Formation of Binary Ionic Compounds

Binary Ionic Compound

Binary- twoIonic- ionsCompound- joined together

Binary Ionic Compound

Solid formed between a metal and a nonmetal

The oppositely charged ions together have lower energy

Lattice Energy

The change in energy that takes place when separated gaseous ions are packed together to form an ionic solid

Lattice Energy

The energy released when an ionic solid is formed

M+(g) + X-

(g) --> MX(s)

Sign will be negative b/c process is exothermic

Energy Changes

Look at the formation of an ionic solid from its elements

Keep in mind that energy is a STATE FUNCTION!!

Li(s) + ½ F2(g) --> LiF(s)

Sublimation of solid LiLi(s) --> Li(g)

Enthalpy for sublimation is 161 kJ/mol

Li(s) + ½ F2(g) --> LiF(s)

Ionization of Li atoms to form Li+

Li(g) --> Li+(g) + e-

Ionization is 520 kJ/mol

Li(s) + ½ F2(g) --> LiF(s)

Dissociation of F2 molecules to F atoms

½ F2(g) --> F(g)

154 kJ/mol Divide by two = 77 kJ/mol

Li(s) + ½ F2(g) --> LiF(s)

Formation of F- ionsElectron affinityF(g) + e- --> F-

(g)

Electron affinity = - 328 kJ/mol

Li(s) + ½ F2(g) --> LiF(s)

Formation of LiF(s)

Lattice energyLi+(g) + F-

(g) --> LiF(s)

Lattice energy = -1047 kJ/mol

Li(s) + ½ F2(g) --> LiF(s)

Sum of these 5 processes yields the desired overall reaction

-617 kJ (per mole of LiF)

Energy Diagram

Summarizes processNotice that most of the

processes are endothermic and unfavorable

Energy Diagram

However, the large lattice energy makes the whole process worthwhile

K(s) + ½ Cl2(g) --> KCl(s)

Sublimation of K = +64 kJ Ionization of K = +419 kJ Bond energy of Cl2 = +240 kJ

e- affinity of Cl = -349 kJ Lattice energy = -690 kJ

K(s) + ½ Cl2(g) --> KCl(s)

Net energy of formation equals the sum of the energy changes

Hfo = -436 kJ

Lattice Energy Calculations

Lattice energy is important in contributing to the stability of the compounds

Lattice Energy Calculations

Modified from Coulomb’s Law

Lattice energy = k(Q1Q2/r)k = constant that depends

on structure of solid

Lattice energy = k(Q1Q2/r)

Q1 and Q2 = charges of ions

r = shortest distance between ions

Lattice EnergyThe higher the charge on

each ion, the greater the lattice energy will be

Lattice EnergyThis value counteracts the

higher endothermic ionization energies, thus resulting in a more stable energetically stable crystal

Li(s) + ½ Br2(g) --> LiBr(s)

Ionization of Li = +520 kJ/mol e- affinity for Br = -324 kJ/mol sublimation of Li = +161 kJ/mol lattice energy = -787 kJ/mol bond energy Br2 = +193 kJ/mol

Li(s) + ½ Br2(g) --> LiBr(s)

Hfo = -334 kJ