An enthalpy cycle for the formation of monovalent aluminium halides is given below.

Use the data given below (kJ mol^-1) to show which oxidation state is favoured in each case.
I₁ = 577 kJ mol^-1 I₂ = 1816 kJ mol^-1 I₃ = 2743 kJ mol^-1
ΔH_sub = 324 kJ mol^-1
ΔH (AlF) = -910 kJ mol^-1
ΔH (AlI) = -696 kJ mol^-1
ΔH (AlF₃) = -6380 kJ mol^-1
ΔH (AlI₃) = -4706 kJ mol^-1

Question

An enthalpy cycle for the formation of monovalent aluminium halides is given below.

Use the data given below (kJ mol^-1) to show which oxidation state is favoured in each case.
I₁ = 577 kJ mol^-1 I₂ = 1816 kJ mol^-1 I₃ = 2743 kJ mol^-1
ΔH_sub = 324 kJ mol^-1
ΔH (AlF) = -910 kJ mol^-1
ΔH (AlI) = -696 kJ mol^-1
ΔH (AlF₃) = -6380 kJ mol^-1
ΔH (AlI₃) = -4706 kJ mol^-1

Quizplus · Accepted Answer

To determine which oxidation state is favored in each case, we need to compare the enthalpies of formation. The enthalpy of formation is the energy change when one mole of a substance is formed from its constituent elements in their standard state.

For option A, we need to form aluminum(III) fluoride and aluminum(III) iodide. The enthalpies of formation for these compounds are not given, but the enthalpies of formation for aluminum fluoride and aluminum iodide are given. We can use these values in conjunction with the enthalpy of sublimation of aluminum to calculate the enthalpies of formation for the aluminum halides. However, since we do not have enough information to directly compare the enthalpies of formation for aluminum(III) fluoride and aluminum(III) iodide, we cannot determine which oxidation state is favored.

For option B, we need to form aluminum(I) fluoride and aluminum(III) iodide. The enthalpy of formation for aluminum(I) fluoride is not given, but we can use the enthalpy of formation for aluminum fluoride and the enthalpy of sublimation of aluminum to calculate it. The enthalpy of formation for aluminum(III) iodide is given. Comparing the enthalpies of formation for aluminum(I) fluoride and aluminum(III) iodide, we see that aluminum(III) iodide is favored since its formation releases more energy.

For option C, we need to form aluminum(I) fluoride and aluminum(I) iodide. The enthalpies of formation for these compounds are not given, but we can use the enthalpy of sublimation of aluminum to calculate their enthalpies of formation. However, since we do not have enough information to directly compare the enthalpies of formation for aluminum(I) fluoride and aluminum(I) iodide, we cannot determine which oxidation state is favored.

For option D, we need to form aluminum(III) fluoride and aluminum(I) iodide. The enthalpies of formation for these compounds are given, and we can compare them directly to see which oxidation state is favored. The enthalpy of formation for aluminum(III) fluoride is much more negative than that of aluminum(I) fluoride, indicating that aluminum(III) fluoride is more stable. The enthalpy of formation for aluminum(I) iodide is more negative than that of aluminum(III) iodide, indicating that aluminum(I) iodide is more stable. Therefore, option D is the best choice since it results in the favored oxidation states for both halides.

An Enthalpy Cycle for the Formation of Monovalent Aluminium Halides