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Ch. 9 and 10 Chemical Bonding

Ch. 9 and 10 Chemical Bonding. Brady & Senese, 5th Ed. A general comparison of metals and nonmetals. Figure 9.1. Types of Chemical Bonding. 1. Metal with nonmetal:. electron transfer and ionic bonding. 2. Nonmetal with nonmetal:. electron sharing and covalent bonding.

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Ch. 9 and 10 Chemical Bonding

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  1. Ch. 9 and 10 Chemical Bonding Brady & Senese, 5th Ed

  2. A general comparison of metals and nonmetals. Figure 9.1

  3. Types of Chemical Bonding 1. Metal with nonmetal: electron transfer and ionic bonding 2. Nonmetal with nonmetal: electron sharing and covalent bonding 3. Metal with metal: electron pooling and metallic bonding

  4. The three models of chemical bonding. Figure 9.2

  5. The A group number gives the number of valence electrons. Place one dot per valence electron on each of the four sides of the element symbol. Place single dots (electrons) around the symbol and pair the dots when necessary . . . : . . . : N . N . . N N : . . : . Lewis Electron-Dot Symbols For main group elements - Example: Nitrogen, N, is in Group 5A and therefore has 5 valence electrons.

  6. Periodic Trends in Lattice Energy Coulomb’s Law charge A X charge B electrostatic force a distance2 energy = force X distance therefore charge A X charge B electrostatic energy a distance cation charge X anion charge a DH0lattice electrostatic energy a cation radius + anion radius

  7. Figure 9.7 Trends in lattice energy.

  8. Covalent bond formation in H2. Figure 9.11

  9. Distribution of electron density of H2. Figure 9.12

  10. back to previous slide

  11. PROBLEM: Using the periodic table, but not Tables 9.2 and 9.3, rank the bonds in each set in order of decreasing bond length and bond strength: (a) S - F, S - Br, S - Cl (b) C = O, C - O, C O PLAN: (a) The bond order is one for all and sulfur is bonded to halogens; bond length should increase and bond strength should decrease with increasing atomic radius. (b) The same two atoms are bonded but the bond order changes; bond length decreases as bond order increases while bond strength increases as bond order increases. Bond length: C - O > C = O > C O Bond strength: C O > C = O > C - O SAMPLE PROBLEM 9.2 Comparing Bond Length and Bond Strength SOLUTION: (a) Atomic size increases going down a group. (b) Using bond orders we get Bond length: S - Br > S - Cl > S - F Bond strength: S - F > S - Cl > S - Br

  12. Strong covalent bonding forces within molecules Weak intermolecular forces between molecules Figure 9.14 Strong forces within molecules and weak forces between them.

  13. Covalent bonds of network covalent solids. Figure 9.15

  14. Figure 9.16 Using bond energies to calculate H0rxn. DH0rxn = DH0reactant bonds broken + DH0product bonds formed Enthalpy, H BOND BREAKING DH01 = + sum of BE DH02 = - sum of BE BOND FORMATION DH0rxn

  15. 2[-BE(C O)]= -1598kJ DH0rxn= -818kJ Using bond energies to calculate H0rxn of methane. Figure 9.17 BOND BREAKING 4BE(C-H)= +1652kJ 2BE(O2)= + 996kJ DH0(bond breaking) = +2648kJ BOND FORMATION 4[-BE(O-H)]= -1868kJ Enthalpy,H DH0(bond forming) = -3466kJ

  16. PROBLEM: Use Table 9.2 (button at right) to calculate DH0rxn for the following reaction: CH4(g) + 3Cl2(g) CHCl3(g) + 3HCl(g) PLAN: Write the Lewis structures for all reactants and products and calculate the number of bonds broken and formed. SAMPLE PROBLEM 9.3 Calculating Enthalpy Changes from Bond Energies SOLUTION: bonds broken bonds formed

  17. SAMPLE PROBLEM 9.3 Calculating Enthalpy Changes from Bond Energies continued bonds broken bonds formed 4 C-H = 4 mol(413 kJ/mol) = 1652 kJ 3 C-Cl = 3 mol(-339 kJ/mol) = -1017 kJ 3 Cl-Cl = 3 mol(243 kJ/mol) = 729 kJ 1 C-H = 1 mol(-413 kJ/mol) = -413 kJ 3 H-Cl = 3 mol(-427 kJ/mol) = -1281 kJ DH0bonds broken = 2381 kJ DH0bonds formed = -2711 kJ DH0reaction = DH0bonds broken + DH0bonds formed = 2381 kJ + (-2711 kJ) = - 330 kJ

  18. Table 9.5 Heats of Combustion(DHcomb) of Some Carbon Compounds Two-Carbon Compounds One-Carbon Compounds Ethane (C2H6) Ethanol (C2H6O) Methane (CH4) Methanol (CH4O) Structural Formulas Sum of C-C and C-H Bonds 7 6 4 3 Sum of C-O and O-H Bonds 0 2 0 2 DHcomb(kJ/mol) -1560 -1367 -890 -727 DHcomb(kJ/g) -51.88 -29.67 -55.5 -22.7

  19. Figure 9.19 The Pauling electronegativity (EN) scale.

  20. Figure 9.20 Electronegativity and atomic size.

  21. PROBLEM: (a) Use a polar arrow to indicate the polarity of each bond: N-H, F-N, I-Cl. (b) Rank the following bonds in order of increasing polarity: H-N, H-O, H-C. SAMPLE PROBLEM 9.4 Determining Bond Polarity from EN Values PLAN: (a) Use Figure 9.19(button at right) to find EN values; the arrow should point toward the negative end. (b) Polarity increases across a period. SOLUTION: (a) The EN of N = 3.0, H = 2.1; F = 4.0; I = 2.5, Cl = 3.0 N - H F - N I - Cl (b) The order of increasing EN is C < N < O; all have an EN larger than that of H. H-C < H-N < H-O

  22. Figure 9.21 Electron density distributions in H2, F2, and HF.

  23. 3.0 DEN 2.0 0.0 Figure 9.22 Boundary ranges for classifying ionic character of chemical bonds.

  24. Figure 9.23 Percent ionic character of electronegativity difference (DEN).

  25. Steps for converting a molecular formula into a Lewis structure: • The atom requiring the most additional valence e- is placed centrally. If multiple atoms require the same number, place the larger one centrally. • Sum up the total valence e- (adjust for polyatomic ion charge if needed) • Connect the surrounding atoms to the central atom with single bonds. • Place the remaining e- as lone pairs (starting with the surrounding atoms) to satisfy the octet rule. • If you run out of e- before the octet rule has been satisfied (where applicable) convert one or more lone pairs to multiple bonds. Common octet rule exceptions: • H, Be and B form covalent compounds with less than 8 valence e-. They are stable with 2,4 and 6 valence e- respectively. • Elements in the 3rd row and below may sometimes exceed the octet rule having as many as 10 or 12 valence e-. • Compounds with an odd number of valence e- are not able to satisfy the octet rule. These compounds are called radicals.

  26. Molecular formula For NF3 Atom placement : : N 5e- : F : : F : : Sum of valence e- F 7e- X 3 = 21e- N Total 26e- : F : : Remaining valence e- Lewis structure

  27. PROBLEM: Write a Lewis structure for CCl2F2, one of the compounds responsible for the depletion of stratospheric ozone. PLAN: Follow the steps outlined in Figure 10.1 . : : : Cl : : : : Cl C F : : : : F : SAMPLE PROBLEM 10.1 Writing Lewis Structures for Molecules with One Central Atom SOLUTION: Cl Cl C F F

  28. PROBLEM: Write the Lewis structure for methanol (molecular formula CH4O), an important industrial alcohol that is being used as a gasoline alternative in car engines. SAMPLE PROBLEM 10.2 Writing Lewis Structure for Molecules with More than One Central Atom SOLUTION: Hydrogen can have only one bond so C and O must be next to each other with H filling in the bonds. There are 4(1) + 4 + 6 = 14 valence e-. C has 4 bonds and O has 2. O has 2 pair of nonbonding e-. H : H C O H : H

  29. H H H H C C C C H H H H N : N : . . N : N : N : N : . . . . SAMPLE PROBLEM 10.3 Writing Lewis Structures for Molecules with Multiple Bonds. PROBLEM: Write Lewis structures for the following: (a) Ethylene (C2H4), the most important reactant in the manufacture of polymers (b) Nitrogen (N2), the most abundant atmospheric gas PLAN: For molecules with multiple bonds, there is a Step 5which follows the other steps in Lewis structure construction. If a central atom does not have 8e-, an octet, then two e- (either single or nonbonded pair)can be moved in to form a multiple bond. SOLUTION: (a) There are 2(4) + 4(1) = 12 valence e-. H can have only one bond per atom. : (b) N2 has 2(5) = 10 valence e-. Therefore a triple bond is required to make the octet around each N.

  30. A - central atom X -surrounding atom E -nonbonding valence electron-group integers VSEPR - Valence Shell Electron Pair Repulsion Theory Each group of valence electrons around a central atom is located as far away as possible from the others in order to maximize distance These repulsions maximize the space that each object attached to the central atom occupies. The result is five electron-group arrangements of minimum energy seen in a large majority of molecules and polyatomic ions. The electron-groups are defining the object arrangement,but the molecular shape is defined by the relative positions of the atomic nuclei. Because valence electrons can be bonding or nonbonding, the same electron-group arrangement can give rise to different molecular shapes. AXmEn

  31. linear tetrahedral trigonal planar trigonal bipyramidal octahedral Figure 10.2 Electron-group repulsions and the five basic molecular shapes.

  32. Figure 10.3 The single molecular shape of the linear electron-group arrangement. Examples: CS2, HCN, BeF2

  33. Class Shape Figure 10.4 The two molecular shapes of the trigonal planar electron-group arrangement. Examples: SO2, O3, PbCl2, SnBr2 Examples: SO3, BF3, NO3-, CO32-

  34. 1220 Effect of Double Bonds 1160 real Effect of Nonbonding(Lone) Pairs Factors Affecting Actual Bond Angles Bond angles are consistent with theoretical angles when the atoms attached to the central atom are the same and when all electrons are bonding electrons of the same order. 1200 larger EN 1200 ideal greater electron density Lone pairs repel bonding pairs more strongly than bonding pairs repel each other. 950

  35. The three molecular shapes of the tetrahedral electron-group arrangement. Figure 10.5 Examples: CH4, SiCl4, SO42-, ClO4- NH3 PF3 ClO3 H3O+ H2O OF2 SCl2

  36. Lewis structures and molecular shapes. Figure 10.6

  37. Figure 10.7 The four molecular shapes of the trigonal bipyramidal electron-group arrangement. PF5 AsF5 SOF4 SF4 XeO2F2 IF4+ IO2F2- XeF2 I3- IF2- ClF3 BrF3

  38. Figure 10.8 The three molecular shapes of the octahedral electron-group arrangement. SF6 IOF5 BrF5 TeF5- XeOF4 XeF4 ICl4-

  39. A summary of common molecular shapes with two to six electron groups. Figure 10.9

  40. PROBLEM: Draw the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) PF3 and (b)COCl2. SOLUTION: (a) For PF3 - there are 26 valence electrons, 1 nonbonding pair SAMPLE PROBLEM 10.6 Predicting Molecular Shapes with Two, Three, or Four Electron Groups The shape is based upon the tetrahedral arrangement. The F-P-F bond angles should be <109.50 due to the repulsion of the nonbonding electron pair. The final shape is trigonal pyramidal. <109.50 The type of shape is AX3E

  41. 124.50 1110 SAMPLE PROBLEM 10.6 Predicting Molecular Shapes with Two, Three, or Four Electron Groups continued (b) For COCl2, C has the lowest EN and will be the center atom. There are 24 valence e-, 3 atoms attached to the center atom. C does not have an octet; a pair of nonbonding electrons will move in from the O to make a double bond. Type AX3 The shape for an atom with three atom attachments and no nonbonding pairs on the central atom is trigonal planar. The Cl-C-Cl bond angle will be less than 1200 due to the electron density of the C=O.

  42. PROBLEM: Determine the molecular shape and predict the bond angles (relative to the ideal bond angles) of (a) SbF5 and (b) BrF5. SOLUTION: (a) SbF5 - 40 valence e-; all electrons around central atom will be in bonding pairs; shape is AX5 - trigonal bipyramidal. SAMPLE PROBLEM 10.7 Predicting Molecular Shapes with Five or Six Electron Groups (b) BrF5 - 42 valence e-; 5 bonding pairs and 1 nonbonding pair on central atom. Shape is AX5E, square pyramidal.

  43. Electric field OFF Electric field ON The orientation of polar molecules in an electric field.

  44. PROBLEM: From electronegativity (EN) values (button) and their periodic trends, predict whether each of the following molecules is polar and show the direction of bond dipoles and the overall molecular dipole when applicable: PLAN: Draw the shape, find the EN values and combine the concepts to determine the polarity. SOLUTION: (a) NH3 SAMPLE PROBLEM 10.9 Predicting the Polarity of Molecules (a) Ammonia, NH3 (b) Boron trifluoride, BF3 (c) Carbonyl sulfide, COS (atom sequence SCO) The dipoles reinforce each other, so the overall molecule is definitely polar. ENN = 3.0 ENH = 2.1 molecular dipole bond dipoles

  45. SAMPLE PROBLEM 10.10 Predicting the Polarity of Molecules continued (b) BF3 has 24 valence e- and all electrons around the B will be involved in bonds. The shape is AX3, trigonal planar. F (EN 4.0) is more electronegative than B (EN 2.0) and all of the dipoles will be directed from B to F. Because all are at the same angle and of the same magnitude, the molecule is nonpolar. 1200 (c) COS is linear. C and S have the same EN (2.0) but the C=O bond is quite polar(DEN) so the molecule is polar overall.

  46. both C are sp3 hybridized s-sp3 overlaps to s bonds sp3-sp3 overlap to form a s bond relatively even distribution of electron density over all s bonds The s bonds in ethane(C2H6). Figure 11.9

  47. overlap in one position - s p overlap -  electron density The s and p bonds in ethylene (C2H4). Figure 11.10

  48. overlap in one position - s p overlap -  The s and p bonds in acetylene (C2H2). Figure 11.11

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