The lattice energy

The lattice energy is the energy necessary in order to decompose, at the temperature T=0 K, an ionic crystal in its fundamental atomic component by carrying them at a distance infinite. Ionic crystals are constituted of the combination of positive (cations) and negative (anions) ions. The first are main elements of Group IA (alkaline metals), IIA (earth metals alkaline), part of the IIIA and transitions metals with the lowest oxidation number.  The seconds are from elements of the groups VIA and VIIA. Two ions (A and B) with opposite charges and placed at a distance RAB are attracted with an electrostatic force given by the Coulomb relation:

\displaystyle F_{AB}(R) = \frac{e^2}{4\pi\epsilon_0} \frac{Z_AZ_B}{R^2_{AB}} \hfill (1)

(where e0=8.85×10-12C2J-1m-1 it is the vacuum permittivity), by integrating the previous relation it is possible to derive the energy:

\displaystyle V_{AB}^a = \int_{\inf}^R F_{AB}(r) dr = - \frac{e^2}{4\pi\epsilon_0} \frac{Z_AZ_B}{R_{AB}} \hfill (2)

Beyond the attractive force defined by the Coulombic force, there is also a repulsive one defined by the Born expression:

\displaystyle V_{AB}^r = \frac{B}{4\pi\epsilon_0 R_{AB}^n} \hfill (3)

or, alternatively:

\displaystyle V_{AB}^r = b e^{\frac{R}{\rho}} \hfill (4)

where B, {\rho}, n are parameters obtained experimentally using compressibility measurements.  Therefore, the total interaction energy of two ions in the vacuum is given by the sum of repulsive and attractive terms, as follows:

\displaystyle V_{AB} = - \frac{e^2}{4\pi\epsilon_0} \frac{Z_AZ_B}{R_{AB}} + \frac{B}{4\pi\epsilon_0 R_{AB}^n} \hfill (5)

The energy of an ionic pair in a crystal is obtained by adding the interaction energies between all the possible ionic pairs. In this case, the energy of each ionic pair interaction (for a binary compound) becomes:

\displaystyle V_{AB} = - A \frac{e^2}{4\pi\epsilon_0} \frac{Z_AZ_B}{R_{AB}} + \frac{B}{4\pi\epsilon_0 R_{AB}^n} \hfill (6)

The electrostatic term is now multiplied by the factor A, called Madelung constant, that depends on the type of crystalline reticulum.

Equation 6 can be further simplified eliminating the parameter B. In fact, in the crystal the forces acting on particles to the equilibrium distance (RAB ) are  zero:

\displaystyle \frac{V_{AB}}{dr} = 0 = A \frac{e^2}{4\pi\epsilon_0} \frac{Z_AZ_B}{R_{c}^2} - n\frac{B}{4\pi\epsilon_0 R_{c}^{n+1}} \hfill (7)

from which the value of the constant B is obtained:

\displaystyle B = A \frac{e^2}{4\pi\epsilon_0} \frac{Z_AZ_B}{n} R_c^{n-1} \hfill (8)

by replacing B in equation 6 the new relation is obtained:

\displaystyle V_{AB} = - A \frac{e^2}{4\pi\epsilon_0} \frac{Z_AZ_B}{R_{c}} \left( 1 - \frac{1}{n} \right) \hfill (9)

The term n depends on the electronic configuration of the elements. In the table below, the values of n are reported for various electronic configurations.

Table 1: Value of n for various electronic configurations.

Electronic Configuration 


He (Li+, Be2+)


Ne, Na+, Mg2+, Al3+


Cu+, Ar, K+, Ca2+, Zn2+


Ag+, Kr, Rb+, Sr2+, Cd2+


Au+, Xe, Cs+, Ba2+, Hg2+


If you have found interesting and useful my article, do not forget to press “Like it” and subscribe for updates on new ones!

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.