1. Chemical Bonding I: Basic Concepts Chapter 9

2. Intermolecular Forces 11.2 Intermolecular forces are attractive forces between molecules. Intramolecular forces hold atoms together in a molecule. Intermolecular vs Intramolecular 41 kJ to vaporize 1 mole of water (inter) 930 kJ to break all O-H bonds in 1 mole of water (intra) Generally, intermolecular forces are much weaker than intramolecular forces. “Measure” of intermolecular force boiling point melting point  H vap  H fus  H sub

3. 9.1 Valence electrons are the outer shell electrons of an atom. The valence electrons are the electrons that particpate in chemical bonding. 1A 1ns 1 2A 2ns 2 3A 3ns 2 np 1 4A 4ns 2 np 2 5A 5ns 2 np 3 6A 6ns 2 np 4 7A 7ns 2 np 5 Group# of valence e - e - configuration

4. 9.2 Li + F Li + F - The Ionic Bond 1s 2 2s 1 1s 2 2s 2 2p 5 1s 2 1s 2 2s 2 2p 6 [He][Ne] Li Li + + e - e - + FF - F - Li + + Li + F -

5. 9.1

6. A covalent bond is a chemical bond in which two or more electrons are shared by two atoms. Why should two atoms share electrons? FF + 7e - FF 8e - F F F F Lewis structure of F 2 lone pairs single covalent bond 9.4

7. 8e - H H O ++ O HH O HHor 2e - Lewis structure of water Double bond – two atoms share two pairs of electrons single covalent bonds O C O or O C O 8e - double bonds Triple bond – two atoms share three pairs of electrons N N 8e - N N triple bond or 9.4

8. Bond Type Bond Length (pm) C-CC-C 154 CCCC 133 CCCC 120 C-NC-N 143 CNCN 138 CNCN 116 Lengths of Covalent Bonds Bond Lengths Triple bond < Double Bond < Single Bond 9.4

10. 1.Draw skeletal structure of compound showing what atoms are bonded to each other. Put least electronegative element in the center. 2.Count total number of valence e -. Add 1 for each negative charge. Subtract 1 for each positive charge. 3.Complete an octet for all atoms except hydrogen 4.If structure contains too many electrons, form double and triple bonds on central atom as needed. Writing Lewis Structures 9.6

11. Write the Lewis structure of nitrogen trifluoride (NF 3 ). Step 1 – N is less electronegative than F, put N in center FNF F Step 2 – Count valence electrons N - 5 (2s 2 2p 3 ) and F - 7 (2s 2 2p 5 ) 5 + (3 x 7) = 26 valence electrons Step 3 – Draw single bonds between N and F atoms and complete octets on N and F atoms. Step 4 - Check, are # of e - in structure equal to number of valence e - ? 3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons 9.6

12. Write the Lewis structure of the carbonate ion (CO 3 2- ). Step 1 – C is less electronegative than O, put C in center OCO O Step 2 – Count valence electrons C - 4 (2s 2 2p 2 ) and O - 6 (2s 2 2p 4 ) -2 charge – 2e - 4 + (3 x 6) + 2 = 24 valence electrons Step 3 – Draw single bonds between C and O atoms and complete octet on C and O atoms. Step 4 - Check, are # of e - in structure equal to number of valence e - ? 3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons 9.6 Step 5 - Too many electrons, form double bond and re-check # of e - 2 single bonds (2x2) = 4 1 double bond = 4 8 lone pairs (8x2) = 16 Total = 24

13. 9.7 Two possible skeletal structures of formaldehyde (CH 2 O) HCOH H CO H An atom’s formal charge is the difference between the number of valence electrons in an isolated atom and the number of electrons assigned to that atom in a Lewis structure. formal charge on an atom in a Lewis structure = 1 2 total number of bonding electrons () total number of valence electrons in the free atom - total number of nonbonding electrons - The sum of the formal charges of the atoms in a molecule or ion must equal the charge on the molecule or ion.

14. HCOH C – 4 e - O – 6 e - 2H – 2x1 e - 12 e - 2 single bonds (2x2) = 4 1 double bond = 4 2 lone pairs (2x2) = 4 Total = 12 formal charge on C = 4 -2 -2 -½ x 6 = -1 formal charge on O = 6 -2 -2 -½ x 6 = +1 formal charge on an atom in a Lewis structure = 1 2 total number of bonding electrons () total number of valence electrons in the free atom - total number of nonbonding electrons - +1 9.7

15. C – 4 e - O – 6 e - 2H – 2x1 e - 12 e - 2 single bonds (2x2) = 4 1 double bond = 4 2 lone pairs (2x2) = 4 Total = 12 H CO H formal charge on C = 4 -0 -0 -½ x 8 = 0 formal charge on O = 6 -4 -4 -½ x 4 = 0 formal charge on an atom in a Lewis structure = 1 2 total number of bonding electrons () total number of valence electrons in the free atom - total number of nonbonding electrons - 00 9.7

16. Formal Charge and Lewis Structures 9.7 1.For neutral molecules, a Lewis structure in which there are no formal charges is preferable to one in which formal charges are present. 2.Lewis structures with large formal charges are less plausible than those with small formal charges. 3.Among Lewis structures having similar distributions of formal charges, the most plausible structure is the one in which negative formal charges are placed on the more electronegative atoms. Which is the most likely Lewis structure for CH 2 O? HCOH +1 H CO H 00

17. Predicting Molecular Geometry 1.Draw Lewis structure for molecule. 2.Count number of lone pairs on the central atom and number of atoms bonded to the central atom. 3.Use VSEPR to predict the geometry of the molecule. What are the molecular geometries of SO 2 and SF 4 ? SO O AB 2 E bent S F F F F AB 4 E distorted tetrahedron 10.1

18. Valence shell electron pair repulsion (VSEPR) model: Predict the geometry of the molecule from the electrostatic repulsions between the electron (bonding and nonbonding) pairs. AB 2 20 Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry 10.1 linear B B

19. Cl Be 2 atoms bonded to central atom 0 lone pairs on central atom 10.1

20. AB 2 20linear Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry VSEPR AB 3 30 trigonal planar 10.1

22. AB 2 20linear Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry VSEPR AB 3 30 trigonal planar 10.1 AB 4 40 tetrahedral

23. 10.1

24. AB 2 20linear Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry VSEPR AB 3 30 trigonal planar 10.1 AB 4 40 tetrahedral AB 5 50 trigonal bipyramidal trigonal bipyramidal

25. 10.1

26. AB 2 20linear Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry VSEPR AB 3 30 trigonal planar 10.1 AB 4 40 tetrahedral AB 5 50 trigonal bipyramidal trigonal bipyramidal AB 6 60 octahedral

27. 10.1

29. bonding-pair vs. bonding pair repulsion lone-pair vs. lone pair repulsion lone-pair vs. bonding pair repulsion >>

30. Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry VSEPR AB 3 30 trigonal planar AB 2 E21 trigonal planar bent 10.1

31. Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry VSEPR AB 3 E31 AB 4 40 tetrahedral trigonal pyramidal 10.1

32. Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry VSEPR AB 4 40 tetrahedral 10.1 AB 3 E31tetrahedral trigonal pyramidal AB 2 E 2 22tetrahedral bent H O H

33. Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry VSEPR 10.1 AB 5 50 trigonal bipyramidal trigonal bipyramidal AB 4 E41 trigonal bipyramidal distorted tetrahedron

34. Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry VSEPR 10.1 AB 5 50 trigonal bipyramidal trigonal bipyramidal AB 4 E41 trigonal bipyramidal distorted tetrahedron AB 3 E 2 32 trigonal bipyramidal T-shaped Cl F F F

35. Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry VSEPR 10.1 AB 5 50 trigonal bipyramidal trigonal bipyramidal AB 4 E41 trigonal bipyramidal distorted tetrahedron AB 3 E 2 32 trigonal bipyramidal T-shaped AB 2 E 3 23 trigonal bipyramidal linear I I I

36. Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry VSEPR 10.1 AB 6 60 octahedral AB 5 E51 octahedral square pyramidal Br FF FF F

37. Class # of atoms bonded to central atom # lone pairs on central atom Arrangement of electron pairs Molecular Geometry VSEPR 10.1 AB 6 60 octahedral AB 5 E51 octahedral square pyramidal AB 4 E 2 42 octahedral square planar Xe FF FF

38. 10.1

39. A resonance structure is one of two or more Lewis structures for a single molecule that cannot be represented accurately by only one Lewis structure. OOO + - OOO + - OCO O -- OCO O - - OCO O - - 9.8 What are the resonance structures of the carbonate (CO 3 2 -) ion?

40. Exceptions to the Octet Rule The Incomplete Octet HHBe Be – 2e - 2H – 2x1e - 4e - BeH 2 BF 3 B – 3e - 3F – 3x7e - 24e - FBF F 3 single bonds (3x2) = 6 9 lone pairs (9x2) = 18 Total = 24 9.9

41. Exceptions to the Octet Rule Odd-Electron Molecules N – 5e - O – 6e - 11e - NO N O The Expanded Octet (central atom with principal quantum number n > 2) SF 6 S – 6e - 6F – 42e - 48e - S F F F F F F 6 single bonds (6x2) = 12 18 lone pairs (18x2) = 36 Total = 48 9.9

42. Covalent share e - Polar Covalent partial transfer of e - Ionic transfer e - Increasing difference in electronegativity Classification of bonds by difference in electronegativity DifferenceBond Type 0Covalent  2 Ionic 0 < and <2 Polar Covalent 9.5

43. Classify the following bonds as ionic, polar covalent, or covalent: The bond in CsCl; the bond in H 2 S; and the NN bond in H 2 NNH 2. Cs – 0.7Cl – 3.03.0 – 0.7 = 2.3Ionic H – 2.1S – 2.52.5 – 2.1 = 0.4Polar Covalent N – 3.0 3.0 – 3.0 = 0Covalent 9.5

44. H F F H Polar covalent bond or polar bond is a covalent bond with greater electron density around one of the two atoms electron rich region electron poor region e - riche - poor ++ -- 9.5

45. Electronegativity is the ability of an atom to attract toward itself the electrons in a chemical bond. Electron Affinity - measurable, Cl is highest Electronegativity - relative, F is highest X (g) + e - X - (g) 9.5

47. Dipole Moments and Polar Molecules 10.2 H F electron rich region electron poor region    = Q x r Q is the charge r is the distance between charges 1 D = 3.36 x 10 -30 C m

48. 10.2

50. Which of the following molecules have a dipole moment? H 2 O, CO 2, SO 2, and CH 4 O H H dipole moment polar molecule S O O CO O no dipole moment nonpolar molecule dipole moment polar molecule C H H HH no dipole moment nonpolar molecule

51. Does CH 2 Cl 2 have a dipole moment? 10.2

52. Intermolecular Forces Dipole-Dipole Forces Attractive forces between polar molecules Orientation of Polar Molecules in a Solid 11.2

53. Intermolecular Forces Ion-Dipole Forces Attractive forces between an ion and a polar molecule 11.2 Ion-Dipole Interaction

54. 11.2

55. Intermolecular Forces Dispersion Forces Attractive forces that arise as a result of temporary dipoles induced in atoms or molecules 11.2 ion-induced dipole interaction dipole-induced dipole interaction

56. Intermolecular Forces Dispersion Forces Continued 11.2 Polarizability is the ease with which the electron distribution in the atom or molecule can be distorted. Polarizability increases with: greater number of electrons more diffuse electron cloud Dispersion forces usually increase with molar mass.

57. S O O What type(s) of intermolecular forces exist between each of the following molecules? HBr HBr is a polar molecule: dipole-dipole forces. There are also dispersion forces between HBr molecules. CH 4 CH 4 is nonpolar: dispersion forces. SO 2 SO 2 is a polar molecule: dipole-dipole forces. There are also dispersion forces between SO 2 molecules. 11.2

58. Intermolecular Forces Hydrogen Bond 11.2 The hydrogen bond is a special dipole-dipole interaction between they hydrogen atom in a polar N-H, O-H, or F-H bond and an electronegative O, N, or F atom. A H … B A H … A or A & B are N, O, or F

59. Hydrogen Bond 11.2

60. Why is the hydrogen bond considered a “special” dipole-dipole interaction? Decreasing molar mass Decreasing boiling point 11.2

61. Types of Crystals Molecular Crystals Lattice points occupied by molecules Held together by intermolecular forces Soft, low melting point Poor conductor of heat and electricity 11.6

62. Types of Crystals Metallic Crystals Lattice points occupied by metal atoms Held together by metallic bonds Soft to hard, low to high melting point Good conductors of heat and electricity 11.6 Cross Section of a Metallic Crystal nucleus & inner shell e - mobile “sea” of e -

63. Types of Crystals Ionic Crystals Lattice points occupied by cations and anions Held together by electrostatic attraction Hard, brittle, high melting point Poor conductor of heat and electricity CsClZnSCaF 2 11.6

64. 9.3 Lattice energy (E) increases as Q increases and/or as r decreases. cmpd lattice energy MgF 2 MgO LiF LiCl 2957 3938 1036 853 Q= +2,-1 Q= +2,-2 r F < r Cl Electrostatic (Lattice) Energy E = k Q+Q-Q+Q- r Q + is the charge on the cation Q - is the charge on the anion r is the distance between the ions Lattice energy (E) is the energy required to completely separate one mole of a solid ionic compound into gaseous ions.

65. Types of Crystals Covalent Crystals Lattice points occupied by atoms Held together by covalent bonds Hard, high melting point Poor conductor of heat and electricity 11.6 diamond graphite carbon atoms

66. Types of Crystals 11.6

67. Hybridization – mixing of two or more atomic orbitals to form a new set of hybrid orbitals. 1.Mix at least 2 nonequivalent atomic orbitals (e.g. s and p). Hybrid orbitals have very different shape from original atomic orbitals. 2.Number of hybrid orbitals is equal to number of pure atomic orbitals used in the hybridization process. 3.Covalent bonds are formed by: a.Overlap of hybrid orbitals with atomic orbitals b.Overlap of hybrid orbitals with other hybrid orbitals 10.4

70. Predict correct bond angle

71. Formation of sp Hybrid Orbitals 10.4

72. Formation of sp 2 Hybrid Orbitals 10.4

73. # of Lone Pairs + # of Bonded Atoms HybridizationExamples 2 3 4 5 6 sp sp 2 sp 3 sp 3 d sp 3 d 2 BeCl 2 BF 3 CH 4, NH 3, H 2 O PCl 5 SF 6 How do I predict the hybridization of the central atom? Count the number of lone pairs AND the number of atoms bonded to the central atom 10.4

74. Valence Bond Theory and NH 3 N – 1s 2 2s 2 2p 3 3 H – 1s 1 If the bonds form from overlap of 3 2p orbitals on nitrogen with the 1s orbital on each hydrogen atom, what would the molecular geometry of NH 3 be? If use the 3 2p orbitals predict 90 0 Actual H-N-H bond angle is 107.3 0 10.4

75. 10.5

77. Sigma (  ) and Pi Bonds (  ) Single bond 1 sigma bond Double bond 1 sigma bond and 1 pi bond Triple bond 1 sigma bond and 2 pi bonds How many  and  bonds are in the acetic acid (vinegar) molecule CH 3 COOH? C H H CH O OH  bonds = 6 + 1 = 7  bonds = 1 10.5

78. The enthalpy change required to break a particular bond in one mole of gaseous molecules is the bond energy. H 2 (g) H (g) +  H 0 = 436.4 kJ Cl 2 (g) Cl (g) +  H 0 = 242.7 kJ HCl (g) H (g) +Cl (g)  H 0 = 431.9 kJ O 2 (g) O (g) +  H 0 = 498.7 kJ OO N 2 (g) N (g) +  H 0 = 941.4 kJ N N Bond Energy Bond Energies Single bond < Double bond < Triple bond 9.10

79. Average bond energy in polyatomic molecules H 2 O (g) H (g) +OH (g)  H 0 = 502 kJ OH (g) H (g) +O (g)  H 0 = 427 kJ Average OH bond energy = 502 + 427 2 = 464 kJ 9.10

80. Bond Energies (BE) and Enthalpy changes in reactions  H 0 = total energy input – total energy released =  BE(reactants) –  BE(products) Imagine reaction proceeding by breaking all bonds in the reactants and then using the gaseous atoms to form all the bonds in the products. 9.10

81. H 2 (g) + Cl 2 (g) 2HCl (g)2H 2 (g) + O 2 (g) 2H 2 O (g)

82. Use bond energies to calculate the enthalpy change for: H 2 (g) + F 2 (g) 2HF (g)  H 0 =  BE(reactants) –  BE(products) Type of bonds broken Number of bonds broken Bond energy (kJ/mol) Energy change (kJ) HH1436.4 FF 1156.9 Type of bonds formed Number of bonds formed Bond energy (kJ/mol) Energy change (kJ) HF2568.21136.4  H 0 = 436.4 + 156.9 – 2 x 568.2 = -543.1 kJ 9.10