Chemistry रसायनशास्त्र

Quick summary

Chemistry is the study of matter, its properties and the changes it undergoes. This chapter classifies matter, explains physical and chemical properties and SI units, and introduces the laws of chemical combination, Dalton's atomic theory, atomic and molecular mass, and the mole concept. The mole is the base for every calculation in chemistry.

Key points

• Matter is classified into pure substances (elements and compounds) and mixtures (homogeneous and heterogeneous).
• Pure substances have a fixed composition and fixed properties; mixtures do not.
• Physical properties (colour, melting point, density) can be measured without changing the substance. Chemical properties (burning, reacting with acid) involve a chemical change.
• There are 7 SI base units: metre (m), kilogram (kg), second (s), kelvin (K), ampere (A), mole (mol), candela (cd).
• Law of conservation of mass (Lavoisier): mass is neither created nor destroyed in a chemical reaction.
• Law of definite proportions (Proust): a compound always contains the same elements in the same proportion by mass, whatever its source.
• Law of multiple proportions (Dalton): when two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in a simple whole-number ratio (e.g. NO and NO₂: 16 g : 32 g oxygen per 14 g N = 1 : 2).
• Gay Lussac's law of gaseous volumes: gases react and are produced in simple whole-number ratios by volume at the same temperature and pressure (N₂ + 3H₂ → 2NH₃ gives 1 : 3 : 2).
• Avogadro's law: equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules.
• Dalton's atomic theory (1808): matter is made of tiny indivisible atoms; atoms of one element are identical; compounds form when atoms combine in simple ratios.
• Atomic masses are measured on the carbon-12 scale: 1 u = 1/12th of the mass of one C-12 atom.

Important terms

• Atomic mass unit (u) – 1/12th the mass of one carbon-12 atom.
• Average atomic mass – weighted average of the masses of the natural isotopes of an element (chlorine ≈ 35.5 u).
• Molecular mass – sum of the atomic masses of all atoms in a molecule (H₂O = 18 u).
• Formula mass – used for ionic compounds which have no separate molecules (NaCl = 58.5 u).
• Mole – amount of substance containing 6.022 × 10²³ particles (atoms, molecules or ions).
• Molar mass – mass of one mole of a substance in grams; numerically equal to atomic/molecular mass (H₂O = 18 g mol⁻¹).
• Molar volume – volume of one mole of a gas at STP = 22.4 dm³ (at the new IUPAC STP of 1 bar it is 22.71 L).

Formulas

• Number of moles n = mass (g) ÷ molar mass (g mol⁻¹)
• Number of particles = n × Nₐ, where Nₐ = 6.022 × 10²³ mol⁻¹ (Avogadro's number)
• Moles of a gas n = volume at STP (dm³) ÷ 22.4 dm³ mol⁻¹
• Average atomic mass = Σ (isotopic mass × fractional abundance)
• Temperature: K = °C + 273.15 ; °F = (9/5)°C + 32

5 important questions

Q1. State and explain Avogadro's law.

Equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules. Example: 2 volumes of H₂ + 1 volume of O₂ give 2 volumes of steam, so 2 molecules of H₂ and 1 molecule of O₂ give 2 molecules of H₂O. This is why oxygen must be diatomic (O₂).

Q2. What is the difference between 12 g of carbon and 12 u of carbon?

12 u is the mass of ONE carbon-12 atom. 12 g is the mass of ONE MOLE of carbon atoms, i.e. 6.022 × 10²³ atoms.

Q3. Calculate the number of moles of hydrogen in 0.448 L of hydrogen gas at STP.

n = 0.448 dm³ ÷ 22.4 dm³ mol⁻¹ = 0.02 mol.

Q4. Calculate the number of atoms in (a) 254 u of iodine (b) 254 g of iodine. (I = 127 u)

(a) 254 u ÷ 127 u = 2 atoms.
(b) 254 g ÷ 127 g mol⁻¹ = 2 mol; atoms = 2 × 6.022 × 10²³ = 1.2044 × 10²⁴ atoms.

Q5. State the law of multiple proportions with an example.

When two elements combine to form two or more compounds, the masses of one element that combine with a fixed mass of the other bear a simple whole-number ratio. In CO and CO₂, 12 g of carbon combines with 16 g and 32 g oxygen, ratio 1 : 2.

Exam tip

Always write units (u, g, mol, dm³) in every step of a mole problem. Remember: "u" is for one particle, "g" is for one mole.

Quick summary

Analytical chemistry is the branch of chemistry that deals with separation, identification and quantitative determination of the composition of substances. This chapter covers qualitative and quantitative analysis, handling numbers (scientific notation, significant figures, accuracy, precision, errors), finding empirical and molecular formulae, stoichiometry and limiting reagent, and ways of expressing concentration.

Key points

• Qualitative analysis tells WHAT is present; quantitative analysis tells HOW MUCH is present.
• Chemical methods of quantitative analysis include gravimetric (by weighing) and volumetric (titration) analysis.
• Large and small numbers are written in scientific notation: N × 10ⁿ, with 1 ≤ N < 10 (e.g. 0.00016 = 1.6 × 10⁻⁴).
• Accuracy = closeness of a measurement to the true value. Precision = closeness of repeated measurements to each other.
• Errors are expressed as absolute error, relative error and percentage error; mean deviation shows precision.
• Significant figures are the meaningful digits in a measurement: all certain digits plus one uncertain digit.
• In addition/subtraction, the answer keeps the least number of decimal places; in multiplication/division, the least number of significant figures.
• Empirical formula = simplest whole-number ratio of atoms; molecular formula = actual number of atoms = n × empirical formula.
• Stoichiometry is the calculation of masses/volumes of reactants and products using the balanced equation.
• The limiting reagent is the reactant used up first; it decides how much product forms.
• Molarity depends on temperature (volume changes with temperature); molality does not.

Important terms

• Least count – the smallest value that an instrument can measure correctly.
• Rounding off – dropping extra digits: if the digit removed is more than 5, raise the previous digit by 1; if less than 5, leave it.
• Percentage composition – mass of each element in 100 g of the compound.
• Mole fraction (x) – moles of a component ÷ total moles of all components.
• Molarity (M) – moles of solute per litre (dm³) of solution.
• Molality (m) – moles of solute per kilogram of solvent.

Formulas

• Absolute error = |observed value − true value|
• Relative error = absolute error ÷ true value ; percentage error = relative error × 100
• Mass percent = (mass of solute ÷ mass of solution) × 100
• Mole fraction xA = nA ÷ (nA + nB); xA + xB = 1
• Molarity M = moles of solute ÷ volume of solution in L (unit mol L⁻¹ or M)
• Molality m = moles of solute ÷ mass of solvent in kg (unit mol kg⁻¹)
• n (in molecular formula) = molar mass ÷ empirical formula mass

Significant figure rules (quick list)

1. All non-zero digits are significant (285 → 3).
2. Zeros between non-zero digits are significant (2.05 → 3).
3. Leading zeros are not significant (0.0034 → 2).
4. Trailing zeros after a decimal point are significant (2.200 → 4).
5. In scientific notation, count only the digits of N (1.50 × 10⁴ → 3).

5 important questions

Q1. Distinguish between accuracy and precision.

Accuracy is how close a measured value is to the true value. Precision is how close repeated readings are to each other. Readings can be precise but not accurate (e.g. a faulty balance giving the same wrong value every time).

Q2. Calculate the molarity of NaOH when 4 g of it is dissolved to make 250 mL of solution.

Moles = 4 ÷ 40 = 0.1 mol. Volume = 0.250 L. M = 0.1 ÷ 0.250 = 0.4 M.

Q3. A compound contains 4.07 % H, 24.27 % C and 71.65 % Cl. Molar mass = 98.96 g mol⁻¹. Find its empirical and molecular formula.

Moles: H = 4.07/1.008 = 4.04; C = 24.27/12.01 = 2.02; Cl = 71.65/35.45 = 2.02. Divide by 2.02 → H 2 : C 1 : Cl 1. Empirical formula CH₂Cl (mass 49.48). n = 98.96/49.48 = 2. Molecular formula C₂H₄Cl₂.

Q4. What is a limiting reagent? Explain with an example.

The reactant that is completely consumed first and limits the product. In 2NO + O₂ → 2NO₂ with 8 mol NO and 7 mol O₂: 8 mol NO needs only 4 mol O₂, so NO is the limiting reagent and 3 mol O₂ is left over; 8 mol NO₂ forms.

Q5. Why does molarity of a solution depend on temperature but molality does not?

Molarity uses volume of solution, which expands or contracts with temperature. Molality uses mass of solvent, and mass does not change with temperature.

Exam tip

In empirical formula problems, make a neat table: element, %, atomic mass, moles, simplest ratio. Examiners give step marks for the table.

Quick summary

Chemicals made in the laboratory or obtained from nature are usually impure. This chapter explains common techniques to purify and separate substances: crystallization, distillation (simple, fractional, under reduced pressure), solvent extraction and chromatography. The method chosen depends on the physical properties of the substances, such as solubility, boiling point and adsorption.

Key points

• Crystallization purifies solid compounds. The impure solid is dissolved in a minimum amount of hot solvent, filtered hot, and cooled slowly; pure crystals separate and impurities stay in the solution (mother liquor).
• A good crystallization solvent dissolves the compound well when hot and poorly when cold, does not react with it, and dissolves the impurities easily (or not at all).
• Fractional crystallization separates two solids that have different solubilities in the same solvent; the less soluble one crystallizes first.
• Simple distillation separates a liquid from non-volatile impurities, or two liquids whose boiling points differ a lot.
• Fractional distillation separates liquids whose boiling points are close (e.g. acetone and benzene, petroleum fractions). A fractionating column gives many vaporisation–condensation cycles.
• Distillation under reduced pressure is used for liquids that decompose at or below their boiling point. Lowering the pressure lowers the boiling point. In the soap industry, glycerol is separated from soap this way.
• Solvent extraction: an organic compound in water is shaken with an immiscible organic solvent in which it is more soluble (e.g. ether, chloroform); the layers are separated and the solvent evaporated.
• Chromatography (Greek chroma = colour), discovered by Tswett in 1903 for plant pigments, separates components by their different distribution between a stationary phase and a mobile phase.
• Adsorption chromatography – column chromatography and thin layer chromatography (TLC), with silica gel or alumina as adsorbent.
• Partition chromatography – paper chromatography; water held in the paper is the stationary phase.
• Colourless spots are seen under UV light, in an iodine chamber (spots turn brown) or with a spraying reagent.

Important terms

• Saturated solution – a solution that cannot dissolve any more solute at that temperature.
• Residue / filtrate – the solid left on the filter paper / the clear liquid that passes through.
• Stationary phase – the fixed phase (solid adsorbent or liquid held on paper) on which components are held back.
• Mobile phase – the moving solvent that carries the components.
• Chromatogram – the developed plate, paper or column showing separated components.
• Retention factor (Rf) – a measure of how far a component moves compared to the solvent.

Formulas

• Rf = distance travelled by the solute from the base line ÷ distance travelled by the solvent from the base line
• Rf has no unit and is always less than 1. A component held strongly by the stationary phase has a small Rf.

5 important questions

Q1. Distinguish between simple distillation and fractional distillation.

Simple distillation separates liquids with a large difference in boiling points (or a liquid from non-volatile solids) and uses no column. Fractional distillation separates liquids with close boiling points and uses a fractionating column, e.g. acetone and benzene, crude oil fractions.

Q2. Why is distillation under reduced pressure used? Give one example.

Some liquids decompose at or before their normal boiling point. Reducing pressure (with a water pump or vacuum pump) makes them boil at a lower temperature without decomposing. Example: separating glycerol from soap in the soap industry.

Q3. Why should the mixture be spotted above the level of the mobile phase in TLC/paper chromatography?

If the spot is dipped in the solvent, the mixture dissolves into the solvent in the jar instead of moving up the plate, so no separation takes place.

Q4. Compare TLC and paper chromatography.

TLC: stationary phase is a thin layer (about 0.2 mm) of silica gel or alumina on a glass plate; works by adsorption; faster. Paper chromatography: stationary phase is water held in special paper; works by partition. In both, colourless spots are seen by UV light, iodine or spray reagents, and Rf values are calculated.

Q5. What will happen if the upper outlet of the condenser is connected to the tap?

The condenser will not fill completely with water; air remains in the jacket, cooling is poor and vapours escape without condensing. Water must enter at the lower inlet and leave from the upper outlet.

Exam tip

Learn one example for each technique and practise drawing the labelled distillation set-up and a TLC chromatogram — diagrams are asked often.

Quick summary

This chapter traces how our picture of the atom developed: discovery of the electron, proton and neutron, Rutherford's nuclear model, Bohr's model of the hydrogen atom and its line spectrum, and finally the quantum mechanical model with orbitals, quantum numbers and electronic configuration.

Key points

• Electron discovered by J. J. Thomson (1897, cathode rays). Rutherford's gold foil (α-particle scattering) experiment (1911) showed a small, heavy, positive nucleus; he identified the proton in 1919. Neutron discovered by James Chadwick (1932).
• Charge on electron = −1.6 × 10⁻¹⁹ C; mass of electron ≈ 9.1 × 10⁻³¹ kg. An electron is about 1837 times lighter than a hydrogen atom.
• Isotopes have the same atomic number, different mass number (¹²C, ¹⁴C). Isobars have the same mass number, different atomic number (⁴⁰Ar, ⁴⁰Ca). Isotones have the same number of neutrons (¹⁴C, ¹⁶O). Isoelectronic species have the same number of electrons (Ne, Na⁺, O²⁻).
• Rutherford's model could not explain the stability of the atom: a revolving electron should lose energy and fall into the nucleus.
• Bohr's postulates: electrons move in fixed circular orbits (stationary states) with fixed energy; energy is emitted or absorbed only when an electron jumps between orbits (ΔE = hν); angular momentum is quantised (mvr = nh/2π).
• Hydrogen line spectrum series: Lyman (to n = 1, ultraviolet), Balmer (to n = 2, visible), Paschen (n = 3), Brackett (n = 4), Pfund (n = 5) – the last three in infrared.
• Bohr's model failed for multi-electron atoms, finer lines of the spectrum, Zeeman and Stark effects, and chemical bonding.
• de Broglie (1924): matter has dual nature; λ = h/mv. Heisenberg: position and momentum of an electron cannot both be known exactly.
• The quantum mechanical model (Schrödinger equation) describes electrons by orbitals – regions of high probability of finding the electron.
• Four quantum numbers: principal (n, size and energy), azimuthal (l = 0 to n−1, shape: s, p, d, f), magnetic (mₗ = −l to +l, orientation) and spin (mₛ = +½ or −½).
• s orbital is spherical; p orbitals are dumb-bell shaped (px, py, pz); d orbitals are mostly double dumb-bell.
• Electrons fill orbitals by the Aufbau principle (lower energy first: 1s 2s 2p 3s 3p 4s 3d 4p…), Pauli's exclusion principle (no two electrons with all four quantum numbers same; max 2 per orbital) and Hund's rule (pairing starts only after each orbital of a subshell has one electron).
• Half-filled and completely filled subshells are extra stable: Cr = [Ar] 4s¹ 3d⁵ and Cu = [Ar] 4s¹ 3d¹⁰.

Important terms

• Atomic number (Z) – number of protons in the nucleus.
• Mass number (A) – number of protons + neutrons.
• Stationary state – a Bohr orbit in which the electron does not radiate energy.
• Ground state – the lowest energy state of the atom (n = 1 for H).
• Orbital – three-dimensional region around the nucleus where the probability of finding an electron is maximum.
• Node – region where the probability of finding the electron is zero.

Formulas

• Neutrons = A − Z
• Energy of a photon E = hν = hc/λ ; h = 6.626 × 10⁻³⁴ J s, c = 3 × 10⁸ m s⁻¹
• Bohr radius rₙ = n² a₀ (for H), a₀ = 52.9 pm; for hydrogen-like ions rₙ = 52.9 × n²/Z pm
• Energy of nth orbit Eₙ = −2.18 × 10⁻¹⁸ (Z²/n²) J
• ΔE = 2.18 × 10⁻¹⁸ (1/n_f² − 1/n_i²) J
• de Broglie λ = h/mv ; Heisenberg Δx × Δp ≥ h/4π
• Maximum electrons in a shell = 2n²; orbitals in a shell = n²; electrons in a subshell = 2(2l + 1)

5 important questions

Q1. State Pauli's exclusion principle and Hund's rule.

Pauli: no two electrons in the same atom can have the same set of all four quantum numbers, so an orbital holds at most two electrons with opposite spins. Hund: in orbitals of equal energy (e.g. three 2p), electrons first occupy each orbital singly with parallel spins; pairing starts only after that.

Q2. Explain the anomalous configuration of Cr and Cu.

Expected Cr: [Ar] 4s² 3d⁴, actual [Ar] 4s¹ 3d⁵. Expected Cu: [Ar] 4s² 3d⁹, actual [Ar] 4s¹ 3d¹⁰. One 4s electron moves to 3d because half-filled (d⁵) and fully filled (d¹⁰) subshells have extra stability (symmetry and exchange energy).

Q3. Calculate the energy of the first orbit of He⁺.

For He⁺, Z = 2, n = 1. E₁ = −2.18 × 10⁻¹⁸ × (2²/1²) = −8.72 × 10⁻¹⁸ J.

Q4. Using quantum numbers, find the maximum electrons in the M shell.

M shell: n = 3, so l = 0, 1, 2 (3s, 3p, 3d). Orbitals = 1 + 3 + 5 = 9 = n². Electrons = 2 × 9 = 18 = 2n² (3s², 3p⁶, 3d¹⁰).

Q5. An atom has 29 electrons and 35 neutrons. Find protons, mass number and configuration.

Protons = 29 (copper). Mass number = 29 + 35 = 64. Configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d¹⁰.

Exam tip

Practise configurations of Z = 1 to 30, including Cr, Cu and ions like Fe²⁺ ([Ar] 3d⁶) and Fe³⁺ ([Ar] 3d⁵) — electrons are removed from 4s first.

Quick summary

Atoms join together by chemical bonds to reach a more stable (lower energy) state. This chapter explains ionic and covalent bonds (Kossel–Lewis approach and octet rule), Lewis structures and formal charge, shapes of molecules by VSEPR theory, valence bond theory with hybridization, molecular orbital theory, bond parameters, dipole moment and resonance.

Key points

• Octet rule: atoms gain, lose or share electrons to get 8 electrons in the outer shell (like a noble gas).
• Ionic (electrovalent) bond: complete transfer of electrons from a metal to a non-metal (Na → Na⁺, Cl → Cl⁻ in NaCl). Higher lattice enthalpy means a more stable ionic solid.
• Covalent bond: sharing of electron pairs (single, double, triple bonds). Coordinate bond: both shared electrons come from one atom (NH₃ → H⁺ gives NH₄⁺).
• Limitations of octet rule: incomplete octet (BeCl₂, BF₃), expanded octet (PCl₅, SF₆), odd-electron molecules (NO, NO₂); it cannot explain shape or energy of molecules.
• VSEPR theory: electron pairs around the central atom repel and stay as far apart as possible. Repulsion order: lone pair–lone pair > lone pair–bond pair > bond pair–bond pair.
• Shapes: BeCl₂ linear (180°), BF₃ trigonal planar (120°), CH₄ tetrahedral (109°28'), NH₃ pyramidal (107°), H₂O bent/V-shaped (104°35'), PCl₅ trigonal bipyramidal, SF₆ octahedral.
• Valence bond theory: a covalent bond forms by overlap of half-filled atomic orbitals with opposite spins. Head-on (axial) overlap gives a sigma (σ) bond; sideways overlap gives a pi (π) bond. σ is stronger than π.
• Hybridization: mixing of atomic orbitals of nearly equal energy to form equal new orbitals. sp (linear, 180°, C₂H₂, BeCl₂), sp² (trigonal planar, 120°, C₂H₄, BF₃), sp³ (tetrahedral, 109°28', CH₄, C₂H₆).
• Molecular orbital theory: atomic orbitals combine to form bonding MOs (lower energy) and antibonding MOs (higher energy, marked *). Electrons fill MOs by Aufbau, Pauli and Hund's rules.
• MO theory explains why O₂ is paramagnetic (two unpaired electrons in π*2p) — Lewis theory cannot.
• Bond order higher → bond shorter and stronger. Bond order 0 means the molecule does not exist (He₂).
• Resonance: when one Lewis structure is not enough (O₃, CO₃²⁻), the real molecule is a hybrid of several structures.
• Dipole moment shows polarity. CO₂ (linear) and BF₃ have zero dipole moment though their bonds are polar; H₂O and NH₃ have a net dipole moment.

Important terms

• Lattice enthalpy – energy needed to completely separate one mole of a solid ionic compound into gaseous ions.
• Formal charge – charge given to an atom in a Lewis structure, to choose the best structure.
• Bond length – equilibrium distance between nuclei of two bonded atoms.
• Bond enthalpy – energy needed to break one mole of bonds of a type in gaseous molecules.
• Bond angle – angle between two bonds around the central atom.
• Paramagnetic / diamagnetic – has unpaired electrons, attracted by magnet / all electrons paired, weakly repelled.

Formulas

• Formal charge = (valence electrons) − (non-bonding electrons) − ½ (bonding electrons)
• Bond order = ½ (Nb − Na), where Nb = electrons in bonding MOs and Na = electrons in antibonding MOs
• Dipole moment μ = charge (Q) × distance (r); unit Debye (1 D = 3.33564 × 10⁻³⁰ C m)

5 important questions

Q1. Distinguish between sigma and pi bonds.

Sigma: formed by head-on overlap along the internuclear axis; strong; can exist alone; free rotation possible. Pi: formed by sideways overlap of p orbitals; weaker; forms only after a sigma bond; no free rotation. A double bond = 1σ + 1π; triple bond = 1σ + 2π.

Q2. Explain the geometry of methane by hybridization.

C (ground state 2s² 2p²) promotes one 2s electron to 2p, then one 2s and three 2p orbitals mix to give four equal sp³ orbitals pointing to corners of a tetrahedron. Each overlaps with the 1s orbital of H forming four C–H σ bonds. Shape tetrahedral, H–C–H angle 109°28'.

Q3. Why is the bond angle 107° in NH₃ but 104°35' in H₂O, though both are sp³?

NH₃ has one lone pair; H₂O has two. Lone pair–lone pair repulsion is the greatest, so in water the two bond pairs are pushed closer, giving a smaller angle than in NH₃.

Q4. Write the MO configuration of O₂ and find its bond order and magnetic nature.

O₂ (16 electrons): σ1s² σ1s² σ2s² σ2s² σ2pz² π2px² π2py² π2px¹ π2py¹. Nb = 10, Na = 6, bond order = ½(10 − 6) = 2. Two unpaired electrons → paramagnetic.

Q5. CO₂ has polar bonds but zero dipole moment. Explain.

CO₂ is linear (O=C=O). The two C=O bond dipoles are equal and point in opposite directions, so they cancel and the net dipole moment is zero.

Exam tip

Make a table of molecule → electron pairs → hybridization → shape → bond angle. Practise Lewis structures of H₂O, CO₂, CH₄, C₂H₄, SO₂ and MO diagrams of H₂, He₂, O₂.

Quick summary

Redox reactions are reactions in which oxidation and reduction take place together. This chapter explains oxidation and reduction in three ways — classical (oxygen/hydrogen), electronic, and oxidation number — and shows how to balance redox equations by the oxidation number method and the ion-electron (half-reaction) method. It ends with electrode potential.

Key points

• Classical idea: oxidation = addition of oxygen or removal of hydrogen; reduction = removal of oxygen or addition of hydrogen.
• Electronic idea: oxidation is loss of electrons, reduction is gain of electrons.
• An oxidising agent (oxidant) accepts electrons and is itself reduced. A reducing agent (reductant) gives electrons and is itself oxidised.
• In terms of oxidation number: oxidation = increase in oxidation number; reduction = decrease.
• Rules for oxidation number: free element = 0; monoatomic ion = its charge; O is usually −2 (−1 in peroxides like H₂O₂, −½ in superoxides like KO₂, +2 in OF₂); H is +1 (−1 in metal hydrides like NaH); F is always −1; alkali metals +1, alkaline earth metals +2.
• Sum of oxidation numbers = 0 in a neutral molecule and = charge in an ion.
• Stock notation writes the oxidation number in Roman numerals: FeCl₃ = iron(III) chloride, MnO₂ = Mn(IV)O₂.
• Not every reaction is redox: BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl has no change in oxidation numbers.
• In bleaching powder CaOCl₂ = Ca(OCl)Cl, one Cl is +1 and the other is −1.
• Oxidation number method: find the change in oxidation number, make increase = decrease, then balance O with H₂O and H with H⁺ (acidic) or OH⁻ (basic).
• Ion-electron method: split into oxidation and reduction half-reactions, balance each (atoms, then O with H₂O, H with H⁺, charge with electrons), equalise electrons and add.
• A redox reaction can be arranged to produce electricity in a galvanic cell. Standard electrode potential (E°) is measured against the standard hydrogen electrode (E° = 0.00 V).

Important terms

• Oxidation number – the charge an atom appears to have when electrons are counted by fixed rules.
• Half-reaction – the oxidation part or the reduction part of a redox reaction shown separately.
• Disproportionation – one element in a substance is oxidised and reduced at the same time.
• Standard electrode potential (E°) – potential of an electrode at 1 M concentration, 1 bar and 298 K.

Formulas

• In a compound: Σ (oxidation numbers) = 0 ; in an ion: Σ = charge on ion
• Example: C in H₂CO₃: 2(+1) + x + 3(−2) = 0 → x = +4
• Electrons lost by reductant = electrons gained by oxidant

5 important questions

Q1. Define oxidation and reduction in terms of electron transfer with an example.

Oxidation is loss of electrons; reduction is gain of electrons. In 2Na + Cl₂ → 2NaCl, each Na loses one electron (oxidised) and each Cl gains one electron (reduced).

Q2. Find the oxidation number of Cr in K₂Cr₂O₇.

2(+1) + 2x + 7(−2) = 0 → 2 + 2x − 14 = 0 → x = +6.

Q3. Balance in acidic medium by ion-electron method: Cr₂O₇²⁻ + Fe²⁺ → Cr³⁺ + Fe³⁺.

Oxidation: Fe²⁺ → Fe³⁺ + e⁻ (× 6).
Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O.
Add: Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O.

Q4. In which reaction does carbon change oxidation state from −4 to +4?

Combustion of methane: CH₄ + 2O₂ → CO₂ + 2H₂O. In CH₄ carbon is −4; in CO₂ it is +4.

Q5. Write formulas: mercury(II) chloride, tin(IV) oxide, chromium(III) oxide, thallium(I) sulphate.

HgCl₂, SnO₂, Cr₂O₃, Tl₂SO₄.

Exam tip

In balancing, always check both atoms and total charge on each side at the end. Write the steps clearly — each step carries marks.

Quick summary

The modern periodic table arranges elements in order of increasing atomic number. Its 7 periods and 18 groups follow directly from electronic configuration, which also divides the table into s, p, d and f blocks. This chapter explains these blocks and the periodic trends in atomic radius, ionic radius, ionization enthalpy, electron gain enthalpy, electronegativity, valency and metallic character.

Key points

• Modern periodic law: properties of elements are a periodic function of their atomic numbers.
• Period number = principal quantum number (n) of the valence shell. Elements of a group have the same outer electronic configuration, so similar chemical properties.
• s-block (groups 1, 2; ns¹⁻²): reactive metals, low ionization enthalpy, form ionic compounds, fixed valency.
• p-block (groups 13–18; ns² np¹⁻⁶): metals, metalloids and non-metals; group 17 = halogens (ns² np⁵), group 18 = noble gases (ns² np⁶).
• d-block (groups 3–12; (n−1)d¹⁻¹⁰ ns¹⁻²): transition metals; variable oxidation states, coloured ions (when d-orbitals are partly filled), catalytic activity. Cu²⁺ (3d⁹) salts are coloured; Zn²⁺ (3d¹⁰) salts are colourless.
• f-block (lanthanoids and actinoids; (n−2)f orbitals filling): placed separately at the bottom of the table.
• Effective nuclear charge (Zeff) is the actual pull felt by valence electrons = Z − screening by inner electrons.
• Atomic radius decreases across a period (Zeff rises) and increases down a group (new shells added).
• A cation is smaller than its parent atom; an anion is larger. In isoelectronic ions, more protons → smaller ion (F⁻ 133 pm > Na⁺ 98 pm).
• Ionization enthalpy increases across a period and decreases down a group. Second ionization enthalpy is always greater than the first (removing an electron from a positive ion is harder).
• Electronegativity increases across a period and decreases down a group. Fluorine is the most electronegative element (4.0 on Pauling scale).
• Metallic character decreases across a period and increases down a group (K > Mg > Al > B). In period 3: Al is a metal, Si a metalloid, P a non-metal.

Important terms

• Screening (shielding) effect – inner electrons reduce the nuclear attraction on outer electrons.
• Ionization enthalpy – energy needed to remove the most loosely held electron from an isolated gaseous atom.
• Electron gain enthalpy – enthalpy change when an electron is added to an isolated gaseous atom (can be measured experimentally).
• Electronegativity – tendency of a bonded atom to attract the shared electron pair to itself (a relative number, not measured directly).
• Valency – combining capacity of an atom; for s and p elements it equals the number of valence electrons or 8 minus that number.
• Representative elements – s and p block elements.

Key facts

• Period lengths: 2, 8, 8, 18, 18, 32, 32 elements.
• Group number of p-block element = 10 + number of valence electrons (e.g. Ge, 4s² 4p² → group 14).
• Electronegativity values (Pauling): Li 1.0, Be 1.5, B 2.0, C 2.5, N 3.0, O 3.5, F 4.0.
• Atomic radii: Cl 99 pm < Br 114 pm < I 133 pm (increase down group 17).
• Ionization enthalpy: Li 520 kJ mol⁻¹, F 1681 kJ mol⁻¹.

5 important questions

Q1. Why do elements of the same group have similar chemical properties?

Chemical properties depend on valence electrons. All elements of a group have the same outer electronic configuration (e.g. group 1 = ns¹), so they react in similar ways.

Q2. Why is the ionization enthalpy of F (1681 kJ mol⁻¹) much higher than that of Li (520 kJ mol⁻¹)?

Both are in period 2. Across the period nuclear charge increases from 3 to 9 while electrons enter the same shell, so Zeff increases and atomic size decreases. The valence electron of F is held much more tightly.

Q3. Explain the screening effect with an example.

Inner-shell electrons stand between the nucleus and valence electrons and reduce the attraction felt by them. In Na (2, 8, 1), the 10 inner electrons screen the 3s electron, so it feels a small effective charge and is easily removed.

Q4. Why does Cu form coloured salts while Zn forms colourless salts?

Cu²⁺ has 3d⁹ configuration with an unpaired electron; d–d transitions absorb visible light, giving colour. Zn²⁺ is 3d¹⁰ (completely filled), no d–d transition is possible, so its salts are colourless.

Q5. Write the outer configuration of Ge (period 4, group 14) and Po (period 6, group 16).

Ge: 4s² 4p² (n = 4 from period; 2 + 2 = 4 valence electrons for group 14). Po: 6s² 6p⁴ (n = 6; 6 valence electrons for group 16).

Exam tip

Learn trends as arrows: across a period → size ↓, IE ↑, EN ↑, metallic ↓; down a group → the opposite. Always give the reason (Zeff or number of shells).

Quick summary

This chapter covers hydrogen and the s-block elements: alkali metals (group 1) and alkaline earth metals (group 2). It explains hydrogen's position, isotopes and preparation, then the trends in properties of groups 1 and 2, the diagonal relationship, and some important compounds such as sodium carbonate, sodium hydroxide, calcium carbonate, hydrogen peroxide and lithium aluminium hydride.

Key points

• Hydrogen (1s¹) resembles alkali metals (one valence electron, forms H⁺) and also halogens (one electron short of a noble gas, forms H⁻, diatomic). So its position in the periodic table is unique.
• Hydrogen is the most abundant element in the universe.
• Isotopes of hydrogen: protium ¹H (1p, 0n), deuterium ²H or D (1p, 1n), tritium ³H or T (1p, 2n). Tritium is radioactive.
• Lab preparation: Zn + 2HCl → ZnCl₂ + H₂ ; also Zn + 2NaOH → Na₂ZnO₂ + H₂.
• Industrial preparation from steam: (1) C + H₂O → CO + H₂ (water gas, 1270 K); (2) water-gas shift: CO + H₂O → CO₂ + H₂ (673 K, iron chromate catalyst); (3) CO₂ removed by scrubbing with sodium arsenite solution. Electrolysis of dilute acidified water also gives H₂.
• Group 1: Li, Na, K, Rb, Cs, Fr (ns¹). Group 2: Be, Mg, Ca, Sr, Ba, Ra (ns²). Francium is radioactive; Li is the smallest alkali metal atom.
• Down the group, atomic size increases; ionization enthalpy and electronegativity decrease; reactivity increases.
• Alkali metals have large negative standard reduction potentials, so they are strong reducing agents.
• Alkali metals dissolve in liquid ammonia giving blue, conducting solutions (ammoniated electrons).
• Diagonal relationship: Li resembles Mg (both form nitrides directly, carbonates decompose on heating, LiCl and MgCl₂ are deliquescent). Be resembles Al.
• BeCl₂ is covalent (very small, highly polarising Be²⁺) while MgCl₂ is ionic. BeO is amphoteric.
• Na⁺ and K⁺ are important in body fluids; K⁺ is high in cell sap.

Important terms

• Washing soda – Na₂CO₃·10H₂O; made by the Solvay (ammonia-soda) process.
• Caustic soda – NaOH; made by electrolysis of brine.
• Hard water – water containing Ca²⁺ and Mg²⁺ salts; does not lather with soap.
• Lewis acid – electron pair acceptor (e.g. BeCl₂).
• LiAlH₄ – lithium aluminium hydride, a strong reducing agent used in organic chemistry.

Formulas

Key reactions:

• Solvay process: NH₃ + H₂O + CO₂ → NH₄HCO₃ ; NH₄HCO₃ + NaCl → NaHCO₃↓ + NH₄Cl ; 2NaHCO₃ → Na₂CO₃ + CO₂ + H₂O
• Magnesium burnt in air: 2Mg + O₂ → 2MgO and 3Mg + N₂ → Mg₃N₂
• BeO is amphoteric: BeO + 2HCl → BeCl₂ + H₂O ; BeO + 2NaOH → Na₂BeO₂ + H₂O
• H₂O₂ from 50 % H₂SO₄: 2HSO₄⁻ → HO₃SOOSO₃H (peroxodisulphuric acid) + 2e⁻ ; hydrolysis gives 2HSO₄⁻ + 2H⁺ + H₂O₂
• Removal of temporary hardness by washing soda: Ca(HCO₃)₂ + Na₂CO₃ → CaCO₃↓ + 2NaHCO₃

5 important questions

Q1. Why is hydrogen's position in the periodic table not fixed?

Its configuration 1s¹ is like alkali metals (ns¹): it forms H⁺ and acts as a reducing agent. It is also one electron short of helium, like halogens (ns² np⁵): it forms H⁻ (NaH) and exists as diatomic H₂. Hence it resembles both groups.

Q2. Describe the industrial production of dihydrogen from steam.

Steam is passed over coke at 1270 K with nickel catalyst: C + H₂O → CO + H₂ (water gas). Water-gas shift: CO + H₂O → CO₂ + H₂ at 673 K with iron chromate catalyst. CO₂ is removed by scrubbing with sodium arsenite solution, leaving H₂.

Q3. Explain the diagonal relationship between Li and Mg with two properties.

Li and Mg have similar size and polarising power. (1) Both burn in N₂ to form nitrides, Li₃N and Mg₃N₂. (2) Both form carbonates which decompose on heating to oxide and CO₂ (other alkali metal carbonates are stable).

Q4. Lithium floats on water calmly while sodium floats and catches fire. Why?

Both are lighter than water. Sodium reacts much more vigorously; the heat released ignites the hydrogen formed. Lithium, though it has the most negative E°, reacts slowly (kinetically), so not enough heat is produced to ignite H₂.

Q5. Why is BeCl₂ covalent while MgCl₂ is ionic?

Be²⁺ is very small with high charge density, so it strongly polarises the chloride ion's electron cloud, giving covalent character. Mg²⁺ is larger and less polarising, so MgCl₂ is ionic.

Exam tip

Write the Solvay process equations step by step — this question is very common. Learn the three isotopes of hydrogen with p, n, e counts.

Quick summary

This chapter studies the p-block elements of group 13 (boron family), group 14 (carbon family) and group 15 (nitrogen family). It covers their electronic configuration, trends in properties, oxidation states and inert pair effect, catenation, allotropy of carbon and phosphorus, and structures and uses of important compounds such as BCl₃, AlCl₃, diborane, boric acid, borax, silicones, SiO₂, ammonia, nitric acid and orthophosphoric acid.

Key points

• Group 13: B, Al, Ga, In, Tl (ns² np¹). Group 14: C, Si, Ge, Sn, Pb (ns² np²). Group 15: N, P, As, Sb, Bi (ns² np³).
• Down each group atomic radius increases (Ga is slightly smaller than Al due to poor shielding by 3d electrons), ionization enthalpy generally decreases and metallic character increases.
• Inert pair effect: down the group the ns² electrons become reluctant to take part in bonding, so the lower oxidation state becomes more stable. Tl⁺ is more stable than Tl³⁺; Pb²⁺ more stable than Pb⁴⁺ (PbCl₄ is less stable than PbCl₂).
• So Ga³⁺ salts are stable and Ga⁺ is a reducing agent, while Tl³⁺ salts are good oxidising agents (they readily become Tl⁺).
• Group 15 oxidation states range from −3 to +5; the +5 state becomes less stable and +3 more stable from N to Bi.
• Catenation (self-linking) is maximum in carbon due to strong C–C bonds; it decreases down group 14.
• Carbon forms strong pπ–pπ multiple bonds; heavier members do not. CO₂ is the most acidic oxide of group 14.
• Allotropes of carbon: diamond (sp³, hard, non-conductor), graphite (sp², layers, soft, conducts electricity), fullerenes (C₆₀, bucky ball). Allotropes of phosphorus: white (P₄, reactive, glows in dark, stored under water) and red (polymeric, stable).
• BCl₃ is trigonal planar (sp²) and a Lewis acid (electron deficient). AlCl₃ exists as a dimer Al₂Cl₆ where Al is sp³.
• Diborane (B₂H₆) has 4 terminal B–H bonds and 2 bridging B–H–B bonds (3-centre-2-electron "banana bonds").
• Boric acid H₃BO₃ is a weak monobasic Lewis acid. Borax Na₂B₄O₇·10H₂O is made by boiling colemanite with Na₂CO₃ solution; used in the borax bead test.
• Silicones are polymers with –R₂SiO– units; water repellent, heat stable — used as sealants, lubricants, in cosmetics and surgical implants. SiO₂ has a 3D network of SiO₄ tetrahedra.
• Ammonia is made by Haber's process; its lone pair makes it a good complexing agent (Lewis base). HNO₃ is made by Ostwald's process.

Important terms

• Allotropy – existence of an element in two or more forms with different physical properties.
• Inert pair effect – reluctance of ns² electrons of heavy p-block elements to participate in bonding.
• Catenation – property of atoms of an element to link with each other forming chains or rings.
• Inorganic graphite – boron nitride (BN), with a layered structure like graphite.
• Borate test – borate heated with ethyl alcohol and conc. H₂SO₄ gives ethyl borate, which burns with a green-edged flame.

Formulas

Key reactions:

• Borax from colemanite: Ca₂B₆O₁₁ + 2Na₂CO₃ → 2CaCO₃ + Na₂B₄O₇ + 2NaBO₂
• Haber's process: N₂ + 3H₂ ⇌ 2NH₃ (about 700 K, 200 atm, iron oxide catalyst with K₂O and Al₂O₃)
• Ostwald's process: 4NH₃ + 5O₂ → 4NO + 6H₂O (Pt/Rh gauze, 500 K) ; 2NO + O₂ → 2NO₂ ; 3NO₂ + H₂O → 2HNO₃ + NO
• Oxidation states of carbon: +4 in CO₂, +2 in CO, −4 in CH₄

5 important questions

Q1. Explain the structure and bonding of diborane.

B₂H₆ has two B atoms, each sp³ hybridised. Four terminal H atoms form normal 2-centre-2-electron B–H bonds in one plane. Two bridging H atoms lie above and below this plane, each forming a 3-centre-2-electron B–H–B bond (banana bond). Bridging B–H bonds are longer than terminal ones.

Q2. Why is PbCl₄ less stable than PbCl₂?

Due to the inert pair effect, the 6s² electrons of Pb do not easily take part in bonding. So +2 is the stable oxidation state for lead and PbCl₄ readily decomposes to PbCl₂ and Cl₂.

Q3. Distinguish between diamond and graphite.

Diamond: each C is sp³ bonded to 4 others in a 3D network; very hard; transparent; bad conductor. Graphite: each C is sp² bonded to 3 others in flat hexagonal layers held by weak forces; soft and slippery; good conductor (free π electrons); used as lubricant and electrodes.

Q4. What are silicones? Give their uses.

Silicones are organosilicon polymers with repeating –R₂SiO– units (R = methyl or phenyl), made by hydrolysis of alkyl chlorosilanes followed by condensation. They are water-repellent, heat resistant and chemically inert. Uses: sealants, greases, electrical insulators, water-proofing fabric, cosmetics and surgical implants.

Q5. Why is ammonia a good complexing agent?

The nitrogen in NH₃ has a lone pair of electrons which it donates to metal ions, forming coordinate bonds, e.g. Cu²⁺ + 4NH₃ → [Cu(NH₃)₄]²⁺ (deep blue).

Exam tip

Practise drawing structures of diborane, AlCl₃ dimer, orthophosphoric acid and resonance structures of HNO₃ — each is a regular 2–3 mark question.

Quick summary

This chapter studies the gaseous and liquid states. It explains intermolecular forces, the gas laws (Boyle, Charles, Gay-Lussac, Avogadro), the ideal gas equation and its uses, Dalton's law of partial pressures, kinetic molecular theory, deviation of real gases, liquefaction and critical constants, and properties of liquids — vapour pressure, surface tension and viscosity.

Key points

• Intermolecular forces: dispersion (London) forces, dipole–dipole, dipole–induced dipole and hydrogen bonding. Their balance with thermal energy decides whether a substance is solid, liquid or gas.
• Gases have no fixed shape or volume, are highly compressible, mix completely (diffusion) and exert pressure equally in all directions.
• Boyle's law: at constant T and n, P ∝ 1/V, so P₁V₁ = P₂V₂.
• Charles' law: at constant P and n, V ∝ T (in kelvin), so V₁/T₁ = V₂/T₂. Absolute zero = 0 K = −273.15 °C.
• Gay-Lussac's law: at constant V and n, P ∝ T, so P₁/T₁ = P₂/T₂.
• Avogadro's law: at constant T and P, V ∝ n.
• Combining these gives the ideal gas equation PV = nRT.
• Dalton's law: total pressure of a mixture of non-reacting gases = sum of partial pressures.
• Kinetic theory assumes gas molecules are point masses, have no attraction for each other, move randomly in straight lines and have perfectly elastic collisions; average kinetic energy ∝ absolute temperature.
• Real gases deviate from ideal behaviour at high pressure and low temperature, because molecules have volume and attract each other.
• A gas can be liquefied only below its critical temperature.
• Vapour pressure rises with temperature; a liquid boils when its vapour pressure equals the outside pressure. At 1 atm this is the normal boiling point.
• Surface tension and viscosity both decrease with rise in temperature.

Important terms

• Partial pressure – pressure a gas in a mixture would exert if it alone occupied the container.
• Diffusion – mixing of gases by random motion of molecules.
• Critical temperature (Tc) – the temperature above which a gas cannot be liquefied however high the pressure.
• Surface tension – force acting per unit length perpendicular to a line on the liquid surface (unit N m⁻¹).
• Viscosity – resistance of a liquid to flow (unit poise in CGS; N s m⁻² or Pa s in SI).
• Compressibility factor (Z) – Z = PV/nRT; Z = 1 for an ideal gas.

Formulas

• PV = nRT ; R = 8.314 J K⁻¹ mol⁻¹ = 0.0821 L atm K⁻¹ mol⁻¹ = 0.083 L bar K⁻¹ mol⁻¹
• Combined gas law: P₁V₁/T₁ = P₂V₂/T₂
• Molar mass M = mRT/PV ; density d = PM/RT, so M = dRT/P
• Dalton: P(total) = p₁ + p₂ + p₃ + … ; partial pressure p₁ = x₁ × P(total) (x = mole fraction)
• T (K) = t (°C) + 273.15 ; 1 atm = 101325 Pa = 1.01325 bar = 760 torr

5 important questions

Q1. State Boyle's law. A gas occupies 22.4 dm³ at 105 kPa. Find its volume at 210 kPa at the same temperature.

At constant temperature, volume of a fixed amount of gas is inversely proportional to pressure. V₂ = P₁V₁/P₂ = (105 × 22.4)/210 = 11.2 dm³.

Q2. A bubble of methane rises from the bottom of the sea. What happens to its size? Why?

As it rises, the water pressure on it decreases. By Boyle's law, volume increases as pressure decreases, so the bubble grows bigger.

Q3. Calculate the volume of 2 mol of an ideal gas at 27 °C and 1 bar. (R = 0.083 L bar K⁻¹ mol⁻¹)

T = 300 K. V = nRT/P = (2 × 0.083 × 300)/1 = 49.8 L.

Q4. State Dalton's law of partial pressure. A mixture has 0.4 mol N₂ and 0.6 mol O₂ at 2 bar. Find the partial pressure of O₂.

Total pressure of a mixture of non-reacting gases equals the sum of their partial pressures. x(O₂) = 0.6/1.0 = 0.6. p(O₂) = 0.6 × 2 = 1.2 bar.

Q5. Why do real gases deviate from ideal behaviour?

Kinetic theory assumes (1) molecules have no volume and (2) no attraction between them. At high pressure the volume of molecules is not negligible, and at low temperature intermolecular attraction becomes important, so PV ≠ nRT.

Exam tip

Always convert °C to kelvin before using any gas law, and choose the value of R that matches your pressure and volume units.

Quick summary

Adsorption is the gathering of molecules of a substance on the surface of a solid or liquid. This chapter explains adsorption and its types, factors affecting it, the Freundlich isotherm and uses, then catalysis (homogeneous, heterogeneous, adsorption theory) and finally colloids — their classification, preparation, purification, properties, coagulation, emulsions and applications.

Key points

• Adsorption is a surface phenomenon; absorption is a bulk phenomenon (water spreads throughout a sponge). When both happen together it is called sorption.
• Adsorption occurs because surface particles have unbalanced (residual) forces. It is exothermic, and the entropy decreases.
• Physisorption: weak van der Waals forces, low enthalpy (20–40 kJ mol⁻¹), reversible, multilayer, favoured by low temperature.
• Chemisorption: chemical bonds form, high enthalpy (80–240 kJ mol⁻¹), irreversible, unilayer, highly specific.
• Factors affecting adsorption of gases on solids: nature of gas (easily liquefiable gases like NH₃, SO₂ are adsorbed more), surface area (finely divided solids adsorb more), pressure (increases adsorption) and temperature (physisorption falls with rise in temperature).
• Uses: gas masks (charcoal), decolourising raw sugar with animal charcoal, chromatography, silica gel for removing moisture, heterogeneous catalysis.
• Catalyst changes rate of reaction without being used up. Homogeneous: catalyst and reactants in the same phase. Heterogeneous: in different phases (Fe in Haber process). Promoters increase activity (K₂O, Al₂O₃ in Haber process); inhibitors slow a reaction.
• Colloidal particle size: 1 nm to 1000 nm (10⁻⁹ m to 10⁻⁶ m). True solution particles are smaller; suspension particles larger.
• Lyophilic sols (solvent-loving, e.g. gum, starch, gelatin in water) are stable and reversible; lyophobic sols (solvent-hating, e.g. gold, Fe(OH)₃ sol) are unstable and irreversible.
• Preparation: Bredig's arc method (metal sols such as gold, silver, platinum), peptization, chemical methods. Purification: dialysis.
• Properties: Tyndall effect (scattering of light), Brownian motion (zig-zag motion), electrical charge, electrophoresis and coagulation.
• Hardy–Schulze rule: greater the charge on the oppositely charged ion, greater its coagulating power (Al³⁺ > Ba²⁺ > Na⁺ for a negative sol).
• Emulsions: oil in water (milk) and water in oil (butter, cold cream).

Important terms

• Adsorbate / adsorbent – substance that gets adsorbed / surface on which it is adsorbed.
• Desorption – removal of the adsorbed substance from the surface.
• Adsorption isotherm – graph of amount adsorbed (x/m) against pressure at constant temperature.
• Dispersed phase / dispersion medium – the colloidal particles / the medium in which they are spread.
• Gel – liquid dispersed in a solid (jelly, fruit jam, cheese).
• Coagulation – settling of colloidal particles when their charge is neutralised.

Formulas

• Freundlich adsorption isotherm: x/m = k P^(1/n) (n > 1), where x = mass of gas adsorbed on mass m of adsorbent, P = pressure, k and n are constants
• Log form: log (x/m) = log k + (1/n) log P (a straight line graph of log x/m against log P)

5 important questions

Q1. Distinguish between physisorption and chemisorption.

Physisorption: van der Waals forces, low enthalpy (20–40 kJ mol⁻¹), reversible, multilayer, not specific, decreases with temperature. Chemisorption: chemical bond, high enthalpy (80–240 kJ mol⁻¹), irreversible, unilayer, specific, first increases with temperature.

Q2. Why is a finely divided substance a better adsorbent?

Adsorption takes place on the surface. Finely dividing a solid greatly increases its surface area, so more sites are available and more adsorbate is held.

Q3. Distinguish between lyophilic and lyophobic sols.

Lyophilic: strong attraction for medium, prepared directly by mixing, reversible, stable, not easily coagulated (gum, starch). Lyophobic: little attraction, prepared by special methods, irreversible, unstable, easily coagulated by electrolytes (gold sol, As₂S₃ sol).

Q4. What is the Tyndall effect?

When a beam of light passes through a colloidal sol, its path becomes visible because colloidal particles scatter light. True solutions do not show it. Example: sunlight beam visible in a dusty room.

Q5. Identify dispersed phase and medium: milk, fog, blood, printing ink.

Milk: liquid fat in liquid water (emulsion). Fog: liquid in gas. Blood: solid particles (blood cells/proteins) in liquid. Printing ink: solid pigment in liquid.

Exam tip

Draw the electrophoresis apparatus and the Freundlich graph (x/m vs P and log form) — both are common diagram questions.

Quick summary

Many reactions do not go to completion; they reach a state where forward and backward reactions occur at the same rate. This chapter explains reversible reactions, physical and chemical equilibrium, the law of mass action, equilibrium constants Kc and Kp and their relation, homogeneous and heterogeneous equilibria, uses of K, Le Chatelier's principle and its industrial use in the Haber process.

Key points

• Irreversible reactions go in one direction only (burning of fuel). Reversible reactions go both ways (N₂ + 3H₂ ⇌ 2NH₃).
• Equilibrium is reached only in a closed system and is dynamic: forward and backward reactions continue at equal rates, so concentrations stay constant.
• Physical equilibria: ice ⇌ water at 0 °C, water ⇌ vapour, solid ⇌ solution (saturated solution).
• Law of mass action (Guldberg and Waage): rate of a reaction is proportional to the product of active masses (molar concentrations) of reactants, each raised to its stoichiometric coefficient.
• For aA + bB ⇌ cC + dD, Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ.
• Pure solids and pure liquids are not written in K expressions (their concentration is constant). Example: CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kc = [CO₂].
• K depends only on temperature, not on initial concentration, pressure or catalyst.
• Large K (> 10³) → products dominate; small K (< 10⁻³) → reactants dominate.
• Reaction quotient Q: if Q < K reaction moves forward; Q > K moves backward; Q = K at equilibrium.
• Le Chatelier's principle: if a system at equilibrium is disturbed, it shifts so as to reduce the effect of the change.
• Adding a reactant or removing a product shifts equilibrium forward. Increasing pressure shifts it to the side with fewer gas moles. For an exothermic reaction, rise in temperature shifts it backward. A catalyst does not shift equilibrium; it only helps reach it faster.
• Haber process (exothermic, 4 gas moles → 2): high pressure about 200 atm, about 700 K, iron oxide catalyst with K₂O and Al₂O₃ promoters, and removal of NH₃.

Important terms

• Chemical equilibrium – state at which rates of forward and backward reactions are equal.
• Homogeneous equilibrium – all species in the same phase (all gases).
• Heterogeneous equilibrium – species in more than one phase.
• Equilibrium constant – ratio of product of equilibrium concentrations of products to that of reactants, each raised to its coefficient.

Formulas

• Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ ; Kp = (pC)ᶜ(pD)ᵈ / (pA)ᵃ(pB)ᵇ
• Kp = Kc (RT)^Δn, where Δn = moles of gaseous products − moles of gaseous reactants
• If Δn = 0, Kp = Kc (e.g. H₂ + I₂ ⇌ 2HI)
• K for reverse reaction = 1/K ; K for equation multiplied by n = Kⁿ

5 important questions

Q1. State the law of mass action.

At constant temperature, the rate of a chemical reaction is directly proportional to the product of the molar concentrations of the reactants, each raised to the power equal to its stoichiometric coefficient in the balanced equation.

Q2. Write Kc for (i) 2SO₂ + O₂ ⇌ 2SO₃ (ii) N₂O₄ ⇌ 2NO₂.

(i) Kc = [SO₃]² / [SO₂]²[O₂]. (ii) Kc = [NO₂]² / [N₂O₄].

Q3. Find the relation between Kp and Kc for A(g) + B(g) ⇌ 2C(g) + D(g).

Δn = (2 + 1) − (1 + 1) = 1. Kp = Kc (RT)¹, so Kp/Kc = RT.

Q4. For BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq), what is the effect of (a) adding Ba²⁺ (b) removing SO₄²⁻ (c) adding BaSO₄(s)?

(a) Equilibrium shifts left, more BaSO₄ precipitates. (b) Shifts right, more BaSO₄ dissolves. (c) No effect, because a pure solid does not appear in the K expression.

Q5. Using Le Chatelier's principle, give suitable conditions for the manufacture of NH₃.

N₂ + 3H₂ ⇌ 2NH₃, ΔH negative. High pressure (fewer gas moles on product side) ~200 atm; low temperature favours NH₃ but slows the rate, so optimum ~700 K; iron oxide catalyst with K₂O and Al₂O₃ to reach equilibrium quickly; remove NH₃ as it forms.

Exam tip

For Kp–Kc questions, count only gaseous moles for Δn. For Le Chatelier questions, always state the direction AND the reason.

Quick summary

Nuclear chemistry studies changes in the nucleus of an atom. This chapter covers classification of nuclides, why some nuclei are stable (even–odd rule, N/Z ratio, magic numbers, binding energy), radioactivity and its rate law, half-life, modes of decay, nuclear reactions (transmutation, artificial radioactivity, fission and fusion), and uses of radioisotopes such as radiocarbon dating and medicine.

Key points

• Nuclides are classified as isotopes (same Z), isobars (same A), isotones (same neutrons), mirror nuclei (proton and neutron numbers interchanged, e.g. ³H and ³He) and as stable or unstable.
• Nuclei with even numbers of both protons and neutrons are most stable; odd–odd nuclei are the least stable.
• N/Z ratio: light stable nuclei have N/Z ≈ 1; it rises up to about 1.5 for heavy nuclei. Nuclei above the stability belt (too many neutrons) are β⁻ emitters; below it (too many protons) emit positrons (β⁺) or capture electrons.
• Magic numbers 2, 8, 20, 28, 50, 82, 126 (of protons or neutrons) give extra stability; ²⁰⁸Pb has two magic numbers (82 p and 126 n).
• Mass defect: actual nuclear mass is less than the sum of masses of its nucleons; the lost mass becomes binding energy (E = mc²). Higher binding energy per nucleon means a more stable nucleus (maximum near ⁵⁶Fe).
• Radioactive decay is a first-order process: rate ∝ number of nuclei present. The decay constant does not depend on temperature or pressure.
• α-decay: A decreases by 4, Z by 2. β⁻-decay: A unchanged, Z increases by 1 (a neutron becomes a proton). γ-emission: no change in A or Z.
• Nuclear fission: a heavy nucleus (U-235) splits into two medium nuclei with neutrons and huge energy; a chain reaction is used in nuclear reactors for electricity.
• Nuclear fusion: light nuclei join to form a heavier one (source of the Sun's energy). It needs extremely high temperature (millions of kelvin), which is hard to achieve and contain.
• Radiocarbon dating: ¹⁴C (half-life 5730 years) forms from ¹⁴N in the upper atmosphere; living things keep a constant ¹⁴C/¹²C ratio, which falls after death. Measuring it gives the age of old wood or fossils.
• Medical uses: ⁶⁰Co as external radiation source for cancer, ¹³¹I for thyroid, ³²P for leukemia.

Important terms

• Radioactivity – spontaneous emission of radiation by unstable nuclei.
• Half-life (t½) – time in which half of the radioactive nuclei decay.
• Transmutation – changing one element into another by bombarding with particles (Rutherford: ¹⁴N + α → ¹⁷O + p).
• Artificial radioactivity – a stable nucleus made radioactive by bombardment (²⁷Al + α → ³⁰P + n; ³⁰P then emits a positron).
• Curie (Ci) – 1 Ci = 3.7 × 10¹⁰ disintegrations per second; 1 Becquerel (Bq) = 1 dps.

Formulas

• Rate law: −dN/dt = λN ; integrated: N = N₀ e^(−λt) or λ = (2.303/t) log (N₀/N)
• Half-life: t½ = 0.693/λ
• Mass defect Δm = calculated mass of nucleons − observed nuclear mass
• Binding energy = Δm × 931.5 MeV (Δm in u); 1 u = 1.66 × 10⁻²⁷ kg ; E = mc²
• After n half-lives, N = N₀ (½)ⁿ

5 important questions

Q1. ⁶⁰Co decays with a half-life of 5.27 years. Find its decay constant.

λ = 0.693/t½ = 0.693/5.27 = 0.1315 ≈ 0.132 y⁻¹.

Q2. Why are odd–odd nuclides the least stable?

In a nucleus, protons pair with protons and neutrons with neutrons; paired nucleons have opposite spins, which adds stability. In odd–odd nuclei one proton and one neutron are unpaired, so the nucleus is least stable.

Q3. Explain nuclear fission with an example.

When U-235 absorbs a slow neutron it splits into two medium nuclei such as Ba-141 and Kr-92, releasing 3 neutrons and about 200 MeV energy. The neutrons cause more fissions — a chain reaction, controlled in a nuclear reactor.

Q4. How is the age of an old wood sample found?

Measure its ¹⁴C activity and compare it with that of fresh wood. Using t = (2.303/λ) log (A₀/A), with λ = 0.693/5730 y⁻¹, the age is calculated.

Q5. Why is it difficult to use fusion to produce energy?

Fusion needs temperatures of millions of degrees to overcome repulsion between nuclei. No material container can hold such hot plasma, and controlling the reaction is very difficult, so practical fusion reactors are not yet available.

Exam tip

In nuclear equations, total mass numbers and total atomic numbers must be equal on both sides — check both before moving on.

Quick summary

Organic chemistry is the chemistry of carbon compounds. This chapter shows how organic molecules are drawn (structural, condensed, bond-line and 3D formulae), how they are classified, how they are named by IUPAC rules, the types of structural isomerism, and the theory behind organic reactions — bond cleavage, reagents (electrophiles and nucleophiles), reactive intermediates and electronic effects.

Key points

• Carbon forms millions of compounds because of catenation (forming long chains and rings) and its ability to form single, double and triple bonds.
• Representation: complete (dash) formula, condensed formula (CH₃CH₂CH₂CH₃), bond-line (zig-zag) formula where each corner/end is a carbon, and 3D forms (wedge-dash, sawhorse, Newman projections).
• Classification by carbon skeleton: acyclic (open chain) and cyclic (alicyclic, aromatic, heterocyclic). By functional group: alcohols (–OH), aldehydes (–CHO), ketones (>C=O), carboxylic acids (–COOH), amines (–NH₂), ethers (–O–), esters (–COOR), haloalkanes (–X).
• IUPAC name = prefix (substituents) + word root (number of carbons: meth, eth, prop, but, pent…) + primary suffix (-ane, -ene, -yne) + secondary suffix (functional group: -ol, -al, -one, -oic acid).
• Choose the longest chain containing the functional group; number it so the functional group (then multiple bond, then substituents) gets the lowest number; write substituents in alphabetical order.
• Structural isomers have the same molecular formula but different structures: chain, position, functional group and metamerism.
• Homolytic cleavage: each atom takes one electron → free radicals. Heterolytic cleavage: one atom takes both electrons → carbocation and carbanion.
• Electrophiles are electron-loving (H⁺, NO₂⁺, carbocations, BF₃). Nucleophiles are nucleus-loving, electron-pair donors (OH⁻, CN⁻, NH₃, H₂O).
• Stability of carbocations and free radicals: 3° > 2° > 1° > methyl.
• Inductive effect: permanent shift of σ electrons due to an electronegative atom (–I by –Cl, –NO₂; +I by alkyl groups). It weakens with distance.
• Resonance (mesomeric) effect: delocalisation of π electrons (+R: –OH, –NH₂; −R: –NO₂, –CHO).
• Electromeric effect: temporary complete shift of π electrons in presence of a reagent.
• Hyperconjugation: delocalisation of σ electrons of C–H bonds on a carbon next (α) to a carbocation or double bond; more α-hydrogens → more stable.

Important terms

• Functional group – an atom or group of atoms that decides the chemical properties of a compound.
• Homologous series – compounds with the same functional group, differing by a –CH₂– unit.
• Free radical – an atom or group with an unpaired electron (CH₃•).
• Carbocation – a carbon with only six electrons and a positive charge (sp², planar).
• Carbanion – a carbon with a negative charge and a lone pair.
• α-carbon – the carbon atom directly attached to the functional group or charged carbon.

Formulas

Word roots and suffixes (IUPAC):

• General formulas: alkane CₙH₂ₙ₊₂, alkene CₙH₂ₙ, alkyne CₙH₂ₙ₋₂
• Degree of a carbon: primary (1°) bonded to one other C, secondary (2°) to two, tertiary (3°) to three, quaternary (4°) to four

5 important questions

Q1. Distinguish between homolytic and heterolytic cleavage.

Homolytic: shared pair splits equally, each atom gets one electron, free radicals form, favoured by light/heat and non-polar conditions. Heterolytic: the more electronegative atom takes both electrons, ions (carbocation and carbanion) form, favoured in polar solvents.

Q2. Write the structure of 5-ethyl-3-methylheptane.

Heptane chain C1–C7 with a methyl on C3 and an ethyl on C5: CH₃–CH₂–CH(CH₃)–CH₂–CH(C₂H₅)–CH₂–CH₃. (Name is correct: substituents alphabetical, ethyl before methyl.)

Q3. Arrange in order of stability: CH₃⁺, (CH₃)₂CH⁺, (CH₃)₃C⁺. Justify.

(CH₃)₃C⁺ > (CH₃)₂CH⁺ > CH₃⁺. More alkyl groups give more +I effect and more hyperconjugation (9, 6 and 0 α-hydrogens), which spread the positive charge and stabilise the cation.

Q4. What is the inductive effect? Give an example.

Permanent displacement of σ-bond electrons towards a more electronegative atom along a carbon chain. In CH₃–CH₂–Cl, Cl pulls electrons (–I effect), giving C1 a partial positive charge; the effect fades by the third carbon.

Q5. Write all structural isomers of C₄H₁₀O that are alcohols.

Butan-1-ol CH₃CH₂CH₂CH₂OH; butan-2-ol CH₃CH(OH)CH₂CH₃; 2-methylpropan-1-ol (CH₃)₂CHCH₂OH; 2-methylpropan-2-ol (CH₃)₃COH. (Chain and position isomers.)

Exam tip

For IUPAC names, first draw the structure, find the longest chain with the functional group, then number. Most marks are lost by numbering from the wrong end.

Quick summary

Hydrocarbons are compounds of carbon and hydrogen only. This chapter covers alkanes (saturated), alkenes and alkynes (unsaturated) and aromatic hydrocarbons like benzene — their isomerism, conformations, preparation, physical and chemical properties and uses, along with the structure and aromaticity of benzene, directive influence of groups and the toxicity of some aromatic compounds.

Key points

• Alkanes (CₙH₂ₙ₊₂) have only C–C single bonds (sp³). Natural sources: petroleum and natural gas. Ethane has staggered (more stable) and eclipsed conformations.
• Branching lowers boiling point: n-pentane > isopentane (2-methylbutane) > neopentane (2,2-dimethylpropane).
• Alkanes undergo substitution (halogenation by free radical mechanism in light), combustion, and cracking. Higher alkanes are used as lubricants and fuels.
• Alkenes (CₙH₂ₙ) have a C=C bond (sp², 1σ + 1π) and show geometrical (cis–trans) isomerism.
• Alkenes are prepared by dehydrohalogenation of alkyl halides with alcoholic KOH (1-chloropropane → propene) and by dehydration of alcohols with conc. H₂SO₄.
• Alkenes give addition reactions: H₂ (Ni), X₂, HX, H₂O. Addition of HBr follows Markovnikov's rule (H goes to the carbon having more H atoms); in presence of peroxide the anti-Markovnikov product forms.
• Bromine water test: alkenes and alkynes decolourise red-brown bromine water. Ozonolysis breaks the double bond into carbonyl compounds. Ethene is used to make polythene.
• Alkynes (CₙH₂ₙ₋₂) have C≡C (sp, linear). Acetylene (ethyne) is made from calcium carbide and water: CaC₂ + 2H₂O → C₂H₂ + Ca(OH)₂.
• Hydrogen on a triple-bonded carbon is acidic: acetylene reacts with sodium. Hydration of alkynes with dil. H₂SO₄ and HgSO₄ gives carbonyl compounds (ethyne → ethanal; propyne → propanone). Acetylene is used for oxy-acetylene welding.
• Benzene C₆H₆: planar hexagon, all C–C bonds equal (139 pm), between single (154 pm) and double (134 pm), due to resonance of 6 delocalised π electrons.
• Hückel rule: a planar, cyclic, fully conjugated molecule with (4n + 2) π electrons is aromatic (benzene: n = 1, 6 π electrons).
• Benzene undergoes electrophilic substitution: nitration, halogenation, sulphonation, Friedel–Crafts alkylation and acylation.
• Directive influence: –OH, –NH₂, –CH₃, –Cl are ortho/para directing; –NO₂, –CHO, –COOH, –CN are meta directing.
• Benzene and some polynuclear hydrocarbons (e.g. benzopyrene) are toxic and carcinogenic.

Important terms

• Conformations – different arrangements of atoms obtained by rotation about a C–C single bond.
• Markovnikov's rule – in addition of HX to an unsymmetrical alkene, the negative part goes to the carbon with fewer hydrogens.
• Ozonolysis – reaction with ozone followed by Zn/H₂O to split an alkene into aldehydes/ketones.
• Aromaticity – special stability of planar cyclic molecules with (4n + 2) delocalised π electrons.

Formulas

Key reactions:

• Ethane from ethyl bromide: C₂H₅Br + H₂ (Zn/HCl) → C₂H₆ + HBr ; from Grignard reagent: C₂H₅MgI + H₂O → C₂H₆ + Mg(OH)I
• Chlorination: CH₄ + Cl₂ (light) → CH₃Cl + HCl
• Markovnikov: CH₃–CH=CH₂ + HBr → CH₃–CHBr–CH₃ (2-bromopropane)
• Propene + cold conc. H₂SO₄, then water → propan-2-ol
• Nitration: C₆H₆ + HNO₃ (conc. H₂SO₄) → C₆H₅NO₂ + H₂O (electrophile NO₂⁺)
• Friedel–Crafts: C₆H₆ + CH₃Cl (anhydrous AlCl₃) → C₆H₅CH₃ + HCl

5 important questions

Q1. Arrange the isomers of C₅H₁₂ in increasing order of boiling point and name them.

Neopentane (2,2-dimethylpropane) < isopentane (2-methylbutane) < n-pentane (pentane). More branching makes molecules more compact, reducing surface area and van der Waals forces, so boiling point falls.

Q2. State Markovnikov's rule with an example.

When HX adds to an unsymmetrical alkene, H goes to the double-bonded carbon that already has more hydrogens and X to the other. Propene + HBr → 2-bromopropane (major).

Q3. Why is the hydrogen of acetylene acidic?

In H–C≡C–H, carbon is sp hybridised with 50 % s-character, so it is more electronegative and holds bonding electrons strongly. The C–H bond is polarised and H can leave as H⁺, e.g. with sodium it forms sodium acetylide and H₂.

Q4. How many monochlorination products does 2-methylpropane give?

Two: 1-chloro-2-methylpropane (ClCH₂–CH(CH₃)₂) and 2-chloro-2-methylpropane ((CH₃)₃C–Cl), because it has two kinds of hydrogen (9 primary, 1 tertiary).

Q5. Explain the aromatic character of benzene using Hückel's rule.

Benzene is cyclic and planar; each carbon is sp² with one unhybridised p-orbital, giving a continuous ring of overlap. It has 6 π electrons = 4n + 2 with n = 1. So it is aromatic and unusually stable, preferring substitution over addition.

Exam tip

Learn reagents with conditions (alc. KOH, Zn/HCl, HgSO₄/H₂SO₄, AlCl₃) — reaction-completion questions carry easy marks.

Quick summary

This chapter connects chemistry with daily life. It covers the chemistry of food (nutrients, food quality, rancidity, browning, preservation, healthy fats), compounds with medicinal properties (analgesics, antipyretics, antimicrobials, traditional Indian remedies) and cleansing agents (soaps and synthetic detergents, and how they remove dirt).

Key points

• Nutrients are carbohydrates, lipids (fats), proteins, vitamins, minerals and water. Cooking breaks down complex molecules and softens food, so it is easier to digest.
• Food quality changes because of enzymes, reactions with air and microorganisms. Some changes are useful (curd setting, bread dough rising).
• Rancidity: fats and oils develop bad smell and taste on keeping. Hydrolytic rancidity releases fatty acids such as butyric acid; oxidative rancidity is a free radical reaction with oxygen. Antioxidants such as vitamin E (tocopherol, natural) and BHT (synthetic) prevent it; vitamin E is added to packed edible oils.
• Browning of cut fruits/vegetables (apple, potato) happens when an enzyme oxidises polyphenols (released from cut cells) to quinones, which then form brown products called tannins. Reducing agents such as SO₂ or ascorbic acid (vitamin C, e.g. lemon juice), or a change of pH, slow it.
• Mono- and poly-unsaturated fats are healthier; saturated and trans fats raise the risk of heart disease.
• Analgesics relieve pain; antipyretics reduce fever. Aspirin (acetyl salicylic acid) and paracetamol are both.
• Antiseptics are applied to living tissue (Dettol, whose active ingredient is chloroxylenol, 4.8 % w/v; trichlorophenol; thymol from oil of thyme; turmeric). Disinfectants are used on non-living surfaces (p-chlorobenzyl phenol in all-purpose cleaners). Antibiotics target bacteria inside the body (penicillin, salvarsan — arsenic based, prontosil, sulphapyridine).
• Traditional knowledge: turmeric heals wounds — India won the turmeric patent case in 1997, led by Dr R. A. Mashelkar (CSIR). Neem, tulsi and wintergreen are other traditional remedies.
• Soaps are sodium or potassium salts of long-chain fatty acids, made by saponification (alkaline hydrolysis of oils/fats with NaOH or KOH), giving glycerol as by-product.
• Soaps do not work well in hard water because Ca²⁺ and Mg²⁺ form insoluble scum.
• Synthetic detergents work even in hard water: anionic (sodium alkylbenzenesulphonate), cationic (quaternary ammonium salts) and non-ionic (polyethylene glycol ethers/esters).
• Cleansing action: soap molecules have a polar head (water-loving) and long non-polar tail (oil-loving). Tails dissolve in grease, heads face water, forming micelles that lift dirt off and wash it away.

Important terms

• Saponification – alkaline hydrolysis of fats or oils to make soap and glycerol.
• Antioxidant – a substance that prevents oxidation (vitamin E, BHA, BHT).
• Micelle – a cluster of soap molecules with tails inside (holding oil) and heads outside.
• Antimicrobial – umbrella term for drugs that inhibit or kill microbes (bacteria, fungi, viruses).

Formulas

• Saponification: fat/oil (triglyceride) + 3NaOH → glycerol + 3 RCOONa (soap)
• Soap in hard water: 2RCOONa + Ca²⁺ → (RCOO)₂Ca↓ (scum) + 2Na⁺
• Functional groups: aspirin – carboxylic acid and ester; paracetamol – phenolic –OH and amide

5 important questions

Q1. Why do cut apples turn brown? How can this be slowed?

Cutting damages cells and releases polyphenols. An enzyme helps oxygen of air oxidise them to quinones, which form brown tannins. Applying lemon juice (vitamin C and acid) or using SO₂ slows the reaction.

Q2. Why is vitamin E added to packed edible oil?

Vitamin E is an antioxidant. It stops oxidative rancidity (a free radical reaction with oxygen), so the oil does not develop a bad smell or taste.

Q3. Distinguish between antiseptics and disinfectants.

Antiseptics are mild, used on living tissue (wounds, skin), e.g. Dettol, thymol, turmeric. Disinfectants are non-selective and used on non-living surfaces like floors, instruments and sanitary ware, e.g. p-chlorobenzyl phenol in cleaners.

Q4. Explain the cleansing action of soap.

A soap molecule has a polar –COO⁻Na⁺ head and a long hydrocarbon tail. Tails dissolve in grease, heads stay in water, forming micelles around the dirt. On rinsing, the micelles with trapped dirt are washed away.

Q5. Why do detergents work in hard water but soaps do not?

Soaps form insoluble calcium and magnesium salts (scum) with hard water ions, wasting soap. Calcium/magnesium salts of detergents are soluble, so detergents lather even in hard water.

Exam tip

Learn one example each for analgesic, antiseptic, disinfectant, antibiotic and each type of detergent — one-mark questions come straight from these.