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.