Ionic compounds
Properties: hard, brittle, high mp and bp, often water soluble, electrolytes in solution or melt, usually crystalline. All of this suggests large bond energies.
Necessary conditions for an ionic compound: cation (implies neutral species has a low IP) and anion (implies neutral has a high EA)
We develop a theory of ionic bonding using simple electrostatics. Assume
that the ions behave as point charges (i.e., the ions have no volume).
Then build a lattice from condensation of gas phase ions and use Coulomb's
law to calculate the energy.
Ion Pair
q_{+}, q_{–} are the ionic charges in coulombs
r is the distance between ions in meters
_{0} is the permitivitty of free space
Ion square
Since (in this example) q_{+}=q_{–}
ion cube
Continue this over the entire lattice to give:
N_{0} = Avogadros number
Z^{+}, Z^{–} are the ionic charges in units of e (i.e.,
+1, +2, –1, –2, etc.)
r is the interionic distance
M is called the Madelung constant.
M depends solely on geometry, i.e., the kind of lattice but the internuclear
distances are factored out. Madelung constants are tabulated for all the common lattice types.
E_{Coul} is negative (all terms are positive except Z^{–}), i.e., the bonding is favorable. In fact, too favorable: the most stable situation is when r = 0.
Why? Initial point charge assumption does not allow for nuclearnuclear repulsion at close distances.
Correct this empirically: add a term that is repulsive at small r but negligible at larger distances. This means either an exponential term or a 1/r^{n} term with large n.
Born repulsion:
n is known as the Born exponent, B is an arbitrary constant
The lattice energy becomes
Need to find two new quantities: B and n.
Evaluate B at equilbrium because (r_{o} is the equilibrium internuclear distance measured experimentally)
So, at r = r_{o}
To find n, we must move away from the equilibrium position. This means compressing the sample, i.e. measure the compressibility, :
(SI units) 
(Å) 
(kJ/mole) 

AgCl 



AgBr 



AgI 


CaF_{2} 


CaCl_{2} 


CaBr_{2} 


LiF 



LiCl 



LiBr 



LiI 



MgO 



NaCl 



NaBr 



NaI 



KF 



KCl 



KBr 



KI 





Sodium chloride (NaCl) 

Cesium chloride (CsCl) 

Zinc blende (ZnS) 

Wurtzite (ZnS) 

Fluorite (CaF_{2}) 

Rutile (TiO_{2}) 

Anatase (TiO_{2}) 

Cadmium iodide (CdI_{2}) 

Quartz (SiO_{2}) 

Corundum (Al_{2}O_{3}) 

Sample calculation for NaCl:
= 8.854×10^{–12} SI units
e = 1.602×10^{–19} coulombs
M = 1.74756 (NaCl structure)
d = 5.628×10^{–10} m giving r_{o} = 2.814×10^{–10} m
= 4.18×10^{–11} SI
n is found by
This compares to 769.4 kJ/mole experimental (2.4% error)