Oxidation numbers

Sometimes it is not easy to decide if a reaction is a redox one or not. To help keep track of the movement of electrons in reactions chemists use the idea of oxidation numbers. Oxidation numbers make it easy to spot redox reactions and to help us work out exactly what is happening in a chemical reaction.

Ionic compounds typically contain metal and non-metal ions. These ions are formed when the metal atoms are oxidised (lose electrons) and the non-metal atoms are reduced (gain electrons). The number of electrons lost or gained by the metal and non-metal atoms is simply the number needed to gain a stable octet of electrons (from gcse chemistry- 8 electrons in the outer electron shell). For example sodium chloride consists of sodium ions (Na⁺) and chloride ions (Cl^-). The oxidation number for these two ions is simply the number of electrons lost or gained to complete a stable octet of electron. So for sodium its oxidation number is +1 and for the chloride ion its oxidation state or number would be -1.

For other ionic compounds such as those which contain a group 2 metal such as Mg or Ca the oxidation state of the metal will be +2, for group 3 metals such as Al the oxidation state is always +3.

Hydrogen usually loses 1 electron in its reactions and so will have an oxidation number of +1. However hydrogen can form compounds with metals such as sodium and lithium where it gains an electron from the metal. This H^- ion is called a hydride ion and it has an oxidation state or oxidation number of -1.

More oxidation numbers

The halogens

Oxidation numbers in covalent compounds.

In deciding on oxidation numbers above we assumed that electrons were transferred from on atom to another and that ions were produced. However in covalent compounds the electrons are shared and not transferred. However recall that oxidation numbers are only used to help us keep track of what is happening in a chemical reaction. So we will simply assume or pretend that the covalent compounds are in fact ionic- even though they are NOT!

In working out oxidation numbers for the elements in a covalent compound we will assume that the most electronegative element in the covalent bond basically has the electrons in the covalent bond transferred to it. This will mean that the more electronegative element in the covalent bond will have a negative oxidation number and the least electronegative element will have a positive oxidation number e.g.

• The most electronegative element here is nitrogen and it makes 3 covalent bonds to the hydrogen atoms. So despite the fact that this is a covalent compound we will assume that the electrons in the covalent bonds are transferred to the most electronegative element, which in this case will be the nitrogen atom. So this means that the nitrogen atom will gain 3 electrons, one electron from each hydrogen atom. This means the oxidation number of nitrogen will be -3.
• Each hydrogen atom will lose 1 electron so the oxidation state of hydrogen will be +1.

Example 2- What is the oxidation number for each atom in a molecule of sulfur dioxide (S0₂)?

• The most electronegative element is present in this molecule is oxygen and it needs to gain 2 electrons to complete its octet; so it will gain 2 electrons from the sulfur atom and end up with an oxidation number of -2.
• The molecule is neutral and since the sulfur atom is bonded to two oxygen atoms and each one takes 2 electrons then the sulfur will end up with an oxidation number of +4.

Example 3 - What is the oxidation state of each atom in the sulfate ion, SO₄^2-?

• Each oxygen atom will have an oxidation number of -2. Since there are x4 oxygen atoms this will give a total of -8.
• The molecule has a charge of -2; so the oxidation state of the sulfur atom must be +6. Since -8+6=2-.

Example 4- What is the oxidation number of chlorine in the ClO₃^- molecule?

• The oxidation number of the oxygen atom is -2. There are 3 oxygen atoms giving a total charge of -6.
• The molecule has a charge of -1, so the oxidation number of the chlorine must be +5 since -6+5=-1.

Example 5- What is the oxidation state of aluminium in aluminium oxide (Al₂O₃)?

• The oxidation state of the oxygen ion will be -2. Since there are x3 oxide ions this gives a total charge of -6.
• The "molecule" is neutral so the total oxidation number for the two aluminium ions must be +6. So simply divide by 2 to get the oxidation number for each aluminium ion, so the answer is +3.

Oxidation numbers and transition metals

The transition metals form a variety of compounds where the metal has variable oxidation states or numbers. I am sure you will have seen the names of many transition metals written with the oxidation number of the metal present in Roman numerals e.g.

• iron (III) oxide - here the iron has an oxidation number of +3
• iron(II) oxide - here the iron has an oxidation number of +2
• copper(I) chloride - here the copper has an oxidation number of +1
• copper (II) chloride - here the copper has an oxidation number of +2

You may also see the oxidation state of the metal written in Roman numerals after the name of the compound e.g.
• Iron (III) oxide (Fe₂O₃)- here the iron has an oxidation state of +3. The Roman numerals after the iron in the compounds name are telling you the oxidation state of the iron.
• Copper(II) sulfate pentahydrate (CuSO₄·5H₂O)- here the copper has an oxidation state of +2.
• potassium dichromate (VI) (Cr₂O₇^2-) here the metal chromium has an oxidation number of +6
• potassium manganate (VII) (MnO₄^-)- here the manganese has an oxidation number of +7

The common oxidation states for the first row transition metals are shown in the table below. The oxidation state of the metal will depend on the reaction conditions and the reagents used.

The lower oxidation states of the transition metals consist of simple ions such as M²⁺ and M³⁺. Transition metals with a low oxidation number can act as reducing agents and donate electrons to other substances, e.g.

Fe²⁺ → Fe³⁺ + e

Cr²⁺ → Cr³⁺ + e

• The highest oxidation state for the metal vanadium is +5, this is found for example in the VO₂⁺ ion.
• The highest oxidation state for chromium is +6, this is found for example in the chromate ion (CrO₄^2-) and dichromate ions Cr₂O₇^2-.
• The highest oxidation state for manganese is +7, this is found for example in the MnO₄^- ion.

Ions which contain metals in high oxidation states are good electron acceptors; that is they are good oxidising agents.

The oxidation states of vanadium

The transition metal vanadium has a wide range of oxidation states ranging from +2, +3, +4 and +5. Each of these oxidation states has a different colour. You can view all the different colours of the various oxidation states of vanadium, start by dissolving the compound ammonium vanadate (V) in sulfuric acid. This compound has the formula NH₄VO₃contains vanadium ions in the +5 oxidation state. Now when it dissolves in the sulfuric acid it converts the vanadate (V) ions into the dioxovanadium (V) ions (VO₂⁺) which are yellow in colour (see image below). An equation for this reaction is shown below, note that the ammonium ion and hydrogen sulfate ions are spectator ions and for simplicity I have removed them from the equation below:

VO_3(aq)^- + 2H ⁺ → VO_2(aq)⁺ + H₂O_(aq)

Zn_(s) → Zn²⁺_(aq) + 2e

The oxidation states of chromium metal

Chromium metal is another transition metal with variable oxidation states, it has 3 oxidation states: +2, +3 and +6; that is Cr²⁺, Cr³⁺ and Cr⁶⁺, with the Cr³⁺ ion being the most stable. Now chromium metal reacts with concentrated acid in the absence of air to yield the beautiful blue chromous ion, Cr²⁺. We can write a simplified equation for this reaction as:

Cr_(s) + 2H⁺ → Cr²⁺_(aq) + H_2(g)

4Cr²⁺_(aq) + O_2(g) + 4H⁺_(aq) → 4Cr³⁺_(aq) + 2H₂O_(l)

The highest oxidation state for chromium is +6 and this is found in two ions, the yellow chromate ion CrO₄^2- and the orange dichromate ion, Cr₂O₇^2- ion. These two ions are in equilibrium with each other and the concentration of each ion in the equilibrium mixture depends upon the pH of the solution, this is shown by the equation below:

CrO₄^2- + 2H⁺ ⇌ Cr₂O₇^2- + H₂O_(l)