1. CHE1101, Chapter 8 Learn, 1 The Basics of Chemical Bonding

2. CHE1101, Chapter 8 Learn, 2 Chemical Bonds Three basic types of bonds –Ionic Electrostatic attraction between ions. –Covalent Sharing of electrons. –Metallic Metal atoms bonded to several other atoms.

3. CHE1101, Chapter 8 Learn, 3 Lewis Symbols G. N. Lewis developed a method to denote potential bonding electrons by using one dot for every valence electron around the element symbol. When forming compounds, atoms tend to gain, lose, or share electrons until they are surrounded by eight valence electrons (the octet rule).

4. CHE1101, Chapter 8 Learn, 4 Ionic Bond Formation Atoms tend to lose (metals) or gain (nonmetals) electrons to make them isoelectronic to the noble gases.

5. CHE1101, Chapter 8 Learn, 5 Energetics of Ionic Bonding As we saw in the last chapter, it takes 496 kJ/mol to remove electrons from sodium.

6. CHE1101, Chapter 8 Learn, 6 Energetics of Ionic Bonding We get 349 kJ/mol back by giving electrons to chlorine.

7. CHE1101, Chapter 8 Learn, 7 Energetics of Ionic Bonding But these numbers don’t explain why the reaction of sodium metal and chlorine gas to form sodium chloride is so exothermic!

8. CHE1101, Chapter 8 Learn, 8 Energetics of Ionic Bonding There must be a third piece to the puzzle. What is as yet unaccounted for is the electrostatic attraction between the newly formed sodium cation and chloride anion.

9. CHE1101, Chapter 8 Learn, 9 Lattice Energy This third piece of the puzzle is the lattice energy: –The energy required to completely separate a mole of a solid ionic compound into its gaseous ions. The energy associated with electrostatic interactions is governed by Coulomb’s law: E el =  Q1Q2dQ1Q2d

10. CHE1101, Chapter 8 Learn, 10 Lattice Energy Lattice energy, then, increases with the charge on the ions. It also increases with decreasing size of ions.

11. CHE1101, Chapter 8 Learn, 11 Energetics of Ionic Bonding By accounting for all three energies (ionization energy, electron affinity, and lattice energy), we can get a good idea of the energetics involved in such a process.

12. CHE1101, Chapter 8 Learn, 12 Practice Exercise Which of the following orderings of lattice energy is correct for these ionic compounds? a) NaCl > MgO > CsI > ScN b) ScN > MgO > NaCl > CsI c) NaCl > CsI > ScN > MgO d) MgO > NaCl > ScN > CsI e) ScN > CsI > NaCl > MgO

13. CHE1101, Chapter 8 Learn, 13 Practice Exercise Which substance do you expect to have the greatest lattice energy? a) MgF 2 b) CaF 2 c) ZrO 2

14. CHE1101, Chapter 8 Learn, 14 Electron Configuration of s- and p-Block Elements The energetics of ionic bond formation explains why many ions adopt noble-gas electron configurations eg. Na always forms a +1 charge Even though lattice energy increases with increasing ionic charge, we never find Na 2+ The second electron removed would have to come from an inner shell Removing electrons from the inner shell would cost more energy than could be compensated for by the increased lattice energy

15. CHE1101, Chapter 8 Learn, 15 Electron Configuration of Transition Metals Ionization energies increase rapidly for each successive electron removed; the lattice energies compensate for loss of up to only three electrons in main block elements. Most transition metals have more than three electrons beyond a noble-gas core Transition metals generally do not form ions that have a noble-gas configuration The driving force for transition elements is often formation of an empty, filled, or half filled subshell

16. CHE1101, Chapter 8 Learn, 16 Determine the empirical formula expected for a compound containing Ca and F.

17. CHE1101, Chapter 8 Learn, 17 Covalent Bonding In covalent bonds, atoms share electrons. There are several electrostatic interactions in these bonds: –attractions between electrons and nuclei, –repulsions between electrons, and –repulsions between nuclei. For a bond to form, the attractions must be greater than the repulsions.

18. CHE1101, Chapter 8 Learn, 18 Covalent Bonding In a covalent “single” bond, 2 electrons are “shared” between 2 atoms Covalently bound species are different than ionic –exist as individual, discrete species (vs. 3-D crystal lattice structure for ionic) –tend to exhibit much lower melting and boiling points (vs. ionic)

19. CHE1101, Chapter 8 Learn, 19 Covalent Bonding

20. CHE1101, Chapter 8 Learn, 20 Lewis Structures Sharing electrons to make covalent bonds can be demonstrated using Lewis structures. We start by trying to give each atom the same number of electrons as the nearest noble gas by sharing electrons. The simplest examples are for hydrogen, H 2, and chlorine, Cl 2, shown below.

21. CHE1101, Chapter 8 Learn, 21 Electrons on Lewis Structures Lone pairs: electrons located on only one atom in a Lewis structure Bonding pairs: shared electrons in a Lewis structure; they can be represented by two dots or one line

22. CHE1101, Chapter 8 Learn, 22 Multiple Bonds Some atoms share only one pair of electrons. These bonds are called single bonds. Sometimes, two pairs need to be shared. These are called double bonds. There are even cases where three bonds are shared between two atoms. These are called triple bonds.

23. CHE1101, Chapter 8 Learn, 23 Multiple Bonds single bond – 2 electrons (1 pair of electrons) are shared between 2 atoms double bond – 4 electrons (2 pairs of electrons) are shared between 2 atoms triple bond – 6 electrons (3 pairs of electrons) are shared between 2 atoms

24. CHE1101, Chapter 8 Learn, 24

25. CHE1101, Chapter 8 Learn, 25 a double bond is shorter and stronger than a single bond a triple bond is shorter and stronger than a double bond Multiple Bonds

26. CHE1101, Chapter 8 Learn, 26 Polar Covalent Bonds The electrons in a covalent bond are not always shared equally. Fluorine pulls harder on the electrons it shares with hydrogen than hydrogen does. Therefore, the fluorine end of the molecule has more electron density than the hydrogen end.

27. CHE1101, Chapter 8 Learn, 27 Cl 2 nonpolar bond – electrons are shared equally in the bond polar bond – electrons are NOT shared equally HCl

28. CHE1101, Chapter 8 Learn, 28 electronegativity increases from left right electronegativity decreases from top bottom Electronegativity – the ability of an element to attract electron density to itself in a molecule

29. CHE1101, Chapter 8 Learn, 29

30. CHE1101, Chapter 8 Learn, 30 Arrange the following in order of increasing electronegativity: Na, F, O, K, Al, Si, Mg

31. CHE1101, Chapter 8 Learn, 31 Electronegativity and Polar Covalent Bonds When two atoms share electrons unequally, a polar covalent bond results. Electrons tend to spend more time around the more electronegative atom. The result is a partial negative charge (not a complete transfer of charge). It is represented by δ –. The other atom is “more positive,” or δ +.

32. CHE1101, Chapter 8 Learn, 32 Polar Covalent Bonds When two atoms share electrons unequally, a bond dipole results. The dipole moment, , produced by two equal but opposite charges separated by a distance, r, is calculated:  = Q r It is measured in debyes (D).

33. CHE1101, Chapter 8 Learn, 33 Polar Covalent Bonds The greater the difference in electronegativity, the more polar is the bond.

34. CHE1101, Chapter 8 Learn, 34 Polar Covalent Bonds We can use the difference in electronegativity between two atoms to gauge the polarity of the bond the atoms form

35. CHE1101, Chapter 8 Learn, 35 Rank the following in order of increasing bond polarity: H─FH─BrF─FNa─Cl a)H─F < H─Br < F─F < Na─Cl b)F─F < H─F < H─Br < Na─Cl c)H─Br < H─F < F─F < Na─Cl d)F─F < H─Br < H─F < Na─Cl e)Na─Cl < H─F < H─Br < F─F

36. CHE1101, Chapter 8 Learn, 36 Rank the following in order of increasing bond polarity: H─FH─BrF─FNa─Cl a)H─F < H─Br < F─F < Na─Cl b)F─F < H─F < H─Br < Na─Cl c)H─Br < H─F < F─F < Na─Cl d)F─F < H─Br < H─F < Na─Cl e)Na─Cl < H─F < H─Br < F─F

37. CHE1101, Chapter 8 Learn, 37 Metallic Bonds The relatively low ionization energy of metals allows them to lose electrons easily. The simplest theory of metallic bonding involves the metal atoms releasing their valence electrons to be shared as a pool by all the atoms/ions in the metal. –An organization of metal cation islands in a sea of electrons –Electrons delocalized throughout the metal structure Bonding results from attraction of cation for the delocalized electrons.

38. CHE1101, Chapter 8 Learn, 38 Metallic Bonding

39. CHE1101, Chapter 8 Learn, 39 a) H 2 O b) C 6 H 12 O 6 c) NO 2 d) Al e) CaCO 3 Which of the following compounds has ionic bonding?

40. CHE1101, Chapter 8 Learn, 40 a) H 2 O b) C 6 H 12 O 6 c) NO 2 d) Al e) CaCO 3 Which of the following compounds has ionic bonding?

41. CHE1101, Chapter 8 Learn, 41 Write the Lewis Structure for Arsenic.

42. CHE1101, Chapter 8 Learn, 42 Writing Lewis Structures (Covalent Molecules) 1.Sum the valence electrons from all atoms, taking into account overall charge. –If it is an anion, add one electron for each negative charge. –If it is a cation, subtract one electron for each positive charge. PCl 3 Keep track of the electrons: 5 + 3(7) = 26

43. CHE1101, Chapter 8 Learn, 43 Writing Lewis Structures 2.Write the symbols for the atoms, show which atoms are attached to which, and connect them with a single bond (a line representing two electrons). Keep track of the electrons: 26 − 6 = 20 Heavier element is often the central element

44. CHE1101, Chapter 8 Learn, 44 Writing Lewis Structures 3.Complete the octets around all atoms bonded to the central atom. Keep track of the electrons: 26 − 6 = 20; 20 − 18 = 2

45. CHE1101, Chapter 8 Learn, 45 Writing Lewis Structures 4.Place any leftover electrons on the central atom. Keep track of the electrons: 26 − 6 = 20; 20 − 18 = 2; 2 − 2 = 0

46. CHE1101, Chapter 8 Learn, 46 Writing Lewis Structures 5. If there are not enough electrons to give the central atom an octet, try multiple bonds. Do NOT (under any circumstance…..ever) form a multiple bond to a halogen or hydrogen

47. CHE1101, Chapter 8 Learn, 47 Writing Lewis Structures Then assign formal charges. –For each atom, count the electrons in lone pairs and half the electrons it shares with other atoms. –Subtract that from the number of valence electrons for that atom: the difference is its formal charge.

48. CHE1101, Chapter 8 Learn, 48 Writing Lewis Structures The best Lewis structure… –…is the one with the fewest charges. –…puts a negative charge on the most electronegative atom.

49. CHE1101, Chapter 8 Learn, 49 Draw the Lewis structure for F 2 bonding pair of e - – e - that hold two atoms (bonding pair) together nonbonding pair of e - – e - that are NOT holding (lone pair) 2 atoms together

50. CHE1101, Chapter 8 Learn, 50 Draw the Lewis structure for H 2 O

51. CHE1101, Chapter 8 Learn, 51 Draw the Lewis structure for ethene, C 2 H 4

52. CHE1101, Chapter 8 Learn, 52 Draw the Lewis structure for NO +

53. CHE1101, Chapter 8 Learn, 53 The Best Lewis Structure? Following our rules, this is the Lewis structure we would draw for ozone, O 3. However, it doesn’t agree with what is observed in nature: Both O to O connections are the same.

54. CHE1101, Chapter 8 Learn, 54 Resonance One Lewis structure cannot accurately depict a molecule like ozone. We use multiple structures, resonance structures, to describe the molecule.

55. CHE1101, Chapter 8 Learn, 55 Resonance In truth, the electrons that form the second C—O bond in the double bonds below do not always sit between that C and that O, but rather can move among the two oxygens and the carbon. They are not localized; they are delocalized.

56. CHE1101, Chapter 8 Learn, 56 Resonance The organic compound benzene, C 6 H 6, has two resonance structures. It is commonly depicted as a hexagon with a circle inside to signify the delocalized electrons in the ring. Localized electrons are specifically on one atom or shared between two atoms; Delocalized electrons are shared by multiple atoms.

57. CHE1101, Chapter 8 Learn, 57 Determine the Lewis structure for NO 2 - What are your bond expectations for nitrite ?

58. CHE1101, Chapter 8 Learn, 58 A single Lewis structure can NOT be drawn to describe the “real” nitrite species go to lab and measure the actual bond lengths in a real nitrite anion The N-O bonds in nitrite are identical (in every sense; same length; same strength)

59. CHE1101, Chapter 8 Learn, 59 The Real molecule is somewhere in between these two extremes

60. CHE1101, Chapter 8 Learn, 60 Draw the Lewis structure for HNO 3

61. CHE1101, Chapter 8 Learn, 61 Exceptions to the Octet Rule There are three types of ions or molecules that do not follow the octet rule: –ions or molecules with an odd number of electrons, –ions or molecules with less than an octet, –ions or molecules with more than eight valence electrons (an expanded octet).

62. CHE1101, Chapter 8 Learn, 62 Odd Number of Electrons Though relatively rare and usually quite unstable and reactive, there are ions and molecules with an odd number of electrons.

63. CHE1101, Chapter 8 Learn, 63 Draw the Lewis structure for BF 3

64. CHE1101, Chapter 8 Learn, 64 Fewer Than Eight Electrons Consider BF 3 : –Giving boron a filled octet places a negative charge on the boron and a positive charge on fluorine. –This would not be an accurate picture of the distribution of electrons in BF 3.

65. CHE1101, Chapter 8 Learn, 65 Fewer Than Eight Electrons Therefore, structures that put a double bond between boron and fluorine are much less important than the one that leaves boron with only 6 valence electrons.

66. CHE1101, Chapter 8 Learn, 66 Fewer Than Eight Electrons The lesson is: If filling the octet of the central atom results in a negative charge on the central atom and a positive charge on the more electronegative outer atom, don’t fill the octet of the central atom.

67. CHE1101, Chapter 8 Learn, 67 Draw the Lewis structure for PF 5 octet expansion – some atoms can exceed 8 valence electrons (usually P & S)

68. CHE1101, Chapter 8 Learn, 68 More Than Eight Electrons The only way PCl 5 can exist is if phosphorus has 10 electrons around it. It is allowed to expand the octet of atoms on the third row or below. –Presumably d orbitals in these atoms participate in bonding.

69. CHE1101, Chapter 8 Learn, 69 Draw the Lewis structure for PO 4 3 -

70. CHE1101, Chapter 8 Learn, 70 More Than Eight Electrons Even though we can draw a Lewis structure for the phosphate ion that has only 8 electrons around the central phosphorus, the better structure puts a double bond between the phosphorus and one of the oxygens.

71. CHE1101, Chapter 8 Learn, 71 More Than Eight Electrons This eliminates the charge on the phosphorus and the charge on one of the oxygens. The lesson is: When the central atom is on the third row or below and expanding its octet eliminates some formal charges, do so.

72. CHE1101, Chapter 8 Learn, 72 Metallic Bonds The low ionization energy of metals allows them to lose electrons easily. The simplest theory of metallic bonding involves the metal atoms releasing their valence electrons to be shared by all atoms/ions in the metal. –An organization of metal cation islands in a sea of electrons –Electrons delocalized throughout the metal structure Bonding results from attraction of the cations for the delocalized electrons.

73. CHE1101, Chapter 8 Learn, 73 Metallic Bonding

74. CHE1101, Chapter 8 Learn, 74 Practice Problems

75. CHE1101, Chapter 8 Learn, 75 Predict the formula of the compound that forms between calcium and chlorine. Draw the Lewis dot symbols of the elements. Ca ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ Transfer all the valence electrons from the metal to the nonmetal, adding more of each atom as you go, until all electrons are lost from the metal atoms and all nonmetal atoms have eight electrons. Ca ∙ ∙ Cl ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙ Ca 2+ CaCl 2 Cl Using Lewis Theory to Predict Chemical Formulas of Ionic Compounds

76. CHE1101, Chapter 8 Learn, 76 Use Lewis symbols to predict the formula of an ionic compound made from reacting a metal, M, that has two valence electrons with a nonmetal, X, that has five valence electrons M3X2M3X2

77. CHE1101, Chapter 8 Learn, 77 Assuming that the separation between cations and anions in the lattice is nearly identical, which species would have the greatest lattice energy? A. sodium chloride B. calcium chloride C. calcium nitride D. sodium oxide E. calcium oxide Highest charges (+2 and -3) Your Turn!

78. CHE1101, Chapter 8 Learn, 78 Which combination below will have the most negative lattice energy? A. Low-charge ions separated by large distances B. Low-charge ions separated by small distances C. High-charged ions separated by large distances D. High-charged ions separated by small distances As q increases,  H increases As r increases,  H decreases Your Turn!

79. CHE1101, Chapter 8 Learn, 79 First examine the ion charges and order by sum of the charges Ca 2+ & O 2-, K + & Br ─, K + & Cl ─, Sr 2+ & O 2─ (KBr, KCl) < (CaO, SrO) Then examine the ion sizes of each group and order by radius; larger < smaller (KBr, KCl) same cation, Br ─ > Cl ─ (same Group) KBr < KCl < (CaO, SrO) (CaO, SrO) same anion, Sr 2+ > Ca 2+ (same Group) KBr < KCl < SrO < CaO Order the following ionic compounds in order of increasing magnitude of lattice energy: CaO, KBr, KCl, SrO

80. CHE1101, Chapter 8 Learn, 80 First examine the ion charges and order by sum of the charges Mg 2+ & S 2-, Na + & Br ─, Li + & Br ─, Sr 2+ & S 2─ (NaBr, LiBr) < (MgS, SrS) Then examine the ion sizes of each group and order by radius; larger < smaller (NaBr, LiBr) same anion, Na + > Li + (same Group) NaBr < LiBr < (MgS, SrS) (MgS, SrS) same anion, Sr 2+ > Mg 2+ (same Group) NaBr < LiBr < SrS < MgS Order the following ionic compounds in order of increasing magnitude of lattice energy: MgS, NaBr, LiBr, SrS

81. CHE1101, Chapter 8 Learn, 81 KBr (734 ºC) CaCl 2 (772 ºC) MgF 2 (1261 ºC) KBr CaCl 2 MgF 2 Which ionic compound below has the highest melting point?

82. CHE1101, Chapter 8 Learn, 82 Your Turn! What is the correct electron configuration for Cs and Cs + ? A. [Xe] 6s 1, [Xe] B. [Xe] 6s 2, [Xe] 6s 1 C. [Xe] 5s 1, [Xe] D. [Xe] 6s 1, [Xe] 6s 2 E. [Xe] 6p 2, [Xe] 6p 1

83. CHE1101, Chapter 8 Learn, 83 What charges are aluminum, phosphorus, sulfur, strontium, and rubidium most likely to have when they become ions? A. -3, +3, +2, -2, -1, respectively B. -5, -3, -2, +2, +1, respectively C. +3, -3, -2, +2, +1, respectively D. +13, +3, -2, +2, +2, respectively E. +3, +5, -2, -2, +1, respectively Your Turn!

84. CHE1101, Chapter 8 Learn, 84 Which of the following compounds will have the highest-magnitude lattice energy value? KBr NaBr LiBr RbBr All are the same.

85. CHE1101, Chapter 8 Learn, 85 Which of the following compounds will have the highest-magnitude lattice energy value? KBr NaBr LiBr RbBr All are the same.

86. CHE1101, Chapter 8 Learn, 86 a)Al(s) + 3/2 Cl 2 (g)  AlCl 3 (s) b)Al(s)  Al(g) c)3/2 Cl 2 (g)  3 Cl(g) d)Al 3+ (g) + 3 Cl – (g)  AlCl 3 (s) e)Cl(g) + e –  Cl – (g) Which of the following represents the lattice energy for AlCl 3 ?

87. CHE1101, Chapter 8 Learn, 87 a)Al(s) + 3/2 Cl 2 (g)  AlCl 3 (s) b)Al(s)  Al(g) c)3/2 Cl 2 (g)  3 Cl(g) d)Al 3+ (g) + 3 Cl – (g)  AlCl 3 (s) e)Cl(g) + e –  Cl – (g) Which of the following represents the lattice energy for AlCl 3 ?

88. CHE1101, Chapter 8 Learn, 88 Your Turn! Which species is most likely to have multiple bonds? A. CO B. H 2 O C. PH 3 D. BF 3 E. CH 4

89. CHE1101, Chapter 8 Learn, 89 Your Turn! Which species is most likely covalently bonded? A. CsCl B. NaF C. CaF 2 D. CO E. MgBr 2

90. CHE1101, Chapter 8 Learn, 90 Your Turn! What is the primary driving force behind the formation of covalent bonds? A. The energy released when two electrons attract each other B. The energy gained when two electrons attract each other C. The energy released when nuclei attract electrons D. The energy released by two nuclei attracting each other E. The energy gained by the repulsion between nuclei

91. CHE1101, Chapter 8 Learn, 91 Your Turn! What force(s) limit(s) the distance between nuclei in a covalent bond? A. The repulsive forces between nuclei B. The attractive forces between nuclei C. The attractive forces between electrons D. The attractive forces between nuclei and electrons E. The octet rule

92. CHE1101, Chapter 8 Learn, 92 Which of the following has the more electronegative atom written first? Br, Cl S, F Si, P O, S

93. CHE1101, Chapter 8 Learn, 93 Which of the following has the more electronegative atom written first? Br, Cl S, F Si, P O, S

94. CHE1101, Chapter 8 Learn, 94 Determine the electronegativity of each element N = 3.0; O = 3.5 Subtract the electronegativities, large minus small (3.5) − (3.0) = 0.5 If the difference is 2.0 or larger, then the bond is ionic; otherwise it’s covalent –difference (0.5) is less than 2.0, therefore covalent If the difference is 0.5 to 1.9, then the bond is polar covalent; otherwise it’s covalent –difference (0.5) is 0.5 to 1.9, therefore polar covalent Determine whether an N―O bond is ionic, covalent, or polar covalent

95. CHE1101, Chapter 8 Learn, 95 Which situation below results in the largest dipole moment? A. +1 and -1 charges separated by 6 Å B. +1 and -1 charges separated by 8 Å C. +2 and -2 charges separated by 4 Å D. +2 and -2 charges separated by 6 Å E. +2 and -2 charges separated by 8 Å μ = q × r dipole moment increases as both q and r increase. Your Turn!

96. CHE1101, Chapter 8 Learn, 96 Your Turn! Which of the following species has the least polar bond? A. HCl B. HF C. HI D. HBr

97. CHE1101, Chapter 8 Learn, 97 Predict the type of bonding (covalent or ionic) in the following: magnesium chloride, carbon tetrachloride, iron(III) oxide, sulfur dioxide, carbon disulfide A. ionic, ionic, ionic, covalent, covalent B. covalent, ionic, covalent, ionic, ionic C. ionic, covalent, covalent, covalent, ionic D. ionic, covalent, ionic, covalent, covalent Your Turn!

98. CHE1101, Chapter 8 Learn, 98 Example: H 2 CO 3 CO 3 2– oxoanion, so C central, and O s around, H + attached to two O s 1 C = 1  4e – = 4 e – 3 O = 3  6e – = 18 e – 2 H = 2  1e – = 2 e – Total = 24 e – single bonds – 10 e – 14 e – O lone pairs – 14 e – 0 e –  But C only has 6 e –

99. CHE1101, Chapter 8 Learn, 99 Example: H 2 CO 3 (cont.) Too few electrons on C Must convert one of lone pairs on O to second bond to C Form double bond between C and O

100. CHE1101, Chapter 8 Learn, 100 Example: N 2 F 2 2 N = 2  5e – = 10 e – 2F = 2  7e – = 14 e – Total = 24 e – single bonds – 6 e – 18 e – F lone pairs – 12 e – 6 e – N electrons – 6 e – 0 e – Skeletal Structure Complete terminal atom octets Put remaining electrons on central atom

101. CHE1101, Chapter 8 Learn, 101 Example: N 2 F 2 (cont.) Not enough electrons to complete octets on nitrogen Must form double bond between nitrogen atoms to satisfy both octets

102. CHE1101, Chapter 8 Learn, 102 Example: BBr 3  B has only six electrons  Does not form double bond  Has incomplete octet 1 B = 1  3 e – = 3 e – 3 Br = 3  7e – = 21 e – Total = 24 e – single bonds – 6 e – 18 e – Br lone pairs – 18 e – 0 e –

103. CHE1101, Chapter 8 Learn, 103 Example: PCl 5 1 P = 1  5 e – = 5 e – 5 Cl = 5  7e – = 35 e – Total = 40 e – single bonds – 10 e – 30 e – Cl lone pairs – 30 e – 0 e –  P has 10 electrons  Third period element  Can expand its shell

104. CHE1101, Chapter 8 Learn, 104 Draw the best Lewis structure for XeI 2. How many lone pairs of electrons are on Xe? 0 1 2 3 4

105. CHE1101, Chapter 8 Learn, 105 Draw the best Lewis structure for XeI 2. How many lone pairs of electrons are on Xe? 0 1 2 3 4

106. CHE1101, Chapter 8 Learn, 106 What type of bond exists between the C and O atoms in CH 2 O (i.e., single, double, etc.)? A. single bond B. double bond C. triple bond D. There is no bond between the C and the O because the oxygen is bonded to a H: C-H-H-O C HH O 106 Your Turn!

107. CHE1101, Chapter 8 Learn, 107 What is the best Lewis structure for CS 2 ?

108. CHE1101, Chapter 8 Learn, 108 What is the best Lewis structure for CS 2 ?

109. CHE1101, Chapter 8 Learn, 109 CO 2 SeOF 2 NO 2 − H 3 PO 4 SO 3 2− P 2 H 4 Practice – Draw Lewis Structures of the Following

110. CHE1101, Chapter 8 Learn, 110 CO 2 SeOF 2 NO 2 − H 3 PO 4 SO 3 2− P 2 H 4 16 e − 26 e − 18 e − 26 e − 32 e − 14 e − Practice – Draw Lewis Structures of the Following

111. CHE1101, Chapter 8 Learn, 111 1 S = 1  6e – = 6 e – 4 O = 4  6e – = 24 e – 2 H = 2  1e – = 2 e – Total = 32 e – single bonds – 12 e – 20 e – O lone pairs – 20 e – 0 e – Example: H 2 SO 4 n = 3, has empty d orbitals Could expand its octet Could write structure with double bonds.

112. CHE1101, Chapter 8 Learn, 112 FC = # valence e –  [# bonds to atom + # unshared e – ] Structure 1 FC S = 6 – (4 + 0) = 2 FC H = 1 – (1 + 0) = 0 FC O(s) = 6 – (1 + 6) = –1 FC O(d) = 6 – (2 + 4) = 0 Structure 2 FC S = 6 – (6 + 0) = 0 FC H = 1 – (1 + 0) = 0 FC O(s) = 6 – (2 + 4) = 0 FC O(d) = 6 – (2 + 4) = 0 +2

113. CHE1101, Chapter 8 Learn, 113 H 2 SO 4 No formal charges on any atom in structure 2 When several Lewis structures are possible, Most Stable Lewis Structure has: 1. Lowest sum of total formal charges 2. All FC   1  3. Any negative FC on most electronegative element

114. CHE1101, Chapter 8 Learn, 114 Example: CO 2 1 C = 1  4e – = 4 e – 2 O = 2  6e – = 12 e – Total = 16 e – single bonds – 4 e – 12 e – O lone pairs – 12 e – 0 e –  C only has four electrons  Need two extra bonds to O to complete octet  3 ways you can do this  Which of these is correct?  Need another criteria  Come back to this

115. CHE1101, Chapter 8 Learn, 115 CO 2 Which Structure is Best Use formal charge to determine which structure is best FC C = 4 – (4 + 0) = 0 FC O(single) = 6 – (1 + 6) = –1 FC O(double) = 6 – (2 + 4) = 0 FC O(triple) = 6 – (3 + 2) = +1 Central structure best All FC’s = 0 11 +1

116. CHE1101, Chapter 8 Learn, 116 CO 2 SeOF 2 NO 2 − H 3 PO 4 SO 3 2− P 2 H 4 Practice – Assign formal charges

117. CHE1101, Chapter 8 Learn, 117 CO 2 SeOF 2 NO 2 − all 0 Se = +1 H 3 PO 4 SO 3 2− P 2 H 4 P = +1 rest 0 S = +1 all 0 Practice – Assign formal charges

118. CHE1101, Chapter 8 Learn, 118 What is the formal charge on each atom in the structure below? A. O: -2N: 0S: -2 B. O: 0N: 0S: 0 C. O: -1N: 0S: -1 D. O: -2N: +1S: +2 E. O: -1N: 0S: 0 ONS Your Turn!

119. CHE1101, Chapter 8 Learn, 119 Multiple Lewis structures for a single molecule –No single Lewis structure is correct –Structure not accurately represented by any one Lewis structure –Actual structure = “average” of all possible structures A “resonance hybrid” –Double headed arrow between resonance structures used to denote resonance What are Resonance Structures?

120. CHE1101, Chapter 8 Learn, 120 Lewis structures assume electrons are localized between 2 atoms. In resonance structures, electrons are delocalized. –They are smeared out over all the atoms –They can move around the entire molecule to give equivalent bond distances. Resonance Hybrid A way to depict resonance delocalization. Resonance Structures

121. CHE1101, Chapter 8 Learn, 121 1 N = 1  5e – = 5 e – 3 O = 3  6e – = 18 e –  1 charge = 1 e – Total = 24 e – single bonds – 6 e – 18 e – O lone pairs –18 e – 0 e – C only has 6 electrons so a double bond is needed Example: CO 3 2–

122. CHE1101, Chapter 8 Learn, 122 All structures have the same formal charges on C and O’s –FC = –1 on singly bonded O’s –FC = O on doubly bond O and C Three Equivalent Resonance Structures

123. CHE1101, Chapter 8 Learn, 123 Drawing Resonance Structures 1.Draw the first Lewis structure that maximizes octets. 2.Assign formal charges. 3.Move electron pairs from atoms with (−) formal charge toward atoms with (+) formal charge. 4.If (+) formal charge atom is in second row, only move in electrons if you can move out electron pairs from a multiple bond. 5.If (+) formal charge atom is in the third row or below, keep bringing in electron pairs to reduce the formal charge, even if you get an expanded octet. +1 −1 +1 −1

124. CHE1101, Chapter 8 Learn, 124 Drawing Resonance Structures 1.Draw the first Lewis structure that maximizes octets. 2.Assign formal charges. 3.Move electron pairs from atoms with (−) formal charge toward atoms with (+) formal charge. 4.If (+) formal charge atom in second row, only move in electrons if you can move out electron pairs from a multiple bond. 5.If (+) formal charge atom In third row or below, keep bringing in electron pairs to reduce the formal charge, even if you get expanded octet. −1 +2

125. CHE1101, Chapter 8 Learn, 125 Evaluating Resonance Structures Better structures have fewer formal charges. Better structures have smaller formal charges. Better structures have the negative formal charge on the more electronegative atom. better resonance structure −1 +2

126. CHE1101, Chapter 8 Learn, 126 CO 2 SeOF 2 NO 2 − all 0 +1 H 3 PO 4 SO 3 2− P 2 H 4 Practice – Identify Structures with Better or Equal Resonance Forms and Draw Them +1

127. CHE1101, Chapter 8 Learn, 127 Practice – Identify Structures with Better or Equal Resonance Forms and Draw Them none −1 +1 none H 3 PO 4 SO 3 2− P 2 H 4 CO 2 SeOF 2 NO 2 −

128. CHE1101, Chapter 8 Learn, 128 1 C = 1  4e – = 4 e – 1 N = 1  5e – = 5 e – 1 O = 1  6e – = 6 e – –1 charge = 1 e – Total = 16 e – single bonds – 4 e – 12 e – lone pairs –12 e – 0 e – Example: NCO –

129. CHE1101, Chapter 8 Learn, 129 FC N = 5 – 2 – 3 = 0 FC C = 4 – 0 – 4 = 0 FC O = 6 – 6 – 1 = –1 FC N = 5 – 4 – 2 = –1 FC C = 4 – 0 – 4 = 0 FC O = 6 – 4 – 2 = 0 FC N = 5 – 6 – 1 = –2 FC C = 4 – 0 – 4 = 0 FC O = 6 – 2 – 3 = +1 OK Best Not Accept- able 22 +1+1 11 11 Example: NCO –

130. CHE1101, Chapter 8 Learn, 130 Determine the formal charge of chlorine in the most stable Lewis structure for ClO 2 –. a)–2 b)–1 c) 0 d)+1 e)+2

131. CHE1101, Chapter 8 Learn, 131 Determine the formal charge of chlorine in the most stable Lewis structure for ClO 2 –. a)–2 b)–1 c) 0 d)+1 e)+2

132. CHE1101, Chapter 8 Learn, 132 a)S: –2C: –4 N: –3 b)S: +6C: +4 N: +5 c)S: –2C: –4 N: –3 d)S: +1C: 0 N: –1 e)S: 0C: 0 N: –1 Which of the following sets gives the correct formal charges for the elements in the Lewis structure below?

133. CHE1101, Chapter 8 Learn, 133 a)S: –2C: –4 N: –3 b)S: +6C: +4 N: +5 c)S: –2C: –4 N: –3 d)S: +1C: 0 N: –1 e)S: 0C: 0 N: –1 Which of the following sets gives the correct formal charges for the elements in the Lewis structure below?

134. CHE1101, Chapter 8 Learn, 134 Based on formal charges, which structure of hydrogen cyanide is the most stable? a) b) c) d)

135. CHE1101, Chapter 8 Learn, 135 Based on formal charges, which structure of hydrogen cyanide is the most stable? a) b) c) d)

136. CHE1101, Chapter 8 Learn, 136 Draw the best Lewis structure for NO 3 –. How many equivalent resonance structures can be drawn? a) 1 b) 2 c) 3 d) 4 e) 5

137. CHE1101, Chapter 8 Learn, 137 Draw the best Lewis structure for NO 3 –. How many equivalent resonance structures can be drawn? a) 1 b) 2 c) 3 d) 4 e) 5

138. CHE1101, Chapter 8 Learn, 138 Which of the following is false? A. Not all resonance structures for a given molecule are equivalent B. Resonance structures indicate that the molecule rapidly switches between multiple electron distributions C. Resonance structures are necessary when a single Lewis structure is not sufficient to depict the electron distribution in a molecule D. Formal charges are calculated the same way in resonance structures as in other structures 138 Your Turn!

139. CHE1101, Chapter 8 Learn, 139 Which of the structures below exhibit resonance (hint: there is more than one)? A. NO 2 B. H 2 O C. N 3 – D. N 2 O (nitrogen is central atom) E. CH 3 CH 2 CH Your Turn!