1. Molecular Geometries and Bonding Chapter 9 Molecular Geometries and Bonding Theories Chemistry, The Central Science, 10th edition Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice-Hall, Inc.

2. Molecular Geometries and Bonding 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. Molecular Geometries and Bonding 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. Molecular Geometries and Bonding 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. This molecule has four electron domains.

5. Molecular Geometries and Bonding 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.”

6. Molecular Geometries and Bonding Electron-Domain Geometries These are the electron-domain geometries for two through six electron domains around a central atom. (p. 319)

7. Molecular Geometries and Bonding 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 that number of electron domains.

8. Molecular Geometries and Bonding 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.

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

10. Molecular Geometries and Bonding Linear Electron Domain In this 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.

11. Molecular Geometries and Bonding 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.

12. Molecular Geometries and Bonding 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.

13. Molecular Geometries and Bonding SAMPLE EXERCISE 9.1 Using the VSEPR Model Use the VSEPR model to predict the molecular geometry of (a) O 3, (b) SnCl 3 –. Solve: (a) We can draw two resonance structures for O 3 : Because of resonance, the bonds between the central O atom and the outer O atoms are of equal length. In both resonance structures the central O atom is bonded to the two outer O atoms and has one nonbonding pair. Thus, there are three electron domains about the central O atoms. (Remember that a double bond counts as a single electron domain.) The best arrangement of three electron domains is trigonal planar (Table 9.1). Two of the domains are from bonds, and one is due to a nonbonding pair, so the molecule has a bent shape with an ideal bond angle of 120° (Table 9.2).

14. Molecular Geometries and Bonding The central Sn atom is bonded to the three Cl atoms and has one nonbonding pair. Therefore, the Sn atom has four electron domains around it. The resulting electron-domain geometry is tetrahedral (Table 9.1) with one of the corners occupied by a nonbonding pair of electrons. The molecular geometry is thus trigonal pyramidal (Table 9.2), like that of NH 3. Answers: (a) tetrahedral, bent; (b) trigonal planar, trigonal planar PRACTICE EXERCISE Predict the electron-domain geometry and the molecular geometry for (a) SeCl 2, (b) CO 3 2–. SAMPLE EXERCISE 9.1 continued (b) The Lewis structure for the SnCl 3 – ion is

15. Molecular Geometries and Bonding 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.

16. Molecular Geometries and Bonding 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

17. Molecular Geometries and Bonding Trigonal Bipyramidal Electron Domain There are two distinct positions in this geometry:  Axial  Equatorial

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

19. Molecular Geometries and Bonding Trigonal Bipyramidal Electron Domain There are four distinct molecular geometries in this domain:  Trigonal bipyramidal  Seesaw  T-shaped  Linear

20. Molecular Geometries and Bonding Octahedral Electron Domain All positions are equivalent in the octahedral domain. There are three molecular geometries:  Octahedral  Square pyramidal  Square planar

21. Molecular Geometries and Bonding SAMPLE EXERCISE 9.2 Molecular Geometries of Molecules with Expanded Valence Shells Use the VSEPR model to predict the molecular geometry of (a) SF 4, (b) IF 5. Solve: (a) The Lewis structure for SF 4 is The sulfur has five electron domains around it: four from the S—F bonds and one from the nonbonding pair. Each domain points toward a vertex of a trigonal bipyramid. The domain from the nonbonding pair will point toward an equatorial position. The four bonds point toward the remaining four positions, resulting in a molecular geometry that is described as seesaw-shaped:

22. Molecular Geometries and Bonding The iodine has six electron domains around it, one of which is from a nonbonding pair. The electron-domain geometry is therefore octahedral, with one position occupied by the nonbonding electron pair. The resulting molecular geometry is therefore square pyramidal (Table 9.3): SAMPLE EXERCISE 9.2 continued (b) The Lewis structure of IF 5 is (There are three lone pairs on each of the F atoms, but they are not shown.)

23. Molecular Geometries and Bonding SAMPLE EXERCISE 9.2 continued PRACTICE EXERCISE Predict the electron-domain geometry and molecular geometry of (a) ClF 3, (b) ICl 4 –. Answers: (a) trigonal bipyramidal, T-shaped; (b) octahedral, square planar

24. Molecular Geometries and Bonding 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.

25. Molecular Geometries and Bonding Larger Molecules This approach makes sense, especially because larger molecules tend to react at a particular site in the molecule.

26. Molecular Geometries and Bonding Predict the approximate values for the H—O—C and O—C—C bond angles in vinyl alcohol. 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: Solution Analyze: We are given a molecular structure and asked to determine two bond angles in the structure. Plan: To predict a particular bond angle, we consider the middle atom of the angle and determine the number of electron domains surrounding that atom. The ideal angle corresponds to the electron-domain geometry around the atom. The angle will be compressed somewhat by nonbonding electrons or multiple bonds. Solve: For the H—O—C bond angle, there are four electron domains around the middle O atom (two bonding and two nonbonding). The electron-domain geometry around O is therefore tetrahedral, which gives an ideal angle of 109.5°. The H—O—C angle will be compressed somewhat by the nonbonding pairs, so we expect this angle to be slightly less than 109.5°. To predict the O—C—C bond angle, we must examine the leftmost C atom, which is the central atom for this angle. There are three atoms bonded to this C atom and no nonbonding pairs, and so it has three electron domains about it. The predicted electron-domain geometry is trigonal planar, resulting in an ideal bond angle of 120°. Because of the larger size of the domain, however, the O—C—C bond angle should be slightly greater than 120°.

27. Molecular Geometries and Bonding SAMPLE EXERCISE 9.3 continued Answers: 109.5°, 180° PRACTICE EXERCISE Predict the H—C—H and C—C—C bond angles in the following molecule, called propyne:

28. Molecular Geometries and Bonding 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.

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

30. Molecular Geometries and Bonding Polarity

31. Molecular Geometries and Bonding SAMPLE EXERCISE 9.4 Polarity of Molecules Predict whether the following molecules are polar or nonpolar: (a) BrCl, (b) SO 2, (c) SF 6. Solution Analyze: We are given the molecular formulas of several substances and asked to predict whether the molecules are polar. Plan: If the molecule contains only two atoms, it will be polar if the atoms differ in electronegativity. If it contains three or more atoms, its polarity depends on both its molecular geometry and the polarity of its bonds. Thus, we must draw a Lewis structure for each molecule containing three or more atoms and determine its molecular geometry. We then use the relative electronegativities of the atoms in each bond to determine the direction of the bond dipoles. Finally, we see if the bond dipoles cancel each other to give a nonpolar molecule or reinforce each other to give a polar one. Solve: (a) Chlorine is more electronegative than bromine. All diatomic molecules with polar bonds are polar molecules. Consequently, BrCl will be polar, with chlorine carrying the partial negative charge: The actual dipole moment of BrCl, as determined by experimental measurement, is µ = 0.57 D. (b) Because oxygen is more electronegative than sulfur, SO 2 has polar bonds. Three resonance forms can be written for SO 2 :

32. Molecular Geometries and Bonding Answers: (a) polar because polar bonds are arranged in a trigonal-pyramidal geometry, (b) nonpolar because polar bonds are arranged in a trigonal-planar geometry SAMPLE EXERCISE 9.4 continued For each of these, the VSEPR model predicts a bent geometry. Because the molecule is bent, the bond dipoles do not cancel and the molecule is polar: Experimentally, the dipole moment of SO 2 is µ = 1.63 D. PRACTICE EXERCISE Determine whether the following molecules are polar or nonpolar: (a) NF 3, (b) BCl 3. (c) Fluorine is more electronegative than sulfur, so the bond dipoles point toward fluorine. The six S—F bonds are arranged octahedrally around the central sulfur: Because the octahedral geometry is symmetrical, the bond dipoles cancel, and the molecule is nonpolar, meaning that µ = 0.

33. Molecular Geometries and Bonding 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. Molecular Geometries and Bonding 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.

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

36. Molecular Geometries and Bonding Hybrid Orbitals Consider beryllium:  In its ground electronic state, it would not be able to form bonds because it has no singly-occupied orbitals.

37. Molecular Geometries and Bonding 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.

38. Molecular Geometries and Bonding 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.

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

40. Molecular Geometries and Bonding 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.

41. Molecular Geometries and Bonding Hybrid Orbitals Using a similar model for boron leads to…

42. Molecular Geometries and Bonding Hybrid Orbitals …three degenerate sp 2 orbitals.

43. Molecular Geometries and Bonding Hybrid Orbitals With carbon we get…

44. Molecular Geometries and Bonding Hybrid Orbitals …four degenerate sp 3 orbitals.

45. Molecular Geometries and Bonding Hybrid Orbitals For geometries involving expanded octets on the central atom, we must use d orbitals in our hybrids.

46. Molecular Geometries and Bonding Hybrid Orbitals This leads to five degenerate sp 3 d orbitals… …or six degenerate sp 3 d 2 orbitals.

47. Molecular Geometries and Bonding Hybrid Orbitals Once you know the electron-domain geometry, you know the hybridization state of the atom.

48. Molecular Geometries and Bonding SAMPLE EXERCISE 9.5 Hybridization Indicate the hybridization of orbitals employed by the central atom in (a) NH 2 –, (b) SF 4 (see Sample Exercise 9.2). Solution Analyze: We are given two chemical formulas—one for a polyatomic anion and one for a molecular compound—and asked to describe the type of hybrid orbitals surrounding the central atom in each case. Plan: To determine the hybrid orbitals used by an atom in bonding, we must know the electron-domain geometry around the atom. Thus, we first draw the Lewis structure to determine the number of electron domains around the central atom. The hybridization conforms to the number and geometry of electron domains around the central atom as predicted by the VSEPR model. Answers: (a) tetrahedral, sp 3 ; (b) octahedral, sp 3 d 2 (b) The Lewis structure and electron-domain geometry of SF 4 are shown in Sample Exercise 9.2. There are five electron domains around S, giving rise to a trigonal-bipyramidal electron-domain geometry. With an expanded octet of ten electrons, a d orbital on the sulfur must be used. The trigonal-bipyramidal electron-domain geometry corresponds to sp 3 d hybridization (Table 9.4). One of the hybrid orbitals that points in an equatorial direction contains a nonbonding pair of electrons; the other four are used to form the S—F bonds. Solve: (a) The Lewis structure of NH 2 – is Because there are four electron domains around N, the electron-domain geometry is tetrahedral. The hybridization that gives a tetrahedral electron-domain geometry is sp 3 (Table 9.4). Two of the sp 3 hybrid orbitals contain nonbonding pairs of electrons, and the other two are used to make two-electron bonds with the hydrogen atoms. PRACTICE EXERCISE Predict the electron-domain geometry and the hybridization of the central atom in (a) SO 3 2–, (b) SF 6.

49. Molecular Geometries and Bonding 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.

50. Molecular Geometries and Bonding Sigma (  ) Bonds Sigma bonds are characterized by  Head-to-head overlap.  Cylindrical symmetry of electron density about the internuclear axis.

51. Molecular Geometries and Bonding Pi (  ) Bonds Pi bonds are characterized by  Side-to-side overlap.  Electron density above and below the internuclear axis.

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

53. Molecular Geometries and Bonding Multiple Bonds In a multiple bond one of the bonds is a  bond and the rest are  bonds.

54. Molecular Geometries and Bonding Multiple Bonds In a molecule like formaldehyde (shown at left) an sp 2 orbital on carbon overlaps in  fashion with the corresponding orbital on the oxygen. The unhybridized p orbitals overlap in  fashion.

55. Molecular Geometries and Bonding 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.

56. Molecular Geometries and Bonding 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. Solve: The C atom has three electron domains around it, which suggests a trigonal-planar geometry with bond angles of about 120°. This geometry implies sp 2 hybrid orbitals on C (Table 9.4). These hybrids are used to make the two C—H and one C—O  bonds to C. There remains an unhybridized 2p orbital on carbon, perpendicular to the plane of the three sp 2 hybrids. Solution Analyze: We are asked to describe the bonding in formaldehyde in terms of orbital overlaps. Plan: Single bonds will be of the  type, whereas double bonds will consist of one  bond and one  bond. The ways in which these bonds form can be deduced from the geometry of the molecule, which we predict using the VSEPR model.

57. Molecular Geometries and Bonding SAMPLE EXERCISE 9.6 continued The O atom also has three electron domains around it, and so we will assume that it has sp 2 hybridization as well. One of these hybrids participates in the C—O  bond, while the other two hybrids hold the two nonbonding electron pairs of the O atom. Like the C atom, therefore, the O atom has an unhybridized 2p orbital that is perpendicular to the plane of the molecule. The unhybridized 2p orbitals on the C and O atoms overlap to form a C—O π bond, as illustrated in Figure 9.27. Answers: (a) approximately 109° around the left C and 180° on the right C; (b) sp 3, sp; (c) five  bonds and two π bonds PRACTICE EXERCISE 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. Figure 9.27 Formation of  and  bonds in formaldehyde, H 2 CO.

58. Molecular Geometries and Bonding 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.

59. Molecular Geometries and Bonding 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.

60. Molecular Geometries and Bonding 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.

61. Molecular Geometries and Bonding Resonance The organic molecule benzene has six  bonds and a p orbital on each carbon atom.

62. Molecular Geometries and Bonding 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.

63. Molecular Geometries and Bonding SAMPLE EXERCISE 9.7 Delocalized Bonding Describe the bonding in the nitrate ion, NO 3 –. Does this ion have delocalized  bonds? Solution Analyze: Given the chemical formula for a polyatomic anion, we are asked to describe the bonding and determine whether the ion has delocalized  bonds. Plan: Our first step in describing the bonding in NO 3 – is to construct appropriate Lewis structures. If there are multiple resonance structures that involve the placement of the double bonds in different locations, that suggests that the  component of the double bonds is delocalized. In each of these structures the electron-domain geometry at nitrogen is trigonal planar, which implies sp 2 hybridization of the N atom. The sp 2 hybrid orbitals are used to construct the three N—O  bonds that are present in each of the resonance structures. Solve: In Section 8.6 we saw that NO 3 – has three resonance structures:

64. Molecular Geometries and Bonding SAMPLE EXERCISE 9.7 continued The unhybridized 2p orbital on the N atom can be used to make  bonds. For any one of the three resonance structures shown, we might imagine a single localized N––O  bond formed by the overlap of the unhybridized 2p orbital on N and a 2p orbital on one of the O atoms, as shown in Figure 9.30(a). Because each resonance structure contributes equally to the observed structure of NO 3 –, however, we represent the  bonding as spread out, or delocalized, over the three N—O bonds, as shown in Figure 9.30(b). Figure 9.30 Localized and delocalized  bonds in NO 3 –.

65. Molecular Geometries and Bonding SAMPLE EXERCISE 9.7 continued Answer: SO 3 and O 3, as indicated by the presence of two or more resonance structures involving  bonding for each of these molecules PRACTICE EXERCISE Which of the following molecules or ions will exhibit delocalized bonding: SO 3, SO 3 2–, H 2 CO, O 3, NH 4 + ?

66. Molecular Geometries and Bonding Molecular Orbital (MO) Theory Though valence bond theory effectively conveys most observed properties of ions and molecules, there are some concepts better represented by molecular orbitals.

67. Molecular Geometries and Bonding Molecular Orbital (MO) Theory In MO theory, we invoke the wave nature of electrons. If waves interact constructively, the resulting orbital is lower in energy: a bonding molecular orbital.

68. Molecular Geometries and Bonding Molecular Orbital (MO) Theory If waves interact destructively, the resulting orbital is higher in energy: an antibonding molecular orbital.

69. Molecular Geometries and Bonding MO Theory In H 2 the two electrons go into the bonding molecular orbital. The bond order is one half the difference between the number of bonding and antibonding electrons.

70. Molecular Geometries and Bonding MO Theory For hydrogen, with two electrons in the bonding MO and none in the antibonding MO, the bond order is 1212 (2 - 0) = 1

71. Molecular Geometries and Bonding MO Theory In the case of He 2, the bond order would be 1212 (2 - 2) = 0 Therefore, He 2 does not exist.

72. Molecular Geometries and Bonding MO Theory For atoms with both s and p orbitals, there are two types of interactions:  The s and the p orbitals that face each other overlap in  fashion.  The other two sets of p orbitals overlap in  fashion.

73. Molecular Geometries and Bonding MO Theory The resulting MO diagram looks like this. There are both  and  bonding molecular orbitals and  * and  * antibonding molecular orbitals.

74. Molecular Geometries and Bonding MO Theory The smaller p-block elements in the second period have a sizeable interaction between the s and p orbitals. This flips the order of the s and p molecular orbitals in these elements.

75. Molecular Geometries and Bonding Second-Row MO Diagrams