# Oxidation number

How to find oxidation numbers, and a brief introduction to oxidation-reduction (redox) reactions.

## What are oxidation numbers?

Chemists use oxidation numbers (or oxidation states) to keep track of how many electrons an atom has. Oxidation numbers don’t always correspond to real charges on molecules, and we can calculate oxidation numbers for atoms that are involved in covalent (as well as ionic) bonding.
Let's make sense of oxidation numbers with a few examples!

## Guidelines for determining the oxidation number

Oxidation numbers are usually written with the sign ($+$ or $-$) first, then the magnitude, which is the opposite of charges on ions. Chemists use the following guidelines to determine oxidation numbers:
Step $1.~$Atoms in their elemental state have an oxidation number of $0$.
Step $2.~$Atoms in monatomic (i.e. single atom) ions have an oxidation number equal to their charge.
Step $3.~$In compounds: fluorine is assigned a oxidation number; oxygen is usually assigned a oxidation number (except in peroxide compounds where it is , and in binary compounds with fluorine where it is positive); and hydrogen is usually assigned a $+1$ oxidation number except when it exists as the hydride ion, , in which case rule $2$ wins.
A peroxide is a compound that contains an oxygen-oxygen single bond. If you aren't sure if you have a peroxide based on the molecular formula, try drawing the most likely Lewis structure. In general, peroxides are not especially common because the oxygen-oxygen bond is a pretty weak (and reactive) single bond, so I would not recommend invoking them unless you have a good reason!
Step $4.~$In compounds, all other atoms are assigned an oxidation number so that the sum of the oxidation numbers on all the atoms in the species equals the charge on the species.

## Determining the oxidation state in $\text H_2$ and $\text H_2 \text O$

When using the guidelines, we usually start with rule $1$ and move downward in the list until we are done. By following that order, we determine the oxidation states for the easy atoms first, and use the process of elimination for trickier examples.
Example: What is the oxidation state of the hydrogen atom in $\text H_2$ and $\text H_2 \text O$?
$\text H_2$: We can use rule $1$ here. Since $\text H_2$ is in its elemental state, it has an oxidation number of $0$.
$\text H_2 \text O$: Rules $1$ and $2$ don’t apply to this situation, so we can skip them. Rule $3$ tells us that hydrogen usually has the oxidation number of $+1$, except in hydrides. We don’t have a hydride here, but even if we aren’t sure if our hydrogen has a $+1$ or $-1$ oxidation state, we can use rule $4$ to check.
Rule $4$ tells us that all the oxidation numbers in a compound have to add up to charge on the compound, and rule $3$ says that oxygen usually has an oxidation number of $-2$. Water is a neutral compound, so we need our oxidation numbers to add up to $0$. If we assign a $-1$ oxidation number to hydrogen because we think it might be a hydride, we get for the sum of oxidation numbers:
That doesn't look right! If instead we use the oxidation number $+1$ for hydrogen, we get for the sum: $(-2 \times 1)+(+1 \times 2) = 0$, which is what we would expect for a neutral molecule. Thus, using rules $3$ and $4$ we can assign the hydrogen in $\text H_2 \text O$ an oxidation state of $+1$.
Concept check: What is the oxidation state of sulfur in the sulfate ion, $\text{SO}_4^{2-}$?
A quick skim of the guidelines tells us that we don't have the oxidation state of sulfur listed. That means we will have to use rules $3$ and $4$ to figure out the oxidation state of sulfur.
Rule $3$ tells us that the oxidation state of oxygen is $-2$. We can then use rule $4$ to calculate the oxidation state of sulfur:
\begin{aligned} \text{Net charge on SO}_4^{2-} &= (\text{oxidation number of O} \times 4) + \text{oxidation number of S} \\ -2 &= (-2 \times 4) + \text{oxidation number of S} \\ -2 &= -8 + \text{oxidation number of S} \end{aligned}
Solving the equation for the oxidation number of sulfur, we get the following answer:
$\text{oxidation number of S in } \text{SO}_4^{2-} = +6$

## Using oxidation numbers: Oxidation and reduction

Now we can follow instructions to find oxidation numbers. That’s cool, but why do oxidation states matter? After all, they are practically imaginary!
A thin ribbon of silver magnesium metal burns with white flame to produce a white-grey solid, magnesium oxide.
When pure magnesium metal is burned, the magnesium gets oxidized to form magnesium oxide. Photo from Wikimedia Commons, CC BY-SA 3.0
Chemists are usually interested in keeping track of electrons because we want to know when electrons get transferred from one atom to another. In the above example, we can see that the oxidation state of the hydrogen is different in $\text H_2$ compared to $\text{H}_2 \text O$. If we compare the oxidation states, a chemist might say that hydrogen in water has fewer electrons compared to elemental hydrogen.
Here is the reaction to make water from hydrogen gas and oxygen gas:
$2\text H_2(g)+ \text O_2(g) \rightarrow 2\text{H}_2 \text O (l)$
We will see that this reaction not only breaks and makes new chemical bonds, but also involves a loss of electrons from $\text H_2$. Chemists have specific terms to describe processes that involve the loss or gain of electrons:
Oxidation is the loss of one or more electrons by an atom. When the oxidation number of an element increases, that means there is a loss of electrons and that element is being oxidized. In the above reaction, $\text H_2$ is being oxidized because it loses electrons to form $\text{H}_2 \text O$. The loss of electrons is evident from the oxidation number changing from $0$ to $+1$: since the hydrogen atom lost an electron (which has a negative charge), the oxidation number increased.
Reduction is the gain of one or more electrons by an atom. When the oxidation number of an element decreases, that means there is a gain of electrons and that element is being reduced. In the reaction to make water, $\text O_2$ is being reduced because the oxidation number of each oxygen atom decreased from $0$ to $-2$ from the gain of $2$ negatively charged electrons.
Some ways to remember oxidation and reduction include:
$1.~$OIL RIG: "Oxidation Is Loss" and "Reduction Is Gain"
$2.~$LEO the lion says GER: "Loss of Electron is Oxidation" and "Gain of Electrons is Reduction"
The terms oxidizing agent and reducing agent are closely related to oxidation and reduction.
A reducing agent, or reductant, loses electrons and is therefore oxidized in a chemical reaction. In our example, $\text H_2$ is acting as a reducing agent in the reaction since it is getting oxidized and causing the reduction of $\text O_2$.
An oxidizing agent, or oxidant, gains electrons and gets reduced in a chemical reaction. $\text O_2$ is acting as an oxidizing agent in our reaction because it is responsible for the oxidation of the reducing agent, $\text H_2$.

## Oxidation numbers and redox reactions

Chemical reactions that involve the transfer of electrons are called oxidation-reduction or redox reactions. Oxidation state changes are a sign that an electron transfer is occurring. All redox reactions involve both a reduction and an oxidation.
Many reaction types such as combustion, single replacement, and some synthesis and decomposition reactions fall into the category of redox reactions. Oxidation numbers and redox reactions are important concepts in more advanced chemistry classes as well, such as biochemistry, electrochemistry, organic chemistry, and inorganic chemistry to name a few. We aren't going to go into more detail about redox reactions in this article, but keep in mind being comfortable with oxidation numbers will come in handy!

## Summary

Oxidation numbers are used by chemists to keep track of electrons within a compound. We can use guidelines to assign oxidation numbers to atoms in a compound. Changes in oxidation state during a reaction tell us that there is a transfer of electrons. Reactions that involve the transfer of electrons are called redox reactions, and they include a reduction (a gain of electrons) and an oxidation (a loss of electrons). The substance that is reduced is called the oxidizing agent, and the substance that is oxidized is called the reducing agent.