Calculating lattice energies: Ionic crystals

Given a crystal lattice, the lattice energy is defined to be the energy needed to separate a crystal into its component atoms, molecules, or ions at zero-temperature.

Here we will consider a simple ionic crystal such as NaCl, which is composed of Na$^+$ and Cl$^-$ ions and whose underlying crystal lattice is fcc. Let us begin by looking at a simple one-dimensional ionic crystal, such as that shown in the figure below:

Figure: A simple one-dimensional crystal

Let this one-dimensional crystal contain $N_A$ positive ions and $N_A$ negative ions, so that the total number of particles is $2N_A$. If the positions of the atoms are ${\bf R}_1,...,{\bf R}_{2N_A}$, then the lattice energy is just the total Coulomb energy of the crystal:

$$V_{\rm tot} = {e^2 \over 4\pi\epsilon_0} \sum_{i=1}^{2N_A}\sum_{j=i+1}^{2N_A} {1 \over \vert{\bf R}_i-{\bf R}_j\vert}$$

This expression can be simplified considerably using the regularity of the lattice. Let $R_0$ be the distance between nearest neighbor particles on the lattice. The separation between a given ion and the nearest atom of the same charge will be $2R_0$. Based on this, we decompose $V_{\rm tot}$ into repulsive and attractive contributions:

$$V_{\rm tot} = V_{++} + V_{--} + V_{+-}$$

where $V_{++}$ is the sum of all Coulomb repulsion terms between positive charges, $V_{--}$ is the sum of all Coulomb repulsion terms between negative charges, and $V_{+-}$ is the total Coulomb attraction between opposite charges.

Consider first the Coulomb attraction term. Because of the regularity of the lattice, we can write $V_{+-}$ as

$$V_{+-} = N_A\times({\rm Potential\ energy\ of\ all\ negative\ charges\ with\ one\ positive\ charge)}$$

$$+ N_A\times({\rm Potential\ energy\ of\ all\ positive\ charges\ with\ one\ negative\ charge)}$$

Based on the given distances, this can be written as

$$V_{+-} = -{N_A e^2 \over 4\pi\epsilon_0} \left[{1 \over R_0} + {1 \over 3R... ...\right] + \left[{1 \over R_0} + {1 \over 3R_0} + {1 \over 5R_0} + \cdots\right]$$

$$= -{N_A e^2 \over 2\pi\epsilon_0} \left[{1 \over R_0} + {1 \over 3R_0} + {1 \over 5R_0} + \cdots\right]$$

$$= -{N_A e^2 \over 2\pi\epsilon_0 R_0}\sum_{n=1}^{N_A} {1 \over 2n-1}$$

For the repulsion terms, we can write

$$V_{++} = N_A\times({\rm Potential\ energy\ of\ all\ positive\ charges\ with\ one\ positive\ charge})$$

Note that with each positive charge, there are $N_A-1$ charges that can interact with it. Thus, $V_{++}$ becomes

$$V_{++} = {N_A e^2 \over 4\pi\epsilon_0} \left[{1 \over 2R_0} + {1 \over 4R_0} + {1 \over 6R_0} + cdots\right]$$

$$= {N_A e^2 \over 4\pi\epsilon_0 R_0}\sum_{n=1}^{N_A-1}{1 \over 2n}$$

The same expression results for $V_{--}$. Hence, the total potential energy is

$$V_{\rm tot} = {N_A e^2 \over 4\pi\epsilon_0R_0} \left[\sum_{n=1}^{N_A-1} {1 \over n} - \sum_{n=1}^{N_A} {2 \over 2n-1}\right]$$

$$= {N_A e^2 \over 4\pi\epsilon_0 R_0}M$$

where $M$ is a pure number known as the Madelung constant. Here, we have derived an expression for the Madelung constant of a one-dimensional ionic crystal:

$$M = \sum_{n=1}^{N_A-1} {1 \over n} - \sum_{n=1}^{N_A} {2 \over 2n-1}$$

However, each ionic crystal in any number of dimensions will have a unique Madelung constant that depends on the particular crystal structure. Since $M$ is a pure number, however, it is independent of the unit cell dimensions. For an NaCl crystal, for example, $M=1.7476$, and any crystal that has the same structure as NaCl will have the same Madelung constant. Finally, note that the Madelung constant is really derived in the limit $N_A\longrightarrow\infty$.

Example: What is the lattice energy per mole for an NaCl crystal given that the distance between Na$^+$ and Cl$^-$ is 2.82 Å?

Using the expression for $V_{\rm tot}$ with $M=1.7476$, we have

$$V_{\rm tot} = {N_0 e^2 M\over 4\pi \epsilon_0 R_0}$$

$$= {(6.022\times 10^{23}\;{\rm mol}^{-1})(1.602\times 10^{-19}\;{\rm... ...es 10^{-12}\;{\rm C}^2{\rm J}^{-1}{\rm m}^{-1}) (2.82\times 10^{-14}\;{\rm m})}$$

$$= 861\;{\rm kJ/mol}$$