To find the oxidation number, first remember that the total charge must be preserved. The total charge of the perchlorate ion is .

Each oxygen will have an oxidation number of . The oxidation of chlorine must be used to balance this negative charge.

The rules for oxidation state note that certain atoms will have constant oxidation values, while others are more likely to vary. Oxygen, for example, will always have an oxidation state of , alkali metals will always have an oxidation state of , and hydrogen will (almost) always have an oxidation state of .

All of the given compounds are neutral, meaning that their oxidation numbers must sum up to zero. Using our listed rules to look at these molecules, we can try to find any missing values.

Chromium in potassium dichromate is the only atom in this set to have an oxidation state of .

The sum of the oxidation numbers of each atom in a molecule must equal the net charge of the molecule.

Oxygen will always have an oxidation number of . In we can find the oxidation number of manganese with the following equation:

Solving this out, we can see that the charge on manganese must be .

In the manganese ions, there is only one atom and the net charge is . The oxidation of this manganese atom must be to balance the ionic charge.

The difference between these oxidation numbers is:

Hydrogen generally has an oxidation number of  in most compounds. The exception is when it is bonded to an alkali metal (group 1 of the periodic table). When hydrogen bonds to one of these elements, it can have an oxidation number of . Alkali metals so readily lose their outermost electron that hydrogen must accept it, and thereby acquire a negative charge.

Potassium is an alkali metal, and will have an oxidation number of . Hydrogen will balance this by acquiring an oxidation number of to keep the molecule neutral. Our answer is .

Hydrogen maintains an oxidation number of in all other answer options.

The oxidation state of oxygen is always , and the oxidation state of magnesium is . We know that the oxidation numbers must cancel out to create a neutral molecule.

There are four oxygen atoms, so the total charge from oxygen is . If we add  from the magnesium atom and  from the oxygen atoms, we get . To get the oxidation state of sulfur, we must add a number to  that totals to zero.

The only number that does this is .

The question states that an oxidized atom will increase its oxidation state, which means that the atom will lose electrons and become more positively charged. Since the atom will lose electrons, the number of electrons will be less than the number of protons. Electrons are negatively charged and protons are positively charged; therefore, the presence of more protons in an oxidized atom will make the atom more positive. An oxidized atom will usually become a cation because its number of electrons will decrease.

Reduced atoms will gain electrons and become more negatively charged (decrease oxidation state). Notice that the reactions given in the passage are not redox reactions because the atoms in both reactions retain their oxidation states.

The atom that is oxidized will have a higher oxidation number as a reactant than as a product. Since reactant phosphorus (P₄) is in elemental form, it has an oxidation number of 0. As a product, phosphorus (PO₄^3-) has an oxidation number of +5.

Total charge = (P) + 4(O) = –3

–3 = (P) + 4(–2) = (P) + (–8)

p = +5

Because it is now more positive, we say that the phosphorus has been oxidized.

The oxidizing reagent is the reactant, not the atom, that is responsible for receiving electrons from another atom in order to oxidize it. The oxygen gas goes from having an oxidation number of 0 to –2. This means that oxygen gas is reduced, but because it oxidized another atom by taking its electrons, we call it the oxidizing reagent.

Reactant oxygen is elemental, and has oxidation number 0.

Product oxygen is in H₂O, with net charge of 0.

Total charge = 2(H) + (O) = 0

2(+1) + (O) = 2 + (O) = 0

O = –2

Decrease in oxidation number indicates reduction, and also identifies the oxidizing agent.

In the reaction, solid potassium (K) is initially in elemental form, so it has an oxidation state of 0. The potassium ions (K⁺) have a charge of +1 in the product. Since the potassium has lost an electron, it has been oxidized.

Since each hydrogen in reactants (NH₃) has an oxidation state of +1, and the hydrogen gas (H₂) has a charge of 0 as a product, the hydrogen has been reduced. Since it was reduced, we conclude that NH₃ is the oxidizing reagent for the reaction.

In this reaction, the oxidation number of oxygen goes from , in diatomic gaseous oxygen, to , in both carbon dioxide and water. This indicates that it has gained electrons; a gain of electrons indicates that oxygen has been reduced. Since it is reduced, it is the oxidizing agent.

The oxidation number of carbon goes from  in cyclohexane to  in carbon dioxide. This indicates that it has lost electrons; a loss of electrons indicates that carbon has been oxidized. Since it is oxidized, cyclohexane is the reducing agent.