CheMentor ISC Chemistry Notes - Class XI (Complete)

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Table of Contents (Class XI)

1. Some Basic Concepts

Importance of chemistry, laws of chemical combination, Dalton's theory, mole concept, stoichiometry, equivalents, and volumetric terms.

Quick Links: Introduction & Matter Measurement & Analysis Laws & Dalton's Theory Mole Concept & Molar Mass Formulae & Composition Equivalents & Volumetric Terms Stoichiometry Practice Micro Tests

1.1 Introduction & Study of Matter

Key Points:

Chemistry: Study of matter, its properties, structure, composition, and the changes it undergoes.
Importance & Scope: Crucial in medicine, agriculture, industry, environment, materials science, daily life.
Matter Classification: Physical (Solid, Liquid, Gas) and Chemical (Elements, Compounds, Mixtures).
Elements, Atoms, Molecules: Basic definitions revisited.

1.2 Measurement & Dimensional Analysis

Key Points:

Precision vs Accuracy: Precision = closeness of various measurements for the same quantity. Accuracy = agreement of a particular value to the true value.
Significant Figures: Meaningful digits in a measured or calculated quantity. Rules for determining and using in calculations (addition/subtraction, multiplication/division).
SI Units: International System of Units (Base units: meter, kilogram, second, Kelvin, mole, ampere, candela). Derived units.
Dimensional Analysis (Factor-Label Method): Technique for converting units using conversion factors. Useful for problem-solving.

Q: How many significant figures are in 0.0250 cm?

A: Three. Leading zeros (0.0) are not significant. Non-zero digits (2, 5) are significant. Trailing zero after decimal point (0) is significant.

1.3 Laws of Chemical Combination & Dalton's Theory

Key Points:

Dalton's Atomic Theory (Postulates): Matter from indivisible atoms; atoms of same element identical, different elements differ; atoms combine in fixed ratios; atoms rearranged in reactions.
Limitations: Atoms are divisible (subatomic particles); isotopes exist; atoms not always combine in simple ratios (organic); doesn't explain bonding forces.
Laws of Chemical Combination:
- Law of Conservation of Mass (Lavoisier): Mass is neither created nor destroyed in a chemical reaction.
- Law of Definite Proportions (Proust): A given compound always contains exactly the same proportion of elements by mass.
- Law of Multiple Proportions (Dalton): If two elements form >1 compound, the masses of one element combining with a fixed mass of the other are in simple whole number ratios.
- Law of Reciprocal Proportions (Richter): If two elements combine separately with a fixed mass of a third element, the ratio of masses in which they combine is same or simple multiple of ratio in which they combine with each other.
- Gay Lussac's Law of Gaseous Volumes: Gases combine/are produced in simple volume ratios at constant T & P.
Understand statements, explanations, and simple problems based on these laws.

Critical Concept Check: How does the existence of isotopes contradict Dalton's theory?

A: Dalton postulated that all atoms of a given element are identical, including having the same mass. Isotopes are atoms of the same element (same atomic number) but have different masses due to different numbers of neutrons, contradicting this postulate.

1.4 Mole Concept, Atomic & Molecular Mass

Key Points:

Atomic Mass Unit (amu/u): Defined as 1/12th the mass of a Carbon-12 isotope atom. Standard for relative masses.
Atomic Mass (Isotopic Mass): Mass of an atom of an isotope relative to 1/12th mass of C-12.
Average Atomic Mass: Weighted average of isotopic masses based on natural abundance. (This is the value usually found in periodic tables).
Molecular Mass: Sum of atomic masses of all atoms in a molecule.
Mole Concept:
- 1 mole = Amount containing Avogadro's Number (N_A = 6.022 x 10²³) of particles.
- Molar Mass = Mass of 1 mole of substance in grams (numerically equal to atomic/molecular mass in amu).
- Gram Atomic Mass (GAM) = Mass of 1 mole of atoms.
- Gram Molecular Mass (GMM) = Mass of 1 mole of molecules.
Molar Volume of Gas at STP: 1 mole of any gas occupies 22.4 L at STP (Standard Temp=273K, Std Pressure=1 atm). [Note: IUPAC STP uses 1 bar pressure, molar vol ≈ 22.7 L, but 22.4 L at 1 atm is commonly used in calculations unless specified otherwise].
Numericals: Calculating moles from mass/volume/particles and vice versa.

Critical Concept Check: What is the mass of a single water molecule in grams?

A: Molar mass of H₂O = 18 g/mol. 1 mole contains 6.022 x 10²³ molecules. Mass of 1 molecule = Molar Mass / Avogadro's Number = 18 g / (6.022 x 10²³) ≈ 2.99 x 10⁻²³ g.

1.5 Empirical & Molecular Formulae

Key Points:

Percentage Composition: % by mass of each element in a compound.
Empirical Formula (EF): Simplest whole number ratio of atoms in a compound.
Molecular Formula (MF): Actual number of atoms of each element in a molecule. MF = (EF) × n, where n = Molecular Mass / Empirical Formula Mass.
Calculation from % Composition: (Same steps as Class X).
Numericals based on finding EF and MF.

Q: Glucose has MF C₆H₁₂O₆. What is its Empirical Formula?

A: The ratio C:H:O is 6:12:6. The simplest whole number ratio is obtained by dividing by 6, which gives 1:2:1. Empirical Formula = CH₂O.

1.6 Equivalents & Volumetric Terms

Key Points:

Equivalent Weight (Eq. Wt.): Mass of substance that combines with or displaces 1 part H, 8 parts O, or 35.5 parts Cl by mass. Standard: C=12.00.
- Eq. Wt. of Element = Atomic Mass / Valency.
- Eq. Wt. of Acid = Molecular Mass / Basicity (no. of replaceable H⁺).
- Eq. Wt. of Base = Molecular Mass / Acidity (no. of replaceable OH⁻).
- Eq. Wt. of Salt = Formula Mass / Total positive or negative charge on ions.
- Eq. Wt. of Oxidizing/Reducing Agent = Molecular Mass / Change in Oxidation Number per molecule (or electrons gained/lost per molecule).
Variable Equivalent Weight: Possible if element has variable valency.
Gram Equivalent Weight (GEW): Eq. Wt. expressed in grams. Mass of 1 gram equivalent.
Relationship: Gram Molecular Mass = n × Gram Equivalent Weight (where n = Valency/Basicity/Acidity/Total charge/O.N change).
Volumetric Terms:
- Percentage: w/w (mass solute/mass soln × 100), w/v (mass solute(g)/vol soln(mL) × 100).
- Normality (N): No. of gram equivalents of solute per litre of solution. (N = GEW / Vol(L)).
- Molarity (M): No. of moles of solute per litre of solution. (M = Moles / Vol(L)).
- Molality (m): No. of moles of solute per kilogram of solvent. (m = Moles / Mass Solvent(kg)).
- Mole Fraction (χ): Ratio of moles of one component to total moles in solution. (χ_A = n_A / (n_A+n_B+...)).
Know formulae, normality/molarity equations, simple calculations. (Experimental details for Eq Wt determination not required).

Critical Concept Check: Which concentration term is independent of temperature: Molarity or Molality? Why?

A: Molality is independent of temperature. Molarity involves the volume of solution, which changes with temperature (liquids expand/contract). Molality involves the mass of the solvent, which does not change with temperature.

1.7 Stoichiometry & Limiting Reagent

Key Points:

Stoichiometry: Quantitative relationships between reactants and products in a balanced chemical equation.
Calculations based on balanced equation: Mass-mass, Mass-volume, Volume-volume relationships (revisited from Class X).
Limiting Reagent (Reactant): Reactant that is completely consumed first in a reaction, thereby limiting the amount of product formed. The other reactant(s) are in excess.
Calculations involving identifying the limiting reagent and determining the amount of product formed based on it.

Critical Concept Check: 3g of H₂ reacts with 28g of N₂ to form NH₃ according to N₂ + 3H₂ → 2NH₃. Which is the limiting reagent?

A: Moles H₂ = 3g / 2g/mol = 1.5 mol. Moles N₂ = 28g / 28g/mol = 1.0 mol. Stoichiometric ratio N₂ : H₂ = 1 : 3. For 1.0 mol N₂, we need 3.0 mol H₂. We only have 1.5 mol H₂. Therefore, H₂ is the limiting reagent.

Try Yourselves - Chapter 1 (Class XI)

1. State Law of Multiple Proportions.

2. How many moles in 22g CO₂? (C=12,O=16)

3. How many molecules in 22g CO₂?

4. Calculate Eq. Wt. of H₂SO₄. (S=32)

5. Define Molarity.

6. EF of Benzene (C₆H₆)?

7. Vol of 2 moles N₂ at STP?

8. Define Limiting Reagent.

Micro Tests - Chapter 1 (Class XI)

Micro Test XI1.1 (Laws & Dalton - 10 Marks)

Test XI1.1

Time: 10 Min

Answer the following.

1.State the Law of Conservation of Mass.[2]
2.State the Law of Definite Proportions.[2]
3.State Gay Lussac's Law of Gaseous Volumes.[2]
4.Give one postulate of Dalton's Atomic Theory.[2]
5.Give one limitation of Dalton's Theory.[2]

Micro Test XI1.2 (Mole Concept Calc - 10 Marks)

Test XI1.2

Time: 12 Min

Calculate (N_A=6x10²³, Molar Vol=22.4L, C=12, H=1, O=16, Na=23):

1.Number of moles in 88g of CO₂.[2]
2.Mass of 0.25 moles of NaOH.[2]
3.Volume of 3 moles of CH₄ gas at STP.[2]
4.Number of atoms in 12g of Carbon.[2]
5.Number of molecules in 5.6 L of O₂ gas at STP.[2]

Micro Test XI1.3 (Eq Wt & Conc Terms - 10 Marks)

Test XI1.3

Time: 12 Min

Calculate/Define (H=1, O=16, S=32, Ca=40):

1.Equivalent weight of Ca(OH)₂.[2]
2.Equivalent weight of H₃PO₄ (assuming complete neutralization).[2]
3.Define Normality.[2]
4.Define Molality.[2]
5.What mass of H₂SO₄ (Mol mass 98) is needed for 250mL of 0.1 M solution?[2]

Micro Test XI1.4 (Stoichiometry & Limiting Reagent - 10 Marks)

Test XI1.4

Time: 15 Min

Calculate (N=14, H=1, O=16):

1.Reaction: 2H₂ + O₂ → 2H₂O. Mass of water from 4g H₂?[3]
2.Reaction: N₂ + 3H₂ → 2NH₃. Volume of N₂ needed to react with 30L H₂?[2]
3.Reaction: N₂ + 3H₂ → 2NH₃. If 14g N₂ reacts with 6g H₂, identify limiting reagent.[3]
4.Using Q3 data, calculate mass of NH₃ formed.[2]

2. Structure of Atom

Discovery of fundamental particles, atomic models (Thomson, Rutherford, Bohr), dual nature, quantum mechanical model (orbitals, quantum numbers), electron configuration rules.

Quick Links: Fundamental Particles & Early Models Atomic Number, Mass No, Isotopes, Isobars Dual Nature & Bohr's Model Quantum Mechanical Model Electron Configuration Rules Practice Micro Tests

2.1 Discovery of Particles & Early Models

Key Points:

Subatomic Particles:
- Electrons (e⁻): Discovered via Cathode Ray experiments (J.J. Thomson). Negative charge (-1.602x10⁻¹⁹ C), mass ≈ 9.11x10⁻³¹ kg. Properties of cathode rays (travel straight, neg charge, produce fluorescence, heating effect, X-rays).
- Protons (p⁺): Discovered via Anode Ray/Canal Ray experiments (Goldstein). Positive charge (+1.602x10⁻¹⁹ C), mass ≈ 1.672x10⁻²⁷ kg. Properties of anode rays depend on gas in tube.
- Neutrons (n⁰): Discovered by Chadwick (bombarding Be with α-particles). No charge, mass ≈ 1.674x10⁻²⁷ kg (slightly heavier than proton).
Thomson's Model (Plum Pudding/Watermelon): Atom as a sphere of positive charge with electrons embedded. Limitation: Couldn't explain scattering experiment or spectral lines.
Rutherford's Model (Nuclear Model): Based on α-particle scattering experiment (most α passed through, few deflected, very few bounced back).
- Conclusions: Atom mostly empty space, positive charge concentrated in tiny, dense nucleus, electrons orbit nucleus.
- Limitations: Couldn't explain atom stability (orbiting electron should radiate energy & spiral into nucleus), couldn't explain line spectra.
Electromagnetic Wave Theory Limitations: Black body radiation, Photoelectric effect (explained by Planck's Quantum Theory: E=hν).

Critical Concept Check: What was the key observation in the alpha scattering experiment that led Rutherford to conclude the existence of a nucleus?

A: The deflection of a small fraction of alpha particles through large angles, and the bouncing back (deflection > 90°) of very few alpha particles. This indicated collision with a very small, dense, positively charged region within the atom (the nucleus), as the positively charged alpha particles were strongly repelled.

2.2 Atomic Number, Mass No, Isotopes, Isobars

Key Points:

Atomic Number (Z): Number of protons in the nucleus. Determines the element. (Z = p⁺ = e⁻ in neutral atom).
Mass Number (A): Total number of protons (Z) + neutrons (n⁰) in the nucleus. (A = Z + n⁰).
Isotopes: Atoms of the same element (same Z) with different mass numbers (A) due to different numbers of neutrons (n⁰). (e.g., ¹H, ²H, ³H; ¹²C, ¹⁴C; ³⁵Cl, ³⁷Cl).
Isobars: Atoms of different elements (different Z) with the same mass number (A). (e.g., ⁴⁰Ar, ⁴⁰K, ⁴⁰Ca).

Q: An atom has 17 protons and 20 neutrons. Write its symbol using A and Z notation. Is it an isotope of ³⁵Cl?

A: Z = No. of protons = 17. Element is Chlorine (Cl). A = p⁺ + n⁰ = 17 + 20 = 37. Symbol: ³⁷₁₇Cl. Yes, it is an isotope of ³⁵₁₇Cl (same Z=17, different A).

2.3 Dual Nature & Bohr's Model

Key Points:

Dual Nature of Matter & Radiation:
- Light: Wave-particle duality (Wave nature - diffraction; Particle nature - photoelectric effect, Planck's E=hν).
- Matter (de Broglie Hypothesis): Moving particles exhibit wave-like properties. λ = h / mv (h=Planck's const, m=mass, v=velocity). Significance mainly for microscopic particles.
- Heisenberg's Uncertainty Principle: Impossible to determine simultaneously the exact position (Δx) and exact momentum (Δp) of a small moving particle like an electron. Δx ⋅ Δp ≥ h / 4π. Rules out fixed orbits.
Bohr's Atomic Model (for H-like species):
- Postulates: Electrons revolve in fixed circular orbits (energy levels/shells) without radiating energy; energy emitted/absorbed only during electron jumps between orbits (ΔE = hν); Angular momentum is quantized (mvr = nh/2π).
- Merits: Explained stability, H-spectrum lines (using Rydberg's formula: 1/λ = R [1/n₁² - 1/n₂²]).
- Limitations: Failed for multi-electron atoms; couldn't explain splitting of spectral lines (Zeeman/Stark effect); ignored dual nature/uncertainty principle; orbits assumed circular.

Critical Concept Check: How did Bohr's model explain the line spectrum of Hydrogen?

A: Bohr proposed that electrons exist only in specific energy levels. When an electron absorbs energy, it jumps to a higher level (excited state). When it falls back to a lower level, it emits energy as a photon of specific frequency (and wavelength), corresponding to the energy difference between the levels (ΔE = hν). Since only specific jumps are allowed, only specific frequencies/wavelengths are emitted, resulting in discrete lines (line spectrum) rather than a continuous spectrum.

2.4 Quantum Mechanical Model

Key Points:

Based on dual nature and uncertainty principle. Treats electron as a wave.
Schrödinger Wave Equation: Mathematical equation describing electron wave behavior. Solution (Ψ - wave function) gives probability of finding electron. |Ψ|² = Probability density.
Orbitals: 3D region around nucleus with high probability (>90%) of finding electron. Replaces fixed orbits.
Quantum Numbers: Describe energy, shape, orientation, spin of electron/orbital.
- Principal (n): Shell number (1, 2, 3...), determines size & energy level.
- Azimuthal/Angular Momentum (l): Subshell (0 to n-1). l=0(s), l=1(p), l=2(d), l=3(f). Determines shape.
- Magnetic (m_l): Orbital orientation (-l to +l including 0). Determines no. of orbitals in subshell (s=1, p=3, d=5, f=7).
- Spin (m_s): Electron spin (+½ or -½).
Shapes of Orbitals: s (spherical), p (dumbbell - p_x, p_y, p_z along axes), d (double dumbbell/cloverleaf - d_xy, d_yz, d_zx, d_x²-y², d_z²). (No derivations needed).
Nodes/Nodal Planes: Regions where probability of finding electron is zero.

Critical Concept Check: What is the difference between an orbit (Bohr model) and an orbital (Quantum model)?

A: An orbit is a fixed, well-defined circular path around the nucleus where an electron was thought to revolve (Bohr). An orbital is a 3-dimensional region of space around the nucleus where the probability of finding an electron is maximum (Quantum Mechanical Model). Orbitals have specific shapes (s, p, d) and do not represent fixed paths.

2.5 Electron Configuration Rules

Key Points:

Aufbau Principle: Electrons fill orbitals starting from lowest energy level upwards. Order generally: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d... (Follow (n+l) rule: lower n+l fills first; if n+l same, lower n fills first).
Pauli Exclusion Principle: No two electrons in an atom can have the same set of all four quantum numbers. An orbital can hold max 2 electrons, and they must have opposite spins (↑↓).
Hund's Rule of Maximum Multiplicity: Pairing of electrons in orbitals of the same subshell (degenerate orbitals like p, d, f) does not occur until each orbital is singly occupied with parallel spin.
Electronic Configuration: Representation using s, p, d, f notation (e.g., N (Z=7): 1s² 2s² 2p³).
Stability of Half-filled & Completely Filled Orbitals: Extra stability due to symmetrical distribution and higher exchange energy (e.g., Cr [Ar] 3d⁵ 4s¹ instead of 3d⁴ 4s²; Cu [Ar] 3d¹⁰ 4s¹ instead of 3d⁹ 4s²).

Critical Concept Check: Write the electronic configuration of Iron (Fe, Z=26) using orbital notation.

A: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶. Or using noble gas core: [Ar] 3d⁶ 4s². (Note: 4s fills before 3d according to Aufbau).

Try Yourselves - Chapter 2 (Class XI)

1. Particle discovered by Chadwick?

2. Define Isobars.

3. State Heisenberg's Principle.

4. Limitation of Bohr's model?

5. Shape of 'p' orbital?

6. How many orbitals in d-subshell?

7. State Hund's Rule.

8. Electronic config of Cl (Z=17)?

Micro Tests - Chapter 2 (Class XI)

Micro Test XI2.1 (Particles & Models - 10 Marks)

Test XI2.1

Time: 10 Min

Answer the following.

1.Charge & Relative Mass of electron?[2]
2.Who discovered the proton?[1]
3.Describe Thomson's atomic model.[2]
4.Key conclusion from α-scattering about atom's structure?[2]
5.One limitation of Rutherford's model?[1]
6.Define Isotopes.[2]

Micro Test XI2.2 (Bohr & Dual Nature - 10 Marks)

Test XI2.2

Time: 10 Min

Answer the following.

1.State one postulate of Bohr's model about energy levels.[2]
2.What phenomenon did Bohr's model successfully explain for Hydrogen?[1]
3.State one limitation of Bohr's model.[1]
4.State de Broglie's equation.[2]
5.State Heisenberg's Uncertainty Principle.[2]
6.What does Planck's equation E=hν relate?[2]

Micro Test XI2.3 (Quantum Numbers & Orbitals - 10 Marks)

Test XI2.3

Time: 10 Min

Answer the following.

1.What does the Principal Quantum Number (n) signify?[1]
2.What does the Azimuthal Quantum Number (l) signify?[1]
3.What are the possible values of 'l' for n=3?[2]
4.What are the possible values of 'm_l' for l=2?[2]
5.What is the shape of an s orbital?[1]
6.What is the shape of a p orbital?[1]
7.How many 'd' orbitals are there in a d-subshell?[1]
8.What are the two possible values for spin quantum number (m_s)?[1]

Micro Test XI2.4 (Electron Configuration - 10 Marks)

Test XI2.4

Time: 10 Min

Answer the following.

1.State Aufbau Principle.[2]
2.State Pauli Exclusion Principle.[2]
3.State Hund's Rule of Maximum Multiplicity.[2]
4.Write electronic configuration of Silicon (Si, Z=14).[2]
5.Write electronic configuration of Chromium (Cr, Z=24), noting exception.[2]

3. Classification & Periodicity

Understand historical context (Mendeleev), Modern Periodic Law/Table, division into blocks, IUPAC nomenclature (Z>100), and detailed trends in periodic properties with reasons.

Quick Links: Historical & Modern Law Periodic Table Structure Trends: Radii & Valency Trends: IE, EA, EN Other Trends & Anomalies Practice Micro Tests

3.1 Historical Context & Modern Periodic Law

Key Points:

Significance of Classification: Systematic study of elements and prediction of properties.
Mendeleev's Periodic Law: Properties are periodic function of Atomic *Masses*.
Mendeleev's Table: Arranged by increasing At. Mass into Periods/Groups.
- Merits: Grouping similar elements, predicting undiscovered elements (eka-aluminium/Ga, eka-silicon/Ge), correcting atomic masses.
- Defects: Position of H, Isotopes, Anomalous pairs (Ar/K, Co/Ni, Te/I).
Modern Periodic Law (Moseley): Properties are periodic function of Atomic *Numbers* (Z). Resolves Mendeleev's defects.

3.2 Long Form Periodic Table & Nomenclature

Key Points:

Structure: 7 Periods (horizontal rows, n=shell no.), 18 Groups (vertical columns, similar valence config & properties).
Blocks: Based on differentiating electron entry:
- s-block: Groups 1 & 2 (outermost e⁻ in s-subshell).
- p-block: Groups 13 to 18 (outermost e⁻ in p-subshell).
- d-block: Groups 3 to 12 (Transition elements, e⁻ enters penultimate d-subshell).
- f-block: Lanthanides & Actinides (Inner transition, e⁻ enters anti-penultimate f-subshell).
IUPAC Nomenclature (Z > 100): Uses roots for digits 0-9 (nil, un, bi, tri, quad, pent, hex, sept, oct, enn). Assemble roots + suffix '-ium'. (e.g., Z=101: Unnilunium (Unu), Z=118: Ununoctium (Uuo)).

Q: What is the IUPAC name and symbol for element Z=104?

A: 1=un, 0=nil, 4=quad. Name: Unnilquadium. Symbol: Unq.

3.3 Periodic Trends: Radii & Valency

Key Points:

Atomic Radius: Half the internuclear distance in homonuclear molecule (covalent radius) or distance to outermost shell.
- Across Period (L→R): Decreases (↑Nuclear Charge pulls e⁻ closer).
- Down Group (T→B): Increases (New shells added, ↑Shielding).
Ionic Radius:
- Cation: Smaller than parent atom (lost e⁻, ↑effective nuclear charge).
- Anion: Larger than parent atom (gained e⁻, ↓effective nuclear charge, ↑repulsion).
- Trends: Follow similar trends to atomic radius across period/down group for isoelectronic species.
Valency: Combining capacity, often equal to group number (Gp 1, 2) or 8 - group number (Gp 13-17) or related to valence electrons. Shows periodicity.

Critical Concept Check: Why is the radius of Cl⁻ larger than Cl, but Na⁺ smaller than Na?

A: Cl gains an electron to form Cl⁻. The nuclear charge (17+) remains the same, but it now has to hold 18 electrons. The electron-electron repulsion increases, and the effective nuclear charge per electron decreases, causing the electron cloud to expand → Cl⁻ > Cl. Na loses an electron to form Na⁺. The nuclear charge (11+) remains, but it now holds only 10 electrons. The effective nuclear charge per electron increases, pulling the remaining electrons closer → Na⁺ < Na.

3.4 Periodic Trends: IE, EA, EN

Key Points:

Ionisation Enthalpy (IE): Energy required to remove outermost electron from isolated gaseous atom. (IE₁, IE₂, ...).
- Factors: Size (↓Size, ↑IE), Nuclear Charge (↑Charge, ↑IE), Shielding (↑Shielding, ↓IE), Penetration effect (s>p>d>f), Stable Config (Half/Full filled harder to remove e⁻).
- Across Period: Generally Increases.
- Down Group: Generally Decreases.
Electron Gain Enthalpy (EA / Δ_egH): Enthalpy change when electron added to isolated gaseous atom. Usually exothermic (negative value). More negative = higher affinity.
- Factors: Size, Nuclear Charge, Stable Config.
- Across Period: Generally becomes more negative (↑affinity). Halogens most negative. Noble gases positive (unstable anion).
- Down Group: Generally becomes less negative (↓affinity). *Exception: EA of Cl > F*.
Electronegativity (EN): Ability of atom in a covalent bond to attract shared electron pair. (Pauling, Mulliken scales).
- Factors: Size, Nuclear Charge.
- Across Period: Increases.
- Down Group: Decreases.

Critical Concept Check: Why is the electron gain enthalpy of Fluorine less negative than that of Chlorine, despite Fluorine being more electronegative?

A: Fluorine atom is very small. The existing electrons in its compact 2p subshell cause significant electron-electron repulsion when an extra electron is added. This repulsion slightly counteracts the attraction from the nucleus, resulting in less energy being released (less negative EA) compared to Chlorine, which has a larger 3p subshell where the incoming electron experiences less repulsion.

3.5 Other Trends & Anomalies

Key Points:

Periodicity of Valence / Oxidation States: Valency typically relates to group number. Oxidation state trends seen across periods (e.g., max O.S. often = group number for p-block).
Anomalous Properties of 2nd Period Elements: Elements Li, Be, B, C, N, O, F differ significantly from heavier members of their groups due to: Small size, High IE/EN, Absence of d-orbitals, Ability to form pπ-pπ multiple bonds.
Diagonal Relationship: Similarity in properties between elements placed diagonally in 2nd and 3rd periods (Li-Mg, Be-Al, B-Si). Due to similar ionic size / charge/radius ratio / electronegativity.
Nature of Oxides:
- Across Period: Basic → Amphoteric → Acidic. (e.g., Na₂O(basic), Al₂O₃(amphoteric), Cl₂O₇(acidic)).
- Down Group: Basicity of oxides increases (metallic character increases).

Q: Give one reason for the diagonal relationship between Lithium and Magnesium.

A: Similar ionic radii (Li⁺ ≈ Mg²⁺) / Similar electronegativity / Similar charge/radius ratio.

Try Yourselves - Chapter 3 (Class XI)

1. Basis of Modern Periodic Law?

2. Element Z=17 belongs to which block?

3. IUPAC name for Z=112?

4. Trend in Atomic Radius down Group 2?

5. Trend in IE across Period 3?

6. Which has higher EA: O or S?

7. Why F differs from Cl, Br, I?

8. Nature of Na₂O?

Micro Tests - Chapter 3 (Class XI)

Micro Test XI3.1 (Laws & Table Structure - 10 Marks)

Test XI3.1

Time: 8 Min

Answer the following.

1.State Modern Periodic Law.[2]
2.Basis of Mendeleev's classification?[1]
3.How many groups and periods in long form table?[2]
4.Elements of Group 1 & 2 belong to which block?[1]
5.Elements of Group 13-18 belong to which block?[1]
6.IUPAC symbol for element Z=109?[1]
7.IUPAC name for element Z=102?[2]

Micro Test XI3.2 (Trends - Radius, Valency, Metallic - 10 Marks)

Test XI3.2

Time: 10 Min

Answer the following.

1.Trend in Atomic Radius across Period 2 (Li to Ne)? Reason?[3]
2.Trend in Atomic Radius down Group 1 (Li to Cs)? Reason?[3]
3.Why is K⁺ smaller than K?[2]
4.What is the general trend in valency across Period 3 (Na to Ar)?[1]
5.Trend in metallic character down Group 14?[1]

Micro Test XI3.3 (Trends - IE, EA, EN - 10 Marks)

Test XI3.3

Time: 10 Min

Answer the following.

1.Define Ionisation Enthalpy.[2]
2.Trend in IE across Period 3? Reason?[2]
3.Trend in EA down Group 17? Reason?[2]
4.Define Electronegativity.[2]
5.Which element has highest EN?[1]
6.Why Noble gases have positive EA?[1]

Micro Test XI3.4 (Anomalies & Oxides - 10 Marks)

Test XI3.4

Time: 10 Min

Answer the following.

1.Give one reason for anomalous behaviour of 2nd period elements.[2]
2.Name the element showing diagonal relationship with Beryllium (Be).[1]
3.What is the reason for diagonal relationships?[2]
4.Trend in nature of oxides across Period 3 (Na₂O to Cl₂O₇)?[3]
5.Nature of MgO? Basic/Acidic/Amphoteric?[1]
6.Nature of SO₃? Basic/Acidic/Amphoteric?[1]

4. Chemical Bonding & Structure

Understand Lewis approach, ionic/covalent/coordinate bonds, VSEPR theory, hybridisation, MOT (qualitative), resonance, H-bonding.

Quick Links: Kossel-Lewis & Octet Rule Ionic Bonding Covalent Bonding & Parameters Polarity, Fajan's Rules VSEPR Theory Hybridisation Molecular Orbital Theory (MOT) Coordinate Bond, Resonance, H-Bond Practice Micro Tests

4.1 Kossel-Lewis Approach & Octet Rule

Key Points:

Valence Electrons: Outermost shell electrons involved in bonding.
Lewis Symbols: Element symbol surrounded by dots representing valence electrons.
Kossel-Lewis Approach: Atoms combine to achieve stable noble gas configuration (octet/duplet) by transferring (ionic) or sharing (covalent) electrons.
Octet Rule: Tendency of atoms to have 8 electrons in valence shell.
Applications: Explains formation of simple ionic (e.g., NaCl) and covalent (e.g., Cl₂) bonds.

4.2 Ionic Bonding

Key Points:

Formation: Transfer of e⁻ (Metal→Non-metal). Lewis structures (NaCl, Li₂O, MgO, CaO, MgF₂, Na₂S).
Conditions: Low IE (metal), High EA (non-metal), High Lattice Enthalpy, Large EN difference.
Lattice Enthalpy: Energy released when 1 mole of ionic solid formed from gaseous ions. (Born-Haber cycle concept - qualitative).
Variable Electrovalency: Due to Inert Pair Effect (reluctance of ns² e⁻ to participate, esp. in heavier p-block elements like Pb²⁺/Pb⁴⁺, Sn²⁺/Sn⁴⁺) or unstable core.
Characteristics: Crystalline solids, High MP/BP, Conductors (molten/aq), Soluble in polar solvents.

Q: Explain Inert Pair Effect with example.

A: Reluctance of the outermost 's' electrons (ns²) to participate in bonding due to poor shielding by inner d/f electrons, especially in heavier p-block elements. Example: Lead (Pb) can show +2 (using only p e⁻) and +4 (using s and p e⁻) oxidation states, but Pb(II) is more stable than Pb(IV) due to inert pair effect.

4.3 Covalent Bonding & Parameters

Key Points:

Formation: Mutual sharing of electrons. Lewis structures (H₂, Cl₂, O₂, N₂, CH₄, NH₃, H₂O, C₂H₄, C₂H₂, CO₂ etc.).
Conditions: Small EN difference, high IE/EA (non-metals).
Bond Parameters:
- Bond Length: Avg distance between nuclei of bonded atoms. (Factors: size, multiplicity - single>double>triple).
- Bond Angle: Angle between orbitals containing bonding electron pairs around central atom.
- Bond Enthalpy: Energy required to break 1 mole of specific type of bond in gaseous state. (Higher = Stronger bond).
- Bond Order: No. of bonds between two atoms (1 for single, 2 for double, 3 for triple). (↑Order, ↑Enthalpy, ↓Length).
Sigma (σ) & Pi (π) Bonds:
- σ Bond: Formed by head-on/axial overlap (s-s, s-p, p-p). Stronger. First bond formed.
- π Bond: Formed by sideways/lateral overlap of p-orbitals. Weaker. Formed after σ bond in multiple bonds.
- Single bond = 1σ. Double bond = 1σ + 1π. Triple bond = 1σ + 2π.
- Examples: H₂(σ), O₂(1σ, 1π), N₂(1σ, 2π), C₂H₄(5σ, 1π), C₂H₂(3σ, 2π).
Variable Covalency: Due to availability of vacant d-orbitals for expanding octet (e.g., P in PCl₅ (valency 5), S in SF₆ (valency 6)). Cl can show 1, 3, 5, 7.
Formal Charge: Hypothetical charge on atom assuming equal sharing (Valence e⁻ - Non-bonding e⁻ - ½ Bonding e⁻). Helps choose most stable Lewis structure.

Critical Concept Check: Why can Phosphorus form PCl₅ but Nitrogen cannot form NCl₅?

A: Nitrogen (Period 2) has only s and p orbitals in its valence shell (n=2) and lacks vacant d-orbitals. It can only form a maximum of 4 covalent bonds (e.g., in NH₄⁺). Phosphorus (Period 3) has vacant 3d orbitals available. It can promote one of its 3s electrons to a 3d orbital, allowing it to form 5 covalent bonds, as in PCl₅ (sp³d hybridization).

4.4 Polarity & Fajan's Rules

Key Points:

Polar Covalent Bond: Unequal sharing due to EN difference → Partial charges (δ⁺, δ⁻).
Dipole Moment (μ): Measure of polarity. μ = charge (q) × distance (d). Vector quantity. Symmetrical molecules (CO₂, CCl₄, BF₃) have zero net dipole moment even with polar bonds. H₂O, NH₃ have net dipole moment (lone pairs contribute).
Fajan's Rules (Covalent Character in Ionic Bonds): Covalent character increases with:
- Small Cation size.
- Large Anion size.
- High charge on Cation &/or Anion.
- Cation with non-noble gas config (pseudo-noble gas config - e.g., Cu⁺).
Consequences: Predicts relative covalent character (e.g., LiCl more covalent than NaCl; AlCl₃ more covalent than MgCl₂). Affects solubility, melting point.
Deviation from Octet Rule:
- Incomplete Octet: Central atom < 8 valence e⁻ (e.g., BeCl₂, BF₃).
- Expanded Octet: Central atom > 8 valence e⁻ (requires d-orbitals, e.g., PCl₅, SF₆).
- Odd Electron Molecules: (e.g., NO, NO₂).

Q: Which is expected to have more covalent character: NaCl or MgCl₂? Why?

A: MgCl₂. According to Fajan's rules, higher charge on the cation increases covalent character. Mg²⁺ has a higher charge (+2) than Na⁺ (+1). (Also Mg²⁺ is smaller than Na⁺, further increasing covalent character).

4.5 VSEPR Theory

Key Points:

Valence Shell Electron Pair Repulsion Theory: Predicts geometry of simple molecules based on minimizing repulsion between electron pairs (bonding pairs - BP, lone pairs - LP) in valence shell of central atom.
Repulsion Order: LP-LP > LP-BP > BP-BP.
Predicting Shape: Count total e⁻ pairs around central atom → Determine basic geometry (Linear, Trigonal Planar, Tetrahedral, Trigonal Bipyramidal, Octahedral) → Account for lone pairs distorting the ideal angles/shape.
Examples (Shape & Bond Angle approx.):
- BeCl₂ (2BP, 0LP): Linear, 180°.
- BF₃ (3BP, 0LP): Trigonal Planar, 120°.
- CH₄ (4BP, 0LP): Tetrahedral, 109.5°.
- NH₃ (3BP, 1LP): Pyramidal (based on tetrahedral), ~107°.
- H₂O (2BP, 2LP): Bent/V-Shape (based on tetrahedral), ~104.5°.
- PCl₅ (5BP, 0LP): Trigonal Bipyramidal, 120° & 90°.
- SF₆ (6BP, 0LP): Octahedral, 90°.

Critical Concept Check: Why is the bond angle in H₂O (~104.5°) smaller than in NH₃ (~107°), even though both have a tetrahedral electron geometry?

A: Water (H₂O) has two lone pairs on the central Oxygen atom, while Ammonia (NH₃) has only one lone pair on Nitrogen. Lone pair - lone pair repulsion is stronger than lone pair - bond pair repulsion, which is stronger than bond pair - bond pair repulsion. The two lone pairs in water exert a greater repulsive force, pushing the O-H bonds closer together, resulting in a smaller bond angle compared to ammonia where there is less lone pair repulsion.

4.6 Hybridisation

Key Points:

Concept: Mixing of atomic orbitals of slightly different energies to form new hybrid orbitals of equivalent energy and shape.
Purpose: Explains observed bond angles and shapes of molecules that VBT alone cannot easily explain.
Types (sp, sp², sp³ only primarily, maybe d-involvement):
- sp: Mixing 1s + 1p → 2 sp orbitals. Geometry: Linear, Angle: 180°. (e.g., BeCl₂, C₂H₂).
- sp²: Mixing 1s + 2p → 3 sp² orbitals. Geometry: Trigonal Planar, Angle: 120°. (e.g., BF₃, C₂H₄).
- sp³: Mixing 1s + 3p → 4 sp³ orbitals. Geometry: Tetrahedral, Angle: 109.5°. (e.g., CH₄, NH₃, H₂O).
- sp³d: Trigonal Bipyramidal (e.g., PCl₅).
- sp³d²: Octahedral (e.g., SF₆).
Relate hybridisation type to number of sigma bonds + lone pairs around central atom.

Q: What is the hybridisation of Carbon atoms in Ethene (C₂H₄)? What is the geometry around each Carbon?

A: Each Carbon atom forms 3 sigma bonds (2 C-H, 1 C-C) and 1 pi bond (C=C). Hybridisation involving 3 sigma regions = sp². Geometry around each Carbon = Trigonal Planar (bond angles ~120°).

4.7 Molecular Orbital Theory (MOT)

Key Points: (Qualitative Idea for H₂ to Ne₂)

Atomic orbitals combine to form molecular orbitals (MOs) spanning the molecule.
Types of MOs: Bonding MOs (lower energy, stable) & Antibonding MOs (higher energy, unstable, denoted by *).
Filling MOs: Follow Aufbau, Pauli, Hund's rules.
Energy Level Diagrams: Order of filling for O₂, F₂, Ne₂ (σ1s, σ*1s, σ2s, σ*2s, σ2p_z, π2p_x=π2p_y, π*2p_x=π*2p_y, σ*2p_z). Order reverses slightly for B₂, C₂, N₂ (π2p before σ2p).
Bond Order (BO): ½ [No. of e⁻ in Bonding MOs - No. of e⁻ in Antibonding MOs]. (BO=1 single, 2 double, 3 triple). Higher BO = Stronger/Shorter bond.
Magnetic Properties: Paramagnetic (unpaired e⁻, attracted by magnet), Diamagnetic (all e⁻ paired, repelled). Explains paramagnetism of O₂ (has 2 unpaired e⁻ in π*2p MOs).
Relative Stabilities: Compare BO for species like O₂, O₂⁺, O₂⁻, O₂²⁻ and N₂, N₂⁺, N₂⁻, N₂²⁻.

Critical Concept Check: Using MOT, explain why He₂ molecule does not exist.

A: Helium atom (He) has config 1s². In hypothetical He₂ molecule (4 electrons total), the MO config would be (σ1s)² (σ*1s)². Bond Order = ½ [Bonding e⁻ - Antibonding e⁻] = ½ [2 - 2] = 0. A bond order of zero indicates no net bond formation, hence He₂ is unstable and does not exist.

4.8 Coordinate Bond, Resonance, H-Bond

Key Points:

Coordinate (Dative) Bond: Both shared e⁻ from one atom (donor) to another (acceptor). Examples: NH₄⁺, H₃O⁺, O₃, formation of oxy-acids of chlorine (HClO, HClO₂, HClO₃, HClO₄).
Resonance: When a molecule/ion cannot be represented by a single Lewis structure, but is a hybrid of multiple contributing structures (resonance structures). Actual structure has lower energy (more stable) than any single structure. Examples: Ozone (O₃), Carbon Dioxide (CO₂), Carbonate ion (CO₃²⁻), Nitrate ion (NO₃⁻). (Draw contributing structures).
Hydrogen Bonding (H-Bond): Special dipole-dipole attraction between H atom bonded to highly electronegative atom (F, O, N) and a lone pair on another nearby F, O, N atom.
- Types: Intermolecular (between molecules, e.g., HF, H₂O, alcohols) & Intramolecular (within same molecule, e.g., o-nitrophenol).
- Conditions: H bonded to F/O/N; F/O/N atom must have lone pair(s).
- Consequences: Affects properties like boiling point (unusually high for H₂O, HF, NH₃), solubility, viscosity, density of ice < water.

Q: Why does water (H₂O) have a much higher boiling point than H₂S?

A: Water molecules form strong intermolecular Hydrogen bonds due to the high electronegativity of Oxygen and the presence of lone pairs. H₂S has weaker dipole-dipole forces (S is less electronegative, H-bonds negligible). More energy is needed to overcome the strong H-bonds in water, resulting in a higher boiling point.

Try Yourselves - Chapter 4 (Class XI)

1. Condition for ionic bond formation?

2. Define Lattice Enthalpy.

3. Difference between Sigma & Pi bond?

4. Shape of CH₄ using VSEPR?

5. Hybridisation in BeCl₂?

6. Bond order of N₂?

7. Is O₂ paramagnetic or diamagnetic? Why?

8. Give example of intermolecular H-bond.

Micro Tests - Chapter 4 (Class XI)

Micro Test XI4.1 (Ionic & Covalent Basics - 10 Marks)

Test XI4.1

Time: 10 Min

Answer the following.

1.Define Ionic Bond.[1]
2.Define Covalent Bond.[1]
3.State one condition favouring ionic bond formation.[1]
4.Draw Lewis structure for NaCl formation.[2]
5.Draw Lewis structure for Cl₂ molecule.[2]
6.Define Bond Length.[1]
7.Define Bond Enthalpy.[1]
8.What is Bond Order?[1]

Micro Test XI4.2 (Sigma/Pi, Polarity, Fajan - 10 Marks)

Test XI4.2

Time: 10 Min

Answer the following.

1.How many sigma (σ) and pi (π) bonds in N₂ molecule?[2]
2.How many sigma (σ) and pi (π) bonds in Ethene (C₂H₄)?[2]
3.Define Dipole Moment.[1]
4.Is CO₂ molecule polar or non-polar? Why?[2]
5.State one condition from Fajan's rules that increases covalent character.[1]
6.Which is more covalent: LiF or LiI? Why?[2]

Micro Test XI4.3 (VSEPR & Hybridisation - 10 Marks)

Test XI4.3

Time: 12 Min

Predict shape & hybridisation:

1.BeCl₂[2]
2.BF₃[2]
3.NH₃[2]
4.H₂O[2]
5.PCl₅[2]

Micro Test XI4.4 (MOT, Resonance, H-Bond - 10 Marks)

Test XI4.4

Time: 12 Min

Answer the following.

1.According to MOT, what is Bond Order?[1]
2.Calculate Bond Order for O₂ (Total e⁻=16). Config: ...(σ2p_z)² (π2p_x)²(π2p_y)² (π*2p_x)¹(π*2p_y)¹[2]
3.Is O₂ paramagnetic or diamagnetic based on MOT?[1]
4.Define Resonance.[1]
5.Give one example molecule showing resonance.[1]
6.Define Hydrogen Bond.[1]
7.What are the two types of H-Bonding?[2]
8.Why does ice float on water?[1]

5. Chemical Thermodynamics

Concepts of system/surroundings, properties, processes, First Law (Internal Energy, Enthalpy, Work, Heat, Cp, Cv), Hess's Law, Enthalpy changes, Second Law (Spontaneity, Entropy, Gibbs Free Energy), Third Law statement.

Quick Links: Basic Concepts First Law & Enthalpy Heat Capacity & Enthalpy Changes Hess's Law Second Law & Entropy Gibbs Free Energy & Spontaneity Third Law Practice Micro Tests

5.1 Basic Concepts: System, Surroundings, Properties, Processes

Key Points:

System: Part of universe under observation.
Surroundings: Rest of the universe outside the system.
Types of System:
- Open: Exchanges both energy and matter with surroundings (e.g., beaker with reaction).
- Closed: Exchanges energy but not matter (e.g., sealed flask with reaction).
- Isolated: Exchanges neither energy nor matter (e.g., ideal thermos flask).
Properties:
- Extensive: Depend on amount of substance (e.g., Mass, Volume, Internal Energy, Enthalpy, Entropy).
- Intensive: Independent of amount of substance (e.g., Temperature, Pressure, Density, Molarity).
State Functions: Properties whose value depends only on the initial and final state of the system, not the path taken (e.g., P, V, T, U, H, S, G).
Path Functions: Properties whose value depends on the path followed (e.g., Work (w), Heat (q)).
Processes:
- Isothermal: Temperature constant (ΔT=0).
- Adiabatic: No heat exchange (q=0).
- Isobaric: Pressure constant (ΔP=0).
- Isochoric: Volume constant (ΔV=0).
- Reversible: Process occurs infinitesimally slowly, can be reversed by infinitesimal change; system always near equilibrium. Max work done.
- Irreversible: Process occurs rapidly, cannot be reversed easily.
- Cyclic: System returns to initial state after series of changes (ΔU=0, ΔH=0).
Thermodynamic Equilibrium: System state where macroscopic properties (T, P, Composition) do not change with time.

Critical Concept Check: Is density an intensive or extensive property? Explain.

A: Density is an intensive property. Density = Mass/Volume. Although both mass and volume are extensive (depend on amount), their ratio (density) is independent of the amount of substance. A small drop of water has the same density as a large pool of water (at the same temperature).

5.2 First Law of Thermodynamics, Internal Energy, Enthalpy

Key Points:

First Law (Law of Conservation of Energy): Energy can neither be created nor destroyed, only converted from one form to another.
Internal Energy (U or E): Total energy stored within a system (KE + PE of molecules). State function. Absolute value cannot be determined, only change (ΔU).
Mathematical Statement (First Law): ΔU = q + w
- ΔU = Change in internal energy.
- q = Heat absorbed by the system (+ve) / Heat released by the system (-ve).
- w = Work done *on* the system (+ve) / Work done *by* the system (-ve). [Note: Chemistry convention often uses ΔU = q - w where w is work done *by* system]. Be consistent with sign convention used.
Work (Pressure-Volume): Work done by gas expansion = -P_extΔV (irreversible). Work done in reversible isothermal expansion = -2.303 nRT log(V₂/V₁).
Enthalpy (H): Thermodynamic property, H = U + PV. State function. Represents total heat content at constant pressure.
Enthalpy Change (ΔH): Heat absorbed or released at constant pressure. ΔH = ΔU + PΔV (for constant P).
- ΔH positive = Endothermic reaction.
- ΔH negative = Exothermic reaction.

Critical Concept Check: For the reaction N₂(g) + 3H₂(g) → 2NH₃(g) at constant T & P, how does ΔH relate to ΔU?

A: We know ΔH = ΔU + PΔV. For ideal gases, PΔV = Δn_gRT, where Δn_g is the change in moles of gas (products - reactants). Here, Δn_g = 2 - (1+3) = -2. So, ΔH = ΔU + (-2)RT, or ΔH = ΔU - 2RT. Since RT is positive, ΔH will be more negative (or less positive) than ΔU.

5.3 Heat Capacity & Enthalpy Changes

Key Points:

Heat Capacity (C): Heat required to raise temperature of a substance by 1°C (or 1K). Extensive property.
Specific Heat Capacity (c or s): Heat required to raise temperature of 1 gram of substance by 1°C (or 1K). Intensive property. q = mcΔT.
Molar Heat Capacity (C_m): Heat required to raise temperature of 1 mole of substance by 1°C (or 1K). Intensive property. q = nC_mΔT.
C_p (Molar heat capacity at constant pressure): C_p = (ΔH/ΔT)_p.
C_v (Molar heat capacity at constant volume): C_v = (ΔU/ΔT)_v.
Relationship: C_p - C_v = R (for 1 mole ideal gas). C_p > C_v (heat supplied at const P also does expansion work).
Enthalpy Changes (ΔH): Heat change at constant pressure. Standard state (°) refers to 1 bar pressure, specified temp (usually 298K).
- Δ_fH° (Formation): Enthalpy change when 1 mole of compound formed from elements in standard states.
- Δ_cH° (Combustion): Enthalpy change when 1 mole of substance completely burnt in excess O₂.
- Δ_solH° (Solution): Enthalpy change when 1 mole solute dissolves in specified amount solvent.
- Δ_dilH° (Dilution): Enthalpy change when solution diluted.
- Δ_neutH° (Neutralization): Enthalpy change when 1 mole H₂O formed by neutralization of strong acid & strong base (~ -57.1 kJ/mol). Lower for weak acid/base (some heat used for ionization).
- Δ_dissH° / Δ_bondH° (Bond Dissociation): Enthalpy change to break 1 mole of specific bonds in gaseous state. Always positive. Bond Enthalpy = Average value.
- Δ_atomH° (Atomisation): Enthalpy change to form 1 mole gaseous atoms from element in standard state.
- Δ_subH° (Sublimation): Enthalpy change for Solid → Gas.
- Δ_fusH° (Fusion), Δ_vapH° (Vaporisation).
- Δ_ionH° (Ionisation): Ionisation Enthalpy.
Calorific Value: Heat produced by complete combustion of 1 gram of fuel.

Q: Why is the enthalpy of neutralization of a strong acid and strong base constant (~ -57.1 kJ/mol)?

A: Strong acids and bases are fully ionized in solution. The net ionic reaction for neutralization is always H⁺(aq) + OH⁻(aq) → H₂O(l). Since the actual reaction is the same (formation of 1 mole of water from its ions), the enthalpy change is constant, regardless of the specific strong acid and strong base used.

5.4 Hess's Law of Constant Heat Summation

Key Points:

Statement: The total enthalpy change for a reaction is the same, whether the reaction occurs in one step or in several steps. (Based on enthalpy being a state function).
Application: Allows calculation of enthalpy changes for reactions that cannot be measured directly, by combining enthalpy changes of known related reactions (treating thermochemical equations algebraically).
Calculations involving bond enthalpies: Δ_rH ≈ Σ(Bond Enthalpies of Reactants) - Σ(Bond Enthalpies of Products).
Simple numerical problems based on Hess's Law and bond enthalpies.

Critical Concept Check: How can Hess's Law be used to find the enthalpy of formation (Δ_fH°) of CO(g), given Δ_fH°(CO₂) and Δ_cH°(CO)?

A: Target reaction: C(s) + ½O₂(g) → CO(g) ; Δ_fH°(CO) = ?
Given: (1) C(s) + O₂(g) → CO₂(g) ; ΔH₁ = Δ_fH°(CO₂)
(2) CO(g) + ½O₂(g) → CO₂(g) ; ΔH₂ = Δ_cH°(CO)
Reverse equation (2): (3) CO₂(g) → CO(g) + ½O₂(g) ; ΔH₃ = -ΔH₂
Add equations (1) and (3): [C(s) + O₂(g)] + [CO₂(g)] → [CO₂(g)] + [CO(g) + ½O₂(g)]
Simplify: C(s) + ½O₂(g) → CO(g)
By Hess's Law: Δ_fH°(CO) = ΔH₁ + ΔH₃ = Δ_fH°(CO₂) - Δ_cH°(CO).

5.5 Second Law & Entropy

Key Points:

Inadequacy of First Law: Doesn't predict spontaneity/direction of process.
Spontaneous Process: Occurs on its own without external aid (may need initiation). Non-spontaneous requires external work.
Entropy (S): Measure of randomness or disorder of a system. State function. Units: J K⁻¹ mol⁻¹.
Second Law Statement (Entropy): Entropy of the universe (system + surroundings) always increases for a spontaneous process (ΔS_total = ΔS_sys + ΔS_surr > 0). For reversible process at equilibrium, ΔS_total = 0.
Physical Significance: Nature tends towards higher disorder. Gas > Liquid > Solid (Entropy increases). ↑Temp, ↑Volume, ↑No. of moles → ↑Entropy.
Entropy Change Calculation: ΔS = q_rev / T (for reversible process at const T).

Critical Concept Check: Can the entropy of a system decrease during a spontaneous process?

A: Yes, the entropy of the *system* can decrease (e.g., freezing of water, condensation of gas), but for the process to be spontaneous, the entropy increase in the *surroundings* must be greater, resulting in an overall increase in the entropy of the *universe* (ΔS_total > 0).

5.6 Gibbs Free Energy & Spontaneity

Key Points:

Gibbs Free Energy (G): Thermodynamic state function combining enthalpy and entropy. G = H - TS. Represents maximum available useful work.
Gibbs Free Energy Change (ΔG): Criterion for spontaneity at constant Temperature and Pressure.
- ΔG < 0 (negative): Process is spontaneous.
- ΔG > 0 (positive): Process is non-spontaneous (reverse is spontaneous).
- ΔG = 0: System is at equilibrium.
Relationship: ΔG = ΔH - TΔS. Spontaneity depends on balance between enthalpy (tendency towards lower energy) and entropy (tendency towards higher disorder), and temperature.
Effect of T on Spontaneity:
- ΔH -, ΔS + : Always spontaneous (ΔG always -).
- ΔH +, ΔS - : Never spontaneous (ΔG always +).
- ΔH -, ΔS - : Spontaneous at low T (enthalpy driven).
- ΔH +, ΔS + : Spontaneous at high T (entropy driven).
Standard Free Energy Change (ΔG°): ΔG for reaction with reactants/products in standard states.
ΔG° and Equilibrium Constant (K): ΔG° = -2.303 RT log K_eq. Relates thermodynamic favourability to extent of reaction.

Q: For a reaction, ΔH = +100 kJ/mol and ΔS = +200 J K⁻¹ mol⁻¹. Is the reaction spontaneous at 27°C (300K)?

A: ΔG = ΔH - TΔS. Convert ΔS to kJ: 200 J/K = 0.2 kJ/K.
ΔG = (+100 kJ) - (300 K)(+0.2 kJ/K) = +100 kJ - 60 kJ = +40 kJ.
Since ΔG is positive, the reaction is non-spontaneous at 300K.

5.7 Third Law of Thermodynamics

Key Points:

Statement: The entropy of a perfectly crystalline solid approaches zero as the absolute zero of temperature (0 K) is approached.
Significance: Allows calculation of absolute entropies of substances.
(Self-explanatory, no calculations expected).

Try Yourselves - Chapter 5 (Class XI)

1. Define Isolated System.

2. State First Law mathematically.

3. What is Enthalpy (H)?

4. Is Cp or Cv greater for a gas?

5. State Hess's Law.

6. What is Entropy (S)?

7. Condition for spontaneity using ΔG?

8. Relationship ΔG = ?

Micro Tests - Chapter 5 (Class XI)

Micro Test XI5.1 (Concepts & First Law - 10 Marks)

Test XI5.1

Time: 10 Min

Answer the following.

1.Define Extensive property + example.[2]
2.Define State Function + example.[2]
3.State First Law of Thermodynamics.[2]
4.Write mathematical form of First Law.[1]
5.What is the sign of 'q' when heat is released by system?[1]
6.What is the sign of 'w' when work is done BY system?[1]
7.What is Enthalpy (H)?[1]

Micro Test XI5.2 (Enthalpy Changes & Hess - 10 Marks)

Test XI5.2

Time: 12 Min

Define/State:

1.Standard Enthalpy of Formation (Δ_fH°).[2]
2.Standard Enthalpy of Combustion (Δ_cH°).[2]
3.Enthalpy of Neutralization. Approx value for strong acid/base?[2]
4.Hess's Law of Constant Heat Summation.[2]
5.Given: A→B, ΔH=+50; C→B, ΔH=-30. Find ΔH for A→C.[2]

Micro Test XI5.3 (Entropy & Second Law - 10 Marks)

Test XI5.3

Time: 10 Min

Answer the following.

1.What does the First Law not predict?[1]
2.Define Entropy (S).[2]
3.State the Second Law in terms of entropy of universe.[2]
4.Sign of entropy change (ΔS) for melting ice?[1]
5.Sign of entropy change for condensation of steam?[1]
6.Sign of entropy change for N₂(g) + 3H₂(g) → 2NH₃(g)?[1]
7.What happens to entropy as temperature increases?[1]
8.What is the unit of Entropy?[1]

Micro Test XI5.4 (Gibbs Energy & Spontaneity - 10 Marks)

Test XI5.4

Time: 10 Min

Answer the following.

1.Define Gibbs Free Energy (G).[2]
2.What is the criterion for spontaneity using ΔG (const T, P)?[2]
3.Write the Gibbs-Helmholtz equation relating ΔG, ΔH, ΔS.[1]
4.If ΔH is (-) and ΔS is (+), the reaction is spontaneous at...?[1]
5.If ΔH is (+) and ΔS is (+), the reaction is spontaneous at...?[1]
6.What is the value of ΔG at equilibrium?[1]
7.State the Third Law of Thermodynamics.[2]

6. Equilibrium

Concepts of physical/chemical equilibrium, Law of Mass Action, K_c/K_p, Le Chatelier's principle. Ionic equilibrium: Acids/bases (concepts), Ostwald's law, pH, buffers, common ion effect, hydrolysis, K_sp.

Quick Links: Chemical Equilibrium Basics Equilibrium Constant (Kc, Kp) Le Chatelier's Principle Ionic Eq: Acids/Bases & Ionisation Ionic Product of Water & pH Buffers & Common Ion Effect Salt Hydrolysis Solubility Product (Ksp) Practice Micro Tests

6.1 Chemical Equilibrium: Basics

Key Points:

Reversible Reactions: Proceed in both forward and backward directions (⇌).
Irreversible Reactions: Proceed mainly in one direction (→).
Physical Equilibrium: Equilibrium involving physical changes (phase transitions). Examples: Solid ⇌ Liquid (Melting point), Liquid ⇌ Vapour (Boiling point), Solid ⇌ Vapour (Sublimation point). Characteristics: Occurs in closed system, dynamic, properties constant.
Chemical Equilibrium: State in reversible reaction where rate of forward reaction = rate of backward reaction. Concentrations of reactants/products become constant (not necessarily equal).
Characteristics: Dynamic nature (reactions continue), attainable from either direction, occurs in closed system, catalyst affects rate but not equilibrium position.

6.2 Equilibrium Constant (K_c, K_p)

Key Points:

Law of Mass Action (Guldberg & Waage): Rate of reaction is proportional to product of active masses (molar concentrations) of reactants raised to stoichiometric coefficients.
Equilibrium Constant (K_c): Ratio of product of molar concentrations of products to reactants, each raised to stoichiometric coefficient, at equilibrium. For aA + bB ⇌ cC + dD, K_c = [C]^c[D]^d / [A]^a[B]^b. (Value depends only on temperature).
Equilibrium Constant (K_p): Ratio using partial pressures of gaseous reactants/products. K_p = (P_C)^c(P_D)^d / (P_A)^a(P_B)^b.
Relationship: K_p = K_c(RT)^Δn_g. Where Δn_g = (moles of gaseous products) - (moles of gaseous reactants). Derivation required.
Characteristics of K: Constant at given T, independent of initial conc/pressure/catalyst. Large K favours products, Small K favours reactants.
Simple calculations involving K_c and K_p.

Critical Concept Check: For the reaction H₂(g) + I₂(g) ⇌ 2HI(g), what is the relationship between K_p and K_c?

A: Calculate Δn_g = (moles gaseous products) - (moles gaseous reactants) = 2 - (1+1) = 0. Since Δn_g = 0, K_p = K_c(RT)⁰ = K_c(1). Therefore, K_p = K_c for this reaction.

6.3 Le Chatelier's Principle

Key Points:

Statement: If a change of condition (concentration, temperature, pressure) is applied to a system in equilibrium, the system will shift in a direction that tends to counteract the effect of the change.
Effect of Changes:
- Concentration: ↑Reactant conc → Shifts Right (favours products). ↑Product conc → Shifts Left (favours reactants).
- Pressure (for gases): ↑Pressure → Shifts towards side with fewer moles of gas. ↓Pressure → Shifts towards side with more moles of gas. No effect if Δn_g = 0.
- Temperature: ↑Temperature → Favours endothermic direction (absorbs heat). ↓Temperature → Favours exothermic direction (releases heat). (Value of K changes).
- Catalyst: Increases rate of both forward/backward reactions equally. Equilibrium reached faster, but position (K value) unchanged.
- Inert Gas Addition: At constant Volume: No effect on partial pressures/concentrations → No shift. At constant Pressure: Volume increases, partial pressures decrease → Shifts towards side with more moles of gas.
Applications (Maximizing Yield):
- Haber's Process (N₂ + 3H₂ ⇌ 2NH₃ + Heat): High P, Low T (compromise ~450°C), Catalyst (Fe).
- Contact Process (2SO₂ + O₂ ⇌ 2SO₃ + Heat): High P (~2 atm), Low T (compromise ~450°C), Catalyst (V₂O₅).
- Dissociation of N₂O₄ (N₂O₄ ⇌ 2NO₂ - Endothermic): Low P, High T.
- Ester Hydrolysis (Ester + H₂O ⇌ Acid + Alcohol): Add excess water or remove products.

Q: For the exothermic reaction A(g) + 2B(g) ⇌ C(g), what conditions favour maximum yield of C?

A: 1. High Pressure (Shifts right, 3 moles gas → 1 mole gas). 2. Low Temperature (Forward reaction is exothermic). 3. Increase concentration of A or B / Remove C as it forms.

6.4 Ionic Equilibrium: Acids/Bases & Ionisation

Key Points:

Electrolytes/Non-electrolytes: Revisited. Strong vs Weak electrolytes.
Degree of Ionisation (α): Fraction of total substance ionized at equilibrium.
Acid-Base Concepts:
- Arrhenius: Acid → H⁺ donor in water. Base → OH⁻ donor in water.
- Brønsted-Lowry: Acid → Proton (H⁺) donor. Base → Proton acceptor. (Concept of conjugate acid-base pairs).
- Lewis: Acid → Electron pair acceptor. Base → Electron pair donor.
Multistage Ionisation: Polybasic acids (H₂SO₄, H₃PO₄) ionize in steps, each with different K_a (K_a1 > K_a2 > ...).
Acid/Base Strength: Based on degree of ionisation (α) or dissociation constant (K_a for acids, K_b for bases). Larger K_a/K_b = Stronger acid/base. pK_a = -log K_a (Smaller pK_a = Stronger acid).
Ostwald's Dilution Law (for weak electrolytes): α² ≈ K⋅V or α ≈ √(K/C). Degree of ionisation increases on dilution. (Derivation required). Problems based on α, K_a/K_b, concentration.

Critical Concept Check: Identify the conjugate acid-base pairs in: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻

A: NH₃ (Base) accepts H⁺ to form NH₄⁺ (Conjugate Acid). Pair: NH₄⁺ / NH₃.
H₂O (Acid) donates H⁺ to form OH⁻ (Conjugate Base). Pair: H₂O / OH⁻.

6.5 Ionic Product of Water & pH

Key Points:

Self-Ionisation of Water: H₂O + H₂O ⇌ H₃O⁺ + OH⁻.
Ionic Product of Water (K_w): K_w = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25°C (298K). Constant at given temp.
pH Scale: pH = -log₁₀[H₃O⁺]. pOH = -log₁₀[OH⁻].
Relationship: pH + pOH = pK_w = 14 (at 25°C).
Acidity/Basicity & pH: Acidic (pH<7), Neutral (pH=7), Basic (pH>7).
pH Indicators: Weak organic acids/bases changing colour over specific pH range. Choice depends on pH change at equivalence point of titration.
Numericals involving pH, pOH, [H⁺], [OH⁻], K_w.

Q: Calculate the pH of a 0.01 M HCl solution.

A: HCl is a strong acid, fully ionized. HCl → H⁺ + Cl⁻. [H⁺] = [HCl] = 0.01 M = 10⁻² M. pH = -log₁₀[H⁺] = -log₁₀(10⁻²) = -(-2) = 2.

6.6 Buffers & Common Ion Effect

Key Points:

Buffer Solution: Resists change in pH upon addition of small amounts of acid or alkali.
- Acidic Buffer: Weak acid + its salt with strong base (e.g., CH₃COOH + CH₃COONa).
- Basic Buffer: Weak base + its salt with strong acid (e.g., NH₄OH + NH₄Cl).
Buffer Action (Mechanism): Added H⁺ reacts with conjugate base; added OH⁻ reacts with weak acid/base. Use examples to explain.
Henderson-Hasselbalch Equation:
- Acidic: pH = pK_a + log ([Salt]/[Acid])
- Basic: pOH = pK_b + log ([Salt]/[Base])
Common Ion Effect: Suppression of ionisation of a weak electrolyte by adding a strong electrolyte containing a common ion.
- Example: Adding CH₃COONa (strong) to CH₃COOH (weak) suppresses CH₃COOH ionisation due to common CH₃COO⁻ ion. Adding NH₄Cl to NH₄OH suppresses NH₄OH ionisation.
- Application: Qualitative analysis (precipitating Grp II sulphides in acidic medium, Grp III hydroxides in ammoniacal buffer).

6.7 Salt Hydrolysis

Key Points:

Definition: Reaction of cation or anion (or both) of a salt with water to produce acidity or alkalinity (change in pH from 7). Reverse of neutralization.
Types & pH of Solutions:
- Salt of Strong Acid + Strong Base (e.g., NaCl): No hydrolysis. Solution neutral (pH=7).
- Salt of Strong Acid + Weak Base (e.g., NH₄Cl, CuSO₄): Cation hydrolysis (NH₄⁺ + H₂O ⇌ NH₄OH + H⁺). Solution acidic (pH<7). pH = 7 - ½(pK_b + log C).
- Salt of Weak Acid + Strong Base (e.g., CH₃COONa, Na₂CO₃): Anion hydrolysis (CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻). Solution alkaline (pH>7). pH = 7 + ½(pK_a + log C).
- Salt of Weak Acid + Weak Base (e.g., CH₃COONH₄): Both ions hydrolyze. Solution pH depends on relative K_a, K_b (approx neutral if K_a≈K_b). pH = 7 + ½(pK_a - pK_b).
pH calculation formulae and qualitative explanations needed.

6.8 Solubility Product (K_sp)

Key Points:

Solubility Product (K_sp): Product of molar concentrations of constituent ions in a saturated solution of a sparingly soluble salt, each raised to power of its stoichiometric coefficient. (Applies to equilibrium between undissolved salt and its ions).
Example: For AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq), K_sp = [Ag⁺][Cl⁻]. For CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq), K_sp = [Ca²⁺][F⁻]².
Solubility (s): Molar concentration of saturated solution.
Relationship between K_sp and s:
- For AB type (AgCl): K_sp = s².
- For AB₂ type (CaF₂): K_sp = (s)(2s)² = 4s³.
- For A₂B type (Ag₂CrO₄): K_sp = (2s)²(s) = 4s³.
Application in Qualitative Analysis: Predicting precipitation. If Ionic Product (IP) > K_sp → Precipitation occurs. If IP < K_sp → No precipitation/Dissolution occurs. If IP = K_sp → Solution is saturated (equilibrium).
Numericals involving K_sp, solubility, prediction of precipitation.

Q: The K_sp of AgCl is 1.8 x 10⁻¹⁰. Calculate its molar solubility (s).

A: AgCl ⇌ Ag⁺ + Cl⁻. K_sp = [Ag⁺][Cl⁻] = (s)(s) = s².
s = √K_sp = √(1.8 x 10⁻¹⁰) = 1.34 x 10⁻⁵ mol/L.

Try Yourselves - Chapter 6 (Class XI)

1. Define Chemical Equilibrium.

2. Write K_c expression for PCl₅ ⇌ PCl₃ + Cl₂.

3. State Le Chatelier's Principle.

4. Define Brønsted-Lowry base.

5. pH of 10⁻⁴ M NaOH soln?

6. Define Buffer Solution.

7. What is Common Ion Effect?

8. Define Solubility Product (K_sp).

Micro Tests - Chapter 6 (Class XI)

Micro Test XI6.1 (Chemical Equilibrium - 10 Marks)

Test XI6.1

Time: 10 Min

Answer the following.

1.Define dynamic equilibrium.[2]
2.State the Law of Mass Action.[2]
3.Write expression for K_p for N₂O₄(g) ⇌ 2NO₂(g).[2]
4.State Le Chatelier's Principle.[2]
5.Effect of catalyst on equilibrium constant (K)?[1]
6.Effect of increasing pressure on H₂(g)+I₂(g)⇌2HI(g)?[1]

Micro Test XI6.2 (Ionic Eq - Acids/Bases/pH - 10 Marks)

Test XI6.2

Time: 10 Min

Answer the following.

1.Define Lewis Acid.[2]
2.Identify conjugate base of H₂SO₄.[1]
3.Define Ionic Product of Water (K_w).[2]
4.What is the value of K_w at 25°C?[1]
5.Define pH.[1]
6.Calculate pH of 10⁻³ M HNO₃.[1]
7.Calculate [H⁺] in solution with pOH = 11.[2]

Micro Test XI6.3 (Ionic Eq - Buffers, Hydrolysis - 10 Marks)

Test XI6.3

Time: 10 Min

Answer the following.

1.Define Buffer Solution.[2]
2.Give example of an acidic buffer.[1]
3.Define Common Ion Effect.[2]
4.Define Salt Hydrolysis.[2]
5.Will aq. solution of NH₄Cl be acidic, basic or neutral?[1]
6.Will aq. solution of CH₃COONa be acidic, basic or neutral?[1]
7.Will aq. solution of NaCl be acidic, basic or neutral?[1]

Micro Test XI6.4 (Ionic Eq - Ksp & Calculations - 10 Marks)

Test XI6.4

Time: 12 Min

Answer the following.

1.Define Solubility Product (K_sp).[2]
2.Write K_sp expression for Mg(OH)₂.[2]
3.If solubility of AgBr is 's' mol/L, what is its K_sp?[1]
4.If solubility of PbCl₂ is 's' mol/L, what is its K_sp?[1]
5.When does precipitation occur in terms of Ionic Product (IP) and K_sp?[2]
6.Calculate solubility (s) of BaSO₄ if K_sp = 1.0 x 10⁻¹⁰.[2]

7. Redox Reactions

Understand different concepts of oxidation/reduction, calculating oxidation numbers, balancing redox equations using both methods (O.N. and Ion-Electron).

Quick Links: Concepts of Oxidation & Reduction Oxidation Number (O.N.) Balancing: Oxidation Number Method Balancing: Ion-Electron Method Practice Micro Tests

7.1 Concepts of Oxidation & Reduction

Key Points:

Classical Concept:
- Oxidation: Addition of Oxygen / Electronegative element; Removal of Hydrogen / Electropositive element.
- Reduction: Removal of Oxygen / Electronegative element; Addition of Hydrogen / Electropositive element.
Electronic Concept:
- Oxidation: Loss of Electron(s) (LEO - Loss Electrons Oxidation). Increase in positive charge / Decrease in negative charge.
- Reduction: Gain of Electron(s) (GER - Gain Electrons Reduction). Decrease in positive charge / Increase in negative charge.
Oxidation Number Concept:
- Oxidation: Increase in Oxidation Number (O.N.).
- Reduction: Decrease in Oxidation Number (O.N.).
Redox Reaction: Reaction involving simultaneous oxidation and reduction.
Oxidizing Agent (Oxidant): Substance that causes oxidation (gets reduced itself).
Reducing Agent (Reductant): Substance that causes reduction (gets oxidized itself).

Critical Concept Check: In the reaction Zn + CuSO₄ → ZnSO₄ + Cu, identify the species oxidized, reduced, oxidizing agent, and reducing agent.

A: Zn (O.N. 0) → Zn²⁺ in ZnSO₄ (O.N. +2): Oxidation Number Increased → Zn is Oxidized. Cu²⁺ in CuSO₄ (O.N. +2) → Cu (O.N. 0): Oxidation Number Decreased → Cu²⁺ (in CuSO₄) is Reduced.
Zn causes reduction of Cu²⁺ → Zn is the Reducing Agent.
Cu²⁺ causes oxidation of Zn → CuSO₄ (or Cu²⁺) is the Oxidizing Agent.

7.2 Oxidation Number (O.N.)

Key Points:

Definition: Apparent charge assigned to an atom in a molecule or ion, assuming electrons in a covalent bond belong entirely to the more electronegative element.
Rules for Assignment:
1. O.N. of element in free state = 0 (e.g., O₂ , P₄, Na).
2. O.N. of monoatomic ion = its charge (e.g., Na⁺ = +1, Cl⁻ = -1, Mg²⁺ = +2).
3. O.N. of Oxygen = -2 (usually). Exceptions: Peroxides (O₂²⁻, O.N.=-1, e.g., H₂O₂), Superoxides (O₂⁻, O.N.=-½, e.g., KO₂), OF₂ (O.N.=+2).
4. O.N. of Hydrogen = +1 (usually, with non-metals). Exception: Metal Hydrides (O.N.=-1, e.g., NaH, CaH₂).
5. O.N. of Fluorine = -1 (always). Other Halogens = -1 (usually, except when bonded to O or more electronegative halogen).
6. Sum of O.N.s in neutral molecule = 0.
7. Sum of O.N.s in polyatomic ion = charge on the ion.
Calculation of O.N. in various compounds/ions (e.g., K₂Cr₂O₇, KMnO₄, H₂SO₄, S₂O₃²⁻, Fe₃O₄).

Q: Calculate the Oxidation Number of Cr in K₂Cr₂O₇.

A: K is Gp 1 (+1), O is -2. Let O.N. of Cr be x. Sum of O.N.s = 0.
2(+1) + 2(x) + 7(-2) = 0
+2 + 2x - 14 = 0
2x = +12
x = +6.

7.3 Balancing: Oxidation Number Method

Steps:

Write skeleton equation.
Assign O.N.s and identify atoms undergoing change.
Calculate total increase & decrease in O.N. per formula unit.
Equalize total increase and decrease by multiplying species with suitable integers.
Balance atoms other than O and H.
Balance O atoms by adding H₂O molecules.
Balance H atoms by adding H⁺ ions (acidic medium) or OH⁻ ions (basic medium - add H₂O to balance H, then add equal OH⁻ to both sides to neutralize H⁺ if needed).
Verify balanced equation (atoms and charges).

7.4 Balancing: Ion-Electron (Half-Reaction) Method

Steps:

Write skeleton ionic equation.
Split into two half-reactions (oxidation and reduction).
Balance atoms other than O and H in each half-reaction.
Balance O atoms by adding H₂O.
Balance H atoms by adding H⁺ (acidic) or H₂O/OH⁻ (basic).
- Basic Medium Hint: Balance H using H⁺ first as if acidic, then add equal OH⁻ to both sides to neutralize H⁺ into H₂O.
Balance charge by adding electrons (e⁻) to appropriate side.
Equalize electrons gained and lost by multiplying half-reactions by suitable integers.
Add the balanced half-reactions. Cancel common species (electrons must cancel).
Verify balanced equation (atoms and charges).

Critical Concept Check: What is the fundamental difference between the two balancing methods?

A: The Oxidation Number method focuses on tracking the change in oxidation numbers of specific atoms and balancing the total increase and decrease. The Ion-Electron method focuses on separating the overall reaction into oxidation and reduction half-reactions, balancing mass and charge (using electrons explicitly) in each half separately, and then combining them. Both methods achieve the same balanced equation.

Try Yourselves - Chapter 7 (Class XI)

1. Define oxidation (electronic concept).

2. Define reducing agent.

3. O.N. of S in H₂SO₄?

4. O.N. of Mn in KMnO₄?

5. O.N. of O in H₂O₂?

6. Name the two methods for balancing redox eq.

7. In Ion-electron method, how balance O atoms (acidic)?

8. In Ion-electron method, how balance charge?

Micro Tests - Chapter 7 (Class XI)

Micro Test XI7.1 (Concepts & Definitions - 10 Marks)

Test XI7.1

Time: 8 Min

Define/Identify:

1.Oxidation (in terms of electrons).[2]
2.Reduction (in terms of O.N.).[2]
3.Oxidizing Agent.[2]
4.Reducing Agent.[2]
5.In Mg + Cl₂ → MgCl₂, which is oxidized?[1]
6.In Mg + Cl₂ → MgCl₂, which is the oxidant?[1]

Micro Test XI7.2 (Oxidation Number Calculation - 10 Marks)

Test XI7.2

Time: 10 Min

Calculate O.N. of underlined element:

1.S in SO₄²⁻[2]
2.Cr in Cr₂O₇²⁻[2]
3.Mn in MnO₂[2]
4.N in HNO₃[2]
5.Cl in HClO₄[2]

Micro Test XI7.3 (Balancing Steps - O.N. Method - 10 Marks)

Test XI7.3

Time: 10 Min

Describe steps in O.N. method:

1.First step?[1]
2.How identify oxidized/reduced species?[2]
3.How equalize change in O.N.?[2]
4.How balance Oxygen atoms (acidic)?[2]
5.How balance Hydrogen atoms (acidic)?[1]
6.Final check?[2]

Micro Test XI7.4 (Balancing Steps - Ion-Electron - 10 Marks)

Test XI7.4

Time: 10 Min

Describe steps in Ion-Electron method (acidic):

1.First step after writing ionic eq?[1]
2.How balance atoms other than O & H?[1]
3.How balance O atoms?[1]
4.How balance H atoms?[1]
5.How balance charge in each half-reaction?[2]
6.How equalize electrons between half-reactions?[2]
7.Final step before verification?[1]
8.What must cancel out when adding half-reactions?[1]

8. Organic: Basic Principles

Introduction to organic chemistry, classification, IUPAC nomenclature, isomerism types, electron displacement effects, reaction intermediates, reaction types, and qualitative/quantitative analysis methods.

Quick Links: Introduction & Characteristics Classification IUPAC Nomenclature Isomerism Electron Displacement Effects Bond Fission & Reagents Reaction Types Purification & Analysis Practice Micro Tests

8.1 Introduction & Characteristics

Key Points:

Vital Force Theory (Berzelius): Initially believed organic compounds could only be formed by living organisms due to a 'vital force'. Disproved by Wöhler's synthesis of Urea (organic) from Ammonium cyanate (inorganic).
Separate Study Reasons: Vast number of compounds, unique properties due to Catenation and Tetravalency.
Importance: Fuels, polymers, drugs, dyes, food, biomolecules etc.
Characteristics of Carbon Atom: Tetravalency (forms 4 covalent bonds), Catenation (self-linking ability), tendency to form Multiple Bonds (C=C, C≡C, C=O etc.).

8.2 Classification of Organic Compounds

Key Points: Based on structure & functional group.

Structure Based:
- Acyclic / Open Chain / Aliphatic: Straight or branched chains (e.g., Ethane, Isobutane).
- Cyclic / Closed Chain / Ring: Atoms form rings.
  - Homocyclic (Carbocyclic): Ring contains only Carbon atoms.
    - Alicyclic: Properties similar to aliphatic (e.g., Cyclohexane).
    - Aromatic: Special stability, contain benzene ring (e.g., Benzene, Toluene).
  - Heterocyclic: Ring contains Carbon and at least one heteroatom (O, N, S etc.). (e.g., Pyridine, Furan).
Functional Group Based: Atom or group determining chemical properties.
- Homologous Series: Series differing by -CH₂ group, same functional group, similar properties (e.g., Alkanes, Alcohols, Aldehydes, Ketones, Carboxylic Acids, Amines, Ethers, Esters, Halides etc. - Recognize common functional groups).

Q: Classify Cyclohexane based on structure.

A: Cyclic → Homocyclic → Alicyclic.

8.3 IUPAC Nomenclature

Key Points: Systematic naming rules.

Components: Prefix (substituents) + Word Root (no. of C in main chain) + Primary Suffix (saturation: -ane, -ene, -yne) + Secondary Suffix (principal functional group).
Rules (Summary):
1. Longest continuous carbon chain selection.
2. Numbering chain (lowest number to principal functional group > multiple bond > substituent).
3. Identify & name substituents (prefixes like methyl-, ethyl-, chloro-, bromo-, nitro-).
4. Alphabetical order for multiple different substituents.
5. Use di-, tri-, tetra- for identical substituents (not considered for alphabetizing).
6. Correct placement of locants (numbers).
Naming Aliphatic (straight/branched alkanes, -enes, -ynes, compounds with one functional group - alcohols, aldehydes, ketones, acids, halides, amines), Alicyclic (cycloalkanes), and simple Aromatic compounds (benzene derivatives - mono/di-substituted).

Q: IUPAC name for CH₃-CH(OH)-CH₃?

A: Longest chain = 3C (Prop). Suffix = -ane. Functional group = -OH at C2. Name: Propan-2-ol.

8.4 Isomerism

Key Points: Same molecular formula, different properties (due to different structures/spatial arrangements).

Structural Isomerism: Different connectivity of atoms.
- Chain: Different C-skeleton (e.g., n-Butane, Isobutane).
- Position: Different position of functional group/multiple bond (e.g., Propan-1-ol, Propan-2-ol; But-1-ene, But-2-ene).
- Functional: Different functional groups (e.g., Ethanol C₂H₅OH & Dimethyl ether CH₃OCH₃ - MF C₂H₆O).
- Metamerism: Different alkyl groups on either side of polyvalent functional group (e.g., Ethers, Ketones: Diethyl ether C₂H₅OC₂H₅ & Methyl propyl ether CH₃OC₃H₇ - MF C₄H₁₀O).
- Tautomerism: Dynamic equilibrium between two isomers differing in position of proton & double bond (e.g., Keto-enol tautomerism).
Stereoisomerism: Same connectivity, different spatial arrangement.
- Geometrical (cis-trans): Restricted rotation (C=C, cyclic). Requires two different groups on *each* C of the restricted bond. Cis (same side), Trans (opposite side). Syn/Anti for oximes etc. Examples: But-2-ene (cis & trans).
- Optical: Rotate plane-polarised light. Requires Chiral center (Carbon bonded to 4 different groups). Non-superimposable mirror images (Enantiomers - d/l forms). Examples: Lactic acid, Tartaric acid. Racemic mixture (equimolar d+l, inactive). Meso form (internal compensation, inactive, e.g., meso-Tartaric acid).

Critical Concept Check: Why does But-1-ene not show geometrical isomerism, but But-2-ene does?

A: But-1-ene (CH₂=CH-CH₂-CH₃): One carbon of the double bond (C1) has two identical groups (two H atoms). Geometrical isomerism requires each carbon of the C=C bond to have two different groups attached. But-2-ene (CH₃-CH=CH-CH₃): Each carbon of the double bond (C2 and C3) has two different groups attached (a H and a CH₃ group). Thus, it exists as cis and trans isomers.

8.5 Electron Displacement Effects

Key Points: Permanent or temporary effects influencing electron density distribution.

Inductive Effect (I): Permanent polarisation of sigma (σ) bond due to electronegativity difference. Transmitted along chain, decreases with distance.
- -I Effect: Electron withdrawing groups (pull e⁻ density) (e.g., -NO₂, -CN, -COOH, -F, -Cl, -OH).
- +I Effect: Electron donating groups (push e⁻ density) (e.g., Alkyl groups -CH₃, -C₂H₅).
Electromeric Effect (E): Temporary, complete transfer of shared pi (π) electron pair to one atom under influence of attacking reagent. Occurs in multiple bonds (C=C, C=O). Reverses when reagent removed.
Resonance (Mesomeric Effect - R/M): Permanent delocalisation of pi (π) electrons through conjugated system (alternating single/multiple bonds or atoms with lone pairs). Represented by resonance structures, actual molecule is resonance hybrid.
- +R Effect: Groups donating e⁻ to conjugated system (e.g., -OH, -NH₂, -OR).
- -R Effect: Groups withdrawing e⁻ from conjugated system (e.g., -NO₂, -CN, -CHO, -COOH).
Hyperconjugation: Delocalisation of sigma (σ) electrons of C-H bond of alkyl group adjacent to unsaturated system (C=C, C⁺) or atom with unshared p-orbital. Stabilizes carbocations, alkenes. (No-bond resonance).

Q: Classify the -CH₃ group based on inductive effect.

A: Alkyl groups like methyl (-CH₃) are electron donating via the inductive effect, so it shows a +I effect.

8.6 Bond Fission & Reaction Intermediates/Reagents

Key Points:

Covalent Bond Fission:
- Homolytic Fission (Homolysis): Symmetrical breaking, each atom gets one electron → Forms Free Radicals (neutral species with unpaired e⁻). Favoured by non-polar conditions, heat, light (UV), peroxides. (A:B → A• + B•)
- Heterolytic Fission (Heterolysis): Unsymmetrical breaking, one atom takes both electrons → Forms Ions (Carbocation + Carbanion). Favoured by polar conditions. (A:B → A⁺ + :B⁻ or :A⁻ + B⁺)
Reaction Intermediates: Short-lived, highly reactive species formed during reaction.
- Free Radicals (e.g., •CH₃): Neutral, unpaired e⁻. Involved in substitution reactions of alkanes.
- Carbocations (e.g., CH₃⁺): Positive charge on C, electron deficient (6e⁻). Stability: 3° > 2° > 1° > Methyl.
- Carbanions (e.g., :CH₃⁻): Negative charge on C, electron rich (8e⁻, including lone pair). Stability: Methyl > 1° > 2° > 3°.
Attacking Reagents:
- Electrophiles (E⁺): Electron-loving species. Electron deficient, attack electron-rich centers. (e.g., H⁺, NO₂⁺, Cl⁺, BF₃, AlCl₃).
- Nucleophiles (Nu:⁻): Nucleus-loving species. Electron rich (lone pair or neg charge), attack electron-deficient centers (positive charge). (e.g., OH⁻, Cl⁻, CN⁻, H₂O:, NH₃:).

8.7 Types of Organic Reactions (Brief)

Key Points: (Definitions & Examples)

Substitution: Replacement of an atom/group by another. (e.g., Alkane halogenation, Benzene nitration).
Addition: Adding atoms/groups across a multiple bond (π bond breaks). (e.g., Alkene/Alkyne + H₂/Br₂/HX).
Elimination: Removal of atoms/groups from adjacent carbons to form multiple bond. (e.g., Dehydration of alcohols, Dehydrohalogenation of alkyl halides).
Mechanisms: Free radical (e.g., alkane halogenation), Polar (Electrophilic/Nucleophilic Substitution/Addition - S_N1, S_N2, E1, E2 concepts introduced briefly).

8.8 Purification & Analysis

Key Points:

Purification Methods: Basic principles of Crystallisation, Sublimation, Distillation (Simple, Fractional, Steam, Vacuum), Chromatography (brief idea).
Qualitative Analysis (Detection of Elements):
- Lassaigne's Test (Sodium Fusion Test): Organic compound fused with Na metal → converts covalently bonded N, S, Halogens into ionic NaCN, Na₂S, NaX.
- Test for N: Fuse filtrate + FeSO₄ + conc H₂SO₄ → Prussian blue colour (Fe₄[Fe(CN)₆]₃).
- Test for S: Fuse filtrate + Sodium nitroprusside → Violet colour. OR + Lead acetate → Black ppt (PbS).
- Test for Halogens (X): Fuse filtrate + dil HNO₃ + AgNO₃ → Ppt (AgCl-White, AgBr-Pale Yellow, AgI-Yellow). Differentiate solubility in NH₄OH.
- Test for C & H: Heating compound with CuO → CO₂(lime water milky), H₂O(anhydrous CuSO₄ blue).
Quantitative Analysis (Estimation):
- C & H (Liebig's Method): Known mass compound burnt in O₂, products absorbed in anhyd CaCl₂ (for H₂O) and KOH soln (for CO₂). Calculate % from mass increase.
- Nitrogen (Kjeldahl's Method): Compound + conc H₂SO₄ → (NH₄)₂SO₄. + NaOH → NH₃ evolved. NH₃ absorbed in std acid, titrated. Calculate %N. (Not for nitro/azo compounds). Dumas method alternative.
- Halogens (Carius Method): Compound + fuming HNO₃ + AgNO₃ → AgX ppt weighed. Calculate %X.
- Sulphur (Carius Method): Compound + fuming HNO₃ → H₂SO₄. + BaCl₂ → BaSO₄ ppt weighed. Calculate %S.
- Phosphorus: Compound + fuming HNO₃ → H₃PO₄. + Magnesia mixture → Mg₂P₂O₇ ppt weighed. Calculate %P.
Numericals based on estimation methods included.

Try Yourselves - Chapter 8 (Class XI)

1. Define Catenation.

2. Classify Toluene (Methylbenzene).

3. IUPAC name of CH₃COCH₃?

4. Define Metamerism.

5. Effect shown by -NO₂ group?

6. Type of fission forming free radicals?

7. Is H₂O an electrophile or nucleophile?

8. Which test detects Nitrogen in organic compound?

Micro Tests - Chapter 8 (Class XI)

Micro Test XI8.1 (Classification & Nomenclature - 10 Marks)

Test XI8.1

Time: 10 Min

Answer the following.

1.Define Functional Group.[1]
2.Give example of an Alicyclic compound.[1]
3.Give example of a Heterocyclic compound.[1]
4.IUPAC name of CH₂=CH-CHO?[2]
5.IUPAC name of CH₃-CH₂-O-CH₃?[2]
6.Draw structure of Butan-2-one.[1]
7.Draw structure of 2-Methylpropan-1-ol.[2]

Micro Test XI8.2 (Isomerism - 10 Marks)

Test XI8.2

Time: 12 Min

Define/Identify/Draw:

1.Define Chain Isomerism + Example.[2]
2.Define Position Isomerism + Example.[2]
3.Define Functional Isomerism + Example.[2]
4.Draw cis and trans isomers of But-2-ene.[2]
5.What condition is necessary for optical isomerism?[1]
6.What is a racemic mixture?[1]

Micro Test XI8.3 (Electronic Effects & Intermediates - 10 Marks)

Test XI8.3

Time: 10 Min

Define/Identify:

1.Inductive Effect.[1]
2.Resonance Effect.[1]
3.Type of effect shown by -Cl group?[1]
4.Type of effect shown by -OH group?[1]
5.Homolytic Fission leads to?[1]
6.Heterolytic Fission leads to?[1]
7.Define Electrophile + Example.[2]
8.Define Nucleophile + Example.[2]

Micro Test XI8.4 (Analysis - 10 Marks)

Test XI8.4

Time: 10 Min

Answer the following.

1.Name the test used to detect N, S, Halogens.[1]
2.What is the colour confirming Nitrogen in Lassaigne's test?[1]
3.Reagent used to test for Sulphur in Lassaigne's filtrate?[1]
4.How distinguish AgCl, AgBr, AgI ppts?[2]
5.Name the method for estimating Carbon & Hydrogen.[1]
6.Name the method for estimating Nitrogen (commonly used).[1]
7.Name the method for estimating Halogens.[1]
8.What compound absorbs CO₂ in Liebig's method?[1]
9.What compound absorbs H₂O in Liebig's method?[1]

9. Hydrocarbons

Detailed study of alkanes, alkenes, alkynes (nomenclature, isomerism, preparation, physical/chemical properties, mechanisms). Introduction to aromatic hydrocarbons (benzene structure, aromaticity, electrophilic substitution mechanism, directive influence).

Quick Links: Alkanes Alkenes Alkynes Aromatic Hydrocarbons (Benzene) Practice Micro Tests

9.1 Alkanes (Paraffins)

Key Points: (C_nH_2n+2, saturated, sp³ hybridised)

Nomenclature & Isomerism: Review IUPAC naming, chain isomerism (starts from C4).
Conformation (Ethane): Different spatial arrangements via rotation around C-C single bond.
- Sawhorse Projections: View molecule obliquely.
- Newman Projections: View along C-C bond. Eclipsed (H atoms aligned, max repulsion, least stable), Staggered (H atoms max distance apart, min repulsion, most stable), Skew (intermediate).
Preparation:
- From Carboxylic Acids: Decarboxylation (RCOONa + NaOH
- From Alcohols/Alkyl Halides: Wurtz reaction (2RX + 2Na
- From Aldehydes/Ketones: Grignard reagent (RMgX) followed by hydrolysis; Clemmensen/Wolff-Kishner reduction.
Physical Properties: Non-polar, insoluble in water, soluble in organic solvents. BP/MP increase with molar mass. Branching lowers BP. C1-C4 gases, C5-C17 liquids, >C17 solids.
Chemical Properties (Less Reactive):
- Halogenation (Free Radical Substitution): Via homolytic fission (Initiation, Propagation, Termination steps). Requires UV light/heat. (e.g., CH₄ + Cl₂ → CH₃Cl + HCl etc.).
- Combustion: Complete → CO₂ + H₂O. Incomplete → C (soot) + CO + H₂O.
- Pyrolysis (Cracking): Decomposition at high temp in absence of air → smaller alkanes/alkenes.
- Other: Controlled oxidation (→ alcohols/aldehydes), Isomerisation, Aromatisation (e.g., n-hexane → benzene).
Uses: Fuels (LPG, CNG, gasoline), solvents, starting materials.

Critical Concept Check: Explain the mechanism steps (Initiation, Propagation, Termination) for chlorination of methane.

A: 1. Initiation: Cl₂ 2. Propagation: Cl• + CH₄ → HCl + •CH₃ (Chlorine radical abstracts H from methane); •CH₃ + Cl₂ → CH₃Cl + Cl• (Methyl radical reacts with Cl₂, regenerating Cl• - chain continues).
3. Termination: Radicals combine to end chain. Cl• + Cl• → Cl₂; •CH₃ + •CH₃ → C₂H₆; Cl• + •CH₃ → CH₃Cl.

9.2 Alkenes (Olefins)

Key Points: (C_nH_2n, unsaturated C=C, sp² hybridised)

Nomenclature & Isomerism: Review IUPAC (suffix -ene, lowest number to C=C). Position isomerism (But-1-ene, But-2-ene). Geometrical (cis-trans) isomerism possible (e.g., But-2-ene).
Structure of Double Bond: One sigma (σ) bond + one pi (π) bond. Planar geometry around C=C. Restricted rotation.
Preparation:
- Dehydration of Alcohols: C₂H₅OH
- Dehydrohalogenation of Alkyl Halides: C₂H₅Br + KOH(alcoholic)
- From Vicinal Dihalides: CH₂Br-CH₂Br + Zn
- Kolbe's Electrolysis: Electrolysis of potassium succinate.
- Partial reduction of Alkynes: Using Lindlar's catalyst (Pd/CaCO₃/quinoline) → cis-alkene. Using Na/liq NH₃ (Birch) → trans-alkene.
Physical Properties: Similar trends to alkanes but slightly lower MP/BP for similar mass. Slightly polar due to sp² C.
Chemical Properties (Addition Reactions across C=C):
- Addition of H₂ (Hydrogenation): C₂H₄ + H₂
- Addition of Halogens (X₂): C₂H₄ + Br₂ → C₂H₄Br₂ (Decolorizes Br₂ water - Test for unsaturation).
- Addition of Hydrogen Halides (HX): C₂H₄ + HBr → C₂H₅Br. Follows Markownikoff's Rule for unsymmetrical alkenes (H adds to C with more H's, X adds to other C - 'rich get richer'). Anti-Markownikoff (Peroxide Effect) occurs with HBr in presence of peroxide (Br adds to C with more H's). Mechanism involves carbocation (Mark.) or free radical (Anti-Mark.).
- Addition of H₂SO₄: Follows Markownikoff's rule. Product hydrolyzes to alcohol.
- Addition of Water (Hydration): Direct (high T/P) or indirect (via H₂SO₄). Follows Markownikoff's rule.
Oxidation:
- Combustion: Complete → CO₂ + H₂O.
- With cold, dilute, alkaline KMnO₄ (Baeyer's Reagent): → Diol (e.g., Ethene glycol). Purple colour disappears - Test for unsaturation.
- With hot alkaline KMnO₄: Cleavage of C=C bond → Ketones/Acids/CO₂ depending on substitution.
- Ozonolysis (O₃ followed by Zn/H₂O): Cleavage of C=C bond → Aldehydes and/or Ketones. Useful for locating double bond.
Polymerisation: n(CH₂=CH₂) → [-CH₂-CH₂-]_n (Polyethene).
Uses: Making polymers (polythene), ethanol, antifreeze (ethene glycol), ripening fruits.

Critical Concept Check: Explain Markownikoff's rule using stability of carbocations during addition of HBr to Propene.

A: Propene is CH₃-CH=CH₂. H⁺ from HBr adds first to form a carbocation. If H⁺ adds to C1 (CH₂), it forms CH₃-C⁺H-CH₃ (a secondary carbocation, 2°). If H⁺ adds to C2 (CH), it forms CH₃-CH₂-C⁺H₂ (a primary carbocation, 1°). Secondary carbocations are more stable than primary carbocations (due to hyperconjugation/inductive effect from alkyl groups). Therefore, the more stable 2° carbocation forms preferentially. Br⁻ then attacks this carbocation, leading to the major product CH₃-CHBr-CH₃ (2-Bromopropane). This aligns with Markownikoff's rule (H adds to C with more H's).

9.3 Alkynes (Acetylenes)

Key Points: (C_nH_2n-2, unsaturated C≡C, sp hybridised)

Nomenclature & Isomerism: Review IUPAC (suffix -yne, lowest number to C≡C). Position isomerism (But-1-yne, But-2-yne). No geometrical isomerism around triple bond (linear).
Structure of Triple Bond: One sigma (σ) bond + two pi (π) bonds. Linear geometry (180°).
Preparation:
- From Calcium Carbide: CaC₂ + 2H₂O → C₂H₂ + Ca(OH)₂.
- Dehydrohalogenation of Vicinal/Geminal Dihalides: Using strong base like alcoholic KOH followed by NaNH₂.
- Kolbe's Electrolysis: Electrolysis of potassium maleate/fumarate.
Physical Properties: Similar trends to alkanes/alkenes. Lower members gases.
Chemical Properties:
- Acidic Character (Terminal Alkynes): H atom attached to sp-hybridised C is acidic (due to high s-character). Reacts with strong bases (NaNH₂) or ammoniacal AgNO₃ (Tollen's reagent → White ppt Ag₂C₂) / Cu₂Cl₂ (→ Red ppt Cu₂C₂). Test to distinguish terminal alkynes.
- Addition Reactions (Occur twice):
  - + H₂: C₂H₂ → C₂H₄ → C₂H₆ (requires catalyst).
  - + Halogens (X₂): C₂H₂ + Br₂ → C₂H₂Br₂ → C₂H₂Br₄ (Decolorizes Br₂ water).
  - + Hydrogen Halides (HX): Follows Markownikoff's rule. C₂H₂ + HCl → CH₂=CHCl (Vinyl chloride); CH₂=CHCl + HCl → CH₃-CHCl₂ (1,1-Dichloroethane).
  - + Water (Hydration - Kucherov's reaction): C₂H₂ + H₂O
- Oxidation: Combustion → CO₂ + H₂O. With KMnO₄ → Cleavage/Oxalic acid etc. Ozonolysis → Carboxylic acids.
- Polymerisation: Linear (→ polyacetylene). Cyclic (Ethyne passed through red hot iron tube → Benzene).
Distinguishing Test (Alkane/ene/yne): Baeyer's reagent (Alkene/yne decolorize, Alkane no). Ammoniacal AgNO₃/Cu₂Cl₂ (Terminal Alkyne gives ppt, Alkene/Alkane no).
Uses: Oxy-acetylene flame (welding/cutting), making organic chemicals (acetaldehyde, acetic acid, polymers like PVC).

Q: Why is ethyne acidic while ethene and ethane are not?

A: In ethyne (C₂H₂), the carbon atoms are sp hybridized. sp hybrid orbitals have 50% s-character, making them highly electronegative. This pulls the electrons of the C-H bond strongly towards carbon, making the hydrogen atom relatively positive and easily removable as H⁺ (acidic). In ethene (sp²) and ethane (sp³), the carbons have less s-character (33% and 25% respectively), are less electronegative, and the C-H bond is much less polar, hence hydrogens are not acidic.

9.4 Aromatic Hydrocarbons (Benzene)

Key Points:

Aromaticity (Hückel's Rule): Cyclic, planar, completely conjugated system with (4n+2) π electrons (n=0,1,2...). Benzene (C₆H₆) fits this (n=1).
Benzene Structure: Hexagonal planar ring. All C-C bonds intermediate length between single/double due to Resonance (delocalization of 6 π electrons over ring). Kekulé structures are contributing resonance forms.
Preparation:
- From Sodium Benzoate: Decarboxylation with Sodalime (NaOH+CaO).
- From Phenol: Reduction with Zinc dust.
- From Ethyne: Cyclic polymerisation (red hot Fe tube).
Physical Properties: Colourless liquid, characteristic smell, insoluble in water, good organic solvent.
Chemical Properties (Electrophilic Aromatic Substitution - EAS):
- Mechanism: Generation of electrophile (E⁺) → Attack of E⁺ on π-electron cloud → Formation of carbocation intermediate (sigma complex - resonance stabilized) → Loss of H⁺ to restore aromaticity.
- Nitration: Benzene + Conc HNO₃ + Conc H₂SO₄ (Nitrating mixture)
- Halogenation: Benzene + Cl₂/Br₂ + Anhydrous FeCl₃/FeBr₃ (Lewis acid catalyst) → Chlorobenzene/Bromobenzene + HCl/HBr. (E⁺ = Cl⁺/Br⁺).
- Sulphonation: Benzene + Conc H₂SO₄ (or Oleum)
- Friedel-Crafts Alkylation: Benzene + Alkyl halide (R-X) + Anhydrous AlCl₃ → Alkylbenzene (Toluene) + HX. (E⁺ = R⁺ carbocation).
- Friedel-Crafts Acylation: Benzene + Acyl halide (RCOCl) or Acid Anhydride + Anhydrous AlCl₃ → Acylbenzene (Acetophenone) + HCl. (E⁺ = RCO⁺ acylium ion).
Directive Influence of Substituents (Monosubstituted Benzene):
- Activating Groups (Electron Donating, o/p-directing): Increase e⁻ density at ortho/para positions, make ring more reactive towards EAS. (e.g., -OH, -NH₂, -OR, -R (alkyl)).
- Deactivating Groups (Electron Withdrawing, m-directing): Decrease e⁻ density, esp. at o/p positions, make ring less reactive towards EAS. Direct incoming group to meta position. (e.g., -NO₂, -CN, -CHO, -COR, -COOH, -SO₃H).
- Halogens (-X): Deactivating (due to -I effect) but o/p-directing (due to +R effect).
Addition Reactions (Under drastic conditions, destroys aromaticity): + H₂ (Ni/Pt, Heat, Pressure) → Cyclohexane. + Cl₂ (UV light) → Benzene hexachloride (BHC/Gammexane).
Oxidation: Resistant usually. Vigorous oxidation (V₂O₅, Heat) → Maleic anhydride. Combustion → CO₂ + H₂O (sooty flame).
Pyrolysis: High temp → Bi-phenyl.
Carcinogenicity/Toxicity: Benzene is carcinogenic.
Uses: Solvent, starting material for dyes, drugs, detergents, explosives, polymers.

Critical Concept Check: Why is Friedel-Crafts acylation preferred over alkylation for introducing a straight-chain alkyl group longer than ethyl onto benzene?

A: Friedel-Crafts alkylation often suffers from carbocation rearrangements and polyalkylation. For example, reacting benzene with n-propyl chloride and AlCl₃ gives mainly isopropylbenzene (cumene) due to rearrangement of the primary n-propyl carbocation to the more stable secondary isopropyl carbocation. Acylation introduces an acyl group (RCO-) which does not rearrange. The resulting ketone can then be reduced (e.g., using Clemmensen or Wolff-Kishner reduction) to the desired straight-chain alkylbenzene without rearrangement.

Try Yourselves - Chapter 9 (Class XI)

1. Draw staggered Newman projection of ethane.

2. Name the product: Ethene + HBr (Peroxide).

3. Test to distinguish But-1-yne from But-2-yne?

4. Define Aromaticity (Hückel's Rule).

5. Reagents for nitration of benzene?

6. Is -OH group activating or deactivating?

7. Is -CHO group o/p or m-directing?

8. Product of cyclic polymerisation of ethyne?

Micro Tests - Chapter 9 (Class XI)

Micro Test XI9.1 (Alkanes & Conformation - 10 Marks)

Test XI9.1

Time: 10 Min

Answer the following.

1.General formula for alkanes?[1]
2.Define conformation.[1]
3.Draw Sawhorse projection of eclipsed ethane.[2]
4.Which conformer of ethane is most stable? Why?[2]
5.Write equation for decarboxylation of sodium propanoate.[2]
6.Write first propagation step for bromination of methane.[2]

Micro Test XI9.2 (Alkenes - Prep & Reactions - 10 Marks)

Test XI9.2

Time: 12 Min

Write balanced equations/products:

1.Preparation of Ethene from Ethanol.[2]
2.Addition of Cl₂ to Propene.[2]
3.Addition of HBr to Propene (Markownikoff).[2]
4.Addition of HBr to Propene (Peroxide).[2]
5.Reaction of Ethene with Baeyer's Reagent.[1]
6.Products of ozonolysis of But-2-ene (followed by Zn/H₂O).[1]

Micro Test XI9.3 (Alkynes - Prep & Reactions - 10 Marks)

Test XI9.3

Time: 12 Min

Write balanced equations/products:

1.Preparation of Ethyne from CaC₂.[1]
2.Acidic nature: Reaction of Ethyne with NaNH₂.[2]
3.Addition of excess Br₂ to Ethyne.[2]
4.Addition of H₂O to Ethyne (Kucherov). Product?[2]
5.Cyclic Polymerisation of Ethyne.[2]
6.Distinguishing test for terminal alkynes?[1]

Micro Test XI9.4 (Benzene - Reactions - 10 Marks)

Test XI9.4

Time: 12 Min

Write equations/products/reagents:

1.Nitration of Benzene.[2]
2.Halogenation (Chlorination) of Benzene.[2]
3.Friedel-Crafts Alkylation of Benzene with CH₃Cl.[2]
4.Friedel-Crafts Acylation of Benzene with CH₃COCl.[2]
5.Is -NO₂ group o/p or m-directing?[1]
6.Is -CH₃ group activating or deactivating?[1]

10. Practicals (XI)

Paper II: Focus on basic lab techniques, molarity-based titrations, qualitative analysis (specific ions), inorganic prep, chromatography. Includes Project Work and File assessment.

Quick Links: Lab Techniques Titration (Molarity) Qualitative Analysis Inorganic Prep Chromatography Practice (Conceptual) Micro Tests

10.1 Basic Laboratory Techniques

Skills:

Cutting glass tube (scoring with file, gentle pressure).
Bending glass tube (heating middle in luminous flame, rotating uniformly, bending slowly to desired angle).
Drawing out a glass jet (heating middle strongly in Bunsen flame until soft, pulling ends apart steadily).
Boring a cork (selecting correct borer size, moistening, using twisting pressure).

Critical Concept Check: Why is a luminous (yellow) flame used for bending glass, while a non-luminous (blue) flame is used for heating strongly?

A: A luminous flame is cooler and less intense, allowing the glass to heat more slowly and evenly over a larger area, which is necessary for smooth bending without melting or collapsing. A non-luminous flame is hotter and provides more concentrated heat, suitable for strong heating required for operations like drawing a jet or performing reactions.

10.2 Titration (Acid-Base, Molarity)

Focus: Determine molarity and concentration (g/L) of a solution using volumetric analysis.

Titrations Involved:
- Sodium Carbonate (Na₂CO₃) vs Dilute HCl/H₂SO₄ (Methyl Orange indicator).
- Sodium Hydroxide (NaOH)/Potassium Hydroxide (KOH) vs Dilute HCl/H₂SO₄ (Methyl Orange or Phenolphthalein indicator - usually Phenolphthalein for strong base/strong acid).
Procedure: Correct rinsing of pipette (with solution to be measured) & burette (with titrant solution). Accurate filling/reading of burette (remove air bubble, read bottom of meniscus for colourless, top for coloured). Accurate pipetting. Use of indicator (2-3 drops). Titration with constant swirling until sharp endpoint colour change. Repeat for concordant readings (usually values agreeing within 0.1 mL, best practice is identical values).
Observation Table: Standard format (Initial burette reading, Final burette reading, Volume of titrant used = Difference). Concordant volume used for calculation.
Calculations:
- Balanced chemical equation is essential.
- Use Molarity Equation: M₁V₁/n₁ = M₂V₂/n₂ (M=Molarity, V=Volume, n=stoichiometric coefficient).
- Calculate unknown Molarity.
- Calculate Concentration (g/L) = Molarity (mol/L) × Molar Mass (g/mol).
- Calculate % Purity = (Pure Mass / Impure Mass) × 100. (Pure mass calculated from titration).
- Calculate Water of Crystallisation (x) by relating molar mass of hydrated and anhydrous salt.
Indicator Choice: Methyl Orange (pH range ~3.1-4.4, Red→Yellow); Phenolphthalein (pH range ~8.2-10, Colourless→Pink). Choice depends on pH at equivalence point.

Critical Concept Check: In the titration of NaOH vs HCl using phenolphthalein, what is the colour change at the endpoint when adding acid from the burette?

A: Initially, the NaOH solution in the flask with phenolphthalein is pink. As HCl is added from the burette, neutralization occurs. At the endpoint (when all NaOH is just neutralized), the addition of one extra drop of HCl makes the solution slightly acidic, and the phenolphthalein indicator turns from Pink to Colourless.

10.3 Qualitative Analysis

Focus: Identification of ONE anion and ONE cation from a given single salt (from specified list). Systematic procedure involving preliminary tests and confirmatory tests.

Anions List: CO₃²⁻, NO₂⁻, S²⁻, SO₃²⁻, SO₄²⁻, NO₃⁻, CH₃COO⁻, Cl⁻, Br⁻, I⁻, C₂O₄²⁻, PO₄³⁻.
Cations List: NH₄⁺, Pb²⁺, Cu²⁺, Al³⁺, Fe³⁺, Zn²⁺, Mn²⁺, Ni²⁺, Co²⁺, Ba²⁺, Sr²⁺, Ca²⁺, Mg²⁺.
General Scheme (Anions):
- Dil. H₂SO₄ Test: Add to salt → Observe gas (CO₂, SO₂, H₂S, NO₂ from NO₂⁻). Perform gas tests.
- Conc. H₂SO₄ Test: Add to salt, warm → Observe gas/fumes (HCl, Br₂, I₂, NO₂, CH₃COOH vinegar smell). Perform gas tests/confirmatory tests (AgNO₃ for halides, FeSO₄ ring test for NO₃⁻).
- Independent Tests (using original soln or soda extract): BaCl₂ test for SO₄²⁻ (white ppt insol in acid). Tests for PO₄³⁻ (amm. molybdate→yellow ppt), C₂O₄²⁻ (CaCl₂→white ppt, +KMnO₄ decolorized on warming).
- Note: Prepare Sodium Carbonate Extract (boil salt with Na₂CO₃ soln, filter) for testing anions (except CO₃²⁻) to remove interfering cations.
General Scheme (Cations): Group separation not required, use specific tests.
- Test for NH₄⁺: Warm with NaOH → NH₃ gas test.
- Using NaOH/NH₄OH: Add reagent dropwise then excess (See Ch 4 for observations for Pb²⁺, Cu²⁺, Al³⁺, Fe³⁺, Zn²⁺, Ca²⁺, Mg²⁺). Confirmatory tests may be needed.
- Flame Test: For Ba²⁺ (Apple green), Sr²⁺ (Crimson red), Ca²⁺ (Brick red).
Note: Chromyl chloride test not required. Insoluble salts like BaSO₄, PbSO₄ etc. not given.

Q: A salt gives reddish-brown fumes with conc. H₂SO₄ on heating. Adding FeSO₄ solution carefully to the salt solution + conc H₂SO₄ along the sides forms a brown ring. Identify the anion.

A: The anion is Nitrate (NO₃⁻).

10.4 Preparation of Inorganic Compounds

Focus: Prepare reasonably pure crystalline samples.

(a) Potash Alum (KAl(SO₄)₂·12H₂O) or Mohr's Salt (FeSO₄·(NH₄)₂SO₄·6H₂O):
- Type: Double Salts.
- Method: Dissolve stoichiometric amounts of constituent salts (e.g., K₂SO₄ + Al₂(SO₄)₃ or FeSO₄ + (NH₄)₂SO₄) in minimum hot water, maybe acidify slightly. Filter if needed. Allow solution to cool slowly for crystallization. Filter crystals, wash gently, dry.
(b) Crystalline Ferrous Sulphate (FeSO₄·7H₂O) or Copper Sulphate (CuSO₄·5H₂O):
- Method: React excess metal/oxide/carbonate with appropriate dilute acid (e.g., Fe filings + dil H₂SO₄ or CuO + dil H₂SO₄). Warm if needed. Filter off excess solid. Concentrate filtrate by heating gently until crystallization point reached (test by cooling a drop on glass rod). Cool undisturbed. Filter crystals, wash, dry.

10.5 Paper Chromatography

Focus: Separation based on differential partitioning.

Principle: Components distribute differently between stationary phase (water adsorbed on cellulose paper) and mobile phase (organic solvent mixture).
Procedure:
- Draw baseline (pencil) near bottom edge of chromatography paper strip.
- Apply concentrated spot of mixture (leaf extract, flower extract, ink) on baseline using capillary tube, allow to dry.
- Suspend paper in sealed chromatography jar containing solvent (mobile phase) below the baseline. Ensure spot is above solvent level.
- Allow solvent to run up the paper by capillary action (developing the chromatogram).
- Remove paper when solvent front nears top, mark solvent front immediately. Dry.
- Observe separated coloured spots.
R_f Value (Retardation Factor):
- Calculation: R_f = (Distance moved by spot centre from baseline) / (Distance moved by solvent front from baseline).
- Value ≤ 1. Characteristic for a component under specific conditions (paper, solvent, temp).
Separation: Components with higher solubility in mobile phase / lower adsorption on stationary phase travel further (higher R_f).

Q: In paper chromatography, solvent front moved 10 cm and a green spot moved 4 cm from the baseline. Calculate the R_f value for the green component.

A: R_f = (Distance moved by spot) / (Distance moved by solvent front) = 4 cm / 10 cm = 0.4.

Try Yourselves - Chapter 10 (Class XI Practicals - Conceptual)

1. Why rinse burette with titrant?

2. Indicator for NaOH vs HCl titration?

3. Confirmatory test for SO₄²⁻?

4. Test for NH₄⁺?

5. What is R_f value?

6. Principle of chromatography?

7. Why use soda extract?

8. What is concordant reading?

Micro Tests - Chapter 10 (Class XI Practicals)

Micro Test XI10.1 (Lab Techniques & Titration - 10 Marks)

Test XI10.1

Time: 10 Min

Answer the following.

1.Which flame type for bending glass?[1]
2.Why rinse pipette with solution to be measured?[1]
3.Name indicator for Na₂CO₃ vs HCl titration.[1]
4.Endpoint colour change for Q3 (acid in burette).[1]
5.Write molarity equation M₁V₁/n₁ = ...[1]
6.20mL of 0.1M NaOH reqd 25mL HCl. Molarity of HCl?[3]
7.Convert Molarity from Q6 to g/L (HCl=36.5).[2]

Micro Test XI10.2 (Qualitative Analysis - Anions - 10 Marks)

Test XI10.2

Time: 10 Min

Identify anion based on test:

1.Add dil H₂SO₄ → Gas turns lime water milky.[1]
2.Add dil H₂SO₄ → Gas with rotten egg smell.[1]
3.Add conc H₂SO₄, warm → Pungent gas giving white fumes with NH₃.[2]
4.Add conc H₂SO₄, warm → Red-brown vapour.[1]
5.Add conc H₂SO₄, warm → Violet vapour.[1]
6.Original soln + BaCl₂ → White ppt insoluble in dil HCl.[2]
7.Brown Ring Test positive.[1]
8.Why use soda extract?[1]

Micro Test XI10.3 (Qualitative Analysis - Cations - 10 Marks)

Test XI10.3

Time: 10 Min

Identify cation from observation:

1.Salt soln + NaOH → Reddish-brown ppt, insol in excess.[1]
2.Salt soln + NH₄OH → Gelatinous white ppt, soluble in excess.[1]
3.Salt soln + NaOH → Pale blue ppt, insol in excess.[1]
4.Salt soln + NH₄OH → Pale blue ppt, soluble in excess (deep blue soln).[2]
5.Flame test gives Apple Green colour.[1]
6.Flame test gives Crimson Red colour.[1]
7.Salt + NaOH, warm → Gas turns moist red litmus blue.[1]
8.Salt soln + dil HCl → White ppt, soluble in hot water.[2]

Micro Test XI10.4 (Prep & Chromatography - 10 Marks)

Test XI10.4

Time: 10 Min

Answer the following.

1.Mohr's salt is what type of salt?[1]
2.Name the two simple salts making Potash Alum.[2]
3.How are crystals generally obtained from a saturated solution?[1]
4.Reactants to prepare crystalline FeSO₄?[1]
5.What is the stationary phase in paper chromatography?[1]
6.What is the mobile phase in paper chromatography?[1]
7.Define R_f value.[2]
8.If component A has R_f=0.8 and B has R_f=0.3, which is more soluble in mobile phase?[1]

10. Practicals (XI)

Paper II: Focus on basic lab techniques, molarity-based titrations, qualitative analysis (specific ions), inorganic prep, chromatography. Includes Project Work and File assessment.

Quick Links: Lab Techniques Titration (Molarity) Qualitative Analysis Inorganic Prep Chromatography Practice (Conceptual) Micro Tests

10.1 Basic Laboratory Techniques

Skills:

Cutting glass tube (scoring with file, gentle break).
Bending glass tube (heating middle in luminous flame, rotating, bending slowly).
Drawing out a glass jet (heating middle strongly, pulling ends apart).
Boring a cork (using appropriate sized borer, twisting action).

10.2 Titration (Acid-Base, Molarity)

Focus: Determine molarity and concentration (g/L) of a solution.

Titrations Involved:
- Sodium Carbonate (Na₂CO₃) vs Dilute HCl/H₂SO₄ (Methyl Orange indicator).
- Sodium Hydroxide (NaOH)/Potassium Hydroxide (KOH) vs Dilute HCl/H₂SO₄ (Methyl Orange or Phenolphthalein indicator - usually Pht).
Procedure: Rinsing pipette/burette, filling burette (no air bubbles), pipetting accurately, adding indicator, titration (swirling) until endpoint (colour change), recording readings (initial, final, titre), repeating for concordant readings (±0.1 mL, exact same value preferred).
Observation Table: Standard format (Initial, Final, Difference). Concordant value used.
Calculations:
- Use Molarity Equation: M₁V₁/n₁ = M₂V₂/n₂ (where M=Molarity, V=Volume, n=stoichiometric coefficient from balanced equation).
- Calculate unknown Molarity.
- Calculate Concentration (g/L) = Molarity × Molar Mass.
- Calculate % Purity or Water of Crystallisation if required.
Indicators: Methyl Orange (Acidic: Red, Alkaline: Yellow, Endpoint: Orange/Peach). Phenolphthalein (Acidic: Colourless, Alkaline: Pink, Endpoint: Colourless).

Critical Concept Check: Why is methyl orange suitable for titrating Na₂CO₃ (weak base part) with a strong acid, while phenolphthalein is generally not preferred?

A: Na₂CO₃ hydrolyzes to give a weakly alkaline solution. The equivalence point for its titration with a strong acid occurs at a slightly acidic pH (due to formation of H₂CO₃). Methyl orange changes colour in the acidic range (pH 3.1-4.4), covering this equivalence point well. Phenolphthalein changes colour in the alkaline range (pH 8.2-10) and would change colour too early, before complete neutralization.

10.3 Qualitative Analysis

Focus: Identification of ONE anion and ONE cation from a given single salt.

Anions (Systematic Tests):
- Dilute Acid Group (CO₃²⁻, NO₂⁻, S²⁻, SO₃²⁻): Add dil H₂SO₄/HCl → Observe gas (CO₂, NO₂, H₂S, SO₂). Perform confirmatory tests for gas.
- Conc. Acid Group (Cl⁻, Br⁻, I⁻, NO₃⁻, CH₃COO⁻): Add conc H₂SO₄ → Observe gas/fumes (HCl, Br₂, I₂, NO₂, Vinegar smell). Perform confirmatory tests (AgNO₃ for halides, FeSO₄ ring test for NO₃⁻, Ester test for acetate).
- Special Group (SO₄²⁻, PO₄³⁻, C₂O₄²⁻): Test original solution/soda extract. BaCl₂ test for SO₄²⁻. Tests for PO₄³⁻, C₂O₄²⁻ involve specific reagents (Ammonium molybdate, CaCl₂).
Cations (Systematic Group Analysis - using H₂S, (NH₄)₂S, etc. is NOT required, use specific tests):
- Group 0 (NH₄⁺): Warm with NaOH → NH₃ gas test.
- Group I (Pb²⁺): Add dil HCl → White ppt (PbCl₂), soluble in hot water.
- Group II (Cu²⁺, Pb²⁺): Add NaOH/NH₄OH (See Ch 4). Confirmatory tests (e.g., K₄[Fe(CN)₆] for Cu²⁺ - Choc ppt).
- Group III (Al³⁺, Fe³⁺): Add NaOH/NH₄OH. (NaOH test for Al³⁺ solubility). Confirmatory (e.g., KSCN for Fe³⁺ - blood red).
- Group IV (Zn²⁺, Mn²⁺, Ni²⁺, Co²⁺): Add NaOH/NH₄OH. (NaOH/NH₄OH test for Zn²⁺ solubility). Specific tests if needed.
- Group V (Ba²⁺, Sr²⁺, Ca²⁺): Flame test. Add (NH₄)₂CO₃ → White ppt.
- Group VI (Mg²⁺): Add NH₄Cl, NH₄OH, then (NH₄)₂HPO₄ → White ppt.
Note: Na₂CO₃ extract must be used for anion tests (except CO₃²⁻). Chromyl chloride test not performed. Insoluble salts not given.

Critical Concept Check: Why is Sodium Carbonate extract prepared for testing most anions?

A: Many salts are insoluble, or the cation present might interfere with the anion tests (e.g., forming precipitates with reagents like AgNO₃ or BaCl₂). Boiling the salt with Na₂CO₃ solution precipitates the interfering cation as its insoluble carbonate (e.g., Pb²⁺ → PbCO₃↓). The desired anion remains dissolved in the filtrate as its soluble sodium salt, allowing for clear testing without cation interference.

10.4 Preparation of Inorganic Compounds

Focus: Prepare crystals of given salts.

(a) Potash Alum (K₂SO₄·Al₂(SO₄)₃·24H₂O) / Mohr's Salt (FeSO₄·(NH₄)₂SO₄·6H₂O):
- Double salts prepared by mixing equimolar amounts of constituent salts in water, concentrating the solution, and allowing crystallization on cooling.
(b) Crystalline FeSO₄ / CuSO₄:
- Prepared by reacting the metal/oxide/carbonate with dilute H₂SO₄, filtering (if needed), concentrating the filtrate, and cooling to crystallize. (Fe + dil H₂SO₄ → FeSO₄ + H₂; CuO + dil H₂SO₄ → CuSO₄ + H₂O).

10.5 Paper Chromatography

Focus: Separation technique based on differential distribution between stationary (paper) and mobile (solvent) phases.

Preparation of Chromatogram: Spotting the mixture (leaf/flower extract, ink) on baseline of chromatography paper strip, suspending strip in closed jar with solvent below baseline, allowing solvent to run up (developing).
Separation: Components move at different rates depending on solubility in mobile phase and adsorption onto stationary phase.
R_f Value (Retardation Factor): Ratio of distance travelled by component spot to distance travelled by solvent front. R_f = (Distance travelled by solute) / (Distance travelled by solvent). Characteristic value for a given substance, solvent, paper under specific conditions.
Determination: Measure distances from baseline, calculate R_f for separated components.

Critical Concept Check: What does a higher R_f value indicate about a component's interaction with the stationary and mobile phases?

A: A higher R_f value (closer to 1) indicates that the component travelled further up the paper. This means it is *more soluble* in the mobile phase (solvent) and/or *less strongly adsorbed* onto the stationary phase (paper), allowing it to move along faster with the solvent front.

Try Yourselves - Chapter 10 (Class XI Practicals - Conceptual)

[Practice questions covering titration calculations, qualitative analysis deductions, chromatography principles would go here]

Micro Tests - Chapter 10 (Class XI Practicals)

Micro Test XI10.1 (Lab Techniques & Titration - 10 Marks)

[Test questions on glass work, titration procedure, indicators, molarity calculations]

Micro Test XI10.2 (Qualitative Analysis - Anions - 10 Marks)

[Test questions on identifying anions based on dil/conc acid tests, confirmatory tests]

Micro Test XI10.3 (Qualitative Analysis - Cations - 10 Marks)

[Test questions on identifying cations based on specific reagent tests (NaOH, NH₄OH, flame test etc.)]

Micro Test XI10.4 (Prep & Chromatography - 10 Marks)

[Test questions on principle of double salt prep, chromatography setup, Rf calculation, interpretation]

Chapters (XI)

Interactive ISC Chemistry Notes - Class XI

Table of Contents (Class XI)

Solutions

Test XI1.1

Solutions XI1.1

Test XI1.2

Solutions XI1.2

Test XI1.3

Solutions XI1.3

Test XI1.4

Solutions XI1.4

Solutions

Test XI2.1

Solutions XI2.1

Test XI2.2

Solutions XI2.2

Test XI2.3

Solutions XI2.3

Test XI2.4

Solutions XI2.4

Solutions

Test XI3.1

Solutions XI3.1

Test XI3.2

Solutions XI3.2

Test XI3.3

Solutions XI3.3

Test XI3.4

Solutions XI3.4

Solutions

Test XI4.1

Solutions XI4.1

Test XI4.2

Solutions XI4.2

Test XI4.3

Solutions XI4.3

Test XI4.4

Solutions XI4.4

Solutions

Test XI5.1

Solutions XI5.1

Test XI5.2

Solutions XI5.2

Test XI5.3

Solutions XI5.3

Test XI5.4

Solutions XI5.4

Solutions

Test XI6.1

Solutions XI6.1

Test XI6.2

Solutions XI6.2

Test XI6.3

Solutions XI6.3

Test XI6.4

Solutions XI6.4

Solutions

Test XI7.1

Solutions XI7.1

Test XI7.2

Solutions XI7.2

Test XI7.3

Solutions XI7.3

Test XI7.4

Solutions XI7.4

Solutions

Test XI8.1

Solutions XI8.1

Test XI8.2

Solutions XI8.2

Test XI8.3

Solutions XI8.3

Test XI8.4

Solutions XI8.4

Solutions

Test XI9.1

Solutions XI9.1

Test XI9.2

Solutions XI9.2