Oxidation Number

8 MCQs9-step worked example

Source: NCERT Redox ReactionsPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

Oxidation number assignment is a bookkeeping system — not a measure of actual charge on an atom. NEET regularly tests whether you can assign oxidation numbers correctly in compounds, ions, and polyatomic species. The confusion almost always comes from exceptions to the default rules, and from elements that take multiple oxidation states in the same compound.

The core rules (NCERT Class 11 Chemistry, Chapter 8, Part 2):

Free element: oxidation number = 0 (Na, O₂, P₄, S₈ — all zero).
Monoatomic ion: oxidation number = ion charge (Na⁺ = +1, Cl⁻ = −1).
Oxygen is −2, EXCEPT in peroxides (−1), superoxides (−1/2), OF₂ (+2), and O₂F₂ (+1).
Hydrogen is +1 with non-metals, −1 with metals (NaH, CaH₂).
Fluorine is always −1 (most electronegative element — no exceptions).
Sum rule: oxidation numbers in a neutral compound sum to 0; in a polyatomic ion, sum to ion charge.

Where aspirants lose marks:

Peroxide vs. oxide confusion. Seeing "Na₂O₂" and assigning O as −2 gives Na as +2 (wrong — Na is +1, O is −1 in the peroxide).
Fractional oxidation states. In Fe₃O₄, average iron oxidation state is +8/3. NEET may ask you to identify this or recognise that Fe₃O₄ is a mixed oxide (FeO + Fe₂O₃, so Fe is both +2 and +3).
Elements in multiple states. In Na₂S₄O₆ (tetrathionate), the four sulfur atoms are not all equivalent — two are 0, two are +5, but the average is +2.5. Questions may test whether you treat them as equivalent or identify the structural reality.
OF₂ and O₂F₂. Fluorine's −1 rule overrides oxygen's −2 default. In OF₂, oxygen is +2. Many aspirants default to O = −2 here and get a nonsensical fluorine value.

The method that doesn't fail: Start by assigning atoms with fixed oxidation states (F, alkali metals, alkaline earths). Then assign H and O using their rules and exceptions. Solve for the unknown using the sum rule. Always verify the sum.

Practice MCQs

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

What is the oxidation number of each atom in a free element (e.g., O₂, P₄, S₈)?

MCQ 2Easy RecallPractice

What is the oxidation number of fluorine in all its compounds?

MCQ 3Easy RecallPractice

What is the oxidation number of hydrogen in calcium hydride (CaH₂)?

MCQ 4Direct ApplicationPractice

What is the oxidation number of oxygen in Na₂O₂?

MCQ 5Direct ApplicationPractice

What is the oxidation number of oxygen in OF₂?

MCQ 6Direct ApplicationPractice

What is the oxidation number of Cr in K₂Cr₂O₇?

MCQ 7Concept TrapPractice

Fe₃O₄ can be written as FeO·Fe₂O₃. What are the oxidation states of iron in Fe₃O₄?

MCQ 8CalculationPractice

In the reaction 2KMnO₄ + 3H₂SO₄ + 5H₂C₂O₄ → 2MnSO₄ + K₂SO₄ + 8H₂O + 10CO₂, what is the change in oxidation number of Mn per atom?

Worked Example

1
Given
- Compound: Na₂O₂ - Na is an alkali metal (Group 1) - The compound is a peroxide (contains O₂²⁻ ion)
2
Required
Oxidation number of Na and O in Na₂O₂.
3
Concept
Alkali metals always have oxidation number +1. Oxygen is −1 in peroxides (not the default −2). The sum of all oxidation numbers in a neutral compound must equal zero.
4
Formula / Rule
Sum rule: Σ(oxidation numbers × atom count) = 0 for a neutral compound.
5
Assignment
- Na: +1 (alkali metal rule, no exceptions) - O: −1 (peroxide exception to the −2 default)
6
Verification
Sum = 2(+1) + 2(−1) = +2 − 2 = 0 ✓ Note: The subscript 2 on both Na and O are exact counting integers — they do not affect any significant-figure analysis.
7
Final answer
Na = +1, O = −1 in Na₂O₂.
8
Common trap
Assigning O = −2 (the default) gives Na = +2, which is impossible for sodium — Na is ALWAYS +1 in compounds. If your sum rule gives Na anything other than +1, re-check whether the compound is a peroxide or superoxide.
9
Similar NEET-style question
Assign oxidation numbers to all elements in KO₂ (potassium superoxide). [Answer: K = +1, O = −1/2. Verification: +1 + 2(−1/2) = 0 ✓] ---

Before solving, remember these

Law

Rules for assigning oxidation number

1) Free element: 0. 2) Monatomic ion: charge. 3) O: -2 (peroxide -1, OF₂ +2). 4) H: +1 (-1 in metal hydride). 5) Sum = charge of species. 6) Group I/II metals: fixed +1/+2.

-- NCERT, p. 6

Formulas

Cell EMF

E_{cell}^{\circ} = E_{cath}^{\circ} - E_{anode}^{\circ}

Both as reduction potentials. E°_cell > 0 → spontaneous.

Symbol	Quantity	SI Unit
E°_cell	standard cell EMF	V
E°_red	reduction potential	V

Valid when

Standard conditions (1 M, 1 bar, 298 K)
Both half-reactions as reductions

Faraday's law of electrolysis

m = \frac{M I t}{n F}

Mass deposited at electrode. M = molar mass; I = current; t = time; n = electrons per ion.

Symbol	Quantity	SI Unit
m	mass deposited	g
M	molar mass	g/mol
I	current	A
t	time	s
n	electrons per ion	-
F	Faraday	C/mol

Valid when

Steady current
Single product

ΔG from EMF

Δ G = - n F E

Connection between thermodynamics and electrochemistry. F = 96485 C/mol.

Symbol	Quantity	SI Unit
ΔG	Gibbs energy change	J
n	electrons transferred	-
F	Faraday 96485	C/mol
E	cell EMF	V

Valid when

Single redox process

Molar conductivity

Λ_{m} = \frac{1000 κ}{C}

Molar conductivity from specific conductance. Increases with dilution as more ions are free.

Symbol	Quantity	SI Unit
Λ_m	molar conductivity	S cm^2/mol
κ	specific conductance	S/cm
C	molarity	mol/L

Valid when

Aqueous solution
Single electrolyte

Nernst equation

E = E^{\circ} - \frac{0.0591}{n} lo g Q

Cell potential at non-standard conditions. n = electrons transferred. At equilibrium E=0, Q=K.

Symbol	Quantity	SI Unit
E	cell potential	V
E°	standard	V
n	electrons	-
Q	reaction quotient	-

Valid when

298 K (else use RT/F)
Single redox process

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

Trap

Nernst n electrons cumulative error

Category: Negative Marking

Multi-step Nernst problem: identify electrons, write Q correctly, plug into 0.0591/n. Each sub-step has factor errors.

When it triggers

Cell EMF problem at non-standard conditions.

How to avoid

Step-by-step: (1) write balanced redox; (2) count n electrons; (3) compute Q from concentrations; (4) plug into Nernst. Verify by checking limits: at standard conditions Q=1, log Q=0, E=E°.

Trap

KMnO4 oxidation product depends on medium

Category: Inorganic Exception

Student assumes Mn²⁺ is the product regardless of medium. Acidic: → Mn²⁺ (5e⁻). Neutral/weakly basic: → MnO₂ (3e⁻). Strongly basic: → MnO₄²⁻ (1e⁻).

When it triggers

Question gives KMnO4 oxidation in unspecified or specific medium.

How to avoid

Always check medium. In acidic: Mn(+7) → Mn(+2). In neutral: → Mn(+4) (MnO₂). In basic: → Mn(+6) (manganate). The number of electrons (n) in Nernst calculations depends accordingly.

Mistake

Uses n=1 in Faraday's law for ions like Cu²⁺ (actual n=2) or Al³⁺ (n=3).

Root cause: concept gap

Correction

n = number of electrons per ion to deposit. Cu²⁺ + 2e⁻ → Cu: n=2. Al³⁺ + 3e⁻: n=3. m = MIt/(nF).

Mistake

Uses incorrect electron count n in Nernst equation.

Root cause: formula misuse

Correction

n = electrons transferred per balanced redox equation. For Cu²⁺ + 2e⁻ → Cu: n=2. For Mn(VII) → Mn(II): n=5.

Past Year Questions

17 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.

NEET 2025

If the rate constant of a reaction is 0.03 s–1, how much time does it take for 7.2 mol L–1 concentration of the reactant to get reduced to 0.9 mol L–1? (Given: log 2 = 0.301)

121.0 s

269.3 s

323.1 s

4210 s

NTA Answer: Option 2(final)

NEET 2025

Consider the following compounds : KO , H O and H SO 2 2 2 2 4 The oxidation state of the underlined elements in them are, respectively,

1+4, –4, and +6

2+1, –1, and +6

3+2, –2, and +6

4+1, –2, and +4

NTA Answer: Option 2(final)

NEET 2025

Match List-I with List-II List-I List-II (Example) (Type of Solution) A. Humidity I. Solid in solid B. Alloys II. Liquid in gas C. Amalgams III. Solid in gas D. Smoke IV. Liquid in solid Choose the correct answer from the options given below:

1A-III, B-II, C-I, D-IV

2A-II, B-IV, C-I,D-III

3A-II, B-I, C-IV, D-III

4A-III, B-I, C-IV, D-II

NTA Answer: Option 3(final)

NEET 2024Revised key

Given below are two statements: Statement I : The boiling point of three isomeric pentanes follows the order n-pentane > isopentane > neopentane Statement II : When branching increases, the molecule attains a shape of sphere. This results in smaller surface area for contact, due to which the intermolecular forces between the spherical molecules are weak, thereby lowering the boiling point. In the light of the above statements, choose the most appropriate answer from the options given below:

1Both Statement I and Statement II are correct.

2Both Statement I and Statement II are incorrect.

3Statement I is correct but Statement II is incorrect.

4Statement I is incorrect but Statement II is correct.

NTA Answer: Option 1(revised_final)

NEET 2024Revised key

Match List I with List II. List I List II (Conversion) (Number of Faraday required) A. 1 mol of H O to O I. 3F 2 2 B. 1 mol of MnO− 4 to Mn2+ II. 2F C. 1.5 mol of Ca from molten CaCl III. 1F 2 D. 1 mol of FeO to Fe O IV. 5F 2 3 Choose the correct answer from the options given below:

1A-II, B-IV, C-I, D-III

2A-III, B-IV, C-I, D-II

3A-II, B-III, C-I, D-IV

4A-III, B-IV, C-II, D-I

NTA Answer: Option 1(revised_final)

NEET 2024Revised key

Mass in grams of copper deposited by passing 9.6487 A current through a voltmeter containing copper sulphate solution for 100 seconds is (Given : Molar mass of Cu : 63 g mol–1, 1 F = 96487 C)

13.15 g

20.315 g

331.5 g

40.0315 g

NTA Answer: Option 2(revised_final)

NEET 2024Revised key

The products A and B obtained in the following reactions, respectively, are 3ROH + PCl → RCl + A 3 ROH + PCl → RCl + HCl + B 5

1POCl and H PO 3 3 3

2POCl and H PO 3 3 4

3H PO and POCl 3 4 3

4H PO and POCl 3 3 3

NTA Answer: Option 4(revised_final)

NEET 2023

The conductivity of centimolar solution of KCl at 25°C is 0.0210 ohm–1 cm–1 and the resistance of the cell containing the solution at 25°C is 60 ohm. The value of cell constant is

13.28 cm–1

21.26 cm–1

33.34 cm–1

41.34 cm–1

NTA Answer: Option 2(final)

NEET 2023

The number of  bonds,  bonds and lone pair of electrons in pyridine, respectively are:

112, 3, 0

211, 3, 1

312, 2, 1

411, 2, 0

NTA Answer: Option 2(final)

NEET 2022

At 298 K, the standard electrode potentials of Cu2+ / Cu, Zn2+ / Zn, Fe2+ / Fe and Ag+ / Ag are 0.34 V, –0.76 V, –0.44 V and 0.80 V, respectively. On the basis of standard electrode potential, predict which of the following reaction cannot occur?

12CuSO (aq) + 2Ag(s) → 2Cu(s) + Ag SO (aq) 4 2 4

2CuSO (aq) + Zn(s) → ZnSO (aq) + Cu(s) 4 4

3CuSO (aq) + Fe(s) → FeSO (aq) + Cu(s) 4 4

4FeSO (aq) + Zn(s) → ZnSO (aq) + Fe(s) 4 4

NTA Answer: Option 1(final)

NEET 2022

Which of the following statement is not correct about diborane?

1Both the Boron atoms are sp2 hybridised.

2There are two 3-centre-2-electron bonds.

3The four terminal B-H bonds are two centre two electron bonds.

4The four terminal Hydrogen atoms and the two Boron atoms lie in one plane.

NTA Answer: Option 1(final)

NEET 2022

Given below are half cell reactions: - 13 - M n O −4 + 8 H + + 5 e − → M n 2 + + 4 H 2 O E ºM n 2 + /M n O −4 = − 1 .5 1 0 V 1 2 O 2 + 2 H + + 2 e − → H 2 O Eº =+1.223V O2/H2O Will the permanganate ion, M n O –4 liberate O from water in the presence of an acid? 2

1No, because E cell =−2.733V

2Yes, because E c e ll = + 0 .2 8 7 V

3No, because E c e ll = – 0 .2 8 7 V

4Yes, because E c e ll = + 2 .7 3 3 V

NTA Answer: Option 2(final)

NEET 2022

If radius of second Bohr orbit of the He+ ion is 105.8 pm, what is the radius of third Bohr orbit of Li2+ ion?

1158.7 Å

2158.7 pm

315.87 pm

41.587 pm

NTA Answer: Option 2(final)

NEET 2021

Which one among the following is the correct option for right relationship between C and C for one mole P V of ideal gas?

1C = RC V P

2C + C = R P V

3C – C = R P V

4C = RC P V

NTA Answer: Option 3(final)

NEET 2021

The molar conductance of NaCl, HCl and CH COONa 3 at infinite dilution are 126.45, 426.16 and cm2 mol–1 91.0 S respectively. The molar conductance of CH COOH at infinite dilution is. 3 Choose the right option for your answer. cm2 mol–1

1540.48 S cm2 mol–1

2201.28 S cm2 mol–1

3390.71 S cm2 mol–1

4698.28 S

NTA Answer: Option 3(final)

NEET 2021

Which one of the following polymers is prepared by addition polymerisation?

1Dacron

2Teflon

3Nylon-66

4Novolac

NTA Answer: Option 2(final)

NEET 2021

The molar conductivity of 0.007 M acetic acid is 20 cm2 mol–1. S What is the dissociation constant of acetic acid? Choose the correct option. Λ° =350Scm2 mol−1  H+   Λ° mol−1 =50Scm2  CHCOO−  3 2.50×10−5 molL−1

11.75×10−4 molL−1

22.50×10−4 molL−1

31.75×10−5 molL−1

4Answer

NTA Answer: Option 4(final)

How NEET usually asks this

Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.

Compute Λ_m from κ and C. Or compute Λ°_m of weak electrolyte from strong electrolyte values via Kohlrausch.

Direct ApplicationMedium

Common distractors

inverts c multiplier

Uses Λ = κC instead of κ/C

forgets 1000 factor

Drops 1000 in unit conversion

Apply Nernst equation to compute non-standard cell EMF given Q. n = electrons transferred.

Multi StepMedium

Common distractors

wrong n electrons

Uses incorrect electron count from half-reactions

Test yourself on this topic with real past-paper questions:

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