Final answer:
Magnesium in Mg + S has an oxidation state of 0, which becomes +2 in the form of the Mg²⁺ ion. Sulfur has an oxidation state of 0, which becomes -2 as S²⁻. Hence, Mg is oxidized, and S is reduced in this reaction.
Step-by-step explanation:
The question asks us to determine the oxidation states of reactants and products in the reaction Mg + S → Mg²⁺ +S²⁻.
Magnesium (Mg) in its elemental form has an oxidation state of 0, as it is a pure element. Sulfur (S) also has an oxidation state of 0 for the same reason. Upon reacting, magnesium becomes a magnesium ion (Mg²⁺) with an oxidation state of +2, indicating it has lost two electrons. Sulfur becomes a sulfide ion (S²⁻) with an oxidation state of -2, indicating it has gained two electrons. The magnesium atom is oxidized, and the sulfur atom is reduced in this process.
Considering other provided examples, in each reaction, we can determine which atoms are oxidized and reduced by looking at the changes in their oxidation states. For instance, in the reaction Mg(s) + NiCl₂(aq) → MgCl₂(aq) + Ni(s), magnesium is oxidized from 0 to +2, and nickel is reduced since it goes from a positive oxidation state in NiCl₂ to 0 in nickel metal. Similarly, in Mg(s) + Cl₂(g) → MgCl₂(s), magnesium loses two electrons (oxidation), and chlorine gains electrons (reduction).