1. As a general rule, the repulsion decreases as follows:
lone pair – lone pair > lone pair – bond pair > bond pair – bond pair
repulsion repulsion repulsion
Electrons in a bond pair are held by the attractive forces of the two nuclei in the two atoms. These electrons have less spatial distribution than lone pair electrons, which are associated with only one particular atom.

2. Consequently, the lone pair electrons occupy more space and experience a greater repulsion from the neighboring lone pairs and bond pairs.

3. Consequently, the lone pair electrons occupy more space and experience a greater repulsion from the neighboring lone pairs and bond pairs.
Examples:
1. Sulfur dioxide, SO2

4. Consequently, the lone pair electrons occupy more space and experience a greater repulsion from the neighboring lone pairs and bond pairs.
Examples:
1. Sulfur dioxide, SO2
The Lewis structure is

5. Consequently, the lone pair electrons occupy more space and experience a greater repulsion from the neighboring lone pairs and bond pairs.
Examples:
1. Sulfur dioxide, SO2
The Lewis structure is
That is ignoring the lone pairs on
the surrounding atoms.

6. Note that it is essential to retain the lone pair on the central atom.

7. Note that it is essential to retain the lone pair on the central atom
Note that it is essential to retain the lone pair on the central atom. This particular example has resonance.

8. Note that it is essential to retain the lone pair on the central atom
Note that it is essential to retain the lone pair on the central atom. This particular example has resonance. The second structure is

9. Note that it is essential to retain the lone pair on the central atom
Note that it is essential to retain the lone pair on the central atom. This particular example has resonance. The second structure is
That is ignoring the lone pairs on the surrounding atoms.

10. Note that it is essential to retain the lone pair on the central atom
Note that it is essential to retain the lone pair on the central atom. This particular example has resonance. The second structure is
That is ignoring the lone pairs on the surrounding atoms.
If we take the average of the two resonance structures, then we have
where represents 1.5 bonds.

11. There are three charge clouds off the central S atom.

12. There are three charge clouds off the central S atom
There are three charge clouds off the central S atom. To get these as far apart as possible, requires the structure of SO2 to be bent (or nonlinear).

13. There are three charge clouds off the central S atom
There are three charge clouds off the central S atom. To get these as far apart as possible, requires the structure of SO2 to be bent (or nonlinear).

14. There are three charge clouds off the central S atom
There are three charge clouds off the central S atom. To get these as far apart as possible, requires the structure of SO2 to be bent (or nonlinear).
If SO2 were linear repulsion between bond pairs would be minimized, but the angle between the lone pair and each of the two S O bond pairs would be 90o.

15. There are three charge clouds off the central S atom
There are three charge clouds off the central S atom. To get these as far apart as possible, requires the structure of SO2 to be bent (or nonlinear).
If SO2 were linear repulsion between bond pairs would be minimized, but the angle between the lone pair and each of the two S O bond pairs would be 90o. Since lone pair-bond pair repulsion is greater than bond pair-bond pair repulsion, SO2 must be bent.

16. There are three charge clouds off the central S atom
There are three charge clouds off the central S atom. To get these as far apart as possible, requires the structure of SO2 to be bent (or nonlinear).
If SO2 were linear repulsion between bond pairs would be minimized, but the angle between the lone pair and each of the two S O bond pairs would be 90o. Since lone pair-bond pair repulsion is greater than bond pair-bond pair repulsion, SO2 must be bent. Note that we have no means to predict the exact OSO angle.

17. 2. ammonia, NH3

18. 2. ammonia, NH3
The Lewis structure is

19. 2. ammonia, NH3
The Lewis structure is
The underlying symmetry is tetrahedral.

20. Four charge clouds (three bond pairs and one lone pair) point to the vertices of a tetrahedron.

21. Four charge clouds (three bond pairs and one lone pair) point to the vertices of a tetrahedron.
No bonds in place yet.

22. Four charge clouds (three bond pairs and one lone pair) point to the vertices of a tetrahedron.

23. Four charge clouds (three bond pairs and one lone pair) point to the vertices of a tetrahedron.
However, the shape of a molecule is determined by the positions of the nuclei.

24. Four charge clouds (three bond pairs and one lone pair) point to the vertices of a tetrahedron.
However, the shape of a molecule is determined by the positions of the nuclei. So the shape of NH3 is trigonal pyramidal.

25. Four charge clouds (three bond pairs and one lone pair) point to the vertices of a tetrahedron.
However, the shape of a molecule is determined by the positions of the nuclei. So the shape of NH3 is trigonal pyramidal. Because the lone pair repels the bond pairs more strongly than the bond pair-bond pair repulsions, the three N H bonds in ammonia are pushed closer together, with the result that the bond angle HNH is smaller than the tetrahedral angle.

26. 3. water, H2O

27. 3. water, H2O
The Lewis structure is

28. 3. water, H2O
The Lewis structure is
that is

30. Four charge clouds (two bond pairs and two lone pairs) point to the vertices of a tetrahedron (underlying symmetry is tetrahedral).

31. Four charge clouds (two bond pairs and two lone pairs) point to the vertices of a tetrahedron (underlying symmetry is tetrahedral). But the shape is determined by the position of the nuclei, so the molecule is bent.

32. Four charge clouds (two bond pairs and two lone pairs) point to the vertices of a tetrahedron (underlying symmetry is tetrahedral). But the shape is determined by the position of the nuclei, so the molecule is bent.

33. We would expect a further decrease in the bond angle HOH (due to lone pair-lone pair repulsion and lone pair-bond pair repulsion) from the regular tetrahedral angle of 109o 28’. The bond angle in water is 105o (from experiment – not from VSEPR theory).

34. 4. xenon tetrafluoride, XeF4

35. 4. xenon tetrafluoride, XeF4 The valence electron count for xenon is taken to be 8 electrons.

36. 4. xenon tetrafluoride, XeF4 The valence electron count for xenon is taken to be 8 electrons. So the Lewis structure is

37. 4. xenon tetrafluoride, XeF4 The valence electron count for xenon is taken to be 8 electrons. So the Lewis structure is or, in more simplified form (for the purpose of getting the shape)

38. 4. xenon tetrafluoride, XeF4 The valence electron count for xenon is taken to be 8 electrons. So the Lewis structure is or, in more simplified form (for the purpose of getting the shape) Note it is essential to retain the lone pairs on Xe.

39. possibility one

40. possibility two

41. possibility two
Which of the two structures is to be preferred?

42. possibility two
Which of the two structures is to be preferred?
The key interaction to consider is the lone pair-lone pair repulsion.

43. possibility two
Which of the two structures is to be preferred?
The key interaction to consider is the lone pair-lone pair repulsion.
You usually want to make the angle between the lone pairs as large as possible, so possibility two is energetically preferred.

44. So XeF4 is square planar.

45. 5. bromine pentafluoride, BrF5

46. 5. bromine pentafluoride, BrF5 The Lewis structure is

47. 5. bromine pentafluoride, BrF5 The Lewis structure is or in more compact form (for the purpose of getting the shape)

48. There are a total of six charge clouds off the central atom (five bond pairs and one lone pair).

50. There are a total of six charge clouds off the central atom (five bond pairs and one lone pair).

51. Arrange these to point to the vertices of a regular octahedron.
There are a total of six charge clouds off the central atom (five bond pairs and one lone pair).
Arrange these to point to the vertices of a regular octahedron.

52. There are a total of six charge clouds off the central atom (five bond pairs and one lone pair).
Arrange these to point to the vertices of a regular octahedron. So the shape of the BrF5 molecule is square pyramidal.

53. Note in the examples with lone pairs on the central atom, these lone pairs play an important role in predicting the correct shape of the molecule.

54. Summary of rules for applying VSEPR Theory to predict the geometry of simple molecules

55. Summary of rules for applying VSEPR Theory to predict the geometry of simple molecules
1. Write the Lewis structure of the molecule: (i) Draw in chemical bonds from the central atom. (ii) Draw in any lone pairs on the central atom.

56. Summary of rules for applying VSEPR Theory to predict the geometry of simple molecules
1. Write the Lewis structure of the molecule: (i) Draw in chemical bonds from the central atom. (ii) Draw in any lone pairs on the central atom. 2. To predict basic shape, note that lone pairs repel one another and bond pairs more strongly than bond pairs repel one another.

57. 3. As a good approximation, treat double and triple bonds as if they were single electron pairs between neighboring atoms. Finally, remember that there is no way to predict bond angles accurately in cases involving lone pairs on the central atom.

58. Exercises Predict the shape of the following: 1. CO2 2. BCl3 3. SiH4
4. AsCl5
5. PF6-

59. Exercises Predict the shape of the following: 6. CO32- 7. NO2-
8. ClO3-
9. BrF3
10. ICl4-

60. Exercises 1. linear 2. trigonal planar 3. tetrahedral
4. trigonal bipyramidal
5. octahedral