1. 1 Take out CLICKERS: “GO 41 GO” -or- “CH 41 CH” NEXT LECTURE:Monday, Nov. 9; Bring CLICKERS READ BLBM, Chapters 8 and 9 HOMEWORK #12:Ch. 8: #; Ch. 9: # (Due in Recitation Thursday, November 12) HOUR EXAM #4:Monday, November 16, 6:30-7:30 PM OPTIONAL EXAM:Monday, December 7, 6:30-7:30 PM (Multiple Choice) There are 4 exams; ONLY one can replace your lowest exam score… You must have taken Exams 1-4 or have valid excuse for missing one… FINAL EXAM:Monday, December 14, 7-9 PM Chemistry 177 November 6, 2009

2. 2 (a)Create a molecular skeleton… Rule of thumb – most EN elements to outside (few bonds), e.g., F, Cl, O, and H; less EN elements to inside (more bonds), e.g. B, C, N, Si, P. (b) Count total number of valence electron pairs, including net charge. # valence electron pairs = # valence electrons/2; (Bond or Lone pairs) (c) Draw a single bond pair (  -bond) between each pair of connected atoms. (d) Complete the octet around the outside (“terminal”) atoms first using lone pairs, then central atoms (SEE Rule (e) below…) Utilize multiple bonds if central atoms do not achieve octet; Check for resonance, if there are several possibilities… (e) Calculate formal charges at each element Sum of formal charges = total charge on molecule. Molecular Geometry: Lewis Structures (Rules) Chapter 8: Basic Concepts of Chemical Bonding Best Choice: All formal charges close to 0; and consistent with ENs… There are EXCEPTIONS!

3. 3 (1) Count the number of valence electrons and electron pairs (Bond and Lone Pairs) (2) Draw a Lewis Structure (consider resonance structures, if needed…) (3) Identify regions of electron density around the central atoms (4) Calculate formal charges at each atom (5) Identify the shape of the molecule (VSEPR) (6) Classify bond types using electronegativity differences (  EN) (7) Classify the molecule as polar or nonpolar (need shape) (8) Identify the numbers of  and  bonds and their locations (9) Identify the bond orders for each bond (10) Determine the oxidation states of each atom Top 10 List of Fun Things to do with a Molecular Formula… Chapter 8: Basic Concepts of Chemical Bonding

4. 4 Nitrate Ion: (NO 3 )  5 + 3(6) + 1 = 24 e  (12 e  pairs) 8 LONE PAIRS Formal Charges: N: 5  (4  1) = +1 O: 6  (6 + 1  1) =  1 (2x) O: 6  (4 + 2  1) = 0 N O OO  N O OO  N O OO 4 BOND PAIRS  Molecular Geometry: Lewis Structures (Resonance) Chapter 8: Basic Concepts of Chemical Bonding

5. 5 Aromatic Compounds: Benzene (C 6 H 6 ) 6(4) + 6(1) = 30 e  (15 e  pairs) Molecular Geometry: Lewis Structures (Resonance) Chapter 8: Basic Concepts of Chemical Bonding

6. 6 Bond TypeBond Order Bond Distance (pm) Bond TypeBond Order Bond Distance (pm) 154147 134124 120110 143 138136 116122 143(Nitrate)121 123149 113121 Molecular Geometry: Bond Distances / Bond Order Chapter 8: Basic Concepts of Chemical Bonding

7. 7 Boron Trifluoride: BF 3 Molecular Geometry: Lewis Structures (Violating the Octet Rule) Chapter 8: Basic Concepts of Chemical Bonding

8. 8 Phosphorus Pentafluoride: PF 5 Molecular Geometry: Lewis Structures (Violating the Octet Rule) Chapter 8: Basic Concepts of Chemical Bonding

9. 9 Nitrogen Dioxide: NO 2 Molecular Geometry: Lewis Structures (Violating the Octet Rule) Chapter 8: Basic Concepts of Chemical Bonding

10. 10 Molecular Geometry: Lewis Structures – Formal Charges Isocyanate Ion: (OCN)  Chapter 8: Basic Concepts of Chemical Bonding

11. 11  H 0 for reactions involving molecular species can be well estimated using average bond enthalpies for chemical bonds in the molecules… Bond Enthalpy = Cl 2 (g)  2Cl  (g);  H 0 = D(Cl  Cl) = 242 kJ/mol (bonds) HCl(g)  H  (g) + Cl  (g);  H 0 = D(H  Cl) = 431 kJ/mol CH 4 (g)  H  (g) +  CH 3 (g);  H 0 = D(H  CH 3 ) = 427 kJ/mol CH 4 (g)  C(g) + 4 H  (g);  H 0 = 1660 kJ/mol Bond Enthalpies = Bond Dissociation Energies Chapter 8: Basic Concepts of Chemical Bonding

12. 12 REACTANTS  PRODUCTS H0H0 REACTANTS  GASEOUS ATOMS  PRODUCTS   Chapter 8: Basic Concepts of Chemical Bonding Bond Enthalpies N 2 (g) + 3 H 2 (g)  2 NH 3 (g)

13. 13 2 C 2 H 2 (g) + 5 O 2 (g)  4 CO 2 (g) + 2 H 2 O(g)  H 0 =  2511 kJ Example: Use average bond enthalpies to estimate the enthalpy of combustion of acetylene gas, C 2 H 2 (g). Chapter 8: Basic Concepts of Chemical Bonding Bond Enthalpies