10.11: Lattice Energy

Discussion Questions

• How is lattice energy estimated using Born-Haber cycle?
• How is lattice energy related to crystal structure?

The Lattice energy, \(U\), is the amount of energy required to separate a mole of the solid (s) into a gas (g) of its ions.

\[\ce{M_{a} L_{b} (s) \rightarrow a M^{b+} (g) + b X^{a-} (g) } \label{eq1}\]

This quantity cannot be experimentally determined directly, but it can be estimated using a Hess Law approach in the form of Born-Haber cycle. It can also be calculated from the electrostatic consideration of its crystal structure. As defined in Equation \ref{eq1}, the lattice energy is positive, because energy is always required to separate the ions. For the reverse process of Equation \ref{eq1}:

\[\ce{ a M^{b+} (g) + b X^{a-} (g) \rightarrow M_{a}L_{b}(s) }\]

the energy released is called energy of crystallization (\(E_{cryst}\)). Therefore,

\[U_{lattice} = - E_{cryst}\]

Values of lattice energies for various solids have been given in literature, especially for some common solids. Some are given here.

Table \(\PageIndex{1}\): Comparison of Lattice Energies (U in kJ/mol) of Some Salts

Solid	U	Solid	U	Solid	U	Solid	U
LiF	1036	LiCl	853	LiBr	807	LiI	757
NaF	923	NaCl	786	NaBr	747	NaI	704
KF	821	KCl	715	KBr	682	KI	649
MgF2	2957	MgCl2	2526	MgBr2	2440	MgI2	2327

The following trends are obvious at a glance of the data in Table \(\PageIndex{1}\):

• As the ionic radii of either the cation or anion increase, the lattice energies decrease.
• The solids consists of divalent ions have much larger lattice energies than solids with monovalent ions.

How is lattice energy estimated using Born-Haber cycle?

Estimating lattice energy using the Born-Haber cycle has been discussed in Ionic Solids. For a quick review, the following is an example that illustrate the estimate of the energy of crystallization of NaCl.

Hsub of Na = 108 kJ/mol (Heat of sublimation)
D of Cl2 = 244 (Bond dissociation energy)
IP of Na(g) = 496 (Ionization potential or energy)
EA of Cl(g) = -349 (Electron affinity of Cl)
Hf of NaCl = -411 (Enthalpy of formation)

The Born-Haber cycle to evaluate E_lattice is shown below:

     -----------Na+ + Cl(g)--------
         ­                       |
         |                       |-349
         |496+244/2              ¯
         |                 Na+(g) + Cl-(g)
         |                       |
   Na(g) + 0.5Cl2(g)             |
         ­                       |
         |108                    |
         |                       |Ecryst= -788 
   Na(s) + 0.5Cl2(l)             |
         |                       |
         |-411                   |
         ¯                       ¯
     -------------- NaCl(s) --------------

E_cryst = -411-(108+496+244/2)-(-349) kJ/mol
= -788 kJ/mol.

Discussion
The value calculated for U depends on the data used. Data from various sources differ slightly, and so is the result. The lattice energies for NaCl most often quoted in other texts is about 765 kJ/mol.

Compare with the method shown below

Na(s) + 0.5 Cl2(l) ® NaCl(s)	- 411	Hf
Na(g) ® Na(s)	- 108	-Hsub
Na^+(g) + e ® Na(g)	- 496	-IP
Cl(g) ® 0.5 Cl2(g)	- 0.5 * 244	-0.5*D
Cl^-(g) ® Cl(g) + 2 e	349	-EA
Add all the above equations leading to
Na^+(g) + Cl^-(g) ® NaCl(s)	-788 kJ/mol = Ecryst

Lattice Energy is Related to Crystal Structure

There are many other factors to be considered such as covalent character and electron-electron interactions in ionic solids. But for simplicity, let us consider the ionic solids as a collection of positive and negative ions. In this simple view, appropriate number of cations and anions come together to form a solid. The positive ions experience both attraction and repulson from ions of opposite charge and ions of the same charge.

As an example, let us consider the the NaCl crystal. In the following discussion, assume r be the distance between Na⁺ and Cl^- ions. The nearest neighbors of Na⁺ are 6 Cl^- ions at a distance 1r, 12 Na⁺ ions at a distance 2r, 8 Cl^- at 3r, 6 Na⁺ at 4r, 24 Na⁺ at 5r, and so on. Thus, the energy due to one ion is

\[ E = \dfrac{Z^2e^2}{4\pi\epsilon_or} M \label{6.13.1}\]

The Madelung constant, \(M\), is a poorly converging series of interaction energies:

\[ M= \dfrac{6}{1} - \dfrac{12}{2} + \dfrac{8}{3} - \dfrac{6}{4} + \dfrac{24}{5} ... \label{6.13.2}\]

with

• \(Z\) is the number of charges of the ions, (e.g., 1 for NaCl),
• \(e\) is the charge of an electron (\(1.6022 \times 10^{-19}\; C\)),
• \(4\pi \epsilon_o\) is 1.11265x10^-10 C²/(J m).

The above discussion is valid only for the sodium chloride (also called rock salt) structure type. This is a geometrical factor, depending on the arrangement of ions in the solid. The Madelung constant depends on the structure type, and its values for several structural types are given in Table 6.13.1.

A is the number of anions coordinated to cation and C is the numbers of cations coordinated to anion.

Table \(\PageIndex{2}\): Madelung Constants

Compound	Crystal Lattice	M	A : C	Type
NaCl	NaCl	1.74756	6 : 6	Rock salt
CsCl	CsCl	1.76267	6 : 6	CsCl type
CaF2	Cubic	2.51939	8 : 4	Fluorite
CdCl2	Hexagonal	2.244
MgF2	Tetragonal	2.381
ZnS (wurtzite)	Hexagonal	1.64132
TiO2 (rutile)	Tetragonal	2.408	6 : 3	Rutile
bSiO2	Hexagonal	2.2197
Al2O3	Rhombohedral	4.1719	6 : 4	Corundum
A is the number of anions coordinated to cation and C is the numbers of cations coordinated to anion.

Madelung constants for a few more types of crystal structures are available from the Handbook Menu. There are other factors to consider for the evaluation of energy of crystallization, and the treatment by M. Born led to the formula for the evaluation of crystallization energy \(E_{cryst}\), for a mole of crystalline solid.

\[ E_{cryst} = \dfrac{N Z^2e^2}{4\pi \epsilon_o r} \left( 1 - \dfrac{1}{n} \right)\label{6.13.3a} \]

where N is the Avogadro's number (6.022x10^-23), and n is a number related to the electronic configurations of the ions involved. The n values and the electronic configurations (e.c.) of the corresponding inert gases are given below:

n =	5	7	9	10	12
e.c.	He	Ne	Ar	Kr	Xe

The following values of n have been suggested for some common solids:

n =	5.9	8.0	8.7	9.1	9.5
e.c.	LiF	LiCl	LiBr	NaCl	NaBr

Example \(\PageIndex{1}\)

Estimate the energy of crystallization for \(\ce{NaCl}\).

Solution

Using the values giving in the discussion above, the estimation is given by Equation \ref{6.13.3a}:

\[ \begin{align*} E_cryst &= \dfrac{(6.022 \times 10^{23} /mol (1.6022 \times 10 ^{-19})^2 (1.747558)}{ 4\pi \, (8.854 \times 10^{-12} C^2/m ) (282 \times 10^{-12}\; m} \left( 1 - \dfrac{1}{9.1} \right) \\[4pt] &= - 766 kJ/mol \end{align*}\]

Discussion

Much more should be considered in order to evaluate the lattice energy accurately, but the above calculation leads you to a good start. When methods to evaluate the energy of crystallization or lattice energy lead to reliable values, these values can be used in the Born-Haber cycle to evaluate other chemical properties, for example the electron affinity, which is really difficult to determine directly by experiment.

Exercise \(\PageIndex{1}\)

Which one of the following has the largest lattice energy? LiF, NaF, CaF₂, AlF₃

Answer

Skill: Explain the trend of lattice energy.

Exercise \(\PageIndex{2}\)

Which one of the following has the largest lattice energy? LiCl, NaCl, CaCl₂, Al₂O₃

Answer

Corrundum Al₂O₃ has some covalent character in the solid as well as the higher charge of the ions.

Exercise \(\PageIndex{3}\)

Lime, CaO, is know to have the same structure as NaCl and the edge length of the unit cell for CaO is 481 pm. Thus, Ca-O distance is 241 pm. Evaluate the energy of crystallization, E_cryst for CaO.

Answer

Energy of crystallization is -3527 kJ/mol

Skill: Evaluate the lattice energy and know what values are needed.

Exercise \(\PageIndex{4}\)

Assume the interionic distance for NaCl₂ to be the same as those of NaCl (r = 282 pm), and assume the structure to be of the fluorite type (M = 2.512). Evaluate the energy of crystallization, E_cryst .

Answer

-515 kJ/mol

Discussion: This number has not been checked. If you get a different value, please let me know.

Contributors and Attributions

• Chung (Peter) Chieh (Professor Emeritus, Chemistry @ University of Waterloo)