# 7.7.1: Lattice Energy

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##### 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}:

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

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

$U_{lattice} = - E_{cryst} \nonumber$

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 Elattice 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) --------------


Ecryst = -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

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

Skill: Explain the trend of lattice energy.

##### Exercise $$\PageIndex{2}$$

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

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, Ecryst for CaO.

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, Ecryst .