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The rules for oxidation numbers work by adding up the formula: oxygen is −2, hydrogen is +1, and the total equals the charge. Most of the time that is enough. But sometimes the answer is 8/3, or 2.5, or +10 for an element that cannot go above +6. Those are the questions where the formula is hiding the structure. This post shows how to read oxidation numbers from a structure, with four worked examples.
Is this on the syllabus?
The Syllabus for JEE (Main) 2026 lists "oxidation number, rules for assigning oxidation number and balancing of redox reactions" in Unit 7, and the NEET (UG) 2026 syllabus uses almost the same words. The 2026-27 reprint of the NCERT Class 11 chapter Redox Reactions keeps its box "The Paradox of Fractional Oxidation Number" in Section 7.3, which is the starting point for this post. Check the syllabus for your exam year.
What does an oxidation number really count?
The IUPAC Recommendations of 2016 define oxidation state "using ionic approximation of bonds" and give two algorithms, for a compound described by a Lewis formula or a bond graph. The 2014 IUPAC technical report behind it names the two: one based on composition, the other on topology (the structure).
Composition (the formula rule). Assign standard values and solve for the unknown. If an element appears in several sites, you get the average.
Topology (the structure rule). For each bond, give both electrons to the more electronegative atom. A bond between two atoms of the same element is split equally, so it adds nothing to either. Each atom's oxidation number is its valence electrons minus the electrons it ends up with.
When an element sits in one kind of site, the two rules agree. When it sits in two or more, only the structure tells you the individual values.
Worked example 1: S₄O₆²⁻, by counting bonds
Formula rule: 4x + 6(−2) = −2, so x = 2.5.
Structure: the tetrathionate ion is O₃S–S–S–SO₃. NCERT's box gives the answer: the two end sulphurs are +5 and the two middle ones 0. Here is why, by the structure rule.
Each end sulphur is bonded to three oxygens and one sulphur. In the usual Lewis structure it has two S=O double bonds and one S–O⁻ single bond, five bonding pairs to oxygen in all. Oxygen is more electronegative, so sulphur loses a count of 5: +5.
The S–S bonds are between identical atoms, so they count for nothing.
Each middle sulphur is bonded only to sulphurs: 0.
Check the total: 5 + 0 + 0 + 5 + 6(−2) = −2. It matches the charge.
The average, 10/4 = 2.5, is what the formula rule gave.
Worked example 2: Fe₃O₄, from the mixed oxide
Formula rule: 3x = 8, so x = 8/3.
NCERT's Class 12 chapter on the d- and f-block elements lists Fe₃O₄ in its table of 3d metal oxides between the +3 and +2 rows and marks it as a mixed oxide. So the iron atoms are +2 and +3. With three of them adding to +8, the only split is 2 + 3 + 3: one Fe(II) and two Fe(III). The same table gives the same answer for Mn₃O₄.
Worked example 3: CrO₅, where the formula rule is impossible
Formula rule with every oxygen at −2: x − 10 = 0, so x = +10.
That cannot be right. NCERT gives chromium's configuration as 3d⁵4s¹, and its Table 4.3 lists chromium's oxidation states only up to +6. So some oxygens must not be −2.
In the structure, chromium is bonded to one oxide oxygen and two peroxide groups (O–O). PubChem names the compound chromium(VI) oxide peroxide. The NCERT redox chapter states that in peroxides such as H₂O₂ and Na₂O₂, each oxygen is assigned −1. So:
x + (−2) + 4(−1) = 0, giving x = +6.
The O–O bond is between identical atoms, so by the structure rule each peroxide oxygen takes only one bond's worth of electrons from chromium: −1, not −2.
Worked example 4: Pb₃O₄, from a reaction
Formula rule: 3x = 8, so x = 8/3 again. NCERT calls Pb₃O₄ a mixed oxide and gives the reaction
Pb₃O₄ + 8HCl → 3PbCl₂ + Cl₂ + 4H₂O
Two chloride ions are oxidised to Cl₂: two electrons are given up.
Lead must take those two electrons. In PbCl₂, every lead is +2.
Total lead oxidation number goes from +8 to +6, a drop of 2. So one lead atom went from +4 to +2.
Pb₃O₄ therefore has two Pb(II) and one Pb(IV): 2 + 2 + 4 = 8.
Common mistakes and how to avoid them
Reporting a fraction as the answer for each atom. Say it is an average, and give the real values if asked.
Taking oxygen as −2 every time. Check for O–O bonds. Peroxide oxygen is −1.
Ignoring the maximum state. If your answer exceeds the s plus d electron count for a 3d metal, recheck the structure.
Counting same-element bonds. S–S and O–O bonds add nothing.
Our guides to redox balancing and transition metal oxidation states cover the formula rules and electron counts in more depth. If drawing a Lewis structure is the weak step, the Fermi Concept Library covers every concept in chemistry, each with its own lesson.
A practice routine
Redo the four examples with only the formula and a blank page.
Draw C₃O₂ and Br₃O₈ and check NCERT's values from its box (+2, 0, +2 and +6, +4, +6).
Find the oxidation number of oxygen in H₂O₂ and of sulphur in S₂O₃²⁻ by the formula rule, then say whether the answer is an average.
For any fraction, write one line on which atoms differ and why.
Try it now: find the average oxidation number of carbon in C₃O₂, then draw O=C=C=C=O and assign each carbon separately.
