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Chemistry 177 November 6, 2009. 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
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Chemistry 177 November 6, 2009 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
Chapter 8: Basic Concepts of Chemical Bonding Molecular Geometry:Lewis Structures (Rules) • 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…) • Utilizemultiple 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. There are EXCEPTIONS! Best Choice: All formal charges close to 0; and consistent with ENs…
Chapter 8: Basic Concepts of Chemical Bonding Top 10 List of Fun Things to do with a Molecular Formula… • Count the number of valence electrons and electron pairs (Bond and Lone Pairs) • Draw a Lewis Structure (consider resonance structures, if needed…) • Identify regions of electron density around the central atoms • Calculate formal charges at each atom • Identify the shape of the molecule (VSEPR) • Classify bond types using electronegativity differences (EN) • Classify the molecule as polar or nonpolar (need shape) • Identify the numbers of s and p bonds and their locations • Identify the bond orders for each bond • Determine the oxidation states of each atom
O O 4 BOND PAIRS N N O O O O - - - Chapter 8: Basic Concepts of Chemical Bonding Molecular Geometry:Lewis Structures (Resonance) Nitrate Ion: (NO3) O N O O 8 LONE PAIRS 5 + 3(6) + 1 = 24 e (12 e pairs) Formal Charges: N: 5 (41) = +1 O: 6 (6 + 11) = 1 (2x) O: 6 (4 + 21) = 0
Chapter 8: Basic Concepts of Chemical Bonding Molecular Geometry:Lewis Structures (Resonance) Aromatic Compounds: Benzene (C6H6) 6(4) + 6(1) = 30 e (15 e pairs)
Chapter 8: Basic Concepts of Chemical Bonding Molecular Geometry: Bond Distances / Bond Order
Chapter 8: Basic Concepts of Chemical Bonding Molecular Geometry:Lewis Structures (Violating the Octet Rule) Boron Trifluoride: BF3
Chapter 8: Basic Concepts of Chemical Bonding Molecular Geometry:Lewis Structures (Violating the Octet Rule) Phosphorus Pentafluoride: PF5
Chapter 8: Basic Concepts of Chemical Bonding Molecular Geometry:Lewis Structures (Violating the Octet Rule) Nitrogen Dioxide: NO2
Chapter 8: Basic Concepts of Chemical Bonding Molecular Geometry: Lewis Structures – Formal Charges Isocyanate Ion: (OCN)
Chapter 8: Basic Concepts of Chemical Bonding Bond Enthalpies = Bond Dissociation Energies DH0 for reactions involving molecular species can be well estimated using average bond enthalpies for chemical bonds in the molecules… Bond Enthalpy = Cl2(g) 2Cl(g); DH0 = D(ClCl) = 242 kJ/mol (bonds) HCl(g) H(g) + Cl(g); DH0 = D(HCl) = 431 kJ/mol CH4(g) H(g) + CH3(g); DH0 = D(HCH3) = 427 kJ/mol CH4(g) C(g) + 4 H(g); DH0 = 1660 kJ/mol
Chapter 8: Basic Concepts of Chemical Bonding Bond Enthalpies DH0 REACTANTS PRODUCTS N2(g) + 3 H2(g) 2 NH3(g) (1) (2) REACTANTS GASEOUS ATOMS PRODUCTS
Chapter 8: Basic Concepts of Chemical Bonding Bond Enthalpies Example: Use average bond enthalpies to estimate the enthalpy of combustion of acetylene gas, C2H2(g). 2 C2H2(g) + 5 O2(g) 4 CO2(g) + 2 H2O(g) DH0= 2511 kJ