Use the following to calculate delta H^o of MgF_2 M_g(g) rightarrow Mg(g) delta

Question

Use the following to calculate delta H^o of MgF_2 M_g(g) rightarrow Mg(g) delta H^o = 1.48 kJ F_z(g) rightarrow 2F(g) delta H^o = 159 kJ M_g(g) rightarrow Mg^-(g) + e^- delta H^o = 738 kJ M_g^-(g) rightarrow M_g^2-(g) + e^- delta H^o = 1450 kJ F(g) - e^- rightarrow F(g) delta H^o = -328 kJ M_g(s) + F_2(g) rightarrow M_gF_2(s) delta H^o = -1123 kJ Compared with the lattice energy of LiF(1050kJ/mol) or the lattice energy you calculated for NaCl in problem 9.30 does the relative magnitude of the value for MgF_2 you? Explain. Example the following about problem 9.31. Why do you use the electron affinity of F twice, but two separate ionization energies for Mg? Why is the lattice energy of this to much larger than the 788 kJ/mol that you calculated for NaCl

Explanation / Answer

5b) Lattice energy is proportional to the charges of the ions and inversely proportional to the size of ions.

Lattice energy = |z+| |z-| / (r+ +r-)

While comparing MgF2 and NaCl the charge of Mg is +2 and Na is +1, thus the lattice energy is doubled.

but the changes in sizes of ions is lesscomparatively as the sizes of ions are quite small.

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