CBSEClass11Chemistry

NCERTExemplarSolutions

Chapter11

Thep-BlockElements

MultipleChoiceQuestion(Type-1)

1. Theelementwhichexistsinliquidstateforawiderangeoftemperatureandcanbe

usedformeasuringhightemperatureis

(i) B

(ii) Al

(iii) Ga

(iv) In

Ans.(iii)Ga

Explanation:GalliumhasdifferentstructureconsistingGa2moleculewithlowestmelting

point.Itexistsasaliquidfrom30°Cto2000°Candhenceitisusedinhightemperature

thermometry.

2. WhichofthefollowingisaLewisacid?

(i) AlCl3

(ii) MgCl2

(iii) CaCl2

(iv) BaCl2

Ans.(i)AlCl3

Explanation:Lewisacidsarethespeciesinwhichoctateisnotcompleteandreadytoaccept

electrons.InAlCl3,Alissurroundedby6electronsandallthreeClatomsaresurroundedby

8electrons,therefore,AlCl3iselectronaccepter.Itisacovalentcompound.

3. Thegeometryofacomplexspeciescanbeunderstoodfromtheknowledgeoftypeof

Exercise

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hybridizationoforbitalsofcentralatom.Thehybridizationoforbitalsofcentralatom

in[Be(OH)4]–andthegeometryofthecomplexarerespectively

(i) sp3,tetrahedral

(ii) sp3,squareplanar

(iii) sp3d2,octahedral

(iv) dsp2,squareplanar

Ans.(i)sp3,tetrahedral

Explanation:In[B(OH)4]-,Boronissurroundedby4bondspairandhasnolonepair.

Geometryandhybridizationofcentralatomisbasedonbondpairsandlonepairsaround

thecentralatom.[B(OH)4]-istetrahedralandBoronissp3hybridized

4. Whichofthefollowingoxidesisacidicinnature?

(i) B2O3

(ii) Al2O3

(iii) Ga2O3

(iv) In2O3A

Ans.(i)B2O3

Explanation:Ingroup13(boronfamily)onmovingdownthegroup,acidiccharacter

decreasesandbasiccharacterincreases,therefore,B2O3isacidicinnature.

5. Theexhibitionofhighestco-ordinationnumberdependsontheavailabilityofvacant

orbitalsinthecentralatom.Whichofthefollowingelementsisnotlikelytoactas

centralatomin ?

(i) B

(ii) Al

(iii) Ga

(iv) In

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Ans.(i)B

Explanation:Borondoesnothaved-orbital.TheelementMinthecomplexion has

coordinationnumber6.Boroncanhavemaximumcoordinationnumber4.Thus,Bcannot

formthiscomplex.

6. Boricacidisanacidbecauseitsmolecule

(i) containsreplaceableH+ion

(ii) givesupaproton

(iii) acceptsOH–fromwaterreleasingproton

(iv) combineswithprotonfromwatermolecule

Ans.(iii)acceptsOH–fromwaterreleasingproton

Explanation:BoricacidisLewisacidandacceptselectron.Itreactswithwaterandaccepts

OH-andreleaseH+ionandthus,actsasweakmonobasicacid.

7. Catenationi.e.,linkingofsimilaratomsdependsonsizeandelectronicconfiguration

ofatoms.ThetendencyofcatenationinGroup14elementsfollowstheorder:

(i) C>Si>Ge>Sn

(ii) C>>Si>Ge≈Sn

(iii) Si>C>Sn>Ge

(iv) Ge>Sn>Si>C

Ans.(ii)C>>Si>Ge≈Sn

Explanation:Onmovingdownthegroupthesizeoftheatomincreasesandthebondenergy

decreasesandpropertyofcatenationdecreases.Ingroup14carbonshowsmaximum

catenation.C>>Si>Ge Sn.

8. Siliconhasastrongtendencytoformpolymerslikesilicones.Thechainlengthof

siliconepolymercanbecontrolledbyadding

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(i) MeSiCl3

(ii) Me2SiCl2

(iii) Me3SiCl

(iv) Me4Si

Ans.(iii)Me3SiCl

Explanation:Thechainlengthofthepolymercanbecontrolledbyadding(CH3)3SiClwhich

blockstheendsasshownbelow:

9. Ionizationenthalpy( iH1kJmol–1)fortheelementsofGroup13followstheorder.

(i) B>Al>Ga>In>Tl

(ii) B<Al<Ga<In<Tl

(iii) B<Al>Ga<In>Tl

(iv) B>Al<Ga>In<Tl

Ans.(iv)B>Al<Ga>In<Tl

Explanation:Ionizationenthalpy( iH)decreasesdownthegroupasthesizeofatom

increasesandscreeningeffect.AltoGa:Ionizationenthalpyincreasesslightlybecauseboth

nuclearchargeandscreeningeffectincreasesbutduetopoorshieldingbyd-electrons

effectivenuclearchargeonvalenceelectronismoreasaresultionizationenthalpy

increases.OnmovingGatoInagainionizationenthalpydecreasesduetoshieldingeffectof

d-electrons.OnmovingfromIntoTlionizationenthalpyagainincreasesbecauseofpoor

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shieldingeffectof4f~electrons.Thatiswhyeffectivenuclearchargeincreasesand

ionizationenthalpyincreases.

10. Inthestructureofdiborane

(i) Allhydrogenatomslieinoneplaneandboronatomslieinaplaneperpendiculartothis

plane.

(ii) 2boronatomsand4terminalhydrogenatomslieinthesameplaneand2bridging

hydrogenatomslieintheperpendicularplane.

(iii) 4bridginghydrogenatomsandboronatomslieinoneplaneandtwoterminalhydrogen

atomslieinaplaneperpendiculartothisplane.

(iv) Alltheatomsareinthesameplane.

Ans.(ii)2boronatomsand4terminalhydrogenatomslieinthesameplaneand2bridging

hydrogenatomslieintheperpendicularplane.

Explanation:Thefourterminalhydrogenatomsandthetwoboronatomslieinoneplane.

Aboveandbelowthisplane,therearetwobridginghydrogenatoms.ThefourterminalB-H

bondsareregulartwo-centre-two-electronbondswhilethetwobridge(B-H-B)bondsare

differentandcanbedescribedintermsofthree-centre-two-electronbondsasshownin

figure:

11. AcompoundX,ofboronreactswithNH3onheatingtogiveanothercompoundY

whichiscalledinorganicbenzene.ThecompoundXcanbepreparedbytreating

BF3withLithiumaluminumhydride.ThecompoundsXandYarerepresentedbythe

formulas.

(i) B2H6,B3N3H6

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(ii) B2O3,B3N3H6

(iii) BF3,B3N3H6

(iv) B3N3H6,B2H6

Ans.(i)B2H6,B3N3H6

Explanation:B2H6reactswithammoniaandgivesB2H6.2NH3whichisformulatedas

[BH2(NH3)2]+[BH4]

-andonheatinggivesB3N3H6borazinealsocalledborazole.

3B2H6+6NH3 3[BH2(NH3)2]+[BH4]

- 2B3N3H6+12H2

B2H6canbepreparedbyreductionofBF3withLiAlH4.

4BF3+3LiAlH4 2B2H6,+3LiF+3AIF3.

12. Quartzisextensivelyusedasapiezoelectricmaterial,itcontains_____.

(i) Pb

(ii) Si

(iii) Ti

(iv) Sn

Ans.(ii)Si

Explanation:Quartzisoneofthecrystallineformofsilicaandathightemperaturecanbe

convertedintoothercrystallineforms.Itisextensivelyusedasapiezoelectricmaterial.

13. Themostcommonlyusedreducingagentis

(i) AlCl3

(ii) PbCl2

(iii) SnCl4

(iv) SnCl2

Ans.(iv)SnCl2

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Explanation:InSnCl2,Sn2+canbeeasilyoxidizedtoSn4+because+4oxidationstateofSnis

morestablethan+2oxidationstate.Hence,SnCl2actsasareducingagent,

e.g.,SnCl2+2FeCl3 SnCl4+2FeCl2,

14. Dryiceis

(i) SolidNH3

(ii) SolidSO2

(iii) SolidCO2

(iv) SolidN2

Ans.(iii)SolidCO2

Explanation:SolidCO2iscalleddryicebecauseitisusedformakingicebathfororganic

reactioninlaboratory.ItispreparedbycoolingCO2gasathighpressure.

15. Cement,theimportantbuildingmaterialisamixtureofoxidesofseveralelements.

Besidescalcium,ironandSulphur,oxidesofelementsofwhichofthegroup(s)are

presentinthemixture?

(i) group2

(ii) groups2,13and14

(iii) groups2and13

(iv) groups2and14

Ans.(ii)groups2,13and14

Explanation:Cementismanufacturedbycombiningsubstanceswhicharelime(CaO),clay

containsSilica(SiO)andoxidesofAl,Mgandiron.

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Multiple Choice Questions (Type-II)

Inthefollowingquestionstwoor

moreoptionsmaybecorrect.16.Thereasonforsmallradius

ofGacomparedtoAlis_______.

(i) poorscreeningeffectofandforbitals

(ii) increaseinnuclearcharge

(iii) presenceofhigherorbitals

(iv) higheratomicnumber

Ans.(i),(ii)

Explanation:OnmovingdownthegroupfromAltoGa,theatomicradiidecreasesdueto

shieldingeffectofd-electrons.Thiseffectispoorhence,effectivenuclearchargeincreases.

17. ThelinearshapeofCO2isdueto_____.

(i) sp3hybridizationofcarbon

(ii) sphybridizationofcarbon

(iii) pπ–pπbondingbetweencarbonandoxygen

(iv) sp2hybridizationofcarbon

Ans.(ii),(iii)

Explanation:HybridizationofCinCO2isspandthestructureislinear.Carbonislinkedto

twooxygenatomsthroughdoublebonds,oneissigmaandotheroneispibond(p -p

bonding).

18. Me3SiClisusedduringpolymerizationoforganosiliconsbecause

(i) thechainlengthoforganosiliconpolymerscanbecontrolledbyaddingMe3SiCl

(ii) Me3SiClblockstheendterminalofsiliconepolymer

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(iii) Me3SiClimprovesthequalityandyieldofthepolymer

(iv) Me3SiClactsasacatalystduringpolymerization

Ans.(i),(ii)

Explanation:Thechainlengthofthepolymercanbecontrolledbyadding(CH3)3SiClwhich

blockstheends.

19. Whichofthefollowingstatementsarecorrect?

(i) Fullereneshavedanglingbonds

(ii) Fullerenesarecage-likemolecules

(iii) Graphiteisthermodynamicallymoststableallotropeofcarbon

(iv) Graphiteisslipperyandhardandthereforeusedasadrylubricantinmachines

Ans.(ii)and(iii)

Explanation:Fullerenesarecagelikemolecules.C-60moleculehastheshapelikeasoccer

ballandcalledBuckminsterfullerene.

Graphiteisthermodynamicallymoststableallotropeofcarbonand,therefore, of

graphiteistakenaszero. valuesofdiamondanfullerene,C-60are1.90and38.1kJ

mol-1,respectively.

20. Whichofthefollowingstatementsarecorrect.AnsweronthebasisofFig.11.1.

(i) Thetwobirdgedhydrogenatomsandthetwoboronatomslieinoneplane;

(ii) OutofsixB–Hbondstwobondscanbedescribedintermsof3centre2-electronbonds.

(iii) OutofsixB-HbondsfourB-Hbondscanbedescribedintermsof3centre2electron

bonds;

(iv) ThefourterminalB-Hbondsaretwocentre-twoelectronregularbonds.

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Ans.(i),(ii),(iv)

Explanation:Thefourterminalhydrogenatomsandthetwoboronatomslieinoneplane.

Aboveandbelowthisplane,therearetwobridginghydrogenatoms.ThefourterminalB-H

bondsareregulartwocentre-twoelectronbondswhilethetwobridge(B-H-B)bondsare

differentandcanbedescribedintermsofthreecentre-twoelectronbonds.

21. Identifythecorrectresonancestructuresofcarbondioxidefromtheonesgiven

below:

(i) O–C≡O(ii) O=C=O

(iii) –O≡C–O+

(iv) –O–C≡O+

Ans.(ii),(iv)

Explanation: Resonanceexplains

delocalizedelectronswithincertainmoleculesorpolyatomicionswherethebondingcannot

beexpressedbyonesingleLewisformula.

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Short Answer Type

22. DrawthestructuresofBCl3.NH3andAlCl3(dimer).

Ans.BCl3iscovalentinnatureanditisLewisacidbecauseoctaneofBoronisnotcompleted.

Boronneedsapairofelectrons.Itissp2hybridizedandtriangularplanar.Suchelectron

deficientmoleculeshavetendencytoacceptapairofelectronstoachievestableelectronic

configurationandthus,behaveasLewisacids.ThetendencytobehaveasLewisacid

decreaseswiththeincreaseinthesizedownthegroup.BCl3easilyacceptsalonepairof

electronsfromammoniatoformBCl3NH3

AlCl3achievesstabilitybyformingadimer.Intrivalentstate,mostofthecompoundsbeing

covalentarehydrolyzedinwater.Forexample,thetrichloridesonhydrolysisinwaterform

tetrahedralM(OH)4-.ElementMissp3hybridizedandhastetrahedralshape.

23. ExplainthenatureofboricacidasaLewisacidinwater.

Ans.BoricacidactsasLewisacidinwaterbyacceptingapairofelectronsfromahydroxyl

ion:

B(OH)3+2HOH→[B(OH)4]–+H3O+

24. Drawthestructureofboricacidshowinghydrogenbonding.Whichspeciesis

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presentinwater?Whatisthehybridizationsofboroninthisspecies?

Ans.IthasalayerstructureinwhichplanarH3BO3unitsarejoinedbyhydrogenbonds

forminghexagonalrings.Boricacidisaweakmonobasicacid.Itisnotaprotonicacidbut

actsasaLewisacidbyacceptingelectronsfromahydroxylion.Inwater,boricacidis

presentintheformof[B(OH)4]-species.Thehybridizationsofboroninthisspeciesissp3.

B(OH)3+2HOH [B(OH)4]-+H3O+

25. ExplainwhythefollowingcompoundsbehaveasLewisacids?

(i) BCl3

(ii) AlCl3

Ans.Inboththespecies,thecentralatomi.e.,BoronandAluminumaresurroundedby6

electronsandeachClatominbothissurroundedby8electrons.BCl3andAlCl3areelectron-

deficientandreadytoacceptapairofelectrons.Hence,theyactasLewisacids.

26. Givereasonsforthefollowing:

(i) CCl4isimmiscibleinwater,whereasSiCl4iseasilyhydrolyzed.

(ii) Carbonhasastrongtendencyforcatenationcomparedtosilicon.

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Ans.(i)CCl4isacovalentcompoundandcannotformH—bondwithpolarH2OandCdoes

nothavedorbitaltoaccommodatethelonepairofelectronsfromoxygenatomofwater

molecule.Othertetrachloride’sareeasilyhydrolyzedbywaterbecausethecentralatomcan

accommodatethelonepairofelectronsfromoxygenatomofwatermoleculeindorbital.

HydrolysiscanbeunderstoodbytakingtheexampleofSiCl4.Itundergoeshydrolysisby

initiallyacceptinglonepairofelectronsfromwatermoleculeindorbitalsofSi,finally

leadingtotheformationofSi(OH)4asshown:

(ii) Carbonatomshavethetendencytolinkwithoneanotherthroughcovalentbondstoform

chainsandrings.Thispropertyiscalledcatenation.ThisisbecauseC—Cbondsarevery

strong.Downthegroupthesizeincreasesandelectronegativitydecreases,and,thereby,

tendencytoshowcatenationdecreases.Thiscanbeclearlyseenfrombondenthalpies

values.TheorderofcatenationisC>>Si>Ge Sn.Leaddoesnotshowcatenation.

Bond Bondenthalpy/kJmol-1

C-C 348'

Si-Si 297

Ge—Ge 260

Sn-Sn 240

27. Explainthefollowing:

(i) CO2isagaswhereasSiO2isasolid.

(ii) SiliconformsSiF62–ionwhereascorrespondingfluorocompoundofcarbonisnot

known.

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Ans.InCO2,carbonatomundergoessphybridization.Twosphybridizedorbitalsofcarbon

atomoverlapwithtwoporbitalsofoxygenatomstomaketwosigmabondswhileothertwo

electronsofcarbonatomareinvolvedinpπ-pπbondingwithoxygenatom.Thisresultsinits

linearshape[withbothC-Obondsofequallength(115pm)]withnodipolemoment.

Silicondioxideisacovalent,three-dimensionalnetworksolidinwhicheachsiliconatomis

covalentlybondedinatetrahedralmannertofouroxygenatoms.Eachoxygenatominturn

covalentlybondedtoanothersiliconatomsasshownindiagram.Eachcomerissharedwith

anothertetrahedron.Theentirecrystalmaybeconsideredasgiantmoleculeinwhicheight

memberedringsareformedwithalternatesiliconandoxygenatoms.

(iii) Siliconhasvacantdorbitalinitsvalenceshellduetowhichitcanaccommodate6

electronsfrom6fluorineatomswhereascarbondoesnothavedorbitalandcannotexpand

itscovalencebeyondfour.

28. The+1-oxidationstateingroup13and+2oxidationstateingroup14becomesmore

andmorestablewithincreasingatomicnumber.Explain.

Ans.Onmovingdown,thegroupingroup13and14,loweroxidationstatebecomesmore

stableascomparedtohigheroxidationstate,becauseofinertpaireffect.Ininertpaireffect

Velectronsofvalenceshelldonotparticipateinbondingonly‘p’electronsparticipatein

bonding.Asthesizeofatomincreases,moreenergyisneededbyVelectronstoparticipatein

bondingfromvalenceshell.Ingroup13valenceshellconfigurationisYis1Yipx(n=2to6),

whenelectronsofboth‘s’and‘p’orbitalsparticipatetheyshow+3oxidationstatebut,ifonly

‘p’electronsparticipatethentheyshow+1oxidationstate.Ingroup14valenceshell

configurationisns2np2(n=2to6),whenelectronsofboth‘s’and‘p’orbitalsparticipatethen

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+4oxidationstateandifonly‘p’electronsparticipate,then+2oxidations.

29. Carbonandsiliconbothbelongtothegroup14,butinspiteofthestoichiometric

similarity,thedioxides,(i.e.,carbondioxideandsilicondioxide),differintheir

structures.Comment.

Ans.InCO2,carbonatomundergoessphybridization.Twosphybridizedorbitalsofcarbon

atomoverlapwithtwoporbitalsofoxygenatomstomaketwosigmabondswhileothertwo

electronsofcarbonatomareinvolvedinp

-p bondingwithoxygenatom.Thisresultsinitslinearshape[withbothC-Obondsofequal

length(115pm)]withnodipolemoment.Carboncangivedoublebondwithoxygenduetoits

smallsize.

Silicondioxideisacovalent,three-dimensionalnetworksolidinwhicheachsiliconatomis

covalentlybondedinatetrahedralmannertofouroxygenatoms.Eachoxygenatominturn

covalentlybondedtoanothersiliconatomsasshownindiagram.Eachcomerissharedwith

anothertetrahedron.Theentirecrystalmaybeconsideredasgiantmoleculeinwhicheight

memberedringsareformedwithalternatesiliconandoxygenatoms.Siliconcannotgive

doublebondwithoxygenduetolargesizeandlesselectronegativity.

30. Ifatrivalentatomreplacesafewsiliconatomsinthree-dimensionalnetworkof

silicondioxide,whatwouldbethetypeofchargeonoverallstructure?

Ans.InSiO2structure,ifatrivalentatomreplacesSiatom,thenholesarecreated.

Theseholeswillmakethecrystal,conductorofelectricity,calledp-typeconduction.The

crystalsaswholeareelectricallyneutral.

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31. WhenBCl3istreatedwithwater,ithydrolysesandforms[B(OH)4]–onlywhereas

AlCl3inacidifiedaqueoussolutionforms[Al(H2O)6]3+ion.Explainwhatisthe

hybridizationsofboronandaluminuminthesespecies?

Ans.Intrivalentstate,mostofcompoundsbeingcovalentarehydrolyzedinwater.For

example,thetrichloridesonhydrolysisinwaterformtetrahedral[M(OH)4]species;the

hybridizationstateofelementMissp3.Aluminumchlorideinacidifiedaqueoussolution

formsoctahedral[AI(H2O)6]+3Ion.Inthiscomplexion,the3dorbitalsofAIareinvolvedand

thehybridizationsstateofAIissp3d2.

32. Aluminumdissolvesinmineralacidsandaqueousalkalisandthusshows

amphotericcharacter.Apieceofaluminumfoilistreatedwithdilutehydrochloricacid

ordilutesodiumhydroxidesolutioninatesttubeandonbringingaburning

matchsticknearthemouthofthetesttube,apopsoundindicatestheevolutionof

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hydrogengas.Thesameactivitywhenperformedwithconcentratednitricacid,

reactiondoesn’tproceed.Explainthereason.

Ans.Aluminumisamphotericinnature,itreactswithacidandbasetogivesaltandH2gas.

Itburnswithpopsound.

2Al+6HCl 2AlCl3+3H2(g)

2Al+2NaOH+2H2O 2NaAlO2+3H2(g)

WhenAlreactswithconc.HNO3,athinlayerofAl2O3onthesurfaceofAlmetalwhich

protectfurtherreaction.Thislayeriscalledprotectivelayer.

2Al+6HNO3conc. Al2O3+6NO2+3H2O

33. Explainthefollowing:

(i) Galliumhashigherionisationenthalpythanaluminum.

(ii) BorondoesnotexistasB3+ion.

(iii) Aluminumforms[AlF6]3–ionbutborondoesnotform[BF6]3–ion.

(iv) PbX2ismorestablethanPbX4.

(v) Pb4+actsasanoxidizingagentbutSn2+actsasareducingagent.

(vi) Electrongainenthalpyofchlorineismorenegativeascomparedtofluorine.

(vii) Tl(NO3)3actsasanoxidizingagent.

(viii) Carbonshowscatenationpropertybutleaddoesnot.

(ix) BF3doesnothydrolyze.

(x) Whydoestheelementsilicon,notformagraphitelikestructurewhereascarbondoes.

Ans.(i)TheionizationenthalpyvalueofGaishigherthanAlduetoinabilityofd-andf-

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electrons,whichhavescreeningeffect,tocompensatetheincreaseinnuclearcharge.

(ii) Boronhassmallsizeandsumof iH1+ iH2+ iH3isveryhigh.Borondoesnotform

B3+iontherefore,givecovalentcompounds.

(iii) Alhasvacant‘dorbitalsandcanexpanditsco-ordinationno.andforms[AlF6]3-.Onthe

otherhand,Borondoesnothave‘dorbitalsandcannotform[BF6]3-andcannotexpandits

covalencebeyond4andthus,gives[BF4]-.

(iv) Duetoinertpaireffect,+2oxidationstateismorestablethan+4oxidationstate.

(v) Pb4+bygaining2electronschangesintoPb2+whichismorestableduetoinertpaireffect.

Sn2+islessstablethanSn4+bylosingelectrons.Therefore,Pb4+actsasanoxidizingagent

whileSn2+actsasareducingagent.

(vi) ThesizeofFatomisverysmallandincomingelectronsfeelinterelectronicrepulsionand

electrongainenthalpyofFislessnegativeascomparedtoCl.

(vii) Duetoinertpaireffect,Tlismorestablein+1oxidationstatethanthatof+3oxidation

state.Therefore,Tl(NO3)3actsasstrongoxidizingagent.

(viii) Carbonatomshavethetendencytolinkwithoneanotherthroughcovalentbondsto

formchainsandrings.Thispropertyiscalledcatenation.ThisisbecauseC—Cbondsare

verystrong.Downthegroupthesizeincreasesandelectronegativitydecreases,and,thereby,

tendencytoshowcatenationdecreases.Thiscanbeclearlyseenfrombondenthalpies

values.TheorderofcatenationisC>>Si>Ge Sn.Leaddoesnotshowcatenation.

(ix) BF3doesnothydrolyzecompletely.Itformsboricacidandfluoroboricacidthisis

becausetheHFformedreactswithH3BO3.

BF3+3H20 H3BO3+3HF}x4

H3BO3+3HF H+[BF4]-+3H2O}x3

4BF3+3H20 H3BO3+3[BF4]-+3H+

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(x) Ingraphite,carbonissp2hybridized.Carbonhasatendencytoformmultiplep -p

bondsduetoitssmallsizeandhighestelectronegativityingroup14.Siliconduetoitslarge

sizeandlesselectronegativitycannotformmultiplebonds.Thus,siliconcannotform

graphitelikestructure.

34. IdentifythecompoundsA,XandZinthefollowingreactions:

(i) A+2HCl+5H2O→2NaCl+X

Ans.(i)Na2B4O7(A)+2HCl+5H20 2NaCl+4B(OH)3(X)

H3BO3(X) HBO2 B2O3(Z)

Na2B4O7(A),H3BO3(X)andB2O3(Z)

35. Completethefollowingchemicalequations:

Z+3LiAlH4→X+3LiF+3AlF3

X+6H2O→Y+6H2

3X3O2 B2O33H2O

Ans.4BF3+3LiAlH4 2B2H6+3LiF+3A1F3

B2H6(g)+6H20(l) 2B(OH)3(aq)+6H2(g)

B2H6+3O2 B2O3+3H2O

Z=BF3,X=B2H6andY=B(OH)3orH3BO3

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Matching Type

Inthefollowingquestionsmorethanonecorrelationispossiblebetweenoptionsof

ColumnIandColumnII.Makeasmanycorrelationsasyoucan.

36.MatchthespeciesgiveninColumnIwiththepropertiesmentionedinColumnII.

ColumnI ColumnII

(i) BF4- (a) Oxidationstateofcentralatomis+4

(ii) AlCl3 (b) Strongoxidizingagent

(iii) SnO (c) Lewisacid

(iv) PbO2 (d) Canbefurtheroxidized

(e) Tetrahedralshape

Ans.(i) (e);(ii) (c);(iii) (d);(iv) (a),(b)

Column

IColumnII

(i) BF4- InBF4,Bissp

3hybridizedandsurroundedby4bondpairsandhasnolonepair.

(ii)

ALCL3InAlCl3octetofAlisnotcomplete,electrondeficientcompound.

(iii) SnO InSnOoxidationstateofSnis+2andcanbechangedinto+4.

(iv)

PbO4

InPbO2,oxidationstateofPbis+4,itislessstableduetoinertpaireffectand+2

oxidationstateismorestable.Pb4+changesintoPb2+actsasastrongoxidizing

agent.

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37. MatchthespeciesgiveninColumnIwithpropertiesgiveninColumnII.

ColumnI ColumnII

(i)Diborane (a) Usedasafluxforsolderingmetals

(ii)Galluim (b) Crystallineformofsilica

(iii) Borax (c) Bananabonds

(iv) Aluminosilicate(d) Lowmelting,highboiling,usefulfor

measuringhightemperatures

(v) Quartz(e) Usedascatalystinpetrochemical

industries

Ans.(i) (c);(ii) (d);(iii) (a);(iv) (e);(v) (b)

ColumnI ColumnII

(i) Diborane

InB2H6,EachBatomusessp3hybridsforbonding.Outofthefour

sp3hybridsoneachBatom,oneiswithoutanelectronshowninbroken

lines.TheterminalB-Hbondsarenormal2-centre-2-electronbondsbut

thetwobridgebondsare3-centre-2-electronbonds.The3-centre-2-

electronbridgebondsarealsoreferredtoasbananabonds.

(ii) Gallium

Galliumwithunusuallylowmeltingpoint(303K),couldexistinliquid

stateduringsummer.Itshighboilingpoint(2676K)makesitauseful

materialformeasuringhightemperatures.

(iii) BoraxBoraxisusedasafluxforsolderingmetals,forheat,scratchandstain

resistantglazedcoatingtoearthenwares.

(iv)

Zeolitesarealuminosilicatesandwidelyusedasacatalystin

petrochemicalindustriesforcrackingofhydrocarbonsandisomerization,

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Aluminosilicate e.g.,ZSM-5(Atypeofzeolite)usedtoconvertalcoholsdirectlyinto

gasoline.

(v) QuartzQuartz,cristobaliteandtridymitearesomeofthecrystallineformsof

silica.

38. MatchthespeciesgiveninColumnIwiththehybridizationsgiveninColumnII.

ColumnI ColumnII

(i) Boronin[B(OH)4]– (a) Sp2

(ii) Aluminumin[Al(H2O)6]3+ (b) sp3

(iii) BoroninB2H6 (c) sp3d2

(iv) CarboninBuckminsterfullerene

(v) SiliconinSiO44–

(vi) Germaniumin[GeCl6]2–

Ans.(i) (b);(ii) (c);(iii) (b);(iv) (a);(v) (b);(vi) (c)

ColumnI ColumnII

(i) Boronin[B(OH)4J- Boroniscentralatomandissurroundedby4bondpairsonly.

(ii) Aluminumin[Al

(H2O)6]3+

In[Al(H2O)6]3+,coordinationnumberofAlis6andgeometryis

octahedral.

(iii) BoroninB2H6InB2H6,eachBatomusessp

3hybridorbitalsforbonding.Outof

thefour,sp3hybridsoneachBatom,oneiswithoutanelectron.

(iv) Carbonin

Buckminsterfullerene

Allthecarbonatomsareequalandtheyundergosp2hybridization.

Eachcarbonatomformsthreesigmabondswithotherthree

carbonatoms.

Thebasicstructuralunitofsilicatesis inwhichsiliconatom

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(v) Siliconin

isbondedtofouroxygenatomsintetrahedronfashion.

(vi) Germaniumin

[GeCl6]2-

In[GeCl6]2-,Gehascoordinationnumber6andithasoctahedral

geometryandcentralatomGeissp3d2hybridized.

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Assertion and Reason Type

InthefollowingquestionsastatementofAssertion(A)followedbyastatementof

Reason(R)isgiven.Choosethecorrectoptionoutofthechoicesgivenbeloweach

question.

39.Assertion(A):Ifaluminumatomsreplaceafewsiliconatomsinthreedimensional

networkofsilicondioxide,theoverallstructureacquiresanegativecharge.

Reason(R):Aluminumistrivalentwhilesiliconistetravalent.

(i) BothAandRarecorrectandRisthecorrectexplanationofA.

(ii) BothAandRarecorrectbutRisnotthecorrectexplanationofA.

(iii) BothAandRarenotcorrect

(iv) AisnotcorrectbutRiscorrect.

Ans.(iv)AisnotcorrectbutRiscorrect.

Explanation:AlistrivalentandSiistetravalentandReasoniscorrect.Herethesiliconis

dopedwithgr.13elements.

40. Assertion(A):Silicon’sarewaterrepellinginnature.

Reason(R):Silicon’sareorganosiliconpolymers,whichhave(–R2SiO–)asrepeating

unit.

(i) AandRbotharecorrectandRisthecorrectexplanationofA.

(ii) BothAandRarecorrectbutRisnotthecorrectexplanationofA.

(iii) AandRbotharenottrue.

(iv) AisnottruebutRistrue.

Ans.(ii)BothAandRarecorrectbutRisnotthecorrectexplanationofA.

Explanation:Siliconesarethegroupofsiliconpolymers,whichhave(R2SiO)asarepeating

unit.Siliconesbeingsurroundedbynon-polaralkylgroupsarewaterrepellinginnaturesoA

andRbotharecorrectbutRisnotthecorrectexplanationofA.

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Long Answer Type

41.DescribethegeneraltrendsinthefollowingpropertiesoftheelementsinGroups13

and14.

(i) Atomicsize

(ii) Ionizationenthalpy

(iii) Metalliccharacter

(iv) Oxidationstates

(v) Natureofhalides

Ans.Group13

(i) Atomicsize:Onmovingdownthegroupthesizeofatomincreasesbecauseforeach

successivememberofthegrouponeextrashellofelectronsisaddedbutinsomecases

deviationisseen.ThesizeofGaislessthanthatofAlduetothepoorshieldingeffectof10

electronspresentindorbital.

(ii) Ionizationenthalpy:Theionizationenthalpyvaluesasexpectedfromthegeneraltrends

donotdecreasesmoothlydownthegroup.ThedecreasefromBtoAlisassociatedwith

increasedinsize.TheobserveddiscontinuityintheionizationenthalpyvaluesbetweenAl

andGa,andbetweenInandTlareduetoinabilityofd-andf-electrons,whichhavelow

screeningeffect,tocompensatetheincreaseinnuclearcharge.

(iii)Metalliccharacter:Boronisnonmetallicandallotherelementsaremetallic.Metallic

characterincreasesfromBtoAlbutfromAltoTlitdecreasesduetopoorshieldingeffectof

d-electronsandf-electrons.

(iv) Oxidationstates:Allelementsshow+3oxidationstatebutonmovingdownthegroup

duetoinertpaireffect+3oxidationstatedecreasesand+1oxidationstateprogressively

increasesintheorderofAl<Ga<In<Tl.

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(v) NatureofHalides:Theseelementsreactwithhalogenstoformtrihalides(exceptTlI3).

2E(s)+3X2(g) 2EX3(s)(X=F,Cl,Br,I)

HalidesofBoronandAluminumareelectrondeficientandactasLewisacids.Lewisacidic

characterofhalidesofborondecreasesinthefollowingorder:

BF3<BCl3<BBr3<BI3

Group-14

(i) Atomicsize:Ingroup14atomicradiusisreferredascovalentradius.Thereisa

considerableincreaseincovalentradiusfromCtoSi,thereafterfromSitoPbasmall

increaseinradiusisobserved.Thisisduetothepresenceofcompletelyfilleddandforbitals

inheaviermembersduetoscreeningeffect.

(ii) lonizationEnthalpy:Thefirstionizationenthalpyofgroup14membersishigherthan

thecorrespondingmembersofgroup13.Theinfluenceofinnercoreelectronsisvisiblehere

also.Ingeneral,theionizationenthalpydecreasesdownthegroup.SmalldecreaseinDiH

fromSitoGetoSnandslightincreaseinDiHfromSntoPbistheconsequenceofpoor

shieldingeffectofinterveningdandforbitalsandincreaseinsizeoftheatom.

(iii)Metalliccharacter:Ingroup14,onmovingdownthemetalliccharacterincreases.

Carbonisnon-metal,SiandGearemetalloidandSnandPbaremetals.

(iv) Oxidationstates:Thecommonoxidationstatesexhibitedbytheseelementsare+4and

+2.Carbonalsoexhibitsnegativeoxidationstates.Since,thesumofthefirstfourionization

enthalpiesisveryhigh,compoundsin+4oxidationstatearegenerallycovalentinnature.In

heaviermembers,thetendencytoshow+2oxidationstateincreasesinthesequenceGe<Sn

<Pb.

Itisduetotheinabilityofns2electronsofvalenceshelltoparticipateinbonding.The

relativestabilitiesofthesetwooxidationstatesvarydownthegroup.Carbonandsilicon

mostlyshow+4oxidationstate.Germaniumformsstablecompoundsin+4stateandonly

fewcompoundsin+2state.

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(v) NatureofHalides:TheseelementscanformhalidesofformulaMX2andMX4(X=F,Cl,

Br,I).Exceptcarbon,allothermembersreactdirectlywithhalogenundersuitable

conditionstomakehalides.Mostofthecompounds(MX4)arecovalentinnature.Thecentral

metalatominthesehalidesundergoessp3hybridizationandthemoleculeistetrahedralin

shape.ExceptionsareSnF4andPbF1,whichareionicinnature.PbI4doesnotexistbecause

Pb—Ibondinitiallyformedduringthereactiondoesnotreleaseenoughenergytounpair

6s2electronsandexciteoneofthemtohigherorbitaltohavefourunpairedelectronsaround

leadatom.HeaviermembersGetoPbareabletomakehalidesofformulaMX2.

42. Accountforthefollowingobservations:

(i) AlCl3isaLewisacid.

(ii) ThoughfluorineismoreelectronegativethanchlorineyetBF3isaweakerLewis

acidthanBCl3

(iii) PbO2isastrongeroxidizingagentthanSnO2

(iv) The+1-oxidationstateofthalliumismorestablethanits+3state.

Ans.(i)InAlCl3,theoctetofAlisincompleteasithas6electronsandacceptspairof

electrons.ElectronacceptersareLewisacids.

(ii) InBF3,boronhasavacant2porbitalandfluorinehasoneofthe2porbitalcompletely

filledandunutilized.Bothhavesameenergyandcanoverlapeffectivelytogivepπ—pπ

bond.Thistypeofbondingcalledbackbonding.BackbondingisstronginFanddecreases

downtheGp17asthesizeofhalogensincreases.ThrustofelectronsofBoronisgoingtobe

fulfilledbybackbonding.

(iii) Itismorestableandactsasoxidizingagent.Onmovingdown,agroup,inertpaireffect

increasesanditisverystrongincaseofPbtherefore,PbO2isstrongeroxidizingagent.

(iv) Itisduetoinertpaireffect.

43. Whenaqueoussolutionofboraxisacidifiedwithhydrochloricacid,awhite

crystallinesolidisformedwhichissoapytotouch.Isthissolidacidicorbasicin

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nature?Explain.

Ans.AnaqueoussolutionofboraxisacidifiedwithHCltogiveboricacid.

Na2B4O7+2HCl+5H20 2NaCl+4H3B03

Boricacidisaweakmonobasicacid.ItisnotaprotonicacidbutactsasaLewisacidby

acceptingelectronsfromahydroxylion:

B(OH)3+2HOH [B(OH)4]-+H3O

+

44. Threepairsofcompoundsaregivenbelow.Identifythatcompoundineachofthe

pairswhichhasgroup13elementinmorestableoxidationstate.Givereasonforyour

choice.Statethenatureofbondingalso.

(i) TlCl3,TlCl

(ii) AlCl3,AlCl

(iii) InCl3,InCl

Ans.(i)TlClismorestablethanTlCl3becausemovingdownthegrouploweroxidationstate

ismorestableduetoinertpaireffect.

(ii) AlCl3ismorestablebecauseitdoesnotshowinertpaireffect.Itisacovalentcompound

andactsasaLewisacid.

(iii) InClismorestableduetoinertpaireffectandloweroxidationstate+1ismorestable.In

showsboththeoxidationstates+3and+1.

45. BCl3existsasmonomerwhereasAlCl3isdimerizedthroughhalogenbridging.Give

reason.ExplainthestructureofthedimerofAlCl3also.

Ans.BCl3doesnotdimerizebecauseofitssmallsize.BCl3donotexistasadimer.Boron

cannotaccommodatefourlargesizedchlorideions.AlCl3existasadimerinwhichAlusesits

vacant3p-orbitalbycoordinatewithCltocompletetheiroctetbyformingdimer.

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AlCl3achievesstabilitybyformingadimer.

46. BoronfluorideexistsasBF3butboronhydridedoesn’texistasBH3.Givereason.In

whichform,doesitexist?Explainitsstructure.

Ans.InBF3,lonepairoffluorinegivesbacksupportofelectronstoboronatompπ—pπback

bonding).Thisdelocalizationofelectronsreducesthedeficiencyofelectronsandthus

reducesLewisacidiccharacterandincreasesstabilityofBF3.InBH3,thereisnolonepairof

electronsonHatom,therefore,BH3dimerizestogiveB2H6.4terminalHatomsand2boron

atomslieinoneplaneandaboveandbelowtheplanetherearetwobridgingHatoms.

47. (i)Whataresilicones?Statetheusesofsilicones.

(ii) Whatareboranes?Givechemicalequationforthepreparationofdiborane.

Ans.(i)Siliconesareagroupoforganosiliconpolymers,whichhave(R2SiO)asarepeating

unit.Thestartingmaterialsforthemanufactureofsiliconesarealkylorarylsubstituted

siliconchlorides, whereRisalkylorarylgroup.Whenmethylchloridereacts

withsiliconinthepresenceofcopperasacatalystatatemperature573Kvarioustypesof

methylsubstitutedchloramineofformulaMeSiCl3,Me2SiCl2,Me3SiClwithsmallamountof

Me4Siareformed.Hydrolysisofdimethyldichlorosilane,(CH3)2SiCl2followedby

condensationpolymerizationyieldsstraightchainpolymers.

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Uses:Theyareusedassealant,greases,electricalinsulatorsandforwaterproofingof

fabrics.Beingbiocompatibletheyarealsousedinsurgicalandcosmeticplants.

(ii) BoranesarethebinarycompoundsofboronandhydrogenlikeaAlkanes.Theseare

covalenthydridesofformulaB2H6calleddiborane.

Preparationofdiborane:

4BF3+3LiAlH4 2B2H6+3LiF+3A1F3

48. Acompound(A)ofboronreactswithNMe3togiveanadduct(B)whichon

hydrolysisgivesacompound(C)andhydrogengas.Compound(C)isanacid.

IdentifythecompoundsA,BandC.Givethereactionsinvolved.

Ans.Compound[A]ofboronreactswithNMe3andgivesanadduct[B]thuscompound[A]is

Lewisacid.Since]B]onhydrolysisgivesanacid[C]andH2gas,therefore[A]isB2H6,[B]is

anadduct2BH3NMe3and[C]isboricacid.Reactionsareasfollows:

B2H6(A)+2NMe3 2BH3NMe3(B)

BH3NMe3+3H20 H3BO3(C)+NMe3+6H2

49. Anonmetallicelementofgroup13,usedinmakingbulletproofvestsisextremely

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hardsolidofblackcolour.Itcanexistinmanyallotropicformsandhasunusuallyhigh

meltingpoint.ItstrifluorideactsasLewisacidtowardsammonia.Theelementexhibits

maximumcovalenceoffour.Identifytheelementandwritethereactionofits

trifluoridewithammonia.ExplainwhydoesthetrifluorideactasaLewisacid.

Ans.Ingroup13,boronisonlynon-metallicandextremelyhardandalsousedformaking

bulletproofvests.Boronexistsinmanyallotropicforms.Itusuallyshowshighmeltingpoint

anddoesnothave‘dorbital.Itcanshowmaximumcovalenceof4byusing2sand2p

orbitals.Intrivalenthalidesofboron,octetofboronisnotcompletedhenceactsasLewis

acid.ItreactswithLewisbaseandformsadduct.

50. Atetravalentelementformsmonoxideanddioxidewithoxygen.Whenairispassed

overheatedelement(1273K),producergasisobtained.Monoxideoftheelementisa

powerfulreducingagentandreducesferricoxidetoiron.Identifytheelementand

writeformulasofitsmonoxideanddioxide.Writechemicalequationsforthe

formationofproducergasandreductionofferricoxidewiththemonoxide.

Ans.ProducergasisamixtureofCO(g)andN2(g).Carbonistetravalentandgivesmonoxide

(CO)anddioxide(CO2)withoxygen.

2C(s)+02(g)+4N2(g) 2CO(g)+

Fe2O3(S)+3CO(g 2Fe(s)+3C02(g)

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