1. Chapter 9 Molecular Geometry and Bonding Theories

2. 9.1 Molecular Shapes

3. Molecular Shapes
The shape of a molecule plays an important role in its reactivity.
By noting the number of bonding and nonbonding electron pairs we can easily predict the shape of the molecule. 3

4. What Determines the Shape of a Molecule?
Simply put, electron pairs, whether they be bonding or nonbonding, repel each other.
By assuming the electron pairs are placed as far as possible from each other, we can predict the shape of the molecule. 4

5. Electron Domains
We can refer to the electron pairs as electron domains.
In a double or triple bond, all electrons shared between those two atoms are on the same side of the central atom; therefore, they count as one electron domain.
The central atom in this molecule, A, has four electron domains. 5

6. 9.2 The VSEPR Model

7. Valence Shell Electron Pair Repulsion Theory (VSEPR)
“The best arrangement of a given number of electron domains is the one that minimizes the repulsions among them.” 7

8. Electron-Domain Geometries
P.344 GIST: One of the common shapes for AB4 molecules is square planar. All 5 atoms lie in the same plane with the B atoms on the corner of a square and A in the middle. Which shape could lead to square planar geometry upon removal of one or more atoms? 8

9. Electron-Domain Geometries
All one must do is count the number of electron domains in the Lewis structure.
The geometry will be that which corresponds to the number of electron domains. 9

10. Molecular Geometries
The electron-domain geometry is often not the shape of the molecule, however.
The molecular geometry is that defined by the positions of only the atoms in the molecules, not the nonbonding pairs. 10

11. Molecular Geometries
Within each electron domain, then, there might be more than one molecular geometry. 11

12. Linear Electron Domain
In the linear domain, there is only one molecular geometry: linear.
NOTE: If there are only two atoms in the molecule, the molecule will be linear no matter what the electron domain is. 12

13. Trigonal Planar Electron Domain
There are two molecular geometries:
Trigonal planar, if all the electron domains are bonding,
Bent, if one of the domains is a nonbonding pair. 13

14. Nonbonding Pairs and Bond Angle
Nonbonding pairs are physically larger than bonding pairs.
Therefore, their repulsions are greater; this tends to decrease bond angles in a molecule. 14

15. Multiple Bonds and Bond Angles
Double and triple bonds place greater electron density on one side of the central atom than do single bonds.
Therefore, they also affect bond angles. 15

16. Tetrahedral Electron Domain
There are three molecular geometries:
Tetrahedral, if all are bonding pairs,
Trigonal pyramidal if one is a nonbonding pair,
Bent if there are two nonbonding pairs. 16

17. Exercise 9.1
1) Use the VSEPR model to predict the molecular geometry of
A) O3
B) SnCl3-
2) Predict the electron-domain geometry and the molecular geometry for
A) SeCl2
B) CO32-
3) Draw the structure for NO3- and describe the bond angles.

18. Trigonal Bipyramidal Electron Domain
There are two distinct positions in this geometry:
Axial
Equatorial 18

19. Trigonal Bipyramidal Electron Domain
Lower-energy conformations result from having nonbonding electron pairs in equatorial, rather than axial, positions in this geometry. 19

20. Trigonal Bipyramidal Electron Domain
There are four distinct molecular geometries in this domain:
Trigonal bipyramidal
Seesaw
T-shaped
Linear 20

21. Octahedral Electron Domain
All positions are equivalent in the octahedral domain.
There are three molecular geometries:
Octahedral
Square pyramidal
Square planar 21

22. Exercise 9.2
1) Use the VSEPR model to predict the molecular geometry of
A) SF4
B) IF5
2) Predict the electron-domain geometry and molecular geometry of
A) ClF3
B) ICl4-
3) For an AB4 atom, why is a tetrahedron geometry preferred over a square planar geometry?

23. Larger Molecules
In larger molecules, it makes more sense to talk about the geometry about a particular atom rather than the geometry of the molecule as a whole. 23

24. Larger Molecules
This approach makes sense, especially because larger molecules tend to react at a particular site in the molecule. 24

25. Sample Exercise 9.3 Predicting Bond Angles
Eyedrops for dry eyes usually contain a water-soluble polymer called poly(vinyl alcohol), which is based on the unstable organic molecule
called vinyl alcohol:
Predict the approximate values for the H—O—C and O—C—C bond angles in vinyl alcohol.

26. Sample Exercise 9.3 Predicting Bond Angles
Predict the H—C—H and C—C—C bond angles in the following molecule, called propyne:
Practice Exercise

27. 9.3 Molecular Shape and Molecular Polarity

28. Polarity In Chapter 8 we discussed bond dipoles.
But just because a molecule possesses polar bonds does not mean the molecule as a whole will be polar. 28

29. Polarity
By adding the individual bond dipoles, one can determine the overall dipole moment for the molecule. 29

30. Polarity 30

31. For ABn molecules where B is the same atom, the following geometries will be nonpolar:
Linear (AB2)
Trigonal planar (AB3)
Tetrahedral (AB4)
Square planar (AB4E2)
Tirgonal bipyramidal (AB5)
Octahedral (AB6)

32. Exercise 9.4
Predict whether the following molecules are polar or nonpolar:
BrCl
SO2
SF6
NF3
BCl3
The molecule O=C=S has a Lewis structure analogous to CO2 and is a linear molecule. Will it necessarily have a zero dipole moment like CO2?

33. 9.4 Covalent Bonding and Orbital Overlap

34. Overlap and Bonding
We think of covalent bonds forming through the sharing of electrons by adjacent atoms.
In such an approach this can only occur when orbitals on the two atoms overlap. 34

35. Overlap and Bonding
Increased overlap brings the electrons and nuclei closer together while simultaneously decreasing electron-electron repulsion.
However, if atoms get too close, the internuclear repulsion greatly raises the energy.
p.357 GIST: If you could put pressure on the hydrogen molecule so that its bond length decreased, would its bond strength increase or decrease? 35

36. 9.5 Hybrid Orbitals

37. Hybrid Orbitals
But it’s hard to imagine tetrahedral, trigonal bipyramidal, and other geometries arising from the atomic orbitals we recognize. 37

38. Hybrid Orbitals Consider beryllium:
In its ground electronic state, it would not be able to form bonds because it has no singly-occupied orbitals. 38

39. Hybrid Orbitals
But if it absorbs the small amount of energy needed to promote an electron from the 2s to the 2p orbital, it can form two bonds. 39

40. Hybrid Orbitals
Mixing the s and p orbitals yields two degenerate orbitals that are hybrids of the two orbitals.
These sp hybrid orbitals have two lobes like a p orbital.
One of the lobes is larger and more rounded as is the s orbital. 40

41. Hybrid Orbitals
These two degenerate orbitals would align themselves 180 from each other.
This is consistent with the observed geometry of beryllium compounds: linear. 41

42. Hybrid Orbitals
With hybrid orbitals the orbital diagram for beryllium would look like this.
The sp orbitals are higher in energy than the 1s orbital but lower than the 2p.
P.358 GIST: Suppose that two unhybridized 2p orbitals on Be were used to make the Be-F bonds in BeF2. Would the two bonds be equivalent to each other? What would be the expected F-Be-F bond angle? 42

43. Hybrid Orbitals
Using a similar model for boron leads to…
p.358 GIST: In an sp2 hybridized atom, what is the orientation of the unhybridized p orbital relative to the three sp2 hybrid orbitals? 43

44. Hybrid Orbitals
…three degenerate sp2 orbitals. 44

45. Hybrid Orbitals
With carbon we get… 45

46. Hybrid Orbitals
…four degenerate
sp3 orbitals. 46

47. Hybrid Orbitals
For geometries involving expanded octets on the central atom, we must use d orbitals in our hybrids. 47

48. Hybrid Orbitals This leads to five degenerate sp3d orbitals…
…or six degenerate sp3d2 orbitals. 48

49. Hybrid Orbitals
Once you know the electron-domain geometry, you know the hybridization state of the atom. 49

50. Exercise 9.5
1) Indicate the hybridization of orbitals employed by the central atom in:
A) NH2-
B) SF4
2) Predict the electron-domain geometry and the hybridization of the central atom in:
A) SO32-
B) SF6

51. 9.6 Multiple Bonds

52. Valence Bond Theory
Hybridization is a major player in this approach to bonding.
There are two ways orbitals can overlap to form bonds between atoms. 52

53. Sigma () Bonds Sigma bonds are characterized by Head-to-head overlap.
Cylindrical symmetry of electron density about the internuclear axis. 53

54. Pi () Bonds Pi bonds are characterized by Side-to-side overlap.
Electron density above and below the internuclear axis. 54

55. Single Bonds
Single bonds are always  bonds, because  overlap is greater, resulting in a stronger bond and more energy lowering. 55

56. Multiple Bonds
In a multiple bond one of the bonds is a  bond and the rest are  bonds. 56

57. Multiple Bonds
In a molecule like formaldehyde (shown at left) an sp2 orbital on carbon overlaps in  fashion with the corresponding orbital on the oxygen.
The unhybridized p orbitals overlap in  fashion. 57

58. Multiple Bonds
In triple bonds, as in acetylene, two sp orbitals form a  bond between the carbons, and two pairs of p orbitals overlap in  fashion to form the two  bonds. 58

59. Formaldehyde has the Lewis structure
Sample Exercise 9.6 Describing  and  Bonds in a Molecule
Formaldehyde has the Lewis
structure
Describe how the bonds in formaldehyde are formed in terms of overlaps of appropriate hybridized and unhybridized orbitals.
p.363 GIST: The molecule diazine has the formula N2H2 and the following Lewis structure. Would you expect it to be linear? If not would you expect it to be planar?

60. Consider the acetonitrile molecule:
Sample Exercise 9.6 Describing  and  Bonds in a Molecule
Consider the acetonitrile
molecule:
(a) Predict the bond angles around each carbon atom; (b) describe the hybridization at each of the carbon atoms; (c) determine the total number of and bonds in the molecule.
Practice Exercise

61. Delocalized Electrons: Resonance
When writing Lewis structures for species like the nitrate ion, we draw resonance structures to more accurately reflect the structure of the molecule or ion. 61

62. Delocalized Electrons: Resonance
In reality, each of the four atoms in the nitrate ion has a p orbital.
The p orbitals on all three oxygens overlap with the p orbital on the central nitrogen. 62

63. Delocalized Electrons: Resonance
This means the  electrons are not localized between the nitrogen and one of the oxygens, but rather are delocalized throughout the ion. 63

64. Resonance
The organic molecule benzene has six  bonds and a p orbital on each carbon atom. 64

65. Resonance
In reality the  electrons in benzene are not localized, but delocalized.
The even distribution of the electrons in benzene makes the molecule unusually stable. 65

66. Sample Exercise 9.7 Delocalized Bonding
Describe the bonding in the nitrate ion, NO3–. Does this ion have delocalized  bonds?

67. Practice
Which of the following molecules or ions will exhibit delocalized bonding: SO3, SO32-, H2CO, O3, NH4+

68. 9.7 Molecular Orbitals

69. Bond Order
Single bond = 1
Double bond = 2
Triple bond = 3

70. Integrative Exercise
Elemental sulfur is a yellow solid that consists of S8 molecules. The structure is an eight-membered ring (Figure 7.30). Heating elemental sulfur to high temperatures produces gaseous S2 molecules.
A) With respect to electronic structure, which element in the second row if most similar to sulfur?
B) Use VSEPR to predict the S-S-S bond angles in S8 and the hybridization of S in S8.
D) Use average bond enthalpies (Table 8.4) to estimate the enthalpy change for the reaction just described. Is the reaction exothermic or endothermic?