Oxidation number

What is an oxidation number?

In plain words: in a covalent molecule such as water no electron is fully handed over, but we can pretend it is. Give each shared pair to the atom that pulls harder, count the charge each atom is left with, and you have its oxidation number. Then any rise in that number is oxidation and any fall is reduction, whether the compound is ionic or covalent.

Why oxidation numbers are needed

In 2H₂(g) + O₂(g) → 2H₂O(l), hydrogen goes from a neutral state in H₂ to a positive state in water and oxygen to a dinegative state, as if electrons passed from H to O. In fact the charge transfer is only partial, an electron shift, as in many reactions of covalent compounds (NCERT page 239). To keep track of such shifts, oxidation numbers assume a complete transfer from the less electronegative atom to the more electronegative one (see electronegativity). This is book-keeping only, but it describes every redox reaction simply:

• 2H₂ + O₂ → 2H₂O: H 0 → +1, O 0 → −2;
• H₂ + Cl₂ → 2HCl: H 0 → +1, Cl 0 → −1;
• CH₄ + 4Cl₂ → CCl₄ + 4HCl: C −4 → +4, Cl 0 → −1, H stays +1.

(NCERT’s equations 7.19 and 7.22 print hydrogen as H₂(s); it is a gas, H₂(g).)

Rules for assigning oxidation numbers

NCERT gives six rules (page 240):

1. Every atom of an element in the free state has oxidation number 0: H₂, O₂, Cl₂, O₃, P₄, S₈, Na, Mg, Al.
2. A one-atom ion has the oxidation number of its charge: Na⁺ +1, Mg²⁺ +2, Fe³⁺ +3, Cl⁻ −1, O²⁻ −2. In their compounds, alkali metals are +1, alkaline earth metals +2 and aluminium +3 (page 240); strictly, these are rules for ordinary compounds (reasoned).
3. Oxygen is −2 in most compounds, with two kinds of exception: −1 in peroxides (H₂O₂, Na₂O₂) and −½ in superoxides (KO₂, RbO₂), where oxygen atoms are bonded to each other; and +2 in OF₂ and +1 in O₂F₂, where it is bonded to fluorine.
4. Hydrogen is +1, except when it is bonded to metals in binary compounds, such as LiH, NaH and CaH₂, where it is −1 (page 240). More generally, hydrogen is −1 whenever its partner is less electronegative, as for the B–H bonds of BH₄⁻ (reasoned from the definition above).
5. Fluorine is −1 in all its compounds. Chlorine, bromine and iodine are −1 as halide ions, but positive when combined with oxygen, as in oxoacids and oxoanions.
6. The sum rule:

The algebraic sum of the oxidation number of all the atoms in a compound must be zero.

NCERT Chemistry Part II (Class 11), Section 7.3, page 240

For a polyatomic ion the sum equals the charge on the ion: in CO₃²⁻, x + 3(−2) = −2 gives carbon +4. When an element has several atoms in one species, as in Na₂S₂O₃ or Cr₂O₇²⁻, the rules give the average oxidation number of those atoms (page 240).

Patterns. Metals have positive oxidation numbers in their compounds; non-metals can be positive or negative, and transition elements usually show several positive states. The highest oxidation number of a representative element is the group number for groups 1 and 2 and the group number minus 10 for groups 13 to 17, so it generally rises across a period (page 240). Strictly, a metal can also be 0 in some compounds, as nickel in the neutral carbonyl Ni(CO)₄ (reasoned). Oxygen and fluorine are exceptions: fluorine is never positive, and oxygen reaches only +2 (in OF₂). The term oxidation state is used interchangeably: in CO₂ the oxidation state, and the oxidation number, of carbon is +4.

NCERT’s table (page 241) gives HClO for chlorine. That is an error: in HClO, hypochlorous acid, chlorine is only +1 (+1 + x − 2 = 0). The +7 compound is perchloric acid, HClO₄ (+1 + x − 8 = 0).

Stock notation

In NCERT’s Stock notation, the oxidation number is expressed by a Roman numeral in parenthesis after the symbol of the metal in the molecular formula (page 241). Aurous and auric chloride become Au(I)Cl and Au(III)Cl₃; stannous and stannic chloride become Sn(II)Cl₂ and Sn(IV)Cl₄. Beyond the NCERT text: today the numeral is usually written after the element’s name, as in iron(III) oxide, or as a superscript on the symbol, Fe^III (Wikipedia, “Oxidation state”). In an exam on NCERT’s Stock notation, write NCERT’s form, Au(I)Cl. The numeral shows at once which is the oxidised and which the reduced form: Hg₂(I)Cl₂ is the reduced form of Hg(II)Cl₂. In NCERT Problem 7.3, HAuCl₄, Tl₂O, FeO, Fe₂O₃, CuI, CuO, MnO and MnO₂ become HAu(III)Cl₄, Tl₂(I)O, Fe(II)O, Fe₂(III)O₃, Cu(I)I, Cu(II)O, Mn(II)O and Mn(IV)O₂.

Types of redox reactions

NCERT Section 7.3.1 (pages 242–245) sorts redox reactions into four types.

1. Combination, A + B → C. A free element is common but not required: NO + NO₂ → N₂O₃ is a combination and comproportionation, with N +2 and +4 becoming +3. Combustion in dioxygen is always redox. NCERT lists these examples here, although methane’s combustion gives two products and so is not strictly A + B → C:

• C(s) + O₂(g) → CO₂(g): C 0 → +4, O 0 → −2;
• 3Mg(s) + N₂(g) → Mg₃N₂(s): Mg 0 → +2, N 0 → −3;
• CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l): C −4 → +4, while hydrogen stays +1.

2. Decomposition, the opposite: a compound breaks into two or more products. A redox decomposition need not produce an element: NH₄NO₃ → N₂O + 2H₂O changes nitrogen oxidation states although all products are compounds.

• 2H₂O(l) → 2H₂(g) + O₂(g)
• 2NaH(s) → 2Na(s) + H₂(g)
• 2KClO₃(s) → 2KCl(s) + 3O₂(g): Cl +5 → −1, O −2 → 0, potassium unchanged.

Not every decomposition is redox. In CaCO₃(s) → CaO(s) + CO₂(g), Ca stays +2, C +4 and O −2 (page 242).

3. Displacement, X + YZ → XZ + Y: an atom or ion in a compound is replaced by one of another element.

Metal displacement is used in metallurgy to obtain pure metals. In each case the reducing metal is a better reducing agent than the metal it displaces (page 242):

• CuSO₄(aq) + Zn(s) → Cu(s) + ZnSO₄(aq)
• V₂O₅(s) + 5Ca(s) → 2V(s) + 5CaO(s)
• TiCl₄(l) + 2Mg(s) → Ti(s) + 2MgCl₂(s)
• Cr₂O₃(s) + 2Al(s) → Al₂O₃(s) + 2Cr(s)

Hydrogen displacement. All alkali metals and calcium, strontium and barium displace hydrogen from cold water: 2Na + 2H₂O → 2NaOH + H₂, and Ca + 2H₂O → Ca(OH)₂ + H₂. Iron reacts with steam; magnesium reacts with hot water and more readily with steam (page 243). With steam, magnesium gives its oxide, Mg + H₂O(g) → MgO + H₂, and iron gives Fe₃O₄: 3Fe + 4H₂O(g) → Fe₃O₄ + 4H₂. NCERT’s equations 7.35 and 7.36 write the water as a liquid and give Mg(OH)₂, which magnesium forms with hot water, and Fe₂O₃, which iron does not form with steam.

Many metals, even some that do not react with steam, such as cadmium and tin, displace hydrogen from acids. Zn, Mg and Fe with hydrochloric acid are used to make dihydrogen in the laboratory: Zn + 2HCl → ZnCl₂ + H₂. The rate of hydrogen evolution reflects the reactivity of the metal: slowest for the least active, iron, and fastest for the most reactive, magnesium; much less active metals, such as silver and gold, do not react even with hydrochloric acid (page 243). Strictly, surface films and temperature also affect the rate, so compare clean surfaces under the same conditions (reasoned).

Halogen displacement. The oxidising power of the halogens falls from fluorine to iodine. Fluorine is so reactive that it even displaces the oxygen of water, 2H₂O + 2F₂ → 4HF + O₂, so its displacement reactions are not carried out in water. Chlorine displaces bromide and iodide, and bromine displaces iodide:

• Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂
• Cl₂ + 2I⁻ → 2Cl⁻ + I₂
• Br₂ + 2I⁻ → 2Br⁻ + I₂

Bromine and iodine colour a layer of CCl₄ shaken with the solution, which is the basis of the layer test for Br⁻ and I⁻ (page 243). Recovering a halogen from its halide, 2X⁻ → X₂ + 2e⁻, is an oxidation. Chemical oxidants can do this for Cl⁻, Br⁻ and I⁻ (page 243).

NCERT slip (page 244): NCERT says there is no way to convert F⁻ ions to F₂ by chemical means, and that the only way is to oxidise them electrolytically. Ordinary oxidants in water cannot do it (reasoned), but beyond the NCERT text, in 1986 Karl Christe made fluorine chemically, from K₂MnF₆ and SbF₅; industrial fluorine is still made by electrolysis (Wikipedia, “Fluorine”). In an exam, give NCERT’s answer: electrolysis.

4. Disproportionation.

In a disproportionation reaction an element in one oxidation state is simultaneously oxidised and reduced.

NCERT Chemistry Part II (Class 11), Section 7.3.1, page 244

The element must be able to exist in at least three oxidation states, and the reacting substance holds it in an intermediate one, so that both a higher and a lower state can form. Examples (page 244):

• 2H₂O₂ → 2H₂O + O₂: O −1 → −2 and 0;
• P₄ + 3OH⁻ + 3H₂O → PH₃ + 3H₂PO₂⁻: P 0 → −3 and +1;
• S₈ + 12OH⁻ → 4S²⁻ + 2S₂O₃²⁻ + 6H₂O: S 0 → −2 and average +2 in thiosulphate (whose individual S assignments are +5 and −1);
• Cl₂ + 2OH⁻ → ClO⁻ + Cl⁻ + H₂O: Cl 0 → +1 and −1.

The last reaction makes household bleach: hypochlorite oxidises coloured stains to colourless compounds. Bromine and iodine “follow the same trend as exhibited by chlorine” (NCERT page 244). Beyond the NCERT text: their hypohalites differ in stability, and the hypoiodite ions formed disproportionate at once to iodide and iodate (Wikipedia, “Iodine”). Fluorine does not follow this pattern. With alkali it gives 2F₂ + 2OH⁻ → 2F⁻ + OF₂ + H₂O, in which fluorine is −1 in both products and it is oxygen that is oxidised. Being the most electronegative element, fluorine cannot take a positive oxidation state, so it cannot disproportionate (page 244).

Fractional oxidation numbers

Sometimes the rules give a fraction: carbon 4/3 in C₃O₂, bromine 16/3 in Br₃O₈, and sulphur 2.5 in Na₂S₄O₆ (NCERT page 245). NCERT explains that electrons are never shared or transferred in fractions, so the fraction is the average oxidation state of atoms that are in different states (page 245); give that in an exam. Strictly, a fraction can also describe equivalent atoms over which the electrons are spread, as in the superoxide ion O₂⁻ (reasoned).

• Carbon suboxide, O=C=C=C=O: the two end carbons are +2 and the middle one 0; average 4/3.
• Br₃O₈: in NCERT’s structure the two end bromines are +6 and the middle one +4 (page 245); average 16/3.
• Tetrathionate, S₄O₆²⁻: the two end sulphurs are +5 and the two middle ones 0; average 2.5.

Mixed oxides such as Fe₃O₄, Mn₃O₄ and Pb₃O₄ give fractional averages for the metal in the same way. Fe₃O₄ contains one Fe²⁺ and two Fe³⁺, average 8/3. In O₂⁺ and O₂⁻, though, the equivalent oxygens are assigned average +½ and −½ (page 245).

Balancing redox equations

NCERT Section 7.3.2 (pages 246–249) gives two methods. Either may be used.

(a) Oxidation number method.

1. Write the correct formula of every reactant and product.
2. Assign oxidation numbers and find the atoms whose numbers change.
3. Find the increase and the decrease per atom and per formula, and multiply so that the total increase equals the total decrease. If no oxidation balances a reduction in the complete reaction, check for an omitted species or external electron source, as well as incorrect formulae or oxidation numbers.
4. In water, add H⁺ (acidic solution) or OH⁻ (basic solution) to the side that makes the total ionic charges equal.
5. Add H₂O to balance hydrogen, then check oxygen. If oxygen balances, the equation is balanced.

Examples 6 and 7 below are worked by this method (NCERT Problems 7.8 and 7.9).

(b) Half reaction (ion–electron) method. Each half reaction is balanced separately, then the two are added. NCERT’s example (pages 247–248) is the oxidation of Fe²⁺ by dichromate in acid:

1. Skeleton: Fe²⁺ + Cr₂O₇²⁻ → Fe³⁺ + Cr³⁺.
2. Split: oxidation Fe²⁺ → Fe³⁺; reduction Cr₂O₇²⁻ → Cr³⁺.
3. Balance atoms other than O and H: Cr₂O₇²⁻ → 2Cr³⁺.
4. In acid, add H₂O for oxygen and H⁺ for hydrogen: Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O.
5. Add electrons to balance charge. The left side has +12 and the right +6, so 6e⁻ go on the left: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. The iron half becomes Fe²⁺ → Fe³⁺ + e⁻, multiplied by 6.
6. Add and cancel the electrons.
7. Check atoms and charges.

Charge check: 12 − 2 + 14 = +24 on the left and 18 + 6 = +24 on the right.

In basic solution, balance first as in acid, then add as many OH⁻ to both sides as there are H⁺. Wherever H⁺ and OH⁻ stand on the same side, combine them into H₂O (page 248; see Example 8).

Redox reactions as the basis for titrations

Just as acid–base titrations use a pH-sensitive indicator, redox titrations find the strength of a reductant or oxidant with a redox-sensitive indicator (NCERT Section 7.3.3, page 249):

• Permanganate as its own indicator. MnO₄⁻ is intensely coloured. After the last of the reductant (Fe²⁺ or C₂O₄²⁻) is oxidised, the first lasting tinge of pink appears at an MnO₄⁻ concentration as low as 10⁻⁶ mol L⁻¹, so the colour overshoots the equivalence point, where oxidant and reductant match in mole stoichiometry, only slightly. The acid used is dilute H₂SO₄, not HCl: permanganate would also oxidise Cl⁻ to Cl₂, so too much KMnO₄ would be used.
• An added indicator. Dichromate is not a self-indicator, but just after the equivalence point it oxidises diphenylamine to an intense blue.
• Iodine and thiosulphate. An oxidant such as Cu²⁺ frees iodine from iodide, 2Cu²⁺ + 4I⁻ → Cu₂I₂(s) + I₂. The iodine, kept in solution by KI as KI₃, is titrated with thiosulphate: I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻. Starch gives an intense blue with iodine that vanishes when the last iodine is used up.

The volumes then give the amounts through the balanced equation (see reactions in solution), as in Example 9.

Limitations of the oxidation number concept

The idea of redox has kept changing: from adding oxygen, to losing electrons, to changing oxidation numbers. Oxidation numbers are a formal device; the fractional averages above show that they need not match the real state of each atom. NCERT Section 7.3.4 pictures oxidation as decreasing and reduction as increasing electron density. Strictly, this is a qualitative picture: changes of calculated partial charge alone are not a universal redox test, whereas oxidation state is a defined formal assignment (reasoned).

Solved examples

Common mistakes to avoid

Quick revision summary

• Oxidation number: the charge an atom would have if every shared pair went to the more electronegative atom.
• Oxidation is an increase in oxidation number, reduction a decrease; redox reactions change oxidation numbers.
• Rules: element 0; simple ion = charge; O −2 (peroxide −1, superoxide −½, with F positive); H +1 (metal hydride −1); F −1; the sum is 0, or the charge of an ion.
• Common maximum oxidation-number pattern: group number for groups 1–2 and group number −10 for groups 13–17, with exceptions including oxygen and fluorine.
• Types: combination, decomposition, displacement (metal, hydrogen, halogen), disproportionation.
• Fractional values can describe averages or equivalent delocalised sites; structural information is needed for individual assignments, which remain formal.
• Balance by oxidation numbers or by half reactions: total increase = total decrease.
• Redox titrations: KMnO₄ as its own indicator, Cr₂O₇²⁻ with diphenylamine, iodine with starch.

Questions students ask

What is the difference between oxidation number and valency?

Valency is an atom’s combining capacity, a whole number without a sign that counts the bonds it forms or the electrons it loses or gains, while oxidation number is the charge the atom would carry if each shared pair went wholly to the more electronegative atom, so it has a sign and can be zero or, as an average, fractional. Carbon has valency 4 in CH₄, CH₂Cl₂ and CCl₄, but oxidation numbers −4, 0 and +4, because it takes the bonding electrons from the less electronegative hydrogen and gives them to the more electronegative chlorine. In N₂, likewise, each nitrogen has valency 3 (a triple bond) but oxidation number 0.

Are oxidation number and oxidation state the same?

Yes, in NCERT and in JEE and NEET questions oxidation number and oxidation state mean the same thing: the charge an atom would carry if every shared pair went to the more electronegative atom, such as +6 for sulphur in H₂SO₄. NCERT itself calls oxidation state “a term that is often used interchangeably with the oxidation number” (page 240). Roman numerals in names such as iron(III) chloride are a notation for this value, not a separate physical property. Neither is the same as formal charge, which splits each shared pair equally between the two atoms.

Is oxidation number the same as the charge on an atom?

Oxidation number equals the real charge only in simple cases, such as a one-atom ion like Fe³⁺ or Cl⁻; in a covalent molecule it is a bookkeeping charge, because the shared electrons are only drawn towards the more electronegative atom, never fully handed over. In water, for example, oxygen is −2 and each hydrogen +1, but the actual partial charges are much smaller, and in SF₆ sulphur is +6 although SF₆ does not contain a bare S⁶⁺ ion. By convention the sign is written first for an oxidation number (+3) and last for an ionic charge (3+).

What are the oxidation states of nitrogen?

Nitrogen shows every oxidation state from −3 to +5: −3 in NH₃, −2 in N₂H₄, −1 in NH₂OH, 0 in N₂, average +1 in N₂O (its two N atoms are inequivalent), +2 in NO, +3 in HNO₂, +4 in NO₂ and +5 in HNO₃. The limits follow from its five valence electrons: it can take three more to complete an octet, giving −3, or lose all five to more electronegative atoms, giving +5, the group number (15) minus 10. In hydrazoic acid, HN₃, the rules even give an average of −1/3, which here hides different states of the three nitrogen atoms.

What is the difference between disproportionation and comproportionation?

In disproportionation one element in a single intermediate oxidation state is both oxidised and reduced, while in comproportionation, the reverse, the same element in two different oxidation states reacts to give one intermediate state. For example, 2H₂S + SO₂ → 3S + 2H₂O takes sulphur from −2 and +4 to 0, and IO₃⁻ + 5I⁻ + 6H⁺ → 3I₂ + 3H₂O takes iodine from +5 and −1 to 0. NCERT does not name comproportionation, but it is a redox reaction like any other: in the sulphur example the 4 electrons lost by two H₂S are the 4 gained by one SO₂.

What does KMnO₄ change into in acidic, neutral and basic solution?

Potassium permanganate, with manganese at +7, is reduced to different products in different media: to nearly colourless Mn²⁺ (+2, five electrons per MnO₄⁻) in acidic solution, to a brown precipitate of MnO₂ (+4, three electrons) in neutral or weakly basic solution, and to green manganate, MnO₄²⁻ (+6, one electron), in strongly basic solution with suitable reduction conditions. The reductant and conditions also matter, so medium alone does not guarantee a unique product. The electrons taken per formula unit, 5, 3 or 1, are what coaching books call the n-factor, so the equivalent mass of KMnO₄ (molar mass 158.0 g mol^⁻¹) is 31.6, 52.7 or 158.0 g per mole of electron equivalents for those respective reduction products.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Find the oxidation number of (a) P in NaH₂PO₄; (b) Mn in K₂MnO₄; (c) B in NaBH₄; (d) S in H₂S₂O₇ (from NCERT Exercise 7.1).

   Show answerHide answer

   (a) 1 + 2 + x − 8 = 0, x = +5. (b) 2 + x − 8 = 0, x = +6. (c) 1 + x − 4 = 0, x = +3 (hydrogen is −1 bonded to boron, the less electronegative element). (d) 2 + 2x − 14 = 0, x = +6.

2. Q2. Find the average oxidation number of iron in Fe₃O₄ and of carbon in CH₃COOH, and explain each result (from NCERT Exercise 7.2).

   Show answerHide answer

   Fe₃O₄: 3x = 8, x = 8/3; it is a mixed oxide with one Fe²⁺ and two Fe³⁺. CH₃COOH: 2x + 4 − 4 = 0, x = 0; in fact the CH₃ carbon is −3 and the COOH carbon +3.

3. Q3. Balance H₂O₂ + Fe²⁺ → Fe³⁺ + H₂O in acidic solution (NCERT Exercise 7.18 (c)).

   Show answerHide answer

   Reduction: H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O (O −1 → −2). Oxidation: Fe²⁺ → Fe³⁺ + e⁻, × 2. Sum: H₂O₂ + 2Fe²⁺ + 2H⁺ → 2Fe³⁺ + 2H₂O.