Chapter 8

Chapter 8 PowerPoint PPT Presentation

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8.1 What is a Bond?. A force that holds atoms together.Why?We will look at it in terms of energy.Bond energy the energy required to break a bond.Why are compounds formed?Because it gives the system the lowest energy.. Ionic Bonding. An atom with a low ionization energy reacts with an atom with high electron affinity.The electron moves.Opposite charges hold the atoms together..

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Chapter 8

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1. Chapter 8 Bonding

2. 8.1 What is a Bond? A force that holds atoms together. Why? We will look at it in terms of energy. Bond energy the energy required to break a bond. Why are compounds formed? Because it gives the system the lowest energy.

3. Ionic Bonding An atom with a low ionization energy reacts with an atom with high electron affinity. The electron moves. Opposite charges hold the atoms together.

4. Coulomb's Law E= 2.31 x 10-19 J•nm(Q1Q2)/r Q is the charge. r is the distance between the centers. If charges are opposite, E is negative, indicating attractive force. Therefore, the ion pair has lower energy than the separated ions. Same charge, positive E, requires energy to bring them together.

5. What about covalent compounds? The electrons in each atom are attracted to the nucleus of the other. The electrons repel each other. The nuclei repel each other. They reach a distance with the lowest possible energy. The distance between is the bond length.

12. Covalent Bonding Electrons are shared by atoms. In between covalent and ionic are polar covalent bonds. The electrons are not shared evenly. One end is slightly positive, the other negative. Indicated using small delta ?.

17. 8.2 Electronegativity The ability of an electron to attract shared electrons to itself. Developed by Linus Pauling. Imaginary molecule HX Expected H-X energy = H-H energy + X-X energy 2 ? = (H - X)actual - (H - X)expected

18. Electronegativity ? is known for almost every element. Gives us relative electronegativities of all elements. Tends to increase left to right, decreases as you go down a group. Noble gases aren’t discussed. Difference in electronegativity between atoms tells us how polar.

19. Practice Predict the order of increasing electronegativity in each group. Mg,Ca,Sr C,F,Ge Se,O,Te P,O,N

21. Practice Rank the following bonds in order of increasing ionic character. P-S, Sr-S, C-N, I-I, Na-Cl

22. 8.3 Dipole Moments(dipolar) A molecule with a center of negative charge and a center of positive charge is dipolar (two poles), or has a dipole moment. Center of charge doesn’t have to be on an atom. Will line up in the presence of an electric field.

23. How It is drawn

24. Which Molecules Have Them? Any two atom molecule with a polar bond. With three or more atoms there are two considerations. There must be a polar bond. Geometry can’t cancel it out.

25. Geometry and polarity Three shapes will cancel them out. Linear

26. Geometry and polarity Three shapes will cancel them out. Planar triangles

27. Geometry and polarity Three shapes will cancel them out. Tetrahedral

28. Geometry and polarity Others don’t cancel Bent

29. Geometry and polarity Others don’t cancel Trigonal Pyramidal

30. Practice Predict which bond in each of the following groups will be the most polar. C-H, Si-H, Sn-H Al-Br, Ga-Br, In-Br, Tl-Br C-O or Si-O O-F or O-Cl

31. 8.4 Ions Atoms tend to react to form noble gas configuration. Metals lose electrons to form cations. Nonmetals gain electrons to form anions. Nonmetals can can share electrons in covalent bonds when two non metals react.(more later)

32. Ionic Compounds We mean the solid crystal. Ions align themselves to maximize attractions between opposite charges, and to minimize repulsion between like ions. Crystals stabilize ions that would be unstable as a gas. React to achieve noble gas configuration

33. Size of ions Ion size increases down a group. Cations are smaller than the atoms they came from. Anions are larger. Across a row they get smaller, and then suddenly larger. First 3/4 of table are cations. Second 1/4 are anions.

34. Periodic Trends Across the period nuclear charge increases so they get smaller.

35. Size of Isoelectronic ions Iso - same Iso electronic ions have the same # of electrons Al+3 Mg+2 Na+1 Ne F-1 O-2 and N-3 All have 10 electrons. All have the configuration 1s22s22p6

36. Size of Isoelectronic ions Positive ions have more protons so they are smaller.

37. Practice What noble gas has the same electron configuration as each of the ions in the following compounds: Cesium Sulfide, strontium fluoride. Give three ions that are isoelectronic with krypton.

38. Practice For each group, place the atoms or ions in order of increasing size. V, V+2, V+3, V+5 Na+, K+, Rb+, Cs+

39. 8.5 Forming Ionic Compounds Lattice energy - the energy associated with making a solid ionic compound from its gaseous ions. M+(g) + X-(g) ? MX(s) Energy change is exothermic. The formation of the ionic compounds is endothermic until the formation of the lattice.

40. Na(s) + ½F2(g) ? NaF(s) First sublime Na Na(s) ? Na(g) ?H = 109 kJ/mol Ionize Na(g) Na(g) ? Na+(g) + e- (Ionization energy) ?H = 495 kJ/mol Break F-F Bond ½F2(g) ? F(g) (Bond energy) ?H = 77 kJ/mol Add electron to F F(g) + e- ? F-(g) (Electron affinity) ?H = -328 kJ/mol

41. Na(s) + ½F2(g) ? NaF(s) Na+(g) + F-(g) ? NaF(s) (Lattice energy) ?H = -928 kJ/mol Total??Hf ?H = -575 kJ/mol

43. Calculating Lattice Energy Lattice Energy = k(Q1Q2 / r) k is a constant that depends on the structure of the crystal. Q’s are charges. r = shortest distance between centers of the cations and anions. Lattice energy is greater with more highly charged ions. A decreased distance between ions (smaller ions) will have more energy than a larger distance.

44. Practice Which compound in the following pairs has the most exothermic lattice energy? Justify your answer. LiF, CsF NaBr, NaI BaCl2, BaO Na2SO4, CaSO4 Li2O, Na2S

45. 8.6 Partial Ionic Character There are probably no totally ionic bonds between individual atoms. Calculate % ionic character. Compare measured dipole of X-Y bonds to the calculated dipole of X+Y- the completely ionic case. % dipole = Measured X-Y x 100 Calculated X+Y- In the gas phase.

47. How do we deal with it? Ionic bond = generally, >50% ionic character, electronegativity difference of 1.6. Difficult to calculate for polyatomic ions. An ionic compound will be defined as any substance that conducts electricity when melted. Also use the generic term salt.

48. 8.7 The Covalent Bond The forces that cause a group of atoms to behave as a unit. Why? Due to the tendency of atoms to achieve the lowest energy state. The bond is a human invention. It is a method of explaining the energy change associated with forming molecules. Bonds don’t exist in nature, but are useful. We have a model of a bond.

49. What is a Model? Explains how nature operates on the microscopic level. Derived from observations on the macroscopic level. Models can be wrong, because they are based on speculations and oversimplification. Become more complicated with age. You must understand the assumptions in the model, and look for weaknesses.

50. 8.8 Covalent Bond Energies We use an average of the individual bond dissociation energies to calculate the bond energy. Each C-H bond has a different energy. CH4 ? CH3 + H ?H = 435 kJ/mol CH3 ? CH2 + H ?H = 453 kJ/mol CH2 ? CH + H ?H = 425 kJ/mol CH ? C + H ?H = 339 kJ/mol Each bond is sensitive to its environment.

51. Averages Have made a table of the averages of different types of bonds pg. 372 single bond one pair of electrons is shared. double bond two pair of electrons are shared. triple bond three pair of electrons are shared. More bonds, shorter bond length.

52. Using Bond Energies We can find ?H for a reaction. It takes energy to break bonds, and end up with atoms (+). We get energy when we use atoms to form bonds (-). If we add up the energy it took to break the bonds, and subtract the energy we get from forming the bonds we get the ?H. ?H = ?D(bonds broken) - ?D(bonds formed)

53. Find the energy for this CH4(g) + 2Cl2(g) + 2F2(g) --> CF2Cl2(g) + 2HF(g) + 2HCl(g) Ex8.5 C-H 413 kJ/mol C-Cl 339 kJ/mol Cl-Cl 239 kJ/mol H-F 565 kJ/mol F-F 154 kJ/mol H-Cl 427 kJ/mol C-F 485 kJ/mol

54. Find the energy for this CH4(g) + 2Cl2(g) + 2F2(g) --> CF2Cl2(g) + 2HF(g) + 2HCl(g) C-H 413 kJ/mol C-Cl 339 kJ/mol Cl-Cl 239 kJ/mol H-F 565 kJ/mol F-F 154 kJ/mol H-Cl 427 kJ/mol C-F 485 kJ/mol Ans -1194kJ/mol

55. Find the energy for this

56. Find the energy for this

57. 8.9 Localized Electron Model Simple model, easily applied. A molecule is composed of atoms that are bound together by sharing pairs of electrons using the atomic orbitals of the bound atoms. Three Parts: Valence electrons using Lewis structures. Prediction of geometry using VSEPR. Description of the types of orbitals (Chapter 9).

58. 8.10 Lewis Structure Shows how the valence electrons are arranged. One dot for each valence electron. A stable compound has all its atoms with a noble gas configuration. Hydrogen follows the duet rule. The rest follow the octet rule. Bonding pair is the one between the symbols.

59. Rules Sum the valence electrons. Use a pair to form a bond between each pair of atoms. Arrange the rest to fulfill the octet rule (except for H and the duet). Do not worry where electrons come from, they are now part of the molecule. A line can be used instead of a pair.

60. Practice Write the Lewis structures for the following structures. POCl3 SO42- XeO4 PO43-

61. 8.11 Exceptions to the octet BH3 B typically will have 6 electrons around it. Boron will often form molecules that obey the octet rule. SF6 Sulfur has 12 electrons around it. Exceeding the rule. 3s and 3p are filled with the other 4 electrons occupying the 3d orbital.

62. Exceptions to the octet When we must exceed the octet, extra electrons go on central atom. ClF3 XeF4 ICl4- BeCl2

63. Comments About the Octet Rule 2nd row elements C, N, O, F observe the octet rule. 2nd row elements B and Be often have fewer than 8 electrons around themselves - they are very reactive. 3rd row and heavier elements CAN exceed the octet rule using empty valence d orbitals. When writing Lewis structures, satisfy octets first, then place electrons around elements having available d orbitals.

64. 8.12 Resonance Sometimes there is more than one valid structure for an molecule or ion. NO3- Use double arrows to indicate it is the “average” of the structures. It doesn’t switch between them. NO2- Localized electron model is based on pairs of electrons, doesn’t deal with odd numbers.

65. Formal Charge For molecules and polyatomic ions that exceed the octet there are several different structures. Use charges on atoms to help decide which. Trying to use the oxidation numbers to put charges on atoms in molecules doesn’t work.

66. Formal Charge The difference between the number of valence electrons on the free atom and that assigned in the molecule. We count half the electrons in each bond as “belonging” to the atom. SO4-2 Molecules try to achieve as low a formal charge as possible. Negative formal charges should be on the most electronegative elements. Experiment can only determine correct bonding.

67. Examples Not as good Better XeO3 NO4-3 ClO3-

68. 8.13 VSEPR Lewis structures tell us how the atoms are connected to each other. They don’t tell us anything about shape. The shape of a molecule can greatly affect its properties. Valence Shell Electron Pair Repulsion Theory allows us to predict geometry.

70. VSEPR Molecules take a shape that puts electron pairs as far away from each other as possible. Have to draw the Lewis structure to determine electron pairs. Electron pairs are placed as far apart as possible. Nonbonding Lone pairs take more space. Multiple bonds count as one pair.

71. VSEPR The number of pairs determines bond angles underlying structure The number of atoms determines actual shape (BDVD)


73. Actual shape from unshared pairs

74. Actual Shape

76. Practice Predict the structure and bond angles for the following molecules: POCl3, NF3, PCl2-, SO3, ICl5, XeCl4, SeCl6 Which are polar?

77. Practice Write the Lewis structures and predict whether each is polar or nonpolar. COS, XeF2, CF2Cl2, SeF6, H2CO

78. How well does it work? Does an outstanding job for such a simple model. For non-ionic compounds, predictions are almost always accurate. Like all simple models, it has exceptions.

79. Chapter 9 Orbitals and Covalent Bond

80. Atomic Orbitals Don’t Work to explain molecular geometry. In methane, CH4, the shape is tetrahedral. The valence electrons of carbon should be two in s, and two in p. The p orbitals would have to be at right angles. The atomic orbitals change when making a molecule.

81. 9.1 Hybridization We blend the s and p orbitals of the valence electrons and end up with the tetrahedral geometry. We combine one s orbital and 3 p orbitals. The atoms are responding as needed to give the minimum energy for the molecule. sp3 hybridization has tetrahedral geometry.

83. In terms of energy

84. How we get to hybridization - CH4 We know the geometry from experiment. Four bonds of equal length and strength. We know the orbitals of the central atom. Hybridizing atomic orbitals can explain the geometry. So if the geometry requires a tetrahedral shape, it is sp3 hybridized. This includes bent and trigonal pyramidal molecules because one of the sp3 lobes holds the lone pair.


86. sp2 hybridization C2H4 Trigonal planar. 120° Double bond acts as one pair. This results in 3 effective pairs surrounding the carbon atoms. One s and two p orbitals hybridize into 3 identical orbitals of equal length and energy to make sp2 orbitals. This leaves one p orbital unhybridized.

87. sp2 hybridization

88. In terms of energy

89. Two types of Bonds Sigma bonds (?) form from the overlap of orbitals along the internuclear axis. Pi bond (?) occupies the space above and below internuclear axis. Between adjacent unhybridized p orbitals. The double bond always consists of one ? bond and one ? bond. C-C double bond (BDVD)

90. ? and ? bonds (YDVD)

91. ? and ? bonds (YDVD)

92. sp hybridization CO2 Each carbon has two hybridized orbitals 180º apart. Also 2 unhybridized p orbitals. p orbitals are at right angles (Fig. 9.17) Makes room for two p bonds and two sigma bonds.

93. In terms of energy


95. CO2 C can make two s and two p O can make one s and one p (Fig. 9.19)

96. dsp3 PCl5 Five pairs of electrons around the central atom. Trigonal bypyramidal. Only ? bonds no ? bonds. The model predicts that we must use the d orbitals. Five electron pairs require dsp3 hybridization. (Fig. 9.21) There is some controversy about how involved the d orbitals are.

97. d2sp3 SF6 Six pairs of electrons around the central atom. Octahedral shape. (Fig. 9.23)

98. How do we figure this out? Use the Localized Electron Model. Draw the Lewis structure(s). Determine the arrangement of electron pairs (VSEPR model). Specify the necessary hybrid orbitals based upon the pairs of electrons around the central atom.

99. What is the expected hybridization of the central atom for the following molecules? POCl3, NF3, PCl2-, SO3, ICl5, XeCl4, SeCl6 HOCN, COS, XeF2, CF2Cl2, SeF6, H2CO

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