General Chemical Bonding

University:
Harvard University
Course:
CHEM 20 | Organic Chemistry
Academic year:

2023
Views:

145

Pages:

96
Author:

Dan Thompson

MWTFSSPage NoYOUVADate:Chemical BondingBond : Force of attraction between twoor more species.Chemical Bond t The attractive forces whichis responsible for holdingC ions, atoms, molecule) together in adifferent chemical species is called chemical BondAtoms are more reactive than moleculesbecause they have higher energy thanmoleculesAtoms Combines together to form malecules and attains stability by develosingenergyStability Can be attained by :-is octet Completion :- when atoms Combinesthey have a tendencyto achieve nearest noble gas Configura-tion so they can have 8 e in itsvalence shell and Complete its octet[ H- duplet]21 By decreasing Energy Atoms Combinestoget to acquiresstability by minimising their potentialEnergywhen atoms approaches to each otherthe force of attraction is dominatingover force of repulsion MTWTFPage No.YOUVADate:e repulsioneABNNattractionNANB(+)(+)force of repulsioneneBforce of repulsioneA nBforce of attractionNA eBforce of attractionGraph of P.E v/s Internuclear distance:-P.ET-0.roDrOrFR> FAFAYFRinternucleardistanceP.EminimumFA=FRBond Energyt It is the amt of energyreleased during bond formationorIt is the amt of energy required tobreak a bond MFPage NoYOUVADateminTBond length It is the internuclear distancewhere bond formation takesplaceQ NO - 36, Page NO-119 (JEE MAINS)Q:- The figure below is the plot of P.E vsinternuctear distance (d) of H2 moleculein the electronic ground state whatis the value of net potential energy EoCas shown in figure) in KJ most for d=doat which the e-e- repulsion andnucleus - nucleus repulsion energies areabsent? As reference, the potential Energyof H atom is taken as zero when the eand nucleus are infinitely far apartPotentialPennaydEdointernucleardistanceAnsJP.E =- Kq,92ro-9x109 x (16x10-99) 2X2 x 6x1023529 x 10-10- 5265 KJmole MWFPage NoYOUVADateTypes of chemical Bondweak bondStrong Bond(ERe > 42 KJ /mole)(ERe <42KJ/mle(i) Jonic Bond(ii) Covalent Bond'iii) Coordinate Bond(iv) metallic BondHydrogen Bondweak forces(8-42 KJ/mole)(<8 KJ/mole)Int 80 molecular(i) Ion- dipoleIntermoleH Bondingcular H'ii) Ion: induced(weak)Bonding(8-25) KJ/mol(Strong)dipole(25-mm)'iii) Dipole-Dipule(iv, Dipole induceddipole(V) Instaneous dibate-induced dipole(i) Dipote-Dipole ( Keesom)ii) Dipole- Induced dipoleVanderwaal'sC Debye)forces.'iii) Instaneous dipole-induced dipoleChondon forces / dispersion forces) MWTPage NoYOUVADateIonic Bond \ Electrovalent Bonds It is anelectrostaticforces of attraction between oppositily chargedions, produced by the complete transferof e- from electropositive elemenent toelectronegative element. (IA - III A)(VA- VITA)NaNatg)+ leLow I.FCI +le(1-(g)High E.G.FNatry'+ cicglNa+ (s) High L.E.CICondition to form ionic Bond f(i) large cation $ small anion.(ii)D.E.N >, 2.1I(iii)L.E of electropositive element.liys High E.G.E of electronegative element(x )High Lattice EnergyValues of E.N ace to Pauling scaleH (2.1)Li(1) Be(1.5) B(2) ((2.5) (30) 0(3.5) F(4)Ng. mg.,2. A1 1.5 Si 1.8 (2.18) S(2.5) Cl (3)K(8)Rb(78)Cs (07) MWTFSPage No.YOUVADateProperties of Ionic Compound"1 Ionic Bonds are non directional innature.2. Ionic Compounds don't Conduct electricityin solid state but do soin moltenor aqueous medium3. Ionic Compounds don't have molecularformula It has only empirical formula41 Electrovalency! :- Total no of e- lost orgained by an atom isknown as its electrovalency,orIt is the magnitude of charge persenton ion in ionic CompoundegNaclNa+ + ciE.V of Na = 1Ev of Cl = 1eg.mgclzmg+2 + 2ciE.V of mg=2E.V ofCl=1gmgomg+ + 0-2E.V of mg it 2E.g. 20.2 M W I F S SPage NoYOUVADate:egAIF3A1+3 + 3Fa.v of A1 =3E.V of F = 1Covalent Bond 1/4 The bond which is formedby mutual sharing of ebetween two atome(i) Single Bondeg:-H- HABH-ClABAB'iii Double BondA:X Beg0=0xABA =B MWTFPage No.YOUVADate:Triple bond t.A :XxBegNENABAEBNoteonly valence e takes part in bondformationACovalency J The total no of Covalent bondformed by one element withanother element is called CovalencyorIt is also defined as total no of unpaire present in an atom inits groundstate or excited statesegt- 5B - is2202 2p' (ground state)ALAL1ground stateal111Covalency = 3 M T W T F S SPage NoYOUVADate21 C = 1st 202 2p26AL11ground staten441Covalency = 431N =71s2 2p3ALaaris2s2p.ground stateCovalency = 3But nitrogen can extend its Covalencymaximum to H. This is due to formationof \ io Coordinate Het bondH- N - HHHJ of=AL ALAN2sCovalency= = 2.5) 9F= is 202 2p5AL202p.Covalency I MTWTFSSPage No.YOUVADate:6) P= [Ne] 352 3p310AL dan G.S2)Coralency 3.z) (Ist &.S) In 111 Acovalency 571S = [Ne] 382 3p410AL1411Covalency = 2Ist 9.5AL144covalency = 422nd asTn 191 11Covalency 681Cl(Ne) 3s2 3/51710IL11/14/1,Covalenly/1L ALAA 1111Ist as Covalency = 3^L 111 A12nd G-5 Covalency = 5 M W TPage NoYOUVADate:3rd E.JN TaxaCovalency = 7Bond pair $ lone pair e fThe e which participate in bond formationis Called Bond pair eThe e which don't participate in bondformation is called lone pair e -Coordinate Bond 1 In this bond shared pairof e is Contributed byonly / atom.orIt is interaction between Lewis acid&lewis baselewis acide pair acceptorLewis basee pair donar.Types of Coordinate Bond11 Sigma (o) type Coordinate Bond / If theIst bond isCoordinate bond between two atoms then thebond is called o type Coordinate bondHtH- - HM WPage No :YOUVADate:2) Pi(II) type Coordinate bonds Coordinate If thebond between two atom is not firstthen it is called II type Coordinatebond for esT type: Co:FegFto type)1 N :BtoFFBE B Feg+eg 03/ MwPage No.:YOUVADate:Selection Criteria for central atom)is central atom should be least inno and less electronegative21 Surrounding atom must completeits octet while H must completeits duplet31 If central atom belongs to 2ndperiod then they never extend itsoctet due to absence of d subshellus If Central atom belongs to 3rdperiod then they can extend theiro'ctet or Cavalency due to presenceof vacant d orbitalHow to draw structure 1/1(in Identify central atom(a) less in no'b' which can form multiple bondii)place surrounding atom around thecentral atom.(iii) Connect surrounding atom with centralatom & form single bond I double bondtriple bond by which octet of surroun-If oxygen $ hydrogen both are presentding atom Can be completed.(iv)together in a molecule then Connectthem with central atom in the formof -OH bond.V If oxygen is present in an anionthen place -ve charge on oxygen(vi) In other Condition of anion cation MTSPage No.YOUVADate:place charge an central atomvii) Finally Check whether bonds central not atomCan form such no of or2g Pcl3P:C1:eg2t NH3:-HHIHeg 13/- Co2CO: WSPage NoYOUVADateeg SFySF:: Feg:- CIF 3:FCl: Feg SF6F.:F:FSFF : Page No.YOUVADate:6) So3S7) PclsP8) XeF2as XeFXe101 SO2 M W 1 F SPage No.:YOUVADate:+11)NO2N121 Bcl3:CI:13) PockP:C1 :14]So cl2 Page No.:YOUVADate:15) SOCIAS16) NOCICI:17)N204N18)N2O5N.191Becl2Be Page NoPageYOUVAYOUVADate:Date:20)NF315)SOCI2F :161NOCI21co217)N20,N221po 318)N2O5N.231Bo3191Becl2 PageYOUVADate:241 NO2N55 C10226 NH2HH7J=C10,1.9Cl11:0:28) soIS TSPage No.Page NoYOUVAYOUVADate:Date:295C10301-31 NOHH:N(odd e species)+321 PH.HIIHPH:OH4331 C1O2+ MTWTFSSPage No.YOUVADate341NO2351 No::0:361 XeoFyI.IXe371so32S WFPage No.:YOUVADate:385 C1207OO11110=ClCIO11391AIF3:: FF401SHHH.HC.CHH41S2 HgHHCHHHH421CH2 2H C E CH FPage No.:YOUVADate:431 NO F3F::F:44) POF3PF:cless stable)(more stable)Due to formalcharge451 SNF3NH461CH2SF:F: TWTF5Page NoYOUVADate:47H3POiPPHOHHHo0/1HXoxy acid of 'p' haveone&5 Covalencyof P481H3PO3b11PHofoh491Sncl2SnSo 1 SF2:FF :0600 MWTFPage No.YOUVADateLewis Dot Structure fLewis or electronic structures are thediagrams that represents the valence eof atoms in O molecule.Acc to Lewis all atoms including surroundingatoms and central atom must Complete itsoctet while H complete its duplet.-ve charge is carried by electronegativeatom while (+ve) charge is carried byless electro negative atomeg SO2eg 2 soSeg 3 NO CCI:eg 4 xeo WPage NoYOUVADate:5 NO 2N61co32C3-7) Po!NO3asNO2+ MPage No.YOUVADate1o101 N3N N:N:0 611] CIF3CI F.Limitationof LDS125 Pcl3:C1 :Lewis dot structure (LOS):-n no of total valence en2 total valence e after octetcompletion ( 2e in case of H)my = m2-n,no of bondsn32eg So2 =>n,=18243 = n,-n,= 6bonds = 6 = 32 WTFPage No:YOUVADate:2)SO3 :-n,- 24n= 32n 8bonds is 431CHn = 88+ (4x2)= 16 (Duplet in H)CN3 8H\Hbonds = 4H41sock6+6+7+7 = 2632o2nE63bonds - 3SCICI51 So, Cl2n 6+6+6+7+7 = 32n,=408OA°.0bonds = 4's 'ClCI6J Pocl3n 5+6+7X3= 32On = 40n3 = 8Ibonds = 4P(1CIC / M W T F SPage No.:YOUVADate:eg Nocl3n 5+6+7x3= 32n,= 40ON3 8bonds- 4NC110CICesson 6+ 4x6 + 2 = 32n 40ng 8no of bonds - 4@So--oeg CIF377 783-28n32=4No of bonds 2(less than no of side atoms.is LOS of CF3 is not possible.eg SFn6+7x6 = 48n 56n3= 8bonds = 4( less than no of side atoms): LDS of SF6 is not possible WPage No.YOUVADateesxe F2n8+7X2 = 22m 24n3 2bonds - \Cless than no of side atoms): LDS not possibleeg BF3n,= 3+7x3 'c 24m2 = 32n 3 = 8bonds --H(Not possible 0.0 only 3 v e in 3)egNO3n,20n 32m 3 . 8bonds - 4egNo2m 18n2 24n3 6bonds = 33--gPon=32n,40nz 8bonds = 4 M W T F SPage No.YOUVADate:eg xeoFyn 42m, 48n3 = 6bonds = 3Cless than no of side atoms)LDS Not possiblees NO2n,- 17n 24n 73bonds - 3.5(Not possible)Limitations of Lewis Dot structureU we Can't draw LDS for CIF3, Pcls,IF7, NO2 etc.21 LDS didn't explain the geometry orshape of moleculeException of octet Rule fis Hypervalent Species It is also called etrich speciesmolecules in which Central atom hasmore than eight e in its valence shellis known as hypervalent. MTWTFSPage NoYOUVADateIn such molecules central atom generallybelongs to thired period and onwards.egPcls=10 eSF6=12 e -SFu=10 e --xeoFu14 exe FloeXeE6the-2)IFz)-10e-IF72)14e21 Hypovalent species Ce- deficient )molecules in which Central atom hasless than eight e is known as hypoNatentSuch molecules acts as an lowis acidegis Bcl321BBr33)BI3CtBrIB- CIB BrB-I1\tC1IBr91AICI351 Sncl261BF3cl,aF1AI -clSr:B-FciC1:F0)71Beel,CI - Be - CID. FPage NoYOUVADate3) odd e speciesmolecules in which central atom has oddno of e in its valence shell is knownas odd e species.These molecules are generally paramag noticin nature due to presence of unpaired ein the valence shell of central atom .egNON-O:7eegNO,te-NegCIO2lle:efC103aCl.II13e- WTFPage NoYOUVADate:Formal charge Formal charge on anatom in a moleculilionis defined as difference in no of valencee of that atom in free state andno of e assigned to that atom inthe Combined state C Generally Lewisstructure )Formal charge = Valence e- - - Lane pairin free atom1/2 (shared pair e-)2F =V - L- 1 S2(6) =+1egO36-6-1(2)6-4-1(4)-I0efso301021111SSS0230%30F.C ofSS=it+2O,O012002-002-12=O3>-,= - 103 M W T F S SPage No.:YOUVADate:If C.A belongs to 3rd or below periodthen double bond is preffered overCoordinate bondCoordinate bond Can be replaced bya single bond followed by at \ chargeon donar and -ve charge on acceptorHtHFH++OegNorNHNBH/PHHH17Htoeg: C O:F.C one = u-2-1(6)= - 12F.C on O = 6-2-1(6) = +12020Dipole moment of co is fromO0 [ from more electronegativeto less electro negative elementT type Coordinate bond -> back bonding. MWTFSPage No.YOUVADateResonance f of when a molecule cannot beCompletely represented by asingle lewis structure but its Chracteristicsproperties Can be described by the two ormore different structure , with similarenergy position of nuclei bonding &non bonding pairs of electrons then thelew structure is said to be resonancehybrid of these structureThis phenomenon is called resonance $different contributing structures are calledresonating structure or Caronical structuresResonance stabilises the molecule as energyof resonance hybrid is less than the energyof any Canenical structureAResorance averages the bond chractersticsas a whole*The Canonical forms has no relalexistance2-eg(i)coC--C-C0overall charge = - 2charge = -2/3Bond order = 4A.RB.O = 4/3 M W T F SPage No.:YOUVADate:11)-8-SCO-2/3-2/3:O-2/3-8ResonancehybridPossible R. structure: 3-2)No: NNo :BO = 3=1.52-dNO1/2031 soS0-1/21/2B-O = 6- 1.5 = Total bond4No of bonds in which they distributed areCharge - -2=overall charge42no of distribution MWTFSPage No.:YOUVADateBond orders No of bonds between twoatomsQ,:- Cal B.O of given speciesB.OH2H-H1N2 NEN3O20=024CO2OF C=025HCNH - -CENH-C = 1C-N 3Q2:- Compare the B-O of given speciesNO2 No: c10 co(i)NO20- N 1=0B-O= 3 = 1.52'ii)NO30=OB-O= I'D 4 =1.33is'iii) ciouB.O = 7 = 1.75H(iv)CO:CEo: B.O= 3B-O order :-COciou>NO2>NO3 M W : F SPage No.YOUVADate:StrengthNotef B.O 2 1 & Bond length 2 B.D.EB.L2 B.EB-OBond orderB.1Bond lengthB.D.EBond dissociation EnergyB-Er Bond Energya3f Bond length order1d N-O (NO;-)dN-O (NOg)O=N-0 -OF N - oI0B-O = 3/2B.O = 4/32dc-o (co) 2 dc-o (co2) < dc.o (co32):C o :O=C=00=0-0-10B.O = 3.B-O 2B-O: 4/3=1.3Q4 :- Bond strength orderso > soo :oi i6o- S=0So- B.O 6:-oB.O= 443 MTWTFSSPage No.:YOUVADate2CI - OCIO > c103"6"C1=0of CI=C11.Oo--7/4=1.755/3-1.66Valence Bond theory According to valence bondtheory as many numberof unpaired electrons are present either intheir ground state or in excited state, sameno of Covalent bonds will be formed and iflone pair is present on central atom (wherecentral atom belongs to 2nd period only) thenit can form Coordinate bond which dependon circmstanceseg 15 S [Ne]AL hunn3s3p3deg. H2S SF2, Sel, etcs+(Ne)AL3s3p3deg SF, SO2, Socl2, SOF, , so 2 etcS**(Ne)N3s3p3d2-eg SF6, So, So2 Cl2, SOF, H SO , 50 2etc M T W T F SPage NoYOUVADateeg2t 7 N (He]ALnnn2s 2peg NH2 3 NF3 NCI, NOCI etcNHeALcoordinate covalentBond bondif NOF3Q1 Check which of the following existsSF6OF6SF4existOFdon'texistPelsNCISCLF3Fcl,Reason Due to the absen ce of 2d subshell01SF 2SFXSF4SF3SF6SFsSurphur Can only form bonds witheven no(Reasondue to V.BT) M W T F SPage No:YOUVADate:osIciICI2XIcl3IF4IclsIF6IF7a forms bonds only in odd noReason -) due to V.B.T. .asSF2+SE +SFLxSF,+s+ forms bond only in odd NoReason V.B.Tof Icl2ICI2XIF4IFIFCIF6I forms bonds only in even NoReasons- due to V.B.T. M T W T F SPage No.YOUVADate:Types of Covalent Bonds :These are two types of covalent bonds -bond & TI bondisa bond (sigmart It is formed by headon or axial overlappi-mg of orbitals2) T1-bond (pie) :- It is formed by sidewiseor lateral overlappingof atomic orbitalsSigma (o) bond is stronger than II bondas overlapping is more electron densityis moreTypes of G bond 28-28 > 3s-33'ii)2p-2p - 2s-2p > 2s-2s'iii)is- is > 2p-2p(N,is-is 7 1s-2p > 18-2s M W T F S SPage No.:YOUVADate:Relative Strength of TI - bond i'(i) If shell No is differentB.S 2IShell Noii) If shell NC is sameB.S 2 directional nature oforbital(Shell No factor is dominating)Cl 2pT1 2pT, > 2pt 3dTI > 2pT1- 3pt, >588.201amore stableI912Iless Stable highlyunstableas SH exists but SizHu doesnt why ?Asthe4tobyC=C- Si=sitoHto2pTi- 2PF3pTi-3pX WPage No:YOUVADate:as At Room temp N2 exists but P2 Can't ?AMNENPEPI2pr-2pT,3pr-3pT,asAt Room temp co2 molecule exists in more -meric forms while Sig molecule in polymeric form not in monomeric forms why ?As2pTi-2pTO=C=00 = Si=02pti3PTIpolymeric forms0Si- 0- Si-o- SiSi0SiOSr0S: 0 Si o SiS.Si0O0Si MTWTFSSPage No.:YOUVADate:o-Si0-Si-0-O-00-si-0-oOS.O-F2 is the second highest strongestoxidising agent(yeog)4-Li is most strong reducing agentvinpB.S orderICSi-C=CSi-Si-2- -p-NIN-S-P=P-3ceCI4 M T W T F S SPage No :YOUVADate:Small size her ILip - Lip repulsionis morehighly unstableQ) compare B-S / B.D.E ( B.E order -F2, C1, Br, I2expected F. F > CI-CI > Br-Br> I-IReal :- CI-CI > Br-Br > F- F> I-I[oncy 01121 fre 3113 Boss ItofspeciesPTI PnPTI-dTIisCO22O2fPO - 3-OI3)So,2HjSo2I15jC10O36)C10O37sIO3O28)soI M W T F SPage No.YOUVADate:asN2OS101GH2 WPage NoYOUVADate:Need of Hybridisation Acc to NBT inmethane all C-Hb.L should not same. and location of 4thhydrogen is not fixedCsripp.ssHHHB.L of Three C-Hshould be same4 one of the 4 C-H should havedifferent B.L. aBut in actual all B.L of C-H bonds inCHu are equal and u hydrogens arearranged in at corners of tetrahedralTo explain this Concept hybridisation isused.CHuHC =is 2s2 2p 2611LAL11H- - C- His2s2p.HALa111is2s2psDps(H,) M W T F S SPage No.:YOUVACHypothetical phenomenon):-Date:Hybridisation It is intermixing of atomicorbitals having almostsamel Comparable energyIn hybridisation C Properties)LNo of hybrid orbitals form=No of atomic orbitals participate inhybridisation.2L b bond is formed by the hybridorbitals while F bond is formedby unhybrid \ pure atomic orbital.eg. C2Hg SHu SH2 2 (draw structure)inclassroom) if needed3f Lone pair if present will participatein hybridisationLAny kind of orbital ( filled, hay filled /vacant) Can participate in hybridisationEf Hybrid orbitals are more reactive innature as Compare to atomic orbitalsas they are more directional in nature6fHybrid orbital formed are havingthe same sppd Chracter (if any)length and strength C only in equivalenthybridisation)No of hybrid orpitalsNo of b bond formedrequiredby that atom +orno of lone pairSteric Number)present on that atom7Bonds formed by hybrid orbitals aremore stronger as compare to atomicorbitals M T W T F S SPage No.YOUVADate:A.O (atomic orbitalss98 L e densityp2% e densityd.Inp8fon increasing the 5% chractu in hybridorbital -spherical Nature of H.OLength ofH.OJsize ofH.Oenergy of H.OIB.EBond LengthBond StrengthE.N of hybrid atomBond anglest plst2 Hless distanceB.L(more repulsi6thon)HBond angleTmore distan CoB-LCless1repulsion)Bond angle L MTWTFSSPage NoYOUVADate:'c'% S% pC-CB.AE.NBond lengthsp325751.54 A109°28'2.5sp233-3366.661.34 A°1202.75 0228sp50301.20A°180°3.25Types of hybridisationH.O=3 (sp2)s + p + p+ + p + d + dt d + d + d.HO 2 (sp)H.O=Y (sp3)H.O = 5 (sp3d)H.O -- 6 (sp3d2)H.O=7 (sp3d3)01 Comment on hybridisation of given speciestI SFy :-.0S4+1 - sp3d WTFSPage NoYOUVADate:it2Poc'sP(uto) = 4CIsp33SO2(2+1) = 3Osp24NH3(1+3) = 4H.Hsp3SHO(2+2) 4Isp3F6Ne FF - xe -F (4+2) = 6Fsp'd27XeOZF2(5+0) - sp3d:0 Xe-0F F8xeoofO"xe(3+1) = sp3(6+0)9xeofxesb'd2 MTWTFSSPage No.:YOUVAitDate:1110 :- JogI(4+0) = sp311O411PclsaP-CI(5+0) = sp'dC1C112snckCl Snaecl (3+1) = 4Isp3CI13OF2Rt2)=4 sp31=11ofIMsoS3+1=4= sp3F15IF7F7- sp3d3I11FFF16Becl2C Be - CI 2 sp. W T F S SPage NoYOUVADate:17NO20- in - o-1+2 = 3=sp2C118PcleCI -P-cl4+0 = sp3C1+F19BEuF- BOEF4+0 = sp'F+20Is+I-I-I2+2-4 sp321Br:Br-Br-B32+3 = sp3d22Hot4-4+0 = sp3iH+23C103ci=o3+1 = 4= sp30024C1022+2 = 4= sp3Ht(33)NH+HNH4+0 us sp3 MTWIFSSPage No.YOUVADate:261 IsI - I-3+2=5sp3d, 427) SHG4-c-c-H(4+0) = usp'28) SH4-car(3+0) = spb29j C2H2 4CEC-H (2+0) 2= sp30) N205(3+0) = sp23y Benzene (grs)Isp2spr.H - C01/142sp2CspringHH1sp2sp2 17 M W T F SPage NoYOUVADate:Note Hybrid orbitals always form o bondexcept Benzyne (gHu)In Benzyne all catoms are sp2 hybri-disedC11^sp2H-CocII bondCCCsp2HCInDrawing of structures using hybridisationH.OHybarrangement ofElectronB.A.planer None pair inspacegeometryplaner.2spLinear180°P(1)3sp2Trigonal120Pplaner.(3)4)sp3tetrahedre109°28'N.P.(6)aesa5.sp'de.Trigonal90°16) N.Pbibyramidasabipyramida=1800/1ese180°(3) M W T F S SPage No.:YOUVADate:6 spd2octahedralorSBP90° (12)SquareN.Pbibyramida180°(3)7sp'd?Pentae $e) = 72 (5)N.Pgone=114 (5)bipyraasa=180 (1)midal.esa=(10)900sp hybridisationst prO+2 8180°B - A B5 = 7,2.3-0Geometry = linearsp2 hybridication, st pxt by & 3 8Bond angle! 120° 3 angles0120°Geometry t Trigonal planerTotal No of atoms =4in a pain M W T F SPage No.YOUVADate:sp3 :s + pxt by of p2XIBond anyle 109°28'109°28No of angles 6x, m xyX, mx3X244X, mx2X3from planerX2 m x 3X3mx3X2 mxymax. no of bond in a plane 2max no of atoms in a plane: 3Geometry - Tetrahedra(10) how many planets tX,CX3X,CX2X1 CX46 planes.X3CX4X2CX3X, CX4x,x2x34 planesX, X3 X4X, Xy X2X2 X3 X 4 M T W I F S SPage No.:YOUVADate:sp3ds + pxt by + p2 + d2 22axidequitorial.35 GeometryTrigonal pipyramdae(Non planer)Angle between1821485] 180°184283120°(1)2341 $ 3 (3)38490°185285(6)385max. no of atoms in a plane - 4sp3d? f2Geometry5.9SquareXbipyramida/43octabudral(New)69.0° ( 12 angles180° ( 3)No of atoms in a plane = 5 M T W T F SPage NoYOUVADate:=sp3d3/s+ pactpy + pat d22 + dazy 2(Non planer'Bond angle90° o = 1077 = 5114 = 5180 = INo of atoms in a plane = 6Geometrypenta gonal bipyramideQ) Find max no of atoms in one planeof o(cHs)4toH-cAns -5CCH3Q1 Solve above for(C(N)4As-s5CEN"IfNIII M T W T F S SPage No.:YOUVADateValence shell electron pair repulsions(VSEPR) fElectron pairs of valence shell of centra Iatom of a molecule arrange themselvesin 0 manner to minimise muture electronicrepulsionmagnitude of different electronic repulsionis1lone-pair lone pair 7 lone pair- bond pair )bond pair- - bond pairElectronic geometry = Sigma (o) bond +lone pairMolecular geometry shape \ structure(Total no of a bond + e.p) - e.p.2)Electron geometry - e. pdon't Cut up & - e.p asit would change the geometry00gtoNNHtoelectronic geometryTetrahedrup.molecular geometry z PyramidalSAN at 1.b.b. MT W T F S SPage No.YOUVADate:.esH2OtoS.N 2lp + 2b.p.molecullar geometry = shape 1 Denteg! sp3de.p-bip repulsionHpropulsimatat 90 - 3ao 2.(4 lip - bip)H.p.l.ppiplip6 lip bip MTFSPage No.:YOUVADate:sp3dIn sp3d2 hypridisation axial x equitorialterms are not defined I'm sp3dzhybridisating lip Can be placed at19.5any position but USEPR should byfollowXAcc to YSEPR10Acc to VSEPR16F0012Q1115+xeXeXe12FFF(A)X (B)'c' M T W TESSPage NoYOUVADate:tosp3d3l.p is placed will at be minimum axial position thererepulsion-IOFSxeotsException- XeFfS.N = 7 -Geometry should be pentagonahapyramidalReal geometryCapped octahedralororShapedistortedoctahedralTables MWTPage No.:YOUVADateAtomic orbitals Combines in hybridisationssp3st pre+ by t prsp3dst pt x by + pato + d22sp3d2s + psetpy+ dacing? + d2?sp3dss+ psc + by + dx2y2 + dz2 day.Q)which of the following at orbitalsis lare participat, in hyb of centralatom inIOF3sp2dx2-y2d22a match the ColumnIA.II BSoy 2-PS:2503Qp23XeO2F 2Rday4SF2C12Sdr2-y25XeFuT d2.2 M W T F SPage No.:YOUVADate:AP, 00BP,OCP,CDP,00aPOSTPcls (g)Cavalent sp3d (TBP)Hy brid. sation in Jonic solid 1 liquid 1/Cation partAnionic part+is Pcl5(S)PclyPoli2)PBr(s)PBr +PBS Brdon't existdue tosteric repulsion31PT+P Is (s)I i PI (x)4) C12 of (s)C102C10If two centralare present dwideit by 2. ( C1206) 2CTO351 N204 (s)No++NO31NO2 Xe F6 (covalent in gaseous phase)M T weak vandlywast forcesIYOUNGDate:N2O5 (s)No+2NO2did cation521121 anionCNO2 NO3)N2O3 (s)No+NO2NO, NO2)*I2C'6 (e)(ioniccod)Ic'2IclyIIC'3YeF (s)-XeF+xeta F(don't sit intocrystal lattice)size of anionscation should besimilar to formcrystalI2 (e) :-I3+I'ionicI MFSPage No.YOUVADateQ) what will be the product when solidN205 react with molten Nacl.AsjN20,(5) NO2 T No,Na++ccNa NO2 NO2C1CP) Comment on hybridisation shape xmax no of atoms in anionic part ofI2clo (e)I2cl61e)Icht + IclyIC13ClC14+2=6sp2d 2/9I5-Cl M T W T F S SPage No :YOUVADateQ) what is the bond angle dyjerine betweenanionic part of solid molecule and cationicpart of N20, (s) I2 (liquid) anionicN205 (s)NO2+NO3NO2 NO3sp2(1200)I2I3++ I'm180°difference = 60° MTWTFSSPage NoYOUVADate:Bond parametersisBond ordersIt is the total no of Bonds present betweentwo atoms.A-B B.O-1A=B B.O=2A E B 3.0=3eg.H2 N2, O2H-H NEN 0=00eg.co32IfB.O = 4/3 =1.33C6egPo 311B.O. 5. 1.25Pueg. Compare B.Oco32 po 3 sou2-B.O 1.33 1.25 / 1.5B.O so > co32's po 3ab9 b c02H- 6-C-0tH-O c=ogd:-/( = zz B-Oz 1.5beIa= I M T W T F SPage NoYOUVADate:25 Bona length:- It is the internuclear distancebetween two atoms (bonded ).1Bond lengthBond orderA-BA = BBgco32- so 2- Po 3B-o1.33 1.5 1.25B.LPo 3- > Co3 > so [ S-o]2Bond length 2 At SizeIf B.O is same then B.L 2 A.SegH-F < H-CI L H-Br L H- I3% S chracterBond length thLI% S chrocter MWFSPage NoYOUVADate:comparison of Bond angles in molecules.1) Hybridisationsp > sp27sp3 3.eg.Co2 > BE3 > CHB.A 2 1 Schractereg.NoLNo+2spasp.esC2H2 > CItty > E2Hg21 No of lone pair If Hbd is samethenB.A 2No of lone pain on central atunnef CH NH3 H2O41)o)c115h r 1016hrB2hH170,70,203a)NO2NO2NO3sp2sp.sp2o+itN=O 05 N=6Noof1032 Page NoYOUVADate3NHNH37NH2(IONNN16toIto2H.O3or)oz>0,01Sick, 4 >Pcl3SciCI\eISiPS4C340, )O>ofa)BCl3= BBr, = BI3 (sym)aB>I1BisBBrI4aBrozIO2=0301cclu =Sicl = GeclyatC\CS,ccareC1 CIac.C,CGINote SIf in a different molecules hyb issame and molecule is symetrical thenbond angle will be same in all FPage NoYOUVADate:3) Electro negativity of central atometIf in a given mole hyb & lone pair ofe are same thenBond a angle2Electro-vity ofErect Central atom..1v.)SesHHepNCI,Pcl37Ascl3(+)NAsg>c?11a4CICIC1 C,ciciesOCI,2Scl,SO,C00PH3ASH>3SbH3CP)H2O < HotasSo2 L So,SHoa TilSSPage NoYOUVADate:usSize of side aton EN of side atomIf in 0 different molecules, hyb, lip 8 E.Nof central atom same thenB.AdSize of Side aton (steric Crowdiy)E.N of side atomin:g 274>3. MWTPage No.:YOUVADate:Ha)'scto2F3.ITQ)3(cisicHC,17273i -of-6Pof PPIC/11oO172as AsH > AsF,1(i)AsAs10, bFh r0,7 O2 M W T F $ SPage NoYOUVADate:Bentls Rules According to Bentls rule)1 more EN element attached to thathybrid orbital of C.A in which . I.Schracter is less.2) lone pain is attached at that hybridorbital of central atom which Containsmore 0/0 S chractersp3dst poct by + dazage + do 2 + p2A90°% schracter L120% Schroder n% schroady Tes, B.A% Schroctis Tes E.N Tes-0/2 S chracts TesB.LL1/ S Chroch TesB-S1esequal and by % S chrocter in equitorial isIn sp3d hyb all hybrid orbitals are notmore than that at axial position.is axial lengths is more than equitorialbond length M T W T F S SPage No.:YOUVADate:In sp3d? all bond length are identialwhen symetrical molecule is consideredIn sp3d3 hyb b equitorial bond lengthsare greater than axial bonds in sym.moleculeS more720S lessEquitorial B.A J (720) P chroacterAxial B.A T (900) Schroaches.eg.SFy sp3dFFFeg. Xeo F2 Fxe =0FFCIeg SF2CL2SOC I/F M W T F S SPage NoYOUVADate:esSOF, ? sp3d$ 1.0ofxeoFyBt+toXeFFOeg.IF3sp3dFIDFes :- xeo2Oxeto20eg SF, (CH3)2sp3dH3S:HC3FepXeF5sp'd3FFFXeF MWTPage NoYOUVADateeg. IFS : sp3d2AFIFFg Comparison of Bond length)up. PF3C12aCICPFCI1Fa> byc%.scmat.sizeasCom Pare C-H B.L44SA4R'/C.2/c3/c\Cto/112)RinIn inin1hFhBond angle)3>2>3d C-HCM 7 CH3F ) CM2F2 ) CHF3 WPage NoYOUVADate(0) Compare B.L11H2O2O2F2x,4y0-/sr'n'FB.Lx y y .2)h-F - C-C-FSLabSanhnFFB.L a>b.31N2H4, N2F4, 4FFtoN- NN-IVh ,SL."17Fsn Fxyy4f In which of following B.L are not CI identialF4(i) 0=0=01BP-cl12/ 1toC ,hhC, M T W T F S SPage NoYOUVADateas In which of the following molecule All B.Lare identicalev SF4 (b) CIF3 (c) xeF2 (d, not(P) ComfareDrago's RulesNH3 = 202 2p3PH3= 3s2 sp3intromizingNo intermixingHybridisation(gap betweenpossible.3s-3p ies)C No hybridisatinsp3sp/37 sp3takes ploa)Pure orbited(s)Nspsppures P/Aorbitalbyh 02HhIIs(p)(1070)(=94°)Conditions for Dragols rule)11 Central atom should belong to third ordownward period2)Central atom must lone pain of is3) EN of surrounding atom should be 2.5 MWTFSPage NoYOUVACDate:exNPH,ASHSbH3toto9492.591to107°H2OH25ty seH2Te9291690.5104.5a) Compare Lowis base strength in NH3PH3.(100%S)N(25%S)PItoh hhismolecule doesn't exist1)d- orbital contractiontdSPF5PHS3/03d( gap decreases)gap Tes M W T F S SPage No.:YOUVADate:molecule existmolecule doesnotexistPFSPH,SF6SHGSE,SH,XeF2xelt,XeFuXetlyXeF6XeHIE7IH72) Element doesn't break octet Rule)molecule existmolecule doesn't existSFCOF,SF4OF,PclsNCI5CIF,Fclz3) Steric crowdingmoleculeexistmolecule don't existIpclfPisr PIj2SFself3SecifSclf M T W T F S SPage No :YOUVADate(not)[SiE]](sick 2-Br,IBF iBelu BBr= BIInpInterhalogen Compoundsmolecule existmolecule dont existCIFFCl3, FC's, Ecl2, FBr3,FI3CIF, CF3, HF5GE7BrF, Br F BrFsBrF7IF, JF3, IFg, IFBrciBrcls, Brcl7Icl, Icl3 (dimer)Icls, ICI,IBrIBr, IBrs, I Brgos which of the following molecule doesn't exist ?TA CIF3(B) Jch 23.etc) BrF, X'D CI3