结构化学课件problems2

第7-9章习题课Keypoints/conceptsLatticeofcrystalstructure:translationsymmetryalatticepoint=astructuremotif--unitcellCrystalsystems(7),BravaisLattice(14)Symmetryoperations(point&translation)Crystallographicpointgroups(32),spacegroups(230),millerindexofcrystalplaneetc.X-raydiffraction,Laueequation,Bragg’sLaw,reciprocallattice,Ewaldsphere,structuralfactor,systemabsence,generalprocessofx-raycrystalstructuredetermination.Close-packingofspheres(ccp/A1,hcp/A3,bcp/A2)inmetalsandioniccompounds,coordinationofcations.Crystalstructuresofsometypicalioniccompounds.Example:p224,7.2--conceptoflatticeAstructuremotif=alatticepointEachlatticepointhasidenticalsurroundings.AlatticefulfillsTranslationsymmetry.Keypointistofindthestructuremotif(basis)thatfulfillstranslationsymmetry!p.227,7.26SnF4Sn:(0,0,0),(1/2,1/2,1/2)F:(0,1/2,0),(1/2,0,0)(0,0,0.237),(0,0,-0.237)1)I-body-centredtetragonal2latticepointwithinaunitcell.Theblackdots(2Sn)andredballs(4F)aredefinedbythecoordinatesgiven!Other4Fatomscanbeobtainedbytranslationoperation(2LP).EachSnislocatedinadistortedoctahedralhole.R(Sn-F)1=0.237x793=187.9pm;R(Sn-F)2=0.5x404=202pmSimplecubiccrystal–example:CsClp.2869.13

Eachlatticepointcontains2atoms,Cl(0,0,0),Cs(1/2,1/2,1/2);Onlyonelatticepointwithinaunitcell.Thestructuralfactoris

Therefore,allpossiblediffractionsareobservablewithoutsystemabsence!However,Ifh+k+l=2n,Strongdiffraction

Ifh+k+l=2n+1,Weakdiffraction

Face-centeredcubiccrystal–generalcasep227,7.21Latticepoints(LPs):(0,0,0),(1/2,1/2,0),(0,1/2,1/2),(1/2,0,1/2)Supposeeachlatticepointcontainsnatoms,{(xj,yj,zj)}(j=1,…,n)EachunitcellcontainsN=4natoms.E.g.,AnatomA(xi,yi,zi)inoneLPhasotherthreeequivalentAatomswithinthesameunitcell!Thenthestructurefactoris

Whenh,k,lareneitherallevennorallodd,systemabsence!Whenh,k,lareallevenorallodd,diffractionobservable!Sumupoverallatomswithinaunitcell!NowsumupoverallatomswithinaLP!Face-centeredcubiccrystal–

Specialcase:NaClp227,7.23Unitcellcontains4latticepoints,or4NaClEachlatticepointcorrespondstoaNaCl.PutClat(0,0,0),thenaneighboringNaat(1/2,0,0).Whenh,k,lareallevenorallodd,diffractionsobservable!Case1:ifh=2n(notewealsohavel=2nandk=2n)

Strongdiffraction!Case2:ifh=2n+1(notewealsohavel=2n+1,k=2n+1)

Weakdiffraction!Latticepoints:(0,0,0)+,(1/2,1/2,0)+,(0,1/2,1/2)+,(1/2,0,1/2)+EachlatticepointcontainstwoCatoms(structuremotif=2C)C1--(0,0,0),C2--(1/4,1/4,1/4)Specialcase:DiamondOh7-Fd3m

Sideviewtopview41dglideplane1/8,3/8,5/8,8/7Face-centredcubiccrystal--Face-centeredcubiccrystal–

Specialcase:DiamondStep1:

Systemabsenceofface-centeredcrystal–h,k,lareeitherallevennorallodd!Observable:(111),(200),(220),(311),(222),(400),(331),(420),(422)Step2:

Twocarbonatomswithinalatticepoint:(0,0,0),(1/4,1/4,1/4).Whenh,k,lareallevenorallodd,Ifh+k+l=4n+2,(200),(222),(420),(442)…systemabsence!

Step3:observable:(111),(220),(311),(400),(331),(422)Ifh+k+l=4n,Fhkl=8fC,(220),(400)…strongestdiffraction!

Face-centeredcubiccrystal–

Specialcase:ZnS(sphalerite)Unitcellcontains4latticepoints,or4ZnSEachlatticepointcorrespondstoaZnS.S--(0,0,0),Zn--(1/4,1/4,1/4).Whenh,k,lareallevenorallodd,diffractionsobservable,

(111),(200),(220),(311),(222),(400),(331),(420),(422),…Case1:ifh+k+l=4n,e.g.,(220),(400),(440)…Strongdiffraction!Case2:ifh+k+l=4n+2,e.g.,(200),(222),(420),(442)…..

Weakdiffraction!Body-centercrystal–p227,7.22

Ageneralcase:eachlatticepointcontainsnatomsThetotalnumberofatomswithinaunitcellis2n;Forjthatominastructuremotif(alatticepoint):(xj,yj,zj)

Itsbody-centerequivalentpointis:(0.5+xj,0.5+yj,0.5+zj)

Whileh+k+l=2n+1,

SystemabsenceBody-centercrystal–example(p.227,7.24)Eachlatticepoint(LP)contains2atoms,A(0,0,0),B(x,y,z);ThusinanotherLP,A(1/2,1/2,1/2),B(x+1/2,y+1/2,z+1/2)Thestructuralfactoris

Whileh+k+l=2n+1,

SystemabsenceExample:钨为立方晶系，其粉末衍射线指标为:110,200,211,220,310,222,321,400,….属何种点阵类型？若X射线

=154.4pm，220衍射角43.6°，试计算晶胞参数。Answer:h+k+l=2n+1

systemabsence!CubicB-centred.Bragglaw:

2dhkl

sin

=

dhkl=/2sin

since

220=43.6°cellparameter:

p.226,7.18;Theanswerinp.317iswrong!P227,7.27Bragg’slaw,d-spacingNumberofS8inaunitcelloforthorhombiccrystal:Fororthohombiccrystal,dhkl=1/[(h/a)2+(k/b)2+(l/c)2]1/2p.200AccordingtoBragg’sLaw,wehavesinq=

/2dhkl=[(h/a)2+(k/b)2+(l/c)2]1/2×

/2=[(2/1048)2+(2/1292)2+(4/2455)2]1/2×154.18/2=[(1/1048)2+(1/1292)2+(2/2455)2]1/2×154.18=0.2273

q=13.1°Note:thevalueof

isnotgiveninquestion7.27!p.257,8.8indexingofdiffractiondata!Tametal’sx-raydiffractiondata(sin2q)isknown.1)

Indexing:sin2q1:sin2q2:sin2q3:sin2q4:sin2q5:sin2q6:sin2q7:sin2q8:sin2q9….=1:2:3:4:5:6:7:8:9:…=2:4:6:8:10:12:14:16:18:…

h+k+l=2n+1systemabsence,body-centeredcubiclatticeObserved:(110)(200)(211)(220)(310)(222)(321)(400)(330)…Cellparameter:dhkl=a/(h2+k2+l2)1/2a=dhkl(h2+k2+l2)1/2AccordingtoBragg’sLaw,wehavedhkl=

/2sinq

a=

(h2+k2+l2)1/2/2sinq

Choosingthe(330)diffraction,wehavea=

(h2+k2+l2)1/2/2sinq=154.1x(18)1/2/(2x0.98907)=330.5pmp.259,8.21Statistically,theprobabilityforaCuatomtoappearatalatticepointisequaltothepercentage(x)ofCudoping.ThustheAuCualloybelongstoface-centeredcubiclatticesystem.Eachunitcellcontains4latticepointsandeachofthemisametalatom(Au1-xCux).Theorderedphasebelongstosimpletetragonallattice;eachunitcell/latticepointcontainsAuCu.Au(0,0,0),Cu(1/2,1/2,1/2);Cellparameterfortheorderedphase:a=272.23pm,c=385pmThefirstobserveddiffractionis(001)d001=1/(h/a)2+(k/a)2+(l/c)2)1/2=cAccordingtoBragg’sLaw,wehavesinq

=

/2d001=

/2c=0.2

q

=11.5

p.259,8.21Fortherandomlydopingphase,itsfirstobservablediffractionis(111).d001=a/(h2+k2+l2)1/2=a/31/2sinq

=

/2d111=31/2

/2a=0.3464q

=22.3

Example:银为立方晶系，用CuK

射线(

=154.18pm)作粉末衍射，在hkl类型衍射中，hkl奇偶混合的系统消光。衍射线经指标化后，选取333衍射线，

=78.64°，试计算晶胞参数。已知Ag的密度为10.507g·cm-3，相对原子质量为107.87，问晶胞中有几个Ag原子。试写出Ag原子的分数坐标Answer:whenh,k,lareneitheralloddnoralleven,

systemabsence!CubicF-centred.Bragglaw:

2dhkl

sin

=

dhkl=/2sin

cellparameter:atomsinaunitcell:Thuseachatomcorrespondstoonelatticepoint,atomiccoordinates:(0,0,0)(1/2,1/2,0)(1/2,0,1/2)(0,1/2,1/2)Example:金属Mg是由Mg原子按A3型堆积而成，已知Mg的原子半径是160pm，求晶胞参数。Answer:Hexagonalclose-packingmode

a=b=2R=320pm,Theheightofatetrahedron(edgelength2R)is

Example:p.286,9.14CellparametersofNaCl-typeKBr,LiBr,KF,andLiFare658,550,534,402pm,pleasederivetheionicradiiofK,Li,F,Br.Answer:1.BranionisthelargestandLicationisthesmallest!ForLiBr,theanionsadoptccpstructurewithLicationsoccupyingtheoctahedralholes.Thus,ForKBr,theKcationislarge,thus,theanionsmaynotcloselycontactwitheachother.Instead,thecationsandanionscloselycontactwitheachother.Thus,Similarly,forKF,ForLiF,theultimatecaseisLi-Fcloselycontactwitheachother,

Example:p.288,9.32Thecellparametersofmonocliniccrystalofbiphenyl:a=824pm,b=573pm,c=951pm,

=94.5

;

=1.16gcm-3.Answer:1)Numberofmoleculeswithinacell,2)hisodd,h0lsystemabsence.glideplane(a/2)parellelto(010).kisodd,0k0systemabsence,21screwaxisisparalleltob-axis.LikelytobeP21/c3)ThemoleculeisD2dsymmetric.Onemoleculeiscenteringaround(0,0,0),anotheriscenteringaround(1/2,1/2,0)Example:p.288,9.30SiP2O7.Answer:1)EachSi(4+)issurroundedby6Oxygenanions.

s(Si-O1)=+4/6=+2/3

s(P-O1)=2-(4/6)=+4/3

s(P-O2)=5-(4/3)x3=+1

Z(O2)=2x(+1)=2

P2O74-inSiP2O7solidisstable!s(P-O1)>s(P-O2)

R(P-O1)<R(P-O2)3)FreeP2O74-anionisunstable!s(P-O2)=+5/4Z(O2)=2x(+5/4)=2.5>2!面心立方点阵的倒易点阵为体心立方点阵，反之亦然！若面心立方点阵单胞的边长为a,其三个正交的单位矢量为注意到面心立方点阵的素原胞为棱方单胞，其基矢为:其体积为：则其倒易点阵的基矢为：此为体心立方点阵素单胞的三个基矢，该体心立方点阵单胞的边长为2/a.

故单胞参数为a的面心立方点阵,其倒易点阵为体心立方点阵(单胞参数为2/a)！第5-6章习题课Keypoints/conceptsHybridorbitaltheoryandVSEPRHMO,HMOtreatmentofp-conjugatedsystemsandgraphicalmethodtopredefinecoefficientsofHMOofp-conjugatedsystems.Symmetryrulesformolecularreactions.ChemicalbondsinElectron-deficientmolecules(boranesandcarboranesetc),styxmethod.Chemicalbondingincoordinationcomplexes,crystal-fieldtheory,ligand-fieldtheory,18eruleanditsapplicationinmetalclustercompounds.VSEPRandHybridorbitaltheoryp.149,5.8,5.12SCl3+:S+(5Ve),eachClprovides1etoformaS-Clbond;thus

SCl3+hasalonepaironS.PyramidalstructurelikePCl3.Ssp3hybridization.S(3+)ICl4-:I-(8Ve),eachClprovides1etoformaI-Clbond;thusICl4-hastwolone-pairsonIwithasquareplanarstructure.Iatomadoptssp3d2hybridization.I(3+).ICl3:I(7Ve),eachClprovides1etoformaI-Clbond;thusICl3hastwolone-pairsonIwithaplanartrigonalstructure.Iatomadoptssp3dhybridization.I(3+),non-polarSO3:S(6ve),sp2hybridization,trigonalplanar;threeS-Osigmabond,andaP46

SO32-:S(6ve),sp3hybridization,PyramidalstructurelikePCl3;threeS-Osigmabond.CO32-:C(4ve),sp2hybridization,trigonalplanar;aP46

MOtheoryp.149,5.12SupposethebondingMOofABis

From2e,8e,18erulestoelectroncountingrulesformetalclustercompoundsandstyxmethodforelectron-deficientmolecules.2e,8e,18erulesgovernthestabilityofsuchmolecules/groupsAXnandaccountsfortheirgeometriesincombinationwithVSEPRorhybridorbitaltheory.Yet,thereareexceptionsoftheserules,e.g.,PCl5,andPtCl42-,whosestructurescanbewell-understoodwithuseoftheVSEPRandhybridorbitaltheory.Theelectroncountingrulefortransition-metalclustercompoundsisactuallyanaturalextensionofthe18erule,whichstatesthatwhenthevalenceelectronsprovidedbytheligandsofaTMcenterarenotenoughtofulfillthe18erule,valenceelectronsfromneighboringTMcenter(s)shouldbeinvolvedtomaintaintheTMcentertohave18VE.Forelectron-deficientmolecules/ions,the(styx)methodisalsoanaturalextensionofthe8e/2erule.Incasethatthetotalvalenceelectronsofamoleculeareinadequatetomaintaintheessentialnumberof2c-2ebonds,formationofseveral3c-2e(ornc-ne)bondscanmakeallcomponentatomsfulfillingthe8e/2erule.BBHBBBBHHBBstyxStyxmethodx=m-st=n-sy=(2s-m)/2

psetsofstyxpisomersGeneralruleforstabilityofmolecules/ionsThemolecules/ionsshouldhavealargeHOMO-LUMOgap.The2e/8e/18erulesworkwellbecausemoleculesadoptingsuchelectronconfigurationsalwayshavealargeHOMO-LUMOgap.Followingthisgeneralrule,theWade’s(n+1)ruleaccountsforthestabilityofcloso-boranes/carboranes,whoseskeletalbondingelectronsareactuallydelocalizedoverthewholemolecule.Similarly,theHuckel(4n+2)ruleaccountsforthestabilityofcyclicp-conjugatedsystemsthathavedelocalizedelectrons.Examplesp.185,6.17,6.24,Ni(CO)4:1)Ni(10ve)+4(CO)(8ve)=18ve;sp3hybridization.tetrahedralstructure.2)Ni(3d10),3dorbitalssplitinto(3dz2,3dx2-y2)4(3dxy,3dyz,3dxz)63)nod-dexcitationcanbeobservedinsuchacase.Fe5C(CO)15:g=5x8+4+15x2=74;b=(18n-g)/2=8Squarepyramidwith8Fe-Febonds.Ru6C(CO)16:g=6x8+4+16x2=84;b=(18n-g)/2=12Octahedronwith12Ru-Rubonds![Rh6C(CO)15]2-:g=6x9+4+15x2+2=90;b=(18n-g)/2=9trigonalprismwith9Rh-Rhbonds![Ni8C(CO)16]4-:g=8x10+4+16x2+4=120,b=(18n-g)/2=12cube-shapedwith12Ni-Nibonds.[Ni8C(CO)16]2-:g=8x10+4+16x2+4=118,b=(18n-g)/2=13squareantiprism.(actuallysynthesized!)Examplesp.185,6.26,Fe6(CO)18:g=6x8+18x2=84;b=(18n-g)/2=12Octahedronwith12Fe-Febonds.[Fe4RhC(CO)14]-:g=4x8+9+4+14x2+1=74;b=(18n-g)/2=8squarepyramidwith8M-Mbonds![Co6N(CO)15]-:g=6x9+5+15x2+1=90;b=(18n-g)/2=9trigonalprismwith9Rh-Rhbonds!Ni8(CO)8(PPh3)12:g=8x10+8x2+12x2=120,b=(18n-g)/2=12cube-shapedwith12Ni-Nibonds.(wrongligandintextbook!)Ni8(PPh)6(CO)8:g=8x10+6x4+8x2=120,b=(18n-g)/2=12cube-shapedwith12Ni-Nibonds.ThePPhligandism4.Examplesp.185,6.25,Fe2(m2-CO)3(CO)6:1)g=2x8+9x2=34;b=(18n-g)/2=1ThereexistsoneFe-Febond.TheCO(m2)formstwoFe-CbondswiththeFeatoms,adoptingC=Odoublebond.Onthecontrary,aterminalCOformsadonativeFe-COs-bondand,meanwhile,getbackdonationfromFe(3d

)arisingfromthe3d(Fe)2

*(CO)bonding,whichresultsinapartiallyweakenedC

Otriplebond.ThustheC-ObondinaterminalCOisstrongthanthatinabridgingCO(m2).Whyaretherethreem2-COligands?Tofulfillthe18erule,allvalenceAOsofaFecentershouldbeinvolved,demandingeachFecenterbeing6-foldcoordinated(excludingFe-Febond).ThusforFe2(CO)9,threem2-COligandsarerequiredtokeepbothFecentersbeing6-foldcoordinated.ExampleRe2(CO)4(h5-C5H5)2

Re-Rebondorder=?Whyaretheretwom2-COligands?g=2x7+4x2+2x5=32;b=(18n-g)/2=2Re-Rebondorderis2.2)ForaTMcomplexfulfillingthe18erule,allthevalenceAOsofaTMcentershouldbeinvolvedinthechemicalbondings.AssuchthecoordinationnumberofaRecentershouldbe6(excludingtheRe-Rebond).AsaC5H5ligandmakesuseof3valenceAOsofRe,thereshouldbe3COligandscoordinatedtoaRecenter.Thus,twom2-COligandsarerequiredtokeepbothRecenterbeing6-foldcoordinated!Linearcarbonchainwith2ncarbonatoms:HOMOandLUMO2nep,noccupiedp-MOs;soHOMO(m=n),LUMO(m=n+1).Linearcarbonchainwith2ncarbonatoms:HOMOIf2n=4l,If2n=4l+2,Similarly,HOMOisasymmetric!Similarly,HOMOissymmetric!Linearcarbonchainwith2ncarbonatoms:LUMOIf2n=4l,If2n=4l+2,Similarly,LUMOissymmetric!Similarly,LUMOisasymmetric!Linearcarbonchainwith2ncarbonatoms:HOMOandLUMOIf2n=4l,If2n=4l+2,HOMO:symmetric!asymmetric!LUMO:asymmetric!LUMO:HOMO:symmetric!AsymmetricwithrespecttoCsoperation!Symmetric!Simplifiedmotifofsymm.&asymm.p-MOs:pp-AOofterminalcarbonatomLinearcarbonchainwith(2n+1)pp-orbitals.symmetricNep=2n+1ndoublyoccupiedMOs(m=1,..,n),andasinglyoccupiedMO(SOMO,m=n+1).

SOMO=p/2,ESOMO=a,

non-bonding!If2n+1=4l+1(e.g.,5),

If2n+1=4l+3(e.g.,7),

asymmetric[1,n]sigmatropicshift:stereochemicalrules[1,5]-s-shift[1,7]-s-shiftSupposethetransitionstateofasigmatropicshiftreactionisacombinationofHatomanda[n]polyenewithodd-numberpporbitals.NotetheSOMOofsuch[n]polyeneissymmetricwhenn=4k+1,andasymmetricwhenn=4k+3.QM-predictedTransitionstatesof[1,n]-sigmatropicshift[1,5]-sigmatropicshift[1,7]-sigmatropicshiftSuprafacialmodeantarafacialmoden=4k+1n=4k+3SuprafacialmodeantarafacialmodeCS-symmetryrequired!C2-symmetryrequired!TheH-transferdemandsbothendsofSOMOofsuch[n]polyene(n=odd)canoverlapwiththeH1sorbital.i.e.,

根据Hückel近似，写出下列分子的p电子分子轨道久期行列式：

Answer:Thesequlardeterminantofmoleculeais

Thesequlardeterminantofmoleculebis

Thesequlardeterminantofmoleculecis

C5H5anditsanion:symmetricMOs,boundaryconditionAsymmetricMOs:

C5H5anditsanion:asymmetricMOs,boundaryconditionThreesymmetricandtwoasymmetricMOs!s1isthelowestoccupiedMO(LOMO)withEs1=a+2bEas1=Es2,Eas2=Es3(doublydegenerateMOs)C5H5anditsanion:2)ForC5H5anion,6eintotal.Thes1,s2andas1Mosareoccupied.Etotal=6a+4b[1+2cos(2p/5)]Forthelocalizedsystem:Eloc=6a+4b

Edeloc=Etotal–Eloc=8bcos(2p/5)

2.47b1)ForC5H5,5eintotal.Etotal=2(a+2b)+3[a+2bcos(2p/5)]=5a+4b+6bcos(2p/5)Forthelocalizedsystem:Eloc=2x2(a+b)+a=5a+4b

Edeloc=Etotal–Eloc=6bcos(2p/5)

1.85bCyclicsystemwith2ncarbonatoms:HOMOandLUMOSymmetricMOsCyclicsystemwith2ncarbonatoms:HOMOandLUMOAsymmetricMOsSothep-conjugatedmoleculehas1LOMO(s),1HUMO(as)and(n-1)doublydegenerateMOs.HOMO--doublydegenerated!For2n=4l+2,thetwodegenerateHOMOsarefullyoccupied!Cyclicsystemwith2ncarbonatoms:HOMOandLUMOFor2n=4l,thetwodegenerateHOMOsaresinglyoccupied!e.g.,cyclobutadiene,theHOMOsaresinglyoccupied!Thus,Etotal=2(a+2b)+2a=

4(a+b)Forlocalizedsystem:Eloc=4(a+b)

Edeloc=Etotal–Eloc=0Nodelocalizationstability!Antiaromatic!Cyclicsystemwith2n+1carbonatoms:HOMOandLUMOSymmetricMOsCyclicsystemwith2n+1carbonatoms:HOMOandLUMOAsymmetricMOsSothep-conjugatedmoleculehas1LOMO,andndoubly-degenerateMOs(oneissymmetric,oneisasymmetric).HOMO--doublydegenerated!For2n+1=4l+1,thetwodegenerateHOMOshold3electrons!Cyclicsystemwith2n+1carbonatoms:HOMOandLUMOFor2n+1=4l+3,theHOMOshave1e!e.g.,C3H3,theHOMOsaresinglyoccupied!Thus,Etotal=2(a+2b)+(a-b)=

3(a+b)Forlocalizedsystem:Eloc=2(a+b)+a=3a+2b

Edeloc=Etotal–Eloc=bForC3H3cation,Etotal=2(a+2b)Forlocalizedsystem:Eloc=2(a+b)Edeloc=Etotal–Eloc=2bEnhanceddelocalizationfulfillingtheHückelrule!Cyclicsystemwith2n+1carbonatoms:HOMOandLUMOe.g.,C7H7cation,

Etotal=2x(a+2b)+4(a+2bcos(2p/7))=6a+4b(1+2cos(2p/7))Forlocalizedsystem:Eloc=3x2(a+b)Delocalizationenergy:

Edeloc=Etotal–Eloc=8bcos(2p/7)-2b=

2.99bHMO---MorecomplexsystemsThesequlardeterminantofthismolecule:is

ForsymmetricMOsForasymmetricMOItistheantibondingp-MOofaC=Cbond!cos=-1C2H4+Br2

CH2Br-CH2Br

isnotanelementaryrxn!P.150,5.29LUMO(Br2)–su*HOMO(Br2)–pg*LUMO(C2H4)–pg*HOMO(C2H4)–puSymmetrypatible!Overlap=0Symmetrycompatible!Wrongdirectionofelectrontransfer!eLUMO(Br2)andHOMO(C2H4)arepatible!TheinteractionbetweenHOMO(Br2)andLUMO(C2H4),thoughbeingpatible,leadstoelectrontransferpatiblewiththerelativeelectron-negativityofthereactants!Thereactioncannotbeanelementaryrxn.Pleasederivethep-MOsofcyclopentadieneandexplainwhythereactionoftwocyclopentadienesgivesriseto[4+2]cycloadditionproduct,butnotthe[2+2]or[4+4]cycloadditionproduct.Answer:1)Thep-MOsofcyclopentadienearesimilartothoseof1,3-butadiene;theMOcoefficientsaregiveninthediagram.ThusTheboundaryconditionisThefourp-MOsareHOMOLUMO2)HOMOLUMOFor[4+2]cycloadditionoftwocyclopentadienes,the1,4-siteofmoleculeainteractswiththe1,2-siteofmoleculeb.The1,4-siteofHOMO(a)iscompatiblewiththe1,2-siteofLUMO(b);meanwhile,the1,4-siteofLUMO(a)iscompatiblewiththe1,2-siteofHOMO(b).Sothe[4+2]cycloadditionissymmetry-allowed.B)For[4+4]cycloaddition,the1,4-siteofmoleculeashouldinteractwiththe1,4-siteofmoleculeb.However,the1,4-siteofHOMO(a)isnotcompatiblewiththe1,4-siteofLUMO(b);likewise,the1,4-siteofLUMO(a)isnotcompatiblewiththe1,4-siteofHOMO(b).Sothe[4+4]cycloadditionissymmetry-forbidden.Similarly,the[2+2]cycloadditionissymmetry-forbiddenandcouldnotoccur!PossiblestructuresofXeOnFm(n,m=1,2,3)?P.150,5.19AnystableXeOnFm(n,m=0,1,2,3)compoundshouldhaveevennumberofvalenceelectrons.Som=even.

Ifn=0,a)m=2,XeF2,Xehasthreelonepairs,linearstructureofDh;b)m=4,XeF4,Xehastwolonepairs,planarsquareD4h;c)m=6,XeF6,Xehasonelonepair,thu