Consider the following reaction: 2Al(s) + 3I₂(s) → 2AlI₃(s). In Al(s) the aluminum is present as uncharged atoms. In I₂(s) each iodine atom is uncharged. How does Al become Al³⁺ to form AlI₃(s)? How does I become I⁻ to form AlI₃(s)? (Select all that apply)
a) Al gains three electrons.
b) Al loses three electrons.
c) I gains one electron.
d) I loses one electron.