AP Chem

1. AP Chem STUFF YOU PROBABLY FORGOT!!!!

2. Chapter 2

3. Atomic Size Size Increases going down a group. Because electrons are added further from the nucleus, there is less attraction. Size Decreases going across a period. Electrons are held closer.

4. Trends in Ionization Energy IE increases across a period because Z* increases. Metals lose electrons more easily than nonmetals. Metals are good reducing agents. Nonmetals lose electrons with difficulty.

5. Trends in Ionization Energy IE decreases down a group Because size increases. Electrons are held less closely

6. Electron affinity Trends Cont • In general electrons are more difficult to add going down a group. Because • Nuclear charge increases • Atomic radius increases down a group decreasing electron affinity

7. Electronegativity A measure of the ability of an atom in a chemical compound to attract electrons. Most electronegative is Fluorine. Least electronegative are alkali and alkaline metals Most electronegative are the halogens nitrogen and oxygen.

8. Quantum Mechanics Quantum Numbers • The use of wavefunctions generates four quantum numbers. • Principal Quantum Number (n) • This is the same as Bohr’s n • Allowed values: 1, 2, 3, 4, … (integers) 8

9. Quantum Mechanics Quantum Numbers • Secondary (Azimuthal) Quantum Number (l) • Allowed values: 0, 1, 2, 3, 4, . , (n – 1) (integers) • Each l represents an orbital type 9

10. Quantum Mechanics Quantum Numbers • Magnetic Quantum Number (ml ). • This quantum number depends on l. • Allowed values: -l +l by integers. • Magnetic quantum number describes the orientation of the orbital in space. 10

11. Quantum Mechanics Quantum Numbers • Spin Quantum Number (s) • Allowed values: -½  +½. • Electrons behave as if they are spinning about their own axis. • This spin can be either clockwise or counter clockwise. 11

12. Quantum Mechanics Orbitals and Quantum Numbers 12

13. Chapter 3

14. Percent composition • Percent of each element in a compound. TO Calculate: • Find the mass of each element • Divide that by the molar mass • Multiply by a 100.

15. Limiting Reactant Problems Balanced chemical equation. Determine the number of moles of each reactant. Divide each number of moles of each reactant by the coefficient in the balanced equation. The smallest number is the limiting reactant. Proceed with solving problem. 15

16. Percent Yield = Actual yield x 100 Theoretical yield 16

17. Chapter 4

18. Acid/Base definitions #1: Arrhenius (traditional) • Arrhenius Acids contain H+ or hydronium H3O+ ions as the only positive ions • Arrhenius Bases contain OH- ions  problem: some bases don’t have hydroxide ions

19. Acid/Base Definitions Definition #2: Brønsted – Lowry Acids – any substance that donates a H+ ion (proton donor) Bases – any substance that accepts a H+ ion (proton acceptor) A “proton” is really just a hydrogen atom that has lost it’s electron!

20. Acids & Base Definitions Lewis acid - a substance that accepts an electron pair Definition #3 – Lewis Lewis base - a substance that donates an electron pair

21. Conjugate Pairs

22. 5 Solubility Rules • 1) All nitrates, hydrogen carbonates, acetates, and chlorates are soluble • 2) All alkali metal and NH4+ compounds are soluble

23. 5 Solubility Rules • 3) All Cl-are soluble except Ag+, Hg22+,and Pb2+ • 4) All SO42- are soluble except Ca2+, Ba2+, Sr2+, and Pb2+

24. 5 Solubility Rules • 5) Everything else is insoluble!

25. Chapter 5 Gases

26. Pressure Column height measures Pressure of atmosphere • 1 standard atmosphere (atm) * = 760 mm Hg (or torr) * = 29.92 inches * = 14.7 pounds/in2 (psi) * = about 34 feet of water = 101.3 kPa * (SI unit is PASCAL) * Memorize these!

27. Combined Gas Law If you should only need one of the other gas laws, you can cover up the item that is constant and you will get that gas law! = P1 V1 P2 Boyle’s Law Charles’ Law Gay-Lussac’s Law V2 T1 T2

28. And now, we pause for this commercial message from STP OK, so it’s really not THIS kind of STP… STP in chemistry stands for Standard Temperature and Pressure Standard Pressure = 1 atm (or an equivalent) Standard Temperature = 0 deg C (273 K) STP allows us to compare amounts of gases between different pressures and temperatures

29. Ideal Vs Real Gases • No gas behaves in a truly ideal fashion. • Ideal gas law fails under conditions of high pressure and low temperature. • WHY? Gas molecules DO take up space and DO interact through attractive forces with one another.

30. Remember Kinetic Molecular Theory • Theory tells why the things happen. • explains why ideal gases behave the way they do. • Assumptions that simplify the theory, but don’t work in real gases. • The particles are so small we can ignore their volume. • The particles are in constant motion and their collisions cause pressure.

31. Kinetic Molecular Theory • The particles do not affect each other, neither attracting or repelling. • The average kinetic energy is proportional to the Kelvin temperature. • Appendix 2 shows the derivation of the ideal gas law and the definition of temperature. • We need the formula KE = 1/2 mv2

32. Root Mean Square • m is kg for one particle, so Na m is kg for a mole of particles. We will call it M • Where M is the molar mass in kg/mole, and R has the units 8.3145 J/Kmol. • The velocity will be in m/s

33. Effusion • Passage of gas through a small hole, into a vacuum. • The effusion rate measures how fast this happens. • Graham’s Law the rate of effusion is inversely proportional to the square root of the mass of its particles.

34. Chapter 6 Thermodynamics

35. 6.1 Fluid Work (p. 233-4) • Gases do work via volume change according to the formula below. The system is often a gas-filled piston. • When a gas expands (+DV), work is negative. • When a gas compresses (-DV), work is positive. • Units of gas expansion/compression work are conveniently L · atm. 1 L · atm = 101.3 J. w = -PDV

36. 6.2 Enthalpy • At constant pressure, a system’s enthalpy change equals the energy flow as heat: • For a reaction at constant pressure, DH is given by: • If DH is positive, the reaction is endothermic. • If DH is negative, the reaction is exothermic. DH = qP DHreaction = Hproducts - Hreactants

37. 6.3 Calculating DHreaction • There are three quantitative relationships between a chemical equation and DH: • 1) If a chemical equation is multiplied by some integer, DH is also multiplied by the same integer. • 3A + B  2C DH • 2 · (3A + B  2C) 2 · (DH) • 6A + 2B 4C 2DH • 2) If a chemical equation is reversed, then DH changes sign. • A + B  C + D DH • C + D  A + B -DH

38. 6.3 Calculating DHreaction • Hess’ Law • 3) If a chemical reaction can be expressed as the sum of a series of steps, then DH for the overall equation is the sum of the DH’sfor each step. • 3A + B  C DH1 • C  2DDH2 • 3A + B  2D DH1 + DH2 • Since enthalpy is a state function, DH is dependent only on initial and final states.

39. 6.4 Standard Enthapy of Reaction, 1111111 • The enthalpy change for a given reaction is calculated by subtracting the enthalpies of formation of the reactants from the enthalpies of formation of the products: • Elements in their standard states aren’t included since their is zero. = Snp [products] - Snr [reactants]

40. Chapter 8

41. Percent Ionic Character Ionic Bond % IC > 50 % Polar Covalent % IC 5 - 50 % Nonpolar Covalent % IC < 5 % where xA is the larger electronegativity and xB is the smaller value. Watch significant figures!!!

42. Energy Changes & Chemical Bonding Most chemical reactions can be explained in terms of the rearrangements of bonds. • Energy must be added to break existing bonds in the reacting substances. • Energy is released when new bonds are formed in the products. • For the reactions of covalent compounds, the net energy of the reaction, ∆Hrxn, = (sum of energy added to break bonds) - (sum of energy released when bonds form).

43. Rules for Formal Charges • Assign each atom half of the electrons in each pair it shares. • Also give each atom all electrons from unshared pairs it has. • Subtract the number of electrons assigned to each atom from the number of valence electrons for an atom of the element.

44. Formal Charges For each oxygen (4 electrons assigned from unshared e- + 2 e- from the bonds) = 6 total Formal charge = 6 - 6 = 0 For carbon 4 e- assigned from the bonds = 4 total Formal charge = 4 - 4 = 0 O=C=O 0 0 0 -1 +1 0 O C=O Structure 1 Structure 2 For the single-bond oxygen (6 e- from unshared e- + 1e- from bond) = 7 total Formal charge = 6 - 7 = -1 For the triple-bond oxygen (2 e- from unshared e- + 3e- from bonds) = 5 total Formal charge = 6 - 5 = +1 For carbon 4 e- from the bonds = 4 total Formal charge = 4 - 4 = 0

45. VSEPR shapes Coordination Electron pairs General Number Bonding Unshared Formula Shape 2 2 0 AB2 Linear 3 3 0 AB3 Trigonal planar 2 1 AB2 Bent 4 4 0 AB4 Tetrahedral 3 1 AB3 Trigonal pyramidal 2 2 AB2 Bent 1 3 AB Linear

46. VSEPR shapes Coordination Electron pairs General Number Bonding Unshared Formula Shape 5 5 0 AB5Trigonalbipyramidal 4 1 AB4 Seesaw 3 2 AB3 T-shaped 2 3 AB2 Linear 6 6 0 AB6 Octahedral 5 1 AB5 Square pyramidal 4 2 AB4 Square Planar

47. Other hybrid orbitals d orbitals can also be involved in the formation of hybrid orbitals. Hybrid Shape sp Linear sp2 Trigonal planar sp3 Tetrahedral sp3d Trigonal bipyramidal sp3d 2 Octahedral sp3d 3 Pentagonal bipyramidal

48. MO diagram for NO π *2px π *2pz π *2py 2px 2py 2pz 2px 2py 2pz π 2px π 2pz π 2py Ó *2s 2s 2s Ó2s Ó *1s 1s 1s Ó1s N NO O # bonds = 10 bonding e- - 5 antibonding e- = 2.5 2

49. Chapter 9

50. A sigma () bondcenters along the internuclear axis. A pi () bondoccupies the space above and below the internuclear axis. 50