Adsorption is the accumulation of molecules of a substance on the surface of a solid or liquid, rather than in its bulk. This topic covers types of adsorption, factors affecting it, Freundlich and Langmuir adsorption isotherms, catalysis, and colloids with their properties and applications.
महत्त्वाचे मुद्दे
The substance that collects on the surface is the adsorbate; the surface is the adsorbent (charcoal, silica gel, alumina).
Absorption is uniform distribution throughout the bulk (water in a sponge). Sorption means both happen together.
Adsorption is exothermic (ΔH negative) and decreases entropy (ΔS negative). So it is favoured at lower temperature.
Physisorption: weak van der Waals forces, low heat (about 20–40 kJ/mol), reversible, multilayer, not specific, decreases with rise in temperature.
Chemisorption: chemical bonds form, high heat (about 80–240 kJ/mol), usually irreversible, monolayer, specific, needs activation energy (may first increase with temperature).
Factors: nature and surface area of the adsorbent, nature of gas (easily liquefiable gases with higher critical temperature, like NH₃ and SO₂, are adsorbed more), pressure and temperature.
Freundlich isotherm (empirical): x/m = k P^(1/n), with n > 1. Good at intermediate pressure, fails at high pressure.
Langmuir isotherm (theoretical): assumes a monolayer, uniform sites, no interaction between adsorbed molecules, and dynamic equilibrium between adsorption and desorption.
Catalysis: homogeneous (same phase) and heterogeneous (different phases, e.g. Fe in the Haber process). Heterogeneous catalysis works by adsorption of reactants on the surface.
Colloids: particle size about 1 nm to 1000 nm. Types: lyophilic (solvent-loving, stable, e.g. starch, gelatin) and lyophobic (solvent-hating, less stable, e.g. gold sol, As₂S₃ sol).
Properties of colloids: Tyndall effect (scattering of light), Brownian motion, electrophoresis, coagulation.
Hardy–Schulze rule: the higher the charge on the coagulating ion, the greater its coagulating power. For a negative sol: Al³⁺ > Ba²⁺ > Na⁺.
महत्त्वाच्या संज्ञा
Adsorption (अधिशोषण) – concentration of a substance at a surface.
Desorption – removal of the adsorbed substance from the surface.
Adsorption isotherm – graph of amount adsorbed against pressure at constant temperature.
Gold number – milligrams of a protective colloid needed to prevent coagulation of 10 mL of a gold sol by 1 mL of 10% NaCl. Smaller gold number means better protection.
Emulsion – colloid of one liquid in another (milk is oil-in-water).
सूत्रे
Freundlich: x/m = k P^(1/n); log (x/m) = log k + (1/n) log P (straight line; slope 1/n, intercept log k)
For solutions: x/m = k C^(1/n)
Langmuir: x/m = aP/(1 + bP)
Linear form: P/(x/m) = 1/a + (b/a) P
At low P: x/m ≈ aP (first order); at high P: x/m ≈ a/b (constant, surface full)
५ महत्त्वाचे प्रश्न
Q1. Distinguish between physisorption and chemisorption (any four points).
Physisorption uses weak van der Waals forces, chemisorption uses chemical bonds. Physisorption has low enthalpy (about 20–40 kJ/mol), chemisorption high. Physisorption is reversible and multilayer; chemisorption is usually irreversible and monolayer. Chemisorption is highly specific.
Q2. Why is adsorption exothermic?
When molecules stick to a surface, their freedom decreases, so ΔS is negative. For the process to be spontaneous (ΔG = ΔH − TΔS negative), ΔH must be negative. Also, residual surface forces are satisfied, releasing energy.
Q3. In a Freundlich plot of log(x/m) against log P, the slope is 0.5 and the intercept is 0.3. Find x/m at P = 4 atm.
1/n = 0.5; log k = 0.3 → k = 10^0.3 ≈ 2.0
log (x/m) = 0.3 + 0.5 × log 4
log 4 = 0.602, so 0.5 × 0.602 = 0.301
log (x/m) = 0.601 → x/m ≈ 4.0
Q4. Derive the Langmuir adsorption isotherm.
Let θ = fraction of surface covered. Rate of adsorption = k₁ P (1 − θ); rate of desorption = k₂ θ.
At equilibrium: k₁ P (1 − θ) = k₂ θ.
θ = k₁P/(k₂ + k₁P) = bP/(1 + bP), where b = k₁/k₂.
Amount adsorbed x/m ∝ θ, so x/m = aP/(1 + bP).
Discuss low and high pressure limits.
Q5. Explain the Tyndall effect and electrophoresis.
Tyndall effect: when a beam of light passes through a colloid, the path becomes visible because colloidal particles scatter light. True solutions do not show it.
Example: sunlight beam in a dusty room, headlight beam in fog.
Electrophoresis: movement of colloidal particles towards an electrode under an electric field.
It shows that colloidal particles carry charge (Fe(OH)₃ sol positive, As₂S₃ sol negative).
परीक्षेसाठी टीप
Langmuir derivation and the physisorption vs chemisorption table are the most asked. Draw the Freundlich isotherm curve and its log–log straight line.
Phase equilibria deals with systems in which two or more phases are in equilibrium. Gibbs' phase rule connects the number of phases, components and degrees of freedom. This topic covers the phase rule, phase diagrams of one-component systems (water and sulphur) and two-component systems (simple eutectic like Pb–Ag), and the Clausius–Clapeyron equation.
महत्त्वाचे मुद्दे
Phase (P): a homogeneous, physically distinct and mechanically separable part of a system. A mixture of gases is always one phase. Each solid is a separate phase.
Component (C): the smallest number of independent chemical species needed to express the composition of every phase.
Degrees of freedom (F): the number of variables (T, P, concentration) that can be changed independently without changing the number of phases.
Gibbs' phase rule: F = C − P + 2.
For CaCO₃(s) ⇌ CaO(s) + CO₂(g): P = 3, C = 2, so F = 1 (univariant).
Water system (C = 1): areas (ice, water, vapour) are bivariant (F = 2); curves are univariant (F = 1); the triple point where all three phases meet is invariant (F = 0). Triple point of water ≈ 0.0098 °C (273.16 K) and 4.58 mm Hg.
The ice–water (fusion) curve of water slopes towards the pressure axis (negative slope), because ice is less dense than water. So increase in pressure lowers the melting point of ice.
Sulphur system: has two solid forms, rhombic and monoclinic, and four phases in all. It has more than one triple point; a metastable triple point also exists. The transition temperature of rhombic to monoclinic is about 95.6 °C.
Condensed (reduced) phase rule for solid–liquid systems at constant pressure: F′ = C − P + 1.
Eutectic point: the lowest temperature at which a liquid mixture of two components can exist; the eutectic mixture melts at a fixed temperature. In the Pb–Ag system it is used in Pattinson's process for desilverisation of lead.
महत्त्वाच्या संज्ञा
Phase (प्रावस्था) – uniform, separable part of a system.
Triple point – point where three phases are in equilibrium (F = 0 in a one-component system).
Metastable equilibrium – state that is stable only if not disturbed (supercooled water).
Eutectic mixture – mixture of two solids with the lowest melting point possible for that pair.
Transition temperature – temperature at which one solid form changes into another.
F = C − P + 2. P is the number of phases, C the number of components, and F the degrees of freedom (independent variables like T, P, concentration that can be varied without changing the number of phases).
Q2. Why does the ice–water curve of the water system have a negative slope?
Ice has a larger volume than the same mass of liquid water. On melting ΔV is negative, so by Clapeyron's equation dP/dT is negative. Hence increasing pressure lowers the melting point of ice.
Q3. Find P, C and F for NH₄Cl(s) ⇌ NH₃(g) + HCl(g) when NH₃ and HCl come only from NH₄Cl.
Phases: solid NH₄Cl and one gas mixture → P = 2
NH₃ and HCl are always in 1:1 ratio, so composition of both phases can be expressed by NH₄Cl alone → C = 1
F = C − P + 2 = 1 − 2 + 2 = 1
System is univariant.
Q4. The vapour pressure of water is 760 mm at 373 K. Find the vapour pressure at 363 K. (ΔH_vap = 40.7 kJ/mol, R = 8.314 J K⁻¹ mol⁻¹)
log (P₂/P₁) = (ΔH/2.303R) × (T₂ − T₁)/(T₁T₂), with P₁ = 760 at T₁ = 373, T₂ = 363
A solution is a homogeneous mixture of two or more substances. This topic covers ways of expressing concentration, Raoult's law, ideal and non-ideal solutions, and colligative properties — relative lowering of vapour pressure, elevation of boiling point, depression of freezing point and osmotic pressure — with their use in finding molar mass, and the van't Hoff factor.
महत्त्वाचे मुद्दे
Concentration terms: molarity (mol L⁻¹, changes with temperature), molality (mol kg⁻¹ solvent, does not change with temperature), mole fraction, normality, mass percent, ppm.
Raoult's law: partial vapour pressure of a component = its mole fraction × vapour pressure of the pure component (p₁ = x₁p₁°).
Ideal solutions obey Raoult's law at all compositions; ΔH(mix) = 0 and ΔV(mix) = 0 (benzene + toluene).
Positive deviation: A–B attraction weaker than A–A and B–B; vapour pressure higher than expected (ethanol + water, acetone + CS₂). Forms minimum-boiling azeotropes.
Colligative properties depend only on the number of solute particles, not on their nature. They are used to find the molar mass of non-volatile solutes.
Relative lowering of vapour pressure = mole fraction of solute.
Elevation of boiling point ΔTb = Kb·m; depression of freezing point ΔTf = Kf·m. For water Kb = 0.52 K kg mol⁻¹ and Kf = 1.86 K kg mol⁻¹.
Osmosis: flow of solvent through a semipermeable membrane from dilute to concentrated solution. Osmotic pressure π = CRT (van't Hoff equation). Isotonic solutions have equal π (0.9 % NaCl is isotonic with blood).
Reverse osmosis: applying pressure greater than π pushes solvent out of a solution — used in water purifiers and desalination.
van't Hoff factor (i): i > 1 for dissociation (NaCl, i ≈ 2), i < 1 for association (acetic acid dimerises in benzene, i ≈ 0.5). Such solutes give abnormal molar masses.
महत्त्वाच्या संज्ञा
Molality (मोललता) – moles of solute per kg of solvent.
Ebullioscopic constant (Kb) – elevation of boiling point for a 1 molal solution.
Cryoscopic constant (Kf) – depression of freezing point for a 1 molal solution.
Azeotrope – mixture that boils at constant temperature with the same composition in liquid and vapour.
सूत्रे
(p° − p)/p° = x₂ ≈ n₂/n₁ (dilute solution)
ΔTb = Kb × m = Kb × (W₂ × 1000)/(M₂ × W₁)
ΔTf = Kf × m
π = CRT = (n₂/V)RT (R = 0.0821 L atm K⁻¹ mol⁻¹)
With dissociation/association: ΔT = i × K × m; π = iCRT
Degree of dissociation α = (i − 1)/(n − 1), n = ions per formula unit
५ महत्त्वाचे प्रश्न
Q1. Why is molality preferred over molarity for colligative properties?
Molality uses mass of solvent, which does not change with temperature. Molarity uses volume, which expands or contracts with temperature.
Q2. What is reverse osmosis? Give one use.
When pressure greater than the osmotic pressure is applied on the solution side, pure solvent flows out through the membrane. It is used to make drinking water from sea water and in RO water purifiers.
Q3. 6 g of urea (M = 60) is dissolved in 250 g of water. Find the freezing point of the solution (Kf = 1.86 K kg mol⁻¹).
Moles of urea = 6/60 = 0.1 mol
m = 0.1 × 1000/250 = 0.4 mol kg⁻¹
ΔTf = 1.86 × 0.4 = 0.744 K
Freezing point = 273.15 − 0.744 = 272.41 K (−0.744 °C)
Q4. Calculate the osmotic pressure of a 0.1 M glucose solution at 300 K. Explain osmosis.
π = CRT = 0.1 × 0.0821 × 300 = 2.46 atm
Explain semipermeable membrane, direction of flow, and the Berkeley–Hartley method of measuring π.
Q5. Explain positive and negative deviations from Raoult's law.
Draw vapour pressure–composition curves for both. Give the cause (intermolecular forces), signs of ΔH(mix) and ΔV(mix), examples and the type of azeotrope formed. Add one example each: ethanol–water (95.6 % ethanol, minimum-boiling, 351 K) and HNO₃–water (68 % HNO₃, maximum-boiling, 393.5 K).
परीक्षेसाठी टीप
Write units with every number in colligative numericals and convert grams of solvent to kg. Most marks are lost on that conversion.
The first law tells us energy is conserved but not whether a process will happen by itself. The second law answers this through entropy. This topic covers spontaneous processes, statements of the second law, the Carnot cycle and efficiency, entropy and its calculation, Gibbs and Helmholtz free energy, the Gibbs–Helmholtz and Clausius–Clapeyron equations, and the third law.
महत्त्वाचे मुद्दे
Spontaneous process: takes place by itself without outside help (heat flows from hot to cold, gas expands into vacuum). All natural processes are irreversible.
Kelvin–Planck statement: no engine working in a cycle can convert all the heat taken from a source into work.
Clausius statement: heat cannot flow by itself from a colder body to a hotter body.
Carnot cycle: four reversible steps — isothermal expansion, adiabatic expansion, isothermal compression, adiabatic compression. Efficiency depends only on the two temperatures.
Entropy (S): a measure of randomness (disorder) of a system; a state function. dS = dq(rev)/T.
For an isolated system, ΔS > 0 for a spontaneous process and ΔS = 0 at equilibrium. The entropy of the universe always increases.
Entropy increases on melting, boiling, dissolving, and when moles of gas increase.
Gibbs free energy: G = H − TS. At constant T and P, ΔG < 0 spontaneous, ΔG = 0 equilibrium, ΔG > 0 non-spontaneous.
Helmholtz free energy: A = U − TS; criterion at constant T and V.
ΔG° = −RT ln K = −2.303 RT log K links free energy to the equilibrium constant.
Trouton's rule: ΔS(vap) ≈ 88 J K⁻¹ mol⁻¹ for many normal liquids; water (about 109) deviates due to H-bonding.
Third law (Nernst / Planck): the entropy of a perfect crystal at 0 K is zero. It allows calculation of absolute entropies.
महत्त्वाच्या संज्ञा
Entropy (एंट्रॉपी) – measure of disorder; unit J K⁻¹ mol⁻¹.
Free energy – part of energy available to do useful work.
Reversible process – carried out in infinitely small steps, always near equilibrium.
Q1. A Carnot engine works between 500 K and 300 K. Find its efficiency.
η = 1 − 300/500 = 1 − 0.6 = 0.4, i.e. 40 %.
Q2. State the third law of thermodynamics.
The entropy of a perfectly crystalline substance is zero at absolute zero (0 K), because there is only one possible arrangement of particles.
Q3. The enthalpy of vaporisation of water is 40.7 kJ mol⁻¹ at 373 K. Find the entropy of vaporisation and comment.
ΔS = ΔH/T = 40 700/373 = 109.1 J K⁻¹ mol⁻¹
This is higher than Trouton's value (about 88) because H-bonds in liquid water make it more ordered, so more disorder is gained on boiling.
Q4. Describe the Carnot cycle and derive its efficiency.
Draw the P–V diagram, write work for each step (w₁ = nRT₂ ln V₂/V₁, adiabatic terms cancel, w₃ = nRT₁ ln V₄/V₃), use V₂/V₁ = V₃/V₄ from adiabatic relations, and obtain η = (T₂ − T₁)/T₂. Conclude that efficiency is always less than 1.
Q5. Explain Gibbs free energy as a criterion of spontaneity.
Derive ΔG = −TΔS(total) at constant T and P, give the three conditions, the four-case table above and one example (melting of ice above 273 K).
परीक्षेसाठी टीप
Remember: entropy of the universe decides spontaneity; ΔG is just the system-only shortcut at constant T and P. Always convert kJ to J before dividing by T.
Electrochemistry studies the relation between chemical reactions and electricity. This topic covers conductance of electrolytes, Kohlrausch's law, electrochemical cells, electrode potential, the Nernst equation, EMF and its relation with Gibbs energy, and types of electrodes.
महत्त्वाचे मुद्दे
Electrolytic conductance is due to movement of ions. It increases with temperature (unlike metals).
Specific conductance (κ) is the conductance of a 1 cm cube of solution. Molar conductance (Λm) is the conductance of all ions from one mole of electrolyte.
On dilution, κ decreases (fewer ions per cm³) but Λm increases.
For strong electrolytes (KCl, HCl), Λm increases slightly on dilution and follows the Debye–Hückel–Onsager relation Λm = Λ°m − b√C.
For weak electrolytes (CH₃COOH), Λm rises sharply at high dilution because the degree of dissociation increases. Λ°m cannot be found by extrapolation.
Kohlrausch's law: at infinite dilution, each ion makes its own independent contribution to the molar conductance. It helps find Λ° of weak electrolytes and their degree of dissociation.
Galvanic (voltaic) cell converts chemical energy into electrical energy. Example: Daniell cell, Zn | Zn²⁺ || Cu²⁺ | Cu. Oxidation at the anode (negative), reduction at the cathode (positive).
A salt bridge completes the circuit and keeps the solutions electrically neutral.
Standard hydrogen electrode (SHE) is the reference with E° = 0.00 V.
Nernst equation gives the electrode or cell potential at any concentration.
ΔG = −nFE. A positive EMF means a spontaneous cell reaction.
Types of electrodes: metal–metal ion, gas electrode (SHE), metal–insoluble salt (calomel, Ag/AgCl), redox electrode (Pt in Fe²⁺/Fe³⁺).
Calomel electrode (Hg | Hg₂Cl₂ | KCl) is a commonly used secondary reference electrode.
महत्त्वाच्या संज्ञा
Electrolyte (विद्युत अपघट्य) – substance that conducts electricity in solution or molten state by ions.
Cell constant – l/A of the conductivity cell (unit cm⁻¹).
Electrode potential – potential developed between a metal and its ion solution.
EMF – potential difference between two electrodes when no current flows.
Transport number – fraction of total current carried by an ion.
सूत्रे
κ = G × (l/A) = conductance × cell constant (S cm⁻¹)
Λm = (κ × 1000)/C (S cm² mol⁻¹, C in mol/L)
Kohlrausch: Λ°m = ν₊ λ°₊ + ν₋ λ°₋
Degree of dissociation: α = Λm/Λ°m; Ka = Cα²/(1 − α)
Nernst (at 298 K): E = E° − (0.0591/n) log Q
For Mⁿ⁺/M: E = E° + (0.0591/n) log [Mⁿ⁺]
ΔG° = −nFE°; ΔG° = −2.303 RT log K; F = 96500 C mol⁻¹
log K = nE°/0.0591 (298 K)
५ महत्त्वाचे प्रश्न
Q1. State Kohlrausch's law and give one application.
At infinite dilution, the molar conductance of an electrolyte is the sum of the independent contributions of its cation and anion. Application: finding Λ° of a weak electrolyte like acetic acid from Λ° values of strong electrolytes: Λ°(CH₃COOH) = Λ°(CH₃COONa) + Λ°(HCl) − Λ°(NaCl).
Q2. What is the function of a salt bridge?
It connects the two half-cells and completes the internal circuit. It keeps both solutions electrically neutral by allowing ions to move, and it reduces liquid junction potential.
Q3. Calculate the EMF of the cell Zn | Zn²⁺ (0.01 M) || Cu²⁺ (1 M) | Cu at 298 K. E°(Zn²⁺/Zn) = −0.76 V, E°(Cu²⁺/Cu) = +0.34 V.
Chemical kinetics studies the rate of chemical reactions, the factors affecting the rate, and the mechanism. This topic covers rate, rate law, order and molecularity, integrated rate equations for zero and first order reactions, half-life, methods to find order, the Arrhenius equation and collision theory.
महत्त्वाचे मुद्दे
Rate of reaction = change in concentration of a reactant or product per unit time. Units: mol L⁻¹ s⁻¹.
For aA → bB: rate = −(1/a) d[A]/dt = (1/b) d[B]/dt.
Rate law: rate = k[A]^m[B]^n, found by experiment, not from the balanced equation.
Order = m + n (sum of powers in rate law). It can be 0, a whole number or a fraction.
Molecularity = number of molecules colliding in an elementary step. It is always a whole number (1, 2, rarely 3), never zero.
Zero order: rate does not depend on concentration (e.g. some reactions on a metal surface).
First order: rate ∝ [A]. Half-life does not depend on initial concentration. Radioactive decay is first order.
Pseudo-first-order: a higher-order reaction behaves like first order because one reactant is in large excess (acid hydrolysis of ester, inversion of cane sugar).
Arrhenius equation: k = A e^(−Ea/RT). The rate usually becomes roughly double for every 10 °C rise (temperature coefficient about 2–3).
Activation energy (Ea): minimum extra energy needed for reactants to form products. A catalyst lowers Ea and speeds up the reaction without being used up.
Collision theory: reaction happens only when molecules collide with energy ≥ threshold energy and with proper orientation.
महत्त्वाच्या संज्ञा
Rate constant (वेग स्थिरांक, k) – rate when all concentrations are 1 mol/L.
Half-life (t½) – time for concentration to fall to half its initial value.
Elementary reaction – reaction that occurs in a single step.
Rate-determining step – slowest step of a mechanism.
Activated complex – unstable high-energy arrangement at the top of the energy barrier.
सूत्रे
Order
Integrated equation
t½
Unit of k
Zero
[A]₀ − [A] = kt
[A]₀/2k
mol L⁻¹ s⁻¹
First
k = (2.303/t) log ([A]₀/[A])
0.693/k
s⁻¹
Second (equal conc.)
1/[A] − 1/[A]₀ = kt
1/(k[A]₀)
L mol⁻¹ s⁻¹
Arrhenius: k = A e^(−Ea/RT); log k = log A − Ea/(2.303RT)
Two temperatures: log (k₂/k₁) = (Ea/2.303R) × (T₂ − T₁)/(T₁T₂)
For order n, t½ ∝ 1/[A]₀^(n−1)
५ महत्त्वाचे प्रश्न
Q1. Distinguish between order and molecularity.
Order is the sum of the powers of concentrations in the experimental rate law; it may be zero or fractional. Molecularity is the number of molecules taking part in an elementary step; it is a whole number and cannot be zero. Order applies to the overall reaction, molecularity to each elementary step.
Q2. Show that the half-life of a first order reaction is independent of initial concentration.
k = (2.303/t) log ([A]₀/[A]). At t = t½, [A] = [A]₀/2. So t½ = (2.303/k) log 2 = 0.693/k. No [A]₀ appears, so t½ is independent of the initial concentration.
Q3. A first order reaction is 20% complete in 10 minutes. Find k and t½.
[A]₀ = 100, [A] = 80
k = (2.303/10) log (100/80) = 0.2303 × log 1.25
log 1.25 = 0.0969 → k = 0.02232 min⁻¹
t½ = 0.693/0.02232 = 31.05 min
Q4. The rate constant of a reaction doubles when temperature rises from 300 K to 310 K. Calculate Ea.
Q5. Derive the integrated rate equation for a first order reaction.
A → products; rate = −d[A]/dt = k[A].
Separate: d[A]/[A] = −k dt.
Integrate: ln [A] = −kt + I. At t = 0, [A] = [A]₀, so I = ln [A]₀.
ln ([A]₀/[A]) = kt → k = (2.303/t) log ([A]₀/[A]).
Plot of log [A] vs t is a straight line with slope −k/2.303.
परीक्षेसाठी टीप
Learn the table of zero, first and second order formulas with units of k. In numericals, keep time units the same throughout and write log values clearly.
Transition elements are d-block elements whose atoms or common ions have partly filled d orbitals. They lie in Groups 3 to 12. This topic covers their electronic configuration and general properties: variable oxidation states, colour, magnetic behaviour, complex formation, catalytic activity, alloy and interstitial compound formation, and a comparison of the 3d, 4d and 5d series.
महत्त्वाचे मुद्दे
General configuration: (n − 1)d¹⁻¹⁰ ns¹⁻². Four series: 3d (Sc–Zn), 4d (Y–Cd), 5d (La, Hf–Hg) and 6d.
Zn, Cd and Hg have full d¹⁰ shells in the atom and common ions, so they are usually not counted as true transition elements.
Variable oxidation states: ns and (n − 1)d electrons have close energies, so different numbers can be lost. Mn shows +2 to +7; the maximum is seen in the middle of the series.
Colour: due to d–d transitions. An electron absorbs visible light and jumps between split d levels; the complementary colour is seen. Ions with d⁰ (Sc³⁺, Ti⁴⁺) or d¹⁰ (Zn²⁺, Cu⁺) are colourless.
Magnetic properties: unpaired electrons make ions paramagnetic. Spin-only magnetic moment μ = √(n(n + 2)) BM.
Complex formation: small size, high charge and vacant d orbitals help them form complexes like [Cu(NH₃)₄]²⁺.
Catalytic activity: due to variable oxidation states and surface adsorption. Fe in the Haber process, V₂O₅ in the contact process, Ni in hydrogenation of oils.
Interstitial compounds: small atoms (H, C, N) fit in the gaps of the metal lattice (e.g. steel, TiC). They are hard and have high melting points.
Alloys form easily because transition metal atoms have similar sizes (brass = Cu + Zn, bronze = Cu + Sn).
Atomic radii decrease slowly across the series and then become almost constant. 4d and 5d elements of the same group have nearly the same size due to lanthanide contraction (Zr ≈ Hf).
They have high melting points, high density and high enthalpy of atomisation due to strong metallic bonding with d electrons.
महत्त्वाच्या संज्ञा
Transition element (संक्रमण मूलद्रव्य) – element with incomplete d subshell in the atom or common ion.
d–d transition – jump of an electron from a lower to a higher d level after splitting.
Paramagnetism – weak attraction to a magnetic field due to unpaired electrons.
Diamagnetism – weak repulsion when all electrons are paired.
Interstitial compound – compound with small atoms trapped in metal lattice holes.
सूत्रे
Spin-only magnetic moment: μ = √(n(n + 2)) BM (n = number of unpaired electrons)
Q1. Why are Zn²⁺ salts white while Cu²⁺ salts are coloured?
Zn²⁺ is 3d¹⁰; all d orbitals are full, so no d–d transition is possible and no visible light is absorbed. Cu²⁺ is 3d⁹ with one unpaired electron; d–d transition absorbs light in the visible region, so its salts are blue.
Q2. Why do transition elements show variable oxidation states?
The (n − 1)d and ns orbitals have nearly the same energy. So electrons from both can be lost in different numbers, giving several oxidation states. Example: Fe shows +2 and +3.
Q3. Calculate the spin-only magnetic moment of Fe³⁺ (Z = 26).
Fe = [Ar] 3d⁶ 4s²; Fe³⁺ = [Ar] 3d⁵
5 unpaired electrons (Hund's rule)
μ = √(5 × 7) = √35
μ ≈ 5.92 BM
Q4. An ion of a 3d element has μ = 3.87 BM. How many unpaired electrons does it have? Suggest one such ion.
μ = √(n(n + 2)) → 3.87² = n(n + 2)
14.98 ≈ 15 = n² + 2n → n² + 2n − 15 = 0
(n + 5)(n − 3) = 0 → n = 3
Example: Cr³⁺ (3d³) or Co²⁺ high-spin (3d⁷)
Q5. Discuss the general characteristics of transition elements.
Metallic nature, high melting points and density.
Variable oxidation states.
Coloured ions due to d–d transitions.
Paramagnetism due to unpaired electrons.
Complex formation, catalytic activity, alloy and interstitial compound formation.
Give one example for each property.
परीक्षेसाठी टीप
Magnetic moment numericals are easy marks: write the ion's configuration, count unpaired electrons, then apply √(n(n + 2)).
Nitro compounds contain the –NO₂ group bonded to carbon through nitrogen. They may be aliphatic (nitroalkanes, R–NO₂) or aromatic (nitroarenes, Ar–NO₂). This topic covers their structure, preparation, the acidity of α-hydrogen in nitroalkanes, reduction under different conditions, and the effect of the nitro group on the benzene ring.
महत्त्वाचे मुद्दे
Structure: the –NO₂ group is a resonance hybrid of two equal structures; both N–O bonds have the same length (about 1.21 Å). N carries a positive charge, so the group is strongly electron-withdrawing (−I and −R).
Nitro compounds are isomeric with alkyl nitrites (R–O–N=O). Nitroalkanes have high boiling points and are polar; nitrites are low boiling and are hydrolysed to alcohols.
Preparation of nitroalkanes: vapour-phase nitration of alkanes (HNO₃ at about 400 °C) and alkyl halide + AgNO₂ (gives mainly nitroalkane; NaNO₂/KNO₂ give more nitrite).
Preparation of nitrobenzene: benzene + nitrating mixture (conc. HNO₃ + conc. H₂SO₄) at about 50–60 °C. The electrophile is the nitronium ion, NO₂⁺.
Acidity of α-H: 1° and 2° nitroalkanes dissolve in NaOH because the α-H is acidic; the anion (nitronate) is resonance-stabilised. 3° nitroalkanes have no α-H and do not dissolve.
Tautomerism: nitro (keto-like) form ⇌ aci-nitro (nitronic acid) form.
Nitrous acid test: 1° nitroalkane gives nitrolic acid (red colour in NaOH), 2° gives pseudonitrole (blue colour), 3° no reaction.
acidic (Sn/HCl or Fe/HCl) → aniline
neutral (Zn/NH₄Cl) → N-phenylhydroxylamine
alkaline (Zn/NaOH, or LiAlH₄) → azoxybenzene, azobenzene, hydrazobenzene
electrolytic in conc. H₂SO₄ → p-aminophenol
Reduction of nitrobenzene depends on the medium:
Effect on the ring: –NO₂ deactivates the ring and directs incoming electrophiles to the meta position (nitration of nitrobenzene gives m-dinitrobenzene).
–NO₂ at ortho/para positions activates aryl halides towards nucleophilic substitution.
महत्त्वाच्या संज्ञा
Nitronium ion (नायट्रोनियम आयन) – NO₂⁺, the electrophile in nitration.
It withdraws electrons by −I and −R effects. Resonance structures show positive charge at ortho and para positions, so these become electron-poor. The meta position is less deactivated, so electrophiles attack there.
Q2. How will you distinguish between nitroethane and 2-nitropropane?
With nitrous acid followed by NaOH: nitroethane (1°) forms nitrolic acid that gives a red solution; 2-nitropropane (2°) gives pseudonitrole, which is blue.
Q3. Explain the mechanism of nitration of benzene.
Generation of NO₂⁺ from the nitrating mixture.
Attack of NO₂⁺ on the π system → arenium ion (σ-complex), resonance stabilised.
Loss of H⁺ (taken by HSO₄⁻) → nitrobenzene; aromaticity restored.
Q4. Discuss the reduction of nitrobenzene in different media.
Write equations for acidic, neutral, alkaline, catalytic and electrolytic reduction, name each product and give one use (aniline for dyes, p-aminophenol for paracetamol).
Q5. Explain the acidic nature of 1° and 2° nitroalkanes.
Show removal of α-H by OH⁻, resonance of the nitronate ion, nitro–aci-nitro tautomerism and why 3° nitroalkanes are not acidic.
परीक्षेसाठी टीप
The table of reduction products is asked almost every year. Learn it with the reagent written beside each product.
An active methylene compound has a –CH₂– group placed between two electron-withdrawing groups such as C=O or –COOR. Its hydrogens are unusually acidic, so it forms a stable carbanion that is very useful in synthesis. This topic covers ethyl acetoacetate (acetoacetic ester) and diethyl malonate (malonic ester): preparation, keto–enol tautomerism and synthetic uses.
महत्त्वाचे मुद्दे
Why active: the carbanion formed after losing H⁺ is stabilised by resonance over both C=O groups. pKa of ethyl acetoacetate ≈ 11, diethyl malonate ≈ 13, compared with about 50 for an alkane.
Ethyl acetoacetate (EAA), CH₃COCH₂COOC₂H₅, is made by Claisen condensation: two molecules of ethyl acetate with sodium ethoxide, then acidification.
Keto–enol tautomerism of EAA: CH₃COCH₂COOC₂H₅ ⇌ CH₃C(OH)=CHCOOC₂H₅. At room temperature it contains about 92–93 % keto and 7–8 % enol form. The enol is stabilised by intramolecular H-bonding and conjugation.
Evidence for the keto form: reacts with HCN, NaHSO₃, 2,4-DNP. Evidence for enol form: violet colour with FeCl₃, decolourises bromine water, reacts with Na to give H₂.
Diethyl malonate (DEM), CH₂(COOC₂H₅)₂, made from chloroacetic acid → cyanoacetic acid → esterification with ethanol.
Alkylation: sodium salt + alkyl halide → mono- or di-alkyl derivative.
Q1. Why are the methylene hydrogens of ethyl acetoacetate acidic?
They lie between two carbonyl groups. When one is removed, the negative charge on carbon spreads onto both oxygens by resonance, giving a stable enolate ion.
Q2. What is the difference between ketonic and acidic hydrolysis of EAA?
Ketonic hydrolysis (dilute acid/alkali) gives acetone (or a methyl ketone) with CO₂ and ethanol. Acidic hydrolysis (conc. alcoholic KOH) breaks a C–C bond and gives two molecules of acetic acid (or acids).
Q3. Discuss keto–enol tautomerism of ethyl acetoacetate with evidence.
Write both forms, the percentage, the factors stabilising the enol (conjugation, intramolecular H-bond) and the chemical tests for each form listed above.
Q4. How will you synthesise (a) butanone, (b) succinic acid, (c) n-butyric acid using EAA or DEM?
Show each step with reagents: sodium salt formation, alkylation (CH₃I; ClCH₂COOEt; C₂H₅Br on DEM), hydrolysis and decarboxylation.
Q5. Describe the preparation of diethyl malonate and its use in making barbituric acid.
ClCH₂COOH → (NaCN) NC–CH₂–COOH → (C₂H₅OH/H⁺) CH₂(COOC₂H₅)₂. Condensation with urea in sodium ethoxide gives barbituric acid, the parent of barbiturate drugs.
परीक्षेसाठी टीप
For synthesis questions, first count the carbons in the target and decide which group comes from the alkyl halide. Then write the route step by step.
Coordination compounds contain a central metal atom or ion bonded to a fixed number of ions or molecules called ligands. This topic covers Werner's theory, important terms, IUPAC naming, isomerism, the effective atomic number (EAN) rule, and valence bond theory (VBT) with its limitations.
महत्त्वाचे मुद्दे
Example: [Co(NH₃)₆]Cl₃. The part in square brackets is the coordination sphere; Cl⁻ ions outside are counter ions.
Werner's theory (1893): metals show two valencies. Primary valency = oxidation state, ionisable, satisfied by negative ions. Secondary valency = coordination number, non-ionisable, satisfied by ligands, and directed in space (gives geometry).
[Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅SO₄]Br are ionisation isomers.
EAN rule (Sidgwick): EAN = Z − oxidation state + 2 × CN. Many stable complexes have EAN equal to the next noble gas number.
VBT: metal uses hybrid orbitals: sp³ (tetrahedral), dsp² (square planar), d²sp³ (inner orbital octahedral, low spin), sp³d² (outer orbital octahedral, high spin).
Strong field ligands (CN⁻, CO) cause pairing of electrons; weak field ligands (F⁻, H₂O) do not.
Limitations of VBT: does not explain colour, gives no quantitative idea of magnetic data beyond spin, does not separate strong and weak ligands clearly.
महत्त्वाच्या संज्ञा
Ligand (लिगंड) – ion or molecule that donates an electron pair to the metal.
Denticity – number of donor atoms in one ligand.
Chelation – ring formation by a polydentate ligand.
Inner orbital complex – uses (n − 1)d orbitals (d²sp³).
Outer orbital complex – uses nd orbitals (sp³d²).
सूत्रे
Oxidation state x: sum of charges = charge on complex. In [Fe(CN)₆]⁴⁻: x + 6(−1) = −4 → x = +2
Naming rules: cation first; ligands in alphabetical order before metal; prefixes di, tri, tetra (bis, tris for complex names); anionic ligands end in -o (chlorido, cyanido); anionic complex ends in -ate (ferrate, cuprate); oxidation state in Roman numerals.
Complex
Hybridisation
Shape
Magnetic
[Ni(CO)₄]
sp³
tetrahedral
diamagnetic
[Ni(CN)₄]²⁻
dsp²
square planar
diamagnetic
[Fe(CN)₆]³⁻
d²sp³
octahedral
paramagnetic (1 unpaired)
[CoF₆]³⁻
sp³d²
octahedral
paramagnetic (4 unpaired)
५ महत्त्वाचे प्रश्न
Q1. Write IUPAC names of K₄[Fe(CN)₆] and [Co(NH₃)₅Cl]Cl₂.
Q2. What are ambidentate ligands? Give two examples.
Ligands that have two different donor atoms but bind through only one at a time. NO₂⁻ binds through N (nitro) or O (nitrito); SCN⁻ binds through S (thiocyanato) or N (isothiocyanato). They give linkage isomers.
Q3. Calculate the EAN of Fe in K₄[Fe(CN)₆] (Z of Fe = 26).
Oxidation state of Fe: x + 6(−1) = −4 → x = +2
Coordination number = 6
EAN = 26 − 2 + 2 × 6 = 36
36 = atomic number of krypton, so the complex is stable.
Q4. Explain the structure of [Ni(CN)₄]²⁻ using VBT.
Ni²⁺ = 3d⁸ (two unpaired electrons in free ion).
CN⁻ is a strong field ligand; the two unpaired 3d electrons pair up, freeing one 3d orbital.
One 3d, one 4s and two 4p orbitals mix → dsp² hybridisation.
Four CN⁻ donate electron pairs → square planar shape.
No unpaired electrons → diamagnetic.
Q5. State the main postulates of Werner's theory.
Metal shows primary (ionisable) and secondary (non-ionisable) valencies.
Primary valency = oxidation state; satisfied by anions.
Secondary valency = coordination number; satisfied by ligands (neutral or negative).
Secondary valencies are directed in space and decide geometry.
Redox reactions are reactions in which oxidation and reduction happen together. This topic covers the electronic concept of oxidation and reduction, oxidation number rules, oxidising and reducing agents, balancing redox equations by the oxidation number and ion-electron (half-reaction) methods, equivalent weight, and redox titrations.
महत्त्वाचे मुद्दे
Oxidation: loss of electrons, or increase in oxidation number. Reduction: gain of electrons, or decrease in oxidation number. Memory aid: OIL RIG (Oxidation Is Loss, Reduction Is Gain).
Oxidation number rules: element in free state = 0; monoatomic ion = its charge; O = −2 (but −1 in peroxides, +2 in OF₂); H = +1 (but −1 in metal hydrides like NaH); F = −1 always; sum in a neutral molecule = 0; sum in an ion = its charge.
Disproportionation: the same element is both oxidised and reduced. Example: 2H₂O₂ → 2H₂O + O₂; Cl₂ + 2NaOH → NaCl + NaClO + H₂O.
Balancing (oxidation number method): find changes in oxidation numbers, equalise increase and decrease, then balance O with H₂O and H with H⁺ (acid) or OH⁻ (base).
Ion-electron method: split into oxidation and reduction half-reactions, balance each for atoms and charge (using electrons), multiply to equalise electrons, and add.
Equivalent weight of an oxidant or reductant = molar mass ÷ number of electrons gained or lost per formula unit.
KMnO₄ in acidic medium gains 5 electrons (MnO₄⁻ → Mn²⁺); in neutral/weakly basic medium gains 3 (→ MnO₂).
K₂Cr₂O₇ in acid gains 6 electrons per formula unit (Cr₂O₇²⁻ → 2Cr³⁺).
Redox titrations: KMnO₄ vs oxalic acid or Mohr's salt (KMnO₄ is a self-indicator: first permanent pink colour is the end point). K₂Cr₂O₇ vs Fe²⁺ uses diphenylamine as indicator. Iodometry uses starch.
महत्त्वाच्या संज्ञा
Oxidation number (ऑक्सिडीकरण अंक) – charge an atom would have if all bonds were ionic.
Half-reaction – oxidation or reduction part of a redox reaction shown separately.
Self-indicator – titrant whose own colour shows the end point (KMnO₄).
Normality – gram equivalents of solute per litre of solution.
Disproportionation – simultaneous oxidation and reduction of one element.
सूत्रे
Equivalent weight = Molar mass / n (n = electrons transferred per formula unit)
Q1. Find the oxidation number of Cr in K₂Cr₂O₇ and of S in H₂SO₄.
K₂Cr₂O₇: 2(+1) + 2x + 7(−2) = 0 → 2x = 12 → x = +6. H₂SO₄: 2(+1) + x + 4(−2) = 0 → x = +6.
Q2. What is a disproportionation reaction? Give an example.
A redox reaction in which the same element in one substance is oxidised and reduced at the same time. Example: 2H₂O₂ → 2H₂O + O₂; oxygen goes from −1 to −2 (in H₂O) and to 0 (in O₂).
Q3. Balance by the ion-electron method: MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺ (acidic medium).
Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
Oxidation: Fe²⁺ → Fe³⁺ + e⁻ (multiply by 5)
5Fe²⁺ → 5Fe³⁺ + 5e⁻
Add: MnO₄⁻ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
Check charge: left 17+, right 17+ ✓
Q4. Find the equivalent weight of KMnO₄ in acidic medium and the normality of a 0.02 M solution. (KMnO₄ = 158 g/mol)
Mn changes from +7 to +2 → n = 5
Equivalent weight = 158/5 = 31.6 g
Normality = Molarity × n = 0.02 × 5 = 0.1 N
Q5. 20 mL of 0.1 N oxalic acid needs 25 mL of KMnO₄ for complete titration. Find the normality and strength of KMnO₄ in g/L.
N₁V₁ = N₂V₂ → 0.1 × 20 = N₂ × 25
N₂ = 2/25 = 0.08 N
Strength = N × Eq. wt = 0.08 × 31.6
= 2.528 g/L
End point: colourless to permanent light pink (KMnO₄ acts as self-indicator; titration done hot, about 60 °C, in dilute H₂SO₄).
परीक्षेसाठी टीप
Always check both atoms and total charge at the end of balancing. In titration problems write N₁V₁ = N₂V₂ with labels for each side.
Titrimetric (volumetric) analysis finds the amount of a substance by measuring the volume of a standard solution that reacts completely with it. This topic covers basic terms, primary and secondary standards, the four types of titrations (acid–base, redox, precipitation, complexometric), theory of indicators, titration curves, and calculations.
महत्त्वाचे मुद्दे
Titration: slow addition of a solution of known concentration (titrant, in burette) to the solution being analysed (titrand / analyte, in conical flask) until the reaction is complete.
Equivalence point: exact amount of titrant needed has been added (theoretical). End point: point at which the indicator changes colour (practical). A good indicator makes them almost equal.
Primary standard: pure, stable, non-hygroscopic, high molar mass, easily available — e.g. oxalic acid (H₂C₂O₄·2H₂O), anhydrous Na₂CO₃, K₂Cr₂O₇, borax. A secondary standard (NaOH, KMnO₄, HCl) must be standardised against a primary standard.
Acid–base titrations: indicators are weak organic acids or bases whose two forms have different colours (Ostwald's theory). Phenolphthalein (pH 8.3–10, colourless → pink) and methyl orange (pH 3.1–4.4, red → yellow).
Choice of indicator: strong acid–strong base: either; weak acid–strong base: phenolphthalein; strong acid–weak base: methyl orange; weak acid–weak base: no sharp end point.
Redox titrations: KMnO₄ (self-indicator, in dilute H₂SO₄, colourless → pale pink), K₂Cr₂O₇ (diphenylamine indicator), iodometry and iodimetry (starch indicator, blue colour).
Precipitation titrations: argentometric; Mohr's method (K₂CrO₄ indicator, red-brown Ag₂CrO₄ at the end), Volhard's method (Fe³⁺ indicator, back titration with KSCN), Fajans' method (adsorption indicators like fluorescein).
Complexometric titrations: EDTA (hexadentate) forms 1:1 complexes with metal ions; Eriochrome Black T indicator (wine red → blue) at pH 10 buffer; used to find hardness of water.
Titration curves: plot of pH against volume of titrant; the steep jump near equivalence decides which indicator can be used.
महत्त्वाच्या संज्ञा
Standard solution (प्रमाणित द्रावण) – solution whose concentration is exactly known.
Normality (N) – gram equivalents of solute per litre.
Back titration – excess reagent is added and the unused part is titrated.
Equivalent weight of KMnO₄ in acidic medium = M/5 = 158/5 = 31.6
Equivalent weight of oxalic acid dihydrate = 126/2 = 63
Titration
Indicator
Colour change
HCl vs NaOH
phenolphthalein
colourless → pink
HCl vs Na₂CO₃
methyl orange
yellow → orange-red
KMnO₄ vs oxalic acid
self
colourless → pink
Iodometry
starch
blue → colourless
EDTA vs Ca²⁺/Mg²⁺
Eriochrome Black T
wine red → blue
५ महत्त्वाचे प्रश्न
Q1. Distinguish between end point and equivalence point.
Equivalence point is when chemically equivalent amounts of titrant and analyte have reacted. End point is when the indicator visibly changes colour. The difference between them is the titration error.
Q2. Why is KMnO₄ called a self-indicator?
MnO₄⁻ is deep purple and its reduced form Mn²⁺ is almost colourless. The first extra drop after the end point leaves the solution pale pink, so no separate indicator is needed.
Q3. 25 mL of NaOH solution required 20 mL of 0.1 N HCl for neutralisation. Find the normality and strength of NaOH.
N₁V₁ = N₂V₂ → N(NaOH) × 25 = 0.1 × 20
N(NaOH) = 2/25 = 0.08 N
Strength = 0.08 × 40 = 3.2 g L⁻¹
Q4. Explain Ostwald's theory of acid–base indicators.
Phenolphthalein as a weak acid HIn ⇌ H⁺ + In⁻ (colourless / pink). Le Chatelier: in acid HIn dominates, in base In⁻. Methyl orange as a weak base. Indicator range pH = pKIn ± 1. Limitation: does not explain the colour itself (quinonoid theory does).
Q5. Explain complexometric titration with EDTA and its use to find hardness of water.
Structure and hexadentate nature of EDTA, 1:1 complex, pH 10 ammonia buffer, EBT indicator colour change, calculation of total hardness as ppm of CaCO₃.
परीक्षेसाठी टीप
Always write the balanced reaction before using N₁V₁ = N₂V₂, and check whether the question gives normality or molarity.
Chromatography is a group of methods for separating the components of a mixture. Separation happens because the components are distributed differently between a stationary phase and a mobile phase that moves over it. This topic covers the principle and classification, paper, thin-layer and column chromatography, Rf value, and an introduction to gas chromatography and HPLC.
महत्त्वाचे मुद्दे
Russian botanist M. S. Tswett (1906) first separated plant pigments on a column of chalk (CaCO₃); the name means "colour writing".
Stationary phase: a solid or a liquid held on a solid support. Mobile phase: a liquid or gas that moves through it.
Adsorption chromatography: components are held on the surface of a solid (column chromatography with silica gel or alumina, TLC).
Partition chromatography: components divide between two liquids, one held stationary on a support (paper chromatography — water held in cellulose fibres is the stationary phase).
Other types: ion-exchange, gel (size-exclusion), affinity, gas chromatography (GC), HPLC.
Paper chromatography: spot near the bottom of Whatman filter paper, dip in solvent (ascending, descending or radial). Colourless spots are made visible with a locating reagent: ninhydrin for amino acids (purple), iodine vapour, UV light.
TLC: glass plate coated with a thin layer of silica gel G (gypsum binder). Faster and sharper than paper; used to check purity and follow reactions.
Column chromatography: adsorbent packed in a glass column; mixture added at top; solvents of increasing polarity (eluent) wash out components one by one (elution). Less strongly adsorbed component comes out first.
Rf value is constant for a substance under fixed conditions (same solvent, paper, temperature), so it helps identify it. It always lies between 0 and 1.
Gas chromatography: mobile phase is an inert carrier gas (He, N₂); used for volatile substances; detected by flame ionisation or thermal conductivity detectors. Result is shown by retention time.
HPLC: liquid mobile phase pumped at high pressure through a column of very fine particles; fast, sensitive; used for drugs and food analysis.
महत्त्वाच्या संज्ञा
Eluent – solvent used to wash components through a column.
Retention factor, Rf (धारण गुणांक) – ratio of distance moved by the substance to distance moved by the solvent front.
Retention time – time taken by a component to pass through a GC/HPLC column.
सूत्रे
Rf = distance travelled by the solute from the origin / distance travelled by the solvent front from the origin
Type
Stationary phase
Mobile phase
Principle
Paper
water on cellulose
organic solvent
partition
TLC
silica gel / alumina
organic solvent
adsorption
Column
silica gel / alumina
solvent of increasing polarity
adsorption
GC
liquid on solid support
inert gas
partition
HPLC
fine particles in column
liquid under pressure
adsorption/partition
५ महत्त्वाचे प्रश्न
Q1. In a paper chromatogram the solvent front moved 7.5 cm and a spot moved 3.0 cm. Find Rf.
Rf = 3.0/7.5 = 0.40.
Q2. Why is Rf always less than 1?
A solute can at most move as far as the solvent; it is slowed down by the stationary phase, so its distance is smaller than that of the solvent front.
Q3. Explain the principle and procedure of thin-layer chromatography.
Preparation and activation of the plate, spotting, development in a closed jar, drying, locating spots (iodine chamber/UV), calculating Rf. Applications: purity check, monitoring reactions, identifying drugs and dyes. Advantages over paper chromatography.
Q4. Describe column chromatography with a labelled diagram.
Packing the column, loading the sample, elution with solvents in order of increasing polarity (hexane → benzene → chloroform → ethyl acetate → methanol), collecting fractions, separation of chlorophyll pigments as an example.
Q5. Write a note on gas chromatography.
Carrier gas, injection port, column in oven, detector and recorder; principle of partition; retention time; uses (essential oils, alcohol in blood, pesticide residues). Limitation: only volatile, thermally stable compounds.
परीक्षेसाठी टीप
A neat diagram of a TLC plate showing origin, spots and solvent front, with the Rf formula, scores full marks in short notes.
Amines are organic derivatives of ammonia in which one or more hydrogen atoms are replaced by alkyl or aryl groups. This topic covers classification, nomenclature, methods of preparation, basic character, chemical reactions, distinction between 1°, 2° and 3° amines, and diazonium salts.
Nitrogen in amines is sp³ with one lone pair; the shape is pyramidal.
Preparation: reduction of nitro compounds (Sn/HCl or H₂/Ni), reduction of nitriles and amides (LiAlH₄), Gabriel phthalimide synthesis (gives pure 1° aliphatic amines), Hofmann bromamide degradation (amide → amine with one carbon less).
Basicity: the lone pair on N accepts a proton. Aliphatic amines are stronger bases than NH₃ (+I effect of alkyl groups). Aromatic amines like aniline are weaker bases than NH₃ because the lone pair is delocalised into the benzene ring.
In water, the order for methylamines is (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃, due to a balance of +I effect, solvation (H-bonding) and steric hindrance.
Hinsberg's test (benzenesulphonyl chloride) distinguishes 1°, 2° and 3° amines.
Carbylamine test: only 1° amines (aliphatic and aromatic) give a foul-smelling isocyanide with CHCl₃ and alcoholic KOH.
Reaction with nitrous acid (HNO₂): 1° aliphatic amines give alcohol and N₂ gas; 1° aromatic amines at 0–5 °C give diazonium salts; 2° amines give yellow oily N-nitrosoamines.
Diazonium salts (C₆H₅N₂⁺Cl⁻) are very useful: Sandmeyer reaction (CuCl/HCl → chlorobenzene; CuCN → benzonitrile), replacement by −OH (warm water), −I (KI), −H (H₃PO₂), and coupling with phenol or aniline to form azo dyes.
The −NH₂ group in aniline is strongly ortho/para directing; aniline with bromine water gives a white precipitate of 2,4,6-tribromoaniline.
महत्त्वाच्या संज्ञा
Amine (अमाइन) – compound with −NH₂, −NH− or −N< group attached to carbon.
Diazotisation – conversion of a 1° aromatic amine into a diazonium salt with NaNO₂ + HCl at 0–5 °C.
Coupling reaction – diazonium salt joins with phenol/aniline to give coloured azo compounds.
pKb – measure of base strength; lower pKb means stronger base.
Acetylation – replacing H of −NH₂ by CH₃CO−, used to protect the amino group.
Q1. Why is aniline a weaker base than methylamine?
In aniline the lone pair on N is delocalised into the benzene ring by resonance, so it is less available to accept H⁺. In methylamine the CH₃ group pushes electrons (+I) to N, increasing availability. So methylamine is the stronger base.
Q2. What is the carbylamine test? Which amines give it?
When a primary amine is warmed with chloroform and alcoholic KOH, a foul-smelling isocyanide (carbylamine) is formed. Only primary amines give it; secondary and tertiary amines do not.
Q3. Explain Hinsberg's test to distinguish 1°, 2° and 3° amines.
Reagent: benzenesulphonyl chloride (C₆H₅SO₂Cl) with aqueous KOH.
1° amine forms N-alkylbenzenesulphonamide which has an acidic H on N; it dissolves in alkali.
2° amine forms N,N-dialkylbenzenesulphonamide with no H on N; insoluble in alkali.
3° amine does not react (no H on N).
Q4. Explain Hofmann bromamide degradation with mechanism outline.
Amide + Br₂ + NaOH → primary amine with one carbon less.
Example: CH₃CONH₂ → CH₃NH₂.
Steps: N-bromoamide forms; base removes H; the alkyl group migrates from C to N as Br⁻ leaves (isocyanate forms); hydrolysis gives amine and CO₂ (as carbonate).
Used to step down a carbon chain.
Q5. What are diazonium salts? Give their preparation and any three reactions.
Formula ArN₂⁺X⁻, e.g. benzenediazonium chloride.
Prepared by diazotisation of aniline with NaNO₂ and HCl at 0–5 °C (unstable above this).
With CuCl/HCl → chlorobenzene (Sandmeyer).
With KI → iodobenzene.
With warm water → phenol.
With phenol (alkaline) → orange azo dye (coupling).
परीक्षेसाठी टीप
Name reactions — Gabriel, Hofmann, Sandmeyer, carbylamine, coupling — are favourite short questions. Write each with a balanced equation and the condition.
Heterocyclic compounds are cyclic compounds in which the ring contains at least one atom other than carbon (a heteroatom) such as N, O or S. This topic covers classification and nomenclature, and the structure, aromaticity, preparation and reactions of the five-membered rings furan, pyrrole and thiophene and the six-membered ring pyridine.
Numbering starts at the heteroatom (position 1). In five-membered rings, positions 2 and 5 are called α, and 3 and 4 are β.
Aromaticity of pyrrole, furan, thiophene: four carbon atoms give 4 π electrons, and the heteroatom gives one lone pair (2 electrons) to the ring. Total 6 π electrons → aromatic by Hückel's rule.
Order of aromaticity: benzene > thiophene > pyrrole > furan (O is most electronegative and holds its lone pair most tightly).
These five-membered rings are π-excessive: they undergo electrophilic substitution more easily than benzene, mainly at the 2-position (α), because the intermediate formed by attack at C-2 is more resonance-stabilised.
Pyrrole is a very weak base, because its N lone pair is part of the aromatic sextet. Its N–H is weakly acidic.
Pyridine: N is sp² and its lone pair is in an sp² orbital in the plane of the ring, not part of the π sextet. So pyridine is basic (weaker than aliphatic amines, stronger than pyrrole).
Pyridine is π-deficient: electrophilic substitution is difficult and occurs at the 3-position (β) under harsh conditions; nucleophilic substitution occurs at 2 and 4 (Chichibabin reaction with NaNH₂ gives 2-aminopyridine).
Paal–Knorr synthesis: 1,4-dicarbonyl compounds give furan (with acid/P₂O₅), pyrrole (with NH₃) or thiophene (with P₂S₅).
Industrial note: furan from furfural (obtained from pentosans in agricultural waste like corn cobs and bran).
महत्त्वाच्या संज्ञा
Heteroatom – atom other than carbon in a ring (N, O, S).
Heterocycle (विषमचक्रीय संयुग) – ring compound containing a heteroatom.
π-excessive ring – ring with more π electron density per atom than benzene (pyrrole).
π-deficient ring – ring with less π electron density than benzene (pyridine).
Chichibabin reaction – amination of pyridine at C-2 with sodamide.
Pyrrole is cyclic and planar. Each carbon gives one p electron (4 total) and nitrogen gives its lone pair (2), making 6 π electrons in a continuous ring of p orbitals. 6 = 4n + 2 with n = 1, so it is aromatic.
Q2. Why is pyridine more basic than pyrrole?
In pyridine the N lone pair lies in an sp² orbital outside the π system, so it is free to accept H⁺. In pyrrole the lone pair is part of the aromatic sextet; using it would destroy aromaticity. So pyridine is much more basic.
Q3. Why does electrophilic substitution in pyrrole occur at the 2-position?
Attack at C-2 gives a carbocation intermediate with three resonance structures (charge spread over C-3, C-5 and N).
Attack at C-3 gives an intermediate with only two resonance structures.
More delocalisation means a more stable intermediate and lower activation energy.
So 2-substitution is preferred (e.g. nitration, sulphonation, acylation).
Q4. Compare the reactivity of pyrrole, furan, thiophene and benzene towards electrophiles.
All three five-membered rings are π-excessive (6 electrons over 5 atoms).
Pyrrole: N is least electronegative of N and O; donates lone pair best → most reactive.
Furan: O holds lone pair more tightly → less reactive than pyrrole.
Thiophene: S is large; 3p–2p overlap is weaker, but it is most aromatic → least reactive of the three.
Organometallic compounds contain at least one direct bond between a carbon atom and a metal (or metalloid like B, Si). This topic covers classification, the Grignard reagent (preparation and many reactions), organolithium and organozinc compounds, metal carbonyls, ferrocene, and the 18-electron rule, with industrial uses.
महत्त्वाचे मुद्दे
A compound is organometallic only if a metal–carbon bond is present. Sodium ethoxide (C₂H₅ONa) is not organometallic (Na is bonded to O).
Classification by bond type: ionic (Na, K compounds), σ-bonded (Grignard reagent, R₂Zn, (C₂H₅)₄Pb), π-bonded (ferrocene, Zeise's salt), and metal carbonyls.
Grignard reagent (RMgX), discovered by Victor Grignard (Nobel Prize 1912). Prepared from an alkyl halide and Mg turnings in dry ether.
The C–Mg bond is polar; carbon carries a partial negative charge. So R acts as a carbanion-like nucleophile and a strong base.
Grignard reagent must be kept away from water, alcohols and acids, because it reacts to give alkanes.
Organolithium (RLi) are more reactive than Grignard reagents; organozinc (R₂Zn, Frankland) are less reactive. Organozinc is used in the Reformatsky reaction.
Metal carbonyls: CO bonds to metals by σ-donation from C plus π-back-bonding from metal d orbitals into CO π* orbitals. Examples: Ni(CO)₄ (tetrahedral), Fe(CO)₅ (trigonal bipyramidal), Cr(CO)₆ (octahedral).
18-electron rule: stable organometallic complexes often have 18 valence electrons around the metal (like a noble gas). Ni(CO)₄: 10 + 4 × 2 = 18.
Ferrocene Fe(C₅H₅)₂ is a 'sandwich' compound: Fe²⁺ between two cyclopentadienyl rings; 18 electrons; aromatic-like reactions (Friedel–Crafts acylation).
Uses: Mond process (Ni(CO)₄ to purify nickel), Wilkinson's catalyst [RhCl(PPh₃)₃] for hydrogenation, Ziegler–Natta catalyst (TiCl₄ + Al(C₂H₅)₃) for polythene, Grignard reagents in synthesis.
महत्त्वाच्या संज्ञा
Organometallic compound (कार्बधातू संयुग) – compound with a direct metal–carbon bond.
Hapticity (η) – number of ligand atoms bonded to the metal (ferrocene rings are η⁵).
Back-bonding – metal gives d electrons back to empty ligand π* orbitals.
Sandwich compound – metal between two flat aromatic rings.
Dry ether – water-free ether; it stabilises RMgX by coordination.
18-electron count = metal group number electrons + 2 per CO (or as per ligand) ± charge
५ महत्त्वाचे प्रश्न
Q1. What are organometallic compounds? Is CH₃COONa organometallic?
Compounds containing at least one direct metal–carbon bond, like CH₃MgBr and ferrocene. CH₃COONa is not organometallic, because Na is bonded to oxygen, not carbon.
Q2. Why must a Grignard reagent be prepared in dry ether?
RMgX reacts instantly with water to form an alkane, so moisture destroys it. Dry ether also coordinates with Mg through its oxygen lone pairs and stabilises the reagent.
Q3. How will you prepare ethanol, propan-2-ol and 2-methylpropan-2-ol from CH₃MgI?
Ethanol (1°): CH₃MgI + HCHO, then hydrolysis.
Propan-2-ol (2°): CH₃MgI + CH₃CHO, then hydrolysis → CH₃CH(OH)CH₃.
2-methylpropan-2-ol (3°): CH₃MgI + CH₃COCH₃, then hydrolysis.
In each case the R group adds to the carbonyl carbon, then H₂O/H⁺ gives the alcohol.
Q4. Apply the 18-electron rule to Fe(CO)₅ and Cr(CO)₆.
Fe is in Group 8 → 8 valence electrons; 5 CO × 2 = 10; total = 18 ✓
Cr is in Group 6 → 6 electrons; 6 CO × 2 = 12; total = 18 ✓
In a molecular rearrangement, an atom or group moves from one atom to another inside the same molecule, giving a product with a new carbon skeleton or a new functional group position. This topic covers the classification of rearrangements and the mechanisms of the pinacol–pinacolone, Beckmann, Hofmann, Curtius, benzilic acid, Claisen and Fries rearrangements.
महत्त्वाचे मुद्दे
Classification by the atom receiving the group: electron-deficient carbon (pinacol, Wagner–Meerwein, benzilic acid), electron-deficient nitrogen (Beckmann, Hofmann, Curtius, Lossen), electron-deficient oxygen (Baeyer–Villiger), and aromatic/sigmatropic types (Claisen, Fries).
Migratory aptitude (rough order): aryl > H ≈ alkyl; among alkyls 3° > 2° > 1° > methyl. Groups move with their electron pair (1,2-shift).
Wagner–Meerwein: carbocation rearranges to a more stable carbocation by an alkyl or hydride shift (seen in SN1/E1 reactions).
Beckmann rearrangement: ketoxime + acid (H₂SO₄, PCl₅) → N-substituted amide. The group anti to the –OH migrates. Cyclohexanone oxime → caprolactam (monomer of nylon-6).
Hofmann rearrangement (bromamide): RCONH₂ + Br₂ + NaOH → RNH₂ (one carbon less) through an N-bromoamide and an isocyanate intermediate.
Curtius rearrangement: acyl azide (RCON₃) on heating → isocyanate (R–N=C=O) + N₂ → amine on hydrolysis.
Q4. Explain the Beckmann rearrangement with mechanism and stereochemistry.
Protonation of oxime –OH, loss of water with simultaneous anti migration, nitrilium ion, attack by water, tautomerisation to amide. Show how syn and anti oximes give different amides. Mention caprolactam for nylon-6.
Q5. Write notes on the benzilic acid and Fries rearrangements.
Equations, mechanism outline (OH⁻ attack and 1,2-phenyl shift for benzilic; acylium ion and AlCl₃ complex for Fries), and the effect of temperature on the Fries ortho/para ratio.
परीक्षेसाठी टीप
Remember "anti group migrates" for Beckmann and "one carbon less" for Hofmann; these two lines are the most tested facts.
Photochemistry studies chemical reactions that are started by the absorption of light (UV or visible). This topic covers the difference between thermal and photochemical reactions, the laws of photochemistry, quantum yield and the reasons for high and low values, photosensitisation, fluorescence and phosphorescence (Jablonski diagram), and chemiluminescence.
महत्त्वाचे मुद्दे
Thermal reactions take place in the dark and depend strongly on temperature. Photochemical reactions need light; their rate depends on light intensity and is only slightly affected by temperature. ΔG can even be positive for a photochemical reaction.
Grotthuss–Draper law: only light that is absorbed can cause a photochemical change.
Stark–Einstein law of photochemical equivalence: each molecule taking part in the primary process absorbs one quantum of light (one photon).
Einstein: energy of one mole of photons, E = Nₐhc/λ. For λ = 400 nm, E ≈ 299 kJ mol⁻¹.
Quantum yield (φ) = number of molecules reacted / number of photons absorbed.
High quantum yield (φ > 1): chain reactions. H₂ + Cl₂ → 2HCl has φ about 10⁴–10⁶. Decomposition of HI has φ = 2.
Low quantum yield (φ < 1): deactivation of excited molecules by collision, fluorescence, recombination of fragments (e.g. H₂ + Br₂ has a low φ at room temperature).
Photosensitisation: a substance absorbs light and passes the energy to another molecule that cannot absorb it directly. Example: Hg vapour sensitises the dissociation of H₂; chlorophyll in photosynthesis.
Fluorescence: emission of light from the excited singlet state (S₁ → S₀); stops almost at once when light is removed (about 10⁻⁸ s). Fluorite, fluorescein.
Phosphorescence: emission from the triplet state (T₁ → S₀) after intersystem crossing; continues after the light is removed (ZnS).
Chemiluminescence: light given out by a chemical reaction (oxidation of luminol; firefly light is bioluminescence).
महत्त्वाच्या संज्ञा
Photon – one quantum of light energy, E = hν.
Quantum yield (क्वांटम उत्पन्न) – molecules reacted per photon absorbed.
Intersystem crossing – non-radiative change from a singlet to a triplet state.
सूत्रे
E (one photon) = hν = hc/λ (h = 6.626 × 10⁻³⁴ J s, c = 3 × 10⁸ m s⁻¹)
One einstein = Nₐhc/λ = 0.1196/λ J mol⁻¹ (λ in m)
φ = moles reacted / einsteins absorbed
Process
Transition
Lifetime
Fluorescence
S₁ → S₀
about 10⁻⁹–10⁻⁷ s
Phosphorescence
T₁ → S₀
10⁻³ s to seconds
५ महत्त्वाचे प्रश्न
Q1. Define quantum yield. Why is it very high for H₂ + Cl₂?
φ = molecules reacted per photon absorbed. One photon splits Cl₂ into two Cl atoms, which start a chain (Cl + H₂ → HCl + H; H + Cl₂ → HCl + Cl) that repeats thousands of times, so φ is about 10⁴–10⁶.
Q2. Distinguish between fluorescence and phosphorescence.
Fluorescence comes from the singlet excited state and stops immediately when light is cut off. Phosphorescence comes from the triplet state after intersystem crossing and continues for some time after the light is removed.
Q3. Calculate the energy of one einstein of light of wavelength 400 nm.
Grotthuss–Draper law with an example; Stark–Einstein law; primary and secondary processes; why many reactions do not show φ = 1 (secondary reactions). Give the HI example: HI + hν → H + I; H + HI → H₂ + I; I + I → I₂, so φ = 2.
Q5. Explain the Jablonski diagram.
Draw S₀, S₁, S₂ and T₁ levels with vibrational sub-levels. Mark absorption, vibrational relaxation, internal conversion, fluorescence, intersystem crossing and phosphorescence with straight (radiative) and wavy (non-radiative) arrows.
परीक्षेसाठी टीप
Learn the H₂–Cl₂ chain and HI decomposition steps; they are used to explain both high quantum yield and quantum yield = 2.