level 2 8. effect of addition of CaCl2 on freezing point of H2O 1. (i) conditions required for burning. 9. (i) components of gunpowder (ii) utilisation of heat energy supplied (ii) solubility of the components in water and CS2 (iii) products obtained on oxidation of constituents 2. (i) process of welding (ii) changes in pressure during the process of gunpowder ( iii) effect of change in pressure on melting point (iv) constituent of gun power acting as oxygen provider 3. (i) d urable characteristics of the metal for proper 10. (i) predicting the order of vapour pressure of given casting compounds (ii) change observed in silver during phase transition (ii) changes occurring when metal is preserved in water 4. (i) comparison of reactivity of Ag and Mg (ii) conditions required for the ions to get displaced (iii) properties of sodium in amalgam (iv) effect of water on sodium amalgam 5. (i) structure of khus khus mats to exhibit capillary action 11. (i) relation between vapour pressure and boiling point (ii) m ethods of separation of liquids varying in (ii) changes accruing in water due to large surface area of mats b­ oiling points (iii) relation between vapour pressure and boiling point (iii) changes observed in the surrounding (iv) m ethods of separation of liquids varying in 6. (i) m etal present in paints which gets darkened b­ oiling points (black) on long exposure to polluted air 12. (i) requisite to support combustion (ii) nature of H2O2 (ii) products formed on decomposition of N2O (iii) change in the colour of the compound when (iii) volume of one of the product in air which washed with H2O2 ­supports combustion (iv) comparison of volumes of the component in air 7. (i) principle involved in chromatography (ii) changes observed in chromatogram B and product of N2O which supports combustion (iii) reason for the change

13. (i) requisite for displacement of one metal by the Nature of Matter 1.27 other rate. The sweat absorbs heat energy equal to the (ii) relative positions of Ag, Cu and Fe in activity latent heat of vaporisation from the body and gets series evaporated, thus, keeping the body cool. 14. (i) energy changes involved in the reaction 19. Steel is an alloy, i.e., a mixture in which iron is the between H2SO4 and H2O major component and it retains its property in steel. Hence, pieces of steel will liberate hydrogen and (ii) conditions required for controlling the reaction effervescence will be observed. However, rust being a compound of iron does not exhibit the chemical 15. (i) effect of light on AgBr property of iron. Rust dissolves in hydrochloric acid (ii) comparison of reactivity of Ag and Au due to the formation ferric chloride and no efferves- cence will be observed in that test tube. 16. Water droplets are found on the inner surface of one test tube while the other test tube is completely 20. When a mixture of O2, H2 and CO2 is present, dry. In the case of water, adhesive forces are more the mixture can first be passed through KOH as it than cohesive forces but in case of mercury cohesive will dissolve the CO2. This CO2 can be obtained forces are more than adhesive forces. Due to stronger back from KOH by adding dilute HCl. The left adhesive force, water droplets stick to the test tube. over mixture contains O2 and H2. As the molecu- lar weight difference is high, therefore, they can be 17. Due to very low temperature in the freezer, the separated by diffusion, where H2 will come out first. water vapour gets deposited as crystals of ice. Therefore, we get crystals of ice inside the box if 21. Since the critical temperature of CO2 is high, the kept in freezer for longer time. intermolecular forces of attraction among CO2 molecules are more when compared to N2. Hence, 18. People wear cotton clothes to keep themselves cool CO2 will liquefy easily. in summer. Cotton fabric absorb sweat more due to greater adhesive forces between cotton and water 22. Yes, water can be boiled below its boiling point and also allows the sweat to be evaporated at a faster as the boiling point of the water decreases with decrease in pressure. HINTS AND EXPLANATION Level 3 1. (i) movement of water when water undergoes freezing (iii) effect addition of ethylene glycol to water of a (ii) effect of pressure on the melting point of ice radiator 2. (i) comparison of vapour pressure of solids which 5. (i) relation between intermolecular forces of undergo sublimation to normal solids a­ ttraction and boiling point (ii) cause of sublimation (ii) relation between critical temperature and (iii) requisite of vapour pressure of solid to sublime b­ oiling point 3. (i) method of separation (iii) relation between intermolecular forces of (ii) changes in the boiling point of water on ­attraction and boiling point a­ ddition of water-soluble salt (iv) relation between critical temperature and (iii) method of separation ­boiling point (iv) method of separation (v) changes in the boiling point of water on 6. Fractional distillation is employed to separate m­ iscible liquids which have a difference in their ­addition of water-soluble salt boiling points. In this process of separation, the (vi) method of separation liquid with a low boiling point, i.e., a high vapour pressure, distils. Therefore, from the given data, it 4. (i) boiling point of ethylene glycol can be said that the boiling point of B is less than (ii) changes observed in the radiator of a car during that of C which is less than that of A. For liquids with high boiling points the intermolecular forces summer and winter

1.28 Chapter 1 9. The solid remains as it is. Water boils at 100°C. Transmission of heat takes place as long as the of attraction are also high, and hence, they have ­temperature of the solid is below 0°C. As soon high critical temperatures. Therefore, the critical as the temperature of the solid reaches 100°C, temperature of A is greater than that of C which is transmission of heat from water to the solid stops. greater than that of B. Hence, it does not melt. 7. After sometime one ice block will disappear and 10. Mercury shows a convex meniscus as it has higher the other one remains intact. The ice block which cohesive forces than adhesive forces with glass. is made up of sea water melts because of lowering Water, on the other hand, shows a concave menis- of melting point due to presence of dissolved salts. cus as the adhesive forces between water and glass are higher than the cohesive forces between water 8. Yes, the melting point of pure solid is always con- molecules. stant. The purity of gold can be measured with the help of its melting point. However, if the gold is not pure then it does not have a sharp melting point, because the presence of impurities reduces the melting point of the pure gold. HINTS AND EXPLANATION

2CChhaapptteerr ASSAttrtotruomucmctiutciucrree Remember Before beginning this chapter, you should be able to: • understand basic concept of matter and atom. • identify presence of subatomic particles. Key Ideas After completing this chapter, you should be able to: • understand the concept of divisibility of atom and the characteristics of subatomic particles. •  study the stability of nucleus and nuclear reactions. • conceptuliaze the model of atom based on the discovery of electrons. • learn the planetary model of atom based on the concept of nucleus. • study the concept of stationary orbits based on quantisation of energy and approach towards the modern atomic model. F I G U R E 2 . 1   Figure Caption

2.2 Chapter 2 INTRODUCTION The concept that atoms are the fundamental building blocks of matter dates back to very ancient times. However, the ideas regarding atoms of those times had no experimental evidence and remained as mere speculation. Therefore, these ideas had to lay dormant for a long period until John Dalton proposed his atomic theory on the basis of certain observations and experimental results. The basic principle of his theory was that it regarded an atom as the ultimate particle of matter. Dalton’s atomic theory has been successful in giving a convincing explanation for the various laws of chemical combination, such as, the law of conservation of mass, the law of definite proportions and the law of multiple proportions. However, Dalton’s idea that the atom is an indivisible particle of matter has been disproved by later discovery of radioactivity. Several series of experiments on radioactivity which were carried out later proved the presence of various subatomic particles in an atom. Atoms are found to be mainly composed of three types of fundamental particles, namely, positively-charged protons, neutral particles known as neutrons and negatively-charged electrons. The discovery of these fundamental particles paved the way for further research on the internal structure of an atom which obviously explains the enormous diversity of chemistry involved in a wide range of chemical reactions. DISCOVERY OF FUNDAMENTAL PARTICLES The electron was the first fundamental particle that was discovered. The credit for the discovery of the electron goes to J.J. Thomson based on his experiments carried out in a discharge tube. Sir William Crookes was the first scientist who designed the discharge tube which was called Crooke’s discharge tube or cathode ray tube. It is a long glass tube having two metal plates connected to the oppositely charged poles of a battery. The pressure inside the discharge tube can be adjusted by means of an exhaust pump. This discharge tube was later slightly modified by J.J. Thomson. When high voltage (HV) was applied between the cathode and the anode with a small hole at the centre of a partially evacuated tube at a pressure of 0.01 mm of Hg, a bright spot of light was formed on the zinc sulphide screen kept at the opposite end of the discharge tube. This was caused by the rays which originated from the cathode called cathode rays. Cathode Anode Cathode rays Exhaust pump _+ HV source FIGURE 2.1  Cathode ray tube J. J. Thomson conducted some experiments with a discharge tube for studying the properties of cathode rays.

Atomic Structure 2.3 Gas at low Bright spot Cathode anode pressure Fluorescent –+ material Cath_ode+ Perforation HV source FIGURE 2.2 J.J. Thomson’s cathode ray tube TABLE 2.1  Experiments involved in discovery of fundamental particles Experiments Properties based on observation Placing a small object in between the cathode and anode Formation of a shadow of the object on the opposite side of the cathode Metal object Shadow of object Cathode rays travel in straight lines • Anode Cathode • –+ HV source Rotation of light paddle wheel. Small particles having mass and kinetic energy Placing a light paddle wheel between cathode and anode Light paddle wheel – • + • • Cathode Anode HV source Bending of rays towards the positive plate Negatively-charged particles Passing through electric field Deflection perpendicular to the applied Cathode anode Fluorescent magnetic field material – (ZnS) –+ + • Bright spot HV Schematic diagram Passing through magnetic field applied perpendicular to the path of the cathode rays Cathode Anode S Fluorescent N material (ZnS) –+ Schematic diagram • Bright spot HV The above experiments were carried out with different No change in properties gases in the discharge tube. The properties do not depend on the nature of gas taken in the discharge tube. Specific charge (e/m value) remains the same.