Transcript Chapter 10
Chemical Bonding II:
Molecular Geometry and
Hybridization of Atomic
Orbitals
Chapter 10
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Molecules are 3-dimensional objects
=> geometry helps us predict physical and
chemical properties
•
•
•
•
melting point
boiling point
density
reactivity
Lewis structure shows the number of valence
electrons on the central atom of the molecule
Electron pairs in the valence shell repel each other
=> Try to get as far away from each other as
possible
Valence Shell Electron Pair Repulsion Model
(VESPR Model)
10.1
Predict geometries based on Lewis Structures
Systematically go through all possible
geometries.
Notation:
A = Central atom
=> atom we predict geometry surrounding
B = atoms attached to the central atom
E = lone pairs on the central atom
(cannot forget lone pairs !!!)
start simple
===>
First Case: A - B
more complex
2 atoms linear
Valence shell electron pair repulsion (VSEPR) model:
Predict the geometry of the molecule from the electrostatic
repulsions between the electron (bonding and nonbonding) pairs.
B’s
E’s
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
AB2
2
0
linear
linear
B
B
10.1
Be
Cl
Cl
0 lone pairs on central atom
Cl
Be
Cl
2 atoms bonded to central atom
10.1
VSEPR
B’s
E’s
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
AB2
2
0
linear
linear
0
trigonal
planar
trigonal
planar
AB3
3
10.1
F
B
F
F
10.1
VSEPR
B’s
E’s
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
AB2
2
0
linear
linear
trigonal
planar
tetrahedral
AB3
3
0
trigonal
planar
AB4
4
0
tetrahedral
H
C
H
H
H
10.1
VSEPR
B’s
E’s
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
AB2
2
0
linear
linear
AB3
3
0
trigonal
planar
trigonal
planar
AB4
4
0
tetrahedral
tetrahedral
0
trigonal
bipyramidal
trigonal
bipyramidal
AB5
5
F
F
P
F
F
F
10.1
VSEPR
B’s
E’s
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
Arrangement of
electron pairs
Molecular
Geometry
AB2
2
0
linear
linear
trigonal
planar
AB3
3
0
trigonal
planar
AB4
4
0
tetrahedral
tetrahedral
AB5
5
0
trigonal
bipyramidal
trigonal
bipyramidal
AB6
6
0
octahedral
octahedral
Think of
axis
system
z
y
x
F
F
F
S
F
F
F
10.1
10.1
Lone
pairs
Lone pairs take up more room than bonding pairs.
lone-pair vs. lone pair
> lone-pair vs. bonding > bonding-pair vs. bonding
repulsion
pair repulsion
pair repulsion
VSEPR
B’s
E’s
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
AB3
3
0 120°
AB2E
2
1
Arrangement of
electron pairs
Molecular
Geometry
trigonal
planar
trigonal
planar
trigonal
planar
bent
118.5°
for SO2
VSEPR
B’s
E’s
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
AB4
4
0
AB3E
3
1
Arrangement of
electron pairs
Molecular
Geometry
tetrahedral
tetrahedral
tetrahedral
trigonal
pyramidal
VSEPR
B’s
E’s
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
AB4
4
0
Arrangement of
electron pairs
Molecular
Geometry
tetrahedral
tetrahedral
AB3E
3
1
tetrahedral
trigonal
pyramidal
AB2E2
2
2
tetrahedral
bent
O
H
H
VSEPR
B’s
E’s
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
AB5
5
0
trigonal
bipyramidal
trigonal
bipyramidal
AB4E
4
1
trigonal
bipyramidal
distorted
tetrahedron
Arrangement of
electron pairs
Molecular
Geometry
VSEPR
B’s
E’s
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
AB5
5
0
AB4E
4
1
AB3E2
3
2
Arrangement of
electron pairs
Molecular
Geometry
trigonal
bipyramidal
trigonal
bipyramidal
trigonal
bipyramidal
trigonal
bipyramidal
distorted
tetrahedron
T-shaped
F
F
Cl
F
B’s
Class
AB5
# of atoms
bonded to
central atom
5
VSEPR
E’s
# lone
pairs on
central atom
0
AB4E
4
1
AB3E2
3
2
AB2E3
2
3
Arrangement of
electron pairs
Molecular
Geometry
trigonal
bipyramidal
trigonal
bipyramidal
trigonal
bipyramidal
trigonal
bipyramidal
distorted
tetrahedron
trigonal
bipyramidal
T-shaped
linear
I
I
I
VSEPR
B’s
E’s
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
AB6
6
0
octahedral
octahedral
AB5E
5
1
octahedral
square
pyramidal
F
F
F
Arrangement of
electron pairs
Molecular
Geometry
Br
F
F
VSEPR
B’s
E’s
Class
# of atoms
bonded to
central atom
# lone
pairs on
central atom
AB6
6
0
octahedral
octahedral
AB5E
5
1
octahedral
AB4E2
4
2
octahedral
square
pyramidal
square
planar
Arrangement of
electron pairs
Molecular
Geometry
F
F
Xe
F
F
3
4
5
6
Predicting Molecular Geometry
1. Draw Lewis structure for molecule.
2. Count number of lone pairs on the central atom and
number of atoms bonded to the central atom.
3. Use VSEPR to predict the geometry of the molecule.
What are the molecular geometries of SO2 and SF4?
O
S
AB2E
bent
F
O
F
S
F
AB4E
F
distorted
tetrahedron
Dipole Moments and Polar Molecules
electron poor
region
electron rich
region
H
F
d+
d-
m=Qxr
Q is the charge
r is the distance between charges
1 D = 3.36 x 10-30 C m
10.2
10.2
10.2
Which of the following molecules have a dipole moment?
H2O, CO2, SO2, and CH4
S
O
dipole moment
polar molecule
O
C
O
no dipole moment
nonpolar molecule
Element
H
C
S
O
EN
2.1
2.5
2.5
3.5
dipole moment
polar molecule
H
H
C
H
H
no dipole moment
nonpolar molecule
A non-polar molecule can have polar bonds.
A polar molecule must have polar bonds.
Does CH2Cl2 have
a dipole moment?
10.2
10.2
How does Lewis theory explain the bonds in H2 and F2?
Sharing of two electrons between the two atoms.
H
H
F
F
Bond Dissociation Energy Bond Length Overlap Of
H2
1s1
436.4 kJ/mole
74 pm
2 1s
F2 1s22s22p5 150.6 kJ/mole
142 pm
2 2p
Valence bond theory – bonds are formed by sharing
of e- from overlapping atomic orbitals.
10.3
Zero Energy
is defined as
infinite
separation of
atoms
=> no
interaction
As atoms get closer, the
electron of one atom is
attracted to the nucleus of
the other atom.
Change in electron
density as two hydrogen
atoms approach each
other.
10.3
Valence Bond Theory and NH3
N – 1s22s22p3
3 H – 1s1
If the bonds form from overlap of 3 2p orbitals on nitrogen
with the 1s orbital on each hydrogen atom, what would
the molecular geometry of NH3 be?
If use the
3 2p orbitals
predict 900
Actual H-N-H
bond angle is
107.30
How do we solve this dilemma??
Hybridization – mixing of two or more atomic
orbitals to form a new set of hybrid orbitals.
1. Mix at least 2 nonequivalent atomic orbitals (e.g. s
and p). Hybrid orbitals have very different shape
from original atomic orbitals. Can not mix only p’s
2. Number of hybrid orbitals is equal to number of
pure atomic orbitals used in the hybridization
process. 1 s + 2 p => 3 hybrid orbitals Conservation
in number of
orbitals.
3. Covalent bonds are formed by:
a. Overlap of hybrid orbitals with atomic orbitals
N
H
b. Overlap of hybrid orbitals with other hybrid
C
C or N
orbitals
1s+3p
4 sp3
Not all orbitals are along an axis.
Angle between orbitals is 109.5°
10.4
Predict correct
bond angle
Since non-bonding
orbitals are larger than
bonding orbitals, expect
H - N - H angle to be
less than 109.5°.
Formation of sp Hybrid Orbitals
1s+2p
2 sp
10.4
Formation of sp2 Hybrid Orbitals
1s+2p
3 sp2
How do I predict the hybridization of the central atom?
Count the number of lone pairs AND the number
of atoms bonded to the central atom
# of Lone Pairs
+
# of Bonded Atoms Hybridization
AB2
Examples
2
sp
BeCl2
AB3 or AB2E
3
sp2
BF3
AB4, AB3E or
AB2E2
4
sp3
5 Allows for
expanded
octet
6
sp3d
PCl5
sp3d2
SF6
AB5, AB4E,
AB3E2, or AB2E3
AB6, AB5E, or
AB4E2
CH4, NH3, H2O
NOTE: Not all orbitals need to be used.
How do we get multiple bonds?
For carbon, if orbitals are sp3, we can not get
enough overlap between 2 carbon atoms.
=> do not use all p orbitals for hybridization.
pz
sp2
sp2
sp2
pz
H
H
sp2
C=C
H
sp2
sp2 sp2
H
sp2
sp2
Sigma bond (s) – electron density between the 2 atoms
Pi bond (p) – electron density above and below plane of
nuclei
of the bonding atoms
All electron density is
above or below the
plane of the molecule
in a π bond.
For triple bond, leave 2 p orbitals alone.
π
s
180°
H
C
C
H
Sigma (s) and Pi Bonds (p)
1 sigma bond
Single bond
Double bond
1 sigma bond and 1 pi bond
Triple bond
1 sigma bond and 2 pi bonds
How many s and p bonds are in the acetic acid
(vinegar) molecule CH3COOH?
H
C
O
H
C
O
H
s bonds = 6 + 1 = 7
p bonds = 1
H
10.5
Carbon Hybridization
Ground
state
2s2
2p2
Promotion
of electron
2s1
2p3
All single
bonds => sp3
One double
bond => sp2
Two double
bonds => sp
sp3
H
H
C=C
sp2
p
H
H
sp
p
H
C
C
O=C=O
H
Experiments show O2 is paramagnetic
O
O
No unpaired e-
Should be diamagnetic
Molecular orbital theory – bonds are formed from
interaction of atomic orbitals to form molecular
orbitals.
10.6
Remember that electrons have a wave nature.
Constructive
Destructive
Energy levels of bonding and antibonding molecular
orbitals in hydrogen (H2).
A bonding molecular orbital has lower energy and greater
stability than the atomic orbitals from which it was formed.
An antibonding molecular orbital has higher energy and
lower stability than the atomic orbitals from which it was
formed.
10.6
10.6
Same
configuration
of orbitals
=> fill with
allowed
number of
electrons.
Molecular Orbital (MO) Configurations
1. The number of molecular orbitals (MOs) formed is
always equal to the number of atomic orbitals
combined. Keep same number of orbitals.
2. The more stable the bonding MO, the less stable the
corresponding antibonding MO. Energy shifts have same
magnitude.
3. The filling of MOs proceeds from low to high
energies.
4. Each MO can accommodate up to two electrons.
5. Use Hund’s rule when adding electrons to MOs of
the same energy.
Same
as
before
(Ch 7).
6. The number of electrons in the MOs is equal to the
sum of all the electrons on the bonding atoms.
10.7
1
bond order =
2
(
Number of
electrons in
bonding
MOs
-
Number of
electrons in
antibonding
MOs
Anti-bonding orbital
)
Bonding orbital
bond
order
½
1
½
0
No He2
molecule !!
10.7
Delocalized molecular orbitals are not confined between
two adjacent bonding atoms, but actually extend over three
or more atoms.
Benzene – C6H6
H
H
H
H
C
C
C
C
C
C
H
H
H
H
H
Kekulé structures
C
C
C
C
C
C
H
H
H
Electron density above and below the plane of the
benzene molecule.
10.8
10.8