Oxidation Number: Definition and Rules for Assignment

Directly counting electrons transferred works cleanly for simple ionic reactions like $\text{Zn} + \text{Cu}^{2+}$, but many redox reactions involve covalently bonded species — $\text{KMnO}_4$, $\text{K}_2\text{Cr}_2\text{O}_7$, $\text{H}_2\text{SO}_4$ — where no whole electron is ever fully transferred to or from a single atom in the way an ionic bond forms. Oxidation number (also called oxidation state) is the bookkeeping device that lets the same 'loss/gain of electrons' language be applied uniformly to every compound, ionic or covalent.

The oxidation number of an atom in a species is the hypothetical charge that atom would carry if every bond it forms were treated as fully ionic, with the shared pair of electrons in each bond assigned entirely to the more electronegative atom. It is a formal accounting convention, not a measurement of real, physically localized charge — but it is extremely useful precisely because it can be assigned to any atom in any compound by a short, fixed set of rules, without needing to know the real bonding picture in detail.

The rules (given in the reference table alongside this section) are applied in the order that resolves the fewest-unknowns case first: fix the oxidation numbers of atoms with an invariant value (Group 1, Group 2, F, usually H and O), then solve for the one remaining unknown atom using the requirement that the oxidation numbers of all atoms in a neutral formula must sum to zero, or to the ionic charge for a polyatomic ion.

Worked example — Mn in $\text{KMnO}_4$: K is a Group 1 metal, so its oxidation number is fixed at $+1$. Oxygen is $-2$ in this compound (no peroxide bond present), and there are four oxygen atoms, contributing $4 \times (-2) = -8$. Since $\text{KMnO}_4$ is a neutral formula, the sum of all oxidation numbers must equal $0$: $(+1) + x + (-8) = 0 \Rightarrow x = +7$. So Mn is in the $+7$ oxidation state — the highest possible for manganese, which is exactly why $\text{KMnO}_4$ is such a powerful oxidizing agent (its manganese atom has room only to be reduced, never oxidized further).

Worked example — Cr in $\text{K}_2\text{Cr}_2\text{O}_7$: two K atoms contribute $2(+1)=+2$; seven O atoms contribute $7(-2)=-14$. Neutral formula: $2 + 2x - 14 = 0 \Rightarrow 2x = 12 \Rightarrow x = +6$. Chromium is $+6$ in dichromate. …