2021 H1 Chemical Bonding Tutorial Tutors

DISCUSSIONQUESTIONS Metalliclatticestructures1. [2015/P1/Q4]Themeltingpointofcalcium,839°C,ismuchhigherthanthemeltingpointofsodium,98°C.Whichstatementismostrelevantinexplainingthisdifference?A Thecalciumatomislargerthanthesodiumatom.B Thecalciumatomisheavierthanthesodiumatom.C Thecalciumion,Ca 2+ ,hasahigherchargethanthesodiumion,Na + .D Thecalciumion,Ca 2+ ,containsmoreelectronsthanthesodiumion,Na + .ThebestanswerisC.OptionDshouldhavetheword“valence”electronsinordertobecorrect.Examiners’ comments:Thisisoneofthequestionswhichcandidatesfoundchallenging.Manycandidatesfeltthenumberofelectronsintheionswasthedecidingfactorratherthanthechargeontheions. AfL:GetstudentstosuggestwhatiswrongwiththeoptionD. HingeQuestion:Whataffectsthestrengthofmetallicbonding? Ans:1)Numberofvalenceelectronscontributedtotheseaofdelocalisedelectrons2)Cationicradius Dot crossdiagram, VSEPRtheory, Shapesof molecules2. N2016/3/5(modified)(a) OutlinetheprinciplesoftheValenceShell ElectronPairRepulsion(VSEPR)Theory.Central Atom’sValenceShell ElectronPairs(bothbondpairsandlonepairs)arrangethemselvesinspacetominimiserepulsionandmaximiseseparation.LonePair−LonePairrepulsion>LonePair−BondPairrepulsion>BondPair−BondPairrepulsion (b) Although they have similar formulae, the following three compounds have differently shapedmolecules. BF3 NF3 ClF3 (i) Predictandexplaintheirshapes,usingdiagramswhereappropriate.BF3 NF3 ClF3 BondPairs 3 3 3LonePairs 0 1 2StericNumber 3 4 5Shape Trigonal Planar Trigonal Pyramidal T-shapedDiagram 1

(ii) Label thepartial charges,δ+andδ-,presentononeB-Fbond.Statethecauseofthese partial charges.Hence,state,withareason,whetherornotBF3 isapolarmolecule. δ+ δ- B FF is more electronegative than B and hence pull the shared electrons closer towardsitself, giving rise tothedipole. BF3 hasnonet dipolemoment. Hence, thereisnodipolemoment. AfL:Hingequestion(tutorsmaydothisquestionbeforegoingthroughQ3): Whichdot-crossdiagramisinaccuratelyrepresented? Reasonfor myanswer isbecause(youmaychoosemorethan1answer): 1.Diagram1shouldhavealternatedotcrossonO.2.Diagram2shouldhavenodativebonds.3.Diagram3havemissingvalenceelectronsonNa + .4.Diagram4havemissingvalenceelectrons.5.Diagram5shouldhaveOgainingtheadditional electron. 2

Ans: Incorrect diagram:1, 2,4, 5 Correct reason: 1, 2, 4, 5 3. N2010/3/4c(a) Drawdot–and–crossdiagramstoshowthebondinginthemoleculesofNO2 andO3 .Each molecule contains a dative covalent bond. In the NO2 molecule, the central atom isnitrogen.In each case you should distinguish carefully between electrons originating fromthe centralatomandthosefromthetwoothermostatoms.Includeall lonepairsinyourdiagram. (b) Suggest a value for the bond angle in each of the above two molecules, giving reasons foryourchoice.● In O3 , there are 2 bond pairs and 1 lone pair around the central Oatom. Sincelone–pair – bond pair repulsion isgreater thanbondpair –bondpair repulsion,thebondangleisabout 112°(ie.109°<bondangle<120°)● InNO2 ,thereisalone–electronand2bondpairsaroundthecentral Natom.Thusthelone–electronwill repel thebondpair toasmaller extent thanalonepair.ThusthebondangleinNO2 is134 o .(ie.120°<bondangle<180°) ● Note: Givenvaluesareactual valuesfor NO2 andO3 . NO2 (shouldhavegreater bondanglethanO3 . Allowanyvaluebetweenthat givefor O3 and170 o ); O3 (accept anyvaluebetween110–120 o ). AfL:HingequestionAccordingtoVSEPR,Lp-lprepulsion>lp-bprepulsion>bp-bprepulsionHowwilltheloneelectron-bondpairrepulsioncomparewiththerest? Ans:Lp-lprepulsion>lp-bprepulsion>bp-bprepulsion>lonee–bprepulsion ● Examiners’ comments:● Estimatingthebondangleinthesetwomoleculesismoredifficult.AsNO2hasanunpairedelectronitisnotsoreadilycoveredbytheValenceShell ElectronPairRepulsiontheory,andthereforethelogical approachistobeginwithozone.Thecentral oxygenissurroundedbytwobondsandaloneelectronpair:thebasicangleistherefore120 o ,butallowingforthefactthatalonepairisclosertotheoxygenthanabondpair,orissaidto‘occupyagreatervolume’,thisanglewill bediminishedtosomethinglessthan120 o ;avaluebetween110 o and120 o wasallowed.MovingtotheNO2 molecule,asingleunpaired 3

electronwill belesseffectiveindiminishingthebasicangle,andsothebondangleforNO2 shouldbegreaterthanthatforO3 .ExaminersallowedanyvaluebetweenthatgivenforO3 and170 o (c) The compound FO2 does not exist but ClO2 does. By considering the possible types ofbondinginthetwocompounds, suggest reasonsfor thisdifference.(Assumethatthehalogenatomoccupiesacentral positionineachofthesemolecules.) [6]ApossiblestructureforXO2 (whereX=ForCl)isO=X=O or O=X→O,inwhichtherearemorethan8electronsaroundthecentral Xatominbothstructures.AsClis in period 3andhas energetically accessible d-orbitalsto expand its octet structure,ClO2 exists. The above structures do not exist for FO2 since fluorine lacks energeticallyaccessibled-orbitalstoexpanditsoctetstructure. Another possiblestructurefor FO2 is O←F→O.However,whilethereareonly7electronsaround the central F atom, fluorine is too electronegative to formdative bonds. Thus,theabovestructuredoesnotexisttoo. Examiners’ comments:CandidateswereaskedtosuggestareasonwhythecompoundClO2 exists,butFO2 doesnot.VerymanycandidatespointedoutthattheClO2 moleculewouldinvolvechlorineexpandingitsoctetofelectrons,whichitisabletodothroughthe3dorbital whichisaccessibleforelectronoccupation.Veryfewcandidateswentfurtherwiththediscussion.Iffluorinecannotexpanditsoctet,thentheotheralternativewouldbeforthefluorinetoprovidetwodativebonds:itdoesnotdothisbecauseitistooelectronegativefordativebonding. AfL:Hingequestion(N2017/P3/1c):*ChallengingQnstostretchstudents’ thinking ThecentralatominClF3 issurroundedbyfivepairsofelectronsarrangedinatrigonalbipyramidalshape.AtrigonalbipyramidalarrangementisshowninFig1.1. ThreedifferentmoleculararrangementsofClF3 arepossible. 4

(i) Drawcleardiagramsofthesethreemoleculararrangements,eachshowingthefivepairsofelectrons.Statewhicharrangement,ifany, wouldresultinamoleculewithnodipolemoment. [2] X: Y: Z(nodipolemoment): (ii) ApplytheprinciplesofVSEPRtheorytodiscusstherelativestabilitiesofthesethreepossiblearrangements. [2]VSEPRTheorystatesthatLone-pair-lonepairrepulsion>lonepair-bondpairrepulsion>bondpair-bondpairrepulsion.Thestrongertherepulsion,thefurtherapart(orthelargertheangle)is.ForstructureX,lonepairsare120 o apart.Lonepair-bondpairangleis90°.Fitsthetheory.ForstructureY, lonepairsare90 o apart.Lonepair-bondpairangleis120°and90°.Doesnotfitthetheory.ForstructureZ,lonepair-bondpairangleis90 o .Bondpairsare120 o apart.Doesnotfitthetheory.XismorestablethanZthanY. (BasedontheorythatT-shapedistheshapefor3bondpairsand2lonepairs) Sigmaandpi bonds4. Whichoneofthecompoundsisionicandcontainsbothsigmaandpi bonds?A Fe(OH)3 B HClO C H2 S D NaCN AandDisionic.However,onlyC Ξ N - containsbothsigmaandpi bonds. 5

DativeBonding5. N2016/31/1(modified)(a) Carbonmonoxidereactswithboronhydride. BH3 , at highpressuretogivethecompoundH3 BCO,in which carbon is bonded to both boron and oxygen. Drawa dot-and-crossdiagramtoshowthebondinginH3 BCO,clearlyindicatinganydativebondsitcontains. (b) Explainwhythesetwomoleculesformaproductwhentheyreactinthemolarratio1:1. BinBH3 containsoneemptyorbitalandCinCOcontains1lonepair of electrons.Hence,theCatomdonatesitslonepairelectronstoBtoformdative/coordinatebonding. (c) By considering the numbers of bonding and non–bonding electron pairs, drawdiagrams toshow the likely shapes of carbon monoxide, boron hydride and the product. In yourdiagrams,clearlystatethevaluesofthebondangles.CO BH3 Product 6

(d) Suggestthestructural formulaeofpossiblesimilarproductsfrom(i) berylliumdifluorideandcarbonmonoxide (ii) borontrifluorideandammonia AfL:HingeQuestion Ans:specieswithoutdativebond(1);correctreason:4 Physical Propertiesrelatingtostructure: MeltingandBoilingPoints6.Thefollowingtableliststheboilingpointsofsomeorganiccompounds. 7

Compound Formula Mr Boilingpoint/°CA CH3 CH2 CH2 CH2 OH 74.0 118B CH3 CH2 CH2 CH2 CH3 72.0 36C 191 130 With reference to the type and extent of relevant intermolecular forces, explain the differenc