Use the following to calculate ΔH°latice of MgF₂:
Mg(s) → Mg(g) ΔH° = 148kJ F₂(g) → 2F(g) ΔH° = 159kJ
M(g) → Mg⁺ (g) + e⁻ ΔH° = 738kJ
M⁺(g) → Mg²⁺ (g) + e⁻ ΔH° = 1450kJ
F(g) + e⁻ → F⁻(g) ΔH° = -328kJ
Mg(s) + F₂(g) → MgF₂(s) ΔH° = -1123KJ
Compared with the lattice energy of LiF (1050 kJ/mol) or the lattice energy you calculated for NaCl in Problem 9.30, does the relative magnitude of the value for MgF _2 surprise you? Explain.

Respuesta :

Yes, Since magnesium ions have a charge that is two times greater than that of lithium and sodium ions, MgF2 has higher lattice energy than LiF and NaCl.

Lattice energy: MgF2 (s) → Mg2+(g) + 2 F–(g)

Use Hess’s law:                                      

∆Hº Mg(s) → Mg(g)                                  

Mg(s) → Mg(g) ΔH° = 148kJ F₂(g) → 2F(g) ΔH° = 159kJ

M(g) → Mg⁺ (g) + e⁻ ΔH° = 738kJ

M⁺(g) → Mg²⁺ (g) + e⁻ ΔH° = 1450kJ

F(g) + e⁻ → F⁻(g) ΔH° = -328kJ

Mg(s) + F₂(g) → MgF₂(s) ΔH° = -1123KJ

(Reaction is reversed and the sign of ∆Hº changed.)  

MgF2 (s) → Mg2+(g) + 2F–(g)            

2962 kJ

As ion charge rises, lattice energy rises as well. The lattice energy increases as the ion charge variable is increased. Accordingly, ions with higher charge intensities will result in ionic compounds with higher lattice energies.

Learn more about lattice energy here:

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