IB Chemistry SLTopic 3 — Classifying the ElementsPaper 1 & 2Core skill~12 min read
Oxidation States
An oxidation state is an accounting trick. You pretend every bond in a substance is fully ionic, hand each electron to whichever atom pulls harder, and write down the charge each atom would end up with. Nobody thinks this is what really happens — but it works, and it turns “is this redox?” from a judgement call into arithmetic.
📘 What you need to know
The oxidation state of an atom is the charge it would carry if all the bonding were completely ionic.
Uncombined elements are 0. A monatomic ion equals its charge.
In compounds: group 1 is +1, group 2 is +2, fluorine is always −1, hydrogen is usually +1, oxygen is usually −2.
The states in a neutral compound sum to zero; in a polyatomic ion they sum to the charge.
The more electronegative atom takes the negative value.
Oxidation is an increase in oxidation state; reduction is a decrease.
Roman numerals (Stock notation) show the oxidation state in names, e.g. iron(III) oxide, manganate(VII).
The number line
Every element sits somewhere on a scale that runs from strongly negative to strongly positive. Movement along that scale is the definition of oxidation and reduction, which is why oxidation states make redox questions so much easier.
Manganese in the manganate(VII) ion sits at +7, the highest state any first-row transition metal reaches. That is exactly why it is such an aggressive oxidising agent — there is nowhere left to go but down.
The rules, in the order you should use them
Rule
Detail
Example
1. Uncombined elements
Always 0, however many atoms are in the molecule
Zn, O2, P4 are all 0
2. Monatomic ions
Equal to the charge on the ion
Ca2+ is +2, Br− is −1
3. Fixed values
Group 1 = +1, group 2 = +2, fluorine = −1
Na in NaCl is +1
4. Hydrogen
+1, except −1 in metal hydrides
+1 in HCl, −1 in NaH
5. Oxygen
−2, except −1 in peroxides and +2 in OF2
−2 in H2O, −1 in H2O2
6. The sum
Zero in a neutral compound, equal to the charge in an ion
In SO42− everything sums to −2
Use the rules in that order and the exceptions look after themselves. Fluorine outranks oxygen, which is why oxygen is forced to +2 in OF2 — there is no more electronegative element than fluorine, so it never gives ground.
The standard method
🧩 Finding an unknown oxidation state
Write down what you know. Usually oxygen at −2 and hydrogen at +1.
Multiply by the number of atoms of each known element.
Write the target total — zero for a compound, the charge for an ion.
Solve for the unknown, remembering to divide by the number of those atoms.
Sanity check: is the answer plausible for that element? Nothing goes above +7.
That final division is the single most common slip in this topic. Students correctly reach +12 and then write it down as the oxidation state of chromium.
Fractional values
Sometimes the arithmetic gives you a fraction. In Fe3O4, four oxygens give −8, so the three irons must total +8, and each one comes out at +8/3.
No atom actually has two-thirds of an electron missing. What the fraction really means is that the iron atoms are in different environments — in this case some are +2 and some are +3 — and the calculation has given you the average. A single atom always has a whole-number oxidation state.
Naming with Roman numerals
When an element can have more than one oxidation state, the name must say which one, using a Roman numeral in brackets straight after the element name.
FeO is iron(II) oxide; Fe2O3 is iron(III) oxide.
KMnO4 is potassium manganate(VII).
Cu2O is copper(I) oxide; CuO is copper(II) oxide.
You are not expected to use Stock notation for non-metals. SO2 is sulfur dioxide, not sulfur(IV) oxide. Save the Roman numerals for metals with variable oxidation states.
Worked examples
WORKED EXAMPLE
Find the oxidation state of nitrogen in the nitrate ion, NO3−.
Step 1: what you know
Oxygen is −2, and there are three of them.
3 × (−2) = −6Step 2: the target total
The ion has a charge of −1, so everything sums to −1.
Step 3: solveN + (−6) = −1, so N = +5Nitrogen is +5only one nitrogen atom here, so no final division is needed
WORKED EXAMPLE
In the reaction Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s), identify what is oxidised and what is reduced.
Step 1: assign states before and afterZn: 0 → +2Cu: +2 → 0Step 2: check the sulfate
S stays at +6 and O at −2 throughout, so sulfate is a spectator.
Step 3: read the direction of change
Zinc’s state went up; copper’s went down.
Zinc is oxidised, copper is reduced — and zinc is the reducing agenta species whose oxidation state does not change is not involved in the redox
WORKED EXAMPLE
Name the compound Fe2(SO4)3, showing your reasoning.
Step 1: use the ion you know
The sulfate ion is SO42−, and there are three of them.
3 × (−2) = −6Step 2: balance the charge
The compound is neutral, so the two iron ions must total +6.
+6 ÷ 2 = +3 eachStep 3: write the name
Iron is in the +3 state, so the numeral is III.
Iron(III) sulfatetreating a familiar polyatomic ion as one lump is nearly always faster than working atom by atom
💡 Exam tip
Write oxidation states with the sign before the number: +2, not 2+. Ionic charges are written the other way round, as 2+.
Show the arithmetic. “7 × −2 = −14” written out is worth marks even if the final answer slips.
If an element appears more than once, remember to divide at the end.
For redox questions, tabulate the states before and after. It makes the changes impossible to miss.
Learn the common polyatomic ions — sulfate, nitrate, carbonate, manganate(VII) — and their charges.
“Oxidation state” and “oxidation number” mean the same thing. Either is accepted.
⚠ Common mix-up
Confusing oxidation state with ionic charge. In SO42− the sulfur is +6 even though the ion carries 2−.
Forgetting the final division when there are two or more of the unknown atom.
Giving oxygen −2 in a peroxide. In H2O2 it is −1.
Assigning a non-zero state to an uncombined element. O2, Cl2 and P4 are all 0.
Thinking a fractional state means a fractional electron. It is an average across atoms in different environments.
Reversing oxidation and reduction. Increase is oxidation — the numbers go the same way as the name suggests.
Up next: Ionisation Energy Trends Across a Period (HL) — we return to ionisation energy and look at the two places where the neat trend breaks, because those two dips are the best evidence we have that subshells exist at all.
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