1. Bonding: General Concepts

2. Copyright © Houghton Mifflin Company. All rights reserved. 13a–2 A computer representation of K 3 C 60, a superconducting substance formed by reacting potassium with buckminster fullerine (C 60 ) Source: Photo Researchers, Inc.

3. Copyright © Houghton Mifflin Company. All rights reserved. 13a–3 Two forms of carbon; graphite and diamond. Source: Grant Hellman

4. Copyright © Houghton Mifflin Company. All rights reserved. 13a–4 Quartz grows in beautiful, regular crystals SiO 2 Vs CO 2

5. Copyright © Houghton Mifflin Company. All rights reserved. 13a–5 Figure 13.1: (a) The interaction of two hydrogen atoms (b) Energy profile as a function of the distance between the nuclei of the hydrogen atoms.

6. Copyright © Houghton Mifflin Company. All rights reserved. 13a–6 Figure 13.1: (a) The interaction of two hydrogen atoms (b) Energy profile as a function of the distance between the nuclei of the hydrogen atoms.

7. Copyright © Houghton Mifflin Company. All rights reserved. 13a–7 WHERE DO THE ELECTRONS GO? Are they shared equally? Are they more on one atom than the other?

8. Copyright © Houghton Mifflin Company. All rights reserved. 13a–8 WHERE DO THE ELECTRONS GO? Are they shared equally? Are they more on one atom than the other?

9. Copyright © Houghton Mifflin Company. All rights reserved. 13a–9 WHERE DO THE ELECTRONS GO? Are they shared equally? Are they more on one atom than the other? ANSWER: It depends who is pulling harder

10. Copyright © Houghton Mifflin Company. All rights reserved. 13a–10 WHERE DO THE ELECTRONS GO? Are they shared equally? Are they more on one atom than the other? ANSWER: It depends who is pulling harder (“Electro negativity”)

11. Copyright © Houghton Mifflin Company. All rights reserved. 13a–11

12. Copyright © Houghton Mifflin Company. All rights reserved. 13a–12 The HCL molecule has a dipole moment

13. Copyright © Houghton Mifflin Company. All rights reserved. 13a–13 Pauling and his electronegativity

14. Copyright © Houghton Mifflin Company. All rights reserved. 13a–14 Pauling and his electronegativity

15. Copyright © Houghton Mifflin Company. All rights reserved. 13a–15 Figure 13.3: The Pauling electronegativity values as updated by A.L. Allred in 1961. Expect A-B bond energy to be average of A-A and B-B If it is not electro negativities must be different (Stronger bond – more ionic – larger EN difference)

16. Copyright © Houghton Mifflin Company. All rights reserved. 13a–16 Figure 13.3: The Pauling electronegativity values as updated by A.L. Allred in 1961. (cont’d) Arbitrarily set F as 4

17. Copyright © Houghton Mifflin Company. All rights reserved. 13a–17

18. Copyright © Houghton Mifflin Company. All rights reserved. 13a–18

19. Copyright © Houghton Mifflin Company. All rights reserved. 13a–19 The HCL molecule has a dipole moment

20. Copyright © Houghton Mifflin Company. All rights reserved. 13a–20

21. Copyright © Houghton Mifflin Company. All rights reserved. 13a–21

22. Copyright © Houghton Mifflin Company. All rights reserved. 13a–22

23. Copyright © Houghton Mifflin Company. All rights reserved. 13a–23

24. Copyright © Houghton Mifflin Company. All rights reserved. 13a–24 models Now lets consider more than two atoms in a molecule

25. Copyright © Houghton Mifflin Company. All rights reserved. 13a–25 Linear molecules

26. Copyright © Houghton Mifflin Company. All rights reserved. 13a–26 Planar molecules

27. Copyright © Houghton Mifflin Company. All rights reserved. 13a–27 Tetrahedral molecules

28. Copyright © Houghton Mifflin Company. All rights reserved. 13a–28 MODELS These are only models

29. Copyright © Houghton Mifflin Company. All rights reserved. 13a–29 MODELS These are only models But…. Models are very useful for describing properties.

30. Copyright © Houghton Mifflin Company. All rights reserved. 13a–30 MODELS These are only models But…. Models are very useful for describing properties. Newton: particles

31. Copyright © Houghton Mifflin Company. All rights reserved. 13a–31 MODELS These are only models But…. Models are very useful for describing properties. Newton: particles Huygens: waves

32. Copyright © Houghton Mifflin Company. All rights reserved. 13a–32 models

33. Copyright © Houghton Mifflin Company. All rights reserved. 13a–33 Ball-and-stick model of a protein segment illustrating the alpha helix. Source: Photo Researchers, Inc.

34. Copyright © Houghton Mifflin Company. All rights reserved. 13a–34 The concept of individual bonds makes it much easier to deal with complex molecules such as DNA. Source: Photo Researchers, Inc.

35. Copyright © Houghton Mifflin Company. All rights reserved. 13a–35

36. Copyright © Houghton Mifflin Company. All rights reserved. 13a–36 Polarity of Molecules Dipole Moments of Polyatomic Molecules Example: in CO 2, each C-O dipole is canceled because the molecule is linear. In H 2 O, the H-O dipoles do not cancel because the molecule is bent. OCO

37. Copyright © Houghton Mifflin Company. All rights reserved. 13a–37 Skeletal Structure Hydrogen atoms are always terminal atoms. Central atoms are generally those with the lowest electronegativity. Carbon atoms are always central atoms. Generally structures are compact and symmetrical.

38. Copyright © Houghton Mifflin Company. All rights reserved. 13a–38 Skeletal Structure Identify central and terminal atoms in the molecule C 2 H 6 O (ethyl alcohol or ethanol). C H H H H H H O C NOTE: Terminal atoms are all bonded to only one other atoms. Central atoms are bonded to two or more other atoms Now where do the electrons go?

39. Copyright © Houghton Mifflin Company. All rights reserved. 13a–39 Writing Lewis Structures All the valence e - of atoms must appear. Usually, the e - are paired. Usually, each atom requires an octet. –H only requires 2 e -. Multiple bonds may be needed. –Readily formed by C, N, O, S, and P.

40. Copyright © Houghton Mifflin Company. All rights reserved. Lewis Structures Draw Lewis structures for: HF: H 2 O: NH 3 : CH 4 : H F  or H F  H O H  or H O H  H N H H  or H N H H  H C H H H  or H C H H H

41. Copyright © Houghton Mifflin Company. All rights reserved. Lewis Structures Draw Lewis structures for: HF: H 2 O: NH 3 : CH 4 : H F  or H F  H O H  or H O H  H N H H  or H N H H  H C H H H  or H C H H H

42. Copyright © Houghton Mifflin Company. All rights reserved. Exceptions to the Octet Rule Molecules with an odd number of electrons. Molecules in which an atom has less than an octet of electrons. Molecules in which an atom has more than an octet of electrons.

43. Copyright © Houghton Mifflin Company. All rights reserved. Resonance Forms Lewis structures that differ only in the placement of electrons are resonance forms. For O 3 : Experimentally, it is found that both bonds are 0.128 nm long. The Lewis structure of O 3 must show both resonance forms. O O O   O O O  = =

44. Copyright © Houghton Mifflin Company. All rights reserved. 13a–44 Resonance: Delocalized Electron-Pair Bonding Resonance structures have the same relative atom placement but a difference in the locations of bonding and nonbonding electron pairs. O 3 can be drawn in 2 ways - Neither structure is actually correct but can be drawn to represent a structure which is a hybrid of the two - a resonance structure. Note: Resonance structure exists full time: it is not an equilibrium mixture of interconverting contributing structures. E.g. Mule is not a horse 50% of the time and a donkey 50% of the time. Curved line is not a solid line.

45. Copyright © Houghton Mifflin Company. All rights reserved. 13a–45 SAMPLE PROBLEM 10.4Writing Resonance Structures SOLUTION: PROBLEM:Write resonance structures for the nitrate ion, NO 3 -. Nitrate has 1(5) + 3(6) + 1 = 24 valence e - N does not have an octet; a pair of e - will move in to form a double bond.

46. Copyright © Houghton Mifflin Company. All rights reserved. 13a–46 Draw the resonance hybrid structure of NO 3 - The three resonance structures Hybrid structure. Averaging leads to partial N-O double bond character (solid + dotted lines)

47. Copyright © Houghton Mifflin Company. All rights reserved. Resonance Forms C2H6OC2H6O 20 ve’s Ethyl alcohol Methyl ether H C C O H H H H H H C O C H H H H H   

48. Copyright © Houghton Mifflin Company. All rights reserved. 13a–48 Benzene Fredrich August von Kekule (German chemist) said that he discovered the ring-shaped chemical structure of benzene because of a strange, reptilian dream he had in 1865: "I turned my chair to the fire and dozed. Again the atoms were gamboling before my eyes.... My mental eye... could not distinguish larger structures, of manifold conformation; long rows, sometimes more closely fitted together; all twining and twisting in snakelike motion. But look! What was that? One of the snakes had seized hold of its own tail, and the form whirled mockingly before my eyes. As if by a flash of lighting I awoke... " ( From "Creativity, Beyond the Myth of Genius" by Robert Weisberg published by W. H. Freeman 1992.) Although some scholars now believe that Kekule's dream was a hoax to avoid sharing credit for the discovery of the hexagonal shape of benzene, it still makes a wonderful story.

49. Copyright © Houghton Mifflin Company. All rights reserved. Resonance Forms C6H6C6H6 30 ve’s C C C H H H H H H C C H H H H H H 

50. Copyright © Houghton Mifflin Company. All rights reserved. Nonequivalent Forms The molecule N 2 O has three resonance forms: Which one contributes most to the overall electronic structure? N N O   N N O ==  N N O  

51. Copyright © Houghton Mifflin Company. All rights reserved. 13a–51 Formal Charge: Selecting the Best Resonance Structure An atom “owns” all of its nonbonding electrons and half of its bonding electrons. Formal charge of atom = # valence e - - (# unshared electrons + 1/2 # shared electrons) For O A # valence e - = 6 # nonbonding e - = 4 # bonding e - = 4 X 1/2 = 2 Formal charge = 0 For O B # valence e - = 6 # nonbonding e - = 2 # bonding e - = 6 X 1/2 = 3 Formal charge = +1 For O C # valence e - = 6 # nonbonding e - = 6 # bonding e - = 2 X 1/2 = 1 Formal charge = -1

52. Copyright © Houghton Mifflin Company. All rights reserved. Formal Charges The formal charge on an atom assumes that the bonding electrons are shared equally. Formal charge = # valence electrons - # nonbonding electrons - ½ # bonding electrons Formal charges in N 2 O: N N O   N N O ==  N N O   -2 +1 +10 0

53. Copyright © Houghton Mifflin Company. All rights reserved. Formal Charges The most favored resonance form is the one that has: The smallest formal charges. No formal charges with the same sign on adjacent atoms. Negative formal charges on the most electronegative atoms.

54. Copyright © Houghton Mifflin Company. All rights reserved. Formal Charges The formal charge on an atom assumes that the bonding electrons are shared equally. Formal charge = # valence electrons - # nonbonding electrons - ½ # bonding electrons Formal charges in N 2 O: N N O   N N O ==  N N O   -2 +1 +10 0 Most favorable

55. Copyright © Houghton Mifflin Company. All rights reserved. 13a–55 EXAMPLE: NCO - has 3 possible resonance forms - Resonance (continued) ABC formal charges -20+10000 Forms B and C have negative formal charges on N and O; this makes them more important than form A. Form C has a negative charge on O which is the more electronegative element, therefore C contributes the most to the resonance hybrid.

56. Copyright © Houghton Mifflin Company. All rights reserved. Electron-Deficient Atoms Molecules containing Be and B can have less than an octet for these atoms. BeF 2 : The small Be atom can’t accomodate a large formal charge. F Be F   Most Favorable F Be F  = = 0 0 0 +1 -2 +1

57. Copyright © Houghton Mifflin Company. All rights reserved. Electron-Deficient Atoms BF 3 :  F B F F  0 0 0 0 F B F F  =  F B F F  = F B F F  +1 0 0 Most Favorable

58. Copyright © Houghton Mifflin Company. All rights reserved. Electron-Rich Atoms An atom that has a d subshell in the valence electron shell can accomodate more than an octet of electrons. Nitrogen can’t accommodate more than 8 electrons. 1s 2s 2p 3s 3p 4s 3d 0 E     N in NF 3

59. Copyright © Houghton Mifflin Company. All rights reserved. Electron-Rich Atoms An atom that has a d subshell in the valence electron shell can accomodate more than an octet of electrons. P can accommodate more than 8 electrons 1s 2s 2p 3s 3p 4s 3d 0 E   P in PF 5

60. Copyright © Houghton Mifflin Company. All rights reserved. Electron-Rich Atoms PF 5 P F F F FF 

61. Copyright © Houghton Mifflin Company. All rights reserved. 13a–61 Exceptions to the Octet Rule Expanded octets. P Cl P Cl Cl Cl Cl S F F F F F F Cl

62. Copyright © Houghton Mifflin Company. All rights reserved. 13a–62 Molecular Shapes Lewis structures give atomic connectivity: they tell us which atoms are physically connected to which. The shape of a molecule is determined by its bond angles.

63. Copyright © Houghton Mifflin Company. All rights reserved. 13a–63 Molecular Shapes In order to predict molecular shape, we assume the valence electrons repel each other. Therefore, the molecule adopts whichever 3D geometry minimizes this repulsion. We call this process Valence Shell Electron Pair Repulsion (VSEPR) theory.

64. Copyright © Houghton Mifflin Company. All rights reserved. 13a–64 The VSEPR Model Predicting Molecular Geometries

65. Copyright © Houghton Mifflin Company. All rights reserved. 13a–65 The VSEPR Model Predicting Molecular Geometries

66. Copyright © Houghton Mifflin Company. All rights reserved. 13a–66 Molecular Shapes experimentally we find all Cl-C-Cl bond angles are 109.5 . Therefore, the molecule cannot be planar. All Cl atoms are located at the vertices of a tetrahedron with the C at its center.

67. Copyright © Houghton Mifflin Company. All rights reserved. 13a–67 The VSEPR Model Predicting Molecular Geometries

68. Copyright © Houghton Mifflin Company. All rights reserved. 13a–68 The VSEPR Model Predicting Molecular Geometries To determine the electron pair geometry: draw the Lewis structure count the total number of electron pairs around the central atom arrange the electron pairs in one of the above geometries to minimize e  -e  repulsion multiple bounds count as one bonding pair

69. Copyright © Houghton Mifflin Company. All rights reserved. 13a–69 The VSEPR Model Predicting Molecular Geometries

70. Copyright © Houghton Mifflin Company. All rights reserved. 13a–70 The VSEPR Model Predicting Molecular Geometries

71. Copyright © Houghton Mifflin Company. All rights reserved. 13a–71 Molecular Shapes AB 2 Linear AB 3 Trigonal planar AB 4 Tetrahedral AB 5 Trigonal bipyramidal AB 6 Octahedral AB 2 E Angular or Bent AB 3 E Trigonal pyramidal AB 2 E 2 Angular or Bent AB 4 E Irregular tetrahedral (see saw) AB 3 E 2 T-shaped AB 2 E 3 Linear AB 5 E Square pyramidal AB 4 E 2 Square planar

72. Copyright © Houghton Mifflin Company. All rights reserved. 13a–72 The VSEPR Model The valence electrons in a molecule are the bonding pairs of electrons as well as the lone pairs. There are 11 shapes that are important to us: Number of atoms, formula Shapes (3 atoms, AB 2 ) linear or bent (4 atoms, AB 3 ) trigonal planar, trigonal bipyramidal, or T-shaped (5 atoms, AB 4 ) tetrahedral, square planar, or see-saw (6 atoms, AB 5 ) trigonal bipyramidal or square pyramidal (7 atoms, AB 6 ) octahedral

73. Copyright © Houghton Mifflin Company. All rights reserved. 13a–73 The VSEPR Model

74. Copyright © Houghton Mifflin Company. All rights reserved. 13a–74 The VSEPR Model The Effect of Nonbonding Electrons and Multiple Bonds on Bond Angles By experiment, the H-X-H bond angle decreases on moving from C to N to O: Since electrons in a bond are attracted by two nuclei, they do not repel as much as lone pairs. Therefore, the bond angle decreases as the number of lone pairs increase.

75. Copyright © Houghton Mifflin Company. All rights reserved. 13a–75 The VSEPR Model The Effect of Nonbonding Electrons and Multiple Bonds on Bond Angles Similarly, electrons in multiple bonds repel more than electrons in single bonds.

76. Copyright © Houghton Mifflin Company. All rights reserved. 13a–76 The VSEPR Model Molecules with Expanded Valence Shells

77. Copyright © Houghton Mifflin Company. All rights reserved. 13a–77 The VSEPR Model Molecules with Expanded Valence Shells

78. Copyright © Houghton Mifflin Company. All rights reserved. 13a–78 The VSEPR Model Molecules with More than One Central Atom In acetic acid, CH 3 COOH, there are three central atoms. We assign the geometry about each central atom separately. Number of electron domains Electron-domain geometry Predicted bond angles Tetrahedral Trigonal planar Tetrahedral 109.5 o 120 o 109.5 o COHH H HO 434 C

79. Copyright © Houghton Mifflin Company. All rights reserved. 13a–79 The VSEPR Model Molecules with More than One Central Atom In acetic acid, CH 3 COOH, there are three central atoms. We assign the geometry about each central atom separately.

80. Copyright © Houghton Mifflin Company. All rights reserved. 13a–80 Polarity of Molecules If two charges, equal in magnitude and opposite in sign, are separated by a distance r, then a dipole is established. The dipole moment, , is given by  = Qr where Q is the magnitude of charge. Dipole Moments of Polyatomic Molecules In a polyatomic molecule, each bond is treated as a dipole. The orientation of these individual dipole moments determines whether the molecule has an overall dipole moment.

81. Copyright © Houghton Mifflin Company. All rights reserved. 13a–81 Dipole Moment Nonpolar Polar.. HH O C OO Bond dipoles Overall dipole moment = 0 Bond dipoles Overall dipole moment The overall dipole moment of a molecule is the sum of its bond dipoles. In CO 2 the bond dipoles are equal in magnitude but exactly opposite each other. The overall dipole moment is zero. In H 2 O the bond dipoles are also equal in magnitude but do not exactly oppose each other. The molecule has a nonzero overall dipole moment. Coulomb’s law  = Q r Dipole moment, 

82. Copyright © Houghton Mifflin Company. All rights reserved. 13a–82 Figure 13.6: The carbon dioxide molecule Nonpolar Molecules –Dipole moments are symmetrical and cancel out.

83. Copyright © Houghton Mifflin Company. All rights reserved. 13a–83 Determining Molecular Polarity Nonpolar Molecules –Dipole moments are symmetrical and cancel out. BF 3 F F F B Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

84. Copyright © Houghton Mifflin Company. All rights reserved. 13a–84 Determining Molecular Polarity Polar Molecules –Dipole moments are asymmetrical and don’t cancel. net dipole moment H2OH2O H H O Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

85. Copyright © Houghton Mifflin Company. All rights reserved. 13a–85 CHCl 3 H Cl Determining Molecular Polarity polar molecules have... –asymmetrical shape (lone pairs) or –asymmetrical atoms net dipole moment Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

86. Copyright © Houghton Mifflin Company. All rights reserved. 13a–86 Figure 13.5: (a) The structure and charge distribution of the ammonia molecule. (b) The dipole moment of the ammonia molecule oriented in an electric field.

87. Copyright © Houghton Mifflin Company. All rights reserved. 13a–87

88. Copyright © Houghton Mifflin Company. All rights reserved. 13a–88 Sulfur has a partial positive charge

89. Copyright © Houghton Mifflin Company. All rights reserved. 13a–89 Hydrogen atoms have a partial positive charge

90. Copyright © Houghton Mifflin Company. All rights reserved. 13a–90 Hydrogen atoms and a small partial negative charge on the carbon

91. Copyright © Houghton Mifflin Company. All rights reserved. 13a–91.. Polar Bonds H Cl Polar A molecule has a zero dipole moment when bond dipoles cancel one another. HH O Polar FF B F Nonpolar H H H N Polar Nonpolar FF Cl F F F Xe FF Cl C Nonpolar Polar Cl H C H H

92. Copyright © Houghton Mifflin Company. All rights reserved. Covalent Bond Energy Covalent bond energy is measured by the energy required to break the bond. The bond enthalpy, D(X-Y) is the average  H for breaking one mole of X-Y bonds in the gas phase: D(C-O) =  H = 358 kJ When one mole of X-Y bonds is formed, the enthalpy change is  D(X-Y).  C O  C + O  

93. Copyright © Houghton Mifflin Company. All rights reserved. Bond Enthalpies and Bond Lengths As bond order increases, the bond enthalpy increases and the bond length decreases. D(C  C) = 348 kJ 0.154 nm D(C=C) = 614 kJ 0.134 nm D(C  C) = 839 kJ 0.120 nm D(C  O) = 358 kJ 0.143 nm D(C=O) = 799 kJ 0.123 nm D(C  O) = 1072 kJ 0.113 nm

94. Copyright © Houghton Mifflin Company. All rights reserved. 13a–94

95. Copyright © Houghton Mifflin Company. All rights reserved. Bond Enthalpies and  H rxn Consider the reaction of H 2 and O 2 to form H 2 O:  H for breaking bonds = 2 D(H–H) + D(O=O)  H for forming bonds = 4  -D(O–H)  H rxn = 2 D(H–H) + D(O=O) - 4 D(O–H)  H =  D(bonds broken) -  D(bonds formed) H H + H H + O O =  H O H + H O H  

96. Copyright © Houghton Mifflin Company. All rights reserved. Bond Enthalpies and  H rxn Estimate  H for the combustion of CH 4 :  H = 4 D(C–H) + 2 D(O=O) - 2 D(C=O)- 4 D(O–H) = [ 4(413) + 2(495) - 2(799)- 4(463) ] kJ = -808 kJ +  + H C H H H 2 O O =  O C O ==  2 H O H 

97. Copyright © Houghton Mifflin Company. All rights reserved. 13a–97 Comparing fuels Natural gas: CH 4 + 2O 2 → CO 2 + 2 H 2 O ΔH=-808 kJ/mol Coal: C + O 2 → CO 2 ΔH=-393.5 kJ/mol Oil: C 20 H 42 + 30½O 2 → 20CO 2 + 21 H 2 O ΔH=-13315 kJ/mol ΔH=-666 kJ/mol.CO 2

98. Copyright © Houghton Mifflin Company. All rights reserved. 13a–98 Comparing fuels Production of 1 GigaJoule of energy releases: Natural gas: 10 9 J x 0.044 kg/mol ÷ 808,000 J/mol = 54.5 kg CO 2 Coal: 112 kg CO 2 Oil: 66 kg CO 2

99. Copyright © Houghton Mifflin Company. All rights reserved. 13a–99 N 3 high energy density

100. Copyright © Houghton Mifflin Company. All rights reserved. 13a–100 ATP energy

101. Copyright © Houghton Mifflin Company. All rights reserved. 13a–101 ATP energy Repulsion weakens these bonds Resonance!