What an Oxidation Number Means
When hydrogen burns in oxygen to give water, $\ce{2H2(g) + O2(g) -> 2H2O(l)}$, there is no obvious ionic transfer; the bonds in water are covalent. Yet the H atom plainly moves from a neutral state in $\ce{H2}$ to a more positive condition in $\ce{H2O}$, while the O atom moves from neutral in $\ce{O2}$ to a more negative one. NCERT §7.3 captures this by assuming a complete transfer of each bonding electron pair from the less electronegative atom to the more electronegative atom, purely for book-keeping. Under that fiction we can label every atom with a charge.
The number so obtained is the oxidation number (ON): the apparent charge an atom would carry if every shared pair were handed entirely to the more electronegative element. NCERT states it formally as the value that "denotes the oxidation state of an element in a compound, ascertained according to a set of rules formulated on the basis that electron pair in a covalent bond belongs entirely to the more electronegative element." The transfer is real only in part — it is better described as an electron shift — but treating it as total gives a clean, reproducible scheme.
Because identifying the more electronegative partner by eye is awkward in larger species, NCERT replaces that judgement with a fixed set of priority rules. Master the rules and the electronegativity reasoning is done for you.
The actual charge separation in $\ce{H2O}$ is partial, but the oxidation-number scheme pretends the electron pairs move fully to oxygen. That assumption yields H = +1 and O = −2 every time.
The Rules for Assigning ON
NCERT lays out six rules, applied roughly in the order of priority below. Higher rules win when two would conflict — the charge balance in rule 6 is what you usually solve for last, after the fixed values are placed.
| # | Rule | Examples |
|---|---|---|
| 1 | An atom of a free (uncombined) element has ON = 0. | $\ce{H2}$, $\ce{O2}$, $\ce{Cl2}$, $\ce{O3}$, $\ce{P4}$, $\ce{S8}$, $\ce{Na}$, $\ce{Mg}$, $\ce{Al}$ all 0 |
| 2 | For a monatomic ion, ON equals the charge on the ion. | $\ce{Na+}$ = +1, $\ce{Mg^2+}$ = +2, $\ce{Fe^3+}$ = +3, $\ce{Cl-}$ = −1, $\ce{O^2-}$ = −2 |
| 3 | Oxygen is −2 in most compounds (peroxide / superoxide / O–F exceptions aside). | $\ce{CO2}$, $\ce{H2O}$, $\ce{SO4^2-}$ → O = −2 |
| 4 | Hydrogen is +1, except in binary metal hydrides where it is −1. | $\ce{HCl}$ → H = +1; $\ce{LiH}$, $\ce{NaH}$, $\ce{CaH2}$ → H = −1 |
| 5 | Fluorine is always −1; other halides (Cl, Br, I as halide ions) are −1, but positive when bonded to O. | $\ce{NaF}$ → F = −1; $\ce{HClO}$ → Cl = +1 |
| 6 | The algebraic sum of all ON equals 0 (neutral compound) or the ionic charge (polyatomic ion). | $\ce{CO3^2-}$ → sum = −2 |
Rules 1 and 2 are absolute. Rules 3 to 5 are reliable defaults that hold in the overwhelming majority of compounds, and rule 6 is the equation you actually solve. Alkali metals are fixed at +1 and alkaline-earth metals at +2 in all their compounds; aluminium is +3 throughout. The defaults of rules 3 to 5 carry the famous exceptions that NEET loves to probe, so they deserve a table of their own.
Exceptions: Peroxides, OF2, Hydrides
Three places break the comfortable −2 / +1 habit. NCERT names each explicitly, and each has surfaced in examination items disguised as a "trick" species. The logic is consistent: whenever the usual junior partner is bonded to something even more electronegative, or to itself, the assigned value moves.
| Species / class | Atom | ON | Reason |
|---|---|---|---|
| Peroxides ($\ce{H2O2}$, $\ce{Na2O2}$) | O | −1 | O atoms are bonded to each other; the O–O bond contributes 0, so each O is −1. |
| Superoxides ($\ce{KO2}$, $\ce{RbO2}$) | O | −½ | An O–O linkage with one extra electron shared over two O atoms. |
| Oxygen difluoride ($\ce{OF2}$) | O | +2 | F is more electronegative than O, so O loses; two F at −1 force O to +2. |
| Dioxygen difluoride ($\ce{O2F2}$) | O | +1 | Two F at −1 distributed over two O atoms gives each O = +1. |
| Metal hydrides ($\ce{LiH}$, $\ce{NaH}$, $\ce{CaH2}$) | H | −1 | The metal is less electronegative than H, so H takes the electron pair. |
The −2 default fails more often than students expect
Plugging O = −2 into a peroxide or into $\ce{OF2}$ is the single most common error in ON questions. In $\ce{H2O2}$ the answer for O is −1, not −2; in $\ce{OF2}$ it is +2; in $\ce{KO2}$ it is −½ (a fractional value). Likewise hydrogen is +1 only with non-metals — in $\ce{NaH}$ it is −1. Read the formula and ask: is oxygen bonded to oxygen, or to fluorine? Is hydrogen bonded to a metal?
Quick screen: O–O present → peroxide/superoxide; O–F present → oxygen positive; H–metal binary → hydrogen −1.
Worked Calculations
Every ON calculation is the same algebra: place the fixed values, let the unknown be $x$, multiply by the number of atoms, and set the sum equal to 0 (compound) or the charge (ion). Three high-frequency species follow.
Find the oxidation number of sulphur in sulphuric acid, $\ce{H2SO4}$.
Each H is +1 (bonded to a non-metal) and each O is −2. Let S be $x$. Setting the sum to zero:
$$2(+1) + x + 4(-2) = 0 \;\Rightarrow\; 2 + x - 8 = 0 \;\Rightarrow\; x = +6$$
Sulphur is in its +6 oxidation state, the highest available to a group-16 element.
Find the oxidation number of chromium in the dichromate ion, $\ce{Cr2O7^2-}$.
Each O is −2. With two Cr atoms each at $x$, the sum must equal the ionic charge, −2:
$$2x + 7(-2) = -2 \;\Rightarrow\; 2x - 14 = -2 \;\Rightarrow\; 2x = 12 \;\Rightarrow\; x = +6$$
Each chromium is +6. This is exactly the reduction half that NEET 2023 built a balancing question around, $\ce{Cr2O7^2- + 6e- -> 2Cr^3+}$, a drop of three units per Cr.
Find the oxidation number of manganese in the permanganate ion, $\ce{MnO4-}$.
Each O is −2; let Mn be $x$. The sum equals the ionic charge, −1:
$$x + 4(-2) = -1 \;\Rightarrow\; x - 8 = -1 \;\Rightarrow\; x = +7$$
Manganese is +7, its maximum state. When permanganate oxidises iodide to iodate in neutral or faintly alkaline medium it falls to $\ce{MnO2}$, i.e. +7 → +4, the change tested in NEET 2022.
A change in oxidation number is precisely what defines oxidation and reduction. See how this connects to electron loss and gain in Redox: Electron Transfer.
Fractional and Average Oxidation Numbers
NCERT is explicit that "if two or more than two atoms of an element are present in the molecule or ion, the oxidation number of the atom of that element will then be the average of the oxidation number of all the atoms of that element." Each atom individually still carries a whole number; the value the rules return is the mean. When the atoms occupy different structural positions, that mean can be a fraction.
Find the average oxidation number of sulphur in the tetrathionate ion, $\ce{S4O6^2-}$.
Each O is −2; let the average S be $x$ over four atoms. Sum equals the charge, −2:
$$4x + 6(-2) = -2 \;\Rightarrow\; 4x - 12 = -2 \;\Rightarrow\; 4x = 10 \;\Rightarrow\; x = +\tfrac{10}{4} = +2.5$$
The average is +2.5. Structurally the four sulphurs are not equivalent — two bridging atoms sit near 0 and two terminal atoms near +5 — but the rules deliver only the mean. A fractional ON is always a signal that the atoms are in chemically distinct sites.
The same pattern explains other classic fractions: iron in $\ce{Fe3O4}$ averages +8/3, and carbon in propane sits at a non-integer mean across its three positions. None of these conflicts with the whole-number rule for an individual atom; they are bulk averages.
The rules report only the mean +2.5; the individual sulphur atoms occupy whole-number states determined by their bonding environment.
Oxidation State and Stock Notation
NCERT treats oxidation state and oxidation number as interchangeable. In $\ce{CO2}$ the oxidation state of carbon is +4, which is also its oxidation number, and oxygen is −2 in both senses. The oxidation number simply denotes the oxidation state of an element in a compound.
For metals, the state is often written by Alfred Stock's convention: a Roman numeral in parentheses after the metal symbol records its oxidation number. So aurous and auric chlorides become $\ce{Au(I)Cl}$ and $\ce{Au(III)Cl3}$, while stannous and stannic chlorides become $\ce{Sn(II)Cl2}$ and $\ce{Sn(IV)Cl4}$. The notation makes the oxidised-versus-reduced status immediate — $\ce{Hg2(I)Cl2}$ is the reduced form of $\ce{Hg(II)Cl2}$.
| Compound | Metal ON | Stock notation |
|---|---|---|
| $\ce{HAuCl4}$ | Au = +3 | $\ce{HAu(III)Cl4}$ |
| $\ce{FeO}$ | Fe = +2 | $\ce{Fe(II)O}$ |
| $\ce{Fe2O3}$ | Fe = +3 | $\ce{Fe2(III)O3}$ |
| $\ce{MnO}$ | Mn = +2 | $\ce{Mn(II)O}$ |
| $\ce{MnO2}$ | Mn = +4 | $\ce{Mn(IV)O2}$ |
Redox Language Built on ON
Once oxidation number is in hand, the entire vocabulary of redox follows from a single rule: watch which way the number moves. NCERT defines each term against the change in ON, which is why ON is the gateway concept of the whole chapter.
| Term | Definition in terms of ON |
|---|---|
| Oxidation | An increase in the oxidation number of the element. |
| Reduction | A decrease in the oxidation number of the element. |
| Oxidising agent (oxidant) | A reagent that increases the ON of another element (and is itself reduced). |
| Reducing agent (reductant) | A reagent that lowers the ON of another element (and is itself oxidised). |
| Redox reaction | Any reaction involving a change in oxidation number of the interacting species. |
Take NCERT's own illustration, $\ce{2Cu2O(s) + Cu2S(s) -> 6Cu(s) + SO2(g)}$. Assigning oxidation numbers shows copper falling from +1 to 0 (reduced) and sulphur rising from −2 to +4 (oxidised); $\ce{Cu2O}$ is the oxidant and the sulphur of $\ce{Cu2S}$ is the reductant. The whole analysis is just ON arithmetic. This same lens turns "is this a redox reaction?" into a mechanical test, which is exactly how NEET 2024 framed a question — the only non-redox option was the one with no ON change at all, the metathesis $\ce{BaCl2 + Na2SO4 -> BaSO4 + 2NaCl}$.
Oxidation Number in one screen
- ON is the apparent charge assuming each bonding pair goes fully to the more electronegative atom; oxidation state is the same thing.
- Free element = 0; monatomic ion = its charge; O usually −2; H usually +1; F always −1.
- Exceptions: peroxides O = −1, superoxides O = −½, $\ce{OF2}$ O = +2, $\ce{O2F2}$ O = +1, metal hydrides H = −1.
- Sum of ON = 0 for a neutral compound, = ionic charge for a polyatomic ion; solve for the unknown.
- Multiple inequivalent atoms of one element give an average, which may be fractional (e.g. S in $\ce{S4O6^2-}$ = +2.5).
- Oxidation = ON increases, reduction = ON decreases; a redox reaction is any change in ON.