Oxidation Numbers and Naming

Spec Mapping — OCR H432 Module 2.1.5 — Redox, covering oxidation as electron loss / oxidation-number increase, reduction as electron gain / oxidation-number decrease, the rules for assigning oxidation numbers in elements, ions and compounds, Roman-numeral nomenclature for variable-oxidation-state elements, and disproportionation as a redox process in which a single element is simultaneously oxidised and reduced (refer to the official OCR H432 specification document for exact wording).

The oxidation number is the central bookkeeping device of inorganic chemistry. It converts the messy reality of partly-ionic, partly-covalent bonding into a tractable integer that you can subtract across a reaction to see which atoms have gained electrons (reduced) and which have lost them (oxidised). Mastery of oxidation numbers makes everything that follows in the OCR specification more tractable — half-equation balancing (next lesson), transition-metal redox titrations (Module 5.3), standard electrode potentials and electrochemical cells (Module 5.2), the qualitative tests for anions, and the entire mechanism vocabulary of organic redox in Module 4 (alcohol-to-aldehyde-to-carboxylic acid, haloalkane substitution). This lesson formalises the rules, drills them across the species OCR most often examines (oxyanions, peroxides, hydrides, mixed-valence salts), introduces IUPAC Roman-numeral nomenclature, and explores disproportionation as the most-examined application.

Key Definition: an oxidation number is the hypothetical charge an atom would carry if every bond to a different element were treated as fully ionic, with the more electronegative atom taking both bonding electrons. It is therefore a formal accounting device, not a physical charge — the actual partial charges on atoms (computed from quantum-mechanical electron density) are smaller in magnitude and rarely integer.

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OIL RIG and what it really means

The traditional mnemonic — OIL RIG (Oxidation Is Loss, Reduction Is Gain) — refers to electrons. Equivalently:

• Oxidation ↔ loss of electrons ↔ increase in oxidation number.
• Reduction ↔ gain of electrons ↔ decrease in oxidation number.
• Redox reaction — any reaction in which at least one oxidation number changes.
• Oxidising agent — the species that causes oxidation by being reduced (it gains electrons; its oxidation number falls).
• Reducing agent — the species that causes reduction by being oxidised (it loses electrons; its oxidation number rises).

A useful mental picture: oxidation number is a signed integer attached to each atom; a redox reaction is a transfer of electrons from one atom to another; the oxidising agent is the receiver, the reducing agent the donor.

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The eight rules for assigning oxidation numbers

Apply these in order; later rules defer to earlier ones in case of conflict.

1. Uncombined elements (in their standard state) have oxidation number 0. Examples: Na(s), Fe(s), O₂(g), Cl₂(g), S₈(s), P₄(s), Hg(l).
2. Monatomic ions have an oxidation number equal to their ionic charge. Na⁺ is +1; O²⁻ is −2; Al³⁺ is +3.
3. Sum rule: oxidation numbers sum to 0 in a neutral compound and to the overall charge in a polyatomic ion.
4. Group 1 metals are always +1; Group 2 always +2; Aluminium always +3 in its compounds.
5. Hydrogen is +1 in almost all compounds, except in metal hydrides (NaH, CaH₂, LiAlH₄) where it is −1 because the metal is the less electronegative partner.
6. Oxygen is −2 in almost all compounds, except:
   • in peroxides (H₂O₂, Na₂O₂, BaO₂) where it is −1;
   • in superoxides (KO₂, NaO₂) where it is −½ (a non-integer, signalling the resonance-delocalised superoxide radical);
   • in compounds with the more electronegative fluorine (OF₂, O₂F₂) where it is +2 or +1 respectively.
7. Fluorine is always −1 in its compounds (because nothing is more electronegative than F).
8. Other halogens are −1 in binary halides; positive when bonded to oxygen or to a more electronegative halogen above them (Cl is positive in ClO⁻, ClF₃; Br is positive in HBrO).

A higher-priority rule beats a lower one. If a question hands you H₂O₂, apply rule 6 (peroxide oxygen = −1) and use the sum rule (3) to deduce H = +1, not rule 5 followed by rule 6.

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Worked examples — assigning oxidation numbers

Example 1 — sulfur in H₂SO₄

Let O.S. of S = $x$x. Apply rules 5, 6, 3 in sequence: $2(+1) + x + 4(-2) = 0 \implies 2 + x - 8 = 0 \implies x = +6$2(+1)+x+4(−2)=0⟹2+x−8=0⟹x=+6 Sulfur is in its highest oxidation state — sulfuric acid is therefore named sulfuric(VI) acid in systematic nomenclature.

Example 2 — manganese in $\text{MnO}_4^-$MnO4−​

$x + 4(-2) = -1 \implies x = +7$x+4(−2)=−1⟹x=+7 Mn(VII) — the highest oxidation state of manganese, and a powerful oxidising agent in acidic solution.

Example 3 — chromium in $\text{Cr}_2\text{O}_7^{2-}$Cr2​O72−​

$2x + 7(-2) = -2 \implies 2x = +12 \implies x = +6$2x+7(−2)=−2⟹2x=+12⟹x=+6 Cr(VI) — the second of the great A-Level oxidising agents (dichromate in acid).

Example 4 — nitrogen in ammonium nitrate

NH₄NO₃ contains two nitrogen atoms in different oxidation states. Treat them separately:

• In NH₄⁺: $x + 4(+1) = +1 \implies x = -3$x+4(+1)=+1⟹x=−3.
• In NO₃⁻: $x + 3(-2) = -1 \implies x = +5$x+3(−2)=−1⟹x=+5.

The same is true for $(\text{NH}_4)_2\text{Cr}_2\text{O}_7$(NH4​)2​Cr2​O7​ — N is −3 in the cation, Cr is +6 in the anion. When this salt is heated, the ammonium reduces the dichromate to Cr₂O₃ in the classic "volcano" demonstration — an internal redox.

Example 5 — chlorine in NaClO₃

Cl in NaClO₃: $x + 3(-2) = -1 \implies x = +5$x+3(−2)=−1⟹x=+5. Hence sodium chlorate(V) — the V indicates Cl(+5).

Example 6 — sulfur in thiosulfate $\text{S}_2\text{O}_3^{2-}$S2​O32−​ (a classic pitfall)

By the sum rule: $2x + 3(-2) = -2 \implies x = +2$2x+3(−2)=−2⟹x=+2. This is the average oxidation number that A-Level exam papers accept. Structurally the two sulfurs are inequivalent — the central S is roughly +5 and the terminal S roughly −1 — but OCR accepts +2 as the formal answer. The lesson: oxidation number is a bookkeeping device, not a quantum-mechanical observable.

Example 7 — oxygen in OF₂

By rule 7, F is −1. By rule 3: $x + 2(-1) = 0 \implies x = +2$x+2(−1)=0⟹x=+2. Oxygen positive because F is the only thing more electronegative than O.

Example 8 — oxygen in H₂O₂

H is +1 (rule 5), peroxide O is −1 (rule 6). Check: $2(+1) + 2(-1) = 0$2(+1)+2(−1)=0 ✓.

Example 9 — vanadium in VO₂⁺ (vanadyl ion)

V in $\text{VO}_2^+$VO2+​: O is −2 (rule 6), so $x + 2(-2) = +1 \implies x = +5$x+2(−2)=+1⟹x=+5. V(+5) — vanadium in its highest oxidation state, the oxidation level present in commercial vanadium chemistry (the catalyst in the Contact Process for sulfuric acid manufacture is V₂O₅, where V is also +5).

Example 10 — phosphorus in $\text{P}_4\text{O}_{10}$P4​O10​

By rule 1, free P₄ has P at 0. In P₄O₁₀, each O is −2 (10 of them give −20). The sum rule needs $4x + 10(-2) = 0 \implies x = +5$4x+10(−2)=0⟹x=+5. Phosphorus(V), as expected for the maximum group-15 oxidation state.

Example 11 — carbon in HCN and in $\text{CN}^-$CN−

In HCN, H is +1 and N is −3 (more electronegative than C, but less so than O). The sum rule gives $1 + x + (-3) = 0 \implies x = +2$1+x+(−3)=0⟹x=+2. C(+2) in HCN. In CN⁻: $x + (-3) = -1 \implies x = +2$x+(−3)=−1⟹x=+2 — same C oxidation state, as expected since the cyanide ion is just HCN minus a proton with no electron-transfer step.

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Roman-numeral nomenclature

When an element can exhibit several oxidation states (notably the d-block transition metals and the higher main-group elements), IUPAC nomenclature places a Roman numeral in brackets directly after the element name to indicate the oxidation state. There is no space between name and bracket.

Formula	Systematic name	Comment
FeCl₂	iron(II) chloride	Fe(+2) — pale green in solution
FeCl₃	iron(III) chloride	Fe(+3) — yellow/brown in solution
Cu₂O	copper(I) oxide	Cu(+1) — red solid
CuO	copper(II) oxide	Cu(+2) — black solid
KMnO₄	potassium manganate(VII)	Mn(+7) — purple
K₂Cr₂O₇	potassium dichromate(VI)	Cr(+6) — orange
K₂CrO₄	potassium chromate(VI)	Cr(+6) — yellow
NaClO	sodium chlorate(I)	Cl(+1) — household bleach
NaClO₃	sodium chlorate(V)	Cl(+5) — weedkiller
NaClO₄	sodium chlorate(VII)	Cl(+7) — strong oxidant
HNO₂	nitric(III) acid (nitrous acid)	N(+3)
HNO₃	nitric(V) acid (nitric acid)	N(+5)
H₂SO₃	sulfuric(IV) acid (sulfurous acid)	S(+4)
H₂SO₄	sulfuric(VI) acid (sulfuric acid)	S(+6)

Traditional names ("ferrous", "cuprous", "hypochlorite", "perchlorate") survive in older textbooks and industrial contexts but are deprecated by IUPAC and OCR. Always use the Roman-numeral systematic name in exam answers.

The chlorine oxyanion series

Formula	Cl O.S.	Systematic name	Traditional name
Cl⁻	−1	chloride	chloride
Cl₂	0	chlorine	chlorine
ClO⁻	+1	chlorate(I)	hypochlorite
ClO₂⁻	+3	chlorate(III)	chlorite
ClO₃⁻	+5	chlorate(V)	chlorate
ClO₄⁻	+7	chlorate(VII)	perchlorate

This series is the classic OCR exam pivot for Roman-numeral nomenclature — make sure you can navigate it both directions (formula → name and name → formula).

Nitrogen oxidation states

Nitrogen ranges from −3 (NH₃, NH₄⁺) through 0 (N₂) to +5 (HNO₃) — one of the widest oxidation-state ranges in the periodic table:

Compound	N O.S.	Notes
NH₃ / NH₄⁺	−3	Ammonia / ammonium
N₂H₄	−2	Hydrazine (rocket fuel)
NH₂OH	−1	Hydroxylamine
N₂	0	Atmospheric nitrogen
N₂O	+1	Nitrous oxide / "laughing gas"
NO	+2	Nitric oxide / signalling molecule
HNO₂ / NO₂⁻	+3	Nitric(III) acid / nitrite
NO₂ / N₂O₄	+4	Nitrogen dioxide
HNO₃ / NO₃⁻	+5	Nitric(V) acid / nitrate

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Tracking oxidation-number change in a reaction

flowchart TD
    A[Write balanced equation] --> B[Assign O.N. to every atom on LHS]
    B --> C[Assign O.N. to every atom on RHS]
    C --> D{Any atom changed O.N.?}
    D -- "No" --> E[Not a redox reaction]
    D -- "Yes" --> F[Atom whose O.N. increases = OXIDISED]
    D -- "Yes" --> G[Atom whose O.N. decreases = REDUCED]
    F --> H[Species containing oxidised atom = REDUCING AGENT]
    G --> I[Species containing reduced atom = OXIDISING AGENT]
    H --> J[Electrons lost = sum of increases]
    I --> K[Electrons gained = sum of decreases]
    J --> L{Electrons balance?}
    K --> L
    L -- "Yes" --> M[Equation is consistent]
    L -- "No" --> N[Scale species to balance electrons]

Worked example — Mg with HCl

$\text{Mg(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)}$Mg(s)+2HCl(aq)→MgCl2​(aq)+H2​(g)

Atom	LHS O.N.	RHS O.N.	Change	Role
Mg	0	+2	+2	Oxidised (reducing agent)
H	+1	0	−1	Reduced (oxidising agent — the H⁺ ion)
Cl	−1	−1	0	Spectator

Electron balance: Mg loses 2 e⁻ per atom; each H gains 1 e⁻, two Hs per Mg → 2 e⁻ gained. Balanced.

Worked example — Cr₂O₇²⁻ oxidising Fe²⁺ in acid

This is the key OCR Year 13 redox titration but the bookkeeping works at A-Level Year 12:

• Cr: +6 → +3, Δ = −3 per Cr, −6 per Cr₂O₇²⁻ unit.
• Fe: +2 → +3, Δ = +1 per Fe.

Need 6 Fe²⁺ per Cr₂O₇²⁻ to balance electrons. Balancing O via H₂O and H via H⁺ in acidic solution:

$\text{Cr}_2\text{O}_7^{2-} + 6\text{Fe}^{2+} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}$Cr2​O72−​+6Fe2++14H+→2Cr3++6Fe3++7H2​O

You will return to this method-of-balancing in the next lesson, where it is formalised as the half-equation method.

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Disproportionation