10.11: Lattice Energy
 Page ID
 392374
Discussion Questions
 How is lattice energy estimated using BornHaber 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 BornHaber 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.
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 
MgF_{2}  2957  MgCl_{2}  2526  MgBr_{2}  2440  MgI_{2}  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 BornHaber cycle?
Estimating lattice energy using the BornHaber 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 BornHaber cycle to evaluate E_{lattice} is shown below:
Na^{+} + Cl(g)   349 496+244/2 ¯  Na^{+}(g) + Cl^{}(g)   Na(g) + 0.5Cl_{2}(g)   108   E_{cryst}= 788 Na(s) + 0.5Cl_{2}(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 Cl_{2}(l) ® NaCl(s)   411  H_{f} 
Na(g) ® Na(s)   108  H_{sub} 
Na^{+}(g) + e ® Na(g)   496  IP 
Cl(g) ® 0.5 Cl_{2}(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 = E_{cryst} 
Lattice Energy is Related to Crystal Structure
There are many other factors to be considered such as covalent character and electronelectron 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} C^{2}/(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.



A : C  Type 

NaCl  NaCl  1.74756  6 : 6  Rock salt 
CsCl  CsCl  1.76267  6 : 6  CsCl type 
CaF_{2}  Cubic  2.51939  8 : 4  Fluorite 
CdCl_{2}  Hexagonal  2.244  
MgF_{2}  Tetragonal  2.381  
ZnS (wurtzite)  Hexagonal  1.64132  
TiO_{2} (rutile)  Tetragonal  2.408  6 : 3  Rutile 
bSiO_{2}  Hexagonal  2.2197  
Al_{2}O_{3}  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 BornHaber 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, CaF_{2}, AlF_{3}
 Answer

Skill: Explain the trend of lattice energy.
Exercise \(\PageIndex{2}\)
Which one of the following has the largest lattice energy? LiCl, NaCl, CaCl_{2}, Al_{2}O_{3}
 Answer

Corrundum Al_{2}O_{3} 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, CaO 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 NaCl_{2} 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)