Chapter 9 Molecular Geometries and Bonding Theories

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

2. 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. 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. Electron Domains • We can refer to the electron pairs as electrondomains. • 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. 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. Electron-Domain Geometries These are the electron-domain geometries for two through six electron domains around a central atom.

7. 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 • 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 Within each electron domain, then, there might be more than one molecular geometry.

10. 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. 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. 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. 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.

14. 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

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

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

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

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

19. 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.

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

21. Solve: (a) We can draw two resonance structures for O3: 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). SAMPLE EXERCISE 9.1 Using the VSEPR Model Use the VSEPR model to predict the molecular geometry of (a) O3, (b) SnCl3–. Solution Analyze: We are given the molecular formulas of a molecule and a polyatomic ion, both conforming to the general formula ABn and both having a central atom from the p block of the periodic table. Plan: To predict the molecular geometries of these species, we first draw their Lewis structures and then count the number of electron domains around the central atom. The number of electron domains gives the electron-domain geometry. We then obtain the molecular geometry from the arrangement of the domains that are due to bonds. As this example illustrates, when a molecule exhibits resonance, any one of the resonance structures can be used to predict the molecular geometry.

22. SAMPLE EXERCISE 9.1continued (b) The Lewis structure for the SnCl3– ion is 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 NH3. PRACTICE EXERCISE Predict the electron-domain geometry and the molecular geometry for (a) SeCl2, (b) CO32–. Answers:(a) tetrahedral, bent; (b) trigonal planar, trigonal planar

23. Solve: (a) The Lewis structure for SF4 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: SAMPLE EXERCISE 9.2 Molecular Geometries of Molecules with Expanded Valence Shells Use the VSEPR model to predict the molecular geometry of (a) SF4, (b) IF5. Solution Analyze: The molecules are of the ABn type with a central atom from the p block of the periodic table. Plan: We can predict their structures by first drawing Lewis structures and then using the VSEPR model to determine the electron-domain geometry and molecular geometry.

24. (b) The Lewis structure of IF5 is 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 9.2continued Comment: The experimentally observed structure is shown on the previous slide on the right, and we can infer that the nonbonding electron domain occupies an equatorial position, as predicted. The axial and equatorial S––F bonds are slightly bent back away from the nonbonding domain, suggesting that the bonding domains are “pushed” by the nonbonding domain, which is larger and has greater repulsion (Figure 9.7). (There are three lone pairs on each of the F atoms, but they are not shown.) Comment: Because the domain for the nonbonding pair is larger than the other domains, the four F atoms in the base of the pyramid are tipped up slightly toward the F atom on top. Experimentally, it is found that the angle between the base and top F atoms is 82°, smaller than the ideal 90° angle of an octahedron.

25. SAMPLE EXERCISE 9.2continued PRACTICE EXERCISE Predict the electron-domain geometry and molecular geometry of (a) ClF3, (b) ICl4–. Answers:(a) trigonal bipyramidal, T-shaped; (b) octahedral, square planar

26. 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: 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°. Predict the approximate values for the H—O—C and O—C—C bond angles in 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.

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

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.

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

30. Polarity

31. 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, SO2 has polar bonds. Three resonance forms can be written for SO2: SAMPLE EXERCISE 9.4 Polarity of Molecules Predict whether the following molecules are polar or nonpolar: (a) BrCl, (b) SO2, (c) SF6. 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.

32. SAMPLE EXERCISE 9.4continued 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 SO2 is µ = 1.63 D. (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. PRACTICE EXERCISE Determine whether the following molecules are polar or nonpolar: (a) NF3, (b) BCl3. 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

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

36. 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. 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. 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. 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. 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. Hybrid Orbitals Using a similar model for boron leads to…

42. Hybrid Orbitals …three degenerate sp2 orbitals.

43. Hybrid Orbitals With carbon we get…

44. Hybrid Orbitals …four degenerate sp3 orbitals.

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

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

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

48. Solve: (a) The Lewis structure of NH2– 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 sp3 (Table 9.4). Two of the sp3 hybrid orbitals contain nonbonding pairs of electrons, and the other two are used to make two-electron bonds with the hydrogen atoms. SAMPLE EXERCISE 9.5 Hybridization Indicate the hybridization of orbitals employed by the central atom in (a) NH2–, (b) SF4 (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. (b) The Lewis structure and electron-domain geometry of SF4 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 sp3d 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. PRACTICE EXERCISE Predict the electron-domain geometry and the hybridization of the central atom in (a) SO32–, (b) SF6. Answers: (a) tetrahedral, sp3; (b) octahedral,sp3d2

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