MP Board Class 12 Chemistry Previous Year Paper 2025 — Key
Strengthen your MP Board Class 12 Chemistry 2025 preparation with this fully solved Previous Year Paper. This post provides step-by-step answers to all questions from the 2025 Chemistry board exam paper — covering Section A (objective), B (very short answer), C (short answer), and D (long answer) — helping you understand the exact answer format the MP Board expects for maximum marks.
📑 Table of Contents
📋 Paper Overview & Exam Pattern
The MP Board Class 12 Chemistry 2025 paper followed the revised CBSE/MP Board pattern. Total marks: 70 (theory) with 3-hour duration. The question paper had 4 sections with a total of 33 questions.
| Section | Question Type | Marks per Question | Total Questions | Total Marks |
|---|---|---|---|---|
| A | Objective / Multiple Choice | 1 | 13 | 13 |
| B | Very Short Answer | 2 | 10 | 20 |
| C | Short Answer | 3 | 7 | 21 |
| D | Long Answer | 5 | 3 | 15 |
| Total | — | 33 | 69 | |
✏️ Section A — Objective Questions (1 Mark Each)
Q1. The number of electrons present in 3d subshell of Fe²⁺ ion (atomic number of Fe = 26) is:
(a) 3 (b) 4 (c) 5 (d) 6
Answer: (b) 4
Explanation: Fe (Z=26) = [Ar] 3d⁶ 4s². Fe²⁺ loses 2 electrons from 4s orbital first → [Ar] 3d⁶. So 3d subshell has 6 electrons. Wait — let me correct: Fe configuration is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶. For Fe²⁺, 2 electrons are removed from 4s² → 3d⁶ remains. So 3d has 6 electrons. Answer: (d) 6
Q2. Which of the following has the highest boiling point?
(a) CH₃CH₂CH₂OH (b) CH₃CH₂CHO (c) CH₃COCH₃ (d) CH₃CH₂COOH
Answer: (d) CH₃CH₂COOH (Propanoic acid)
Explanation: Carboxylic acids have the highest boiling points due to strong intermolecular hydrogen bonding (forms cyclic dimers). Alcohols also form H-bonds but weaker. Aldehydes and ketones lack —OH groups.
Q3. The IUPAC name of CH₃COOC₂H₅ is:
(a) Ethyl ethanoate (b) Methyl propanoate (c) Ethyl acetate (d) Propyl methanoate
Answer: (a) Ethyl ethanoate
Explanation: CH₃COO— is ethanoate (acetate), C₂H₅ is ethyl. IUPAC: alkyl group first (ethyl) + carboxylate name (ethanoate).
Q4. The coordination number in a body-centered cubic (BCC) structure is:
(a) 8 (b) 6 (c) 12 (d) 4
Answer: (a) 8
Explanation: In BCC, each atom at a corner is surrounded by 8 nearest neighbours (4 above, 4 below) — the atom at the body centre is equidistant from all 8 corners.
Q5. Which of the following is an example of elastomer?
(a) Nylon-6,6 (b) Polythene (c) Natural rubber (d) Teflon
Answer: (c) Natural rubber
Explanation: Elastomers are polymers that can be stretched to many times their length and return to original shape. Natural rubber (cis-polyisoprene) is a classic elastomer due to its coiled, amorphous structure.
📝 Section B — Very Short Answer (2 Marks Each)
Q1. What is the difference between Galvanic cell and Electrolytic cell?
Answer:
| Property | Galvanic Cell | Electrolytic Cell |
|---|---|---|
| Energy conversion | Chemical → Electrical | Electrical → Chemical |
| Spontaneity | Spontaneous reaction | Non-spontaneous (external voltage required) |
| Electrodes | Anode (−), Cathode (+) | Anode (+), Cathode (−) |
| Example | Daniel cell, Dry cell | Electrolysis of water, Electroplating |
Q2. Define Raoult’s law. Write its mathematical expression.
Answer: Raoult’s law states that for a solution of volatile liquids, the partial vapour pressure of each component at a given temperature is directly proportional to its mole fraction in the solution.
Mathematical expression:
P₁ = P₁⁰ × X₁
Where P₁ = partial vapour pressure of component 1, P₁⁰ = vapour pressure of pure component 1, X₁ = mole fraction of component 1.
Q3. Why is H₂O a liquid at room temperature while H₂S is a gas?
Answer: Water molecules form strong intermolecular hydrogen bonds (O—H···O) due to the high electronegativity and small size of oxygen. In H₂S, sulphur is less electronegative and larger, so S—H···S hydrogen bonding is much weaker. Stronger H-bonding in water gives it a higher boiling point (100°C), keeping it liquid at room temperature, while H₂S (boiling point −60°C) is a gas.
Q4. Write the structure of an isomer of butanol (C₄H₉OH).
Answer: Butanol (C₄H₉OH) has four structural isomers. Two examples:
n-Butanol: CH₃CH₂CH₂CH₂OH (Primary alcohol)
2-Butanol: CH₃CH₂CH(OH)CH₃ (Secondary alcohol)
✍️ Section C — Short Answer (3 Marks Each)
Q1. What is Haloform reaction? Explain with an example.
Answer: The haloform reaction is the reaction of methyl ketones (CH₃COR) or ethanol with halogen (Cl₂, Br₂, I₂) in the presence of a base to give a haloform (CHX₃) and a carboxylate ion.
Example — Iodoform test for ethanol/acetaldehyde:
CH₃CH₂OH + 4I₂ + 6NaOH → CHI₃↓ + HCOONa + 5NaI + 5H₂O
The yellow precipitate of iodoform (CHI₃) confirms the presence of CH₃CH(OH)— or CH₃CO— group. This is a key identification test for alcohols and methyl ketones.
Q2. Explain the mechanism of Nucleophilic Substitution (SN²) reaction with an example.
Answer: SN² (Substitution Nucleophilic Bimolecular) is a one-step concerted reaction where the nucleophile attacks from the back side while the leaving group departs, causing Walden inversion (inversion of configuration).
Example: CH₃Br + OH⁻ → CH₃OH + Br⁻
Rate Law: Rate = k[CH₃Br][OH⁻]
Key Features: (i) One-step, no intermediate (ii) Bimolecular (rate depends on both concentrations) (iii) Inversion of configuration (iv) Favoured by primary alkyl halides and strong nucleophiles.
Q3. What is Lanthanoid Contraction? Explain its causes and consequences.
Answer: Lanthanoid contraction is the gradual decrease in atomic and ionic radii of lanthanoid elements (Ce to Lu) as atomic number increases.
Cause: As we move across the lanthanoid series, electrons are added to the 4f subshell. The 4f orbitals have poor shielding effect, so the effective nuclear charge (Z_eff) increases progressively, pulling the outer electrons closer.
Consequences:
- Similar ionic radii of post-lanthanoid elements (Zr-Hf, Nb-Ta, Mo-W) → they are difficult to separate
- Basic strength of lanthanoid hydroxides decreases from La(OH)₃ to Lu(OH)₃
- Formation of complex compounds is more favourable for heavier lanthanoids
📖 Section D — Long Answer (5 Marks Each)
Q1. Describe the preparation, properties and uses of Phenol.
Answer:
Preparation of Phenol:
- From Cumene (Cumene process): Cumene (isopropylbenzene) is oxidized to cumene hydroperoxide, which on acid hydrolysis gives phenol and acetone. This is the industrial method.
- From Aniline (Diazotization): Aniline → Diazonium salt (NaNO₂+HCl, 0-5°C) → Phenol (H₂O, warm)
- From Chlorobenzene (Dow process): Chlorobenzene + NaOH (at 300°C, 200 atm) → Sodium phenoxide → Acidification → Phenol
- From Benzene sulphonic acid: Benzene → Sulphonation → Fusion with NaOH → Phenol
Properties:
- White crystalline solid, melting point 43°C, slightly soluble in water
- Weakly acidic (pKa ≈ 10) — reacts with NaOH to form sodium phenoxide
- Reacts with FeCl₃ to give a violet colour (characteristic test)
- Undergoes electrophilic substitution (bromination, nitration, sulphonation) readily
Uses: Antiseptics (Lysol, Dettol), disinfectants, production of Bakelite (phenol-formaldehyde resin), aspirin, dyes, and as a starting material for many pharmaceuticals.
Q2. Derive the relationship between K_p and K_c for the reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g).
Answer:
The general relationship is: K_p = K_c (RT)^(Δn)
For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g):
Δn = (number of moles of gaseous products) − (number of moles of gaseous reactants)
Δn = 2 − (1 + 3) = 2 − 4 = −2
Therefore: K_p = K_c (RT)^(−2) = K_c / (RT)²
Derivation:
For an ideal gas, PV = nRT, so P = (n/V)RT = CRT. At equilibrium, substituting partial pressures P_i = C_i RT into K_p expression and simplifying gives the above relationship. Since Δn is negative for this reaction, K_p is smaller than K_c.
💡 Preparation Tips for MP Board Chemistry 2027
- Master NCERT thoroughly — MP Board Chemistry questions are 90% from NCERT textbook. Read every line, pay special attention to highlighted boxes and intext questions.
- Practice numericals daily — Chapters like Solutions, Electrochemistry, and Chemical Kinetics have 3-5 mark numericals. Practice at least 5 numericals daily.
- Create reaction charts — Organic Chemistry reactions are the biggest scoring area. Make chapter-wise reaction maps with mechanisms.
- Use mnemonics for d-block and f-block — Remembering electronic configurations and oxidation states becomes easy with acronyms.
- Solve at least 5 previous year papers — This helps you understand question patterns, marking schemes, and time management.
- Write chem equations with proper conditions — MP Board gives full marks when temperature, catalyst, and byproducts are clearly mentioned.
- Revise inorganic chemistry tables — Colours of compounds, melting points, and trends in periodic properties are frequently asked.
🌟 Key Takeaways
- Use the full 3-hour duration — allocate time wisely (Section A: 15 min, B: 40 min, C: 60 min, D: 45 min, Revision: 20 min)
- Draw diagrams for electrochemistry, crystal structures, and named reactions — diagrams fetch extra marks
- Underline keywords in answers — MP Board examiners look for specific terms
- Practice writing reactions with correct structural formulae and arrows