Chem 14A

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Chem 14A

• Monday, July 16, 2007

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Lecture Outline

• Integrated problems combining Lewis structures
  and VSEPR shape
• Molecular Orbital Theory

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Integrated Problems

• Typical Question Draw the Lewis structure and
  determine the molecular geometry of a compound
• Some problems will also ask for bond angles based
  on the shape
• Give individual hybridization or shape for each
  atom in the molecule, ie sp2, sp3 etc.

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Integrated Problems

• Lets try a few of these problems
• I3- write the Lewis structure and find the vsepr
  shape
• I3- valence electrons (7x3) 1 22e-

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Integrated Problems

• Now that we have the Lewis structure, lets
  determine the molecular shape
• The central iodine has 3 LP 2 bonds 5
  connections

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Summary of Molecular Shapes
Electron Pair Geometry
Lone electron pairs
Shape of Molecule
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Linear
0
Linear
0
Trigonal planar
Trigonal planar
3
1
V-shaped
0
Tetrahedral
4
Tetrahedral
1
2
V-shaped
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Total valence electron pairs
Electron Pair Geometry
Lone electron pairs
Shape of Molecule
0
Trig. bipyramid.
1
See-saw
Trigonal bipyramidal
5
2
T-shaped
3
Linear
0
Octahedral
6
Octahedral
1
Square pyramid
2
Square planar
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Integrated Problems

• A molecule with 5 connections falls under dsp3
  hybridization (5 orbitals mixed)
• Using the chart, 5 connections or total electron
  pairs gives us an overall shape of trigonal
  bipyramidal
• The electronic geometry, taking into account the
  shape contribution from the lone pairs, is linear

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Integrated Problems

• Lets try another example
• Draw the Lewis structure and vsepr shape for IF5
• Total valence electrons 7x6 42e-
• Important Notice the bond angles in the
  structure are slightly smaller than expected-
  extra repulsion

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Integrated Problems

• For larger molecules with more than one central
  atom, you may be asked to identify the
  hybridization and vsepr shape for each atom in
  the molecule
• For example, lets look at acetone

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Integrated Problems
SOLUTION
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Quick Exercise I.

• Draw the Lewis structure and vsepr shape for SF4
• What bond angles do you expect for this molecule?
• Draw the Lewis structure and vsepr shape for
  CH3CH2OH. Label the hybridization and shape for
  each atom.

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Answers Quick Exercise I.

• Each carbon is sp3 tetrahedral
• The oxygen is sp3 with an overall tetrahedral
  shape,
• bent or v-shape electronic geometry

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Transitioning from atoms to molecules

• Weve discussed how atomic orbitals hybridize
• Hybridized atomic orbitals then overlap with
  either single atomic orbitals such as
• H 1s or other hybrid orbitals to form bonds, s
  or p
• The resulting orbitals become transformed in
  molecular orbitals

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MOLECULAR ORBITAL THEORY
electrons occupy orbitals each of which spans the
entire molecule
molecular orbitals each hold up to two electrons
and obey Hunds rule, just like atomic orbitals
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H2 molecule
1s orbital on Atom A
1s orbital on Atom B
the H2 molecules molecular orbitals can be
constructed from the two 1s atomic orbitals
1sA 1sB MO1
constructive interference
1sA 1sB MO2
destructive interference
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ADDITION OF ORBITALS
builds up electron density in overlap region
1sA 1sB MO1
A
B
combine them by addition
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ADDITION OF ORBITALS
builds up electron density in overlap region.
1sA 1sB MO1
A
B
what do we notice?
electron density between atoms
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SUBTRACTION OF ORBITALS
results in low electron density in overlap
region..
1sA 1sB MO2
A
B
subtract
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SUBTRACTION OF ORBITALS
results in low electron density in overlap
region..
1sA 1sB MO2
A
B
what do we notice?
no electron density between atoms
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COMBINATION OF ORBITALS
1sA 1sB MO1
builds up electron density between nuclei
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COMBINATION OF ORBITALS
ANTI-BONDING
1sA 1sB MO2
results in low electron density between nuclei
1sA 1sB MO1
BONDING
builds up electron density between nuclei
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THE MOs FORMED BY TWO 1s ORBITALS
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1sA 1sB MO2
s1s
sigma anti-bonding s1s
s1s
1sA 1sB MO1
sigma bonding s1s
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COMBINING TWO 1s ORBITALS
Energy of a 1s orbital in a free atom
Energy of a 1s orbital in a free atom
A
B
E
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Energy of a 1s orbital in a free atom
Energy of a 1s orbital in a free atom
A
B
E
s1s
1sA1sB MO
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1sA-1sB MO
s1s
Energy of a 1s orbital in a free atom
Energy of a 1s orbital in a free atom
A
B
E
s1s
1sA1sB MO
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COMBINING TWO 1s ORBITALS
s1s
A
B
1sB
1sA
E
s1s
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bonding in H2
H
H
H2
s1s
E
1s
1s
s1s
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H
H
H2
s1s
E
1s
1s
s1s
the electrons are placed in the s1s molecular
orbitals
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H
H
H2
s1s
E
1s
1s
s1s
H2 (s1s)2
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He2
1s2
atomic configuration of He
He
He
He2
s1s
E
1s
1s
s1s
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He2 (s1s)2(s1s)2
He
He
He2
s1s
E
1s
1s
s1s
bonding effect of the (s1s)2 is cancelled by the
antibonding effect of (s1s)2
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BOND ORDER
net number of bonds existing after the
cancellation of bonds by antibonds
the electronic configuration is.
He2
(s1s)2(s1s)2
the two bonding electrons were cancelled out by
the two antibonding electrons
BOND ORDER 0
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BOND ORDER
measure of bond strength and molecular stability
If of bonding electrons gt of antibonding
electrons
the molecule is predicted to be stable
Bond order

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BOND ORDER
measure of bond strength and molecular stability
If of bonding electrons gt of antibonding
electrons
the molecule is predicted to be stable

Bond order


of bonding electrons(nb)
of antibonding electrons (na)

1/2

(nb - na)
1/2
high bond order indicates high bond energy and
short bond length
H2,H2,He2
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s1s s1s Magnetism Bond order Bond energy
(kJ/mol) Bond length (pm)
H2
He2
He2
H2
E
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s1s s1s Magnetism Bond order Bond energy
(kJ/mol) Bond length (pm)
H2
He2
He2
H2 Dia- 1 436 74
E
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s1s s1s Magnetism Bond order Bond energy
(kJ/mol) Bond length (pm)
H2 Para- ½ 225 106
He2
He2
H2 Dia- 1 436 74
E
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s1s s1s Magnetism Bond order Bond energy
(kJ/mol) Bond length (pm)
H2 Para- ½ 225 106
He2 Para- ½ 251 108
He2
H2 Dia- 1 436 74
E
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First row diatomic molecules and ions
s1s s1s Magnetism Bond order Bond energy
(kJ/mol) Bond length (pm)
H2 Para- ½ 225 106
He2 Para- ½ 251 108
He2 0
H2 Dia- 1 436 74
E
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second period
HOMONUCLEAR DIATOMICS
Li2
Li 1s22s1
both the 1s and 2s overlap to produce s bonding
and anti-bonding orbitals
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ENERGY LEVEL DIAGRAM FOR DILITHIUM
s2s
Li2
2s
2s
E
s2s
s1s
1s
1s
s1s
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ELECTRONS FOR DILITHIUM
s2s
Li2
2s
2s
E
s2s
s1s
1s
1s
s1s
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Electron configuration for DILITHIUM
s2s
Li2
(s1s)2(s1s)2(s2s)2
2s
2s
E
s2s
Bond Order ?
1s
1s
s1s
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Electron configuration for DILITHIUM
s2s
Li2
(s1s)2(s1s)2(s2s)2
2s
2s
nb 4
na 2
E
s2s
Bond Order 1
single bond.
1s
1s
s1s
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Electron configuration for DILITHIUM
s2s
Li2
(s1s)2(s1s)2(s2s)2
2s
2s
the s1s and s1s orbitals can be ignored when
both are FILLED!
E
s2s
1s
1s
omit the inner shell
s1s
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only valence orbitals contribute to molecular
bonding
Li2 (s2s)2
Li
Li
Li2
s2s
E
2s
2s
s2s
The complete configuration is (s1s)2(s1s)2
(s2s)2
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Be2
Be
Be
Be2
s2s
E
2s
2s
s2s
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Electron configuration for DIBERYLLIUM
Be2
Be
Be
Be2
s2s
E
2s
2s
s2s
Bond order 0
(s2s)2(s2s)2
Configuration
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Electron configuration for DIBERYLLIUM
Be2
Be
Be
Be2
(s2s)2(s2s)2
s2s
nb 2
E
2s
2s
na 2
s2s
Bond Order
1/2(2 - 2) 0
1/2(nb - na)
Now B2...
No bond!!!
The molecule is not stable!
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B2
the Boron atomic configuration is
1s22s22p1
we expect B to use 2p orbitals to
form molecular orbitals
addition and subtraction
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s-molecular orbitals
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p molecular orbitals
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ENERGY LEVEL DIAGRAM
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s2p
p2p
p2p
s2p
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expected orbital splitting
2p
E
This pushes the s2p up
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MODIFIED ENERGY LEVEL DIAGRAM
E
Notice that the s2p and p2p have changed
places!!!!
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Electron configuration for B2
B is He 2s22p1
s2p
p2p
s2p
2p
2p
p2p
E
Place electrons from 2s into s2s and s2s
s2s
2s
2s
s2s
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s2p
p2p
s2p
2p
2p
p2p
E
Place electrons from 2p into p2p and p2p
s2s
2s
2s
Remember HUNDs RULE
s2s
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ELECTRONS ARE UNPAIRED
Abbreviated configuration
(s2s)2(s2s)2(p2p)2
E
Complete configuration
(s1s)2(s1s)2(s2s)2(s2s)2(p2p)2
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Electron configuration for B2
s2p
(s2s)2(s2s)2(p2p)2
p2p
na 2
nb 4
s2p
2p
2p
p2p
E
Bond order
1/2(nb - na)
s2s
1/2(4 - 2) 1
Molecule is predicted to be stable and
paramagnetic.
2s
2s
s2s
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A SUMMARY OF THE MOs
Emphasizing nodal planes
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ELECTRONIC CONFIGURATION OF THE HOMONUCLEAR
DIATOMICS
B2
C2
N2
O2
F2
Li2
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O2
F2
B2
C2
N2
Li2
E
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Second row diatomic molecules
s2p p2p s2p p2p s2s s2s Magnetism Bond
order Bond E. (kJ/mol) Bond length(pm)
B2
C2
N2
O2
F2
E
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Second row diatomic molecules
s2p p2p s2p p2p s2s s2s Magnetism Bond
order Bond E. (kJ/mol) Bond length(pm)
B2 Para- 1 290 159
C2
N2
O2
F2
E
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Second row diatomic molecules
s2p p2p s2p p2p s2s s2s Magnetism Bond
order Bond E. (kJ/mol) Bond length(pm)
B2 Para- 1 290 159
C2 Dia- 2 620 131
N2
O2
F2
E
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Second row diatomic molecules
s2p p2p s2p p2p s2s s2s Magnetism Bond
order Bond E. (kJ/mol) Bond length(pm)
B2 Para- 1 290 159
C2 Dia- 2 620 131
N2 Dia- 3 942 110
O2
F2
E
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Second row diatomic molecules
NOTE SWITCH OF LABELS
s2p p2p p2p s2p s2s s2s Magnetism Bond
order Bond E. (kJ/mol) Bond length(pm)
B2 Para- 1 290 159
C2 Dia- 2 620 131
N2 Dia- 3 942 110
O2 Para- 2 495 121
F2
E
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Second row diatomic molecules
NOTE SWITCH OF LABELS
s2p p2p p2p s2p s2s s2s Magnetism Bond
order Bond E. (kJ/mol) Bond length(pm)
B2 Para- 1 290 159
C2 Dia- 2 620 131
N2 Dia- 3 942 110
O2 Para- 2 495 121
F2 Dia- 1 154 143
E
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s2p p2p p2p s2p s2s s2s
O2
O2
O2
O22-
E
O2 O2 O2 O22-
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s2p p2p p2p s2p s2s s2s
O2
O2
O2
O22-
E
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s2p p2p p2p s2p s2s s2s
O2
O2
O2
O22-
E
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s2p p2p p2p s2p s2s s2s
O2
O2
O2
O22-
E
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s2p p2p p2p s2p s2s s2s
O22-
O2
O2
O2
E
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s2p p2p p2p s2p s2s s2s
O22-
O2
O2
O2
E
O2 B.O. (8 - 4)/2 2
O2 B.O. (8 - 3)/2 2.5
O2 B.O. (8 - 5)/2 1.5
O22- B.O. (8 - 6)/2 1
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s2p p2p p2p s2p s2s s2s
O22-
O2
O2
O2
E
O2 B.O. 2
BOND ENERGY ORDER
O2 B.O. 2.5
O2 gtO2 gtO2 gt O22-
O2 B.O. 1.5
O22- B.O. 1
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OXYGEN
How does the Lewis dot picture correspond to MOT?
12 valence electrons
BO 2 but PARAMAGNETIC
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Quick Exercise II.

• Draw the molecular orbital configuration of N2
  and N22
• Give the expected bond order for each molecule
• Indicate paramagnetism or diamagnetism
• Which molecule will be more stabile?

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N2
s2p p2p s2p p2p s2s s2s

• Molecule is paramagnetic
• BO is 2.5

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N22
s2p p2p s2p p2p s2s s2s

• Molecule is diamagnetic
• BO 2