Consider an ionic compound, mx2 , composed of generic metal m and generic, gaseous halogen x .

the enthalpy of formation of mx2 is δ∘f=−985 kj/mol.

the enthalpy of sublimation of m is δsub=135 kj/mol.

the first and second ionization energies of m are ie1=731 kj/mol and ie2=1403 kj/mol.

the electron affinity of x is δea=−335 kj/mol. (refer to the hint).

the bond energy of x2 is be=207 kj/mol.

determine the lattice energy of mx2 .

Explanation:

One way to calculate the lattice energy is to use Hess's Law .

The lattice energy U is the energy released when the gaseous ions combine to form a solid ionic crystal:

M²⁺(g) + 2X⁻(g) ⟶ MgX₂(s); U = ?

We must generate this reaction rom the equations given.

(1) M(s) + X₂ (g) ⟶ MX₂(s);      ΔHf⁰ = -985 kJ·mol⁻¹

(2) M(s) ⟶ M(g);                  ΔHsub =   135 kJ·mol⁻¹

(3) M(g) ⟶M⁺(g) +e⁻                   IE₁ =   731 kJ·mol⁻¹

(4) M⁺(g) ⟶ M²⁺(g) + e⁻             IE₂ = 1403 kJ·mol⁻¹

(5) X(g) + e⁻ ⟶ X⁻(g)                 EA =  -335 kJ·mol⁻¹

(6) X₂(g) ⟶ 2X(g)                      BE =   207 kJ·mol⁻¹

Now, we put these equations together to get the lattice energy.  Underlined species have been cancelled.

                                                        E/kJ  

(7)   M²⁺(g) + e⁻⟶ M⁺(g)                 -1403

(8)   M⁺(g) + e⁻ ⟶ M(g)                     -731

(9)   M(g) ⟶ M(s)                              -135

(10) M(s) + X₂(g) ⟶ MX₂(s)              -985

(11)  2X(g) ⟶ X₂(g)                           -207

(12) 2X⁻(g) ⟶ 2X(g) + 2e⁻               +670

     M²⁺(g) +  2X⁻(g) ⟶ MX₂(s)       -2791

The lattice energy of MX₂ is .

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