1Solutions
Colligative properties, Raoult's law
Electrochemistry
Nernst equation, Conductance, Batteries
Chemical Kinetics
Rate law, Arrhenius equation, Half-life
Chemical Equilibrium
Le Chatelier's principle, Kp, Kc
Surface Chemistry
Adsorption, Catalysis, Colloids
General Principles of Metallurgy
Extraction, Refining, Thermodynamic principles
p-Block Elements
Group 15, 16, 17, 18
d and f Block Elements
Transition metals, Lanthanoids, Actinoids
Coordination Compounds
Werner's theory, Crystal field theory, Isomerism
Haloalkanes and Haloarenes
Reactions, SN1, SN2 mechanisms
Alcohols, Phenols, Ethers
Preparation, Properties, Reactions
Aldehydes, Ketones, Carboxylic Acids
Nucleophilic addition, Oxidation, Reduction
Amines
Classification, Basicity, Diazotisation
Biomolecules
Carbohydrates, Proteins, Nucleic acids
Polymers
Addition, Condensation polymers, Rubber
Chemistry in Everyday Life
Drugs, Food additives, Cleansing agents
Chapter 1: Solutions
Key Formulas:
• Molarity (M) = moles of solute/L of solution
• Molality (m) = moles of solute/kg of solvent
• Raoult's Law: Psolution = P°solvent × xsolvent
• ΔTb = Kb · m, ΔTf = Kf · m
• Π = CRT (osmotic pressure)
Q1. Calculate the molality of a solution containing 5.85 g of NaCl in 500 g of water.
Solution
Molar mass of NaCl = 58.5 g/mol
Moles of NaCl = 5.85/58.5 = 0.1 mol
Mass of solvent = 500 g = 0.5 kg
Molality = 0.1/0.5 = 0.2 mol/kg = 0.2 m
Q2. The boiling point of a 0.1 m aqueous solution of a non-electrolyte is 100.18°C. Find the Kb of water. (Kb = 0.512 K kg/mol)
Solution
ΔTb = 100.18 − 100 = 0.18°C
Kb = ΔTb/m = 0.18/0.1 = 1.8 K kg/mol
Wait, given Kb = 0.512, so: ΔTb = Kb × m = 0.512 × 0.1 = 0.0512°C
Actual ΔT = 0.18 suggests non-ideal behavior or i > 1.
Q3. Calculate the osmotic pressure of a 0.05 M glucose solution at 27°C.
Solution
Π = CRT
= 0.05 × 0.0821 × 300
= 1.23 atm
Q4. 18 g of glucose (C₆H₁₂O₆) is dissolved in 1 kg of water. Calculate the freezing point depression.
Solution
Molar mass of glucose = 180 g/mol
Moles = 18/180 = 0.1 mol
Molality = 0.1/1 = 0.1 m
ΔTf = Kf × m = 1.86 × 0.1 = 0.186°C
Q5. A solution of urea (mol. mass = 60) boils at 100.18°C at 1 atm. Calculate the freezing point of the same solution.
Solution
ΔTb = 0.18°C
m = ΔTb/Kb = 0.18/0.52 = 0.346
ΔTf = Kf × m = 1.86 × 0.346
= 0.644°C
Freezing point = 0 − 0.644 = −0.644°C
Chapter 2: Electrochemistry
Key Formulas:
• Nernst: E = E° − (RT/nF) ln Q
• E°cell = E°cathode − E°anode
• ΔG° = −nFE°cell
• K = enFE°/RT
• Λm = κ/c (molar conductivity)
Q1. Calculate the emf of the cell: Zn | Zn₂₁(0.1M) || Cu₂₁(1M) | Cu. Given E°(Zn₂₁/Zn) = −0.76V, E°(Cu₂₁/Cu) = +0.34V.
Solution
E°cell = 0.34 − (−0.76) = 1.10V
Ecell = E° − (0.0591/2) log([Zn₂₁]/[Cu₂₁])
= 1.10 − (0.0591/2) log(0.1/1)
= 1.10 − (0.02955)(-1)
= 1.10 + 0.02955
= 1.13V
Q2. Calculate ΔG° for the reaction: Zn + Cu₂₁ → Zn₂₁ + Cu. E°cell = 1.10V.
Solution
ΔG° = −nFE°cell
= −(2)(96500)(1.10)
= −212300 J/mol
= −212.3 kJ/mol
Q3. The resistance of a 0.5 M KCl solution is 4 Ω. The cell constant is 1.2 cm⁻¹. Find the molar conductivity.
Solution
κ = 1/R × cell constant = (1/4) × 1.2 = 0.3 S/cm
Λm = κ/c = 0.3/0.5 = 0.6 S cm²/mol
= 600 S cm²/mol
Q4. What is the equilibrium constant for the reaction: Cu + 2Ag₁ → Cu₂₁ + 2Ag? (E°cell = 0.46V)
Solution
log K = nE°/(0.0591)
= (2)(0.46)/0.0591
= 0.92/0.0591 = 15.57
K = 1015.57 = 3.7 × 1015
Q5. How long does it take to deposit 1 g of copper from CuSO₄ solution using a current of 2 A? (M = 63.5, F = 96500)
Solution
W = (M × I × t)/(n × F)
1 = (63.5 × 2 × t)/(2 × 96500)
t = (2 × 96500)/(63.5 × 2)
t = 193000/63.5
= 3039.4 s ≈ 50.7 minutes
Chapter 3: Chemical Kinetics
Key Formulas:
• Rate = k[A]m[B]n
• First order: k = (2.303/t) log(a/(a−x))
• t1/2 = 0.693/k (first order)
• Arrhenius: k = Ae−Ea/RT
• ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂)
Q1. For a first order reaction, the half-life is 10 minutes. How much time will it take for 75% completion?
Solution
75% completion means 25% remains, so x = 0.75a
t = (2.303/k) log(a/(a−x))
k = 0.693/10 = 0.0693 min⁻¹
t = (2.303/0.0693) log(a/0.25a)
= 33.23 × log(4)
= 33.23 × 0.6021
= 20 minutes
Q2. The rate constant of a reaction is 1.5 × 10−3 s⁻¹ at 300K and 4.5 × 10−3 s⁻¹ at 310K. Calculate Ea.
Solution
ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂)
ln(4.5/1.5) = (Ea/8.314)(1/300 − 1/310)
ln(3) = (Ea/8.314)(10/(300 × 310))
1.0986 = (Ea/8.314)(1.075 × 10−4)
Ea = (1.0986 × 8.314)/(1.075 × 10−4)
= 84.9 kJ/mol
Q3. The rate of a reaction triples when the concentration of a reactant is doubled. What is the order of the reaction with respect to that reactant?
Solution
Rate₂/Rate₁ = (2[Reactant])n/[Reactant]n
3 = 2n
n = log 3/log 2 = 1.58 ≈ 1.6
Q4. A first order reaction takes 40 minutes for 30% decomposition. Calculate t1/2.
Solution
k = (2.303/t) log(100/70)
= (2.303/40) log(1.4286)
= (2.303/40)(0.1549)
= 0.00888 min⁻¹
t1/2 = 0.693/k = 0.693/0.00888 = 78 minutes
Q5. For a reaction A + B → Products, the rate law is Rate = k[A][B]2. If concentration of A is doubled and B is halved, how does the rate change?
Solution
Rate₁ = k[A][B]²
Rate₂ = k[2A][(B/2)]² = k · 2A · B²/4 = kAB²/2
Rate₂/Rate₁ = 1/2
The rate becomes half of the original rate.
Chapter 4: Chemical Equilibrium
Q1. For the reaction: N₂(g) + 3H₂(g) ⇔ 2NH₃(g), Kc = 0.5 at 400K. Find Kp.
Solution
Kp = Kc(RT)Δn
Δn = 2 − (1 + 3) = −2
Kp = 0.5 × (0.0821 × 400)−2
= 0.5/(32.84)²
= 4.65 × 10−4
Q2. At equilibrium, [SO₂] = 0.6M, [O₂] = 0.4M, [SO₃] = 0.8M for the reaction: 2SO₂ + O₂ ⇔ 2SO₃. Find Kc.
Solution
Kc = [SO₃]²/([SO₂]²[O₂])
= (0.8)²/((0.6)² × 0.4)
= 0.64/(0.36 × 0.4)
= 0.64/0.144
= 4.44
Q3. Explain Le Chatelier's principle with an example.
Solution
Le Chatelier's Principle: If a system at equilibrium is disturbed, the system adjusts itself to counteract the disturbance and re-establish equilibrium.
Example: N₂(g) + 3H₂(g) ⇔ 2NH₃(g) + Heat
• Increase pressure → equilibrium shifts to right (fewer moles)
• Increase temperature → shifts left (endothermic direction)
• Remove NH₃ → shifts right (produces more NH₃)
Q4. For a reaction A ⇔ B, Kc = 4. If 0.4 mol of A is taken in 1L flask, find the equilibrium concentration of B.
Solution
Initial: [A] = 0.4, [B] = 0
Change: −x, +x
Equilibrium: 0.4 − x, x
Kc = x/(0.4 − x) = 4
x = 4(0.4 − x) = 1.6 − 4x
5x = 1.6
x = 0.32 M
[B] = 0.32 M
Chapter 5: Surface Chemistry
Q1. Distinguish between physisorption and chemisorption.
Solution
Physisorption:
• Weak van der Waals forces
• Low enthalpy (20-40 kJ/mol)
• Reversible, no specificity
Chemisorption:
• Strong chemical bonds
• High enthalpy (80-240 kJ/mol)
• Irreversible, highly specific
Q2. What is the difference between a sol and a gel? Give one example of each.
Solution
Sol: Solid particles dispersed in liquid (e.g., starch sol, gold sol)
Gel: Liquid trapped in a solid network (e.g., gelatin, cheese)
Q3. Why does physisorption decrease with increase in temperature?
Solution
Physisorption involves weak van der Waals forces. As temperature increases, the kinetic energy of adsorbed molecules increases, causing them to overcome the weak attractive forces and desorb from the surface. Hence, physisorption decreases with increasing temperature.
Q4. What type of catalysis is involved in: (i) Haber's process (ii) Hydrogenation of oils (iii) Decomposition of H₂O₂?
Solution
(i) Haber's process: Heterogeneous catalysis (Fe catalyst)
(ii) Hydrogenation: Heterogeneous catalysis (Ni catalyst)
(iii) H₂O₂ decomposition: Homogeneous catalysis (MnO₂ or catalase enzyme)
Q5. What are emulsions? Give two examples and mention the type of each.
Solution
An emulsion is a colloidal dispersion of one liquid in another immiscible liquid.
O/W (oil-in-water): Milk, butter, vanishing cream
W/O (water-in-oil): Cold cream, mayonnaise
Chapter 6: General Principles of Metallurgy
Q1. What is the principle behind the extraction of aluminium by Hall-Heroult process?
Solution
Electrolytic reduction of purified alumina (Al₂O₃) dissolved in molten cryolite (Na₃AlF₆).
At cathode: Al₃⁷ + 3eⁿ → Al
At anode: 2O₂⁷ − 4eⁿ → O₂
Cryolite lowers the melting point from 2050°C to 950°C.
Q2. Define the following terms: (a) Roasting (b) Calcination (c) Slag
Solution
Roasting: Heating sulphide ores in excess air to convert them to oxides (e.g., 2ZnS + 3O₂ → 2ZnO + 2SO₂)
Calcination: Heating ores in limited air to remove volatile impurities (e.g., ZnCO₃ → ZnO + CO₂)
Slag: Fusible product formed when flux reacts with impurities (e.g., CaSiO₃ in iron extraction)
Q3. Why is zinc extracted by electrolysis and not by carbon reduction?
Solution
Zinc is a more reactive metal (E° = −0.76V). Carbon cannot reduce ZnO effectively because zinc has a high affinity for oxygen. Electrolytic reduction provides sufficient driving force to extract zinc from its oxide.
Q4. Explain the Ellingham diagram. What is its significance in metallurgy?
Solution
The Ellingham diagram plots ΔG° vs Temperature for formation of oxides.
Significance:
• Lower line = more stable oxide
• Metal with lower ΔG° can reduce oxide above it
• At intersection, both metals have same affinity for O₂
• Helps choose reducing agent and extraction temperature
Chapter 7: p-Block Elements
Q1. Why does PCl₅ exist but NCl₅ does not?
Solution
Nitrogen cannot expand its octet (no d-orbitals in the second period), so NCl₅ is not possible. Phosphorus has vacant 3d-orbitals and can expand its octet to form PCl₅ using sp₃d hybridization.
Q2. Arrange the following in increasing order of acidity: HClO, HClO₂, HClO₃, HClO₄
Solution
Increasing acidity: HClO < HClO₂ < HClO₃ < HClO₄
Reason: As the number of oxygen atoms increases, the O-H bond becomes more polar due to the electron-withdrawing inductive effect, making H+ release easier.
Q3. Complete and balance: Cu + conc. HNO₃ →
Solution
Cu + 4HNO₃(conc.) → Cu(NO₃)₂ + 2NO₂ + 2H₂O
(Dilute HNO₃ gives NO instead of NO₂)
Q4. Why is H₂S a better reducing agent than H₂O?
Solution
H₂S has a longer bond length (S-H vs O-H) and weaker bond dissociation energy. The S-H bond is easier to break, making H₂S more easily oxidized and thus a better reducing agent than H₂O.
Q5. What happens when (i) SO₂ gas is passed through water (ii) Cl₂ reacts with cold dilute NaOH?
Solution
(i) SO₂ + H₂O ⇔ H₂SO₃ (sulphurous acid)
(ii) Cl₂ + 2NaOH(dilute, cold) → NaCl + NaOCl + H₂O
(gives hypochlorite, bleaching agent)
Chapter 8: d and f Block Elements
Q1. Why do transition elements show variable oxidation states?
Solution
Transition elements have (n−1)d and ns orbitals with small energy difference. Electrons from both orbitals can participate in bonding, leading to variable oxidation states. For example, Mn shows +2, +3, +4, +5, +6, +7.
Q2. Why is Cu₁ ion unstable in aqueous solution while Cu₂ is stable?
Solution
Cu₁ (3d₄) has higher hydration enthalpy than Cu₂ (3d₃)
But Cu₂ has more than double the charge, giving much higher hydration enthalpy (~2100 kJ/mol vs ~600 kJ/mol), making Cu₂ more stable in solution.
Q3. What is lanthanoid contraction? Mention its consequences.
Solution
Lanthanoid contraction is the gradual decrease in atomic and ionic radii of lanthanoids with increasing atomic number, due to poor shielding by 4f electrons.
Consequences:
• Similarity of 2nd and 3rd transition series elements
• Separation of lanthanoids is difficult
• Basicity of lanthanoid hydroxides decreases across the series
Q4. Why does Zn₂ not show coloured ions?
Solution
Zn₂ has the electronic configuration 3d₄ (completely filled d-orbitals). Since there are no unpaired electrons, d-d transitions are not possible, and hence Zn₂ does not show coloured ions.
Chapter 9: Coordination Compounds
Q1. Give the IUPAC name of [Co(NH₃)₄Cl₂]Cl.
Solution
Complex ion: [Co(NH₃)₄Cl₂]+
Co is in +3 oxidation state (since 2Clⁿ + 4NH₃ = +2, so Co = +3)
Tetraamminedichloridocobalt(III) chloride
Q2. What is the coordination number and oxidation state of metal in K₄[Fe(CN)₆]?
Solution
Coordination number = 6 (6 CNⁿ ligands)
Oxidation state: 4(+1) + x + 6(−1) = 0
x = +2
Coordination number = 6, Oxidation state = +2
Q3. Explain Crystal Field Theory. Draw the d-orbital splitting in an octahedral field.
Solution
In CFT, metal-ligand bonding is electrostatic. In an octahedral field, d-orbitals split into two sets:
t₄⁴ (lower): dxy, dyz, dxz
eg (higher): dz², dx²−y²
Energy difference = Δ₀ (crystal field splitting energy)
If Δ₀ > P (pairing energy) → low spin (strong field)
If Δ₀ < P → high spin (weak field)
Q4. What is the difference between a double salt and a coordination compound? Give one example of each.
Solution
Double salt: Dissociates into simple ions in solution (e.g., KCl·MgCl₂·6H₂O)
Coordination compound: Retains identity in solution (e.g., [Co(NH₃)₆]Cl₃)
Chapter 10: Haloalkanes and Haloarenes
Q1. Distinguish between SN1 and SN2 mechanisms.
Solution
SN1: Two steps, rate depends on [substrate] only, racemization, follows 3° > 2° > 1°
SN2: One step, rate depends on [substrate][nucleophile], inversion of configuration, follows 1° > 2° > 3°
Q2. Why is chlorobenzene less reactive than chloroethane towards nucleophilic substitution?
Solution
In chlorobenzene, the C-Cl bond has partial double bond character due to resonance with the benzene ring. This makes the bond stronger and harder to break compared to chloroethane where C-Cl is a pure single bond.
Q3. What is the Wurtz reaction? Write the equation for the preparation of butane.
Solution
2CH₃CH₂Cl + 2Na &xrightarrow;dry ether; CH₃CH₂CH₂CH₃ + 2NaCl
Butane
Q4. What is the Reimer-Tiemann reaction?
Solution
Phenol + CHCl₃ + NaOH → o-Hydroxybenzaldehyde (salicylaldehyde)
Used to introduce −CHO group at ortho position of phenol.
Chapter 11: Alcohols, Phenols, Ethers
Q1. Arrange in increasing order of acidic strength: ethanol, phenol, water.
Solution
Increasing acidity: ethanol < water < phenol
Ethanol: pKa ~ 16
Water: pKa = 15.7
Phenol: pKa = 10
Phenol is most acidic due to resonance stabilization of phenoxide ion.
Q2. How is ethyl methyl ether prepared by Williamson synthesis?
Solution
CH₃ONa + CH₃CH₂Br → CH₃OCH₂CH₃ + NaBr
Sodium methoxide + Bromoethane → Ethyl methyl ether
Q3. What happens when phenol reacts with bromine water?
Solution
Phenol + 3Br₂(aq) → 2,4,6-Tribromophenol (white precipitate) + 3HBr
The −OH group activates the ring at ortho and para positions.
Q4. What is the Kolbe's reaction?
Solution
Phenol + NaOH → Sodium phenoxide
Sodium phenoxide + CO₂ (under pressure) → Sodium salicylate
Acidification → Salicylic acid
Used to prepare salicylic acid from phenol.
Chapter 12: Aldehydes, Ketones, Carboxylic Acids
Q1. How is acetaldehyde prepared from ethanol?
Solution
CH₃CH₂OH &xrightarrow[PCC]{}; CH₃CHO
Or: 2CH₃CH₂OH + O₂ &xrightarrow[Cu, 573K]{}; 2CH₃CHO + 2H₂O
Q2. Distinguish between aldehyde and ketone using Tollen's test.
Solution
Aldehyde + 2[Ag(NH₃)₂]+ + 3OHⁿ → RCOOⁿ + 2Ag ↓ (silver mirror) + 4NH₃ + 2H₂O
Ketones do NOT give this test.
Q3. Why are carboxylic acids more acidic than phenols?
Solution
Carboxylate ion (RCOOⁿ) is stabilized by equivalent resonance between two oxygen atoms, making it much more stable than phenoxide ion. Hence, carboxylic acids are stronger acids (pKa ~ 5) compared to phenols (pKa ~ 10).
Q4. What is the Cannizzaro reaction?
Solution
Aldehydes without α-hydrogen undergo self-oxidation and reduction in conc. NaOH:
2HCHO + NaOH → CH₃OH + HCOONa
(One molecule is reduced to alcohol, other is oxidized to carboxylate)
Q5. How is benzaldehyde prepared from toluene?
Solution
C₆H₃CH₃ &xrightarrow[CrO₂Cl₂]{}; C₆H₃CHO
(Etard reaction: Toluene + chromyl chloride gives benzaldehyde)
Chapter 13: Amines
Q1. Arrange in decreasing order of basicity: CH₃NH₂, (CH₃)₂NH, (CH₃)₃N, C₆H₃NH₂
Solution
(CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > C₆H₃NH₂
2° amine is most basic (more +I effect than 1°, less steric hindrance than 3°).
Aniline is least basic due to resonance delocalization of lone pair.
Q2. What is the Hinsberg test? How does it distinguish between 1°, 2°, and 3° amines?
Solution
1° amine + C₆H₃SO₂Cl → N-alkylbenzenesulphonamide (soluble in NaOH)
2° amine + C₆H₃SO₂Cl → N,N-dialkylbenzenesulphonamide (insoluble in NaOH)
3° amine + C₆H₃SO₂Cl → No reaction
Q3. Write the reaction for the diazotisation of aniline.
Solution
C₆H₃NH₂ + NaNO₂ + 2HCl &xrightarrow[0-5°C]{}; C₆H₃N₂ⁿClⁿ + NaCl + 2H₂O
Benzene diazonium chloride
Q4. What is the Gabriel phthalimide synthesis?
Solution
Phthalimide + KOH → Potassium phthalimide
Potassium phthalimide + R-X → N-alkylphthalimide
Hydrolysis → Primary amine (RNH₂)
Used to prepare pure 1° amines only.
Chapter 14: Biomolecules
Q1. What is the difference between glucose and fructose?
Solution
Glucose: Aldohexose (contains −CHO group)
Fructose: Ketohexose (contains >C=O group)
Both have same molecular formula C₆H₁₂O₆ but different structures.
Q2. Name the four types of bonds that hold the secondary structure of proteins.
Solution
1. Hydrogen bonds (primary stabilizing force)
2. Van der Waals forces
3. Electrostatic interactions
4. Disulphide bridges (covalent bonds between cysteine residues)
Q3. What is the difference between DNA and RNA?
Solution
DNA: Deoxyribose sugar, double helix, bases A,T,G,C
RNA: Ribose sugar, single strand, bases A,U,G,C
DNA stores genetic information; RNA is involved in protein synthesis.
Q4. What are enzymes? Give their characteristics.
Solution
Enzymes are biological catalysts (mostly proteins) that speed up biochemical reactions.
Characteristics:
• Highly specific (lock and key model)
• Work at mild conditions (body temp, neutral pH)
• Extremely efficient
• Affected by temperature, pH, inhibitors
Chapter 15: Polymers
Q1. Distinguish between addition and condensation polymerization.
Solution
Addition: Monomers add without loss of small molecules (e.g., polythene from ethene)
Condensation: Monomers combine with loss of small molecules like H₂O (e.g., Nylon-6,6 from hexamethylenediamine + adipic acid)
Q2. What is vulcanization of rubber? Why is it done?
Solution
Vulcanization is the process of heating natural rubber with sulfur to form cross-links between polymer chains.
Purpose:
• Increases tensile strength
• Improves elasticity
• Makes it resistant to abrasion
• Prevents it from becoming sticky at high temperatures
Q3. Write the name and structure of the polymer formed from: (i) Caprolactam (ii) Phenol + Formaldehyde
Solution
(i) Caprolactam → Nylon-6 (addition of ring)
(ii) Phenol + Formaldehyde → Bakelite (cross-linked thermosetting polymer)
Q4. What is biodegradable polymer? Give an example.
Solution
A polymer that can be broken down by natural processes (bacteria, enzymes) into harmless products.
Example: PHBV (Poly-β-hydroxybutyrate-co-β-hydroxyvalerate) - used in medical applications.
Chapter 16: Chemistry in Everyday Life
Q1. Classify drugs based on chemical structure.
Solution
1. Sulphonamides (e.g., Sulphadiazine)
2. Quinolones (e.g., Ciprofloxacin)
3. Heterocyclic (e.g., Ofloxacin)
4. Alkaloids (e.g., Morphine)
5. Antibiotics (e.g., Penicillin)
Q2. What is the difference between an analgesic and an antibiotic?
Solution
Analgesic: Pain-relieving drug (e.g., Aspirin, Paracetamol)
Antibiotic: Drug that kills or inhibits bacterial growth (e.g., Penicillin, Ampicillin)
Q3. What are the two types of detergents? Give an example of each.
Solution
Anionic: Sodium lauryl sulphate (CH₃(CH₂)₁₀OSO₃ⁿNa₁+)
Cationic: Cetyltrimethylammonium bromide
Non-ionic: Polyethylene glycol stearate
Q4. What is the difference between tranquilizers and analgesics?
Solution
Tranquilizers: Used for mental diseases like anxiety, tension, depression (e.g., Equanil, Valium)
Analgesics: Used to relieve pain without causing loss of consciousness (e.g., Aspirin, Ibuprofen)