IIIIII IV. Lewis Diagrams (p. 170 – 175) Unit 6 – Molecular Structure.
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Transcript of IIIIII IV. Lewis Diagrams (p. 170 – 175) Unit 6 – Molecular Structure.
![Page 1: IIIIII IV. Lewis Diagrams (p. 170 – 175) Unit 6 – Molecular Structure.](https://reader037.fdocuments.us/reader037/viewer/2022110115/55142e61550346d8488b5e60/html5/thumbnails/1.jpg)
I II III
IV. Lewis Diagrams
(p. 170 – 175)
Unit 6 – Molecular Structure
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A. Octet Rule
Remember… Most atoms form bonds in order to
have 8 valence electrons.
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Hydrogen 2 valence e-
Groups 1,2,3 get 2,4,6 valence e-
Expanded octet more than 8
valence e- (e.g. S, P, Xe)
Radicals odd # of valence e-
Exceptions:
A. Octet Rule
F B FFH O HN O
Very unstable!!
F F
F S F
F F
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B. Drawing Lewis Diagrams Find total # of valence e-.
Arrange atoms - singular atom is usually in the middle.
Form bonds between atoms (2 e-).
Distribute remaining e- to give each atom an octet (recall exceptions).
If there aren’t enough e- to go around, form double or triple bonds.
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B. Drawing Lewis Diagrams CF41 C × 4e- = 4e-
4 F × 7e- = 28e-
32e- FF C F
F
- 8e-
24e-
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B. Drawing Lewis Diagrams BeCl21 Be × 2e- = 2e-
2 Cl × 7e- = 14e-
16e-
Cl Be Cl - 4e-
12e-
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B. Drawing Lewis Diagrams CO2
1 C × 4e- = 4e-
2 O × 6e- = 12e-
16e-
O C O - 4e-
12e-
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C. Polyatomic Ions
To find total # of valence e-:
Add 1e- for each negative charge.
Subtract 1e- for each positive charge. Place brackets around the ion and label the
charge.
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C. Polyatomic Ions
ClO4-
1 Cl × 7e- = 7e-
4 O × 6e- = 24e-
31e- OO Cl O
O
+ 1e-
32e-
- 8e-
24e-
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NH4+
1 N × 5e- = 5e-
4 H × 1e- = 4e-
9e- HH N H
H
- 1e-
8e-
- 8e-
0e-
C. Polyatomic Ions
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D. Resonance Structures
Molecules that can’t be correctly represented by a single Lewis diagram.
Actual structure is an average of all the possibilities.
Show possible structures separated by a double-headed arrow.
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D. Resonance Structures
OO S O
OO S O
OO S O
SO3