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CBSEClass–12SubjectChemistry

NCERTSolutions

Chapter–01

TheSolidState

1.Definetheterm'amorphous'.Giveafewexamplesofamorphoussolids.

Ans.Amorphous solids are the solidswhose constituent particles are irreguraly arranged

andhave short rangeorder.These solidsare isotropic innatureandmelt overa rangeof

temperature. Therefore, amorphous solids are sometimes called pseudo solids or super

cooledliquids.Theydonothavedefiniteheatoffusion.Whencutwithasharp-edgedtool,

theycutintotwopieceswithirregularsurfaces.Examplesofamorphoussolidsincludeglass,

rubber,andplastic.

2.Whatmakes a glass different from a solid such as quartz?Underwhat conditions

couldquartzbeconvertedintoglass?

Ans.The arrangement of the constituent particles makes glass different from quartz. In

glass,theconstituentparticleshaveshort-rangeorder,butinquartz,theconstituentparticles

havelongrangeorder.

Quartzcanbeconvertedintoglassbyheatingandthencoolingitrapidly.

3.Classifythefollowingasamorphousorcrystallinesolids:

Polyurethane, naphthalene, benzoic acid, teflon, potassium nitrate, cellophane,

polyvinylchloride,fibreglass,copper.

Ans.Amorphoussolids:Polyurethane,teflon,cellophane,polyvinylchloride,fibreglass

Crystallinesolids:Naphthalene,benzoicacid,potassiumnitrate,copper

4.(i)Whatismeantbytheterm'coordinationnumber'?

(ii)Whatisthecoordinationnumberofatoms:

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(a)inacubicclose-packedstructure?

(b)inabody-centredcubicstructure?

Ans.(i)Thenumberofnearestneighboursofanyconstituentparticlepresentinthecrystal

latticeiscalleditscoordinationnumber.

(ii)Thecoordinationnumberofatoms

(a)inacubicclose-packedstructureis12,and

(b)inabody-centredcubicstructureis8

5.Howcanyoudeterminetheatomicmassofanunknownmetalifyouknowitsdensity

andthedimensionofitsunitcell?Explain.

Ans.Byknowing the density of anunknownmetal and the dimension of its unit cell, the

atomicmassofthemetalcanbedetermined.

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

atomicmassofoneatomofthemetaland‘z’bethenumberofatomsintheunitcell.

Now,densityoftheunitcell=Massoftheunitcell/Volumeoftheunitcell

d=zxm/a3xNA……………………(i)

[Sincemassoftheunitcell=Numberofatomsintheunit ofoneatom

andmassofoneatom=atomicmass/NA]

[Volumeoftheunitcell=(Edgelengthofthecubicunitcell)3]

Fromequation(i),wehave:

m=dxa3xNA/z……………………..(ii)

ifitisnotacubicunitcellthendimensionsofunitcellwillbea,b,candvolumeofunitcell=

axbxc

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som=dx(abc)xNA/z

6. 'Stability of a crystal is reflected in themagnitudeof itsmeltingpoint'. Comment.

Collectmeltingpointsofsolidwater,ethylalcohol,diethyletherandmethanefroma

databook.Whatcanyousayabouttheintermolecularforcesbetweenthesemolecules?

Ans.Higherthemeltingpoint,greateristheintermolecularforceofattractionandgreateris

the stability. A substancewith highermelting point ismore stable than a substancewith

lowermeltingpoint.

Themeltingpointsofthegivensubstancesare:

Solidwater=273K

Ethylalcohol=158.8K

Diethylether=156.85K

Methane=89.34K

Now, on observing the values of the melting points, it can be said that among the given

substances,theintermolecularforceinsolidwateristhestrongestandthatinmethaneisthe

weakest.Thisisbecaseofexistanceofhydrogenbonding(stronginteraction)insolidwater

(ice)andvanderwallforceswhichareweakerforceexistinmethane.

7.Howwillyoudistinguishbetweenthefollowingpairsofterms:

(i)Hexagonalclose-packingandcubicclose-packing?

(ii)Crystallatticeandunitcell?

(iii)Tetrahedralvoidandoctahedralvoid?

Ans.A2-Dhexagonalclose-packingcontainstwotypesoftriangularvoids(shadedaandb)

asshowninfigure1.Letuscallthis2-DstructureaslayerA.Now,particlesarekeptinthe

voidspresentinlayerA(itcanbeeasilyobservedfromfigures2and3thatonlyoneofthe

voidswillbeoccupiedintheprocess,i.e.,eitheraorb).Letuscalltheparticlesorspheres

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presentinthevoidsoflayerAaslayerB.Now,twotypesofvoidsarepresentinlayerB(c

andd).UnlikethevoidspresentinlayerA,thetwotypesofvoidspresentinlayerBarenot

similar. Void c is surrounded by 4 spheres and is called the tetrahedral void. Void d is

surroundedby6spheresandiscalledtheoctahedralvoid.

Now,thenextlayercanbeplacedoverlayerBin2ways.

Case1:Whenthethirdlayer(layerC)isplacedoverthesecondone(layerB)insucha

mannerthatthespheresoflayerCoccupythetetrahedralvoidsc.

Inthiscasewegethexagonalclose-packing.Thisisshowninfigure4.Infigure4.1,layerBis

present over the voids a and layer C is present over the voids c. In figure 4.2, layer B is

presentoverthevoidsbandlayerCispresentoverthevoidsc.Itcanbeobservedfromthe

figurethatinthisarrangement,thespherespresentinlayerCarepresentdirectlyabovethe

spheres of layer A. Hence, we can say that the layers in hexagonal close-packing are

arrangedinanABAB…..pattern.

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Case2:Whenthethirdlayer(layerC)isplacedoverlayerBinsuchamannerthatthe

spheresoflayerCoccupytheoctahedralvoidsd.

Inthiscasewegetcubicclose-packing.Infigure5.1,layerBispresentoverthevoidsaand

layerCispresentoverthevoidsd.Infigure5.2,layerBispresentoverthevoidsbandlayer

C is present over the voids d. It can be observed from the figure that the arrangement of

particlesinlayerCiscompletelydifferentfromthatinlayersAorB.Whenthefourthlayer

is kept over the third layer, the arrangement of particles in this layer is similar to that in

layer A. Hence, we can say that the layers in cubic close-packing are arranged in an

ABCABC…..pattern.

8.Howmanylatticepointsarethereinoneunitcellofeachofthefollowinglattice?

(i)Face-centredcubic

(ii)Face-centredtetragonal

(iii)Body-centred

Ans.(i) Thereare14 (8 from the corners+6 from the faces) latticepoints in face-centred

cubic.

(ii) There are 14 (8 from the corners + 6 from the faces) lattice points in face-centred

tetragonal.

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

9.Explain

(i)Thebasisofsimilaritiesanddifferencesbetweenmetallicandioniccrystals.

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(ii)Ionicsolidsarehardandbrittle.

Ans.(i)Thebasisofsimilaritiesbetweenmetallicandioniccrystalsisthatboththesecrystal

types areheldby the electrostatic force of attraction. Inmetallic crystals, the electrostatic

forceactsbetweenthepositivemetalions(Kernels)andtheelectrons.Inioniccrystals,itacts

between the oppositely-charged ions (cation and anion). Hence, both have high melting

points.

Thebasisofdifferencesbetweenmetallicand ioniccrystals is that inmetalliccrystals, the

electronsarefreetomoveandso,metalliccrystalscanconductelectricity.However,inionic

crystals,theionsarenotfreetomove.Asaresult,theycannotconductelectricity.However,

inmoltenstateorinaqueoussolution,theydoconductelectricity.

(ii)Theconstituentparticlesofioniccrystalsareions.Theseionsareheldtogetherinthree-

dimensional arrangements by the electrostatic force of attraction. Since the electrostatic

force of attraction is very strong, the charged ions are held in fixed positions. This is the

reasonwhyioniccrystalsarehardandbrittle.

10.Calculatetheefficiencyofpackingincaseofametalcrystalfor

(i)simplecubic

(ii)body-centredcubic

(iii)face-centredcubic(withtheassumptionsthatatomsaretouchingeachother).

Ans.(i)Simplecubic

Inasimplecubiclattice,theparticlesarelocatedonlyatthecornersofthecubeandtouch

eachotheralongtheedge.

Lettheedgelengthofthecubebe‘a’andtheradiusofeachparticleber.

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So,wecanwrite:a=2r

Now,volumeofthecubicunitcell=

= =

Weknowthatthenumberofparticlesperunitcellis1.

Therefore,volumeoftheoccupiedbyatomsinunitcell=4/3x3.14xr3

Hence,packingefficiency=(volumeoccupiedbyparticlesinunitcell/volumeofcubicunit

cell)x100

=(4/3x3.14xr3/a3)x100

anda=2r

=52.4%

(ii)Body-centredcubic

Itcanbeobservedfromtheabovefigurethat theatomat thecentre is incontactwiththe

othertwoatomsdiagonallyarranged.

FromΔFED,wehave:

b2=a2+a2⇒b2=2a2⇒b=1.414a

Again,from ,wehave:

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c2=a2+b2⇒c2=a2+2a2⇒c2=3a2(Sinceb2=2a2)

c=1.73a

Lettheradiusoftheatomber.

Lengthofthebodydiagonal,c=4r

=1.73a=4r

Volumeofthecube,a3=(4r/1.73)3

Abody-centredcubiclatticecontains2atoms.

So,volumeoftheoccupiedbyatomsincubiclattice=2x4/3x3.14xr3

Therefore, Packing efficiency = Volume occupied by two spheres in the unit cell / Total

volumeoftheunitcellx100

=(2x4/3x3.14xr3)/a3x100

anda3=(4r/1.73)3

packingeffiency=68%

(iii)Face-centredcubic

Lettheedgelengthoftheunitcellbe‘a’andthelengthofthefacediagonalACbeb.

FromΔABC,wehave:

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b=1.414a

Letrbetheradiusoftheatom.

Now,fromthefigure,itcanbeobservedthat:

b=4r

Now,volumeofthecube,a3

Weknowthatthenumberofatomperunitcellis4.

So,volumeoftheoccupiedbyatomsofunitcell=4x4/3x3.14xr3

Packingefficiency=(4x4/3x3.14xr3)/a3x100

and1.414a=4r

=74%

11. Silver crystallizes in fcc lattice. If edge length of the cell is and

densityis10.5g ,calculatetheatomicmassofsilver.

Ans.Itisgiventhattheedgelength,

Density,

Asthelatticeisfcctype,thenumberofatomsperunitcell,z=4

Wealsoknowthat,NA=6.023x1023

Usingtherelation:

d=(zxm)/(a3xNA)

m=dxa3xNA/z

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12.AcubicsolidismadeoftwoelementsPandQ.AtomsofQareatthecornersofthe

cubeandPatthebody-centre.Whatistheformulaofthecompound?Whatarethe

coordinationnumbersofPandQ?

Ans.ItisgiventhattheatomsofQarepresentatthecornersofthecube.

Therefore,numberofatomsofQinoneunitcell=8x1/8=1

ItisalsogiventhattheatomsofParepresentatthebody-centre.

Therefore,numberofatomsofPinoneunitcell=1

ThismeansthattheratioofthenumberofPatomstothenumberofQatoms,P:Q=1:1

Hence,theformulaofthecompoundisPQ.

ThecoordinationnumberofbothPandQis8.

13. Niobium crystallises in body–centred cubic structure. If density is 8.55 g ,

calculateatomicradiusofniobiumusingitsatomicmass93u.

Ans.Itisgiventhatthedensityofniobium,d=8.55gcm-3

Atomicmass,M=93g

Asthelatticeisbcctype,thenumberofatomsperunitcell,z=2

Wealsoknowthat,NA=6.023x1023

Applyingtherelation:

d=(zxm)/(a3xNA)

a3=(2x93)/(8.55x6.023x1023)

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=

So,

Forbody-centredcubicunitcell:

4r=1.73a

r=(1.73x3.306x10-8)/4

14.Iftheradiusoftheoctachedralvoidisrandradiusoftheatomsinclosepackingis

R,deriverelationbetweenrandR.

Ans.

AspherewithcentreO,isfittedintotheoctahedralvoidasshownintheabovefigure.Itcan

beobservedfromthefigurethat isright-angled

Now,applyingPythagorastheorem,wecanwrite:

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15.Coppercrystallisesintoafcclatticewithedgelength .Showthatthe

calculateddensityisinagreementwithitsmeasuredvalueof8.92gcm-3.

Ans.Edgelength,

Asthelatticeisfcctype,thenumberofatomsperunitcell,z=4

Atomicmass,M=63.5g

Wealsoknowthat,NA=6.023X1023

Applyingtherelation:

d=(zxM)/(NAxa3)

=4x63.5/6.023x1023x(3.61x10-8)3

=8.96gcm-3

The measured value of density is given as . Hence, the calculated density

isinagreementwithitsmeasuredvalue.

16.AnalysisshowsthatnickeloxidehastheformulaNi0.98O1.00.Whatfractionsofnickel

existas and ions?

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Ans.TheformulaofnickeloxideisNi0.98O1.00.

Therefore,theratioofthenumberofNiatomstothenumberofOatoms,Ni:O=0.98:1.00=

98:100

Now,totalchargeon100O2-ions= =-200

Letthenumberof ionsbex.

So,thenumberof ionsis98–x.

Now,totalchargeon ions=x(+2)

=+2x

And,totalchargeon ions=(98–x)(+3)

=294–3x

Since,thecompoundisneutral,wecanwrite:

2x+(294–3x)+(–200)=0

⇒–x+94=0

⇒x=94

Therefore,numberof ions=94

And,numberof ions=98–94=4

Hence,fractionofnickelthatexistsas =94/98

=0.959

And,fractionofnickelthatexistsas =4/98

=0.041

Alternatively,fractionofnickelthatexistsas =1–0.959

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=0.041

17. What is a semiconductor? Describe the two main types of semiconductors and

contrasttheirconductionmechanism.

Ans.Semiconductors are substances having conductance in the intermediate range of 10–

6to104ohm–1m–1.

Thetwomaintypesofsemiconductorsare:

(i)n-typesemiconductor

(ii)p-typesemiconductor

n-type semiconductor: The semiconductor whose increased conductivity is a result of

negatively-chargedelectronsiscalledann-typesemiconductor.Whenthecrystalofagroup

14 element such as Si orGe is dopedwith a group 15 element such as P orAs, ann-type

semiconductorisgenerated.

SiandGehavefourvalenceelectronseach.Intheircrystals,eachatomformsfourcovalent

bonds.On theotherhand,PandAs contain fivevalenceelectrons each.WhenSi orGe is

dopedwithPorAs,thelatteroccupiessomeofthelatticesitesinthecrystal.Fouroutoffive

electronsareusedintheformationoffourcovalentbondswithfourneighbouringSiorGe

atoms.The remaining fifth electronbecomesdelocalisedand increases the conductivityof

thedopedSiorGe.

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p-type semiconductor: The semiconductor whose increased in conductivity is a result of

electronholeiscalledap-typesemiconductor.Whenacrystalofgroup14elementssuchas

SiorGe isdopedwithagroup13element suchasB,Al,orGa (whichcontainsonly three

valenceelectrons),ap-typeofsemiconductorisgenerated.

WhenacrystalofSi isdopedwithB, the threeelectronsofBareused in the formationof

threecovalentbondsandanelectronholeiscreated.Anelectronfromtheneighboringatom

cancomeand fill this electronhole,but indoing so, itwould leaveanelectronholeat its

original position. The process appears as if the electron hole has moved in the direction

opposite to that of the electron that filled it. Therefore, when an electric field is applied,

electronswillmovetowardthepositively-chargedplatethroughelectronholes.However,it

will appear as if the electron holes are positively-charged and are moving toward the

negatively-chargedplate.

18. Non-stoichiometric cuprous oxide, can be prepared in laboratory. In this

oxide,coppertooxygenratioisslightlylessthan2:1.Canyouaccountforthefactthat

thissubstanceisap-typesemiconductor?

Ans. In the cuprous oxide prepared in the laboratory, copper to oxygen ratio is

slightlylessthan2:1.Thismeansthatthenumberof ionsisslightlylessthantwicethe

numberofO2−ions.Thisisbecausesome ionshavebeenreplacedby ions.Every

ionreplacestwo ions,therebycreatingholes.Asaresult,thesubstanceconducts

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electricity with the help of these positive holes. Hence, the substance is a p-type

semiconductor.

19.Ferricoxidecrystallises inahexagonalclose-packedarrayofoxide ionswith two

outofeverythreeoctahedralholesoccupiedbyferricions.Derivetheformulaofthe

ferricoxide.

Ans.LetthenumberofoxideO2-ionsbex.

So,numberofoctahedralvoids=x

Itisgiventhattwooutofeverythreeoctahedralholesareoccupiedbyferricions.

So,numberofferric ions=2/3x

Therefore,ratioofthenumberofFe3+ionstothenumberofO2−ions,Fe3+:O2−=2/3x:x

=2/3:1

=2:3

Hence,theformulaoftheferricoxideis .

20.Classifyeachofthefollowingasbeingeitherap-typeorann-typesemiconductor:

(i)GedopedwithIn

(ii)BdopedwithSi.

Ans.(i)Ge(agroup14element)isdopedwithIn(agroup13element).Therefore,aholewill

becreatedandthesemiconductorgeneratedwillbeap-typesemiconductor.

(ii)B (a group13element) isdopedwithSi (a group14element).Thus, a electronwillbe

createdandthesemiconductorgeneratedwillbean-typesemiconductor.

21.Gold(atomicradius=0.144nm)crystallisesinaface-centredunitcell.Whatisthe

lengthofasideofthecell?

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Ans.Foraface-centredunitcell4r=1.414a

Itisgiventhattheatomicradius,r=0.144nm

So,4(0.144)/1.414=a

=0.407nm

Hence,lengthofasideofthecell=0.407nm

22.Intermsofbandtheory,whatisthedifference

(i)Betweenaconductorandaninsulator

(ii)Betweenaconductorandasemiconductor

Ans. (i) The valence band of a conductor is partially-filled or it overlaps with a higher

energy,unoccupiedconductionband.

Ontheotherhand,inthecaseofaninsulator,thevalencebandisfully-filledandthereisa

largegapbetweenthevalencebandandtheconductionband.

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(ii)Inthecaseofaconductor,thevalencebandispartially-filledoritoverlapswithahigher

energy, unoccupied conduction band. So, the electrons can flow easily under an applied

electricfield.

On theotherhand, thevalencebandof a semiconductor is filled and there is a small gap

betweenthevalencebandandthenexthigherconductionband.Therefore,someelectrons

canjumpfromthevalencebandtotheconductionbandandconductelectricity.

23.Explainthefollowingtermswithsuitableexamples:

(i)Schottkydefect

(ii)Frenkeldefect

(iii)Interstitialsand

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(iv)F-centres

Ans.(i)Schottkydefect:Schottkydefectisbasicallyavacancydefectshownbyionicsolids.

In this defect, an equal number of cations and anions are missing to maintain electrical

neutrality.Itdecreasesthedensityofasubstance.SignificantnumberofSchottkydefectsis

presentinionicsolids.Forexample,inNaCl,thereareapproximately106Schottkypairsper

cm3atroomtemperature.Ionicsubstancescontainingsimilar-sizedcationsandanionsshow

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

(ii)Frenkeldefect: Ionic solids containing largedifferences in the sizes of ions show this

typeofdefect.Whenthesmallerion(usuallycation)isdislocatedfromitsnormalsitetoan

interstitialsite,Frenkeldefectiscreated.Itcreatesavacancydefectaswellasaninterstitial

defect.Frenkeldefectisalsoknownasdislocationdefect.IonicsolidssuchasAgCl,AgBr,AgI,

andZnSshowthistypeofdefect.

(iii) Interstitials: Interstitial defect is shown by non-ionic solids. This type of defect is

createdwhensomeconstituentparticles (atomsormolecules)occupyan interstitial siteof

thecrystal.Thedensityofasubstanceincreasesbecauseofthisdefect.

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(iv)F-centres:When the anionic sites of a crystal are occupied by unpaired electrons, the

ionic sitesare calledF-centres.Theseunpairedelectrons impart colour to thecrystals.For

example,whencrystalsofNaClareheatedinanatmosphereofsodiumvapour,thesodium

atomsaredepositedonthesurfaceofthecrystal.TheClionsdiffusefromthecrystaltoits

surfaceandcombinewithNaatoms,formingNaCl.Duringthisprocess,theNaatomsonthe

surface of the crystal lose electrons. These released electrons diffuse into the crystal and

occupythevacantanionicsites,creatingF-centres.

24.Aluminiumcrystallisesinacubicclose-packedstructure.Itsmetallicradiusis125

pm.

(i)Whatisthelengthofthesideoftheunitcell?

(ii)Howmanyunitcellsaretherein1.00 ofaluminium?

Ans.(i)Forcubicclose-packedstructure:

4r=1.414a

so,a=4(125pm)/1.414

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=353.55pm

=354pm(approximately)

(ii)Volumeofoneunitcell=(354pm)3

=4.4×107

=4.4×107×

=4.4×

soin4.4x10-23cm3wehave1unitcell

thereforein1cm3wehave=1/(4.4x10-23)unitcell

=2.27×1022

25. If NaCl is doped with of SrCl2, what is the concentration of cation

vacancies?

Ans.ItisgiventhatNaClisdopedwith %of .

Thismeansthat100molofNaClisdopedwith10−3molof .

Therefore,1molofNaClisdopedwith=

= molof

Cationvacanciesproducedbyone ion=1

Therefore,Concentrationofthecationvacancies

Producedby molof ions=10-5x6.023x1023

=6.023x1018

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Hence,theconcentrationofcationvacanciescreatedby is6.023x1018

26.Explainthefollowingwithsuitableexamples:

(i)Ferromagnetism

(ii)Paramagnetism

(iii)Ferrimagnetism

(iv)Antiferromagnetism

(v)12-16and13-15groupcompounds.

Ans.(i)Ferromagnetism:Thesubstancesthatarestronglyattractedbyamagneticfieldare

calledferromagneticsubstances.Ferromagneticsubstancescanbepermanentlymagnetised

even in the absence of amagnetic field. Some examples of ferromagnetic substances are

iron,cobalt,nickel,gadolinium,and .

In solid state, themetal ions of ferromagnetic substances are grouped together into small

regionscalleddomainsandeachdomainactsasatinymagnet.Inanun-magnetisedpieceof

a ferromagnetic substance, the domains are randomly-oriented and so, their magnetic

momentsgetcancelled.However,when thesubstance isplaced inamagnetic field,all the

domains get oriented in thedirectionof themagnetic field.As a result, a strongmagnetic

effectisproduced.Thisorderingofdomainspersistsevenaftertheremovalofthemagnetic

field.Thus,theferromagneticsubstancebecomesapermanentmagnet.

Schematicalignmentofmagneticmomentsinferromagneticsubstances

(ii) Paramagnetism: The substances that are attracted by a magnetic field are called

paramagneticsubstances.

Paramagneticsubstancesgetmagnetisedinamagneticfieldinthesamedirection,butlose

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magnetismwhen themagnetic field is removed. To undergo paramagnetism, a substance

must have one or more unpaired electrons. This is because the unpaired electrons are

attractedbyamagneticfield,therebycausingparamagnetism.

(iii)Ferrimagnetism:The substances inwhich themagneticmomentsof thedomainsare

aligned in parallel and anti-parallel directions, in unequal numbers, are said to have

ferrimagnetism. Examples include (magnetite), ferrites such as and

.

Ferrimagnetic substances are weakly attracted by a magnetic field as compared to

ferromagneticsubstances.Onheating,thesesubstancesbecomeparamagnetic.

Schematicalignmentofmagneticmomentsinferrimagneticsubstances

(iv)Antiferromagnetism:Antiferromagneticsubstanceshavedomainstructuressimilarto

ferromagnetic substances, but are oppositely-oriented. The oppositely-oriented domains

cancelouteachother'smagneticmoments.

Schematicalignmentofmagneticmomentsinantiferromagneticsubstances

(v) 12-16 and 13-15 group compounds: The 12-16 group compounds are prepared by

combininggroup12andgroup16elementsandthe13-15groupcompoundsarepreparedby

combining group 13 and group 15 elements. These compounds are prepared to stimulate

averagevalenceoffourasinGeorSi.Indium(III)antimonide(IrSb),aluminiumphosphide

(AlP), and gallium arsenide (GaAs) are typical compounds of groups 13-15. GaAs

semiconductors have a very fast response time and have revolutionised the designing of

semiconductor devices. Examples of group 12-16 compounds include zinc sulphide (ZnS),

cadmium sulphide (CdS), cadmium selenide (CdSe), andmercury (II) telluride (HgTe). The

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bonds in these compounds are not perfectly covalent. The ionic character of the bonds

dependsontheelectronegativitiesofthetwoelements.

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