This topic explains how an atom is built and how its electrons are arranged. It starts with the failure of classical ideas, moves to Bohr's model, and then to the quantum mechanical model with the de Broglie relation, Heisenberg's uncertainty principle, the Schrödinger equation and quantum numbers. It ends with rules for filling electrons in orbitals.
Key points
Bohr's model (1913): electrons move in fixed circular orbits (stationary states) without losing energy. Angular momentum is quantised: mvr = nh/2π.
Energy is absorbed or emitted only when an electron jumps between orbits: ΔE = E₂ − E₁ = hν.
Bohr's model works for hydrogen and one-electron ions (He⁺, Li²⁺) but fails for multi-electron atoms, the Zeeman and Stark effects, and fine spectral lines.
de Broglie (1924): every moving particle has wave nature, λ = h/mv. This is called dual nature of matter.
Heisenberg's uncertainty principle: position and momentum of an electron cannot both be known exactly at the same time: Δx·Δp ≥ h/4π. So the fixed orbits of Bohr are not possible.
Schrödinger wave equation describes the electron as a wave. Its solution ψ (wave function) has no direct meaning, but ψ² gives the probability of finding the electron. The region of high probability is an orbital.
Four quantum numbers describe each electron: n (principal), l (azimuthal), m (magnetic), s (spin).
Shapes: s orbital is spherical, p orbitals are dumb-bell shaped (px, py, pz), d orbitals are mostly double dumb-bell (five d orbitals).
Aufbau principle: electrons fill orbitals in increasing order of energy (n + l rule). Order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d…
Pauli exclusion principle: no two electrons in an atom have the same set of all four quantum numbers. So an orbital holds at most 2 electrons with opposite spins.
Hund's rule: in degenerate orbitals, electrons first occupy separately with parallel spins, then pair up.
Half-filled and fully filled subshells are extra stable. Examples: Cr = [Ar] 3d⁵ 4s¹ and Cu = [Ar] 3d¹⁰ 4s¹.
Important terms
Orbit – fixed circular path of the electron in Bohr's model.
Orbital (कक्षक) – 3D region around the nucleus where the probability of finding the electron is maximum.
Node – region where the probability of finding the electron is zero. Radial nodes = n − l − 1; angular nodes = l.
Degenerate orbitals – orbitals with the same energy, like the three 2p orbitals.
Wave function (ψ) – mathematical function that describes the electron wave.
Quantisation – energy or angular momentum can have only certain fixed values.
Formulas
Angular momentum: mvr = nh/2π
Energy of nth orbit of hydrogen: Eₙ = −13.6/n² eV (for H-like ions: −13.6 Z²/n² eV)
Maximum electrons in a shell = 2n²; in a subshell = 2(2l + 1)
Quantum number
Values
Tells us
n
1, 2, 3 …
size and energy of shell
l
0 to n − 1
subshell and shape (s, p, d, f)
m
−l to +l
orientation in space
s
+½ or −½
spin of electron
5 important questions
Q1. State Heisenberg's uncertainty principle.
It is impossible to measure exactly both the position and the momentum of a small particle like an electron at the same time. Mathematically Δx·Δp ≥ h/4π. It is important only for very small particles.
Q2. Write the electronic configuration of Cr (Z = 24) and explain why it is unusual.
Cr = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s¹. The expected 3d⁴ 4s² changes to 3d⁵ 4s¹ because a half-filled d subshell has extra stability (symmetry and more exchange energy).
Q3. Calculate the de Broglie wavelength of an electron moving with velocity 1.0 × 10⁶ m/s. (m = 9.1 × 10⁻³¹ kg)
λ = h/mv
λ = 6.626 × 10⁻³⁴ / (9.1 × 10⁻³¹ × 1.0 × 10⁶)
Denominator = 9.1 × 10⁻²⁵
λ = 7.28 × 10⁻¹⁰ m ≈ 7.3 Å
Q4. Calculate the energy needed to excite a hydrogen electron from n = 1 to n = 2.
E₁ = −13.6/1² = −13.6 eV
E₂ = −13.6/2² = −3.4 eV
ΔE = E₂ − E₁ = −3.4 − (−13.6) = 10.2 eV
In joules: 10.2 × 1.602 × 10⁻¹⁹ = 1.63 × 10⁻¹⁸ J
Q5. Explain the four quantum numbers. Give the set for the last electron of sodium.
n gives the main shell and energy; l gives the subshell and shape; m gives orientation; s gives spin direction.
Na = 1s² 2s² 2p⁶ 3s¹. Last electron is in 3s.
So n = 3, l = 0, m = 0, s = +½.
Add a short note on Pauli's principle, which uses these four numbers.
Exam tip
Learn the n + l rule with two examples (4s before 3d) and practise at least one de Broglie and one Bohr-energy numerical — they are asked almost every year.
Periodic properties are properties of elements that change in a regular way across a period and down a group of the periodic table. This topic covers the modern periodic law, effective nuclear charge and Slater's rules, and the trends in atomic radius, ionisation energy, electron affinity and electronegativity.
Key points
Modern periodic law (Moseley): properties of elements are a periodic function of their atomic numbers.
The long form table has 18 groups and 7 periods, divided into s, p, d and f blocks.
Shielding (screening) effect: inner electrons reduce the pull of the nucleus on outer electrons.
Effective nuclear charge Z* = Z − σ, where σ is the screening constant (from Slater's rules). Z* increases across a period.
Atomic radius decreases across a period (Z* increases) and increases down a group (new shells are added).
A cation is smaller than its parent atom; an anion is larger. In isoelectronic ions, size falls as nuclear charge rises: O²⁻ > F⁻ > Na⁺ > Mg²⁺.
Ionisation energy (IE) increases across a period and decreases down a group. IE₂ > IE₁ always.
Exceptions: IE of Be > B and N > O, because of stable fully filled (2s²) and half-filled (2p³) configurations.
Electron affinity generally becomes more negative (more energy released) across a period. Chlorine has the highest electron affinity, more than fluorine, because F is very small and its electrons repel the incoming electron.
Electronegativity increases across a period and decreases down a group. F is the most electronegative element (4.0 on Pauling scale).
Lanthanide contraction makes 4d and 5d elements of the same group nearly equal in size (Zr ≈ Hf).
Diagonal relationship: Li–Mg, Be–Al and B–Si show similar properties.
Important terms
Period (आवर्त) – horizontal row of the periodic table.
Group – vertical column; elements have similar valence shell configuration.
Ionisation energy – minimum energy needed to remove the outermost electron from an isolated gaseous atom.
Electron affinity – energy released when an electron is added to an isolated gaseous atom.
Electronegativity – tendency of a bonded atom to attract shared electrons towards itself.
Isoelectronic species – atoms or ions with the same number of electrons.
Formulas
Effective nuclear charge: Z* = Z − σ
Slater's rules (for an ns or np electron): electrons in the same (ns, np) group contribute 0.35 each (0.30 for 1s); electrons in (n − 1) shell contribute 0.85 each; electrons in (n − 2) and lower shells contribute 1.00 each. For a d or f electron, all inner electrons contribute 1.00.
Mulliken electronegativity: χ = (IE + EA)/2
Pauling: difference in electronegativity, (χA − χB) = 0.208 √Δ, where Δ (in kcal/mol) = actual bond energy of A–B minus the average of A–A and B–B bond energies.
5 important questions
Q1. Define effective nuclear charge.
It is the actual positive charge felt by an outer electron after the shielding by inner electrons. Z* = Z − σ. It increases from left to right in a period.
Q2. Why is the first ionisation energy of nitrogen more than that of oxygen?
Nitrogen has a half-filled, stable 2p³ configuration. Oxygen (2p⁴) has one paired electron in a 2p orbital; repulsion between the paired electrons makes it easier to remove. So IE(N) > IE(O).
Q3. Calculate the effective nuclear charge for the 4s electron of potassium (Z = 19).
Chemical bonding explains how atoms join to form molecules. This topic covers ionic and covalent bonds, lattice energy and the Born–Haber cycle, Fajans' rules, VSEPR theory, valence bond theory with hybridisation, and molecular orbital (MO) theory with bond order.
Key points
Ionic bond: formed by complete transfer of electrons from a metal to a non-metal (NaCl). Favoured by low ionisation energy, high electron affinity and high lattice energy.
Covalent bond: formed by sharing of electrons (H₂, Cl₂, CH₄).
Lattice energy is found indirectly using the Born–Haber cycle (based on Hess's law).
Fajans' rules: covalent character in an ionic bond increases when the cation is small and highly charged, the anion is large, and the cation has a pseudo-noble gas (18-electron) configuration. Example: LiCl is more covalent than NaCl; AlCl₃ is more covalent than NaCl.
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.
Hybridisation: mixing of atomic orbitals of similar energy to give equal new orbitals. sp (linear, 180°), sp² (trigonal planar, 120°), sp³ (tetrahedral, 109°28′), sp³d (trigonal bipyramidal), sp³d² (octahedral).
MO theory: atomic orbitals combine to give bonding (lower energy) and antibonding (higher energy) molecular orbitals.
Bond order = ½ (Nb − Na). Higher bond order means a stronger and shorter bond. Bond order zero means the molecule does not exist (He₂).
O₂ has two unpaired electrons in π* orbitals, so it is paramagnetic. Valence bond theory cannot explain this; MO theory can.
For molecules up to N₂, the σ2p orbital lies above the π2p orbitals (s–p mixing). For O₂ and F₂, σ2p lies below π2p.
Important terms
Lattice energy – energy released when one mole of an ionic solid forms from its gaseous ions.
Polarisation (ध्रुवीकरण) – distortion of the electron cloud of an anion by a cation.
Hybrid orbital – new orbital formed by mixing atomic orbitals.
Sigma (σ) bond – formed by head-on overlap; stronger.
Pi (π) bond – formed by sideways overlap of p orbitals.
Paramagnetic – attracted by a magnetic field due to unpaired electrons.
Formulas
Bond order = ½ (Nb − Na), Nb = electrons in bonding MOs, Na = in antibonding MOs
Born–Haber cycle for NaCl: ΔHf = S + ½D + IE + EA + U (S = sublimation, D = dissociation energy of Cl₂, IE = ionisation energy, EA = electron gain enthalpy, U = lattice energy, with signs)
Number of hybrid orbitals = number of σ bonds + lone pairs on central atom
Molecule
Bond order
Magnetic nature
H₂
1
diamagnetic
He₂
0
does not exist
N₂
3
diamagnetic
O₂
2
paramagnetic
O₂⁻
1.5
paramagnetic
O₂²⁻
1
diamagnetic
5 important questions
Q1. State Fajans' rules.
Covalent character of an ionic compound increases when (1) the cation is small, (2) the cation has a high charge, (3) the anion is large, and (4) the cation has an 18-electron outer shell instead of a noble gas shell.
Q2. Why is the bond angle in H₂O smaller than in CH₄?
Both are sp³ hybridised. Oxygen has two lone pairs. Lone pair–lone pair repulsion is the strongest, so it pushes the O–H bonds closer. The angle becomes about 104.5° instead of 109.5°.
Q3. Draw the MO energy diagram of O₂ and calculate its bond order.
This topic explains two special kinds of bonding. Metallic bonding holds metal atoms together and explains why metals conduct electricity, shine and can be beaten into sheets. Hydrogen bonding is a weak attraction between a hydrogen atom bonded to F, O or N and another electronegative atom. It explains the high boiling point of water, the structure of ice and many properties of biological molecules.
Key points
Free electron (electron sea) model (Drude–Lorentz): metal atoms lose their valence electrons, which move freely through a lattice of positive kernels. The attraction between kernels and the electron cloud is the metallic bond.
The electron sea model explains electrical and thermal conductivity, lustre, malleability and ductility (layers of kernels can slide without breaking the bond).
Valence bond (Pauling) model: the metallic bond is pictured as resonance among many covalent and ionic structures. It cannot explain conductivity well.
Band theory (MO model): orbitals of N atoms combine to give N molecular orbitals that lie so close together that they form a continuous band. The filled band is the valence band; the empty or partly filled band above it is the conduction band. The gap between them is the band gap (forbidden zone).
Conductors: valence and conduction bands overlap, or the band is half-filled (Na). Conductivity decreases as temperature rises.
Insulators: large band gap (diamond about 5–6 eV), so electrons cannot jump.
Semiconductors: small band gap (Si about 1.1 eV, Ge about 0.7 eV). Conductivity increases with temperature.
n-type semiconductor: Si doped with a Group 15 element (P, As) – extra electrons carry current. p-type: Si doped with a Group 13 element (B, Ga) – holes carry current.
Hydrogen bond: X–H····Y, where X and Y are highly electronegative small atoms (F, O, N). Strength about 10–40 kJ mol⁻¹, weaker than a covalent bond but stronger than van der Waals forces.
Intermolecular H-bond: between two molecules (H₂O, HF, alcohols) → raises boiling point, viscosity and solubility.
Intramolecular H-bond (chelation): inside one molecule (o-nitrophenol, salicylaldehyde) → lowers boiling point and water solubility, increases volatility.
Important terms
Kernel – metal atom without its valence electrons (positive ion core).
Band gap – energy gap between the valence band and conduction band.
Doping – adding a small, controlled amount of impurity to a semiconductor.
Hydrogen bond (हायड्रोजन बंध) – electrostatic attraction between H attached to F/O/N and a lone pair on another F/O/N.
Formulas
Material
Band gap
Effect of heating on conductivity
Metal (conductor)
none / overlap
decreases
Semiconductor (Si)
about 1.1 eV
increases
Insulator (diamond)
about 5–6 eV
very poor at all temperatures
Boiling points: H₂O (100 °C) > H₂S (about −60 °C), because H₂O molecules are H-bonded.
Boiling point order of hydrogen halides: HF > HI > HBr > HCl (HF is high due to H-bonding).
5 important questions
Q1. Why is ice lighter than water?
In ice, each water molecule is H-bonded to four others in a tetrahedral open cage structure with large empty spaces. On melting some H-bonds break, molecules come closer, and density increases. So ice (less dense) floats on water.
Q2. Why is o-nitrophenol more volatile than p-nitrophenol?
o-Nitrophenol has an intramolecular H-bond, so its molecules do not associate. p-Nitrophenol forms intermolecular H-bonds and its molecules associate. Association raises the boiling point, so the ortho isomer is more volatile and can be steam distilled.
Q3. Explain the free electron theory of metallic bonding and how it explains metallic properties.
Metal atoms give their valence electrons to a common "sea"; positive kernels sit in fixed lattice points.
Bond = attraction between kernels and mobile electrons; non-directional.
Conductivity: mobile electrons move under an applied voltage.
Lustre: free electrons absorb and re-emit light.
Malleability and ductility: layers slide; the electron sea still holds them.
Limitation: cannot explain why some elements are semiconductors.
Q4. Explain conductors, insulators and semiconductors on the basis of band theory.
Draw three diagrams: overlapping bands (metal), small gap (semiconductor), large gap (insulator).
Explain why a semiconductor conducts more when heated (electrons jump the small gap).
Add n-type and p-type doping with examples.
Q5. What is hydrogen bonding? Give its types and effects with examples.
Definition and conditions (H attached to small, highly electronegative atom).
s-block elements are Group 1 (alkali metals) and Group 2 (alkaline earth metals). p-block elements are Groups 13 to 18. This topic covers their general properties, trends, anomalous behaviour of the first element, the inert pair effect and some important compounds.
Key points
s-block: valence configuration ns¹ (Group 1) or ns² (Group 2). They are soft, reactive metals and strong reducing agents.
Alkali metals have the lowest ionisation energies in their periods. They give characteristic flame colours: Li crimson red, Na golden yellow, K lilac (violet).
Alkali metals dissolve in liquid ammonia to give a blue solution due to ammoniated (solvated) electrons. The solution is conducting and paramagnetic.
Lithium is anomalous because of its very small size and high polarising power. It resembles magnesium (diagonal relationship). Both form nitrides with N₂ and their carbonates decompose on heating.
Beryllium is anomalous and resembles aluminium. BeCl₂ is covalent.
p-block: valence configuration ns² np¹⁻⁶. Includes metals, metalloids and non-metals.
Inert pair effect: down Groups 13 and 14, the ns² electrons are less willing to take part in bonding. So the lower oxidation state becomes more stable: Tl⁺ more stable than Tl³⁺, Pb²⁺ more stable than Pb⁴⁺.
Boron forms electron-deficient compounds. Diborane (B₂H₆) has two 3-centre–2-electron (banana) B–H–B bonds.
Catenation (self-linking) is highest in carbon and decreases down Group 14.
Nitrogen exists as N₂ with a strong triple bond, so it is fairly inert. Phosphorus exists as P₄.
Halogens are strong oxidising agents. F₂ is the strongest oxidising agent.
Noble gases (Group 18) have full ns² np⁶ shells. Xenon forms compounds like XeF₂, XeF₄ and XeF₆.
Diagonal relationship – similarity between an element and the element diagonally below-right of it.
Inert pair effect – reluctance of ns² electrons to take part in bonding in heavy p-block elements.
Electron-deficient compound – a compound with too few electrons for normal 2-electron bonds, like B₂H₆.
Catenation – ability of an element to form chains with itself.
Allotropy – existence of an element in two or more physical forms (diamond and graphite).
Formulas
Important reactions:
2Na + 2H₂O → 2NaOH + H₂
6Li + N₂ → 2Li₃N
2NaBH₄ + I₂ → B₂H₆ + 2NaI + H₂
B₂H₆ + 6H₂O → 2H₃BO₃ + 6H₂
Xe + F₂ → XeF₂ (with excess Xe, heat)
Shapes of xenon fluorides (VSEPR): XeF₂ linear (sp³d), XeF₄ square planar (sp³d²), XeF₆ distorted octahedral (sp³d³)
5 important questions
Q1. What is the inert pair effect? Give one example.
In heavy p-block elements the ns² pair does not easily take part in bonding. So the oxidation state two less than the group valency becomes stable. Example: Pb²⁺ is more stable than Pb⁴⁺.
Q2. Why do alkali metals give colour to the flame?
Heat excites the valence electron to a higher level. When it comes back, energy is given out as visible light. Na gives golden yellow, K gives lilac.
Q3. Explain the structure of diborane.
B₂H₆ has 12 valence electrons, not enough for 8 normal bonds.
Each B is sp³ hybridised.
Four terminal B–H bonds are normal 2-centre–2-electron bonds and lie in one plane.
Two bridging H atoms lie above and below this plane.
Each B–H–B bridge is a 3-centre–2-electron (banana) bond.
Q4. Explain the anomalous behaviour of lithium and its diagonal relationship with magnesium.
Reasons: very small atom and ion, high polarising power, no d orbitals.
Li is harder and has higher melting point than other alkali metals.
Like Mg, Li forms a nitride (Li₃N) directly with N₂.
Li₂CO₃ and MgCO₃ decompose on heating to give oxides and CO₂.
LiCl and MgCl₂ are deliquescent and partly covalent.
Q5. Discuss the structure of XeF₄ using VSEPR theory.
Xe has 8 valence electrons; four form bonds with F, four remain as two lone pairs.
Total 6 electron pairs → sp³d² hybridisation, octahedral arrangement.
Two lone pairs occupy opposite (trans) positions to reduce repulsion.
Shape: square planar.
Exam tip
Short questions on flame colours, inert pair effect and diagonal relationship are very common. Learn diborane and xenon fluoride structures with neat diagrams.
This topic studies the structure and bonding of important compounds of p-block elements. It covers hydrides of Group 15, diborane and borazine, allotropes of carbon, carbides, silicates, oxides and oxyacids of phosphorus and sulphur, and interhalogen compounds. Exams usually ask for structures with reasons.
Key points
Hydrides of Group 15 (NH₃, PH₃, AsH₃, SbH₃): all pyramidal. Bond angle falls down the group: NH₃ 107° > PH₃ about 93.5° > AsH₃ about 92° > SbH₃ about 91.5°. Basic strength and thermal stability also fall down the group; reducing power increases.
Diborane (B₂H₆): electron-deficient. It has 4 terminal B–H bonds (normal 2c–2e) and 2 bridging B–H–B bonds called three-centre two-electron (3c–2e) "banana" bonds. Each B is roughly sp³ hybridised.
Borazine (B₃N₃H₆): "inorganic benzene", a flat six-membered ring of alternating B and N. It is more reactive than benzene because B–N bonds are polar.
Allotropes of carbon: diamond (each C sp³, 3D network, hard, insulator), graphite (sp², layers, soft, conductor, lubricant), fullerene C₆₀ (football shape, 20 hexagons and 12 pentagons).
Carbides: ionic (CaC₂, Al₄C₃), covalent (SiC, B₄C – very hard) and interstitial (TiC, WC – hard, high melting).
Silicates: basic unit is the SiO₄⁴⁻ tetrahedron. Types depend on how many O atoms are shared: orthosilicates, pyrosilicates (Si₂O₇⁶⁻), cyclic, chain (pyroxenes), sheet (mica, talc) and 3D (quartz, feldspar, zeolites).
Oxides of phosphorus: P₄O₆ and P₄O₁₀, both based on the P₄ tetrahedron with O bridges; P₄O₁₀ has four extra terminal P=O bonds.
Oxyacids of phosphorus: basicity = number of P–OH groups. H₃PO₄ tribasic, H₃PO₃ dibasic (one P–H), H₃PO₂ monobasic (two P–H). P–H bonds make H₃PO₃ and H₃PO₂ reducing agents.
Interhalogens: AX (ClF), AX₃ (ClF₃, T-shaped), AX₅ (BrF₅, square pyramidal), AX₇ (IF₇, pentagonal bipyramidal). More reactive than halogens because the A–X bond is weaker.
S₄N₄: cage structure; an eight-membered S–N ring folded so that S atoms come close in pairs.
Important terms
Electron-deficient compound – has fewer electrons than needed for normal 2-electron bonds (B₂H₆).
Basicity of an acid – number of replaceable (ionisable) H atoms.
Allotropy (अपरूपता) – existence of an element in two or more forms with different physical properties.
Formulas
Compound
Hybridisation
Shape
NH₃
sp³
pyramidal
ClF₃
sp³d
T-shaped
BrF₅
sp³d²
square pyramidal
IF₇
sp³d³
pentagonal bipyramidal
SiO₄⁴⁻
sp³
tetrahedral
Number of 3c–2e bonds in B₂H₆ = 2; total valence electrons = 12.
5 important questions
Q1. Why is H₃PO₃ dibasic?
Its structure has one P=O, two P–OH and one P–H bond. Only H attached to O can ionise, so it gives two H⁺ ions and is dibasic.
Q2. Why is graphite a good conductor but diamond is not?
In graphite each C uses three electrons for σ bonds (sp²); the fourth electron is delocalised over the layer and can move. In diamond all four electrons are in σ bonds (sp³), so no free electrons remain.
Q3. Describe the structure and bonding of diborane.
Two BH₂ units in one plane; two bridging H atoms above and below the plane.
Give the six classes, the number of shared O atoms per tetrahedron, one formula and one mineral for each. Mention that zeolites are used as ion exchangers and molecular sieves.
Q5. Compare the hydrides of Group 15 elements.
Shape, bond angle trend with reason (decreasing electronegativity of central atom, less bond-pair repulsion), basic strength, thermal stability, reducing nature and boiling point trend (NH₃ high due to H-bonding).
Exam tip
Practise drawing B₂H₆, borazine, P₄O₁₀ and the oxyacids of phosphorus. A clear labelled structure earns most of the marks.
This topic compares the main theories of acids and bases: Arrhenius, Brønsted–Lowry, Lux–Flood, solvent system and Lewis. It also covers the relative strength of acids and bases, the levelling effect of solvents, and Pearson's Hard and Soft Acids and Bases (HSAB) principle.
Key points
Arrhenius (1884): acid gives H⁺ in water; base gives OH⁻ in water. Limited to aqueous solutions.
Brønsted–Lowry (1923): acid = proton donor; base = proton acceptor. Every acid has a conjugate base and every base a conjugate acid. A strong acid has a weak conjugate base.
Lux–Flood: acid = oxide ion acceptor, base = oxide ion donor. Example: CaO + SiO₂ → CaSiO₃ (CaO base, SiO₂ acid). Useful for high-temperature melts.
Solvent system concept: an acid increases the concentration of the solvent's characteristic cation; a base increases its characteristic anion. In liquid NH₃: 2NH₃ ⇌ NH₄⁺ + NH₂⁻, so NH₄Cl is an acid and NaNH₂ a base.
Lewis (1923): acid = electron-pair acceptor (BF₃, AlCl₃, H⁺, metal ions); base = electron-pair donor (NH₃, H₂O, Cl⁻). Most general theory, but cannot order acid strengths well.
Levelling effect: water levels all strong acids (HCl, HNO₃, HClO₄) to H₃O⁺, so they look equally strong. In a less basic solvent like acetic acid, their differences appear (differentiating solvent).
Strength of oxyacids rises with more O atoms on the central atom: HClO < HClO₂ < HClO₃ < HClO₄.
Strength of hydracids in a group rises down the group: HF < HCl < HBr < HI (weaker H–X bond).
HSAB (Pearson, 1963): hard acids prefer hard bases and soft acids prefer soft bases.
Hard species: small, high charge, not easily polarised (H⁺, Li⁺, Mg²⁺, Al³⁺, F⁻, OH⁻, H₂O, NH₃).
B in BF₃ has only six electrons in its valence shell and an empty p orbital. It accepts an electron pair (for example from NH₃) to complete its octet.
Q3. Explain the Lewis concept with its merits and limitations.
Definition with examples of cations, electron-deficient molecules and molecules with empty d orbitals (SiF₄) as acids.
Merits: covers reactions without protons, explains complex formation.
Limitations: no quantitative scale of strength, catalytic behaviour not explained, includes too many reactions.
Q4. Explain the HSAB principle and its applications.
Classification table of hard, soft and borderline acids and bases.
Rule: hard–hard and soft–soft combinations are more stable.
Applications: occurrence of ores (Al, Mg as oxides/carbonates – hard; Cu, Hg, Ag as sulphides – soft), stability of complexes ([HgI₄]²⁻ is more stable than [HgF₄]²⁻), Kornblum's rule, the symbiosis effect.
Limitation: qualitative only.
Q5. Explain the levelling and differentiating effect of solvents.
Water as a levelling solvent (all strong acids → H₃O⁺). Acetic acid as a differentiating solvent showing HClO₄ > HBr > H₂SO₄ > HCl > HNO₃. Liquid NH₃ levels even weak acids like acetic acid.
Exam tip
Learn the hard/soft lists as a table; one question on HSAB applications appears in most papers.
This topic gives the basic ideas needed to understand organic reactions. It covers hybridisation of carbon, bond fission, reactive intermediates (carbocations, carbanions, free radicals, carbenes), electronic effects (inductive, resonance, hyperconjugation, electromeric) and types of reagents and reactions.
Key points
Carbon shows sp³ (alkanes), sp² (alkenes, benzene) and sp (alkynes) hybridisation. More s-character means shorter bonds and higher electronegativity of carbon.
Homolytic fission: each atom keeps one electron of the bond → free radicals. Favoured by light, heat and non-polar solvents.
Heterolytic fission: one atom takes both electrons → ions (carbocation and carbanion). Favoured by polar solvents.
Carbocation: carbon with a positive charge and six electrons, sp², planar. Stability: 3° > 2° > 1° > CH₃⁺. Benzyl and allyl cations are extra stable by resonance.
Carbanion: carbon with negative charge, usually sp³, pyramidal. Stability: CH₃⁻ > 1° > 2° > 3° (opposite of carbocation).
Free radical: carbon with an unpaired electron; stability 3° > 2° > 1° > CH₃•.
Carbene: neutral carbon with two bonds and two non-bonded electrons (:CH₂). Singlet and triplet forms exist.
Inductive effect (I): permanent shift of σ electrons along a chain due to an electronegative or electropositive group. −I groups: −NO₂, −CN, −COOH, −F, −Cl. +I groups: alkyl groups.
Resonance (mesomeric) effect: delocalisation of π electrons or lone pairs through a conjugated system. +R: −OH, −NH₂, −OCH₃. −R: −NO₂, −CHO, −CN.
Hyperconjugation: delocalisation of σ electrons of an α C–H bond into an adjacent empty p orbital or π bond. It explains why more alkyl groups stabilise carbocations and alkenes.
Electromeric effect: temporary complete shift of π electrons in presence of an attacking reagent.
Electrophiles are electron-seeking (H⁺, NO₂⁺, BF₃, AlCl₃). Nucleophiles are nucleus-seeking, electron-rich (OH⁻, CN⁻, NH₃, H₂O).
Main reaction types: substitution, addition, elimination, rearrangement.
Important terms
Reactive intermediate – short-lived species formed during a reaction, like a carbocation.
Nucleophile (केंद्रकस्नेही) – species that gives an electron pair to an electron-poor centre.
Electrophile – species that accepts an electron pair.
Conjugation – alternate single and double bonds, allowing delocalisation.
Acidity order of acetic acids – affected by −I groups; more −I means stronger acid.
Q1. Distinguish between homolytic and heterolytic fission.
In homolytic fission the bond breaks evenly and each atom keeps one electron, giving free radicals. In heterolytic fission one atom takes both electrons, giving a cation and an anion. Homolysis needs light/heat; heterolysis is helped by polar solvents.
Q2. Why is trichloroacetic acid stronger than acetic acid?
Three Cl atoms pull electrons away (−I effect). This weakens the O–H bond and spreads out the negative charge of the carboxylate ion, making it more stable. So H⁺ is lost more easily.
Q3. Explain the structure and stability of carbocations.
Carbon has 6 electrons and a positive charge; it is sp² hybridised.
Three bonds lie in one plane at 120°; an empty p orbital is perpendicular.
Alkyl groups stabilise by +I effect and hyperconjugation.
Stability: 3° > 2° > 1° > CH₃⁺ (tert-butyl cation has 9 α-H, so 9 hyperconjugative forms).
Allyl and benzyl cations are stabilised by resonance.
Q4. Explain hyperconjugation with an example.
Also called no-bond resonance (Baker–Nathan effect).
σ electrons of the C–H bond next to a carbocation or double bond overlap with the empty p orbital or π bond.
Example: ethyl cation CH₃–CH₂⁺ has 3 α-H, so 3 hyperconjugative structures.
Explains: stability of alkenes (more substituted alkene more stable), stability of carbocations and radicals.
Q5. Classify reagents as electrophiles and nucleophiles with examples.
Electrophiles: positive ions (H⁺, NO₂⁺, Br⁺) and neutral Lewis acids (BF₃, AlCl₃, SO₃).
Nucleophiles: negative ions (OH⁻, CN⁻, Cl⁻, RO⁻) and neutral molecules with lone pairs (H₂O, NH₃, ROH).
Questions on stability orders and acid strength are frequent. Always give the reason (inductive, resonance or hyperconjugation) in one line with the order.
Ionic equilibria deals with the equilibrium between ions and un-ionised molecules of weak electrolytes in solution. This topic covers Ostwald's dilution law, ionic product of water, pH, common ion effect, buffer solutions and the Henderson equation, salt hydrolysis, and solubility product with its uses.
Common ion effect: adding a salt with a common ion suppresses the ionisation of a weak electrolyte (CH₃COONa reduces ionisation of CH₃COOH). Used in qualitative analysis (NH₄Cl + NH₄OH in Group III).
Buffer solution: resists change in pH when a little acid or base is added. Acidic buffer: weak acid + its salt with strong base (CH₃COOH + CH₃COONa). Basic buffer: weak base + its salt with strong acid (NH₄OH + NH₄Cl).
Buffer capacity is greatest when [salt] = [acid], i.e. pH = pKa.
Salt hydrolysis: salt of weak acid + strong base gives a basic solution (CH₃COONa); salt of strong acid + weak base gives an acidic solution (NH₄Cl); salt of strong acid + strong base does not hydrolyse (NaCl).
Solubility product (Ksp): product of ion concentrations in a saturated solution of a sparingly soluble salt, each raised to its stoichiometric power.
Precipitation occurs when ionic product > Ksp.
Important terms
Degree of dissociation (α) – fraction of total molecules that ionise.
Buffer (बफर द्रावण) – solution that keeps its pH nearly constant.
Hydrolysis constant (Kh) – equilibrium constant for the reaction of a salt's ion with water.
Applications: precipitation of Group II sulphides in acidic medium (low [S²⁻] due to common ion H⁺); salting out of soap with NaCl; purification of common salt by passing HCl gas.
Exam tip
Always write the formula first, then substitute with units, then the log step. Examiners give marks for each step even if the final digit is off.
Stereochemistry studies the three-dimensional arrangement of atoms in molecules. Isomers with the same structure but different spatial arrangement are called stereoisomers. This topic covers optical isomerism, chirality, R/S configuration, geometrical (cis–trans, E/Z) isomerism and conformations of ethane and butane.
Key points
Stereoisomers have the same molecular formula and same bonding sequence but differ in spatial arrangement.
Optical activity: the ability to rotate plane-polarised light. Measured with a polarimeter. Dextrorotatory (+, d) rotates to the right; laevorotatory (−, l) to the left.
A carbon bonded to four different groups is a chiral centre (asymmetric carbon). Example: lactic acid, CH₃CH(OH)COOH.
Enantiomers are non-superimposable mirror images. They have the same physical properties except the direction of rotation of light.
Diastereomers are stereoisomers that are not mirror images. They have different physical properties.
A meso compound has chiral centres but also a plane of symmetry, so it is optically inactive. Example: meso-tartaric acid.
A racemic mixture is a 50:50 mix of enantiomers; it shows no rotation (external compensation).
Number of optical isomers = 2ⁿ (n = chiral centres), when the molecule is not symmetric. Tartaric acid has only 3 (d, l, meso).
CIP rules (Cahn–Ingold–Prelog): give priority by higher atomic number. Keep the lowest priority group away; if 1 → 2 → 3 is clockwise, it is R; anticlockwise is S.
Geometrical isomerism occurs due to restricted rotation around C=C or in rings. Cis: same groups on the same side; trans: opposite sides. E/Z uses CIP priorities.
Conformations arise by rotation about a C–C single bond. Ethane: staggered (most stable) and eclipsed (least stable).
Q1. What is a meso compound? Why is it optically inactive?
A meso compound has two or more chiral centres but has an internal plane of symmetry. The rotation by one half is cancelled by the other half (internal compensation). Example: meso-tartaric acid.
Q2. Distinguish between enantiomers and diastereomers.
Enantiomers are non-superimposable mirror images with identical physical properties except the sign of rotation. Diastereomers are not mirror images and differ in melting point, boiling point and solubility. Cis and trans-2-butene are diastereomers.
Q3. A solution of 0.2 g/mL of a sugar in a 2 dm tube shows rotation +13.3°. Find its specific rotation.
[α] = α/(l × c)
= 13.3 / (2 × 0.2)
= 13.3 / 0.4
= +33.25°
Q4. Assign R or S to (+)-lactic acid drawn with H pointing back, and explain the steps.
Hydrocarbons are compounds of carbon and hydrogen only. This topic covers alkanes, alkenes, alkynes and aromatic hydrocarbons (arenes): their preparation, important reactions and mechanisms, Markovnikov's rule, aromaticity and electrophilic substitution in benzene.
Key points
Alkanes (CₙH₂ₙ₊₂) are saturated, sp³, and mostly give substitution reactions.
Preparation of alkanes: Wurtz reaction, Kolbe's electrolysis, decarboxylation of sodium salts with soda lime, hydrogenation of alkenes.
Halogenation of methane is a free radical chain reaction: initiation, propagation, termination.
Alkenes (CₙH₂ₙ) have a C=C (one σ + one π bond), sp². They give electrophilic addition reactions.
Markovnikov's rule: in addition of HX to an unsymmetrical alkene, H goes to the carbon that already has more hydrogens (because the more stable carbocation forms).
Anti-Markovnikov (peroxide/Kharasch effect): HBr in presence of peroxide adds the opposite way by a free radical mechanism. Works only with HBr.
Ozonolysis of alkenes gives aldehydes/ketones and helps locate the double bond.
Alkynes (CₙH₂ₙ₋₂) have C≡C, sp, linear. Terminal alkynes are weakly acidic (sp carbon has 50% s-character) and form acetylides with sodium or ammoniacal AgNO₃.
Aromaticity (Hückel's rule): a cyclic, planar, fully conjugated molecule with (4n + 2) π electrons is aromatic. Benzene has 6 π electrons (n = 1).
Benzene gives electrophilic substitution (not addition), keeping its stable aromatic ring: nitration, halogenation, sulphonation, Friedel–Crafts alkylation and acylation.
Mechanism: formation of electrophile → attack on ring to form a σ-complex (arenium ion) → loss of H⁺ to restore aromaticity.
Directive influence: −OH, −NH₂, −CH₃, −Cl are ortho/para directors; −NO₂, −CHO, −COOH, −CN are meta directors. Halogens are o/p directing but deactivating.
Important terms
Saturated hydrocarbon (संतृप्त) – contains only single C–C bonds.
Electrophilic addition – addition started by attack of an electrophile on a π bond.
When an unsymmetrical reagent like HX adds to an unsymmetrical alkene, the negative part (X) goes to the carbon having fewer hydrogen atoms. Propene + HBr gives 2-bromopropane as the main product.
Q2. Why are terminal alkynes acidic?
The carbon of ≡C–H is sp hybridised with 50% s-character. It holds electrons closer and is more electronegative. So the C–H bond is polarised and H⁺ can be removed by strong bases like NaNH₂.
Q3. Explain the mechanism of nitration of benzene.
Learn three mechanisms well: free radical halogenation, electrophilic addition of HBr (with peroxide effect), and nitration of benzene. Write each step with arrows and intermediates.
Alkyl halides (haloalkanes, R–X) and aryl halides (haloarenes, Ar–X) are compounds in which a halogen is bonded to an sp³ carbon or to a benzene ring. This topic covers their classification, preparation, nucleophilic substitution (SN1 and SN2) and elimination (E1 and E2) mechanisms, and why aryl halides are much less reactive.
Key points
Classification: 1°, 2° and 3° alkyl halides depending on how many carbons are attached to the C bearing X. Also allylic, benzylic, vinylic and aryl halides.
Preparation of R–X: from alcohols (HX/ZnCl₂, PCl₅, PCl₃, SOCl₂ – SOCl₂ is best because by-products SO₂ and HCl are gases), from alkenes (HX addition), Finkelstein reaction (R–Cl + NaI in acetone → R–I), Swarts reaction (R–Br + AgF → R–F).
Preparation of Ar–X: electrophilic halogenation of benzene (Cl₂/FeCl₃) and the Sandmeyer reaction (ArN₂⁺Cl⁻ + CuCl → ArCl).
SN2: one step, bimolecular, rate = k[RX][Nu⁻]. Backside attack → Walden inversion of configuration. Reactivity: CH₃X > 1° > 2° > 3°. Favoured by polar aprotic solvents.
SN1: two steps through a carbocation, rate = k[RX]. Gives racemisation (with some inversion). Reactivity: 3° > 2° > 1° > CH₃X. Favoured by polar protic solvents.
E2: one step, anti-periplanar H and X, strong base (alc. KOH). E1: through carbocation.
Saytzeff (Zaitsev) rule: in elimination, the more substituted alkene is the main product (2-bromobutane → but-2-ene mainly).
Aryl halides are less reactive towards nucleophiles because (1) the C–X bond gets partial double-bond character by resonance, (2) the C is sp² and holds the bond more tightly, (3) the phenyl cation is unstable, (4) electron-rich ring repels the nucleophile.
Electron-withdrawing groups (–NO₂) at ortho/para positions make aryl halides react (2,4-dinitrochlorobenzene hydrolyses easily). Benzyne (elimination–addition) mechanism occurs with NaNH₂.
Nucleophile (केंद्रकस्नेही) – electron-rich species that attacks a positive centre.
Leaving group – group that departs with the bond pair (I⁻ > Br⁻ > Cl⁻ > F⁻).
Walden inversion – inversion of configuration in SN2.
Formulas
Feature
SN1
SN2
Steps
two
one
Rate law
k[RX]
k[RX][Nu]
Intermediate
carbocation
none (transition state)
Stereo result
racemisation
inversion
Best substrate
3°
CH₃ / 1°
Rearrangement
possible
no
5 important questions
Q1. Why is SOCl₂ preferred for converting alcohols to alkyl chlorides?
R–OH + SOCl₂ → R–Cl + SO₂↑ + HCl↑. Both by-products are gases and escape, so pure alkyl chloride is left.
Q2. Why does tert-butyl bromide undergo SN1 faster than methyl bromide?
It forms a stable 3° carbocation (stabilised by +I and hyperconjugation of nine C–H bonds). Methyl cation is very unstable. Bulky groups also block SN2 at the tertiary carbon.
Q3. Explain the SN2 mechanism with stereochemistry.
Example: OH⁻ + CH₃Br → CH₃OH + Br⁻.
Nucleophile attacks from the side opposite to Br; five groups in the transition state (C in trigonal bipyramid form).
Second-order kinetics; inversion of configuration (umbrella flip), e.g. (R)-2-bromobutane → (S)-butan-2-ol.
Effect of substrate, nucleophile, leaving group and solvent.
Q4. Explain the E2 mechanism and Saytzeff rule.
One-step removal of β-H by base and loss of X⁻ in anti arrangement. Example: 2-bromobutane + alc. KOH → but-2-ene (major, about 80 %) + but-1-ene (minor). More substituted alkene is more stable due to hyperconjugation.
Q5. Why are aryl halides less reactive than alkyl halides towards nucleophilic substitution?
Draw resonance structures of chlorobenzene showing C=Cl⁺ character, explain sp² carbon, instability of phenyl cation and repulsion by π electrons. Show how –NO₂ groups at o/p positions activate the ring (Meisenheimer complex).
Exam tip
A comparison table of SN1 and SN2 is a guaranteed 3–4 mark answer. Always draw the transition state for SN2.
Alcohols (R–OH) and phenols (Ar–OH) contain a hydroxyl group; ethers (R–O–R′) have an oxygen between two carbon groups. This topic covers classification, preparation, acidity, distinction tests and key reactions such as oxidation, pinacol–pinacolone rearrangement, Reimer–Tiemann and Kolbe reactions, and cleavage of ethers by HI.
Preparation: hydration of alkenes, hydrolysis of alkyl halides, reduction of aldehydes/ketones (H₂/Ni, NaBH₄, LiAlH₄), Grignard reagent with HCHO (1°), other aldehydes (2°) and ketones (3°).
Lucas test (conc. HCl + anhydrous ZnCl₂): 3° alcohols give turbidity at once, 2° in about 5 minutes, 1° not at room temperature.
Glycols: cleaved by HIO₄ or Pb(OAc)₄ into two carbonyl compounds. Pinacol–pinacolone rearrangement: pinacol + acid → pinacolone through a carbocation and 1,2-methyl shift.
Phenols are more acidic than alcohols: the phenoxide ion is stabilised by resonance; alkoxide is not. Phenol (pKa about 10) reacts with NaOH; ethanol (pKa about 16) does not.
Electron-withdrawing groups (–NO₂) increase phenol acidity; electron-donating groups (–CH₃) decrease it. Picric acid is strongly acidic.
Reactions of phenol: bromine water gives white precipitate of 2,4,6-tribromophenol; Reimer–Tiemann (CHCl₃ + NaOH → salicylaldehyde, via dichlorocarbene); Kolbe's reaction (sodium phenoxide + CO₂, 125 °C, pressure → salicylic acid); FeCl₃ gives violet colour.
Ethers: prepared by Williamson synthesis (R–ONa + R′–X → R–O–R′, SN2; use 1° halide). Fairly unreactive; cleaved by hot conc. HI.
Anisole + HI → phenol + CH₃I (never iodobenzene, because the Ar–O bond does not break).
Epoxides (three-membered cyclic ethers) open easily with acid or base because of ring strain.
Important terms
Phenoxide ion – C₆H₅O⁻, resonance-stabilised conjugate base of phenol.
Dichlorocarbene (:CCl₂) – electrophile in Reimer–Tiemann reaction.
Hydroxyl group (हायड्रॉक्सिल गट) – –OH functional group.
Formulas
Test / reaction
Reagent
Result
Lucas test
conc. HCl + ZnCl₂
3° fast, 2° slow, 1° none
Ferric chloride
neutral FeCl₃
phenol: violet
Bromine water
Br₂(aq)
phenol: white ppt
Iodoform
I₂ + NaOH
CH₃CH(OH)– alcohols: yellow ppt
5 important questions
Q1. Why is phenol acidic but ethanol is almost neutral?
After losing H⁺, phenoxide ion spreads its negative charge over the ring by resonance and becomes stable. Ethoxide has no resonance and +I effect of ethyl increases its charge, so ethanol hardly ionises.
Q2. How will you distinguish between 1°, 2° and 3° alcohols?
Use the Lucas test (turbidity speed) or Victor Meyer's test (red, blue, colourless). Oxidation products also differ: aldehyde, ketone, no reaction.
Q3. Explain the pinacol–pinacolone rearrangement with mechanism.
Protonation of one –OH, loss of water → 3° carbocation.
1,2-shift of a methyl group to the cationic carbon → more stable oxocarbenium ion.
Loss of H⁺ → pinacolone (3,3-dimethylbutan-2-one).
Q4. Explain the Reimer–Tiemann reaction with mechanism.
CHCl₃ + OH⁻ → :CCl₂ (dichlorocarbene). Phenoxide attacks :CCl₂ at the ortho position; the –CHCl₂ group is hydrolysed to –CHO. Product after acidification: salicylaldehyde (o-hydroxybenzaldehyde), with a little para isomer.
Q5. Describe the Williamson synthesis and the action of HI on ethers.
Show the SN2 step; explain why tert-butyl halide gives alkene instead. Give HI cleavage of diethyl ether (cold: C₂H₅I + C₂H₅OH; hot excess: 2C₂H₅I) and of anisole.
Exam tip
Name reactions (Reimer–Tiemann, Kolbe, Williamson, pinacol) carry easy marks — write the equation, reagents and conditions for each.
Aldehydes (R–CHO) and ketones (R–CO–R′) contain the carbonyl group (C=O). The polar C=O makes the carbon electrophilic, so the typical reaction is nucleophilic addition. This topic covers structure, preparation, nucleophilic addition, aldol, Cannizzaro, benzoin, Perkin, Knoevenagel, Wittig and Mannich reactions, and oxidation and reduction tests.
Key points
Carbonyl carbon is sp² hybridised, trigonal planar (about 120°). C=O is polar: C is δ+, O is δ−.
Preparation: oxidation of alcohols (1° → aldehyde with PCC, 2° → ketone), Rosenmund reduction (RCOCl + H₂ / Pd–BaSO₄ → RCHO), from nitriles with Grignard reagent (R–C≡N + R′MgX → ketone after hydrolysis), Friedel–Crafts acylation (aromatic ketones), ozonolysis of alkenes.
Reactivity order: HCHO > other aldehydes > ketones (alkyl groups push electrons by +I and block the nucleophile).
Aldol condensation: aldehydes/ketones with α-H, dilute NaOH → β-hydroxy carbonyl compound, which loses water on heating to give an α,β-unsaturated compound.
Cannizzaro reaction: aldehydes with no α-H (HCHO, C₆H₅CHO) + conc. NaOH → one molecule oxidised to acid salt, one reduced to alcohol (disproportionation).
Benzoin condensation: 2 C₆H₅CHO with KCN (alcoholic) → benzoin.
Tests: Tollens' (silver mirror) and Fehling's (red Cu₂O) are given by aliphatic aldehydes, not by ketones. Benzaldehyde gives Tollens' but not Fehling's. Iodoform test for CH₃CO– group.
Reductions: Clemmensen (Zn–Hg / conc. HCl) and Wolff–Kishner (NH₂NH₂ / KOH, heat) convert C=O to CH₂.
Important terms
α-hydrogen – H on the carbon next to C=O; it is acidic due to resonance-stabilised enolate.
Carbonyl group (कार्बोनिल गट) – C=O group.
Ylide – neutral molecule with adjacent opposite charges (Ph₃P⁺–CH₂⁻).
Formulas
Reaction
Needs
Product
Aldol
α-H, dil. NaOH
β-hydroxy aldehyde/ketone
Cannizzaro
no α-H, conc. NaOH
alcohol + carboxylate
Benzoin
ArCHO, CN⁻
α-hydroxy ketone
Perkin
ArCHO, (CH₃CO)₂O, CH₃COONa
cinnamic acid
Wittig
Ph₃P=CR₂
alkene
5 important questions
Q1. Why are aldehydes more reactive than ketones towards nucleophilic addition?
Ketones have two alkyl groups that reduce the positive charge on carbon by +I effect and cause more steric hindrance. Aldehydes have only one, so their carbonyl carbon is more electrophilic and easier to reach.
Q2. Give a test to distinguish acetaldehyde from acetone.
Tollens' reagent: acetaldehyde gives a silver mirror, acetone does not. (Fehling's solution: red precipitate with acetaldehyde only.)
Q3. Explain the mechanism of aldol condensation of acetaldehyde.
OH⁻ removes α-H → enolate ion (resonance stabilised).
Enolate carbon attacks carbonyl C of second CH₃CHO → alkoxide.
Alkoxide takes H⁺ from water → 3-hydroxybutanal (aldol).
On heating: loss of H₂O → but-2-enal (crotonaldehyde).
Q4. Explain the Cannizzaro reaction with mechanism.
OH⁻ adds to C=O of HCHO/C₆H₅CHO → tetrahedral anion.
Hydride ion (H⁻) transfer to a second aldehyde molecule (rate-determining).
Carboxylic acids contain the –COOH group. Their derivatives are acid chlorides, anhydrides, esters and amides, in which the –OH is replaced by another group. This topic covers acidity and the effect of substituents, preparation, important reactions (HVZ, decarboxylation, reduction), and the relative reactivity of acid derivatives towards nucleophilic acyl substitution.
Key points
Acidity: RCOOH ⇌ RCOO⁻ + H⁺. The carboxylate ion is stabilised by resonance (two equal C–O bonds), so carboxylic acids are much more acidic than alcohols and phenols.
Effect of substituents: electron-withdrawing groups increase acidity, electron-donating groups decrease it. Cl₃C–COOH > Cl₂CH–COOH > ClCH₂–COOH > CH₃COOH. HCOOH is stronger than CH₃COOH.
The nearer the –Cl is to –COOH, the stronger the acid (2-chlorobutanoic > 3-chloro > 4-chloro).
Carboxylic acids form hydrogen-bonded dimers, giving high boiling points.
Preparation: oxidation of 1° alcohols/aldehydes, hydrolysis of nitriles (R–CN + 2H₂O → RCOOH + NH₃), Grignard reagent + CO₂ then H₃O⁺, oxidation of alkylbenzenes (toluene → benzoic acid with KMnO₄).
Hell–Volhard–Zelinsky (HVZ): RCH₂COOH + Br₂ / red P → RCHBrCOOH (α-halogenation).
Decarboxylation: sodium salt + soda lime (NaOH + CaO), heat → alkane with one C less. Kolbe electrolysis gives R–R.
Reduction: LiAlH₄ reduces –COOH to –CH₂OH.
Derivatives reactivity (towards nucleophiles): acid chloride > anhydride > ester > amide. Better leaving group (Cl⁻) means higher reactivity; amide is least reactive due to strong resonance of N lone pair.
Q1. Why is chloroacetic acid stronger than acetic acid?
Cl has a −I effect. It pulls electron density away from the –COO⁻ group and spreads the negative charge, making the anion more stable. So ClCH₂COOH loses H⁺ more easily.
Q2. What is the HVZ reaction?
α-Halogenation of a carboxylic acid with Cl₂ or Br₂ in the presence of red phosphorus. Example: CH₃COOH + Cl₂ (red P) → ClCH₂COOH + HCl.
Q3. Explain the effect of substituents on the acidity of carboxylic acids with examples.
Inductive effect: −I groups (Cl, NO₂, CN) raise acidity, +I groups (alkyl) lower it.
Number and position of substituents (mono-, di-, trichloroacetic acid; α vs β vs γ).
Aromatic acids: p-nitrobenzoic acid > benzoic acid > p-methylbenzoic acid. Ortho effect: almost all ortho-substituted benzoic acids are stronger than benzoic acid.
Q4. Explain the mechanism of acid-catalysed esterification.
Protonation of C=O → attack by R′OH → tetrahedral intermediate → proton transfer → loss of H₂O → loss of H⁺ → ester. Isotope labelling (¹⁸O in alcohol ends up in the ester) shows the C–OH bond of the acid breaks.
Q5. Why is an acid chloride more reactive than an amide?
Compare leaving-group ability (Cl⁻ weak base, good leaving group; NH₂⁻ strong base, poor leaving group) and resonance donation (strong from N, weak from Cl) to the carbonyl carbon. Show resonance structures.
Exam tip
For acidity questions always mention the stability of the anion and the inductive/resonance effect by name.
This topic explains the behaviour of gases. It covers the gas laws and ideal gas equation, the postulates of the kinetic theory of gases, molecular speeds, deviations of real gases from ideal behaviour, the van der Waals equation and critical phenomena.
Key points
Boyle's law: at constant temperature, PV = constant.
Charles' law: at constant pressure, V/T = constant.
Avogadro's law: equal volumes of gases at same T and P contain equal numbers of molecules.
Ideal gas equation: PV = nRT. R = 8.314 J K⁻¹ mol⁻¹ = 0.0821 L atm K⁻¹ mol⁻¹.
Kinetic theory postulates: gas molecules are tiny points with negligible volume; they move randomly in straight lines; collisions are perfectly elastic; there are no attractive forces between molecules; average kinetic energy is proportional to absolute temperature.
Kinetic gas equation: PV = ⅓ mN c², where c is the root mean square speed.
Three molecular speeds: most probable (α), average (ū) and root mean square (c). Ratio α : ū : c = 1 : 1.128 : 1.224.
Maxwell distribution: at higher temperature the curve becomes flatter and its peak moves to higher speed.
Real gases deviate from ideal behaviour at high pressure and low temperature. Compressibility factor Z = PV/nRT. For an ideal gas Z = 1.
The ideal postulates fail because real molecules have a finite volume and they attract each other.
van der Waals equation corrects both: (P + an²/V²)(V − nb) = nRT. 'a' measures attraction; 'b' measures the volume of molecules.
Critical temperature (Tc): the temperature above which a gas cannot be liquefied however high the pressure.
Boyle temperature: temperature at which a real gas behaves ideally over a range of pressure; T_B = a/Rb.
Important terms
Ideal gas (आदर्श वायू) – gas that obeys PV = nRT at all T and P.
Compressibility factor (Z) – PV/nRT; shows deviation from ideal behaviour.
Critical pressure (Pc) – pressure needed to liquefy a gas at its critical temperature.
Critical volume (Vc) – volume of one mole of gas at Tc and Pc.
Mean free path – average distance a molecule travels between two collisions.
Formulas
PV = nRT
Kinetic gas equation: PV = ⅓ mNc²
c (rms) = √(3RT/M); ū = √(8RT/πM); α = √(2RT/M) (M in kg/mol, R = 8.314)
Q1. Why do real gases deviate from ideal behaviour?
Two postulates of kinetic theory are wrong for real gases. Real molecules occupy some volume, and they attract each other. These effects become large at high pressure and low temperature.
Q2. Define critical temperature. Give the relation with van der Waals constants.
Critical temperature is the temperature above which a gas cannot be liquefied by pressure alone. Tc = 8a/27Rb.
Q3. Calculate the rms speed of O₂ molecules at 27 °C.
T = 300 K, M = 32 g/mol = 0.032 kg/mol
c = √(3RT/M) = √(3 × 8.314 × 300 / 0.032)
= √(7482.6 / 0.032) = √233831
c ≈ 483.6 m/s
Q4. 2 mol of a gas occupy 10 L at 300 K. Find the pressure (ideal gas). (R = 0.0821 L atm K⁻¹ mol⁻¹)
P = nRT/V
= 2 × 0.0821 × 300 / 10
= 49.26 / 10
P = 4.93 atm
Q5. Derive the van der Waals equation (outline).
Volume correction: molecules have own volume; free volume = V − nb (b = excluded volume per mole, four times the actual volume of molecules).
Pressure correction: molecules near the wall are pulled inward, so observed P is less than ideal pressure.
Correction is proportional to (n/V)², written an²/V²; ideal P = P + an²/V².
Put both in PV = nRT → (P + an²/V²)(V − nb) = nRT.
At low P and high T the corrections become small, and it reduces to PV = nRT.
Exam tip
Always convert temperature to kelvin and molar mass to kg/mol before using speed formulas. The van der Waals derivation is a favourite long question.
Liquids have a definite volume but no definite shape. Their molecules are close together and held by intermolecular forces, but they can still move. This topic covers vapour pressure, surface tension, viscosity, their measurement, the effect of temperature, and liquid crystals.
Key points
Intermolecular forces in liquids: London (dispersion) forces, dipole–dipole forces and hydrogen bonding.
Vapour pressure is the pressure of vapour in equilibrium with the liquid at a given temperature. It increases with temperature.
A liquid boils when its vapour pressure equals the outside pressure. Water boils at 100 °C at 1 atm, and at a lower temperature on mountains.
Surface tension (पृष्ठताण): the force acting per unit length on the surface, at right angles to any line on it. Surface molecules are pulled inward, so a liquid tries to reduce its surface area. That is why drops are spherical.
Surface tension decreases with rise in temperature and becomes zero at the critical temperature.
Surface tension is measured by the stalagmometer (drop count or drop weight method) or the capillary rise method.
Viscosity: internal resistance to flow of a liquid, due to friction between layers moving at different speeds.
Viscosity of a liquid decreases with rise in temperature (for gases it increases).
Viscosity is measured with an Ostwald viscometer by comparing flow times of the liquid and water.
Parachor is an additive and constitutive property linking surface tension, molar mass and densities; it helps in finding structures.
Liquid crystals are a state between solid and liquid. They flow like liquids but have some order like crystals. Types: smectic, nematic and cholesteric. They are used in LCD displays and in temperature indicators.
Liquids with strong hydrogen bonds (water, glycerol) have high surface tension, high viscosity and high boiling points.
Important terms
Vapour pressure – pressure of vapour in equilibrium with its liquid at a fixed temperature.
Surface tension (γ) – force per unit length on the surface; SI unit N m⁻¹.
Surface energy – work needed to increase surface area by one unit; J m⁻², equal in value to surface tension.
Coefficient of viscosity (η) – force per unit area needed to keep unit velocity gradient between layers. SI unit Pa s (N s m⁻²); CGS unit poise.
Fluidity – reciprocal of viscosity, 1/η.
Formulas
Newton's law of viscous flow: F = η A (dv/dx)
Ostwald viscometer: η₁/η₂ = (d₁ t₁)/(d₂ t₂), d = density, t = flow time
Stalagmometer (drop number): γ₁/γ₂ = (n₂ d₁)/(n₁ d₂), n = number of drops of a fixed volume
Drop weight method: γ₁/γ₂ = m₁/m₂ (m = mass of one drop)
Capillary rise: γ = ½ h r d g (contact angle zero)
Parachor: [P] = M γ^(1/4) / (D − d), M = molar mass, D = density of liquid, d = density of vapour
1 poise = 0.1 Pa s
5 important questions
Q1. Why are liquid drops spherical?
Surface tension makes a liquid reduce its surface area to the minimum. For a given volume, a sphere has the smallest surface area. So free drops take a spherical shape.
Q2. What is the effect of temperature on viscosity of a liquid?
As temperature rises, molecules get more kinetic energy and intermolecular attraction becomes less effective. Layers slide more easily, so viscosity decreases.
Q3. In an Ostwald viscometer, water takes 120 s and a liquid takes 90 s to flow. Density of liquid = 0.8 g/cm³, water = 1.0 g/cm³, η(water) = 1.0 cP. Find η of the liquid.
η₁/η₂ = (d₁ t₁)/(d₂ t₂)
η₁ = 1.0 × (0.8 × 90)/(1.0 × 120)
= 72/120
η₁ = 0.60 cP
Q4. In a stalagmometer, a fixed volume gives 25 drops of water and 40 drops of a liquid of density 0.9 g/cm³. γ(water) = 72 dyne/cm. Find γ of the liquid.
γ₁/γ₂ = (n₂ d₁)/(n₁ d₂), liquid = 1, water = 2
γ₁ = 72 × (25 × 0.9)/(40 × 1.0)
= 72 × 22.5/40 = 72 × 0.5625
γ₁ = 40.5 dyne/cm
Q5. Describe the determination of viscosity by Ostwald's viscometer.
Clean, dry viscometer fixed vertically in a constant-temperature bath.
Fill a fixed volume of water, suck it above the upper mark, note time t₂ for flow between two marks.
Repeat with the liquid to get t₁.
Find densities d₁ and d₂.
Calculate η₁ = η₂ (d₁ t₁)/(d₂ t₂). Mention the diagram: bulb, two marks, capillary, lower reservoir.
Exam tip
Both viscometer and stalagmometer numericals use simple ratios. Write the formula, then substitute carefully; keep units of density the same for both liquids.
Solids have a definite shape and volume because their particles are held in fixed positions. This topic covers crystalline and amorphous solids, laws of crystallography, unit cells and crystal systems, Miller indices, Bragg's law of X-ray diffraction, packing in cubic lattices, calculation of density, and defects in crystals.
Key points
Crystalline solids: regular 3D arrangement (long-range order), sharp melting point, anisotropic (NaCl, quartz). Amorphous solids: short-range order only, melt over a range, isotropic (glass, rubber).
Laws of crystallography: law of constancy of interfacial angles, law of rational indices, law of symmetry.
Symmetry elements of a cube: 9 planes, 13 axes and 1 centre of symmetry — 23 elements in all.
Unit cell: smallest repeating unit of a crystal lattice, described by edge lengths a, b, c and angles α, β, γ.
Seven crystal systems (cubic, tetragonal, orthorhombic, monoclinic, triclinic, hexagonal, rhombohedral) and 14 Bravais lattices.
Atoms per unit cell (Z): simple cubic 1, body-centred cubic (bcc) 2, face-centred cubic (fcc) 4.
Miller indices (h k l): take intercepts in units of a, b, c → take reciprocals → clear fractions to smallest whole numbers.
Bragg's law: X-rays reflected from parallel planes interfere constructively when nλ = 2d sinθ.
For cubic crystals, the ratio of interplanar distances d₁₀₀ : d₁₁₀ : d₁₁₁ helps decide sc, bcc or fcc.
NaCl structure: fcc, coordination number 6:6, 4 NaCl units per cell. CsCl: bcc-type, 8:8, 1 unit per cell.
Defects: Schottky (equal cation and anion vacancies, density falls; NaCl, KCl) and Frenkel (ion moves to an interstitial site, density unchanged; AgCl, ZnS). AgBr shows both.
Important terms
Crystal lattice (स्फटिक जालक) – regular 3D arrangement of points representing particles.
Coordination number – number of nearest neighbours of a particle.
Interplanar spacing (d) – distance between parallel lattice planes.
Formulas
Density ρ = Z × M / (a³ × Nₐ) (a in cm, ρ in g cm⁻³, Nₐ = 6.022 × 10²³ mol⁻¹)
Radius relations: sc a = 2r; bcc √3 a = 4r; fcc √2 a = 4r
Lattice
Z
Coordination no.
Packing
Simple cubic
1
6
52.4 %
bcc
2
8
68 %
fcc
4
12
74 %
5 important questions
Q1. Find the Miller indices of a plane with intercepts 2a, 3b and c.
Intercepts: 2, 3, 1 → reciprocals: ½, ⅓, 1 → multiply by 6: 3, 2, 6. Miller indices = (3 2 6).
Q2. Distinguish between Schottky and Frenkel defects.
Schottky: pair of cation and anion missing, density decreases, found in ionic solids with similar-sized ions (NaCl). Frenkel: smaller ion leaves its site and sits in an interstitial position, density unchanged, found where ions differ greatly in size (AgCl, ZnS).
Q3. NaCl has an fcc unit cell with edge length 5.64 Å. Calculate its density (M = 58.5 g mol⁻¹).
Two parallel planes distance d apart; X-rays strike at angle θ.
The ray reflected from the lower plane travels an extra path = 2d sinθ.
Constructive interference when the path difference = whole number of wavelengths: nλ = 2d sinθ.
Example: λ = 1.54 Å, first-order reflection at θ = 30° → d = 1.54/(2 × 0.5) = 1.54 Å.
Q5. Explain the seven crystal systems with their axial lengths and angles.
Make a table: cubic (a = b = c; all 90°), tetragonal (a = b ≠ c; all 90°), orthorhombic (a ≠ b ≠ c; all 90°), monoclinic (a ≠ b ≠ c; α = γ = 90°, β ≠ 90°), triclinic (all different), hexagonal (a = b ≠ c; α = β = 90°, γ = 120°), rhombohedral (a = b = c; α = β = γ ≠ 90°). Give one example each.
Exam tip
Learn Z, coordination number and packing for sc, bcc and fcc as one table. Density and Bragg's law numericals are favourites.
Thermodynamics studies energy changes in physical and chemical processes. This topic covers system and surroundings, state functions, the first law, internal energy and enthalpy, heat capacities, work in reversible and irreversible processes, Hess's law, and an introduction to entropy, the second law and Gibbs free energy.
Key points
System: the part studied. Surroundings: everything else. Systems are open, closed or isolated.
State functions depend only on initial and final states (U, H, S, G, P, V, T). Path functions depend on the path (q, w).
Processes: isothermal (T constant), adiabatic (q = 0), isobaric (P constant), isochoric (V constant), reversible and irreversible.
First law: energy can neither be created nor destroyed. ΔU = q + w (IUPAC sign: work done on the system is positive).
Enthalpy: H = U + PV. At constant pressure, ΔH = q_p. At constant volume, ΔU = q_v.
ΔH = ΔU + Δn(g)RT for reactions involving gases.
Cp − Cv = R for one mole of an ideal gas. γ = Cp/Cv.
Reversible isothermal expansion of an ideal gas gives maximum work. ΔU = 0, so q = −w.
Hess's law: the total enthalpy change of a reaction is the same whether it occurs in one step or several steps.
Kirchhoff's equation gives the effect of temperature on enthalpy of reaction.
Second law: heat does not flow by itself from a colder to a hotter body; the entropy of the universe increases in a spontaneous process.
Gibbs free energy: G = H − TS. At constant T and P, ΔG < 0 means spontaneous, ΔG = 0 means equilibrium, ΔG > 0 non-spontaneous.
Third law: entropy of a perfect crystal at 0 K is zero.
Important terms
Internal energy (अंतर्गत ऊर्जा, U) – total energy stored in a system.
Enthalpy (H) – heat content at constant pressure.
Entropy (S) – measure of randomness or disorder.
Heat capacity – heat needed to raise the temperature by 1 K.
Extensive property – depends on amount (mass, volume, U, H). Intensive property – does not (T, P, density).
Formulas
ΔU = q + w
Work against constant external pressure: w = −P_ext ΔV
Q1. State the first law of thermodynamics and write its mathematical form.
Energy can neither be created nor destroyed, only changed from one form to another. The energy of an isolated system is constant. ΔU = q + w.
Q2. What are state functions? Give two examples.
Properties whose value depends only on the present state of the system, not on how it was reached. Examples: internal energy and enthalpy. Heat and work are not state functions.
Q3. Calculate the work done when 2 mol of an ideal gas expands reversibly and isothermally at 300 K from 10 L to 20 L.
w = −2.303 nRT log(V₂/V₁)
= −2.303 × 2 × 8.314 × 300 × log 2
= −2.303 × 4988.4 × 0.3010
= −3458 J ≈ −3.46 kJ (work done by the gas)
Q4. For N₂(g) + 3H₂(g) → 2NH₃(g), ΔH = −92.4 kJ at 298 K. Find ΔU.
Δn(g) = 2 − (1 + 3) = −2
ΔnRT = −2 × 8.314 × 298 = −4955 J = −4.96 kJ
ΔU = ΔH − ΔnRT = −92.4 − (−4.96)
ΔU = −87.44 kJ
Q5. Explain the criteria of spontaneity in terms of ΔG.
G = H − TS; ΔG = ΔH − TΔS at constant T and P.
ΔH negative, ΔS positive → ΔG always negative, spontaneous at all T.
ΔH positive, ΔS negative → never spontaneous.
ΔH negative, ΔS negative → spontaneous at low T.
ΔH positive, ΔS positive → spontaneous at high T (e.g. melting of ice above 0 °C).
Exam tip
Use the IUPAC sign convention throughout and say so in the answer. Write R = 8.314 J K⁻¹ mol⁻¹ and convert J to kJ at the end.