⚗️ CHEMISTRY · MARCH 2026
ANSWER KEY – FULL SOLUTIONS & EXPLANATIONS
📋 Questions 1 – 5 (Multiple choice)
1 Electrolysis of molten NaCl
In the electrolysis of molten NaCl, the substance liberated at the cathode is : (a) Cl₂ (b) Na (c) H₂ (d) O₂
✅ Answer: (b) Na
Na⁺ ions are reduced to Na metal at the cathode: Na⁺ + e⁻ → Na.
Na⁺ ions are reduced to Na metal at the cathode: Na⁺ + e⁻ → Na.
2 Half‑life (first order)
The half-life of a first-order reaction depends on : (a) Initial concentration (b) Temperature only (c) Rate constant (d) Both (a) and (c)
✅ Answer: (c) Rate constant
t₁/₂ = 0.693/k ; independent of initial concentration.
t₁/₂ = 0.693/k ; independent of initial concentration.
3 Chelating ligand
Which of the following is a chelating ligand ? (a) NH₃ (b) H₂O (c) Cl⁻ (d) C₂O₄²⁻
✅ Answer: (d) C₂O₄²⁻ (oxalate)
Bidentate ligand with two donor oxygen atoms, forms chelate rings.
Bidentate ligand with two donor oxygen atoms, forms chelate rings.
4 Least reactive toward SN1
Which is least reactive towards nucleophilic substitution (SN1) ? (a) Benzyl chloride (b) Methyl chloride (c) Chlorobenzene (d) Allyl chloride
✅ Answer: (c) Chlorobenzene
Delocalisation of lone pair on Cl gives partial double bond character; very stable C–Cl bond.
Delocalisation of lone pair on Cl gives partial double bond character; very stable C–Cl bond.
5 Polysaccharide
Which of the following is a polysaccharide ? (a) maltose (b) sucrose (c) fructose (d) cellulose
✅ Answer: (d) cellulose
Cellulose is a polymer of glucose; others are di‑ or monosaccharides.
Cellulose is a polymer of glucose; others are di‑ or monosaccharides.
✍️ Questions 6 – 15 (short answers, 2 scores each)
6 Saline with medicine
Why saline water is mixed with medicine before injection?
Answer: Saline (0.9% NaCl) is isotonic with blood plasma. Mixing ensures same osmotic pressure, preventing RBCs from shrinking (crenation) or bursting (hemolysis).
7 ΔG° for Daniel cell
E° = 1.1 V. Calculate ΔG° for Zn(s)+Cu²⁺→Zn²⁺+Cu(s).
Answer: ΔG° = –nFE° = –2 × 96500 × 1.1 = –212300 J/mol = –212.3 kJ/mol.
8 Activation energy & catalyst
(i) Define activation energy. (ii) Effect of catalyst on rate.
(i) Minimum extra energy required for reactants to form products.
(ii) Catalyst provides alternative path with lower activation energy, increasing rate; itself not consumed.
(ii) Catalyst provides alternative path with lower activation energy, increasing rate; itself not consumed.
9 Rate constant of N₂O₅
[N₂O₅]₀ = 1.24×10⁻² M, after 60 min = 0.20×10⁻² M. Find k (300 K).
Answer: k = (2.303/60) log(1.24×10⁻² / 0.20×10⁻²) = (2.303/60) log(6.2) = (2.303/60)×0.7924 ≈ 0.0304 min⁻¹.
10 Products X and Y
(i) CH₃CH₂OH + PCl₃ → X + POCl₃ + HCl (ii) CH₃Br + AgF → Y + AgBr
X = CH₃CH₂Cl (chloroethane); Y = CH₃F (fluoromethane).
11 SN1 vs SN2 (two differences)
- Kinetics: SN1 first‑order (rate ∝ substrate); SN2 second‑order (rate ∝ [substrate][nucleophile]).
- Stereochemistry: SN1 racemisation; SN2 inversion.
12 (i) poisonous gas from CHCl₃ (ii) chlorobenzene → 1‑chloro‑2‑methylbenzene
(i) In light, chloroform slowly oxidised by air to an extremely poisonous gas called ?
Phosgene (COCl₂).
(ii) Friedel‑Crafts alkylation: C₆H₅Cl + CH₃Cl → o‑ClC₆H₄CH₃ (anhyd. AlCl₃).
(ii) Friedel‑Crafts alkylation: C₆H₅Cl + CH₃Cl → o‑ClC₆H₄CH₃ (anhyd. AlCl₃).
13 Industrial preparation of methanol
How methanol is prepared industrially ? Write equation.
From water gas (CO + H₂): CO + 2H₂ → CH₃OH (ZnO/Cr₂O₃, 300 °C, 200 atm).
14 Distinguish HCHO and CH₃CHO
One chemical test to distinguish formaldehyde and acetaldehyde.
Iodoform test: CH₃CHO gives yellow precipitate (CHI₃) with I₂/NaOH; HCHO does not react.
15 Nitrobenzene → 2,4,6‑Tribromo aniline
How will you convert nitrobenzene to 2,4,6‑Tribromo aniline?
1. Reduction (Sn/HCl) → aniline. 2. Bromine water → 2,4,6‑tribromoaniline.
🔬 Questions 16 – 26 (any eight, 3 scores each)
16 Molar mass of protein
200 cm³ solution, 1.26 g protein, π=2.57×10⁻³ bar at 300 K. R=0.083. Find M.
M = wRT/(πV) = (1.26×0.083×300) / (2.57×10⁻³×0.200) = 31.374 / 5.14×10⁻⁴ ≈ 61039 g/mol.
17 Rusting of iron (electrochemical)
Anode: Fe → Fe²⁺ + 2e⁻. Cathode: O₂ + 2H₂O + 4e⁻ → 4OH⁻. Fe²⁺ + 2OH⁻ → Fe(OH)₂, further oxidized to Fe₂O₃·xH₂O (rust).
18 Order vs molecularity, pseudo order
(i) Differences: Order is experimental sum of exponents; molecularity is number of species in rate step. Order can be fractional, molecularity integer.
(ii) Pseudo order: Reaction appears lower order due to excess reagent (e.g., ester hydrolysis with excess water follows first order).
(ii) Pseudo order: Reaction appears lower order due to excess reagent (e.g., ester hydrolysis with excess water follows first order).
19 Reasons (Zn,Cd,Hg; complexes; Sc³⁺/Ti³⁺ colour)
(i) d¹⁰ configuration, no partially filled d‑orbitals.
(ii) Small size, high charge, vacant d‑orbitals.
(iii) Sc³⁺ = d⁰ (no d‑d transition); Ti³⁺ = d¹ (one d‑d transition gives colour).
(ii) Small size, high charge, vacant d‑orbitals.
(iii) Sc³⁺ = d⁰ (no d‑d transition); Ti³⁺ = d¹ (one d‑d transition gives colour).
20 K₂Cr₂O₇ from chromite ore
1. Fusion with Na₂CO₃ + air → Na₂CrO₄. 2. Leach with water. 3. Acidify → Na₂Cr₂O₇. 4. Treat with KCl → K₂Cr₂O₇ crystals.
21 [Ni(CN)₄]²⁻, [Ni(CO)₄] & formula
(i) Different structures but same magnetic behaviour? (ii) Tetraamineaquachloridocobalt(III) chloride formula.
(i) Both diamagnetic: [Ni(CN)₄]²⁻ square planar (low‑spin d⁸), [Ni(CO)₄] tetrahedral (d¹⁰, all paired).
(ii) [Co(NH₃)₄(H₂O)Cl]Cl₂.
(ii) [Co(NH₃)₄(H₂O)Cl]Cl₂.
22 d‑orbital splitting (tetrahedral) & VBT limitation
(i) In tetrahedral field, e set (dₓ²₋ᵧ², d₂²) lower, t₂ set (dₓᵧ, dᵧ₂, d₂ₓ) higher, Δt smaller than Δo.
(ii) VBT cannot explain colour or magnetic details quantitatively.
(ii) VBT cannot explain colour or magnetic details quantitatively.
23 Phenol + bromine water / Zn dust / conc. HNO₃
Bromine water: 2,4,6‑tribromophenol. Zinc dust: benzene. Conc. HNO₃: o‑ and p‑nitrophenol.
24 Complete the table
| SL No | Reactant | Reagent | Major product | Name of reaction |
|---|---|---|---|---|
| 1 | RCOCl | H₂/Pd/BaSO₄ | RCHO | Rosenmund reduction |
| 2 | CH₃COOH | Cl₂/Red P | Cl–CH₂–COOH | HVZ reaction |
| 3 | CH₃CHO | Zn(Hg)/HCl | CH₃CH₃ | Clemmensen reduction |
25 Hofmann degradation & Hinsberg test
(i) Hofmann bromamide degradation.
(ii) Primary amine: sulphonamide soluble in alkali. Secondary amine: precipitate insoluble. Tertiary amine: no reaction.
(ii) Primary amine: sulphonamide soluble in alkali. Secondary amine: precipitate insoluble. Tertiary amine: no reaction.
26 Globular vs fibrous proteins; denaturation
Globular: spherical, compact, e.g., enzymes. Fibrous: thread‑like, structural, e.g., keratin. Denaturation: loss of 3D structure (biological activity) due to heat/pH etc., primary structure intact.
📐 Questions 27 – 31 (any four, 4 scores each)
27 Henry’s law & positive deviation plot
Law: p = K_H · x (solubility ∝ pressure). Applications: carbonated beverages, scuba diving (He/O₂ to avoid bends).
Plot: total pressure curve lies above ideal line (Raoult); sketch with p_A, p_B above dashed line.
Plot: total pressure curve lies above ideal line (Raoult); sketch with p_A, p_B above dashed line.
28 Daniel cell Nernst & conductivity
Cell reaction: Zn + Cu²⁺ → Zn²⁺ + Cu; Nernst: E = E° – (0.0591/2) log([Zn²⁺]/[Cu²⁺]).
Conductivity (κ) decreases with dilution; molar conductivity (Λₘ) increases with dilution.
Conductivity (κ) decreases with dilution; molar conductivity (Λₘ) increases with dilution.
29 Werner’s postulates & [Co(NH₃)₃(NO₂)₃] isomers
Two postulates: Primary & secondary valency; fixed coordination number with definite geometry. Isomers: fac (same ligands on one face) and mer (ligands in a meridional plane).
30 Hydroboration products A, B & cumene to phenol
3CH₃–CH=CH₂ + (H–BH₂)₂ → A → (H₂O₂/OH⁻) B.
A = (CH₃CH₂CH₂)₃B (tripropylborane); B = CH₃CH₂CH₂OH (propan‑1‑ol).
Cumene → phenol: cumene + O₂ → cumene hydroperoxide; then H⁺ → phenol + acetone.
Cumene → phenol: cumene + O₂ → cumene hydroperoxide; then H⁺ → phenol + acetone.
31 Aldol condensation & conversions
Aldol eg: 2CH₃CHO → CH₃CH(OH)CH₂CHO (3‑hydroxybutanal) then dehydration.
(a) Benzene → benzaldehyde: Gatterman‑Koch (CO+HCl, AlCl₃/CuCl).
(b) Ethanoic acid → ethanol: LiAlH₄ reduction.
(a) Benzene → benzaldehyde: Gatterman‑Koch (CO+HCl, AlCl₃/CuCl).
(b) Ethanoic acid → ethanol: LiAlH₄ reduction.