# Assigning & Calculating Oxidation States

Show/Hide Sub-topics (Chemical Reactions | O Level)

### Rules to Assign Oxidation Numbers

The following table summarises the rules to assign oxidation numbers (O.N.):

Rules Examples
All elements have an oxidation state of 0. O.N. of Li = 0
Simple ions are given an oxidation number that is equal to its charge. O.N. of $\text{Li}^{+} = +1$
Oxygen in compounds usually have an oxidation number of $-2$, except in $\text{H}_{2}\text{O}_{2}$, where its O.N. is $-1$. O.N. of O in $\text{SO}_{2} =-2$

O.N. of O in $\text{H}_{2}\text{O}_{2}=-1$

Hydrogen in compounds usually has an O.N. of $+1$, except in metal hydrides, where it is $-1$. O.N. of H in $\text{H}_{2}\text{O}= +1$

O.N. of H in $\text{ZnH}_{2}=-1$

Group I elements have O.N of $+1$ in their compounds. O.N. of $\text{K}$ in $\text{K}_{2}\text{O} = +1$
Group II elements have O.N of $+2$ in their compounds. O.N. of $\text{Ca}$ in $\text{CaCl}_{2} = +2$
Group III elements have O.N. of $+3$ in their compounds. O.N. of $\text{Al}$ in $\text{Al}_{2}\text{O}_{3} = +3$
Fluoride ion always has an O.N. of $-1$. O.N of $\text{F}$ in $\text{HF} =-1$
The sum of the O.N. of all the atoms must add up to 0 in a compound. For $\text{H}_{2}\text{SO}_{4}$, the O.N. of H, S and O are +1, +4 and -2 respectively. The sum of O.N. of all atoms is $2(+1) + 4 + 4(-1) = 0$
The sum of the O.N. of all atoms in a particle is equal to the charge on the particle. For sulphate ion, $\text{SO}_{4}^{2-}$, the sum of O.N. of S and 4 O atoms must add up to the overall charge of the particle, $-2$.

### Calculation of Oxidation States

For simple ionic compounds, the oxidation state is simply the charge on the simple ion. E.g.: (Using the rules above)

• Magnesium Chloride, $\text{MgCl}_{2}$, is an ionic compound, which consists of $\text{Mg}^{2+}$ and $\text{Cl}^{-}$ ions.
• The oxidation state of magnesium in $\text{MgCl}_{2}$ is +2, and the oxidation state of chlorine is -1.

For covalent compounds and more complicated compounds, the atoms in those compounds are given oxidation states by assuming that they are ionic compounds. E.g. of a covalent compound: (Using the rules above)

• $\text{NO}_{2}$ is a covalent compound, which we will assume to be ionic.
• The oxygen atoms would be $\text{O}^{2-}$ ions.
• Hence, the nitrogen would have to be $\text{N}^{4+}$ ions.
• Therefore, the oxidation state of nitrogen is +4.

An example of a complicated compound: (Using the rules above)

• $\text{KMnO}_{4}$ is a complicated compound as it has a mixture of ionic and covalent bonds.
• We assume that the compound is ionic.
• The potassium and oxygen must be $K^{+}$ and $O^{2-}$ ions respectively.
• Hence, the manganese would have to be a $\text{Mn}^{7+}$ ion so that all the charges on the ions add up to zero. 