oxidation numbers

Oxidation numbers

Sometimes it is not easy to decide if a reaction is a redox one or not. To help keep a 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.

recap of the terms oxidation and reduction, reducing agents and oxidising agents

Ionic compounds contain metal and non-metal ions. These ions are formed when the metal is oxidised (loses electrons) and the non-metal is reduced (gains 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 - 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 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 would be -1.

For other ionic compounds such as those which contain a group 2 metal such as Mg or Ca, here 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 of -1.

Oxygen usually accepts 2 electrons to complete its outer shell of electrons and form a stable octet. This will produce the oxide ion, O2-. Here the oxygen has an oxidation number of -2. However oxygen can for example react with the alkali metals to form compounds where the oxidation state of the oxygen is -1. These compounds are called metal peroxides e.g. lithium peroxide (Li2O2) and potassium peroxide (K2O2) contain the Na+ and K+ cations. The compounds formed are neutral, that is they have no charge and since each metal peroxide contains 2 metal ions each with a + charge and there are 2 oxygen ions present, each ion must have a -1 charge, or an oxidation state of -1. Potassium can react with oxygen to form potassium superoxide (K02). In a superoxide the oxidation state of the oxygen is -1/2.

Chlorine has only to gain 1 electron to achieve an octet of electrons so it has an oxidation state of -1 unless it is bonded to fluorine or oxygen, in this case the oxidation number can vary.

Oxidation numbers in covalent compounds.

In deciding on oxidation numbers above we assumed that electrons were transferred from on atom to another and ions were produced. However in covalent compounds the electrons are shared and not transferred. However 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! 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.

Example 1 - what is the oxidation states of nitrogen and hydrogen atoms in ammonia (NH3)?

ammonia molecule

This makes sense since the ammonia molecule is neutral. The -3 oxidation number of the nitrogen will be cancelled out by the three +1 oxidation numbers from each hydrogen.

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

sulfur dioxide molecule

Since the molecule is neutral we can be confident that these oxidation numbers are correct, since 2x the oxidation number of oxygen added to the oxidation number of the sulfur atom is 0.

Example 3 - What is the oxidation state of each atom in sulfate ion, SO42-?

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

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

Oxidation numbers and transition metals

The transition metals form a variety of compounds where the metal has different oxidation 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.

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

Sc Ti V Cr Mn Fe Co Ni Cu
+2 +2 +2 +2 +2 +2 +2 +2 +2
+3 +3 +3 +3 +3 +3 +3
+4 +4 +4
+6 +6

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 higher oxidation states for the transition metals are only found in compounds where the metal is bonded to a element with a high electronegativity value, such as oxygen e.g. 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. To view all the colours of the various oxidation states of vanadium start by dissolving the compound ammonium vanadate(V) in sulfuric acid. This compound contains vandanium ions in the +5 oxidation state, when it dissolves in the acid it is converts the vanadate (V) ions into dioxovanadium(V) ions (VO2+) which are yellow in colour (see image below).

VO3(aq)- + 2H + → VO2(aq)+ + H2O(aq)
When this yellow solution is shaken with powdered zinc it is reduced to form blue oxovanadium(IV) ions, VO2+. If more zinc is added then green vanadium(III) ions are formed, V3+( and if even more zinc is added then it will reduce the green V3+( ions to form the violet V2+(aq) ion. The zinc is the reducing agent in these reactions and supplies all the electrons necessary to reduce the V5+ ion to the V2+ ion. The zinc itself is oxidised according to the equation below:
Zn(s) → Zn2+(aq) + 2e
The colour changes are shown in the image below:

colours of the oxidation states of vanadium

The oxidation states of chromium metal

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

Cr(s) + 2H+ Cr2+(aq) + H2(g)
Cr2+(aq) ions are good reducing agents and if the above reaction is carried out in the presence of air then the blue Cr2+(aq) ions are oxidised by the oxygen in the air to form the green Cr3+(aq) ion.
4Cr2+(aq) + O2(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 CrO42- and the orange dichromate ion, Cr2O72- ion. these two ions are in equilibrium with each other and the concentration of each ion in the equilibrium mixture depends upon the pH.

CrO42- + 2H+Cr2O72- + H2O(l)
In basic or alkaline solutions the yellow chromate ion (CrO42-) is the main species present while the addition of acid to the equilibrium mixture will cause the yellow solution to turn orange as the concentration of the dichromate ion (CrO42-) increases. The Cr2O72- ion in acidic conditions is an excellent oxidising agent while the chromate ion, present in basic solutions is a much weaker oxidising agent.
This is shown in the image below:

colours of chromium ions

Key Points

Practice questions

Check your understanding - Questions on oxidation numbers