Oxidation numbersEdexcel A-Level Chemistry: Revision notes

The oxidation number of an element in a species is the charge it would have if all the bonding were completely ionic. It is a bookkeeping tool that shows where electrons have been gained or lost. It is written with the sign before the number (+2, −1), unlike an ionic charge (2+, 1−).

Key rules:

• An uncombined element has oxidation number 0 (Na, O₂, S₈)
• A simple monatomic ion has an oxidation number equal to its charge (Mg²⁺ is +2)
• The oxidation numbers in a neutral compound add up to 0; in an ion they add up to the charge on the ion
• In compounds: Group 1 is +1, Group 2 is +2, fluorine is −1, hydrogen is +1, oxygen is −2, other halogens usually −1

Set the sum of oxidation numbers equal to the overall charge and solve for the unknown.

Worked example 1: Mn in MnO₄⁻. Mn + 4(−2) = −1, so Mn = +7.

Worked example 2: Cr in Cr₂O₇²⁻. 2Cr + 7(−2) = −2, so 2Cr = +12 and Cr = +6.

Worked example 3: S in SO₃²⁻. S + 3(−2) = −2, so S = +4.

In a compound with more than one atom of the unknown element, divide the total by the number of atoms. For S₂O₃²⁻ this gives an average oxidation number of +2.

Oxygen is usually −2 and hydrogen +1, but there are important exceptions:

• In peroxides (H₂O₂, Na₂O₂) oxygen is −1, as each O is bonded to the other O
• In metal hydrides (NaH, CaH₂) hydrogen is −1, because it is bonded to a less electronegative metal
• In OF₂ oxygen is +2, because fluorine is the most electronegative element and is always −1

Check for these before assuming the usual values. In H₂O₂, 2(+1) + 2(O) = 0 gives O = −1.

Oxidation numbers are written as Roman numerals in the names of compounds to show the oxidation number of an element that can have several values: iron(II) chloride is FeCl₂, iron(III) chloride is FeCl₃. Sulfate(VI) shows S is +6 in SO₄²⁻, and manganate(VII) shows Mn is +7 in MnO₄⁻.

To write a formula from oxidation numbers, balance the total positive and negative:

• Iron(III) oxide: Fe is +3, O is −2, so 2 × (+3) balances 3 × (−2): Fe₂O₃
• Copper(II) nitrate: Cu is +2, NO₃⁻ is −1, so Cu(NO₃)₂
• Vanadium(V) oxide: V₂O₅

A reaction is redox if any element changes its oxidation number. An increase is oxidation; a decrease is reduction. They always occur together.

If no element changes oxidation number the reaction is not redox (e.g. acid plus carbonate).

Disproportionation is a redox reaction in which one element in a single species is simultaneously oxidised and reduced. Example: 2H₂O₂ → 2H₂O + O₂. Oxygen goes from −1 to −2 (reduced) and from −1 to 0 (oxidised).

To prove disproportionation, state the oxidation number of the element in the reactant, then in each product, and identify which one increased and which decreased.