Final answer:
The third ionization of aluminum occurs when it loses its three valence electrons, leading to the formation of an Al³+ cation, correctly represented by option A (Al³+).
Step-by-step explanation:
The question is asking which option correctly represents the third ionization of aluminum. To address this, we must consider the process in which aluminum atoms lose electrons to form cations. Aluminum has an atomic number of 13, which means it has 13 protons in its nucleus. In its neutral state, aluminum also has 13 electrons. The valence electron configuration for aluminum is 3s²3p¹. The third ionization of aluminum occurs when it loses three valence electrons, resulting in the formation of an Al³+ cation. Option A (Al³+) correctly represents this third ionization state.
The process can be shown as a half-reaction where the aluminum atom loses three electrons:
Al → Al³+ + 3e-