Chapter 8. Basic Concepts of Chemical Bonding

Transcription

1 Chapter 8 Basic Concepts of Chemical Bonding

2 Chemical Bonds An attractive force that holds two atoms together in a more complex unit Three basic types of bonds Ionic Electrons are transferred from one atom or group of atoms to another Electrostatic attraction between ions Covalent Electrons are shared between atoms Metallic Metal atoms bonded to several other atoms 2015 Pearson Education, Inc. 2

3 Chemical Bonds Fundamental concepts in bonding Not all electrons in an atom participate in chemical bonding Certain arrangements of electrons are more stable than others Form if energy of the aggregate is lower than the separate atoms 2015 Pearson Education, Inc. 3

4 Chemical Bonds: Valence Electrons Valence electrons participate in chemical bonding Outermost electron shell in a representative element or noble gas 4

5 Chemical Bonds: Valence Electrons How do you visually represent valence electrons on an atom? Lewis symbol Chemical symbol of an element surrounded by its valence electrons Methodology for writing Lewis symbols Write the chemical symbol Begin adding dots, representing electrons, at any side Place the first four dots separately on the four sides of the chemical symbol Begin pairing dots if further dots are added 5

6 Chemical Bonds Fundamental concepts in bonding Not all electrons in an atom participate in chemical bonding Certain arrangements of electrons are more stable than others Form if energy of the aggregate is lower than the separate atoms 2015 Pearson Education, Inc. 6

7 Chemical Bonds: Octet Rule Certain arrangements of electrons are more stable than others Noble gases is considered to be the most stable of all valence electron configurations Has a completely filled outermost subshell Most unreactive of all elements Atoms without full outer shell form compounds to complete shell Atoms tend to gain, lose or share electrons until they are surrounded by eight valence electrons octet rule 7

8 Chemical Bonds: Octet Rule Atoms lose or gain electrons until they reach an electron configuration that is the same as a noble gas Process involving the fewer number of electrons will always be the more energetically favorable process 8

9 Chemical Bonds: Octet Rule Atoms lose or gain electrons until they reach an electron configuration that is the same as a noble gas Process involving the fewer number of electrons will always be the more energetically favorable process 9

10 Chemical Bonds Fundamental concepts in bonding Not all electrons in an atom participate in chemical bonding Certain arrangements of electrons are more stable than others Form if energy of the aggregate is lower than the separate atoms 2015 Pearson Education, Inc. 10

11 Chemical Bonds: Energy Fundamental concepts in bonding Form if energy of the aggregate is lower than the separate atoms Copyright Cengage Learning. All rights reserved 11

12 Ionic Bond Ionic Bond Electron transfer in the ionic bond produces ions Ions are isoelectronic with noble gas Isoelectronic species are ions, or atoms and ions, having the same number and configuration of electrons Na + : 1s 2 2s 2 2p 6 Cl - : 1s 2 2s 2 2p 6 3s 2 3p 6 Ne: 1s 2 2s 2 2p 6 Ar: 1s 2 2s 2 2p 6 3s 2 3p Pearson Education, Inc. 12

13 Ionic Bond Ionic Bond Ionic bonds tend to form between metals and nonmetals Ionic bonds result from the attraction of positive and negative ions Ratio ions combine achieves charge neutrality for the resulting compound 2015 Pearson Education, Inc. 13

14 Ionic Bond: Chemical Formula Guidelines for ionic formulas Ion charges are not shown in the chemical formula Charges are needed to determine the chemical formula Positive ion symbol is always written first Subscripts give the combining ratio for the ions to form a electrically neutral compound 14

15 Which pure substance is held together by ionic bonds? a. LiH b. HCl c. H 2 O d. NH 3 15

16 Which pure substance is held together by ionic bonds? a. LiH b. HCl c. H 2 O d. NH 3 16

17 Which is isoelectronic with Cl? a. Ca 2+ b. Na + c. F d. Se 2 17

18 Which is isoelectronic with Cl? a. Ca 2+ b. Na + c. F d. Se 2 18

19 What measures the strength of attraction of the ions in the solid state? Lattice Energy 19

20 Ionic Bond Energetics Lattice energy Energy needed to transform one mole of an ionic compound to its gaseous ions Q1Q 2 Electrosta tic potential energy d energy of interaction between a pair of ions k = proportionality constant = 8.99 x 10 9 J-m/C 2 Q = ion charge * X C r = shortest distance between the centers of the cations and anions 20

21 Ionic Bond Energetics Lattice energy increases with: decreasing size of ions increasing charge on the ions 2015 Pearson Education, Inc. 21

22 Example Predicting Relative Lattice Energies Arrange these ionic compounds in order of increasing magnitude of lattice energy: CaO, KBr, KCl, SrO. Solution KBr and KCl have lattice energies of smaller magnitude than CaO and SrO because of their lower ionic charges (1+, 1 compared to 2+, 2 ). Order of increasing magnitude of lattice energy: KBr < KCl < SrO < CaO Between KBr and KCl, KBr has a lattice energy of lower magnitude due to the larger ionic radius of the bromide ion relative to the chloride ion. Actual lattice energy values: Between CaO and SrO, the magnitude of the lattice energy of SrO is lower due to the larger ionic radius of the strontium ion relative to the calcium ion.

23 Is there another way to calculate lattice energy? 23

24 Born-Haber Cycle Ionic Bond Energetics: Determine the lattice energies of ionic solids Born-Haber Cycle 2015 Pearson Education, Inc. 24

25 Born-Haber Cycle Ionic Bond Energetics: Determine the lattice energies of ionic solids Cycle that show how ionic compounds are formed from their constituent elements Series of summed reactions Allow lattice enthalpies to be calculated theoretically using empirical data Born-Haber Cycle 2015 Pearson Education, Inc. 25

26 Born-Haber cycle procedure Start with the metal and nonmetal elements: Na(s) and Cl 2 (g) Make gaseous atoms: Na(g) and Cl(g) Make ions: Na + (g) and Cl (g) Combine the ions: NaCl(s) Ionic Bond Energetics: Born-Haber Cycle 2015 Pearson Education, Inc. 26

27 Ionic Bond Energetics: Calculating lattice energy o DH fo [NaCl(s)] = DH fo [Na(g)] + DH fo [Cl(g)] + I 1 (Na) + E(Cl) - DH lattice o DH o rxn = n p DH fo (products) n r DH (reactants) Born-Haber Cycle o DH lattice = -DH fo [NaCl(s)] + DH fo [Na(g)] + DH fo [Cl(g)] + I 1 (Na) + E(Cl) 2015 Pearson Education, Inc. 27

28 Covalent Bond In covalent bonds, atoms share electrons Electrostatic interactions in these bonds: attractions between electrons and nuclei repulsions between electrons repulsions between nuclei For a bond to form, the attractions must be greater than the repulsions 2015 Pearson Education, Inc. 28

29 Covalent Bond Overlap of orbitals containing valence electrons occur with bonding 2015 Pearson Education, Inc. Produces orbital common to both atoms Electrons move throughout this new orbital 29

30 Covalent Bond Distance between bonded atoms is called it s bond length Measured from nucleus to nucleus 2015 Pearson Education, Inc. 30

31 Electrons are shared in a covalent bond Are the electrons shared equally? 31

32 Covalent Bonds: Polarity Electrons not always shared equally Why Not? 2015 Pearson Education, Inc. 32

33 Covalent Bonds: Polarity Electronegativity Measures how strongly atoms attract shared electrons 2015 Pearson Education, Inc. 33

34 Covalent Bonds: Polarity What happens if electrons are unequally shared? Polar covalent bond results Electrons gravitate towards the electronegative atom Result is a partial negative charge Represented by δ Partial charges are less than +1 or -1 Other atom is more positive, or δ+ 34

35 Covalent Bonds: Polarity Greater the difference in electronegativity, the more polar the bond 2015 Pearson Education, Inc. 35

36 CONCEPT CHECK! If lithium and fluorine react, which has more attraction for an electron? Why? In a bond between fluorine and iodine, which has more attraction for an electron? Why? 36

37 Covalent Bonds: Types Electrons shared equally form nonpolar covalent bond Atoms have equal electronegativity Electrons shared unequally form polar covalent bond Atoms have different electronegativity 2015 Pearson Education, Inc. nonpolar covalent bond polar covalent bond 37

38 Is a Compound Ionic or Covalent? Simplest approach: Metal + nonmetal is ionic Nonmetal + nonmetal is covalent Electronegativity difference can be used Exceptions: Higher oxidation number of a metal can give covalent bonding Properties of compounds best way to determine i.e. Lower melting points mean covalent bonding Copyright Cengage Learning. All rights reserved 38

39 CONCEPT CHECK! Which of the following bonds would be the least polar yet still be considered polar covalent? Mg O C O O O Si O N O 39

40 CONCEPT CHECK! Which of the following bonds would be the most polar without being considered ionic? Mg O C O O O Si O N O 40

41 Does having different electronegative atoms in a molecule affect it? Yes How? 41

42 Covalent Bonds: Dipole Moment Bond Polarity Measure of inequality in the sharing of electrons in a bond Copyright Cengage Learning. All rights reserved 42

43 EXERCISE! Arrange the following bonds from most to least polar: a) N F O F C F b) C F N O Si F c) Cl Cl B Cl S Cl 43

44 Covalent Bonds: Dipole Moment Dipole Separation of charge between two covalently bonded atoms Dipole moment,, Measure of the polarity of molecule Polar molecule if centers of positive and negative charge do not overlap 2015 Pearson Education, Inc. Copyright Cengage Learning. All rights reserved 44

45 Covalent Bonds: Strength Bond energy (bond enthalpy) Measures strength of a bond Energy required to break a bond Example: Bond energy for a Cl Cl bond, D(Cl Cl), is measured to be 239 kj/mol. Write reactions for breaking one mole of those bonds: Cl Cl 2 Cl 45

46 Covalent Bonds: Strength Average bond energies Averages over many different compounds Bonds in nature do not have exact bond energy Enthalpies are positive Breaking bond is endothermic Strength of bond single bond < double bond < triple bond 46

47 Covalent Bonds: Bond Energy to Enthalpy of Reaction Estimate DH rxn from energy of bonds broken and bonds formed Breaking bonds is endothermic (positive sign) Forming bonds is exothermic, (negative sign) DH rxn = nd(bonds broken) nd(bonds formed) D = bond energy per mole of bond, always positive n = moles of type of bond 47

48 Example Calculate the enthalpy of the reaction CH 4 (g) + Cl 2 (g) CH 3 Cl(g) + HCl(g) H H H C H H + Cl Cl C + H H Cl Given: One C H bond and one Cl Cl bond are broken One C Cl and one H Cl bond are formed Solve DH = [D(C H) + D(Cl Cl)] [D(C Cl) + D(H Cl)] = [(413 kj) + (242 kj)] [(328 kj) + (431 kj)] = (655 kj) (759 kj) = 104 kj 48 Cl H

49 Example Calculate the enthalpy of the reaction CH 4 (g) + Cl 2 (g) CH 3 Cl(g) + HCl(g) H H H C H H + Cl Cl C + H H Cl Given: One C H bond and one Cl Cl bond are broken One C Cl and one H Cl bond are formed Solve DH = [D(C H) + D(Cl Cl)] [D(C Cl) + D(H Cl)] = [(413 kj) + (242 kj)] [(328 kj) + (431 kj)] = (655 kj) (759 kj) = 104 kj 49 Cl H

50 Example Calculate the enthalpy of the reaction CH 4 (g) + Cl 2 (g) CH 3 Cl(g) + HCl(g) H H H C H H + Cl Cl C + H H Cl Given: One C H bond and one Cl Cl bond are broken One C Cl and one H Cl bond are formed Solve DH = [D(C H) + D(Cl Cl)] [D(C Cl) + D(H Cl)] = [(413 kj) + (242 kj)] [(328 kj) + (431 kj)] = (655 kj) (759 kj) = 104 kj 50 Cl H

51 Covalent Bonds: Bond Enthalpy and Bond Length Average bond length for different bond types As bond number increases, the bond length decreases 51

52 Stop 52

53 Sharing electrons to make covalent bonds can be demonstrated using Lewis structures 53

54 Lewis Structures Lewis structure is a grouping of Lewis symbols that show bonding Lewis Symbol Lewis Structure Chemical Formula Cl 2 54

55 Lewis Structures Bonding pairs (bonding electrons): shared electrons in a Lewis structure; represented by two dots or one line Lone pairs (non bonding electrons): electrons located on only one atom in a Lewis structure 55

56 Lewis Structures Bonding pairs (bonding electrons): shared electrons in a Lewis structure represented by two dots or one line Lone pairs (non bonding electrons): electrons located on only one atom in a Lewis structure 56

57 Lewis Structures Electrons needed to achieve a noble gas configuration determine number of bond Single bonds Atoms share only one pair of electrons Double bonds Two pairs of electrons shared Triple bonds Three pairs of electrons are shared between two atoms 57

58 Procedure for drawing Lewis structures 58

59 Writing Lewis Structures PCl 3 Valence electrons: P: 5 Cl: 7 Keep track of the electrons: 5 + 3(7) = Sum the valence electrons from all atoms Take into account overall charge If an anion, add one electron for each negative charge If a cation, subtract one electron for each positive charge 59

60 Writing Lewis Structures 2. Form a single bond between the atoms Keep track of the electrons: 26 6 = 20 60

61 Writing Lewis Structures 3. Complete octet around all atoms bonded to the central atom Keep track of the electrons: 26 6 = 20; = 2 61

62 Writing Lewis Structures 4. Place any leftover electrons on the central atom Keep track of the electrons: 26 6 = 20; = 2; 2 2 = 0 62

63 Writing Lewis Structures 5. Check that all atoms have 8 electrons Keep track of the electrons: 26 6 = 20; = 2; 2 2 = 0 63

64 Writing Lewis Structures 6. If each atom does not contain 8 electrons, try multiple bonds Valence Electrons :

65 EXAMPLE WRITING LEWIS STRUCTURES FOR COVALENT COMPOUNDS Write the Lewis structure for CO 2. Calculate the total number of electrons for the Lewis structure by summing the valence electrons of each atom in the molecule. Distribute the electrons among the atoms, giving octets (or duets for hydrogen) to as many atoms as possible. Begin with the bonding electrons, and proceed to lone pairs on terminal atoms, and finally to lone pairs on the central atom. SOLUTION Total number of electrons for Lewis structure = = 4 + 2(6) = 16 Work with bonding electrons first. Proceed to lone pairs on terminal atoms next. 65

66 EXAMPLE WRITING LEWIS STRUCTURES FOR COVALENT COMPOUNDS Continued If any atoms lack octets, form double or triple bonds as necessary to give them octets. Move lone pairs from the oxygen atoms to bonding regions to form double bonds. SKILLBUILDER Write the Lewis structure for CO. Answer: 66

67 EXAMPLE WRITING LEWIS STRUCTURES FOR POLYATOMIC IONS Write the Lewis structure for the NH 4+ ion. Calculate the total number of electrons for the Lewis structure by summing the number of valence electrons for each atom and subtracting 1 for the positive charge. Next, place 2 electrons between each pair of atoms. Since the nitrogen atom has an octet and all of the hydrogen atoms have duets, the placement of electrons is complete. Write the entire Lewis structure in brackets indicating the charge of the ion in the upper right corner. SOLUTION 67

68 EXAMPLE WRITING LEWIS STRUCTURES FOR POLYATOMIC IONS Continued SKILLBUILDER Writing Lewis Structures for Polyatomic Ions Write the Lewis structure for the ClO ion. Answer: 68

69 Writing Lewis Structures What happens if more than one possible Lewis structure? O C O O C O All atoms in both molecules contain 8 electrons Which structure is best? 69

70 Writing Lewis Structures O C O O C O Assign formal charge Charge on atom if all electrons in a covalent bond are shared equally Formal charge = valence electrons ½ (bonding electrons) all nonbonding electrons Sum of the formal charges equal the overall charge on the molecule 70

71 Writing Lewis Structures Assign formal charge Formal charge = valence electrons ½ (bonding electrons) all nonbonding electrons Best Lewis structure One in which atoms have formal charges closest to zero Puts a negative formal charge on the most electronegative atom 71

72 Writing Lewis Structures Best Lewis structure? One in which atoms have formal charges closest to zero Puts a negative formal charge on the most electronegative atom 72

73 Example Assigning Formal Charges Assign formal charges to each atom in the resonance forms of the cyanate ion (OCN ). Which resonance form is likely to contribute most to the correct structure of OCN? Solution The sum of all formal charges for each structure is 1, as it should be for a 1 ion. Structures A and B have the least amount of formal charge and are therefore preferable over structure C. Structure A is preferable to B because it has the negative formal charge on the more electronegative atom. We thus expect structure A to make the biggest contribution to the resonance forms of the cyanate ion.

74 RESONANCE STRUCTURES 74

75 Best Lewis Structure Lewis structure for ozone, O 3 Bond length observed in nature does not match Lewis structure Both O to O connections are the same 75

76 Best Lewis Structure How should you depict ozone as a Lewis Structure? Use Resonance Structures 76

77 Resonance structures Multiple structures to describe a molecule Resonance Same arrangement of atoms with electrons in different locations Electrons distributed over area in resonance 77

78 Resonance The organic compound benzene, C 6 H 6, has two resonance structures 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. 78

79 EXAMPLE WRITING RESONANCE STRUCTURES Write the Lewis structure for the NO 3 ion. Include resonance structures. Sum the valence electrons (adding one electron to account for the 1 charge) to determine the total number of electrons in the Lewis structure. SOLUTION Place 2 electrons between each pair of atoms. Distribute the remaining electrons, first to terminal atoms. 79

80 EXAMPLE WRITING RESONANCE STRUCTURES Continued Since there are no electrons remaining to complete the octet of the central atom, form a double bond by moving a lone pair from one of the oxygen atoms into the bonding region with nitrogen. Enclose the structure in brackets and write the charge at the upper right. Notice that you can form the double bond with either of the other two oxygen atoms as well. Since the three Lewis structures are equally correct, write the three structures as resonance structures. 80

81 EXAMPLE WRITING RESONANCE STRUCTURES Continued SKILLBUILDER Writing Resonance Structures Write the Lewis structure for the NO 2 ion. Include resonance structures. Answer: 81

82 OCTET RULE EXCEPTIONS 82

83 Exceptions to the Octet Rule 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 83

84 Odd Number of Electrons Rare and usually quite unstable and reactive there are ions and molecules with an odd number of electrons 84

85 Exceptions to the Octet Rule 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 85

86 Fewer Than Eight Electrons Second period elements before carbon can make stable compounds with 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 Not an accurate picture of the distribution of electrons in BF 3 86

87 Exceptions to the Octet Rule 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 87

88 More Than Eight Electrons Elements in period 3 or higher can use d-orbitals to make more than four bonds Example: PF 5, Phosphorus Pentafluoride 88

89 Example Writing Lewis Structures for Compounds Having Expanded Octets Write the Lewis structure for XeF 2. Solution Begin by writing the skeletal structure. Since fluorine is so electronegative, put the fluorine atoms in terminal positions. Calculate the total number of electrons for the Lewis structure by summing the number of valence electrons for each atom. F Xe F Total number of electrons for Lewis structure = (number of valence e in Xe) + 2(number of valence e in F) = 8 + 2(7) = 22 (4 of 22 electrons are used) Place two bonding electrons between each pair of atoms. Distribute the remaining electrons to give octets to as many atoms as possible, beginning with terminal atoms and finishing with the central atom. Arrange additional electrons (beyond an octet) around the central atom, giving it an expanded octet of up to 12 electrons. (16 of 22 electrons are used) (22 of 22 electrons are used)

90 Chapter 9 Sections 1-3 MOLECULAR STRUCTURE 90

91 Molecular Shapes Lewis Structures show bonding and lone pairs do not denote shape Use Lewis Structures to determine shapes Molecular geometry describes the three dimensional shape of a molecule 91

92 Molecular Shapes Molecular geometry determines Physical properties Polarity Solubility Boiling and melting point Chemical properties Chemical reactivities Square planar molecule cisplatin [Pt(NH 3 ) 2 Cl 2 ]» cis isomer treats certain cancers» trans isomer can not treat cancer 92

93 Predict molecular geometry of simple molecules by using valence shell electron pair repulsion theory (VSEPR Theory) 93

94 Valence-Shell Electron-Pair Repulsion (VSEPR) Model Electrons try get as far apart as possible The balloon analogy demonstrates the maximum distances that minimize repulsions 2015 Pearson Education, Inc. 94

95 Valence-Shell Electron-Pair Repulsion (VSEPR) Model Bonding and nonbonding electron pairs repel each other Assume electron pairs are placed as far as possible 2015 Pearson Education, Inc. 95

96 Valence-Shell Electron-Pair Repulsion (VSEPR) Model Assume electron pairs are placed as far as possible 2015 Pearson Education, Inc. 96

97 Valence-Shell Electron-Pair Repulsion (VSEPR) Model Nonbonding pairs Larger than bonding pairs Repulsions are greater Compress bond angles 97

98 The basic principle of the VSEPR model is that bonding electron pairs in a molecule will be as far apart as possible, but nonbonding pairs can approach more closely. a. True b. False 98

99 The basic principle of the VSEPR model is that bonding electron pairs in a molecule will be as far apart as possible, but nonbonding pairs can approach more closely. a. True b. False 99

100 Valence-Shell Electron-Pair Repulsion (VSEPR) Direction in which electrons point called electron domains Domains may have Two electrons Single covalent bond Nonbonding electron pair Four electrons Double bond Six electrons Triple covalent bond 2015 Pearson Education, Inc. 100

101 Valence-Shell Electron-Pair Repulsion (VSEPR) Count electron domains around central atom 2 electron domains 3 electron domains 4 electron domains 2015 Pearson Education, Inc. 101

102 Valence-Shell Electron-Pair Repulsion (VSEPR) Model Multiple bonds Greater repulsive force than single bonds 102

103 Valence-Shell Electron-Pair Repulsion (VSEPR) Applying VSEPR theory 1. Draw a Lewis structure 2. Determine number of electron domains 3. Predict electron domain arrangement 4. Establish molecular geometry by bonding electrons placement 103

104 Valence-Shell Electron-Pair Repulsion (VSEPR): Molecular Geometries Determined the electron-domain geometry Look at each electron domain to see what molecular geometries are possible Use arrangement of bonded atoms to determine molecular geometry 2015 Pearson Education, Inc. 104

105 What molecular geometries are possible 105

106 VSEPR: Electron Group Possible electron group geometries 106

107 VSEPR: Molecular Geometry 2 electron domains 1 molecular geometry 3 electron domains 2 molecular geometries 4 electron domains 3 molecular geometries 107

108 VSEPR: Molecular Geometry 5 electron domains 4 molecular geometry 2015 Pearson Education, Inc. 108

109 VSEPR: Molecular Geometry 6 electron domains 3 molecular geometry 109

110 VSEPR: Molecular Geometry Molecules with multiple central atoms Consider each atom separately, then combine results 2015 Pearson Education, Inc. 110

111 EXAMPLE PREDICTING GEOMETRY USING VSEPR THEORY Predict the electron and molecular geometry of PCl Draw a Lewis structure for the SOLUTION molecule. PCl 3 has 26 electrons. 2. Determine the total number of electron groups around the central atom. Lone pairs, single bonds, double bonds, and triple bonds each count as one group. The central atom (P) has four electron groups. 3. Determine the number of bonding groups and the number of lone pairs around the central atom. These should sum to the result from Step 2. Bonding groups include single bonds, double bonds, and triple bonds. Three of the four electron groups around P are bonding groups, and one is a lone pair. 111

112 EXAMPLE PREDICTING GEOMETRY USING VSEPR THEORY Continued 4. Determine the electron geometry and molecular geometry. The electron geometry is tetrahedral (four electron groups), and the molecular geometry the shape of the molecule is trigonal pyramidal (four electron groups, three bonding groups, and one lone pair). 112

113 EXAMPLE PREDICTING GEOMETRY USING VSEPR THEORY Continued SKILLBUILDER Predict the molecular geometry of ClNO (N is the central atom). Answer: 113

114 Does molecular shape affect the polarity of a molecule? Yes How? 114

115 Molecular Polarity Bond Polarity Measure of inequality in the sharing of electrons in a bond Copyright Cengage Learning. All rights reserved 115

116 Molecular Polarity: Nonpolar Nonpolar molecules have symmetrical distribution of electronic charge 2015 Pearson Education, Inc. 116

117 Molecular Polarity: Polar Polar molecules have unsymmetrical electronic charge 2015 Pearson Education, Inc. 117

118 Molecular Polarity: Polarity To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE 118

119 Molecular Polarity: Comparison A NONPOLAR molecule A POLAR molecule 2015 Pearson Education, Inc. 119

120 120