3.1-3.2 - Oxidation Numbers | Edexcel A-Level Chemistry

Oxidation numbers track electron ownership in compounds and ions. Learn to calculate them from a species’ total charge, recognise peroxides and metal hydrides, and distinguish an assigned oxidation number from a real ionic charge.

The Oxidation-Number Model

Suppose you meet these three formulae in the same topic:

$\ce{NaCl}$
$\ce{HCl}$
$\ce{ClO3-}$

Chlorine is not doing the same electronic job in all three. In sodium chloride, chloride is a real $\ce{Cl-}$ ion. In hydrogen chloride, the bond is covalent but polar. In $\ce{ClO3-}$ , chlorine is surrounded by oxygen atoms and the whole species has a $-1$ charge. Oxidation numbers let chemists compare these cases using one consistent accounting method.

Oxidation number

The charge an atom would have if each bond were treated as completely ionic: assign bonding electrons to the more electronegative atom and share them equally between identical atoms.

For a simple ion, it is the actual ionic charge: sodium in $\ce{Na+}$ has oxidation number $+1$ ; chloride in $\ce{Cl-}$ has oxidation number $-1$ . In a covalent molecule or polyatomic ion, the oxidation number is usually not a real charge sitting on the atom. It is the charge the atom is treated as having when bonding electrons are assigned for bookkeeping.

This is why water can be written as:

Species	Atom considered	Oxidation number	What this does and does not mean
$\ce{H2O}$	$H$	$+1$	Hydrogen is treated as electron-poor compared with oxygen.
$\ce{H2O}$	$O$	$-2$	Oxygen is treated as electron-rich compared with hydrogen.

The water molecule does not contain two free $\ce{H+}$ ions and one free $\ce{O^2-}$ ion. Oxidation numbers are a model for electron accounting, not a claim that every covalent substance is ionic.

Assigned charge, not real charge

Oxygen in $\ce{H2O}$ is assigned $-2$ even though water is a neutral covalent molecule with no oxide ions. Oxidation numbers are assigned charges used for accounting; their total still has to match the overall charge of the species.

Oxidation-Number Rules

The most important rule is the total rule:

Total oxidation-number rule

\sum(\text{oxidation numbers of all atoms}) = \text{overall charge of the species}

The sum is $0$ for a neutral element or compound and equals the ion charge for a polyatomic ion.

So in $\ce{H2SO4}$ , all the assigned oxidation numbers must add to $0$ . In $\ce{SO4^2-}$ , they must add to $-2$ .

Use the fixed rules first, then use algebra for the unknown:

Situation	Usual oxidation number	Examples
Uncombined element	$0$	$\ce{Na}$ , $\ce{O2}$ , $\ce{Cl2}$ , $\ce{S8}$
Simple monatomic ion	ionic charge	$\ce{Mg^2+}$ is $+2$ ; $\ce{Br-}$ is $-1$
Group 1 metal in a compound	$+1$	$\ce{Na2O}$ , $\ce{KCl}$
Group 2 metal in a compound	$+2$	$\ce{MgCl2}$ , $\ce{CaO}$
Aluminium in a compound	$+3$	$\ce{Al2O3}$
Fluorine in a compound	$-1$	$\ce{HF}$ , $\ce{NaF}$
Oxygen in most compounds	$-2$	$\ce{H2O}$ , $\ce{CO2}$ , $\ce{SO4^2-}$
Oxygen in a peroxide	$-1$	$\ce{H2O2}$ , $\ce{Na2O2}$
Hydrogen with non-metals	$+1$	$\ce{HCl}$ , $\ce{H2O}$ , $\ce{NH4+}$
Hydrogen in a metal hydride	$-1$	$\ce{NaH}$ , $\ce{CaH2}$

The order matters when there is an exception. In $\ce{H2O2}$ , oxygen is not $-2$ ; it is a peroxide, so each oxygen is $-1$ . In $\ce{NaH}$ , hydrogen is not $+1$ ; it is a metal hydride, so hydrogen is $-1$ .

Worked mini-example: carbon dioxide

Find the oxidation number of carbon in $\ce{CO2}$ .

Oxygen is usually $-2$ , and there are two oxygen atoms:

x + 2(-2) = 0

x - 4 = 0

x = +4

Carbon has oxidation number $+4$ in $\ce{CO2}$ . The chemical sense check is that oxygen is more electronegative than carbon, so carbon is assigned a positive oxidation number.

Compounds And Ions

Most calculations are not about memorising a table; they are about setting up the total correctly. Write the unknown as $x$ , insert the fixed oxidation numbers, and make the sum equal to the overall charge.

Worked example 1: sulfur in sulfuric acid

Calculate the oxidation number of sulfur in $\ce{H2SO4}$ .

Decision points:

$\ce{H2SO4}$ is neutral, so the total is $0$ .
Hydrogen is $+1$ with a non-metal.
Oxygen is $-2$ in this compound.

Set up the equation:

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

Simplify:

2 + x - 8 = 0

x = +6

Sulfur has oxidation number $+6$ in $\ce{H2SO4}$ .

Worked example 2: nitrogen in nitrate

Calculate the oxidation number of nitrogen in $\ce{NO3-}$ .

This time the species is an ion, so the sum must equal $-1$ , not $0$ .

x + 3(-2) = -1

x - 6 = -1

x = +5

Nitrogen has oxidation number $+5$ in $\ce{NO3-}$ . A common error is to make the total $0$ out of habit; that would give $+6$ , which does not match the ion charge.

Worked example 3: nitrogen in ammonium

Calculate the oxidation number of nitrogen in $\ce{NH4+}$ .

Hydrogen is $+1$ because it is bonded to a non-metal, and the total is $+1$ because this is an ion:

x + 4(+1) = +1

x + 4 = +1

x = -3

Nitrogen has oxidation number $-3$ in $\ce{NH4+}$ .

Guided practice: Calculate the oxidation number of the named element.

Species	Element	Route cue	Answer
$\ce{CO3^2-}$	$C$	total is $-2$	$+4$
$\ce{PO4^3-}$	$P$	total is $-3$	$+5$
$\ce{ClO3-}$	$\ce{Cl}$	total is $-1$	$+5$

A correct final oxidation number can earn full calculation marks. Showing the total-charge equation makes errors easier to diagnose; when a question explicitly asks for that equation, include it as a separately requested representation.

Peroxides And Metal Hydrides

Peroxides and metal hydrides require different oxygen or hydrogen values from their usual ones. Check the chemical environment before inserting a familiar value into the total-charge equation.

Peroxides

In hydrogen peroxide and ionic metal peroxides, the two linked oxygen atoms each have oxidation number $-1$ . The peroxide ion is $\ce{O2^2-}$ , not the superoxide ion $\ce{O2-}$ . The common examples at this level are $\ce{H2O2}$ and compounds such as $\ce{Na2O2}$ .

For $\ce{H2O2}$ :

2(+1) + 2x = 0

2x = -2

x = -1

Each oxygen is $-1$ . If you forced oxygen to be $-2$ , the total would be $2(+1) + 2(-2) = -2$ , which is impossible for a neutral molecule.

For $\ce{Na2O2}$ :

2(+1) + 2x = 0

x = -1

Each oxygen is again $-1$ .

Metal hydrides

In a metal hydride, hydrogen has oxidation number $-1$ . This is different from hydrogen in acids, water, ammonia or ammonium ions, where hydrogen is usually $+1$ .

For $\ce{CaH2}$ , calcium is Group 2, so it is $+2$ :

+2 + 2x = 0

2x = -2

x = -1

Hydrogen has oxidation number $-1$ in $\ce{CaH2}$ .

Oxygen bonded to fluorine

Fluorine has oxidation number $-1$ in compounds. Therefore oxygen in $\ce{OF2}$ is $+2$ : $x+2(-1)=0$ . This further example shows why “oxygen is always −2” is not a universal rule. The explicitly required peroxide and metal-hydride exceptions remain essential.

Common-error contrast

Formula	Tempting wrong route	Correct route
$\ce{H2O2}$	oxygen is always $-2$	peroxide: oxygen is $-1$
$\ce{NaH}$	hydrogen is always $+1$	metal hydride: hydrogen is $-1$
$\ce{CaH2}$	two hydrogens must total $+2$	calcium is $+2$ , so two hydrides total $-2$

Why hydrogen can be +1 or −1

Hydrogen is $+1$ in $\ce{HCl}$ but $-1$ in $\ce{NaH}$ not because the formula looks different, but because hydrogen is assigned $+1$ when bonded to a more electronegative non-metal, but in a metal hydride the metal is assigned a positive oxidation number and hydrogen is treated as hydride, $\ce{H-}$ .

Non-Equivalent Atoms

Sometimes the same element appears in more than one chemical environment. When individual oxidation numbers are required, calculate the different environments separately; an average does not identify each atom’s oxidation number.

Ammonium nitrate, $\ce{NH4NO3}$ , is a good test. It is made from $\ce{NH4+}$ and $\ce{NO3-}$ , so there are two different nitrogen environments.

For nitrogen in $\ce{NH4+}$ :

x + 4(+1) = +1

x = -3

For nitrogen in $\ce{NO3-}$ :

x + 3(-2) = -1

x = +5

So the two nitrogen atoms in ammonium nitrate have different oxidation numbers: $-3$ in $\ce{NH4+}$ and $+5$ in $\ce{NO3-}$ .

This matters because oxidation numbers are attached to atoms in their chemical environments, not just to element symbols in a whole empirical formula. Later redox lessons use changes in oxidation number, so the starting assignment has to identify which atom is being tracked.

One element, two environments

It is wrong to say that nitrogen in $\ce{NH4NO3}$ must have one oxidation number because it is one element. Oxidation number depends on the atom's environment: nitrogen in $\ce{NH4+}$ ( $-3$ ) and nitrogen in $\ce{NO3-}$ ( $+5$ ) are calculated separately.

Checking Oxidation Numbers

To check a proposed oxidation number, substitute it into the total-charge equation. A wrong total identifies an incorrect assignment, an overlooked exception or a misread formula. For example, $\ce{VO2^+}$ contains two oxygen atoms and has overall charge +1; $\ce{VO^2+}$ contains one oxygen atom and has charge +2. The subscript and superscript have different meanings.

Use this route whenever you are asked to calculate or deduce an oxidation number:

Identify whether the species is neutral or an ion.
Write the total: $0$ for a neutral species, or the ion charge for an ion.
Insert fixed oxidation numbers, checking for peroxides and metal hydrides.
Use $x$ for the unknown atom.
Solve the equation.
Sense-check the sign and total.

Worked response

Calculate the oxidation number of manganese in $\ce{MnO4^-}$ .

The ion has overall charge $-1$ . Oxygen is $-2$ :

x + 4(-2) = -1

x - 8 = -1

x = +7

Manganese has oxidation number $+7$ .

Recap: Oxidation number is an assigned charge used for electron bookkeeping. The calculation skill is to make the sum of assigned oxidation numbers equal the charge on the species, while applying the named exceptions for peroxides and metal hydrides. The next lesson uses these numbers to decide whether oxidation, reduction or disproportionation has happened.

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3.1-3.2 - Oxidation Numbers