Lewis Structures of Simple Molecules

1
Lewis Structures of Simple Molecules
H
H
..
CH4
..
H
C
O
H
C
Methane
H
H
Ethyl Alcohol (Ethanol)
..
..
..
..
O
K
..
Cl
..
..
..
..
..
..
O
O
..
CF4
..
KClO3
Potassium Chlorate
Carbon Tetrafluoride
2
Resonance Delocalized
Electron-Pair Bonding - I
Ozone O3
..
..
..
..
..
O
..
O
..
..
..
..
..
O
O
O
O
..
II
I
Resonance Hybrid Structure
..
O
..
..
..
..
O
O
One pair of electrons resonances between
the two locations!!
3
Resonance Delocalized
Electron-Pair Bonding - II
H
H
C
C
H
H
C
C
C
C
H
H
C
C
H
H
C
C
H
H
H
C
C
C
H
H
C
H
H
C
H
C
C
H
C
Benzene
Resonance Structure
H
4
Lewis Structures for Octet Rule Exceptions
..
..
..
..
..
..
F
..
..
..
..
Cl
F
..
..
F
Each chlorine atom has 8 electrons associated.
Boron has only 6!
Each fluorine atom has 8 electrons associated.
Chlorine has 10 electrons!
.
..
..
..
..
..
..
N
..
..
..
..
Cl
Cl
Be
..
O
O
Each chlorine atom has 8 electrons associated.
The beryllium has only 4 electrons.
NO2 is an odd electron atom. The nitrogen has 7
electrons.
5
Resonance Structures - Expanded Valence Shells
..
..
..
..
..
..
..
..
..
..
..
..
..
..
F
F
F
F
..
..
..
..
..
..
..
S
F
F
P
..
..
F
..
F
..
..
..
..
..
..
..
..
F
F
..
F
p 10e-
S 12e-
Sulfur hexafluoride
Phosphorous pentafluoride
..
..
..
..
..
Resonance Structures
..
..
..
..
..
Sulfuric acid
S 12e-
6
Lewis Structures of Simple Molecules
. .
Sulfate
. .
-2
O
. .
. .
Resonance Structures-V
. .
. .
S
O
O
. .
. .
-2
. .
. .
O
o
o o
o o
O
Plus 4 others for a total of 6
x o
o
o
o o
x
S
O
O
x o
o
o o
. .
x
o o
-2
o o
O
. .
. .
x x
. .
. .
o o
O
o o
S
O
O
o o
. .
. .
O
. .
. .
. .
x Sulfur electrons
o Oxygen electrons
7
Fig. 9.14
8
Figure 9.15 Electronegatives of the elements.
9
The Periodic Table of the Elements
2.1
He
0.9
1.5
2.0
2.5
3.0
3.5
4.0
Ne
Electronegativity
0.9
1.2
Ar
1.5
1.8
2.1
2.5
3.0
0.8
1.0
1.3
1.5
1.6
1.6
1.5
1.8
1.8
1.8
1.9
1.6
1.6
1.8
2.0
2.4
2.8
Kr
0.8
1.0
1.2
1.4
1.6
1.8
1.9
2.2
2.2
2.2
1.9
1.7
Xe
2.5
2.1
1.9
1.8
1.7
0.7
0.9
1.1
1.5
1.7
1.9
2.2
2.2
2.4
1.9
Rn
2.2
2.0
1.9
1.8
1.8
1.3
2.2
0.7
0.9
1.1
Ce Pr Nd Pm
Yb Lu
1.1
1.1
1.1
1.1
1.2
1.2
1.2
1.2
1.2
1.2
1.2
1.2
1.2
1.3
1.3
1.5
1.7
1.3
1.3
1.3
1.3 1.3
1.3
1.3
1.3
1.5
1.3
Th Pa U Np
No Lr
10
Fig. 9.16
11
Fig. 9.17
12
Determining Bond Polarity from
Electronegativity Values
Problem (a)Indicate the polarity of the
following bonds with a polarity arrow O -
H, O - Cl, C - N, P - N, N - S, C - Br, As - S
(b) rank those bonds in order of increasing
polarity. Plan (a) We use Fig. 9.16 to find the
EN values, and point the arrow toward the
negative end. (b) Use the EN values. Solution
a) the EN of O 3.5 and of H 2.1 O - H
the EN of O 3.5 and of Cl 3.0 O
- Cl the EN of C 2.5 and of P 2.1
C - P the EN of P
2.1 and of N 3.0 P - N the EN of N
3.0 and of S 2.1 N - S the
EN of C 2.5 and of Br 2.8
C - Br the EN of As 2.0 and of O
3.5 As - O
b) C - Br lt C - P lt O - Cl lt P - N lt N - S lt O -
H lt As - O 0.3 lt 0.4 lt 0.5 lt
0.9 lt 0.9 lt 1.4 lt 1.5
13
Fig. 9.18
14
Percent Ionic Character as a Function
ofElectronegativity Difference (?En)
Fig. 9.19
15
The Charge Density of LiF
Fig. 9.20
16
Figure 9.15 Electronegatives of the elements.
17
The Periodic Table of the Elements
2.1
He
0.9
1.5
2.0
2.5
3.0
3.5
4.0
Ne
Electronegativity
0.9
1.2
Ar
1.5
1.8
2.1
2.5
3.0
0.8
1.0
1.3
1.5
1.6
1.6
1.5
1.8
1.8
1.8
1.9
1.6
1.6
1.8
2.0
2.4
2.8
Kr
0.8
1.0
1.2
1.4
1.6
1.8
1.9
2.2
2.2
2.2
1.9
1.7
Xe
2.5
2.1
1.9
1.8
1.7
0.7
0.9
1.1
1.5
1.7
1.9
2.2
2.2
2.4
1.9
Rn
2.2
2.0
1.9
1.8
1.8
1.3
2.2
0.7
0.9
1.1
Ce Pr Nd Pm
Yb Lu
1.1
1.1
1.1
1.1
1.2
1.2
1.2
1.2
1.2
1.2
1.2
1.2
1.2
1.3
1.3
1.5
1.7
1.3
1.3
1.3
1.3 1.3
1.3
1.3
1.3
1.5
1.3
Th Pa U Np
No Lr
18
Fig. 9.16
19
Fig. 9.17
20
Determining Bond Polarity from
Electronegativity Values
Problem (a)Indicate the polarity of the
following bonds with a polarity arrow O -
H, O - Cl, C - N, P - N, N - S, C - Br, As - S
(b) rank those bonds in order of increasing
polarity. Plan (a) We use Fig. 9.16 to find the
EN values, and point the arrow toward the
negative end. (b) Use the EN values. Solution
a) the EN of O 3.5 and of H 2.1 O - H
the EN of O 3.5 and of Cl 3.0 O
- Cl the EN of C 2.5 and of P 2.1
C - P the EN of P
2.1 and of N 3.0 P - N the EN of N
3.0 and of S 2.1 N - S the
EN of C 2.5 and of Br 2.8
C - Br the EN of As 2.0 and of O
3.5 As - O
b) C - Br lt C - P lt O - Cl lt P - N lt N - S lt O -
H lt As - O 0.3 lt 0.4 lt 0.5 lt
0.9 lt 0.9 lt 1.4 lt 1.5
21
Fig. 9.18
22
Percent Ionic Character as a Function
ofElectronegativity Difference (?En)
Fig. 9.19
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The Charge Density of LiF
Fig. 9.20
28
VSEPR Valence Shell Electron Pair Repulsion
A way to predict the shapes of molecules
Pairs of valence electrons want to get as far
away from each other as possible in
3-dimensional space.
29
Balloon Analogy for the MutualRepulsion of
Electron Groups
Two
Three Four Five
Six
Number of Electron Groups
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AX2 Geometry - Linear
..
..
..
..
Molecular Geometry Linear Arrangement
..
..
Cl
Cl
Be
BeCl2
1800
Gaseous beryllium chloride is an example of a
molecule in which the central atom - Be does not
have an octet of electrons, and is electron
deficient.
Other alkaline earth elements also have the same
valence electron configuration, and the same
geometry for molecules of this type. Therefore
this geometry is common to group II elements.
..
..
..
..
C
O
O
CO2
1800
Carbon dioxide also has the same geometry, and is
a linear molecule, but in this case, the bonds
between the carbon and oxygens are double bonds.
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The Two Molecular Shapes of the Trigonal
Planar Electron-Group Arrangement
34
AX3 Geometry - Trigonal Planar
..
..
..
..
All of the boron Family(IIIA) elements have the
same geometry. Trigonal Planar !
..
..
F
F
BF3
B
1200
Boron Trifluoride
..
..
..
F
AX2E SO2
..
-
..
..
..
..
O
..
..
1200
NO3-
..
..
N
..
..
..
..
O
O
The AX2E molecules have a pair of Electrons where
the third atom would appear in the space around
the central atom, in the trigonal
planar geometry.
1200
Nitrate Anion
35
The Three Molecular Shapes of the
Tetrahedral Electron-Group Arrangement
36
AX4 Geometry - Tetrahedral
H
Methane
H
109.50
CH4
C
H
H
C
H
H
H
H
All molecules or ions with four electron groups
around a central atom adopt the tetrahedral
arrangement
H
H
109.50
109.50
H
H

all angles are the same!
Ammonia is in a tetrahedral shape, but it has
only an electron pair in one location, so the
smaller angle!
Ammonium Ion
H
37
The Four Molecular Shapes of the
Trigonal Bipyramidal Electron- Group Arrangement
38
AX5 Geometry - Trigonal Bipyramidal
..
..
..
..
..
..
F
I
..
..
..
86.20
..
..
I
..
1800
Br
F
..
..
..
..
..
..
I
F
..
..
AX3E2 - BrF3
AX2E3 - I3-
..
..
..
..
Cl
..
Cl
..
..
..
P
Cl
..
AX5 - PCl5
..
..
Cl
..
..
..
..
Cl
39
The Three Molecular Shapes of the Octahedral
Electron-Group Arrangement
40
AX6 Geometry - Octahedral
..
..
..
..
..
..
..
..
..
..
F
F
..
..
..
..
F
F
..
..
..
F
F
..
..
..
..
..
S
I
..
..
..
F
F
..
..
..
..
..
F
F
..
..
..
..
F
AX5E Iodine
Pentafluoride
AX6 Sulfur Hexafluoride
..
..
..
..
..
F
..
F
..
Xe
..
..
..
..
..
F
F
..
..
Xenon Tetrafluoride
Square planar shape
41
Using VSEPR Theory to Determine Molecular Shape
1) Write the Lewis structure from the molecular
formula to see the relative placement of
atoms and the number of electron groups. 2)
Assign an electron-group arrangement by counting
all electron groups around the central atom,
bonding plus nonbonding. 3) Predict the ideal
bond angle from the electron-group arrangement
and the direction of any deviation caused by
the lone pairs or double bonds. 4) Draw and
name the molecular shape by counting bonding
groups and non-bonding groups separately.
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Hybrid Orbital Model
44
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The sp Hybrid Orbitals in Gaseous BeCl2
46
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The sp3 Hybrid Orbitals in NH3 and H2O
50
The sp3d Hybrid Orbitals in PCl5
51
The sp3d2 Hybrid Orbitals in SF6
Sulfur Hexafluoride -- SF6
52
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Figure 10.26 Sigma and pi bonds.
54
Figure 10.27 Bonding in ethylene.
55
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Figure 10.28 Bonding in acetylene.
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Restricted Rotation of ?-Bonded Molecules
A) Cis - 1,2 dichloroethylene B)
trans - 1,2 dichloroethylene
59
Postulating the Hybrid Orbitals in a Molecule
Problem Describe how mixing of atomic orbitals
on the central atoms leads to the hybrid
orbitals in the following a) Methyl amine,
CH3NH2 b) Xenon tetrafluoride, XeF4 Plan
From the Lewis structure and molecular shape, we
know the number and arrangement of electron
groups around the central atoms, from which we
postulate the type of hybrid orbitals involved.
Then we write the partial orbital diagram for
each central atom before and after the orbitals
are hybridized.
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Postulating the Hybrid Orbitals in a Molecule
Problem Describe how mixing of atomic orbitals
on the central atoms leads to the hybrid
orbitals in the following a) Methyl amine,
CH3NH2 b) Xenon tetrafluoride, XeF4 Plan
From the Lewis structure and molecular shape, we
know the number and arrangement of electron
groups around the central atoms, from which we
postulate the type of hybrid orbitals involved.
Then we write the partial orbital diagram for
each central atom before and after the orbitals
are hybridized. Solution a) For CH3NH2 The
shape is tetrahedral around the C and N
atoms. Therefore, each central atom is sp3
hybridized. The carbon atom has four half-filled
sp3 orbitals
2s
2p
sp3
Isolated Carbon Atom
Hybridized Carbon Atom
61
The N atom has three half-filled sp3 orbitals and
one filled with a lone pair.
2s
sp3
2p
..
H
C
H
N
H
H
H
62
b) The Xenon atom has filled 5 s and 5 p
orbitals with the 5 d orbitals empty.
Isolated Xe atom
5 d
5 s
5 p
Hybridized Xe atom
5 d
sp3d2
63
b) continuedFor XeF4. for Xenon, normally it has
a full octet of electrons,which would mean an
octahedral geometry, so to make the compound,
two pairs must be broken up, and bonds made to
the four fluorine atoms. If the two lone pairs
are on the equatorial positions, they will be at
900 to each other, whereas if the two polar
positions are chosen, the two electron groups
will be 1800 from each other. Thereby
minimizing the repulsion between the two electron
groups.
F
F
F
F
Xe
Xe
1800
F
F
F
F
Square planar
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That's all, folks!!