Oxidation number

Definition. The oxidation number (oxidation state) of an atom in a molecule or ion is the charge that atom would bear if all its bonds to atoms of different elements were counted as completely ionic — the shared electron pair assigned wholly to the more electronegative partner — while bonds between two atoms of the same element contribute nothing, since neither atom is more electronegative than the other.

Assigning it: a small set of priority rules. Rather than drawing Lewis structures every time, chemists apply a short checklist, used in this order whenever two rules could otherwise conflict: An atom in its free, uncombined element — Na(s), $\text{O}_2$, $\text{P}_4$ — is always $0$; there is no more-electronegative partner to steal electrons. A monatomic ion simply carries its actual charge: $\text{Na}^+$ is $+1$, $\text{Al}^{3+}$ is $+3$. Fluorine, the most electronegative element of all, is always $-1$ in every compound. Oxygen is usually $-2$, but drops to $-1$ in peroxides such as $\text{H}_2\text{O}_2$ (two oxygens share the "extra" negative charge), to $-\tfrac12$ in superoxides such as $\text{KO}_2$, and rises to $+2$ in the one compound where it bonds to something even more electronegative, $\text{OF}_2$. Hydrogen is $+1$ next to non-metals but flips to $-1$ in metal hydrides like NaH, where hydrogen is the more electronegative partner. Alkali metals are fixed at $+1$ and alkaline earth metals at $+2$ in their compounds. Finally, the numbers must add up: the oxidation numbers of all atoms in a neutral molecule sum to zero, and in a polyatomic ion they sum to the ion's charge. This last rule is the one that actually solves problems — the others just supply the known values you plug in.

Reading redox off the numbers. Once every atom has a number, spotting a redox reaction is mechanical: an atom whose oxidation number increases has been oxidised; one whose number decreases has been reduced. If not a single oxidation number changes across a reaction, it is not a redox reaction at all — a fast way to classify reactions without thinking about electrons directly.

Why it is not the same as valency. Valency is a fixed, essentially positive count of an atom's combining capacity. Oxidation number is signed, can change from compound to compound, and can even be a fraction — $-\tfrac12$ for oxygen in $\text{KO}_2$, or an average of $+\tfrac{8}{3}$ for iron in $\text{Fe}_3\text{O}_4$ (a mixed-valence oxide really containing $\text{Fe}^{2+}$ and $\text{Fe}^{3+}$ in a 1:2 ratio). Nitrogen's valency of 3 stays put, but its oxidation number ranges from $-3$ in $\text{NH}_3$ to $+5$ in $\text{HNO}_3$ depending on what it is bonded to — the same atom, wildly different bookkeeping charges. The figure below shows exactly this kind of spread for a single element, manganese, across three compounds you will meet again in the redox unit.

Question. Compare the oxidation number of sulphur in the sulphite ion, $\text{SO}_3^{2-}$, and the sulphate ion, $\text{SO}_4^{2-}$, and say which ion has sulphur in the higher oxidation state.

Step 1. For sulphite, let sulphur's oxidation number be $x$. Oxygen is $-2$ (three oxygens present), and the ion carries a $-2$ charge, so the sum rule gives $x + 3(-2) = -2$, i.e. $x - 6 = -2$, so $x = +4$.

Step 2. For sulphate, let sulphur's oxidation number be $y$. Oxygen is again $-2$, now with four oxygens, and the ion still carries $-2$ charge: $y + 4(-2) = -2$, i.e. $y - 8 = -2$, so $y = +6$.

Step 3. Compare the two results directly: $+6$ is greater than $+4$, so sulphur sits in a higher oxidation state in the sulphate ion than in the sulphite ion. Since going from sulphite to sulphate raises sulphur's oxidation number from $+4$ to $+6$, that conversion is an oxidation — confirmed by arithmetic alone, without ever drawing a Lewis structure.

Competitive stretch

Iron in magnetite, $\text{Fe}_3\text{O}_4$, gives an average oxidation number rather than a single clean integer. Treat the formula as an oxide with three iron atoms and four oxygens: $3x + 4(-2) = 0$, so $3x = 8$ and $x = +\tfrac{8}{3} \approx +2.67$. This fractional value is not a mistake — magnetite genuinely contains iron in two different real oxidation states, one $\text{Fe}^{2+}$ and two $\text{Fe}^{3+}$ per formula unit, and the sum rule only reports their average, $\tfrac{(+2) + (+3) + (+3)}{3} = +\tfrac{8}{3}$, which matches.

• Forgetting the peroxide and superoxide exceptions for oxygen and assigning $-2$ everywhere by reflex — this produces an impossible or wrong answer for $\text{H}_2\text{O}_2$, $\text{Na}_2\text{O}_2$, or $\text{KO}_2$.
• Confusing oxidation number with formal charge or with a real, measured charge distribution — it is a bookkeeping convention built on the assumption of 100% ionic bonding, not a physical quantity you could measure directly.
• Treating oxidation number as identical to valency. They agree in many simple ionic compounds but diverge for elements with variable states: chlorine in NaCl is $-1$ with valency 1, but chlorine in $\text{HClO}_4$ has oxidation number $+7$.
• Setting the sum equal to zero out of habit even when the species is a polyatomic ion — remember to set the sum equal to the ion's actual charge instead.

Ask the tutor: Want to practise finding oxidation numbers in trickier formulas like $\text{Na}_2\text{S}_2\text{O}_3$ or $\text{K}_2\text{Cr}_2\text{O}_7$, or need the peroxide exception explained again with a diagram? Ask your AI tutor for guided practice.