CBSE Class 12 Chemistry Sample Paper 2026 (with Solutions) — Free PDF
This is a full-length CBSE Class 12 Chemistry paper built to the current board blueprint: 33 questions across five sections for 70 marks in 3 hours, the re
TL;DR: This is a full-length CBSE Class 12 Chemistry paper built to the current board blueprint: 33 questions across five sections for 70 marks in 3 hours, t…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
This is a full-length CBSE Class 12 Chemistry paper built to the current board blueprint: 33 questions across five sections for 70 marks in 3 hours, the remaining 30 marks being the practical component assessed in school. Section A is objective, Section D is case-based and Section E carries the long answers. Every question is followed by a complete worked solution, with the reasoning and the calculation set out step by step rather than just a final value.
Board: CBSE · Class: Class 12 · Subject: Chemistry · Total: 70 marks · Time: 3 hours
Section A — Objective (1 mark each) — 16 marks
Questions 1–14 are multiple choice. Questions 15 and 16 are Assertion–Reason: mark (a) if both A and R are true and R explains A, (b) if both are true but R does not explain A, (c) if A is true and R is false, (d) if A is false and R is true.
Q1. The molal elevation constant of a solvent depends on (1 mark)
Answer: The nature of the solvent alone. Kb is a property of the solvent — it is independent of the solute added.
Q2. Which of the following aqueous solutions has the highest boiling point at the same molality: glucose, NaCl, or CaCl₂? (1 mark)
Answer: CaCl₂. Elevation depends on the number of particles; CaCl₂ gives three ions (i ≈ 3) against two for NaCl and one for glucose.
Q3. For a first-order reaction, the half-life is (1 mark)
Answer: Independent of the initial concentration, t½ = 0.693/k. This constancy is the standard test for first-order kinetics.
Q4. The rate constant of a reaction doubles when the temperature rises by 10 K. This is explained by (1 mark)
Answer: The Arrhenius equation, k = Ae^(−Ea/RT): a small rise in T sharply increases the fraction of molecules with energy above Ea.
Q5. The standard EMF of a cell with E°(cathode) = +0.34 V and E°(anode) = −0.76 V is (1 mark)
Answer: +1.10 V. E°cell = E°cathode − E°anode = 0.34 − (−0.76) = 1.10 V. This is the Daniell cell.
Q6. The oxidation state of chromium in the dichromate ion, Cr₂O₇²⁻, is (1 mark)
Answer: +6. Let x be the oxidation state: 2x + 7(−2) = −2, so 2x = 12 and x = +6.
Q7. Which of the following is a transition element: Zn, Sc, or Ca? (1 mark)
Answer: Sc. A transition element has a partly filled d subshell in its ground state or a common oxidation state; Zn is d¹⁰ throughout and Ca has no d electrons.
Q8. The coordination number of the central metal in [Co(en)₃]³⁺ is (1 mark)
Answer: Six. Ethylenediamine is bidentate, so three molecules occupy six coordination sites.
Q9. The IUPAC name of CH₃CH₂CHO is (1 mark)
Answer: Propanal. A three-carbon chain ending in −CHO takes the suffix -al.
Q10. Which reagent distinguishes an aldehyde from a ketone? (1 mark)
Answer: Tollens' reagent (ammoniacal silver nitrate). An aldehyde reduces it to a silver mirror; a ketone does not.
Q11. The compound formed when phenol is treated with bromine water is (1 mark)
Answer: 2,4,6-tribromophenol, a white precipitate. The −OH group is strongly activating and ortho/para directing, so all three positions brominate.
Q12. SN1 reactions proceed fastest with (1 mark)
Answer: Tertiary alkyl halides. The rate-determining step forms a carbocation, and a tertiary carbocation is the most stabilised by hyperconjugation and inductive effects.
Q13. The monomer of natural rubber is (1 mark)
Answer: Isoprene (2-methyl-1,3-butadiene). Natural rubber is its cis-1,4 polymer.
Q14. Insulin belongs to which class of biomolecules? (1 mark)
Answer: Proteins — it is a polypeptide hormone made of two chains linked by disulphide bridges.
Q15. Assertion (A): Transition metals form coloured compounds. Reason (R): They have partly filled d orbitals allowing d–d electronic transitions. (1 mark)
Answer: (a) Both true and R explains A. Absorption of visible light promotes a d electron to a higher d level, and the transmitted light gives the colour.
Q16. Assertion (A): The boiling point of an aqueous solution is higher than that of pure water. Reason (R): The vapour pressure of the solution is lower than that of pure water. (1 mark)
Answer: (a) Both true and R explains A. A non-volatile solute lowers vapour pressure, so a higher temperature is needed to reach atmospheric pressure.
Section B — Very Short Answer (2 marks each) — 10 marks
Answer all 5 questions. Each carries 2 marks.
Q1. Define molarity and molality, and state which one changes with temperature and why. (2 marks)
Answer: Molarity is the number of moles of solute per litre of solution; molality is the number of moles of solute per kilogram of solvent. Molarity changes with temperature because the volume of the solution expands or contracts, while molality does not, since mass is temperature-independent.
Q2. State Raoult's law for a solution of a non-volatile solute in a volatile solvent. (2 marks)
Answer: The relative lowering of vapour pressure equals the mole fraction of the solute: (p° − p)/p° = x_solute. Equivalently, the vapour pressure of the solution equals the vapour pressure of the pure solvent multiplied by its mole fraction, p = p°·x_solvent.
Q3. Write the cell reaction and the Nernst equation for the Daniell cell. (2 marks)
Answer: Cell reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). Nernst equation at 298 K: Ecell = E°cell − (0.059/2)·log([Zn²⁺]/[Cu²⁺]), with n = 2 electrons transferred.
Q4. What are lanthanoid contraction and one of its consequences? (2 marks)
Answer: Lanthanoid contraction is the steady decrease in atomic and ionic radii across the lanthanoid series, caused by the poor shielding of the nuclear charge by 4f electrons. As a consequence the second and third transition series have almost identical radii — for example Zr and Hf — which makes them chemically very hard to separate.
Q5. Distinguish between an addition polymer and a condensation polymer, with one example of each. (2 marks)
Answer: An addition polymer forms by repeated addition of monomers containing double bonds, with no loss of any molecule — for example polythene from ethene. A condensation polymer forms by reaction between bifunctional monomers with the elimination of a small molecule such as water — for example nylon-6,6 from adipic acid and hexamethylenediamine.
Section C — Short Answer (3 marks each) — 21 marks
Answer all 7 questions. Each carries 3 marks.
Q1. A solution containing 18 g of glucose (molar mass 180 g/mol) in 500 g of water. Calculate its molality and the boiling point elevation. (Kb for water = 0.52 K kg/mol) (3 marks)
Answer: Moles of glucose = 18/180 = 0.1 mol. Molality = 0.1 mol / 0.5 kg = 0.2 mol/kg. Elevation ΔTb = Kb·m = 0.52 × 0.2 = 0.104 K, so the solution boils at about 100.10 °C.
Q2. The rate constant of a first-order reaction is 2.303 × 10⁻³ s⁻¹. Calculate the time required for the reaction to be 90% complete. (3 marks)
Answer: For a first-order reaction t = (2.303/k)·log([A]₀/[A]). When 90% is complete, [A] = 0.1[A]₀, so the ratio is 10 and log 10 = 1. Therefore t = 2.303/(2.303 × 10⁻³) × 1 = 1000 s, that is about 16.7 minutes.
Q3. Explain why transition metals show variable oxidation states, and give the common oxidation states of manganese. (3 marks)
Answer: In transition metals the (n−1)d and ns orbitals are very close in energy, so electrons from both can take part in bonding, and the energy needed to remove successive electrons rises only gradually. That allows several stable oxidation states differing by one unit. Manganese shows +2, +3, +4, +6 and +7, with +2 the most stable in aqueous solution and +7 in KMnO₄, its strongest oxidising form.
Q4. Write the IUPAC names and draw the structures of the geometrical isomers of [Pt(NH₃)₂Cl₂]. (3 marks)
Answer: The complex is square planar, so it has cis and trans isomers. In the cis isomer the two chloride ligands occupy adjacent corners, and it is named cis-diamminedichloridoplatinum(II) — this is the anticancer drug cisplatin. In the trans isomer the chlorides are diagonally opposite, giving trans-diamminedichloridoplatinum(II), which is biologically inactive.
Q5. How do you convert ethanol to ethanoic acid, and ethanoic acid to ethanamide? Give the reagents and equations. (3 marks)
Answer: Ethanol is oxidised to ethanoic acid by acidified potassium dichromate: CH₃CH₂OH + 2[O] →(K₂Cr₂O₇/H⁺) CH₃COOH + H₂O. Ethanoic acid is converted to ethanamide by treating it with ammonia to give ammonium ethanoate, which on heating loses water: CH₃COOH + NH₃ → CH₃COONH₄ →(heat) CH₃CONH₂ + H₂O.
Q6. Explain the mechanism of the SN2 reaction of bromomethane with hydroxide ion, and state its stereochemical outcome. (3 marks)
Answer: The hydroxide ion attacks the carbon from the side directly opposite the bromine, so bond formation and bond breaking happen together in a single step through a five-coordinate transition state in which the carbon is partially bonded to both. There is no intermediate. Because attack is from the back, the three remaining bonds turn inside out and the configuration is inverted — the Walden inversion. The rate depends on both reactants, so it is second order overall.
Q7. Define enzymes and explain, with one example, why enzyme catalysis is highly specific. (3 marks)
Answer: Enzymes are biological catalysts, almost all globular proteins, that greatly accelerate reactions in living cells under mild conditions. Their specificity arises from the active site, a small pocket whose shape and chemical groups fit only a particular substrate, as a key fits one lock. Maltase, for instance, hydrolyses maltose but not sucrose, because only maltose fits the active site correctly.
Section D — Case-based questions (4 marks each) — 8 marks
Answer both case-based questions. Each has sub-parts totalling 4 marks.
Q1. CASE STUDY 1 — Seawater freezes below 0 °C, and salt is spread on icy roads in cold countries to melt the ice. A student prepares a solution by dissolving 5.85 g of NaCl (molar mass 58.5 g/mol) in 1000 g of water. (Kf for water = 1.86 K kg/mol) (i) Which colligative property explains road salting? (ii) Calculate the molality. (iii) Calculate the expected freezing point, assuming complete dissociation. (4 marks)
Answer: (i) Depression of freezing point — the dissolved solute lowers the temperature at which ice and liquid are in equilibrium, so the ice melts. (ii) Moles of NaCl = 5.85/58.5 = 0.1 mol in 1 kg of water, so molality = 0.1 mol/kg. (iii) NaCl dissociates completely into two ions, so the van 't Hoff factor i = 2. ΔTf = i·Kf·m = 2 × 1.86 × 0.1 = 0.372 K, and the solution freezes at about −0.37 °C.
Q2. CASE STUDY 2 — Aspirin, an ester of salicylic acid, is one of the most widely used medicines. It is prepared by treating salicylic acid with acetic anhydride. (i) Name the functional groups present in salicylic acid. (ii) Name the type of reaction used to make aspirin. (iii) Why is aspirin taken after food rather than on an empty stomach? (iv) Name the class of drug aspirin belongs to. (4 marks)
Answer: (i) Salicylic acid contains a carboxylic acid group (−COOH) and a phenolic hydroxyl group (−OH) on the benzene ring. (ii) Acetylation — an esterification in which the phenolic −OH is converted to an ester group by acetic anhydride. (iii) Aspirin is acidic and irritates the stomach lining, and can cause bleeding, so it is taken after food to reduce contact with the mucous membrane. (iv) It is an analgesic and antipyretic — a non-narcotic pain reliever that also reduces fever.
Section E — Long Answer (5 marks each) — 15 marks
Answer all 3 questions. Each carries 5 marks.
Q1. (a) Define conductivity and molar conductivity, and give their units. (b) Explain how each varies with dilution for a strong electrolyte. (c) State Kohlrausch's law and one of its applications. (5 marks)
Answer: (a) Conductivity κ is the conductance of a solution held between two electrodes of unit area separated by unit distance, with units S m⁻¹. Molar conductivity Λm = κ/c is the conducting power of all the ions from one mole of electrolyte, with units S m² mol⁻¹. (b) On dilution, conductivity DECREASES, because the number of ions per unit volume falls. Molar conductivity INCREASES, because the total volume containing one mole of electrolyte rises and the ions move more freely; for a strong electrolyte it rises slowly and approaches a limiting value Λ°m. (c) Kohlrausch's law states that the limiting molar conductivity of an electrolyte is the sum of the individual contributions of its cation and anion. It allows Λ°m of a weak electrolyte such as acetic acid — which cannot be found by extrapolation — to be calculated from strong-electrolyte values, and hence its degree of dissociation and dissociation constant.
Q2. (a) Give the structure of the amino acid glycine at physiological pH and explain why amino acids are amphoteric. (b) What is a peptide bond? (c) Distinguish between the primary and secondary structure of a protein. (d) What is denaturation? (5 marks)
Answer: (a) At physiological pH glycine exists as a zwitterion, ⁺H₃N−CH₂−COO⁻: the carboxyl group has donated its proton to the amino group. Amino acids are amphoteric because they carry both an acidic −COOH and a basic −NH₂ group, so they react with both acids and bases. (b) A peptide bond is the amide linkage −CO−NH− formed when the carboxyl group of one amino acid condenses with the amino group of the next, with the loss of water. (c) Primary structure is the sequence of amino acids in the chain, held by covalent peptide bonds. Secondary structure is the local folding of that chain into an α-helix or β-pleated sheet, held by hydrogen bonds between the backbone C=O and N−H groups. (d) Denaturation is the loss of secondary and tertiary structure — by heat, acid or heavy metal ions — leaving the primary structure intact; the protein loses its biological activity, as when egg white coagulates on boiling.
Q3. (a) Explain the difference between an ideal and a non-ideal solution, with one example of each type of deviation. (b) A solution of two liquids shows a positive deviation from Raoult's law — comment on the sign of ΔHmix and ΔVmix. (c) What is an azeotrope, and why can such a mixture not be separated by fractional distillation? (5 marks)
Answer: (a) An ideal solution obeys Raoult's law at all compositions, with solute–solvent interactions equal in strength to the solute–solute and solvent–solvent interactions, so ΔHmix = 0 and ΔVmix = 0 — benzene and toluene, for example. A non-ideal solution disobeys it: a mixture of ethanol and acetone shows positive deviation, while chloroform and acetone show negative deviation because hydrogen bonding between unlike molecules is stronger. (b) In positive deviation the new interactions are WEAKER than the original ones, so energy is absorbed and the volume expands: ΔHmix is positive and ΔVmix is positive. (c) An azeotrope is a mixture that boils at a constant temperature and distils unchanged in composition, because at that composition the vapour has the same composition as the liquid. Fractional distillation separates components by exploiting the difference between vapour and liquid composition, and since no such difference exists at the azeotropic point, the method fails there.
FAQs
What is the total marks for Class 12 Chemistry board exam?
70 marks for theory and 30 marks for practicals; total 100 marks.
Which chapters have highest weightage?
Organic chemistry reactions, chemical kinetics, and coordination chemistry are high-weightage topics.
Are chemical equations required in answers?
Yes, balanced chemical equations are essential for mechanism and reaction-based questions.
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