Oxidation Number Calculation: From Intuition to Precision
Imagine you're watching a tug-of-war between two atoms in a molecule. Each atom has a certain "pull" on the shared electrons — chemists call this electronegativity. The oxidation number is like a scorecard that tells us: If the more electronegative atom took all the shared electrons, what charge would each atom end up with?
This isn't a real charge — it's a bookkeeping tool. Real molecules don't have these exact charges. But this imaginary scorecard helps us track where electrons go during chemical reactions, especially in redox (reduction-oxidation) processes.
The Core Idea
Oxidation number (also called oxidation state) is the hypothetical charge an atom would have if all bonds to atoms of different elements were 100% ionic — meaning the more electronegative atom keeps all the shared electrons.
For an atom bonded to another atom of the same element (like O₂ or N₂), the electrons are shared equally. So the oxidation number is zero — no one "wins" the tug-of-war.
The Rules (Your Toolkit)
These rules are applied in order — rule 1 overrides rule 2, and so on. Memorise them in this sequence:
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Free elements (uncombined, like Fe, O₂, H₂, S₈) have oxidation number = 0.
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Monatomic ions have oxidation number = their charge.
Example: Na⁺ = +1, Cl⁻ = −1, Mg²⁺ = +2.
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Fluorine is always −1 in compounds (it's the most electronegative element — it always "wins").
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Oxygen is usually −2, except:
- In peroxides (like H₂O₂) it's −1
- In OF₂ (with fluorine) it's +2 (fluorine wins)
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Hydrogen is usually +1 when bonded to non-metals, −1 when bonded to metals (like NaH, CaH₂).
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The sum of oxidation numbers in a neutral compound = 0.
In a polyatomic ion, the sum = the ion's charge.
Never apply rule 6 before rules 1–5. The sum rule is your check, not your starting point.
How to Calculate: A Step-by-Step Example
Let's find the oxidation number of sulphur in H₂SO₄ (sulphuric acid).
Step 1: Write the known oxidation numbers.
Hydrogen: +1 (rule 5, bonded to non-metal oxygen)
Oxygen: −2 (rule 4, not a peroxide)
Step 2: Let the unknown be x (for sulphur).
Step 3: Apply the sum rule (rule 6). The compound is neutral, so:
2(+1)+x+4(−2)=0
Step 4: Solve:
2+x−8=0
Sulphur in H₂SO₄ has oxidation number +6.
Another Example: A Polyatomic Ion
Find the oxidation number of chromium in Cr₂O₇²⁻ (dichromate ion).
Oxygen: −2 (rule 4)
Let chromium = x
Sum of oxidation numbers = charge of ion (−2):
2x+7(−2)=−2
2x−14=−2
When you get a fractional oxidation number (like +2.5 in Fe₃O₄), it means the compound has two different oxidation states for the same element. Fe₃O₄ actually contains Fe²⁺ and Fe³⁺ in a 1:2 ratio.
Common Traps to Avoid
| Mistake | Why it's wrong |
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