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Solutions to problems at the end of Chapter 13

# Solutions to problems at the end of Chapter 13 - Solutions...

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Solutions to problems at the end of Chapter 13: 29. ΔH f = E sublimation (K) + IE 1 (K) + ½E bond (Cl 2 ) + EA(Cl) + Lattice Energy = 64 + 419 + 119.5 + (-349) + (-690) = -436.5 kJ/mol 30. ΔH f = E sublimation (Mg) + IE 1 (Mg) + IE 2 (Mg) + E bond (F 2 ) + 2EA(F) + Lattice Energy = 150 + 735 + 1445 + 154 + 2(-328) + (-2913) = -1085 kJ/mol 31. ΔH f = E sublimation (Li) + IE 1 (Li) + ½E bond (I 2 ) + EA(I) + Lattice Energy Given: IE 1 = 520 kJ/mol E bond (I 2 ) = 151 kJ/mol EA(I) = -295 kJ/mol Lattice E = -753 kJ/mol ΔH f (LiI) = -272 kJ.mol -272 = E sub (Li) + 520 + ½(151) + (-295) + (-753) E sub (Li) = 180.5 kJ/mol 32. Remember that elements typically form ions in order to have a filled valence octet (i.e. look like a noble gas in terms of electron configuration). Although the lattice energy might increase if you were to ionize elements to +2, +3, +4 or higher oxidation states, the ionization energies associated with removing these extra electrons would also be extremely high. 33.a)
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