Oxidation Numbers — Part 2
Using the Algebra Rules to find unknown oxidation numbers in compounds and polyatomic ions.
The sum of oxidation numbers in a neutral compound = 0
In H₂O: (+1×2) + (−2) = 0 ✓
The sum in a polyatomic ion = the ion's charge
In SO₄²⁻: sum must equal −2
The Constants (memorise these)
Oxygen
−2
except in peroxides (−1)
Hydrogen
+1
except in metal hydrides (−1)
The Algebra Board — choose a compound
Carbon dioxide — the gas you breathe out. But what's carbon's 'tax bill' here?
What is the oxidation number of Carbon (C) in CO₂?
The Method — A Universal Recipe
Write the algebraic equation
Let the unknown element = x. Write: (known contributions) + x = target sum.
Multiply counts × values
E.g. 4 oxygen atoms × (−2) = −8. Do this for every known element.
Set the target
Neutral compound → target = 0. Ion → target = ion charge (e.g. −1 for NO₃⁻).
Solve for x
Simple algebra: add/subtract both sides to isolate x.
Write with sign
Always write the sign: +6, not just 6. Cambridge demands this.
⚠️ Examiner Traps
✗ Forgetting the sign
Always write +6, never just 6. Cambridge will mark "+6" correct and "6" wrong.
✗ Assuming oxygen is always −2
In peroxides (H₂O₂), oxygen is −1. The question will usually tell you if this applies.
✗ Confusing ion charge with oxidation number
The ion charge is the TARGET of the sum. The oxidation number of each element is calculated inside.
✗ Getting the algebra sign wrong
If N − 6 = −1, then N = −1 + 6 = +5. Not +1 − 6 = −5!
Cambridge IGCSE Chemistry 0620 · Syllabus 6.4.9c & 6.4.9d · Supplement
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