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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 Between the Extremes: Electronegativity and Bond Polarity. 9.5 An Introduction to Metallic Bonding.

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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 Between the Extremes: Electronegativity and Bond Polarity 9.5 An Introduction to Metallic Bonding

  3. Figure 9.1 A general comparison of metals and non-metals

  4. Types of Chemical Bonding 1. Metal with non-metal electron transfer and ionic bonding 2. Non-metal with non-metal electron sharing and covalent bonding (localized) 3. Metal with metal electron pooling and metallic bonding (delocalized)

  5. The three models of chemical bonding Figure 9.2

  6. The 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. Lewis electron-dot symbols for elements in Periods 2 and 3 Figure 9.3

  8. General Rules For a metal, the total number of dots equals the maximum number of electrons it loses to form a cation. For a non-metal, the number of unpaired dots equals the number of electrons that become paired either through electron gain or electron sharing. The number of unpaired dots equals either the negative charge of the anion an atom forms or the number of covalent bonds it forms.

  9. The Ionic Bonding Model Involves the transfer of electrons from metal to non-metal to form ions that come together in a solid ionic compound The Octet Rule When atoms bond, they lose, gain or share electrons to attain a filled outer shell of eight (or two) electrons In ionic bonding, the total number of electrons lost by the metal atoms equals the total number of electrons gained by the non-metal atoms.

  10. 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 two sodiums are needed for each oxygen. SOLUTION: 2 Na+

  11. + F- 1s22s22p6 Li+ Li 1s 2s 2p 1s 2s 2p F- + + F 1s 2s 2p 1s 2s 2p . : . Li : F : Li+ + : F - : + : : Figure 9.4 Three ways to represent the formation of Li+ and F- through electron transfer 1. Electron configurations Li 1s22s1 + F 1s22s22p5 Li+ 1s2 2. Orbital diagrams 3. Lewis electron-dot symbols

  12. Ionic Bonding and Lattice Energy The electron transfer process is an endothermic process, but ionic compound formation is an exothermic process. Li(g) Li+ + e- IE1 = 520 kJ F(g) + e- F-(g) EA = -328 kJ Li(g) + F(g) Li+(g) + F-(g) IE1 + EA = 192 kJ But ∆Hfo for solid LiF = -617 kJ/mol! Li+(g) + F-(g) LiF(g) ∆Ho = -755 kJ (an exothermic process due to the attraction of oppositely charged ions)

  13. Even more energy is released when the gaseous ions coalesce into a crystalline solid. Thus…. Li+(g) + F-(g) LiF(s) ∆Holattice of LiF = lattice energy = -1050 kJ The lattice energy is the enthalpy change that occurs when gaseous ions coalesce into an ionic solid. How do we measure lattice energy experimentally? Use Hess’s law in a Born-Haber cycle

  14. The Born-Haber cycle for lithium fluoride Figure 9.6

  15. Working the Numbers STEP 1: Enthalpy of Li atomization = 161 kJ STEP 2: 1/2 the bond energy of F2(g)= 0.5(159 kJ) = 79.5 kJ STEP 3: IE1 for Li(g) = 520 kJ STEP 4: EA of F(g) = -328 kJ The enthalpy change for the overall process, ∆Hfo, = -617 kJ Only the lattice energy is unknown, and it is equal to the enthalpy change of the overall process minus the sum of the above four steps = -1050 kJ

  16. Central Point Ionic solids exist only because the lattice energy drives the energetically unfavorable electron transfer.

  17. Periodic Trends in Lattice Energy Coulomb’s Law charge A x charge B electrostatic force a distance2 But energy = force x distance. Therefore, charge A x charge B electrostatic energy a distance cation charge x anion charge a DHolattice electrostatic energy a cation radius + anion radius

  18. Trends in lattice energy Figure 9.7

  19. Effect of Ionic Charge on Lattice Energy Compare LiF and MgO: Li+ and Mg2+ have similar radii, and F- and O2- have similar radii. ∆Holattice (LiF) = -1050 kJ/mol ∆Holattice (MgO) = -3923 kJ/mol The nearly four-fold larger value for MgO reflects the difference in the product of the charges (12vs 22) in the numerator of the electrostatic energy equation (monovalent vs divalent ions).

  20. Does the ionic model explain the properties of ionic compounds?

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

  22. Solid ionic compound Molten ionic compound Ionic compound dissolved in water Electrical Conductance and Ion Mobility Figure 9.9

  23. Table 9.1 Melting and Boiling Points of Some Ionic Compounds Compound mp (oC) bp (oC) 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

  24. Vaporizing an ionic compound Figure 9.10

  25. The Covalent Bonding Model Each atom in a covalent bond “counts” the shared electrons as belonging entirely to itself. An electron pair that is part of an atom’s valence shell but not involved in bonding is called a lone pair, or unshared pair. Bond order: the number of electron pairs being shared between any two bonded atoms single bond (H2) - bond order of 1 double bond (H2C=CH2) - bond order of 2 triple bond (N2) - bond order of 3

  26. Covalent bond formation in H2 Figure 9.11

  27. The attractive and repulsive forces in covalent bonding Figure 9.12

  28. Properties of Covalent Bonds Bond energy (bond enthalpy or bond strength): the energy required to overcome the mutual attraction between the bonded nuclei and the shared electrons. Bond breakage is an endothermic process; bond energy is always positive. Bond formation is an exothermic process.

  29. Bond Length For a given pair of atoms, a higher bond order results in a shorter bond length and a higher bond energy. A shorter bond is a stronger bond.

  30. internuclear distance (bond length) covalent radius internuclear distance (bond length) covalent radius internuclear distance (bond length) covalent radius internuclear distance (bond length) covalent radius Bond length and covalent radius Figure 9.13

  31. (b) C = O, C - O, C O 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 PROBLEM: Using the periodic table, rank the bonds in each set in order of decreasing bond length and bond strength: (a) S - F, S - Br, S - Cl PLAN: (a) Bond order =1 for all and sulfur is bonded to halogens; bond length should increase and bond strength should decrease with increasing atomic radius. (b) Similar atoms (C) are bonded but bond order changes; bond length decreases as bond order increases, and bond strength increases as bond order increases. SOLUTION: (a) Atomic size increases moving 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

  32. Properties of Covalent Compounds Weak forces between molecules, not the strong covalent bonds within each molecule, are responsible for the physical properties of covalent compounds. Covalent compounds have relatively low melting and boiling points. Most covalent compounds are poor electrical conductors.

  33. Strong covalent bonding forces within molecules Weak intermolecular forces between molecules Strong forces within molecules, weak forces between them Figure 9.14

  34. Network Covalent Solids No separate molecules; held together by covalent bonds that extend throughout the sample quartz: melts at 1550 oC. diamond: melts at 3550 oC. These examples illustrate the strength of covalent bonds.

  35. Covalent bonds of network covalent solids Figure 9.15

  36. The Concept of Electronegativity (EN) Defined as the relative ability of a bonded atom to attract shared electrons (not the same as EA) Bond energy of H2 = 432 kJ/mol Bond energy of F2 = 159 kJ/mol Bond energy of HF = 565 kJ/mol, not 296 kJ/mol The stronger-than-expected HF bond is due to unequal sharing of electrons, with F bearing a partial negative charge and H bearing a partial positive charge. The attraction between the partial charges strengthens the bond.

  37. The Pauling electronegativity (EN) scale Figure 9.16

  38. Trends in Electronegativity In general, electronegativity is inversely related to atomic size. For main-group elements, EN generally increases up a group and across a period. Non-metals are more electronegative than metals. The least electronegative (most electropositive) non- radioactive element is Cs (lower left-hand corner of the Periodic Table).

  39. Electronegativity and atomic size Figure 9.17

  40. Electronegativity and Oxidation Number (a) The more electronegative atom in a bond is assigned all of the shared electrons; the less electronegative atom is assigned none of the shared electrons. (b) Each atom in a bond is assigned all of its unshared electrons. (c) The oxidation number is given by: O.N. = # valence e- - (# shared e- + # unshared e-) e.g.: HCl: Cl more electronegative than H; has 7 valence electrons; has an O.N. of 7 - 8 = -1 H has 1 valence electron; has an O.N. of 1 - 0 = +1

  41. Polar Covalent Bonds and Bond Polarity Covalent bonds involving atoms with different electronegativities: generate partial (+) and (-) charges; defined as polar covalent bonds (e.g., HCl) Polar covalent bonds: depicted by a polar arrow ( ) that points toward the negative pole H2 and F2: examples of nonpolar covalent bonds

  42. SAMPLE PROBLEM 9.3 Determining Bond Polarity from EN Values 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. PLAN: (a) Use Figure 9.16 to find EN values; the arrow should point toward the negative end. (b) EN 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

  43. Partial Ionic Character of Polar Covalent Bonds Related directly to the electronegativity difference (∆EN) between the bonded atoms The greater the ∆EN, the larger the partial charges and the higher the partial ionic character (PIC). Thus LiF has more PIC than HF; HF has more PIC than F2.

  44. 3.0 2.0 DEN 0.0 Boundary ranges for classifying the ionic character of chemical bonds Figure 9.18

  45. Percent ionic character as a function of electronegativity difference (DEN) Figure 9.19

  46. Charge density of the LiF molecule (an ionic compound) Li F No bond has 100% ionic character; electron sharing occurs to some extent Figure 9.20

  47. Ionic-To-Covalent Bonding Continuum Across a Period Consider bonding between a metal and non-metal in Period 3 NaCl, MgCl2, AlCl3, SiCl4, PCl3, S2Cl2, and Cl2 Increasing covalent character (decreasing ionic character) from NaCl to Cl2 Underlying factor: As ∆EN becomes smaller, the bond becomes more covalent.

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