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The octet rule: bonding and lone electron pairs fill the valence levels of each atom in a molecule. Why ionic compounds are brittle and high melting and conduct electricity only when molten or in aqueous solution.Ĭovalent bonding model: nonmetal atoms share electrons to form a covalent bond. How Coulomb’s law accounts for periodic trends in lattice energy. Ionic compound formation in hypothetical steps (Born-Haber cycle) and how they can be used to calculate lattice energies. The reasons that lattice energy is responsible for the stability of solid ionic compounds.
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The key features of ionic bonding: electron transfer to form ions and their electrostatic attraction to form an ionic solid. The differences in atomic properties that lead to each of the basic types of bonding. The heat of formation, Δ H f 0 may be expressed as, Δ H f 0 = ΔH sub + D + IE + E A + U Δ H f 0 = 146.4 + 155.8 + 2186 – 2 x 332.6 – 2922.5 Δ H f 0= 1099.5 KJ/mole Ans.This study guide contains over 3 hours of recorded lecture topics and example problems that follow the material covered in chapter 9 of the textbook Chemistry, The Molecular Nature of Matter and Change by Silberberg. Question 3) Calculate the enthalpy of formation of MgF 2 from the following data- Enthalpy of sublimation of Mg = 146.4 KJ/mole Enthalpy of dissociation of Fluorine = 155.8 KJ/ mole Ionisation energy of Mg (IE 2 ) = 2186.0 KJ/mole Electron gain enthalpy of Fluorine = -322.6 KJ/mole Lattice Enthalpy of MgF 2 = – 2922.5 KJ/mole Solution ) U = – 2870.8 KJ/mole Lattice energy of CaCl 2 = 2870.8 KJ/mole Ans. Question 2) – Calculate the lattice enthalpy of CaCl 2, given that the enthalpy of – Enthalpy of sublimation for Ca (s) -> Ca(g) = 121 KJ/mole Enthalpy of dissociation of Cl 2 (g) -> 2Cl(g) = 242.8 KJ/ mole Ionisation energy of Ca(g) -–>Ca ++ = 2422 KJ/mole Electron gain enthalpy of 2Cl -–> 2 Cl – = 2 x -355 = -710 KJ/mole Enthalpy of formation of CaCl2= -795 KJ/mole Solution )Ĭa (s) + ΔH sub -> Ca (g) ΔH sub = 121 KJ/moleĬl 2 (g) + D -> 2Cl(g) D = 242.8KJ/molĬa(g) + IE -–> Ca ++ (g) + 2e – IE= 2422 KJ/moleĢCl(g) +2 e -–> 2Cl – (g) +E A E A = – 2x 355 = 710 KJ/moleĬa ++ (g) + 2Cl – (g) -> CaCl 2 (s) According to Born – Haber cycle- Δ H f 0 = ΔH sub + D + IE + E A +U K + (g) + Cl – (g) -> KCl (s) Δ H f 0 = -438 KJ/mole According to Born – Haber cycle- Δ H f 0 = ΔH sub + D/2 + IE + E A +U Source : Born – Haber cycle- Question 1) – Calculate the lattice enthalpy of KCl from the following data at standard states- Enthalpy of sublimation of K = 89 KJ/mole Enthalpy of dissociation of chlorine = 244 KJ/ mole Ionisation energy of K = 425 KJ/mole Electron gain enthalpy of chlorine = -355 KJ/mole Enthalpy of formation of KCl = -438 KJ/mole Solution )