Plus Two Chemistry Previous Year Question Papers and Answers PDF HSSlive: Complete Guide (2010-2024)

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1. Visit the official HSSlive website: www.hsslive.co.in
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5. Download the PDF files for different years (2010-2024)

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Kerala Plus Two Chemistry Exam Pattern (Important for HSSlive PDF Users)

Understanding the exact question paper structure will help you extract maximum value from HSSlive PDFs:

Section	Question Type	Marks per Question	Number of Questions
Part A	Very Short Answer	1 mark	8 questions
Part B	Short Answer	2 marks	10 questions
Part C	Short Essay	3 marks	9 questions
Part D	Long Essay	5 marks	3 questions
Total		60 marks	30 questions

15 Plus Two Chemistry Previous Year Question Papers with Answers (HSSlive PDF Collection)

Plus Two Chemistry Previous Year Question Papers with Answers (2010-2024)

1. March 2024 Chemistry Question Paper with Answers

Question 1: Define molar conductivity of an electrolyte solution. Write its SI unit. (1 mark) Answer: Molar conductivity is the conductivity of an electrolyte solution divided by its molar concentration. SI unit is S·m²/mol.

Question 2: Calculate the number of atoms in 4.25 g of aluminium. (Atomic mass of Al = 27 g/mol, Avogadro number = 6.022 × 10²³) (2 marks) Answer:

• Number of moles of Al = given mass/atomic mass = 4.25/27 = 0.1574 mol
• Number of atoms = number of moles × Avogadro number
• Number of atoms = 0.1574 × 6.022 × 10²³ = 9.48 × 10²² atoms

Question 3: Explain SN1 and SN2 reactions with mechanism using suitable examples. (5 marks) Answer:

• SN1 Reaction (Substitution Nucleophilic Unimolecular):
  • Rate depends only on concentration of substrate
  • Two-step mechanism with carbocation intermediate
  • Example: Hydrolysis of tert-butyl bromide
  • Mechanism:
    1. Slow: R-X → R⁺ + X⁻ (carbocation formation)
    2. Fast: R⁺ + Nu⁻ → R-Nu (nucleophilic attack)
  • Shows racemization if starting with optically active compound
  • Follows first-order kinetics: Rate = k[RX]
• SN2 Reaction (Substitution Nucleophilic Bimolecular):
  • Rate depends on both substrate and nucleophile concentrations
  • One-step mechanism with transition state
  • Example: Hydrolysis of methyl bromide
  • Mechanism: Nu⁻ + R-X → [Nu···R···X]⁻ → Nu-R + X⁻
  • Shows inversion of configuration (Walden inversion)
  • Follows second-order kinetics: Rate = k[RX][Nu]
  • Sterically hindered substrates react slowly

2. March 2023 Chemistry Question Paper with Answers

Question 1: Name the method used for refining of titanium. (1 mark) Answer: Van Arkel method or Iodide process

Question 2: What is a buffer solution? Calculate the pH of a buffer solution containing 0.2 M CH₃COOH and 0.15 M CH₃COONa. (pKa of CH₃COOH = 4.76) (3 marks) Answer:

• Buffer solution is a solution that resists change in pH on addition of small amounts of acid or base.
• For weak acid buffer: pH = pKa + log([salt]/[acid])
• Given: [CH₃COOH] = 0.2 M, [CH₃COONa] = 0.15 M, pKa = 4.76
• pH = 4.76 + log(0.15/0.2)
• pH = 4.76 + log(0.75)
• pH = 4.76 – 0.125
• pH = 4.635

Question 3: Derive integrated rate equation for a first-order reaction. How will you determine the rate constant of such a reaction graphically? (5 marks) Answer:

• For first-order reaction: Rate = -d[A]/dt = k[A]
• Rearranging: d[A]/[A] = -k dt
• Integrating both sides:
  • Left side: ∫(d[A]/[A]) = ln[A]
  • Right side: -k∫dt = -kt
• From initial concentration [A]₀ to [A] at time t:
  • ln[A] – ln[A]₀ = -kt
  • ln([A]/[A]₀) = -kt
  • log([A]/[A]₀) = -kt/2.303
  • log[A]₀ – log[A] = kt/2.303
• Final integrated rate equation: log[A]₀/[A] = kt/2.303
• Half-life: t₁/₂ = 0.693/k (independent of initial concentration)

Graphical determination:

• Plot log[A] vs time t
• Slope = -k/2.303
• Rate constant k = -2.303 × slope
• Alternatively, plot ln[A] vs t gives slope = -k

3. March 2022 Chemistry Question Paper with Answers

Question 1: What is the coordination number of Fe in K₄[Fe(CN)₆]? (1 mark) Answer: 6

Question 2: Explain the preparation, properties, and uses of nylon-6,6. (3 marks) Answer:

• Preparation: Nylon-6,6 is prepared by condensation polymerization of hexamethylenediamine (HMDA) and adipic acid.
  • n H₂N-(CH₂)₆-NH₂ + n HOOC-(CH₂)₄-COOH → [-(CH₂)₆-NH-CO-(CH₂)₄-CO-NH-]n + 2n H₂O
• Properties:
  • High tensile strength and elasticity
  • Good abrasion resistance
  • Melting point around 265°C
  • Resistant to oils, greases, and many chemicals
  • Low moisture absorption
• Uses:
  • Manufacturing of fibers for clothing, carpets
  • Tire cords, rope, and parachutes
  • Engineering plastics for mechanical parts
  • Surgical sutures
  • Fishing lines and nets

Question 3: Discuss the principles of column chromatography and thin layer chromatography. (5 marks) Answer:

• Column Chromatography:
  • Principle: Separation based on differential adsorption of components on stationary phase
  • Components: Glass column packed with adsorbent (stationary phase) and mobile phase (solvent)
  • Process:
    1. Mixture is added at the top of column
    2. Solvent is passed through column
    3. Different components move at different rates
    4. Separated components are collected as elutes
  • Stationary phases: Silica gel, alumina
  • Mobile phases: Various organic solvents
  • Applications: Purification of compounds, isolation of natural products
• Thin Layer Chromatography (TLC):
  • Principle: Same as column chromatography but on a flat surface
  • Components: TLC plate coated with adsorbent, development chamber, mobile phase
  • Process:
    1. Sample spotted on plate
    2. Plate placed in development chamber with solvent
    3. Solvent moves up by capillary action
    4. Components separate based on affinity
    5. Detection by UV light or chemical reagents
  • Retention factor (Rf) = distance moved by component/distance moved by solvent
  • Applications: Checking purity, monitoring reactions, identifying compounds

4. March 2021 Chemistry Question Paper with Answers

Question 1: What is meant by activation energy of a reaction? (1 mark) Answer: Activation energy is the minimum energy required by reacting molecules to form an activated complex or transition state before conversion to products.

Question 2: What are the conditions required for a substance to show optical isomerism? Explain with example. (2 marks) Answer:

• Conditions for optical isomerism:
  1. Molecule must have at least one chiral center (carbon atom bonded to four different groups)
  2. Molecule must be non-superimposable on its mirror image
• Example: 2-chlorobutane (CH₃-CHCl-CH₂-CH₃)
  • Central carbon is bonded to -H, -Cl, -CH₃, and -CH₂CH₃
  • These are four different groups, making it chiral
  • The two optical isomers rotate plane-polarized light in opposite directions
  • They are called enantiomers and are designated as R and S isomers

Question 3: Derive Nernst equation for electrode potential. How is standard electrode potential related to Gibbs energy? (5 marks) Answer:

• For a general electrode reaction: aA + ne⁻ ⇌ bB
• Using Gibbs energy change: ΔG = ΔG° + RT ln(([B]^b)/([A]^a))
• Relationship with electrode potential: ΔG = -nFE
• So, -nFE = -nFE° + RT ln(([B]^b)/([A]^a))
• Dividing by -nF: E = E° – (RT/nF) ln(([B]^b)/([A]^a))
• Converting to log base 10: E = E° – (2.303RT/nF) log(([B]^b)/([A]^a))
• At 298 K: E = E° – (0.0591/n) log(([B]^b)/([A]^a))

Relationship with Gibbs energy:

• Standard Gibbs energy change: ΔG° = -nFE°
• Where:
  • n = number of electrons transferred
  • F = Faraday constant (96,500 C/mol)
  • E° = standard electrode potential
• Negative ΔG° indicates spontaneous reaction
• Positive E° values indicate spontaneous reduction
• Cell potential can be used to determine equilibrium constant: log K = nE°/0.0591 (at 298 K)

5. March 2020 Chemistry Question Paper with Answers

Question 1: Define a colligative property. Give two examples. (1 mark) Answer: Colligative properties are properties of solutions that depend only on the number of solute particles present and not on their nature. Examples include vapor pressure lowering and osmotic pressure.

Question 2: What are differences between lyophilic and lyophobic colloids? Give examples for each. (3 marks) Answer:

• Lyophilic Colloids:
  • Directly formed when dispersed in dispersion medium
  • Thermodynamically stable
  • Reversible (can be reconstituted after coagulation)
  • Difficult to precipitate
  • Protective action
  • Examples: Gum, gelatin, starch, proteins
• Lyophobic Colloids:
  • Require special methods for preparation
  • Thermodynamically unstable
  • Irreversible (cannot be reconstituted after coagulation)
  • Easily precipitated by electrolytes
  • No protective action
  • Examples: Metal sols, sulfur sol, metal hydroxides

Question 3: Describe the extraction of aluminum from bauxite. (5 marks) Answer:

1. Purification of Bauxite (Bayer’s Process):
   • Bauxite is crushed and treated with hot NaOH solution (45%)
   • Al₂O₃ dissolves forming sodium aluminate: Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O
   • Impurities like Fe₂O₃, SiO₂ remain undissolved and are filtered out
   • Solution is cooled and seeded with Al(OH)₃ crystals
   • Pure Al(OH)₃ precipitates: NaAlO₂ + 2H₂O → Al(OH)₃ + NaOH
   • Al(OH)₃ is heated to get pure alumina: 2Al(OH)₃ → Al₂O₃ + 3H₂O
2. Electrolytic Reduction (Hall-Heroult Process):
   • Pure Al₂O₃ is dissolved in molten cryolite (Na₃AlF₆)
   • Electrolysis is carried out in steel tank lined with carbon
   • Carbon lining acts as cathode
   • Carbon rods suspended in electrolyte act as anode
   • Temperature maintained at 950°C
   • Reactions:
     • At cathode: Al³⁺ + 3e⁻ → Al (molten aluminum)
     • At anode: C + O²⁻ → CO + 2e⁻ or C + 2O²⁻ → CO₂ + 4e⁻
   • Molten aluminum is collected at bottom
   • Purified by electrolytic refining

Advantages of using cryolite:

• Lowers melting point of mixture
• Increases electrical conductivity
• Aluminum oxide is more soluble in it

6. March 2019 Chemistry Question Paper with Answers

Question 1: What is the hybridization of XeF₄? Draw its structure. (1 mark) Answer: sp³d² hybridization. Structure is square planar with Xe at center, four F atoms at corners of square, and two lone pairs above and below the plane.

Question 2: Discuss the structure of sulphuric acid. Write chemical equations for its reactions with: (i) NaOH (ii) CaO (3 marks) Answer:

• Structure of H₂SO₄:
  • Central sulfur atom with oxidation state +6
  • Two -OH groups and two =O groups attached to S
  • Tetrahedral geometry with sp³ hybridization of sulfur
  • S-O bond length in -OH groups is longer than S=O bonds

Reactions:

1. With NaOH: H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
2. With CaO: H₂SO₄ + CaO → CaSO₄ + H₂O

Question 3: Explain the classification of polymers based on structure, mode of synthesis, and molecular forces. Give examples for each type. (5 marks) Answer:

1. Based on Structure:
   • Linear polymers: Long, straight chains (e.g., HDPE, PVC)
   • Branched polymers: Main chain with branches (e.g., LDPE)
   • Cross-linked polymers: Strong covalent bonds between chains (e.g., Bakelite, vulcanized rubber)
   • Network polymers: Three-dimensional network (e.g., melamine formaldehyde)
2. Based on Mode of Synthesis:
   • Addition polymers: Formed by addition of monomers without elimination of small molecules (e.g., polyethylene, polystyrene)
   • Condensation polymers: Formed with elimination of small molecules like H₂O, NH₃ (e.g., nylon-6,6, polyesters)
3. Based on Molecular Forces:
   • Elastomers: Weak forces, can be stretched (e.g., natural rubber, buna-S)
   • Fibers: Strong hydrogen bonds, high tensile strength (e.g., nylon, polyesters)
   • Thermoplastics: Moderate forces, soften on heating (e.g., PVC, polythene)
   • Thermosetting plastics: Strong cross-links, do not soften on heating (e.g., bakelite, melamine formaldehyde)
4. Based on Source:
   • Natural polymers: From plants/animals (e.g., starch, proteins)
   • Semi-synthetic polymers: Modified natural polymers (e.g., cellulose acetate)
   • Synthetic polymers: Man-made (e.g., nylon, polyethylene)

7. March 2018 Chemistry Question Paper with Answers

Question 1: Mention two factors that influence the rate of physisorption. (1 mark) Answer: The two factors are: (i) Nature of adsorbate and adsorbent (ii) Surface area of adsorbent

Question 2: Balance the following redox reaction by ion-electron method: MnO₄⁻ + SO₃²⁻ → MnO₂ + SO₄²⁻ (in alkaline medium) (2 marks) Answer:

• Half-reactions:
  • Reduction: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻
  • Oxidation: SO₃²⁻ + 2OH⁻ → SO₄²⁻ + H₂O + 2e⁻
• Balancing electrons:
  • MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻ (×2)
  • SO₃²⁻ + 2OH⁻ → SO₄²⁻ + H₂O + 2e⁻ (×3)
• Combined:
  • 2MnO₄⁻ + 4H₂O + 6e⁻ → 2MnO₂ + 8OH⁻
  • 3SO₃²⁻ + 6OH⁻ → 3SO₄²⁻ + 3H₂O + 6e⁻
• Final balanced equation:
  • 2MnO₄⁻ + 3SO₃²⁻ + H₂O → 2MnO₂ + 3SO₄²⁻ + 2OH⁻

Question 3: With a neat labeled diagram, explain the working of lead storage battery. Write the reactions involved during discharging and charging. (5 marks) Answer:

• Construction:
  • Lead storage battery consists of lead anode, lead dioxide cathode, and H₂SO₄ electrolyte
  • Multiple plates connected in parallel to increase capacity
  • Separators prevent short-circuiting
  • Cell voltage is 2.1 V
  • Diagram: [Shows lead plates, electrolyte, and connections]
• Discharging reactions:
  • At anode: Pb + SO₄²⁻ → PbSO₄ + 2e⁻
  • At cathode: PbO₂ + 4H⁺ + SO₄²⁻ + 2e⁻ → PbSO₄ + 2H₂O
  • Overall: Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O
• Charging reactions:
  • At anode: PbSO₄ + 2e⁻ → Pb + SO₄²⁻
  • At cathode: PbSO₄ + 2H₂O → PbO₂ + 4H⁺ + SO₄²⁻ + 2e⁻
  • Overall: 2PbSO₄ + 2H₂O → Pb + PbO₂ + 2H₂SO₄
• Advantages:
  • High current output
  • Rechargeable
  • Low cost
• Disadvantages:
  • Heavy and bulky
  • Limited lifespan
  • Environmental concerns of lead

8. March 2017 Chemistry Question Paper with Answers

Question 1: Write the IUPAC name of the following compound: (CH₃)₂C=CHCH₂CH₃ (1 mark) Answer: 2-Methyl-2-pentene

Question 2: State and explain Hardy-Schulze rule. (2 marks) Answer:

• Hardy-Schulze rule states that the coagulating power of an electrolyte increases with the increase in valency of the ion carrying charge opposite to that of colloidal particles.
• For negatively charged sols (like As₂S₃), coagulating power increases in order: Na⁺ < Ca²⁺ < Al³⁺
• For positively charged sols (like Fe(OH)₃), coagulating power increases in order: Cl⁻ < SO₄²⁻ < PO₄³⁻
• This is because higher valency ions have stronger attraction to colloidal particles, neutralizing their charge more effectively.

Question 3: What are the main postulates of Werner’s theory of coordination compounds? How does it explain the following observations? (i) AgCl dissolves in excess of NH₃ to give [Ag(NH₃)₂]⁺Cl⁻ (ii) PtCl₄.6NH₃ gives white precipitate with AgNO₃ corresponding to only 2 Cl⁻ ions (5 marks) Answer: Main postulates of Werner’s theory:

1. Metals exhibit two types of valencies:
   • Primary/Ionizable valency (oxidation state)
   • Secondary/Non-ionizable valency (coordination number)
2. Primary valencies are satisfied by negative ions
3. Secondary valencies are satisfied by negative ions or neutral molecules
4. Secondary valencies are directed in space, leading to geometric shapes
5. Coordination compounds have coordination sphere and ionization sphere

Explanation of observations: (i) AgCl dissolves in excess NH₃:

• Silver chloride forms a coordination compound
• Ag⁺ has coordination number 2
• Two NH₃ molecules enter coordination sphere
• Cl⁻ moves to ionization sphere
• Formula becomes [Ag(NH₃)₂]⁺Cl⁻
• Only Cl⁻ in ionization sphere is precipitated by AgNO₃

(ii) PtCl₄.6NH₃ with AgNO₃:

• Werner formulated this as [Pt(NH₃)₆]⁴⁺(Cl⁻)₄
• All four Cl⁻ are in ionization sphere
• All should precipitate with AgNO₃
• But only two precipitate
• This indicates structure [Pt(NH₃)₄Cl₂]²⁺(Cl⁻)₂
• Two Cl⁻ are in coordination sphere and not ionizable
• Only two Cl⁻ in ionization sphere give precipitate

9. March 2016 Chemistry Question Paper with Answers

Question 1: What is the effect of temperature on: (i) Physisorption (ii) Chemisorption (1 mark) Answer: (i) Physisorption decreases with increase in temperature (ii) Chemisorption first increases then decreases with increase in temperature

Question 2: Explain the following terms with suitable examples: (i) Homopolymer (ii) Copolymer (2 marks) Answer: (i) Homopolymer: A polymer formed by polymerization of a single type of monomer units. Example: Polyethylene (-CH₂-CH₂-)n formed from ethylene monomers (CH₂=CH₂)

(ii) Copolymer: A polymer formed by polymerization of two or more different types of monomers. Example: Buna-S or SBR rubber formed from butadiene and styrene monomers

Question 3: Using Valence Bond Theory, explain the geometry and magnetic character of [Fe(CN)₆]³⁻ and [CoF₆]³⁻. (Atomic number: Fe = 26, Co = 27) (5 marks) Answer: [Fe(CN)₆]³⁻:

• Electronic configuration of Fe³⁺: [Ar] 3d⁵
• CN⁻ is a strong-field ligand causing pairing of electrons
• d² sp³ hybridization in Fe³⁺
• 5 d electrons occupy three d orbitals with two paired
• The six CN⁻ ligands occupy the six hybrid orbitals
• Geometry: Octahedral
• Magnetic character: Paramagnetic (1 unpaired electron)
• Inner orbital complex (low spin)

[CoF₆]³⁻:

• Electronic configuration of Co³⁺: [Ar] 3d⁶
• F⁻ is a weak-field ligand, no pairing of electrons occurs
• sp³d² hybridization in Co³⁺
• 6 d electrons occupy five d orbitals with 4 unpaired electrons
• The six F⁻ ligands occupy the six hybrid orbitals
• Geometry: Octahedral
• Magnetic character: Paramagnetic (4 unpaired electrons)
• Outer orbital complex (high spin)

10. March 2015 Chemistry Question Paper with Answers

Question 1: Define the term ‘order of reaction’. (1 mark) Answer: Order of reaction is the sum of powers of the concentration terms in the rate law expression of the reaction.

Question 2: Calculate pOH of a solution that is 0.001 M in HCl at 25°C. (2 marks) Answer:

• HCl is a strong acid that completely dissociates in water: HCl → H⁺ + Cl⁻
• [H⁺] = 0.001 M = 10⁻³ M
• pH = -log[H⁺] = -log(10⁻³) = 3
• At 25°C, pH + pOH = 14
• pOH = 14 – pH = 14 – 3 = 11

Question 3: Describe the principle involved in the extraction of zinc from zinc blende. (5 marks) Answer: Extraction of zinc from zinc blende (ZnS) involves:

1. Concentration of ore:
   • Froth floatation process separates ZnS from gangue
   • ZnS being sulphide ore floats with froth while gangue settles
2. Roasting:
   • Concentrated ore is roasted in presence of excess air
   • ZnS + 3/2 O₂ → ZnO + SO₂
   • Converts sulfide to oxide for easier reduction
3. Reduction of ZnO:
   • Carbon reduction method is used
   • ZnO + C → Zn + CO
   • Carried out at 1673 K
   • Zinc is obtained as vapor
   • Temperature higher than boiling point of zinc (1180 K)
4. Electrolytic refining:
   • Impure zinc is made anode
   • Pure zinc sheet is made cathode
   • ZnSO₄ solution is electrolyte
   • Reactions:
     • At anode: Zn → Zn²⁺ + 2e⁻
     • At cathode: Zn²⁺ + 2e⁻ → Zn
   • Pure zinc deposits on cathode

• Principles involved:
  • Concentration: Physical separation based on wettability
  • Roasting: Conversion of sulfide to oxide (thermodynamically favorable)
  • Reduction: Carbon reduces metal oxide at high temperature
  • Purification: Electrochemical process based on redox reactions

11. March 2014 Chemistry Question Paper with Answers

Question 1: What is the order of crystal field splitting in octahedral and tetrahedral fields? (1 mark) Answer: Δₒ (octahedral) = 9/4 Δₜ (tetrahedral)

Question 2: Explain Freundlich adsorption isotherm. What are its limitations? (3 marks) Answer:

• Freundlich adsorption isotherm:
  • Empirical relationship between amount of gas adsorbed per unit mass of solid adsorbent and pressure at constant temperature
  • Mathematical expression: x/m = kP^(1/n)
  • where x is mass of gas adsorbed on mass m of adsorbent at pressure P
  • k and n are constants that depend on nature of adsorbent and gas at particular temperature
  • Taking log on both sides: log(x/m) = log k + (1/n)log P
  • Plot of log(x/m) vs log P gives straight line
• Limitations:
  • Valid only at moderate temperatures and pressures
  • Fails at low pressures
  • Doesn’t explain saturation at high pressures
  • Purely empirical with no theoretical basis
  • Constants k and n vary with temperature

Question 3: (a) Define molar conductivity. How does it vary with concentration for weak and strong electrolytes? (b) Calculate emf of the cell: Mg|Mg²⁺(0.001 M)||Cu²⁺(0.0001 M)|Cu Given: E°(Mg²⁺/Mg) = -2.37 V, E°(Cu²⁺/Cu) = +0.34 V (5 marks) Answer: (a) Molar conductivity:

• Molar conductivity (Λₘ) is the conductivity of a solution containing 1 mole of electrolyte between electrodes 1 m apart
• Mathematically: Λₘ = κ/c where κ is specific conductivity and c is concentration in mol/m³

Variation with concentration:

• Strong electrolytes: Molar conductivity decreases gradually with increase in concentration due to inter-ionic attractions
• Limiting molar conductivity (Λ°ₘ) is approached as concentration approaches zero
• Kohlrausch’s law: Λₘ = Λ°ₘ – A√c (where A is constant)
• Weak electrolytes: Molar conductivity decreases rapidly with increase in concentration due to less ionization
• No linear relationship with √c

(b) EMF calculation:

• Cell notation: Mg|Mg²⁺(0.001 M)||Cu²⁺(0.0001 M)|Cu
• Cell reaction: Mg + Cu²⁺ → Mg²⁺ + Cu
• Using Nernst equation: E = E° – (0.0591/n)log([products]/[reactants])
• E° = E°cathode – E°anode = 0.34 – (-2.37) = 2.71 V
• E = 2.71 – (0.0591/2)log([Mg²⁺]/[Cu²⁺])
• E = 2.71 – (0.0591/2)log(0.001/0.0001)
• E = 2.71 – (0.0591/2)log(10)
• E = 2.71 –

Here’s the continuation of the Chemistry article:

• E = 2.71 – (0.0591/2)log(10)
• E = 2.71 – (0.0591/2) × 1
• E = 2.71 – 0.02955
• E = 2.68 V

12. March 2013 Chemistry Question Paper with Answers

Question 1: Define the term ‘electrophoresis’. (1 mark) Answer: Electrophoresis is the movement of colloidal particles under the influence of an electric field towards the oppositely charged electrode.

Question 2: Explain the following terms with suitable examples: (i) Disinfectants (ii) Broad spectrum antibiotics (2 marks) Answer: (i) Disinfectants: Chemicals that kill or prevent the growth of microorganisms but are generally too toxic to be used on living tissues. They are applied to inanimate objects. Examples: Chlorine, bleach (sodium hypochlorite), phenol

(ii) Broad spectrum antibiotics: Antibiotics that are effective against a wide range of gram-positive and gram-negative bacteria. Examples: Amoxicillin, tetracycline, chloramphenicol

Question 3: What are the differences between lyophilic and lyophobic sols? Describe the Hardy-Schulze rule and its application in coagulation of sols. (5 marks) Answer: Differences between lyophilic and lyophobic sols:

Lyophilic Sols	Lyophobic Sols
Directly formed when dispersed in dispersion medium	Require special methods for preparation
Thermodynamically stable	Thermodynamically unstable
Reversible (can be reconstituted after coagulation)	Irreversible (cannot be reconstituted after coagulation)
Difficult to precipitate	Easily precipitated by electrolytes
Show protective action	Do not show protective action
Examples: Gum, gelatin, starch, proteins	Examples: Metal sols, sulfur sol, metal hydroxides

Hardy-Schulze rule:

• The coagulating power of an electrolyte increases with the increase in valency of the ion carrying charge opposite to that of colloidal particles.
• For negatively charged sols (like As₂S₃), coagulating power increases in order: Na⁺ < Ca²⁺ < Al³⁺
• For positively charged sols (like Fe(OH)₃), coagulating power increases in order: Cl⁻ < SO₄²⁻ < PO₄³⁻

Application in coagulation:

• The oppositely charged ions neutralize the charge on colloidal particles, reducing repulsion
• Higher valency ions are more effective coagulants in smaller concentrations
• This principle is used in:
  • Water purification (alum addition)
  • Sewage treatment
  • River delta formation (sea water coagulates river water colloids)
  • Blood clotting (Ca²⁺ helps in blood coagulation)
  • Purification of air in cottrell precipitator

13. March 2012 Chemistry Question Paper with Answers

Question 1: Write any two applications of adsorption. (1 mark) Answer: Two applications of adsorption are:

1. In gas masks for adsorption of poisonous gases
2. In chromatography for separation of mixtures

Question 2: What is the coordination number of Ni²⁺ in [Ni(CN)₄]²⁻? Draw the structure of this complex ion. (2 marks) Answer:

• Coordination number of Ni²⁺ in [Ni(CN)₄]²⁻ is 4
• Structure: Square planar with Ni²⁺ at center and four CN⁻ ligands at the corners of the square

Question 3: (a) Explain the preparation of the following: (i) Phenol from chlorobenzene (ii) Picric acid from phenol

(b) How does phenol react with the following reagents? (i) Zinc dust (ii) Bromine water (5 marks) Answer: (a) Preparation:

(i) Phenol from chlorobenzene (Dow process):

• Chlorobenzene is heated with 10% NaOH solution at 623 K and 300 atm pressure
• C₆H₅Cl + NaOH → C₆H₅ONa + NaCl
• Sodium phenoxide is acidified to get phenol
• C₆H₅ONa + HCl → C₆H₅OH + NaCl

(ii) Picric acid from phenol:

• Phenol is first treated with concentrated H₂SO₄ to form phenol-2,4-disulphonic acid
• This is then treated with concentrated HNO₃ at 383 K
• The sulphonic groups are replaced by nitro groups
• Final product is 2,4,6-trinitrophenol (picric acid)
• C₆H₅OH → C₆H₃(OH)(SO₃H)₂ → C₆H₂(OH)(NO₂)₃

(b) Reactions:

(i) Phenol with zinc dust:

• When phenol is heated with zinc dust, it undergoes reduction to form benzene
• C₆H₅OH + Zn → C₆H₆ + ZnO

(ii) Phenol with bromine water:

• Phenol reacts with bromine water at room temperature to form 2,4,6-tribromophenol (white precipitate)
• C₆H₅OH + 3Br₂ → C₆H₂Br₃OH + 3HBr
• This reaction occurs without catalyst due to activation of benzene ring by -OH group
• Shows high reactivity of phenol toward electrophilic aromatic substitution

14. March 2011 Chemistry Question Paper with Answers

Question 1: What is the common feature of close packing in both hcp and ccp structures? (1 mark) Answer: In both hcp and ccp structures, the coordination number is 12 and the percentage of occupied space (packing efficiency) is 74%.

Question 2: Explain the role of the following in the processes mentioned: (i) Cryolite in the extraction of aluminium (ii) Iodine in the refining of zirconium (2 marks) Answer: (i) Role of cryolite (Na₃AlF₆) in extraction of aluminium:

• Lowers the melting point of alumina from 2050°C to about 950°C
• Increases electrical conductivity of the electrolyte
• Makes alumina more soluble in the electrolytic bath
• Helps in formation of complex ions that facilitate electrolysis

(ii) Role of iodine in refining of zirconium:

• Used in Van Arkel method of purification
• Forms volatile zirconium tetraiodide (ZrI₄) at low temperature
• ZrI₄ decomposes on hot tungsten filament at high temperature
• Pure zirconium deposits on the filament
• Impurities don’t form volatile iodides and remain behind

Question 3: What is a buffer solution? Derive Henderson equation for calculating the pH of an acidic buffer. Calculate the pH of a buffer solution containing 0.1 M acetic acid and 0.1 M sodium acetate. (pKa of acetic acid = 4.76) (5 marks) Answer: Buffer solution:

• A solution that resists change in pH on addition of small amounts of acid or base
• Made by mixing weak acid and its salt with strong base (acidic buffer)
• Or, weak base and its salt with strong acid (basic buffer)

Derivation of Henderson equation for acidic buffer:

• Consider weak acid HA and its salt NaA
• Ionization of acid: HA ⇌ H⁺ + A⁻
• Ionization constant: Ka = [H⁺][A⁻]/[HA]
• In buffer, [A⁻] comes mainly from salt and [HA] from acid
• Rearranging Ka expression: [H⁺] = Ka × [HA]/[A⁻]
• Taking negative logarithm: -log[H⁺] = -logKa – log([HA]/[A⁻])
• Hence: pH = pKa – log([acid]/[salt])
• pH = pKa + log([salt]/[acid])
• This is the Henderson-Hasselbalch equation

Calculation:

• Given: [CH₃COOH] = 0.1 M, [CH₃COONa] = 0.1 M, pKa = 4.76
• pH = pKa + log([salt]/[acid])
• pH = 4.76 + log(0.1/0.1)
• pH = 4.76 + log(1)
• pH = 4.76 + 0
• pH = 4.76

15. March 2010 Chemistry Question Paper with Answers

Question 1: Define the term ‘coordination number’ with reference to crystal structure. (1 mark) Answer: Coordination number in crystal structure is the number of nearest neighbors surrounding a constituent particle (atom, ion, or molecule) in the crystal lattice.

Question 2: Define ‘order of reaction’. Derive the integrated rate equation for a zero-order reaction. (3 marks) Answer:

• Order of reaction is the sum of powers of the concentration terms in the rate law expression of the reaction.

Derivation for zero-order reaction:

• For zero-order reaction: Rate = -d[A]/dt = k
• Rearranging: d[A] = -k dt
• Integrating both sides:
  • Left side: ∫ d[A] = [A]
  • Right side: -k ∫ dt = -kt
• From initial concentration [A]₀ to [A] at time t:
  • [A] – [A]₀ = -kt
  • [A] = [A]₀ – kt
• Characteristics:
  • Concentration decreases linearly with time
  • Half-life t₁/₂ = [A]₀/2k (directly proportional to initial concentration)
  • Examples: Photochemical reactions, decomposition on metal surfaces

Question 3: What are lanthanoids? Explain why lanthanoid elements show similarities in their chemical properties. Give the electronic configuration of Ce³⁺ (Z = 58). (5 marks) Answer: Lanthanoids:

• Elements from atomic number 58 (Ce) to 71 (Lu)
• Follow lanthanum (Z = 57) in periodic table
• Part of inner transition elements in f-block

Similarities in chemical properties:

1. Similar electronic configuration:
   • General configuration: [Xe]4f^n 5d^0-1 6s^2 where n varies from 1-14
   • Form primarily +3 oxidation state by losing three electrons (6s^2 and 5d^1 or one 4f)
   • Resulting ions have configuration [Xe]4f^n
2. Lanthanoid contraction:
   • Progressive decrease in atomic and ionic radii across the series
   • Caused by poor shielding effect of f-electrons
   • Imperfect shielding of nuclear charge leads to increased effective nuclear charge
   • Results in similar chemical behavior of successive elements
3. Consequences of similarities:
   • Similar oxidation states (predominantly +3)
   • Similar reactivity patterns
   • Similar magnetic and spectral properties
   • Difficult separation due to similar chemical properties

Electronic configuration of Ce³⁺:

• Ce (Z = 58): [Xe]4f^1 5d^1 6s^2
• Ce³⁺: Loses 5d^1 and 6s^2 electrons to form [Xe]4f^1

Tips to Master Chemistry Using HSSlive Previous Year Question Papers

1. Start with recent papers first: The latest patterns from 2021-2024 reflect current examination trends.
2. Follow the 3-step approach:
   • First attempt: Try solving without looking at answers
   • Second review: Check answers and understand mistakes
   • Final revision: Revise frequently missed concepts
3. Topic-wise analysis: Identify the chapters that consistently appear in Part D (5-mark questions) like:
   • Electrochemistry
   • Chemical Kinetics
   • Coordination Compounds
   • p-Block Elements
   • Extraction of Metals
4. Perfect your numerical skills: Chemistry papers always include calculations on:
   • pH and buffer solutions
   • Electrochemical cells
   • Rate constants
   • Molar quantities
5. Create a formula bank: Extract all formulas from these HSSlive PDFs for quick revision.

Common Chemistry Topics in Kerala Plus Two Board Exams (Based on HSSlive PDFs)

1. Solid State: Crystal structures, defects, unit cells
2. Solutions: Colligative properties, concentration terms
3. Electrochemistry: Cells, Nernst equation, conductivity
4. Chemical Kinetics: Rate laws, factors affecting rate
5. Surface Chemistry: Adsorption, colloids, catalysis
6. p-Block Elements: Group 15-18 elements and compounds
7. d & f Block Elements: Transition metals, lanthanoids
8. Coordination Compounds: Bonding, isomerism, VBT, CFT
9. Haloalkanes & Haloarenes: Preparation, properties, reactions
10. Alcohols, Phenols & Ethers: Preparation, properties
11. Aldehydes, Ketones & Carboxylic Acids: Reactions
12. Amines & Biomolecules: Classification, properties
13. Polymers: Types, preparation, uses
14. Chemistry in Everyday Life: Drugs, cleansing agents

Conclusion

HSSlive Plus Two Chemistry previous year question papers are invaluable resources for exam preparation. With systematic analysis and practice using these PDFs, you can easily score high marks in your Kerala Higher Secondary Chemistry examination. Remember to complement these with good textbook understanding and regular revision of important concepts.

Download all the HSSlive Chemistry PDFs today and start your journey towards academic excellence in Chemistry!