1. DOT STRUCTURES!

2. One way of doing dot structures is to follow a few simple rules:

3. 1. Count the total number of valence electrons from each atom 2. Arrange the molecule: place the outer atoms around the central atom. 3. Place two electrons between each atom; add the rest of the electrons so that each atom has an octet. 4. If one or more of the atoms cannot get an octet with the available electrons, then rearrange the electrons into double or triple bonds as needed for the octets.

4. EXAMPLE: SO 3 sulfur trioxide

5. step 1: How many total valence electrons does SO 3 have?

6. 6 + 6 + 6 + 6 = 24 step 1: How many total valence electrons does SO 3 have?

7. 24

9. So, distribute 24 valence electrons around SO 3 so that each atom has an octet!

10. 24 electrons step 2: arrange the outer atoms around the central atom

11. S 24 electrons step 2: arrange the outer atoms around the central atom

12. SO 24 electrons step 2: arrange the outer atoms around the central atom

13. SO O 24 electrons step 2: arrange the outer atoms around the central atom

14. OSO O 24 electrons step 2: arrange the outer atoms around the central atom

15. OO O step 3: put electrons in the bonds 24 electrons S

16. OO O 1 23 electrons left S step 3: put electrons in the bonds

17. OO O 1 2 22 electrons left S step 3: put electrons in the bonds

18. OO O 1 2 3 21 electrons left S step 3: put electrons in the bonds

19. OO O 1 2 3 4 20 electrons left S step 3: put electrons in the bonds

20. OO O 1 2 3 4 5 19 electrons left S step 3: put electrons in the bonds

21. OO O 1 2 3 4 5 6 18 electrons left S step 3: put electrons in the bonds

22. OO O 1 2 3 4 5 6 18 electrons left S

23. OO O then, distribute the rest to form the octets 1 2 3 4 5 6 18 electrons left S

24. OO O 1 2 3 4 5 6 87 16 electrons left S

25. OO O 1 2 3 4 5 6 87 9 10 14 electrons left S

26. OO O 1 2 3 4 5 6 87 9 10 11 12 electrons left S

27. OO O 1 2 3 4 5 6 87 9 10 11 12 14 13 10 electrons left S

28. OO O 1 2 3 4 5 6 87 9 10 11 12 14 13 15 16 8 electrons left S

29. OO O 1 2 3 4 5 6 87 9 10 11 12 14 13 15 16 17 18 6 electrons left S

30. OO O 1 2 3 4 5 6 87 9 10 11 12 14 13 15 16 17 18 19 20 4 electrons left S

31. OO O 1 2 3 4 5 6 87 9 10 11 12 14 13 15 16 17 18 19 20 21 22 2 electrons left S

32. OO O 1 2 3 4 5 6 87 9 10 11 12 14 13 15 16 17 18 19 20 21 22 2324 S

33. OO O 1 2 3 4 5 6 87 9 10 11 12 14 13 15 16 17 18 19 20 21 22 2324 S step 4: check to see that each atom has an octet

34. OO O oxygen does not have an octet 1 2 3 4 5 6 87 9 10 11 12 14 13 15 16 17 18 19 20 21 22 2324 S

35. OO O sulfur can share these with oxygen and still count them in its octet 1 2 3 4 5 6 87 9 10 11 12 14 13 15 16 17 18 19 20 21 22 2324 S

36. OO O 1 2 3 4 5 6 87 9 10 11 12 14 13 15 16 17 18 19 20 21 22 2324 S

37. OO O 1 2 3 4 5 6 87 9 10 11 12 14 13 15 16 17 18 19 20 21 22 2324 S

38. OO O 1 2 3 4 5 6 87 9 10 11 12 14 13 15 16 17 18 19 20 21 22 2324 S

39. OO O 1 2 3 4 5 6 87 9 10 11 12 14 13 15 16 17 18 19 20 21 22 2324 S

40. OO O 1 2 3 4 5 6 87 9 10 11 12 14 13 15 16 17 18 19 20 21 22 2324 S

41. OO O 1 2 3 4 5 6 87 9 10 11 12 14 13 15 16 17 18 19 20 21 22 2324 S

42. OO O 1 2 3 4 5 6 87 9 10 11 12 14 13 15 16 17 18 19 20 21 22 2324 S

43. OO O 1 2 3 4 5 6 87 9 10 11 12 14 13 15 16 17 18 19 20 21 22 2324 S

44. OO O 1 2 3 4 5 6 87 9 10 11 12 14 13 15 16 17 18 19 20 21 22 2324 S

45. OO O 1 2 3 4 5 6 87 9 10 11 12 14 13 15 16 17 18 19 20 21 22 2324 S

46. OO O 1 2 3 4 5 6 87 9 10 11 12 14 13 15 16 17 18 19 20 21 22 2324 oxygen now has an octet S

47. OO O 1 2 3 4 5 6 87 9 10 11 12 14 13 15 16 17 18 19 20 21 22 2324 sulfur still has its octet S

48. OO O 1 2 3 4 5 6 87 9 10 11 12 14 13 15 16 17 18 19 20 21 22 2324 S

49. OO O 1 2 3 4 5 6 87 9 10 11 12 14 13 15 16 17 18 19 20 21 22 2324 S

50. OO O S

51. OO O S

52. (don’t forget resonance!) OO O S

53. (don’t forget shapes!) OO O S

54. O S O O

55. SO 3 O S O O

56. ! O S O O