UNIT 2 CHEMISTRY · PEARSON AAQ · C1.2.1
Oxidation Numbers
FREE CHEMISTRY REVISION · TOPIC 19
Keep the atom counts, signs and overall charge consistent.
Use a reliable sum rule to assign oxidation numbers in elements, compounds and ions, with clear worked calculations and appropriate exceptions.
Pearson BTEC Level 3 AAQ Applied Science
Unit 2: Principles and Applications of Chemistry · C1.2.1
What you will learn
Assign oxidation numbers to elements in free elements, compounds and ions. Use the sum rule, apply atom-count multipliers and keep positive and negative signs throughout. This is the foundation for deriving formulas in C1.2.2 and recognising redox changes later in C1.2.3.
1. What an oxidation number means
An oxidation number (oxidation state) is a formal electron-accounting value assigned to an atom. In the ionic approximation, electrons in a bond between unlike atoms are assigned to the more electronegative partner; electrons in a bond between identical atoms are divided equally. The resulting formal value helps track electron distribution.
For a monatomic ion, oxidation number equals its charge: Na⁺ is +1, Mg²⁺ is +2 and Cl⁻ is −1. In covalent substances it need not be a real ionic charge. Sulfur can be assigned +6 in H₂SO₄ without claiming the molecule contains a free S⁶⁺ ion. It is also different from partial charge δ and from valency, which is normally stated without a sign.
Write the sign explicitly: +3, not merely 3 when an oxidation number is requested. Charge notation usually places magnitude before sign, such as Al³⁺, whereas the oxidation-number value is written +3. Roman numerals in names such as sulfur(VI) identify +6; they do not count atoms.
2. The rules to apply in a sensible order
| Situation | Rule | Example |
|---|---|---|
| Free element | Every atom has oxidation number 0 | Na(s), P₄, S₈, Cl₂, Ar: 0 |
| Monatomic ion | Equal to the ion charge | Al³⁺: +3; S²⁻: −2 |
| Neutral compound | Sum over all atoms = 0 | MgCl₂: (+2) + 2(−1) = 0 |
| Polyatomic ion | Sum over all atoms = overall ion charge | SO₄²⁻: x + 4(−2) = −2 |
| Group 1 / Group 2 in ordinary compounds | +1 / +2 respectively | Na₂O: Na +1; MgO: Mg +2 |
| Fluorine in compounds | −1 | OF₂: each F −1 |
| Oxygen in most compounds | −2, with stated exceptions | SO₂: each O −2 |
| Hydrogen in most compounds | +1; −1 in metal hydrides | HCl: H +1; NaH: H −1 |
| Cl, Br and I in common binary compounds | Usually −1; check O/F compounds | NaCl: Cl −1; HClO: Cl +1 |
Use the definite rules and the sum rule before guessing a central value. Aluminium is normally +3 in the common compounds considered here. Phosphorus, sulfur and chlorine have variable oxidation numbers, so the formula or a stated oxidation state must guide you. Group number alone is not a universal oxidation number.
3. A repeatable calculation method
1. Identify whether the species is neutral or an ion, and write its total charge.
2. Assign known oxidation numbers, checking exceptions.
3. Let x represent the unknown oxidation number per atom.
4. Multiply each oxidation number by the number of those atoms.
5. Set the sum equal to the overall charge and solve.
6. Substitute the result back into the original sum and include its sign.
For H₂SO₄: H is +1 and O is −2. The species is neutral, so 2(+1) + x + 4(−2) = 0. Hence x − 6 = 0 and S is +6. In SO₄²⁻ the same sulfur value follows from x + 4(−2) = −2. The total charge on an ion is not automatically the oxidation number of its central atom.
4. Exceptions: apply them only where they belong
Peroxides: the peroxide group O₂²⁻ has oxygen at −1 per atom. In Na₂O₂, 2(+1) + 2x = 0 gives x = −1. This does not make oxygen −1 in ordinary Na₂O, where it is −2. Recognise the peroxide identity rather than assuming any formula containing two O atoms is a peroxide: CO₂ contains ordinary O at −2.
Metal hydrides: hydrogen is −1 in compounds such as NaH or MgH₂, paired with more electropositive metals. In water and common acids H is +1. Oxygen bonded to fluorine: F remains −1, so O is +2 in OF₂. These are useful exceptions to understand; they are not reasons to abandon the normal rules for the specified Period 3 oxides and acids.
Halogens with oxygen: chlorine is not always −1. In HClO, (+1) + x + (−2) = 0 gives Cl +1; in HClO₄ the result is +7. Elemental Cl₂ still has oxidation number 0, not −1. Bonding to the same element does not change the free-element rule.
5. Charges, brackets and atom counts
In OH⁻, oxygen is −2 and hydrogen +1, giving total −1. In Al(OH)₃ there are three complete hydroxide groups: 3 O and 3 H. Their total is 3[−2 + 1] = −3, which balances Al at +3. Parentheses multiply every atom inside them; they do not change the oxidation numbers of O or H.
A leading coefficient does not change an atom’s oxidation number: chlorine is 0 in Cl₂ and in 3Cl₂. If a species has two unknown atoms of the same element in equivalent environments, write 2x rather than x. The sum method may give an average value for more complicated mixed-state compounds; do not assume an average proves every atom is equivalent. Such cases are beyond the main Period 3 examples here.
Worked examples
Example 1 · Free elements and simple ions
Na(s), P₄ and Cl₂ each have oxidation number 0 for their atoms. For Na⁺ use +1; for Cl⁻ use −1. The presence of a subscript 2 or 4 does not create an ionic charge.
Example 2 · Sulfur in SO₂
Neutral species: x + 2(−2) = 0. Therefore x − 4 = 0 and S = +4. Check (+4) − 4 = 0. The two oxygens contribute −4 together, not −2 together.
Example 3 · Phosphorus in P₄O₁₀
Write 4x + 10(−2) = 0. Thus 4x = 20 and P = +5 per atom. Check 4(+5) + 10(−2) = 0. Do not report +20: that is the combined phosphorus contribution.
Example 4 · Sulfur in sulfate, SO₄²⁻
Use the actual ion charge: x + 4(−2) = −2. Add 8 to both sides: x = +6. Setting the right side to zero would incorrectly give +8.
Example 5 · Chlorine in ClO⁻
The ion has total charge −1: x + (−2) = −1. Therefore Cl = +1. Check (+1) + (−2) = −1. Chlorine is positive here because oxygen takes the usual negative assignment.
Example 6 · Chlorine in HClO₄
The neutral acid gives (+1) + x + 4(−2) = 0. Hence x − 7 = 0 and Cl = +7. Do not treat HClO₄ as though its H were −1; it is not a metal hydride.
Example 7 · Aluminium in Al(OH)₃
Each OH group contributes −1, so x + 3(−1) = 0 and Al = +3. Expanding gives x + 3(−2) + 3(+1) = 0, the same result.
Example 8 · Recognise the peroxide exception
For Na₂O₂, 2(+1) + 2x = 0 gives O = −1. For Na₂O, 2(+1) + x = 0 gives O = −2. The different formulas describe different substances, not interchangeable ways of writing sodium oxide.
Common misconceptions and exam tips
“All oxidation numbers in a compound must be zero.” The sum is zero for a neutral compound; individual values need not be.
“The ion charge belongs entirely to the central atom.” Sum the contributions from every atom.
“Chlorine is always −1.” Check oxygen-containing species and free Cl₂.
“A subscript is a charge.” It counts atoms.
“+6 means a free 6+ ion exists.” Oxidation numbers are formal assignments.
“Oxygen is −2 without exception.” Recognise peroxide and fluorine cases.
Show a one-line algebraic sum even when the answer seems obvious. Preserve minus signs in brackets, distinguish per-atom value from total contribution, and finish with a charge-sum check. Keep the original formula unchanged while calculating.
Your revision summary
Free elements are 0; monatomic ions use their charge. Sum all atom contributions to zero for a neutral species or to the total charge for an ion. Use atom-count multipliers, explicit signs and the appropriate exceptions. Oxidation numbers describe formal electron accounting and do not automatically represent real ionic charges.
Check your understanding
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