CBSE Class 12 Chemistry Board Paper 4 — MCQs with Answers
Original paper · CBSE patternWritten by our subject team, not a reprint of an official sample paper.
- Questions
- 20
- Time
- 30 min
- Marking
- +1, no negative marking
- Mix
- 4 Easy · 8 Medium · 8 Hard
Board Paper 4 — CBSE Class 12 Chemistry is a full 20-question objective paper for Class 12 Chemistry, written in the CBSE board pattern and marked the way the board marks: +1 for a correct answer and 0 for a wrong one — no negative marking, exactly like the real board paper. It follows Section A of a real CBSE paper: Q1–Q16 are single-correct MCQs, Q17–Q18 are case-based questions built on a short source, and Q19–Q20 are assertion–reason items using the board's own four option strings.
The difficulty sits deliberately a notch above the real board — 4 Easy · 8 Medium · 8 Hard — because a paper you can clear comfortably tells you nothing on exam day. You get 30 minutes — about 90 seconds a question, the pace the real paper demands. 8 of the questions are solved in full below, with the correct option and a worked explanation for each; the remaining 12 are timed and scored when you sign in free.
Sit it under exam conditions rather than open-book. The score you get on a timed full-length paper is the only honest signal of whether your Chemistry revision is holding together across chapters — which is precisely what the board tests and what chapter-wise practice cannot tell you.
8 solved questions from this paper
Answer and worked explanation shown for each. The remaining 12 are timed and scored when you sign in free.
- Q1Easy
For a first-order reaction, which of the following plots gives a straight line?
A.[A] against tB.ln[A] against t✓ CorrectC.1/[A] against tD.[A]² against tAnswer: B. ln[A] against t
Explanation: The integrated first-order rate law is ln[A] = ln[A]₀ − kt, which is a straight line of slope −k when ln[A] is plotted against time. A straight line of [A] against t is the signature of a ZERO-order reaction, and 1/[A] against t is the second-order plot.
- Q2Easy
A solution contains 46 g of ethanol (molar mass 46 g mol⁻¹) and 54 g of water (molar mass 18 g mol⁻¹). The mole fraction of ethanol is:
A.0.46B.0.75C.0.33D.0.25✓ CorrectAnswer: D. 0.25
Explanation: Mole fraction uses amounts, not masses: ethanol is 46/46 = 1.0 mol and water is 54/18 = 3.0 mol, so the total is 4.0 mol and x(ethanol) = 1.0/4.0 = 0.25. Working from masses gives 46/100 = 0.46, which is the mass fraction, and 0.75 is the mole fraction of water rather than of ethanol.
- Q3Easy
Which of the following carbonyl compounds is the most reactive towards nucleophilic addition?
A.HCHO✓ CorrectB.CH₃CHOC.CH₃COCH₃D.C₆H₅COCH₃Answer: A. HCHO
Explanation: Nucleophilic addition is easier when the carbonyl carbon is more positive and less crowded. Methanal has only hydrogen atoms attached, so nothing releases electrons into the carbonyl and nothing blocks the approaching nucleophile. Each added alkyl group both feeds electron density in and adds bulk, and the phenyl group of acetophenone also delocalises the carbonyl, making it the least reactive of the four.
- Q4Medium
During the electrolysis of aqueous sodium chloride with inert electrodes, hydrogen rather than sodium is liberated at the cathode because:
A.sodium reacts with chlorine as soon as it is formedB.sodium ions are not attracted towards the cathodeC.the reduction of water to H₂ occurs at a much less negative potential than the reduction of Na⁺✓ CorrectD.chloride ions coat the cathode and block the discharge of sodium ionsAnswer: C. the reduction of water to H₂ occurs at a much less negative potential than the reduction of Na⁺
Explanation: At a cathode, the species with the HIGHER (less negative) reduction potential is discharged first. Reduction of water, 2H₂O + 2e⁻ → H₂ + 2OH⁻, occurs near −0.83 V, far above the −2.71 V required for Na⁺ + e⁻ → Na, so water wins and hydrogen is evolved. Sodium ions are certainly attracted to the cathode; they simply are not discharged there in an aqueous medium.
- Q5Medium
Which pair of ions has almost identical radii as a direct consequence of the lanthanoid contraction?
A.Zr⁴⁺ and Hf⁴⁺✓ CorrectB.Ti⁴⁺ and Zr⁴⁺C.La³⁺ and Lu³⁺D.Sc³⁺ and Y³⁺Answer: A. Zr⁴⁺ and Hf⁴⁺
Explanation: The steady shrinkage of the 4f series cancels the size increase expected on going from the second to the third transition series, so zirconium and hafnium end up almost the same size and are notoriously hard to separate. La³⁺ and Lu³⁺ are the two ENDS of that contraction and therefore differ appreciably, while Ti⁴⁺ and Zr⁴⁺ are separated by a normal period jump that no 4f block intervenes in.
- Q6Medium
1-Bromopropane is heated with alcoholic KOH and the product is then treated with HBr in the presence of benzoyl peroxide. The final product is:
A.2-bromopropaneB.propan-1-olC.propan-2-olD.1-bromopropane✓ CorrectAnswer: D. 1-bromopropane
Explanation: Alcoholic KOH eliminates HBr to give propene. Adding HBr in the presence of a peroxide follows the free-radical anti-Markovnikov path, so bromine attaches to the terminal carbon and the starting material is regenerated. Without the peroxide the addition would be Markovnikov and would give 2-bromopropane, which is exactly what the peroxide effect reverses.
- Q7Hard
A 0.10 m aqueous solution of a weak electrolyte AB boils at 100.0728 °C. Taking Kb for water as 0.52 K kg mol⁻¹, the degree of dissociation of AB is:
A.40%✓ CorrectB.70%C.14%D.60%Answer: A. 40%
Explanation: Without dissociation the elevation would be Kb m = 0.52 × 0.10 = 0.052 K, but the observed elevation is 0.0728 K, so the van't Hoff factor is i = 0.0728/0.052 = 1.40. For AB ⇌ A⁺ + B⁻ two particles come from one, so i = 1 + α, giving α = 0.40, that is 40%. Dividing i by 2 instead of subtracting 1 gives the tempting but wrong figure of 70%.
- Q8Hard
If the activation energy of a reaction were zero, its rate constant would:
A.double for every 10 K rise in temperatureB.be independent of temperature✓ CorrectC.decrease as the temperature is raisedD.become zero at every temperatureAnswer: B. be independent of temperature
Explanation: The Arrhenius equation is k = Ae^(−Ea/RT); putting Ea = 0 makes the exponential term e⁰ = 1, so k = A, a quantity with no temperature term in it. Every colliding pair would already have enough energy, so warming the mixture could not increase the fraction that reacts. Note that k equals the frequency factor A, which is certainly not zero — a zero barrier means a fast reaction, not a stalled one.
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Attempt this paper free →More Class 12 Chemistry board papers
- Board Paper 1 — 20 questions
- Board Paper 2 — 20 questions
- Board Paper 3 — 20 questions
- Board Paper 4 — you are here
- Board Paper 5 — 20 questions
Prefer chapter-by-chapter revision first? Class 12 Chemistry notes & chapter MCQs →
Questions students ask about this paper
Is this an official CBSE sample paper?
No — and that is deliberate. This is an original paper written in the CBSE board pattern by our subject team, not a copy of CBSE's official Sample Question Paper. It means you get fresh questions you have not already seen on a dozen other sites, with instant scoring and worked solutions rather than a PDF.
How is this Class 12 Chemistry paper marked?
+1 for a correct answer and 0 for a wrong one — no negative marking, exactly like the real board paper. So attempt every question — there is no penalty for a wrong answer on a board paper, and leaving a question blank can only cost you.
How long should this paper take?
30 minutes for 20 questions, roughly 90 seconds each. The timer runs whether or not you are watching it, which is the point: board marks are lost to pacing at least as often as to gaps in knowledge.
Do I need to pay to attempt it?
No. Sign in free with Google and the full paper opens with a live timer, automatic scoring and a worked solution for every question — including the ones not shown on this page.