Transcript Chapter 10 - Aureus Student Portal

Petrucci • Harwood • Herring • Madura
Ninth
Edition
GENERAL
CHEMISTRY
Principles and Modern Applications
Chapter 10: Chemical Bonding I:
Basic Concepts
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General Chemistry: Chapter 10
Slide 1 of 50
10-1 Lewis Theory: An Overview
• Valence e- play a fundamental role in
chemical bonding.
• e- transfer leads to
ionic bonds (metal-non metals)
• Sharing of e- leads to covalent bonds
(two non-metals)
• e- are transferred of shared to give
each atom a noble gas configuration
– the octet.
Slide 2 of 50
General Chemistry: Chapter 10
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Lewis Symbols
• A chemical symbol represents the nucleus
and the core e- (inner shell)
• Dots around the symbol represent valence e.
•
• Si •
•
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I •
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Ar
••
••
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• Se
•
•
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Bi
•
•
•
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Sb
•
•
•
••
• Al •
•
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As
•
•
•
••
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P
• •
•
••
N
• •
•
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Lewis Structure
(Dot and cross diagram)
• A lewis structure is a combination of Lewis
symbols that represents either the transfer or
the sharing of electrons in a chemical bond.
EXAMPLE 10-2
Writing Lewis Structures of Ionic Compounds. Write Lewis structures for the
following compounds: (a) BaO; (b) MgCl2 ; (c) aluminum oxide.
••
O
• •
••
Ba
2+
•• 2O
••
••
•
Ba •
••
BaO
Note the use of the “fishhook” arrow to denote a single electron
movement. A “double headed” arrow means that two electrons
move.
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10-2 Covalent Bonding: An
Introduction
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General Chemistry: Chapter 10
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Coordinate Covalent Bonds
A covalent bond in which a single atom contributes both
of the electrons to a shared pair.
+
H
Cl
H
H
N
H
•• Cl
••
••
H
••
N
••
H
H
H
Note
 the “double headed” arrow showing that two electrons
move.
Any atom with electron pairs (non-metals: O, S, P) can form
coordinate covalent bonds.
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General Chemistry: Chapter 10
Slide 7 of 50
Multiple Covalent Bonds
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C
•
•
O
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•
C
•
C
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O
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••
••
•
O
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••
••
•
O
••
•
O
•
••
••
O
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General Chemistry: Chapter 10
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O
••
••
•
C
• •
•
••
•
O•
••
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Multiple Covalent Bonds
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N
N
General Chemistry: Chapter 10
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•
N
•
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•
N
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N
•
•
N
•
••
••
••
•
•N
•
••
•
N•
•
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Paramagnetism of Oxygen
Failure of the Lewis Structure Model
Explain by:
Molecular Orbital Theory
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General Chemistry: Chapter 10
Slide 10 of 50
Practice Examples
10-3 Polar Covalent Bonds and
Electrostatic Potential Maps
A covalent bond in which electrons are not shared equally between two
atoms
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Electronegativity (EN)
It is an atoms ability to compete for electrons with other atoms to which it is
bonded.
The lower the EN, the more metallic the element is
The higher the EN, the more nonmetallic the element is
+
A
(less electro n eg ativ e)
B
(m o re electro n eg ativ e)
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General Chemistry: Chapter 10
Slide 13 of 50
Percent Ionic Character
Electronegativity difference, DEN =
the absolute value of the difference in EN values of the bonded atoms
DEN >1.5 (Ionic bond)
0.3<DEN <1.5 (polar covalent bond)
DEN<0.3 (covalent bond)
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General Chemistry: Chapter 10
Slide 14 of 50
Practice Questions
10-4 Writing Lewis Structures
• All the valence e- of atoms must appear.
• Usually, the e- are paired.
• Usually, each atom requires an octet.
– H only requires 2 e-.
• Multiple bonds may be needed.
– Readily formed by C, N, O, S, and P.
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Skeletal Structure
• Identify central and terminal atoms.
H
H
H
C
C
H
H
O
H
Skeletal Structure
• Hydrogen atoms are always terminal atoms.
• Central atoms are generally those with the lowest
electronegativity.
– O is only central in peroxo (-O-O-) or a hydroxy go (-O-H)
• Carbon atoms are always central atoms (organic
molecules)
• Generally structures are compact and symmetrical.
H
O
H
H
O
O
P
O
O
H
H
O
P
O
H
O
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Slide 18 of 50
Strategy for
Writing Lewis
Structures
Slide 19 of 50
General Chemistry: Chapter 10
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EXAMPLE10-7
Writing a Lewis Structure for a Polyatomic Ion. Write the Lewis structure for the
nitronium ion, NO2+.
Step 2:
Identify the central and terminal atoms
Step 3:
Plausible structure:
Step 4:
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Add
e-
to terminal atoms:
General Chemistry: Chapter 10
O—N—O
••
••
O—N—O
••
••
••
Total valence e- = 5 + 6 + 6 – 1 = 16 e-
••
Step 1:
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EXAMPLE 10-7
Step 5:
Determine e- left over:
Step 6:
Use multiple bonds to satisfy octets.
16 – 4 – 12 = 0
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••
••
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O—N—O
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General Chemistry: Chapter 10
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O=N=O
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+
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Practice Problems
Formal Charge
These are apparent charges on certain atoms in a Lewis structure that arise when atoms
have not contributed equal # of electrons to the covalent bonds joining them
FC = # of valence e- in free atom - # of valence lone-pair e- - 1/2( # of bond-pair
e-)
+
•• ••
O=N=O
•• ••
1
FC(O) = 6 - 4 –
2
FC(N) = 5 - 0 –
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(4) = 0
1
2
(8) = +1
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Alternative Lewis Structure
FC(O≡) = 6 - 2 –
FC(N) = 5 - 0 –
1
2
1
+
N
•• -
O
••
(6) = +1
(8) = +1
2
FC(O—) = 6 - 6 –
+
O
••
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O—N—O
••
••
1
(2) = -1
2
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General Chemistry: Chapter 10
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Alternative Lewis Structures
• Sum of FC is the overall charge.
• FC should be as small as possible.
• Negative FC usually on most electronegative
elements.
• FC of same sign on adjacent atoms is unlikely.
+
+
••-
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••
••
O≡N—O••
General Chemistry: Chapter 10
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