1 Take out CLICKERS: “GO 41 GO” -or- “CH 41 CH” NEXT LECTURE:Monday, Nov. 9; Bring CLICKERS...
-
Upload
adele-nichols -
Category
Documents
-
view
212 -
download
0
Transcript of 1 Take out CLICKERS: “GO 41 GO” -or- “CH 41 CH” NEXT LECTURE:Monday, Nov. 9; Bring CLICKERS...
1
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
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 areEXCEPTIONS!
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
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
5
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
6
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
7
Boron Trifluoride: BF3
Molecular Geometry: Lewis Structures (Violating the Octet Rule)
Chapter 8: Basic Concepts of Chemical Bonding
8
Phosphorus Pentafluoride: PF5
Molecular Geometry: Lewis Structures (Violating the Octet Rule)
Chapter 8: Basic Concepts of Chemical Bonding
9
Nitrogen Dioxide: NO2
Molecular Geometry: Lewis Structures (Violating the Octet Rule)
Chapter 8: Basic Concepts of Chemical Bonding
10
Molecular Geometry: Lewis Structures – Formal Charges
Isocyanate Ion: (OCN)
Chapter 8: Basic Concepts of Chemical Bonding
11
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
12
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)
13
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