Formation of Binary
Ionic Compounds
Binary Ionic Compound
Binary-
two
Ionic- ions
Compound- 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
 Li(s)
of solid Li
--> Li(g)
 Enthalpy for sublimation is
161 kJ/mol
Li(s) + ½ F2(g) --> LiF(s)
 Ionization
form Li+
 Li(g)
of Li atoms to
+
Li
e
--> (g) +
 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
F
of ions
 Electron affinity
 F(g) + e --> F (g)
 Electron affinity =
- 328 kJ/mol
Li(s) + ½ F2(g) --> LiF(s)
 Formation
of LiF(s)
 Lattice energy
+
 Li (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
process
 Notice 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 Energy
 The
higher the charge on
each ion, the greater the
lattice energy will be
Lattice Energy
 This
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