MP Board Class 12 Chemistry Previous Year Paper 2023 — Solved Questions with Answers

Prepare effectively for the MP Board Class 12 Chemistry Board Exam 2027 with this fully solved MP Board Class 12 Chemistry Previous Year Paper 2023. This paper follows the exact pattern set by the Madhya Pradesh Board of Secondary Education (MPBSE). Go through each question, check your answers, and understand the marking scheme to boost your score.

Paper Overview & Exam Pattern

The MP Board Class 12 Chemistry 2023 paper follows a structured pattern. Understanding the marks distribution helps you plan your preparation better. Here is the complete breakdown:

The total marks for the MP Board Class 12 Chemistry paper is 60 marks, with a duration of 3 hours. The question paper is bilingual (Hindi and English) for the convenience of students.

Section A — Objective Questions (1 Mark Each)

This section contains 10 multiple-choice questions. Each question carries 1 mark. Select the correct option for each question.

Q1. Which of the following is an example of a lyophilic colloid?

(a) Gold sol    (b) Fe(OH)₃ sol    (c) Starch solution    (d) As₂S₃ sol

Answer: (c) Starch solution

Q2. What is the coordination number of Na⁺ in NaCl crystal?

(a) 4    (b) 6    (c) 8    (d) 12

Answer: (b) 6 — NaCl has a face-centred cubic (FCC) structure where each Na⁺ is surrounded by 6 Cl⁻ ions.

Q3. Which of the following is a natural polymer?

(a) Nylon-6,6    (b) Teflon    (c) Cellulose    (d) PVC

Answer: (c) Cellulose — Cellulose is a polysaccharide found in plant cell walls, making it a natural polymer.

Q4. The IUPAC name of CH₃-CH₂-CH₂-CHO is:

(a) Butanal    (b) Butanone    (c) Pentanal    (d) Butanoic acid

Answer: (a) Butanal — The chain has 4 carbon atoms with an aldehyde (−CHO) functional group, so the IUPAC name is Butanal.

Q5. Which of the following is the weakest acid?

(a) CH₃COOH    (b) ClCH₂COOH    (c) Cl₃CCOOH    (d) Cl₂CHCOOH

Answer: (a) CH₃COOH — Acetic acid has no electron-withdrawing group, unlike the chlorinated derivatives which are stronger due to the −I effect of chlorine.

Q6. The rate of a reaction is expressed in terms of:

(a) Δ[G]/Δt    (b) Δ[H]/Δt    (c) Δ[C]/Δt    (d) Δ[E]/Δt

Answer: (c) Δ[C]/Δt — Rate of reaction is defined as the change in concentration of reactants or products per unit time.

Section B — Very Short Answer Questions (2 Marks Each)

Attempt any six of the following questions. Each carries 2 marks. Write concise answers.

Q7. Define electrochemical cell. Give one example.

Answer:

An electrochemical cell is a device that converts chemical energy into electrical energy (or vice versa) through redox reactions. Example: Daniel cell (Zn-Cu cell) where Zn acts as anode and Cu acts as cathode.

Q8. What is the difference between emulsion and gel?

Answer:

Q9. Write the IUPAC name of [Co(NH₃)₅Cl]Cl₂.

Answer: Pentaamminechloridocobalt(III) chloride

Explanation: The complex has 5 ammine (NH₃) ligands, one chloro ligand, and the oxidation state of Co is +3. The counter ions are 2 chloride ions.

Q10. What is the order of a reaction if its half-life is independent of initial concentration?

Answer: First-order reaction

For a first-order reaction, the half-life (t₁/₂ = 0.693/k) is independent of the initial concentration. This is a key identifying feature of first-order kinetics.

Section C — Short Answer Questions (3 Marks Each)

Attempt any six questions. Each carries 3 marks. Provide detailed but concise answers with supporting equations where needed.

Q11. Explain the mechanism of aldol condensation with an example.

Answer:

Aldol condensation is a reaction in which two molecules of an aldehyde or ketone (with α-hydrogen) react in the presence of a dilute base to form a β-hydroxy aldehyde or β-hydroxy ketone (aldol product), followed by dehydration to form an α,β-unsaturated carbonyl compound.

Example: Acetaldehyde (CH₃CHO) undergoes aldol condensation:

Step 1: Formation of enolate ion — OH⁻ abstracts α-hydrogen from CH₃CHO.

Step 2: Enolate attacks second molecule of CH₃CHO → 3-hydroxybutanal (aldol).

Step 3: On heating, 3-hydroxybutanal loses water to form crotonaldehyde (CH₃-CH=CH-CHO).

Q12. State and explain Faraday’s laws of electrolysis.

Answer:

First Law: The mass of a substance liberated at an electrode during electrolysis is directly proportional to the quantity of electricity (charge) passed through the electrolyte.

m ∝ Q    i.e., m = ZQ where Z = electrochemical equivalent.

Second Law: When the same quantity of electricity is passed through different electrolytes, the masses of substances liberated are proportional to their chemical equivalent weights.

m₁/m₂ = E₁/E₂ where E = equivalent weight = atomic mass / valency.

Q13. Distinguish between SN¹ and SN² reaction mechanisms.

Answer:

Section D — Long Answer Questions (5 Marks Each)

Attempt any four questions. Each carries 5 marks. Write comprehensive answers with proper chemical equations and diagrams (describe in text).

Q14. Explain the valence bond theory (VBT) for the formation of coordination compounds. Illustrate with [Co(NH₃)₆]³⁺.

Answer:

Valence Bond Theory (VBT), proposed by Linus Pauling, explains the bonding in coordination compounds through hybridisation of atomic orbitals of the central metal ion.

Key points of VBT:

• The central metal ion provides vacant orbitals for accepting lone pairs from ligands.
• These orbitals undergo hybridisation to form a set of equivalent hybrid orbitals.
• The number of hybrid orbitals equals the coordination number.
• Each ligand donates a lone pair to form a coordinate covalent bond.
• Magnetic properties depend on the number of unpaired electrons.

Application to [Co(NH₃)₆]³⁺:

Co³⁺ has electronic configuration [Ar]3d⁶. In the presence of NH₃ (strong field ligand), the 3d electrons pair up. The metal ion uses d²sp³ hybridisation (two 3d, one 4s, three 4p orbitals) to form six hybrid orbitals. These accept lone pairs from six NH₃ ligands. The complex is octahedral, diamagnetic (no unpaired electrons), and is an inner orbital or low-spin complex.

Magnetic moment: μ = 0 BM (diamagnetic).

Q15. Describe the preparation, properties, and uses of H₂SO₄ (Sulphuric Acid).

Answer:

Preparation — Contact Process:

Step 1: Burning of sulphur or roasting of sulphide ores: S + O₂ → SO₂

Step 2: Catalytic oxidation of SO₂ to SO₃: 2SO₂ + O₂ ⇌ 2SO₃ (V₂O₅ catalyst, 450°C, 1-2 atm)

Step 3: Absorption of SO₃ in H₂SO₄ to form oleum: SO₃ + H₂SO₄ → H₂S₂O₇ (oleum)

Step 4: Dilution of oleum: H₂S₂O₇ + H₂O → 2H₂SO₄

Properties: Heavy oily liquid (density 1.84 g/mL), strong acid, hygroscopic, powerful oxidising agent (hot conc. H₂SO₄), charring action on carbohydrates.

Uses: Fertiliser industry (phosphoric acid), petroleum refining, car batteries, detergent manufacture, dye and pigment production, metal pickling.

H₂SO₄ is called the “King of Chemicals” due to its wide industrial importance.

Preparation Tips for Chemistry Board Exam 2027

Use these expert tips to score 90%+ in MP Board Class 12 Chemistry based on the 2023 paper pattern:

1. Master NCERT thoroughly — 80% of questions are directly from NCERT Class 12 Chemistry textbook.
2. Focus on name reactions — Aldol condensation, Cannizzaro, Sandmeyer, and Reimer-Tiemann are frequently asked.
3. Practise numericals — Electrochemistry, chemical kinetics, and solution chapters have numerical questions worth 3-5 marks.
4. Learn IUPAC nomenclature — At least one question on naming coordination compounds or organic compounds appears every year.
5. Draw and label diagrams — Electrochemical cells, crystal structures, and distillation setups are common in long answer questions.

Practice this MP Board Class 12 Chemistry Previous Year Paper 2023 regularly and track your progress. Solving previous year papers builds confidence, improves time management, and helps you identify weak areas before the board exam. Good luck with your MP Board Class 12 Chemistry 2027 preparation! 🎯