Chapter 9
AP Chemistry

 Valence Shell Electron Pair
Repulsions
 Electrons are negatively charged, so each pair will repel other pairs such that they spread out in 3-D space to minimize the repulsions.

 Domains are regions about an atom’s shell where electrons are concentrated.
 This is easier to see with a Lewis diagram.
 For example, the carbon atom above has electrons on two sides (even though they are double bonds). So this carbon atom has 2 domains.

 The shapes that molecules take, and thus the angles between bonds, depends on the number of domains.
 2 domains need to be 180 o apart to minimize repulsions.
 3 Domains need to be 120 o apart.
 2 & 3 domains can remain 2-D. Any more domains and it must be 3-D.

# of domains
2
3
4
5
6
Arrangement Domain Geometry linear trigonal planar
Bond Angles
180
120
Tetrahedral 109.5
Trigonal bipyramidal
120 & 90 octahedral 90

 The shape may not match the domain geometry.
 Why?

 In the Lewis
Structure of water, we see 4 domains. Yet when we look at a water molecule, we can only see the bonds, not the nonbonding pairs.
 Look back at the angles.
 4 domains should have an angle of
109.5.
 The water molecule is 104.5.
 These angles are too close to be coincidence.

 There are 2 domains
 There are Zero nonbonding domains.
 The Molecular Geometry is linear
 Example:

Trigonal Planar Domain Geometry option 1
 There are 3 domains
 If there is Zero nonbonding domains, then The Molecular
Geometry is trigonal planar
 Example:

Trigonal Planar Domain Geometry option 2
 There are 3 domains
 If there is 1 nonbonding domain, then The Molecular Geometry is bent
 Example:

Tetrahedral Domain Geometry option 1
 There are 4 domains
 If there is Zero nonbonding domains, then The Molecular
Geometry is tetrahedral
 Example:

Tetrahedral Domain Geometry option 2
 There are 4 domains
 If there is 1 nonbonding domain, then The Molecular Geometry is trigonal pyramidal
 Example:

Tetrahedral Domain Geometry option 3
 There are 4 domains
 If there are 2 nonbonding domains, then The Molecular
Geometry is bent
 Example:

Trigonal Bipyramidal Domain
Geometry option 1
 There are 5 domains
 If there are zero nonbonding domains, then The Molecular
Geometry is trigonal bipyramidal
 Example:

Trigonal Bipyramidal Domain
Geometry option 2
 There are 5 domains
 If there is 1 nonbonding domain, then The Molecular Geometry is
SeeSaw
 Example:

Trigonal Bipyramidal Domain
Geometry option 3
 There are 5 domains
 If there are 2 nonbonding domains, then The Molecular
Geometry is T-Shaped
 Example:

Trigonal Bipyramidal Domain
Geometry option 4
 There are 5 domains
 If there are 3 nonbonding domains, then The Molecular
Geometry is linear
 Example:

Octahedral Domain Geometry option 1
 There are 6 domains
 If there are zero nonbonding domains, then The Molecular
Geometry is octahedral
 Example:

Octahedral Domain Geometry option 2
 There are 6 domains
 If there is 1 nonbonding domain, then The Molecular Geometry is square pyramidal
 Example:

Octahedral Domain Geometry option 3
 There are 6 domains
 If there are 2 nonbonding domains, then The Molecular
Geometry is square planar
 Example:

 CO
2
 CH
4
 XeF
4
 H
2
CO
 H
2
O
 XeF
2
 PCl
5
 ICl
5

Molecular
Geometry
CO
2
CH
4
XeF
4 linear tetrahedral octahedral linear tetrahedral
Square planar
H
2
CO Trigonal planar Trigonal planar
H
2
O tetrahedral bent
XeF
2
Trigonal bipyramidal linear
PCl
5
ICl
5
Domain
Geometry
Trigonal bipyramidal octahedral
Trigonal bipyramidal
Square pyramidal

 The Electron Dot Structure of
Carbon shows four unpaired electrons, but the Orbital Notation only shows 2. Why?

*

*C*

*
 Will carbon make 2 bonds, or 4?

 Bonding usually involves s-orbitals.
For the s-orbital of carbon to bond, one of the electrons has to go somewhere.
 That somewhere is the empty p orbital. In order to make 4 bonds, the carbon will combine its s-orbital with its 3 p-orbitals into a new set of
4 orbitals all of equal energy.
 This new set is called a hybrid and is referred to as an sp 3 hybrid.

3
 On the left are regular porbitals and s-orbital.
 On the right are the 4 hybrized sp 3 -orbitals.

 When there are 2 domains, there is an SP hybrid.
 When there are 3 domains, there is an SP 2 hybrid.
 When there are 4 domains, there is an SP 3 hybrid.
 When there are 5 domains, there is an SP 3 D hybrid.
 When there are 6 domains, there is an SP 3 D 2 hybrid.

 CO
2
 CH
4
 XeF
4
 H
2
CO
 H
2
O
 XeF
2
 PCl
5
 ICl
5

 CO
2
= SP
 CH
4
= SP 3
 XeF
4
= SP 3 D 2
 H
2
CO = SP 2
 H
2
O = SP 3
 XeF
2
= SP 3 D
 PCl
5
= SP 3 D
 ICl
5
= SP 3 D 2

 Earlier, we stated that bonding usually involves an s-orbital. How does that happen?
 When 2 s-orbitals overlap, the electrostatic forces of attraction of the nucleus of one atom will attract the electrons of the other atom and vice versa, forming a bond.
 If two s-orbitals directly overlap then the bond formed is linear between the 2 nuclear centers & is called a sigma (  ) bond.

 While this is a depiction of a sigma bond, a sigma bond is not always formed between two sorbitals.

 Let’s examine a C
2
H
4 molecule.
 From the Lewis
Structure, we expect a double bond. We can also see that carbon has 3 domains, so we expect SP 2 hybridization.

 SP 2 hybridized orbitals bond each carbon atom (and hydrogen atoms) along the axis connecting the atoms, forming  bonds.
 Since SP 2 uses 3 orbitals, we see that there is an unhybridized P-orbital.
 As the  bond forms, the atoms move closer and the p-orbitals of the 2 carbons merge into a 2nd bond called a pi (  ) bond.
 The top and bottom portion are both part of the same  bond.

 Let’s examine a C
2
H
2 molecule.
 From the Lewis Structure, we expect a triple bond. We can also see that carbon has 2 domains, so we expect SP hybridization.

 SP hybridized orbitals bond each carbon atom (and hydrogen atoms) along the axis connecting the atoms, forming  bonds.
 Since SP uses 2 orbitals, there must be 2 unhybridized P-orbitals.
 As the  bond forms, the atoms move closer and the p-orbitals of the 2 carbons merge into 2 pi (  ) bonds.

 How many pi bonds and how many sigma bonds are present (in total) in the molecule below?
QuickTime™ and a
decompressor are needed to see this picture.

 A single bond consist of 1 sigma bond.
 A double bond consist of 1 sigma bond and 1 pi bond.
 A triple bond consist of 1 sigma bond and 2 pi bonds.
 So the answer to the last question is 11 sigma bonds and 1 pi bond.