Rusting, burning, batteries and photosynthesis all move electrons from one substance to another. To follow the electrons, chemists give each atom an oxidation number, a charge it would have if every bond were fully ionic. It builds on Predicting Ion Charges and Polyatomic Ions.
Key idea
Rules, in order of priority: (1) An atom in a pure element is 0. (2) A monatomic ion equals its charge. (3) F is always $-1$; group 1 metals $+1$; group 2 metals $+2$. (4) H is $+1$ with nonmetals ($-1$ in metal hydrides such as NaH). (5) O is usually $-2$ ($-1$ in peroxides such as $\mathrm{H_2O_2}$). (6) The oxidation numbers add up to the overall charge: zero for a neutral compound, the ion's charge for a polyatomic ion.
Why it works
Oxidation numbers are bookkeeping, not real charges. Each bond's electrons are assigned entirely to the more electronegative atom, which is why fluorine (the most electronegative element) is always $-1$ and oxygen nearly always $-2$. Because every electron is assigned to some atom, the oxidation numbers must add up to the actual total charge, and that is how you find the one atom whose number is unknown. Later, a change in an atom's oxidation number will tell you it gained or lost electrons.
Worked example: Sulfur in sulfuric acid
Find the oxidation number of S in $\mathrm{H_2SO_4}$.
- H is $+1$ (two of them: $+2$); O is $-2$ (four of them: $-8$).
- Neutral compound: $+2 + x - 8 = 0$, so $x = +6$.
Answer: $+6$
Worked example: Chromium in dichromate
Find the oxidation number of Cr in $\mathrm{Cr_2O_7^{2-}}$.
- Seven O at $-2$: $-14$.
- The total must be the ion's charge: $2x - 14 = -2$, so $2x = 12$.
- $x = +6$ for each chromium.
Answer: $+6$
Worked example: Nitrogen in ammonium
Find the oxidation number of N in $\mathrm{NH_4^+}$.
- Four H at $+1$: $+4$.
- $x + 4 = +1$, so $x = -3$.
Answer: $-3$
An oxidation number can even be a fraction when it is an average: in $\mathrm{Fe_3O_4}$ the three iron atoms average $+\tfrac{8}{3}$, because the crystal holds one $\mathrm{Fe^{2+}}$ for every two $\mathrm{Fe^{3+}}$.
Common mistake
Giving every atom in a polyatomic ion the ion's charge (calling each O in $\mathrm{SO_4^{2-}}$ "$-2$" and S "$-2$" too), or forgetting that the numbers in a neutral compound add to zero. Check: multiply each oxidation number by the number of atoms of that kind and add; the sum must equal the overall charge.