CBSE Class 12 Chemistry Important Questions & Answers

A practical question bank with solutions, aligned to the CBSE 2026–27 syllabus (Board Exam 2027)

Table of Contents

Introduction

Let’s be honest: most students don’t fail Chemistry because it’s too hard. They lose marks because of small things, like a missing unit, a wrongly named complex, or an amine basicity order written for the wrong solvent. The good news is that these mistakes are fixable, and the fastest way to fix them is to practise the questions the board keeps coming back to.

This article gives you exactly that: a focused set of important questions with clear answers, a section on previous-year question formats, competency-based and case-based practice, extra model questions, step-by-step solutions, and a short list of exam-day tips.

First, the exam pattern at a glance. For 2026–27, the Class 12 Chemistry paper is of 70 marks (theory) plus 30 marks (practical), with a time of 3 hours. The syllabus has 10 units and, according to published summaries of the CBSE syllabus, there is no change from the previous session. The paper has 33 questions across five sections: Section A (16 objective-type questions of 1 mark, including assertion–reason), Section B (5 questions of 2 marks), Section C (7 questions of 3 marks), Section D (2 case-based questions of 4 marks) and Section E (3 long-answer questions of 5 marks). A growing share of the paper tests application and competency rather than recall.

Unit	Marks
Solutions	7
Electrochemistry	9
Chemical Kinetics	7
d- and f-Block Elements	7
Coordination Compounds	7
Haloalkanes and Haloarenes	6
Alcohols, Phenols and Ethers	6
Aldehydes, Ketones and Carboxylic Acids	8
Amines	6
Biomolecules	7
Total	70

Always cross-check the unit-wise marks and the question paper design against the official CBSE syllabus and the latest sample paper on cbseacademic.nic.in before you finalise your revision plan.

Download All Solved Questions as a Printable PDF

Part 1: Important Questions and Answers

Solutions

Q1. What is Raoult’s law? Why do solutions of ethanol and acetone show positive deviation?

For a solution of volatile liquids, the partial vapour pressure of each component is directly proportional to its mole fraction: p_A = x_A · p_A°. In ethanol–acetone, ethanol molecules are held together by hydrogen bonds. Adding acetone breaks some of these bonds, and the new ethanol–acetone attractions are weaker than the ones lost. Molecules therefore escape more easily, the vapour pressure is higher than Raoult’s law predicts, and we see positive deviation (ΔmixH > 0 and ΔmixV > 0).

Q2. Why is osmotic pressure preferred over other colligative properties for finding the molar mass of polymers and proteins?

Polymers have very large molar masses and low solubility, so a solution of reasonable mass concentration has an extremely small mole fraction of solute. Changes in boiling point or freezing point would be too tiny to measure. Osmotic pressure, on the other hand, is measurable even for very dilute solutions, and it is usually measured at room temperature, which protects heat-sensitive biomolecules.

Q3. Define the van’t Hoff factor. What is its value for an associating and a dissociating solute?

i = (observed colligative property) ÷ (calculated colligative property assuming no association or dissociation) = (normal molar mass) ÷ (observed molar mass). For association (e.g. ethanoic acid in benzene) i < 1. For dissociation (e.g. KCl in water) i > 1. For a solute that does neither, i = 1.

Q4. Calculate the boiling point of a solution of 18 g of glucose (M = 180 g/mol) in 1 kg of water. (K_b = 0.52 K kg/mol)

Molality = 18/180 = 0.1 m. ΔT_b = K_b × m = 0.52 × 0.1 = 0.052 K. Boiling point = 373.15 + 0.052 ≈ 373.20 K (100.05 °C).

Q5. State Henry’s law. Why do scuba divers use air diluted with helium?

Henry’s law: the partial pressure of a gas in the vapour phase is proportional to its mole fraction in the solution (p = K_H · x). At depth, the high pressure dissolves more nitrogen in the blood. When the diver surfaces, the dissolved gas comes out as bubbles and blocks capillaries (the bends). Helium is much less soluble in blood than nitrogen, so helium–oxygen mixtures greatly reduce this risk.

Electrochemistry

Q6. Write the Nernst equation for the Daniell cell and find its E° given E°(Zn²⁺/Zn) = −0.76 V and E°(Cu²⁺/Cu) = +0.34 V.

Cell: Zn | Zn²⁺ || Cu²⁺ | Cu. E°cell = 0.34 − (−0.76) = 1.10 V. E_cell = E°cell − (0.0591/2) log ([Zn²⁺]/[Cu²⁺]) at 298 K.

Q7. State Kohlrausch’s law. How is Λ°m of ethanoic acid calculated?

At infinite dilution, the limiting molar conductivity of an electrolyte equals the sum of the limiting molar conductivities of its cation and anion, since each ion migrates independently. A weak electrolyte cannot be extrapolated to infinite dilution directly, so: Λ°m(CH₃COOH) = Λ°m(CH₃COONa) + Λ°m(HCl) − Λ°m(NaCl).

Q8. How much copper is deposited when a current of 2 A passes through CuSO₄ solution for 30 minutes? (Molar mass of Cu = 63.5 g/mol; 1 F = 96500 C)

Q = I × t = 2 × 1800 = 3600 C. Cu²⁺ + 2e⁻ → Cu, so 2 × 96500 C deposits 63.5 g. Mass = (63.5 × 3600)/(2 × 96500) ≈ 1.18 g.

Q9. Why does the conductivity of an electrolyte decrease on dilution while its molar conductivity increases?

Conductivity (κ) depends on the number of ions per unit volume, and dilution reduces that number. Molar conductivity (Λm = κ × 1000/M) is conductivity per unit concentration. On dilution the ions are further apart (less inter-ionic attraction), and for weak electrolytes the degree of dissociation also rises. Therefore Λm increases.

Q10. Explain the electrochemical theory of rusting.

A drop of water containing dissolved CO₂ and O₂ acts as an electrolyte. Anode: Fe → Fe²⁺ + 2e⁻. Cathode: O₂ + 4H⁺ + 4e⁻ → 2H₂O. Fe²⁺ is further oxidised by atmospheric oxygen to Fe³⁺, which forms hydrated iron(III) oxide (Fe₂O₃·xH₂O), i.e. rust.

Chemical Kinetics

Q11. A first-order reaction is 75% complete in 32 minutes. Find its half-life and rate constant.

75% complete means 25% = (1/2)² remains, so two half-lives have passed. t½ = 32/2 = 16 min. k = 0.693/16 = 0.0433 min⁻¹.

Q12. Differentiate between molecularity and order of a reaction.

Molecularity is the number of reacting species colliding in an elementary step. It is always a whole number (1, 2, 3) and applies only to elementary steps. Order is the sum of the powers of concentration terms in the experimentally determined rate law. It can be zero, fractional or whole, and applies to the overall reaction.

Q13. What is activation energy? How can it be found graphically?

Activation energy (E_a) is the minimum extra energy that reacting molecules must have to form the activated complex. From the Arrhenius equation, ln k = ln A − E_a/RT, so a plot of log k against 1/T is a straight line with slope = −E_a/(2.303 R). A catalyst lowers E_a and thus increases the rate without changing ΔH.

d- and f-Block Elements

Q14. Why do transition elements show variable oxidation states and form coloured compounds?

The (n−1)d and ns electrons have very similar energies, so different numbers of electrons can be lost, and successive oxidation states usually differ by one unit. Colour comes from d–d transitions: in a complex, the d orbitals split, and electrons absorb visible light to jump to a higher d level. Ions with d⁰ or d¹⁰ configurations (e.g. Sc³⁺, Zn²⁺) have no d–d transition and are colourless.

Q15. What is lanthanoid contraction? State two consequences.

The steady decrease in atomic and ionic radii from La to Lu, caused by poor shielding of the nuclear charge by 4f electrons. Consequences: (i) the second and third transition series have almost identical radii (Zr ≈ Hf), making their separation difficult; (ii) the basic strength of the hydroxides decreases from La(OH)₃ to Lu(OH)₃.

Q16. Write the ionic equations for the oxidising action of acidified K₂Cr₂O₇ and KMnO₄ on Fe²⁺.

Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

Coordination Compounds

Q17. Write the IUPAC names of (i) [Co(NH₃)₅Cl]Cl₂ and (ii) K₃[Fe(CN)₆].

(i) Pentaamminechloridocobalt(III) chloride. (ii) Potassium hexacyanidoferrate(III).

Q18. Explain why [Fe(CN)₆]⁴⁻ is diamagnetic while [CoF₆]³⁻ is paramagnetic, using Crystal Field Theory.

In [Fe(CN)₆]⁴⁻ the metal is Fe²⁺ (d⁶). CN⁻ is a strong-field ligand, so Δo > pairing energy and the electrons pair up: t₂g⁶ eg⁰ (low spin, 0 unpaired electrons, diamagnetic). In [CoF₆]³⁻ the metal is Co³⁺ (d⁶). F⁻ is a weak-field ligand, so Δo < pairing energy and the electrons occupy all orbitals singly first: t₂g⁴ eg² (high spin, 4 unpaired electrons, paramagnetic).

Q19. What is linkage isomerism? Give an example.

It arises with ambidentate ligands, which can bind through two different donor atoms. Example: [Co(NH₃)₅(NO₂)]²⁺ (bonded through N) and [Co(NH₃)₅(ONO)]²⁺ (bonded through O).

Haloalkanes and Haloarenes

Q20. Distinguish between SN1 and SN2 mechanisms.

SN1 is a two-step reaction through a planar carbocation, shows first-order kinetics, gives racemisation, is favoured by tertiary substrates and polar protic solvents. SN2 is a single-step concerted reaction through a pentacoordinate transition state, shows second-order kinetics, gives inversion of configuration (Walden inversion), and is favoured by primary substrates and polar aprotic solvents.

Q21. Why is chlorobenzene less reactive than chloroethane towards nucleophilic substitution?

The lone pair on chlorine is in resonance with the benzene ring, giving the C–Cl bond partial double-bond character, so it is harder to break. In addition, the carbon is sp² hybridised (more electronegative than sp³), which also shortens and strengthens the bond, and the incoming nucleophile is repelled by the electron-rich ring.

Alcohols, Phenols and Ethers

Q22. How can you distinguish primary, secondary and tertiary alcohols with the Lucas test? Why is phenol more acidic than ethanol?

Lucas reagent is anhydrous ZnCl₂ in conc. HCl. Tertiary alcohols give turbidity immediately, secondary alcohols in about 5 minutes, and primary alcohols show no turbidity at room temperature. Phenol is more acidic because the phenoxide ion is stabilised by resonance (the negative charge is delocalised over the ring), while the ethoxide ion is destabilised by the electron-donating ethyl group.

Q23. Why can’t tert-butyl ethyl ether be made well by reacting sodium ethoxide with tert-butyl bromide? What should be used instead?

A tertiary alkyl halide undergoes elimination (E2) with the strongly basic ethoxide, giving isobutene rather than the ether. The correct pair is sodium tert-butoxide + bromoethane (a primary halide that undergoes SN2).

Aldehydes, Ketones and Carboxylic Acids

Q24. Distinguish between aldol condensation and Cannizzaro reaction.

Aldol condensation occurs with aldehydes and ketones that have at least one α-hydrogen, in the presence of dilute alkali; it gives β-hydroxy carbonyl compounds that dehydrate to α,β-unsaturated carbonyl compounds. The Cannizzaro reaction occurs with aldehydes that have no α-hydrogen (e.g. HCHO, C₆H₅CHO) in concentrated alkali: one molecule is oxidised to the carboxylate and another is reduced to the alcohol.

Q25. Give one chemical test each to distinguish (i) propanal and propanone, (ii) ethanal and propanal.

(i) Tollens’ test: propanal gives a silver mirror; propanone does not. (ii) Iodoform test: ethanal (a methyl ketone-type CH₃CO– group) gives a yellow precipitate of CHI₃ with I₂/NaOH; propanal does not.

Amines

Q26. Arrange ethylamine, diethylamine, triethylamine and ammonia in decreasing order of basic strength in aqueous solution and explain.

(C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₂ > NH₃. The ethyl groups push electrons (+I effect) and make the lone pair more available. But in water, the stability of the conjugate acid depends on solvation through hydrogen bonding, and this decreases from primary to tertiary. The balance of +I effect and solvation puts the secondary amine on top.

Q27. Why does aniline not undergo Friedel–Crafts reaction? How is it distinguished from N-methylaniline?

Aniline is basic and forms a salt with the AlCl₃ catalyst, which puts a positive charge on nitrogen and strongly deactivates the ring. Use the Hinsberg test: aniline (primary) forms a sulphonamide soluble in alkali; N-methylaniline (secondary) forms one that is insoluble in alkali.

Biomolecules

Q28. What are reducing and non-reducing sugars? Give examples.

Sugars with a free aldehydic or ketonic (hemiacetal/hemiketal) group that can reduce Tollens’ and Fehling’s reagents are reducing (glucose, fructose, maltose, lactose). In sucrose, the glycosidic bond joins the anomeric carbons of glucose and fructose, so there is no free anomeric –OH, and sucrose is non-reducing.

Q29. What is denaturation of proteins? Compare DNA and RNA briefly.

Denaturation is the loss of the secondary and tertiary structure (and biological activity) of a protein on heating or a change of pH, with the primary structure intact (e.g. coagulation of egg white). DNA has deoxyribose and the bases A, G, C, T and is double-stranded. RNA has ribose and the bases A, G, C, U and is generally single-stranded.

Part 2: Previous Year Question Formats

Editorial note before publishing: replace each [Year / Set code] below with the verified year and paper set from the official CBSE question paper and marking scheme (cbseacademic.nic.in). The questions below are based on formats and themes that CBSE has repeatedly used; do not publish any year label that you have not checked against the paper.

PYQ 1 [Year / Set code: to be verified]. Write the Nernst equation and calculate the emf of a Zn–Cu cell for given concentrations. Typical weightage: 2–3 marks. Method: see Q6 and Model Question 3.

PYQ 2 [Year / Set code: to be verified]. A first-order reaction takes a stated time to reach a given percentage completion. Calculate k or t½. Typical weightage: 2–3 marks. Method: see Q11 and Model Question 5.

PYQ 3 [Year / Set code: to be verified]. Write the IUPAC name and state the hybridisation and magnetic behaviour of a given coordination entity. Typical weightage: 2–3 marks. Method: see Q17 and Q18.

PYQ 4 [Year / Set code: to be verified]. Distinguish between a given pair of compounds by a simple chemical test. Typical weightage: 2 marks. Method: see Q22, Q25 and Q27.

PYQ 5 [Year / Set code: to be verified]. Give reasons: basicity of amines, acidity of phenol, reactivity of haloarenes. Typical weightage: 2–3 marks. Method: see Q21, Q22 and Q26.

PYQ 6 [Year / Set code: to be verified]. Assertion–reason questions on biomolecules, d-block or amines. Typical weightage: 1 mark each. Method: see Part 3.

Part 3: Competency-Based Questions

Case-Based Question 1 (4 marks)

Iron is protected from rusting by coating it with another metal. Zinc-coated (galvanised) iron lasts a long time even when the coating is scratched. Tin-plated cans, however, start rusting quickly once the tin is scratched. Given: E°(Zn²⁺/Zn) = −0.76 V; E°(Fe²⁺/Fe) = −0.44 V; E°(Sn²⁺/Sn) = −0.14 V.

(i) Why does galvanised iron remain protected even when scratched? (1) (ii) Calculate E°cell for the cell formed by Zn and Fe. (1) (iii) Why does tin-plated iron corrode faster when the tin layer is damaged? (2)

Case-Based Question 2 (4 marks)

Compound A (C₆H₅NO₂) is reduced with Sn/HCl to give compound B. B, on treatment with NaNO₂ and HCl at 273–278 K, gives compound C. C reacts with CuCN to give compound D (C₇H₅N). When B is warmed with chloroform and ethanolic KOH, a foul-smelling compound is produced.

(i) Identify A, B, C and D. (2) (ii) Name the reaction by which D is obtained from C. (1) (iii) Why must the diazotisation be carried out at 273–278 K? (1)

Assertion–Reason Questions (1 mark each)

Choose from: (a) both A and R are true and R explains A; (b) both are true but R does not explain A; (c) A is true, R is false; (d) A is false, R is true.

AR 1. Assertion: Molar conductivity of a weak electrolyte rises sharply at high dilution. Reason: The degree of dissociation increases on dilution.

AR 2. Assertion: Phenol is more acidic than ethanol. Reason: The phenoxide ion is less resonance-stabilised than the ethoxide ion.

AR 3. Assertion: Zn, Cd and Hg are not considered typical transition elements. Reason: They have completely filled d¹⁰ configuration in the ground state and in their common oxidation states.

Part 4: Model Questions (Practice Only)

These are practice questions written in the style of board papers. They are not predicted or guaranteed board questions.

M1 (1 mark). The unit of the rate constant of a second-order reaction is: (a) s⁻¹ (b) mol L⁻¹ s⁻¹ (c) L mol⁻¹ s⁻¹ (d) L² mol⁻² s⁻¹

M2 (1 mark). The number of unpaired electrons in [Mn(H₂O)₆]²⁺ is: (a) 1 (b) 3 (c) 5 (d) 0

M3 (2 marks). Calculate the emf of the cell Zn | Zn²⁺ (0.1 M) || Cu²⁺ (0.01 M) | Cu at 298 K. E°cell = 1.10 V.

M4 (2 marks). Give a chemical test to distinguish pentan-2-one from pentan-3-one.

M5 (3 marks). The rate constant of a first-order reaction is 3.0 × 10⁻³ s⁻¹. How long will it take for 90% of the reactant to be consumed?

M6 (3 marks). Convert: (a) benzoic acid to aniline; (b) ethanenitrile to ethanamine; (c) ethanol to ethanal.

M7 (3 marks). 2 g of a non-volatile solute dissolved in 100 g of benzene lowers the freezing point by 0.40 K. Calculate the molar mass of the solute. (K_f for benzene = 5.12 K kg/mol)

M8 (5 marks). (a) Why is Mn²⁺ more stable than Mn³⁺, whereas Cr²⁺ is a strong reducing agent? (3) (b) Draw the crystal-field splitting diagram for a d⁴ ion in a weak-field octahedral field and state the number of unpaired electrons. (2)

Part 5: Answer Explanations and Step-by-Step Solutions

Case-Based Question 1

(i) Zinc is more reactive than iron (more negative E°), so in the scratched area zinc acts as the anode and corrodes in preference to iron. This is sacrificial protection.

(ii) E°cell = E°cathode − E°anode = (−0.44) − (−0.76) = 0.32 V.

(iii) Tin is less reactive than iron (E° of −0.14 V is less negative than −0.44 V). When the coating is broken, iron becomes the anode and tin the cathode, so iron oxidises faster than it would alone.

Case-Based Question 2

(i) A = nitrobenzene (C₆H₅NO₂); B = aniline (C₆H₅NH₂), formed by reduction; C = benzenediazonium chloride (C₆H₅N₂⁺Cl⁻); D = benzonitrile (C₆H₅CN, formula C₇H₅N). The foul-smelling product is phenyl isocyanide, which confirms B as a primary amine (carbylamine test).

(ii) Sandmeyer reaction (using CuCN).

(iii) Benzenediazonium salts are unstable above about 278 K and decompose to phenol and N₂. A low temperature keeps the diazonium ion intact.

Assertion–Reason Answers

AR 1: (a). Both statements are true; the greater dissociation at high dilution explains the sharp rise in Λm.

AR 2: (c). Assertion is true. Reason is false because it is the phenoxide ion that is stabilised by resonance, not the ethoxide ion.

AR 3: (a). Both are true, and the d¹⁰ configuration explains why they do not behave as typical transition metals.

Model Questions

M1. (c). For an nth-order reaction the unit is (mol L⁻¹)^(1−n) s⁻¹. For n = 2: L mol⁻¹ s⁻¹.

M2. (c). Mn²⁺ is d⁵. H₂O is a weak-field ligand, so the complex is high spin, with all five electrons unpaired.

M3. E = E° − (0.0591/2) log([Zn²⁺]/[Cu²⁺]) = 1.10 − 0.02955 × log(0.1/0.01) = 1.10 − 0.02955 × log 10 = 1.10 − 0.02955 ≈ 1.07 V.

M4. Iodoform test. Pentan-2-one has a CH₃CO– group and gives a yellow precipitate of iodoform with I₂/NaOH. Pentan-3-one does not.

M5. t = (2.303/k) log([R]₀/[R]). For 90% consumption, [R] = 0.1[R]₀, so log 10 = 1. t = 2.303/(3.0 × 10⁻³) = 767.7 s ≈ 768 s (about 12.8 minutes).

M6. (a) C₆H₅COOH → (SOCl₂) C₆H₅COCl → (NH₃) C₆H₅CONH₂ → (Br₂ + NaOH; Hofmann bromamide degradation) C₆H₅NH₂. (b) CH₃CN → (H₂/Ni or LiAlH₄) CH₃CH₂NH₂. (c) CH₃CH₂OH → (PCC, or Cu at 573 K) CH₃CHO.

M7. ΔT_f = K_f × w₂ × 1000 / (M₂ × w₁), so M₂ = (5.12 × 2 × 1000)/(0.40 × 100) = 10240/40 = 256 g/mol.

M8. (a) Mn²⁺ is d⁵, a half-filled and exceptionally stable configuration, so Mn³⁺ (d⁴) readily gains an electron (E° Mn³⁺/Mn²⁺ is about +1.57 V). Cr²⁺ is d⁴ and readily loses an electron to become Cr³⁺ (d³, t₂g³, very stable), so it is a strong reducing agent (E° Cr³⁺/Cr²⁺ is about −0.41 V). (b) With weak-field ligands, Δo is small, so the four electrons go t₂g³ eg¹ (high spin). Number of unpaired electrons = 4.

Part 6: Exam Tips, Common Mistakes and Scoring Strategies

Where marks are won. Electrochemistry (9 marks) and Aldehydes, Ketones and Carboxylic Acids (8 marks) are the two heaviest units. Organic chemistry as a whole (Haloalkanes through Biomolecules) carries 33 marks, nearly half the paper, so a strong grip on named reactions and conversions pays off.

Common mistakes to avoid

Missing or wrong units. Always write units for k, molar conductivity and molality, and show them in each line of the calculation.
Log confusion. Use log₁₀ with 2.303; do not mix it with ln.
Naming errors in complexes. Ligands in alphabetical order (ignoring prefixes like di-, tri-), the oxidation state in Roman numerals, and the suffix -ate for anionic complexes.
Amine basicity. Always state whether the order is for aqueous solution or the gas phase. They are different.
Cannizzaro or aldol? Check for α-hydrogens before you name the reaction.
Incomplete reactions. Write the reagent and condition over each arrow. A product without a reagent often earns only part of the marks.
Calculation slips. Do the arithmetic slowly in numericals; a wrong value early in the question carries forward.

Smart revision habits

Keep a one-page name reactions sheet and a one-page reasoning questions sheet (“why is…”) and revise them every few days.
Practise drawing structures and mechanisms by hand, because typed memory rarely survives exam pressure.
Solve at least five previous-year papers from the official CBSE site in a timed 3-hour setting, and mark them against the official marking scheme.
Learn the key constants: F = 96500 C/mol, R = 8.314 J/mol·K, 0.0591 V (Nernst at 298 K).
In the last week, revise case-based and assertion–reason questions, since they now form a significant share of the paper.

In the exam hall. Use the 15-minute reading time to mark the questions you are sure of. Attempt Section A quickly, then move to the 5-mark questions while your mind is fresh. Keep the last 10 minutes for rechecking units, signs and balanced equations.

FAQs

1. Which chapters are the most important in CBSE Class 12 Chemistry? By marks, Electrochemistry (9), Aldehydes, Ketones and Carboxylic Acids (8), and then Solutions, Chemical Kinetics, d- and f-Block, Coordination Compounds and Biomolecules (7 each). But since every unit is examined, ignoring any one chapter is risky.

2. Are these questions going to come in the board exam? No one can guarantee that. These are important and representative questions on concepts that the board tests regularly. The model questions are practice material only.

3. How many marks do numerical problems carry? Numericals are spread across Solutions, Electrochemistry and Chemical Kinetics, and they typically appear in the 2-, 3- and 5-mark questions. Practising them thoroughly is one of the surest ways to score full marks.

4. How should I prepare for case-based questions? Read the passage twice, underline the data, and answer from the passage as well as from your NCERT understanding. Practise by converting each NCERT in-text example into a short case.

5. Is NCERT enough for the board exam? For the theory paper, yes: NCERT (the textbook and its in-text and exercise questions) is the foundation, and most questions are rooted in it. Use sample papers and previous-year papers on top of it to build speed and exam technique.

This article is meant for exam preparation. Please check the official CBSE website for the latest syllabus, sample papers and notifications.

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