Question

Sodium fluoride adopts the NaCl (FCC) structure type. Estimate the lattice energy, ΔU, for NaF using the Born–Lande equation. The bond length for Na–F is 235 pm, and the Madelung constant is 1.7476. The Born exponent of both the cation and the anion is 7.

Answer 1

The lattice energy of a crystalline solid is usually defined as the energy of formation of the crystals from infinitely-separated ions and as such is invariably positive.

Born–Lande equation

where

N_A = the Avogadro constant; = 6.022140857 × 10²³

M = the Madelung constant, relating to the geometry of the crystal; = 1.7476

z⁺ = the charge number of cation; = +1

z⁻ = the charge number of anion; = -1

q_e = the elementary charge, equal to 1.6022×10⁻¹⁹ C;

π = 3.1416

ε₀ = the permittivity of free space, = 8.854×10⁻¹² C² J⁻¹ m⁻¹;

r₀ = the distance to closest ion; = Na-F bond length = 235 x 10^-12 m

n = the Born exponent, = 7

Substituting the above values into the equation one will get:

= - (6.022140857 × 10²³)( 1.7476)(+1)(-1)( 1.6022×10⁻¹⁹)²/ (4)(3.1416)( 8.854×10⁻¹² ) (235 x 10^-12){1 – 1/7}

= - (-27.0163331 x 10^-15)/ (111.2629056 x 10⁻¹²)(235 x 10⁻¹²){1 – 1/7}

= - (-27.0163331 x 10^-15)/ (111.2629056 x 10⁻¹²)(235 x 10⁻¹²){0.857142857}

=  - (-27.0163331 x 10^-15) / 2.6146782816 x 10^-20{0.857142857}

= - (-10.33256492399818 x 10⁵) {0.857142857}

= - (-8.856484219093789 x 10⁵)

= 8.856484219093789 x 10⁵ J/mol

= 885.648219093789 kJ/mol

lattice energy of NaF = 885.648219093789 kJ/mol