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??

??

4

1s 2s 2p 1s Carbon Hydrogen

How can carbon make four bonds with four hydrogen atoms?? How can those bonds be at 109.5o (tetrahedral)?

Energy

Energy

• Answer: Hybrid orbitals: the sublevels in an atom’s outer shell recombine into new orbitals of equal energy with different shapes and angles.

1s sp3 hybrid orbital

Energy

Energy

• Hybridization: when hybrid orbitals form.

• Total energy the same, but redistributed

6

1s 2s 2p 1s Carbon Hydrogen

Energy

Energy

1s sp3 hybrid orbital

Energy

Energy

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

s, p, p, ptrigonal pyramidal

• 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

MultipleBonds

• 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.OCN-

2.NO2-

3.NO3-

4.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) HCO3-

c) 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-