Final answer:
The oxidation number of manganese in potassium permanganate (KMnO4) is +7, determined by balancing the charges of oxygen's typical -2 oxidation state, the compound's neutral charge, and the permanganate anion's -1 charge.
Step-by-step explanation:
The oxidation number of manganese in the compound potassium permanganate (KMnO4) is a topic related to Chemistry, specifically to the field of polyatomic transition metal ions and redox chemistry. In potassium permanganate, that is, KMnO4, manganese (Mn) has an oxidation state that can be determined by considering the typical oxidation state of oxygen, the overall charge of the compound, and the rules for calculating oxidation numbers.
Oxygen usually has an oxidation number of -2, and since there are four oxygen atoms, their total contribution to the oxidation state is -8. Given that the compound as a whole is electrically neutral, but the permanganate anion MnO4- has a charge of -1, we deduce that the total positive oxidation state must counterbalance the negative charge contributed by the oxygen atoms and the extra negative charge of the ion itself. This results in the manganese atom having an oxidation number of +7.
The electronic configuration of manganese in its +7 oxidation state has no electrons in the 4s and 3d orbitals, symbolized by [Ar] 4s03d0. This also corresponds to manganese's highest possible oxidation state among the fourth period transition metals.