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WWU CHEM 121 - Review for Final

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Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Figure 10.26: Sigma and pi bonds.Figure 10.27: Bonding in ethylene.Slide 18Figure 10.28: Bonding in acetylene.Slide 20Slide 21Slide 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 48Resonance:Delocalized Electron-Pair Bonding - IICCCCCCCCCCCCCCCCCCHHHHH HHHHHHHHHHHHHResonance StructureBenzeneResonance Structures - Expanded Valence Shells....SFFFFFF........ ........................Sulfur hexafluoride......PFFFFF........................Phosphorous pentafluorideOSOO OH H.......... ..........OSOO OH H.. .... ..........Sulfuric acidS = 12e-p = 10e-S = 12e-Resonance StructuresUsing VSEPR Theory to Determine Molecular Shape1) Write the Lewis structure from the molecular formula to see the relative placement of atoms and the number of electron groups.2) Assign an electron-group arrangement by counting all electron groups around the central atom, bonding plus nonbonding.3) Predict the ideal bond angle from the electron-group arrangement and the direction of any deviation caused by the lone pairs or double bonds.4) Draw and name the molecular shape by counting bonding groups and non-bonding groups separately.Hybrid Orbital ModelThe sp Hybrid Orbitals in Gaseous BeCl2The sp3 Hybrid Orbitals in NH3 and H2OThe sp3d Hybrid Orbitals in PCl5The sp3d2 Hybrid Orbitals in SF6Sulfur Hexafluoride -- SF6Figure 10.26: Sigma and pi bonds.Figure 10.27: Bonding in ethylene.Figure 10.28: Bonding in acetylene.Restricted Rotation of -Bonded MoleculesA) Cis - 1,2 dichloroethylene B) trans - 1,2 dichloroethylenePostulating the Hybrid Orbitals in a MoleculeProblem: Describe how mixing of atomic orbitals on the central atoms leads to the hybrid orbitals in the following: a) Methyl amine, CH3NH2 b) Xenon tetrafluoride, XeF4Plan: From the Lewis structure and molecular shape, we know the number and arrangement of electron groups around the central atoms,from which we postulate the type of hybrid orbitals involved. Then we write the partial orbital diagram for each central atom before and after the orbitals are hybridized.Postulating the Hybrid Orbitals in a MoleculeProblem: Describe how mixing of atomic orbitals on the central atoms leads to the hybrid orbitals in the following: a) Methyl amine, CH3NH2 b) Xenon tetrafluoride, XeF4Plan: From the Lewis structure and molecular shape, we know the number and arrangement of electron groups around the central atoms,from which we postulate the type of hybrid orbitals involved. Then we write the partial orbital diagram for each central atom before and after the orbitals are hybridized.Solution:a) For CH3NH2: The shape is tetrahedral around the C and N atoms.Therefore, each central atom is sp3 hybridized. The carbon atom has four half-filled sp3 orbitals:Isolated Carbon Atom2s 2p sp3Hybridized Carbon AtomThe N atom has three half-filled sp3 orbitals and one filled with a lone pair.2s2psp3CHHHHHN..b) The Xenon atom has filled 5 s and 5 p orbitals with the 5 d orbitals empty. 5 s 5 p5 dHybridized Xe atom:5 dIsolated Xe atomsp3d2b) continued:For XeF4. for Xenon, normally it has a full octet of electrons,which would mean an octahedral geometry, so to make the compound, two pairs must be broken up, and bonds made to the four fluorine atoms. If the two lone pairs are on the equatorial positions, they will be at 900 to each other, whereas if the two polar positions are chosen, the two electron groups will be 1800 from each other. Thereby minimizing the repulsion between the two electron groups.XeFFFF....XeFF FFSquare planar1800Fig. 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 and of P = 2.1: C - P the EN of P = 2.1 and of N = 3.0: P - N the EN of


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