CBSE Class 12 Chemistry Sample Paper 2027 with Solutions — Complete Guide

In short: CBSE Class 12 Chemistry Sample Paper 2027 has 35 questions across 5 sections — 70 marks in 3 hours. Organic Chemistry carries 28 marks (40%), making it the highest-weightage unit. This guide covers complete solutions for reactions, mechanisms, numericals, and name reactions.

Paper Pattern 2027

Section-wise Breakdown

Chapter-wise Weightage 2027

High-Weightage Topics

Section A — MCQs (16 × 1 = 16 Marks)

Physical Chemistry

Q1. The number of atoms in 100 g of an FCC crystal with density 10 g/cm³ and cell edge 200 pm is:

• (a) 3 × 10²⁵
• (b) 5 × 10²⁴
• (c) 1 × 10²⁵
• (d) 2 × 10²⁵

Solution: FCC has 4 atoms per unit cell. Volume of unit cell = (200 × 10⁻¹⁰)³ = 8 × 10⁻²⁴ cm³ Mass of unit cell = density × volume = 10 × 8 × 10⁻²⁴ = 8 × 10⁻²³ g Number of unit cells in 100 g = 100/(8 × 10⁻²³) = 1.25 × 10²⁴ Number of atoms = 4 × 1.25 × 10²⁴ = 5 × 10²⁴

Answer: (b) 5 × 10²⁴

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Q2. The elevation in boiling point of a solution of 6 g of a non-volatile solute in 100 g of water is 0.2°C. The molar mass of solute is (Kb = 0.52 K kg/mol):

• (a) 156 g/mol
• (b) 176 g/mol
• (c) 186 g/mol
• (d) 196 g/mol

Solution: ΔTb = Kb × m 0.2 = 0.52 × (6/M × 1000/100) 0.2 = 0.52 × 60/M M = 0.52 × 60/0.2 = 156 g/mol

Answer: (a) 156 g/mol

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Q3. The standard EMF of a cell with reaction Zn + Cu²⁺ → Zn²⁺ + Cu is 1.10 V. The standard Gibbs energy change is:

• (a) -212.3 kJ/mol
• (b) +212.3 kJ/mol
• (c) -106.15 kJ/mol
• (d) +106.15 kJ/mol

Solution: ΔG° = -nFE° n = 2 (electrons transferred) ΔG° = -2 × 96500 × 1.10 ΔG° = -212300 J/mol = -212.3 kJ/mol

Answer: (a) -212.3 kJ/mol

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Q4. For a first-order reaction, the half-life is 10 minutes. The time required for 75% completion is:

• (a) 15 min
• (b) 20 min
• (c) 25 min
• (d) 30 min

Solution: For first-order: t = (2.303/k) log(a/(a-x)) 75% completion means 25% remains. t = (2.303/k) log(100/25) = (2.303/k) log 4 = (2.303/k) × 0.602

Also, t₁/₂ = 0.693/k, so k = 0.693/10

t = (2.303 × 10/0.693) × 0.602 = 20 min

Answer: (b) 20 min

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Inorganic Chemistry

Q5. Which of the following oxides is amphoteric?

• (a) CaO
• (b) CO₂
• (c) Al₂O₃
• (d) SO₂

Solution: Al₂O₃ reacts with both acids and bases:

• With acid: Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O
• With base: Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O

Answer: (c) Al₂O₃

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Q6. The IUPAC name of [Co(NH₃)₄Cl₂]Cl is:

• (a) Tetraamminedichloridocobalt(III) chloride
• (b) Tetraamminedichlorocobalt(III) chloride
• (c) Dichloridotetraaminecobalt(III) chloride
• (d) Cobalt tetraamine dichloride chloride

Solution: IUPAC naming: Ligands in alphabetical order (ammine before chlorido), then metal with oxidation state, then counter ion.

Answer: (a) Tetraamminedichloridocobalt(III) chloride

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Organic Chemistry

Q7. The major product of the reaction of 2-bromobutane with alcoholic KOH is:

• (a) But-1-ene
• (b) But-2-ene
• (c) Butan-2-ol
• (d) Butane

Solution: Alcoholic KOH causes elimination (E2 reaction). By Saytzeff's rule, more substituted alkene is major product. CH₃-CHBr-CH₂-CH₃ → CH₃-CH=CH-CH₃ (But-2-ene)

Answer: (b) But-2-ene

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Q8. Which of the following is the most acidic?

• (a) Phenol
• (b) p-Nitrophenol
• (c) p-Cresol
• (d) p-Methoxyphenol

Solution: Electron-withdrawing groups (like -NO₂) increase acidity by stabilizing the phenoxide ion. Electron-donating groups (like -CH₃, -OCH₃) decrease acidity. Order: p-Nitrophenol > Phenol > p-Cresol > p-Methoxyphenol

Answer: (b) p-Nitrophenol

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Q9. The product of reaction between acetaldehyde and HCN is:

• (a) Acetaldehyde cyanohydrin
• (b) Acetone cyanohydrin
• (c) Lactic acid
• (d) Glycolic acid

Solution: CH₃CHO + HCN → CH₃CH(OH)CN (Acetaldehyde cyanohydrin) This is a nucleophilic addition reaction.

Answer: (a) Acetaldehyde cyanohydrin

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Q10. Aniline reacts with bromine water to give:

• (a) o-Bromoaniline
• (b) p-Bromoaniline
• (c) 2,4-Dibromoaniline
• (d) 2,4,6-Tribromoaniline

Solution: -NH₂ is a strong activating group. With bromine water (excess), all three positions (2,4,6) get brominated.

Answer: (d) 2,4,6-Tribromoaniline

Section B — Assertion-Reason (4 × 1 = 4 Marks)

Q17. Assertion (A): Copper is extracted by hydrometallurgy. Reason (R): Copper lies below hydrogen in the electrochemical series.

Solution: Copper lies below hydrogen, so it can be displaced by hydrogen or more reactive metals. Hydrometallurgy (leaching + displacement) is used.

Answer: (a) Both A and R are true, R is correct explanation of A

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Section C — Very Short Answer (5 × 2 = 10 Marks)

Q21. Define the following with one example each: (a) Schottky defect (b) Frenkel defect

Solution:

(a) Schottky Defect: A point defect where equal number of cations and anions are missing from their lattice sites, maintaining electrical neutrality.

• Occurs in ionic compounds with high coordination number
• Decreases density
• Example: NaCl, KCl, CsCl

(b) Frenkel Defect: A point defect where an ion (usually smaller cation) leaves its lattice site and occupies an interstitial site.

• Occurs when there's large difference in ion sizes
• Density remains same
• Example: AgCl, AgBr, ZnS

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Q22. Calculate the osmotic pressure of a solution containing 3.42 g of sucrose (M = 342) in 500 mL of solution at 300 K.

Solution: π = CRT = (n/V)RT

n = 3.42/342 = 0.01 mol V = 500 mL = 0.5 L R = 0.0821 L atm K⁻¹ mol⁻¹ T = 300 K

π = (0.01/0.5) × 0.0821 × 300 π = 0.02 × 24.63 π = 0.493 atm

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Q23. Write the reactions involved in the following: (a) Reimer-Tiemann reaction (b) Kolbe's reaction

Solution:

(a) Reimer-Tiemann Reaction: Phenol reacts with chloroform and aqueous NaOH to form salicylaldehyde.

C₆H₅OH + CHCl₃ + 3NaOH → o-HOC₆H₄CHO + 3NaCl + H₂O

(b) Kolbe's Reaction: Sodium phenoxide reacts with CO₂ under pressure to form sodium salicylate.

C₆H₅ONa + CO₂ → o-HOC₆H₄COONa (Sodium salicylate)

On acidification: o-HOC₆H₄COONa + HCl → o-HOC₆H₄COOH + NaCl (Salicylic acid)

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Section D — Short Answer (7 × 3 = 21 Marks)

Q26. (a) Write the mechanism of the following reaction: CH₃CH₂Br + OH⁻ → CH₃CH₂OH + Br⁻

(b) Why is SN1 reaction faster in polar protic solvents?

Solution:

(a) SN2 Mechanism:

This is an SN2 (bimolecular nucleophilic substitution) reaction.

Step 1: OH⁻ attacks the carbon bearing Br from the backside.

Step 2: Transition state forms with partial bonds to both OH and Br.

Step 3: Br⁻ leaves as OH fully bonds.

        δ-    δ+    δ-
HO⁻ + CH₃-CH₂-Br → [HO---CH₂---Br]‡ → HO-CH₂-CH₃ + Br⁻
                    Transition State

Characteristics:

• One-step mechanism
• Inversion of configuration (Walden inversion)
• Rate = k[substrate][nucleophile]

(b) SN1 in Polar Protic Solvents:

• SN1 involves carbocation intermediate
• Polar protic solvents (water, alcohols) stabilize carbocation through solvation
• They also stabilize the leaving group (anion) through hydrogen bonding
• This lowers activation energy and speeds up the reaction

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Q27. Give reasons for the following: (a) Transition metals form coloured compounds. (b) Zn, Cd, Hg are not considered typical transition elements. (c) Cr²⁺ is a strong reducing agent but Mn³⁺ is not.

Solution:

(a) Coloured Compounds:

• Transition metals have partially filled d-orbitals
• d-d transitions occur when electrons absorb visible light and jump to higher d-orbitals
• The complementary colour is observed
• Energy gap corresponds to visible light wavelength

(b) Zn, Cd, Hg:

• They have completely filled d-orbitals (d¹⁰) in ground state and common oxidation states
• No d-d transitions possible → colourless compounds
• Don't show typical transition metal properties (variable oxidation states, catalytic activity)

(c) Cr²⁺ vs Mn³⁺:

• Cr²⁺ (3d⁴) easily loses one electron to become Cr³⁺ (3d³) which has half-filled t₂g orbitals (stable)
• Mn³⁺ (3d⁴) would need to become Mn⁴⁺ (3d³) which requires high energy
• Mn²⁺ (3d⁵) is more stable due to half-filled d-orbitals
• Hence Cr²⁺ is strong reducing agent, Mn³⁺ is not

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Q28. Complete the following reactions: (a) CH₃CHO + 2[Ag(NH₃)₂]⁺ + OH⁻ → (b) C₆H₅NH₂ + NaNO₂ + HCl (0-5°C) → (c) CH₃COOH + SOCl₂ →

Solution:

(a) Tollens' Test (Silver Mirror Test): CH₃CHO + 2[Ag(NH₃)₂]⁺ + 3OH⁻ → CH₃COO⁻ + 2Ag↓ + 4NH₃ + 2H₂O

(Aldehydes reduce Tollens' reagent, silver mirror forms)

(b) Diazotization: C₆H₅NH₂ + NaNO₂ + 2HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O (Benzenediazonium chloride)

(Reaction at 0-5°C to prevent decomposition of diazonium salt)

(c) Acid Chloride Formation: CH₃COOH + SOCl₂ → CH₃COCl + SO₂↑ + HCl↑ (Acetyl chloride)

(Thionyl chloride converts carboxylic acid to acid chloride)

Section E — Long Answer & Case-Based (19 Marks)

Q32. (5 marks) (a) How would you convert: (i) Ethanol to ethanoic acid (ii) Benzene to acetophenone (b) An organic compound A (C₃H₆O) gives positive iodoform test. Identify A and write the reaction.

Solution:

(a) Conversions:

(i) Ethanol to Ethanoic Acid:

CH₃CH₂OH  ──────────────→  CH₃COOH
          Acidified K₂Cr₂O₇
          or alkaline KMnO₄

Ethanol is oxidized to ethanoic acid (acetic acid).

(ii) Benzene to Acetophenone:

C₆H₆ + CH₃COCl  ──────────→  C₆H₅COCH₃ + HCl
                Anhydrous AlCl₃
                (Friedel-Crafts Acylation)

(b) Identification:

• Molecular formula C₃H₆O
• Positive iodoform test → contains CH₃CO- or CH₃CH(OH)- group
• Compound A is Acetone (CH₃COCH₃) or Propan-2-ol (CH₃CHOHCH₃)

Since formula is C₃H₆O (no extra H for alcohol), A is Acetone.

Iodoform Reaction: CH₃COCH₃ + 3I₂ + 4NaOH → CHI₃↓ + CH₃COONa + 3NaI + 3H₂O (Iodoform - yellow precipitate)

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Q33. (5 marks) (a) State and explain Kohlrausch's law of independent migration of ions. (b) The conductivity of 0.001 M acetic acid is 4 × 10⁻⁵ S cm⁻¹. Calculate its molar conductivity and degree of dissociation. (Λ°m for CH₃COOH = 390.5 S cm² mol⁻¹)

Solution:

(a) Kohlrausch's Law: The limiting molar conductivity of an electrolyte is the sum of the individual contributions of the cation and anion.

Λ°m = ν₊λ°₊ + ν₋λ°₋

Where:

• Λ°m = Limiting molar conductivity of electrolyte
• λ°₊, λ°₋ = Limiting molar conductivities of cation and anion
• ν₊, ν₋ = Number of cations and anions per formula unit

Applications:

1. Calculate Λ°m of weak electrolytes (which can't be measured directly)
2. Calculate degree of dissociation of weak electrolytes
3. Determine solubility of sparingly soluble salts

(b) Numerical:

Molar Conductivity: Λm = (κ × 1000)/C Λm = (4 × 10⁻⁵ × 1000)/0.001 Λm = 40 S cm² mol⁻¹

Degree of Dissociation: α = Λm/Λ°m α = 40/390.5 α = 0.102 or 10.2%

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Case Study (4 marks)

Q34. Read the passage and answer:

Polymers in Daily Life

Polymers are large molecules made of repeating units called monomers. Based on their response to heat, polymers are classified as thermoplastics (soften on heating, can be remoulded) and thermosetting plastics (harden permanently on heating). Biodegradable polymers like PHBV and PLA are gaining importance due to environmental concerns.

(i) Give one example each of addition and condensation polymer. (1 mark)

Solution:

• Addition polymer: Polythene (from ethene)
• Condensation polymer: Nylon-6,6 (from adipic acid and hexamethylenediamine)

(ii) Why is Bakelite a thermosetting plastic? (1 mark)

Solution: Bakelite has extensive cross-linking between polymer chains. On heating, these cross-links become permanent, making it hard and infusible. It cannot be softened or remoulded.

(iii) Write the monomers of Nylon-6,6. (1 mark)

Solution:

• Adipic acid: HOOC-(CH₂)₄-COOH
• Hexamethylenediamine: H₂N-(CH₂)₆-NH₂

(iv) Why are biodegradable polymers preferred? (1 mark)

Solution:

• They decompose naturally by microorganisms
• Don't accumulate in environment
• Reduce plastic pollution
• Made from renewable resources

Important Name Reactions

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