1. A Few “Stand-Alone” Topics in the AP Chemistry Curriculum Internal Energy Organic Molecules Spectroscopy Semiconductors Complexes

2. Topic 1: Change in Internal Energy, ΔE For any change to a gas phase system, you will need to: 1.Calculate the “PV work” involved (w) 2.Calculate the heat gained/lost (q) 3.Relate “q” and “w” to total internal energy (ΔE)

3. ΔE is the total internal energy of the system… a combination of all kinetic and potential energies (including bond energies) ΔE = q + w = smΔT + work So when internal energy is being used to do useful work, the more energy lost as heat, the less work can get done.

4. An expanding gas (the system) does work on surroundings:  w is negative To compress a gas, the surroundings does work on the system  w is positive “Work” Sign Convention

5. 5 Work = - P ex ΔV = - P ex ( V f – V i )

6. 6 So: ΔE = q + w = q + - PΔV = smΔT + - PΔV But wait! The units for smΔT will be Joules, but PΔV units will be Liter∙atm. The conversion is: 101.3 J = 1.000 L∙atm

7. Copyright © Cengage Learning. All rights reserved 7 Exercise A gas in a cylinder fitted with a movable piston releases 18.0 kJ of heat as it is compressed from 3.25 L to 0.50 L by a constant external pressure of 4.50 atm. Calculate ΔE.

8. Topic 2: Organic Structures You will need to: 1.Draw organic structures from their “structural” formulas 2.Determine central atom hybridization and sigma/pi bonding among atoms 3.Rank molecules according to properties such as boiling points and ability to hydrogen bond to other molecules

9. Drawing Lewis Structures There are very consistent bonding “patterns” in organic molecules: 1.Carbon: always 4 bonds… never lone pairs 2.Hydrogen, halogens (Cl, F, Br, I): always 1 single bond 3.Oxygen: always 2 bonds, and 2 lone pairs 4.Nitrogen in neutral molecules: 3 bonds, 1 lone pair 5.Nitrogen in cations: 4 bonds, no lone pairs

10. Examples: CH 3 OCH 2 CH 3 CH 3 (CH 2 ) 4 COCH 3 CH 2 BrCHCHCHBr 2

11. Example: NH 2 C(CH 3 ) 2 CO 2 H CH 3 CH 2 CCC(CH 3 ) 3

12. Interpreting Lewis Structures For structures drawn in “line notation”: 1.Each end of a line or “vertex” is a carbon atom unless there is something else written there. 2.Each carbon has enough hydrogens (not shown) to fulfill it’s 4 bond requirement.

13. Examples - Write the molecular formulas of each:

15. Central Atom Hybridization Now go back to each structure and determine the hybridization of each central atom. Review: 2 e - domains: sp 3 e - domains: sp 2 4 e - domains: sp 3

16. Examples: CH 3 OCH 2 CH 3 CH 3 (CH 2 ) 4 COCH 3 CH 2 BrCHCHCHBr 2

17. Example: NH 2 C(CH 3 ) 2 CO 2 H CH 3 CH 2 CCC(CH 3 ) 3

18. Examples - Write the molecular formulas of each: C 6 H 11 OCl C 7 H 11 Br

19. Examples - Write the molecular formulas of each: C 7 H 9 NO

20. Sigma (  ) and Pi (  Bonding Now go back to each structure and determine how many sigma and pi bonds per molecule Review: Single bond: sigma Double bond: 1 sigma and 1 pi Triple bond: 1 sigma and 2 pi

21. Examples: CH 3 OCH 2 CH 3 CH 3 (CH 2 ) 4 COCH 3 CH 2 BrCHCHCHBr 2

22. Example: NH 2 C(CH 3 ) 2 CO 2 H CH 3 CH 2 CCC(CH 3 ) 3

23. Examples - Write the molecular formulas of each: C 6 H 11 OCl C 7 H 11 Br

24. Examples - Write the molecular formulas of each: C 7 H 9 NO

25. Hydrogen Bonding Finally, go back to each structure and determine 1.Which ones will hydrogen bond to another molecule of the same type 2.Which ones will form hydrogen bonds with water 3.Which ones you would expect to have a low solubility in water.

26. Examples: CH 3 OCH 2 CH 3 CH 3 (CH 2 ) 4 COCH 3 CH 2 BrCHCHCHBr 2

27. Example: NH 2 C(CH 3 ) 2 CO 2 H CH 3 CH 2 CCC(CH 3 ) 3

28. Examples - Write the molecular formulas of each: C 6 H 11 OCl C 7 H 11 Br

29. Examples - Write the molecular formulas of each: C 7 H 9 NO

30. Topic 3: Spectroscopy Analytical Techniques used to Identify Unknown Substances You will need to: 1.Understand the BASICS of each type of spectroscopy. 2.Match the type of spectroscopy to the property tested 3.Interpret Mass Spectrometry Data 4.Interpret PES spectra (see ch. 7 for all PES information)

32. Ultraviolet and visible light (UV/VIS) Spectroscopy Purpose : to determine the identities of elements in a substance based on their electron transitions between energy levels.

33. Frequencies of EMR in this range will excite electrons to higher energy levels.

34. These unstable electrons will then “relax” back to lower levels by emitting EMR

35. Ultraviolet and visible light (UV/VIS) spectroscopy Each set of energy levels is unique for a given element. So each element has a characteristic absorption or emission “fingerprint”

36. Visible Light Emission Spectra

37. A sample UV Spectra… not as interesting to look at since we can’t SEE the evidence of these electron transitions.

38. Ultraviolet and visible light (UV/VIS) spectroscopy Important: colorless solutions will not work in a Visible spectrometer. Solutions will only appear colored if it contains atoms whose electrons absorb energy in the visible range.

39. Ultraviolet and visible light (UV/VIS) spectroscopy Of the elements, only transition metal ions absorb in the visible range. Dyes, pigments and indicators are usually large, complex organic molecules. Ok… now on to the next type of spectroscopy…

40. Infrared (IR) spectroscopy This is used to determine what kinds of bonds are in a compound: O-H vs. C-C vs. C=C etc.

41. Infrared (IR) spectroscopy Energy in the infrared region will cause changes in covalent bond vibrations.

42. Infrared (IR) spectroscopy Energy in the infrared region will cause changes in covalent bond vibrations. Think of bonds as “springs” whose “stretches” and “wiggles” change in a characteristic way as IR energy is increased. The next slide is an IR spectra sample, but you will not need to interpret these.

43. Infrared (IR) Spectra Example

44. Mass Spectrometry Used to identify elements in a substance by the atomic masses of the elements in the substance.

45. Mass Spectrometer

48. Mass Spectrometry 51% 12% 17% 18% 2%

49. 51% 12% 17% 18% 2% At. Wt. = (0.51x90)+(0.12x91)+(0.17x92)+(0.18x94)+(0.02x96) = 91.3 amu Zirconium

50. Topic 4- Semiconductors You will need to: 1.Understand the BASICS of semiconductors. 2.Know the difference between “n-type” and “p-type” semiconductors

51. Electron “Mobility” in Solids In a sample of a solid, elemental substance (Cu, Si, S, for ex.)… …the ability for valence electrons to be mobile throughout the crystal is governed by the “band gap” of that substance.

52. The band gap is the energy “barrier” between what’s called the “valence band” and the “conduction band”

53. Electron “Mobility” in Solids When an electron “jumps” from the valence band to the conduction band, it increases electrical conductivity in two ways: – The mobile electron in the conduction band – The electron “hole” left in the valence band

54. Electron “Mobility” in Solids The greater the band gap, the less conductive the solid will be Increasing temperature, though, will decrease the band gap… increasing conductivity

55. Semiconductors Group 14 Metalloids silicon and germanium have only a little “electron mobility” at room temperature, so they are they called semiconductors. Metalloids would need higher temps to be conductive like “real” metals.

56. Doping BUT… silicon can be “doped” with other elements to increase its conductivity at room temperature. Doping is the process of “replacing” a small percentage of the original atoms with different atoms.

57. Doping The greater the concentration of doped material, the greater the conductivity. This allows for great control over the exact degree of conductivity in the solid… … leading to the development of the modern computer chip!!!

58. n-type Semiconductors Silicon is “doped” with a small amount of a Group 15 element like arsenic or phosphorus. These atoms have one more valence electron than silicon.

59. n-type Semiconductors

60. Once bonded to the silicon, the group 15 atom’s 5 th valence electron is now free to “roam” throughout the crystal. This additional “electron mobility” means increased conductivity.

61. p-type Semiconductors Silicon is doped with a small amount of a Group 13 element like boron, which has one less valence electron than silicon.

62. p-type Semiconductors Once bonded to silicon, there is an “electron hole”… into which a neighboring electron can “fall”, creating an electron hole where IT used to be…

63. “n” and “p” are just mnemonics… “n-type” semiconductors have extra electrons that are free to roam around… thus “n” stands for extra negative electrons. “p-type” semiconductors are electron- deficient… thus “p” stands for the positive “holes”.

64. Topic 5 - Complexes You will need to: 1.Understand the BASICS of what a complex (or complex ion) is. 2.Understand how complex formation affects the solubility of a slightly soluble salt.

65. Complexes A complex is a transition metal cation that is bonded to (usually) 2, 4 or 6 ligands. Common ligands are NH 3, H 2 O, OH-, Cl-, and EDTA

66. Complexes can be neutral, positive or negative, depending on combination. What’s the charge of the TM cation in each complex? [Ag(NH 3 ) 2 ] + [Fe(H 2 O) 4 (OH) 2 ] [NiCl 4 ] 2-

67. Complexes The most common way YOU will see complexes on the AP exam is as a way to decrease the concentration of an ion in solution without precipitating it.

68. For example: adding ammonia, NH 3, to a saturated silver chloride solution… AgCl → Ag + (aq) + Cl - (aq) …will complex some of the silver taking “Ag + ” out of solution, making the solution unsaturated. Ag + + 2 NH 3 → [Ag(NH 3 ) 2 ] + (aq) So even though the solubility of AgCl is not pH dependent, it dissolves better in a solution of NH 3 than in pure water.

69. The process of a ligand complexing a transition metal cation is reversible Co 3+ (aq) + 6 CN - (aq)  [CoCN 6 ] 3- (aq) and would have a corresponding complex formation equilibrium constant, K f If two (or more) ligands are “competing” for the TM cation, the process with the greatest K f will dominate.

70. It may be hard to believe, but… 70

71. 71