Only answer question 5 (the second part). Use the following to calculate delta H
ID: 1049680 • Letter: O
Question
Only answer question 5 (the second part). 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^n = 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 NaClExplanation / Answer
5a) In Mg F2 , only one magenisum ion is formed by loss of two electrons successively, thus we need to use two different ionization energies, namely ionization energy 1 and ionization energy 2.
But two fluoride ionsare needed to combine with one Mg+2 ion, so for each fluorideto form we need to supply one electron afinity . Thus to form two F- ions we take twice electron affinity .
5b) Lattice energy is proportional to |z+| |z-| and inversely proportional to (r+ + r-)
Thus in Mg F2 the |z+| |z-| is twice as that of |z+| |z-| for NaCl . [charge of Mg is +2 and Na is +1]
Though there is difference in (r+ + r-) value for the ions, it is ver y small compared to charges, as the radii of ions themselves are very small value.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.