1. I. Covalent Bonds and Reactions
Covalent Bond Model
A Model is a simple explanation that lets us explain complicated things
Chemical Bonds are a model that help us explain why groups of atoms behave as a unit in reactions
Chemical Bond forms only if that arrangement is more stable
Chemical Bond Energy = energy gained by behaving as a unit
Examples
Methane (CH4)
a. 4 C-H covalent bonds arranged in a tetrahedral shape
b. The total energy of methane is 1656 kJ/mol
c. We can estimate any C-H bond as being 414 kJ/mol
Chloroform (CH3Cl)
Total E = 1581 kJ/mol
We can use C-H value
C-Cl bond E = 339 kJ/mol [ (414) = 339]

2. energy required energy released Other bond energies
Every C-H bond is not exactly 414 kJ/mol, but we estimate as such
Other X-Y bond strengths have been estimated
Single bond: 2 atoms share one pair of e- (C-H)
Double bond: 2 atoms share 2 pairs of e-; stronger than single bond
C=O = 736, C-O = 360
Triple bond: 2 atoms share 3 pairs of e-; even stronger
C≡O = 1072
Bond length shortens as more e- are shared between the atoms
Calculating Enthalpy of a Reaction
Enthalpy = DH = sum of energy to break and form bonds in a reaction
a. Bond breaking requires energy (endothermic).
Bond formation releases energy (exothermic).
H = D(bonds broken)  D(bonds formed)
energy required
energy released

5. Find DH for: 1 H F HF
1(436 kJ/mol) + 1(159 kJ/mol) – 2(565 kJ/mol) = -535 kJ/mol = DH
3. Example: CH Cl F > CF2Cl HF + 2HCl
II. Lewis Structures
Localized Electron Bonding Model
A molecule is composed of atoms that are bound together by sharing pairs of electrons using the atomic orbitals of the bound atoms.
Pairs of e- are either
Shared in a bond (bonding pair)
Localized on a single atom (lone pair)
Duet and Octet Rules are in effect: most stable form of any atom
Lewis Structures
Uses : and – to show arrangement of valence electrons in a molecule
Ionic Compounds are straightforward (and not particularly useful):

6. Covalent Compounds
Shared electrons count for both atoms
A line stands for one pair of electrons
Rules for drawing Lewis structures
Add up total valence e- for all atoms: CH4 = 4x1 + 4 = 8 e-
Use one pair for each bond
Arrange leftover e- to satisfy octet/duet
Examples: CCl4, CO2, H2O and Examples: N2, NH3, CF4, NO+

7. 4. Exceptions to the Octet Rule
a. 2nd row elements C, N, O, F always observe the octet rule.
b. 2nd row elements B and Be (have very small size) often have fewer than 8 electrons around themselves - they are very reactive.
c. 3rd row and heavier elements CAN exceed the octet rule using empty valence d orbitals.
d. When writing Lewis structures, satisfy octets first, then place electrons around elements having available d orbitals.
e. Examples: PCl5, ClF3, RnCl2

8. All the lone pairs electrons c. Examples: i. NO+ ii. NO2- iii. CO2
5. Formal Charge
a. The difference between the number of valence electrons (VE) on the free atom and the number assigned to the atom in the molecule.
b. We need:
a. # VE on free neutral atom—from its group in periodic table
b. # VE “belonging” to the atom in the molecule
One e- for each bond
All the lone pairs electrons
c. Examples:
i. NO+
ii. NO2-
iii. CO2

9. III. Resonance Forms: multiple correct Lewis structures for a given molecule
A. Example: NO3-
B. Rules
1. Bracket the set of structures and separate with 
2. Only electron pairs move to interconvert; atoms don’t move
Electron Pushing
C. Resonance Hybrids = The true structure is a mixture of all resonance forms
1. Every resonance form contributes to the overall structure
2. None of the resonance forms are true pictures of the structure
The real structure is not interconverting between resonance structures, but is a composite all of them, all of the time
Double bonds and lone pairs are delocalized over the whole structure
Single bonds holding the skeleton together are localized and never broken
Example: NO2-

10. Ozone and Resonance Hybrids: O3
Using Formal Charge to pick the “best” Resonance Form
1. Rules regarding Formal Charge
Sum of formal charges for a neutral molecule = 0
Sum of formal charges for an ion = overall charge of the ion
More stable resonance forms minimize formal charges
Separation of charge (+ and – in same structure) is highly unlikely
If formal charges can’t be avoided, match charge with electronegativity

11. HCN Example
Two possible resonance forms can be drawn
Structure A minimized formal charge, so will contribute the most to the overall resonance hybrid between the two forms
OCN- Example
1. Structure C is unlikely because of separation of charge, large charge
Structures A and B have the same amount of formal charge (C is bad)
Oxygen is more electronegative than N, so structure A is favored