Based on the reactivities of the elements involved, which reactions will form products that are more stable than the reactants? 2lif + cl2 → 2licl + f2 2lii + cl2 → 2licl + i2 2licl + br2 → 2libr + cl2 2libr + f2 → 2lif + br2 2libr + i2 → 2lii + br2

2 LiI + Cl₂ → 2 LiCl + I₂

2 LiBr + F₂ → 2 LiF + Br₂

Each of the five reactions involve one halogen molecule (F₂, Cl₂, Br₂, and I₂) substituting the ion of another halogen (F⁻, Cl⁻, Br⁻, and I⁻).

Halogen atoms are found in group 17 of the periodic table. They are all non-metal elements. Each of the halogen atom will gain one electron to form an ion of charge -1. However, the tendency to do so decreases down the group.

Atoms of all four elements have the same effective nuclear charge of +7. However, F has the smallest radius. As a result, it has the strongest hold on electrons around it. Its ion F⁻ is more stable than ions of Cl, Br, or I. Similarly, its molecule F₂ is more reactive than Cl₂, Br₂, and I₂.

As a result, the stability of halogen molecules increases down the group:

The stability of halogen ions decreases down the group:

Cl₂ repaces F⁻ (from LiF) in first reaction. F₂ and Cl⁻ are produced. F₂ is less stable than Cl₂. Cl⁻ is less stable than F⁻.

Cl₂ replaces I⁻ (from LiI) in the second reaction. I₂ and Cl⁻ are produced. I₂ is more stable than Cl₂. Cl⁻ is more stable than I⁻.

Br₂ replaces Cl⁻ (from LiCl) in the third reaction. Cl₂ and Br⁻ are produced. Cl₂ is less stable than Br₂. Br⁻ is less stable than Cl⁻.

F₂ replaces Br⁻ (from LiBr) in the fourth reaction. Br₂ and F⁻ are produced. Br₂ is more stable than F₂. F⁻ is more stable than Br⁻.

I₂ replaces Br⁻ (from LiBr) in the fifth reaction. Br₂ and I⁻ are produced. Br₂ is less stable than I₂. I⁻ is less stable than Br⁻.

c. 4.54 x 10–12 j happens to be correct