1. Chapter 10 Chemical Bonding II

2. Lewis Structure  Molecular Structure Structure determines chemical properties

3. Electron domain/group: area where electrons appear in Lewis structures. It can be electron lone pairs, single bonds, double bonds, triple bonds, or single electrons. H 2 O, NH 3, CH 4, O 2, N 2, SCl 2, CCl 4, PCl 3, NO +, NH 4 +, CO, CO 2

4. Valence Shell Electron Pair Repulsion (VSEPR) model The lowest energy arrangement of a given number of electron domains is the one that minimizes the repulsions among them. The shape of AB n molecules or ions depend on the number of electron domains surrounding the central A atom. number of electron domains: 2 to 6

5. Know how to spell the names!

6. How to predict geometry of a molecule? 1)Draw the Lewis structure of the molecule or ion, and count the number of electron domains around the central atom. 2)Determine the electron domain arrangement by arranging the electron domains about the central atom so that the repulsions among them are minimized. 3) Use the arrangement of the bonded atoms to determine the molecular geometry. CO 2

8. BF 3, NO 3 −, H 2 CO

9. Electron domains for multiple bonds exert a greater repulsion force on adjacent electron domains than do electron domains for single bonds. lone pair-lone pair > lone pair-bonding pair > bonding pair- bonding pair

10. SO 2 119° Electron domain arrangement is not necessarily the same as the molecular structure. Bent or V-shaped

11. CH 4

12. NH 3

14. H2OH2O

16. PCl 5

17. SF 4 To minimize repulsion, electron lone pairs are always placed in equatorial positions for trigonal bipyramidal geometry.

18. BrF 3

19. XeF 2

20. SF 6

21. BrF 5

22. XeF 4

24. Polarity of a molecule

25. How to quantify the polarity of a bond? Dipole moment

26. +− Dipole Dipole has a magnitude and a direction — vector Magnitude (length) of a dipole — dipole moment μ = qr q — charge, r — distance between + and − charge

27. H — F dipole dipole moment of a bond ≠ 0 ↔ polar bond dipole moment of a bond = 0 ↔ nonpolar bond

28. The Pauling Electronegativity Values

29. Polarity of a molecule dipole moment of a molecule ≠ 0 ↔ polar molecule dipole moment of a molecule = 0 ↔ nonpolar molecule dipole of a molecule = sum of all the bond dipoles

30. v1v1 v2v2 v = v 1 + v 2 v 3 = v 2

34. SO 3

35. 120° = = Net dipole moment = 0, nonpolar molecule SO 3

36. 109.5° CCl 4 = = Net dipole moment = 0, nonpolar molecule

37. Polarity of a molecule depends on the polarity of its bonds AND the geometry of the molecule.

38. NH 3

40. How are electrons shared in covalent bonds? Valence Bond Theory Molecular Orbital Theory

41. Valence Bond Theory: Orbital Overlap as a Chemical Bond

44. CH 4

48. 4 electron domains sp 3 hybridization tetrahedral arrangement

49. NH 3

50. sp 3 on O H2OH2O

51. C = C Ethylene H H H H ~120° All atoms are in the same plane

55. C = C Ethylene H H H H ~120° All atoms are in the same plane

56. The σ Bonds in Ethylene

57. C = C Ethylene H H H H ~120° All atoms are in the same plane

58. A Carbon-Carbon Double Bond Consists of a σ and a π Bond

61. 3 electron domains sp 2 hybridization trigonal planar arrangement

62. H−C ≡ C−H Acetylene Linear molecule

67. N2N2 :N≡N:

68. 2 electron domains sp hybridization Linear arrangement

69. Single bond: σ double bond: one σ, one π triple bond: one σ, two π

70. C H H CH = O OC ≡ N: :: ¨ ¨

71. PCl 5

73. The Orbitals Used to Form the Bonds in PCl 5

74. 5 electron domains dsp 3 hybridization trigonal bipyramidal arrangement

75. SF 6

77. An Octahedral Set of d 2 sp 3 Orbitals on Sulfur Atom

78. 6 electron domains d 2 sp 3 hybridization octahedral arrangement

80. paramagnetismdiamagnetism paired electronsunpaired electrons

81. Liquid O 2 is paramagnetic

82. How are electrons shared in covalent bonds? Valence Bond Theory Molecular Orbital Theory

83. Atoms → atomic orbitals Molecules → molecular orbitals Molecular Orbital (MO) ≈ Linear Combination of Atomic Orbitals (LCAO)

84. H2H2

87. Electron configuration (σ 1s ) 2 Pauli principle and Hund’s rule apply

88. Bond order = ½ (number of bonding electrons − number of antibonding electrons) If bond order > 0, the molecule is stable If bond order = 0, the molecule is not stable bond order = 1 → single bond bond order = 2 → double bond bond order = 3 → triple bond

89. Electron configuration (σ 1s ) 2 bond order = 1

90. (σ 1s ) 2 (σ 1s * ) 1 bond order = 0.5

91. (σ 1s ) 2 (σ 1s * ) 2 bond order = 0

92. (σ 1s ) 2 (σ 1s * ) 1 bond order = 0.5

93. Li 2

94. (σ 2s ) 2 bond order = 1

95. (σ 2s ) 2 (σ 2s * ) 2 bond order = 0

100. O 2, F 2, Ne 2 : (σ 2s ) (σ 2s * ) (σ 2p ) (π 2p ) (π 2p * ) (σ 2p * ) B 2, C 2, N 2 : (σ 2s ) (σ 2s * ) (π 2p ) (σ 2p ) (π 2p * ) (σ 2p * ) 1)Electron configuration. 2)Bond order → stable molecule/ion? 3)Paramagnetic or Diamagnetic? O 2, F 2, N 2, N 2 −, N 2 +

102. Homonuclear diatomic molecules/ions Heteronuclear diatomic molecules/ions

105. Problems Chapter 10 1-7,33,35,39,42,47,51,53,59,61,63,69, 71,85,86