CBSE Class 12 Chemistry Board Paper 2 — 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 2 — 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
Assuming complete dissociation, which of the following 0.1 m aqueous solutions will have the LOWEST freezing point?
A.KClB.Al₂(SO₄)₃✓ CorrectC.glucoseD.Na₂SO₄Answer: B. Al₂(SO₄)₃
Explanation: Depression of freezing point is ΔTf = iKf m, so at equal molality the solute giving the most particles wins. Al₂(SO₄)₃ releases 2 Al³⁺ and 3 SO₄²⁻, giving i = 5, against 3 for Na₂SO₄, 2 for KCl and 1 for glucose. Choosing Al₂(SO₄)₃ because of its high charges rather than its particle count happens to give the right answer here, but it is the number of ions, not the charge, that colligative properties count.
- Q2Easy
The mass of copper deposited when 2 faradays of charge are passed through an aqueous solution of CuSO₄ is (atomic mass of Cu = 63.5 u):
A.127 gB.31.75 gC.15.9 gD.63.5 g✓ CorrectAnswer: D. 63.5 g
Explanation: The cathode reaction is Cu²⁺ + 2e⁻ → Cu, so 2 mol of electrons — that is 2 F — deposit exactly 1 mol of copper, or 63.5 g. Treating 1 F as one mole of copper doubles the answer to 127 g, while dividing by 2 a second time gives 31.75 g; the electron count in the half-equation is what fixes the ratio.
- Q3Easy
Among Cr, Mn, Fe and Cu, the element that exhibits the highest oxidation state in its compounds is:
A.Mn✓ CorrectB.CrC.FeD.CuAnswer: A. Mn
Explanation: The maximum oxidation state of a 3d element equals the total of its 4s and 3d electrons available for bonding. Manganese is 3d⁵4s² and reaches +7, seen in MnO₄⁻, which is the highest in the first transition series. Chromium is 3d⁵4s¹ and stops at +6 in Cr₂O₇²⁻, while iron rarely goes beyond +6 and copper beyond +2.
- Q4Medium
For the reaction 2N₂O₅(g) → 4NO₂(g) + O₂(g), the rate of formation of NO₂ is 2.8 × 10⁻³ mol L⁻¹ s⁻¹. The rate of disappearance of N₂O₅ is:
A.1.4 × 10⁻³ mol L⁻¹ s⁻¹✓ CorrectB.2.8 × 10⁻³ mol L⁻¹ s⁻¹C.5.6 × 10⁻³ mol L⁻¹ s⁻¹D.7.0 × 10⁻⁴ mol L⁻¹ s⁻¹Answer: A. 1.4 × 10⁻³ mol L⁻¹ s⁻¹
Explanation: Rates are tied together by the stoichiometric coefficients: −(1/2)d[N₂O₅]/dt = +(1/4)d[NO₂]/dt. Hence −d[N₂O₅]/dt = (2/4) × 2.8 × 10⁻³ = 1.4 × 10⁻³ mol L⁻¹ s⁻¹. Dividing by 4 alone gives 7.0 × 10⁻⁴ — the coefficient of N₂O₅ must be put back in, since two molecules of N₂O₅ vanish for every four NO₂ formed.
- Q5Medium
The pair [Co(NH₃)₅(NO₂)]Cl₂ and [Co(NH₃)₅(ONO)]Cl₂ illustrates:
A.ionisation isomerismB.coordination isomerismC.linkage isomerism✓ CorrectD.hydrate isomerismAnswer: C. linkage isomerism
Explanation: Both complexes contain the same set of ligands and the same counter ions; the only difference is that the ambidentate nitrite ligand binds through nitrogen in one and through oxygen in the other. That is the definition of linkage isomerism. Ionisation isomerism would require the ligand and the counter ion to swap places, which does not happen here — the two chlorides stay outside the coordination sphere in both.
- Q6Medium
The best reagent for converting propan-1-ol into 1-chloropropane in high purity is:
A.concentrated HCl aloneB.SOCl₂ in the presence of pyridine✓ CorrectC.Cl₂ in the presence of sunlightD.NaCl and dilute H₂SO₄Answer: B. SOCl₂ in the presence of pyridine
Explanation: Thionyl chloride converts the alcohol to the chloride while the two by-products, SO₂ and HCl, escape as gases, so the product needs almost no purification. Concentrated HCl alone reacts very slowly with a primary alcohol and needs anhydrous ZnCl₂ as catalyst; chlorine in sunlight is free-radical substitution, which attacks several positions at once and gives a mixture.
- Q7Hard
The correct order of decreasing acid strength is:
A.phenol > p-nitrophenol > p-cresol > cyclohexanolB.cyclohexanol > p-cresol > phenol > p-nitrophenolC.p-cresol > p-nitrophenol > phenol > cyclohexanolD.p-nitrophenol > phenol > p-cresol > cyclohexanol✓ CorrectAnswer: D. p-nitrophenol > phenol > p-cresol > cyclohexanol
Explanation: Acidity tracks the stability of the conjugate base. The electron-withdrawing –NO₂ group spreads the phenoxide charge further, so p-nitrophenol is the strongest; the electron-releasing –CH₃ of p-cresol pushes charge back onto oxygen and weakens the acid relative to phenol. Cyclohexanol is an ordinary alcohol whose alkoxide has no ring to delocalise into, so it is the weakest of the four.
- Q8Hard
For the reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), E°cell = 1.10 V at 298 K. The equilibrium constant of the reaction is closest to:
A.1.9 × 10¹⁸B.2.0 × 10³⁷✓ CorrectC.3.7 × 10¹D.5.0 × 10⁻³⁸Answer: B. 2.0 × 10³⁷
Explanation: At equilibrium the Nernst equation reduces to log K = nE°cell/0.059. Two electrons are transferred, so log K = 2 × 1.10/0.059 = 37.3, giving K ≈ 2 × 10³⁷. Forgetting that n = 2 halves the exponent and produces the far smaller 1.9 × 10¹⁸; a positive E°cell must give a very large K, so a negative exponent is ruled out at once.
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- Board Paper 2 — you are here
- Board Paper 3 — 20 questions
- 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.