1. 1 4.1 Valence Electrons 4.2 Octet Rule and Ions Chapter 4 Compounds and Their Bonds

2. 2  The valence electrons are the electrons in the outer shell.  The electrons in the outer shell have the most contact with other atoms and strongly influence the chemical properties of atoms. Valence Electrons

3. 3 Number of Valence Electrons For Group A elements, the number of valence electrons is the number of electrons in the s and p subshells of the outer shell. For Group A elements, the number of valence electrons is the number of electrons in the s and p subshells of the outer shell. In the electron configuration for phosphorus, there are 5 valence electrons in the s and p subshells with the highest number. 5 valence electrons P Group 5A 1s 2 2s 2 2p 6 3s 2 3p 3 In the electron configuration for phosphorus, there are 5 valence electrons in the s and p subshells with the highest number. 5 valence electrons P Group 5A 1s 2 2s 2 2p 6 3s 2 3p 3

4. 4 Valence Electrons for Groups

5. 5 State the number of valence electrons for each. A. Magnesium 1) 22) 63) 8 1) 22) 63) 8 B. Oxygen 1) 22) 43) 6 1) 22) 43) 6 C. Potassium 1) 12) 23) 7 1) 12) 23) 7 Learning Check

6. 6 State the number of valence electrons for each. A. Magnesium 1) 2 Group 2A 1s 2 2s 2 2p 6 3s 2 1) 2 Group 2A 1s 2 2s 2 2p 6 3s 2 B. Oxygen 3) 6Group 6A 1s 2 2s 2 2p 4 3) 6Group 6A 1s 2 2s 2 2p 4 C. Potassium 1) 1Group 1A 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 1) 1Group 1A 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 Solution

7. 7 Electron Dot Structure An electron-dot structure is a convenient way to represent the valence electrons. An electron-dot structure is a convenient way to represent the valence electrons. For example, the two valence electrons for magnesium are placed as single dots on any two sides of the Mg symbol. For example, the two valence electrons for magnesium are placed as single dots on any two sides of the Mg symbol.

8. 8 Electron-Dot Structures Dot structures are used for Group A elements. Dot structures are used for Group A elements. The valence electrons are placed on the sides of the symbol of an element. The valence electrons are placed on the sides of the symbol of an element.

9. 9 A. X is the electron dot formula for 1) Na2) K3) Al B. X is the electron dot formula of 1) B2) N3) P Learning Check

10. 10 A. X is the electron dot formula for 1) Na2) K B. X i s the electron dot formula of 2) N3) P Solution

11. 11  The stability of the noble gases is associated with 8 valence electrons (He has 2). Ne 2, 8 Ar2, 8, 8 Kr2, 8, 18, 8  Atoms can become more stable by acquiring an octet (8 electrons) in the outer shell.  The process of acquiring an octet involves the loss, gain, or sharing of valence electrons. Octet Rule

12. 12 Ionization Energy Ionization energy is the energy it takes to remove a valence electron. Ionization energy is the energy it takes to remove a valence electron. Metals have lower ionization energies and nonmetals have higher ionization energies. Metals have lower ionization energies and nonmetals have higher ionization energies.

13. 13  Metals acquire octets by losing valence electrons.  The loss of electrons converts an atom to an ion that has the electron configuration of the nearest noble gas.  Metals form positive ions because they have fewer electrons than protons. Group 1A metals  ion 1+ Group 2A metals  ion 2+ Group 3A metals  ion 3+ Metals Form Positive Ions

14. 14 Give the ionic charge for each ion. A. 12 p + and 10 e - 1) 02) 2+3) 2- 1) 02) 2+3) 2- B. 50p + and 46 e - 1) 2+2) 4+3) 4- 1) 2+2) 4+3) 4- C. 15 p + and 18e - 2) 3+ 2) 3-3) 5- 2) 3+ 2) 3-3) 5- Learning Check

15. 15 Give the ionic charge for each ion. A. 12 p + and 10 e - 2) 2+ 2) 2+ B. 50p + and 46 e - 2) 4+ 2) 4+ C. 15 p + and 18e - 2) 3- 2) 3- Solution

16. 16 Sodium forms an octet by losing its one valence electron. Sodium forms an octet by losing its one valence electron. Na  – e   Na + 1s 2 2s 2 2p 6 3s 1 1s 2 2s 2 2p 6 (= Ne) 1s 2 2s 2 2p 6 3s 1 1s 2 2s 2 2p 6 (= Ne) A positive ion forms with a +1 charge. A positive ion forms with a +1 charge. Sodium atom Sodium ion 11 p + 11 p + 11 e - 10 e - 11 e - 10 e - 0 1 + 0 1 + Formation of a Sodium Ion, Na +

17. 17 Magnesium forms an octet by losing its two valence electrons. Magnesium forms an octet by losing its two valence electrons. Magnesium atom Magnesium ion   Mg  – 2e   Mg 2+ Mg  – 2e   Mg 2+ 1s 2 2s 2 2p 6 3s 2 1s 2 2s 2 2p 6 (= Ne) 1s 2 2s 2 2p 6 3s 2 1s 2 2s 2 2p 6 (= Ne) A positive ion forms with a +2 charge. A positive ion forms with a +2 charge. 12 p + 12 p + 12 p + 12 p + 12 e- 10 e - 12 e- 10 e - 0 2 + 0 2 + Formation of Mg 2+

18. 18  When nonmetals gain electrons to achieve an octet arrangement, they form negative ions.  The ionic charge of a nonmetal is 3-, 2-, or 1-. Formation of Negative Ions

19. 19 Fluorine forms an octet by adding an electron to its seven valence electrons. Fluorine forms an octet by adding an electron to its seven valence electrons.     1 -     1 - : F  + e  : F :         1s 2 2s 2 2p 5 1s 2 2s 2 2p 6 (= Ne) 1s 2 2s 2 2p 5 1s 2 2s 2 2p 6 (= Ne) A negative ion forms with a -1 charge. A negative ion forms with a -1 charge. Fluorine atom Fluoride ion 9 p + 9 p + 9 p + 9 p + 9 e - 10 e - 9 e - 10 e - 0 1 – 0 1 – Formation of a Fluoride Ion, F -

20. 20 Group Number and Ions The Group number can be used to determine the charge of an ion. The Group number can be used to determine the charge of an ion. The charge of a positive ion is equal to its Group number. The charge of a positive ion is equal to its Group number. Group 3A = 3+ The charge of a negative ion is obtained by subtracting its Group number from 8. The charge of a negative ion is obtained by subtracting its Group number from 8. Group 6A = - (8-6) = 2-

21. 21 Examples of Ionic Charges

22. 22 Some Important Ions in the Body

23. 23 A.How many valence electrons does aluminum have? 1) 2e - 2) 3e - 3) 5e - B.How does aluminum acquire an octet? 1) loses 3e - 2) gains 3e - 3) gains 5e - 1) loses 3e - 2) gains 3e - 3) gains 5e - C. What is the ionic charge of an aluminum ion? 1) 3- 2) 5- 3) 3 + D. The symbol for the aluminum ion is 1) Al 3+ 2) Al 3- 3) Al + Learning Check

24. 24 A.How many valence electrons does aluminum have? 2) 3e - B.How does aluminum acquire an octet? 1) loses 3e - C. What is the ionic charge of an aluminum ion? 3) 3 + 3) 3 + D. The symbol for the aluminum ion is 1) Al 3+ Solution

25. 25 Chapter 4 Compounds and Their Bonds 4.3 Ionic Compounds 4.4 Naming and Writing Ionic Formulas

26. 26  Ionic compounds consist of positive and negative ions.  An ionic bond is an attraction between the positive and negative charges.  In an ionic formula, the total charge of the positive ions is equal to the total charge of the negative ions. total positive charge = total negative charge Ionic Compounds

27. 27 The formulas of ionic compounds are determined from the charges on the ions. The formulas of ionic compounds are determined from the charges on the ions. atoms ions atoms ions     –     – Na  +  F :  Na + : F :  NaF         sodium fluorine sodium fluoride The overall charge of NaF is zero (0). The overall charge of NaF is zero (0). (1+ ) + (1-) = 0 (1+ ) + (1-) = 0 Ionic Formulas

28. 28 Charge Balance In NaCl The formula does not show the charges of the ions in the compound. The formula does not show the charges of the ions in the compound. The symbol of the metal is written first, followed by the symbol of the nonmetal. The symbol of the metal is written first, followed by the symbol of the nonmetal.

29. 29 Charge Balance In MgCl 2

30. 30 Write the formula of the ionic compound that forms from Ba 2+ and Cl . Write the symbols of the positive ion and the negative ion. Ba 2+ Cl  Write the symbols of the positive ion and the negative ion. Ba 2+ Cl  Balance the charges until the positive charge is equal to the negative charge. Balance the charges until the positive charge is equal to the negative charge. Ba 2+ Cl  two Cl - needed Cl  Ba 2+ Cl  two Cl - needed Cl  Write the formula using subscripts for the number of ions for charge balance. BaCl 2 Write the formula using subscripts for the number of ions for charge balance. BaCl 2 Writing a Formula from Charges

31. 31 Write the correct formula for the ionic compound of A. Na + and S 2- 1) NaS 2) Na 2 S3) NaS 2 1) NaS 2) Na 2 S3) NaS 2 B. Al 3+ and Cl - 1) AlCl 3 2) AlCl 3) Al 3 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 1) MgN 2) Mg 2 N 3 3) Mg 3 N 2 Learning Check

32. 32 A. Na + and S 2- 2) Na 2 S 2) Na 2 S B. Al 3+ and Cl - 1) AlCl 3 1) AlCl 3 C. Mg 2+ and N 3- 3) Mg 3 N 2 3) Mg 3 N 2 Solution

33. 33 Names of Ions Positive ions are named like the element. Positive ions are named like the element. Negative ions are named by changing the end of the element name to –ide. Negative ions are named by changing the end of the element name to –ide.

34. 34 Complete the names of the following ions: N 3  O 2  F  _________ __________ _________ _________ __________ _________ P 3  S 2  Cl  _________ __________ _________ _________ __________ _________ Br  Br  _________ _________ Learning Check

35. 35 N 3  O 2  F  nitride oxide fluoride nitride oxide fluoride P 3  S 2  Cl  phosphide sulfide chloride phosphide sulfide chloride Br  bromide Solution

36. 36 The name of a binary ionic compound (two elements) gives the name of the metal ion first and the name of the negative ion second. The name of a binary ionic compound (two elements) gives the name of the metal ion first and the name of the negative ion second.Examples: NaClsodium chloride K 2 Spotassium sulfide CaI 2 calcium iodide Al 2 O 3 aluminum oxide Naming Ionic Compounds with Two Elements

37. 37 Write the names of the following compounds: 1) Na 3 N___________ 2) KBr___________ 3) Al 2 S 3 ___________ 4) MgO___________ Learning Check

38. 38 Complete the names of the following compounds: 1)Na 3 Nsodium nitride 2)KBrpotassium bromide 3)Al 2 S 3 aluminum sulfide 4)MgOmagnesium oxide Solution

39. 39  Most transition elements have two or more positive ions. Ionic Charges of Transition Metals

40. 40 Summary of Common Ions Of the transition metals, silver and zinc are important elements that form only one ion. Of the transition metals, silver and zinc are important elements that form only one ion.

41. 41 A. The formula for the ionic compound of Na + and O 2- is Na + and O 2- is 1) NaO2) Na 2 O3) NaO 2 B. The formula of a compound of aluminum and chlorine is 1) Al 3 Cl2) AlCl 2 3) AlCl 3 C. The formula of Fe 3+ and O 2- is 1) Fe 3 O 2 2) FeO 3 3) Fe 2 O 3 Learning Check

42. 42 A. The formula for the ionic compound of Na + and O 2- is Na + and O 2- is 2) Na 2 O B. The formula of a compound of aluminum and chlorine is 3) AlCl 3 3) AlCl 3 C. The formula of Fe 3+ and O 2- is 3) Fe 2 O 3 Solution

43. 43 Naming Compounds with Transition Metals Transition metals with two different ions use a Roman numeral following the name of the metal to indicate ionic charge. Transition metals with two different ions use a Roman numeral following the name of the metal to indicate ionic charge.

44. 44 Learning Check Select the correct name for each. A. Fe 2 S 3 1) iron sulfide 1) iron sulfide 2) iron(II) sulfide 2) iron(II) sulfide 3) iron (III) sulfide 3) iron (III) sulfide B. CuSO 4 1) copper sulfate 1) copper sulfate 2) copper(I) sulfate 2) copper(I) sulfate 3) copper (II) sulfate 3) copper (II) sulfate

45. 45 Solution Select the correct name for each. A. Fe 2 S 3 3) iron (III) sulfide 3) iron (III) sulfide B. CuSO 4 3) copper (II) sulfate 3) copper (II) sulfate

46. 46 Learning Check The correct formula is A. Copper (I) nitride 1) CuN2) CuN 3 3) Cu 3 N 1) CuN2) CuN 3 3) Cu 3 N B. Lead (IV) oxide 1) PbO 2 2) PbO 3) Pb 2 O 4 1) PbO 2 2) PbO 3) Pb 2 O 4

47. 47 Solution The correct formula is A. Copper (I) nitride 3) Cu 3 N 3) Cu 3 N B. Lead (IV) oxide 1) PbO 2 1) PbO 2

48. 48 4.5 Covalent Bonds 4.6 Naming and Writing Formulas of Covalent Compounds 4.7 Bond Polarity Chapter 4 Compounds and Their Bonds

49. 49  Covalent bonds form between two nonmetals from Groups 4A, 5A, 6A, and 7A.  In a covalent bond, electrons are shared to complete octets. Covalent Bonds

50. 50 Indicate whether a bond between the following is 1) Ionic2) Covalent ____A.sodium and oxygen ____B.nitrogen and oxygen ____C. phosphorus and chlorine ____D.calcium and sulfur ____E.chlorine and bromine Learning Check

51. 51 Indicate whether a bond between the following is 1) Ionic2) Covalent 1 A. sodium and oxygen 2 B. nitrogen and oxygen 2 C. phosphorus and chlorine 2 C. phosphorus and chlorine 1 D. calcium and sulfur 2 E. chlorine and bromine 2 E. chlorine and bromine Solution

52. 52 In hydrogen, two hydrogen atoms share their electrons to form a covalent bond. In hydrogen, two hydrogen atoms share their electrons to form a covalent bond. Each hydrogen atom acquires a stable outer shell of two (2) electrons like helium (He). Each hydrogen atom acquires a stable outer shell of two (2) electrons like helium (He). H  +  H H : H = H  H = H 2 H  +  H H : H = H  H = H 2 hydrogen molecule H 2, A Covalent Molecule

53. 53 Diatomic Elements As elements, the following share electrons to form diatomic, covalent molecules. As elements, the following share electrons to form diatomic, covalent molecules.

54. 54 What is the name of each of the following diatomic molecules? H 2 hydrogen N 2 nitrogen Cl 2 _______________ O 2 _______________ I 2 _______________ Learning Check

55. 55 What are the names of each of the following diatomic molecules? H 2 hydrogen N 2 nitrogen Cl 2 chlorine O 2 oxygen I 2 iodine Solution

56. 56 The compound NH 3 consists of a N atom and three H atoms. The compound NH 3 consists of a N atom and three H atoms.      N  and 3 H   N  and 3 H    By sharing electrons to form NH 3, the electron dot structure is written as By sharing electrons to form NH 3, the electron dot structure is written as H Bonding pairs H Bonding pairs     H : N : H   Lone pair of electrons   Lone pair of electrons Covalent Bonds in NH 3

57. 57 Number of Covalent Bonds Often, the number of covalent bonds formed by a nonmetal is equal to the number of electrons needed to complete the octet. Often, the number of covalent bonds formed by a nonmetal is equal to the number of electrons needed to complete the octet.

58. 58 Dot Structures and Models of Some Covalent Compounds

59. 59 Multiple Bonds  Sharing one pair of electrons is a single bond. X : X or X–X  In multiple bonds, two pairs of electrons are shared to form a double bond or three pairs of electrons are shared in a triple bond. X : : X or X =X X : : : X or X ≡ X

60. 60 In nitrogen, octets are achieved by sharing three pairs of electrons. In nitrogen, octets are achieved by sharing three pairs of electrons. When three pairs of electrons are shared, the multiple bond is called a triple bond. When three pairs of electrons are shared, the multiple bond is called a triple bond. octets octets                  N  +  N   N:::N     triple bond triple bond Multiple Bonds in N 2

61. 61 In the name of a covalent compound, the first nonmetal is named followed by the name of the second nonmetal ending in –ide. In the name of a covalent compound, the first nonmetal is named followed by the name of the second nonmetal ending in –ide.  Prefixes indicate the number of atoms of each element. Naming Covalent Compounds

62. 62 Complete the name of each covalent compound: COcarbon ______oxide CO 2 carbon _______________ PCl 3 phosphorus ___________ CCl 4 carbon _______________ N 2 O______________________ Learning Check

63. 63 Complete the name of each covalent compound: CO carbon monoxide CO 2 carbon dioxide PCl 3 phosphorus trichloride CCl 4 carbon tetrachloride N 2 Odinitrogen monoxide Solution

64. 64 Formulas and Names of Some Covalent Compounds

65. 65 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

66. 66 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

67. 67  Electronegativity is the attraction of an atom for shared electrons.  The nonmetals have high electronegativity values with fluorine as the highest.  The metals have low electronegativity values. Electronegativity

68. 68 Some Electronegativity Values for Group A Elements

69. 69  The atoms in a nonpolar covalent bond have electronegativity differences of 0.3 or less.  Examples: Atoms Electronegativity Type of Difference Bond N-N 3.0 - 3.0 = 0.0 Nonpolar covalent Cl-Br 3.0 - 2.8 = 0.2 Nonpolar covalent H-Si2.1 - 1.8 = 0.3 Nonpolar covalent Nonpolar Covalent Bonds

70. 70  The atoms in a polar covalent bond have electronegativity differences of 0.4 to 1.6.  Examples: Atoms Electronegativity Type of Difference Bond O-Cl 3.5 - 3.0 = 0.5 Polar covalent Cl-C 3.0 - 2.5 = 0.5 Polar covalent O-S 3.5 - 2.5= 1.0 Polar covalent Polar Covalent Bonds

71. 71 Comparing Nonpolar and Polar Covalent Bonds

72. 72 Ionic Bonds  The atoms in an ionic bond have electronegativity differences of 1.7 or more.  Examples: Atoms ElectronegativityType of Difference Bond Cl-K 3.0 – 0.8 = 2.2Ionic N-Na 3.0 – 0.9 = 2.1Ionic S-Cs2.5 – 0.7= 1.8Ionic

73. 73 Range of Bond Types

74. 74 Predicting Bond Type

75. 75 Identify the type of bond between the following as 1) nonpolar covalent 2) polar covalent 3) ionic 3) ionic A. K-N A. K-N B. N-O C. Cl-Cl Learning Check

76. 76 A. K-N 3) ionic B. N-O 2) polar covalent C. Cl-Cl 1) nonpolar covalent Solution

77. 77 4.8 Polyatomic Ions Chapter 4 Compounds and Their Bonds

78. 78 A polyatomic ion is a group of two or more atoms that has an overall ionic charge. A polyatomic ion is a group of two or more atoms that has an overall ionic charge. Some examples of polyatomic ions are Some examples of polyatomic ions are NH 4 + ammoniumOH - hydroxide NO 3 - nitrateNO 2 - nitrite CO 3 2 - carbonatePO 4 3 - phosphate HCO 3 - hydrogen carbonate (bicarbonate) Polyatomic Ions

79. 79 Common Polyatomic Ions

80. 80 Formulas with Polyatomic Ions The formula of an ionic compound containing a polyatomic ion is written to make the overall charge equal zero (0). The formula of an ionic compound containing a polyatomic ion is written to make the overall charge equal zero (0). Na + and NO 3 - NaNO 3 When two or more polyatomic ions are needed, the polyatomic ion is enclosed in parentheses. When two or more polyatomic ions are needed, the polyatomic ion is enclosed in parentheses. polyatomic ion Mg 2+ and NO 3 - Mg(NO 3 ) 2 subscript 2 for charge balance subscript 2 for charge balance

81. 81 Some Compounds with Polyatomic Ions

82. 82 Select the correct formula for each: A. Aluminum nitrate 1) AlNO 3 2) Al(NO) 3 3) Al(NO 3 ) 3 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 ) 1) CuNO 3 2) Cu(NO 3 ) 2 3) Cu 2 (NO 3 ) C. Iron (III) hydroxide 1) FeOH2) Fe 3 OH3) Fe(OH) 3 1) FeOH2) Fe 3 OH3) Fe(OH) 3 D. Tin(IV) hydroxide 1) Sn(OH) 4 2) Sn(OH) 2 3) Sn 4 (OH) 1) Sn(OH) 4 2) Sn(OH) 2 3) Sn 4 (OH) Learning Check

83. 83 Select the correct formula for each: A. Aluminum nitrate 3) Al(NO 3 ) 3 3) Al(NO 3 ) 3 B. Copper(II) nitrate 2) Cu(NO 3 ) 2 2) Cu(NO 3 ) 2 C. Iron (III) hydroxide 3) Fe(OH) 3 3) Fe(OH) 3 D. Tin(IV) hydroxide 1) Sn(OH) 4 1) Sn(OH) 4 Solution

84. 84  For compounds with polyatomic ions, the positive ion is named first followed by the name of the polyatomic ion. NaNO 3 sodium nitrate K 2 SO 4 potassium sulfate Al(HCO 3 ) 3 aluminum bicarbonate or aluminum hydrogen carbonate (NH 4 ) 3 PO 4 ammonium phosphate Naming Compounds with Polyatomic Ions

85. 85 Match each formula with the correct name: A. Na 2 CO 3 1) magnesium sulfite MgSO 3 2) magnesium sulfate MgSO 3 2) magnesium sulfate MgSO 4 3) sodium carbonate MgSO 4 3) sodium carbonate B. Ca(HCO 3 ) 2 1) calcium carbonate CaCO 3 2) calcium phosphate CaCO 3 2) calcium phosphate Ca 3 (PO 4 ) 2 3) calcium bicarbonate Ca 3 (PO 4 ) 2 3) calcium bicarbonate Learning Check

86. 86 A. Na 2 CO 3 3) sodium carbonate MgSO 3 1) magnesium sulfite MgSO 3 1) magnesium sulfite MgSO 4 2) magnesium sulfate MgSO 4 2) magnesium sulfate B. Ca(HCO 3 ) 2 3) calcium bicarbonate CaCO 3 1) calcium carbonate CaCO 3 1) calcium carbonate Ca 3 (PO 4 ) 2 2) calcium phosphate Ca 3 (PO 4 ) 2 2) calcium phosphate Solution

87. 87 Summary of Naming Compounds

88. 88 Naming Rules

89. 89 Naming Rules (continued)

90. 90 Learning Check Name each of the following compounds: A.Mg(NO 3 ) 2 B.CuCl 2 C.N 2 O 4 D.Fe 2 (SO 4 ) 3 E.Ba 3 (PO 4 ) 2

91. 91 Solution Name each of the following compounds: A.Mg(NO 3 ) 2 magnesium nitrate B.CuCl 2 copper(II) chloride C.N 2 O 4 dinitrogen tetroxide D.Fe 2 (SO 4 ) 3 iron(III) sulfate E.Ba 3 (PO 4 ) 2 barium phosphate

92. 92 Learning Check Write the correct formula for each: A.potassium sulfide B.calcium carbonate C.sodium phosphite D.iron(III) oxide E.iron (II) nitrate

93. 93 Solution Write the correct formula for each: A.potassium sulfideK 2 S B.calcium carbonateCaCO 3 C.sodium phosphiteNa 3 PO 3 D.iron(III) oxideFe 2 O 3 E.iron (II) nitrateFe(NO 3 ) 2

94. 94 Chapter 4 Compounds and Their Bonds 4.9 Shapes of Molecules 4.10Polar and Nonpolar Molecules

95. 95 VSEPR The shape of a molecule is predicted from the geometry of the electrons pairs around the central atom. The shape of a molecule is predicted from the geometry of the electrons pairs around the central atom. In the valence-shell electron-pair repulsion theory (VSEPR), the electron pairs are arranged as far apart as possible to give the least amount of repulsion of the negatively charged electrons. In the valence-shell electron-pair repulsion theory (VSEPR), the electron pairs are arranged as far apart as possible to give the least amount of repulsion of the negatively charged electrons.

96. 96 Two Electron Pairs In a molecule of BeCl 2, there are two bonding pairs around the central atom Be. (Be is an exception to the octet rule.) In a molecule of BeCl 2, there are two bonding pairs around the central atom Be. (Be is an exception to the octet rule.) The arrangement of two electron pairs to minimize their repulsion is 180° or opposite each other. The arrangement of two electron pairs to minimize their repulsion is 180° or opposite each other. The shape of the molecule is linear. The shape of the molecule is linear.

97. 97 Two Electron Pairs with Double Bonds The electron-dot structure for CO 2 consists of two double bonds to the central atom C. The electron-dot structure for CO 2 consists of two double bonds to the central atom C. Because the electrons in a double bond are held together, a double bond is counted as a single unit. Because the electrons in a double bond are held together, a double bond is counted as a single unit. Repulsion is minimized when the double bonds are placed opposite each other at 180° to give a linear shape. Repulsion is minimized when the double bonds are placed opposite each other at 180° to give a linear shape.

98. 98 Three Electron Pairs In BF 3, there are 3 electron pairs around the central atom B. (B is an exception to the octet rule.) In BF 3, there are 3 electron pairs around the central atom B. (B is an exception to the octet rule.) Repulsion is minimized by placing three electron pairs in a plane at angles of 120°, which is a trigonal planar arrangement. Repulsion is minimized by placing three electron pairs in a plane at angles of 120°, which is a trigonal planar arrangement. The shape with three bonded atoms is trigonal planar. The shape with three bonded atoms is trigonal planar.

99. 99 Two Bonding Pairs and A Nonbonding Pair In SO 2, there are 3 electron units around the central atom S. In SO 2, there are 3 electron units around the central atom S. Two electron units are bonded to atoms and one electron pair is a nonbonding pair. Two electron units are bonded to atoms and one electron pair is a nonbonding pair. Repulsion is minimized by placing three electron pairs in a plane at angles of 120°, which is trigonal planar. Repulsion is minimized by placing three electron pairs in a plane at angles of 120°, which is trigonal planar. The shape with two bonded atoms is bent. The shape with two bonded atoms is bent.

100. 100 Learning Check The shape of a molecule of N 2 O (N N O) is 1) linear 2) trigonal planar 3) bent (120°)

101. 101 Solution The shape of a molecule of N 2 O (N N O) is 1) linear The electron-dot structure uses 5 e for each N and 6 e for O (16 e total) has octets using two double bond to the central N and one nonbonding pair. The shape with two bonded atoms is linear. : N :: N :: O : : N :: N :: O :

102. 102 Four Electron Pairs In CH 4, there are 4 electron pairs around the central atom C. In CH 4, there are 4 electron pairs around the central atom C. Repulsion is minimized by placing four electron pairs at angles of 109°, which is a tetrahedral arrangement. Repulsion is minimized by placing four electron pairs at angles of 109°, which is a tetrahedral arrangement. The shape with four bonded atoms is called tetrahedral. The shape with four bonded atoms is called tetrahedral.

103. 103 Three Bonding Atoms and One Nonbonding Pair In NH 3, there are 4 electron pairs around the N. In NH 3, there are 4 electron pairs around the N. Three pairs are bonded to atoms and one is a nonbonding pair. Three pairs are bonded to atoms and one is a nonbonding pair. Repulsion is minimized by placing four electron pairs at angles of 109°, which is a tetrahedral arrangement. Repulsion is minimized by placing four electron pairs at angles of 109°, which is a tetrahedral arrangement. The shape with three bonded atoms is pyramidal. The shape with three bonded atoms is pyramidal.

104. 104 Two Bonding Atoms and Two Lone Pairs In H 2 O, there are 4 electron pairs around O. In H 2 O, there are 4 electron pairs around O. Two pairs are bonded to atoms and two are nonbonding pairs. Two pairs are bonded to atoms and two are nonbonding pairs. Repulsion is minimized by placing four electron pairs at angles of 109° called a tetrahedral arrangement. Repulsion is minimized by placing four electron pairs at angles of 109° called a tetrahedral arrangement. The shape with two bonded atoms is called bent. The shape with two bonded atoms is called bent.

105. 105 Some Steps Using VSEPR to Predict Shape Draw the electron dot structure. Draw the electron dot structure. Count the electron pairs around the central atom. Count the electron pairs around the central atom. Arrange the electron pairs to minimize repulsion. Arrange the electron pairs to minimize repulsion. Determine the shape using the number of bonded atoms in the electron arrangement. Determine the shape using the number of bonded atoms in the electron arrangement.

106. 106 Summary of Electron Arrangements and Shapes

107. 107 Learning Check 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 D. PO 4 3-

108. 108 Solution Use VSEPR theory to determine the shape of the following molecules or ions. 1) tetrahedral2) pyramidal 3) bent A. PF 3 2) pyramidal B.H 2 S3) bent C. CCl 4 1) tetrahedral D. PO 4 3- 1) tetrahedral

109. 109 Polar Molecules A polar molecule contains polar bonds. A polar molecule contains polar bonds. The separation of positive and negative charge is called a dipole. The separation of positive and negative charge is called a dipole. In a polar molecule, dipoles do not cancel. In a polar molecule, dipoles do not cancel.  +  -  +  - H–Cl Cl–N–Cl dipole Cl dipole Cl dipoles do not cancel

110. 110 Nonpolar Molecules A nonpolar molecule contains nonpolar bonds A nonpolar molecule contains nonpolar bonds Cl–ClH–H Cl–ClH–H or a symmetrical arrangement of polar bonds. O=C=O Cl Cl–C–Cl Cl–C–Cl Cl Cl dipoles cancel

111. 111 Learning Check Identify each of the following molecules as 1) polar or 2) nonpolar. Explain. A. PBr 3 B. HBr C. Br 2 D. SiBr 4

112. 112 Solution Identify each of the following molecules as 1) polar or 2) nonpolar. Explain. A. PBr 3 1) polar; pyramidal B. HBr1) polar; polar bond C. Br 2 2) nonpolar, nonpolar bond D. SiBr 4 2) nonpolar; dipoles cancel