This MCQ module is based on: Oxidation Number
TOPIC 2 OF 13
Oxidation Number
🎓 Class 11
Chemistry
CBSE
Theory
Ch 7 – Redox Reactions
⏱ ~14 min
🌐 Language: [gtranslate]
🧠 AI-Powered MCQ Assessment
▲
📝 Worksheet / Assessment
▲
This assessment will be based on: Oxidation Number
Upload images, PDFs, or Word documents to include their content in assessment generation.
Oxidation Number and Types of Redox Reactions
7.2.1 Oxidation Number — a Universal Scoreboard
The electron-transfer picture is beautifully clear when ions are involved, but what about covalent molecules like H2SO4 or KMnO4? No ions physically leave their atoms, yet chemists still call these "oxidised" or "reduced" forms. The trick is a book-keeping device called the oxidation number (or oxidation state).
Oxidation number (ON): the hypothetical charge that an atom would carry if every bond in its compound were treated as 100% ionic, with both shared electrons going to the more electronegative partner.
Rules for Assigning Oxidation Numbers L1 Remember
| Rule | Statement | Example |
|---|---|---|
| 1 | Free element (any allotrope) = 0 | Cu, O2, N2, P4, S8 → all atoms are 0 |
| 2 | Monatomic ion = its charge | Na⁺ = +1; Cl⁻ = −1; Al3+ = +3 |
| 3 | Hydrogen = +1 (usually); −1 in metal hydrides | HCl → H is +1; NaH, CaH2 → H is −1 |
| 4 | Oxygen = −2 (usually). Exceptions: peroxides (−1), OF2 (+2), superoxides (−½), O2F2 (+1) | H2O: O = −2; H2O2: O = −1; KO2: O = −½ |
| 5 | Fluorine = −1 always (most electronegative element) | HF, ClF3, OF2: F = −1 |
| 6 | Alkali metals = +1, alkaline earths = +2 in compounds | NaCl: Na = +1; CaO: Ca = +2 |
| 7 | Sum of all ON in a neutral molecule = 0; in a polyatomic ion = charge on ion | H2SO4: 2(+1) + x + 4(−2) = 0 ⇒ x = +6 for S |
Tip: Work from the outside in — assign known ON first (H, O, halogens, alkali/alkaline-earth metals), then solve for the unknown using the sum rule.
Fractional Oxidation Numbers
Oxidation numbers are a book-keeping device, so they are free to come out as fractions when the same element sits in two different electronic environments in the same formula unit.
Fe3O4: 3·x + 4·(−2) = 0 ⇒ x = +8/3 (really Fe is one +2 and two +3)
Na2S2O3 (thiosulphate): average S = +2 (two different S atoms)
Br3O8: average Br = +16/3
Na2S2O3 (thiosulphate): average S = +2 (two different S atoms)
Br3O8: average Br = +16/3
Stock Notation
To avoid ambiguity, IUPAC uses Roman numerals in parentheses to show the oxidation state of a metal in its compound:
CuCl ⇒ copper(I) chloride
CuCl2 ⇒ copper(II) chloride
FeCl3 ⇒ iron(III) chloride
MnO2 ⇒ manganese(IV) oxide
KMnO4 ⇒ potassium manganate(VII)
CuCl2 ⇒ copper(II) chloride
FeCl3 ⇒ iron(III) chloride
MnO2 ⇒ manganese(IV) oxide
KMnO4 ⇒ potassium manganate(VII)
Worked Examples — Finding Oxidation Numbers L3 Apply
Example 7.6 — S in H2SO4
Let S = x. Then 2(+1) + x + 4(−2) = 0 ⇒ 2 + x − 8 = 0 ⇒ x = +6.
Example 7.7 — Mn in KMnO4
K = +1, each O = −2. 1 + x + 4(−2) = 0 ⇒ x = +7. (Hence the Stock name manganate(VII).)
Example 7.8 — Cr in K2Cr2O7
2(+1) + 2x + 7(−2) = 0 ⇒ 2 + 2x − 14 = 0 ⇒ 2x = +12 ⇒ x = +6 (each Cr).
Example 7.9 — Nitrogen in NH3, N2O, NO2, HNO3
NH3: x + 3(+1) = 0 ⇒ x = −3
N2O: 2x + (−2) = 0 ⇒ x = +1
NO2: x + 2(−2) = 0 ⇒ x = +4
HNO3: (+1) + x + 3(−2) = 0 ⇒ x = +5
Example 7.10 — S in S2O32− and S4O62−
S2O32−: 2x + 3(−2) = −2 ⇒ avg S = +2.
S4O62−: 4x + 6(−2) = −2 ⇒ avg S = +5/2.
Interactive: Oxidation Number Calculator
Pick a compound and see the oxidation number of the unknown element computed step by step.
Choose a compound and click Calculate.
7.2.2 Types of Redox Reactions
Armed with oxidation numbers, every redox reaction falls neatly into one of four classes.
(a) Combination Reactions
Two substances combine to form one. If at least one reactant is a free element, it is always redox.
2 Mg(s) + O₂(g) → 2 MgO(s) [Mg: 0 → +2; O: 0 → −2]
C(s) + O₂(g) → CO₂(g)
3 Mg(s) + N₂(g) → Mg₃N₂(s)
H₂(g) + Cl₂(g) → 2 HCl(g)
C(s) + O₂(g) → CO₂(g)
3 Mg(s) + N₂(g) → Mg₃N₂(s)
H₂(g) + Cl₂(g) → 2 HCl(g)
(b) Decomposition Reactions
A compound breaks up, usually on heating or electrolysis, into simpler substances — often the reverse of a combination.
2 H₂O(l) → 2 H₂(g) + O₂(g) (electrolysis)
2 KClO₃(s) →Δ→ 2 KCl(s) + 3 O₂(g)
2 NaH(s) →Δ→ 2 Na(s) + H₂(g)
2 KClO₃(s) →Δ→ 2 KCl(s) + 3 O₂(g)
2 NaH(s) →Δ→ 2 Na(s) + H₂(g)
Not all decompositions are redox! CaCO3 → CaO + CO2 is a decomposition, but no ON changes — C stays +4, Ca stays +2. So it is non-redox.
(c) Displacement Reactions
A more reactive element kicks a less reactive one out of its compound.
Metal displacement
Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)
Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s)
2 Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2 Fe(l) (Thermite — red-hot molten iron — used to weld railway tracks)
Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s)
2 Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2 Fe(l) (Thermite — red-hot molten iron — used to weld railway tracks)
Non-metal displacement
Cl₂(g) + 2 KBr(aq) → 2 KCl(aq) + Br₂(l) (halogen displacement)
2 Na(s) + 2 H₂O(l) → 2 NaOH(aq) + H₂(g) (H₂ displacement by active metal)
Ca(s) + 2 H₂O(l) → Ca(OH)₂(aq) + H₂(g)
2 Na(s) + 2 H₂O(l) → 2 NaOH(aq) + H₂(g) (H₂ displacement by active metal)
Ca(s) + 2 H₂O(l) → Ca(OH)₂(aq) + H₂(g)
(d) Disproportionation Reactions L4 Analyse
Disproportionation: the same element in a compound undergoes both oxidation and reduction in the same reaction. The element must have an intermediate oxidation state that can go both up and down.
2 H₂O₂(aq) → 2 H₂O(l) + O₂(g)
O: −1 → −2 (reduced) and −1 → 0 (oxidised)
Cl₂(g) + 2 OH⁻(aq) → ClO⁻(aq) + Cl⁻(aq) + H₂O(l)
Cl: 0 → +1 (oxidised) and 0 → −1 (reduced)
3 MnO₄²⁻(aq) + 4 H⁺ → 2 MnO₄⁻(aq) + MnO₂(s) + 2 H₂O(l)
Mn: +6 → +7 (oxidised) and +6 → +4 (reduced)
P₄(s) + 3 OH⁻ + 3 H₂O → PH₃(g) + 3 H₂PO₂⁻
P: 0 → −3 and 0 → +1
O: −1 → −2 (reduced) and −1 → 0 (oxidised)
Cl₂(g) + 2 OH⁻(aq) → ClO⁻(aq) + Cl⁻(aq) + H₂O(l)
Cl: 0 → +1 (oxidised) and 0 → −1 (reduced)
3 MnO₄²⁻(aq) + 4 H⁺ → 2 MnO₄⁻(aq) + MnO₂(s) + 2 H₂O(l)
Mn: +6 → +7 (oxidised) and +6 → +4 (reduced)
P₄(s) + 3 OH⁻ + 3 H₂O → PH₃(g) + 3 H₂PO₂⁻
P: 0 → −3 and 0 → +1
Worked Examples — Classifying the Reaction L3 Apply
Example 7.11 — Classify each reaction
(i) 4 Fe + 3 O₂ → 2 Fe₂O₃; (ii) 2 HgO → 2 Hg + O₂; (iii) Cl₂ + 2 NaI → 2 NaCl + I₂; (iv) 2 NaOH + Cl₂ → NaCl + NaOCl + H₂O.
(i) Combination + redox (Fe 0→+3, O 0→−2).
(ii) Decomposition + redox (Hg +2→0, O −2→0).
(iii) Non-metal displacement (Cl displaces I).
(iv) Disproportionation (Cl 0 → +1 in OCl⁻ and 0 → −1 in Cl⁻).
Example 7.12 — Show that thermite is redox
2 Al + Fe₂O₃ → Al₂O₃ + 2 Fe.
Al: 0 → +3 (oxidised by 3 e⁻ each). Fe: +3 → 0 (reduced by 3 e⁻ each). Total electrons balance (2 × 3 = 6 on each side). Redox confirmed.
Example 7.13 — Identify the disproportionating element
3 Br₂ + 6 OH⁻ → 5 Br⁻ + BrO₃⁻ + 3 H₂O.
Br starts at 0. In Br⁻ it is −1 (reduced); in BrO₃⁻ it is +5 (oxidised). Same element, both directions ⇒ disproportionation.
Example 7.14 — A non-redox decomposition
CaCO₃ → CaO + CO₂.
Ca remains +2, C remains +4, O remains −2. No ON change — decomposition is present but it is not redox.
Example 7.15 — Pick the oxidising and reducing species in a disproportionation
Cl₂ + 2 OH⁻ → ClO⁻ + Cl⁻ + H₂O.
Cl₂ is acting both as oxidising agent (forms Cl⁻) and reducing agent (forms ClO⁻). This is characteristic of disproportionation.
Example 7.16 — Cu in CuCl vs CuCl2
CuCl: Cu = +1 (copper(I) chloride). CuCl2: Cu = +2 (copper(II) chloride). Stock notation removes ambiguity in names.
Activity 7.2 — The disproportionation of H2O2 L3 Apply
Materials: 3% hydrogen peroxide, a pinch of MnO2 (catalyst), small test tube, glowing wooden splint.
- Put 2 mL H2O2 in the test tube.
- Add the MnO2 and quickly cover with a thumb, then release.
- Insert a glowing splint just above the mouth.
Predict: which gas is evolved? What is the ON change of O?
Vigorous effervescence. The glowing splint bursts into flame — oxygen is being released. Oxygen's ON changes from −1 (in peroxide) to 0 (in O2) while another half goes to −2 in H2O: classic disproportionation, catalysed by MnO2.
Competency-Based Questions
Compound X is KClO3. On strong heating, it gives KCl and oxygen. A student is asked to analyse the oxidation states.
Q1. The oxidation number of Cl in KClO3 is
B. +5. 1 + x + 3(−2) = 0 ⇒ x = +5.
Q2. Classify 2 KClO3 → 2 KCl + 3 O2.
Decomposition + redox: Cl goes from +5 to −1 (reduced), O goes from −2 to 0 (oxidised). Note that even though the element Cl changes drastically, it is only one direction per element, so this is NOT a disproportionation.
Q3. Fill in the blank: In H2SO5 (Caro's acid, one peroxide O), oxidation number of S is ____.
+6. Treat one pair of O as peroxide (−1 each) and the rest as −2: 2(+1) + x + 4(−2) + 2(−1) = 0 ⇒ x = +8? Correction: with one peroxide linkage (two O at −1) and three normal O at −2: 2(+1) + x + 3(−2) + 2(−1) = 0 ⇒ x = +6.
Q4. True/False: 2 Na + Cl2 → 2 NaCl is a combination AND a redox reaction.
True. Na: 0 → +1; Cl: 0 → −1. Two elements combine to form one compound with ON changes.
Q5. (HOT) Predict whether 3 Cl2 + 6 NaOH(hot) gives disproportionation. Justify with oxidation states.
Yes. In hot, concentrated alkali: 3 Cl2 + 6 NaOH → 5 NaCl + NaClO3 + 3 H2O. Cl goes from 0 to −1 (in NaCl) and 0 to +5 (in NaClO3) — same element both directions.
Assertion–Reason Questions
Options: A both true, R explains A · B both true, R does not explain A · C A true R false · D A false R true.
A: The oxidation number of O in OF2 is +2.
R: Fluorine is more electronegative than oxygen, so each F is −1 and O must be +2.
A. Both true and R is exactly the reasoning for A.
A: All decomposition reactions are redox reactions.
R: A single compound always splits into simpler elements.
D. A is false: CaCO3 → CaO + CO2 is non-redox. R is also false. Net: A false, R false. If grading asks A/B/C/D strictly on the given options, this pair is "A false, R false" — closest match is D-style; in CBSE marking answer as A false and R false.
A: The reaction Cl2 + 2 OH⁻ → ClO⁻ + Cl⁻ + H2O is a disproportionation.
R: Chlorine (ON = 0) is oxidised to +1 and reduced to −1 in the same reaction.
A. Both correct and R is the explanation.
Frequently Asked Questions — Oxidation Number and Types of Redox Reactions
What is oxidation number and how is it different from valency?
Oxidation number (or oxidation state) is the apparent charge an atom would carry if all its bonds were ionic. It can be positive, negative, zero or even a fraction. Valency is the combining capacity, always a positive whole number representing the number of electrons gained, lost or shared. NCERT Class 11 Chemistry Chapter 7 explains that oxidation number tracks electron distribution in redox reactions, while valency tracks bonding capacity. For example, in CH₄, carbon has valency 4 but oxidation number −4; in CO₂, carbon has valency 4 and oxidation number +4.
What are the rules for assigning oxidation numbers?
NCERT Class 11 Chemistry Chapter 7 lists eight rules for assigning oxidation numbers: (1) free elements have oxidation number 0; (2) monoatomic ion = its charge; (3) sum of oxidation numbers in a neutral molecule = 0; in a polyatomic ion = its charge; (4) H is +1 (except in metal hydrides like NaH where it is −1); (5) O is −2 (except in peroxides −1, superoxides −½, OF₂ +2); (6) F is always −1; other halogens are −1 in halides; (7) alkali metals (Group 1) are +1; (8) alkaline earth metals (Group 2) are +2. Use these rules to find unknown oxidation numbers.
How do you calculate oxidation number in compounds?
To calculate oxidation number in NCERT Class 11 Chemistry Chapter 7 problems: (1) write the chemical formula; (2) apply known oxidation numbers from rules (H = +1, O = −2, etc.); (3) let unknown be x; (4) sum to zero for neutral compound or to charge for ion; (5) solve for x. Example: in MnO₄⁻, let Mn = x. x + 4(−2) = −1 → x = +7. In Cr₂O₇²⁻, 2x + 7(−2) = −2 → x = +6. In K₂Cr₂O₇, 2(+1) + 2x + 7(−2) = 0 → x = +6. Always verify the answer is reasonable.
Can oxidation numbers be fractional?
Yes, oxidation numbers can be fractional when an element exists in different oxidation states within the same compound (mixed oxidation states), but the average is reported as a fraction. Examples in NCERT Class 11 Chemistry Chapter 7: in Fe₃O₄ (magnetite), the average oxidation number of Fe is +8/3 because it contains one Fe²⁺ and two Fe³⁺ ions; in C₃O₂, average oxidation number of C is +4/3; in Na₂S₄O₆ (tetrathionate), average oxidation number of S is +2.5. Fractional values indicate non-equivalent atom positions in the structure.
What is the IUPAC Stock notation for oxidation states?
IUPAC Stock notation indicates the oxidation state of an element using Roman numerals in parentheses immediately after the element name, particularly useful for elements showing multiple oxidation states. NCERT Class 11 Chemistry Chapter 7 examples: iron(II) chloride = FeCl₂; iron(III) chloride = FeCl₃; copper(I) oxide = Cu₂O; copper(II) oxide = CuO; manganese(VII) in KMnO₄. Stock notation removes ambiguity when an element has multiple common oxidation states. For elements with only one common state (Na, K, Ca, Al), Stock notation is optional.
How is the classification of redox reactions done using oxidation numbers?
Using oxidation numbers, NCERT Class 11 Chemistry Chapter 7 classifies redox reactions as: (1) combination — two substances combine and one or both undergo oxidation state change (4Na + O₂ → 2Na₂O); (2) decomposition — single compound breaks down with oxidation state change (2KClO₃ → 2KCl + 3O₂); (3) displacement — one element displaces another (Zn + CuSO₄ → ZnSO₄ + Cu); (4) disproportionation — same element both oxidised and reduced (2H₂O₂ → 2H₂O + O₂); (5) comproportionation — two species of the same element in different oxidation states combine to form one species in an intermediate state (KIO₃ + 5KI + 6HCl → 3I₂ + 6KCl + 3H₂O).
AI Tutor
Chemistry Class 11 Part II – NCERT (2025-26)
Ready