# 167: A Simple Calculation of the Lattice Energy of LiH

Lithium hydride is a white crystalline solid with the face-centered cubic crystal structure (see lattice shown below). The model for LiH(s) proposed in this study constists of the following elements:

1. The bonding in LiH(s) is completely ionic. The lattice sites are occupied by the spherical, two-electron ions, Li+ and H-.

2. The electrons of Li+ and H- occupy hydrogenic 1s atomic orbitals with adjustable scale factors α and β, respectively. Expressed in atomic units the wavefunctions have the form,

Ψ(1,2) = 1s(1)1s(2) = (α3 /π)exp[-α(r1 + r2)]

The scale factor determines how rapidly the wavefunction (and, therefore, the electron density) diminishes as the distance from the nucleus increases. α and β are, therefore, inversely related to the atomic radius. The larger α and β, the smaller the ionic radii are.

3. The average distance of an electron from the nucleus, <r>, in a scaled 1s orbital is 1.5/α. Therefore, it seems reasonable to take 2, or 3/α as the effective ionic radius in the solid. It is easy to show that 94% of the charge is contained within this radius. (See Appendix)

4. Van der Waals interations between the electron clouds of the ions and the quantum mechanical zero-point energy of the lattice are neglected.

To check the validity of this model the lattice energy of LiH(s) will be calculated and compared to the value obtained by a Born-Haber analysis. The lattice energy is defined as the energy required to bring about the following process,

LiH(s) ----> Li+ (g) + H-(g)

The determination of the lattice energy on the basis of the proposed model, therefore, proceeds by calculating the ground state energies of Li+(g) and H-(g). and subtracting from them the ground state energy of LiH(s). Since terms for the kinetic energy of the ions are not included, the calculations refer to absolute zero.

## Li+(g) and H-(g)

The calculations for the ground-state energies of Li+ (g) and H-(g) are similar to that of He. The energy operators consist of five terms: kinetic energy operators for each of the electrons, electron-nuclear potential energy operators for each of the electrons, and an electron-electron potential energy operator.

$\begin{matrix} H_{Li} = - \frac{1}{2r_1} \frac{d^2}{dr_1^2} r_1 - \frac{1}{2r_2} \frac{d^2}{dr_2^2} r_2 - \frac{3}{r_1} - \frac{3}{r_2} + \frac{1}{r_{12}} \\ H_{H} = - \frac{1}{2r_1} \frac{d^2}{dr_1^2} r_1 - \frac{1}{2r_2} r_2 - \frac{1}{r_1} - \frac{1}{r_2} + \frac{1}{r_{12}} \end{matrix}$

When the trial wavefunction and the appropriate energy operator is used in the variational integral,

$E = \int_0^{ \infty} \Psi (1,~2) H \Psi (1,~2) d \tau_1 d \tau_2$

the following expression result (see Appendix for details):

$\begin{matrix} E_{Li} = \alpha^2 - 6 \alpha + \frac{5}{8} \alpha & E_H = \beta^2 - 2 \beta + \frac{5}{8} \beta \end{matrix}$

Minimization of the energy with respect to the scale factors to obtain the ground state energies of the gas-phase ions is the next step.

## Calculation of the energies of the gas phase ions:

Seed value for the cation scale factor: α = 3

Calculate the energy and radius of the gase phase cation: $$E_{Li} ( \alpha ) = \alpha^2 - 5.375 \alpha$$

$\begin{matrix} \alpha = \text{Minimize} \left( E_{Li},~ \alpha \right) & E_{Li} ( \alpha ) = -7.2227 & E_{Li} = E_{Li} ( \alpha ) & R_{Li} ( \alpha ) & R_{Li} = \frac{3}{ \alpha} & R_{Li} = 1.1163 \end{matrix}$

Seed value for the anion scale factor: β = 1

Calculate the energy and radius of the gas phase anion: $$E_{Li} ( \beta ) = \beta^2 - 1.375 \beta$$

$\begin{matrix} \beta = \text{Minimize} \left( E_H,~ \beta \right) & \beta = 0.6875 & E_H ( \beta) = -0.4727 & E_H = E_H ( \beta ) & R_H = \frac{3}{ \beta} & R_H = 4.3636 \end{matrix}$

## Lithium hydride solid - LiH(s)

As noted above, LiH has the face-centered cubic structure shown below.

The ground state energy of LiH(s) consists of three terms: the internal energy of Li+, the internal energy of H-, and the coulombic interaction energy of the ions occupying the lattice sites.

$E_{LiH} = E_{Li} + E_H + E_{coul}$

where

$\begin{matrix} E_{coul} = - \frac{1.748}{R_c + R_a} & \text{for } \frac{R_c}{R_a} >= .414 & E_{coul} = - \frac{1.748}{ \sqrt{2} R_a} & \text{for } \frac{R_c}{R_a} < 0.414 \end{matrix}$

Here 1.748 is the Madelung constant for the face-centered cubic structure for singly charged ions. Rc and Ra are the radii of the cation and anion. (Rc + Ra) is the inter-ionic separation for situations (Rc/Ra >= .414) in which there is cation-anion contact, while 1.414Ra is the inter-ionic separation for those circumstances (Rc/Ra < .414) in which there is only anion-anion contact. On the basis of assumption 3 of the model, Rc and Ra are replaced by 3/α and 3/β, the effective ionic radii of the cation and the anion. The coulombic contribution now has the form

$\begin{matrix} E_{coul} = - \frac{1.478}{ \frac{3}{ \alpha} + \frac{3}{ \beta}} & \text{for } \frac{ \beta}{ \alpha} >= .414 & E_{coul} = - \frac{1.748}{ \frac{ \sqrt{2} 3}{ \beta}} & \text{for} \frac{ \beta}{ \alpha} < .414 \end{matrix}$

Minimization of the energy of the solid simultaneously with respect to α and β is outlined below.

Energy of the solid assuming anion-cation contact. $$f( \alpha,~ \beta) = \alpha^2 - 5.375 \alpha + \beta^2 - 1.375 \beta - \frac{1.748}{ \frac{3}{ \alpha} + \frac{3}{ \beta}}$$

Energy of the solid assuming anion-anion contact and that the cation rattles in the octahedral hole. $$g( \alpha,~ \beta) = \alpha^2 - 5.375 \alpha + \beta^2 - 1.375 \beta - \frac{1.748 \beta}{3 \sqrt{2}}$$

Composite expression for the energy of the solid using a conditional statement. $$E_{LiH} ( \alpha,~ \beta) = \text{if} \left( \frac{ \beta}{ \alpha} \geq .414,~f( \alpha,~ \beta),~ g( \alpha,~ \beta) \right)$$

Minimization of the energy of LiH with respect to the parameters α and β.

$\begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \text{Minimize} \left( E_{Li},~ \alpha,~ \beta \right)$

$\begin{matrix} \alpha = 2.6875 & R_c = \frac{3}{ \alpha} & R_c = 1.1163 & \text{R}_c \text{ (experimental) = 1.134} \\ \beta = 0.8935 & R_a = \frac{3}{ \beta} & R_a = 3.3576 & \text{R}_a \text{ (experimental) = 3.931} \end{matrix}$

$\begin{matrix} E_{LiH} ( \alpha,~ \beta) = -8.0210 & E_{LiH} = E_{LiH} ( \alpha,~ \beta) \end{matrix}$

Comparison of gas-phase and solid-state ion energies (see Appendix for interpretation):

$\begin{matrix} \text{Cation:} & E_{Lis} = \alpha^2 - 5.375 \alpha & E_{Lis} = -7.227 & E_{Li} = -7.227 & \text{Cation energy doesn't change.} \\ \text{Anion:} & E_{Hs} = \beta^2 - 1.375 \beta & E_{Hs} = -0.4032 & E_H = -0.4727 & \text{Anion energy increases.} \end{matrix}$

$\begin{matrix} \text{Coulomb energy in solid state:} & E_{LiH} - E_{Lis} - E_{Hs} = -0.3681 \\ \text{The calculated lattice energy for LiH(s):} & U_{Lattice} = E_{Li} + E_H - E_{LiH} & U_{Lattice} = 0.3257 \end{matrix}$

This result in atomic units is equivalent to a lattice energy expressed in SI units of 856 kJ/mol. A Born-Haber analysis (see below) yields a lattice energy of 912 kJ/mol. Thus, the calculated result of the proposed model is in error by only 6%. The errors for the solid-state ionic radii are 1.6% (cation) and 14.6% (anion). Given the simplicity of the model these comparisons with experimental data are encouraging. For further details on this model see the reference cited below.

$\text{LiH(s)} \xrightarrow{ - \Delta H^o _{form} = 90.4 kJ} \text{Li(s)} + \frac{1}{2} \text{H}_2 \text{(g)} \xrightarrow[ \frac{1}{2} BDE = 218 kJ]{ \Delta H_{sub} = 155 kJ} \text{Li(g) + H(g)} \xrightarrow[EA = -72kJ]{IE = 520 kJ} \text{Li}^+ \text{(g)} + \text{H}^- \text{(g)}$

F. Rioux, "Simple Calculation of the Lattice Energy of Lithium Hydride," Journal of Chemical Education 54, 555 (1977).

## Appendix:

$\int_0^{ \frac{3}{ \alpha}} \left( \sqrt{ \frac{ \alpha^3}{ \pi}} \text{exp} ( - \alpha,~R) \right) 4 \pi r^2 dr = 93.8 \%$

The table below provides a summary of the lattice energy calculation carried out in this tutorial.

$\begin{pmatrix} \text{Property} & \text{Gas Phase} & \text{Solid State} \\ \text{Cation, } \alpha & 2.6875 & 2.6875 \\ \text{Cation Radius} & 1.1163 & 1.163 \\ \text{Cation Energy} & -7.2227 & -7.2227 \\ \text{Anion, } \beta & 0.6875 & 0.8935 \\ \text{Anion Radius} & 4.364 & 3.3576 \\ \text{Anion Energy} & -0.4727 & -0.4302 \\ \frac{ \text{Inter Ion}}{ \text{Coulomb Energy}} & x & -0.3681 \\ \text{Total Energy} & -7.6953 & -8.0210 \\ \text{Lattice Energy} & x & 0.3257 \end{pmatrix}$

From the table it is clear that in the formation of LiH solid, the hydride anion contracts significantly from its gas-phase size. This increases its energy (-0.4302+0.4727=0.0425). The increase in anion energy is more than offset by the attractive inter-ion coulombic energy (-0.3681). In other words, the anion suffers a modest increase in energy by shrinking in size so that it can be on-average closer to the cation, thereby increasing the coulombic attraction between the ions and leading to a stable ionic solid.

Most of the integrals required in the analysis above are now evaluated.

Previous memory of α and β values is cleared: $$\begin{matrix} \alpha = \alpha & \beta = \beta \end{matrix}$$

Trial one-electron wavefunction:

$\Psi (r,~ \beta = \sqrt{ \frac{ \beta^3}{ \pi}} \text{exp} ( - \beta,~r)$

Demonstrate that is is normalized:

$\int_0 ^{ \infty} \Psi (r,~ \beta )^2 4 \pi r^2 dr \text{ assume, } \beta > 0 \rightarrow 1$

Calculate the average value of the electron's distance from the nucleus:

$R( \beta ) = \int_)^{ \infty} \Psi (r,~ \beta) r \Psi (r,~ \beta ) 4 \pi r^2 dr \text{ assume, } \beta > 0 \rightarrow \frac{3}{2 \beta}$

Calculate the average value of the kinetic energy of the electron:

$T( \beta ) = \int_0 ^{ \infty} \Psi (r,~ \beta) - \frac{1}{2r} \frac{d^2}{dr^2} (r \Psi (r,~ \beta )) 4 \pi r^2 dr \text{ assume, } \beta > 0 \rightarrow \frac{ \beta^2}{2}$

Calculate the average value of the electron-nucleus potential energy:

$V( \beta, ~Z) = \int_0^{ \infty} \Psi (r,~ \beta) - \frac{Z}{r} \Psi (r,~ \beta) 4 \pi r^2 dr \text{ assume, } \beta > 0 \rightarrow -Z \beta$

Calculate the average value of the electron-electron potential energy in two steps:

1. The electrostatic potential at r due to electron 1 is:

$\begin{array}{c|c} \Phi (r, ~ \beta _ = \frac{1}{r} \int_0^r \Psi (x,~ \beta)^2 4 \pi x^2 dx + \int_r^{ \infty} \frac{ \Psi (x,~ \beta )^2 4 \pi x^2}{x} dx & ^{ \text{assume,~} \beta >0}_{ \text{simplify}} \rightarrow - \frac{e^{-2 \beta r} + \beta r e^{-2 \beta r} - 1}{r} \end{array}$

2. The electrostatic interaction between the two electrons is:

$\begin{array}{c|c} V_{EE} ( \beta ) = \int_0^{ \infty} \Phi (r,~ \beta) \Psi (r,~ \beta)^2 4 \pi r^2 dr & _{ \text{simplify}}^{ \text{assume, } \beta >0} \rightarrow \frac{5 \beta}{8} \end{array}$

To summarize, the trial wavefunction chosen for two electron systems lead to the following expression for the energy.

$E (Z,~ \beta) = \beta^2 - 2Z \beta + \frac{5}{8} \beta = \beta^2 - 2 \beta \left( Z - \frac{5}{16} \right)$

Minimization of the energy with respect to the variational parameter β yields: $$\beta = Z - \frac{5}{16}$$

Ground state energy:

$E(Z) = - \left( Z - \frac{5}{16} \right)^2$

$R_Z = \frac{3}{Z - \frac{5}{16}}$