solidsaresometimescalledpseudosolidsorsupercooledliquids.Theydonothavedefiniteheatof

fusion.Whencutwithasharp-edgedtool,theycutintotwopieceswithirregularsurfaces.Examplesof

amorphoussolidincludeglass,rubber,andplastic.

Question1.2:

Whatmakesaglassdifferentfromasolidsuchasquartz?Underwhatconditionscouldquartzbe

convertedintoglass?

Answer

Thearrangementoftheconstituentparticlesmakesglassdifferentfromquartz.Inglass,theconstituent

particleshaveshortrangeorder,butinquartz,theconstituentparticleshavebothlongrangeandshort

rangeorders.

Quartzcanbeconvertedintoglassbyheatingandthencoolingitrapidly.

Question1.3:

Classifyeachofthefollowingsolidsasionic,metallic,molecular,network(covalent)oramorphous.

(i)Tetraphosphorusdecoxide(P4O10)(vii)Graphite

(ii)Ammoniumphosphate(NH4)3PO4(viii)Brass

(iii)SiC(ix)Rb

(iv)I2(x)LiBr

(v)P4(xi)Si

Answer

Ionic→(ii)Ammoniumphosphate(NH4)3PO4,(x)LiBr

Metallic→(viii)Brass,(ix)Rb

Molecular→(i)Tetraphosphorusdecoxide(P4O10),(iv)I2,(v)P4.

Covalent(network)→(iii)SiC,(vii)Graphite,(xi)Si

Amorphous→(vi)Plastic

Question1.4:

(i)Whatismeantbytheterm‘coordinationnumber’?

(ii)Whatisthecoordinationnumberofatoms:

(a)inacubicclose-packedstructure?

(b)inabody-centredcubicstructure?

Answer

(i)Thenumberofnearestneighboursofanyconstituentparticlepresentinthecrystallatticeiscalledits

coordinationnumber.

(ii)Thecoordinationnumberofatoms

(a)inacubicclose-packedstructureis12,and

(b)inabody-centredcubicstructureis8

Question1.5:

Howcanyoudeterminetheatomicmassofanunknownmetalifyouknowitsdensityandthedimension

ofitsunitcell?Explain.

Answer

Byknowingthedensityofanunknownmetalandthedimensionofitsunitcell,theatomicmassofthe

metalcanbedetermined.

Let‘a’betheedgelengthofaunitcellofacrystal,‘d’bethedensityofthemetal,‘m’betheatomicmassof

themetaland‘z’bethenumberofatomsintheunitcell.

Fromequations(iii)and(iv),wecandeterminetheatomicmassoftheunknownmetal.

Question1.6:

‘Stabilityofacrystalisreflectedinthemagnitudeofitsmeltingpoint’.Comment.Collectmeltingpointsof

solidwater,ethylalcohol,diethyletherandmethanefromadatabook.

Whatcanyousayabouttheintermolecularforcesbetweenthesemolecules?

Answer

Higherthemeltingpoint,greateristheintermolecularforceofattractionandgreateristhestability.A

substancewithhighermeltingpointismorestablethanasubstancewithlowermeltingpoint.

Themeltingpointsofthegivensubstancesare:

Solidwater→273K

Ethylalcohol→158.8K

Diethylether→156.85K

Methane→89.34K

Now,onobservingthevaluesofthemeltingpoints,itcanbesaidthatamongthegivensubstances,the

intermolecularforceinsolidwateristhestrongestandthatinmethaneistheweakest.

Question1.7:

Howwillyoudistinguishbetweenthefollowingpairsofterms:

(i)Hexagonalclose-packingandcubicclose-packing?

(ii)Crystallatticeandunitcell?

(iii)Tetrahedralvoidandoctahedralvoid?

Answer

i.A2-Dhexagonalclose-packingcontainstwotypesoftriangularvoids(aandb)asshowninfigure1.Let

uscallthis2-DstructureaslayerA.Now,particlesarekeptinthevoidspresentinlayerA(itcanbeeasily

observedfromfigures2and3thatonlyoneofthevoidswillbeoccupiedintheprocess,i.e.,eitheraorb).

LetuscalltheparticlesorspherespresentinthevoidsoflayerAaslayerB.Now,twotypesofvoidsare

presentinlayerB(candd).UnlikethevoidspresentinlayerA,thetwotypesofvoidspresentinlayerB

arenotsimilar.Voidcissurroundedby4spheresandiscalledthetetrahedralvoid.Voiddissurrounded

by6spheresandiscalledtheoctahedralvoid.

Now,the nextlayercanbeplacedoverlayerBin2ways.

Case1:Whenthethirdlayer(layerC)isplacedoverthesecondone(layerB)insuchamannerthatthe

spheresoflayerCoccupythetetrahedralvoidsc.Inthiscasewegethexagonalclose-packing.Thisis

showninfigure4.Infigure4.1,layerBispresentoverthevoidsaandlayerCispresentoverthevoidsc.

Infigure4.2,layerBispresentoverthevoidsbandlayerCispresentoverthevoidsc.Itcanbeobserved

fromthefigurethatinthisarrangement,thespherespresentinlayerCarepresentdirectlyabovethe

spheresoflayerA.Hence,wecansaythatthelayersinhexagonalclose-packingarearrangedinan

ABAB…..pattern.

Case2:Whenthethirdlayer(layerC)isplacedoverlayerBinsuchamannerthatthespheresoflayerC

occupytheoctahedralvoidsd.Inthiscasewegetcubicclose-packing.Infigure5.1,layerBispresentover

thevoidsaandlayerCispresentoverthevoidsd.Infigure5.2,layerBispresentoverthevoidsband

layerCispresentoverthevoidsd.Itcanbeobservedfromthefigurethatthearrangementofparticlesin

layerCiscompletelydifferentfromthatinlayersAorB.Whenthefourthlayeriskeptoverthethird

layer,thearrangementofparticlesinthislayerissimilartothatinlayerA.Hence,wecansaythatthe

layersincubicclosepackingarearrangedinanABCABC…..pattern.

Thesideviewsofhcpandccparegiveninfigures6.1and6.2respectively.

(ii)Thediagrammaticrepresentationoftheconstituentparticles(atoms,ions,ormolecules)presentina

crystalinaregularthree-dimensionalarrangementiscalledcrystallattice.

Aunitcellisthesmallestthree-dimensionalportionofacrystallattice.Whenrepeatedagainandagainin

differentdirections,itgeneratestheentirecrystallattice.

(iii)Avoidsurroundedby4spheresiscalledatetrahedralvoidandavoidsurroundedby6spheresis

calledanoctahedralvoid.Figure1representsatetrahedralvoidandfigure2representsanoctahedral

void.

Question1.8:

Howmanylatticepointsarethereinoneunitcellofeachofthefollowinglattice?

(i)Face-centredcubic

(ii)Face-centredtetragonal

(iii)Body-centred

Answer

(i)Thereare14(8fromthecorners+6fromthefaces)latticepointsinface-centredcubic.

(ii)Thereare14(8fromthecorners+6fromthefaces)latticepointsinface-centredtetragonal.

(iii)Thereare9(1fromthecentre+8fromthecorners)latticepointsinbody-centredcubic.

Question1.9:

Explain

(i)Thebasisofsimilaritiesanddifferencesbetweenmetallicandioniccrystals.

(ii)Ionicsolidsarehardandbrittle.

Answer

(i)Thebasisofsimilaritiesbetweenmetallicandioniccrystalsisthatboththesecrystaltypesareheldby

theelectrostaticforceofattraction.Inmetalliccrystals,theelectrostaticforceactsbetweenthepositive

ionsandtheelectrons.Inioniccrystals,itactsbetweentheoppositely-chargedions.Hence,bothhavehigh

meltingpoints.

Thebasisofdifferencesbetweenmetallicandioniccrystalsisthatinmetalliccrystals,theelectronsare

freetomoveandso,metalliccrystalscanconductelectricity.However,inioniccrystals,theionsarenot

freetomove.Asaresult,theycannotconductelectricity.However,inmoltenstateorinaqueoussolution,

theydoconductelectricity.

(ii)Theconstituentparticlesofioniccrystalsareions.Theseionsareheldtogetherinthree-dimensional

arrangementsbytheelectrostaticforceofattraction.Sincetheelectrostaticforceofattractionisvery

strong,thechargedionsareheldinfixedpositions.Thisisthereasonwhyioniccrystalsarehardand

brittle.

Question1.10:

Calculatetheefficiencyofpackingincaseofametalcrystalfor

(i)simplecubic

(ii)body-centredcubic

(iii)face-centredcubic(withtheassumptionsthatatomsaretouchingeachother).

Answer

(i)Simplecubic

Inasimplecubiclattice,theparticlesarelocatedonlyatthecornersofthecubeandtoucheachother

alongtheedge.

Question1.11:

Silvercrystallisesinfcclattice.Ifedgelengthofthecellis4.07×10−8cmanddensityis10.5gcm−3,

calculatetheatomicmassofsilver.

Answer

Itisgiventhattheedgelength,a=4.077×10−8cm

Density,d=10.5gcm−3

Asthelatticeisfcctype,thenumberofatomsperunitcell,z=4

Wealsoknowthat,NA=6.022×1023mol−1

Usingtherelation:

=107.13gmol−1

Therefore,atomicmassofsilver=107.13u

Question1.12:

AcubicsolidismadeoftwoelementsPandQ.AtomsofQareatthecornersofthecubeandPatthe

body-centre.Whatistheformulaofthecompound?WhatarethecoordinationnumbersofPandQ?

Answer

ItisgiventhattheatomsofQarepresentatthecornersofthecube.

Therefore,numberofatomsofQinoneunitcell=8x(1/8)=1

ItisalsogiventhattheatomsofParepresentatthebody-centre.

Therefore,numberofatomsofPinoneunitcell=1

ThismeansthattheratioofthenumberofPatomstothenumberofQatoms,P:Q=1:1

Hence,theformulaofthecompoundisPQ.

ThecoordinationnumberofbothPandQis8.

Question1.13:

Niobiumcrystallisesinbody-centredcubicstructure.Ifdensityis8.55gcm−3,calculateatomicradiusof

niobiumusingitsatomicmass93u.

Answer

Itisgiventhatthedensityofniobium,d=8.55gcm−3

Atomicmass,M=93gmol−1

Asthelatticeisbcctype,thenumberofatomsperunitcell,z=2

Wealsoknowthat,NA=6.022×1023mol^−1

Applyingtherelation:

Question1.14:

IftheradiusoftheoctachedralvoidisrandradiusoftheatomsinclosepackingisR,deriverelation

betweenrandR.

Answer

Question1.17:

Whatisasemiconductor?Describethetwomaintypesofsemiconductorsandcontrasttheirconduction

mechanism.

Answer

Semiconductorsaresubstanceshavingconductanceintheintermediaterangeof10^-6to10^4ohm^

−1m^−1.

Thetwomaintypesofsemiconductorsare:

(i)n-typesemiconductor

(ii)p-typesemiconductor

n-typesemiconductor:Thesemiconductorwhoseincreasedconductivityisaresultofnegatively-charged

electronsiscalledann-typesemiconductor.Whenthecrystalofagroup14elementsuchasSiorGeis

dopedwithagroup15elementsuchasPorAs,ann-typesemiconductorisgenerated.

SiandGehavefourvalenceelectronseach.Intheircrystals,eachatomformsfourcovalentbonds.Onthe

otherhand,PandAscontainfivevalenceelectronseach.WhenSiorGeisdopedwithPorAs,thelatter

occupiessomeofthelatticesitesinthecrystal.Fouroutoffiveelectronsareusedintheformationoffour

covalentbondswithfourneighbouringSiorGeatoms.Theremainingfifthelectronbecomesdelocalised

andincreasestheconductivityofthedopedSiorGe.

p-typesemiconductor:Thesemiconductorwhoseincreasedinconductivityisaresultofelectronholeis

calledap-typesemiconductor.Whenacrystalofgroup14elementssuchasSiorGeisdopedwitha

group13elementsuchasB,Al,orGa(whichcontainsonlythreevalenceelectrons),ap-typeof

semiconductorisgenerated.

WhenacrystalofSiisdopedwithB,thethreeelectronsofBareusedintheformationofthreecovalent

bondsandanelectronholeiscreated.Anelectronfromtheneighbouringatomcancomeandfillthis

electronhole,butindoingso,itwouldleaveanelectronholeatitsoriginalposition.Theprocessappears

asiftheelectronholehasmovedinthedirectionoppositetothatoftheelectronthatfilledit.Therefore,

whenanelectricfieldisapplied,electronswillmovetowardthepositively-chargedplatethroughelectron

holes.However,itwillappearasiftheelectronholesarepositively-chargedandaremovingtowardthe

negatively-chargedplate.

Question1.20:

Classifyeachofthefollowingasbeingeitherap-typeorann-typesemiconductor:

(i)GedopedwithIn(ii)BdopedwithSi.

Answer

(i)Ge(agroup14element)isdopedwithIn(agroup13element).Therefore,aholewillbecreatedand

thesemiconductorgeneratedwillbeap-typesemiconductor.

(ii)B(agroup13element)isdopedwithSi(agroup14element).So,therewillbeanextraelectronand

thesemiconductorgeneratedwillbeann-typesemiconductor.

Question1.22:

Intermsofbandtheory,whatisthedifference

(i)Betweenaconductorandaninsulator

(ii)Betweenaconductorandasemiconductor

Answer

(i)Thevalencebandofaconductorispartially-filledoritoverlapswithahigherenergy,unoccupied

conductionband.

Ontheotherhand,inthecaseofaninsulator,thevalencebandisfully-filledandthereisalargegap

betweenthevalencebandandtheconductionband.

(ii)Inthecaseofaconductor,thevalencebandispartially-filledoritoverlapswithahigherenergy,

unoccupiedconductionband.So,theelectronscanfloweasilyunderanappliedelectricfield.

Ontheotherhand,thevalencebandofasemiconductorisfilledandthereisasmallgapbetweenthe

valencebandandthenexthigherconductionband.Therefore,someelectronscanjumpfromthevalence

bandtotheconductionbandandconductelectricity.

Question1.23:

Explainthefollowingtermswithsuitableexamples:

(i)Schottkydefect

(ii)Frenkeldefect

(iii)Interstitialsand

(iv)F-centres

Answer

(i)Schottkydefect:Schottkydefectisbasicallyavacancydefectshownbyionicsolids.Inthisdefect,an

equalnumberofcationsandanionsaremissingtomaintainelectricalneutrality.Itdecreasesthedensity

ofasubstance.SignificantnumberofSchottkydefectsispresentinionicsolids.Forexample,inNaCl,there

areapproximately106Schottkypairspercm3atroomtemperature.Ionicsubstancescontaining

similarsizedcationsandanionsshowthistypeofdefect.Forexample:NaCl,KCl,CsCl,AgBr,etc.

(ii)Frenkeldefect:Ionicsolidscontaininglargedifferencesinthesizesofionsshowthistypeofdefect.

Whenthesmallerion(usuallycation)isdislocatedfromitsnormalsitetoaninterstitialsite,Frenkel

defectiscreated.Itcreatesavacancydefectaswellasaninterstitialdefect.Frenkeldefectisalsoknown

asdislocationdefect.IonicsolidssuchasAgCl,AgBr,AgI,andZnSshowthistypeofdefect.

(iii)Interstitials:Interstitialdefectisshownbynon-ionicsolids.Thistypeofdefectiscreatedwhensome

constituentparticles(atomsormolecules)occupyaninterstitialsiteofthecrystal.Thedensityofa

substanceincreasesbecauseofthisdefect.

(iv)F-centres:Whentheanionicsitesofacrystalareoccupiedbyunpairedelectrons,theionicsitesare

calledF-centres.Theseunpairedelectronsimpartcolourtothecrystals.Forexample,whencrystalsof

NaClareheatedinanatmosphereofsodiumvapour,thesodiumatomsaredepositedonthesurfaceof

thecrystal.TheClionsdiffusefromthecrystaltoitssurfaceandcombinewithNaatoms,formingNaCl.

Duringthisprocess,theNaatomsonthesurfaceofthecrystalloseelectrons.Thesereleasedelectrons

diffuseintothecrystalandoccupythevacantanionicsites,creatingF-centres.

Question1.24:

Aluminiumcrystallisesinacubicclose-packedstructure.Itsmetallicradiusis125pm.

(i)Whatisthelengthofthesideoftheunitcell?

(ii)Howmanyunitcellsaretherein1.00cm3ofaluminium?

Answer

Question1.26:

Explainthefollowingwithsuitableexamples:

(i)Ferromagnetism

(ii)Paramagnetism

(iii)Ferrimagnetism

(iv)Antiferromagnetism

(v)12-16and13-15groupcompounds.

Answer

(i)Ferromagnetism:Thesubstancesthatarestronglyattractedbyamagneticfieldarecalled

ferromagneticsubstances.Ferromagneticsubstancescanbepermanentlymagnetisedevenintheabsence

ofamagneticfield.Someexamplesofferromagneticsubstancesareiron,cobalt,nickel,gadolinium,and

CrO2.Insolidstate,themetalionsofferromagneticsubstancesaregroupedtogetherintosmallregions

calleddomainsandeachdomainactsasatinymagnet.Inanunmagnetisedpieceofaferromagnetic

substance,thedomainsarerandomly-orientedandso,theirmagneticmomentsgetcancelled.However,

whenthesubstanceisplacedinamagneticfield,allthedomainsgetorientedinthedirectionofthe

magneticfield.Asaresult,astrongmagneticeffectisproduced.Thisorderingofdomainspersistseven

aftertheremovalofthemagneticfield.Thus,theferromagneticsubstancebecomesapermanentmagnet.

Paramagneticsubstancesgetmagnetisedinamagneticfieldinthesamedirection,butlosemagnetism

whenthemagneticfieldisremoved.Toundergoparamagnetism,asubstancemusthaveoneormore

unpairedelectrons.Thisisbecausetheunpairedelectronsareattractedbyamagneticfield,thereby

causingparamagnetism.

(iii)Ferrimagnetism:Thesubstancesinwhichthemagneticmomentsofthedomainsarealignedin

parallelandanti-paralleldirections,inunequalnumbers,aresaidtohaveferrimagnetism.Examples

includeFe3O4(magnetite),ferritessuchasMgFe2O4andZnFe2O4.

Ferrimagneticsubstancesareweaklyattractedbyamagneticfieldascomparedtoferromagnetic

substances.Onheating,thesesubstancesbecomeparamagnetic.

(v)12-16and13-15groupcompounds:The12-16groupcompoundsarepreparedbycombininggroup

12andgroup16elementsandthe13-15groupcompoundsarepreparedbycombininggroup13and

group15elements.ThesecompoundsarepreparedtostimulateaveragevalenceoffourasinGeorSi.

Indium(III)antimonide(IrSb),aluminiumphosphide(AlP),andgalliumarsenide(GaAS)aretypical

compoundsofgroups13-15.GaAssemiconductorshaveaveryfastresponsetimeandhave

revolutionisedthedesigningofsemiconductordevices.Examplesofgroup12-16compoundsincludezinc

sulphide(ZnS),cadmiumsulphide(CdS),cadmiumselenide(CdSe),andmercury(II)telluride(HgTe).The

bondsinthesecompoundsarenotperfectlycovalent.Theioniccharacterofthebondsdependsonthe

electronegativitiesofthetwoelements.

Question1.1:

Whyaresolidsrigid?

Answer

Theintermolecularforcesofattractionthatarepresentinsolidsareverystrong.Theconstituentparticles

ofsolidscannotmovefromtheirpositionsi.e.,theyhavefixedpositions.However,theycanoscillateabout

theirmeanpositions.Thisisthereasonsolidsarerigid.

Question1.2:

Whydosolidshaveadefinitevolume?

Answer

Theintermolecularforcesofattractionthatarepresentinsolidsareverystrong.Theconstituentparticles

ofsolidshavefixedpositionsi.e.,theyarerigid.Hence,solidshaveadefinitevolume.

Question1.3:

Classifythefollowingasamorphousorcrystallinesolids:

Polyurethane,naphthalene,benzoicacid,teflon,potassiumnitrate,cellophane,polyvinyl

chloride,fibreglass,copper.

Answer

Amorphoussolids

Polyurethane,teflon,cellophane,polyvinylchloride,fibreglass

Crystallinesolids

Naphthalene,benzoicacid,potassiumnitrate,copper

Question1.4:

Whyisglassconsideredasupercooledliquid?

Answer

Similartoliquids,glasshasatendencytoflow,thoughveryslowly.Therefore,glassisconsideredasa

supercooledliquid.Thisisthereasonthatglasswindowsanddoorsareslightlythickeratthebottom

thanatthetop.

Question1.5:

Refractiveindexofasolidisobservedtohavethesamevaluealongalldirections.Commentonthenature

ofthissolid.Woulditshowcleavageproperty?

Answer

Anisotropicsolidhasthesamevalueofphysicalpropertieswhenmeasuredalongdifferentdirections.

Therefore,thegivensolid,havingthesamevalueofrefractiveindexalongalldirections,isisotropicin

nature.Hence,thesolidisanamorphoussolid.Whenanamorphoussolidiscutwithasharpedgedtool,it

cutsintotwopieceswithirregularsurfaces.

Question1.6:

Classifythefollowingsolidsindifferentcategoriesbasedonthenatureofintermolecularforcesoperating

inthem:

Potassiumsulphate,tin,benzene,urea,ammonia,water,zincsulphide,graphite,rubidium,argon,silicon

carbide.

Answer

Potassiumsulphate→Ionicsolid

Tin→Metallicsolid

Benzene→Molecular(non-polar)solid

Urea→Polarmolecularsolid

Ammonia→Polarmolecularsolid

Water→Hydrogenbondedmolecularsolid

Zincsulphide→Ionicsolid

Graphite→Covalentornetworksolid

Rubidium→Metallicsolid

Argon→Non-polarmolecularsolid

Siliconcarbide→Covalentornetworksolid

Question1.7:

SolidAisaveryhardelectricalinsulatorinsolidaswellasinmoltenstateandmeltsat

extremelyhightemperature.Whattypeofsolidisit?

Answer

Thegivenpropertiesarethepropertiesofacovalentornetworksolid.Therefore,thegivensolidisa

covalentornetworksolid.Examplesofsuchsolidsincludediamond(C)andquartz(SiO2).

Question1.8:

Ionicsolidsconductelectricityinmoltenstatebutnotinsolidstate.Explain.

Answer

Inioniccompounds,electricityisconductedbyions.Insolidstate,ionsareheldtogetherbystrong

electrostaticforcesandarenotfreetomoveaboutwithinthesolid.Hence,ionicsolidsdonotconduct

electricityinsolidstate.However,inmoltenstateorinsolutionform,theionsarefreetomoveandcan

conductelectricity.

Question1.9:

Whattypeofsolidsareelectricalconductors,malleableandductile?

Answer

Metallicsolidsareelectricalconductors,malleable,andductile.

Question1.10:

Givethesignificanceofa‘latticepoint’.

Answer

Thesignificanceofalatticepointisthateachlatticepointrepresentsoneconstituentparticleofasolid

whichmaybeanatom,amolecule(groupofatom),oranion.

Question1.11:

Nametheparametersthatcharacterizeaunitcell.

Answer

Thesixparametersthatcharacteriseaunitcellareasfollows.

(i)Itsdimensionsalongthethreeedges,a,b,andc

Theseedgesmayormaynotbeequal.

(ii)Anglesbetweentheedges

Thesearetheangle∝(betweenedgesbandc),β(betweenedgesaandc),andγ(betweenedgesaandb).

Question1.12:

Distinguishbetween

(i)Hexagonalandmonoclinicunitcells

(ii)Face-centredandend-centredunitcells.

Answer

(i)Hexagonalunitcell

Forahexagonalunitcell,

(ii)Face-centredunitcell

Inaface-centredunitcell,theconstituentparticlesarepresentatthecornersandoneatthecentreof

eachface.

End-centredunitcell

Anend-centredunitcellcontainsparticlesatthecornersandoneatthecentreofanytwooppositefaces.

Question1.13:

Explainhowmuchportionofanatomlocatedat(i)cornerand(ii)body-centreofacubicunitcellispart

ofitsneighbouringunitcell.

Answer

(i)Anatomlocatedatthecornerofacubicunitcellissharedbyeightadjacentunitcells.

Therefore,1/8thportionoftheatomissharedbyoneunitcell.

(ii)Anatomlocatedatthebodycentreofacubicunitcellisnotsharedbyitsneighbouringunitcell.

Therefore,theatombelongsonlytotheunitcellinwhichitispresenti.e.,itscontributiontotheunitcellis

1.

Question1.14:

Whatisthetwodimensionalcoordinationnumberofamoleculeinsquareclosepackedlayer?

Answer

Insquareclose-packedlayer,amoleculeisincontactwithfourofitsneighbours.

Therefore,thetwo-dimensionalcoordinationnumberofamoleculeinsquareclosepackedlayeris4.

Question1.15:

Acompoundformshexagonalclose-packedstructure.Whatisthetotalnumberofvoidsin0.5molofit?

Howmanyofthesearetetrahedralvoids?

Answer

Question1.16:

AcompoundisformedbytwoelementsMandN.TheelementNformsccpandatomsofMoccupy1/3rd

oftetrahedralvoids.Whatistheformulaofthecompound?

Answer

TheccplatticeisformedbytheatomsoftheelementN.

Here,thenumberoftetrahedralvoidsgeneratedisequaltotwicethenumberofatomsoftheelementN.

Accordingtothequestion,theatomsofelementMoccupy1/3rdofthetetrahedralvoids.

Therefore,thenumberofatomsofMisequalto2x1/3=2/3rdofthenumberofatomsofN.

Therefore,ratioofthenumberofatomsofMtothatofNisM:N=(2/3):1=2:3

Thus,theformulaofthecompoundisM2N3.

Question1.17:

Whichofthefollowinglatticeshasthehighestpackingefficiency(i)simplecubic(ii)body-centredcubic

and(iii)hexagonalclose-packedlattice?

Answer

Hexagonalclose-packedlatticehasthehighestpackingefficiencyof74%.Thepackingefficienciesof

simplecubicandbody-centredcubiclatticesare52.4%and68%respectively.

Question1.18:

Anelementwithmolarmass2.7×10-2kgmol-1formsacubicunitcellwithedgelength405pm.Ifits

densityis2.7×103kgm−3,whatisthenatureofthecubicunitcell?

Answer

Thisimpliesthatfouratomsoftheelementarepresentperunitcell.Hence,theunitcellisface-centred

cubic(fcc)orcubicclose-packed(ccp).

Question1.19:

Whattypeofdefectcanarisewhenasolidisheated?Whichphysicalpropertyisaffectedbyitandinwhat

way?

Answer

Whenasolidisheated,vacancydefectcanarise.Asolidcrystalissaidtohavevacancydefectwhensome

ofthelatticesitesarevacant.

Vacancydefectleadstoadecreaseinthedensityofthesolid.

Question1.20:

Whattypeofstoichiometricdefectisshownby:

(i)ZnS(ii)AgBr

Answer

(i)ZnSshowsFrenkeldefect.

(ii)AgBrshowsFrenkeldefectaswellasSchottkydefect.

Question1.21:

Explainhowvacanciesareintroducedinanionicsolidwhenacationofhighervalenceisaddedasan

impurityinit.

Answer

Whenacationofhighervalenceisaddedtoanionicsolidasanimpuritytoit,thecationofhighervalence

replacesmorethanonecationoflowervalencesoastokeepthecrystalelectricallyneutral.Asaresult,

somesitesbecomevacant.Forexample,whenSr2+isaddedtoNaCl,eachSr2+ionreplacestwoNa+ions.

However,oneSr2+ionoccupiesthesiteofoneNa+ionandtheothersiteremainsvacant.Hence,

vacanciesareintroduced.

Question1.22:

Ionicsolids,whichhaveanionicvacanciesduetometalexcessdefect,developcolour.Explainwiththe

helpofasuitableexample.

Answer

Thecolourdevelopsbecauseofthepresenceofelectronsintheanionicsites.Theseelectronsabsorb

energyfromthevisiblepartofradiationandgetexcited.Forexample,whencrystalsofNaClareheatedin

anatmosphereofsodiumvapours,thesodiumatomsgetdepositedonthesurfaceofthecrystalandthe

chlorideionsfromthecrystaldiffusetothesurfacetoformNaClwiththedepositedNaatoms.Duringthis

process,theNaatomsonthesurfaceloseelectronstoformNa+ionsandthereleasedelectronsdiffuse

intothecrystaltooccupythevacantanionicsites.Theseelectronsgetexcitedbyabsorbingenergyfrom

thevisiblelightandimpartyellowcolourtothecrystals.

Question1.23:

Agroup14elementistobeconvertedinton-typesemiconductorbydopingitwithasuitableimpurity.To

whichgroupshouldthisimpuritybelong?

Answer

Ann-typesemiconductorconductsbecauseofthepresenceofextraelectrons.Therefore,agroup14

elementcanbeconvertedton-typesemiconductorbydopingitwithagroup15element.

Question1.24:

Whattypeofsubstanceswouldmakebetterpermanentmagnets,ferromagneticorferrimagnetic.Justify

youranswer.

Answer

Ferromagneticsubstanceswouldmakebetterpermanentmagnets.Insolidstate,themetalionsof

ferromagneticsubstancesaregroupedtogetherintosmallregions.Theseregionsarecalleddomainsand

eachdomainactsasatinymagnet.Inanunmagnetisedpieceofaferromagneticsubstance,thedomains

arerandomlyoriented.Asaresult,themagneticmomentsofthedomainsgetcancelled.However,when

thesubstanceisplacedinamagneticfield,allthedomainsgetorientedinthedirectionofthemagnetic

fieldandastrongmagneticeffectisproduced.Theorderingofthedomainspersistsevenafterthe

removalofthemagneticfield.Thus,theferromagneticsubstancebecomesapermanentmagnet.

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