How Do You Determine Oxidation States of Atoms in a Compound?

Determine oxidation states by assigning known values first, then using the rule that their sum equals the species’ overall charge. This method applies at both SL and HL and supports the analysis of electron transfer in Reactivity 3.2.

The Reasoning and Method

An oxidation state is the hypothetical charge assigned to an atom if the compound's bonds are treated as ionic, with bonding electrons assigned to the more electronegative atom. Follow these rules in order:

For example, determine sulfur’s oxidation state in $\text{SO}_4^{2-}$. Let sulfur’s oxidation state be $x$. Each oxygen is $-2$, so:

$$x+4(-2)=-2$$ $$x=+6$$

Therefore, sulfur has oxidation state $+6$.

For $\text{KMnO}_4$, potassium is $+1$ and each oxygen is $-2$:

$$(+1)+x+4(-2)=0$$

Therefore, manganese is $+7$.

A second ionic example is $\text{NH}_4^+$. Hydrogen is $+1$, so nitrogen satisfies:

$$x+4(+1)=+1$$

Hence $x=-3$. Always include the overall ionic charge in the equation. For compounds containing transition elements, do not assume a fixed oxidation state; calculate it from the other atoms because transition elements can have variable oxidation states. This is especially important in redox chemistry.

A common misconception is that oxidation state always equals an atom’s real charge. It does not: oxidation state is a formal electron-accounting model, while ionic charge is an actual overall charge.

Exam Technique

For a deduce question, show the known oxidation states, write the sum equation, and give the answer with the sign before the number, such as $+6$, not $6+$. In redox equations, identify oxidation as an increase in oxidation state and reduction as a decrease.