How to Find Oxidation Numbers — The 7 Rules with Examples
Oxidation number is the bookkeeping tool of redox chemistry. Before you can say what is oxidised, what is reduced, how many electrons move, or how to balance a permanganate equation, you must be able to assign oxidation numbers quickly and correctly. The good news: it is a fixed procedure, seven rules applied in order of priority, and one equation.
The master equation
Sum of oxidation numbers of all atoms = the charge for a polyatomic ion
You assign the "easy" elements from the rules, put x for the unknown, and solve. That is the entire method.
The 7 rules, in priority order
- A free element is 0. Cu, O₂, P₄, S₈, Cl₂, graphite — all zero, regardless of how many atoms are joined together.
- A monatomic ion equals its charge. Na⁺ is +1, Mg²⁺ is +2, Cl⁻ is −1, S²⁻ is −2.
- Fluorine is always −1 in every compound. It is the most electronegative element, so nothing can pull electrons away from it.
- Group 1 metals are +1, Group 2 metals are +2 in their compounds. Aluminium is +3.
- Hydrogen is +1 with non-metals, but −1 in metal hydrides (NaH, CaH₂, LiAlH₄).
- Oxygen is −2 in most compounds, but: −1 in peroxides (H₂O₂, Na₂O₂), −½ in superoxides (KO₂), and +2 in OF₂, because fluorine outranks it under rule 3.
- Other halogens are −1, except when bonded to oxygen or to a more electronegative halogen — Cl in HClO₄ is +7, not −1.
The order matters. If two rules disagree, the one with the lower number wins. That single sentence resolves OF₂, NaH and every other "exception" you have been asked to memorise.
Worked example 1 — Mn in KMnO₄
K is Group 1 → +1 (rule 4). O is normal oxide → −2 (rule 6). Let Mn = x.
(+1) + x + 4(−2) = 0
1 + x − 8 = 0
x = +7
+7 is the highest possible for manganese (group number 7), which is why permanganate is such a powerful oxidising agent — it can only gain electrons.
Worked example 2 — Cr in K₂Cr₂O₇ and in Cr₂O₇²⁻
Neutral salt K₂Cr₂O₇:
2(+1) + 2x + 7(−2) = 0
2 + 2x − 14 = 0 → 2x = 12 → x = +6
Dichromate ion Cr₂O₇²⁻ — the sum equals −2, not 0:
2x + 7(−2) = −2
2x − 14 = −2 → 2x = 12 → x = +6 ✔ same answer, as it must be.
The potassium ions are spectators, so removing them cannot change chromium's oxidation number. If your two answers disagree, you forgot the ionic charge on the right-hand side.
Worked example 3 — The oxygen exceptions
H₂O₂ (hydrogen peroxide): H = +1 (rule 5). 2(+1) + 2x = 0 → x = −1. This is a peroxide linkage O—O.
OF₂ (oxygen difluoride): F = −1 always (rule 3 beats rule 6). x + 2(−1) = 0 → x = +2 for oxygen.
NaH (sodium hydride): Na = +1 (rule 4 beats rule 5), so H = −1.
Worked example 4 — Fractional and average values
Oxidation number is a formal, averaged quantity. It is allowed to come out fractional.
Fe in Fe₃O₄: 3x + 4(−2) = 0 → 3x = 8 → x = +8/3 ≈ +2.67.
No iron atom actually carries +8/3. Fe₃O₄ is FeO·Fe₂O₃ — one Fe(II) and two Fe(III) —
and (2 + 3 + 3)/3 = 8/3. The fraction is the average.
S in S₂O₃²⁻ (thiosulfate): 2x + 3(−2) = −2 → 2x = 4 → x = +2 as an average. Structurally the two sulfur atoms are different (+5 and −1), which is why thiosulfate reacts the way it does with iodine.
S in S₄O₆²⁻ (tetrathionate): 4x + 6(−2) = −2 → 4x = 10 → x = +2.5, again an average.
Worked example 5 — Carbon in glucose, and two different nitrogens
C in C₆H₁₂O₆: 6x + 12(+1) + 6(−2) = 0 → 6x + 12 − 12 = 0 → x = 0. Carbon in glucose has an average oxidation number of zero, even though individual carbons range from −1 to +1 across the molecule.
N in NH₄NO₃: here the averaging trick would be misleading, so split the
ions.
In NH₄⁺: x + 4(+1) = +1 → x = −3
In NO₃⁻: x + 3(−2) = −1 → x = +5
The overall average of +1 is arithmetically true but chemically meaningless. Whenever a
compound is built from two ions, assign each ion separately.
Using oxidation numbers to spot redox
Once assigned, the rest is easy. Compare each element before and after:
Oxidation number decreases → reduction → electrons gained → the species is the oxidising agent
In Zn + Cu²⁺ → Zn²⁺ + Cu: zinc goes 0 → +2 (oxidised, so Zn is the reducing agent), copper goes +2 → 0 (reduced, so Cu²⁺ is the oxidising agent). If an element's oxidation number rises and falls in the same reaction, that is disproportionation.
Common mistakes that cost marks
- Setting a polyatomic ion's sum to zero. For MnO₄⁻ the sum is −1, giving Mn = +7; setting it to 0 gives +8, which manganese cannot reach.
- Assuming oxygen is always −2. Check for peroxide, superoxide and fluorine first.
- Assuming hydrogen is always +1. In a metal hydride it is −1.
- Confusing oxidation number with ionic charge. In CH₄ carbon is −4, but there is no C⁴⁻ ion anywhere — the bonds are covalent. Oxidation number is a formal bookkeeping device, not a real charge.
- Rejecting fractional answers. +8/3 and +2.5 are correct averages, not arithmetic errors.
- Forgetting the subscript multiplies. In K₂Cr₂O₇ there are two chromium atoms, so the unknown enters as 2x.
- Not checking. Add every oxidation number back up including subscripts; it must equal 0 or the ion charge.
Where oxidation numbers appear in exams
| Exam | Typical use |
|---|---|
| CBSE/ICSE Class 11 | Redox reactions chapter; identifying oxidising and reducing agents |
| CBSE/ICSE Class 12 | Electrochemistry, d-block chemistry, balancing by the ion-electron method |
| JEE / NEET | Fractional oxidation states, disproportionation, equivalent weight of oxidants |
| IIT-JAM / CUET-PG | Redox titrations, inorganic qualitative analysis, coordination compounds |
| GATE / CSIR-NET | Transition-metal chemistry, catalytic cycles, organometallic electron counting |
Verify any formula in a second. The Oxidation Number calculator takes a compound or ion — KMnO₄, Cr₂O₇²⁻, Fe₃O₄, NH₄NO₃ — and returns the oxidation state of each element, so you can check your own working rather than trust it.
Open the Oxidation Number Calculator →Redox is one of the highest-scoring chapters once the rules are automatic. ABC Chemistry runs Class 11–12 chemistry coaching at the Gurugram centre and online classes across India, with home tuition available in Delhi-NCR — details at abcchemistry.in.