रसायनशास्त्र Chemistry

थोडक्यात सारांश

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.

महत्त्वाचे मुद्दे

• 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.

महत्त्वाच्या संज्ञा

• 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.

सूत्रे

• 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)
• Radius of nth orbit: rₙ = 0.529 n²/Z Å
• Rydberg formula: 1/λ = R_H Z² (1/n₁² − 1/n₂²), R_H = 1.097 × 10⁷ m⁻¹
• de Broglie: λ = h/mv = h/p (h = 6.626 × 10⁻³⁴ J s)
• Uncertainty: Δx·Δp ≥ h/4π
• Maximum electrons in a shell = 2n²; in a subshell = 2(2l + 1)

५ महत्त्वाचे प्रश्न

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.

परीक्षेसाठी टीप

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.

थोडक्यात सारांश

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.

महत्त्वाचे मुद्दे

• 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.

महत्त्वाच्या संज्ञा

• 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.

सूत्रे

• 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.

५ महत्त्वाचे प्रश्न

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).

• Configuration grouped: (1s²)(2s² 2p⁶)(3s² 3p⁶)(4s¹)
• Same group (4s): 0 other electrons → 0
• n − 1 shell (3s, 3p): 8 × 0.85 = 6.80
• n − 2 and lower: 10 × 1.00 = 10.00
• σ = 16.80, so Z* = 19 − 16.80 = 2.20

Q4. Calculate Z* for a 2p electron of nitrogen (Z = 7).

• Grouping: (1s²)(2s² 2p³)
• Other electrons in same group: 4 × 0.35 = 1.40
• n − 1 shell (1s): 2 × 0.85 = 1.70
• σ = 3.10, Z* = 7 − 3.10 = 3.90

Q5. Discuss the variation of atomic radius and ionisation energy in the periodic table.

• Across a period: nuclear charge rises, shell stays same → radius falls, IE rises.
• Down a group: new shells added, shielding increases → radius rises, IE falls.
• Mention exceptions Be/B and N/O for IE.
• Mention noble gases have the highest IE in each period.
• Mention lanthanide contraction for heavy elements.

परीक्षेसाठी टीप

Slater's rule numericals are easy marks; always write the grouping of electrons first, then the contributions line by line.

थोडक्यात सारांश

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.

महत्त्वाचे मुद्दे

• 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).
• Shapes: CH₄ tetrahedral (109.5°), NH₃ pyramidal (about 107°), H₂O bent (about 104.5°), BF₃ trigonal planar, PCl₅ trigonal bipyramidal, SF₆ 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.

महत्त्वाच्या संज्ञा

• 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.

सूत्रे

• 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

५ महत्त्वाचे प्रश्न

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.

• O₂ has 16 electrons. Configuration: σ1s² σ1s² σ2s² σ2s² σ2pz² π2px² π2py² π2px¹ π2py¹
• Nb = 10, Na = 6
• Bond order = ½ (10 − 6) = 2
• Two unpaired electrons in π* → paramagnetic

Q4. Compare the bond orders of N₂ and N₂⁺.

• N₂ (14 e): Nb = 10, Na = 4, bond order = ½ (10 − 4) = 3
• N₂⁺ (13 e): one bonding electron is removed. Nb = 9, Na = 4
• Bond order = ½ (9 − 4) = 2.5
• So N₂ has the stronger, shorter bond.

Q5. Explain sp³d hybridisation with PCl₅ as an example.

• P (Z = 15): 3s² 3p³. One 3s electron is promoted to 3d → five unpaired electrons.
• One s, three p and one d orbital mix → five sp³d hybrid orbitals.
• Each overlaps with a Cl 3p orbital → five P–Cl σ bonds.
• Shape: trigonal bipyramidal. Three equatorial bonds at 120°, two axial bonds at 90° to the plane.
• Axial bonds are longer than equatorial bonds (more repulsion).

परीक्षेसाठी टीप

Learn the MO diagram once and use it for O₂, N₂ and their ions. Always write the bond order formula, the counting and the magnetic nature together.

थोडक्यात सारांश

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.

महत्त्वाचे मुद्दे

• 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.

महत्त्वाच्या संज्ञा

• 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.

सूत्रे

• 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).

५ महत्त्वाचे प्रश्न

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).
• Types: intermolecular (H₂O, HF, alcohols, carboxylic acid dimers) and intramolecular (o-nitrophenol, salicylaldehyde).
• Effects: high boiling point, high viscosity of glycerol, solubility of alcohols in water, low density of ice, structure of DNA and proteins.

परीक्षेसाठी टीप

Always draw the band diagrams and the H-bond with dotted lines (····). The ortho vs para nitrophenol question is asked very often.

थोडक्यात सारांश

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.

महत्त्वाचे मुद्दे

• 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.

महत्त्वाच्या संज्ञा

• Alkali metals (अल्कली धातू) – Group 1 metals: Li, Na, K, Rb, Cs, Fr.
• 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).

सूत्रे

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)

५ महत्त्वाचे प्रश्न

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.

परीक्षेसाठी टीप

Short questions on flame colours, inert pair effect and diagonal relationship are very common. Learn diborane and xenon fluoride structures with neat diagrams.

थोडक्यात सारांश

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.

महत्त्वाचे मुद्दे

• 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.
• Oxyacids of sulphur: H₂SO₃, H₂SO₄, H₂S₂O₇ (oleum), H₂S₂O₈ (Marshall's acid, peroxodisulphuric acid).
• 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.

महत्त्वाच्या संज्ञा

• 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.

सूत्रे

• Number of 3c–2e bonds in B₂H₆ = 2; total valence electrons = 12.

५ महत्त्वाचे प्रश्न

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.
• 4 terminal B–H bonds: 2c–2e. 2 bridging bonds: 3c–2e (banana bonds).
• Terminal B–H bonds are shorter than bridge bonds.
• Draw the structure with H–B–H angles.

Q4. Classify silicates with examples.

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).

परीक्षेसाठी टीप

Practise drawing B₂H₆, borazine, P₄O₁₀ and the oxyacids of phosphorus. A clear labelled structure earns most of the marks.

थोडक्यात सारांश

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.

महत्त्वाचे मुद्दे

• 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).
• Soft species: large, low charge, easily polarised (Ag⁺, Cu⁺, Hg²⁺, Pt²⁺, I⁻, S²⁻, CN⁻, CO).
• Borderline: Fe²⁺, Co²⁺, Ni²⁺, Cu²⁺, Zn²⁺, Br⁻, pyridine.

महत्त्वाच्या संज्ञा

• Conjugate acid–base pair (संयुग्मी आम्ल–आम्लारी जोडी) – two species that differ by one H⁺ (HCl / Cl⁻).
• Amphoteric (amphiprotic) substance – can act as acid and base (H₂O, HCO₃⁻).
• Levelling solvent – makes acids (or bases) of different strength appear equal.

सूत्रे

५ महत्त्वाचे प्रश्न

Q1. Identify the conjugate pairs: HCl + H₂O → H₃O⁺ + Cl⁻.

Pair 1: HCl (acid) / Cl⁻ (conjugate base). Pair 2: H₂O (base) / H₃O⁺ (conjugate acid).

Q2. Why is BF₃ a Lewis acid?

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.

परीक्षेसाठी टीप

Learn the hard/soft lists as a table; one question on HSAB applications appears in most papers.

थोडक्यात सारांश

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.

महत्त्वाचे मुद्दे

• 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.

महत्त्वाच्या संज्ञा

• 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.

सूत्रे

Useful orders to remember:

• Acid strength: CCl₃COOH > CHCl₂COOH > CH₂ClCOOH > CH₃COOH (−I effect of Cl)
• FCH₂COOH > ClCH₂COOH > BrCH₂COOH > ICH₂COOH
• Carbocation stability: benzyl ≈ allyl, 3° > 2° > 1° > methyl
• Number of hyperconjugative structures = number of α-hydrogens
• Homolysis: Cl–Cl → 2Cl• (light); Heterolysis: (CH₃)₃C–Br → (CH₃)₃C⁺ + Br⁻

५ महत्त्वाचे प्रश्न

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).
• Electrophiles attack electron-rich sites (double bonds, benzene).
• Nucleophiles attack electron-poor sites (carbonyl carbon, alkyl halide carbon).

परीक्षेसाठी टीप

Questions on stability orders and acid strength are frequent. Always give the reason (inductive, resonance or hyperconjugation) in one line with the order.

थोडक्यात सारांश

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.

महत्त्वाचे मुद्दे

• Strong electrolytes ionise almost completely (HCl, NaOH, NaCl). Weak electrolytes ionise partly (CH₃COOH, NH₄OH).
• Ostwald's dilution law: for a weak acid, the degree of dissociation α increases on dilution: α = √(Ka/C).
• Ionic product of water: Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25 °C. Kw increases with temperature.
• pH = −log[H⁺]. Neutral pH = 7 at 25 °C; pH + pOH = 14.
• 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.

महत्त्वाच्या संज्ञा

• 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.

सूत्रे

• Ka = Cα²/(1 − α) ≈ Cα²; α = √(Ka/C); [H⁺] = √(Ka·C)
• pH = −log[H⁺]; pOH = −log[OH⁻]; pH + pOH = 14 (25 °C)
• Henderson equation (acid buffer): pH = pKa + log ([salt]/[acid])
• Basic buffer: pOH = pKb + log ([salt]/[base])
• Salt of weak acid + strong base: Kh = Kw/Ka; pH = 7 + ½pKa + ½log C
• For AB: Ksp = s²; for AB₂ or A₂B: Ksp = 4s³ (s = molar solubility)

५ महत्त्वाचे प्रश्न

Q1. Calculate the pH of 0.01 M HCl.

HCl is a strong acid, so [H⁺] = 0.01 = 10⁻² M. pH = −log 10⁻² = 2.

Q2. Why is an aqueous solution of CH₃COONa basic?

CH₃COO⁻ (conjugate base of a weak acid) reacts with water: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻. Extra OH⁻ makes the solution basic (pH > 7).

Q3. Calculate the pH of 0.1 M acetic acid (Ka = 1.8 × 10⁻⁵).

• [H⁺] = √(Ka × C) = √(1.8 × 10⁻⁵ × 0.1) = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³ M
• pH = −log(1.34 × 10⁻³) = 3 − 0.127 = 2.87
• α = √(Ka/C) = √(1.8 × 10⁻⁴) = 0.0134 (1.34 %)

Q4. Find the pH of a buffer that is 0.2 M CH₃COONa and 0.1 M CH₃COOH (pKa = 4.74). Explain buffer action.

• pH = 4.74 + log (0.2/0.1) = 4.74 + 0.301 = 5.04
• Added H⁺ is removed by CH₃COO⁻ → CH₃COOH; added OH⁻ is removed by CH₃COOH → CH₃COO⁻ + H₂O. So pH hardly changes.

Q5. The Ksp of AgCl is 1.8 × 10⁻¹⁰ at 25 °C. Find its solubility. Give two applications of Ksp.

• AgCl ⇌ Ag⁺ + Cl⁻; Ksp = s² → s = √(1.8 × 10⁻¹⁰) = 1.34 × 10⁻⁵ mol L⁻¹
• 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.

परीक्षेसाठी टीप

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.