1. 4 - 1 CH 104 Chapter 4 CH04 Chapter 5: Compounds & Bonds Valence Electrons & e- Dot Structures Octet Rule & Ions Ionic Compounds & Formulas Covalent Compounds & Formulas Polyatomic Ions Molecular Shapes & Polarity Attractive Forces

2. 4 - 2 CH 104 Chapter 4 Electron arrangement 2 8 18 32 Electrons fill layers around nucleus Low  High Shells = Energy levels 2412 Mg

3. 4 - 3 CH 104 Chapter 4 Inner vs. valence electrons Valence electrons This is where most chemical reactions occur. Inner electrons Not much happens here under normal conditions.

4. 4 - 4 CH 104 Chapter 411H 73Li 42He 94 Be2010Ne 2311 Na 2412 Mg 4018 Ar Octet Rule

5. 4 - 5 CH 104 Chapter 411H 73Li 42He 94 Be2010Ne 2311 Na 2412 Mg 4018 Ar Octet Rule 1s 2, 2s 1 1s 2, 2s 2 1s 1 1s 2 1s 2, 2s 2 2p 6 1s 2, 2s 2 2p 6, 3s 1 1s 2, 2s 2 2p 6, 3s 2 [Ne] 3s 1 1s 2, 2s 2 2p 6, 3s 2 3p 6

6. 4 - 6 CH 104 Chapter 411H 73Li 2311 Na Lewis Structures Show only Valence Electrons H Li Na K

7. 4 - 7 CH 104 Chapter 4 H Li Na K He Be B C O F Ne N Mg Ca Al Ga Si Ge P As S Se ClBrArKr 1 234567 8

8. 4 - 8 CH 104 Chapter 4 2311 Na Ions Metals give e - s to make Cations Na 11 +’s 11 -’s 0 11 +’s 11 -’s 0 11 +’s 10 -’s 1 + 1 + 11 +’s 10 -’s 1 + 1 + Na 1+ 2, 8 = [Ne]

9. 4 - 9 CH 104 Chapter 4 Ions Nonmetals take e - s to make Anions 17 +’s 17 -’s 0 17 +’s 17 -’s 0 17 +’s 18 -’s 1 - 1 - 17 +’s 18 -’s 1 - 1 - Cl 1- 3517 Cl = Cl 1- 2, 8, 8 = [Ar]

10. 4 - 10 CH 104 Chapter 4 Formation of NaCl Na + Cl Na + + Cl + and - ions attract to form an ionic bond. _ e- moves from Metal  Nonmetal Metal Cation Nonmetal Anion Stable octets

11. 4 - 11 CH 104 Chapter 4 Ionic compounds Not individual molecules Form crystal arrays Ions touch many others Formula represents the average ion ratio NaCl sodium chloride NaCl sodium chloride Na Cl

12. Common ions 4 - 6 1+ 2+2+3+3+ 4+4-4+4- 3-3-2-2-1-1- Representative Elements

13. 4 - 13 CH 104 Chapter 4 Ionic Formulas Metal Cations + Nonmetal Anions Na 1+ Cl 1- NaCl Sodium Chloride Al 3+ Cl 1- AlCl 3 Aluminum Chloride Cl 1-

14. 4 - 14 CH 104 Chapter 4 Common ions H Li Na Cs Rb K TlBa Fr He RnAtPoBiPb Be Mg Sr Ca Ra InXeITeSbSn GaKrBrSeAsGe AlArClSPSi BNeFONC HgAuHfLsPtIrOsReWTa CdAgZrYPdRhRuTcMoNb Ac ZnCuTiScNiCoFeMnCrV Gd Cm Tb Bk Sm Pu Eu Am Nd U Pm Np Ce Th Pr Pa Yb No Lu Lr Er Fm Tm Md Dy Cf Ho Es 1+ 2+2+3+3+ 4+4-4+4- 3-3-2-2-1-1- Transition Elements VariableVariable Representative Elements

15. 4 - 15 CH 104 Chapter 4 Information in the table Fe 55.845 Elemental Symbol Electronic Configuration Atomic mass (weight) 26 Iron 2,3 [Ar]3d 6 4s 2 Atomic number Name of the element Oxidation states (Valence) (Valence) Oxidation states (Valence) (Valence)

16. 4 - 16 CH 104 Chapter 4 Transition Metal Ions Fe 2+ Cl 1- FeCl 2 Iron (II) Chloride Fe 3+ Cl 1- FeCl 3 Iron (III) Chloride Cl 1- Ferrous Chloride Ferric Chloride

17. 4 - 17 CH 104 Chapter 4 Ionic compounds Cations Anions Na 1+ Mg 2+ Al 3+ Br 1- O 2- N 3- NaBr Na 2 O Na 3 N MgBr 2 AlBr 3 MgO Al 2 O 3 Mg 3 N 2 AlN Fe 3+ Cu 1+ FeBr 3 Fe 2 O 3 FeN CuBr Cu 2 O Cu 3 N Sodium Bromide Sodium Oxide Sodium Nitride Magnesium Bromide Aluminum Bromide Iron (III) Bromide Ferric Bromide Copper (I) Bromide Cuprous Bromide Magnesium Oxide Aluminum Oxide Iron (III) Oxide Ferric Oxide Copper (I) Oxide Cuprous Oxide Magnesium Nitride Aluminum Nitride Iron (III) Nitride Ferric Nitride Copper (I) Nitride Cuprous Nitride

18. 4 - 18 CH 104 Chapter 4 H Li Na K He Be B C O F Ne N Mg Ca Al Ga Si Ge P As S Se ClBrArKr 1 234567 8 Metals give e-s to nonmetals Nonmetals Share e-s with other nonmetals Nonmetals Share e-s with other nonmetals

19. 4 - 19 CH 104 Chapter 4 Covalent Bonds HH + HHClO + O + + N N N NO O

20. 4 - 20 CH 104 Chapter 4 Covalent Bonds HH Cl N NO O H 2 H-H H 2 Cl 2 Cl-Cl Cl 2 O 2 O=O O 2 N N2N2N2N2

21. 4 - 21 CH 104 Chapter 4 OC OC Covalent Bonds O=C=O C O O Carbon monoxide Carbon dioxide CO 2 May modify rules to improve sound. iemonmono ie - monoxide not monooxide.

22. 4 - 22 CH 104 Chapter 4 CO CO 2 N2O5N2O5 SiO 2 ICl 3 P2O5P2O5 CCl 4 Naming covalent compounds May modify rules to improve the sound. Example Example - use monoxide not monooxide. carbon tetrachloride diphophorous pentoxide iodine trichloride silicon dioxide carbon monoxide carbon dioxide dinitrogen pentoxide

23. 4 - 23 CH 104 Chapter 4 Covalent compounds Discrete molecular units Atoms held together by bonds Covalent compounds exist in all states (CO 2 - gas, H 2 O - liquid, SiO 2 - solid) Formula represents atoms in a molecule Properties of ionic and covalent compounds O=O

24. 4 - 24 CH 104 Chapter 4 Lewis Structures for Molecules & Polyatomic Ions Practice drawing electron dot (Lewis) structures for the following: H2OH2O OH 1- CO 3 2- PO 4 3- Notice options for Resonance Structures:

25. 4 - 25 CH 104 Chapter 4 Polyatomic Ions Na 1+ SO 4 2- Na 2 SO 4 Sodium Sulfate NH 4 1+ PO 4 3- (NH 4 ) 3 PO 4 Ammonium Phosphate NH 4 1+ Na 1+

26. 4 - 26 CH 104 Chapter 4 Polyatomic Ions Ca 2+ C 2 H 3 O 2 1- Calcium Acetate Sn 2+ NO 2 1- Sn(NO 2 ) 2 Tin (II) Nitrite C 2 H 3 O 2 1- Ca(C 2 H 3 O 2 ) 2 NO 2 1- Stannous Nitrite

27. 4 - 27 CH 104 Chapter 4 Naming Practice Aluminum Flouride FeF 3 NF 3 SO 3 Mg 2 C CaCO 3 NaHCO 3 K 2 SO 3 AlF 3 Iron (III) Flouride or Ferric Flouride Nitrogen Triflouride Sulfur Trioxide Potassium Sulfite Magnesium Carbide Calcium Carbonate Sodium Hydrogen Carbonate Sodium Bicarbonate

28. 4 - 28 CH 104 Chapter 4 Bond Polarity, Electronegativity Cl H H Electrons in covalent bonds rarely get shared equally.

29. 4 - 29 CH 104 Chapter 4 polar This unequal sharing results in polar bonds. H Cl Slight positive side Smaller electronegativity Slight positive side Smaller electronegativity Slight negative Larger electronegativity Slight negative Larger electronegativity ++ -- Bond Polarity, Electronegativity

30. Electronegativity Relative ability of atoms to attract e -. At I Br Cl Po Te Se S Bi Sb As P Pb Sn Ge Si FON Tl Na Cs Rb K Ba Mg Sr Ca In Ga Al H LiBeBC 4 - 50

31. 4 - 31 CH 104 Chapter 4 Electronegativity Relative ability of atoms to attract e -. At I Br Cl Po Te Se S Bi Sb As P Pb Sn Ge Si FON Tl Na Cs Rb K Ba Mg Sr Ca In Ga Al H LiBeBC 4 - 50 3.983.443.04 3.16 2.96 2.55 2.20 2.192.58

32. 4 - 32 CH 104 Chapter 4 Electronegativity Relative ability of atoms to attract e -. At I Br Cl Po Te Se S Bi Sb As P Pb Sn Ge Si FON Tl Na Cs Rb K Ba Mg Sr Ca In Ga Al H LiBeBC 4 - 50 4.03.53.0 3.0 2.8 2.5 2.1 2.12.5 2.5 2.1 0.8 0.8 0.7 1.2 1.0 0.9 1.8 1.5 2.0 1.5 2.4 2.1 2.0

33. 4 - 33 CH 104 Chapter 4 Electronegativity Relative ability of atoms to attract electrons. Periodic trends

34. 4 - 34 CH 104 Chapter 4 Cl H H 2.13.0 ---- Polar Covalent ++++ ElectronegativityDifference < 0.5 Nonpolar 0.5-1.7 Polar >1.8 Ionic Bond Polarity, Electronegativity

35. 4 - 35 CH 104 Chapter 4 C Polarity, Shape 3.53.52.5 O=C=O ++++ ---- ---- Polar Covalent Bonds Linear Shape (180 o ) ElectronegativityDifference < 0.5 Nonpolar 0.5-1.7 Polar >1.8 Ionic Nonpolar Compound CO 2 OO

36. 4 - 36 CH 104 Chapter 4 Some common geometries e- directions around e- directions around Shape central atom Example___ e- directions around e- directions around Shape central atom Example___ Linear 2 O=C=O Tetrahedral 4 Trigonal Planar 3

37. 4 - 37 CH 104 Chapter 4 Polarity, Shape (VSEPR) O=C=O ++++ ---- ---- e-’s in 2 directions = 180 o Linear e-’s in 3 directions = 120 o ++++ ---- Trigonal planar Nonpolar Compound Polar Compound

38. 4 - 38 CH 104 Chapter 4 C HH H Cl C HH H 2.1 2.1 2.1 2.5 3.0 ++++ ---- (~109 o ) Tetrahedral Tetrahedral Polarity, Shape (VSEPR) e-’s in 4 directions = 109.5 o

39. 4 - 39 CH 104 Chapter 4 Polarity, Shape (VSEPR) O HH H-O-H2.1 2.1 3.5 ++++ ++++ ---- (105 o ) Tetrahedral Configuration of Electrons Bent Configuration of Atoms (105 o ) Tetrahedral Configuration of Electrons Bent Configuration of Atoms e-’s in 4 directions = 109.5 o

40. 4 - 40 CH 104 Chapter 4 C HH H Cl C HH H O HH H-O-H ++++ ++++ ---- ++++ ---- e-’s in 4 directions = ~109.5 o Tetrahedral Bent 4 directions = ~109.5 o Polarity, Shape (VSEPR)

41. 4 - 41 CH 104 Chapter 4 N HH H ++++ ---- e-’s in 4 directions = 109.5 o Pyramidal N HH H ++++ ++++ (~109.5 o  107 o ) Tetrahedral Configuration of Electrons Trigonal Pyramid Configuration of Atoms (~109.5 o  107 o ) Tetrahedral Configuration of Electrons Trigonal Pyramid Configuration of Atoms Polarity, Shape (VSEPR)

42. 4 - 42 CH 104 Chapter 4 Tetrahedral electron-pair Geometries Tetrahedral Pyramidal Bent

43. 4 - 43 CH 104 Chapter 4 F FB Exceptions to Octet Rule 4.0 4.0 2.0 ++++ ---- ---- Polar Covalent Bonds (120 o ) Trigonal Planar (120 o ) Trigonal Planar Nonpolar Compound BF 3 F 4.0 F B F F ----

44. 4 - 44 CH 104 Chapter 4 Exceptions to Octet Rule Trigonal Bipyramid PCl 5 SF 6 OctahedralOctahedral

45. 4 - 45 CH 104 Chapter 4 Molecular geometry Molecules have specific shapes. Determined by the number of electron pairs around the central species Bonded and unbonded pairs Geometry affects factors like polarity and solubility.

46. 4 - 46 CH 104 Chapter 4 Some common geometries e - pairs around e - pairs around Shape central atomExample Linear 2BeH 2, HF Trigonal plane 3BF 3 Tetrahedral 4CH 4 Pyramidal 4NH 3 Bent 4H 2 O

47. 4 - 47 CH 104 Chapter 4 Geometry and polar molecules For a molecule to be polar - must have polar bonds - must have the proper geometry CH 4 non-polar CH 3 Clpolar CH 2 Cl 2 polar CHCl 3 polar CCl 4 non-polar WHY? WHY?

48. 4 - 48 CH 104 Chapter 4 Properties of ionic & covalent compounds Ionic compounds Exist as 3-D network of ions Held together by electrostatic attraction Ionic compounds are solids at room temp. Formula is simple average Covalent compounds Discrete molecular units Atoms held together by covalent bonds Covalent compounds exist in all states (CO 2 - gas, H 2 O - liquid, SiO 2 - solid) Formula represents atoms in a molecule O=O

49. 4 - 49 CH 104 Chapter 4 Covalent compounds Discrete molecular units Atoms held together by bonds Covalent compounds exist in all states (CO 2 - gas, H 2 O - liquid, SiO 2 - solid) Formula represents atoms in a molecule Properties of ionic and covalent compounds O=O

50. 4 - 50 CH 104 Chapter 4 Attractive Forces Ionic Bonds 150 - 3000 kcal mol 150 - 3000 kcal mol Melting Point NaCl 801 o C Na 2 S 920 o C MgF 2 1248 o C Melting Point NaCl 801 o C Na 2 S 920 o C MgF 2 1248 o C Ionic compounds Held together by electrostatic attraction Ionic compounds are solids at room temp Boiling Point NaCl 1413 o C Boiling Point NaCl 1413 o C

51. 4 - 51 CH 104 Chapter 4 H H O -- ++ ++ H H O -- ++ ++ Attractive Forces Ion-Dipole H H O -- ++ ++ H H O -- ++ ++ H H O -- ++ ++ H H O -- ++ ++ H H O -- ++ ++ H H O -- ++ ++

52. 4 - 52 CH 104 Chapter 4 HCl ++++ ---- HCl ++++ ---- Attractive Forces Dipole-Dipole 0.1 - 1 kcal mol 0.1 - 1 kcal mol Melting Point HCl -114 o C CH 3 F -142 o C Melting Point HCl -114 o C CH 3 F -142 o C HCl ++++ ---- HCl ++++ ---- Boiling Point HCl -85 o C Boiling Point HCl -85 o C

53. 4 - 53 CH 104 Chapter 4 FF ++++ ---- F ++++ ---- F Attractive Forces Induced dipole – Induced dipole (Dispersion Forces) 0.01 kcal mol 0.01 kcal mol Melting Point F 2 -220 o C CH 4 -183 o C Melting Point F 2 -220 o C CH 4 -183 o C F ++++ ----F F ++++ ---- F Boiling Point F 2 -188 o C CH 4 -162 o C Boiling Point F 2 -188 o C CH 4 -162 o C

54. 4 - 54 CH 104 Chapter 4 Dipole-Induced Dipole Attractive Forces I ++++ ---- I H H O -- ++ ++ H H O -- ++ ++ H H O -- ++ ++ H H O -- ++ ++

55. 4 - 55 CH 104 Chapter 4 HH O ++++ ++++ ---- HH O ++++ ++++ ---- HH O ++++ ++++ ---- HH O ++++ ++++ ---- Polar Attraction Attractive Forces Hydrogen Bonds

56. 4 - 56 CH 104 Chapter 4 Hydrogen Bonding of Water Hydrogen Bonds 5 - 10 kcal mol 5 - 10 kcal mol Melting Point H 2 O 0 o C NH 3 -78 o C Melting Point H 2 O 0 o C NH 3 -78 o C Boiling Point H 2 O100 o C NH 3 -33 o C Boiling Point H 2 O100 o C NH 3 -33 o C

57. 4 - 57 CH 104 Chapter 4 Frozen H 2 O: Slow moving molecules H-Bond in patterns Hydrogen Bonding of Water

58. 4 - 58 CH 104 Chapter 4 Freezing Point Depression: Frozen Water: slow molecules form hexagonal rings resulting in crystals Solute particles interfere, causing H 2 O molecules to slow down even more. Thus reducing the temperature.

59. 4 - 59 CH 104 Chapter 4 Boiling and melting points Characteristic physical properties. Boiling point Boiling point The temperature at which a liquid is converted to a gas at atmospheric pressure. Melting point Melting point The temperature at which a solid is converted to a liquid.

60. 4 - 60 CH 104 Chapter 4 Boiling and melting points Chemical Bond Mp Bp N 2 Nonpolar -210-196 O 2 Nonpolar -219-183 NH 3 Polar -78 -33 H 2 O Polar 0 100 Ionic NaCl Ionic 804 1413 Melting and Boiling points Very high for ionic compounds Typically lower for covalent compounds

61. 4 - 61 CH 104 Chapter 4 Polarity and solubility Solubility SolubilityThe maximum amount of a solute that dissolves in a given solvent Depends on the forces of attraction between molecules - intermolecular Types of intermolecular attractions most often encountered Dipole-Dipole Dipole-Dipole Hydrogen bonding Hydrogen bonding Van der Wall forces Van der Wall forces Like dissolves like General rule“Like dissolves like”

62. 4 - 62 CH 104 Chapter 4 62 Flowchart for Naming Ionic Compounds

63. 4 - 63 CH 104 Chapter 4 63 A. The Group number for sulfur is 1) 4A(14) 2) 8A(18) 3) 6A(16) B. The number of valence electrons in sulfur is 1) 4e  2) 6e  3) 8e  C. The change in electrons for an octet requires a 1) gain of 2e  2) loss of 2e  3) a gain of 4e  D. The ionic charge of sulfur is 1) 2 + 2) 2  3) 4  Learning Check

64. 4 - 64 CH 104 Chapter 4 A. The Group number for sulfur is 1) 4A(14) 2) 8A(18) 3) 6A(16) B. The number of valence electrons in sulfur is 1) 4e  2) 6e  3) 8e  C. For Sulfur to form an octet requires a 1) gain of 2e  2) loss of 2e  3) a gain of 4e  D. The ionic charge of sulfur is 1) 2 + 2) 2  3) 4  64 Solution

65. 4 - 65 CH 104 Chapter 4 65 Select the larger atom or ion in the following: 1.A.) Mg or B.) Mg 2+ 2. A.) S or B.) S 2  3. A.) Br or B.) Br  Learning Check

66. 4 - 66 CH 104 Chapter 4 66 Solution Select the larger atom or ion in the following: 1.A.) Mg or B.) Mg 2+ 2. A.) S or B.) S 2  3. A.) Br or B.) Br 

67. 4 - 67 CH 104 Chapter 4 67 Select the correct formula for each of the following ionic compounds: A. Na + and S 2– 1) NaS 2) Na 2 S3) NaS 2 B. Al 3+ and Cl – 1) AlCl 3 2) AlCl 3) Al 3 Cl C. Mg 2+ and N 3– 1) MgN 2) Mg 2 N 3 3) Mg 3 N 2 Learning Check

68. 4 - 68 CH 104 Chapter 4 68 Select the correct formula for each of the following ionic compounds: A. Na + and S 2– 1) NaS 2) Na 2 S3) NaS 2 B. Al 3+ and Cl – 1) AlCl 3 2) AlCl 3) Al 3 Cl C. Mg 2+ and N 3– 1) MgN 2) Mg 2 N 3 3) Mg 3 N 2 Solution

69. 4 - 69 CH 104 Chapter 4 69 Give the names of the following ions: Ba 2+ Al 3+ K + _________ __________ _________ N 3  O 2  F  _________ __________ _________ P 3  S 2  Cl  _________ __________ _________ Learning Check

70. 4 - 70 CH 104 Chapter 4 70 Give the names of the following ions: Ba 2+ Al 3+ K + barium aluminum potassium _____ ____ __________ _________ N 3  O 2  F  nitride oxide fluoride _________ __________ _________ P 3  S 2  Cl  phosphide sulfide chloride _________ __________ _________ Solution

71. 4 - 71 CH 104 Chapter 4 71 Write the names of the following compounds: 1) CaO___________ 2)KBr___________ 3)Al 2 O 3 ___________ 4) MgCl 2 ___________ Learning Check

72. 4 - 72 CH 104 Chapter 4 72 Write the names of the following compounds: 1) CaOcalcium oxide 2)KBrpotassium bromide 3)Al 2 O 3 aluminum oxide 4) MgCl 2 magnesium chloride Solution

73. 4 - 73 CH 104 Chapter 4 73 Learning Check Write the formulas and names for compounds of the following ions: Br – S 2− N 3− Na + Al 3+

74. 4 - 74 CH 104 Chapter 4 74 Solution Br − S 2− N 3− Na + Al 3+ NaBr sodium bromide Na 2 S sodium sulfide Na 3 N sodium nitride AlBr 3 aluminum bromide Al 2 S 3 aluminum sulfide AlN aluminum nitride Write the formulas and names for compounds of the following ions:

75. 4 - 75 CH 104 Chapter 4 75 Learning Check Select the correct name for each. A. Fe 2 S 3 1) iron sulfide 2) iron(II) sulfide 3) iron(III) sulfide B. CuO 1) copper oxide 2) copper(I) oxide 3) copper(II) oxide

76. 4 - 76 CH 104 Chapter 4 76 Solution Select the correct name for each. A. Fe 2 S 3 1) iron sulfide 2) iron(II) sulfide 3) iron(III) sulfideFe 3+ S 2– B. CuO 1) copper oxide 2) copper(I) oxide 3) copper(II) oxideCu 2+ O 2–

77. 4 - 77 CH 104 Chapter 4 77 Learning Check The correct formula for each of the following is: A. copper(I) nitride 1) CuN2) CuN 3 3) Cu 3 N B. lead(IV) oxide 1) PbO 2 2) PbO 3) Pb 2 O 4

78. 4 - 78 CH 104 Chapter 4 78 Solution The correct formula for each of the following is: A. copper(I) nitride 1) CuN2) CuN 3 3) Cu 3 N 3Cu + + N 3– = 3(1+) + (3–) = 0 B. lead(IV) oxide 1) PbO 2 2) PbO 3) Pb 2 O 4 Pb 4+ + 2O 2– = (4+) + 2(2–) = 0

79. 4 - 79 CH 104 Chapter 4 79 Match each formula with the correct name: A. MgS1) magnesium sulfite MgSO 3 2) magnesium sulfate MgSO 4 3) magnesium sulfide B. Ca(ClO 3 ) 2 1) calcium chlorate CaCl 2 2) calcium chlorite Ca(ClO 2 ) 2 3) calcium chloride Learning Check

80. 4 - 80 CH 104 Chapter 4 80 Match each formula with the correct name: A. MgS1) magnesium sulfite MgSO 3 2) magnesium sulfate MgSO 4 3) magnesium sulfide B. Ca(ClO 3 ) 2 1) calcium chlorate CaCl 2 2) calcium chlorite Ca(ClO 2 ) 2 3) calcium chloride Solution

81. 4 - 81 CH 104 Chapter 4 81 Learning Check Name each of the following compounds: A.Mg(NO 3 ) 2 B.Cu(ClO 3 ) 2 C.PbO 2 D.Fe 2 (SO 4 ) 3 E.Ba 3 (PO 3 ) 2

82. 4 - 82 CH 104 Chapter 4 82 Solution Name each of the following compounds: A.Mg(NO 3 ) 2 B.Cu(ClO 3 ) 2 C.PbO 2 D.Fe 2 (SO 4 ) 3 E.Ba 3 (PO 3 ) 2 magnesium nitrate copper(II) chlorate lead(IV) oxide iron(III) sulfate barium phosphite

83. 4 - 83 CH 104 Chapter 4 83 Select the correct formula for each: A. aluminum nitrate 1) AlNO 3 2) Al(NO) 3 3) Al(NO 3 ) 3 B. copper(II) nitrate 1) CuNO 3 2) Cu(NO 3 ) 2 3) Cu 2 (NO 3 ) C. iron(III) hydroxide 1) FeOH2) FeOH 3 3) Fe(OH) 3 D. tin(IV) hydroxide 1) Sn(OH) 4 2) Sn(OH) 2 3) Sn 4 (OH) Learning Check

84. 4 - 84 CH 104 Chapter 4 84 Select the correct formula for each: A. aluminum nitrate 1) AlNO 3 2) Al(NO) 3 3) Al(NO 3 ) 3 B. copper(II) nitrate 1) CuNO 3 2) Cu(NO 3 ) 2 3) Cu 2 (NO 3 ) C. iron(III) hydroxide 1) FeOH2) FeOH 3 3) Fe(OH) 3 D. tin(IV) hydroxide 1) Sn(OH) 4 2) Sn(OH) 2 3) Sn 4 (OH) Solution

85. 4 - 85 CH 104 Chapter 4 85 Learning Check Write the correct formula for each: A.potassium bromate B.calcium carbonate C.sodium phosphate D.iron(III) oxide E.iron(II) nitrite

86. 4 - 86 CH 104 Chapter 4 86 Solution Write the correct formula for each: A.potassium bromate B.calcium carbonate C.sodium phosphate D.iron(III) oxide E.iron(II) nitrite KBrO 3 CaCO 3 Na 3 PO 4 Fe 2 O 3 Fe(NO 2 ) 2

87. 4 - 87 CH 104 Chapter 4 87 Learning Check Name the following compounds: A. Ca 3 (PO 4 ) 2 B. FeBr 3 C. Al 2 S 3 D. Zn(NO 2 ) 2 E. NaHCO 3

88. 4 - 88 CH 104 Chapter 4 88 Solution Name the following compounds: A. Ca 3 (PO 4 ) 2 Ca 2+ PO 4 3− calcium phosphate B. FeBr 3 Fe 3+ Br − iron(III) bromide C. Al 2 S 3 Al 3+ S 2− aluminum sulfide D. Zn(NO 2 ) 2 Zn 2+ NO 2 − zinc nitrite E. NaHCO 3 Na + HCO 3 − sodium hydrogen carbonate or sodium bicarbonate

89. 4 - 89 CH 104 Chapter 4 89 Learning Check Write the formulas for the following: A. calcium nitrate B. iron(II) hydroxide C. aluminum carbonate D. copper(II) bromide E. lithium phosphate

90. 4 - 90 CH 104 Chapter 4 90 Solution Write the formulas for the following: A. calcium nitrateCa 2+, NO 3 − Ca(NO 3 ) 2 B. iron(II) hydroxideFe 2+, OH − Fe(OH) 2 C. aluminum carbonateAl 3+, CO 3 2− Al 2 (CO 3 ) 3 D. copper(II) bromide Cu 2+, Br − CuBr 2 E. lithium phosphateLi +, PO 4 3− Li 3 PO 4

91. 4 - 91 CH 104 Chapter 4 91 What is the name of each of the following diatomic molecules? H 2 _______________ N 2 _______________ Cl 2 _______________ O 2 _______________ I 2 _______________ Learning Check

92. 4 - 92 CH 104 Chapter 4 92 What is the name of each of the following diatomic molecules? H 2 hydrogen N 2 nitrogen Cl 2 chlorine O 2 oxygen I 2 iodine Solution

93. 4 - 93 CH 104 Chapter 4 93 FNO 2, a rocket propellant, has two resonance structures. One is shown below. What is the other resonance structure? Learning Check

94. 4 - 94 CH 104 Chapter 4 94 FNO 2, a rocket propellant, has two resonance structures. One is shown below. What is the other resonance structure? Solution

95. 4 - 95 CH 104 Chapter 4 95 Select the correct name for each compound. A.SiCl 4 1) silicon chloride 2) tetrasilicon chloride 3) silicon tetrachloride B. P 2 O 5 1) phosphorus oxide 2) phosphorus pentoxide 3) diphosphorus pentoxide C.Cl 2 O 7 1) dichlorine heptoxide 2) dichlorine oxide 3) chlorine heptoxide Learning Check

96. 4 - 96 CH 104 Chapter 4 96 Select the correct name for each compound. A.SiCl 4 3) silicon tetrachloride B. P 2 O 5 3) diphosphorus pentoxide C.Cl 2 O 7 1) dichlorine heptoxide Solution

97. 4 - 97 CH 104 Chapter 4 97 Write the name of each covalent compound. CO_____________________ CO 2 _____________________ PCl 3 _____________________ CCl 4 _____________________ N 2 O_____________________ Learning Check

98. 4 - 98 CH 104 Chapter 4 98 Write the name of each covalent compound. CO carbon monoxide CO 2 carbon dioxide PCl 3 phosphorus trichloride CCl 4 carbon tetrachloride N 2 Odinitrogen oxide Solution

99. 4 - 99 CH 104 Chapter 4 Guide to Writing Formulas for Covalent Compounds 99

100. 4 - 100 CH 104 Chapter 4 100 Write the formula for carbon disulfide. STEP 1 Elements are C and S STEP 2 No prefix for carbon means 1 C Prefix di = 2 Formula: CS 2 Writing Formulas of Covalent Compounds

101. 4 - 101 CH 104 Chapter 4 101 Write the correct formula for each of the following: A. phosphorus pentachloride B. dinitrogen trioxide C. sulfur hexafluoride Learning Check

102. 4 - 102 CH 104 Chapter 4 102 Write the correct formula for each of the following: A. phosphorus pentachloride 1P penta = 5ClPCl 5 B. dinitrogen trioxide di = 2N tri = 3 O N 2 O 3 C. sulfur hexafluoride 1S hexa = 6FSF 6 Solution

103. 4 - 103 CH 104 Chapter 4 103 Learning Check Identify each compound as ionic or covalent, and give its correct name. A. SO 2 B. BaCl 2 C. (NH 4 ) 3 PO 4 D. Cu 2 CO 3 E. N 2 O 4

104. 4 - 104 CH 104 Chapter 4 104 Solution Identify each compound as ionic or covalent, and give its correct name. A. SO 2 covalent; sulfur dioxide B. BaCl 2 ionic; barium chloride C. (NH 4 ) 3 PO 3 ionic; ammonium phosphite D. Cu 2 CO 3 ionic; copper(I) carbonate E. N 2 O 4 covalent; dinitrogen tetroxide

105. 4 - 105 CH 104 Chapter 4 105 Learning Check Name the following compounds: A. Ca 3 (PO 4 ) 2 B. FeBr 3 C. SCl 2 D. Cl 2 O

106. 4 - 106 CH 104 Chapter 4 106 Solution Name the following compounds: A. Ca 3 (PO 4 ) 2 ionic Ca 2+, PO 4 3− calcium phosphate B. FeBr 3 ionicFe 3+, Br − iron(III) bromide C. SCl 2 covalent 1S, 2Cl sulfur dichloride D. Cl 2 O covalent 2Cl, 1 O dichlorine oxide

107. 4 - 107 CH 104 Chapter 4 107 Learning Check Write the formulas for the following: A. calcium nitrate B. boron trifluoride C. aluminum carbonate D. dinitrogen tetroxide E. copper(I) phosphate

108. 4 - 108 CH 104 Chapter 4 108 Solution Write the formulas for the following: A. calcium nitrateCa 2+, NO 3 − Ca(NO 3 ) 2 B. boron trifluoride 1B, 3F BF 3 C. aluminum carbonateAl 3+, CO 3 2− Al 2 (CO 3 ) 3 D. dinitrogen tetroxide 2N, 4 O N 2 O 4 E. copper(I) phosphateCu +, PO 4 3− Cu 3 PO 4

109. 4 - 109 CH 104 Chapter 4 109 Use the electronegativity difference to identify the type of bond between the following: nonpolar covalent (NP), polar covalent (P), or ionic (I) A. K–N B. N–O C. Cl–Cl D. B–Cl Learning Check

110. 4 - 110 CH 104 Chapter 4 110 Use the electronegativity difference to identify the type of bond between the following: nonpolar covalent (NP), polar covalent (P), or ionic (I) A. K–N2.2ionic (I) B. N–O0.5 polar covalent (P) C. Cl–Cl0.0nonpolar covalent (NP) D. B–Cl1.0polar covalent (P) Solution

111. 4 - 111 CH 104 Chapter 4 111 Learning Check The shape of a molecule of N 2 O (N N O) is 1) linear 2) trigonal planar 3) bent (120°)

112. 4 - 112 CH 104 Chapter 4 112 Solution The shape of a molecule of N 2 O (N N O) is 1) linear In the electron-dot structure with 16 e –, octets are acquired using two double bonds to the central N atom. The shape of a molecule with two electron groups and two bonded atoms (no lone pairs on N) is linear. two electron groups : N :: N :: O : : N = N=O : linear, 180°

113. 4 - 113 CH 104 Chapter 4 113 Learning Check State the number of electron groups and lone pairs, and use VSEPR theory to determine the shape of the following molecules or ions. 1) tetrahedral 2) pyramidal3) bent A. PF 3 B. H 2 S C. CCl 4

114. 4 - 114 CH 104 Chapter 4 114 Solution State the number of electron groups and lone pairs, and use VSEPR theory to determine the shape of the following molecules or ions. A. PF 3 4 electron groups, 1 lone pair, (2) pyramidal B.H 2 S 4 electron groups, 2 lone pairs, (3) bent C. CCl 4 4 electron groups, 0 lone pairs, (1) tetrahedral

115. 4 - 115 CH 104 Chapter 4 115 Polar Molecules A polar molecule  contains polar bonds  has a separation of positive and negative charge called a dipole, indicated with  + and  –  has dipoles that do not cancel  +  – H–Cl H — N —H dipole H dipoles do not cancel

116. 4 - 116 CH 104 Chapter 4 116 Nonpolar Molecules A nonpolar molecule  contains nonpolar bonds Cl–Cl H–H  or has a symmetrical arrangement of polar bonds

117. 4 - 117 CH 104 Chapter 4 Determining Molecular Polarity Determine the polarity of the H 2 O molecule. Solution: The four electron groups of oxygen are bonded to two H atoms. Thus the H 2 O molecule has a net dipole, which makes it a polar molecule. 117

118. 4 - 118 CH 104 Chapter 4 118 Learning Check Determine the shape of each of the following molecules and whether they are polar or nonpolar. Explain. A. PBr 3 B. HBr C. Br 2 D. SiBr 4

119. 4 - 119 CH 104 Chapter 4 119 Solution Determine the shape of each of the following molecules and whether they are polar or nonpolar. Explain. A. PBr 3 1) pyramidal; polar; dipoles don’t cancel B. HBr1) linear; polar; one polar bond (dipole) C. Br 2 2) linear; nonpolar; nonpolar bond D. SiBr 4 2) tetrahedral; nonpolar; dipoles cancel

120. 4 - 120 CH 104 Chapter 4 120 Learning Check Identify the main type of attractive forces for each: 1) ionic 2) dipole–dipole 3) hydrogen bonds 4) dispersion A. NCl 3 B. H 2 O C. Br–Br D. KCl E. NH 3

121. 4 - 121 CH 104 Chapter 4 121 Solution Identify the main type of attractive forces for each: 1) ionic 2) dipole–dipole 3) hydrogen bonds4) dispersion 2 A. NCl 3 3 B. H 2 O 4 C. Br–Br 1 D. KCl 3 E. NH 3