CHEMISTRY. Chapter 9 The Basics of Chemical Bonding. The Molecular Nature of Matter. Jespersen Brady Hyslop SIXTH EDITION

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2 Chemical Bonds Attractive forces that hold atoms together in a molecule or ionic compounds Changes in these bonding forces are the underlying basis of chemical reactivity During reaction: Breaking bonds Forming bonds Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 2

3 Two Classes of Bonds Ionic Bonding ccurs in ionic solid Electrons transferred from one atom to another Bond generated by attraction of 2 opposite charges. Covalent bonding ccurs in molecules Sharing of electrons Polar and nonpolar bonds Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 3

4 Ionic Bonds Result from attractive forces between opposite charges Na + Cl Metal - nonmetal ionic bonds: Metals have Low ionization energies Easily lose electrons to be stable Non-metals have Very exothermic electron affinities Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 4

5 Ionic Compounds Formed from metal and nonmetal Na + Cl Na + + Cl NaCl(s) e Ionic Bond Attraction between + and ions in ionic compound. Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 5

6 Ionic Compounds Ionic crystals: Exist in 3-dimensional array of cations and anions called a lattice structure Ionic chemical formulas: Always written as empirical formula Smallest ratio of cation to anion Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 6

7 Lattice Energy Potential Energy of system decreases when one mole of solid salt is formed from its gas phase ions. Energy released when ionic lattice forms. Always Lattice = exothermic Lattice q q d Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 7

8 Evaluate Energy of Formation 1. Single step Na(s) + ½Cl 2 (g) NaCl(s) f = kj/mol 2. Stepwise path Na(s) Na(g) (Na, g) = kj/mol ½Cl 2 (g) Cl(g) (Cl, g) = kj/mol Na(g) Na + (g) + e IE(Na) = kj/mol Cl(g) + e Cl (g) EA(Cl) = kj/mol Na + (g) + Cl (g) NaCl(s) lattice = 787 kj/mol Na(s) + ½Cl 2 (g) NaCl(s) f = 411 kj/mol Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 8

Covalent Bond Attraction of valence electrons of one atom by nucleus of other atom Shifting of electron density As distance between nuclei decreases,

9 Covalent Bond Attraction of valence electrons of one atom by nucleus of other atom Shifting of electron density As distance between nuclei decreases, probability of finding either electron near either nucleus increases Pulls nuclei closer together Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 9

As nuclei get close Begin to repel each other Both have high positive charge Covalent Bond Final internuclear distance between two atoms in bond

10 As nuclei get close Begin to repel each other Both have high positive charge Covalent Bond Final internuclear distance between two atoms in bond Balance of attractive and repulsive forces Bond forms since there is a net attraction Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 10

11 Covalent Bond Two quantities characterize this bond Bond Length (bond distance) Distance between 2 nuclei = r A + r B Bond Energy Also bond strength Amount of energy released when bond formed (decreasing PE) or Amount of energy must put in to break bond Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 11

12 Bond Polarity Two atoms of the same element form a bond Equal sharing of electrons NNPLAR BND Two atoms of different elements form a bond Unequal sharing of electrons PLAR BND Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 12

13 Polar Covalent Bond Also known as a polar bond Bond that carries partial + and charges at opposite ends Bond is a dipole Two poles or two charges involved Charge on either end distance between them μ = q r Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 13

14 Table 9.3 Dipole Moments and Bond Lengths for Some Diatomic Molecules Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 14

Electronegativity and Bond Polarity Leads to concept of partial charges + Cl + on = +0.

15 Electronegativity and Bond Polarity Leads to concept of partial charges + Cl + on = on Cl = 0.17 Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 15

16 Electronegativity (EN) Relative attraction of atom for electrons in a bond First proposed by Linus Pauling in Electronegativity cannot be directly measured but can be considered as equivalent to electron affinity Pauling scale, on a relative scale running from around 0.7 to 4.0 Difference in electronegativity = estimate of bond polarity EN = EN 1 EN 2 e.g. N Si F + + Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 16

17 Electronegativity Table Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 17

18 Using Electronegativities DEN= EN A EN B Difference in electronegativity Measure of ionic character of bond Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 18

19 Using Electronegativities Nonpolar Covalent Bond No difference in electronegativity Ionic Character of bond Degree to which bond is polar EN > 1.7 means mostly ionic > 50% ionic More electronegative element almost completely controls electron EN < 0.5 Means almost purely covalent Nonpolar: < 5% ionic 0.5 < EN < 1.7 polar covalent Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 19

20 Your Turn! Which of the following species has the least polar bond? A. Cl B. F C. I D. Br Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 20

21 Your Turn! Which species has mostly covalent bonds? A. CsCl B. NaF C. CaF 2 D. C E. MgBr 2 Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 21

22 Lewis Symbols Electron bookkeeping method Way to keep track of e s Write chemical symbol surrounded by dots for each e Group # 1A 2A 3A 4A Valence e 's e configuration ns 1 ns 2 ns 2 np 1 ns 2 np 2 e Li Be B C Na Mg Al Si Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 22

23 Lewis Symbols Group # 5A 6A 7A 8A Valence e - 's e - configuration ns 2 np 3 ns 2 np 4 ns 2 np 5 ns 2 np 6 N F Ne P S Cl Ar For the representative elements Group number = number of valence e s e Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 23

24 Lewis Symbols Can use to diagram electron transfer in ionic bonding Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 24

25 ctet Rule Works well with Group 1A and 2A metals Al Non-metals and e can't obey Limited to 2 electrons in the n = 1 shell Doesn't work with Transition metals Inner transition metals Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 25

26 Lewis Structures Diatomic Gases: and alogens 2 + : or Each has two electrons through sharing Shared pair of electrons can be written as a line ( ) : or means a covalent bond Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 26

27 Diatomic Gases: F 2 Lewis Structures F + F F F F F Each F has complete octet nly need to form one bond to complete octet The pairs of electrons not included in covalent bond are called lone pairs Same for rest of halogens: Cl 2, Br 2, I 2 Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 27

28 Diatomic Gases: F Lewis Structures + F F F Same for Cl, Br, I Molecules are diatomics of atoms that need only one electron to complete octet Separate molecules Gas in most cases because very weak intermolecular forces Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 28

29 Your Turn! ow many electrons are required to complete the octet around nitrogen, when it forms N 2? A. 2 B. 3 C. 1 D. 4 E. 6 Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 29

30 Method for Drawing Lewis Structure 1. Decide how atoms are bonded Skeletal structure = arrangement of atoms. Central atom Usually given first Usually least electronegative 2. Count all valence electrons (all atoms) 3. Place two electrons between each pair of atoms Draw in single bonds Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 30

31 Method for Drawing Lewis Structure 4. Complete octets of terminal atoms (atoms attached to central atom) by adding electrons in pairs 5. Place any remaining electrons on central atom in pairs 6. If central atom does not have octet Form double bonds If necessary, form triple bonds Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 31

32 Lewis Structures Many nonmetals form more than one covalent bond Needs 4 electrons Forms 4 bonds C C C Needs 3 electrons Forms 3 bonds N N Needs 2 electrons Forms 2 bonds methane ammonia water N Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 32

33 Multiple Bonds Single Bond Bond produced by sharing one pair of electrons between two atoms Many molecules share more than one pair of electrons between two atoms Multiple bonds Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 33

34 Double Bonds Two atoms share two pairs of electrons e.g. C 2 C C C Triple bond Three pairs of electrons shared between two atoms e.g. N 2 N N N N N N Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 34

35 Your Turn! Which species is most likely to have multiple bonds? A. C B. 2 C. P 3 D. BF 3 E. C 4 Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 35

36 Important Compounds of Carbon Alkanes ydrocarbons nly single bonds Isomers Same molecular formula Different physical properties Different connectivity (structure) C C C C butane e.g. C 4 C 3 C 3 C 3 C 2 C 3 Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 36 C methane ethane propane C C C iso-butane

37 ydrocarbons Alkenes Contain at least one double bond C C C C C C ethylene (ethene) butene Alkynes Contain at least one triple bond C C C C acetylene (ethyne) butyne Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 37 C C

38 1 Si = 1 4e = 4 e 4 F = 4 7e = 28 e Total = 32 e single bonds 8 e 24 e F lone pairs 24 e 0 e Ex. SiF 4 Skeletal Structure F F Si F F Complete terminal atom octets F F Si F F Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 38

39 Ex. 2 C 3 C 2 3 oxoanion, so C central, and s around, + attached to two s 1 C = 1 4e = 4 e 3 = 3 6e = 18 e 2 = 2 1e = 2 e Total = 24 e single bonds 10 e 14 e lone pairs 14 e 0 e C C But C only has 6 e Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 39

40 Ex. 2 C 3 (cont.) Too few electrons Must convert one of lone pairs on to second bond to C C Form double bond between C and C Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 40

41 Ex. 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 F N N F Complete terminal atom octets F N N F Remaining electrons on central atom F N N F Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 41

42 Ex. N 2 F 2 (cont.) Not enough electrons to complete octets on nitrogen Must form double bond between nitrogen atoms to satisfy both octets. F N N F F N N F Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 42

43 Expanded ctets Elements after Period 2 in the Periodic Table: Are larger atoms ave d orbitals available - space to expand octets Can accept 18 electrons For Lewis structures: Follow same process as before but add extra electrons to the central atom Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 43

44 Ex. 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 use d orbitals Cl Cl P Cl Cl Cl Cl Cl Cl P Cl Cl Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 44

45 Electron Deficient Structures Boron often has six electrons around it Three pairs Beryllium often has four electrons around it Two pairs Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 45

46 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 Ex. BBr 3 Br Br B Br Br B has only six electrons Does not form double bond as incomplete octet B Br Br Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 46

47 If more than one Lewis structure can be drawn, which is correct? Experiment always decides Concepts such as formal charge and resonance help to make predictions Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 47

48 Experimental evidence The Case of 2 S 4 Bond lengths from X-ray data S bonds (no attached) shorter pm The other two S, longer pm Lewis Structure Lewis structure with two S= is consistent with experimental data - Preferred structure Sulfur has empty 3d orbitals; it could expand its octet Can contain double bonds. Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 48

49 Ex. 2 S 4 1 S = 1 6e = 6 e S 4 = 4 6e = 24 e 2 = 2 1e = 2 e Total = 32 e single bonds 12 e S 20 e lone pairs 20 e 0 e S Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 49

50 Formal Charge (FC) Apparent charge on atom Does not represent real charges FC = # valence e [# bonds around atom + # unshared e ] Indicate formal charges by placing them in circles around atoms Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 50

51 FC = #valence e [#bonds around atom + # unshared e ] S Structure 1 FC S = 6 (4 + 0) = 2 FC = 1 (1 + 0) = 0 FC (s) = 6 (1 + 6) = 1 FC (s,s) = 6 (2 + 4) = 0 S Structure 2 FC S = 6 (6 + 0) = 0 FC = 1 (1 + 0) = 0 FC (s) = 6 (2 + 4) = 0 FC (d) = 6 (2 + 4) = 0 Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 51

52 Best Lewis Structures No formal charges on any atom in structure If charges are unavoidable Those with smallest formal charges are preferred Negative FC on electronegative elements Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 52

1 C = 1 4e = 4 e 2 = 2 6e = 12 e Total = 16 e

53 1 C = 1 4e = 4 e 2 = 2 6e = 12 e Total = 16 e single bonds 4 e 12 e lone pairs 12 e 0 e C only has four electrons Need two extra bonds to to complete octet 3 ways you can do this Ex. C 2 Which of these is correct? Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 53

54 Best Structure for C 2 Use formal charge to determine which structure is best FC C = 4 (4 + 0) = 0 FC (s) = 6 (1 + 6) = 1 FC (d) = 6 (2 + 4) = 0 FC (t) = 6 (3 + 2) = +1 Central structure best All FC s = Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 54

55 Use of Formal Charges to Explain Boron Chemistry BBr 3 Why doesn t a double bond form here? FC B = 3 (3 + 0) = 0 FC Br = 7 (1 + 6) = 0 All FC's = 0 so the molecule has the best possible structure It doesn't need to form double bond Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 55

56 Resonance: Explaining Multiple Equivalent Lewis Structures Delocalization of electrons - a way to describe a molecule with 2 or more equivalent Lewis structures Ex: N 3-1 N = 1 5e = 5 e 3 = 3 6e = 18 e 1 charge = 1 e Total = 24 e single bonds 6 e 18 e lone pairs 18 e 0 e Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 56

57 Ex. N 3 Lewis structure with one bond shorter than other two Experimental observation: All three N bond lengths are same All shorter than N single bonds ave to modify Lewis Structure Electrons cannot distinguish atoms Can write two or more possible structures simply by moving electrons around Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 57

58 What are Resonance Structures? Multiple Lewis structures for single molecule No single Lewis structure is correct Structure not accurately represented by any one Lewis structure Actual structure = "average" of all possible structures Double headed arrow between resonance structures used to denote resonance Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 58

Resonance Structures Lewis structures assume electrons are localized between 2 atoms In resonance structures, electrons are delocalized Move around over all atoms Extended

59 Resonance Structures Lewis structures assume electrons are localized between 2 atoms In resonance structures, electrons are delocalized Move around over all atoms Extended over entire molecule to give equivalent bond distances Resonance ybrid Way to depict resonance delocalization Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 59

60 1 C = 1 4e = 4 e 3 = 3 6e = 18 e 2 charge = 2 e Total = 24 e single bonds 6 e Ex. C e lone pairs 18 e 0 e C only has 6 electrons so a double bond is needed Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 60

Three Equivalent Resonance Structures All have same net formal charges on Carbon and xygens FC = 1 on xygens forming single bonds FC = 0 on xygens forming double bonds with

61 Three Equivalent Resonance Structures All have same net formal charges on Carbon and xygens FC = 1 on xygens forming single bonds FC = 0 on xygens forming double bonds with Carbon C 2- ne Resonance ybrid ne single carbonate ion 2 electrons delocalized All C- bonds, the same length Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 61

62 Resonance Structures Not Always Equivalent Cases in which two or more Lewis Structures for same compound may or may not represent electron distributions of equal energy ow Do We Determine Which are Good Contributors? 1. All octets are satisfied as much as possible 2 FC =0 or minimum value 3 Any negative charges are on the more electronegative atoms. Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 62

63 Ex. NC 1 C = 1 4e = 4 e 1 N = 1 5e = 5 e 1 = 1 6e = 6 e 1 charge = 1 e Total = 16 e single bonds 4 e 12 e /N lone pairs 12 e 0 e Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 63

Ex. NC FC N = 5 3 2 = 0 FC C = 4 4 0 = 0 FC = 6 1 6 = 1 1 Best FC N = 5 2 4 = 1 FC C = 4 4 0 = 0 FC = 6 2 4 = 0 FC N = 5 1 6 =

64 Ex. NC FC N = = 0 FC C = = 0 FC = = 1 1 Best FC N = = 1 FC C = = 0 FC = = 0 FC N = = 2 FC C = = 0 FC = = K Not Acceptable Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 64

65 Resonance Stabilization Actual structure is more stable than either single resonance structure For benzene The extra stability is ~146 kj/mol Resonance energy Extra stabilization energy from resonance Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 65

66 Your Turn! Which of the structures below exhibit resonance? A. N 2 B. 2 C. N 3 D. N 2 (nitrogen is central atom) E. C 3 C 2 C 3 Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 66

67 Coordinate Covalent Bonds Ammonia Normal covalent bonds ne electron from each atom shared between the two 3 + N N Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 67

68 Coordinate Covalent Bond Ammonium Ion + has no electrons N has lone pair Can still get 2 electrons shared between them + N + + N Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 68

69 Coordinate Covalent Bond Both electrons of shared pair come from just one of two atoms nce bond formed, it acts like any other covalent bond Not possible to tell where electrons came from after bond is formed Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 69

70 Coordinate Covalent Bond Especially boron (electron deficient molecule) reacts with nitrogen compounds that contain lone pair of electrons Cl Cl N + B Cl Cl N B Cl Cl Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 70

71 Learning Check Ex. Draw the best Lewis Structure for the following Cl 4 XeF 4 I 3 BrF 5 Jespersen/Brady/yslop Chemistry: The Molecular Nature of Matter, 6E 71

Chapter 8: Concepts of Chemical Bonding

Chapter 8: Concepts of Chemical Bonding Learning Outcomes: Write Lewis symbols for atoms and ions. Define lattice energy and be able to arrange compounds in order of increasing lattice energy based on