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Chapter 9

Chapter 9. Models of Chemical Bonding. Models of Chemical Bonding. 9.1 Atomic Properties and Chemical Bonds. 9.2 The Ionic Bonding Model. 9.3 The Covalent Bonding Model. 9.4 Bond Energy and Chemical Change. 9.5 Between the Extremes: Electronegativity and Bond Polarity.

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Chapter 9

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  1. Chapter 9 Models of Chemical Bonding

  2. Models of Chemical Bonding 9.1 Atomic Properties and Chemical Bonds 9.2 The Ionic Bonding Model 9.3 The Covalent Bonding Model 9.4 Bond Energy and Chemical Change 9.5 Between the Extremes: Electronegativity and Bond Polarity

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

  4. 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

  5. The three models of chemical bonding. Figure 9.2

  6. 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. Pair the dots (electrons) until all of the valence electrons are used. . . . : . . . : 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.

  7. Figure 9.4 Lewis electron-dot symbols for elements in Periods 2 and 3.

  8. PROBLEM: Use partial orbital diagrams and Lewis symbols to depict the formation of Na+ and O2- ions from the atoms, and determine the formula of the compound. O2- Na Na 2s 2p 3s 3p O 2s 2p . . Na : 3s 3p + O : : 2Na+ + O 2- : : . : . Na SAMPLE PROBLEM 9.1 Depicting Ion Formation PLAN: Draw orbital diagrams for the atoms and then move electrons to make filled outer levels. It can be seen that 2 sodiums are needed for each oxygen. SOLUTION: 2 Na+

  9. + F- 1s22s22p6 Li+ Li 1s 2s 2p 1s 2s 2p F- + + F 1s 2s 2p 1s 2s 2p . : . Li : F : Li+ + : F - : + : : Figure 9.5 Three ways to represent the formation of Li+ and F- through electron transfer. Electron configurations Li 1s22s1 + F 1s22s22p5 Li+ 1s2 Orbital diagrams Lewis electron-dot symbols

  10. 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

  11. Figure 9.7 Trends in lattice energy.

  12. Figure 9.8 Electrostatic forces and the reason ionic compounds crack.

  13. Solid ionic compound Molten ionic compound Ionic compound dissolved in water Electrical conductance and ion mobility. Figure 9.9

  14. Table 9.1 Melting and Boiling Points of Some Ionic Compounds Compound mp (0C) bp (0C) CsBr 636 1300 NaI 661 1304 MgCl2 714 1412 KBr 734 1435 CaCl2 782 >1600 NaCl 801 1413 LiF 845 1676 KF 858 1505 MgO 2852 3600

  15. Covalent bond formation in H2. Figure 9.10

  16. Distribution of electron density of H2. Figure 9.11

  17. Table 9.2 back to previous slide

  18. Internuclear distance (bond length) Covalent radius Internuclear distance (bond length) Covalent radius 72 pm 114 pm Internuclear distance (bond length) Covalent radius Internuclear distance (bond length) Covalent radius 100 pm 133 pm Bond length and covalent radius. Figure 9.12

  19. Table 9.3

  20. 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

  21. Strong covalent bonding forces within molecules Weak intermolecular forces between molecules Figure 9.13 Strong forces within molecules and weak forces between them.

  22. Covalent bonds of network covalent solids. Figure 9.14

  23. Figure 9.15 The infrared (IR) spectra of diethyl ether and 2-butanol.

  24. 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

  25. 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

  26. 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

  27. 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

  28. Table 9.4

  29. Figure 9.20 The Pauling electronegativity (EN) scale.

  30. 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

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

  32. 3.0 DEN 2.0 0.0 Figure 9.22 Boundary ranges for classifying ionic character of chemical bonds.

  33. Figure 9.23 Properties of the Period 3 chlorides.

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