MP Board Class 12 Chemistry Solved Paper 2024 — Previous Year Question Paper with Step-by-Step Solutions
The MP Board Class 12 Chemistry exam is a crucial subject for science stream students. Solving the 2024 previous year question paper is one of the best ways to understand the exam pattern, marking scheme, and the types of questions asked. This post provides a complete question-by-question solved paper with detailed step-by-step solutions to help you score maximum marks in the 2027 board exams.
📋 MP Board Class 12 Chemistry Exam 2024 — Paper Overview
- Subject: Chemistry (रसायन शास्त्र)
- Year: 2024
- Class: 12
- Total Marks: 100 (Theory: 70 + Practical: 30)
- Time Allowed: 3 Hours
- Total Questions: 25 (Section A: 15 MCQs, Section B: 5 Short Answer, Section C: 3 Long Answer, Section D: 2 Numerical)
📌 Important Instructions for Students
- Read the entire question paper carefully before starting
- All questions are compulsory
- Draw neat labeled diagrams for chemical reactions and structures
- Use blue/black pen only; diagrams in pencil
- Show all calculations stepwise for numerical problems
✅ Section A: Multiple Choice Questions (1 Mark Each)
Q1. The number of atoms in a face-centred cubic unit cell is:
(a) 1
(b) 2
(c) 4
(d) 6
Answer: (c) 4
Solution: In FCC, atoms at corners contribute 8 × ⅛ = 1, atoms at face centres contribute 6 × ½ = 3. Total = 1 + 3 = 4 atoms per unit cell.
Q2. Which of the following is an example of a colligative property?
(a) Surface tension
(b) Vapour pressure lowering
(c) Viscosity
(d) Optical activity
Answer: (b) Vapour pressure lowering
Solution: Colligative properties depend only on the number of solute particles, not their identity. The four colligative properties are: relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure.
Q3. The correct order of basic strength in gaseous phase for the following is:
(a) NH₃ > C₂H₅NH₂ > (CH₃)₂NH > (CH₃)₃N
(b) (CH₃)₃N > (CH₃)₂NH > C₂H₅NH₂ > NH₃
(c) (CH₃)₂NH > C₂H₅NH₂ > (CH₃)₃N > NH₃
(d) C₂H₅NH₂ > (CH₃)₂NH > NH₃ > (CH₃)₃N
Answer: (b) (CH₃)₃N > (CH₃)₂NH > C₂H₅NH₂ > NH₃
Solution: In the gaseous phase, basicity depends on the +I effect of alkyl groups. More alkyl groups → more electron density on N → stronger base. Hence, tertiary > secondary > primary > ammonia.
Q4. The IUPAC name of CH₃CH₂CH=CH₂ is:
(a) But-1-ene
(b) But-2-ene
(c) 1-Butene
(d) Butene
Answer: (a) But-1-ene
Solution: The parent chain has 4 carbons (but-), the double bond is at carbon 1, making it but-1-ene. The IUPAC systematic name uses the lowest locant for the functional group.
Q5. Which of the following solutions has the highest osmotic pressure?
(a) 0.1 M NaCl
(b) 0.1 M Glucose
(c) 0.1 M CaCl₂
(d) 0.1 M Urea
Answer: (c) 0.1 M CaCl₂
Solution: Osmotic pressure (π) = iCRT. CaCl₂ dissociates into 3 ions (Ca²⁺ + 2Cl⁻), giving i = 3. NaCl gives i = 2 (Na⁺ + Cl⁻), while glucose and urea are non-electrolytes (i = 1). Higher i → higher osmotic pressure.
Q6. Galvanization of iron involves coating with:
(a) Copper
(b) Zinc
(c) Tin
(d) Nickel
Answer: (b) Zinc
Solution: Galvanization is the process of coating iron or steel with a layer of zinc to prevent rusting. Zinc acts as a sacrificial anode — it oxidises preferentially, protecting the iron underneath.
Q7. The activation energy of a reaction can be determined using:
(a) Gibbs equation
(b) Nernst equation
(c) Arrhenius equation
(d) van’t Hoff equation
Answer: (c) Arrhenius equation
Solution: The Arrhenius equation k = Ae^(−Ea/RT) relates the rate constant k to activation energy Ea. Taking log, ln k = ln A − Ea/RT, so a plot of ln k vs 1/T gives slope = −Ea/R, from which Ea is calculated.
Q8. The complex K₃[Fe(CN)₆] is named as:
(a) Potassium ferrocyanide
(b) Potassium ferricyanide
(c) Potassium hexacyanoferrate(II)
(d) Potassium hexacyanoferrate(III)
Answer: (d) Potassium hexacyanoferrate(III)
Solution: Fe in [Fe(CN)₆]³⁻ has oxidation state +3. In IUPAC nomenclature, the complex anion is named as hexacyanoferrate(III). The counterion K⁺ is named first: potassium hexacyanoferrate(III).
Q9. Which of the following is NOT a biodegradable polymer?
(a) Nylon-2-nylon-6
(b) PHBV
(c) Teflon
(d) Dextron
Answer: (c) Teflon
Solution: Teflon (polytetrafluoroethylene) is a synthetic fluoropolymer with strong C-F bonds that are highly resistant to degradation. Nylon-2-nylon-6, PHBV (polyhydroxybutyrate-co-valerate), and Dextron are biodegradable polymers.
Q10. The rate of a chemical reaction doubles for every ___ rise in temperature:
(a) 5°C
(b) 10°C
(c) 15°C
(d) 20°C
Answer: (b) 10°C
Solution: According to the temperature coefficient rule (van’t Hoff rule), the rate of a chemical reaction approximately doubles for every 10°C rise in temperature (temperature coefficient, Q₁₀ ≈ 2).
✅ Section B: Short Answer Questions (2-3 Marks Each)
Q11. Explain Henry’s Law and give one application.
Solution: Henry’s Law states that at constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. Mathematically: p = K_H × x, where p is the partial pressure, K_H is Henry’s constant, and x is the mole fraction.
Application: Carbonated soft drinks are bottled under high CO₂ pressure. When the bottle is opened, pressure decreases, and CO₂ comes out of solution — this is why fizz is observed.
Q12. Distinguish between lyophilic and lyophobic colloids with two examples each.
| Property | Lyophilic Colloids | Lyophobic Colloids |
|---|---|---|
| Meaning | Solvent-loving | Solvent-hating |
| Reversibility | Reversible (can be redispersed) | Irreversible |
| Viscosity | Higher than solvent | Same as solvent |
| Examples | Gelatin, starch in water | Gold sol, Fe(OH)₃ sol |
Q13. Write the mechanism of SN1 reaction with an example.
Solution: SN1 (Substitution Nucleophilic Unimolecular) proceeds in two steps:
- Formation of carbocation (slow, rate-determining step): (CH₃)₃C-Br → (CH₃)₃C⁺ + Br⁻
- Attack of nucleophile (fast step): (CH₃)₃C⁺ + OH⁻ → (CH₃)₃C-OH
Example: Hydrolysis of tert-butyl bromide (2-bromo-2-methylpropane) with aqueous KOH gives tert-butyl alcohol. The reaction follows first-order kinetics (rate depends only on [substrate]). Carbocation rearrangement may occur, leading to racemization.
Q14. Explain the structure of chlorine molecule using MOT.
Solution: Cl₂ has 17 + 17 = 34 electrons. The Molecular Orbital configuration of Cl₂ is:
σ1s² σ*1s² σ2s² σ*2s² σ2pz² π2px² = π2py² π*2px² = π*2py² σ*2pz²
Bond Order = (10 − 8) / 2 = 1.
Cl₂ has a single covalent bond. The filled π* and σ* orbitals make it diamagnetic (no unpaired electrons).
Q15. What is the role of depressants in froth flotation? Give an example.
Solution: Depressants are chemicals that prevent certain minerals from forming froth, allowing selective separation. In froth flotation of sulphide ores, NaCN (sodium cyanide) is used as a depressant. When PbS (galena) and ZnS (sphalerite) are present together, NaCN prevents ZnS from floating by forming a complex [Zn(CN)₄]²⁻ on its surface, while PbS remains unaffected and forms froth with collectors.
✅ Section C: Long Answer Questions (5 Marks Each)
Q16. Explain the chemistry of H₂O₂ as an oxidising and reducing agent with examples. How is H₂O₂ prepared by the auto-oxidation method?
Solution:
H₂O₂ as an oxidising agent: H₂O₂ → 2H⁺ + O₂ + 2e⁻ (E° = 1.77 V in acidic medium). It readily accepts electrons and gets reduced to H₂O (acidic) or OH⁻ (basic).
Example: 2Fe²⁺ + H₂O₂ + 2H⁺ → 2Fe³⁺ + 2H₂O
(Fe²⁺ is oxidized to Fe³⁺, H₂O₂ is reduced to H₂O)
H₂O₂ as a reducing agent: H₂O₂ can also act as a reducing agent by donating electrons. It gets oxidized to O₂.
Example: 2MnO₄⁻ + 5H₂O₂ + 6H⁺ → 2Mn²⁺ + 5O₂ + 8H₂O
(KMnO₄ is reduced to Mn²⁺, H₂O₂ is oxidized to O₂)
Preparation by auto-oxidation: BaO₂ + H₂SO₄ → BaSO₄ + H₂O₂. Barium peroxide reacts with dilute sulphuric acid at 0°C. BaSO₄ precipitates and H₂O₂ is obtained in the filtrate. This is called the auto-oxidation method.
Q17. Describe the principle, construction, and working of a fuel cell. Write the reactions involved in a hydrogen-oxygen fuel cell.
Solution:
Principle: A fuel cell is an electrochemical cell that converts chemical energy of a fuel directly into electrical energy through a redox reaction. Unlike a battery, it operates continuously as long as fuel is supplied.
Construction: The hydrogen-oxygen fuel cell consists of two porous carbon electrodes impregnated with catalysts (Pt, Pd). The electrolyte is a concentrated KOH solution (30%). Hydrogen gas is passed through the anode compartment, and oxygen through the cathode compartment.
Electrode Reactions:
At anode (oxidation): 2H₂(g) + 4OH⁻(aq) → 4H₂O(l) + 4e⁻
At cathode (reduction): O₂(g) + 2H₂O(l) + 4e⁻ → 4OH⁻(aq)
Overall reaction: 2H₂(g) + O₂(g) → 2H₂O(l)
Advantages: High efficiency (~70%), zero pollution (only water is produced), continuous operation.
Q18. Outline the product formed when D-glucose is treated with the following reagents:
- (a) Br₂ water: D-glucose is oxidized to gluconic acid (COOH group at C1). Br₂ water specifically oxidizes the aldehyde group (−CHO) to carboxylic acid (−COOH).
- (b) HNO₃: Both terminal groups are oxidized — D-glucose gives saccharic acid (both C1 and C6 become COOH).
- (c) H₂N-OH (hydroxylamine): D-glucose forms glucose oxime (C=N-OH bond at C1).
- (d) (CH₃CO)₂O (acetic anhydride): D-glucose forms pentaacetate — all five −OH groups are acetylated.
- (e) NH₂OH + H⁺ (phenylhydrazine): D-glucose forms glucosazone (yellow crystalline osazone).
✅ Section D: Numerical Problems (3+4 Marks)
Q19. A 0.5 g sample containing Na₂CO₃·10H₂O and inert impurities is titrated against 0.1 M HCl. The endpoint is reached at 22.4 mL of HCl. Calculate the percentage purity of the sample.
Solution:
Reaction: Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂
Moles of HCl used = M × V(L) = 0.1 × (22.4/1000) = 0.00224 mol
Moles of Na₂CO₃ = ½ × moles of HCl = 0.00112 mol
Molar mass of Na₂CO₃·10H₂O = (2×23) + 12 + (3×16) + 10×(2+16) = 46 + 12 + 48 + 180 = 286 g/mol
Mass of pure Na₂CO₃·10H₂O = 0.00112 × 286 = 0.3203 g
Percentage purity = (0.3203 / 0.5) × 100 = 64.06%
Q20. The rate constant of a first-order reaction is 2.5 × 10⁻³ s⁻¹ at 300 K. Calculate the activation energy if the pre-exponential factor is 1.2 × 10¹³ s⁻¹. (R = 8.314 J/mol·K)
Solution:
Using Arrhenius equation: k = Ae^(−Ea/RT)
ln(k) = ln(A) − Ea/(RT)
ln(2.5 × 10⁻³) = ln(1.2 × 10¹³) − Ea/(8.314 × 300)
ln(2.5) + ln(10⁻³) = ln(1.2) + ln(10¹³) − Ea/(2494.2)
0.9163 − 6.9078 = 0.1823 + 29.9336 − Ea/2494.2
−5.9915 = 30.1159 − Ea/2494.2
Ea/2494.2 = 30.1159 + 5.9915 = 36.1074
Ea = 36.1074 × 2494.2 = 90,028 J/mol ≈ 90.03 kJ/mol
📊 Topic-Wise Weightage Analysis
| Topic | Question Numbers | Total Marks |
|---|---|---|
| Solid State | Q1 | 1 |
| Solutions | Q2, Q5, Q11 | 7 |
| Chemical Kinetics | Q7, Q10, Q20 | 8 |
| Electrochemistry | Q6, Q17 | 7 |
| Surface Chemistry | Q12 | 3 |
| p-Block Elements | Q16 | 5 |
| Coordination Compounds | Q8 | 1 |
| Organic Chemistry (Haloalkanes) | Q4, Q13 | 5 |
| Alcohols, Phenols, Ethers | – | – |
| Biomolecules | Q18 | 5 |
| Chemistry in Everyday Life | Q9 | 1 |
| Metallurgy | Q15 | 3 |
| Stoichiometry & Titration | Q19 | 4 |
🎯 Expert Tips to Score 90%+ in Chemistry Board Exam
- Master NCERT thoroughly — 80% of questions come directly from NCERT textbook
- Practice numerical problems daily — physical chemistry carries 15-20 marks
- Learn named reactions with mechanisms — draw arrow pushing mechanisms for all organic reactions
- Create a formula sheet — colligative properties, rate equations, Nernst equation, Ksp
- Solve at least 10 previous year papers — identify recurring question patterns
- Focus on 3-4 mark questions — they are the easiest to score if concepts are clear
- Use diagrams and flowcharts — metallurgy flowcharts, crystal structures, cell diagrams
- Attempt MCQs first — 15 marks in Section A can be secured in 20 minutes
- Write balanced equations every time — partial equations lose marks
- Practice time management — allocate 60 min for Section A+B, 60 min for C, 30 min for D, 30 min for revision
🔍 Download MP Board Previous Year Papers
Regular practice of previous year question papers is the most effective strategy for board exam preparation. The 2024 Chemistry paper shows that:
- Numerical problems focus on solutions, kinetics, and electrochemistry
- Organic chemistry questions test IUPAC nomenclature, reaction mechanisms, and conversions
- Inorganic chemistry mainly covers coordination compounds, p-block, and d-and f-block elements
- Physical chemistry requires strong numerical aptitude and formula application
Practice this solved paper multiple times, time yourself, and check your answers against the detailed solutions provided. Good luck with your 2027 MP Board Chemistry exam!