previous year question paper

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

by

Are you searching for Kerala Plus One Chemistry previous year question papers and answers in PDF format from HSSlive? You’ve come to the right place! As an experienced Chemistry teacher from Kerala, I’ve compiled this comprehensive resource to help you ace your Chemistry board exams.

Why HSSlive Plus One Chemistry Previous Year Question Papers PDFs Are Essential

Chemistry requires both conceptual clarity and systematic practice. HSSlive.co.in offers the most reliable collection of Plus One Chemistry question papers that:

  • Help you master the exact Kerala Higher Secondary Board examination pattern
  • Reveal frequently tested topics and concepts from past papers
  • Develop effective time management strategies
  • Build confidence through targeted practice
  • Identify your strengths and weak areas in different chapters

How to Download Plus One Chemistry Previous Year Question Papers and Answers PDF from HSSlive

Quick Access Guide:

  1. Visit the official HSSlive website: www.hsslive.co.in
  2. Navigate to “Previous Question Papers” or “Question Bank” section
  3. Select “Plus One” from the class options
  4. Choose “Chemistry” from the subject list
  5. Download the PDF files for different years (2010-2024)

Pro Tip: Create a dedicated folder to organize your HSSlive Chemistry PDFs by year for structured revision.

Kerala Plus One 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 One Chemistry Previous Year Question Papers with Answers (HSSlive PDF Collection)

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

1. March 2024 Plus One Chemistry Previous Year Question Papers with Answers

Question 1: What is Hund’s rule? (1 mark) Answer: Hund’s rule states that electrons occupy orbitals of equal energy singly before pairing, with parallel spins in accordance with Pauli’s exclusion principle.

Question 2: Calculate the pH of a buffer solution containing 0.2 M NH₃ and 0.3 M NH₄Cl. (Given: Kb for NH₃ = 1.8 × 10⁻⁵) (3 marks) Answer: For the buffer NH₃/NH₄Cl: pOH = pKb + log([Salt]/[Base]) pKb = -log(1.8 × 10⁻⁵) = 4.74 pOH = 4.74 + log(0.3/0.2) = 4.74 + log(1.5) = 4.74 + 0.176 = 4.916 pH = 14 – pOH = 14 – 4.916 = 9.084

Question 3: Explain the concept of hybridization with appropriate examples. Illustrate the formation of sp, sp², and sp³ hybrid orbitals. (5 marks) Answer: Hybridization is the mixing of atomic orbitals of similar energies to form new hybrid orbitals of equal energy.

  1. sp hybridization:
    • Mixing of one s and one p orbital
    • Forms two sp hybrid orbitals oriented at 180°
    • Example: BeCl₂, C₂H₂ (acetylene)
    • Linear molecular geometry
    • In acetylene (HC≡CH): Carbon uses sp hybrid orbitals to form sigma bonds with hydrogen and the other carbon, while the unhybridized p orbitals form two pi bonds
  2. sp² hybridization:
    • Mixing of one s and two p orbitals
    • Forms three sp² hybrid orbitals arranged in trigonal planar geometry (120° apart)
    • Example: BF₃, C₂H₄ (ethylene)
    • In ethylene: Each carbon uses sp² hybrids for sigma bonds with hydrogen and the other carbon, while unhybridized p orbitals form a pi bond
  3. sp³ hybridization:
    • Mixing of one s and three p orbitals
    • Forms four sp³ hybrid orbitals arranged in tetrahedral geometry (109.5° apart)
    • Example: CH₄ (methane), NH₃, H₂O
    • In methane: Carbon uses sp³ hybrids to form four sigma bonds with hydrogen atoms

2. March 2023 Plus One Chemistry Previous Year Question Papers with Answers

Question 1: Define electronegativity. How does it vary across a period and down a group in the periodic table? (1 mark) Answer: Electronegativity is the ability of an atom to attract shared electrons towards itself in a chemical bond. It increases across a period (left to right) and decreases down a group.

Question 2: Derive the relationship between Kp and Kc for a gaseous equilibrium. (2 marks) Answer: For a general reaction: aA + bB ⇌ cC + dD Kc = [C]^c[D]^d/[A]^a[B]^b Kp = (PC)^c(PD)^d/(PA)^a(PB)^b

Using ideal gas equation, P = [RT] for each component: Kp = Kc(RT)^Δn Where Δn = (c+d) – (a+b) = change in number of moles of gaseous species

Question 3: Explain the mechanism of SN1 and SN2 reactions with examples. Compare their stereochemical outcomes. (5 marks) Answer: SN1 Reaction (Substitution Nucleophilic Unimolecular):

  • Two-step mechanism:
    1. Slow formation of carbocation intermediate: R-X → R⁺ + X⁻
    2. Fast attack by nucleophile: R⁺ + Nu⁻ → R-Nu
  • Rate = k[R-X] (first-order kinetics)
  • Example: (CH₃)₃C-Br + OH⁻ → (CH₃)₃C-OH + Br⁻
  • Favored by tertiary alkyl halides
  • Stereochemistry: Racemization (mixture of retention and inversion)

SN2 Reaction (Substitution Nucleophilic Bimolecular):

  • One-step concerted mechanism: Nu⁻ + R-X → [Nu···R···X]⁻ → Nu-R + X⁻
  • Rate = k[R-X][Nu⁻] (second-order kinetics)
  • Example: CH₃-Br + OH⁻ → CH₃-OH + Br⁻
  • Favored by primary alkyl halides
  • Stereochemistry: Complete inversion of configuration (backside attack)

Comparison:

  • SN1: Racemization, carbocation intermediate, first-order kinetics
  • SN2: Inversion of configuration, concerted mechanism, second-order kinetics

3. March 2022 Plus One Chemistry Previous Year Question Papers with Answers

Question 1: What are the postulates of Bohr’s model of atom? (1 mark) Answer: Bohr’s model postulates that: (1) Electrons revolve around the nucleus in fixed circular orbits, (2) Angular momentum of electron is quantized as mvr = nh/2π, (3) Electrons emit or absorb energy only when transitioning between orbits, and (4) The energy change during transition is emitted or absorbed as a photon: ΔE = hν.

Question 2: Balance the following redox reaction by ion-electron method: MnO₄⁻ + SO₃²⁻ → MnO₂ + SO₄²⁻ (in basic medium) (3 marks) Answer: Oxidation half-reaction: SO₃²⁻ → SO₄²⁻ SO₃²⁻ + H₂O → SO₄²⁻ + 2H⁺ + 2e⁻ In basic medium: SO₃²⁻ + H₂O → SO₄²⁻ + 2OH⁻ + 2e⁻

Reduction half-reaction: MnO₄⁻ → MnO₂ MnO₄⁻ + 4H⁺ + 3e⁻ → MnO₂ + 2H₂O In basic medium: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻

Balancing electrons: 3(SO₃²⁻ + H₂O → SO₄²⁻ + 2OH⁻ + 2e⁻) 2(MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻)

Final balanced equation: 3SO₃²⁻ + 2MnO₄⁻ + H₂O → 3SO₄²⁻ + 2MnO₂ + 2OH⁻

Question 3: Explain the structure and bonding in benzene. Discuss the concept of resonance and aromaticity. (5 marks) Answer: Structure of Benzene (C₆H₆):

  • Planar hexagonal ring with six carbon atoms
  • Each carbon is sp² hybridized
  • C-C bond length is 1.39 Å (intermediate between single and double bond)
  • C-H bond length is 1.09 Å
  • Bond angles are 120°

Bonding in Benzene:

  • Each carbon forms three sigma bonds: two with adjacent carbons and one with hydrogen
  • Six p-orbitals (one from each carbon) form a delocalized π-electron cloud above and below the plane of the ring
  • Each carbon contributes one electron to the π-system, creating a total of six π-electrons

Resonance:

  • Cannot be represented by a single Lewis structure
  • Two equivalent Kekulé structures with alternating single and double bonds
  • Actual structure is a resonance hybrid of these structures
  • Resonance energy = 152 kJ/mol (extra stability)

Aromaticity:

  • Follows Hückel’s rule: contains 4n+2 π-electrons (where n = 0, 1, 2…)
  • Benzene has 6 π-electrons (n = 1)
  • Planar structure with continuous p-orbital overlap
  • Exhibits enhanced stability
  • Undergoes electrophilic substitution rather than addition reactions

4. March 2021 Plus One Chemistry Previous Year Question Papers with Answers

Question 1: What is common ion effect? Give an example. (1 mark) Answer: Common ion effect is the decrease in solubility of an ionic compound in a solution containing an ion common to the compound. Example: The solubility of AgCl decreases in the presence of NaCl solution because Cl⁻ is the common ion.

Question 2: Describe the principle and working of hydrogen-oxygen fuel cell with a neat diagram. (3 marks) Answer: Principle: A hydrogen-oxygen fuel cell directly converts chemical energy into electrical energy through electrochemical reactions without combustion.

Construction:

  • Anode: Porous carbon containing platinum catalyst
  • Cathode: Similar to anode
  • Electrolyte: Potassium hydroxide solution
  • External circuit connecting electrodes

Working:

  • At anode: H₂ + 2OH⁻ → 2H₂O + 2e⁻
  • At cathode: ½O₂ + H₂O + 2e⁻ → 2OH⁻
  • Overall reaction: H₂ + ½O₂ → H₂O + electrical energy

Advantages:

  • High efficiency (60-70%)
  • Pollution-free operation
  • Continuous operation possible
  • No moving parts

Applications:

  • Space vehicles
  • Electric vehicles
  • Backup power systems

Question 3: What are colloids? Explain different types of colloids with examples. Describe any three methods of preparation of colloids. (5 marks) Answer: Colloids: Heterogeneous mixtures where particles of one substance (dispersed phase) are dispersed in another substance (dispersion medium) with particle size between 1-1000 nm.

Types of Colloids:

Dispersed Phase Dispersion Medium Type Example
Solid Solid Solid Sol Colored glass, gemstones
Solid Liquid Sol Paint, blood, ink
Solid Gas Aerosol Smoke, dust in air
Liquid Solid Gel Cheese, butter, jelly
Liquid Liquid Emulsion Milk, mayonnaise
Liquid Gas Aerosol Fog, clouds, mist
Gas Solid Solid foam Pumice stone, foam rubber
Gas Liquid Foam Whipped cream, soap lather

Methods of Preparation:

  1. Chemical Methods:
    • By double decomposition: AgNO₃ + KI → AgI (colloid) + KNO₃
    • By oxidation: SO₂ + 2H₂S → 3S (colloid) + 2H₂O
    • By reduction: 2AuCl₃ + 3HCHO + 3H₂O → 2Au (colloid) + 3HCOOH + 6HCl
  2. Electrical Disintegration (Bredig’s Arc Method):
    • Metal electrodes immersed in water
    • Electric arc created between electrodes
    • Metal vaporizes and condenses as colloidal particles
  3. Peptization:
    • Converting precipitates into colloidal solution by adding electrolytes
    • Example: Fe(OH)₃ precipitate + FeCl₃ → Fe(OH)₃ colloid

5. March 2020 Plus One Chemistry Previous Year Question Papers with Answers

Question 1: What is the ‘lanthanoid contraction’? What are its consequences? (1 mark) Answer: Lanthanoid contraction is the gradual decrease in atomic and ionic radii across the lanthanoid series due to poor shielding of nuclear charge by 4f electrons. Consequences include: similar properties of elements in the same group, difficulty in separation, and similarity between 4d and 5d elements.

Question 2: Write the IUPAC names of the following compounds: (i) [Fe(CN)₆]³⁻ (ii) K₃[Fe(C₂O₄)₃] (2 marks) Answer: (i) Hexacyanoferrate(III) ion (ii) Potassium trioxalatoferrate(III)

Question 3: What are the main postulates of kinetic molecular theory of gases? How does it explain the gas laws? (5 marks) Answer: Postulates of Kinetic Molecular Theory:

  1. Gases consist of tiny particles (molecules) in constant random motion
  2. The volume of gas molecules is negligible compared to the total volume of the gas
  3. There are no attractive or repulsive forces between gas molecules
  4. Collisions between molecules and with container walls are perfectly elastic
  5. The average kinetic energy of gas molecules is proportional to absolute temperature

Explanation of Gas Laws:

  1. Boyle’s Law (P ∝ 1/V at constant T and n):
    • When volume decreases, molecules collide more frequently with walls
    • Increases the force per unit area (pressure)
    • Mathematically: P ∝ 1/V
  2. Charles’ Law (V ∝ T at constant P and n):
    • Higher temperature means higher kinetic energy of molecules
    • Molecules hit walls with greater force and frequency
    • To maintain constant pressure, volume must increase
    • Mathematically: V ∝ T
  3. Avogadro’s Law (V ∝ n at constant P and T):
    • More molecules (higher n) means more collisions with walls
    • To maintain constant pressure, volume must increase proportionally
    • Mathematically: V ∝ n
  4. Graham’s Law of Diffusion (rate ∝ 1/√M):
    • Kinetic energy = ½mv²
    • At same temperature, KE is constant for all gases
    • Therefore, v ∝ 1/√m
    • Rate of diffusion is proportional to velocity
    • Mathematically: rate ∝ 1/√M

6. March 2019 Plus One Chemistry Previous Year Question Papers with Answers

Question 1: What is meant by ‘activity series’ of metals? (1 mark) Answer: Activity series is an arrangement of metals in decreasing order of their reactivity or tendency to lose electrons. More reactive metals displace less reactive metals from their salt solutions.

Question 2: Explain the following terms with suitable examples: (i) Schottky defect (ii) Frenkel defect (2 marks) Answer: (i) Schottky defect: Equal number of cations and anions missing from their lattice sites, maintaining electrical neutrality. Found in ionic compounds with similar cation and anion sizes. Example: NaCl, KCl, CsCl.

(ii) Frenkel defect: Displacement of an ion (usually smaller cation) from its normal lattice site to an interstitial position. Total number of ions remains the same. Example: AgCl, AgBr, ZnS.

Question 3: What is chemical kinetics? Derive the integrated rate equation for a first-order reaction. What is the half-life of a first-order reaction? (5 marks) Answer: Chemical Kinetics: The branch of chemistry that deals with the study of reaction rates, factors affecting rates, and mechanism of chemical reactions.

Derivation of First-Order Rate Equation: For a reaction: A → Products Rate = -d[A]/dt = k[A] Rearranging: d[A]/[A] = -k dt Integrating between limits [A]₀ at t=0 and [A]t at t=t: ∫([A]₀ to [A]t) d[A]/[A] = -k∫(0 to t) dt ln[A]t – ln[A]₀ = -kt ln([A]t/[A]₀) = -kt [A]t = [A]₀e^(-kt)

Taking logarithm to base 10: log[A]t = log[A]₀ – (k/2.303)t

Half-life of First-Order Reaction: Half-life (t₁/₂) is the time required for the concentration to reduce to half its initial value. At t = t₁/₂, [A]t = [A]₀/2 Substituting in the rate equation: [A]₀/2 = [A]₀e^(-kt₁/₂) 1/2 = e^(-kt₁/₂) ln(1/2) = -k*t₁/₂ t₁/₂ = ln(2)/k = 0.693/k

Characteristics of t₁/₂ for first-order reaction:

  • Independent of initial concentration
  • Inversely proportional to rate constant

Important Chemistry Topics Frequently Asked in Kerala Board Exams

Based on my analysis of past papers, focus on these high-weightage topics:

  1. Chemical Bonding and Molecular Structure
  2. Chemical Equilibrium and Ionic Equilibrium
  3. Redox Reactions and Electrochemistry
  4. States of Matter
  5. Organic Chemistry Fundamentals
  6. Atomic Structure and Periodic Properties
  7. Chemical Kinetics
  8. Thermodynamics
  9. Solutions
  10. Coordination Compounds

Preparation Tips from an Experienced Kerala Chemistry Teacher

  1. Master the Basics: Ensure fundamental concepts are crystal clear before tackling complex problems
  2. Regular Practice: Solve at least 5 years of previous papers to understand question patterns
  3. Focus on Calculations: Practice numerical problems from thermodynamics, equilibrium, and electrochemistry
  4. Revise Organic Reactions: Create flowcharts for important reactions and mechanisms
  5. Time Management: Practice answering questions within time limits (approximately 1 minute per mark)
  6. Use HSSlive Resources: The website offers excellent notes and solved problems
  7. Create Summary Notes: Condense important formulas, equations, and concepts for quick revision
  8. Balance Theory and Numericals: Chemistry exams test both conceptual understanding and problem-solving

Remember, consistent practice with these HSSlive Plus One Chemistry previous year question papers will significantly improve your preparation and boost your confidence for the board examination. Best of luck!

Leave a Comment