Review for final

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Review for final F March 14 1 pm ES 413 Room S
March 16 4 pm BI 234 Office M 1130 am-1230 pm
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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
3
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-
4
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
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The sp Hybrid Orbitals in Gaseous BeCl2
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The sp3 Hybrid Orbitals in NH3 and H2O
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The sp3d Hybrid Orbitals in PCl5
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The sp3d2 Hybrid Orbitals in SF6
Sulfur Hexafluoride -- SF6
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Figure 10.26 Sigma and pi bonds.
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Figure 10.27 Bonding in ethylene.
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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
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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.
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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
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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
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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
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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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Fig. 9.14
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Figure 9.15 Electronegatives of the elements.
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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
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Fig. 9.16
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Fig. 9.17
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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
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Fig. 9.18
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Percent Ionic Character as a Function
ofElectronegativity Difference (?En)
Fig. 9.19
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The Charge Density of LiF
Fig. 9.20
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Figure 9.15 Electronegatives of the elements.
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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
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Fig. 9.16
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Fig. 9.17
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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
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Fig. 9.18
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Percent Ionic Character as a Function
ofElectronegativity Difference (?En)
Fig. 9.19
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The Charge Density of LiF
Fig. 9.20
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That's all, folks!!