1 / 42

Chapter 9

Chapter 9. Models of Chemical Bonding. A general comparison of metals and nonmetals. Figure 9.1. Astatine may be a metalloid but it is radioactive. BASIC CHEMICAL BONDING. Atoms form bonds in order to complete their outer shells of electrons

azure
Download Presentation

Chapter 9

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Chapter 9 Models of Chemical Bonding

  2. A general comparison of metals and nonmetals. Figure 9.1 Astatine may be a metalloid but it is radioactive.

  3. BASIC CHEMICAL BONDING Atoms form bonds in order to complete their outer shells of electrons Main group elements try for 8. (Transition metals are more complex.) Atoms can give up or gain electrons to form ions which then form ionic bonds, or they can share electrons to form covalent bonds (Or in metals, share electrons throughout all atoms)

  4. BASIC CHEMICAL BONDING Atoms/ions seek lower energy states: - ions in a bond have a lower energy than the separated ions Data indicate lowest energy level is for atoms to share electrons at a certain distance = bond length = covalent radii

  5. 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. Cation with anion:ionic bonding(ammonium ion and nitrate ion)

  6. The three models of chemical bonding. Figure 9.2

  7. “NEW” CONCEPT OF CHEMICAL BONDING: LEWIS THEORY Valence electrons have fundamental role in chemical bonding If e-(s) "transferred" ionic bond results If e-(s) "shared" covalent bond results If e-(s) “shared as a pool” metallic bond results Trying to achieve noble gas configuration Lewis symbol: Atomic symbol with valence electrons for neutral atom, correct number of electrons for ion

  8. Figure 2.14 from 4th ed. The relationship between ions formed and the nearest noble gas. (from Chp 2)

  9. The group number digit 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 15 and therefore has 5 valence electrons.

  10. Figure 9.4 Lewis electron-dot symbols for elements in Periods 2 and 3. How will these Lewis symbols look when these atoms form ions?

  11. Lewis Symbols for Respresentative Elements

  12. Lewis Structures For Ions: Put individual ion structures together as group or use coefficient: ...... CaF2 = [:F:]-[Ca]2+[:F:]- or [Ca]2+2[:F:]- ...... Note that Ca:  [Ca]2+ + 2 e- each F + e- --> [F]-

  13. PROBLEM: Use Lewis symbols to depict the formation of Na+ and O2- ions from the atoms, and determine the formula of the compound. Draw in the most correct way when I show you below. Ions should have brackets. . . Na : + O : : 2 [Na]++[ O ]2- : : . : . Na SAMPLE PROBLEM 9.1 Depicting Ion Formation SOLUTION:

  14. Figure 9.5 + [F]- 1s22s22p6 [Li[+ Li 1s 2s 2p 1s 2s 2p [F]- + + F 1s 2s 2p 1s 2s 2p . : . Li : F : [Li]+ + : F - : + : : 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: focus on this way [They should both have brackets.]

  15. Lewis Structures For Ions: Practice drawing Lewis structure for ions and then use them to make ionic compounds of the two elements listed. Be prepared to show your work on the document camera for the class to see. Draw large structures in ink and use brackets where required! Mg and O, Al and Cl, Na and S

  16. LATTICE ENERGY When ionic bonding occurs, energy is released – exothermic – because the positive and negative ions achieve their lowest energy state when surrounded by the opposite charge. However, IE is endothermic and usually larger than EA, so the transfer of electrons is endothermic. In chp 6 we will learn about Hess’ Law, which is how we actually determine the value of lattice energy.

  17. LATTICE ENERGY For now: the energy released when forming an ionic compound arranged in a crystal lattice from the elements is called enthalpy of formation. If we know the value of IE and EA, then we can determine the energy released when the ions are moved into the crystal lattice. Lattice energy is affected by: Ion size: the larger the radius, the lower the lattice energy Ionic charge(s): the greater the charge, the higher the lattice energy

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

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

  20. 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

  21. Covalent Bonding A shared pair of e-s between two atoms Other e-s in valence shell become Lone Pairs Can have single, double or triple bonds (Bond Order)

  22. Covalent Bonding Energy is released when atoms join to form bonds Energy must be absorbed to break bonds - called bond dissociation energy Measured for gaseous species in kJ/mol Increases with multiple bonding Increases with decreased bond length See Bond Energy and Bond Length tables

  23. Formation of a covalent bond between two H atoms. (from Chp 2) Figure 2.13 Covalent bonds form when elements share electrons, which usually occurs between nonmetals.

  24. BOND ENERGIES & DHrxn: DHrxnapprox same as difference between bond breaking energy and bond formation energy REMEMBER THIS IS AN APPROXIMATION BECAUSE WE ARE USING AVERAGES! Estimate DHrxn for N2 + 3 H2 2 NH3 First find kinds of bonds there are Look up average bond energies in table

  25. Draw Lewis structure, look up bond energies, count number of bonds, DHrxn ~ BEprod – BEreact

  26. 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: (NO CALCS, JUST THINK!) (a) S - F, S - Br, S - Cl (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 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

  27. Section 9.4 We can’t really calculate bond energies until we can draw Lewis electron dot structures, which is in chp 10, and learn about enthalpy of reaction, which is in chp 6. We will return to this after chp 6.

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

  29. Covalent bonds of network covalent solids. Figure 9.14

  30. CONCEPT OF ELECTRONEGATIVITY: from Linus Pauling I define it as greediness for electrons in a covalent bond Look at table – a scale with no units Metals are low and nonmetals are high Difference in their electronegativities: If DEN is < .5, bond is covalent (0 to 0.4) If > 2.0, it's ionic, and in between is polar covalent (0.5 to 1.9) Determine something called "% ionic character" using the table in text

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

  32. 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 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

  33. 3.0 DEN 2.0 0.0 Figure 9.21 A. Boundary ranges for classifying ionic character of chemical bonds. B. Gradation in ionic character. > 2.0 Ionic 0.5 to 1.9 Polar covalent 0 to 0.4 Nonpolar covalent Use my simplified version!

  34. Figure 9.21 continued Percent ionic character of electronegativity difference (DEN).

  35. Figure 9.22 Properties of the Period 3 chlorides.

  36. Metallic Bonding (not in text but YOU have to know it) Sea of electrons shared throughout the metal piece, however large it is Explains both electrical and thermal conductivity, since electrons can move throughout the piece A mixture of metals, an alloy, also share electrons throughout Metallic bond strength varies, as you can see from range of melting points, but most metals are solids (only Hg is liquid at room temperature)

  37. Lithium (Li) 180 1347 Tin (Sn) 232 2623 Aluminum (Al) 660 2467 Barium (Ba) 727 1850 Silver (Ag) 961 2155 Copper (Cu) 1083 2570 Uranium (U) 1130 3930 Table 9.7 Melting and Boiling Points of Some Metals Element mp(0C) bp(0C)

  38. Figure 9.26 Melting points of the Group 1A(1) and Group 2A(2) elements.

  39. The reason metals deform. Figure 9.27 metal is deformed

More Related