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 [1581 - 3(414) = 339]

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

Find DH for: 1 H2 + 1 F2 2 HF 1(436 kJ/mol) + 1(159 kJ/mol) – 2(565 kJ/mol) = -535 kJ/mol = DH 3. Example: CH4 + 2Cl2 + 2F2 ----> CF2Cl2 + 2HF + 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):

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+

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

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

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-

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

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