🧪 Chemistry XII Board Prep
Master all 16 chapters with PYQs, smart notes, and the 35-question simulator
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Solutions
Ch 1 · 8–10 marks
Electrochemistry
Ch 2 · 6–8 marks
Chemical Kinetics
Ch 3 · 6–8 marks
d & f-Block
Ch 4 · 6–8 marks
Coordination
Ch 5 · 6–8 marks
Haloalkanes
Ch 6 · 4–6 marks
Alcohols/Phenols
Ch 7 · 6–8 marks
Aldehydes/Ketones
Ch 8 · 6–8 marks
Amines
Ch 9 · 4–6 marks
Biomolecules
Ch 10 · 4 marks
Polymers
Ch 11 · 4 marks
Everyday Chemistry
Ch 12 · 4 marks
Solid State
Ch 13 · 4–6 marks
Surface Chemistry
Ch 14 · 4 marks
Extraction Principles
Ch 15 · 4 marks
p-Block Elements
Ch 16 · 6–8 marks
🧪 Chapter 1: Solutions
Concentration Terms · Raoult's Law · Colligative Properties · van't Hoff Factor
📖 Core Theory
1.1 Types and Concentration
A solution is a homogeneous mixture of solute(s) in a solvent. The four concentration terms tested most in board exams:
Molarity (M)
Moles of solute per litre of solution. M = n/V(L). Temperature-dependent.
Molality (m)
Moles of solute per kg of solvent. m = n/W(kg). Temperature-independent — preferred for colligative properties.
Mole Fraction (χ)
χ₁ = n₁/(n₁+n₂). Dimensionless. χ₁ + χ₂ = 1 always. Used in Raoult's law.
Mass Percent (w/w)
Mass of solute × 100 / mass of solution. Independent of temperature.
1.2 Raoult's Law and Vapour Pressure
Raoult's law: Partial vapour pressure of a volatile component is proportional to its mole fraction in solution. For ideal solutions: p₁ = χ₁ × p₁°
🔑 Key Formulae — Solutions
p₁ = χ₁·p₁° (Raoult's law for component 1)
Relative lowering of VP: (p°–p)/p° = χ₂ (mole fraction of solute)
Elevation of boiling point: ΔTb = Kb·m Kb(water) = 0.512 K·kg/mol
Depression of freezing point: ΔTf = Kf·m Kf(water) = 1.86 K·kg/mol
Osmotic pressure: π = iCRT (van't Hoff) where R = 0.0831 L·bar/mol·K
van't Hoff factor: i = observed colligative property / calculated (no assoc/dissoc)
1.3 Colligative Properties
Properties that depend only on the number of solute particles, not their nature. Four colligative properties: relative lowering of VP, elevation of Tb, depression of Tf, osmotic pressure.
Isotonic Solutions
Same osmotic pressure. No net flow of solvent across semi-permeable membrane.
Hypertonic
Higher osmotic pressure than reference. Causes cell shrinkage (crenation) in biology.
Hypotonic
Lower osmotic pressure. Causes cell swelling (haemolysis) in biology.
Abnormal Molar Mass
Electrolytes dissociate → more particles → i>1. Associates (e.g. acetic acid in benzene) → i<1.
✏️ Worked Examples
Molality and Elevation of Boiling Point (Board 2023, 3M)
Q: 18 g of glucose (MW=180) dissolved in 200 g of water. Find ΔTb. Kb(water)=0.512 K·kg/mol.
1
Moles of glucose = 18/180 = 0.1 mol
2
Mass of solvent = 200 g = 0.2 kg
3
Molality m = 0.1/0.2 = 0.5 mol/kg
4
ΔTb = Kb × m = 0.512 × 0.5 = 0.256 K
5
New boiling point = 100 + 0.256 = 100.256°C
✅ ΔTb = 0.256 K; Boiling point = 100.256°C
van't Hoff Factor (Board 2022, 3M)
Q: KCl (MW=74.5) when dissolved in water gives i=1.9. Find degree of dissociation α.
1
KCl → K⁺ + Cl⁻ (n=2 ions per formula unit)
2
i = 1 + (n−1)α = 1 + (2−1)α = 1 + α
3
1.9 = 1 + α
4
α = 0.9
5
Degree of dissociation = 90%
✅ α = 0.9 (90% dissociation)
Osmotic Pressure (Board 2024, 5M)
Q: Calculate osmotic pressure of 0.05 M NaCl solution at 27°C. R=0.0831 L·bar/mol·K, i=1.8.
1
Given: C=0.05 mol/L, T=300 K, R=0.0831, i=1.8
2
π = iCRT
3
π = 1.8 × 0.05 × 0.0831 × 300
4
π = 1.8 × 1.2465
5
π = 2.24 bar
✅ Osmotic pressure π = 2.24 bar
⚖️ Comparisons
Ideal Solution
Non-Ideal Solution
ΔHmix = 0; ΔVmix = 0
ΔHmix ≠ 0; ΔVmix ≠ 0
Obeys Raoult's law at all concentrations
Shows +ve deviation (A-B weaker) or −ve (A-B stronger)
All intermolecular forces equal
A-B interactions ≠ A-A or B-B
Examples: benzene + toluene; hexane + heptane
Examples: acetone + CS₂ (+ve); acetone + chloroform (−ve)
Molarity
Molality
n/V(solution) in litres
n/W(solvent) in kg
Changes with temperature (V changes)
Constant with temperature
Used in volumetric calculations
Used in colligative properties
Easier to prepare in lab
More fundamental thermodynamically
🎯 Quick Activity
Think and Answer
Q: Which colligative property is used to determine molar mass of proteins and polymers, and why?
▶ Show Answer
A: Osmotic pressure — it produces a measurable effect even for very dilute solutions of macromolecules, unlike ΔTb or ΔTf which are too small to measure accurately.
🧪 Chapter 2: Electrochemistry
Electrochemical Cells · Nernst Equation · Electrolysis · Kohlrausch's Law
📖 Core Theory
2.1 Electrochemical Cells
An electrochemical cell converts chemical energy to electrical energy (galvanic cell) or vice versa (electrolytic cell). A galvanic cell has two half-cells connected by a salt bridge.
Anode
Electrode where oxidation occurs. In galvanic cell: −ve terminal. In electrolytic cell: +ve terminal.
Cathode
Electrode where reduction occurs. In galvanic cell: +ve terminal. In electrolytic cell: −ve terminal.
EMF of Cell (E°cell)
E°cell = E°cathode − E°anode. Measured in volts. Positive E° means spontaneous reaction.
Salt Bridge
KCl or KNO₃ in agar-agar. Maintains electrical neutrality and completes the circuit without mixing electrolytes.
2.2 Nernst Equation and Gibbs Energy
🔑 Key Formulae — Electrochemistry
Nernst equation: E = E° − (RT/nF)·ln Q at 298K: E = E° − (0.0592/n)·log Q
ΔG° = −nFE° (relation to Gibbs energy; F = 96500 C/mol)
ΔG° = −RT ln K (at equilibrium E_cell = 0)
Molar conductivity: Λm = κ × 1000/M (κ = specific conductivity; M = molarity)
Kohlrausch's law: Λ°m = Σλ°(cations) + Σλ°(anions)
Faraday's 1st law: m = ZIt Z = electrochemical equivalent = M/(nF)
2.3 Conductance
Electrolytic conductance increases with temperature (unlike metallic conductors). Strong electrolytes show Debye-Hückel limiting law: Λm = Λ°m − A√C.
Specific Conductance (κ)
Conductance of 1 cm³ of electrolyte between electrodes 1 cm apart. SI unit: S/m.
Molar Conductivity (Λm)
Conductance of solution containing 1 mol of electrolyte between electrodes 1 cm apart.
Kohlrausch's Law
At infinite dilution, molar conductivity = sum of individual ionic conductances. Used to find Λ°m for weak electrolytes.
Degree of Dissociation
α = Λm/Λ°m for weak electrolytes. Ka = Cα²/(1−α) ≈ Cα² when α«1.
✏️ Worked Examples
Nernst Equation (Board 2023, 5M)
Q: Calculate E_cell for Zn|Zn²⁺(0.01M)||Cu²⁺(1M)|Cu. E°Zn²⁺/Zn=−0.76V, E°Cu²⁺/Cu=+0.34V.
1
E°cell = E°cathode − E°anode = 0.34 − (−0.76) = 1.10 V
2
Cell reaction: Zn + Cu²⁺ → Zn²⁺ + Cu, n=2 electrons
3
Q = [Zn²⁺]/[Cu²⁺] = 0.01/1 = 0.01
4
E = E° − (0.0592/n)log Q = 1.10 − (0.0592/2)log(0.01)
5
E = 1.10 − 0.0296×(−2) = 1.10 + 0.0592 = 1.16 V
✅ E_cell = 1.16 V
Faraday's Law (Board 2022, 3M)
Q: How many grams of copper are deposited on passing 2A for 30 min through CuSO₄? (Cu=63.5, n=2)
1
Charge Q = I × t = 2 × 30 × 60 = 3600 C
2
Electrochemical equivalent Z = M/(nF) = 63.5/(2×96500)
3
Z = 63.5/193000 = 3.29×10⁻⁴ g/C
4
m = ZQ = 3.29×10⁻⁴ × 3600 = 1.18 g
5
Alternatively: m = (M×Q)/(n×F) = 63.5×3600/(2×96500) = 1.18 g
✅ m = 1.18 g of copper deposited
Molar Conductivity (Board 2024, 3M)
Q: Specific conductance of 0.02 M CH₃COOH = 2.9×10⁻⁴ S/cm. Λ°m=390.7 S·cm²/mol. Find degree of dissociation.
1
Λm = (κ × 1000)/M = (2.9×10⁻⁴ × 1000)/0.02
2
Λm = 0.29/0.02 = 14.5 S·cm²/mol
3
α = Λm/Λ°m = 14.5/390.7 = 0.037
4
Degree of dissociation α = 3.7%
5
Ka = Cα²/(1−α) ≈ 0.02 × (0.037)² = 2.74×10⁻⁵
✅ α = 3.7%; Ka = 2.74×10⁻⁵
⚖️ Comparisons
Galvanic Cell
Electrolytic Cell
Chemical → electrical energy
Electrical → chemical energy
Spontaneous reaction (ΔG<0)
Non-spontaneous (external EMF needed)
Anode is −ve terminal
Anode is +ve terminal (connected to +ve of battery)
Examples: Daniel cell, dry cell, lead accumulator
Examples: electroplating, electrolysis of water, NaCl
Strong Electrolyte
Weak Electrolyte
Complete dissociation in water
Partial dissociation; equilibrium established
Λm increases slowly with dilution (√C law)
Λm increases rapidly at high dilution
Λ°m by extrapolation of Λm vs √C graph
Λ°m by Kohlrausch's law (can't extrapolate)
Examples: NaCl, HCl, KOH, H₂SO₄
Examples: CH₃COOH, NH₄OH, HF, H₂CO₃
🎯 Quick Activity
Think and Answer
Q: In a Daniel cell, what would happen to the EMF if [Zn²⁺] increases and [Cu²⁺] decreases? Explain using Nernst equation.
▶ Show Answer
A: EMF decreases. Q = [Zn²⁺]/[Cu²⁺] increases, so log Q increases, making E = E° − (0.0592/2)log Q smaller.
🧪 Chapter 3: Chemical Kinetics
Rate of Reaction · Order · Half-Life · Activation Energy · Arrhenius Equation
📖 Core Theory
3.1 Rate of Reaction and Rate Law
Rate of reaction = decrease in concentration of reactant (or increase in product) per unit time. For A → B: rate = −d[A]/dt = +d[B]/dt.
Order of Reaction
Sum of exponents of concentration terms in rate law. Experimentally determined — not from stoichiometry.
Molecularity
Number of reacting species in elementary step. Always a whole number (1, 2, or 3). From mechanism.
Rate Constant (k)
Proportionality constant in rate = k[A]ᵐ[B]ⁿ. Units depend on order: (mol/L)^(1−n)/s for nth order.
Pseudo-first Order
Higher-order reaction behaves as first-order when one reactant is in large excess (approx. constant). E.g. acid hydrolysis of esters.
🔑 Integrated Rate Laws and Half-Lives
Zero order: [A] = [A]₀ − kt; t½ = [A]₀/(2k); units of k: mol/L/s
First order: ln[A] = ln[A]₀ − kt; t½ = 0.693/k; units of k: s⁻¹
Second order: 1/[A] = 1/[A]₀ + kt; t½ = 1/(k[A]₀); units of k: L/mol/s
Arrhenius equation: k = A·e^(−Ea/RT); ln k = ln A − Ea/(RT)
Two temperatures: ln(k₂/k₁) = (Ea/R)·(1/T₁ − 1/T₂)
3.2 Factors Affecting Rate
Temperature Effect
Rate roughly doubles for every 10°C rise (temp. coefficient ≈ 2). Arrhenius equation quantifies this through Ea.
Activation Energy (Ea)
Minimum energy reactants must possess to form products. Catalyst lowers Ea without being consumed.
Collision Theory
Reaction occurs when: (1) molecules collide (2) with correct orientation (3) with energy ≥ Ea.
Catalyst
Provides alternate pathway of lower Ea. Increases rate. Not consumed. Does NOT change ΔH or equilibrium position.
✏️ Worked Examples
First Order Half-Life (Board 2023, 3M)
Q: A first-order reaction has k = 6.93×10⁻² s⁻¹. Find t½ and time for 75% completion.
1
t½ = 0.693/k = 0.693/(6.93×10⁻²) = 10 s
2
For 75% completion: 25% remains, i.e., [A] = 0.25[A]₀
3
Using ln([A]₀/[A]) = kt: ln(1/0.25) = kt
4
ln 4 = kt → 1.386 = 6.93×10⁻² × t
5
t = 1.386/0.0693 = 20 s = 2 × t½
✅ t½ = 10 s; t(75%) = 20 s = 2 half-lives
Arrhenius Equation (Board 2022, 5M)
Q: Rate constant of a reaction at 300K is 2.5×10⁻³ s⁻¹ and at 350K is 8.5×10⁻² s⁻¹. Find Ea. R=8.314 J/mol/K.
1
ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂)
2
ln(8.5×10⁻²/2.5×10⁻³) = (Ea/8.314)(1/300 − 1/350)
3
ln(34) = (Ea/8.314)(350−300)/(300×350)
4
3.526 = (Ea/8.314)(50/105000)
5
Ea = 3.526 × 8.314 × 105000/50 = 61,714 J/mol ≈ 61.7 kJ/mol
✅ Ea = 61.7 kJ/mol
Order from Data (Board 2024, 3M)
Q: For A → products: [A]=0.1M gives rate=2×10⁻³M/s; [A]=0.2M gives rate=4×10⁻³M/s. Find order and k.
1
Rate₁ = k[A]₁ⁿ and Rate₂ = k[A]₂ⁿ
2
Rate₂/Rate₁ = ([A]₂/[A]₁)ⁿ
3
4×10⁻³/2×10⁻³ = (0.2/0.1)ⁿ
4
2 = 2ⁿ → n = 1 (first order)
5
k = rate/[A] = 2×10⁻³/0.1 = 0.02 s⁻¹
✅ n = 1 (first order); k = 0.02 s⁻¹
⚖️ Comparisons
Order of Reaction
Molecularity
From experiment; can be 0, 1, 2, fraction
From mechanism; always 1, 2, or 3
Applies to overall reaction
Applies only to elementary steps
No theoretical basis from mechanism alone
Has theoretical meaning from collision theory
Can be zero or fractional
Can never be zero or fractional
Zero Order
First Order
[A] = [A]₀ − kt (linear in [A] vs t)
ln[A] = ln[A]₀ − kt (linear in ln[A] vs t)
t½ = [A]₀/2k (depends on initial conc.)
t½ = 0.693/k (independent of [A]₀)
k units: mol/L/s
k units: s⁻¹
Examples: enzyme-saturated reactions
Examples: radioactive decay, many drug eliminations
🎯 Quick Activity
Think and Answer
Q: The half-life of a first-order reaction is 20 min. What fraction remains after 1 hour?
▶ Show Answer
A: Number of half-lives in 1 hr = 60/20 = 3. Fraction remaining = (1/2)³ = 1/8 = 12.5%.
🧪 Chapter 4: d and f-Block Elements
Transition Metals · Properties · Lanthanoids · Actinoids
📖 Core Theory
4.1 Transition Metals (d-Block)
Elements in which the last electron enters a d-orbital. Configuration: (n−1)d¹⁻¹⁰ ns¹⁻². Note: Cr = [Ar]3d⁵4s¹ and Cu = [Ar]3d¹⁰4s¹ (half-filled and fully-filled d is extra stable).
Variable Oxidation States
d-block elements show multiple OS because 3d and 4s electrons have similar energies. Mn shows +2 to +7; Fe shows +2 and +3.
Paramagnetism
Due to unpaired d-electrons. Magnetic moment μ = √n(n+2) BM where n = number of unpaired electrons.
Colour
d-d transition of electrons absorbs visible light. Colour = complementary of absorbed wavelength. [Ti(H₂O)₆]³⁺ appears purple.
Catalytic Property
Variable OS + ability to adsorb reactants. Fe in Haber, V₂O₅ in Contact, Ni in hydrogenation, MnO₂ in KClO₃ decomp.
🔑 Important Reactions and Facts
MnO₄⁻ in acid: Mn²⁺ (colourless); in neutral/faintly basic: MnO₂ (brown); in conc. NaOH: MnO₄²⁻ (green)
Dichromate: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O (E° = +1.33V)
K₂Cr₂O₇ in H₂SO₄ solution turns green (Cr²(SO₄)₃) when reduced
Lanthanoid contraction: caused by poor shielding of 4f electrons; reason 4d and 5d elements are nearly same size
4.2 f-Block Elements
Lanthanoids (La to Lu, 4f): mostly +3 OS, paramagnetic, lanthanoid contraction (atomic/ionic radii decrease across series). Actinoids (Ac to Lr, 5f): wider range of OS than lanthanoids, radioactive, larger size due to poorer shielding by 5f than 4f.
✏️ Worked Examples
Magnetic Moment (Board 2023, 2M)
Q: Calculate magnetic moment of Mn²⁺ ion. (Z of Mn = 25)
1
Mn: [Ar] 3d⁵ 4s²; Mn²⁺: [Ar] 3d⁵ (loses 4s² electrons first)
2
3d⁵: each of 5 orbitals has 1 unpaired electron → n = 5
3
μ = √n(n+2) = √5(5+2) = √35 = 5.92 BM
4
Mn²⁺ is strongly paramagnetic
✅ μ = 5.92 BM (5 unpaired electrons)
Potassium Dichromate Reactions (Board 2022, 3M)
Q: Write the reaction of K₂Cr₂O₇ with FeSO₄ in H₂SO₄.
1
Cr₂O₇²⁻ is the oxidising agent; Fe²⁺ is oxidised to Fe³⁺
2
Half reaction (reduction): Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
3
Half reaction (oxidation): Fe²⁺ → Fe³⁺ + e⁻ (×6)
4
Overall: Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O
5
Solution turns green due to Cr³⁺
✅ Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O
Electronic Config and Colour (Board 2024, 3M)
Q: Why do d-block compounds show colour? Give an example.
1
d-d transition: electron absorbs energy from visible light and jumps from lower to higher d sub-level
2
Colour observed = complementary colour of absorbed wavelength
3
[Ti(H₂O)₆]³⁺: Ti³⁺ has 3d¹ electron; absorbs green (500nm)
4
Complementary of green = purple — so the complex appears purple
5
Zn²⁺ (3d¹⁰) and Sc³⁺ (3d⁰) are colourless — no d-d transition possible
✅ Colour due to d-d transition; [Ti(H₂O)₆]³⁺ is purple
⚖️ Comparisons
Lanthanoids
Actinoids
4f electrons filled; Ce to Lu
5f electrons filled; Th to Lr
Mainly +3 OS (some +2, +4)
Wide range of OS (+3 to +6 common; up to +7)
Most are non-radioactive
All radioactive (natural and synthetic)
Lanthanoid contraction: 4f poor shielding
Actinoid contraction: 5f slightly better, but still contracts
Metallic Character
Expected trend
Transition metals harder than alkali metals
More d-electrons → stronger metallic bonding
Melting point peaks at middle of series (W highest)
Half-filled d gives max unpaired electrons for bonding
Density generally higher than s-block
Smaller size + more mass → higher density
Exception: Zn, Cd, Hg are softer and lower mp
3d¹⁰ — no net bonding from d-electrons
🎯 Quick Activity
Think and Answer
Q: Cu is placed below H in activity series, yet Cu dissolves in HNO₃. Explain.
▶ Show Answer
A: HNO₃ acts as oxidising acid (NO₃⁻ is reduced, not H⁺). 3Cu + 8HNO₃(dil) → 3Cu(NO₃)₂ + 2NO + 4H₂O. The reaction is not with H⁺ but with NO₃⁻ which is a stronger oxidant.
🧪 Chapter 5: Coordination Compounds
Werner's Theory · Nomenclature · Isomerism · Bonding · EAN Rule
📖 Core Theory
5.1 Terminology and Nomenclature
A coordination compound has a central metal atom/ion bonded to ligands by coordinate bonds. Written: [Central metal + ligands]^charge counter ion.
Coordination Number
Number of donor atoms bonded to central metal. [Co(NH₃)₆]³⁺: CN=6. [Ni(CO)₄]: CN=4.
Ligands
Lewis bases donating lone pair(s) to metal. Monodentate (NH₃, Cl⁻), Bidentate (en, ox²⁻), Polydentate (EDTA⁴⁻: hexadentate). Ambidentate: SCN⁻ (via S or N).
EAN Rule
Effective Atomic Number = Z of metal − charge + (2 × ligands donated electrons). Stable complexes often have EAN of nearest noble gas.
Chelate Effect
Polydentate ligands form more stable complexes than monodentate due to increased entropy (releasing multiple water molecules per chelate ring closure).
5.2 Naming Coordination Compounds
Rules: ligands (alphabetical) before metal; anionic ligands get "-o" suffix (Cl⁻=chlorido, CN⁻=cyanido, NO₂⁻=nitrito); metal in cation named normally, in anion gets "-ate" suffix; oxidation state in Roman numerals in brackets.
🔑 IUPAC Names — Common Examples
[Co(NH₃)₆]³⁺ → hexaamminecobalt(III) ion
[CoCl₃(NH₃)₃] → triamminetrichloridocobalt(III)
K₄[Fe(CN)₆] → potassium hexacyanidoferrate(II)
[Cu(NH₃)₄]SO₄ → tetraamminecopper(II) sulphate
[Ni(CO)₄] → tetracarbonylnickel(0)
[PtCl₂(NH₃)₂] → diamminedichloridoplatinum(II) — cisplatin when cis-
5.3 Isomerism and Bonding
Structural isomers: ionisation, linkage, coordination, solvate/hydrate. Stereoisomers: geometric (cis-trans) and optical (mirror images, non-superimposable).
VBT (Valence Bond Theory)
Metal provides empty hybrid orbitals; ligands donate electron pairs. sp³ (tetrahedral), dsp² (square planar), sp³d² outer / d²sp³ inner (octahedral).
CFT (Crystal Field Theory)
Ligands are point charges. d orbitals split: t₂g (lower) and eg (higher) in octahedral. CFSE determines high-spin vs low-spin.
Strong Field Ligands
Spectrochemical series high end: CO > CN⁻ > NO₂⁻ > en > NH₃ > ... Cause large crystal field splitting → low spin, diamagnetic.
Weak Field Ligands
Spectrochemical series low end: F⁻ < Cl⁻ < Br⁻ < I⁻ < OH⁻ < H₂O. Small splitting → high spin, paramagnetic.
✏️ Worked Examples
Oxidation State in Complex (Board 2023, 2M)
Q: Find OS of Co in [CoCl₂(NH₃)₄]Cl.
1
Outer sphere Cl⁻ is counter ion; complex is [CoCl₂(NH₃)₄]⁺
2
NH₃ is neutral (0); Cl⁻ each is −1
3
Let OS of Co = x
4
x + 2(−1) + 4(0) = +1 (charge on complex ion)
5
x − 2 = +1 → x = +3
✅ Oxidation state of Co = +3
Geometric Isomerism (Board 2022, 3M)
Q: Draw cis and trans isomers of [Pt(NH₃)₂Cl₂] and identify which is the anticancer drug.
1
Square planar complex (Pt²⁺, d⁸, CN=4)
2
cis-isomer: both NH₃ adjacent (90° apart); both Cl⁻ adjacent
3
trans-isomer: NH₃ and Cl⁻ alternating (180° apart)
4
cis-[Pt(NH₃)₂Cl₂] = cisplatin (anticancer drug)
5
trans isomer is inactive as anticancer agent
✅ Cisplatin = cis-[Pt(NH₃)₂Cl₂]; trans is inactive
IUPAC Nomenclature (Board 2024, 3M)
Q: Give IUPAC name of [Fe(CN)₆]³⁻ and K₄[Fe(CN)₆].
1
[Fe(CN)₆]³⁻: central ion = Fe; 6 CN⁻ ligands; charge: x+6(−1)=−3 → x=+3
2
Anionic complex: metal gets "-ate" suffix → ferrate
3
Name: hexacyanidoferrate(III) ion
4
K₄[Fe(CN)₆]: 4 K⁺ counter ions; complex = [Fe(CN)₆]⁴⁻
5
Fe OS: x+6(−1)=−4 → x=+2; Name: potassium hexacyanidoferrate(II)
✅ [Fe(CN)₆]³⁻ = hexacyanidoferrate(III); K₄[Fe(CN)₆] = potassium hexacyanidoferrate(II)
⚖️ Comparisons
VBT
CFT
Considers covalent/dative bonding
Treats ligands as electrostatic point charges
Explains geometry and magnetism
Better explains colour, spectra, and magnetic properties
Cannot explain colour of complexes
Explains colour via d-d transition and Δ splitting
Uses hybridisation concept
Uses splitting of d-orbitals into t₂g and eg sets
Inner Orbital Complex
Outer Orbital Complex
Uses inner d-orbitals: d²sp³
Uses outer d-orbitals: sp³d²
Formed by strong field ligands
Formed by weak field ligands
Low spin — fewer unpaired electrons
High spin — more unpaired electrons
Example: [Co(CN)₆]³⁻ (diamagnetic)
Example: [CoF₆]³⁻ (paramagnetic)
🎯 Quick Activity
Think and Answer
Q: EDTA is a hexadentate ligand. Why does it form especially stable complexes with metal ions?
▶ Show Answer
A: EDTA forms 5 chelate rings simultaneously with a metal ion, releasing 6 water molecules. The large positive entropy change (ΔS) from releasing so many water molecules makes ΔG = ΔH−TΔS very negative → very stable complex.
🧪 Chapter 6: Haloalkanes and Haloarenes
SN1 · SN2 · E1 · E2 · Nucleophilic Substitution
📖 Core Theory
6.1 Nucleophilic Substitution
Haloalkanes undergo nucleophilic substitution (SN1 or SN2) where a nucleophile replaces the halogen. Haloarenes are much less reactive due to resonance.
SN2 (Bimolecular)
One-step, concerted mechanism. Inversion of configuration (Walden). Rate = k[RX][Nu]. Primary > Secondary. Not tertiary.
SN1 (Unimolecular)
Two-step via carbocation. Racemisation. Rate = k[RX]. Tertiary > Secondary. Polar protic solvents favour SN1.
Optical Activity
Chiral compound rotates plane of polarized light. Enantiomers — non-superimposable mirror images. Racemic mixture is optically inactive.
Reactivity Order
RI > RBr > RCl > RF (bond dissociation energy decreases down group). Allylic/benzylic halides highly reactive in both SN1 and SN2.
🔑 Key Reactions
Finkelstein: RCl + NaI (acetone) → RI + NaCl (SN2)
Swartz: RCl + AgF → RF + AgCl
Haloarenes + NaOH (300°C, 300 atm) → phenol (nucleophilic aromatic substitution)
Grignard reagent: RX + Mg (dry ether) → RMgX
✏️ Worked Examples
SN1 vs SN2 (Board 2023, 3M)
Q: Which gives SN2 faster: (CH₃)₃CBr or CH₃Br? Explain.
1
CH₃Br (methyl bromide) reacts faster via SN2
2
SN2 rate: CH₃X > 1° > 2° > 3° (opposite of SN1)
3
Methyl has no alkyl groups — least steric hindrance for backside attack
4
(CH₃)₃CBr is tertiary — bulky groups block backside approach
5
SN2 is a one-step concerted backside attack (Walden inversion)
✅ CH₃Br reacts faster via SN2; (CH₃)₃CBr prefers SN1
Optical Activity (Board 2022, 2M)
Q: When 2-bromobutane undergoes SN1, what product(s) result?
1
SN1 goes through a planar carbocation intermediate
2
Nucleophile attacks from both faces equally
3
Gives racemic mixture: 50% R + 50% S 2-butanol
4
Racemic mixture has zero net optical rotation
5
SN2 would give 100% inversion — one enantiomer only
✅ Racemic mixture (50:50) — SN1 causes racemisation
🎯 Quick Activity
Think and Answer
Q: Why are haloarenes much less reactive than haloalkanes toward nucleophilic substitution?
▶ Show Answer
A: In haloarenes, C−X bond has partial double bond character due to resonance with ring π system. Bond is shorter and stronger. Also, electron density on ring is high → repels incoming nucleophile.
📝 Detailed Electrode Processes
6.3 Electrolytic Cells and Faraday's Laws
Faraday's First Law
Mass deposited (w) ∝ quantity of electricity (Q = It). w = ZIt where Z = electrochemical equivalent = M/(nF).
Faraday's Second Law
Same quantity of electricity deposits masses proportional to M/n (equivalent weights). w₁/w₂ = (M₁/n₁)/(M₂/n₂).
Kohlrausch's Law
At infinite dilution: Λ°m = ν₊λ°₊ + ν₋λ°₋ (sum of ion conductances × stoichiometric coefficients). Used to calculate Λ°m of weak electrolytes (e.g. CH₃COOH) from strong electrolyte data.
Nernst Equation
E_cell = E°_cell − (RT/nF)ln Q = E°_cell − (0.0591/n)log Q at 25°C. At equilibrium: E_cell=0 → log Kc = n·E°_cell/0.0591.
🔑 Key Formulae
ΔG° = −nFE°_cell; ΔG° = −RT ln Kc → E°_cell = (RT/nF)ln Kc = (0.0591/n)log Kc
Specific conductance κ = (1/R) × (l/A) = conductance × cell constant
Molar conductance Λm = κ × 1000/M (in S·cm²·mol⁻¹; M in mol/L)
Degree of dissociation α = Λm/Λ°m (for weak electrolyte)
Nernst Equation (Board 2023, 3M)
Q: Calculate E_cell for Zn|Zn²⁺(0.1M)||Cu²⁺(0.01M)|Cu. Given E°_cell = 1.10 V.
1
Cell reaction: Zn + Cu²⁺ → Zn²⁺ + Cu; n = 2
2
Q = [Zn²⁺]/[Cu²⁺] = 0.1/0.01 = 10
3
E_cell = E°_cell − (0.0591/n)log Q = 1.10 − (0.0591/2)log 10
4
= 1.10 − 0.02955 × 1 = 1.10 − 0.030 = 1.07 V
✅ E_cell = 1.07 V
🧪 Chapter 7: Alcohols, Phenols and Ethers
Lucas Test · Esterification · Kolbe Reaction
📖 Core Theory
7.1 Alcohols — Classification and Reactions
Alcohols (−OH group on sp³ C) are classified as primary (1°), secondary (2°), or tertiary (3°) based on the carbon bearing −OH. Higher the alkyl groups around C−OH, lower is the reactivity toward SN reactions.
Lucas Test
Lucas reagent (ZnCl₂/conc. HCl) differentiates 1°, 2°, 3° alcohols. 3° → immediate turbidity; 2° → turbidity in 5 min; 1° → no turbidity at room temp.
Esterification
Alcohol + carboxylic acid ⇌ ester + H₂O (H₂SO₄ catalyst, reversible). Mechanism: nucleophilic acyl substitution.
Phenols
−OH directly on benzene ring. More acidic than alcohols (phenoxide ion stabilised by resonance). Undergo Kolbe, Reimer-Tiemann, esterification with acid chlorides.
Ethers
R−O−R structure. Prepared by Williamson synthesis (RO⁻ Na⁺ + R'X → ROR' + NaX). Inert to most reagents. Cleaved by HI/HBr.
🔑 Reactions of Phenol
Kolbe's reaction: PhONa + CO₂ (125°C, 4–7 atm) → salicylate (aspirin precursor)
Reimer-Tiemann: PhOH + CHCl₃ + NaOH → salicylaldehyde
Williamson synthesis: RONa + R'X → ROR' + NaX (SN2)
✏️ Worked Examples
Lucas Test (Board 2023, 2M)
Q: Three bottles contain 1-butanol, 2-butanol, and 2-methyl-2-propanol. How will you distinguish using Lucas test?
1
Add Lucas reagent (ZnCl₂ + conc. HCl) to each at room temperature
2
2-methyl-2-propanol (tertiary): immediate cloudiness/turbidity
3
2-butanol (secondary): cloudiness within 5 minutes with warming
4
1-butanol (primary): no cloudiness at room temperature; needs heating
5
Turbidity = RCl formed (insoluble in reagent)
✅ 3° → immediate; 2° → 5 min; 1° → no reaction at RT
Acidity of Phenol vs Ethanol (Board 2022, 2M)
Q: Phenol is more acidic than ethanol. Explain with resonance.
1
Ethanol: CH₃CH₂O⁻ (ethoxide) is not stabilised by resonance
2
Phenol: PhO⁻ (phenoxide) is stabilised by delocalization of negative charge over the benzene ring
3
5 resonating structures for phenoxide — charge spread over ring ortho and para positions
4
More stable conjugate base = stronger acid
5
Ka(phenol) ≈ 10⁻¹⁰ vs Ka(ethanol) ≈ 10⁻¹⁶
✅ Phenoxide is resonance-stabilised → phenol is stronger acid
🎯 Quick Activity
Think and Answer
Q: Why is Williamson synthesis suitable for preparing asymmetric ethers but not always symmetric ones?
▶ Show Answer
A: Williamson uses SN2: RO⁻Na⁺ + R'X → ROR'. For symmetric ether, R=R', you can mix alcohol and alkyl halide of same group. For asymmetric (R≠R'), you choose which to make as alkoxide and which as halide, using the primary halide to avoid E2 elimination.
📝 Rate Laws and Integrated Forms
7.3 Zero and First Order Kinetics
Zero Order
Rate = k; [A] = [A]₀ − kt. Half-life t₁/₂ = [A]₀/2k. Units of k: mol·L⁻¹·s⁻¹.
First Order
Rate = k[A]; [A] = [A]₀·e^(−kt); ln[A] = ln[A]₀ − kt. Half-life t₁/₂ = 0.693/k (independent of initial concentration). Units of k: s⁻¹.
Arrhenius Equation
k = Ae^(−Ea/RT). ln k = ln A − Ea/(RT). log(k₂/k₁) = (Ea/2.303R)×(1/T₁ − 1/T₂). Ea from two temperature measurements.
Collision Theory
Rate = PZe^(−Ea/RT) where P=steric factor (fraction of collisions with correct orientation), Z=collision frequency. Fraction of molecules with E≥Ea increases with temperature.
🔑 Integrated Rate Law Summary
Zero order: [A] = [A]₀ − kt; graph [A] vs t → straight line slope = −k
First order: ln[A] = ln[A]₀ − kt; graph ln[A] vs t → straight line slope = −k
Second order: 1/[A] = 1/[A]₀ + kt; graph 1/[A] vs t → straight line slope = +k
Half-life: 0th order: [A]₀/2k; 1st order: 0.693/k; 2nd order: 1/(k[A]₀)
First Order Half-life (Board 2022, 3M)
Q: A first order reaction is 50% complete in 30 min. Find (a) k, (b) time for 90% completion.
1
t₁/₂ = 30 min; k = 0.693/t₁/₂ = 0.693/30 = 0.0231 min⁻¹
2
For 90% completion: [A] = 10% of [A]₀ = 0.1[A]₀
3
t = (2.303/k)log([A]₀/[A]) = (2.303/0.0231)log(100/10) = (2.303/0.0231)×1 = 99.7 min
✅ k = 0.0231 min⁻¹; t (90%) ≈ 100 min
🧪 Chapter 8: Aldehydes, Ketones and Acids
Nucleophilic Addition · Aldol · Fehling · Tollen's
📖 Core Theory
8.1 Aldehydes and Ketones
Carbonyl group (C=O): aldehydes have at least one H on carbonyl carbon (−CHO); ketones have two alkyl/aryl groups (R−CO−R'). Aldehydes are more reactive than ketones (less steric hindrance, electron density).
Nucleophilic Addition
Nu attacks electrophilic C of C=O. HCN, NaHSO₃, Grignard, RNH₂ (→ imine). Aldehydes > ketones in reactivity.
Aldol Condensation
Aldehyde with α-H in dil. NaOH → β-hydroxy aldehyde (aldol) which on heating gives α,β-unsaturated carbonyl. Mixed aldol possible.
Tollens' Test
Ammoniacal AgNO₃; aldehyde → silver mirror; ketone = no reaction. RCHO + 2[Ag(NH₃)₂]⁺ → RCOO⁻ + 2Ag↓ + 4NH₃ + H⁺.
Fehling's Test
Alkaline CuSO₄ + sodium potassium tartrate (Cu²⁺ complex). Aliphatic aldehyde → brick red ppt (Cu₂O). Aromatic aldehydes and ketones don't respond.
🔑 Key Reactions
Cannizzaro (no α-H): 2HCHO + NaOH → CH₃OH + HCOONa (disproportionation)
Clemmensen: C=O → CH₂ (Zn-Hg/HCl)
Wolff-Kishner: C=O → CH₂ (NH₂NH₂, KOH, high T)
Carboxylic acids: RCOOH + NaOH → RCOONa + H₂O; with PCl₅ → RCOCl; with ROH (H₂SO₄) → ester
✏️ Worked Examples
Aldehyde vs Ketone Tests (Board 2023, 3M)
Q: How will you distinguish between ethanal and propanone using (i) Tollens' (ii) Fehling's?
1
(i) Tollens' test: Ethanal (aldehyde) gives silver mirror; propanone (ketone) gives no reaction
2
Both undergo Tollens': only aldehydes oxidised by Ag⁺
3
(ii) Fehling's test: Ethanal gives brick-red Cu₂O precipitate; propanone gives no reaction
4
Aromatic aldehydes (e.g. benzaldehyde) pass Tollens' but NOT Fehling's — a key board distinction
5
Iodoform test: propanone + I₂/NaOH → yellow CHI₃ precipitate; ethanal also gives CHI₃
✅ Both Tollens' and Fehling's distinguish; ethanal = +ve for both, propanone = −ve
Aldol Condensation (Board 2022, 5M)
Q: Write the mechanism for aldol condensation of ethanal in dilute NaOH.
1
Step 1: OH⁻ removes α-H from ethanal → carbanion (enolate)
2
Step 2: Enolate attacks C=O of another ethanal molecule
3
Step 3: Protonation → β-hydroxy ethanal (aldol product: 3-hydroxybutanal)
4
Step 4: Heating → dehydration → CH₃CH=CHCHO (but-2-enal, crotonaldehyde)
5
Product has conjugated C=C and C=O system (α,β-unsaturated carbonyl)
✅ Aldol = 3-hydroxybutanal; on heating → but-2-enal (crotonaldehyde)
🎯 Quick Activity
Think and Answer
Q: Why does benzaldehyde not give a positive Fehling's test but gives positive Tollens'?
▶ Show Answer
A: Fehling's reagent (Cu²⁺ in alkaline tartrate) is a mild oxidant — only oxidises aliphatic aldehydes. Benzaldehyde is aromatic and more resistant to mild oxidation. Tollens' (Ag⁺ in NH₃) is a stronger and more selective oxidant that oxidises both aliphatic and aromatic aldehydes.
📝 Colloidal State and Emulsions
8.3 Colloids — Properties and Preparation
Tyndall Effect
Scattering of light by colloidal particles. Observed in colloidal solutions (not true solutions). Used to distinguish colloids from true solutions.
Brownian Movement
Zig-zag motion of colloidal particles due to unequal bombardment by solvent molecules. Prevents settling under gravity (helps stability).
Coagulation
Precipitation of colloid by adding excess electrolyte. Hardy-Schulze rule: higher charge on ion → greater coagulating power. Coagulation value = moles of electrolyte to coagulate 1L of colloid.
Adsorption
Accumulation of substance on surface. Physisorption: weak van der Waals; reversible; multilayer. Chemisorption: strong chemical bond; irreversible; monolayer; specific. High surface area (powders, colloids) enhances adsorption.
🔑 Freundlich Adsorption Isotherm
x/m = k·P^(1/n) where x=mass adsorbed, m=mass adsorbent, P=pressure, k,n=constants
log(x/m) = log k + (1/n)·log P → straight line graph; slope=1/n; intercept=log k
At low P: x/m ∝ P; at high P: x/m = constant (saturation)
Emulsions: O/W (oil in water: milk, vanishing cream) vs W/O (water in oil: butter, cold cream)
Hardy-Schulze Rule (Board 2023, 2M)
Q: Which ion will be most effective in coagulating a negative colloid: Na⁺, Mg²⁺, or Al³⁺?
1
Negative colloid has negative charge on particles
2
Coagulation requires addition of POSITIVE ions (opposite charge)
3
Hardy-Schulze rule: higher positive charge → greater coagulating power
4
Al³⁺ (charge 3) > Mg²⁺ (charge 2) > Na⁺ (charge 1)
✅ Al³⁺ is most effective (highest positive charge; Hardy-Schulze rule)
🧪 Chapter 9: Amines
Basicity · Gabriel Synthesis · Coupling Reaction · Diazonium Salts
📖 Core Theory
9.1 Classification and Basicity
Amines are derivatives of NH₃ where one or more H is replaced by alkyl/aryl group. Classified as 1°, 2°, 3° amines or quaternary ammonium salt.
Basicity Order
In gas phase: 3° > 2° > 1° > NH₃. In aqueous solution: 2° > 1° > 3° > NH₃ (solvation effect). Arylamines much weaker than alkylamines (lone pair delocalised into ring).
Gabriel Synthesis
Phthalimide + KOH → potassium phthalimide + R-X → N-alkylphthalimide → hydrolysis → pure primary amine. Used to avoid secondary/tertiary amine formation.
Hoffmann Bromamide
RCONH₂ + Br₂ + 4NaOH → RNH₂ + Na₂CO₃ + 2NaBr + 2H₂O. Product amine has one carbon less than the amide.
Diazonium Salt
ArNH₂ + NaNO₂ + HCl (0–5°C) → ArN₂⁺Cl⁻. Key intermediate for azo dyes, phenols, halides (Sandmeyer). Aliphatic diazonium salts are too unstable.
🔑 Diazonium Reactions (Sandmeyer)
ArN₂⁺ + CuCl → ArCl + N₂ (Sandmeyer)
ArN₂⁺ + CuBr → ArBr + N₂ (Sandmeyer)
ArN₂⁺ + CuCN → ArCN + N₂
ArN₂⁺ + HBF₄ → ArF + N₂ + BF₃ (Balz-Schiemann)
ArN₂⁺ + H₃PO₂ + H₂O → ArH + N₂ (deamination)
Coupling: ArN₂⁺ + PhOH/ArNH₂ (alk) → Ar−N=N−Ar' (azo dye)
✏️ Worked Examples
Basicity Comparison (Board 2023, 3M)
Q: Arrange in increasing order of basicity: PhNH₂, (CH₃)₃N, (CH₃)₂NH, NH₃.
1
In aqueous solution, basicity is affected by: (1) +I effect of alkyl groups (2) solvation (3) steric hindrance
2
PhNH₂: lone pair delocalised into ring → least basic
3
NH₃: no alkyl groups → benchmark
4
(CH₃)₃N: three +I groups but poor solvation of bulky cation → less basic than (CH₃)₂NH
5
Order: PhNH₂ < NH₃ < (CH₃)₃N < (CH₃)₂NH
✅ PhNH₂ < NH₃ < (CH₃)₃N < (CH₃)₂NH
Coupling Reaction (Board 2022, 3M)
Q: Write the coupling reaction of benzenediazonium chloride with phenol in alkaline medium.
1
Conditions: alkaline (NaOH) medium, 0–5°C, pH 9–10
2
Phenol in alkaline medium → phenoxide ion (more activated toward electrophilic attack)
3
Diazonium ion (electrophile) attacks phenoxide at para position
4
C₆H₅N₂⁺Cl⁻ + C₆H₅OH (NaOH) → p-HO-C₆H₄-N=N-C₆H₅ + HCl
5
Product: p-hydroxyazobenzene (orange azo dye)
✅ p-HO-C₆H₄-N=N-C₆H₅ (para-hydroxyazobenzene — orange dye)
🎯 Quick Activity
Think and Answer
Q: Gabriel synthesis gives only primary amines. Why can't secondary amines be made by this route?
▶ Show Answer
A: Gabriel synthesis uses the nitrogen of phthalimide which has only ONE N-H bond to alkylate. After one alkylation, no more N-H is available. Hydrolysis gives exactly one RNH₂ (primary amine). The method's selectivity is the whole point — it avoids mixtures of 1°/2°/3° amines.
📝 Coordination Chemistry Naming and Isomerism
9.3 IUPAC Naming and Types of Isomerism
IUPAC Naming Rules
1. Name ligands first (alphabetically), then central metal. 2. Ligand names end in -o (anion) or -ine, -o etc. (neutral). 3. Roman numeral = oxidation state. 4. Anionic complex ends in -ate.
Geometric Isomerism
Square planar MA₂B₂: cis (same groups adjacent) vs trans (same groups opposite). Only if two or more different ligands present. MA₄ or MA₂B₂ in octahedral also show cis/trans.
Optical Isomerism
Non-superimposable mirror images (enantiomers). Shown by [Co(en)₃]³⁺ type complexes. Optically active = rotate plane-polarised light; d (+) or l (−).
Ionisation Isomerism
Same formula, different ions in solution. [Co(NH₃)₅Br]SO₄ vs [Co(NH₃)₅(SO₄)]Br. The ions outside brackets dissociate in water.
🔑 Crystal Field Theory
Octahedral field: d orbitals split into t₂g (lower, 3 orbitals, −0.4Δ₀) and eₘ (higher, 2 orbitals, +0.6Δ₀)
CFSE = number of electrons × energy per orbital
Strong field ligands (CO, CN⁻, NO₂⁻): large Δ, low spin; weak field (F⁻, Cl⁻, H₂O): small Δ, high spin
Spectrochemical series (weak→strong): I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < NO₂⁻ < CN⁻ < CO
Oxidation State (Board 2024, 2M)
Q: Find the oxidation state of Co in [Co(NH₃)₄Cl₂]Cl
1
Cl outside bracket: −1 (free Cl⁻ ion). Cl inside bracket: −1 each. NH₃: 0.
2
Overall complex ion charge = +1 (to balance outer Cl⁻)
3
x + 4(0) + 2(−1) = +1 → x − 2 = +1 → x = +3
✅ Oxidation state of Co = +3; Complex is [Co(NH₃)₄Cl₂]⁺·Cl⁻
🧪 Chapter 10: Biomolecules
Carbohydrates · Proteins · Nucleic Acids · Vitamins
📖 Core Theory
10.1 Carbohydrates
Carbohydrates = polyhydroxy aldehydes or ketones (or compounds that give them on hydrolysis). Formula Cₙ(H₂O)ₙ (approximately). Classified by hydrolysis: mono-, di-, polysaccharides.
Monosaccharides
Cannot be hydrolysed. Glucose (aldohexose, open chain: HOCH₂(CHOH)₄CHO), Fructose (ketohexose). D and L forms based on −OH at penultimate C (relative to glyceraldehyde).
Disaccharides
Two monosaccharides linked by glycosidic bond (formed by dehydration). Sucrose (glucose+fructose, non-reducing), Maltose (glucose+glucose, reducing), Lactose (galactose+glucose, reducing).
Polysaccharides
Starch (amylose+amylopectin, α-glycosidic links, food storage), Cellulose (β-glycosidic links, cell wall — humans can't digest). Glycogen (animal storage, highly branched).
Reducing Sugars
Have free −CHO or −OH at anomeric C. Give +ve Fehling's and Tollens'. All monosaccharides + maltose + lactose are reducing. Sucrose is non-reducing.
🔑 Proteins and Nucleic Acids
Amino acids: H₂N−CHR−COOH. 20 standard; essential (must be in diet); non-essential (body synthesises).
Peptide bond: −CO−NH− formed between −COOH of one and −NH₂ of next amino acid
Protein structure: 1° (sequence), 2° (α-helix/β-sheet, H-bonds), 3° (folding, S-S bonds), 4° (multiple subunits)
DNA: deoxyribose + phosphate + bases (A-T, G-C Watson-Crick pairing). Double helix.
RNA: ribose + phosphate + bases (A-U, G-C). Single stranded (mostly). mRNA, tRNA, rRNA.
✏️ Worked Examples
Reducing vs Non-Reducing Sugars (Board 2023, 2M)
Q: Why is sucrose a non-reducing sugar but maltose is reducing?
1
Sucrose: glycosidic bond between C1 of glucose AND C2 of fructose (both anomeric carbons involved)
2
No free anomeric −OH available in sucrose → cannot open to give aldehyde form → non-reducing
3
Maltose: glycosidic bond between C1 of one glucose and C4 of another
4
C1 of the second glucose unit is free → can open to give −CHO → reducing
5
This is why sucrose gives no Fehling's but maltose gives brick-red precipitate
✅ Sucrose: both anomeric C blocked → non-reducing; Maltose: one anomeric C free → reducing
Protein Structure (Board 2022, 3M)
Q: Distinguish between primary and secondary structure of proteins.
1
Primary structure: sequence of amino acids in polypeptide chain linked by peptide bonds
2
Determined by DNA; unique for each protein; determines all higher structures
3
Secondary structure: spatial arrangement of polypeptide backbone
4
α-helix: right-handed coil, H-bonds between C=O and N-H four residues apart
5
β-pleated sheet: H-bonds between adjacent chains; silk fibroin has β-structure
6
Held by hydrogen bonds (intramolecular for α-helix, intermolecular for β-sheet)
✅ 1° = amino acid sequence; 2° = local 3D shape (α-helix/β-sheet) via H-bonds
🎯 Quick Activity
Think and Answer
Q: Starch and cellulose both consist of glucose units but have very different properties. Explain why.
▶ Show Answer
A: Starch has α-glycosidic (1→4) links — allows coiling into compact helix (digestible by amylase). Cellulose has β-glycosidic (1→4) links — allows linear chains to stack via H-bonds into strong fibres (not digestible by humans; cellulase enzyme needed).
🧪 Chapter 11: Polymers
Classification · Addition · Condensation · Rubber · Fibres
📖 Core Theory
11.1 Classification of Polymers
A polymer is a large molecule built from repeating structural units (monomers). Classified by origin (natural/synthetic/semi-synthetic), structure (linear/branched/crosslinked), type of polymerisation, and thermal response.
Addition Polymers
Formed from monomers with C=C double bonds by chain growth. No byproduct. Example: polyethene (-CH₂-CH₂-)ₙ from ethene; PVC from vinyl chloride; Teflon from CF₂=CF₂.
Condensation Polymers
Formed by stepwise reaction with elimination of small molecules (H₂O, HCl etc). Bifunctional monomers. Nylon-6,6: hexamethylenediamine + adipic acid. Dacron/Terylene: ethylene glycol + terephthalic acid.
Thermoplastics
Soften on heating, re-mould. Linear or slightly branched chains. Examples: PVC, polythene, polystyrene, nylon.
Thermosetting
Once set, cannot be re-melted (crosslinked). Bakelite (phenol+formaldehyde), melamine-formaldehyde, urea-formaldehyde.
🔑 Important Polymers
Natural rubber: cis-1,4-polyisoprene (isoprene monomer). Vulcanisation with S improves elasticity/strength by S-S crosslinks.
Nylon-6,6: [−CO(CH₂)₄CO−NH(CH₂)₆NH−]ₙ — condensation polymer (amide bonds)
Nylon-6: from caprolactam (ring-opening polymerisation) — one type of monomer
Buna-S: styrene-butadiene copolymer (synthetic rubber)
Buna-N: acrylonitrile-butadiene copolymer (oil-resistant)
✏️ Worked Examples
Addition vs Condensation (Board 2023, 3M)
Q: Classify: (i) Nylon-6,6 (ii) Teflon (iii) Bakelite.
1
Nylon-6,6: condensation polymer — formed from hexamethylenediamine + adipic acid with elimination of H₂O; amide linkages
2
Teflon: addition polymer — polymerisation of CF₂=CF₂ (tetrafluoroethene); no byproduct
3
Bakelite: condensation polymer AND thermosetting — phenol + formaldehyde, eliminates H₂O, forms crosslinked network
4
Teflon: thermoplastic; Bakelite: thermosetting; Nylon-6,6: thermoplastic
5
Bakelite cannot be re-melted; Teflon and Nylon can be re-shaped on heating
✅ Nylon-6,6 and Bakelite = condensation; Teflon = addition; Bakelite = thermosetting
Vulcanisation (Board 2022, 2M)
Q: What is vulcanisation and how does it improve rubber properties?
1
Natural rubber = cis-1,4-polyisoprene; soft, sticky, loses shape when hot
2
Vulcanisation: heating with sulphur (5–8%) at 150°C
3
Sulphur forms crosslinks between polymer chains (S-S bridges)
4
Result: stronger, harder, more elastic, less sticky, stable over wider temperature range
5
Vehicle tyres: 30% S added for even harder product (ebonite)
✅ Vulcanisation = S crosslinks → improves elasticity, strength, temperature stability
🎯 Quick Activity
Think and Answer
Q: LDPE and HDPE both are polyethene but have different properties. Why?
▶ Show Answer
A: LDPE (low density): formed at high pressure with free radical mechanism → many short branches → chains can't pack closely → low density, flexible, transparent. HDPE: Ziegler-Natta catalyst, low pressure → linear chains, pack closely → high density, rigid, opaque, higher melting point.
🧪 Chapter 12: Chemistry in Everyday Life
Drugs · Drug-Target · Detergents · Food Chemicals
📖 Core Theory
12.1 Drugs and Drug-Target Interaction
A drug is a chemical substance that interacts with a biomolecular target (receptor, enzyme, DNA) and produces a biological response. Drugs are classified by pharmacological effect or target.
Antacids
Neutralise excess HCl in stomach. NaHCO₃, Mg(OH)₂, Al(OH)₃. Should not be taken in excess — may cause alkalosis.
Antihistamines
Block histamine receptors (H₁) — treat allergy, cold. Examples: benadryl (diphenhydramine), ranitidine (also H₂, for ulcer). Do NOT block histamine synthesis.
Antibiotics
Kill or inhibit bacteria. Bactericidal (kill): penicillin, aminoglycosides. Bacteriostatic (inhibit growth): sulphonamides, tetracycline, chloramphenicol. Broad vs narrow spectrum.
Analgesics
Relieve pain. Non-narcotic: aspirin (anti-inflammatory too), paracetamol. Narcotic: morphine, codeine (addictive, controlled). NSAIDs inhibit COX enzyme.
🔑 Detergents and Food Chemicals
Soaps: sodium/potassium salts of long-chain fatty acids. Cannot clean in hard water (Ca²⁺/Mg²⁺ form scum).
Detergents: sodium alkylbenzenesulphonate (ABS) or sodium lauryl sulphate. Work in hard water. Biodegradable if linear chain.
Antioxidants in food: BHA (butylated hydroxy anisole), BHT (butylated hydroxy toluene), vitamin C, vitamin E
Sweeteners: saccharin (350× sweet), aspartame (100×), sucralose (600×, sucrose-based)
✏️ Worked Examples
Broad vs Narrow Spectrum (Board 2023, 2M)
Q: What is the difference between broad and narrow spectrum antibiotics? Give an example of each.
1
Broad spectrum: effective against a wide range of both Gram+ve and Gram−ve bacteria
2
Examples: ampicillin, amoxicillin, chloramphenicol, tetracycline
3
Used when pathogen is unidentified; risk of disrupting normal flora
4
Narrow spectrum: effective against specific types of bacteria only
5
Examples: penicillin G (mainly Gram+ve); streptomycin (Gram−ve)
✅ Broad: works on many bacteria types; Narrow: specific. Broad risk: kills normal flora.
Soaps vs Detergents (Board 2022, 3M)
Q: Why do soaps not work in hard water but detergents do?
1
Soaps are sodium/potassium salts of fatty acids (RCOONa)
2
Hard water contains Ca²⁺ and Mg²⁺ ions
3
Ca²⁺ + 2RCOONa → (RCOO)₂Ca↓ + 2Na⁺ (insoluble scum forms)
4
Scum wastes soap and leaves residue on clothes
5
Detergents use sulphonate/sulphate group: −SO₃Na or −OSO₃Na
6
Calcium/magnesium sulphonates are soluble → no scum; works in hard water
✅ Soaps form insoluble calcium salts (scum) in hard water; detergent calcium salts remain soluble
🎯 Quick Activity
Think and Answer
Q: Aspirin is both an analgesic and an antiplatelet drug. What is the mechanism?
▶ Show Answer
A: Aspirin irreversibly acetylates COX-1 and COX-2 enzymes, blocking prostaglandin and thromboxane synthesis. Prostaglandins mediate pain/inflammation → analgesic. Thromboxane A₂ promotes platelet aggregation → blocking it prevents clotting (antiplatelet use in heart attack prevention, low-dose aspirin).
📝 Aldehydes and Ketones — Reactions
12.3 Nucleophilic Addition and Oxidation-Reduction
Nucleophilic Addition
Carbonyl C⁺ attacked by nucleophile. Aldehydes more reactive than ketones (less steric, more electron-deficient). Nu: HCN, RMgX (Grignard), NaHSO₃, 2,4-DNP, H₂O (gem-diol).
Tollens Test
Silver mirror test: RCHO + [Ag(NH₃)₂]⁺ → RCOOˉ + Ag↓ (silver mirror). Positive for aldehydes, negative for ketones. Also positive for glucose (aldehyde group).
Fehling's Test
Deep blue Cu²⁺ complex + RCHO → brick-red Cu₂O precipitate. Positive for aliphatic aldehydes. Negative for aromatic aldehydes (benzaldehyde) and ketones.
Aldol Condensation
Aldehyde with α-H + dilute base → β-hydroxy aldehyde (aldol); on heating → α,β-unsaturated aldehyde (dehydration). Cross-aldol with two different compounds also possible.
🔑 Key Name Reactions
Cannizzaro Reaction: 2HCHO + NaOH → CH₃OH + HCOONa (no α-H; disproportionation)
Clemmensen Reduction: C=O + Zn-Hg/HCl → CH₂ (acidic medium)
Wolf-Kishner: C=O + NH₂NH₂/KOH/Δ → CH₂ (basic medium)
Rosenmund: RCOCl + H₂/Pd-BaSO₄ → RCHO (selective reduction of acid chloride)
Distinguish Aldehydes (Board 2022, 3M)
Q: How will you distinguish between (i) propanal and propanone (ii) benzaldehyde and propanal?
1
(i) Propanal (aldehyde) vs propanone (ketone): Use Tollens or Fehling's test. Propanal gives silver mirror/brick-red ppt; propanone does not react.
2
(ii) Benzaldehyde (aromatic aldehyde) vs propanal (aliphatic aldehyde): Both give Tollens test positive. Use Fehling's test — propanal gives brick-red ppt; benzaldehyde does NOT (negative).
✅ (i) Tollens/Fehling test; (ii) Fehling's test distinguishes aromatic from aliphatic aldehyde
🧪 Chapter 13: Solid State
Crystal Systems · Packing · Defects · Electrical Properties
📖 Core Theory
13.1 Crystal Systems and Packing
In crystalline solids, particles are arranged in a definite, repeating 3D pattern (unit cell). Amorphous solids have short-range order only (glass, rubber).
Unit Cell
Smallest repeating unit of crystal lattice. Simple cubic (1 atom), BCC (2 atoms), FCC/CCP (4 atoms). Contribution: corner = 1/8, face = 1/2, edge = 1/4, body centre = 1.
Packing Efficiency
% of unit cell volume occupied by atoms. SC = 52.4%, BCC = 68%, FCC/HCP = 74% (maximum; ABCABC or ABABAB stacking).
Crystal Defects
Frenkel defect: cation moves to interstitial (density unchanged); Schottky defect: paired vacancies of cation+anion (density decreases). Both increase with temperature.
Electrical Properties
n-type semiconductor: pentavalent impurity (P, As) donates extra electron. p-type: trivalent impurity (B, Al) creates electron hole. Band gap determines conductor/semiconductor/insulator.
🔑 Key Formulae
Density: ρ = (Z×M)/(a³×Nₐ) where Z=atoms/unit cell, M=molar mass, a=edge length
FCC: 4 atoms; a = 2√2 r; Packing efficiency = 74.05%
BCC: 2 atoms; a = 4r/√3; Packing efficiency = 68.02%
SC: 1 atom; a = 2r; Packing efficiency = 52.36%
✏️ Worked Examples
Density of Unit Cell (Board 2023, 3M)
Q: An element forms BCC with edge length 287 pm. Atomic mass 56 g/mol. Find density. Nₐ=6.022×10²³.
1
BCC: Z = 2 atoms per unit cell
2
a = 287 pm = 287×10⁻¹⁰ cm
3
a³ = (287)³ × 10⁻³⁰ cm³ = 23,639,903 × 10⁻³⁰ = 2.364×10⁻²³ cm³
4
ρ = ZM/(a³Nₐ) = 2 × 56 / (2.364×10⁻²³ × 6.022×10²³)
5
ρ = 112 / 14.24 = 7.87 g/cm³ (this is iron — correct!)
✅ ρ = 7.87 g/cm³ (iron with BCC structure)
Frenkel vs Schottky (Board 2022, 2M)
Q: Distinguish between Frenkel and Schottky defects.
1
Frenkel defect: an ion (usually smaller cation) is displaced from its lattice site to an interstitial site
2
Number of ions unchanged, no change in density; example: AgCl (Ag⁺ moves to interstitial)
3
Schottky defect: equal numbers of cations and anions are missing from lattice sites
4
Density decreases (fewer particles in same volume); example: NaCl, KCl
5
Schottky common in ionic crystals with equal-sized ions; Frenkel when cation much smaller than anion
✅ Frenkel: ion to interstitial (no density change); Schottky: paired vacancies (density decreases)
🎯 Quick Activity
Think and Answer
Q: Why does NaCl crystal have no Frenkel defect but AgCl does?
▶ Show Answer
A: Frenkel defect requires the cation to move to an interstitial void. In NaCl, Na⁺ and Cl⁻ are nearly equal in size, so Na⁺ is too large to fit into the small tetrahedral interstitial sites without seriously distorting the lattice. In AgCl, Ag⁺ is much smaller than Cl⁻ and can easily move to interstitial positions, creating Frenkel defects.
📝 Carboxylic Acids and Derivatives
13.3 Reactions and Acidity
Acidity Order
Carboxylic acids stronger than phenol > water > alcohols. EWG on ring/chain increases acidity (stabilises carboxylate by induction). EDG decreases acidity. CCl₃COOH > CHCl₂COOH > CH₂ClCOOH > CH₃COOH (Cl withdraws e⁻).
Reactions of −COOH
With NaOH: salt + water. With Na₂CO₃: salt + CO₂ + H₂O. With PCl₅/SOCl₂: acid chloride RCOCl. With NH₃: amide (via ammonium salt). With ROH/H⁺: ester (Fischer esterification).
Hell-Volhard-Zelinsky (HVZ)
RCOOH + Cl₂/P → α-chloro acid (RCHClCOOH). Specific to carboxylic acids with α-hydrogen. The −COOH activates α-position.
Decarboxylation
RCOONa + NaOH → RH + Na₂CO₃ (soda lime, heating). Removes −COOH group. Used to descend the homologous series. CH₃COONa + NaOH → CH₄ (methane).
🔑 Derivatives of Carboxylic Acids
Reactivity towards nucleophilic acyl substitution: acid chloride > anhydride > ester > amide
Esterification: RCOOH + R′OH ⇌ RCOOR′ + H₂O (catalyst H⁺; reversible)
Saponification: RCOOR′ + NaOH → RCOONa + R′OH (irreversible; gives soap from fat)
Reduction: RCOOH + LiAlH₄ → RCH₂OH (primary alcohol)
Acidity Comparison (Board 2023, 2M)
Q: Arrange in increasing order of acidity: CH₃COOH, CF₃COOH, CCl₃COOH, CHCl₂COOH
1
More electronegative substituents → stronger inductive withdrawal → better stabilise carboxylate → stronger acid
2
F is more electronegative than Cl; three halogens stronger than two than one
3
CH₃COOH (no halogen, EDG) < CHCl₂COOH < CCl₃COOH < CF₃COOH
✅ CH₃COOH < CHCl₂COOH < CCl₃COOH < CF₃COOH (increasing acidity)
🧪 Chapter 14: Surface Chemistry
Adsorption · Colloids · Emulsions · Catalysis
📖 Core Theory
14.1 Adsorption
Adsorption is accumulation of a substance (adsorbate) on the surface of another (adsorbent). Distinguished from absorption (penetration through bulk). Adsorption is always exothermic (ΔH < 0).
Physisorption
Weak van der Waals forces; low Ea (<40 kJ/mol); reversible; multilayer possible; decreases with temperature; example: N₂ on charcoal.
Chemisorption
Strong chemical bonds (covalent/ionic); high Ea (40–400 kJ/mol); monolayer; may increase then decrease with T (activated process at first); example: H₂ on Ni.
Colloids
Heterogeneous mixture; dispersed phase: 1–1000 nm. Sol (solid in liquid), Aerosol (solid/liquid in gas), Emulsion (liquid in liquid), Gel (liquid in solid), Foam (gas in liquid).
Tyndall Effect
Scattering of light by colloidal particles (visible beam). Absent in true solutions. Used to distinguish colloids from true solutions. Blue colour of sky is Tyndall effect.
🔑 Freundlich Adsorption Isotherm
x/m = k·P^(1/n) where x/m = amount adsorbed per gram, P = pressure
Taking log: log(x/m) = log k + (1/n)log P (straight line, slope = 1/n)
At low pressure: x/m ∝ P; at high pressure: x/m = constant (monolayer)
Coagulation: Hardy-Schulze rule — greater the valency of coagulating ion, greater the coagulating power
✏️ Worked Examples
Physisorption vs Chemisorption (Board 2023, 3M)
Q: Adsorption of a gas on solid surface first increases then decreases with temperature. Explain.
1
At low temperatures: physisorption occurs (van der Waals forces)
2
Rate of physisorption decreases with rising temperature (exothermic process)
3
At higher temperatures: chemisorption begins (needs activation energy to form bonds)
4
Chemisorption increases with temperature up to a maximum
5
At very high temperatures: desorption begins (kinetic energy exceeds adsorption energy)
✅ Low T: physisorption (decreases); intermediate T: chemisorption (increases); very high T: desorption
Tyndall Effect Application (Board 2022, 2M)
Q: Why is the sky blue? Relate to Tyndall effect.
1
Sunlight (white light) passes through atmosphere containing colloidal dust and water droplets
2
Colloidal particles preferentially scatter shorter wavelengths (blue, ~450 nm)
3
This is Tyndall effect — scattering of light by colloidal particles
4
Blue light scattered in all directions → sky appears blue to observer
5
At sunset: light travels longer path → blue scattered away; red/orange remain → sunset colours
✅ Blue light (shorter λ) is scattered by colloidal atmospheric particles (Tyndall effect) → blue sky
🎯 Quick Activity
Think and Answer
Q: Alum (K₂SO₄·Al₂(SO₄)₃·24H₂O) is used to purify muddy water. Explain the chemistry.
▶ Show Answer
A: Muddy water contains negatively charged colloidal clay particles. Alum dissolved in water gives Al³⁺ ions (trivalent cation). By Hardy-Schulze rule, Al³⁺ is very effective at coagulating negatively charged colloids. The Al³⁺ neutralises the charge on clay particles → they coagulate → settle as floc → water becomes clear.
📝 Amines — Basicity and Reactions
14.3 Basic Character and Reactions of Amines
Basicity Order (Aliphatic)
In gas phase: 3° > 2° > 1° > NH₃. In aqueous solution: 2° > 1° > 3° > NH₃ (solvation effect reverses for 3°). Electron-donating groups increase basicity; EWG decrease it.
Aromatic vs Aliphatic
Aniline (C₆H₅NH₂) far less basic than aliphatic amines. Lone pair on N delocalised into benzene ring (resonance) → less available for protonation. Electron-withdrawing groups on ring further decrease basicity.
Diazotisation
ArNH₂ + NaNO₂ + HCl at 0–5°C → ArN₂⁺Cl⁻ (diazonium salt). Must be done below 5°C; diazonium salts are unstable above 5°C.
Coupling Reaction
ArN₂⁺ + ArH (activated) → Ar-N=N-Ar′ (azo dye; orange/yellow colour). Coupling at para position of activated ring (phenol, aniline).
🔑 Distinguishing 1°, 2°, 3° Amines
Hinsberg's test: benzene sulfonyl chloride (C₆H₅SO₂Cl). 1°: sulfonamide soluble in NaOH. 2°: sulfonamide insoluble in NaOH. 3°: no reaction.
Mustard oil reaction (carbylamine test): 1° amine + CHCl₃ + KOH → isocyanide (mustard smell). Negative for 2° and 3°.
With HNO₂ (NaNO₂+HCl): 1° → diazonium (unstable for aliphatic, stable for aromatic); 2° → N-nitrosoamine (yellow oil); 3° → no reaction
Basicity Order (Board 2024, 3M)
Q: Arrange in decreasing basicity: p-nitroaniline, aniline, p-methylaniline, diphenylamine.
1
All are aromatic amines; lone pair on N delocalised into ring
2
CH₃ (EDG) on ring increases electron density → more basic than aniline
3
NO₂ (EWG) further withdraws electrons → less basic than aniline
4
Diphenylamine: lone pair delocalised into TWO rings → least basic
✅ p-methylaniline > aniline > p-nitroaniline > diphenylamine
🧪 Chapter 15: General Principles of Isolation of Metals
Metallurgy · Ores · Refining
📖 Core Theory
15.1 Metallurgy — Principles
Metallurgy is the science of extracting metals from their ores and refining them for use. Key steps: mining, crushing, concentration (beneficiation), extraction, refining.
Concentration Methods
Froth flotation (sulphide ores), Magnetic separation (magnetite/chromite), Gravity separation (gold/tin), Leaching (Al₂O₃ by NaOH in Baeyer's process for Al).
Ellingham Diagram
Plot of ΔG° vs temperature for metal oxide formation. Lower on diagram = more stable oxide = stronger reducing agent. Carbon (as C or CO) is used as reductant when its line is below the metal oxide line.
Hall-Heroult Process
Electrolytic extraction of Al: Al₂O₃ dissolved in molten cryolite (Na₃AlF₆). Cathode: carbon lining (Al deposits). Anode: carbon (oxidised to CO₂). Temperature: ~950°C.
Refining Methods
Electrolytic refining (Cu, Ag, Au, Ni), Zone refining (Ge, Si — semiconductors), Vapour phase (Mond process for Ni: Ni + 4CO → Ni(CO)₄ → Ni), Van Arkel (Ti, Zr: TiI₄ decomp on W wire).
🔑 Extraction Reactions
Iron (Blast furnace): Fe₂O₃ + 3CO → 2Fe + 3CO₂ (at ~700°C)
Copper (froth flotation → roasting → self-reduction): Cu₂S + O₂ → 2Cu + SO₂ (partial oxidation)
Al (Hall-Heroult): Al³⁺ + 3e⁻ → Al (cathode); 2O²⁻ → O₂ + 4e⁻ (anode; C anode burns)
Baeyer's: Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O; then CO₂ + H₂O → Al(OH)₃ → Al₂O₃
✏️ Worked Examples
Froth Flotation (Board 2023, 3M)
Q: How is copper pyrite (CuFeS₂) concentrated by froth flotation?
1
Ore is crushed and ground to fine powder
2
Added to water in a large tank with pine oil (frothing agent) and collectors (xanthates)
3
Air blown through — creates froth; sulphide ore particles are hydrophobic (wetted by oil)
4
Ore particles attach to bubbles and rise to top as froth
5
Gangue (silica, earth) is hydrophilic — sinks to bottom as slurry
6
Froth is skimmed off to get concentrated sulphide ore
✅ Sulphide ore = hydrophobic, attaches to froth and floats; gangue sinks
Mond Process (Board 2022, 3M)
Q: How is nickel refined by Mond process?
1
Impure Ni is heated at 50–60°C with CO gas
2
Ni + 4CO(g) → Ni(CO)₄(g) — volatile nickel tetracarbonyl formed
3
Impurities do not react with CO
4
Ni(CO)₄ vapour is passed to a decomposition chamber at 230°C
5
Ni(CO)₄ → Ni(pure) + 4CO (CO recycled)
6
Very pure Ni deposited; based on reversible reaction Le Chatelier
✅ Ni + 4CO ⇌ Ni(CO)₄; form at 50°C, decompose at 230°C → pure Ni
🎯 Quick Activity
Think and Answer
Q: Why is carbon used as a reducing agent in metallurgy but not for all metals?
▶ Show Answer
A: Carbon works as reductant when ΔG° for C + O₂ → CO₂ (or 2C + O₂ → 2CO) is more negative than for MO → M + O₂. This is shown on the Ellingham diagram. For very stable oxides (Al₂O₃, TiO₂), carbon's line is above the metal oxide line even at high temperatures, so electrolysis must be used instead.
📝 Polymers — Types and Properties
15.3 Classification and Important Polymers
Addition vs Condensation
Addition: monomers with C=C join without losing atoms (polyethylene, PVC, Teflon). Condensation: monomers with two functional groups; small molecule (H₂O, HCl) lost per step (nylon, Dacron, Bakelite).
Thermoplastic vs Thermosetting
Thermoplastic: linear/branched; softens on heating (reversible); recyclable. Examples: polyethylene, PVC, polystyrene. Thermosetting: cross-linked network; does not soften (irreversible). Examples: Bakelite, melamine-formaldehyde.
Copolymers
Two different monomers. Buna-S: styrene + butadiene. Buna-N: acrylonitrile + butadiene. Neoprene: chloroprene. SBR (Buna-S) is the most common synthetic rubber.
Biodegradable Polymers
PHBV (polyhydroxybutyrate-co-hydroxyvalerate): natural, biodegradable; used in surgery. Nylon-2-nylon-6: alternating amino acids; biodegradable. Contrast with non-biodegradable plastics (PVC, polythene).
🔑 Important Named Polymers (Board Favourites)
Nylon-6,6: hexamethylenediamine + adipic acid → [−NH(CH₂)₆NH−CO(CH₂)₄CO−]n; strong fibre
Nylon-6: caprolactam (ring opening) → [−NH(CH₂)₅CO−]n
Dacron/Terylene: ethylene glycol + terephthalic acid; polyester; clothing + bottles
Bakelite: phenol + formaldehyde; cross-linked; electrical switches, combs
Natural rubber: polyisoprene (cis-1,4); vulcanisation with S adds cross-links → harder, elastic
Identify Polymer Type (Board 2023, 3M)
Q: Classify the following: (i) Nylon-6,6 (ii) Polythene (iii) Bakelite. Give reasons.
1
Nylon-6,6: two bifunctional monomers, −H₂O lost per step → condensation polymer; linear → thermoplastic
2
Polythene: CH₂=CH₂ (ethylene) monomers add without losing any atoms → addition polymer; thermoplastic
3
Bakelite: phenol + formaldehyde; condensation; 3D network cross-linking → thermosetting
✅ Nylon-6,6: condensation, thermoplastic; Polythene: addition, thermoplastic; Bakelite: condensation, thermosetting
🧪 Chapter 16: p-Block Elements
Group 15, 16, 17, 18 · Properties · Reactions · Allotropes
📖 Core Theory
16.1 Group 15 — Nitrogen Family
Group 15: N, P, As, Sb, Bi. Valence configuration: ns²np³. Three p-electrons, many possible oxidation states (−3 to +5). Nitrogen unique: forms multiple bonds (N₂, triple bond), unlike heavier congeners.
Oxoacids of N
HNO₂ (nitrous, +3 OS) and HNO₃ (nitric, +5 OS). HNO₃: strong oxidising acid. Dilute → NO with Cu; concentrated → NO₂ with Cu. Aqua regia = 3:1 HCl:HNO₃ (dissolves noble metals).
Allotropes of P
White P (P₄ tetrahedron, reactive, spontaneously flammable in air, stored under water), Red P (polymeric, stable, less reactive), Black P (most stable, layered structure like graphite).
Group 16 — Chalcogens
O, S, Se, Te, Po. O forms 2 bonds; S forms 2, 4, or 6. Allotropes of S: rhombic (stable at RT) and monoclinic (stable 96–119°C). S₈ ring structure.
Group 17 — Halogens
F₂ (most reactive nonmetal, pale yellow gas), Cl₂ (yellow-green), Br₂ (reddish-brown liquid), I₂ (dark violet solid). Only F has OS −1 only; others also +1, +3, +5, +7. HF is weak acid (strong H-bonding).
🔑 Noble Gases and Key Reactions
Group 18: He, Ne, Ar, Kr, Xe, Rn. Octet complete → almost inert. Xe forms compounds: XeF₂, XeF₄, XeF₆, XeO₃, XeOF₄.
Interhalogen compounds: AB, AB₃, AB₅, AB₇ where A is heavier halogen (B is always F or the lighter one)
Bleaching powder: Ca(OCl)Cl — Ca(OH)₂ + Cl₂ (dry) at 40°C. Active agent: OCl⁻ (hypochlorite)
Ozone: O₃ formed at high altitude by UV; shield against UV-B/C. Depleted by Cl· (from CFCs): Cl· + O₃ → ClO· + O₂ (chain reaction)
✏️ Worked Examples
Bleaching Powder (Board 2023, 2M)
Q: Write the reaction of bleaching powder with dilute H₂SO₄ and explain its bleaching action.
1
Bleaching powder: Ca(OCl)Cl (calcium chlorohypochlorite)
2
With dilute H₂SO₄: Ca(OCl)Cl + H₂SO₄ → CaSO₄ + Cl₂ + H₂O
3
Or: 2Ca(OCl)Cl + H₂SO₄ → CaSO₄ + CaCl₂ + 2HOCl
4
HOCl (hypochlorous acid) decomposes: HOCl → HCl + [O]
5
Nascent oxygen [O] oxidises the colouring matter (bleaching action)
✅ Bleaching = nascent [O] from HOCl oxidises coloured matter
Oxoacids of S (Board 2022, 3M)
Q: Why does H₂SO₃ act as a reducing agent while H₂SO₄ acts as oxidising agent?
1
H₂SO₃: S is in +4 oxidation state (can increase to +6)
2
Can be oxidised further → acts as REDUCING agent (loses electrons)
3
Example: H₂SO₃ + Cl₂ + H₂O → H₂SO₄ + 2HCl (S: +4 → +6)
4
H₂SO₄: S is in +6 oxidation state (maximum)
5
Can only be reduced (OS decreases): acts as OXIDISING agent
6
Example: Cu + 2H₂SO₄(conc) → CuSO₄ + SO₂ + 2H₂O (S: +6 → +4)
✅ H₂SO₃: S at +4 → can increase → reductant; H₂SO₄: S at +6 maximum → oxidant
🎯 Quick Activity
Think and Answer
Q: XeF₄ exists but NF₄⁺ does not — explain in terms of atomic properties.
▶ Show Answer
A: Xe is a large atom (period 5) with available d-orbitals for expansion of octet — can form 4, 6 bonds. F is in period 2 with no d-orbitals; nitrogen is also period 2 — cannot expand beyond 4 bonds (octet rule). N in NF₄⁺ would need 4 bonds + a lone pair at N (5 bonds total) which requires d-orbital expansion unavailable in period 2.
📝 Practice Papers
Full-length 70-mark board-pattern papers · Chemistry XII
CHEMISTRY XII — PRACTICE PAPER 1
Time: 3 Hours | Max Marks: 70 | CBSE 2024-25
SECTION A — MCQ (16×1=16)
Q1. The molarity of pure water is: (a)55.5M (b)100M (c)18M (d)1M [Ans: a]
Q2. Henry's law constant (KH) for a gas increases with temperature — this means: (a) Solubility increases (b) Solubility decreases (c) No change (d) Depends on gas [Ans: b]
Q3. Rate law for A+B→C: rate = k[A]²[B]. If [A] doubles, rate becomes: (a) 2× (b) 4× (c) 8× (d) 16× [Ans: b]
Q4. Half-life of first-order reaction: (a) t½ = k/0.693 (b) t½ = 0.693/k (c) t½ = [A]₀/2k (d) t½ = 1/k[A]₀ [Ans: b]
Q5. Which colloid type has solid dispersed in liquid? (a) Emulsion (b) Sol (c) Foam (d) Aerosol [Ans: b]
Q6. Standard electrode potential of SHE (Standard Hydrogen Electrode) is: (a) 1.0V (b) −1.0V (c) 0V (d) 0.5V [Ans: c]
Q7. Which d-block element does NOT show variable oxidation state? (a) Cr (b) Mn (c) Zn (d) Fe [Ans: c]
Q8-16. [Further MCQs on: Coordination compounds (IUPAC), Haloalkanes (SN1 vs SN2), Alcohols (Lucas test), Amines (basicity order), Biomolecules (glucose structure), Polymers (types), D&F block, Aldehyde vs Ketone test] [9M]
SECTION B — SA I (5×2=10)
Q17. Calculate the mole fraction of ethanol in a solution containing 46g ethanol and 36g water. [2M]
Q18. Write the IUPAC name: [Co(NH₃)₄Cl₂]Cl. Give its geometry. [2M]
Q19. Distinguish between SN1 and SN2 mechanisms with one example each. [2M]
Q20. Why are amines more basic than amides? [2M]
Q21. What is adsorption? Distinguish physisorption and chemisorption. [2M]
SECTION C — SA II (7×3=21)
Q22. Derive the integrated rate expression for a first-order reaction. Define half-life. [3M]
Q23. State Kohlrausch's law. How is it used to find limiting molar conductivity of acetic acid? [3M]
Q24. Explain the following: (i) Paramagnetism in transition metals (ii) Coloured compounds of transition elements (iii) High melting points of transition metals. [3M]
Q25. Preparation and chemical properties of phenol (any 3 reactions). [3M]
Q26. What is a biodegradable polymer? Give two examples with monomers. How are polymers classified by backbone chain? [3M]
Q27. (a) What is the function of enzyme in digestion? (b) Differentiate between DNA and RNA. [3M]
Q28. Explain Nessler's reagent test and Baeyer's reagent application in organic chemistry. [3M]
SECTION D — LA (3×5=15) + Case (4M)
Q29. (a) Derive Nernst equation. Calculate EMF: Zn|Zn²⁺(0.1M)||Cu²⁺(0.1M)|Cu, E°cell = 1.10V. (b) What is corrosion? How does galvanisation prevent it? [5M]
Q30. (a) Prepare p-nitroaniline from aniline. Give steps. (b) Compare basicity: NH₃, CH₃NH₂, (CH₃)₂NH, C₆H₅NH₂ in water. [5M]
Q31. (a) Explain Cannizzaro reaction, Aldol condensation with mechanism. (b) Distinguish aldehyde from ketone using Tollens, Fehling and Iodoform. [5M]
Q32. Case-based: A solution shows osmotic pressure 2.47 atm at 27°C. Calculate molar mass of solute (0.5g in 200mL). What is Van't Hoff factor? [4M]
CHEMISTRY XII — PRACTICE PAPER 2
Focus: Organic + Physical chemistry
SECTION A — MCQ (16M)
Q1-16. MCQs on colligative properties (elevation of BP, depression of FP), Freundlich isotherm, EAN rule, Werner's theory, optical isomerism, carbylamine test, biuret test, reducing sugars, natural rubber monomer. [16M]
SECTIONS B, C, D
Q17-32. Cover: Raoult's law deviations, Molar conductivity calculation, Crystal Field Theory (CFT) basics, Halogen acid reactivity order (HF exceptions), Preparation of carboxylic acids (all 4 methods), Mutarotation in glucose, Sabatier-Senderen reaction, Wurtz-Fittig reaction. [54M]
CHEMISTRY XII — PRACTICE PAPER 3 (Challenge)
Board 2024 hardest questions
Q1. Arrhenius equation: ln(k₂/k₁) = Ea/R × (1/T₁ − 1/T₂). Rate doubles from 25°C to 35°C. Find Ea. [4M]
Q2. Complete the reactions (10 organic conversions): Toluene→benzaldehyde; Benzene→aniline; Ethanol→ethylamine; Glucose→lactic acid; etc. [6M]
Q3. Lanthanoid contraction — explain and give 3 consequences. [3M]
Q4. IUPAC names of 5 complex coordination compounds including ambidentate and bridging ligands. [5M]
Q5. Full-length problem: Given electrolytic cell data, calculate quantity of metal deposited (Faraday's law). [4M]
Q6-10. LA questions on all remaining topics: Polymers (condensation vs addition), Hormones (types and functions), Drug-enzyme interaction, Antacids vs antibiotics distinction. [48M total for paper]
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