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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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