Key Fact
Potassium dichromate K₂Cr₂O₇
Powerful oxidising agent in acidic medium. Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. E° = +1.33 V. Used in titrations, leather tanning, oxidation of alcohols.
-- NCERT Class 12 Chemistry, Ch. 4, p. 106K₂Cr₂O₇ KMnO₄
8 MCQs9-step worked example
Source: NCERT The d and f Block ElementsPYQ coverage: NEET 2020, 2022, 2026Official key: NTA-verifiedLast updated: 26 Sep 2026
K₂Cr₂O₇ KMnO₄, explained for NEET
The single trap that costs marks on K₂Cr₂O₇/KMnO₄ questions: assuming Mn²⁺ is the reduction product of KMnO₄ regardless of medium.
KMnO₄ is a versatile oxidising agent, but its reduction product changes with pH:
| Medium | Product | Mn oxidation state | Electrons gained |
|---|---|---|---|
| Acidic (H₂SO₄) | Mn²⁺ (colourless) | +2 | 5 |
| Neutral / weakly basic | MnO₂ (brown ppt) | +4 | 3 |
| Strongly alkaline | MnO₄²⁻ (green, manganate) | +6 | 1 |
Preparation of KMnO₄: Pyrolusite (MnO₂) is fused with KOH in air → K₂MnO₄ (manganate). Manganate is then oxidised electrolytically or disproportionates in acidic/neutral solution → KMnO₄ (NCERT Class 12 Chemistry Chapter 4, page 106).
Preparation of K₂Cr₂O₇: Chromite ore (FeCr₂O₄) is fused with Na₂CO₃ in air → Na₂CrO₄ (yellow chromate). Acidification converts chromate → dichromate (Cr₂O₇²⁻, orange). Treatment with KCl gives K₂Cr₂O₇ (NCERT Class 12 Chemistry Chapter 4, page 105).
K₂Cr₂O₇ as oxidising agent in acidic medium: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. The orange-to-green colour change is a standard titration endpoint.
Watch-out: When a NEET stem says "KMnO₄ in alkaline medium" or doesn't specify acid, do NOT default to Mn²⁺. Read the medium first, then assign the product.
Can you answer these K₂Cr₂O₇ KMnO₄ 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
The final product of KMnO₄ reduction in strongly alkaline medium is:
Show answer and why every option is right or wrong
Answer: C. In strongly alkaline medium, KMnO₄ gains only 1 electron: MnO₄⁻ → MnO₄²⁻ (manganate, green). NCERT Class 12 Chemistry Chapter 4, page 107.
Why A is wrong: A is wrong because Mn²⁺ is the product in acidic medium only (5e⁻ transfer). Selecting Mn²⁺ regardless of medium is the core trap of this topic.
Why B is wrong: B is wrong because MnO₂ forms in neutral or weakly basic medium (3e⁻ transfer), not in strongly alkaline conditions.
Why D is wrong: D is wrong because Mn₂O₇ is the anhydride of permanganic acid, not a reduction product of KMnO₄ in any medium.
MCQ 2Easy RecallPractice
In the preparation of K₂Cr₂O₇ from chromite ore, the intermediate yellow compound formed after fusion with Na₂CO₃ is:
Show answer and why every option is right or wrong
Answer: A. Fusion of FeCr₂O₄ with Na₂CO₃ in air gives Na₂CrO₄ (sodium chromate, yellow). Acidification then converts it to dichromate. NCERT Class 12 Chemistry Chapter 4, page 105.
Why B is wrong: B is wrong because Na₂Cr₂O₇ (dichromate) forms only after acidification of the chromate — it is not the direct fusion product.
Why C is wrong: C is wrong because CrO₃ is chromium trioxide (chromic acid anhydride), formed by treating dichromate with conc. H₂SO₄, not during the ore-fusion step.
Why D is wrong: D is wrong because K₂CrO₄ would require potassium in the fusion mixture. The industrial process uses Na₂CO₃, giving the sodium salt first.
MCQ 3Easy RecallPractice
KMnO₄ is prepared industrially from K₂MnO₄. The conversion of manganate to permanganate is achieved by:
Show answer and why every option is right or wrong
Answer: B. K₂MnO₄ is oxidised to KMnO₄ either electrolytically (at anode) or by disproportionation in acidic/neutral solution: 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O. NCERT Class 12 Chemistry Chapter 4, page 106.
Why A is wrong: A is wrong because H₂ is a reducing agent; converting Mn(+6) to Mn(+7) requires oxidation, not reduction.
Why C is wrong: C is wrong because heating KMnO₄ with conc. HCl produces Cl₂ (KMnO₄ acting as oxidiser). This doesn't describe the manganate-to-permanganate preparation step.
Why D is wrong: D is wrong because Na₂CO₃ is used in the chromite-ore fusion process for K₂Cr₂O₇ preparation, not for the permanganate synthesis.
MCQ 4Direct ApplicationPractice
In acidic medium, the equivalent weight of KMnO₄ (molar mass = 158 g/mol) acting as an oxidising agent is:
Show answer and why every option is right or wrong
Answer: B. In acidic medium, Mn goes from +7 to +2, gaining 5 electrons. Equivalent weight = Molar mass / n-factor = 158/5 = 31.6 g/eq.
Why A is wrong: A is wrong because n = 3 corresponds to neutral medium (MnO₄⁻ → MnO₂). Picking 3 when the stem specifies acidic medium is the medium-dependence trap.
Why C is wrong: C is wrong because n = 1 corresponds to strongly alkaline medium (MnO₄⁻ → MnO₄²⁻). The stem specifies acidic medium.
Why D is wrong: D is wrong because n = 7 would mean reduction from Mn(+7) to Mn(0), which does not occur in any standard KMnO₄ reaction.
MCQ 5Direct ApplicationPractice
K₂Cr₂O₇ in acidic solution oxidises Fe²⁺ to Fe³⁺. The colour change observed in the solution during this reaction is:
Show answer and why every option is right or wrong
Answer: D. Cr₂O₇²⁻ (orange) is reduced to Cr³⁺ (green) in acidic medium: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. The solution changes from orange to green. NCERT Class 12 Chemistry Chapter 4, page 106.
Why A is wrong: A is wrong because this reverses the direction. The dichromate (orange) is consumed as oxidiser, producing Cr³⁺ (green) — not the other way around.
Why B is wrong: B is wrong because 'colourless to brown' describes the colour change when MnO₂ precipitates from KMnO₄ in neutral medium — this is a dichromate reaction in acid.
Why C is wrong: C is wrong because yellow is the colour of chromate (CrO₄²⁻) in basic medium, not dichromate in acid. Violet is associated with Cr³⁺ only in certain crystal-field contexts, not in this acidic titration.
MCQ 6Direct ApplicationPractice
The n-factor of KMnO₄ when it reacts in neutral medium is:
Show answer and why every option is right or wrong
Answer: D. In neutral medium, MnO₄⁻ → MnO₂: Mn goes from +7 to +4, gaining 3 electrons. Therefore n-factor = 3.
Why A is wrong: A is wrong because n = 1 applies only in strongly alkaline medium (MnO₄⁻ → MnO₄²⁻). Confusing alkaline with neutral is a common error.
Why B is wrong: B is wrong because there is no standard KMnO₄ reaction with n = 2. This doesn't correspond to any recognized product.
Why C is wrong: C is wrong because n = 5 applies in acidic medium (MnO₄⁻ → Mn²⁺). The stem specifies neutral medium, where the product is MnO₂, not Mn²⁺ — the core medium-dependence trap.
MCQ 7Concept TrapPractice
A student performing a titration adds KMnO₄ to an alkaline solution of Na₂SO₃ and expects the endpoint to appear when 5 moles of electrons are transferred per mole of KMnO₄. The student's calculation will give:
Show answer and why every option is right or wrong
Answer: A. In alkaline medium, KMnO₄ gains only 1 electron (→ MnO₄²⁻). If the student uses n = 5 (acidic assumption), they underestimate how much KMnO₄ is needed per mole of reductant, calculating a lower titre than actual.
Why B is wrong: B is wrong because the student assumes each KMnO₄ does more work (5e⁻) than it actually does (1e⁻), so the calculated volume needed is less, not more.
Why C is wrong: C is wrong because using the wrong n-factor (5 instead of 1) introduces a systematic error in the equivalence calculation.
Why D is wrong: D is wrong because KMnO₄'s behaviour is strongly medium-dependent — this is the defining property of this topic.
MCQ 8CalculationPractice
20 mL of 0.02 M KMnO₄ in acidic medium exactly reacts with a solution of oxalic acid (C₂O₄²⁻ → CO₂). The millimoles of oxalic acid that reacted are:
Show answer and why every option is right or wrong
Answer: C. KMnO₄ in acid: n-factor = 5. Oxalic acid: C₂O₄²⁻ → 2CO₂, losing 2 electrons, so n-factor = 2. Milliequivalents of KMnO₄ = 20 × 0.02 × 5 = 2.0. At equivalence: meq(oxalic) = 2.0. Millimoles of oxalic acid = 2.0/2 = 1.0 mmol.
Why A is wrong: A is wrong because 0.4 = 20 × 0.02 = millimoles of KMnO₄ without accounting for n-factors. Forgetting that n-factor multiplies in equivalence calculations gives this distractor.
Why B is wrong: B is wrong because 0.8 results from multiplying the KMnO₄ millimoles by oxalate's n-factor instead of going through equivalents: 0.4 × 2 = 0.8. The correct route is 0.4 mmol × 5 = 2.0 meq of KMnO₄, then 2.0/2 = 1.0 mmol of oxalic acid.
Why D is wrong: D is wrong because 2.0 results from using n = 5 for KMnO₄ correctly but then forgetting to divide by oxalic acid's n-factor (2), i.e., treating meq directly as mmol.
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How do you solve a K₂Cr₂O₇ KMnO₄ question? A worked example
- 1
Given
25 mL of 0.04 M KMnO₄ solution in acidic medium is used to oxidise Fe²⁺ to Fe³⁺.
- 2
Required
Find the mass of Fe²⁺ (as FeSO₄, M = 152 g/mol) that reacts completely.
- 3
Concept
At the equivalence point: milliequivalents of oxidiser = milliequivalents of reductant. The medium is acidic, so KMnO₄ → Mn²⁺ (n = 5). Fe²⁺ → Fe³⁺ (n = 1).
- 4
Formula
meq(KMnO₄) = V(mL) × M × n-factor
meq(Fe²⁺) = mmol(FeSO₄) × 1
At equivalence: meq(KMnO₄) = meq(Fe²⁺) - 5
Substitution
meq(KMnO₄) = 25 × 0.04 × 5 = 5.0
mmol(FeSO₄) = 5.0 / 1 = 5.0 mmol - 6
Calculation
Mass of FeSO₄ = 5.0 × 10⁻³ mol × 152 g/mol = 0.76 g
- 7
Final answer
Mass of FeSO₄ = 0.76 g
Note on exact values: The n-factors (5 and 1) are exact integers derived from the balanced half-reactions and do not limit significant figures. - 8
Common trap
If you use n = 3 (neutral medium product MnO₂) instead of n = 5 (acidic medium product Mn²⁺), you get meq = 25 × 0.04 × 3 = 3.0, leading to mass = 0.456 g — a 40% underestimate. Always confirm the medium before assigning the n-factor.
- 9
Similar NEET-style question
"What volume of 0.02 M KMnO₄ in neutral medium is needed to oxidise 10 mL of 0.1 M Na₂C₂O₄?" (Key change: neutral medium → n = 3 for KMnO₄.)
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What to remember before solving K₂Cr₂O₇ KMnO₄ questions
Key Fact
Potassium permanganate KMnO₄
Stronger oxidising agent. Acidic: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (E° = +1.51 V). Neutral/basic: forms MnO₂ (Mn⁺⁴). Self-indicating purple.
-- NCERT Class 12 Chemistry, Ch. 4, p. 107Where do students lose marks on K₂Cr₂O₇ KMnO₄?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
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.
Root cause: concept gap
Correction
Acidic: → Mn²⁺ (5e⁻). Neutral/weakly basic: → MnO₂ (3e⁻). Strongly basic: → MnO₄²⁻ (1e⁻).
More in The d and f Block Elements: 1 exam trap or mistake · 3 formulas · 2 question patterns from its other lessons.
K₂Cr₂O₇ KMnO₄ questions from past NEET papers
3 questions from NEET 2020, 2022, 2026. Answers verified against NTA official keys.
NEET 2026
The green paramagnetic species formed by heating KMnO₄ at 513 K is
1K₂MnO₄
2Mn₃O₄
3MnO
4KO₂
NTA Answer: Option 1(final)
NEET 2022
1+6 to +5
2+7 to +4
3+6 to +4
4+7 to +3
NTA Answer: Option 2(final)
NEET 2020
Identify the incorrect statement.
1The oxidation states of chromium in CrO₄²⁻ and Cr₂O₇²⁻ are not the same.
2Cr2+(d4 ) is a stronger reducing agent than Fe2+(d6 ) in water.
3The transition metals and their compounds are known for their catalytic activity due to their ability to adopt multiple oxidation states and to form complexes.
4Interstitial compounds are those that are formed when small atoms like H, C or N are trapped inside the crystal lattices of metals.
NTA Answer: Option 1(final)
All 15 past-paper questions from The d and f Block Elements →
How does NEET ask about K₂Cr₂O₇ KMnO₄?
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
Use KMnO4 or K2Cr2O7 in acidic/neutral/basic medium for oxidation. Different products by medium.
InterpretationMedium
Common distractors
uses acidic formula in basic medium
Assumes Mn²⁺ regardless of medium
Sources
NCERT refs: Class 12 Chemistry Chapter 4, p.106 | Class 12 Chemistry Chapter 4, p.105
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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