Section 8.13
▪ Covalent bonds are formed by the sharing of electrons; orbitals overlap to allow for this sharing .
▪ The mixing of two or more atomic orbitals of an atom forms hybrid orbitals . This process is called hybridization .
▪ Predicting hybridization is easy – just count the total number of steric numbers (domains).
▪ The AP test does not cover hybridization past 4 domains.

Section 8.13
Molecular Structure: The VSEPR Model
▪ The total number of steric numbers (also known as substituents - bonding plus non-bonding groups) is equal to the number of atomic orbitals that participate in the hybrid orbital.
# of substituents
(steric numbers)
Hybridization Example Molecule
2 sp CO
2
3 sp 2
4 sp 3
CH
3
CH
4
, NH
3
, H
2
O
5
6 sp 3 d sp 3 d 2
PCl
5
, I
3
-
SF
6

Molecular Structure: The VSEPR Model
▪ In sp hybridization, only one p orbital is mixed with the s orbital
▪ Example: BeF
2
- Steric Number: 2
▪ Electron configuration of Be: 1s 2 2s 2
▪ Electron configuration of F:1s 2 2s 2 2p 5

Section 8.13
Molecular Structure: The VSEPR Model
In the ground state, Be has no unpaired electrons – so how can the Be atom form a covalent bond with a fluorine?
Be
F
Be obtains an unpaired electron by moving one electron from the 2s orbital to the 2p orbital resulting in two unpaired electrons, one in a 2s orbital and another in a
2p orbital

Section 8.13
Molecular Structure: The VSEPR Model
▪ The Be atom can now form two covalent bonds with fluorine atoms
Be F
▪ Although we would not expect these bonds to be identical (one is in a 2s electron orbital, the other is in a 2p electron orbital), the structure of BeF
2 is linear and the bond lengths are identical
▪ The 2s and 2p electrons produced a "hybrid" orbital for both electrons

Section 8.13
2
Molecular Structure: The VSEPR Model
▪ In sp 2 hybridization, two p orbitals are mixed with the s orbital to generate three new hybrids
▪ Example: BF
3

Section 8.13
3
Molecular Structure: The VSEPR Model
▪ In sp 3 hybridization, all three p orbitals are mixed with the s orbital to generate four new hybrids
▪ Example: CH
4

Section 8.13
Molecular Structure: The VSEPR Model

Section 8.13
Sigma bond: The first bond made with any other atom
• Made from hybridized orbitals
• s-s, s-p, or p-p head-on overlap between nucleus
• Allows for free rotation
Pi bond: Any 2 nd or 3 rd bond made with any other atom
• Made from leftover p orbitals
• parallel, sideways p-p overlap, nucleus above or below overlap
• Weaker bond than sigma
• Fixed rotation
Single bond 1 sigma bond
Double bond 1 sigma bond and 1 pi bond
Triple bond 1 sigma bond and 2 pi bonds

Section 8.13
How many σ and π bonds are in the acetic acid (vinegar) molecule CH
3
COOH?
H
H
C
H
C O H
σ bonds = 6 + 1 = 7
π bonds = 1

Section 8.13
Molecular Structure: The VSEPR Model
Given the structural formula for propyne:
1. Indicate the hybridization of each carbon atom in the structure above.
2. Indicate the total number of sigma (σ) and pi
(π) bonds in the molecule.

Section 8.13
Molecular Structure: The VSEPR Model
Bond length and strength
▪ The more electrons that are involved in bonding, the shorter the bond length and the stronger the bond (meaning higher bond energy).
Problem: Is the bond length between the two carbon atoms shorter in C
2
H
6
, C
2
H
4
, or C
2
H
2
. Why?
Problem: Use the bonding model to account for the fact that all the bond lengths in SO
3 are identical and are shorter than a sulfur-oxygen single bond.

Section 8.13
Molecular Structure: The VSEPR Model
Problem
A. Draw the Lewis electron-dot structures for CO
3
2, CO
2
CO, including resonance structures where appropriate.
and
Which of the three species has the shortest C-O bond length?
Explain the reason for your answer.
Account for the fact that the carbon-oxygen bond length in
CO
3
2 – is greater than the carbon-oxygen bond length in CO
2
.

Section 8.13
Molecular Structure: The VSEPR Model
Warm Up
▪ Draw the following structures according to shape. Use arrows to indicate bond polarity. Arrows point to the more electronegative atom. Name shape and bond angle.
1. CH
4
2.BF
3
3.CO
2
4. HCN 5. CH
2
Cl
2
6. CH
2
O 7. NH
3

Section 8.13
Molecular Structure: The VSEPR Model
▪ Polar Molecules: Molecules with a somewhat negative end and a somewhat positive end (a dipole moment)
▪ Use an arrow to represent a dipole moment.
▪ Point to the negative charge center with the tail of the arrow indicating the positive center of charge.
Copyright © Cengage Learning. All rights reserved 15

Section 8.13
Molecular Structure: The VSEPR Model
16

Section 8.13
Molecular Structure: The VSEPR Model
No Net Dipole Moment (Dipoles Cancel)
Copyright © Cengage Learning. All rights reserved 17

Section 8.13
Molecular Structure: The VSEPR Model
▪ Polar molecules have a permanent dipole.
▪ Polar molecules line up in the presence of an electric field; nonpolar molecules do not.

Section 8.13
Molecular Structure: The VSEPR Model
▪ Molecules that have a polar bond may or may not show a dipole moment – the shape of the molecule must be considered.

Section 8.13
Molecular Structure: The VSEPR Model

Section 8.13
Molecular Structure: The VSEPR Model
Nonpolar molecules have a symmetrical charge distribution
▪ Diatomic molecules with the same atoms are nonpolar.
▪ Ex: Cl
2
▪ Linear, tetrahedral, trigonal planar shapes must have the same peripheral atoms to be nonpolar
▪ Ex: CH
4
BF
3
CO
2

Section 8.13
Molecular Structure: The VSEPR Model
molecules have an
charge distribution.
▪ Always polar: trigonal pyramidal and bent
Ex: NH
3
, H
2
O
▪ Diatomic molecules with
atoms are polar.
Ex: HCl
▪ Linear, tetrahedral, and trigonal planar shapes have different peripheral atoms to be polar
Ex: HCN, CH
2
Cl
2
, CH
2
O

Section 8.13
Molecular Structure: The VSEPR Model
The melting points of polar substances are higher than the melting points of non-polar substances with similar sizes.
Concept Check: Which of the following would have the higher boiling point?

Section 8.13
Molecular Structure: The VSEPR Model

Section 8.13
Molecular Structure: The VSEPR Model
CONCEPT CHECK!
True or false :
A molecule that has polar bonds will always be polar.
-If true, explain why.
-If false, provide a counter-example.
Answer: False, a molecule may have polar bonds (like
CO
2
) but the individual dipoles might cancel out so that the net dipole moment is zero.
Copyright © Cengage Learning. All rights reserved 25

Section 8.13
Molecular Structure: The VSEPR Model
Let
s Think About It
▪ Draw the Lewis structure for CO
2
.
▪ Does CO
2 contain polar bonds?
▪ Is the molecule polar or nonpolar overall? Why?
Copyright © Cengage Learning. All rights reserved 26

Section 8.13
Molecular Structure: The VSEPR Model
CONCEPT CHECK!
True or false :
Lone pairs make a molecule polar.
-If true, explain why.
-If false, provide a counter-example.
Answer: False, lone pairs do not always make a molecule polar. They might be arranged so that they are symmetrically distributed to minimize repulsions, such as XeF
4
.
Copyright © Cengage Learning. All rights reserved 27

Section 8.13
Molecular Structure: The VSEPR Model
Intramolecular vs. Intermolecular Forces
▪ Intramolecular forces: The forces within individual molecules holding it together
(ex: covalent and ionic bonds)
▪ Intermolecular forces – weak interactions between molecules
Intermolecular forces can cause a condensed state of matter (liquids and solids).
▪ Intermolecular forces are stronger in solids, weaker in liquids, and nearly absent in gases

Section 8.13
Molecular Structure: The VSEPR Model
Three types of Intermolecular forces
(collectively called van der Waals forces)
▪ London Dispersion
▪ Dipole-Dipole
▪ Hydrogen Bonding
Johannes van der Waals

Section 8.13
Molecular Structure: The VSEPR Model
▪ The weakest IM force
▪ Present in all molecules; the only type of IM force present in non-polar substances and Nobel gases.
▪ Caused by instantaneous dipoles
▪ Random movement of electrons can create a momentary nonsymmetrical distribution of charge even in nonpolar molecules

Section 8.13
London Dispersion
Molecular Structure: The VSEPR Model
▪ Instantaneous dipoles can induce a short-lived dipole in a neighboring molecule

Section 8.13
Molecular Structure: The VSEPR Model
▪ Dispersion force increases as the number of electrons in the molecule increases. (higher molecular weight, more electrons.)
(Ex: CCl
4 experiences greater London forces than CH
4
)
▪ Concept Check: Which Nobel gas would you predict to have the lowest boiling point? Why?
▪ Compare the boiling points of the noble gases: helium neon argon krypton xenon radon
- 269 ° C experience fewer dispersion forces
- 246 ° C
- 186 ° C
- 152 ° C
- 108 ° C
- 62 ° C experience more dispersion forces

Section 8.13
Molecular Structure: The VSEPR Model
Dipole-Dipole Forces
▪ Attraction between molecules with dipole moments
(molecules that have a permanent dipole)
▪ Occur in addition to dispersion forces
▪ Molecules orient themselves according to their poles
▪ Maximizes (+,-) interactions
▪ Minimizes (+,+) and (-,-) interactions

Section 8.13
Molecular Structure: The VSEPR Model
▪ Strength increases as polarity increases.
▪ Stronger than dispersion, but yet only 1% the strength of ionic bonds.

Section 8.13
Molecular Structure: The VSEPR Model
Hydrogen Bonding
▪ Special type of
Dipole-Dipole
▪ The strongest of the
Intermolecular forces.

Section 8.13
Molecular Structure: The VSEPR Model
▪ Hydrogen is bound to a highly electronegative atom
(F, O, N) with a lone pair
▪ Important in bonding of molecules such as water and
DNA
▪ Ex: NH
3
, H
2
O, HF can hydrogen bond.
▪ Ex: HCN does not, why?

Section 8.13
Molecular Structure: The VSEPR Model