Chapter 9. Chemical Bonding I: The Lewis Model. Modified by Dr. Cheng-Yu Lai

Transcription

1 Chapter 9. Chemical Bonding I: The Lewis Model Modified by Dr. Cheng-Yu Lai

2 Chemicals Bonds Forces that hold groups of atoms together and make them function as a unit. Ionic bonds transfer of electrons Covalent bonds sharing of electrons Explain how and why atoms attach together to form molecules Explain why some combinations of atoms are stable and others are not Why is water H 2 O, not HO or H 3 O? Can be used to predict the shapes of molecules Can be used to predict the chemical and physical properties of compounds 2014 Pearson Education, Inc.

3 Electronegativity Difference and Bond Type Polar-Covalent bonds Electrons are unequally shared Electronegativity difference between.5 and 1.9 Nonpolar-Covalent bonds Electrons are equally shared Electronegativity difference of 0 to 0.4 Inoic bonds Electrons are unequally shared Electronegativity difference above 2.0

4 Electronegativity The ability of an atom in a molecule to attract shared electrons to itself. Linus Pauling

5 Table of Electronegativities

6 Polar Covalent Bonds A polar covalent bond occurs between nonmetal atoms. is an unequal sharing of electrons. has a moderate electronegativity difference (0.5 to 1.7). Examples: Electronegativity Atoms Difference Type of Bond O-Cl = 0.5 Polar covalent Cl-C = 0.5 Polar covalent O-S = 1.0 Polar covalent 6

7 Polar Covalent Bonds: Electronegativity 2012 Pearson Education, Inc. Chapter 7/7

8 Polar Covalent Bonds and Dipole Moment: The dipole moment is a measure of polarity in a chemical bond or molecule. It is a drawing made on a Lewis Structure composed of an arrow pointing from a "slightly-positive" molecule to a "slightlynegative" molecule. H --- Cl > Partial positive is represented by this symbol Partial negative is represented By this symbol 2012 Pearson Education, Inc.

9 Ionic Bonds An ionic bond occurs between metal and nonmetal ions. is a result of electron transfer. has a large electronegativity difference (1.8 or more). Examples: Electronegativity Atoms Difference Type of Bond Cl-K = 2.2 Ionic N-Na = 2.1 Ionic S-Cs = 1.8 Ionic 9

10 2014 Pearson Education, Inc.

11 Example 9.3 Classifying Bonds as Pure Covalent, Polar Covalent, or Ionic Determine whether the bond formed between each pair of atoms is covalent, polar covalent, or ionic. a. Sr and F b. N and Cl c. N and O Solution a. In Figure 9.8, find the electronegativity of Sr (1.0) and of F (4.0). The electronegativity difference (ΔEN) is ΔEN = = 3.0. Using Table 9.1, classify this bond as ionic. b. In Figure 9.8, find the electronegativity of N (3.0) and of Cl (3.0). The electronegativity difference (ΔEN) is ΔEN = = 0. Using Table 9.1, classify this bond as covalent. c. In Figure 9.8, find the electronegativity of N (3.0) and of O (3.5). The electronegativity difference (ΔEN) is ΔEN = = 0.5. Using Table 9.1, classify this bond as polar covalent. FIGURE 9.8 Electronegativities of the Elements Electronegativity generally increases as we move across a row in the periodic table and decreases as we move down a column.

12 Predicting Bond Types 12

13 Bond Length and Energy It is the energy required to break a bond. It gives us information about the strength of a bonding interaction. Bond Bond type Bond length (pm) Bond Energy (kj/mol) C - C Single C = C Double C C Triple C - O Single C = O Double C - N Single C = N Double C N Triple Bonds between elements become shorter and stronger as multiplicity increases.

14 Bond Energy and Enthalpy bondsbroken H D D bonds formed Energy required Energy released D = Bond energy per mole of bonds Breaking bonds always requires energy Breaking = endothermic Forming bonds always releases energy Forming = exothermic

15 Example 9.11 Calculating ΔH rxn from Bond Energies Hydrogen gas, a potential fuel, can be made by the reaction of methane gas and steam. Use bond energies to calculate ΔH rxn for this reaction. Solution Begin by rewriting the reaction using the Lewis structures of the molecules involved. Determine which bonds are broken in the reaction and sum the bond energies of these. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro 2014 Pearson Education, Inc.

16 Example 9.11 Calculating ΔH rxn from Bond Energies Continued Determine which bonds are formed in the reaction and sum the negatives of their bond energies. Find ΔH rxn by summing the results of the previous two steps. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro 2014 Pearson Education, Inc.

17 Ionic Bonding Electrons are transferred Electronegativity differences are generally greater than 1.7 The formation of ionic bonds is always exothermic!

18 Determination of Ionic Character Electronegativity difference is not the final determination of ionic character Compounds are ionic if they conduct electricity in their molten state

19 Coulomb s Law The energy of interaction between a pair of ions is proportional to the product of their charges, divided by the distance between their centers QQ E (2.31x10 19 J nm) 1 2 r E QQ 1 2 r

20 Example 9.2 Predicting Relative Lattice Energies Arrange these ionic compounds in order of increasing magnitude of lattice energy: CaO, KBr, KCl, SrO. Solution KCl should have lattice energies of smaller magnitude than CaO because of their lower ionic charges (1+, 1 compared to 2+, 2.) Order of increasing magnitude of lattice energy: Actual lattice energy values: KCl < CaO For More Practice 9.2 Which compound has a lattice energy of higher magnitude, NaCl or MgCl 2? Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro 2014 Pearson Education, Inc.

21 Electron-Dot Structures Electron-Dot Structure (Lewis Structure): Represents an atom s valence electrons by dots and indicates by the placement of the dots the way the valence electrons are distributed in a molecule Pearson Education, Inc. Chapter 7/21

22 From Atomic Valance Shell Electrons

23 To Electron Dot Structures H Symbols of atoms with dots to represent the valence-shell electrons only He: Li Be B C N O : F :Ne : Na Mg Al Si P S :Cl :Ar :

24 Octet Rule = atoms tend to gain, lose or share electrons so as to have 8 electrons 2014 Pearson Education, Inc. C would like to N would like to O would like to Gain 4 electrons Gain 3 electrons Gain 2 electrons

25 1. Return your 4 th Exam 2. Course Project please sign your name and number, due on 04/30/14 3. Mastering Chemistry re-opened 2014 Pearson Education, Inc.

26 Valence electrons for Elements Recall that the valence electrons for the elements can be determined based on the elements position on the periodic table. Lewis Dot Symbol 26 Lewis Structure Tutorial :16 PM

27 Lewis Structures Lewis structures show how valence electrons are arranged among atoms in a molecule. Lewis structures Reflect the central idea that stability of a compound relates to noble gas electron configuration. Shared electrons pairs are covalent bonds and can be represented by two dots (:) or by a single line ( - )

28 The Octet Rule Combinations of elements tend to form so that each atom, by gaining, losing, or sharing electrons, has 8 electrons in its outermost occupied energy level. Monatomic chlorine Diatomic chlorine

29 The Octet Rule: The Diatomic Fluorine Molecule F F 1s 2s 2p Each has seven valence electrons 1s 2s 2p F F A Single Bond is when 2 electrons are shared they are represented by a single line in bond diagrams

30 The HONC Rule for Octet Rule Hydrogen (and Halogens) form one covalent bond Oxygen (and sulfur) form two covalent bonds One double bond, or two single bonds Nitrogen (and phosphorus) form three covalent bonds One triple bond, or three single bonds, or one double bond and a single bond Carbon (and silicon) form four covalent bonds. Two double bonds, or four single bonds, or a triple and a single, or a double and two singles

31 The Octet Rule: The Diatomic Oxygen Molecule O O 1s 2s 2p Each has six valence electrons 1s 2s 2p O O A Double bond is when 4 electrons are shared they are represented by two lines in bond diagrams

32 The Octet Rule: The Diatomic Nitrogen Molecule N N 1s 2s 2p Each has five valence electrons 1s 2s 2p N N A Triple bond is when 6 electrons are shared they are represented by three lines in bond diagrams

33 VSEPR Model The Valence Shell Electron Pair Repulsion model predicts the shapes of molecules and ions by assuming that the valence shell electron pairs are arranged as far from one another as possible to minimize the repulsion between them. 33

34 Predict Molecular Shape-VSEPR AX n E m AXE shorthand notation: A - central atom X - terminal atoms n bonding pairs E - lone pair m lone pair electrons Molecular geometry is determined by the arrangement of atoms (or bonding electron pairs only) around the central atom. Electron Pair Geometry is determined by the number and arrangement of all electron pairs (bonding and lone) around the central atom. AX 3 E 0

35 AX n E m Type 1. Central atom S=A 2. Terminal atomes O=X, n=3 3. Calculate Total valence electrons 6+3*6+2 electrons = 26 e SO bonding pairs n = (total valence electrons)/8 = 26 /8= 3 bonding pairs + 2 remainder electrons 5. Lone pair m = (remainder electrons)/2 =2/2=1 lone pairs 6. SO3 2- = AX3E1

36 AX n E m Type NH Central atom N=A 2. Terminal atomes H=X, n=4 3. Calculate Total valence electrons In VSEPR calculation, for H fullfills octet rule, we only need to add one more electron into 1S orbital. But for our calculation here, we make H=7 pseudo valance electrons 3+4*7-1 electrons = 32 e 4. bonding pairs n = (total valence electrons)/8 = 32/8= 4 bonding pairs + 2 remainder electrons 5. Lone pair m = (remainder electrons)/2 =0/2=0 lone pairs 6. NH 4 + = AX4E0

37 More AX n E m Types Example 1: BeCl 2 AX 2 E 0 Example 2: BF 3 AX 3 E 0 Example 3: SO 2 AX 3 E 0 AX 2 E 1 Example 4: CH 4 AX 4 E 0 Example 5: NH 3 AX 4 E 0 AX 3 E 1 Example 6: H 2 O AX 4 E 0 AX 2 E 2

38 More AX n E m Types Example 7: PF 5 AX 5 E 0 Example 8: SF 4 AX 5 E 0 AX 4 E 1 Example 9: BrF 3 AX 5 E 0 AX 3 E 2 Example 10: XeF 2 AX 5 E 0 AX 2 E 3 Example 11: SF 6 AX 6 E 0 Example 12: IF 5 AX 6 E 0 AX 5 E 1 Example 13: XeF 4 AX 6 E 0 AX 5 E 1

39 VSEPR Valence Shell Electron Pair Repulsion Five basic shapes. X + E Overall Structure Forms 2 Linear AX 2 3 Trigonal Planar AX 3, AX 2 E 4 Tetrahedral AX 4, AX 3 E, AX 2 E 2 5 Trigonal bipyramidal AX 5, AX 4 E, AX 3 E 2, AX 2 E 3 6 Octahedral AX 6, AX 5 E, AX 4 E 2 A = central atom X = atoms bonded to A E = nonbonding electron pairs on A

40 AX n E m Geometry in Space A X X AX 2 E 0 A X X AX 3 E 0 X A X X AX 4 E 0 X X A X X AX 5 E 0 X X X A X X AX 6 E 0 X X X X

41 AX2E0

42 AX 3 E 0 AX 2 E 1 Chapter 7/ Pearson Education, Inc.

43 AX4E0 AX3E1 AX2E2

44 AX 5 E 0 AX 5 E 0 AX 3 E 2 AX 2 E 3 AX 4 E 1

45 AX 6 E 0 AX 5 E 1 AX 4 E 2 AX 6 E 0

46 Geometry and Approximate Bond Angle AX 2 E 0 AX 3 E 0 AX 4 E 0 AX 5 E 0 AX 6 E 0

47 AX 2 E 0 AX 3 E 0 AX 4 E 0 AX 5 E 0 AX 6 E 0

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49 Learning Check Example 1: BeCl 2 Example 2: BF 3 Example 3: SO 2 Example 4: CH 4 Example 5: NH 3 Example 6: H 2 O AX 2 E 0 linear AX 3 E 0 triangular planar AX 2 E 1 angular (bent) AX 4 E 0 tetrahedral AX 3 E 1 triangular pyramidal AX 2 E 2 angular (bent)

50 04/28/14 5 th Exam Ch9 and Ch10 CHEM 101 Course Project 04/30/14 Mastering Chemistry 2014-Chem101- Exp11th Heat effect and calorimetry Lab Reports (20 points/per lab) Lab Performance + Attendance (15 points/per lab) Questions (15 points) is check out and make-up lab. Please inform your lab instructor in advance to prepare your dry -lab make up material. A one time make-up session lab will be held on the last week of classes, during the normal lab session. The make-up sessions will comprise a dry-lab performance associated with the missed lab and will include problem solving and theoretical questions.( Please look at the syllabus) Other questions, please discuss with your lab instructor.

51 Geometry and Polarity of Molecules For a molecule to be polar it must 1) have polar bonds, symmetrical shape, and different terminal atoms 2) have polar bonds electronegativity difference - theory bond dipole moments measured 3) have an unsymmetrical shape using vector addition polarity effects the intermolecular forces of attraction

52 Dipole moment is the measured polarity of a polar covalent bond. It is defined as the magnitude of charge (electrons) on the atoms and the distance between the two bonded atoms. H O H : O C O : polar bonds, and unsymmetrical shape causes molecule to be polar polar bonds, but nonpolar molecule because pulls cancel

53 Adding Dipole Moments 53

54 Cl Cl Cl Cl C Cl H H C Cl CCl 4 m = 0.0 D CH 2 Cl 2 m = 2.0 D

55 Geometry and Approximate Bond Angle AX 2 E 0 AX 3 E 0 AX 4 E 0 AX 5 E 0 AX 6 E 0

56 Molecular Geometry Dipole Moment and Polarity CO, PCl 3, BCl 3, GeH 4, CF 4 Which compound is the most polar? Which compounds on the list are non-polar? 56

57 Orbitals Consistent with Molecular Shape Lewis dot + VSEPR gives the correct shape for a molecule. BUT How do atomic orbitals (s, p, d ) produce these shapes? Valence bond theory describes a bond as an overlap of atomic (hybrid) orbitals. 57

58 Hybridization and Molecular Geometry Forms Overall Structure Hybridization of A AX 2 Linear sp AX 3, AX 2 E Trigonal Planar sp 2 AX 4, AX 3 E, AX 2 E 2 Tetrahedral sp 3 AX 5, AX 4 E, AX 3 E 2, AX 2 E 3 Trigonal bipyramidal sp 3 d AX 6, AX 5 E, AX 4 E 2 Octahedral sp 3 d 2 A = central atom X = atoms bonded to A E = nonbonding electron pairs on A

59 Summary - Hybrid Orbitals Hybrid Orbital Geometric Arrangements Number of Orbitals Example sp Linear 2 Be in BeF 2 sp 2 Trigonal planar 3 B in BF 3 sp 3 Tetrahedral 4 C in CH 4 sp 3 d Trigonal bipyramidal 5 P in PCl 5 sp 3 d 2 Octahedral 6 S in SF 6 59

60 Orbitals Consistent with Molecular Shapes Atomic orbitals (AOs) can be hybridized (mixed). Sets of identical hybrid orbitals form identical bonds. # AOs that hybridize = # hybrids orbitals. s + p sp + sp s + p + p sp 2 + sp 2 + sp 2 etc. 60

61 Hybridization and sp 3 Hybrid Orbitals How can the bonding in CH 4 be explained? 4 valence electrons 2 unpaired electrons 2012 Pearson Education, Inc. Chapter 7/61

62 Hybridization and sp 3 Hybrid Orbitals How can the bonding in CH 4 be explained? 4 valence electrons 4 unpaired electrons 2012 Pearson Education, Inc. Chapter 7/62

63 Hybridization and sp 3 Hybrid Orbitals How can the bonding in CH 4 be explained? 4 nonequivalent orbitals 2012 Pearson Education, Inc. Chapter 7/63

64 Hybridization and sp 3 Hybrid Orbitals How can the bonding in CH 4 be explained? 4 equivalent orbitals 2012 Pearson Education, Inc. Chapter 7/64

65 Hybridization and sp 3 Hybrid Orbitals 2012 Pearson Education, Inc. Chapter 7/65

66 Hybridization and sp 3 Hybrid Orbitals 2012 Pearson Education, Inc. Chapter 7/66

67 Energy, E AX 2 E 0, Ex: BeCl 2, sp Hybrid Orbitals 2p 2p 2p Promotion 2s Isolated Be atom 2p 2p 2p 2s Orbital hybridization Two unhybridized p orbitals Two sp hybrid orbitals on Be in BeF 2 sp hybridization occurs around the central atom whenever there are two regions of high e- density. Two equivalent covalent bonds form (180 apart) LINEAR. 67

68 sp Hybrid Orbitals

69 sp 2 Hybrid Orbitals AX 3 E 0, Ex: BF 3 The result is THREE equivalent hybrid orbitals, in a VSEPR basic shape of trigonal planar. 69 p. 396

70 sp 2 Hybrid Orbitals 2012 Pearson Education, Inc. Chapter 7/70

71

72 Orbitals Consistent with Molecular Shapes Describe bonding in PCl 5 using hybrid orbitals. : : : Cl : : : Cl : P : : Cl : : Cl : : Cl : : AX 5 E 0 trigonal bipyramidal We need 5 orbitals. 72

73 sp 3 d Hybrid Orbitals 3d valence shell 3p 3s X hybridization five equal sp 3 d hybrid orbitals P atom (ground state) 73

74 sp 3 d Hybrid Orbitals 3d sp 3 d P atom (hybridized state) 74

75 Orbitals Consistent with Molecular Shapes Describe the bonding in SF 6 using hybrid orbitals. : : : F : : : : : : : S : : F F F : : : F : : : F AX 6 E 0 Octahedral We need 6 orbitals. 75

76 sp 3 d 2 Hybrid Orbitals 3d 3p X hybridization six equal sp 3 d 2 hybrid orbitals 3s X S atom (ground state) 76

77 sp 3 d 2 Hybrid Orbitals 3d sp 3 d 2 S atom (hybridized state) 77

78 Hybridization Mixed Hybrids (#) Remaining Geometry s+p sp (2) p+p Linear s+p+p sp 2 (3) p Triangular planar s+p+p+p sp 3 (4) Tetrahedral Mixed Hybrids (#) Remaining Geometry s+p+p+p+d sp 3 d (5) d+d+d+d Triangular bipyramid s+p+p+p+d+d sp 3 d 2 (6) d+d+d Octahedral 78

79 Hybridization and Molecular Geometry Forms Overall Structure Hybridization of A AX 2 Linear sp AX 3, AX 2 E Trigonal Planar sp 2 AX 4, AX 3 E, AX 2 E 2 Tetrahedral sp 3 AX 5, AX 4 E, AX 3 E 2, AX 2 E 3 Trigonal bipyramidal sp 3 d AX 6, AX 5 E, AX 4 E 2 Octahedral sp 3 d 2 A = central atom X = atoms bonded to A E = nonbonding electron pairs on A

80 Summary - Hybrid Orbitals Hybrid Orbital Geometric Arrangements Number of Orbitals Example sp Linear 2 Be in BeF 2 sp 2 Trigonal planar 3 B in BF 3 sp 3 Tetrahedral 4 C in CH 4 sp 3 d Trigonal bipyramidal 5 P in PCl 5 sp 3 d 2 Octahedral 6 S in SF 6 80

81 Lewis Dot Structure Rules: Count only valence electrons Assemble bonding framework Fill up non-bonding electrons on outer atoms Fill up non-bonding electrons on inner atoms Calculate Formal Charge Minimize Formal Charge

82 To do Lewis Structures: Must be able to recognize polyatomic ions Must be able to identify valence electrons Periodic Table : Column numbers! Must be able to construct Bond framework More complex: H outside Acidic Hs bond to O atoms

83 Hints on Lewis Dot Structures 1. Octet rule is the most useful guideline. 2. Carbon forms 4 bonds. 3. Hydrogen typically forms one bond to other atoms. 4. When multiple bonds are forming, they are usually between C, N, O or S. 5. Nonmetals can form single, double, and triple bonds, but not quadruple bonds. 6. Always account for single bonds and lone pairs before forming multiple bonds. 7. Look for resonance structures.

84 5+(3*7)=26 e - PCl 3 Bonding Pairs Lone Pairs (a.k.a. nonbonding electrons)

85 Drawing Lewis Structures To start: Draw Lewis symbols of all individual atoms in formula 1) Count the total valence electrons for the molecule: To do this, find the number of valence electrons for each atom in the molecule, and add them up. Example: CO 2 Valence electrons: C = 4, (O = 6)x2 16 Total number of valence electrons is important, not where they came from. 85

86 Drawing Lewis Structures 2a) Write the symbols for each atom b) show which atom is attached to which 2 elements: central atom is first more than 2: order of connected atoms as written in formula c) Use a single dash to show bonds between atoms: CO 2 : 2 atoms, C is centrals, both O are attached to C by single bond O C O 86

87 Drawing Lewis Structures 3) Complete octets around all atoms bonded to central atom by adding electrons. CO 2 :.... (H does not get an octet!) : O C O :.... Count electrons now: 16 valence electrons shown, both O have octet, C? 87

88 Drawing Lewis Structures 4) Any leftover valence electrons after step 3? Place them on the central atom. Even if doing so gives that atom more than 8 electrons (there are exceptions to octet rule) CO 2 : no leftover valence electrons 88

89 Drawing Lewis Structures 5) If central atom does not have an octet of electrons after step 4: Try multiple bonds! Use unshared electrons already shown in step 3 and 4 and move them between central atom and other atoms. CO 2 O = C = O 89

90 Drawing Lewis Structures 6) Some handy bond rules to remember for molecules: Hydrogen and the halogens (F, Cl, Br, I) form a single bond. The family oxygen (O, S, Se) forms 2 single bonds (or 1 double bond) The family nitrogen (N, P, As) forms 3 single bonds, or 1 single bond and a double bond or a triple bond. So does boron. The family carbon (C, Si) forms 4 single bonds, or 2 double bonds, or.. A good thing to do is to bond all the atoms together by single bonds, and then add the multiple bonds until the rules above are followed. 90

91 Drawing Lewis Structures Draw the Lewis structure for CHCl 3. # of Valence electrons = x(7) = 32 Draw all atoms, connected by single dash (single bond, 2 electrons). Place electrons around atoms bonded to central atoms for octet. C = central atom Cl H C Cl Cl 91

92 Drawing Lewis Structures Draw the Lewis structure for PCl 3 # valence electrons = 5 + 3x(7) = 26 Draw all atoms, connected by single dash (single bond, 2 electrons). 2 atoms: P is central. Place electrons around atoms bonded to central atoms giving them a total of 8 electrons. Count electrons (have 24 now), place any extra valence electrons on central atom Cl Cl P Cl 92

93 Lewis Structures of Ions 1) Count the total valence electrons for the molecule: To do this, find the number of valence electrons for each atom in the molecule, and add them up. For polyatomic anions, add the charge of the ion to the number of valence electrons. For polyatomic cations, subtract the charge of the ion from the number of valence electrons. PO 4 3- # valence electrons = 5 + 4x(6) + 3 = 32 93

94 Lewis Structures of Ions 2) Draw an arrangement of the atoms for the molecule that follows the rules above, central atom, surrounded by others, connected by single bonds. Hydrogen and the halogens bond once. The family oxygen is in bonds one, two, or three times. The family nitrogen is in bonds two, three, or four times Boron usually bonds four times. The family carbon is in bonds four times. 94

95 Writing Lewis Structures for Polyatomic Ions the procedure is the same, the only difference is in counting the valence electrons for polyatomic cations, take away one electron from the total for each positive charge for polyatomic anions, add one electron to the total for each negative charge

96 Exceptions to the Octet Rule H & Li, lose one electron to form cation Li now has electron configuration like He H can also share or gain one electron to have configuration like He Be shares 2 electrons to form two single bonds B shares 3 electrons to form three single bonds expanded octets for elements in Period 3 or below using empty valence d orbitals some molecules have odd numbers of electrons NO : N O :

97 Some molecules, such as SF 6 and PCl 5 have more than 8 electrons around a central atom in their Lewis structure. Draw an electron-dot structure for SF 6. Step 1: 6 + 4(7) = 34 valence electrons F F F F F F Step 2: S Step 3: S F F F F F F SF6 and PCl5 can violate the octet rule through the use of empty d orbitals: both S and P can utilize empty d orbitals to hold pairs of electrons that help bond halogen atoms.

98 Electron-Dot Structures of Compounds Containing Elements Below the Second Row Draw an electron-dot structure for ICl 3. Step 1: 7 + 3(7) = 28 valence electrons Cl Cl Cl Cl Cl Cl Step 2: I Step 4: I Step 3: Cl Cl I Cl 2014 Pearson Education, Inc.

99 Worked Example 7.3 Drawing an Electron-Dot Structure Draw an electron-dot structure for the deadly gas hydrogen cyanide, HCN. Strategy First, connect the carbon and nitrogen atoms. The only way the carbon can form four bonds and the nitrogen can form three bonds is if there is a carbon nitrogen triple bond. Solution Draw an electron-dot structure for carbon dioxide, CH 3 CN and CH3OH Instructor Resource DVD for Chemistry, 6th Edition John McMurry & Robert C. Fay 2012 Pearson Education, Inc.

100 Electron-Dot Structures of Compounds Containing Elements Below the Second Row Draw an electron-dot structure for O 3. Step 1: 3(6) = 18 valence electrons Step 2: O O O Step 4: O O O Check octet rule O O Step 3: Step 5: O O O O 2014 Pearson Education, Inc.

101 Electron-Dot Structures and Resonance Move a lone pair from this oxygen? Step 4: O O O Or move a lone pair from this oxygen? O O O O O O 2014 Pearson Education, Inc. Resonance Oxygen bond lengths are identical, and intermediate to single and double bonds

102 Resonance we can often draw more than one valid Lewis structure for a molecule or ion in other words, no one Lewis structure can adequately describe the actual structure of the molecule the actual molecule will have some characteristics of all the valid Lewis structures we can draw

103 Resonance Same length Resonance is invoked when more than one valid Lewis structure can be written for a particular molecule. H H H Benzene, C 6 H 6 H H H H H H H H H The actual structure is an average of the resonance structures. The bond lengths in the ring are identical, and between those of single and double bonds.

104 Resonance Bond Length and Bond Energy Resonance bonds are shorter and stronger than single bonds. H H H H H H H H H H H H Resonance bonds are longer and weaker than double bonds.

105 Resonance- Same length Lewis structures often do not accurately represent the electron distribution in a molecule Lewis structures imply that O 3 has a single (147 pm) and double (121 pm) bond, but actual bond length is between, (128 pm) Real molecule is a hybrid of all possible Lewis structures Resonance stabilizes the molecule maximum stabilization comes when resonance forms contribute equally to the hybrid O O + O O O + O

106 Electron Dot Formula for NO 2 +, NO 3 - and O 3

107 Resonance in Polyatomic Ions Resonance in a carbonate ion: The bond lengths in the structures are identical, and between those of single and double bonds. Resonance in an acetate ion: 2014 Pearson Education, Inc.

108 Example 10.1 VSEPR Theory and the Basic Shapes Determine the molecular geometry of NO 3. Solution The molecular geometry of NO 3 is determined by the number of electron groups around the central atom (N). Begin by drawing a Lewis structure of NO 3. NO 3 has 5 + 3(6) + 1 = 24 valence electrons. The Lewis structure has three resonance structures: The hybrid structure is intermediate between these three and has three equivalent bonds. Use any one of the resonance structures to determine the number of electron groups around the central atom. The nitrogen atom has three electron groups. Based on the number of electron groups, determine the geometry that minimizes the repulsions between the groups. Trigonal Planar sp 2 Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro 2014 Pearson Education, Inc.

109 Sigma and Pi Bonds Sigma ( ) bonds exist in the region directly between two bonded atoms. Pi ( ) bonds exist in the region above and below a line drawn between two bonded atoms. Single bond Double Bond Triple Bond 1 sigma bond 1 sigma, 1 pi bond 1 sigma, 2 pi bonds

110 The De-Localized Electron Model Pi bonds ( ) contribute to the delocalized model of electrons in bonding, and help explain resonance H H H H H H H H H H H H Electron density from bonds can be distributed symmetrically all around the ring, above and below the plane.

111 Practice Please draw the lewis structure, identify the bond types and point the types hybridization orbitals in C and N. CH3CN CH3CH2NH2 Identify the pi and sigma bonds in the given molecules. σ σ σ π σ σ σ π, π 111

112 What about multiple bonding! According to valence bond theory hybrid orbitals include: single bonds lone pairs one of the bonds in a multiple bond. The electrons in the unhybridized atomic orbitals are used to form the additional multiple bonds (from Molecular Orbital Theory.) 112

113 Multiple Bonding A (sigma) bond is an overlap of orbitals (hybrids) along the bond axis. A (pi) bond is a overlap of parallel p orbitals, creating an electron distribution above and below the bond axis. () 113

114 Energy Multiple Bonding 2p 2p (unhybridized) 2s sp 2 1s C atom (ground state) 1s (3 sp 2 hybrid + 1 unhybridized p) 114

115 Multiple Bonding σ π 115

116 Sigma and Pi Bonds: Double bonds 1 bond H H H H C C C C H H H H Ethene 1 bond

117 Multiple Bonding 117

118 Practice What are the hybridization and approximate bond angles for each C, N, O in the given molecules? 118

119 Molecular Orbital Theory For Physics / Chem Major Students Atomic Orbital: A wave function whose square gives the probability of finding an electron within a given region of space in an atom. Molecular Orbital: A wave function whose square gives the probability of finding an electron within a given region of space in a molecule Pearson Education, Inc. Chapter 7/119

120 The Octet Rule- Valence Bond Theory The Diatomic Oxygen Molecule O O 1s 2s 2p Each has six valence electrons 1s 2s 2p 1 bond 1 bond O O A Double bond is when 4 electrons are shared they are represented by two lines in bond diagrams BUT NOT 2 bonds

121 Molecular Orbital Theory: The Hydrogen Molecule bonding orbital * antibonding orbital Bond Order = (# bonding e # antibonding e ) Pearson Education, Inc. Chapter 7/121

122 Molecular Orbital Theory: The Hydrogen Molecule Bond Order = = Pearson Education, Inc. Chapter 7/122

123 Molecular Orbital Theory: The Hydrogen Molecule Bond Order: = Pearson Education, Inc = 0 Chapter 7/123

124 Molecular Orbital Theory: Other Diatomic Molecules 2012 Pearson Education, Inc. Chapter 7/124

125 Example Molecular Orbital Theory Draw an MO energy diagram and determine the bond order for the N 2 ion. Do you expect the bond to be stronger or weaker than in the N 2 molecule? Is N 2 diamagnetic or paramagnetic? The N 2 ion has 11 valence electrons (5 for each nitrogen atom plus 1 for the negative charge). Assign the electrons to the molecular orbitals beginning with the lowest energy orbitals and following Hund s rule. Calculate the bond order by subtracting the number of electrons in antibonding orbitals from the number in bonding orbitals and dividing the result by two. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro 2014 Pearson Education, Inc.

126 Formal Charges Formal Charge = # of valence e in free atom 1 2 # of bonding e # of nonbonding e Calculate the formal charge on each atom in O 3. O O O 1 6 (4) 4 = (6) 2 = (2) 6 = Pearson Education, Inc. Chapter 7/126

127 Isomers: Same composition, two different constitutional Lewis structures HCN = atomic compositional structure HCN possesses 10 VE = Lewis compositional structures Two possible Lewis constitutional structures: H-C-N or H-N-C Both need to have 10 VE in their Lewis structure Problem: Try to achieve an acceptable Lewis structure (duet and octet rule followed) for both. 127

128 HCN = atomic compositional structure HCN: 10 VE = Lewis compositional structures Two possible Lewis constitutional structures H-C-N or H-N-C Any acceptable Lewis structure for HCN needs to show 10 VE Try to achieve an acceptable Lewis structure (duet and octet rules obeyed) for all isomeric structures. Two acceptable Lewis structures. Which is better? H C N H N C 128

129 Use formal charges to decide on the stability of isomeric Lewis structures H C N H N C VE (atom) /2 BE (molecule) UE (molecule) FC on atom H C N H N C Important: the net charge of composition HCN = 0, so the sum of the formal charges in any acceptable Lewis structure must be = 0 also. 129

130 Which is more stable? H C N H N C Rule: For two isomeric acceptable Lewis structures, the one with the least separation of formal charges is more stable. H C N Therefore, stable isomer of the pair. is the more 130

131 Electron Dot Formula for CS 2 1. Central atom C=A 2. Terminal atomes S=X, n=2 3. Calculate Total valence electrons 4+2*6 electrons = 16 e 4. bonding pairs n = (total valence electrons)/8 = 16 /8= 2 bonding pairs + 0 remainder electrons 5. Lone pair m = (remainder electrons)/2 =0/2=0 lone pairs 6. CS 2 = AX2E0 Linear S-C-S 7. Hybridization orbital sp 8. Bond angle -180 degree 9. Apply octet rule to all atoms 10. Draw resonance structure 11. Calculate formal charge 12. Which lewis structure is more table

132 Electron Dot Formula for NO 2-1. Central atom N=A 2. Terminal atomes O=X, n=2 3. Calculate Total valence electrons 5+2*6+1 charge electrons = 18 e 4. bonding pairs n = (total valence electrons)/8 = 18 /8= 2 bonding pairs + 2 remainder electrons 5. Lone pair m = (remainder electrons)/2 =2/2=1 lone pair 6. NO 2 - = AX2E1 BENT 7. Hybridization orbital sp 2 8. Bond angle <120 degree 9. Apply octet rule to all atoms 10. Draw resonance structure 11. Calculate formal charge 12. Which lewis structure is more table

133 Example 10.4 Predicting the Shape of Larger Molecules Predict the geometry about each interior atom in methanol (CH 3 OH) and make a sketch of the molecule. Solution Begin by drawing the Lewis structure of CH 3 OH. CH 3 OH contains two interior atoms: one carbon atom and one oxygen atom. To determine the shape of methanol, determine the geometry about each interior atom as follows: Using the geometries of each of these, draw a three-dimensional sketch of the molecule as shown here:

134 Example 9.9 Drawing Resonance Structures and Assigning Formal Charge for Organic Compounds Draw the Lewis structure (including resonance structures) for nitromethane (CH 3 NO 2 ). For each resonance structure, assign formal charges to all atoms that have formal charge. Solution Begin by writing the skeletal structure. For organic compounds, the condensed structural formula (in this case CH 3 NO 2 ) indicates how the atoms are connected. Calculate the total number of electrons for the Lewis structure by summing the number of valence electrons for each atom. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro 2014 Pearson Education, Inc.

135 Example 9.9 Drawing Resonance Structures and Assigning Formal Charge for Organic Compounds Continued Place a dash between each pair of atoms to indicate a bond. Each dash counts for two electrons. (12 of 24 electrons used) Distribute the remaining electrons, first to terminal atoms then to interior atoms. (24 of 24 electrons used) If there are not enough electrons to complete the octets on the interior atoms, form double bonds by moving lone pair electrons from terminal atoms into the bonding region with interior atoms. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro 2014 Pearson Education, Inc.

136 Example 9.9 Drawing Resonance Structures and Assigning Formal Charge for Organic Compounds Continued Draw any necessary resonance structures by moving only electron dots. (In this case, you can form a double bond between the nitrogen atom and the other oxygen atom.) Assign formal charges (FC) to each atom. FC = # valence e (# nonbonding e + ½ # bonding e ) Carbon, hydrogen, and the doubly-bonded oxygen atoms have no formal charge. Nitrogen has a +1 formal charge [5 ½ (8)] and the singly bonded oxygen atom in each resonance structure has a 1 formal charge [6 (6 + ½ (2))]. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro 2014 Pearson Education, Inc.

137 Example 9.8 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 Calculate the formal charge on each atom by finding the number of valence electrons and subtracting the number of nonbonding electrons and one-half the number of bonding electrons. 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 to be preferred over structure C. Structure A is preferable to B because it has the negative formal charge on the more electronegative atom. You therefore expect structure A to make the biggest contribution to the resonance forms of the cyanate ion. Chemistry: A Molecular Approach, 3rd Edition Nivaldo J. Tro 2014 Pearson Education, Inc.