# 168: An Even Simpler LiH Lattice Energy Calculation

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 factor α. 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. It is also assumed that the Li+ and H- ions in the solid have the same α value, energy and radius as they do in the gas phase. Under this assumption the lattice energy is the negative of the inter-ion coulombic energy in the solid state.

To check the validity of this model the calculated lattice energy of LiH(s) will be compared to the value obtained from 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)

We begin with variational calculations for the ground-state energies of Li+ (g) and H- (g). These calculations will yield the ionic radii which will subsequently be used to calculate the LiH lattice energy.

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 the appendix for computational 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 and radii of the ions is shown below.

## 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: $$E_{Li} ( \alpha ) = \alpha^2 - 5.375 \alpha$$

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

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

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

## 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.

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

Assuming a face-centered structure (Madelund constant = 1.748) with cation-anion and anion-anion contact, the coulombic term is as shown below.

$E_{Li} (2.6875) + E_H (0.6875) - \frac{1.748}{R_{Li} + R_H} = -8.0143$

Since it has been assumed that the Li+ (g) and H- (g) ions are the same in the gas phase and the solid state, the lattice energy is the negative of Ecoul.

$\begin{matrix} U_{Lattice} = \frac{1.748}{R_{Li} + R_H} & U_{Lattice} = 0.319 \end{matrix}$

This result in atomic units is equivalent to a lattice energy expressed in SI units of 838 kJ/mol. The Born-Haber analysis shown below yields a lattice energy of 912 kJ/mol. Thus, the calculated result of the proposed model is in error by 8%.

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

Improved results can be obtained by allowing the gas phase ions to change size on the formation of the solid under the influence of the inter-ion coulombic interaction, and by abandoning the simplifying restriction of cation-anion and anion-anion contact. See the reference below for an outline of such a calculation.

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

## Appendix:

Because this is a live Mathcad document and the scale factor, α, has been used above, β will be used for the calculations outlined below.

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:

$R( \beta ) = \int_0^{ \infty} \Psi (r,~ \beta) r \Psi (r,~ \beta ) 4 \pi r^2 dr \text{ assume, } \beta > 0 \rightarrow \frac{3}{2 \beta}$

Demonstrate 94% of electron density is contained within 2<r>: $$\int_0 ^{ \frac{3}{ \beta}} \Psi (r,~ \beta)^2 4 \pi r^2 dr \text{ assume, } \beta >0 \rightarrow (-25)e^{-6} + 2 = 93.8 \% \] Calculate the average value of the kinetic energy of the electron: $T( \beta ) = \int_0 ^{ \infty} \Psi (r,~ \beta) - \frac{1}{2r} \frac{d^2}{dr^2} (r \Psi (r,~ \beta )) 4 \pi r^2 dr \text{ assume, } \beta > 0 \rightarrow \frac{1}{2} \beta^2$ Calculate the average value of the electron-nucleus potential energy: $V( \beta, ~Z) = \int_0^{ \infty} \Psi (r,~ \beta) - \frac{Z}{r} \Psi (r,~ \beta) 4 \pi r^2 dr \text{ assume, } \beta > 0 \rightarrow (- \beta ) Z$ Calculate the average value of the electron-electron potential energy in two steps: 1. The electrostatic potential at r due to electron 1 is: $\begin{array}{c|c} \Phi (r, ~ \beta _ = \frac{1}{r} \int_0^r \Psi (x,~ \beta)^2 4 \pi x^2 dx + \int_r^{ \infty} \frac{ \Psi (x,~ \beta )^2 4 \pi x^2}{x} dx & ^{ \text{assume,~} \beta >0}_{ \text{simplify}} \rightarrow \frac{- \left[ e^{-2 \beta r} + \beta r e^{-2 \beta r} - 1 \right]}{r} \end{array}$ 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:

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

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