Oxidation States — The Exceptions That Trip Students Up
Assigning oxidation numbers looks like the easiest task in inorganic chemistry until a question asks for the oxidation state of chromium in CrO5. Apply the usual rule "oxygen is −2" and you get chromium at +10, which is impossible — chromium only has six valence electrons to lose. The rules are not wrong; the compound is one of a small set of standard exceptions that examiners return to year after year. This article collects those exceptions in one place and works every one out in full, so that the next time an unusual formula appears you know exactly which rule to suspend.
The rules, and the order they apply in
That equation is what you actually solve. The individual rules are just the known values you substitute into it, and they have a strict priority order:
- Free element = 0. O2, P4, S8, Fe metal — all zero.
- Group 1 = +1, group 2 = +2 in all their compounds. No exceptions worth worrying about at school level.
- Fluorine = −1 always. This is the only rule with no exception at all, because fluorine is the most electronegative element. It therefore overrides everything below it.
- Hydrogen = +1, except when bonded to a metal, where it is −1.
- Oxygen = −2, except in peroxides, superoxides and compounds with fluorine.
Read rule 3 carefully. Because fluorine outranks oxygen, a compound containing both makes oxygen positive — that alone answers one of the classic trap questions.
Exception group 1 — oxygen is not always −2
| Compound | Type | Oxidation state of O | Working |
|---|---|---|---|
| H2O2, Na2O2, BaO2 | Peroxide (O–O single bond) | −1 | H2O2: 2(+1) + 2x = 0 → x = −1 |
| KO2, RbO2, CsO2 | Superoxide (O2⁻ ion) | −½ | KO2: (+1) + 2x = 0 → x = −0.5 |
| OF2 | Oxygen fluoride | +2 | x + 2(−1) = 0 → x = +2 |
| O2F2 | Dioxygen difluoride | +1 | 2x + 2(−1) = 0 → x = +1 |
| O2, O3 | Free element | 0 | Rule 1 |
A superoxide with a fractional value of −½ is a genuine, correct answer, not a sign that you have made a mistake. It reflects one extra electron shared over two oxygen atoms.
Exception group 2 — hydrogen in metal hydrides
In NaH, CaH2, LiH and LiAlH4, hydrogen is bonded to a metal that is less electronegative than itself, so the electrons are assigned to hydrogen and it takes −1. In CaH2: Ca is +2, so 2 + 2x = 0 and x = −1. This is also exactly why these hydrides are powerful reducing agents — hydrogen at −1 has nowhere to go but up.
Worked example 1 — CrO5, the question that starts this topic
The naive answer. Cr + 5(−2) = 0 → Cr = +10. Reject it immediately: the oxidation state of a main-group or transition element cannot exceed the number of electrons it can actually lose, and chromium's group number is 6.
The correct structure. CrO5 is a peroxo compound, written CrO(O2)2: one oxygen is a normal doubly bonded oxo group at −2, and the other four are arranged as two peroxide (O–O) linkages, each oxygen at −1.
Cr + 4(−1) + 1(−2) = 0
Cr − 4 − 2 = 0
Cr = +6 ✓ — which is perfectly reasonable, since +6 is chromium's maximum
and the same state it holds in CrO42− and
Cr2O72−.
Worked example 2 — the peroxo acids of sulphur
Caro's acid, H2SO5. Its structure is HO–O–SO2–OH: of the five oxygens, two form a peroxide linkage (−1 each) and three are ordinary (−2 each).
Total oxygen contribution = 2(−1) + 3(−2) = −2 − 6 = −8
2(+1) + S + (−8) = 0 → S − 6 = 0 → S = +6
Marshall's acid, H2S2O8 (peroxydisulphuric acid). Eight oxygens: two peroxidic, six ordinary.
Oxygen total = 2(−1) + 6(−2) = −2 − 12 = −14
2(+1) + 2S + (−14) = 0 → 2S = 12 → S = +6 each
Ignore the peroxide linkage in either compound and you get the impossible answers +8 and +7. Spotting the O–O bond is the entire skill being tested.
Exception group 3 — fractional (average) oxidation states
When the same element sits in two different environments in one formula, the arithmetic gives an average. The average is a correct answer to "what is the oxidation state", but the individual values are what an examiner means by "explain".
Fe3O4 (magnetite).
3x + 4(−2) = 0 → 3x = 8 → x = +8/3 ≈ +2.67
Structurally it is FeO·Fe2O3 — one Fe(II) and two Fe(III). Check the average: (2 + 3 + 3) ÷ 3 = 8/3 ✓. No iron atom is actually at +2.67.
Pb3O4 (red lead). Same arithmetic, 3x = 8, x = +8/3. Structurally 2 × Pb(II) + 1 × Pb(IV): (2 + 2 + 4) ÷ 3 = 8/3 ✓
Tetrathionate, S4O62−.
4x + 6(−2) = −2 → 4x = 10 → x = +2.5
Structurally the two outer sulphurs (each bonded to three oxygens) are +5 and the two
central sulphurs (joined only to sulphur) are 0. Check: 5 + 5 + 0 + 0 = 10 ✓
Thiosulphate, S2O32−.
2x + 3(−2) = −2 → 2x = 4 → x = +2
Structurally it is sulphate with one oxygen replaced by sulphur: the central S is +6 and the
terminal S is −2. Check: 6 + (−2) = +4, average +2 ✓
Azide ion, N3⁻. 3x = −1 → x = −1/3.
Worked example 3 — two oxidation states of the same element in one salt
Ammonium nitrate, NH4NO3. Treat the two ions separately, which is always the safer method for a salt.
NH4⁺: N + 4(+1) = +1 → N = −3
NO3⁻: N + 3(−2) = −1 → N = +5
Average across the formula: (−3 + 5) ÷ 2 = +1. Both answers are defensible, so read the question — "the oxidation state of nitrogen in NH4NO3" usually wants +1, while "the oxidation states of nitrogen" wants −3 and +5.
Bleaching powder, CaOCl2. Same idea: one chlorine is a chloride at −1 and the other is in the hypochlorite group at +1, so the average is 0. This mixture of states in one compound is why bleaching powder both oxidises and releases chloride.
Exception group 4 — carbon in organic compounds
Carbon takes almost any value from −4 to +4, and different carbons in the same molecule take different values. For the whole molecule the arithmetic gives an average; for a single carbon, count its bonds.
Methanol, CH3OH. Whole-molecule route:
C + 3(+1) + (−2) + (+1) = 0 → C + 3 − 2 + 1 = 0 → C = −2
Per-bond route: three C–H bonds (−3) and one C–O bond (+1) → −3 + 1 = −2 ✓ Both routes
agree.
Acetic acid, CH3COOH. Whole-molecule route:
2C + 4(+1) + 2(−2) = 0 → 2C + 4 − 4 = 0 → average C = 0
Per-carbon route: the methyl carbon has three C–H (−3) and one C–C (0) → −3;
the carboxyl carbon has one C=O (+2), one C–O of the hydroxyl (+1) and one C–C (0) →
+3. Average = (−3 + 3) ÷ 2 = 0 ✓ Neither carbon is actually at zero.
Glucose, C6H12O6.
6C + 12(+1) + 6(−2) = 0 → 6C + 12 − 12 = 0 → average C = 0
Dichloromethane, CH2Cl2.
C + 2(+1) + 2(−1) = 0 → C = 0 — a neat reminder that an oxidation state of
zero does not mean "unbonded".
The full exception table, for revision
| Species | Naive answer | Correct answer | Rule that was suspended |
|---|---|---|---|
| O in H2O2 | −2 | −1 | Peroxide linkage |
| O in KO2 | −2 | −½ | Superoxide ion |
| O in OF2 | −2 | +2 | F outranks O |
| H in NaH | +1 | −1 | Bonded to a metal |
| Cr in CrO5 | +10 | +6 | Two peroxo linkages |
| S in H2SO5 | +8 | +6 | One peroxo linkage |
| S in H2S2O8 | +7 | +6 | One peroxo linkage |
| Fe in Fe3O4 | — | +8/3 (Fe²⁺ + 2 Fe³⁺) | Mixed oxidation states |
| S in S4O62− | — | +2.5 (two +5, two 0) | Mixed environments |
| N in NH4NO3 | — | −3 and +5 (average +1) | Two different ions |
| C in CH3COOH | — | −3 and +3 (average 0) | Different carbon environments |
Common mistakes that cost marks
- Applying "O = −2" to a peroxo compound. Whenever the arithmetic produces an oxidation state above an element's group number, stop and look for an O–O bond.
- Letting oxygen outrank fluorine. Fluorine is −1 in every compound without exception; oxygen bends around it, never the reverse.
- Rejecting a fractional answer. +8/3 and +2.5 are real answers. Convert to a fraction rather than rounding — writing +2.67 for iron in Fe3O4 loses the structural meaning.
- Confusing oxidation state with valency or with formal charge. Oxidation state assigns each shared pair entirely to the more electronegative atom; valency counts bonds; formal charge splits every shared pair equally. Three different quantities, three different numbers.
- Averaging a salt when the question wants each ion. Split NH4NO3 and CaOCl2 into their ions first, then decide what is being asked.
- Giving one value for carbon in an organic molecule. Say which carbon. The average is often a number no atom in the molecule actually has.
Where these appear in exams
| Exam | Typical question |
|---|---|
| CBSE / ICSE Class 11 | Redox chapter: assign oxidation numbers, identify oxidising and reducing agents, balance by the oxidation-number method |
| CBSE / ICSE Class 12 | p-block and d-block: variable oxidation states and their stability |
| JEE / NEET | Single-answer questions on CrO5, H2SO5, S4O62−, Fe3O4 |
| IIT-JAM / CUET-PG | Disproportionation, Latimer and Frost diagrams built on these states |
| GATE / CSIR-NET | Non-innocent ligands, formal oxidation states in organometallic complexes |
Check any tricky formula instantly. The Oxidation Number calculator takes a formula or ion and returns the oxidation state of the element you ask about, so you can confirm the CrO5 = +6 and S4O62− = +2.5 results above before you commit them to an answer sheet.
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