Oxidation Number

Oxidation Numbers are needed because for some chemical reactions it is difficult to decide whether oxidation-reduction (redox) is or is not involved.

To begin with, here is a reaction that is easily identified as a redox reaction:

Cl2(aq) + 2Br-(aq)    ®    Br2(aq) + 2Cl-(aq)

It can be separated into two ionic half-equations, in which the transfer of electrons is easily recognised, as follows:

Cl2 + 2e-   ®   2Cl-      Reduction    -    gain of electrons
2Br-   ®   Br2 + 2e-      Oxidation    -    loss of electrons
Now examine the two chemical reactions below. Are these redox reactions?

2CrO42- + 2H+ ® Cr2O72- + H2O

CH4 + Cl2   ®   CH3Cl + HCl

In the above two reactions, covalent bonds in the molecules or polyatomic ions are changing.

The concept of oxidation numbers was conceived to assist in the recognition of redox reactions.

Here is the procedure...

Each of the atoms on both sides of the chemical equation is assigned an oxidation number. Then, each element is examined, as a reactant and a product, to see if it has undergone an increase or decrease in its oxidation number, or if it has remained the same.

Change in Oxidation NumberOxidation or Reduction

There are rules to help in assigning oxidation numbers...

Rule 1

The oxidation number for any atom in its elementary form is zero, as for:

Na, Cu, Cl, O, N2, Cl2, C, S8, P4, O2, and O3.

Rule 2

The oxidation number of any monatomic ion is the same as the charge on the ion, as for:

Na+, Cu2+, Fe3+, H-, and Br-.

Rule 3

The oxidation number of Fluorine is always -1 (except for F and F2, as for rule 1). Fluorine is the most electronegative element.

When it comes to assigning oxidation numbers to atoms that are covalently bonded in a molecule or polyatomic ion, the more electronegative atom of the two bonded atoms involved is considered to gain all of the bonding electrons.

In the example below, the oxidation number of oxygen is +2. This is because each fluorine atom is considered to deprive oxygen of one electron. It can be said that in F2O oxygen is in its +2 oxidation state.

Rule 4

The oxidation number of Oxygen is always -2, except in peroxides (such as hydrogen peroxide (H-O-O-H) where it is -1), and when bonded to fluorine in F2O where it is +2.

In hydrogen peroxide the pair of electrons between the two oxygen atoms is equally shared. Each oxygen atom is considered to gain one electron from each hydrogen atom to acquire an assumed negative charge of -1, and so is assigned an oxidation number of -1. In the peroxide ion, O22-, each oxygen has an oxidation number of -1 (relating to rule 2).

In the water molecule, the more electronegative oxygen atom is considered to gain both pairs of the bonding electrons with hydrogen. In other words, oxygen 'takes' one electron from each of the hydrogen atoms giving it an oxidation number of -2 and each hydrogen an oxidation number of +1.

This assignment imagines each atom in the covalent molecule or polyatomic ion becomes an ion; the oxidation number is the assumed ionic charge. Since covalent bonds are never completely ionic, an oxidation number assigned in these circumstances is not an actual charge.

oxidation numbers+1-2+1

The sum of the oxidation numbers is the charge on the species. For H2O the sum is

-2 + 1 + 1 = 0

As another example, what is the oxidation number of manganese in the MnO4- ion?

You can't possibly know the structure of every molecule or ion, so it is helpful to use the mathematics as above. In MnO4- the overall charge is equal to -1. Oxygen is always -2 (except in peroxides where it is -1 and when bonded to fluorine in F2O where it is +2). There are four oxygen atoms and therefore four lots of -2 making -8. This would make the oxidation number of manganese +7.

? + (4 x -2) = -1      or      +7 -8 = -1

So, is the reaction between methane and chlorine a redox reaction?

To answer the question, assign oxidation numbers to all atoms as shown:

CH4+Cl2  ®  CH3Cl+HCl 
-4 +1 0 -2+1-1 +1-1

The answer is "yes" because there is a change in oxidation number of certain atoms.

In CH4, carbon is oxidised since its oxidation number increases from -4 to -2. In Cl2, chlorine is reduced since its oxidation number decreases from 0 to -1.

What about the change CrO42-  ®  Cr2O72-  ?

2CrO42- + 2H+ ® Cr2O72- + H2O

This is not a redox reaction because the oxidation number for Cr is +6 in both the chromate and dichromate ions.

Systematic Naming of Ions and Ionic Compounds

As for organic compounds, ions and ionic compounds have old names and new names. Old names are often used, so it is helpful to be familiar with both for some ions and ionic compounds. here are some examples.

For metal cations, where the element exists in more than one oxidation state in its compounds, the oxidation number is given (using Roman numerals).

IonNew NameOld Name
Fe2+ Iron(II)Ferrous
Fe3+ Iron(III)Ferric
Cu+ Copper(I)Cuprous
Cu2+ Copper(II)Cupric

For polyatomic anions that are similar, the oxidation number of the lesser electronegative element distinguishes between them (again using Roman numerals).

IonNew NameOld Name
SO42-Sulphate(VI) ionSulphate ion
SO32-Sulphate(IV) ionSulphite ion
NO3-Nitrate(V) ionNitrate ion
NO2-Nitrate(III) ionNitrite ion
MnO4-Manganate(VII) ionPermanganate ion
MnO42-Manganate(VI) ionManganate ion
ClO- Chlorate(I) ionHypochlorite ion
ClO2-Chlorate(III) ionChlorite ion
ClO3-Chlorate(V) ionChlorate ion
ClO4-Chlorate(VII) ionPerchlorate ion
CrO42-Chromate(VI) ionChromate ion
Cr2O72-Dichromate(VI) ionDichromate ion

And finally, a couple of compounds...

CompoundNew NameOld Name
Fe(NO3)2 Iron(II) nitrate(V)Ferrous Nitrate
Ni(CO)4 Tetracarbonyl nickel(0)Nickel tetracarbonyl


  1. Use the ideas above to determine the oxidation numbers in:
  1. OF2
  2. ClF3
  3. manganese in MnO4-
  4. sulphur in (i) SO42-, (ii) SO32-, (iii) S2O32-, (iv) S4O62-, (v) SO2, (vi) SO3
  5. nitrogen in (i) NH3, (ii) N3-, (iii) NH2-,(iv) N2, (v) N2O, (vi) NO, (vii) NO2, (viii) NO2-, (ix) NO3-
  1. Identify the species which are being oxidised or reduced in the following redox reactions:
  1. MnO4-(aq) + 5Fe2+(aq) + 8H+(aq)  ®  Mn2+(aq) + 5Fe3+(aq) + 4H2O(aq)
  2. 3CuO(s) + 2NH3(g)  ®  N2(g) + 3Cu + 3H2O(l)