Chemistry Notes – Set 2: Detailed Guide for UPSC, PCS, SSC Competitive Exams

Class 9: Is Matter Around Us Pure?

Detailed Concepts:

  • Pure Substances vs. Mixtures:
    • Pure Substance: Single type of particle (element or compound) with uniform composition and definite properties (e.g., pure gold, distilled water). Elements (e.g., oxygen) and compounds (e.g., NaCl) are pure.
    • Mixture: Two or more substances physically combined, with variable composition. Classified as:
      • Homogeneous: Uniform composition throughout (e.g., saltwater, air).
      • Heterogeneous: Non-uniform composition, visible boundaries (e.g., sand in water, oil and water).
  • Types of Mixtures:
    • Solution: Homogeneous mixture with solute (dissolved substance) and solvent (dissolving medium). Example: Sugar in water (sugar = solute, water = solvent).
    • Suspension: Heterogeneous mixture with large particles that settle over time (e.g., mud in water). Visible particles, often opaque.
    • Colloid: Intermediate mixture with particles (1–1000 nm) that don’t settle but scatter light (Tyndall effect). Examples: Milk (fat globules in water), fog (water droplets in air).
  • Properties of Colloids:
    • Tyndall Effect: Scattering of light by colloidal particles, visible as a beam (e.g., in fog or milk), not seen in true solutions.
    • Brownian Motion: Random zigzag movement of colloidal particles due to collisions with solvent molecules.
    • Electrophoresis: Movement of charged colloidal particles under an electric field, used in purification.
    • Coagulation: Aggregation of colloidal particles by adding electrolytes (e.g., alum in water purification).
  • Concentration of Solutions:
    • Mass by Mass Percentage: (Mass of solute / Mass of solution) × 100. Example: 10 g salt in 90 g water = (10/100) × 100 = 10% mass/mass.
    • Mass by Volume Percentage: (Mass of solute / Volume of solution) × 100. Example: 5 g sugar in 100 mL solution = (5/100) × 100 = 5% mass/volume.
    • Volume by Volume Percentage: (Volume of solute / Volume of solution) × 100. Example: 20 mL ethanol in 100 mL solution = (20/100) × 100 = 20% vol/vol.
  • Separation Techniques:
    • Evaporation: Recover soluble solid from liquid (e.g., salt from seawater).
    • Filtration: Separate insoluble solids from liquids (e.g., sand from water).
    • Centrifugation: Separate denser particles using high-speed rotation (e.g., cream from milk).
    • Distillation: Separate liquids with different boiling points (e.g., alcohol from water, boiling points 78°C vs. 100°C).
    • Fractional Distillation: Separate miscible liquids with close boiling points (e.g., crude oil fractions).
    • Sublimation: Separate sublimable solids (e.g., naphthalene from salt).
    • Chromatography: Separate components based on differential adsorption (e.g., ink pigments on paper).
    • Separating Funnel: Separate immiscible liquids (e.g., oil and water).
  • Physical vs. Chemical Changes:
    • Physical: No new substance formed, reversible (e.g., dissolving sugar, melting ice).
    • Chemical: New substance formed, usually irreversible (e.g., burning wood, rusting iron).

Formulas:

  • Mass %: (Mass of solute / Mass of solution) × 100.
  • Mass/Volume %: (Mass of solute / Volume of solution) × 100.
  • Volume/Volume %: (Volume of solute / Volume of solution) × 100.
  • Molarity (M): Moles of solute / Volume of solution (L) (introduced conceptually for exams).
  • Tyndall Effect (Qualitative): Observed when particle size is 1–1000 nm.

Applications:

  • Competitive Exams:
    • UPSC/PCS: Questions on separation techniques in environmental contexts (e.g., water purification) or industrial processes (e.g., distillation in petrochemicals).
    • SSC: Objective questions on colloid properties, concentration calculations, or physical vs. chemical changes.
    • Descriptive: Explain chromatography in forensic science or centrifugation in dairy industry.
  • Real-World:
    • Water Purification: Alum causes coagulation of impurities; distillation for drinking water.
    • Industry: Fractional distillation in oil refineries, chromatography in drug analysis.
    • Daily Life: Filtration in coffee making, evaporation in salt production.
  • Exam Tips:
    • Focus on practical applications of separation techniques and colloid properties.
    • Understand concentration calculations for numerical questions in SSC/UPSC.

Diagram (Textual Description):

  • Chromatography Setup: A vertical filter paper strip with a spot of ink near the bottom, dipped in a solvent (e.g., water). Solvent rises, separating ink into colored bands (e.g., blue, red) due to different adsorption rates. Label stationary phase (paper) and mobile phase (solvent).

Class 10: Acids, Bases and Salts

Detailed Concepts:

  • Acids and Bases:
    • Acid: Sour-tasting, releases H⁺ ions in water, turns blue litmus red (e.g., HCl, H₂SO₄, CH₃COOH). Strong acids (e.g., HCl) fully ionize; weak acids (e.g., CH₃COOH) partially ionize.
    • Base: Bitter-tasting, releases OH⁻ ions in water, turns red litmus blue (e.g., NaOH, Ca(OH)₂). Strong bases (e.g., NaOH) fully dissociate; weak bases (e.g., NH₄OH) partially dissociate.
  • pH Scale: Measures [H⁺] concentration (0–14):
    • pH < 7: Acidic (e.g., HCl, pH ≈ 1).
    • pH = 7: Neutral (e.g., pure water).
    • pH > 7: Basic (e.g., NaOH, pH ≈ 13).
  • Indicators: Substances showing pH-dependent color changes:
    • Natural: Litmus (red in acid, blue in base), turmeric (yellow in acid, red in base).
    • Synthetic: Phenolphthalein (colorless in acid, pink in base), methyl orange (red in acid, yellow in base).
  • Chemical Properties:
    • Acids:
      • React with metals to produce H₂ gas (e.g., Zn + 2HCl → ZnCl₂ + H₂).
      • React with carbonates/bicarbonates to produce CO₂ (e.g., Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂).
      • Neutralize bases (e.g., HCl + NaOH → NaCl + H₂O).
    • Bases: Neutralize acids, react with salts (e.g., NH₄OH + HCl → NH₄Cl + H₂O).
  • Salts:
    • Formed by neutralization (acid + base → salt + water).
    • Types by pH:
      • Neutral: Strong acid + strong base (e.g., NaCl).
      • Acidic: Strong acid + weak base (e.g., NH₄Cl).
      • Basic: Weak acid + strong base (e.g., CH₃COONa).
    • Examples:
      • Sodium Chloride (NaCl): From HCl + NaOH, used in food, industry.
      • Sodium Hydroxide (NaOH): From chlor-alkali process (electrolysis of NaCl), used in soap making.
      • Bleaching Powder (Ca(OCl)Cl): From Cl₂ + Ca(OH)₂, used as disinfectant.
      • Baking Soda (NaHCO₃): From NaCl + NH₃ + CO₂ (Solvay process), used in baking, antacids.
      • Washing Soda (Na₂CO₃·10H₂O): From NaHCO₃, used in detergents, glass.
      • Plaster of Paris (CaSO₄·½H₂O): From gypsum (CaSO₄·2H₂O) by heating, used in casts, construction.
  • Chemical Reactions:
    • Chlor-Alkali Process: 2NaCl + 2H₂O → 2NaOH + Cl₂ + H₂ (electrolysis).
    • Plaster of Paris Formation: CaSO₄·2H₂O → CaSO₄·½H₂O + 1½H₂O (heat at 373 K).
  • Applications in Exams: pH calculations, salt properties, and industrial processes are key for competitive exams.

Formulas:

  • pH: pH = –log[H⁺].
  • pOH: pOH = –log[OH⁻], pH + pOH = 14 at 25°C.
  • Neutralization: Acid + Base → Salt + Water (e.g., HCl + NaOH → NaCl + H₂O).
  • Baking Soda Formation: NaCl + H₂O + CO₂ + NH₃ → NaHCO₃ + NH₄Cl.
  • Plaster of Paris: CaSO₄·2H₂O → CaSO₄·½H₂O + 1½H₂O.

Applications:

  • Competitive Exams:
    • UPSC/PCS: Questions on pH in environmental contexts (e.g., acid rain, soil pH) or industrial processes (e.g., Solvay process).
    • SSC: Objective questions on indicators, salt types, or chemical reactions of acids/bases.
    • Descriptive: Discuss applications of NaOH in industry or plaster of Paris in construction.
  • Real-World:
    • Agriculture: pH adjustment in soils using lime (Ca(OH)₂).
    • Medicine: Baking soda as antacid, NaCl in saline solutions.
    • Industry: Bleaching powder in water treatment, washing soda in detergents.
  • Exam Tips:
    • Master pH calculations and neutralization reactions for numericals.
    • Link acid rain (H₂SO₄, HNO₃) to environmental science for mains.

Diagram (Textual Description):

  • pH Scale: A horizontal scale from 0 to 14, color-coded: red (0–6, acidic), green (7, neutral), blue (8–14, basic). Mark indicators (e.g., phenolphthalein: colorless <8.3, pink >8.3; litmus: red <7, blue >7).

Class 11: Chemical Bonding and Molecular Structure

Detailed Concepts:

  • Chemical Bond: Force holding atoms together to achieve stable electron configurations (octet rule for most elements).
  • Types of Bonds:
    • Ionic Bond: Electrostatic attraction between oppositely charged ions, formed by electron transfer (e.g., NaCl: Na⁺ + Cl⁻). Favored by high electronegativity difference (>1.7).
    • Covalent Bond: Electron sharing between atoms (e.g., H₂, Cl₂). Types:
      • Non-Polar: Equal sharing (e.g., O₂, electronegativity difference ≈ 0).
      • Polar: Unequal sharing (e.g., HCl, electronegativity difference 0.4–1.7).
      • Coordinate (Dative): One atom donates both electrons (e.g., NH₄⁺, NH₃ → H⁺).
    • Metallic Bond: Delocalized electrons in a metal lattice (e.g., Cu, Al), responsible for conductivity, malleability.
  • Lewis Dot Structures: Represent valence electrons as dots to predict bonding. Example: CO₂ (O=C=O, each O forms double bond with C, satisfying octet).
  • Valence Shell Electron Pair Repulsion (VSEPR) Theory:
    • Electron domains (bonding pairs + lone pairs) determine molecular geometry by minimizing repulsion.
    • Examples:
      • CH₄: 4 bonding pairs, tetrahedral, 109.5°.
      • NH₃: 3 bonding pairs, 1 lone pair, trigonal pyramidal, ~107°.
      • H₂O: 2 bonding pairs, 2 lone pairs, bent, ~104.5°.
      • CO₂: 2 double bonds, no lone pairs, linear, 180°.
  • Hybridization: Mixing of atomic orbitals to form hybrid orbitals for bonding:
    • sp: Linear (e.g., BeCl₂).
    • sp²: Trigonal planar (e.g., BF₃).
    • sp³: Tetrahedral (e.g., CH₄).
    • sp³d: Trigonal bipyramidal (e.g., PCl₅).
    • sp³d²: Octahedral (e.g., SF₆).
  • Molecular Orbital (MO) Theory:
    • Atomic orbitals combine to form molecular orbitals (bonding and antibonding).
    • Bond order = (Number of bonding electrons – Number of antibonding electrons) / 2.
    • Example: O₂ (bond order = 2, paramagnetic due to 2 unpaired electrons).
  • Hydrogen Bonding: Strong dipole-dipole interaction between H (bonded to N, O, F) and lone pair on N, O, F. Example: H₂O high boiling point (100°C) due to H-bonding.
  • Bond Parameters:
    • Bond Length: Distance between nuclei (shorter for stronger bonds, e.g., C≡C < C=C < C–C).
    • Bond Angle: Angle between bonds (e.g., 109.5° in CH₄).
    • Bond Energy: Energy to break 1 mole of bonds (e.g., C–H ≈ 413 kJ/mol).
  • Dipole Moment: Measure of polarity (μ = q × d). Example: HCl has dipole; CO₂ is non-polar (symmetric).

Formulas:

  • Bond Order: (N_b – N_a) / 2, where N_b = bonding electrons, N_a = antibonding electrons.
  • Formal Charge: Valence electrons – (Lone pair electrons + ½ Bonding electrons).
  • Dipole Moment: μ = charge (q) × distance (d), unit: Debye (D).
  • VSEPR Geometry: Determined by electron domains (e.g., 4 domains = tetrahedral, 3 domains = trigonal planar).

Applications:

  • Competitive Exams:
    • UPSC/PCS: Questions on hybridization, VSEPR, or hydrogen bonding in biological systems (e.g., DNA structure).
    • SSC: Objective questions on bond types, dipole moment, or molecular shapes.
    • Descriptive: Explain role of hydrogen bonding in water’s properties or metallic bonding in conductors.
  • Real-World:
    • Materials: Diamond (covalent network, hard) vs. graphite (layered, lubricant).
    • Biology: H-bonding in DNA base pairing (A-T, C-G).
    • Industry: Conductivity of metals in electrical applications.
  • Exam Tips:
    • Master VSEPR and hybridization for predicting molecular shapes.
    • Link polarity to physical properties (e.g., solubility, boiling points) for mains.

Diagram (Textual Description):

  • VSEPR Geometries: Show five molecules:
    • CH₄: Tetrahedral, 109.5°, 4 bonding pairs.
    • NH₃: Trigonal pyramidal, ~107°, 3 bonding pairs, 1 lone pair.
    • H₂O: Bent, ~104.5°, 2 bonding pairs, 2 lone pairs.
    • CO₂: Linear, 180°, 2 double bonds.
    • SF₆: Octahedral, 90°, 6 bonding pairs. Label bond angles and lone pairs.

Class 12: Chemical Kinetics

Detailed Concepts:

  • Chemical Kinetics: Study of reaction rates and mechanisms, crucial for understanding reaction feasibility and control.
  • Rate of Reaction: Change in concentration of reactant or product per unit time (mol L⁻¹ s⁻¹). Example: For A → B, rate = –Δ[A]/Δt = Δ[B]/Δt.
  • Factors Affecting Rate:
    • Concentration: Higher concentration increases collision frequency, increasing rate.
    • Temperature: Increases kinetic energy, more molecules exceed activation energy (E_a).
    • Catalyst: Lowers E_a, providing an alternative reaction path (e.g., Fe in Haber process).
    • Surface Area: For solids, larger surface area increases rate (e.g., powdered reactants react faster).
  • Rate Law: Expresses rate as a function of reactant concentrations: Rate = k[A]ᵐ[B]ⁿ, where k = rate constant, m, n = order with respect to A, B.
    • Order: Sum of exponents (m + n). Example: Rate = k[A]²[B] has order 3.
    • Molecularity: Number of molecules in rate-determining step (unimolecular, bimolecular, etc.).
  • Rate Constant (k): Specific to temperature, follows Arrhenius equation: k = A e^(–E_a/RT), where A = pre-exponential factor, E_a = activation energy, R = gas constant, T = temperature (K).
  • Integrated Rate Laws:
    • Zero-Order: [A] = [A]₀ – kt, constant rate (e.g., decomposition on catalyst surface).
    • First-Order: ln[A] = ln[A]₀ – kt, common in radioactive decay (e.g., C-14).
    • Second-Order: 1/[A] = 1/[A]₀ + kt, involves two molecules or one bimolecular step.
  • Half-Life (t₁/₂):
    • Zero-Order: t₁/₂ = [A]₀ / (2k).
    • First-Order: t₁/₂ = ln(2) / k ≈ 0.693 / k, independent of initial concentration.
    • Second-Order: t₁/₂ = 1 / (k[A]₀).
  • Collision Theory: Reactions occur when molecules collide with sufficient energy (≥ E_a) and proper orientation. Rate ∝ collision frequency.
  • Transition State Theory: Reactants form a high-energy transition state before products. E_a is energy barrier to reach this state.
  • Applications in Exams: Rate laws, half-life calculations, and Arrhenius equation are key for numerical and conceptual questions.

Formulas:

  • Rate Law: Rate = k[A]ᵐ[B]ⁿ.
  • Arrhenius Equation: k = A e^(–E_a/RT) or ln(k) = ln(A) – E_a/RT.
  • Zero-Order: [A] = [A]₀ – kt, t₁/₂ = [A]₀ / (2k).
  • First-Order: ln([A]₀/[A]) = kt, t₁/₂ = 0.693 / k.
  • Second-Order: 1/[A] = 1/[A]₀ + kt, t₁/₂ = 1 / (k[A]₀).
  • Rate Constant Units:
    • Zero-Order: mol L⁻¹ s⁻¹.
    • First-Order: s⁻¹.
    • Second-Order: L mol⁻¹ s⁻¹.

Applications:

  • Competitive Exams:
    • UPSC/PCS: Questions on reaction rates in industrial processes (e.g., ammonia synthesis) or environmental contexts (e.g., pollutant degradation).
    • SSC: Objective questions on order, half-life, or Arrhenius equation calculations.
    • Descriptive: Explain role of catalysts in industry or temperature effects on reaction rates.
  • Real-World:
    • Industry: Catalysts in Haber process (Fe for NH₃), petroleum cracking.
    • Environment: Kinetics of ozone depletion, pesticide breakdown.
    • Medicine: Drug shelf-life (first-order kinetics).
  • Exam Tips:
    • Master integrated rate laws and half-life calculations for numericals.
    • Link catalysts to environmental/industrial applications for mains.

Diagram (Textual Description):

  • Reaction Profile: A graph of energy vs. reaction coordinate. Reactants at higher energy, transition state at peak (labeled E_a), products at lower energy (exothermic). Show catalyst lowering E_a with a dashed curve.

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