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Chapter 11

Chapter 11. Theories of Covalent Bonding. If you are doing this lecture “online” then print the lecture notes available as a word document, go through this ppt lecture, and do all the example and practice assignments for discussion time. VALENCE BOND THEORY:.

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Chapter 11

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  1. Chapter 11 Theories of Covalent Bonding If you are doing this lecture “online” then print the lecture notes available as a word document, go through this ppt lecture, and do all the example and practice assignments for discussion time.

  2. VALENCE BOND THEORY: • DEVELOPED BY LINUS PAULING, who received the Nobel Prize in 1954 for his work • A view of chemical bonding in which bonds arise from the overlap of atomic orbitals on two atoms to give a bonding orbital of electrons localized between the bonded atoms • RULE: Realize that Valence Bond Theory and all the others don't explain everything

  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 Regular atomic orbital overlap can explain these bonds. Fluorine, F2

  6. VALENCE BOND THEORY: Ha to Hb: 1sa to 1sb overlap radius = 74 pm As overlap increases, strength of bond increases - both electrons are mutually attracted to both atomic nuclei. At optimum distance between nuclei with maximum overlap, a sigma s bond (strong primary bond) forms. Max electron density is along the axis of the bond Ha to Fb: 1sa to 2pb direct overlap or s bond

  7. VALENCE BOND THEORY: F to F: the picture looks like a 2p orbital on one F is overlapping with a 2p orbital on the other F atom, but actually each F is sp3 hybridized & electrons are localized between two atomic nuclei

  8. VALENCE BOND THEORY: We cannot use this direct overlap picture for CH4’s bonding. The 2s and the three 2p orbitals on each C do not fit into the CH4 molecule's 109o bond angles, since the 2p orbitals are at 90° to each other Valence Bond Theory states that HYBRID orbitals of the outermost orbitals on an atom are formed from the atoms’ atomic orbitals

  9. 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.

  10. The sp hybrid orbitals in gaseous BeCl2. Figure 11.2 atomic orbitals on Be hybrid orbitals You have to know how to draw this energy hybrid formation. orbital box diagrams

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

  12. The sp2 hybrid orbitals in BF3. Figure 11.3 You have to know how to draw this energy hybrid formation. Note the three sigma bonds formed between B and each F.

  13. The sp3 hybrid orbitals in CH4. Figure 11.4 You have to know how to draw this energy hybrid formation.

  14. Figure 11.5 The sp3 hybrid orbitals in NH3. You have to know how to draw this energy hybrid formation.

  15. Figure 11.5 continued The sp3 hybrid orbitals in H2O. You have to know how to draw this energy hybrid formation.

  16. VALENCE BOND THEORY Expanded Valence Shells have hybrid orbitals using s, p & d atomic orbitals. Example: PCl5 P: [Ne]3s23p3 dsp3 hybridization results in 5 s bonds and trigonal bipyramidal geometry (You can write these as dsp3 or sp3d)

  17. Figure 11.6 The sp3d hybrid orbitals in PCl5. You have to know how to draw this energy hybrid formation.

  18. The sp3d2hybrid orbitals in SF6. Figure 11.7 You have to know how to draw this energy hybrid formation.

  19. 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

  20. 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: SAMPLE PROBLEM 11.1 Postulating Hybrid Orbitals in a Molecule (a) Methanol, CH3OH (b) Sulfur tetrafluoride, SF4 SOLUTION: The groups around C are arranged as a tetrahedron. (a) CH3OH O also has a tetrahedral arrangement with 2 nonbonding e- pairs.

  21. 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.

  22. VALENCE BOND THEORY There can be more than one central atom, and each has its own hybridization and geometry C2H6 and H2O2 and CH3COOH C2H6: both C's are sp3 hybridized and can rotate around axis of bond. H2O2: both O's are sp3 , etc.

  23. 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

  24. VALENCE BOND THEORY: Multiple Bonds H2CO: the Lewis structures shows a double bond between C and O, but we know it does not have twice the bond dissociation energy of a single C-O bond Pauling proposed that there was only one sigma s bond between any two atoms, and the other multiples were weaker pi p bonds If there are only 3 s bonds around this carbon, it can't be sp3 hybridized - instead we have sp2 hybrid orbitals sp2 hybridization results in only 3 s bonds, and trig planar geometry, with 120° angles p bond is a sideways or parallel overlap of the p atomic orbitals rather than the direct overlap of s bonds

  25. overlap in one position - s p overlap -  electron density The s and p bonds in ethylene (C2H4). Figure 11.10 Proper name is ethene.

  26. VALENCE BOND THEORY Look at acetylene: its geometry is linear. C is forming a triple bond to another C and a single bond to H, so that's only two s bonds Therefore sp hybridization results in only 2 s bonds, and linear geometry There are 2 p bonds from the parallel overlap of the 2p orbitals remaining on both C's

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

  28. 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

  29. H2CO hybrid orbitals and sigma and pi bond formation Remember the C=C double bond has sigman and pi bonds. The C has sigma bonds from its hybrid orbitals to the two H’s and the O. The leftover p orbitals will form the pi bond.

  30. CIS TRANS Figure 11.13 from 4th ed. Restricted rotation of p-bonded molecules in C2H2Cl2. This cis/trans arrangement will be important in chem 2, organic chem and biology!

  31. VALENCE BOND THEORY: RESONANCE Resonance Structures and p Bonding: p resonance structures involve an electron pair used alternately as a p bond or a LP Ozone: O3 O==O--O or O--O==O All are sp2, trig planar, each has 3 sp2 orbitals and a p orbital remaining.

  32. VALENCE BOND THEORY Benzene: C6H6 has carbons with sp2 hybrids and 120o angles, each C has 2 s bonds to other C's, 1 s bond to H, and 1 p bond electron available Get "ring" of delocalized p e-s SUMMARY: draw the Lewis structure; determine arrangement of electron pairs using VSEPR, specify the hybrid orbitals to accommodate the e- pairs

  33. Benzene sigma bond formation between C’s and C-Hs The leftover p orbitals will form alternating pi bonds as shown in sketch.

  34. MOLECULAR ORBITAL THEORY: - explains why H2 forms easily and He2 does not - is an alternate way of viewing e- orbitals in molecules where pure s and pure p orbitals combine to produce orbitals that are delocalized over the molecule - they can have different energies and are assigned electrons just like we do in an atom - Pauli exclusion principle and Hund's rule included Pauling's Valence Bond Theory does not explain everything MO Theory doesn't either, but it does correctly predict the electronic structure of certain molecules that do not follow Lewis's approach, including the paramagnetism of certain molecules, like O2

  35. 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. The number of molecular orbitals produced is always = # of atomic orbitals brought by the combining atoms (only orbitals on different atoms are combined). 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). The electrons of the molecule are placed in bonding or antibonding orbitals of successively higher energy (just like Hund's rule). Atomic orbitals combine most effectively with orbitals of the same type and similar energy (s w/s, n=2 w/ n=2)

  36. Amplitudes of wave functions subtracted. Figure 11.13 An analogy between light waves and atomic wave functions. Amplitudes of wave functions added

  37. 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.

  38. MOLECULAR ORBITAL THEORY BOND ORDER: the number of bonding e- pairs shared by 2 atoms in a molecule Fractional bond orders are possible in MO Theory! Silberberg method: B.O. = ½(# of e- in bonding orbitals - # of e- in antibonding orbitals)

  39. 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

  40. 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

  41. PROBLEM: Use MO diagrams to predict whether H2+ and H2- exist. Determine their bond orders and electron configurations. s s 1s 1s 1s 1s AO of H- AO of H AO of H AO of H s s 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 MO of H2- MO of H2+

  42. s*2s s*2s 2s 2s 2s 2s s2s s2s Figure 11.17 Bonding in s-block homonuclear diatomic molecules. Energy Be2 Li2 Be2 bond order = 0 Li2 bond order = 1

  43. Contours and energies of s and p MOs through combinations of 2p atomic orbitals. Figure 11.18 Or the pz orbitals

  44. Figure 11.19 Relative MO energy levels for Period 2 homonuclear diatomic molecules. with 2s-2p mixing without 2s-2p mixing Memorize this! MO energy levels for O2, F2, and Ne2 MO energy levels for B2, C2, and N2

  45. Figure 11.20 MO occupancy and molecular properties for B2 through Ne2

  46. Figure 11.21 The paramagnetic properties of O2

  47. 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: 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.

  48. 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

  49. MO Theory Practice 1. Draw the bonding and antibonding molecular orbitals for H2. 2. Do Valence Bond Theory (hybridization) and MO Theory for both O2 and O22-. Which theory works better to explain the molecule and ion? 3. For N2, N2+ and N2- compare a. Magnetic character b. Net number of p bonds c. Bond Order d. Bond length e. Bond strength

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