Chemistry 2810 Lecture Notes
5.3
Dr. R. T. Boeré
Page91
Thermodynamics of lattices
5.3.1 Ionic lattices and experimental lattice energy
Ionization energy: In the gas phase, the energy of these processes has been measured, so that for the reaction:
Na → Na + + e − ∆Hrxn = first enthalpy of ionization
Electron affinity: For the opposite process, e.g.:
F + e − → F − ∆Hrxn = enthalpy of electron attachement
In other words, -ve enthalpies of electron attachment implies that the element favours formation of anions. Note that
even the alkali metals have (small) -ve 'Helectron attachment. But the alkaline earths have positive values, while the halogens
have large -ve values, as expected.
Let us consider the imaginary process where one mole of Na+
and Cl- ions in the gas phase at infinite separation are
slowly brought together. The attraction between cations and anions will tend to pull them together, whilst the repulsions of
like charges retards this process.
If this process is continued for an ionic substance like NaCl, energy will be released as the ions approach, and a
crystalline solid will eventually be formed, NaCl(cryst). The energy of this reaction is an enthalpy which is given a common
name, the "lattice energy."
Na+(g) + Cl-(g) à NaCl(cryst) ∆H = -V = -(lattice energy)
We measure lattice energy using the method of thermochemical cycles - the Born-Haber cycle (e.g. for NaCl):
∆ HL.E. = -V
+
+
Na (g)
Cl (g)
NaCl(c)
∆ H1st I.E.
∆ H1srE.A.
Na (g)
∆
∆ HB.E.
Hsubl
Na (s)
∆ H°f
Cl(g)
2
+
Cl2 (g)
∆ HL.E. = -V = ∆ H1st I.E. - ∆ H subl - ∆ H1srE.A. -
∆ HB.E.
2
= -(+496) - (+108) - (-349) - (+243/2) + (-411)
= - 788 kJ mol-1
+ ∆ H°f
5.3.2 Theoretical Calculation of lattice energy from first principles
As a complete alternative to the above method, we can also use the theory of electrostatic attraction to calculate what the
lattice energy should be. Then the degree of agreement between our theoretical values and experiment will be an indication of
how good the assumption of ionic bonding and hence of the electrostatic model really is.
a)
The attractive energy
The energy of attraction is given by Coulomb's law, which in S.I. units can be written:
E AB = N A
( Z e )× ( Z e )
A
B
4πε0rAB
The Z are multipliers of the unit charge e, which is 1.602 10-19 C. NA is Avogadro's number, which merely gives energies
in kJ mol-1, the common unit of enthalpy. e0 is the permittivity of free space, 8.854 10-12C2m-1J-1. If we enter these constants,
and express r in Å, the units used in the tables in SAL, the equation simplifies to (in kJ mol-1):
E AB =
1389Z A Z B
rAB
Chemistry 2810 Lecture Notes
b)
Dr. R. T. Boeré
Page92
The repulsive energy
As we have already seen, ions are NOT rigid spheres. The attractive electrostatic force will be balanced at the point where
the ions start to interpenetrate by repulsions between their respective valence electrons. This repulsion is not purely
electrostatic. It involves a quantum-mechanical interaction between the electrons, and follows a 1/rn law, where n > 2, and is
governed chiefly by the principal quantum number. The expression is:
E repulsive =
N AC '
rAB n
C' is just a constant, and n is known as the Born exponent (the same Born of B-H cycle fame.) It can be estimated for alkali
halides by taking the average of the values for the noble gas configurations of the constituent ions:
Noble Gas
He
Ne
Ar
Kr
Xe
Born exponent
5
7
9
10
12
Note that this is a simplified approach compared to that taken by your textbook. The form of the complete equation will
therefore be different.
c)
The generalized energy expression
The crystal structure is determined by the equilibrium between these attractive and repulsive forces, where:
E AB =
1389 Z A Z B N A C'
−
rAB
rAB n
By evaluating the derivative of V, which must be at a minimum at equilibrium, it is possible to solve for C'. Inserting the
value of C' and simplifying leads to:
E AB =
d)
1389 Z A Z B  1 
1 − 
 n
rAB
Madelung constant and the lattice energy (Born-Meyer equation)
So far all we have considered is two oppositely charged ions on a straight line. We must allow for the 3-D geometrical
array of ions in the lattice, e.g. take the NaCl lattice. Fortunately, this is only a matter of geometry, a point first realized by
Madelung, and the resulting calculations bear his name.
We consider only the interactions of one ion with all the rest. Let it be a Na+ ion. This sodium ion has six nearestneighbors @ r0, if we use this value in the lattice energy equation (the sum of the radii of the ions.) Next are 12 Na+ ions @
Chemistry 2810 Lecture Notes
Dr. R. T. Boeré
Page93
√2 r. Since r is in the electrostatic equation, we now take r as one from here on in. Next are 8 Cl- @ √3. Followed by 6 Na+
@ √4, etc. This leads to the infinite series:
6 12
8
6
24
−
+
−
+
− ........
1
2
3
4
5
This series, or at least its evaluation, is called the Madelung constant.
This series is in fact a very slowly converging one, but it does converge to 1.74756. It is a number ONLY dependent on
the crystal geometry, and has been tabulated for all common lattices. The Madelung constant is simply multiplied by the 1:1
electrostatic formula to get an expression for V. This is known as the Born-Meyer equation. For NaCl, this evaluates as 751
kJ mol-1.
V AB = A ×
1389 Z A Z B  1 
1 − 
 n
rAB
Thus the lattice energy of an ionic solid depends on three factors
(1)
A, defined by structure type
(2)
rAB, determined by the size of the ions
(3)
n, determined by the principle quantum number of the ions
e)
Table of Madelung Constants
The table below summarizes the values obtained for evaluating the infinite series of the Madelung constants for each
lattice type. Note that the values depend only on the lattice type, not the chemical identity of the ions. I have also included
the theoretical range of radius ratios which correspond to each lattice type. Remember that radius ratios are also obtained from
purely geometrical considerations.
Lattice type
A
r+/r_
Zinc blende 1.63805
0.225-0.414
Wurtzite
1.64132
0.225-0.414
NaCl
1.74756
0.414-0.732
CsCl
1.76267
0.732-1.00
b-quartz
2.201
0.225-0.414
Rutile
2.4080
0.414-0.732
CaF2
2.51939
0.732-1.00
f)
Comparison of theoretical and measured lattice energies
Compare this with the solution to V for NaCl above. It is different by only (788 - 751) = 37 kJ, agreement within 5%.
That is very good agreement between experiment and calculation for a relatively crude theory. The textbook contains a table
of calculated and measured lattice energies using an even more sophisticated lattice energy formula than the one discussed
there. Thus the calculated V = 778 kJ mol-1, now only 1% off. For such a small difference in precision, the simple-minded
approach is greatly to be preferred. Clearly the theory developed so far accounts for the great bulk of the energy of ionic
bonding. Where large deviations between the Born-Mayer equation and experimental lattice energies occur, this is evidence
for a strong covalent contribution to the bonding.
If there is good agreement between the lattice energy calculated by the Born-Meyer equation and the measured value, the
compound is considered to be truly ionic. AgCl, on the other hand has Vcalc = 700 kJ mol-1, vs. Vexpl = 912 kJ mol-1.
Chemists conclude from this that there must be a fair covalent contribution to the bonding in AgCl.
Chemistry 2810 Lecture Notes
Dr. R. T. Boeré
Page94
5.3.3. Lattice energy: some applications
There are many important consequences of the crystalline lattice energy for chemical behaviour of ionic compounds. Here
follow some of them:
a)
Effect of relative size of cation and anion on solubility
The best crystal packing, and generally the best lattice energies per mole of ions is usually given by ions of the same
charge and size. Salts with strong lattice energies tend to be insoluble. (We will consider other effects on solubility in the
next chapter.) Thus large organic cations are precipitated by big anions such as BPh4-, large anions by cations such as
NnBu4+.
Salts which are poorly matched by ion size, i.e. at or beyond the limits given by the radius ratio rule, tend to have low
lattice energies, and hence are often highly soluble.
b)
Effect of relative size of cation and anion on thermal stability
For example, MgCO3 is much less stable towards heat than CaCO3, though both lose CO2 when the solids are heated.
MCO3 à MO + CO2(g)
The decomposition parameter 3, the temperature at which there is 1 atmosphere of CO2 in equilibrium with the solid in a
closed vessel, is 300 °C for the Mg and 840 °C for the Ca carbonates. The explanation is that this process converts the large,
non-spherical carbonate ion to the smaller, spherical oxide anion.
c)
Effect of relative size of cation and anion on hydration
The ionic compound Mg(ClO4)2, which has very small cations compared to the large ClO4- anions, is a very strong
desiccant. It literally sucks water out of its environment. Similar reasoning is behind the fact that many transition metals
crystalize as hydrates. The water becomes attached to the metal ion, and increases its effective size.
Copper sulfate forms a pentahydrate, since the Cu2+ ions by themselves are quite small w.r.t. the SO42- ions. Hence, given
a moist environment, the copper ions sheath themselves in four water ions, which increases the effective size of the cation,
making it a better match for the sulfate ion.
CuSO4.5H2O = [Cu(OH2)4][SO4.H2O] à CuSO4 + 5 H2O