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Take out CLICKERS: “GO 41 GO” -or- “CH 41 CH”

NEXT LECTURE: Monday, Nov. 9; Bring CLICKERSREAD 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 177November 6, 2009

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(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 areEXCEPTIONS!

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(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

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Nitrate Ion: (NO3)

5 + 3(6) + 1 = 24 e

(12 e pairs)

8 LONEPAIRS

Formal Charges:N: 5 (41) = +1O: 6 (6 + 11) = 1 (2x)O: 6 (4 + 21) = 0

N

O

OO

N

O

OO

N

O

OO

4 BONDPAIRS

Molecular Geometry: Lewis Structures (Resonance)

Chapter 8: Basic Concepts of Chemical Bonding

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Aromatic Compounds: Benzene (C6H6)

C

C

C

C

C

C

H

H

H

H

H

HC

C

C

C

C

C

H

H

H

H

H

HC

C

C

C

C

C

H

H

H

H

H

H

6(4) + 6(1) = 30 e

(15 e pairs)

Molecular Geometry: Lewis Structures (Resonance)

Chapter 8: Basic Concepts of Chemical Bonding

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Bond Type Bond Order

Bond Distance (pm)

Bond Type Bond Order

Bond Distance

(pm)

154 147

134 124

120 110

143

138 136

116 122

143 (Nitrate) 121

123 149

113 121

Molecular Geometry: Bond Distances / Bond Order

C C

C C

C C

C N

C N

C N

C O

C O

C O

N N

N N

N O

N O

O O

O O

N N

Chapter 8: Basic Concepts of Chemical Bonding

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Boron Trifluoride: BF3

Molecular Geometry: Lewis Structures (Violating the Octet Rule)

Chapter 8: Basic Concepts of Chemical Bonding

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Phosphorus Pentafluoride: PF5

Molecular Geometry: Lewis Structures (Violating the Octet Rule)

Chapter 8: Basic Concepts of Chemical Bonding

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Nitrogen Dioxide: NO2

Molecular Geometry: Lewis Structures (Violating the Octet Rule)

Chapter 8: Basic Concepts of Chemical Bonding

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Molecular Geometry: Lewis Structures – Formal Charges

Isocyanate Ion: (OCN)

Chapter 8: Basic Concepts of Chemical Bonding

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H0 for reactions involving molecular species can be well estimated usingaverage bond enthalpies for chemical bonds in the molecules…

Bond Enthalpy =

Cl2(g) 2Cl(g); H0 = D(ClCl) = 242 kJ/mol (bonds)

HCl(g) H(g) + Cl(g); H0 = D(HCl) = 431 kJ/mol

CH4(g) H(g) + CH3(g); H0 = D(HCH3) = 427 kJ/mol

CH4(g) C(g) + 4 H(g); H0 = 1660 kJ/mol

Bond Enthalpies = Bond Dissociation Energies

Chapter 8: Basic Concepts of Chemical Bonding

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REACTANTS PRODUCTSH0

REACTANTS GASEOUS ATOMS PRODUCTS

0 0 0(1) (2) Sum of in Reactants Sum of in ProductsH H H D D

Chapter 8: Basic Concepts of Chemical Bonding

Bond Enthalpies

N2(g) + 3 H2(g) 2 NH3(g)

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2 C2H2(g) + 5 O2(g) 4 CO2(g) + 2 H2O(g) H0 = 2511 kJ

Example: Use average bond enthalpies to estimate the enthalpy of combustionof acetylene gas, C2H2(g).

Chapter 8: Basic Concepts of Chemical Bonding

Bond Enthalpies