# General Chemistry/Redox Reactions/Oxidation state

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Oxidation states are used to determine the degree of oxidation or reduction that an element has undergone when bonding. The oxidation state of a compound is the sum of the oxidation states of all atoms within the compound, which equals zero unless the compound is ionic.
The oxidation state of an atom within a molecule is the charge it would have if the bonding were completely ionic, even though covalent bonds do not actually result in charged ions.

## Method of notation

Oxidation states are written above the element or group of elements that they belong to (when drawing the molecule), or written with roman numerals in parenthesis when naming the elements.
Examples :
| ${\begin{matrix}_{0}\\Al\end{matrix}}$ | aluminum |
| ${\begin{matrix}_{+3}\\Al\end{matrix}}$ | aluminum(III), an ion |

## Determining oxidation state
### For single atoms or ions

Because oxidation numbers are just the sum of the electrons gained or lost, calculating them for single elements is easy.
Examples :
* ${\begin{matrix}_{1}\\K^{+}\end{matrix}}$
* ${\begin{matrix}_{0}\\Br\cdot \end{matrix}}$
* ${\begin{matrix}_{1}&_{-1}\\Na&Cl\end{matrix}}$
Notice that the oxidation states of ionic compounds are simple to determine.

### For larger molecules

Although covalent bonds do not result in charges, oxidation states are still useful. They label the hypothetical transfer of electrons if the substance were ionic. Determining the oxidation states of atoms in a covalent molecule is very important when analyzing "redox" reactions. When substances react, they may transfer electrons when they form the products, so comparing the oxidation states of the products and reactants allows us to keep track of the electrons.
Examples :
| ${\begin{matrix}_{+1}&_{-1}\\H&Cl\end{matrix}}$ | for hydrogen chloride |
| ${\begin{matrix}_{2(+1)}&_{-2}\\H_{2}&O\end{matrix}}$ | for water |
| ${\begin{matrix}_{+3}&_{2(-2)}\\Cl&O_{2}\end{matrix}}$ | for the chlorite ion (notice the overall charge) |

### Determining Oxidation States

The determination of oxidation states is based on knowing which elements can have only one oxidation state other than the elemental state and which elements are able to form more than one oxidation state other than the elemental state. Let's look at some of the "rules" for determining the oxidation states.
1. The oxidation state of an element is always zero.
2. For metals, the charge of the ion is the same as the oxidation state. The following metals form only one ion: Group IA, Group IIA, Group IIIA (except Tl), Zn$^{2+}$, Cd$^{2+}$.
3. For monatomic anions and cations, the charge is the same as the oxidation state.
4. Oxygen in a compound is −2, unless a peroxide is present. The oxidation state of oxygen in peroxide ion , O$_2$$^{2−}$ is −1.
5. For compounds containing polyatomic ions, use the overall charge of the polyatomic ion to determine the charge of the cation. Here is a convenient method for determining oxidation states. Basically, you treat the charges in the compound as a simple algebraic expression. For example, let's determine the oxidation states of the elements in the compound, KMnO$_4$. Applying rule 2, we know that the oxidation state of potassium is +1. We will assign "x" to Mn for now, since manganese may be of several oxidation states. There are 4 oxygens at −2. The overall charge of the compound is zero: K Mn O4 +1 x 4(-2)
The algebraic expression generated is: 1 + x -8 = 0
Solving for x gives the oxidation state of manganese: x - 7 = 0 x = +7 K Mn O4 +1 +7 4(-2)
Suppose the species under consideration is a polyatomic ion. For example, what is the oxidation state of chromium in dichromate ion, (Cr2O72-)?
As before, assign the oxidation state for oxygen, which is known to be -2. Since the oxidation state for chromium is not known, and two chromium atoms are present, assign the algebraic value of 2x for chromium: Cr2 O7 2- 2x 7(-2)
Set up the algebraic equation to solve for x. Since the overall charge of the ion is -2, the expression is set equal to -2 rather than 0: 2x + 7(-2) = -2
Solve for x: 2x - 14 = -2 2x = 12 x = +6
Each chromium in the ion has an oxidation state of +6. Let's do one last example, where a polyatomic ion is involved. Suppose you need to find the oxidation state of all atoms in Fe2(CO3)3. Here two atoms, iron and carbon, have more than one possible oxidation state. What happens if you don't know the oxidation state of carbon in carbonate ion? In fact, knowledge of the oxidation state of carbon is unnecessary. What you need to know is the charge of carbonate ion (-2). Set up an algebraic expression while considering just the iron ion and the carbonate ion: Fe2 (CO3)3 2x 3(-2) 2x - 6 = 0 2x = 6 x = 3
Each iron ion in the compound has an oxidation state of +3. Next consider the carbonate ion independent of the iron(III) ion: C O3 2- x 3(-2) x - 6 = -2 x = +4
The oxidation state of carbon is +4 and each oxygen is -2.

## Guidelines

Determining oxidation states is not always easy, but there are many guidelines that can help. This guidelines in this table are listed in order of importance. The highest oxidation state that any element can reach is +8 in XeO$_4$.
| Element | Usual Oxidation State |
| Fluorine | Fluorine, being the most electronegative element, will always have an oxidation of -1 (except when it is bonded to itself in F$_2$, when its oxidation state is 0). |
| Hydrogen | Hydrogen always has an oxidation of +1, -1, or 0. It is +1 when it is bonded to a non-metal (e.g. HCl, hydrochloric acid). It is -1 when it is bonded to metal (e.g. NaH, sodium hydride). It is 0 when it is bonded to itself in H$_2$. |
| Oxygen | Oxygen is usually given an oxidation number of -2 in its compounds, such as H$_2$O. The exception is in peroxides (O$_2$$^{-2}$) where it is given an oxidation of -1. Also, in F$_2$O oxygen is given an oxidation of +2 (because fluorine must have -1), and in O$_2$, where it is bonded only to itself, the oxidation is 0. |
| Alkali Metals | The Group 1A metals always have an oxidation of +1, as in NaCl. The Group 2A metals always have an oxidation of +2, as in CaF$_2$. There are some rare exceptions that don't need consideration. |
| Halogens | The other halogens (Cl, Br, I, As) usually have an oxidation of -1. When bonded to another halogen, its oxidation will be 0. However, they can also have +1, +3, +5, or +7. Looking at the family of chlorides, you can see each oxidation state (Cl$_2$ (0), Cl$^-$ (-1), ClO$^-$ (+1), ClO$_2$$^-$ (+3), ClO$_3$$^-$ (+5), ClO$_4$$^-$ (+7)). |
| Nitrogen | Nitrogen (and the other Group 5A elements, such as phosphorus, P) often have -3 (as in ammonia, NH$_3$), but may have +3 (as in NF$_3$) or +5 (as in phosphate, PO$_4$$^{3-}$). |
| Carbon | Carbon can take any oxidation state from -4, as in CH$_4$, to +4, as in CF$_4$. It is best to find the oxidation of other elements first. |
In general, the more electronegative element has the negative number. Using a chart of electronegativities, you can determine the oxidation state of any atom within a compound.

## Periodicity

Oxidation states are another periodic trend. They seem to repeat a pattern across each period.