Oxidation Number

When metals react with non-metals, the electron transfer picture is straightforward: one atom hands over electrons completely, and the other accepts them. But what happens in reactions between non-metals, where atoms share electrons rather than fully transfer them? Consider the formation of water:

$2H_2(g) + O_2(g) \rightarrow 2H_2O(l) \qquad \text{(7.18)}$

There is no complete handover of electrons here. Hydrogen and oxygen share their electron pairs in the water molecule. Yet chemists still find it useful to think of hydrogen as going from a neutral state (in $H_2$) to a “positive” state (in $H_2O$), and oxygen going from neutral (in $O_2$) to a “negative” state (in $H_2O$). The electron shift is partial, better described as electrons being pulled closer to oxygen rather than being fully removed from hydrogen. The same idea applies to many other reactions involving covalent compounds:

$H_2(g) + Cl_2(g) \rightarrow 2HCl(g) \qquad \text{(7.19)}$

$CH_4(g) + 4Cl_2(g) \rightarrow CCl_4(l) + 4HCl(g) \qquad \text{(7.20)}$

To handle situations like these in a clean, systematic way, chemists developed the concept of oxidation number (also called oxidation state). It is a practical bookkeeping tool, not a description of what physically happens to every electron.

The Core Assumption Behind Oxidation Numbers

The oxidation number method works by making a deliberate simplification: it assumes that in every covalent bond, the shared electron pair belongs entirely to the more electronegative atom. This is not what really happens (covalent bonds share electrons), but pretending that it does lets us assign a definite charge-like number to each atom. That number is the oxidation number.

Using this assumption, we can rewrite the reactions above by marking the “charge” each atom would carry if the assumption were true:

$\overset{0}{2H_2}(g) + \overset{0}{O_2}(g) \rightarrow 2\overset{+1 \quad -2}{H_2O}(l) \qquad \text{(7.21)}$

$\overset{0}{H_2}(g) + \overset{0}{Cl_2}(g) \rightarrow 2\overset{+1 \quad -1}{HCl}(g) \qquad \text{(7.22)}$

$\overset{-4 \; +1}{CH_4}(g) + 4\overset{0}{Cl_2}(g) \rightarrow \overset{+4 \quad -1}{CCl_4}(l) + 4\overset{+1 \quad -1}{HCl}(g) \qquad \text{(7.23)}$

Notice how methane ($CH_4$) illustrates the idea well. Carbon is more electronegative than hydrogen, so the bookkeeping system assigns all four shared electron pairs to carbon, giving it a charge of $-4$. Each hydrogen is assigned $+1$. When methane reacts with chlorine to form $CCl_4$, carbon’s oxidation number jumps from $-4$ to $+4$ because chlorine is more electronegative than carbon and pulls all shared pairs to itself.

Key definition: The oxidation number of an element in a compound is the charge that atom would carry if every covalent bond in the molecule were treated as fully ionic, with the electron pair going entirely to the more electronegative partner.

The Six Rules for Assigning Oxidation Numbers

It is not always easy to figure out which atom is more electronegative, especially in complex molecules. So chemists use a set of six standard rules that make the assignment straightforward.

Rule 1: Free Elements Have Oxidation Number Zero

Every atom in the free, uncombined state has an oxidation number of zero. This covers single atoms like $Na$, $Mg$, $Al$ and also molecules of elements: $H_2$, $O_2$, $Cl_2$, $O_3$, $P_4$, $S_8$. Each individual atom in all of these carries zero.

Rule 2: Monoatomic Ions Equal Their Charge

For an ion made of a single atom, the oxidation number is simply the charge on the ion.

Ion	Oxidation number
$Na^+$	$+1$
$Mg^{2+}$	$+2$
$Fe^{3+}$	$+3$
$Cl^-$	$-1$
$O^{2-}$	$-2$

This rule also fixes oxidation numbers for entire groups in the periodic table: all alkali metals (Group 1) are always $+1$ in their compounds, all alkaline earth metals (Group 2) are always $+2$, and aluminium is always $+3$.

Rule 3: Oxygen is Usually $-2$, With Three Exceptions

In most compounds, oxygen carries $-2$. But there are three situations where it differs:

• Peroxides (compounds where oxygen atoms are bonded directly to each other, like $H_2O_2$ and $Na_2O_2$): each oxygen is $-1$.
• Superoxides (like $KO_2$ and $RbO_2$): each oxygen is $-\frac{1}{2}$.
• Compounds with fluorine (like $OF_2$ and $O_2F_2$): fluorine is more electronegative than oxygen, so oxygen gets a positive oxidation number. In $OF_2$, oxygen is $+2$. In $O_2F_2$, each oxygen is $+1$.

Rule 4: Hydrogen is Usually $+1$, Except in Metal Hydrides

Hydrogen is $+1$ in the vast majority of its compounds. The exception comes in binary compounds with metals, called metal hydrides (compounds containing just two elements, one of which is a metal). In $LiH$, $NaH$, and $CaH_2$, hydrogen takes $-1$ because the metal is less electronegative and gives up the electron pair to hydrogen.

Rule 5: Fluorine is Always $-1$; Other Halogens Depend on Their Partner

Fluorine, being the most electronegative element of all, always carries $-1$ in every compound. The other halogens ($Cl$, $Br$, $I$) also take $-1$ when they appear as halide ions in their compounds. However, when chlorine, bromine, or iodine are combined with oxygen (as in oxoacids and oxoanions), they can show positive oxidation numbers because oxygen is more electronegative.

Rule 6: The Sum Rule

For a neutral compound, the oxidation numbers of all atoms must add up to zero. For a polyatomic ion, the sum must equal the charge on the ion.

Example with the carbonate ion ($CO_3^{2-}$): let the oxidation number of carbon be $x$. Each oxygen is $-2$, so the sum is $x + 3(-2) = -2$. Solving: $x - 6 = -2$, giving $x = +4$.

Applying the Rules: Worked Examples

These rules let you find the oxidation number of any element in any compound. Here is how:

Finding the oxidation number of $Cr$ in $Cr_2O_7^{2-}$:

Let the oxidation number of chromium be $x$. There are 2 chromium atoms and 7 oxygen atoms.

$2x + 7(-2) = -2$

$2x - 14 = -2$

$2x = 12$

$x = +6$

Each chromium atom has an oxidation number of $+6$.

Important note about averages: When a molecule or ion contains two or more atoms of the same element, such as $Na_2S_2O_3$ or $Cr_2O_7^{2-}$, the oxidation number calculated by the rules is the average across all atoms of that element. In some cases the individual atoms may actually have slightly different environments, but the rules give a single average value.

Oxidation Numbers Across the Periodic Table

Some useful patterns emerge when we look at oxidation numbers systematically:

• Metals generally show positive oxidation numbers in their compounds.
• Non-metals can show either positive or negative oxidation numbers, depending on what they are bonded to.
• Transition metals typically display several positive oxidation states (like iron, which commonly appears as $Fe^{2+}$ and $Fe^{3+}$).
• The highest oxidation number a representative element can reach follows a clear trend across a period.

For the third period, the pattern looks like this:

Group	1	2	13	14	15	16	17
Element	$Na$	$Mg$	$Al$	$Si$	$P$	$S$	$Cl$
Example compound	$NaCl$	$MgSO_4$	$AlF_3$	$SiCl_4$	$P_4O_{10}$	$SF_6$	$HClO_4$
Highest oxidation number	$+1$	$+2$	$+3$	$+4$	$+5$	$+6$	$+7$

For Groups 1 and 2, the highest oxidation number equals the group number. For Groups 13 through 17, it equals the group number minus 10.

Oxidation Number vs. Oxidation State

These two terms mean the same thing and are used interchangeably. In $CO_2$, saying “the oxidation state of carbon is $+4$” is the same as saying “the oxidation number of carbon is $+4$.” Both describe the charge assigned to an atom under the assumption that all bonds are ionic.

Stock Notation: A Cleaner Way to Name Compounds

When a metal can exist in more than one oxidation state, older names like “aurous chloride” and “auric chloride” can be confusing. The German chemist Alfred Stock introduced a cleaner system: write the Roman numeral of the metal’s oxidation number in parentheses after the metal symbol in the formula.

Solved Example: Stock Notation (Problem 7.3)

Problem: Write the Stock notation for: $HAuCl_4$, $Tl_2O$, $FeO$, $Fe_2O_3$, $CuI$, $CuO$, $MnO$, and $MnO_2$.

Solution:

First, find the oxidation number of the metal in each compound using the rules:

$HAuCl_4$: Hydrogen is $+1$, each chlorine is $-1$ (four chlorines give $-4$). For the compound to be neutral: $+1 + x + (-4) = 0$, so $x = +3$. Gold is $+3$.

$Tl_2O$: Oxygen is $-2$. Two thallium atoms: $2x + (-2) = 0$, so $x = +1$. Thallium is $+1$.

$FeO$: Oxygen is $-2$. So iron is $+2$.

$Fe_2O_3$: Three oxygens give $-6$. Two iron atoms: $2x = +6$, so $x = +3$. Iron is $+3$.

$CuI$: Iodine is $-1$. So copper is $+1$.

$CuO$: Oxygen is $-2$. So copper is $+2$.

$MnO$: Oxygen is $-2$. So manganese is $+2$.

$MnO_2$: Two oxygens give $-4$. So manganese is $+4$.

Now write them in Stock notation:

Compound	Metal oxidation number	Stock notation
$HAuCl_4$	$Au = +3$	$HAu(III)Cl_4$
$Tl_2O$	$Tl = +1$	$Tl_2(I)O$
$FeO$	$Fe = +2$	$Fe(II)O$
$Fe_2O_3$	$Fe = +3$	$Fe_2(III)O_3$
$CuI$	$Cu = +1$	$Cu(I)I$
$CuO$	$Cu = +2$	$Cu(II)O$
$MnO$	$Mn = +2$	$Mn(II)O$
$MnO_2$	$Mn = +4$	$Mn(IV)O_2$

Notice how Stock notation makes the oxidation state immediately visible. Comparing $Fe(II)O$ and $Fe_2(III)O_3$ tells you at a glance that iron is in two different oxidation states. Similarly, $Hg_2(I)Cl_2$ (mercurous chloride) is clearly the reduced form when compared to $Hg(II)Cl_2$ (mercuric chloride).

Redefining Redox Using Oxidation Numbers

With oxidation numbers in hand, we now have a clean, universal way to identify oxidation, reduction, and redox reactions. Here are the definitions:

• Oxidation: An increase in the oxidation number of an element.
• Reduction: A decrease in the oxidation number of an element.
• Oxidising agent (oxidant): A reagent that causes the oxidation number of another element to increase. The oxidant itself gets reduced.
• Reducing agent (reductant): A reagent that causes the oxidation number of another element to decrease. The reductant itself gets oxidised.
• Redox reaction: Any reaction where the oxidation numbers of the reacting species change.

These definitions are powerful because they work for ionic compounds, covalent compounds, and everything in between. You no longer need to argue about whether electrons were “really” transferred; just track the oxidation numbers.

Solved Example: Identifying a Redox Reaction (Problem 7.4)

Problem: Show that the reaction below is a redox reaction. Identify the species oxidised, the species reduced, the oxidant, and the reductant.

$2Cu_2O(s) + Cu_2S(s) \rightarrow 6Cu(s) + SO_2(g)$

Solution:

Step 1: Assign oxidation numbers to every atom.

In $Cu_2O$: oxygen is $-2$, so each copper is $+1$.

In $Cu_2S$: sulphur is $-2$ (by analogy with oxygen in metal compounds), so each copper is again $+1$.

In $Cu$ (free element): oxidation number is $0$.

In $SO_2$: each oxygen is $-2$ (total $-4$), so sulphur must be $+4$.

Writing these on the equation:

$\overset{+1 \quad -2}{2Cu_2O}(s) + \overset{+1 \quad -2}{Cu_2S}(s) \rightarrow \overset{0}{6Cu}(s) + \overset{+4 \quad -2}{SO_2}(g)$

Step 2: Identify changes.

• Copper goes from $+1$ (in both $Cu_2O$ and $Cu_2S$) to $0$ (in free $Cu$). Oxidation number decreases, so copper is reduced.
• Sulphur goes from $-2$ (in $Cu_2S$) to $+4$ (in $SO_2$). Oxidation number increases, so sulphur is oxidised.

Step 3: Identify oxidant and reductant.

Since oxidation numbers change for both copper and sulphur, this is a redox reaction.

• Oxidant (oxidising agent): $Cu_2O$. It provides the copper that helps sulphur increase its oxidation number. Copper in $Cu_2O$ itself gets reduced.
• Reductant (reducing agent): Sulphur in $Cu_2S$. It gets oxidised (its oxidation number rises from $-2$ to $+4$) and in doing so helps copper (in both $Cu_2S$ and $Cu_2O$) decrease from $+1$ to $0$.