This chapter explains how particles (atoms, ions or molecules) are arranged in solids. It covers types of solids, unit cells, cubic lattices, packing efficiency, crystal defects, and the electrical and magnetic properties of solids. Numericals on density and packing efficiency are very common in the board exam.
महत्त्वाचे मुद्दे
Solids have definite shape and volume because of strong interparticle forces.
Crystalline solids have a regular, long-range order, sharp melting point and are anisotropic. Amorphous solids (glass, rubber, plastic) have only short-range order, melt over a range of temperature and are isotropic.
Crystalline solids are of four types: ionic (NaCl), covalent network (diamond, SiO₂), molecular (ice, dry ice) and metallic (Cu, Fe).
A unit cell is the smallest repeating unit of a crystal lattice. There are 7 crystal systems and 14 Bravais lattices.
Particles per cubic unit cell: simple cubic = 1, body-centred cubic (bcc) = 2, face-centred cubic (fcc) = 4.
Contribution of a particle: corner = 1/8, face centre = 1/2, body centre = 1, edge centre = 1/4.
Close packing gives hcp (ABAB…) and ccp/fcc (ABCABC…) structures; coordination number = 12 in both.
In close packing, for N atoms there are N octahedral voids and 2N tetrahedral voids.
Point defects: stoichiometric (vacancy, interstitial, Schottky, Frenkel), impurity and non-stoichiometric (metal deficiency, metal excess) defects.
By band theory, conductors have overlapping/partly filled bands, insulators have a large band gap, semiconductors have a small band gap.
Doping with group 15 element gives n-type; with group 13 element gives p-type semiconductor.
महत्त्वाच्या संज्ञा
Isomorphism – different substances having the same crystal structure (e.g. NaF and MgO).
Polymorphism – one substance existing in more than one crystal form (e.g. diamond and graphite).
Coordination number – number of nearest neighbours of a particle in a crystal.
Void – empty space left between packed spheres (tetrahedral or octahedral).
Schottky defect – equal number of cations and anions missing; density decreases (e.g. NaCl, KCl).
Frenkel defect – a smaller ion (usually cation) leaves its site and sits in an interstitial position; density unchanged (e.g. ZnS, AgCl).
F-centre – an anion vacancy holding a trapped electron; gives colour (NaCl looks yellow when heated in Na vapour).
Doping – adding a tiny amount of impurity to a semiconductor to increase its conductivity.
सूत्रे
Density of unit cell: ρ = (n × M) / (a³ × N_A) n = particles per unit cell, M = molar mass (g mol⁻¹), a = edge length (cm), N_A = 6.022 × 10²³ mol⁻¹; ρ in g cm⁻³.
Radius and edge length: simple cubic r = a/2; bcc r = (√3 × a)/4; fcc r = a/(2√2) = 0.3535a.
Packing efficiency = (volume occupied by particles in unit cell / volume of unit cell) × 100.
Number of unit cells in mass m of a substance = (m × N_A) / (M × n).
1 pm = 10⁻¹⁰ cm; 1 Å = 10⁻⁸ cm.
Unit cell
Particles (n)
Coordination no.
Packing efficiency
Simple cubic
1
6
52.4%
bcc
2
8
68%
fcc
4
12
74%
५ महत्त्वाचे प्रश्न
Q1. Distinguish between crystalline and amorphous solids (any four points).
Crystalline: regular long-range order, sharp melting point, anisotropic, cleave along definite planes, true solids. Amorphous: only short-range order, melt over a temperature range, isotropic, irregular cleavage, called pseudo-solids or supercooled liquids.
Q2. Calculate the number of particles in an fcc unit cell.
8 corners × 1/8 = 1, and 6 face centres × 1/2 = 3. Total = 1 + 3 = 4 particles.
Q3. A metal crystallises in bcc. Edge length = 288 pm, molar mass = 52 g mol⁻¹. Find its density.
n = 2, a = 288 × 10⁻¹⁰ cm, a³ = 2.389 × 10⁻²³ cm³. ρ = (2 × 52) / (2.389 × 10⁻²³ × 6.022 × 10²³) = 104 / 14.39 ≈ 7.23 g cm⁻³.
Q4. Show that packing efficiency of fcc is 74%.
In fcc, √2 a = 4r, so a = 2√2 r. Volume of 4 spheres = 4 × (4/3)πr³ = (16/3)πr³. Volume of cell = a³ = 16√2 r³. Packing efficiency = (16πr³/3) ÷ (16√2 r³) × 100 = π/(3√2) × 100 = 74%.
Q5. What are n-type and p-type semiconductors?
n-type: silicon doped with a group 15 element (P, As); the extra electron carries current. p-type: silicon doped with a group 13 element (B, Al, Ga); the "holes" (electron vacancies) carry current.
परीक्षेसाठी टीप
Always convert pm to cm before using the density formula, and remember the table of n, coordination number and packing efficiency — it gives easy marks.
A solution is a homogeneous mixture of two or more substances. This chapter covers solubility (and Henry's law), vapour pressure and Raoult's law, ideal and non-ideal solutions, and the four colligative properties used to find the molar mass of a solute. It ends with the van't Hoff factor for electrolytes.
महत्त्वाचे मुद्दे
The component present in larger amount is the solvent; the other is the solute.
Solubility of a solid in a liquid usually increases with temperature; solubility of a gas in a liquid decreases with temperature and increases with pressure (Henry's law).
Raoult's law: the partial vapour pressure of a volatile component equals its mole fraction × vapour pressure of the pure component.
Ideal solutions obey Raoult's law at all concentrations (ΔH_mix = 0, ΔV_mix = 0), e.g. benzene + toluene.
Non-ideal solutions show positive deviation (ethanol + water, weaker A–B forces) or negative deviation (chloroform + acetone, stronger A–B forces).
Colligative properties depend only on the number of solute particles, not their nature: relative lowering of vapour pressure, boiling point elevation, freezing point depression and osmotic pressure.
Osmotic pressure is the best method for finding molar mass of polymers and proteins (measurable at room temperature, large values even for dilute solutions).
Solutions with equal osmotic pressure are isotonic; 0.9% NaCl (normal saline) is isotonic with blood.
Reverse osmosis: applying pressure greater than osmotic pressure pushes solvent out of the solution; used to purify sea water.
Electrolytes give more particles, so they show larger colligative effects. This is corrected by the van't Hoff factor i.
महत्त्वाच्या संज्ञा
Molality (m) – moles of solute per kg of solvent (mol kg⁻¹); does not change with temperature.
Mole fraction (x) – moles of a component ÷ total moles.
Ebullioscopic constant (K_b) – boiling point elevation for a 1 molal solution.
Cryoscopic constant (K_f) – freezing point depression for a 1 molal solution (water: 1.86 K kg mol⁻¹).
Semipermeable membrane – lets solvent molecules pass but not solute molecules.
Boiling point elevation: ΔT_b = K_b × m ; M₂ = (1000 × K_b × W₂)/(ΔT_b × W₁)
Freezing point depression: ΔT_f = K_f × m ; M₂ = (1000 × K_f × W₂)/(ΔT_f × W₁)
Osmotic pressure: π = MRT = (W₂RT)/(M₂V) (R = 0.08205 L atm K⁻¹ mol⁻¹ or 8.314 J K⁻¹ mol⁻¹)
With electrolytes: ΔT_b = iK_b m, ΔT_f = iK_f m, π = iMRT
Degree of dissociation: α = (i − 1)/(n − 1), n = ions from one formula unit.
(W₂, M₂ = mass and molar mass of solute; W₁, M₁ = of solvent, in grams.)
५ महत्त्वाचे प्रश्न
Q1. State Raoult's law. What is a positive deviation?
The partial vapour pressure of any volatile component of a solution equals its mole fraction multiplied by the vapour pressure of the pure component. In positive deviation, the observed vapour pressure is higher than predicted because solute–solvent attractions are weaker than solute–solute and solvent–solvent attractions (e.g. ethanol + water).
Q2. 1.8 g of a non-volatile solute in 100 g water lowers the freezing point by 0.186 K. Find molar mass. (K_f = 1.86 K kg mol⁻¹)
Q3. Why is osmotic pressure preferred for molar mass of proteins?
It is measured at room temperature (proteins are not damaged by heating), and even very dilute solutions give a large, easily measured value.
Q4. What are isotonic, hypertonic and hypotonic solutions?
Isotonic: same osmotic pressure. Hypertonic: higher osmotic pressure than the other solution. Hypotonic: lower osmotic pressure. A blood cell shrinks in a hypertonic solution and swells in a hypotonic one.
Q5. Why does 0.1 m NaCl show about double the freezing point depression of 0.1 m glucose?
NaCl dissociates into Na⁺ and Cl⁻, so the number of particles nearly doubles (i ≈ 2). Glucose does not dissociate (i = 1). Colligative properties depend on the number of particles.
परीक्षेसाठी टीप
Write the formula, put units with every value and check the final unit (g mol⁻¹). Remember: freezing point depression uses K_f, boiling point elevation uses K_b — do not mix them up.
This chapter deals with equilibria involving ions in water. It covers theories of acids and bases, ionisation of weak acids and bases (Ostwald's dilution law), the ionic product of water and pH, hydrolysis of salts, buffer solutions, solubility product and the common ion effect.
Arrhenius: acid gives H⁺ in water, base gives OH⁻ in water.
Brønsted–Lowry: acid is a proton donor, base is a proton acceptor. Acid and base differing by one proton form a conjugate acid–base pair (e.g. CH₃COOH / CH₃COO⁻).
Lewis: acid is an electron-pair acceptor (BF₃, AlCl₃, H⁺), base is an electron-pair donor (NH₃, H₂O).
Degree of dissociation (α) of a weak electrolyte increases on dilution (Ostwald's dilution law).
Water ionises slightly: K_w = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 298 K.
Salt hydrolysis: salt of strong acid + strong base (NaCl) gives a neutral solution; strong acid + weak base (NH₄Cl) gives acidic; weak acid + strong base (CH₃COONa) gives basic; weak acid + weak base (CH₃COONH₄) depends on Ka and Kb.
A buffer resists change in pH on adding small amounts of acid or base. Acidic buffer: CH₃COOH + CH₃COONa. Basic buffer: NH₄OH + NH₄Cl.
Buffers are used in blood (pH about 7.4), fermentation, medicines and agriculture.
Solubility product (K_sp) applies to sparingly soluble salts; precipitation occurs when ionic product > K_sp.
Common ion effect: adding a strong electrolyte with a common ion suppresses ionisation of a weak electrolyte or decreases solubility of a sparingly soluble salt.
महत्त्वाच्या संज्ञा
Degree of dissociation (α) – fraction of total molecules that dissociate into ions.
pH – negative logarithm (base 10) of hydronium ion concentration.
Hydrolysis – reaction of ions of a salt with water giving an acidic or basic solution.
Solubility (S) – moles of a salt dissolved per litre of its saturated solution.
Ionic product (IP) – product of ion concentrations (raised to powers) at any moment.
सूत्रे
Ostwald's law (weak acid): Ka = α²c/(1 − α) ≈ α²c, so α = √(Ka/c) and [H₃O⁺] = √(Ka × c)
Q4. What is a buffer? Explain the action of an acetate buffer.
A solution that resists change in pH. In CH₃COOH + CH₃COONa: added H⁺ is taken up by CH₃COO⁻ forming CH₃COOH; added OH⁻ is neutralised by CH₃COOH forming CH₃COO⁻ and water. So pH stays almost the same.
Q5. Solubility of AgBr is 7.2 × 10⁻⁷ mol L⁻¹. Find K_sp.
K_sp = S² = (7.2 × 10⁻⁷)² = 5.18 × 10⁻¹³.
परीक्षेसाठी टीप
For K_sp questions, first write the dissociation equation and the K_sp expression, then put S with the correct powers. Show log steps clearly in pH problems.
Thermodynamics (ऊष्मागतिकी) studies energy changes in physical and chemical processes. This chapter covers system and surroundings, work and heat, the first law, enthalpy and thermochemistry (Hess's law), and then entropy and Gibbs energy to predict whether a process is spontaneous.
महत्त्वाचे मुद्दे
System is the part under study; everything else is surroundings. Systems are open (exchange matter and energy), closed (only energy) or isolated (neither).
State functions depend only on initial and final states (U, H, S, G, P, V, T). Path functions depend on the path (work W and heat Q).
Sign convention (IUPAC, used in the textbook): heat absorbed by system Q is positive; work done ON the system W is positive; work done BY the system is negative.
Expansion against zero external pressure (free expansion) does no work.
Maximum work is obtained in a reversible isothermal expansion.
First law: energy can neither be created nor destroyed; ΔU = Q + W.
At constant volume ΔU = Q_v; at constant pressure ΔH = Q_p.
Exothermic reactions have ΔH negative; endothermic have ΔH positive.
Hess's law: the total enthalpy change of a reaction is the same whether it occurs in one step or many steps.
Second law: the entropy of the universe increases in a spontaneous process.
A process is spontaneous when ΔG < 0, at equilibrium when ΔG = 0, non-spontaneous when ΔG > 0.
महत्त्वाच्या संज्ञा
Internal energy (U) – total energy stored in a system.
Enthalpy (H) – H = U + PV, the heat content at constant pressure.
Standard enthalpy of formation (ΔfH°) – enthalpy change when 1 mol of a compound forms from its elements in their standard states (zero for elements).
Enthalpy of combustion – enthalpy change when 1 mol of a substance burns completely in oxygen.
Bond enthalpy – energy needed to break 1 mol of a particular bond in gaseous molecules.
Entropy (S) – a measure of disorder or randomness.
Gibbs energy (G) – G = H − TS; energy available to do useful work.
ΔH = ΔU + Δn_g RT (Δn_g = moles of gaseous products − moles of gaseous reactants; R = 8.314 J K⁻¹ mol⁻¹)
ΔrH° = Σ ΔfH°(products) − Σ ΔfH°(reactants)
ΔrH° = Σ bond enthalpies of reactants − Σ bond enthalpies of products
ΔS = Q_rev/T (J K⁻¹)
ΔG = ΔH − TΔS ; ΔG° = −2.303 RT log₁₀ K
Temperature of equilibrium: T = ΔH/ΔS
ΔH
ΔS
Spontaneous?
−
+
at all temperatures
+
−
never
−
−
at low temperature
+
+
at high temperature
५ महत्त्वाचे प्रश्न
Q1. Distinguish between state function and path function.
A state function depends only on the initial and final states of the system (e.g. U, H). A path function depends on the route taken between the states (e.g. W, Q).
Q2. 2 mol of an ideal gas expands isothermally and reversibly from 10 L to 20 L at 300 K. Find the work done.
H = U + PV, so ΔH = ΔU + Δ(PV). For ideal gases PV = nRT, so at constant T, Δ(PV) = Δn_g RT. Hence ΔH = ΔU + Δn_g RT.
Q4. State Hess's law and give one use.
The overall enthalpy change of a reaction is the same whether it takes place in one step or several steps. It is used to calculate enthalpies that cannot be measured directly, e.g. ΔfH of CO from combustion data of C and CO.
Q5. For a reaction ΔH = +30 kJ and ΔS = +100 J K⁻¹. Above what temperature is it spontaneous?
T = ΔH/ΔS = 30000/100 = 300 K. Above 300 K, TΔS > ΔH and ΔG becomes negative, so the reaction is spontaneous.
परीक्षेसाठी टीप
Always convert kJ to J (or J to kJ) so that ΔH and TΔS have the same unit before calculating ΔG. Watch the sign of work.
Electrochemistry links chemical reactions with electricity. This chapter covers conductance of electrolyte solutions, Kohlrausch's law, electrolytic cells and Faraday's laws, galvanic cells, electrode potentials and the Nernst equation, the link between cell potential and Gibbs energy, and everyday batteries and fuel cells.
महत्त्वाचे मुद्दे
Metallic conduction is by electrons; electrolytic conduction is by movement of ions.
Conductivity (κ) of a solution decreases on dilution (fewer ions per cm³), but molar conductivity (Λ) increases on dilution.
For strong electrolytes Λ increases slightly and linearly with √c; for weak electrolytes Λ rises sharply at high dilution, so Λ₀ of weak electrolytes is found using Kohlrausch's law.
Electrolytic cell: electrical energy → chemical change (non-spontaneous). Galvanic cell: chemical change → electrical energy (spontaneous).
In every cell, oxidation occurs at the anode and reduction at the cathode. In a galvanic cell the anode is negative; in an electrolytic cell the anode is positive.
Electrolysis of molten NaCl gives Na at cathode and Cl₂ at anode; aqueous NaCl gives H₂ at cathode and Cl₂ at anode (NaOH left in solution).
A salt bridge (KCl or KNO₃ in agar) completes the circuit and keeps the solutions electrically neutral.
Cell notation: anode | anode solution || cathode solution | cathode, e.g. Zn | Zn²⁺(1 M) || Cu²⁺(1 M) | Cu.
The standard hydrogen electrode (SHE) is the reference with E° = 0.000 V.
A higher (more positive) reduction potential means a stronger oxidising agent.
Daily-life cells: dry cell or Leclanché cell (Zn container as anode, graphite rod as cathode), lead storage battery (Pb anode, PbO₂ cathode, H₂SO₄, about 2 V per cell, rechargeable), NICAD cell, mercury battery (about 1.35 V), H₂–O₂ fuel cell.
महत्त्वाच्या संज्ञा
Conductivity (κ) – conductance of a 1 cm cube of solution (S cm⁻¹).
Molar conductivity (Λ) – conductance of the solution containing 1 mol of electrolyte (S cm² mol⁻¹).
Cell constant – l/a, ratio of distance between electrodes to their area (cm⁻¹).
Kohlrausch's law – at infinite dilution, each ion contributes independently to the molar conductivity.
Faraday (F) – charge on 1 mol of electrons = 96500 C mol⁻¹.
Electrochemical series – electrodes arranged in order of their standard reduction potentials.
Fuel cell – a galvanic cell that turns the energy of a fuel (like H₂) directly into electricity.
सूत्रे
κ = (1/R) × (l/a) = cell constant / R
Λ = (1000 × κ)/c (κ in S cm⁻¹, c in mol L⁻¹, Λ in S cm² mol⁻¹)
Kohlrausch: Λ₀ = x λ₀(+) + y λ₀(−) ; degree of dissociation α = Λ/Λ₀
Charge: Q = I × t (C = A × s); moles of electrons = Q/F; mass deposited W = (Q/F) × (molar mass / n)
Q1. Why does molar conductivity increase on dilution?
On dilution the volume containing 1 mol of electrolyte increases. Ions move more freely (less interionic attraction) and for weak electrolytes the degree of dissociation increases. So Λ increases.
Q2. A conductivity cell has cell constant 1.25 cm⁻¹. Resistance of 0.01 M KCl is 500 Ω. Find κ and Λ.
κ = cell constant / R = 1.25/500 = 2.5 × 10⁻³ S cm⁻¹. Λ = 1000κ/c = (1000 × 2.5 × 10⁻³)/0.01 = 250 S cm² mol⁻¹.
Q3. How much charge is needed to deposit 1 mol of Cu from Cu²⁺?
Chemical kinetics is the study of rates of reactions and the factors that change them. This chapter covers average and instantaneous rate, rate law, order and molecularity, integrated rate laws for zero and first order reactions, half-life, collision theory, the Arrhenius equation and the effect of a catalyst.
महत्त्वाचे मुद्दे
Rate of reaction = decrease in reactant concentration (or increase in product concentration) per unit time. Unit: mol L⁻¹ s⁻¹.
The rate law is found by experiment, not from the balanced equation.
Order = sum of powers of concentration terms in the rate law; it can be 0, a whole number or a fraction.
Molecularity = number of reactant molecules taking part in an elementary step; always a whole number (1, 2 or rarely 3), never zero.
For complex reactions, the slowest step is the rate determining step.
For a first order reaction, half-life does not depend on initial concentration. Examples: radioactive decay, decomposition of N₂O₅, H₂O₂.
For a zero order reaction, rate does not depend on concentration; half-life is proportional to initial concentration. Example: decomposition of NH₃ on hot platinum.
Pseudo-first order reactions are really higher order, but one reactant is in large excess (acid hydrolysis of ethyl acetate, inversion of cane sugar).
Collision theory: only collisions with energy ≥ threshold energy and proper orientation give products.
Rate usually roughly doubles for a 10 °C rise in temperature, because more molecules cross the activation energy.
A catalyst gives a new path with lower activation energy; it does not change ΔH or the equilibrium constant.
महत्त्वाच्या संज्ञा
Rate constant (k) – rate of reaction when all concentrations are 1 mol L⁻¹.
Half-life (t½) – time for the reactant concentration to become half.
Activation energy (Ea) – minimum extra energy that reactants must gain to form the activated complex.
Activated complex – unstable high-energy arrangement at the top of the energy barrier.
Elementary reaction – a reaction that occurs in a single step.
सूत्रे
Rate law: rate = k[A]ˣ[B]ʸ ; order = x + y
Units of k: zero order mol L⁻¹ s⁻¹; first order s⁻¹; second order L mol⁻¹ s⁻¹
First order: k = (2.303/t) log₁₀([A]₀/[A]t) ; t½ = 0.693/k
Zero order: [A]t = [A]₀ − kt ; t½ = [A]₀/(2k)
Gas phase first order (A → B + C): k = (2.303/t) log₁₀(P₀/(2P₀ − P))
Arrhenius equation: k = A e^(−Ea/RT) ; log₁₀ k = log₁₀ A − Ea/(2.303RT)
Two temperatures: log₁₀(k₂/k₁) = (Ea/(2.303R)) × ((T₂ − T₁)/(T₁T₂))
Plot of log k vs 1/T is a straight line, slope = −Ea/(2.303R).
५ महत्त्वाचे प्रश्न
Q1. Distinguish between order and molecularity.
Order is found experimentally and can be zero or fractional; it applies to the overall reaction. Molecularity is theoretical, always a whole number (never zero) and applies to an elementary step.
Q2. A first order reaction has k = 6.93 × 10⁻³ s⁻¹. Find its half-life.
t½ = 0.693/k = 0.693/(6.93 × 10⁻³) = 100 s.
Q3. A first order reaction is 75% complete in 60 min. Find k.
Q5. What is a pseudo-first order reaction? Give an example.
A reaction whose true order is more than one but which behaves as first order because one reactant is in large excess. Example: CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH (in acid, water in excess); rate = k'[ester].
परीक्षेसाठी टीप
Learn "75% complete = two half-lives" and "99.9% complete ≈ 10 half-lives" for first order reactions — they save time in MCQs.
This chapter studies the p-block elements of group 16 (oxygen family), group 17 (halogens) and group 18 (noble gases). It covers their electronic configuration, trends in properties, anomalous behaviour of oxygen and fluorine, allotropes of sulfur, oxoacids, and important compounds such as ozone, SO₂, H₂SO₄, chlorine, HCl, interhalogens and xenon compounds.
महत्त्वाचे मुद्दे
Group 16: O, S, Se, Te, Po (ns² np⁴). Group 17: F, Cl, Br, I, At (ns² np⁵). Group 18: He, Ne, Ar, Kr, Xe, Rn (ns² np⁶; He is 1s²).
Down a group: atomic size increases, ionisation enthalpy and electronegativity decrease, metallic character increases.
Oxygen is anomalous: small size, high electronegativity, no d-orbitals, forms hydrogen bonds (H₂O is liquid, H₂S is gas).
Fluorine is anomalous: most electronegative, shows only −1 oxidation state, F–F bond is weak due to repulsion between lone pairs.
Sulfur has two main allotropes: rhombic (α) stable below 369 K and monoclinic (β) stable above 369 K.
Acid strength of oxoacids of chlorine: HOCl < HClO₂ < HClO₃ < HClO₄ (more O atoms → more stable anion).
Ozone (O₃) is made by passing silent electric discharge through dry O₂; it is a strong oxidising agent and protects us from UV rays.
H₂SO₄ is made by the contact process: SO₂ is oxidised to SO₃ over V₂O₅ catalyst (about 720 K, 2 bar); SO₃ is absorbed in H₂SO₄ to give oleum (H₂S₂O₇), which is diluted.
Chlorine bleaches by oxidation (moist Cl₂ gives nascent oxygen) — the bleaching is permanent.
Interhalogens are of types AX, AX₃, AX₅ and AX₇ (ClF, ClF₃, BrF₅, IF₇); they are more reactive than halogens (except F₂).
Noble gases are inert because of a stable full octet; Xe forms compounds with F and O. Neil Bartlett made the first noble gas compound (Xe⁺[PtF₆]⁻).
महत्त्वाच्या संज्ञा
Allotropy – existence of an element in two or more forms with different physical properties.
Oleum – fuming sulphuric acid, H₂S₂O₇.
Aqua regia – 3 parts conc. HCl + 1 part conc. HNO₃; dissolves gold and platinum.
Interhalogen compound – compound formed between two different halogens.
Transition temperature – temperature at which one allotrope changes into another (369 K for sulfur).
Because of its small size, high electronegativity and absence of d-orbitals. So it forms strong hydrogen bonds and pπ–pπ double bonds (O₂ is a gas), and its covalency is limited to 4.
Q2. Explain the manufacture of sulphuric acid by the contact process.
Sulfur is burnt to SO₂. Purified SO₂ is oxidised by air to SO₃ over V₂O₅ catalyst at about 720 K and 2 bar. SO₃ is absorbed in conc. H₂SO₄ to form oleum, which is diluted with water to get H₂SO₄ of required strength.
Q3. Why is HClO₄ a stronger acid than HOCl?
In ClO₄⁻ the negative charge is spread over four oxygen atoms, so it is very stable. A more stable conjugate base means a stronger acid.
Q4. Draw and name the shape of XeF₄.
Xe has 4 bond pairs and 2 lone pairs (sp³d²). The two lone pairs sit opposite each other, so the four F atoms lie in a plane: square planar.
Q5. What happens when chlorine reacts with excess ammonia and with excess chlorine?
Transition (d-block) elements fill their (n−1)d orbitals; inner transition (f-block) elements fill 4f (lanthanoids) or 5f (actinoids) orbitals. This chapter covers electronic configuration, oxidation states, trends in properties, colour and magnetism, KMnO₄ and K₂Cr₂O₇, extraction of iron, and the lanthanoids and actinoids.
महत्त्वाचे मुद्दे
d-block = groups 3 to 12. General configuration: (n−1)d¹⁻¹⁰ ns⁰⁻².
Cr is 3d⁵4s¹ and Cu is 3d¹⁰4s¹ because half-filled and fully-filled d subshells are extra stable.
Transition metals show variable oxidation states (Mn shows +2 to +7) because 3d and 4s energies are close.
Most are hard, have high melting points (strong metallic bonding due to unpaired d electrons) and are good conductors.
Their ions are coloured due to d–d transitions of unpaired electrons; Sc³⁺, Ti⁴⁺, Zn²⁺, Cu⁺ (d⁰ or d¹⁰) are colourless.
Ions with unpaired electrons are paramagnetic.
They are good catalysts (Fe in Haber process, V₂O₅ in contact process, Ni in hydrogenation) and form interstitial compounds and alloys.
KMnO₄: MnO₂ is fused with KOH and an oxidising agent (KClO₃) to give dark green K₂MnO₄, which is oxidised to purple KMnO₄. It is a strong oxidising agent (in acid, MnO₄⁻ → Mn²⁺).
K₂Cr₂O₇: from chromite ore (FeCr₂O₄) → Na₂CrO₄ → Na₂Cr₂O₇ → K₂Cr₂O₇ (with KCl). Chromate (yellow) and dichromate (orange) change into each other with pH.
Iron is extracted from haematite (Fe₂O₃) in a blast furnace using coke and limestone; CO is the main reducing agent; CaSiO₃ is the slag.
Lanthanoids: Ce (58) to Lu (71); common oxidation state +3. Lanthanoid contraction is the steady decrease in size due to poor shielding by 4f electrons.
Actinoids: Th (90) to Lr (103); all radioactive; show more oxidation states than lanthanoids.
महत्त्वाच्या संज्ञा
Transition element – element with partly filled d-orbitals in the atom or in its common ion.
Lanthanoid contraction – decrease in atomic/ionic radii from La to Lu.
Interstitial compound – small atoms (H, C, N) trapped in the metal lattice, e.g. TiC, Fe₃H.
Alloy – mixture of metals (or metal + non-metal), e.g. brass (Cu + Zn), bronze (Cu + Sn).
Slag – fusible waste formed from impurities and flux (CaSiO₃).
Pig iron – impure iron from blast furnace with about 4% carbon.
सूत्रे
Spin-only magnetic moment: μ = √(n(n + 2)) BM, n = number of unpaired electrons.
In the presence of ligands, d-orbitals split into groups of different energy. An unpaired d electron absorbs visible light and jumps to the higher level (d–d transition). The colour we see is the complementary colour of the light absorbed.
Q2. Calculate the spin-only magnetic moment of Fe²⁺ (Z = 26).
Q3. What is lanthanoid contraction? Give one consequence.
The steady decrease in size of atoms and ions from La to Lu, because 4f electrons shield the nuclear charge poorly. Consequence: Zr and Hf have almost the same radius, so they are very hard to separate.
Q4. How is potassium dichromate prepared from chromite ore?
Chromite is fused with Na₂CO₃ in air to give Na₂CrO₄. This is acidified with H₂SO₄ to get Na₂Cr₂O₇. Treating it with KCl gives less soluble K₂Cr₂O₇, which crystallises out.
Q5. Write the reactions in the blast furnace for extraction of iron.
C + O₂ → CO₂ ; CO₂ + C → 2CO ; Fe₂O₃ + 3CO → 2Fe + 3CO₂ ; CaCO₃ → CaO + CO₂ ; CaO + SiO₂ → CaSiO₃ (slag).
परीक्षेसाठी टीप
Write configurations of ions by removing 4s electrons first. Practise μ = √(n(n+2)) for ions of Ti to Cu — it is a sure numerical.
Coordination compounds contain a central metal atom or ion bonded to a fixed number of ligands by coordinate bonds. This chapter covers ligands, Werner's theory, IUPAC naming, the EAN rule, isomerism, stability, bonding by valence bond theory and crystal field theory, and uses of complexes.
महत्त्वाचे मुद्दे
A ligand donates an electron pair to the metal. Monodentate: Cl⁻, NH₃, H₂O, CN⁻. Bidentate: ethylenediamine (en), oxalate (ox²⁻). Hexadentate: EDTA⁴⁻.
Ambidentate ligands can bind through two different atoms: NO₂⁻ (N or O), SCN⁻ (S or N).
Werner's theory: metals have primary valence (ionisable, = oxidation state) and secondary valence (non-ionisable, = coordination number, has a fixed shape).
A double salt (Mohr's salt, potash alum) breaks fully into ions in water; a complex keeps its complex ion intact.
Complexes can be cationic [Co(NH₃)₆]³⁺, anionic [Fe(CN)₆]⁴⁻ or neutral [Ni(CO)₄]; homoleptic (one type of ligand) or heteroleptic.
Uses: EDTA to estimate hardness of water, cisplatin in cancer treatment, cyanide complexes in extraction of gold and silver, haemoglobin (Fe), chlorophyll (Mg), vitamin B₁₂ (Co).
महत्त्वाच्या संज्ञा
Coordination number – number of ligand donor atoms bonded to the metal.
Coordination sphere – the metal and ligands written inside square brackets.
Chelate – ring complex formed by a polydentate ligand; more stable (chelate effect).
Effective atomic number (EAN) – total electrons around the metal after coordination.
Crystal field splitting energy (Δ₀) – energy gap between e_g and t₂g sets.
Q2. What are ambidentate ligands? Give two examples.
Ligands with two different donor atoms that can bind through either one. NO₂⁻ (nitro when bonded through N, nitrito when bonded through O) and SCN⁻ (thiocyanato-S / isothiocyanato-N). They cause linkage isomerism.
Q3. Explain the bonding in [Co(NH₃)₆]³⁺ using VBT.
Co³⁺ is 3d⁶. NH₃ (strong field) pairs the six electrons in three 3d orbitals. Two empty 3d, one 4s and three 4p orbitals form six d²sp³ hybrids which accept six lone pairs from NH₃. Octahedral, inner orbital, diamagnetic.
Q4. Calculate the EAN of Co in [Co(NH₃)₆]³⁺.
EAN = 27 − 3 + 2 × 6 = 36.
Q5. Show geometrical isomers of [Pt(NH₃)₂Cl₂].
It is square planar. In the cis isomer the two Cl are on the same side (adjacent); in the trans isomer they are opposite. The cis isomer (cisplatin) is an anticancer drug.
परीक्षेसाठी टीप
For VBT questions, write the electron boxes for the free ion and then after pairing — examiners give marks for the box diagram and the hybridisation.
Halogen derivatives are compounds in which one or more hydrogen atoms of a hydrocarbon are replaced by halogen atoms (haloalkanes and haloarenes). This chapter covers their classification, naming, preparation, physical properties, optical isomerism, nucleophilic substitution (SN1 and SN2) and elimination reactions, reactions with metals, and the uses and harmful effects of some polyhalogen compounds.
महत्त्वाचे मुद्दे
Haloalkanes are classified as primary (1°), secondary (2°) or tertiary (3°) depending on the carbon carrying the halogen.
Preparation from alcohols: with HX (ZnCl₂ for HCl), PCl₃, PCl₅ or SOCl₂. SOCl₂ is best because the by-products SO₂ and HCl are gases and escape.
From alkenes: HX adds by Markovnikov's rule; HBr with peroxide adds anti-Markovnikov (peroxide effect).
Boiling point increases with size of halogen (R–I > R–Br > R–Cl > R–F) and decreases with branching.
A carbon with four different groups is chiral; such molecules show optical isomerism (dextrorotatory d/+, laevorotatory l/−). A 1:1 mixture of enantiomers is a racemic mixture (optically inactive).
SN2: one step, second order, favoured by 1° halides, gives inversion of configuration.
SN1: two steps through a carbocation, first order, favoured by 3° halides and polar protic solvents, gives racemisation.
Elimination with alcoholic KOH gives alkenes; the more substituted alkene is the main product (Saytzeff rule).
Haloarenes are less reactive towards nucleophiles because the C–X bond has partial double bond character (resonance) and the carbon is sp².
Reactions with metals: Wurtz (2RX + 2Na → R–R), Fittig, Wurtz–Fittig, and Grignard reagent (RX + Mg in dry ether → RMgX).
महत्त्वाच्या संज्ञा
Nucleophile – electron-rich species that attacks a positive carbon (OH⁻, CN⁻, NH₃).
Q1. Why is thionyl chloride preferred for converting alcohols to chloroalkanes?
The by-products SO₂ and HCl are gases and escape, so pure chloroalkane is obtained easily.
Q2. Explain the SN2 mechanism with an example.
CH₃Br + OH⁻ → CH₃OH + Br⁻. OH⁻ attacks carbon from the side opposite to Br. In the transition state C–OH forms while C–Br breaks. Product forms with inversion of configuration. Rate = k[CH₃Br][OH⁻].
Q3. What is a chiral carbon? Identify it in butan-2-ol.
A carbon attached to four different groups. In CH₃–CH(OH)–CH₂–CH₃ the C-2 carbon carries H, OH, CH₃ and C₂H₅, so it is chiral.
Q4. Why are haloarenes less reactive than haloalkanes towards nucleophilic substitution?
Resonance gives the C–X bond partial double bond character; the carbon is sp² (shorter, stronger bond); and the electron-rich ring repels the nucleophile.
Q5. State Saytzeff's rule with an example.
In dehydrohalogenation, the more highly substituted alkene is the major product. 2-Bromobutane with alc. KOH gives mainly but-2-ene.
परीक्षेसाठी टीप
Learn reagents with names: Finkelstein (NaI/acetone), Swarts (AgF), Sandmeyer (CuCl/HCl), Wurtz (Na/dry ether) — conversions carry 2–3 marks each.
Alcohols (R–OH) and phenols (Ar–OH) contain a hydroxyl group; ethers (R–O–R') contain an oxygen between two carbon groups. This chapter covers their classification, IUPAC names, preparation, physical properties, chemical reactions, tests and uses.
महत्त्वाचे मुद्दे
Alcohols are monohydric, dihydric (glycol) or trihydric (glycerol); also 1°, 2° or 3°.
Alcohol preparation: hydration of alkenes (acid), hydroboration–oxidation (anti-Markovnikov product), reduction of aldehydes/ketones (NaBH₄, LiAlH₄), and Grignard reagent with carbonyl compounds.
Phenol preparation: from chlorobenzene (NaOH, 623 K, 300 atm), from benzene sulfonic acid (fuse with NaOH), from diazonium salt (warm with water) and from cumene (air oxidation, then dilute acid; acetone is a by-product).
Alcohols and phenols have high boiling points and some water solubility due to hydrogen bonding.
Acidity order: phenol > water > alcohol. Phenoxide ion is stabilised by resonance. Electron-withdrawing groups (–NO₂) increase acidity of phenol.
Lucas test (anhydrous ZnCl₂ + conc. HCl): 3° alcohol turbid at once, 2° in about 5 minutes, 1° no turbidity at room temperature.
Dehydration with conc. H₂SO₄ at 443 K gives alkene; at 413 K with excess alcohol gives ether.
Phenol reactions: with bromine water gives white precipitate of 2,4,6-tribromophenol; Kolbe's reaction gives salicylic acid; Reimer–Tiemann reaction gives salicylaldehyde; neutral FeCl₃ gives violet colour.
Ethers are made by Williamson synthesis (R–ONa + R'–X → R–O–R'). They are cleaved by HI. Anisole gives ortho and para products in electrophilic substitution.
महत्त्वाच्या संज्ञा
Hydroboration–oxidation – addition of B₂H₆ to alkene followed by H₂O₂/NaOH, giving alcohol.
Denatured spirit – ethanol made unfit to drink by adding methanol/pyridine.
Phenoxide ion – C₆H₅O⁻, formed when phenol loses H⁺.
Symmetrical ether – both groups same (C₂H₅OC₂H₅); unsymmetrical (mixed) – groups differ (CH₃OC₂H₅).
Phenoxide ion is stabilised by resonance (negative charge spread over the ring). Ethoxide ion has no resonance and the ethyl group's +I effect makes it less stable. So phenol loses H⁺ more easily.
Q2. How will you distinguish 1°, 2° and 3° alcohols by the Lucas test?
Shake the alcohol with Lucas reagent at room temperature. 3° gives turbidity immediately, 2° in about 5 minutes, 1° does not give turbidity at room temperature (only on heating).
Q3. Write the preparation of phenol from cumene.
Cumene (isopropylbenzene) is oxidised by air to cumene hydroperoxide. This is treated with dilute acid to give phenol and acetone.
Q4. Explain Williamson synthesis. Why is a 1° alkyl halide used?
Sodium alkoxide reacts with an alkyl halide by SN2 to give an ether. With 3° alkyl halides the alkoxide (a strong base) causes elimination and gives alkene instead, so 1° halides are used.
Q5. What is Kolbe's reaction?
Sodium phenoxide is heated with CO₂ under pressure (about 400 K, 4–7 atm) to give sodium salicylate, which on acidification gives salicylic acid (o-hydroxybenzoic acid), used to make aspirin.
परीक्षेसाठी टीप
Make a one-page chart of name reactions (Kolbe, Reimer–Tiemann, Williamson) with reagents and conditions — conversions are guaranteed questions.
Aldehydes (R–CHO) and ketones (R–CO–R') contain the carbonyl group (C=O); carboxylic acids contain –COOH. This chapter covers their naming, preparation, physical properties, the reactivity of the carbonyl group, important name reactions, laboratory tests, and the acidic nature of carboxylic acids.
महत्त्वाचे मुद्दे
The C=O bond is polar: carbon is partly positive, so carbonyl compounds undergo nucleophilic addition.
Aldehydes are more reactive than ketones (less steric hindrance and less +I effect).
Preparation: oxidation of 1° alcohols (aldehydes) and 2° alcohols (ketones); ozonolysis of alkenes; hydration of alkynes.
Carboxylic acids are made by hydrolysis of nitriles, amides, esters, by oxidation of alkylbenzenes (KMnO₄) and from Grignard reagent + CO₂.
Tests: aldehydes give silver mirror with Tollens' reagent and red precipitate (Cu₂O) with Fehling's solution; ketones do not. Both give orange/yellow precipitate with 2,4-DNP. Methyl ketones give yellow iodoform.
Aldol condensation: aldehydes/ketones with α-H in dilute NaOH give β-hydroxy carbonyl compounds.
Cannizzaro reaction: aldehydes with no α-H (HCHO, C₆H₅CHO) in conc. NaOH give an alcohol and a carboxylic acid salt.
Clemmensen (Zn-Hg/conc. HCl) and Wolff–Kishner (NH₂NH₂, KOH, ethylene glycol) reduce C=O to CH₂.
Carboxylic acids are stronger acids than phenols and alcohols; they give brisk effervescence of CO₂ with NaHCO₃. Electron-withdrawing groups increase acidity (ClCH₂COOH > CH₃COOH).
Decarboxylation: sodium salt heated with soda lime (NaOH + CaO) gives an alkane with one carbon less.
महत्त्वाच्या संज्ञा
α-hydrogen – hydrogen on the carbon next to the carbonyl group.
Nucleophilic addition – a nucleophile adds to the carbonyl carbon, H⁺ to oxygen.
Carboxylate ion – RCOO⁻, stabilised by resonance.
Iodoform test – compounds with CH₃CO– (or CH₃CH(OH)–) give yellow CHI₃ with I₂/NaOH.
Q1. Why are aldehydes more reactive than ketones towards nucleophilic addition?
Ketones have two alkyl groups that push electrons (+I) and reduce the positive charge on carbon, and they also crowd the carbonyl carbon (steric hindrance). Aldehydes have only one, so they react faster.
Q2. What is the Cannizzaro reaction? Which aldehydes give it?
Aldehydes without α-hydrogen, heated with conc. NaOH, undergo self oxidation-reduction: one molecule becomes an alcohol and another a carboxylate salt. Example: 2C₆H₅CHO + NaOH → C₆H₅CH₂OH + C₆H₅COONa.
Q3. How will you distinguish ethanal and propanone?
Tollens' test: ethanal gives a silver mirror; propanone does not. (Fehling's solution: ethanal gives red ppt; propanone does not.)
Q4. Explain aldol condensation with ethanal.
Two molecules of ethanal in dilute NaOH join to form 3-hydroxybutanal (aldol). On heating it loses water to form but-2-enal.
Q5. Why is chloroacetic acid stronger than acetic acid?
The –Cl atom pulls electrons (–I effect), spreading the negative charge of the carboxylate ion and making it more stable. So ClCH₂COOH releases H⁺ more easily.
परीक्षेसाठी टीप
Learn tests in pairs (aldehyde vs ketone, acid vs phenol) — "distinguish between" questions are easy marks if you write the reagent and the observation.
Amines are organic derivatives of ammonia in which one, two or three hydrogens are replaced by alkyl or aryl groups. This chapter covers their classification, naming, preparation, physical properties, basic strength, chemical tests, and the very useful reactions of benzene diazonium salts.
महत्त्वाचे मुद्दे
Classification: primary (RNH₂), secondary (R₂NH) and tertiary (R₃N) amines; aliphatic or aromatic (aniline, C₆H₅NH₂).
Preparation: reduction of nitro compounds (Sn/HCl or Fe/HCl), reduction of nitriles and amides (LiAlH₄), ammonolysis of alkyl halides, Gabriel phthalimide synthesis (pure 1° aliphatic amines only) and Hofmann bromamide degradation (amide → amine with one carbon less).
1° and 2° amines form hydrogen bonds; 3° amines cannot (no H on N), so boiling points: 1° > 2° > 3°.
Amines are basic because of the lone pair on nitrogen.
In water, the observed order is NH₃ < 1° amine < 2° amine, and 3° amine is weaker than 2° — this comes from the combined effect of +I effect, solvation (hydrogen bonding) and crowding.
Aniline is a weaker base than ammonia because its lone pair is delocalised into the benzene ring.
Carbylamine test: 1° amines (aliphatic and aromatic) + CHCl₃ + alcoholic KOH give foul-smelling isocyanide.
Hinsberg test (benzenesulfonyl chloride): 1° gives a sulfonamide soluble in alkali; 2° gives one insoluble in alkali; 3° does not react.
With nitrous acid (NaNO₂ + HCl): 1° aliphatic amines give alcohol + N₂ gas; aniline at 273–278 K gives benzene diazonium chloride.
Diazonium salt reactions: Sandmeyer (CuCl, CuBr, CuCN), with KI (iodobenzene), with HBF₄ (fluorobenzene), with H₃PO₂ (benzene), warming with water (phenol), coupling with phenol (orange dye) or aniline (yellow dye).
Aniline with bromine water gives a white precipitate of 2,4,6-tribromoaniline. Acetylation (acetanilide) reduces the activating effect of –NH₂.
महत्त्वाच्या संज्ञा
Diazotization – conversion of a 1° aromatic amine into a diazonium salt with NaNO₂/HCl at 273–278 K.
In aniline the lone pair on N is spread into the benzene ring by resonance, so it is less available to accept a proton. Ammonia's lone pair is fully available.
Q2. Describe the Hofmann bromamide degradation with an example.
An amide is warmed with Br₂ and KOH to give a 1° amine with one carbon atom less. Example: CH₃CONH₂ → CH₃NH₂ (acetamide to methylamine).
Q3. How will you distinguish 1°, 2° and 3° amines by Hinsberg's reagent?
1° amine forms a sulfonamide that dissolves in KOH. 2° amine forms a sulfonamide that does not dissolve in KOH. 3° amine does not react.
Q4. Why can't Gabriel phthalimide synthesis make aniline?
It needs SN2 attack of phthalimide ion on an alkyl halide. Aryl halides do not undergo this substitution easily, so aromatic amines cannot be made this way.
Q5. Convert aniline into chlorobenzene.
Diazotize aniline with NaNO₂ + HCl at 273–278 K to get benzene diazonium chloride. Treat with CuCl/HCl (Sandmeyer) to get chlorobenzene and N₂.
परीक्षेसाठी टीप
Remember: carbylamine test works only for 1° amines; temperature 273–278 K is essential in diazotization because diazonium salts decompose when warm.
Biomolecules are the organic molecules of living things. This chapter studies carbohydrates (glucose, fructose, sugars, starch, cellulose), proteins (amino acids, peptide bond, structure, enzymes) and nucleic acids (DNA and RNA).
महत्त्वाचे मुद्दे
Carbohydrates are polyhydroxy aldehydes or ketones. Types: monosaccharides (glucose, fructose), oligosaccharides/disaccharides (sucrose, maltose, lactose) and polysaccharides (starch, cellulose, glycogen).
Glucose (C₆H₁₂O₆) is an aldohexose; fructose is a ketohexose.
Evidence for glucose's open-chain structure: with HI it gives n-hexane (straight 6-C chain); forms pentaacetate (5 –OH groups); reacts with HCN and NH₂OH (carbonyl group); bromine water oxidises it to gluconic acid (–CHO group).
Glucose exists mainly in a six-membered ring (pyranose) form, as α and β anomers that differ at C-1. Fructose forms a five-membered ring (furanose).
Glucose and fructose reduce Tollens' and Fehling's reagents (reducing sugars).
Q1. What evidence shows that glucose has an aldehyde group?
Glucose is oxidised by mild bromine water to gluconic acid having the same number of carbons. This shows the carbonyl is an aldehyde (–CHO). It also reduces Tollens' and Fehling's reagents.
Q2. Why is sucrose a non-reducing sugar?
In sucrose, C-1 of glucose and C-2 of fructose are both used in the glycosidic bond, so no free (potential) aldehyde or ketone group is left to reduce Fehling's or Tollens' reagent.
Q3. Explain the secondary structure of proteins.
It is the regular folding of the peptide chain held by hydrogen bonds between C=O and N–H groups. α-helix: chain coils like a spring with H-bonds within the chain. β-pleated sheet: chains lie side by side joined by H-bonds between chains.
Q4. What is denaturation? Give an example.
Loss of a protein's natural 3-D shape and biological activity by heat, pH change or chemicals. Hydrogen bonds break but peptide bonds stay. Example: boiling an egg makes the white albumin solid.
Q5. Give two differences between DNA and RNA.
DNA has deoxyribose sugar and base thymine; RNA has ribose sugar and base uracil. DNA is double stranded; RNA is usually single stranded.
परीक्षेसाठी टीप
Practise drawing Haworth ring structures of α-D-glucose and sucrose, and writing a dipeptide from two amino acids — these are frequent 2–3 mark questions.
Polymers are very large molecules (macromolecules) built from many small repeating units called monomers. This chapter covers the classification of polymers, addition and condensation polymerization, important polymers like rubber, polythene, teflon, nylon, terylene, bakelite and Buna-S, degree of polymerization, and biodegradable polymers.
By polymerization: addition (from unsaturated monomers, no by-product; polythene, PVC, teflon) and condensation (with loss of small molecules like H₂O; nylon, terylene, bakelite).
By number of monomers: homopolymer (one monomer) and copolymer (two or more, e.g. Buna-S, nylon 6,6).
Natural rubber is cis-1,4-polyisoprene. Vulcanisation (heating with sulfur) forms S cross-links, making rubber harder, stronger and less sticky.
LDP: ethene at 1000–2000 atm and 350–570 K with traces of O₂/peroxide (free radical, branched). HDP: ethene with Ziegler–Natta catalyst (Al(C₂H₅)₃ + TiCl₄) at 333–343 K and 6–7 atm (linear, tougher).
Biodegradable polymers are broken down by microbes: PHBV and nylon-2-nylon-6.
Degree of polymerization (DP) = number of monomer units in a polymer molecule. Polymers with H-bonding (nylon 6) need a lower critical DP than those with weak forces (polythene).
महत्त्वाच्या संज्ञा
Monomer – small molecule that joins to form a polymer.
Elastomer – rubber-like polymer that stretches and returns to shape.
Thermosetting polymer – sets permanently on heating, cannot be remoulded.
Copolymer – polymer made from two or more different monomers.
Vulcanisation – heating rubber with sulfur to cross-link the chains.
सूत्रे
Polymer
Monomer(s)
Type
Use
Teflon
Tetrafluoroethylene
Addition
Non-stick pans
Polyacrylonitrile (Orlon)
Acrylonitrile
Addition
Wool substitute
Buna-S
1,3-Butadiene + styrene
Addition copolymer
Tyres, soles
Neoprene
Chloroprene
Addition
Hoses, belts
Nylon 6,6
Hexamethylenediamine + adipic acid
Condensation
Fibres, ropes
Nylon 6
Caprolactam
Condensation (ring opening)
Tyre cords
Terylene (Dacron)
Ethylene glycol + terephthalic acid
Condensation
Clothes
Bakelite
Phenol + formaldehyde
Condensation
Switches, handles
PHBV
3-Hydroxybutanoic acid + 3-hydroxypentanoic acid
Condensation
Packaging, capsules
Nylon-2-nylon-6
Glycine + ε-aminocaproic acid
Condensation
Biodegradable
५ महत्त्वाचे प्रश्न
Q1. Distinguish between addition and condensation polymers.
Addition: formed from unsaturated monomers by repeated addition, no small molecule lost (polythene). Condensation: formed by reaction between two functional groups with loss of small molecules like water (nylon 6,6).
Q2. What is vulcanisation of rubber? Why is it done?
Heating natural rubber with sulfur. Sulfur forms cross-links between chains, so rubber becomes harder, more elastic, less sticky and works over a wider temperature range.
Q3. How is HDP prepared? Write one use.
Ethene is polymerised using Ziegler–Natta catalyst (triethylaluminium + TiCl₄) at 333–343 K and 6–7 atm. It gives a linear, closely packed polymer used for buckets, bottles and pipes.
Q4. Name the monomers of nylon 6,6 and write the reaction.
Hexamethylenediamine H₂N(CH₂)₆NH₂ and adipic acid HOOC(CH₂)₄COOH. They condense with loss of water to form polyamide –[NH(CH₂)₆NH–CO(CH₂)₄CO]ₙ–.
Q5. What are biodegradable polymers? Give two examples.
Polymers that are broken down by microorganisms into harmless products, so they reduce pollution. Examples: PHBV and nylon-2-nylon-6.
परीक्षेसाठी टीप
Learn the monomer table above thoroughly — "name the monomers" and "classify as addition/condensation" are common one-mark and MCQ questions.
Green chemistry is about designing chemical products and processes that reduce or remove hazardous substances and waste, supporting sustainable development. Nanochemistry studies materials at the nanoscale (1–100 nm), where they show special properties. This chapter covers the twelve principles of green chemistry, properties and synthesis of nanoparticles, instruments to study them, and their applications.
महत्त्वाचे मुद्दे
Sustainable development means meeting today's needs without harming the ability of future generations to meet theirs.
The term green chemistry and its twelve principles were given by Paul T. Anastas.
Main principles: prevent waste; atom economy; less hazardous synthesis; design safer chemicals; safer solvents (water, supercritical CO₂); energy efficiency (room temperature and pressure); renewable feedstock; reduce derivatives; catalysis instead of excess reagents; design for degradation; real-time analysis to prevent pollution; safer chemistry to prevent accidents.
Nano means 10⁻⁹. 1 nm = 10⁻⁹ m. Nanomaterials have at least one dimension in the 1–100 nm range.
The idea of nanotechnology was first presented by Richard Feynman (1959); the word "nanotechnology" was used by Norio Taniguchi (1974).
Nanoparticles have very large surface area to volume ratio, so they are excellent catalysts.
Properties such as colour change with size (gold nanoparticles give a ruby red colour, seen in old stained glass, instead of yellow); melting point decreases; mechanical strength and electrical conductivity change.
Sol-gel process (wet chemical method): hydrolysis and condensation of metal alkoxides form a sol, then a gel; drying and heating give nanoparticles.
Characterisation instruments: UV-visible spectroscopy, X-ray diffraction (XRD), scanning electron microscope (SEM), transmission electron microscope (TEM), FTIR.
Applications: medicines and drug delivery, sunscreens (ZnO, TiO₂ block UV), self-cleaning surfaces (lotus effect), catalysts, electronics, water purification (silver nanoparticles kill bacteria), cosmetics, textiles.
महत्त्वाच्या संज्ञा
Green chemistry – chemistry that prevents pollution at its source.
Atom economy – percentage of reactant atoms that end up in the desired product.
Nanomaterial – material with at least one dimension between 1 and 100 nm.
Lotus effect – self-cleaning water-repelling surface due to nanostructures, like a lotus leaf.
Sol – a colloidal suspension of solid particles in a liquid; gel – a rigid connected network.
सूत्रे
% Atom economy = (formula mass of desired product / sum of formula masses of all reactants) × 100
1 nm = 10⁻⁹ m = 10 Å
५ महत्त्वाचे प्रश्न
Q1. Define green chemistry and state any four of its principles.
Green chemistry is the design of chemical products and processes that reduce or eliminate hazardous substances. Principles: prevention of waste, atom economy, use of safer solvents and conditions, use of catalysts rather than stoichiometric reagents (any four, with one line each).
Q2. What is atom economy? Why is it important?
It is the percentage of reactant atoms converted into the desired product. High atom economy means less waste, less cost and less pollution.
Q3. Why do nanoparticles act as good catalysts?
Because of their very small size they have a very large surface area to volume ratio, so many more atoms are on the surface and available for reaction.
Q4. Describe the sol-gel method for preparing nanoparticles.
Metal alkoxide is hydrolysed in a solvent and then undergoes condensation to form a sol. The sol links into a gel network. The gel is aged, dried and heated to get nanoparticles. It works at low temperature and gives uniform particles.
Q5. Give any four applications of nanotechnology.
Targeted drug delivery in medicine; ZnO/TiO₂ nanoparticles in sunscreens; silver nanoparticles for killing bacteria in water purification; self-cleaning glass and fabrics; faster, smaller electronic devices.
परीक्षेसाठी टीप
Learn the twelve principles in order with one example each, and full forms of XRD, SEM, TEM and FTIR — they are often asked as short questions.