Final answer:
In the reaction 2Al₂O₃ → 4Al + 3O₂, aluminum (Al) is reduced from an oxidation number of +3 to 0, transitioning from a compound to its elemental form. Oxygen is oxidized from an oxidation number of -2 in aluminum oxide to 0 in molecular oxygen O₂. The correct option is 4.
Step-by-step explanation:
The question asks about the change in oxidation number of the metal atom in the chemical reaction 2Al₂O₃ → 4Al + 3O₂. When analyzing the changes in oxidation states, we apply certain rules to determine the oxidation numbers before and after the reaction.
In the reactant Al₂O₃, aluminum (Al) has an oxidation number of +3, as it is in a compound. In the product side of the reaction, Al is in its elemental form, which means its oxidation number is 0. Therefore, we can see that Al is reduced from +3 to 0 in this reaction.
Contrary to the given options, oxygen (O) is not reduced from +2 to -1, it is oxidized from an oxidation number of -2 (in Al₂O₃) to 0 in O₂, its elemental form. Moreover, Al is not reduced from +1 to -2 or from 0 to +3 as none of these accurately represent the changes in its oxidation state in the given equation.
The oxidation number of oxygen generally is -2 in compounds (except in peroxides or superoxides), and the oxidation state of an element in its standard state or free form, such as O₂ or Al, is always zero.