Slide 1Slide 2Slide 3Slide 4Slide 5Figure 9.12: Molecular model of nitro-glycerin.Slide 7Slide 8Slide 9Slide 10Slide 11A model of ethylene.A model of acetylene.A model of COCl2.Slide 15Slide 16A model of SCl2.Figure 9.16: Delocalized bonding in sodium metal.Model of CO32-Conceptual Problem 9.103Lewis Structures of Simple MoleculesSlide 22Slide 23Slide 24Slide 25Slide 26Slide 27Figure 9.15: Electronegatives of the elements.Slide 29Slide 30Slide 31Slide 32Slide 33Slide 34Slide 35Slide 36Slide 37Slide 38Slide 39Slide 40Slide 41Slide 42Slide 43Slide 44Slide 45Slide 46Slide 47Slide 48Bond Lengths and Covalent RadiusThe Relation of Bond Order,Bond Length and Bond EnergyBond Bond Order Average Bond Average Bond Length (pm) Energy (kJ/mol)C O 1 143 358C O 2 123 745C O 3 113 1070C C 1 154 347C C 2 134 614C C 3 121 839N N 1 146 160N N 2 122 418N N 3 110 945Table 9.4Figure 9.12: Molecular model of nitro-glycerin. What is the formula for thiscompound?Rules for drawing Lewis structures1. Count up all the valence electrons2. Arrange the atoms in a skeleton3. Have all atoms develop octets (except those around He)Make some Lewis Dot Structures with other elements:CH4 NH3H2OC2H6C2H6OCH2OLook at all these structures and make some bonding rules:The normal number of bonds that common elements make in covalent structures.Element # BondsCNOH, HalogensElement # BondsC 4N 3O 2H, Halogens 1Rules for drawing Lewis structures1. Count up all the valence electrons2. Arrange the atoms in a skeleton3. Have all atoms develop octets (except those around He)4. Satisfy bonding preferences!A model of ethylene.A model of acetylene.A model of COCl2.A model of SCl2.Figure 9.16: Delocalized bonding in sodium metal.Model of CO32-Conceptual Problem 9.103Lewis Structures of Simple Molecules CHH HHClOO OK+KClO3CF4....H C O HHHHHCEthyl Alcohol (Ethanol) Potassium Chlorate Carbon Tetrafluoride..........................CFFFF..................CH4MethaneResonance: Delocalized Electron-Pair Bonding - IOzone : O3........O OO............OOO....IIIOOO..........Resonance Hybrid StructureOne pair of electron’s resonances between the two locations!!Resonance:Delocalized Electron-Pair Bonding - IICCCCCCCCCCCCCCCCCCHHHHH HHHHHHHHHHHHHResonance StructureBenzeneLewis Structures for Octet Rule ExceptionsClFFF..........................BClClCl..............Each fluorine atom has 8 electrons associated. Chlorine has 10 electrons!Each chlorine atom has8 electrons associated. Boron has only 6!Cl ClBe............Each chlorine atom has8 electrons associated. The beryllium has only 4 electrons.NOO...........NO2 is an odd electron atom.The nitrogen has 7 electrons.Resonance Structures - Expanded Valence Shells....SFFFFFF........ ........................Sulfur hexafluoride......PFFFFF........................Phosphorous pentafluorideOSOO OH H.......... ..........OSOO OH H.. .... ..........Sulfuric acidS = 12e-p = 10e-S = 12e-Resonance StructuresLewis Structures of Simple Molecules Resonance Structures-VSO OOOSO OOO. .. . . .. .. .. .. .. .. .. .. .. .. .-2. .. .. .. .. .-2SulfateSOOO Oxxx = Sulfur electrons o = Oxygen electronso oo oo oo oo ox ox xx oo oo oo oo oo *o *-2o oPlus 4 othersfor a total of 6. .. .Fig. 9.14Figure 9.15: Electronegatives of the elements.The Periodic Table of the Elements2.10.9 1.50.9 1.20.8 1.0 1.30.80.70.71.00.91.5 1.6 1.61.5 1.81.21.11.8 1.8 1.9 1.61.4 1.61.51.81.71.91.92.2 2.22.22.22.21.92.41.71.92.0 2.5 3.0 3.54.0HeNeAr1.5 1.8 2.1 2.5 3.01.6 1.8 2.0 2.4 2.8 KrXeRn2.52.12.21.92.01.91.81.71.81.81.1 1.1 1.1 1.11.31.2 1.2 1.2 1.2 1.2 1.2 1.2 1.2 1.21.31.5 1.7 1.3 1.3 1.3 1.3 1.3 1.3 1.31.3 1.50.91.3 2.2Electronegativity1.1Th Pa U NpNo Lr1.3Ce Pr Nd Pm Yb LuFig. 9.16Fig. 9.17Determining Bond Polarity from Electronegativity ValuesProblem: (a)Indicate the polarity of the following bonds with a polarity arrow: O - H, O - Cl, C - N, P - N, N - S, C - Br, As - S (b) rank those bonds in order of increasing polarity.Plan: (a) We use Fig. 9.16 to find the EN values, and point the arrow toward the negative end. (b) Use the EN values.Solution: a) the EN of O = 3.5 and of H = 2.1: O - H the EN of O = 3.5 and of Cl = 3.0: O - Cl the EN of C = 2.5 and of P = 2.1: C - P the EN of P = 2.1 and of N = 3.0: P - N the EN of N = 3.0 and of S = 2.1: N - S the EN of C = 2.5 and of Br = 2.8: C - Br the EN of As = 2.0 and of O = 3.5: As - O b) C - Br < C - P < O - Cl < P - N < N - S < O - H < As - O 0.3 < 0.4 < 0.5 < 0.9 < 0.9 < 1.4 < 1.5Fig. 9.18Percent Ionic Character as a Function ofElectronegativity Difference (En)Fig. 9.19Fig. 9.20The Charge Density of LiFFigure 9.15: Electronegatives of the elements.The Periodic Table of the Elements2.10.9 1.50.9 1.20.8 1.0 1.30.80.70.71.00.91.5 1.6 1.61.5 1.81.21.11.8 1.8 1.9 1.61.4 1.61.51.81.71.91.92.2 2.22.22.22.21.92.41.71.92.0 2.5 3.0 3.54.0HeNeAr1.5 1.8 2.1 2.5 3.01.6 1.8 2.0 2.4 2.8 KrXeRn2.52.12.21.92.01.91.81.71.81.81.1 1.1 1.1 1.11.31.2 1.2 1.2 1.2 1.2 1.2 1.2 1.2 1.21.31.5 1.7 1.3 1.3 1.3 1.3 1.3 1.3 1.31.3 1.50.91.3 2.2Electronegativity1.1Th Pa U NpNo Lr1.3Ce Pr Nd Pm Yb LuFig. 9.16Fig. 9.17Determining Bond Polarity from Electronegativity ValuesProblem: (a)Indicate the polarity of the following bonds with a polarity arrow: O - H, O - Cl, C - N, P - N, N - S, C - Br, As - S (b) rank those bonds in order of increasing polarity.Plan: (a) We use Fig. 9.16 to find the EN values, and point the arrow toward the negative end. (b) Use the EN values.Solution: a) the EN of O = 3.5 and of H = 2.1: O - H the EN of O = 3.5 and of Cl = 3.0: O - Cl the EN of C = 2.5