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. 20K2cr2o7 Kmno4
8 MCQs9-step worked example
Source: NCERT The d and f Block ElementsPYQ coverage: NEET 2021, 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026
Lesson
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 22).
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 20).
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.
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
The final product of KMnO₄ reduction in strongly alkaline medium is:
MCQ 2Easy RecallPractice
In the preparation of K₂Cr₂O₇ from chromite ore, the intermediate yellow compound formed after fusion with Na₂CO₃ is:
MCQ 3Easy RecallPractice
KMnO₄ is prepared industrially from K₂MnO₄. The conversion of manganate to permanganate is achieved by:
MCQ 4Direct ApplicationPractice
In acidic medium, the equivalent weight of KMnO₄ (molar mass = 158 g/mol) acting as an oxidising agent is:
MCQ 5Direct ApplicationPractice
K₂Cr₂O₇ in acidic solution oxidises Fe²⁺ to Fe³⁺. The colour change observed in the solution during this reaction is:
MCQ 6Direct ApplicationPractice
The n-factor of KMnO₄ when it reacts in neutral medium is:
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:
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:
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.) ---
Before solving, remember these
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. 22Formulas
Spin-only magnetic moment
Magnetic moment from n unpaired electrons. 1 unpaired: 1.73 BM; 5: 5.92 BM.
| Symbol | Quantity | SI Unit |
|---|---|---|
| n | unpaired electrons | - |
| mu | magnetic moment | Bohr magneton |
Valid when
- Spin-only contribution (no orbital contribution)
- Octahedral or tetrahedral complex
Spin-only magnetic moment for transition metal
Predicts paramagnetic moment of d-block ion. n unpaired electrons in d-orbitals.
| Symbol | Quantity | SI Unit |
|---|---|---|
| n | unpaired electrons | - |
| mu | magnetic moment | BM |
Valid when
- Spin-only contribution
- Octahedral or tetrahedral complex
- First-row d-block
Common oxidation states (first-row TM)
Catalogues common stable oxidation states across first-row transition metals.
| Symbol | Quantity | SI Unit |
|---|---|---|
| OS | oxidation state | - |
Valid when
- First-row d-block
- Common (not exotic) compounds
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.
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⁻).
Root cause: concept gap
Correction
Lanthanoid contraction (imperfect 4f shielding) makes 5d elements similar in size to 4d. Zr/Hf, Nb/Ta, Mo/W have nearly identical chemistry.
Past Year Questions
10 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.
NEET 2025
The correct order of decreasing basic strength of the given amines is:
1benzenamine > ethanamine > N-methylaniline > N-ethylethanamine
2N-methylaniline > benzenamine > ethanamine > N-ethylethanamine
3N-ethylethanamine > ethanamine > benzenamine > N-methylaniline
4N-ethylethanamine > ethanamine > N-methylaniline > benzenamine
NTA Answer: Option 4(final)
NEET 2024Revised key
1B and D only
2A and E only
3B and C only
4A and D only
NTA Answer: Option 1(revised_final)
NEET 2024Revised key
1d5 to d4 configuration
2d5 to d2 configuration
3d4 to d5 configuration
4d3 to d5 configuration
NTA Answer: Option 3(revised_final)
NEET 2024Revised key
The pair of lanthanoid ions which are diamagnetic is
1Ce4+ and Yb2+
2Ce3+ and Eu2+
3Gd3+ and Eu3+
4Pm3+ and Sm3+
NTA Answer: Option 1(revised_final)
NEET 2023
Which one of the following statements is correct?
1All enzymes that utilise ATP in phosphate transfer require Ca as the cofactor
2The bone in human body is an inert and unchanging substance
3Mg plays roles in neuromuscular function and interneuronal transmission
4The daily requirement of Mg and Ca in the human body is estimated to be 0.2-0.3 g
NTA Answer: Option 4(final)
NEET 2023
1B and D only
2C and D only
3B and C only
4A and E only
NTA Answer: Option 2(final)
NEET 2022
12.57
25.57
33.57
44.57
NTA Answer: Option 2(final)
NEET 2022
Gadolinium has a low value of third ionisation enthalpy because of
1high basic character
2small size
3high exchange enthalpy
4high electronegativity
NTA Answer: Option 3(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 2021
Tritium, a radioactive isotope of hydrogen, emits which of the following particles? (β–)
1Neutron (n)
2Beta (α) (γ)
3Alpha
4Gamma
NTA Answer: Option 2(final)
How NEET usually asks this
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
Effect of lanthanoid contraction on properties of 4d/5d transition metals. Closer atomic/ionic radii, similar properties (Zr/Hf, Nb/Ta).
RecallEasy
Common distractors
predicts large difference due to period shift
Expects 5d to be much larger than 4d
Predict highest stable oxidation state for transition metal. Mn: +7 in MnO4-. Fe: +3 stable. Cu: +2.
InterpretationMedium
Common distractors
misses d0 d10 stability
Doesn't account for closed-shell stability
Sources
NCERT refs: Class 12 Chemistry Chapter 4, p.22 | Class 12 Chemistry Chapter 4, p.20
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