Lesson Explainer: Oxidation Numbers Chemistry

In this explainer, we will learn how to calculate and use oxidation numbers.

In redox reactions, one species is reduced and another species is oxidized. An oxidation involves losing electrons, and a reduction involves gaining electrons. In order to help keep track of how electrons move during a redox reaction, chemists use something called an “oxidation number,” which is largely synonymous with “oxidation state.”

Definition: Oxidation Number

The oxidation number shows the degree of oxidation of an atom in isolation or in a compound in terms of counting electrons. The higher the oxidation number, the greater the degree of oxidation and number of removed electrons.

The oxidation number of an atom shows the degree of oxidation of that element in isolation or in a compound, which is essentially a count of the number of electrons removed. Elements have a positive oxidation state when they are oxidized, and they have a negative oxidation state when they are reduced. Ionic compounds contain some combination of elements with positive and negative oxidation numbers. Their oxidized cations have a positive oxidation number, and their reduced anions have a negative oxidation number. Covalently bonded elements are similarly designated positive or negative oxidation numbers. It is customary to assign a negative oxidation number to the element with the higher electronegativity number and a positive oxidation number to the element with the lower electronegativity number. If an atom is in its elemental form, it has not gained or lost any electrons and its oxidation number is zero. As an example, let’s take a look at the synthesis of magnesium oxide: 2Mg()+O()2MgO()sgs2

In this reaction, we begin with solid magnesium and oxygen gas. Both of these elemental forms have an oxidation number of zero. They have not gained or lost any electrons.

However, the oxidation number of these elements changes as they form magnesium oxide. Magnesium oxide is an ionic compound consisting of an Mg2+ ion and an O2 ion. For neutral magnesium to form a 2+ ion, it needs to lose two electrons. The oxidation number tells us how many electrons have been removed in order to form a chemical bond, so two lost electrons means an oxidation number of +2 for the magnesium ion in MgO.

On the other hand, oxygen gains two electrons to form a 2 ion. Two electrons gained means an oxidation number of 2 for the oxygen ion in MgO. In this case, the oxidation number is the same as the charge of the ions, but oxidation number and charge are distinct values that are not always equal to one another.

We can represent these overall changes by noting the values above the chemical symbols of the reaction equation: 22Mg()+O()MgO()sgs2

There are many rules governing the oxidation states of various elements in different forms. Here are three rules we have touched on so far:

  • The oxidation number of an atom or molecule in its elemental form is zero.
    Examples: Mg and O2
  • The oxidation number of an alkali earth metal in a compound is +2.
    Examples: magnesium in MgO and zinc in ZnI2
  • The oxidation number of oxygen in a compound is 2.
    Examples: oxygen in MgO and oxygen in NaO2
    Exceptions: peroxides such as HO22 where the oxidation number of oxygen is 1, superoxides such as NaO2 where the oxidation number of oxygen is 12, and the dioxygen difluoride compound, which has the chemical formula OF22 and oxygen with an oxidation number of +1

To better understand how to determine the oxidation number of other atoms or ions, let’s take a look at a reaction where the oxidation numbers do not change: MgO+2HClMgCl+HO22

In this reaction, magnesium oxide combines with hydrochloric acid to create magnesium chloride and water. Earlier, we said that the oxidation number is essentially a count of the number of electrons an atom loses to form a chemical bond. This relationship is apparent when looking at the compounds in this reaction.

Earlier, we determined the oxidation numbers of the ions in magnesium oxide. Magnesium loses two electrons to have an oxidation number of +2, while oxygen gains two electrons to have an oxidation number of 2.

The next compound, HCl, is a covalent compound rather than an ionic compound. When looking at compounds with covalent bonds where electrons are shared rather than donated, we do not consider fractions of electrons. Instead, an electron that is shared closer to the other nucleus is considered a full lost electron for the purpose of calculating the oxidation number. In other words, we do not have to consider whether the compound is ionic or covalent because we count shared and donated electrons toward the oxidation number the same way.

For example, to form HCl, hydrogen shares one electron with the electronegative chlorine atom. Hydrogen partially loses a single electron, so its oxidation number is +1. Chlorine partially gains a single electron, so its oxidation number is 1.

The third compound in this reaction, magnesium chloride, is an ionic compound. To form this compound, each chlorine atom completely accepts one electron from the magnesium atom, so chlorine’s oxidation number is 1. Note that chlorine’s oxidation number is 1 in both the covalent compound of HCl and the ionic compound of MgCl2. The magnesium ion in magnesium chloride is a 2+ ion that has given up two electrons, so its oxidation state is +2.

Lastly, each hydrogen atom shares an electron with the oxygen atom to form HO2. Oxygen’s oxidation state is 2, as it has partially gained two electrons. Hydrogen’s oxidation state is +1, as each hydrogen atom has partially given up one electron. Hydrogen is not an alkali earth metal, but like alkali metals such as sodium, it frequently gives up one electron to form compounds where its oxidation number is +1: MgO+HClMgCl+HO222

Here are a few more rules that have appeared in this example:

  • The oxidation state of hydrogen in a compound is +1.
    Examples: HCl and HO2
    Exceptions: Metal hydrides such as AlH3 and MgH2 where the oxidation number is 1
  • The oxidation state of a halogen in a compound is 1.
    Examples: HCl, KI, NaBr, and LiF
    Exceptions: Cl, Br, and I, as they can combine with oxygen or fluorine with a variety of oxidation numbers
  • The oxidation state of an alkali metal (group 1) in a compound is +1.
    Examples: Sodium in NaCl and potassium in KBr

There are a few more useful rules that can help us determine the oxidation number of an unknown atom or compound. The first rule is that the total oxidation number of a neutral compound is zero. If we consider the compound NH3 and are unsure of the oxidation number of nitrogen, we can use the oxidation number of hydrogen and the overall compound to figure it out. Each hydrogen atom has an oxidation number of +1, while the neutral compound as a whole must have an oxidation number of zero. The only way to satisfy both of these conditions is if nitrogen’s oxidation number is 3.

Up to now, we have dealt with neutral compounds, but we can apply several simple rules about ions as well. The oxidation number of an ion is equal to the charge of the ion. For example, the oxidation number of an Fe3+ ion is +3. This rule also applies to polyatomic ions. The oxidation number of a sulfate ion with the chemical formula SO42 is 2. Again, we should note that charge and oxidation number are distinct values that sometimes but not always equal one another.

Let’s summarize these rules:

  • The oxidation number of an ion is the same as the charge of the ion.
    Examples: Na+ has an oxidation number of +1, and S2 has an oxidation number of 2.
  • The sum of the oxidation numbers in a neutral compound is zero. The sum of the oxidation numbers in a polyatomic ion is the charge of the ion.
    Examples: The total oxidation number of HO2 is zero, and the total oxidation number of SO42 is 2.

Example 1: Determining the Oxidation Number of Atoms in a Molecule

Determine the oxidation number of nitrogen in CaN32.

Answer

To determine the oxidation number of nitrogen in this compound, it helps to first identify the oxidation number of calcium. Since calcium is an alkali earth metal that tends to form a 2+ ion, its oxidation number is +2.

This is a neutral compound, so the total oxidation number is zero. Since the compound includes three calcium ions with oxidation numbers of +2, for the compound overall to have an oxidation number of zero, the oxidation number of the nitrogens must total 6. In other words, the oxidation number of each must be 3.

An oxidation number of 3 means that each nitrogen atom gains three extra electrons in order to make the chemical bonds with the calcium atoms in this compound.

The correct answer is 3.

How To: Recalling the Rules for Calculating Oxidation Number

  • The oxidation number of an atom in its elemental form is zero.
    Examples: Cu()s, Mg()s, O()2g, S()8s all have an oxidation number of zero.
  • The oxidation number of an ion is the same as the charge of the ion.
    Examples: Na+ has an oxidation number of +1, and S2 has an oxidation number of 2.
  • The sum of the oxidation numbers in a neutral compound is zero. The sum of the oxidation numbers in a polyatomic ion is the charge of the ion.
    Examples: the total oxidation number of HO2 is zero, that of SO42 is 2, that of NO3 is 1, that of PO43 is 3, that of CO32 is 2, that of OH is 1, and that of NH4+ is +1.
  • The oxidation state of an alkali metal (group 1) in a compound is +1.
    Examples: sodium in NaCl and potassium in KBr
  • The oxidation state of an alkali earth metal (group 2) in a compound is +2.
    Examples: magnesium in MgO and calcium in CaCO3
  • The oxidation state of oxygen in a compound is 2.
    Examples: oxygen in HO2 and oxygen in MgO
    Exceptions: Peroxides such as HO22 where the oxidation number is 1 and oxygen bonded to fluorine such as in OF22 where the oxidation number varies depending on the compound
  • The oxidation state of hydrogen in a compound is +1.
    Examples: HCl and HO2
    Exceptions: metal hydrides such as AlH3 and MgH2 where the oxidation number is 1
  • The oxidation state of a halogen in a compound is 1.
    Examples: HCl, KI, NaBr, and LiF
    Exceptions: Cl, Br, and I, as they can combine with oxygen or fluorine with a variety of oxidation numbers

With these rules in mind, we can determine the oxidation numbers of atoms in many different compounds.

Example 2: Determining the Oxidation Numbers of the Atoms in a Molecule

The oxidation numbers of the atoms in a molecule of carbon dioxide add up to 0.

  1. What is the oxidation number of oxygen in CO2?
  2. What is the oxidation number of carbon in CO2?

Answer

Part 1

The oxidation number of oxygen in a compound is 2, except for peroxides where oxygen is bonded to another oxygen. Carbon dioxide is not a peroxide, so the oxidation number of oxygen here is 2. An oxidation number of 2 indicates that each oxygen atom takes on 2 extra electrons to form this bond. Even though this bond is a polar covalent bond and the electrons are not completely donated, when calculating an oxidation number, we treat each partially shared electron as a whole electron lost or gained. The correct answer is 2.

Part 2

We have already calculated that each oxygen has an oxidation number of 2. In addition, we are told that the oxidation numbers of the atoms in carbon dioxide add up to zero. In fact, the oxidation numbers of the atoms of any neutral compound add up to zero.

If each oxygen has an oxidation number of 2 for a total of 4 and we want the whole compound to have an oxidation number of zero, then carbon’s oxidation number must be +4. This oxidation number suggests that carbon loses four electrons to form this bond. More specifically, carbon partially shares four electrons that are closer to the nucleus of the oxygen atoms than to the nucleus of the carbon atoms. The correct answer is +4.

Oxidation states are useful in letting us track the movement of electrons across a chemical reaction. Let’s take a look at a simple redox reaction to observe how oxidation numbers can change: 2Na()+Cl()2NaCl()sgs2

The reaction of sodium metal and chlorine gas produces sodium chloride. Sodium and chlorine have one oxidation number in their elemental forms and a different oxidation number when they form a compound. As we mentioned earlier, atoms in their elemental form have an oxidation number of zero, so both of the reactants in this reaction have an oxidation number of zero.

What is the oxidation number of the atoms in the product? In order to form this ionic compound, sodium atoms donate an electron to chlorine atoms, forming an Na+ ion and a Cl ion. Since sodium gives up an electron to form a bond, the oxidation number of the sodium atom in sodium chloride is +1. Since chlorine accepts an electron to form a bond, the oxidation number of the chlorine atom in sodium chloride is 1. 22Na()+Cl()NaCl()sgs2

It is worth noting that the oxidation number of an atom changes as it forms and breaks bonds, but the oxidation numbers of the atoms within a certain compound will always be the same from sample to sample. For example, sodium’s oxidation number changes from 0 to +1 in this reaction, but the sodium in any sample of NaCl will always have an oxidation number of +1 as long as it remains in that compound.

If the oxidation numbers change from the beginning of the reaction to the end, it means that electrons are being transferred and a redox reaction is taking place. Oxidation numbers can help us identify what is being oxidized and what is being reduced during a redox reaction.

As we mentioned earlier, reduction involves gaining electrons, and oxidation involves losing electrons. Since the oxidation number tells us how many electrons are lost, oxidizing (losing electrons) increases the oxidation number, while reduction (gaining electrons) reduces it. By comparing an atom’s oxidation number when it is a reactant to that when it is a product, we can determine whether it is oxidized or reduced.

As an example, let’s look at the reaction that takes place when we dip magnesium metal into a zinc chloride solution: Mg()+ZnCl()MgCl()+Zn()saqaqs22

Following the rules we outlined earlier, we can determine what is being reduced and what is being oxidized from the oxidation states of the various atoms. Any elemental form has an oxidation number of 0. The group 2 metals magnesium and zinc have oxidation numbers of +2 if they are in compounds. The halogen chloride has an oxidation number of 1 in compounds. We can visualize these changes using a table.

ElementReactant Oxidation NumberProduct Oxidation Number
Magnesium0+2
Zinc+20
Chlorine11

We can also do so by writing the numbers above the equation: Mg()+ZnCl()MgCl()+Zn()saqaqs22

We can see that magnesium’s oxidation number increases, meaning it is oxidized. In this reaction, magnesium loses electrons. On the other hand, zinc’s oxidation number decreases, meaning it is reduced. In this reaction, zinc gains electrons. Chlorine’s oxidation number does not change, so it is neither oxidized nor reduced.

In this reaction, electrons are transferred from magnesium to zinc. Magnesium loses electrons that go to zinc, causing zinc to reduce. Since the electrons from magnesium reduce the zinc, we call magnesium the “reducing agent.” In any redox reaction, the species that oxidizes and gives electrons to another species is the reducing agent.

Conversely, since zinc takes electrons from magnesium, causing the magnesium to oxidize, we call zinc the “oxidizing agent.” In any redox reaction, the species that oxidizes and takes electrons from another species is the oxidizing agent.

We can apply these definitions to the synthesis of sodium chloride: 22Na()+Cl()NaCl()sgs2

Earlier, we determined that sodium’s oxidation number changes from zero to +1 as it changes from a neutral solid to a positive ion. We also determined that chlorine’s oxidation number changes from zero to 1 as it changes from a neutral gas to a negative ion. In this reaction, sodium is oxidized, as it loses electrons to have a more positive oxidation number. Sodium is the reducing agent. Chlorine is reduced, as it gains electrons to have a more negative oxidation number. Chlorine is the oxidizing agent.

Example 3: Determining the Common Oxidation State of Alkali Metals

What is the common oxidation number of alkali metals?

Answer

Oxidation numbers quantify the number of electrons lost or gained by an element during a chemical reaction. Oxidation numbers are positive when an element effectively loses electrons during a chemical reaction, and they are negative when an element effectively gains electrons during a chemical reaction. Alkali metals are group 1 elements, and they tend to lose a single electron during a chemical reaction. We can use these statements to determine that alkali metals usually have a +1 oxidation number when they are part of a chemical compound. This suggests that +1 is the correct answer for this question.

Key Points

  • Oxidation number, often equivalent to oxidation state, is the degree of the oxidation of an atom or ion.
  • Oxidation numbers show the number of electrons lost or gained during an oxidation or a reduction reaction.
  • There are many rules that govern the oxidation number an atom has, but in general, we can say the following:
    • Atoms in their elemental forms have an oxidation number of 0.
    • The sum of the oxidation numbers in a polyatomic ion is the charge of the ion.
    • Atoms in compounds often have an oxidation number equivalent to the charge of the ion they tend to form, although there are many exceptions.
    • In redox reactions, the oxidation number of an atom changes from the beginning to the end of the reaction. The atom whose oxidation number increases is oxidized, while the atom whose oxidation number decreases is reduced.

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