Review for final F March 14 1 pm ES 413 Room S March 16 4 pm BI 234 Office: M 11:30 am-12:30 pm

1. Review for final F March 14 1 pm ES 413 Room S March 16 4 pm BI 234 Office: M 11:30 am-12:30 pm

2. Resonance: Delocalized Electron-Pair Bonding - II H H C C H H C C C C H H C C H H C C H H H C C C H H C H H C H C C H C Benzene Resonance Structure H

3. Resonance Structures - Expanded Valence Shells .. .. O O .. .. .. .. H O S O H H O S O H .. .. O O .. .. .. .. .. .. .. .. .. .. .. .. .. .. F F F F .. .. .. .. .. .. .. F S F P .. .. F .. F .. .. .. .. .. .. .. .. F F .. F p = 10e- S = 12e- Sulfur hexafluoride Phosphorous pentafluoride .. .. .. .. .. Resonance Structures .. .. .. .. .. Sulfuric acid S = 12e-

4. Using VSEPR Theory to Determine Molecular Shape 1) 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.

6. Hybrid Orbital Model

8. The sp Hybrid Orbitals in Gaseous BeCl2

12. The sp3 Hybrid Orbitals in NH3 and H2O

13. The sp3d Hybrid Orbitals in PCl5

14. The sp3d2 Hybrid Orbitals in SF6 Sulfur Hexafluoride -- SF6

16. Figure 10.26: Sigma and pi bonds.

17. Figure 10.27: Bonding in ethylene.

19. Figure 10.28: Bonding in acetylene.

21. Restricted Rotation of -Bonded Molecules A) Cis - 1,2 dichloroethylene B) trans - 1,2 dichloroethylene

22. Postulating the Hybrid Orbitals in a Molecule Problem: Describe how mixing of atomic orbitals on the central atoms leads to the hybrid orbitals in the following: a) Methyl amine, CH3NH2b) Xenon tetrafluoride, XeF4 Plan: 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.

23. Postulating the Hybrid Orbitals in a Molecule Problem: Describe how mixing of atomic orbitals on the central atoms leads to the hybrid orbitals in the following: a) Methyl amine, CH3NH2b) Xenon tetrafluoride, XeF4 Plan: 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 sp3hybridized. The carbon atom has four half-filled sp3 orbitals: 2s 2p sp3 Isolated Carbon Atom Hybridized Carbon Atom

24. The N atom has three half-filled sp3orbitals and one filled with a lone pair. 2s sp3 2p .. H C N H H H H

25. b) The Xenon atom has filled 5 s and 5 p orbitals with the 5 d orbitals empty. Isolated Xe atom 5 d 5 s 5 p Hybridized Xe atom: sp3d2 5 d

26. .. .. b) 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. F F F F 1800 Xe Xe F F F F Square planar

27. Fig. 9.14

28. Figure 9.15: Electronegatives of the elements.

29. The Periodic Table of the Elements 2.1 He 0.9 1.5 2.0 2.5 3.0 3.5 4.0 Ne Electronegativity 0.9 1.2 1.5 1.8 2.1 2.5 3.0 Ar 0.8 1.0 1.3 1.5 1.6 1.6 1.5 1.8 1.8 1.8 1.9 1.6 1.6 1.8 2.0 2.4 2.8 Kr 0.8 1.0 1.2 1.4 1.6 1.8 1.9 2.2 2.2 2.2 1.9 1.7 1.7 1.8 1.9 2.1 2.5 Xe 0.7 0.9 1.1 1.3 1.5 1.7 1.9 2.2 2.2 2.2 2.4 1.9 1.8 1.8 1.9 2.0 2.2 Rn 0.7 0.9 1.1 Ce Pr Nd Pm Yb Lu 1.1 1.1 1.1 1.2 1.2 1.1 1.2 1.2 1.2 1.2 1.2 1.2 1.2 1.3 1.3 1.5 1.7 1.3 1.3 1.3 1.3 1.3 1.3 1.3 1.3 1.3 1.5 Th Pa U Np No Lr

30. Fig. 9.16

31. Fig. 9.17

32. Determining Bond Polarity from Electronegativity Values Problem: (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 < 1.5

33. Fig. 9.18

34. Percent Ionic Character as a Function ofElectronegativity Difference (En) Fig. 9.19

35. The Charge Density of LiF Fig. 9.20

36. Figure 9.15: Electronegatives of the elements.

37. The Periodic Table of the Elements 2.1 He 0.9 1.5 2.0 2.5 3.0 3.5 4.0 Ne Electronegativity 0.9 1.2 1.5 1.8 2.1 2.5 3.0 Ar 0.8 1.0 1.3 1.5 1.6 1.6 1.5 1.8 1.8 1.8 1.9 1.6 1.6 1.8 2.0 2.4 2.8 Kr 0.8 1.0 1.2 1.4 1.6 1.8 1.9 2.2 2.2 2.2 1.9 1.7 1.7 1.8 1.9 2.1 2.5 Xe 0.7 0.9 1.1 1.3 1.5 1.7 1.9 2.2 2.2 2.2 2.4 1.9 1.8 1.8 1.9 2.0 2.2 Rn 0.7 0.9 1.1 Ce Pr Nd Pm Yb Lu 1.1 1.1 1.1 1.2 1.2 1.1 1.2 1.2 1.2 1.2 1.2 1.2 1.2 1.3 1.3 1.5 1.7 1.3 1.3 1.3 1.3 1.3 1.3 1.3 1.3 1.3 1.5 Th Pa U Np No Lr

38. Fig. 9.16

39. Fig. 9.17

40. Determining Bond Polarity from Electronegativity Values Problem: (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 < 1.5

41. Fig. 9.18

42. Percent Ionic Character as a Function ofElectronegativity Difference (En) Fig. 9.19

47. The Charge Density of LiF Fig. 9.20

48. That's all, folks!!