1. 1 6 5 4 3 2 Mg(s) + C l 2 (g) MgC l 2 (s) Mg(g) + C l 2 (g) Mg(g) + 2C l (g) Mg 2+ (g) + 2C l – (g) 7 Mg + (g) + 2C l (g) Mg 2+ (g) + 2C l (g) Enthalpy of formation of MgC l 2 Mg(s) + C l 2 (g) ——> MgC l 2 (s) Enthalpy of sublimation of magnesium Mg(s) ——> Mg(g) Enthalpy of atomisation of chlorine ½C l 2 (g) ——> C l (g) x2 Ist Ionisation Energy of magnesium Mg(g) ——> Mg + (g) + e¯ 2nd Ionisation Energy of magnesium Mg + (g) ——> Mg 2+ (g) + e¯ Electron Affinity of chlorine C l (g) + e¯ ——> C l ¯(g) x2 Lattice Enthalpy of MgC l 2 Mg 2+ (g) + 2C l ¯(g) ——> MgC l 2 (s) 1 2 3 4 5 7 6 Starter: write the corresponding equations for each enthalpy change.

2. Born-Haber cycle calculations L.O.:  Carry out calculations related to the Born-Haber cycle.

3. Cl¯ Br¯ F¯ O 2- Na + -780-742-918-2478 K + -711-679-817-2232 Rb + -685-656-783 Mg 2+ -2256-3791 Ca 2+ -2259 Lattice Enthalpy Values Smaller ions will have a greater attraction for each other because of their higher charge density. They will have larger Lattice Enthalpies and larger melting points because of the extra energy which must be put in to separate the oppositely charged ions. Units: kJ mol -1 Indentify the trend in the enthalpy values. Explain the reason for this trend.

4. C l ¯ Br¯ F¯ O 2- Na + -780-742-918-2478 K + -711-679-817-2232 Rb + -685-656-783 Mg 2+ -2256-3791 Ca 2+ -2259 Lattice Enthalpy Values Smaller ions will have a greater attraction for each other because of their higher charge density. They will have larger Lattice Enthalpies and larger melting points because of the extra energy which must be put in to separate the oppositely charged ions. Cl¯Cl¯ Na + Cl¯Cl¯ The sodium ion has the same charge as a potassium ion but is smaller. It has a higher charge density so will have a more effective attraction for the chloride ion. More energy will be released when they come together. K+K+

5. Task 1)Construct a Born-Haber cycle for sodium oxide. 2)Calculate the lattice enthalpy for sodium oxide  Ha(Na) = +108 KJ mol -1  Ha(O) = + 249 KJ mol -1  HI1 =+249 KJ mol -1  H EA1 = - 141 KJ mol -1  H EA2 = + 790 KJ mol -1  Hf(Na 2 O) = -414 KJ mol -1