Chemical Bonding II. Molecular Geometry Valence Bond Theory Phys./Chem. Properties Quantum Mechanics Sigma & Pi bonds Hybridization MO theory

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1 Chemical Bonding II Molecular Geometry Valence Bond Theory Phys./Chem. Properties Quantum Mechanics Sigma & Pi bonds ybridization MO theory 1

2 Molecular Geometry 3-D arrangement of atoms 2

3 VSEPR Valence-shell electron-pair repulsion theory. All valence shell e - pairs (or e - domains ) repel each other. (Coulomb s law as applied to the repulsion of valence electrons.) Determines the geometry of e - domains around central atom. 3

4 VSEPR Theory: Rules 1.Single, double, triple bonds and lone pairs are treated as one e - domain (approximation) 2.Apply VSEPR to any one resonance structure 3.Geometry move e - domains as far apart as possible in 3-D space. 4

5 AB 2 (with no lone pairs) Beryllium chloride BeCl 2 e - Be 180 o atoms Cl Be Cl linear geometry 5

6 AB 3 (with no lone pairs) Boron trifluoride B o B Trigonal planar 6

7 AB 4 (with no lone pairs) Methane C 4 C o tetrahedron 7

8 AB 5 (with no lone pairs) Phosphorus Pentachloride PCl 5 90 o axial equatorial 120 o trigonal bipyramid 8

9 AB 6 (with no lone pairs) Sulfur hexafluoride S 6 S All angles 90 o & 180 o octahedral 9

10 Quiz: Name That Shape! AB 2 AB 3 AB 4 AB 5 AB 6 10

11 VSEPR: Lone Pairs Molecules in which the central atom has lone pair(s) AB x E y central atom lone pairs on A surrounding atoms 11

12 AB x E y Geometry is similar to AB, but x nonbonding pairs are treated like bonding pairs to determine geometry of the e - domains. 12

13 VSEPR Bonding e - take up less space than nonbonding e - Bonding e - are focused between the nuclei. Example: water 13

14 VSEPR Decreasing e- pair repulsion lone pair lone pair bonding pair vs vs vs lone pair bonding pair bonding pair repulsion repulsion repulsion 14

15 AB 2 E Sulfur dioxide SO 2 Draw Lewis structure O S O S 3 e - domains O O OSO angle < 120 o 15

16 Shape: e - Pairs vs. Atoms S S O O O O e - domains are trigonal planar molecule is bent linear (always state molecular shape) 16

17 AB 3 E Ammonia N e - tetrahedral trigonal pyramidal 17

18 AB 2 E 2 Water O e - tetrahedral molecule: bent linear 18

19 Comparison AB 4 AB 3 E AB 2 E 2 Methane ammonia water C N O

20 AB 4 E Sulfur tetrafluoride S 4 Draw Lewis structure S Predict: trigonal bipyramid for e - domains. 20

21 AB 4 E: S 4 trigonal bipyramid: 2 choices distorted tetrahedron or seesaw 21

22 AB 3 E 2 Cl 3 A little weird. molecule: T-shaped 22

23 AB 2 E 3 Three possibilities for I 3 - Want nonbonding e - domains farthest apart. 23

24 I 3 - AB 2 E 3 I - molecule: linear I Lone pairs always go on equatorial position(s) 24

25 AB 5 E Br 5 molecule: square pyramidal 25

26 AB 4 E 2 : 2 Possibilities Xe 4 1 Want nonbonding e- pairs farthest apart. 2 26

27 AB 4 E 2 Xe 4 molecule: square planar 27

28 You Try It Predict geometry and approximate bond angles for: AlCl 4 - Xe 2 XeO 2 Text: Tables 10.1 and

29 Bond Polarity d+ d- is more electronegative than. Shift of e - density toward, Thus is polar. 29

30 Overall Molecular Polarity pull test O C O Even though CO 2 has polar bonds, it is nonpolar since the individual bond polarities add to zero. Dipoles are a vector quantities. Symmetrical = nonpolar (Recall the Regents rule about lone pairs on the central atom.) 30

31 Molecular Polarity Molecule Geometry Dipole Strength Linear 1.92 Br Linear 1.08 Water Bent 1.87 SO 2 Bent

32 Molecular Polarity Which has larger polarity? N 3 N 3 N N m = 1.46 D m =0.24 D 32

33 Dipole Moments Predict whether the following molecules are polar. IBr C Cl 2 2 AlCl 3 Remember to distinguish between bond polarity and molecular polarity. 33

34 Shortfall of VSEPR Both 2 & 2 have single bonds, but Bond length Bond energy 2 74 pm 436 kj pm 151 kj 34

35 Quantum Mechanics to the rescue Valence Bond Theory e - in molecule occupy blended atomic orbitals Molecular Orbital Theory molecule has molecular orbitals 35

36 Valence Bond Theory Atomic orbitals (s, p, d ) of the valence electrons hybridize or mix to form new orbitals for the molecule. 36

37 Valence Bond Theory C Consider tetrahedral C 4 carbon atomic orbitals 2s 2p ow can carbon form four equal bonds with four hydrogen atoms using its atomic s and p orbitals? 37

38 C 4 : sp 3 ybridization C carbon atomic orbitals 2s 2p C carbon hybridized orbitals sp 3 (all the same) 38

39 ybridization Analogy s p p p 4 sp 3 hybrid orbitals 39

40 sp 3 ybridization It takes energy to form hybrid orbitals, but this energy is more than compensated by bond formation. 40

41 sp 3 Orbital Shapes One s & three p orbitals change to four sp 3 hybrid orbitals. tetrahedral Each orbital can hold 2 e - 41

42 C 4 Bonding The 4 sp 3 hybrid orbitals of C bond (overlap) with the 4 1s atomic orbitals of the atoms. C 42

43 N 3 sp 3 Bonding N nitrogen atomic orbitals 2s 2p N nitrogen hybridized orbitals sp 3 (all the same) 43

44 N 3 sp 3 Bonding sp 3 lone pair bonding orbitals 44

45 ybridization irst use VSEPR to predict the arrangement of e - pairs, then use hybridization to predict the type of bonding. 45

46 sp ybridization Be Be atomic 2s 2p orbitals Be Be hybrid orbitals sp 2p 46

47 sp ybridization: BeCl 2 Cl Be Cl one atomic p orbital of Cl two sp hybrid orbitals of Be 47

48 sp 2 ybridization B B atomic orbitals 2s 2p B B hybrid orbitals sp 2 2p 48

49 sp 2 ybridization: B 3 one p orbital of B three sp 2 hybrid orbitals of B 49

50 ybridization ybridization for 2 nd Period elements (C, N, O, ) explains the octet rule, since there are 4 hybrid orbitals formed from one s and three p atomic orbitals. s p sp 3 50

51 ybridization: Review 1.Not applied to isolated atoms 2.irst determine VSEPR geometry 3.Mix nonequivalent atomic orbitals of central atom to form hybrid orbitals 51

52 ybridization: Review 4.Requires energy, but energy is more than returned by bond formation 5.Covalent bonds formed by overlap of hybrid-hybrid and/or hybrid-unhybridized orbital 52

53 ybridization: Try It Determine hybridization in: AlBr 3 P 3 gcl 2 53

54 Let s Not orget d Orbitals or elements in the 3 rd Period and higher, hybridization can also include d orbitals. d-orbital hybridization is still being debated! 54

55 d ybridization S 6 VSEPR predicts octahedral geometry S 55

56 ybridization: S 6 S atomic orbitals 3s 3p 3d S hybridized sp 3 d 2 orbitals sp 3 d 2 3d 56

57 ybridization You Try It. What is the VSEPR geometry and the hybridization in PBr 5? 57

58 Summary: ybrid Orbitals 2 e - pairs (sp) linear e - domains (bonding & nonbonding pairs) 3 e - pairs (sp 2 ) trigonal planar 4 e - pairs (sp 3 ) tetrahedral 58

59 Summary: ybrid Orbitals 5 e - pairs (sp 3 d) trigonal bipyramid 6 e - pairs (sp 3 d 2 ) octahedral 59

60 Multiple Bonds In VSEPR, no distinction was made among single, double, triple bond or lone pair. All were counted as an e - domain. C 2 4 C C Each carbon is trigonal planar 60

61 Double Bond: C 2 4 C atomic orbitals 2s 2p C hybrid orbitals sp 2 + p z unhybridized p orbital 61

62 Double Bond: C 2 4 C hybrid orbitals sp 2 + p z trigonal planar dumb bell 62

63 Double Bonds: C 2 4 C C 63

64 Double Bond: C 2 4 Sigma (s) and Pi (p) Bonds (model) 64

65 Pi & Sigma Bonds Sigma (s) bond: covalent bond formed by e - overlap along the axis connecting atoms Pi (p) bond: covalent bond formed by sideways e - overlap above and below the plane connecting atoms (weaker than s) 65

66 Double Bond: C 2 4 C C The double bond is one sigma and one pi bond between the carbon atoms 66

67 Triple Bond: -C=C- C atomic orbitals 2s 2p C hybrid orbitals sp + p y p z unhybridized p orbitals 67

68 Triple Bond: C 2 2 unhybridized p orbitals hybridized sp orbitals ybridize s bonds only 68

69 ybridizing Shortcuts 1.Single bonds: sigma 2.Double bonds: one sigma, one pi 3.Triple bonds: one sigma, two pi 4.ybridize: add the number of sigma bonds plus lone pairs s p d 69

70 You Try It Describe the bonding and hybridization for each atom in: formaldehyde, C 2 O hydrogen cyanide, CN (assign formal charges) 70

71 Delocalized Molecular Orbitals An explanation of resonance. O O O O O O ozone Ozone is a blend of the two resonance structures. 71

72 Delocalized Molecular Orbitals The p component of the double bond is delocalized over the molecule. sp 2 sp 2 sp 2 72

73 Delocalized MOs e - in s bonds localized e - in p bonds-- delocalized benzene 73

74 Sigma bonds (localized) Benzene sp 2 C C sp 2 C C C C 74

75 Benzene sp 2 + p z p bonds (delocalized) 75

76 Benzene Bond order 1.5 Conjugated double bonds are more stable (less reactive) 76

77 Carbonate Ion O C O O 2- + resonance structures C: sp 2 for s bonds (localized) plus a 2p z for p bond (delocalized) O: each has a 2p z orbital (delocalized) 77

78 Carbonate Ion o Additional stability due to delocalization of the pi bond. o C o C o o o 78

79 You Try It Would you predict the NO 3 - anion to have additional stability due to delocalization of the p electrons? 79

80 Polyatomic Ions Many common polyatomic ions have delocalized pi bonds, partially accounting for their additional stability in chemical reactions. CO 3-2, NO 3 -, ClO 3 - etc. 80

81 CaCO 3 Ionic & Covalent Ionic compounds containing polyatomic ions have both ionic and covalent bonds! O 2- Ca 2+ O C O Trigonal planar, sp 2, o

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