Redox Reactions · Lesson 3 of 8

Oxidation Numbers: Rules, Meaning, and Limits

“Use oxidation numbers to track redox changes while distinguishing formal bookkeeping from real atomic charge.”

Learning Objectives

• Assign oxidation numbers using electronegativity and the sum rules. • Apply the exceptions for hydrogen, oxygen, and halogens correctly. • Use Stock notation and oxidation-number changes to identify redox partners. • Distinguish average oxidation numbers from the states of structurally different atoms. • Explain why oxidation number is a useful model rather than a complete description of electron distribution.

Tracking a shift without making ions

Electron transfer is especially clear when a metal becomes an ion. In a covalent molecule, however, electrons are shared. Water formation still oxidises hydrogen and reduces oxygen, even though water is not simply a collection of free hydrogen and oxide ions. We need a consistent way to track the direction of the electron shift without claiming that every covalent bond becomes ionic.

Definition

Oxidation number

A formal number assigned to an atom by treating each bond between different elements as if its electrons belonged entirely to the more electronegative atom. Oxidation state is commonly used with the same meaning.

In $H_{2}$ and $O_{2}$ , identical atoms share electrons equally and each has oxidation number zero. In $H_{2} O$ , the bookkeeping assigns the bonding electrons to oxygen, giving hydrogen $+ 1$ and oxygen $- 2$ . These are oxidation numbers, not measured charges of $+ 1$ and $- 2$ on atoms in a water molecule. The electron distribution is polar, but transfer is not complete.

Water formation as an oxidation-state changeLaTeX

2 H_{2} + O_{2} \to 2 H_{2} O

Hydrogen changes from 0 to +1 and oxygen from 0 to -2. This identifies oxidation and reduction even though the product is covalent.

The same approach works for $H_{2} + C l_{2} \to 2HCl$ and $C H_{4} + 4C l_{2} \to CC l_{4} + 4HCl$ . In the second reaction carbon changes from $- 4$ in methane to $+ 4$ in carbon tetrachloride, while chlorine changes from $0$ to $- 1$ . Hydrogen remains at $+ 1$ . A redox reaction does not require every element to change state.

A consistent set of assignment rules

Most calculations do not require drawing every bond. A small set of rules gives the same bookkeeping result for common species. Apply definite rules first, check for exceptions, and then use the total charge to calculate the unknown value.

Situation	Oxidation-number rule	Illustration
Free element	Each atom is 0.	$H_{2}, O_{2}, O_{3}, P_{4}, S_{8}, Na$
Monatomic ion	Equals the ionic charge.	$M g^{2 +}$ : $+ 2$ ; $C l^{-}$ : $- 1$
Common metals	Group 1: +1; group 2: +2; aluminium: usually +3.	$NaCl, MgO, Al F_{3}$
Oxygen	Usually -2; check peroxides, superoxides, and oxygen fluorides.	$C O_{2}$ : $- 2$
Hydrogen	Usually +1; -1 in binary metal hydrides.	$HCl$ : $+ 1$ ; $NaH$ : $- 1$
Fluorine	-1 in its compounds.	$O F_{2}$ has fluorine $- 1$
Other halogens	Usually -1 as halides; positive states occur with oxygen or more electronegative halogens.	$HCl O_{4}$ has chlorine $+ 7$
Neutral compound	Sum over all atoms is 0.	$H_{2} S O_{4}$ : $2 (+ 1) + x + 4 (- 2) = 0$
Polyatomic ion	Sum over all atoms equals the ion charge.	$C O_{3}^{2 -}$ : $x + 3 (- 2) = - 2$

The number of atoms matters. In $C r_{2} O_{7}^{2 -}$ , there are two chromium atoms, not one. If their average oxidation number is $x$ , the chromium contribution is $2 x$ . When atoms of an element are not equivalent in a structure, the sum method may provide only their average; we will examine that distinction shortly.

Example — An ion with two central atoms

Problem
Calculate the oxidation number of chromium in $C r_{2} O_{7}^{2 -}$ .

1.Oxygen is $- 2$ here, so seven oxygen atoms contribute $7 (- 2) = - 14$ .
2.Let each chromium have oxidation number $x$ . The ion has charge $- 2$ , so $2 x - 14 = - 2$ .
3.Add 14 to both sides: $2 x = 12$ . Divide by 2: $x = + 6$ .
4.The result refers to each chromium atom. The total contribution from both chromium atoms is +12.

Check the oxygen and hydrogen exceptions

A formula can look familiar while containing a different type of bond. A peroxide contains an oxygen–oxygen unit in which each oxygen has oxidation number -1. A superoxide ion has total charge -1 distributed over two equivalent oxygen atoms, giving an average of -1/2 per oxygen. Fluorine is more electronegative than oxygen, so oxygen has a positive oxidation number in oxygen fluorides.

Species	Relevant oxidation number	Reason
$H_{2} O_{2}, N a_{2} O_{2}$	Oxygen: $- 1$	Peroxide oxygen–oxygen unit.
$K O_{2}, Rb O_{2}$	Oxygen: $- \frac{1}{2}$	Superoxide unit has total charge -1.
$O F_{2}$	Oxygen: $+ 2$	Two fluorine atoms each contribute -1.
$O_{2} F_{2}$	Oxygen: $+ 1$ on average	Two fluorines contribute -2 in a neutral molecule.
$LiH, NaH, Ca H_{2}$	Hydrogen: $- 1$	Hydrogen is bonded to an electropositive metal.

Example — Why structure matters in an oxygen-rich acid

Problem
Find sulphur’s oxidation number in $H_{2} S O_{5}$ , whose structure contains one peroxide bond.

1.Do not assign -2 to all five oxygen atoms. The oxygen–oxygen peroxide pair contains two oxygen atoms at -1 each.
2.The other three oxygen atoms contribute $3 (- 2) = - 6$ ; the peroxide pair contributes $2 (- 1) = - 2$ . Total oxygen contribution is $- 8$ .
3.The neutral formula gives $2 (+ 1) + x - 8 = 0$ , so sulphur is $+ 6$ .
4.Assigning every oxygen -2 would give an incorrect +8 for sulphur. The error would come from ignoring the bonding information, not from the sum rule itself.

Representative elements show a useful pattern in their highest commonly encountered oxidation states: +1 and +2 for groups 1 and 2, then generally group number minus 10 for groups 13–17. This is a pattern to interpret compounds, not a replacement for the assignment rules. Transition metals often show several positive oxidation states.

Element	Illustrative compound	Oxidation number of the named element
Na	$NaCl$	+1
Mg	$MgS O_{4}$	+2
Al	$Al F_{3}$	+3
Si	$SiC l_{4}$	+4
P	$P_{4} O_{10}$	+5
S	$S F_{6}$	+6
Cl	$HCl O_{4}$	+7

Name the state and recognise the redox change

Some metals form more than one compound with the same partner. Stock notation specifies the metal’s oxidation state using a Roman numeral. Iron(II) and iron(III) therefore refer to different oxidation states of iron, not different numbers of iron atoms.

For example, $AuCl$ is gold(I) chloride and $AuC l_{3}$ is gold(III) chloride. Similarly, $SnC l_{2}$ is tin(II) chloride and $SnC l_{4}$ is tin(IV) chloride. Mercury(I) chloride, $H g_{2} C l_{2}$ , has mercury at +1 per atom, whereas mercury(II) chloride, $HgC l_{2}$ , has mercury at +2. The first mercury compound is the reduced-state form relative to the second.

Example — Stock notation from the formula

Problem
Assign metal oxidation states in $HAuC l_{4}, T l_{2} O, FeO, F e_{2} O_{3}, CuI, CuO, MnO, Mn O_{2}$ .

1.For $HAuC l_{4}$ , $1 + x + 4 (- 1) = 0$ , giving gold(III). For $T l_{2} O$ , $2 x - 2 = 0$ , giving thallium(I).
2. $FeO$ gives iron(II), while $F e_{2} O_{3}$ gives iron(III). $CuI$ gives copper(I), whereas $CuO$ gives copper(II).
3. $MnO$ contains manganese(II), and $Mn O_{2}$ contains manganese(IV). A Roman numeral states the oxidation number of the metal, not the overall compound charge.

An increase in oxidation number is oxidation; a decrease is reduction. The substance whose changing atom is reduced is the oxidising agent, and the substance whose changing atom is oxidised is the reducing agent. This extends electron accounting to reactions where a literal transfer of whole electrons between free ions is not an adequate picture.

Example — Copper production from oxide and sulphide

Problem
Analyse $2C u_{2} O + C u_{2} S \to 6Cu + S O_{2}$ .

1.Copper is $+ 1$ in both $C u_{2} O$ and $C u_{2} S$ and becomes $0$ in the metal. Copper is reduced.
2.Sulphur is $- 2$ in $C u_{2} S$ and $+ 4$ in $S O_{2}$ . It is oxidised, an increase of six units.
3.Six copper atoms each decrease by one unit, matching the six-unit increase for sulphur.
4.The copper(I) centres accept the electron equivalent and act as oxidising centres. $C u_{2} O$ supplies an oxidant, while sulphide sulphur in $C u_{2} S$ is the reducing centre. Copper in the sulphide is also reduced, so do not assume every atom in a reagent has the same role.

An average can hide different atomic environments

Applying only the sum rule sometimes gives a fraction. That fraction is an average over all atoms of the named element. It does not mean that a particular atom literally donated a fraction of a free electron. For molecules containing non-equivalent atoms, the structure can reveal distinct individual oxidation states hidden by the average.

Three carbons, two environments— In carbon suboxide, the two terminal carbons are each +2 while the central carbon is 0. The average is

4/3

Species	Average from the sum rule	States revealed by structure
$C_{3} O_{2}$	$+ \frac{4}{3}$ for carbon	Terminal carbons +2, central carbon 0.
$B r_{3} O_{8}$	$+ \frac{16}{3}$ for bromine	Terminal bromines +6, central bromine +4.
$S_{4} O_{6}^{2 -}$	$+ \frac{5}{2}$ for sulphur	Two outer sulphurs +5 each; two inner sulphurs 0.
$F e_{3} O_{4}$	$+ \frac{8}{3}$ for iron	One iron(II) and two iron(III) per formula unit.

The sulphur chain in tetrathionate can be viewed as $O_{3} S - S - S - S O_{3}^{2 -}$ : the terminal sulphurs are attached to oxygen, while the two internal sulphurs have only sulphur neighbours. Mixed oxides such as $F e_{3} O_{4}, M n_{3} O_{4}, P b_{3} O_{4}$ similarly contain more than one metal oxidation state. In contrast, the equivalent oxygen atoms in $O_{2}^{-}$ and $O_{2}^{+}$ are described by average values $- \frac{1}{2}$ and $+ \frac{1}{2}$ ; not every fractional result can be split into different integer states on distinguishable atoms.

Structures reveal the individual states— Compare the oxygen attachments. Bonds between atoms of the same element do not change their relative oxidation-number assignment.

Example — Why the same mixed oxide reacts differently

Problem
Explain the contrast between $P b_{3} O_{4} + 8HCl \to 3PbC l_{2} + C l_{2} + 4 H_{2} O$ and $P b_{3} O_{4} + 4HN O_{3} \to 2Pb (N O_{3})_{2} + Pb O_{2} + 2 H_{2} O$ .

1.Treat $P b_{3} O_{4}$ as containing two $PbO$ units and one $Pb O_{2}$ unit. The lead states are +2 and +4, not a single literal +8/3 state on every atom.
2.With HCl, $2PbO + 4HCl \to 2PbC l_{2} + 2 H_{2} O$ is acid–base chemistry. Separately, $Pb O_{2} + 4HCl \to PbC l_{2} + C l_{2} + 2 H_{2} O$ reduces lead(IV) and oxidises chloride.
3.With nitric acid, the lead(II) oxide component undergoes acid–base reaction to give nitrate. Under the stated conditions, $Pb O_{2}$ remains rather than oxidising nitrate, whose nitrogen is already in a high oxidation state.
4.A single overall formula may therefore conceal components with different chemical behaviour. Structural and chemical information improves the formal oxidation-number calculation.

Common mistake

Do not confuse oxidation number, ionic charge, valency, and the number of atoms. They answer different questions. Oxidation number is a signed bookkeeping value; an average from the formula may not describe every individual atom.

Oxidation-number bookkeeping is powerful, but it does not specify a complete reaction mechanism or exact electron density. A broader description views oxidation as a decrease in electron density around the atom or atoms concerned, and reduction as an increase. Use the formal model to track a change, while remembering what the model assumes.

Quick check

What is the chromium oxidation number in $C r_{2} O_{7}^{2 -}$ ?

Quick check

Which assignment is correct in $O F_{2}$ ?

Quick check

A metal changes from oxidation state +2 to +3. What has happened?

Quick check

What does the average sulphur value $+ \frac{5}{2}$ in tetrathionate establish by itself?

Quick check

Why does assigning oxygen -2 everywhere fail for $H_{2} S O_{5}$ ?

Practice Problems

Calculate the named oxidation numbers: P in $Na H_{2} P O_{4}$ and $H_{4} P_{2} O_{7}$ ; Mn in $K_{2} Mn O_{4}$ ; S in $H_{2} S_{2} O_{7}$ .
Find the oxidation numbers of B in $NaB H_{4}$ and O in $Ca O_{2}$ , stating the exceptions used.
Write formulas for mercury(II) chloride, nickel(II) sulphate, tin(IV) oxide, thallium(I) sulphate, iron(III) sulphate, and chromium(III) oxide.
Explain how water of crystallisation affects the sulphur oxidation number in $KAl (S O_{4})_{2} \cdot 12 H_{2} O$ .
Use the structures to distinguish the two carbon oxidation numbers in ethanol, $C H_{3} C H_{2} OH$ , and ethanoic acid, $C H_{3} COOH$ .
Explain why the average iodine oxidation number in $K I_{3}$ does not mean three identical monatomic iodide ions.
Analyse the oxidised and reduced centres in the copper oxide–sulphide reaction, then explain why “one whole reagent is always only an oxidant” can be misleading.

Key Takeaways

Oxidation numbers track formal electron redistribution; they are not generally actual charges in covalent molecules. Assign fixed values and exceptions before applying the total-charge rule. An increase signals oxidation; a decrease signals reduction. Stock numerals specify oxidation states. An average can conceal different atomic environments, as in mixed oxides and tetrathionate. Structure and electron density reveal information that formal oxidation numbers alone cannot provide.

Previous · Lesson 2

Electron Transfer and Competition for Electrons

Next · Lesson 4

Recognising Redox Reaction Patterns

Tracking a shift without making ions

A consistent set of assignment rules

Check the oxygen and hydrogen exceptions

Name the state and recognise the redox change

An average can hide different atomic environments

Quiz

Practice Problems

Key Takeaways