Hybridization
VSEPR Theory Review
• Valence electrons only are
involved in bonding.
• Non-bonding and bonding electron
pairs around the central atom
repel each other.
• This repulsion causes specific
shapes and bond angles for each
molecule.
Question
• How can atoms, that have s, p and d
orbitals, bond in ways that make the
molecule shapes?
• Ex: p orbitals at 90o angles to each
other.
– How can they make a bond angle
of 120o in trigonal planar??
?
Energy
1s
2s
2p
Carbon
1s
Hydrogen
How can carbon make four bonds
with four hydrogen atoms?? How can
those bonds be at 109.5o
(tetrahedral)?
• Answer: Hybrid orbitals: the
sublevels in an atom’s outer shell
recombine into new orbitals of
equal energy with different shapes
and angles.
Energy
1s
sp3 hybrid orbital
• Hybridization: when hybrid
orbitals form.
Energy
Energy
1s
2s
2p
Carbon
1s
sp3 hybrid orbital
1s
Hydrogen
• Total energy the same, but
redistributed
After hybridization
• Total energy is the same.
• Energy redistributed equally among
four new hybrid orbitals.
• Hybrid orbitals are more directional:
in CH4, they point out to the four
corners of the tetrahedral shape.
Shapes of hybrid orbitals
s, p
s, p, p
trigonal
pyramidal
s, p, p, p
• NH3 is also sp3 hybridization: three
orbitals for the H, one orbital for
the non-bonding pair.
• For each compound’s central
atom, draw the orbital notation.
• Then, draw the orbital notation
after hybridization.
• Name the type of hybridization
and the shape of each molecule.
1.SiF4
2.CO2
3.BF3
4.PH3
5.H2O
Homework
• P. 117 # 1 and 2
Multiple
Bonds
• When a double bond forms, the
two bonds are not exactly the
same.
– First: “end on” s-orbital
interaction: σ (sigma) bonds
– Second: “side on” p-orbital
interaction: π (pi) bonds
• Single bonds: always σ bonds.
• Double bonds: one σ + one π bond.
• Triple bonds: one σ bond +two π
bonds
•
•
•
•
•
Count sigma and pi bonds
CHCl3
SCO
SeO2
ClO3-
Practice Problems
• Count the total number of sigma
and pi bonds in each molecule
– Draw a Lewis structure first!
1.H2CO
2.O2
3.CO2
4.HCN (C is the central atom)
5.CSN- (C is the central atom)
6.N3-
Homework
• Draw Lewis structures and count
sigma and pi bonds for:
1.OCN2.NO23.NO34.O3
5.SO3
Delocalization
of Electrons
• CO32- How do you figure out which
oxygen to make the double
bond??
• You can draw the molecule
THREE different WAYS!
• Resonance Structures: Any of
the Lewis structures that can be
drawn if a double bond could be in
more than one place.
• Are any of these structures
correct?
• What would the bond lengths be
like?
– The double bond is shorter than
the single bonds.
• The bond lengths are not actually
different!
But where are the electrons?
• In a π bond (double or triple bonds
only), the electrons can spread
over more than 2 nuclei.
• Why do π bonds delocalize?
– When the electrons spread out,
it gives the molecule a lower
potential energy.
– The molecule is more stable.
• Remember: Resonance structures
are imaginary.
• The electrons are actually being
shared by more than two atoms.
• So we need more than one picture
to show this.
Another Example: NO2-
• If more than one Lewis structure
can be drawn, what actually
happens is part-way between
those Lewis structures.
• Resonance structures have the
same σ bonds, but different π
bonds from one another.
O3 Ozone
Which of the following has more
than one possible Lewis
structure?
a) NH4+
b) HCO3c) C2H2
d) OH-
1. Draw a Lewis structure for NO3-.
2. Does this really explain the shape of the
ion? Why or why not?
3. Circle the electron pair that is delocalized.
4. Is it a sigma or a pi bond?
5. If there is one electron pair being shared
by all the oxygen atoms, what is the
charge of each oxygen atom? (Hint: the
charge of the ion divided by the number
of O atoms.)
6. Repeat 1-5 with CO32-