Chapter 9 Molecular Geometry and Bonding Theories

1. Chapter 9Molecular 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.

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.

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.

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.”

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?

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.

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.

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

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.

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.

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.

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.

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.

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

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

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

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

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.

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

25. 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. Sample Exercise 9.3 Predicting Bond Angles

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

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.

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

30. Polarity

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.

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?

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.

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.

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.

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.

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.

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?

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?

44. Hybrid Orbitals …three degenerate sp2 orbitals.

45. Hybrid Orbitals With carbon we get…

46. Hybrid Orbitals …four degenerate sp3 orbitals.

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

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

49. Hybrid Orbitals Once you know the electron-domain geometry, you know the hybridization state of the atom. http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/hybrv18.swf

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