CBSE Class 12 Chemistry Board Paper 3 — 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 3 — 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
The number of unpaired electrons in the ground state of the Cr³⁺ ion is:
A.4B.3✓ CorrectC.5D.6Answer: B. 3
Explanation: Chromium is [Ar]3d⁵4s¹; forming Cr³⁺ removes the 4s electron first and then two 3d electrons, leaving 3d³. By Hund's rule those three electrons occupy separate orbitals with parallel spins, so three are unpaired. Answering 5 comes from removing three electrons only from the 3d⁵4s¹ set without recognising that the 4s electron goes first.
- Q2Easy
For a reaction the experimentally determined rate law is rate = k[A][B]^½. The overall order of the reaction is:
A.1.5✓ CorrectB.2C.0.5D.1Answer: A. 1.5
Explanation: Overall order is simply the sum of the exponents in the experimental rate law, here 1 + ½ = 1.5. Order need not be a whole number, and it is not read off the balanced equation — that would be molecularity, which must be an integer.
- Q3Easy
Which of the following alcohols gives immediate turbidity on shaking with the Lucas reagent at room temperature?
A.butan-1-olB.butan-2-olC.propan-1-olD.2-methylpropan-2-ol✓ CorrectAnswer: D. 2-methylpropan-2-ol
Explanation: The Lucas reagent, concentrated HCl with anhydrous ZnCl₂, converts an alcohol into its insoluble chloride through a carbocation, so the speed of the turbidity mirrors carbocation stability. A tertiary alcohol such as 2-methylpropan-2-ol reacts at once, a secondary one takes about five minutes, and primary alcohols need heating. Butan-2-ol is the slow secondary case, not the immediate one.
- Q4Medium
The limiting molar conductivities of NaCl, HCl and CH₃COONa are 126.4, 425.9 and 91.0 S cm² mol⁻¹ respectively. The limiting molar conductivity of acetic acid is:
A.425.9 S cm² mol⁻¹B.208.5 S cm² mol⁻¹C.298.5 S cm² mol⁻¹D.390.5 S cm² mol⁻¹✓ CorrectAnswer: D. 390.5 S cm² mol⁻¹
Explanation: Kohlrausch's law lets ionic contributions be added and cancelled: Λ°(CH₃COOH) = Λ°(HCl) + Λ°(CH₃COONa) − Λ°(NaCl), because that combination leaves exactly H⁺ and CH₃COO⁻. Substituting, 425.9 + 91.0 − 126.4 = 390.5 S cm² mol⁻¹. Simply adding the acetate and hydrogen carriers without subtracting NaCl double-counts Na⁺ and Cl⁻ and inflates the value.
- Q5Medium
Which of the following pairs of liquids shows negative deviation from Raoult's law and forms a maximum-boiling azeotrope?
A.ethanol and waterB.acetone and carbon disulphideC.nitric acid and water✓ CorrectD.benzene and tolueneAnswer: C. nitric acid and water
Explanation: Negative deviation means the new A–B interactions are STRONGER than the original A–A and B–B ones, so the vapour pressure drops below the Raoult value and the mixture boils at a maximum. Nitric acid and water hydrogen bond strongly to each other and form a 68% azeotrope boiling at 393.5 K. Ethanol–water and acetone–carbon disulphide both break existing attractions and deviate positively, while benzene–toluene is very nearly ideal.
- Q6Medium
The hybridisation of the central metal ion and the shape of [Fe(CN)₆]³⁻ are:
A.sp³d², octahedralB.sp³, tetrahedralC.d²sp³, octahedral✓ CorrectD.dsp², square planarAnswer: C. d²sp³, octahedral
Explanation: Fe³⁺ is 3d⁵, and the strong field cyanide ligands force the five electrons into three 3d orbitals, freeing two inner 3d orbitals. Those combine with the 4s and three 4p orbitals to give d²sp³ hybridisation and an octahedral inner-orbital complex with one unpaired electron. The sp³d² alternative is octahedral too but uses OUTER 4d orbitals, which is what a weak field ligand such as fluoride would demand.
- Q7Hard
An aqueous solution contains 3.0 g of urea (molar mass 60 g mol⁻¹) per litre. Its osmotic pressure at 300 K, taking R = 0.0821 L atm K⁻¹ mol⁻¹, is:
A.0.62 atmB.1.23 atm✓ CorrectC.2.46 atmD.12.3 atmAnswer: B. 1.23 atm
Explanation: Osmotic pressure is π = CRT with C in mol L⁻¹. The concentration is 3.0/60 = 0.050 mol L⁻¹, so π = 0.050 × 0.0821 × 300 = 1.23 atm. Urea is a non-electrolyte, so no van't Hoff factor is applied; treating it as though it produced two particles would wrongly double the value to 2.46 atm.
- Q8Hard
A steady current of 1.5 A is passed through a solution of AgNO₃ for 20 minutes. The mass of silver deposited is (Ag = 108 g mol⁻¹, F = 96500 C mol⁻¹):
A.2.01 g✓ CorrectB.1.01 gC.4.03 gD.0.02 gAnswer: A. 2.01 g
Explanation: Charge passed Q = It = 1.5 × 1200 s = 1800 C, giving 1800/96500 = 0.01865 mol of electrons. Since Ag⁺ + e⁻ → Ag needs just one electron per atom, the same number of moles of silver deposits: 0.01865 × 108 = 2.01 g. Dividing by 2 as though silver were divalent halves the mass to 1.01 g, and using 20 instead of 1200 seconds loses a factor of 60.
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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 — you are here
- Board Paper 4 — 20 questions
- 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.