Chapter 11

1. Chapter 11 Theories of Covalent Bonding

2. Theories of Covalent Bonding 11.1 Valence Bond (VB) Theory and Orbital Hybridization 11.2 The Mode of Orbital Overlap and the Types of Covalent Bonds 11.3 Molecular Orbital (MO)Theory and Electron Delocalization

3. The Central Themes of VB Theory Basic Principle A covalent bond forms when the orbitals of two atoms overlap and the overlap region, which is between the nuclei, is occupied by a pair of electrons. The two wave functions are in phase so the amplitude increases between the nuclei.

4. The Central Themes of VB Theory Themes A set of overlapping orbitals has a maximum of two electrons that must have opposite spins. The greater the orbital overlap, the stronger (more stable) the bond. The valence atomic orbitals in a molecule are different from those in isolated atoms. There is a hybridization of atomic orbitals to form molecular orbitals.

5. Orbital overlap and spin pairing in three diatomic molecules. Figure 11.1 Hydrogen, H2 Hydrogen fluoride, HF Fluorine, F2

6. Key Points Types of Hybrid Orbitals sp sp2 sp3 sp3d sp3d2 Hybrid Orbitals The number of hybrid orbitals obtained equals the number of atomic orbitals mixed. The type of hybrid orbitals obtained varies with the types of atomic orbitals mixed.

7. The sp hybrid orbitals in gaseous BeCl2. Figure 11.2 atomic orbitals hybrid orbitals orbital box diagrams

8. The sp hybrid orbitals in gaseous BeCl2(continued). Figure 11.2 orbital box diagrams with orbital contours

9. The sp2 hybrid orbitals in BF3. Figure 11.3

10. The sp3 hybrid orbitals in CH4. Figure 11.4

11. Figure 11.5 The sp3 hybrid orbitals in NH3.

12. Figure 11.5 continued The sp3 hybrid orbitals in H2O.

13. Figure 11.6 The sp3d hybrid orbitals in PCl5.

14. The sp3d2hybrid orbitals in SF6. Figure 11.7

16. Step 1 Step 2 Step 3 Figure 10.1 Figure 10.12 Table 11.1 Figure 11.8 The conceptual steps from molecular formula to the hybrid orbitals used in bonding. Molecular shape and e- group arrangement Molecular formula Lewis structure Hybrid orbitals

17. PROBLEM: Use partial orbital diagrams to describe mixing of the atomic orbitals of the central atom leads to hybrid orbitals in each of the following: PLAN: Use the Lewis structures to ascertain the arrangement of groups and shape of each molecule. Postulate the hybrid orbitals. Use partial orbital box diagrams to indicate the hybrid for the central atoms. SAMPLE PROBLEM 11.1 Postulating Hybrid Orbitals in a Molecule (a) Methanol, CH3OH (b) Sulfur tetrafluoride, SF4 SOLUTION: (a) CH3OH The groups around C are arranged as a tetrahedron. O also has a tetrahedral arrangement with 2 nonbonding e- pairs.

18. hybridized C atom hybridized O atom single C atom single O atom hybridized S atom S atom SAMPLE PROBLEM 11.1 Postulating Hybrid Orbitals in a Molecule continued (b) SF4 has a seesaw shape with 4 bonding and 1 nonbonding e- pairs.

19. both C are sp3 hybridized s-sp3 overlaps to s bonds sp3-sp3 overlap to form a s bond relatively even distribution of electron density over all s bonds The s bonds in ethane(C2H6). Figure 11.9

20. overlap in one position - s p overlap -  electron density The s and p bonds in ethylene (C2H4). Figure 11.10

21. overlap in one position - s p overlap -  The s and p bonds in acetylene (C2H2). Figure 11.11

22. Electron density and bond order. Figure 11.12

23. PROBLEM: Describe the types of bonds and orbitals in acetone, (CH3)2CO. PLAN: Use the Lewis structures to ascertain the arrangement of groups and shape at each central atom. Postulate the hybrid orbitals taking note of the multiple bonds and their orbital overlaps. sp3 hybridized sp3 hybridized sp2 hybridized SAMPLE PROBLEM 11.2 Describing the Bond in Molecules SOLUTION: bond bonds

24. CIS TRANS Figure 11.13 Restricted rotation of p-bonded molecules in C2H2Cl2.

25. The Central Themes of MO Theory A molecule is viewed on a quantum mechanical level as a collection of nuclei surrounded by delocalized molecular orbitals. Atomic wave functions are summed to obtain molecular wave functions. If wave functions reinforce each other, a bonding MO is formed (region of high electron density exists between the nuclei). If wave functions cancel each other, an antibonding MO is formed (a node of zero electron density occurs between the nuclei).

26. Amplitudes of wave functions subtracted. Figure 11.14 An analogy between light waves and atomic wave functions. Amplitudes of wave functions added

27. Figure 11.14 Contours and energies of the bonding and antibonding molecular orbitals (MOs) in H2. The bonding MO is lower in energy and the antibonding MO is higher in energy than the AOs that combined to form them.

28. 1s 1s AO of H AO of H Figure 11.15 The MO diagram for H2. Filling molecular orbitals with electrons follows the same concept as filling atomic orbitals. s*1s Energy H2 bond order = 1/2(2-0) = 1 s1s MO of H2

29. s*1s Energy 1s 1s 1s 1s s1s AO of He AO of He+ AO of He AO of He Figure 11.16 MO diagram for He2+ and He2. s*1s Energy s1s MO of He+ MO of He2 He2 bond order = 0 He2+ bond order = 1/2

30. PROBLEM: Use MO diagrams to predict whether H2+ and H2- exist. Determine their bond orders and electron configurations. PLAN: Use H2 as a model and accommodate the number of electrons in bonding and antibonding orbitals. Find the bond order. s s 1s 1s 1s 1s AO of H- AO of H AO of H AO of H s s configuration is (s1s)1 SAMPLE PROBLEM 11.3 Predicting Stability of Species Using MO Diagrams SOLUTION: bond order = 1/2(1-0) = 1/2 bond order = 1/2(2-1) = 1/2 H2+ does exist H2- does exist configuration is (s1s)2(s2s)1 MO of H2- MO of H2+

31. s*2s s*2s 2s 2s 2s 2s s2s s2s Figure 11.18 Bonding in s-block homonuclear diatomic molecules. Energy Be2 Li2 Be2 bond order = 0 Li2 bond order = 1

32. Contours and energies of s and p MOs through combinations of 2p atomic orbitals. Figure 11.19

33. Figure 11.20 Relative MO energy levels for Period 2 homonuclear diatomic molecules. without 2s-2p mixing with 2s-2p mixing MO energy levels for O2, F2, and Ne2 MO energy levels for B2, C2, and N2

34. Figure 11.21 MO occupancy and molecular properties for B2 through Ne2

35. Figure 11.22 The paramagnetic properties of O2

36. PROBLEM: As the following data show, removing an electron from N2 forms an ion with a weaker, longer bond than in the parent molecules, whereas the ion formed from O2 has a stronger, shorter bond: PLAN: Find the number of valence electrons for each species, draw the MO diagrams, calculate bond orders, and then compare the results. N2 N2+ O2 O2+ Bond energy (kJ/mol) 945 841 498 623 Bond length (pm) 110 112 121 112 SAMPLE PROBLEM 11.4 Using MO Theory to Explain Bond Properties Explain these facts with diagrams that show the sequence and occupancy of MOs. SOLUTION: N2 has 10 valence electrons, so N2+ has 9. O2 has 12 valence electrons, so O2+ has 11.

37. 2p 2p 2p 2p s2s s2s SAMPLE PROBLEM 11.4 Using MO Theory to Explain Bond Properties continued N2 N2+ O2 O2+ 2p antibonding e- lost bonding e- lost 2p 2p 2p s2s s2s bond orders 1/2(8-2)=3 1/2(7-2)=2.5 1/2(8-4)=2 1/2(8-3)=2.5

38. s 1s 2px 2py 2p s AO of H AO of F The MO diagram for HF Figure 11.23 Energy MO of HF

39. s*s *p 2p 2p sp p s*s 2s 2s AO of N AO of O ss The MO diagram for NO Figure 11.24 Energy possible Lewis structures MO of NO

40. benzene, C6H6 Figure 11.25 The lowest energy p-bonding MOs in benzene and ozone. ozone,O3

41. Figure 10.1 The steps in converting a molecular formula into a Lewis structure. Place atom with lowest EN in center Molecular formula Step 1 Atom placement Add A-group numbers Step 2 Sum of valence e- Draw single bonds. Subtract 2e- for each bond. Step 3 Give each atom 8e- (2e- for H) Remaining valence e- Step 4 Lewis structure

42. Figure 10.12 The steps in determining a molecular shape. See Figure 10.1 Molecular formula Step 1 Lewis structure Count all e- groups around central atom (A) Step 2 Electron-group arrangement Note lone pairs and double bonds Step 3 Count bonding and nonbonding e- groups separately. Bond angles Step 4 Molecular shape (AXmEn)