2. Midterm WEDNESDAY Ch 8-10.5 Final Exam will be the last day of class. Sorry.

3. Lewis Structures of Simple Molecules C H HH H Cl O OO K+K+ KClO 3 CF 4.. HCOH H H H H C Ethyl Alcohol (Ethanol) Potassium ChlorateCarbon Tetrafluoride.. C F FF F CH 4 Methane

4. Lewis Structures of Simple Molecules Resonance Structures -IIINitrate N O OO N O O O.. N O OO

5. Resonance: Delocalized Electron-Pair Bonding - I Ozone : O 3... OO O OO O I II O O O.. Resonance Hybrid Structure One pair of electron’s resonances between the two locations!!

6. Resonance: Delocalized Electron-Pair Bonding - II C C C CC C C C C C C C C C C CC C H H H H HH H H H H H H H HH H H H Resonance Structure Benzene

8. Postulating the Hybrid Orbitals in a Molecule Problem: Describe how mixing of atomic orbitals on the central atoms leads to the hybrid orbitals in the following: a) Methyl amine, CH 3 NH 2 b) Xenon tetrafluoride, XeF 4 Plan: From the Lewis structure and molecular shape, we know the number and arrangement of electron groups around the central atoms, from which we postulate the type of hybrid orbitals involved. Then we write the partial orbital diagram for each central atom before and after the orbitals are hybridized.

9. Postulating the Hybrid Orbitals in a Molecule Problem: Describe how mixing of atomic orbitals on the central atoms leads to the hybrid orbitals in the following: a) Methyl amine, CH 3 NH 2 b) Xenon tetrafluoride, XeF 4 Plan: From the Lewis structure and molecular shape, we know the number and arrangement of electron groups around the central atoms, from which we postulate the type of hybrid orbitals involved. Then we write the partial orbital diagram for each central atom before and after the orbitals are hybridized. Solution: a) For CH 3 NH 2 : The shape is tetrahedral around the C and N atoms. Therefore, each central atom is sp 3 hybridized. The carbon atom has four half-filled sp 3 orbitals: Isolated Carbon Atom 2s2psp 3 Hybridized Carbon Atom

10. The N atom has three half-filled sp 3 orbitals and one filled with a lone pair. 2s 2p2p sp 3 C H H H H H N..

11. b) The Xenon atom has filled 5 s and 5 p orbitals with the 5 d orbitals empty. 5 s5 p 5 d Hybridized Xe atom: 5 d Isolated Xe atom sp 3 d 2

12. b) continued:For XeF 4. for Xenon, normally it has a full octet of electrons,which would mean an octahedral geometry, so to make the compound, two pairs must be broken up, and bonds made to the four fluorine atoms. If the two lone pairs are on the equatorial positions, they will be at 90 0 to each other, whereas if the two polar positions are chosen, the two electron groups will be 180 0 from each other. Thereby minimizing the repulsion between the two electron groups. Xe F F F F.. Xe F FF F Square planar 180 0

13. Fig. 9.14

14. Figure 9.15: Electronegatives of the elements.

15. Fig. 9.16

16. Fig. 9.17

17. Determining Bond Polarity from Electronegativity Values Problem: (a)Indicate the polarity of the following bonds with a polarity arrow: O - H, O - Cl, C - N, P - N, N - S, C - Br, As - S (b) rank those bonds in order of increasing polarity. Plan: (a) We use Fig. 9.16 to find the EN values, and point the arrow toward the negative end. (b) Use the EN values. Solution: a) the EN of O = 3.5 and of H = 2.1: O - H the EN of O = 3.5 and of Cl = 3.0: O - Cl the EN of C = 2.5 and of P = 2.1: C - P the EN of P = 2.1 and of N = 3.0: P - N the EN of N = 3.0 and of S = 2.1: N - S the EN of C = 2.5 and of Br = 2.8: C - Br the EN of As = 2.0 and of O = 3.5: As - O b) C - Br < C - P < O - Cl < P - N < N - S < O - H < As - O 0.3 < 0.4 < 0.5 < 0.9 < 0.9 < 1.4 < 1.5

18. Fig. 9.18

19. Percent Ionic Character as a Function of Electronegativity Difference (  En) Fig. 9.19

20. Lewis Structures for Octet Rule Exceptions Cl F F F.. B Cl.. Each fluorine atom has 8 electrons associated. Chlorine has 10 electrons! Each chlorine atom has 8 electrons associated. Boron has only 6! Cl Be.. Each chlorine atom has 8 electrons associated. The beryllium has only 4 electrons. N O O... NO 2 is an odd electron atom. The nitrogen has 7 electrons.

21. Resonance Structures - Expanded Valence Shells.. S F F F FF F Sulfur hexafluoride.. P F F F FF Phosphorous pentafluoride O S O OOHH.. O S O OOHH Sulfuric acid S = 12e - p = 10e - S = 12e - Resonance Structures

22. Lewis Structures of Simple Molecules Resonance Structures-V SOO O O SOO O O. -2. -2 Sulfate S O O OO x x x = Sulfur electronso = Oxygen electrons oo o x o x x o o o * -2 o Plus 4 others for a total of 6.

23. Chemical Compounds and Bonds Chemical Bonds - The electrostatic forces that hold the atoms of elements together in the compound. Ionic Compounds - Electrons are transferred from one atom to another to form Ionic Cpds. Covalent Compounds - Electrons are shared between atoms of different elements to form Covalent Cpds. “Cations” - Metal atoms lose electrons to form “ + ” ions. “Anions” - Nonmetal atoms gain electrons to form “ - ” ions. Mono-atomic ions form binary ionic compounds

24. Figure 2.16: Molecular and structural formulas and molecular models for some compounds.

25. Ethanol

26. An ionic compound is composed of cations and anions. Ions are arranged in a repeating three-dimensional pattern, forming a crystal. The formula of an ionic compound gives the smallest possible integer number of ions in the substance (without writing charges) so that the combination is electrically neutral. The formula gives the formula unit of the compounds. A formula unit is not a molecule!

27. Fig. 2.18

28. Figure 2.19: A model of a portion of NaCl.

29. He Ne Ar Kr Xe Rn The Periodic Table of the Elements Most Probable Oxidation State CrMnFeCoNi Mo W Tc Re Ru Os Rh Ir Pd Pt +1 +2 +3+4 +3+_4- 3- 2- 1 0 H Li Na K Rb Cs Fr Sc Y Be Mg Ca Sr Ba Ra La Ac B Al Ga In Tl Ti Rf Hf Zr C Si Ge Sn Pb F Cl Br I At O S Se Te Po N P As Sb Bi Zn Cd Hg + 2+1 Cu Ag Au +5 V Nb Ta Ce Th PrNdPmSmEuGdTbDyHoErTmYbLu PaUNpPuAmCmBkCfEsFmMdNoLr +3 DuSgBoHaMe

30. Fig. 2.20

32. Fig. 2.19

35. What is formula of the ionic compound of Mg 2+ and N 3- ? The common multiple of the charges is 6, so we need three Mg 2+ and two N 3-. The resulting formula is Mg 3 N 2

36. What is the formula of the ionic compound of Ca 2+ and PO 4 3- ? The common multiple of the charges is 6, so we need three Ca 2+ and two PO 4 3-. The resulting formula is Ca 3 (PO 4 ) 2

37. Chemical nomenclature is the systematic naming of chemical compounds. Compounds that are not organic are called inorganic compounds. Carbon monoxide, carbon dioxide, carbonates, and cyanides are also classified as inorganic compounds.

38. Naming Inorganic Compounds 1.Name the cation. 2.Name the anion.

39. 2.Some main-group metals with high atomic number have more than one cation. One cation will have the charge of the group number minus 2; the second cation will have a charge equal to the group number Pb in Group IVA(14) has two ions: Pb 2+ and Pb 4+ Tl in Group IIIA(13) has two ions: Tl + and Tl 3+

40. 3.Most transition metals form more than one cation, of which one is +2. Zn and Cd form only the +2 ion. Ag forms only the +1 ion. 4.Nonmetal main-group elements form one monatomic anion with a charge equal to the group number minus 8. F in Group VIIA(17) forms the F - ion. S in Group VIA(16) forms the S 2- ion. N in Group VA(15) forms the N 3- ion.

41. Naming Monatomic Ions Monatomic cations are named after the element if the element forms only one cation.

42. If more than one cation forms: a.In the Stock system, the charge is written using a Roman numeral and is enclosed in parentheses. Cu 2+ is copper(II). Cu + is copper(I). Fe 3+ is iron(III) Fe 2+ is iron(II) Hg 2+ is mercury(II). The second ion mercury forms is diatomic: Hg 2 2+ is mercury(I).

43. Cr 3+ is chromium(III). Cr 2+ is chromium(II). Mn 2+ is manganese(II). Co 2+ is cobalt(II). Zinc forms only Zn 2+, so it is called zinc ion. Cadmium forms only Cd 2+, so it is called cadmium ion. Silver forms only Ag +, so it is called silver ion.

44. Polyatomic Ion An ion consisting of two or more atoms chemically bonded together and carrying an electrical charge. Table 2.5 lists common polyatomic ions.

45. Cations mercury(I) or mercurousHg 2 2+ ammoniumNH 4 + Anions peroxideO 2 - hydroxideOH - cyanideCN -

46. What are the names of the following ionic compounds? –B–BaO –C–Cr 2 (SO 4 ) 3 BaO is barium oxide. Cr 2 (SO 4 ) 3 is chromium(III) sulfate or chromic sulfate.

47. What are the chemical formulas for the following ionic compounds? –p–potassium carbonate –m–manganese(II) sulfate The ions K + and CO 3 2- form K 2 CO 3 The ions Mn 2+ and SO 4 2- form MnSO 4

48. Binary Molecular Compounds A compound composed of only two elements. Binary compound of a metal and a nonmetal are generally named using ionic rules.

49. Molecular Naming Binary Molecular Compounds We usually name the elements in the order given in the formula. Name the first element using the element name. Name the second element using the element root + -ide suffix.

50. Add a prefix to each name to indicate the number of atoms of that element. The prefix mono- is used only when needed to distinguish two compounds of the same two elements. The final vowel of the prefix is often dropped when followed by an element name that begins with a vowel. Oxygen is the most common example. N 2 O 4 dinitrogen tetroxide (“a” is dropped) NO nitrogen monoxide(only one “o”) (also called nitric oxide)

51. Don’t use these when naming ionic compounds--they’re ONLY for covalent compounds!!

52. Some compounds have common names that differ from their systematic names: H 2 Shydrogen sulfide (the “di” is omitted) H 2 Owater NH 3 ammonia Common names need to be memorized.

53. What are the names of the following compounds? –O–OF 2 –S–S 4 N 4 –B–BCl 3 OF 2 is oxygen difluoride S 4 N 4 is tetrasulfur tetranitride BCl 3 is boron trichloride

54. What are the formulas for the following binary molecular compounds? –c–carbon disulfide –n–nitrogen tribromide –d–dinitrogen tetrafluoride The formula for carbon disulfide is CS 2. The formula for dinitrogen tetrafluoride is N 2 F 4. The formula for nitrogen tribromide is NBr 3.

55. Acids and Corresponding Anions Oxoacids contain hydrogen, oxygen, and a third central atom. To name an acid from its anion name: 1.Change an –ate suffix to –ic. 2.Change an –ite suffix to –ous. 3.Add the word “acid.”” For example: HNO 3 nitric acid H 2 SO 4 sulfuric acid

57. Figure 2.23: Molecular model of nitric acid. What is the formula for nitric acid? HNO 3

58. Bromine has an oxoacid, HBrO 2, bromous acid (compare to HClO 2, chlorous acid). What are the name and formula of the corresponding anion? The anion corresponding to HBrO 2 is bromite, BrO 2 -.

59. Hydrate A compound that contains water molecules weakly bound in the crystals. The formula of a hydrate is written with a dot before the water molecule(s) included. For example: CuSO 4  5H 2 O

60. Hydrates are named using the anhydrous (without water) compound name followed by the prefix for the number of water molecules included and the word “hydrate.” For example: CuSO 4  5H 2 O is named copper(II) sulfate pentahydrate.

61. A compound whose common name is green vitriol has the chemical formula FeSO 4  7H 2 O. What is the chemical name of this compound? FeSO 4  7H 2 O is iron(II) sulfate heptahydrate.

62. Calcium chloride hexahydrate is used to melt snow on roads. What is the chemical formula of the compound? The chemical formula for calcium chloride hexahydrate is CaCl 2  6H 2 O.

63. Organic Compounds An important class of molecular substances; they contain carbon combined with other elements – notably hydrogen, oxygen, and nitrogen. Hydrocarbons contain only carbon and hydrogen.

66. A functional group is a reactive portion of a molecule that undergoes predictable reactions. ExamplesName of Group Functional Group Methyl alcoholAlcohol Dimethyl etherEther Acetic acidCarboxylic acid

67. A chemical equation is the symbolic representation of a chemical reaction in terms of chemical formulas. For example:2Na + Cl 2  2NaCl Reactants are the starting materials; they are written on the left of the equation. Products are the materials at the end of the reaction; they are written on the right of the equation.

68. Because a reaction must accurately describe the chemical reaction, it must be consistent with the law of conservation of mass. When this is not the case, after correct formulas are written for each reactant and product, the coefficients are adjusted so that the same number of each atom is present in both the reactants and the products. This is called balancing the equation.

69. For example, the reaction of sodium with chlorine produced sodium chloride. First, we determine the correct formula for each compound. Sodium is Na. Chlorine is Cl 2. Sodium chloride is NaCl.

70. Second, we write the reaction. Na + Cl 2  NaCl Third, we check the number of each atom on each side of the equation. This equation shows two Cl atoms on the reactant side and only one Cl atom on the product side. To balance the Cl atoms, we insert a coefficient of “2” before NaCl on the product side. Na + Cl 2  2NaCl

71. Now the Na are not balanced: there is one on the reactant side and there are two on the product side. To balance Na, we insert the coefficient “2” before Na on the reactant side. 2Na + Cl 2  2NaCl The reaction is now balanced!

72. Balance the following equation: CS 2 + O 2  CO 2 + SO 2 Tally the number of each atom on each side: C1 on reactant side; 1 on product side S2 on reactant side; 1 on product side O2 on reactant side; 4 on product side Begin by inserting the coefficient “2” before SO 2 on the product side. We leave O 2 until later because it is an element.

73. CS 2 + O 2  CO 2 + 2SO 2 Tally the atoms again: C1 on reactant side; 1 on product side S2 on reactant side; 2 on product side O2 on reactant side; 6 on product side Insert a “3” before O 2 : CS 2 + 3O 2  CO 2 + 2SO 2

74. Tally the atoms again: C1 on reactant side; 1 on product side S2 on reactant side; 2 on product side O6 on reactant side; 6 on product side The reaction is now balanced!

75. Balance the following equation: NH 3 + O 2  NO + H 2 O Tally the number of each atom on each side: N1 on reactant side; 1 on product side H3 on reactant side; 2 on product side O2 on reactant side; 2 on product side Begin by inserting the coefficient “2” before NH 3 on the reactant side and the coefficient “3” before H 2 O on the product side. We leave O 2 until later because it is an element.

76. 2NH 3 + O 2  NO + 3H 2 O Tally the atoms again: N2 on reactant side; 1 on product side H6 on reactant side; 6 on product side O2 on reactant side; 4 on product side To balance N, insert a “2” before NO: 2NH 3 + O 2  2NO + 3H 2 O

77. Tally the atoms again: N2 on reactant side; 2 on product side H6 on reactant side; 6 on product side O2 on reactant side; 5 on product side Since this gives us an odd number oxygens, we double the coefficients on NH 3, NO, and H 2 O and to balance O, insert a “5” before O 2.

78. Tally the atoms again to double check: 4NH 3 + 5O 2  4NO + 6H 2 O N4 on reactant side; 4 on product side H12 on reactant side; 12 on product side O10 on reactant side; 10 on product side The reaction is now balanced!

79. Balance the following equation: C 2 H 5 OH + O 2  CO 2 + H 2 O Tally the number of each atom on each side: C2 on reactant side; 1 on product side H6 on reactant side; 2 on product side O3 on reactant side; 3 on product side Begin by balancing H. Insert the coefficient “3” before H 2 O on the product side. We leave O 2 until later because it is an element.

80. C 2 H 5 OH + O 2  CO 2 + 3H 2 O Tally the number of each atom on each side: C2 on reactant side; 1 on product side H6 on reactant side; 6 on product side O3 on reactant side; 5 on product side To balance C, insert a “2” before CO 2.