NOTES

ELECTRON AFFINITY

It is the enthalpy change accompanying the addition of an electron to a gaseous atom or ion. The trend in electron affinity is the same as electronegativity and ionization energy.

Example:

O + e-  --> O^-  ΔH^o = -142 kJ

O^- + e-  --> O^2-  ΔH^o = +880 kJ      Note that energy has to be supplied to overcome the repulsions between two negative species

Therefore

O + 2e-  -->O^2-  ΔH^o = +639  kJ

LATTICE ENERGY

It is the energy required to break up 1 mole of the crystal into gaseous ions.

Example 1:

NaF(s) + 923 kJ --> Na⁺(g) + F^-(g)

NaCl(s) + 786 kJ --> Na⁺(g) + Cl^-(g)

NaBr(s) + 747 kJ --> Na⁺(g) + Br^-(g)

NaI(s) + 704 kJ --> Na⁺(g) + I^-(g)

Notice that as the halogen increases in size the lattice energy decreases. This happens because the distance between the two nuclei increases making it easier to separate them.

Example 2:  KBr has lower lattice energy than NaCl

Problems:

1. Arrange the following compounds by decreasing lattice energy:  LiF, MgF₂, NaF and KF

BOND ENERGY

Amount of energy it takes to break a bond.

Example: The bond energy of H-H is 436 kJ/mol

CALCULATING ENTHALPY (ΔH^O) FROM BOND ENERGY

The bond energy is the amount of energy it takes to break a bond.

Example: The bond energy of H-H is 436 kJ/mol

ΔH^O = energy absorbed (+) + energy released (-)

Example

H-H bond E = 436 kJ/mol

Br-Br bond E = 193 kJ/mol

H-Br bond E = 366 kJ/mol

Given the reaction H₂ + Br₂ à 2 HBr

E absorbed to break bonds of reactants = 1 mol (436 kJ/mol) + 1 mol (193 kJ/mol) = 629 kJ

E released to form bonds in products = 2 mol (-366 kJ/mol) = -732 kJ

Enthalpy (ΔH^o) = 629 kJ + (-732 kJ) = -103 kJ

Problems

1. What is the enthalpy (ΔH^o) of this reaction, 2H₂(g) + O₂(g) --> 2H₂O(g)

H-H bond E = 436 kJ/mol

O=O bond E = 499 kJ/mol

O-H bond E = 460 kJ/mol

SOME EXCEPTIONS TO THE OCTET RULE

Draw the Lewis structure of the following BF₃, SF₆, PCl₅, and BeCl₂

RESONANCE STRUCTURE

A resonance structure is used to illustrate a special bond length which is in general an intermediate between a single and double bond length.

Example: NO₃^-

Problems:

1. Draw the resonance structure of O₃ ,NO₂^-, and HNO₃

The VSEPR

The Valence Shell Electron Pair Repulsion is used to find the shape of molecules. In the VSEPR a pair site is a lone pair of electron, a single, a double or a triple bond.

First, draw the Lewis structure of the molecule

Second, arrange the pair sites so that they are as far as possible from each other.

Third, consider the shape created by all pair sites excluding the lone e- pairs.

Example:

COORDINATE COVALENT BOND

It is a covalent bond where the two electrons come from only one atom.

Example: H₂O + H⁺ à H₃O⁺

Problem

1. Draw the Lewis structure of the following reaction NH₃ + H⁺ --> NH₄⁺

MOLECULE-ION ATTRACTIONS

It is the attraction between polar molecules and ions.

Example: 

NaCl(s) + H₂O(l) --> Na⁺(aq) +Cl^-(aq)

The ions are said to be hydrated meaning that they are surrounded by many water molecules (the diagram shows only four).

VALENCE BOND THEORY

During bonding orbitals overlap.

Example: H₂, I₂, and HF

When two orbitals overlap to build up e- density along the axis between two nuclei, the resulting localized bond is called a sigma bond (σ). In multiple bonds there is only one sigma bond and the other(s) are called pi bond(s) (π) formed from the overlap of p orbitals.

Example:

ORBITAL HYBRIDIZATION 

The concept of hybrid orbital is use to explain bonding (overlap of orbitals) when unpaired orbitals are not available.

Example: In BeCl₂ the valence configuration of beryllium is 2s² and Cl is 3s² 3p⁵. Chlorine has an unpaired orbital but Be has none. How can the bond between Be and Cl be explained? The solution is hybridization.

Shortcut to hybridization

1. Draw the Lewis structure

2. Add all sigma bonds and lone pairs of electrons from the element

If the total is 2 then it is sp hybridization for this element

If the total is 3 then it is sp2 hybridization for this element

If the total is 4 then it is sp3 hybridization for this element

Example: CO₂, SO₂, H₂O, and NH₃

VAN DER WAALS OR DISPERSION FORCES

In nonpolar molecules electrons can temporally occupy more one side of the molecule than the other creating a dipole and therefore dipole-dipole attractions. Those relatively weak forces are responsible for the formation of the liquid and also solid phase in nonpolar molecules (monatomic or polyatomic). As the molecular size increases van der Waals forces increases. 

Example:

Within the noble gases the boiling point increases moving down the group since a larger molecular mass creates stronger attractions.

home