NCERT Solutions Class 12 Chemistry - All 16 Chapters

All 16 Chapters

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Solutions

Colligative properties, Raoult's law

› 2

Electrochemistry

Nernst equation, Conductance, Batteries

› 3

Chemical Kinetics

Rate law, Arrhenius equation, Half-life

› 4

Chemical Equilibrium

Le Chatelier's principle, Kp, Kc

› 5

Surface Chemistry

Adsorption, Catalysis, Colloids

› 6

General Principles of Metallurgy

Extraction, Refining, Thermodynamic principles

› 7

p-Block Elements

Group 15, 16, 17, 18

› 8

d and f Block Elements

Transition metals, Lanthanoids, Actinoids

› 9

Coordination Compounds

Werner's theory, Crystal field theory, Isomerism

› 10

Haloalkanes and Haloarenes

Reactions, SN1, SN2 mechanisms

› 11

Alcohols, Phenols, Ethers

Preparation, Properties, Reactions

› 12

Aldehydes, Ketones, Carboxylic Acids

Nucleophilic addition, Oxidation, Reduction

› 13

Amines

Classification, Basicity, Diazotisation

› 14

Biomolecules

Carbohydrates, Proteins, Nucleic acids

› 15

Polymers

Addition, Condensation polymers, Rubber

› 16

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: P_solution = P°_solvent × x_solvent
• ΔT_b = K_b · m, ΔT_f = K_f · 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 K_b of water. (K_b = 0.512 K kg/mol)

Solution

ΔT_b = 100.18 − 100 = 0.18°C

K_b = ΔT_b/m = 0.18/0.1 = 1.8 K kg/mol

Wait, given K_b = 0.512, so: ΔT_b = K_b × 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

ΔT_f = K_f × 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

ΔT_b = 0.18°C
m = ΔT_b/K_b = 0.18/0.52 = 0.346

ΔT_f = K_f × m = 1.86 × 0.346
= 0.644°C
Freezing point = 0 − 0.644 = −0.644°C

Q6. 1.0 mol of a non-volatile solute is dissolved in 4.5 moles of water. The vapour pressure of water is 25.0 mm Hg. Find the vapour pressure of the solution.

Solution

P_solution = P° × x_solvent
x_water = 4.5/(4.5 + 1.0) = 4.5/5.5 = 9/11

P_solution = 25.0 × 9/11 = 20.45 mm Hg

Q7. A 0.15 m aqueous solution of NaCl freezes at −0.558°C. What is the van't Hoff factor? (K_f = 1.86 K kg/mol)

Solution

ΔT_f = i × K_f × m
0.558 = i × 1.86 × 0.15
0.558 = i × 0.279
i = 0.558/0.279
i = 2.0 (NaCl gives 2 ions: Na¹ⁿ and Cl¹ⁿ)

Q8. Calculate the molarity of a solution containing 2 g of NaOH dissolved in 500 mL of solution. (Na = 23, O = 16, H = 1)

Solution

Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol
Moles = 2/40 = 0.05 mol
Volume = 500 mL = 0.5 L

Molarity = 0.05/0.5 = 0.1 M

Q9. An aqueous solution of urea boils at 100.12°C. Calculate the freezing point of the solution. (K_b = 0.52, K_f = 1.86)

Solution

ΔT_b = 100.12 − 100 = 0.12°C
m = ΔT_b/K_b = 0.12/0.52 = 0.231 m

ΔT_f = K_f × m = 1.86 × 0.231 = 0.429°C
Freezing point = 0 − 0.429 = −0.429°C

Q10. What is the osmotic pressure of a 5% (w/v) glucose solution at 27°C? (Molar mass of glucose = 180 g/mol)

Solution

5% (w/v) means 5 g glucose in 100 mL solution
Molarity = (5/180) / (100/1000) = 0.0278/0.1 = 0.278 M

Π = CRT = 0.278 × 0.0821 × 300
= 6.85 atm

Q11. Two solutions A (5% urea, mol. mass = 60) and B (5% glucose, mol. mass = 180) are isotonic. Justify this statement.

Solution

For solution A: Molarity = (5/60) / (100/1000) = 0.833 M
For solution B: Molarity = (5/180) / (100/1000) = 0.278 M

Since Π = CRT and both have different molarities, they are NOT isotonic.
Correction: If equal molarities are given, then they would be isotonic. Here, A has 3x the molarity of B, so A has 3x the osmotic pressure.

Q12. Calculate the boiling point of a 0.5 m aqueous solution of KCl. (K_b for water = 0.52 K kg/mol, KCl is fully dissociated)

Solution

KCl → K¹ⁿ + Cl¹ⁿ ∴ i = 2

ΔT_b = i × K_b × m
= 2 × 0.52 × 0.5
= 0.52°C

Boiling point = 100 + 0.52 = 100.52°C

Q13. What happens to the boiling point and osmotic pressure when a solution is diluted?

Solution

When a solution is diluted:
• Concentration of solute particles decreases
• ΔT_b decreases → Boiling point decreases (approaches pure solvent BP)
• Π = CRT, so Π decreases → Osmotic pressure decreases
• Similarly, ΔT_f decreases → Freezing point increases (approaches 0°C)

Q14. A 1 molal aqueous solution of Na₂SO₄ freezes at −2.79°C. Calculate the van't Hoff factor and % dissociation. (K_f = 1.86 K kg/mol)

Solution

ΔT_f = 2.79°C

i = ΔT_f/(K_f × m) = 2.79/(1.86 × 1) = 1.5

For Na₂SO₄ → 2Na¹ⁿ + SO₄₂ⁿ, maximum i = 3
i = 1 + α(n − 1)
1.5 = 1 + α(3 − 1)
0.5 = 2α
α = 0.25 or 25% dissociation

Q15. Why does an aqueous solution of acetic acid have higher boiling point than pure water?

Solution

When acetic acid dissolves in water, it dissociates slightly producing CH₃COOⁿ and H₁ⁿ ions. These solute particles lower the vapour pressure of the solution. According to Raoult's law, a lower vapour pressure means a higher boiling point. Additionally, hydrogen bonding between acetic acid and water molecules further raises the boiling point through colligative effects.

Chapter 2: Electrochemistry

Key Formulas:
• Nernst: E = E° − (RT/nF) ln Q
• E°_cell = E°_cathode − E°_anode
• ΔG° = −nFE°_cell
• K = e^nFE°/RT
• Λ_m = 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

E_cell = 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

Λ_m = 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 = 10^15.57 = 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

Q6. Calculate the cell emf and ΔG° for: Zn | Zn²ⁿ(0.01M) || H₁ⁿ(1M) | H₂(g, 1 atm) | Pt. Given E°(Zn²ⁿ/Zn) = −0.76V.

Solution

E°_cell = E°_cathode − E°_anode = 0.00 − (−0.76) = 0.76V

E_cell = E° − (0.0591/2) log([Zn²ⁿ]/[H₁ⁿ]²)
= 0.76 − (0.02955) log(0.01/1)
= 0.76 − (0.02955)(-2)
= 0.76 + 0.0591
= 0.819V

ΔG° = −nFE° = −(2)(96500)(0.76) = −146.68 kJ/mol

Q7. The specific conductivity of 0.01 M KCl solution is 0.0014 S/cm. Calculate the molar conductivity.

Solution

Λ_m = κ/c
c = 0.01 M = 0.01 mol/L = 0.01/1000 mol/cm³ = 10⁻⁵ mol/cm³

Λ_m = 0.0014/10⁻⁵
= 140 S cm²/mol

Q8. What is the standard emf of a galvanic cell made of a Cd electrode in 1 M CdCl₂ and a hydrogen electrode in 1 M HCl at 298K? (E°(Cd²ⁿ/Cd) = −0.40V)

Solution

Anode: Cd → Cd²ⁿ + 2eⁿ
Cathode: 2H₁ⁿ + 2eⁿ → H₂

E°_cell = E°_cathode − E°_anode
= 0.00 − (−0.40)
= +0.40V

Q9. How many faradays of electricity are required to reduce 1 mole of MnO₄ⁿ to Mn²ⁿ in acidic medium?

Solution

MnO₄ⁿ + 8H₁ⁿ + 5eⁿ → Mn²ⁿ + 4H₂O

Mn goes from +7 to +2, change in oxidation state = 5

Required electricity = 5 F (5 moles of electrons per mole of MnO₄ⁿ)

Q10. Explain why the conductivity of an electrolyte solution decreases with dilution.

Solution

Conductivity (κ) depends on:
• Number of ions per unit volume
• Charge on ions
• Mobility of ions

On dilution:
• Number of ions per unit volume decreases
• Although degree of dissociation increases, the volume effect dominates
• Therefore, κ decreases with dilution

Note: Molar conductivity (Λ_m) increases with dilution.

Q11. A Daniell cell is denoted as Zn | Zn²ⁿ(1M) || Cu²ⁿ(1M) | Cu. What is the maximum work obtainable from this cell? (E°_cell = 1.10V)

Solution

Maximum work = W_max = ΔG° = −nFE°_cell
= −(2)(96500)(1.10)
= −212300 J/mol
= −212.3 kJ/mol

The magnitude 212.3 kJ/mol is the maximum electrical work.

Q12. The resistance of a conductivity cell is 100 Ω when filled with 0.02 M KCl solution. The resistance of the same cell with 0.05 M KCl is 40 Ω. If the conductivity of 0.02 M KCl is 0.0025 S/cm, find the conductivity of 0.05 M KCl.

Solution

Cell constant = κ × R = 0.0025 × 100 = 0.25 cm⁻¹

For 0.05 M KCl:
κ = cell constant/R = 0.25/40
= 0.00625 S/cm

Q13. What type of battery is a lead-acid battery? Write the overall cell reaction during discharge.

Solution

Lead-acid battery is a secondary (rechargeable) cell.

Overall reaction during discharge:
Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(l)

Anode: Pb + SO₄₂ⁿ → PbSO₄ + 2eⁿ
Cathode: PbO₂ + 4H₁ⁿ + SO₄₂ⁿ + 2eⁿ → PbSO₄ + 2H₂O

Q14. Calculate the e.m.f. of the cell in which the following reaction takes place: Ni(s) + 2Ag₁ⁿ(0.002 M) → Ni²ⁿ(0.160 M) + 2Ag(s). (E°_cell = 1.05V)

Solution

E_cell = E° − (0.0591/2) log([Ni²ⁿ]/[Ag₁ⁿ]²)
= 1.05 − (0.02955) log(0.160/(0.002)²)
= 1.05 − (0.02955) log(0.160/0.000004)
= 1.05 − (0.02955) log(40000)
= 1.05 − (0.02955)(4.602)
= 1.05 − 0.136
= 0.914V

Q15. Why does corrosion of iron occur faster in salty water than in pure water?

Solution

Salty water is a better conductor of electricity than pure water due to dissolved ions. During corrosion, electrochemical cells form on the iron surface. The increased conductivity of salty water facilitates faster electron transfer between anodic and cathodic regions, accelerating the rusting process. The ions (Na₁ⁿ, Clⁿ) serve as electrolytes that enhance the galvanic corrosion mechanism.

Chapter 3: Chemical Kinetics

Key Formulas:
• Rate = k[A]^m[B]ⁿ
• First order: k = (2.303/t) log(a/(a−x))
• t_1/2 = 0.693/k (first order)
• Arrhenius: k = Ae^−E_a/RT
• ln(k₂/k₁) = (E_a/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 E_a.

Solution

ln(k₂/k₁) = (E_a/R)(1/T₁ − 1/T₂)
ln(4.5/1.5) = (E_a/8.314)(1/300 − 1/310)
ln(3) = (E_a/8.314)(10/(300 × 310))
1.0986 = (E_a/8.314)(1.075 × 10⁻⁴)
E_a = (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 t_1/2.

Solution

k = (2.303/t) log(100/70)
= (2.303/40) log(1.4286)
= (2.303/40)(0.1549)
= 0.00888 min⁻¹

t_1/2 = 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.

Q6. For a first order reaction, k = 1.54 × 10⁻⁶ s⁻¹ at 300K and 4.6 × 10⁻³ s⁻¹ at 350K. Calculate the activation energy.

Solution

ln(k₂/k₁) = (E_a/R)(1/T₁ − 1/T₂)

ln(4.6 × 10⁻³/1.54 × 10⁻⁶) = (E_a/8.314)(1/300 − 1/350)
ln(2987) = (E_a/8.314)(50/(300 × 350))
8.002 = (E_a/8.314)(4.762 × 10⁻⁴)
E_a = (8.002 × 8.314)/(4.762 × 10⁻⁴)
= 139.6 kJ/mol

Q7. The half-life of a first order reaction is 69.3 s. Calculate the rate constant and the time required for 80% completion.

Solution

k = 0.693/t_1/2 = 0.693/69.3 = 0.01 s⁻¹

For 80% completion, 20% remains: a/(a−x) = 100/20 = 5
t = (2.303/k) log(5)
= (2.303/0.01) × 0.6990
= 160.9 s

Q8. For a second order reaction, the half-life is 0.5 s when the initial concentration is 0.1 M. What is the rate constant?

Solution

For second order reaction:
t_1/2 = 1/(k × a)

0.5 = 1/(k × 0.1)
k = 1/(0.5 × 0.1)
k = 1/0.05
k = 20 M⁻¹s⁻¹

Q9. What is the effect of a catalyst on the rate of reaction, activation energy, and equilibrium constant?

Solution

• Catalyst increases rate by providing an alternate pathway with lower E_a
• E_a is reduced (both forward and reverse)
• Equilibrium constant K is unchanged (catalyst helps reach equilibrium faster, does not shift it)
• Both forward and reverse rates increase equally
• For first order: k = Ae^−E_a/RT, lower E_a means larger k

Q10. The rate of a reaction is found to double when the concentration of the reactant is increased four times. What is the order of the reaction?

Solution

Rate₂/Rate₁ = (4[A])ⁿ/[A]ⁿ
2 = 4ⁿ
n = log 2/log 4 = log 2/(2 log 2)
n = 0.5 (half order)

Q11. The decomposition of N₂O₅ is a first order reaction. The half-life is 1200 s. What fraction of N₂O₅ will decompose after 2400 s?

Solution

2400 s = 2 half-lives

After 1 half-life: remaining = 1/2
After 2 half-lives: remaining = 1/2 × 1/2 = 1/4

Fraction decomposed = 1 − 1/4 = 3/4 or 75%

Q12. Derive the integrated rate equation for a zero order reaction.

Solution

For zero order: Rate = k[A]⁰ = k

−d[A]/dt = k
−d[A] = k · dt

Integrating from [A]₀ to [A] at time t = 0 to t:
−([A] − [A]₀) = kt

[A] = [A]₀ − kt

At t_1/2: [A] = [A]₀/2
t_1/2 = [A]₀/(2k)

Q13. For a reaction A → Products, rate becomes 4 times when temperature increases from 300K to 320K. Calculate E_a.

Solution

k₂/k₁ = 4
ln(k₂/k₁) = (E_a/R)(1/T₁ − 1/T₂)
ln(4) = (E_a/8.314)(1/300 − 1/320)
1.386 = (E_a/8.314)(20/(300 × 320))
1.386 = (E_a/8.314)(2.083 × 10⁻⁴)
E_a = (1.386 × 8.314)/(2.083 × 10⁻⁴)
= 55.4 kJ/mol

Q14. Distinguish between zero order and first order reactions based on their half-life.

Solution

Zero order:
t_1/2 = [A]₀/(2k)
Half-life is directly proportional to initial concentration

First order:
t_1/2 = 0.693/k
Half-life is independent of initial concentration

Q15. What is the difference between an elementary reaction and a complex reaction?

Solution

Elementary reaction: Occurs in a single step, rate law can be written directly from stoichiometry (e.g., A + B → Products, Rate = k[A][B])

Complex reaction: Occurs in multiple steps, rate law is determined experimentally and may differ from stoichiometry (e.g., H₂ + Br₂ → 2HBr proceeds via multiple steps)

Chapter 4: Chemical Equilibrium

Q1. For the reaction: N₂(g) + 3H₂(g) ⇔ 2NH₃(g), K_c = 0.5 at 400K. Find K_p.

Solution

K_p = K_c(RT)^Δn
Δn = 2 − (1 + 3) = −2
K_p = 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 K_c.

Solution

K_c = [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, K_c = 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

K_c = x/(0.4 − x) = 4
x = 4(0.4 − x) = 1.6 − 4x
5x = 1.6
x = 0.32 M

[B] = 0.32 M

Q5. At 25°C, K_c = 100 for the reaction: PCl₅(g) ⇔ PCl₃(g) + Cl₂(g). If 1 mol PCl₅ is placed in a 10 L flask, calculate the equilibrium concentrations.

Solution

Initial [PCl₅] = 1/10 = 0.1 M

PCl₅ ⇔ PCl₃ + Cl₂
0.1 − x x x

K_c = x²/(0.1 − x) = 100
x² = 100(0.1 − x) = 10 − 100x
x² + 100x − 10 = 0
x = (−100 + √(10000 + 40))/2 = (−100 + 101.98)/2
x = 0.099 M

[PCl₅] = 0.1 − 0.099 = 0.001 M
[PCl₃] = [Cl₂] = 0.099 M

Q6. The pH of a 0.1 M solution of a weak acid is 3. Calculate its K_a value.

Solution

pH = 3 ∴ [H₁ⁿ] = 10⁻³ = 0.001 M

HA ⇔ H₁ⁿ + Aⁿ
0.1 − x x x
x = 0.001 M

K_a = x²/(0.1 − x) = (0.001)²/(0.1 − 0.001)
= 10⁻⁶/0.099
= 1.01 × 10⁻⁵

Q7. Calculate the pH of a buffer solution containing 0.4 M CH₃COOH and 0.4 M CH₃COONa. (K_a = 1.8 × 10⁻⁵)

Solution

Henderson-Hasselbalch equation:
pH = pK_a + log([salt]/[acid])

pK_a = −log(1.8 × 10⁻⁵) = 4.74

pH = 4.74 + log(0.4/0.4)
= 4.74 + log(1)
= 4.74 + 0
= 4.74

Q8. What is the solubility product of AgCl if its solubility is 1.35 × 10⁻⁵ mol/L?

Solution

AgCl(s) ⇔ Ag₁ⁿ(aq) + Clⁿ(aq)

[Ag₁ⁿ] = [Clⁿ] = s = 1.35 × 10⁻⁵ M

K_sp = [Ag₁ⁿ][Clⁿ] = s²
= (1.35 × 10⁻⁵)²
= 1.82 × 10⁻¹⁰

Q9. For the equilibrium N₂O₄(g) ⇔ 2NO₂(g), K_p = 0.113 atm at 298K. Calculate K_c.

Solution

K_p = K_c(RT)^Δn
Δn = 2 − 1 = 1

K_c = K_p/(RT) = 0.113/(0.0821 × 298)
= 0.113/24.47
= 4.62 × 10⁻³

Q10. Explain why the pH of pure water is 7 at 25°C but not at other temperatures.

Solution

H₂O ⇔ H₁ⁿ + OHⁿ K_w = [H₁ⁿ][OHⁿ]

At 25°C: K_w = 10⁻¹⁴
[H₁ⁿ] = [OHⁿ] = 10⁻⁷, so pH = 7

At higher T: K_w increases (endothermic dissociation)
e.g., at 60°C, K_w ≈ 10⁻¹³
[H₁ⁿ] = 10^−6.5, pH = 6.5 (still neutral)

pH = 7 only at 25°C; neutrality means [H₁ⁿ] = [OHⁿ], not pH = 7.

Q11. The K_sp of CaF₂ is 3.4 × 10⁻¹¹. Calculate the solubility of CaF₂ in pure water.

Solution

CaF₂(s) ⇔ Ca²ⁿ(aq) + 2Fⁿ(aq)
s 2s

K_sp = [Ca²ⁿ][Fⁿ]² = s(2s)² = 4s³
4s³ = 3.4 × 10⁻¹¹
s³ = 8.5 × 10⁻¹²
s = 2.04 × 10⁻⁴ mol/L

Q12. What is the ionic product of water at 37°C? Is water more or less dissociated at this temperature compared to 25°C?

Solution

At 37°C, K_w = 2.4 × 10⁻¹⁴

Since dissociation of water is endothermic, higher T → larger K_w

Water is more dissociated at 37°C than at 25°C
(K_w = 2.4 × 10⁻¹⁴ > 1.0 × 10⁻¹⁴)

Q13. The K_a of acetic acid is 1.8 × 10⁻⁵. Calculate the percentage dissociation of 0.1 M acetic acid.

Solution

K_a = cα²/(1 − α) ≈ cα² (for small α)

α = √(K_a/c) = √(1.8 × 10⁻⁵/0.1)
= √(1.8 × 10⁻⁴)
= 0.0134

% dissociation = 1.34%

Q14. Explain Le Chatelier's principle applied to the reaction: H₂(g) + I₂(g) ⇔ 2HI(g) + Heat. What happens when (a) temperature is raised, (b) pressure is increased, (c) a catalyst is added?

Solution

(a) Temperature raised → equilibrium shifts left (reverse is endothermic)
(b) Pressure increased → no change (Δn = 0, same moles on both sides)
(c) Catalyst added → no change in equilibrium (both forward and reverse rates increase equally)

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

Q6. What is the Freundlich adsorption isotherm? State its mathematical expression.

Solution

At constant temperature, the extent of adsorption (x/m) is related to pressure:

x/m = k · P^1/n (where 0 < 1/n < 1)

Taking log: log(x/m) = log k + (1/n) log P

At low pressure: x/m ∝ P (linear)
At high pressure: x/m becomes independent of P (saturated)

Q7. What is Langmuir adsorption isotherm? How does it differ from Freundlich isotherm?

Solution

Langmuir isotherm:
θ = KP/(1 + KP)
where θ = fraction of surface covered

Assumes: monolayer adsorption, homogeneous surface, no interaction between adsorbed molecules

Difference:
• Langmuir is valid for monolayer; Freundlich is empirical
• Langmuir gives saturation at high P; Freundlich does not account for saturation
• Langmuir applies to chemisorption; Freundlich applies to physisorption

Q8. What is the Tyndall effect? Why is it observed in colloidal solutions?

Solution

Tyndall effect: Scattering of light by colloidal particles making the beam path visible.

Causes:
• Colloidal particles (1-1000 nm) are comparable to wavelength of visible light
• Particles scatter light in all directions

Conditions:
• Refractive index of dispersed phase ≠ dispersion medium
• Size of particles ≈ wavelength of light

Q9. What is dialysis? Explain its principle and applications.

Solution

Dialysis: Process of separating crystalloids from colloids using a semipermeable membrane.

Principle: Crystalloids (ions, small molecules) pass through the membrane, but colloidal particles cannot.

Applications:
• Purification of colloidal solutions
• Artificial kidney (hemodialysis) removes urea from blood
• Separation of proteins from salts

Q10. Classify the following as lyophilic or lyophobic sols: starch sol, gold sol, protein sol, Fe(OH)₃ sol.

Solution

Lyophilic (solvent-loving):
• Starch sol — reversible, stable
• Protein sol — reversible, stable

Lyophobic (solvent-hating):
• Gold sol — irreversible, unstable, needs stabilizer
• Fe(OH)₃ sol — irreversible, unstable, needs stabilizer

Q11. Explain the mechanism of heterogeneous catalysis using the adsorption theory.

Solution

Steps:
1. Diffusion: Reactant molecules diffuse to catalyst surface
2. Adsorption: Reactants adsorb on catalyst surface (weakens bonds)
3. Reaction: 4. Desorption: Products desorb from surface
5. Diffusion: Products diffuse away from surface

Example: Hydrogenation of ethene on Ni surface

Q12. What is electrophoresis? What does it demonstrate about colloidal particles?

Solution
Electrophoresis: Movement of colloidal particles towards oppositely charged electrode under influence of electric field.

Demonstrates:
• Colloidal particles carry electric charge
• Charge can be positive or negative
• Magnitude of charge determines mobility

Example: Arsenious sulphide sol is negatively charged (moves to anode)

Q13. Why is FeCl₃ used to stop bleeding from a cut?

Solution
Blood is a negatively charged colloidal sol (protein colloid).

FeCl₃ is an electrolyte that provides Fe³ⁿ cations.

The positive Fe³ⁿ ions neutralize the negative charge on blood colloids, causing coagulation (clotting), which stops bleeding.

Q14. What are colloids? How are they classified based on the physical state of dispersed phase and dispersion medium?

Solution
Colloid: Heterogeneous mixture where particle size is 1-1000 nm.

Examples based on states:
• Solid in liquid: Sol (starch sol)
• Liquid in solid: Gel (cheese)
• Gas in liquid: Foam (froth)
• Liquid in liquid: Emulsion (milk)
• Solid in gas: Aerosol (smoke)
• Liquid in gas: Aerosol (fog)

Q15. Explain the role of gold number in protecting colloids.

Solution
Gold number: Milligrams of protective colloid required to prevent coagulation of 10 mL of gold sol by 1 mL of 10% NaCl solution.

• Lower gold number = better protecting power
• Gelatin (gold number = 0.005-0.01) is better than starch (gold number = 15-25)
• Protective colloids adsorb on lyophobic sol and stabilize it

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

Q5. Why is the extraction of copper from its sulphide ore done by self-reduction rather than carbon reduction?

Solution
In copper extraction:
2Cu₂S + 3O₂ → 2Cu₂O + 2SO₂ (roasting)
2Cu₂O + Cu₂S → 6Cu + SO₂ (self-reduction)

Carbon cannot reduce Cu₂O effectively because copper has lower affinity for oxygen compared to carbon. The self-reduction process uses the sulphide ore itself as the reducing agent, which is thermodynamically more favorable.

Q6. Describe the refining of nickel by Mond's process.

Solution
Mond's process (Vapor phase refining):

Step 1: Ni(s) + 4CO(g) → Ni(CO)₄(g) (330-350K)
Step 2: Ni(CO)₄(g) → Ni(s) + 4CO(g) (450-470K)

CO selectively reacts with impure nickel to form volatile nickel carbonyl. Decomposition of the carbonyl yields pure nickel.

Q7. What is zone refining? Name a metal purified by this method.

Solution
Zone refining: Based on the principle that impurities are more soluble in the molten state than in the solid state.

A molten zone is moved along the impure metal rod. Impurities dissolve in the molten zone and travel with it to one end, leaving pure metal behind.

Used to purify: Silicon, Germanium, Gallium (semiconductors)

Q8. What is the role of flux in metallurgical processes? Differentiate between acidic and basic flux.

Solution
Flux: Substance that combines with impurities (gangue) to form fusible slag.

Acidic flux: Reacts with basic impurities
Example: SiO₂ (silica) removes basic gangue like FeO
FeO + SiO₂ → FeSiO₃ (slag)

Basic flux: Reacts with acidic impurities
Example: CaO (lime) removes acidic gangue like SiO₂
SiO₂ + CaO → CaSiO₃ (slag)

Q9. What is the thermodynamic principle behind the extraction of metals? Why is ΔG° important?

Solution
For spontaneous extraction: ΔG° must be negative.

More negative ΔG° of metal oxide formation = more stable oxide = harder to reduce.

A metal with lower (more negative) ΔG° can reduce the oxide of a metal with higher ΔG°.

ΔG° = ΔH° − TΔS°
Since ΔS° is usually negative (gas consumed), ΔG° becomes less negative at higher T. The metal whose oxide line crosses above another can be reduced by that metal.

Q10. Describe the electrolytic refining of impure copper.

Solution
Anode: Impure copper (dissolves)
Cu → Cu²ⁿ + 2eⁿ

Cathode: Pure copper (deposits)
Cu²ⁿ + 2eⁿ → Cu

Electrolyte: Acidified CuSO₄ solution

Impurities:
• More reactive (Fe, Zn) dissolve but don't deposit
• Less reactive (Ag, Au) settle as anode mud

Anode mud contains precious metals recovered separately.

Q11. What is the difference between a mineral and an ore?

Solution
Mineral: Naturally occurring substance from which metal may or may not be extracted profitably.
Example: Bauxite (Al₂O₃·2H₂O) is a mineral of aluminium.

Ore: A mineral from which metal can be extracted economically and conveniently.
Example: Bauxite is an ore of aluminium; clay is not.

All ores are minerals, but not all minerals are ores.

Q12. Why is cryolite added during the extraction of aluminium?

Solution
Cryolite (Na₃AlF₆) serves two purposes:

1. Lowers melting point: Pure Al₂O₃ melts at 2050°C. Adding cryolite reduces it to ~950°C, saving energy.

2. Increases conductivity: Molten Al₂O₃ is a poor conductor. Cryolite increases conductivity, allowing electrolysis to proceed efficiently.

Q13. What is the principle of bessemer process for the manufacture of steel?

Solution
Bessemer process: Air is blown through molten pig iron to oxidize impurities.

Impurities oxidized:
• C → CO₂ (removed as gas)
• Si → SiO₂ (removed as slag)
• Mn → MnO (removed as slag)

The heat generated keeps the metal molten. The process takes about 15-20 minutes and produces steel from pig iron.

Q14. Which metals are extracted by: (a) electrolysis (b) carbon reduction (c) self-reduction? Give one example each.

Solution
(a) Electrolysis: Reactive metals (Al from Al₂O₃, Na from NaCl, Mg from MgCl₂)

(b) Carbon reduction: Moderately reactive (Fe from Fe₂O₃, Zn from ZnO, Sn from SnO₂)

(c) Self-reduction: Least reactive (Cu from Cu₂S, Pb from PbS, Hg from HgS)

Q15. Explain why aluminium thermite reaction can weld railway tracks.

Solution
Fe₂O₃(s) + 2Al(s) → Al₂O₃(l) + 2Fe(l)    ΔH = −850 kJ/mol

• Highly exothermic reaction produces molten iron at ~2500°C
• Molten iron fills the gap between railway tracks
• The heat is sufficient to melt the ends of the tracks, creating a strong weld
• No external heat source needed — reaction is self-sustaining

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)

Q6. Why does phosphorus form P₄ but nitrogen does not form N₄?

Solution
Phosphorus forms P₄ because:
• P has vacant 3d orbitals for sp₃d hybridization
• P-P single bond is stronger than N≡N in terms of total bond energy in P₄

Nitrogen does not form N₄ because:
• N has no d-orbitals (2nd period element)
• Strong N≡N triple bond (946 kJ/mol) is very stable
• N-N single bond (163 kJ/mol) is weak due to lone pair repulsion

Q7. Complete the following reactions: (a) P₄ + NaOH + H₂O → (b) N₂ + Mg →

Solution
(a) P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₃
(Phosphine + Sodium phosphite)

(b) 3Mg + N₂ → Mg₃N₂
(Magnesium nitride)

Q8. Why does H₂O have a higher boiling point than H₂S, H₂Se, and H₂Te?

Solution
H₂O has the highest boiling point due to:
• Hydrogen bonding: Strong intermolecular H-bonds between O-H groups
• H₂S, H₂Se, H₂Te have weaker van der Waals forces

BP order: H₂O (100°C) >> H₂Te (-2°C) > H₂Se (-41°C) > H₂S (-60°C)

Q9. What is the structure of XeF₄ and XeF₆? Explain using VSEPR theory.

Solution
XeF₄:
Xe has 8 valence electrons + 4 from F = 12 electrons = 6 pairs
4 bonding + 2 lone pairs → Square planar geometry

XeF₆:
Xe has 8 valence electrons + 6 from F = 14 electrons = 7 pairs
6 bonding + 1 lone pair → Distorted octahedral (pentagonal bipyramidal with lone pair)

Q10. Why is HNO₃ a stronger acid than HNO₂?

Solution
In HNO₃, nitrogen is in +5 state (more oxidized):
• More O atoms pull electron density from O-H bond
• NO₃ⁿ ion is resonance-stabilized with 3 equivalent structures

In HNO₂, nitrogen is in +3 state:
• Fewer O atoms, less electron withdrawal
• NO₂ⁿ has only 2 resonance structures

More resonance structures → greater stabilization of conjugate base → stronger acid

Q11. How is ammonia manufactured? Write the conditions used.

Solution
Haber's process:
N₂(g) + 3H₂(g) ⇔ 2NH₃(g)    ΔH = −92.4 kJ/mol

Conditions:
• Temperature: 673-723K (compromise between rate and yield)
• Pressure: 150-200 atm (high pressure favors product)
• Catalyst: Finely divided iron with K₂O/Al₂O₃ promoter

Q12. Why does fluorine not exhibit positive oxidation states?

Solution
Fluorine does not show positive oxidation states because:
• It has the highest electronegativity (3.98) of all elements
• It is the most powerful oxidizing agent
• Its small atomic size and high nuclear charge make electron loss impossible
• Fⁿ ion is very stable due to high hydration enthalpy

Fluorine always exhibits −1 oxidation state in compounds.

Q13. What happens when (a) concentrated H₂SO₄ reacts with copper, (b) Cl₂ reacts with NH₃ in excess?

Solution
(a) Cu + 2H₂SO₄(conc.) → CuSO₄ + SO₂ + 2H₂O
(Concentrated H₂SO₄ acts as oxidizing agent)

(b) 3Cl₂ + 8NH₃(excess) → 6NH₄Cl + N₂
(Excess ammonia prevents formation of NCl₃)

Q14. Why are noble gases chemically inert? Name two compounds of xenon with their structures.

Solution
Noble gases are inert because:
• Completely filled ns²np&sup6; configuration
• Very high ionization energy
• Positive electron gain enthalpy

Xenon compounds:
• XeF₄ — square planar
• XeF₂ — linear
• XeOF₄ — square pyramidal
• XeO₃ — pyramidal

Q15. Explain why interhalogen compounds are more reactive than halogen molecules.

Solution
Interhalogen compounds (e.g., ICl, BrF₃) are more reactive because:
• The X-X' bond is polar (different electronegativities)
• The bond is weaker than the corresponding halogen-halogen bond
• Polarity makes the bond more susceptible to nucleophilic attack
• The less electronegative halogen is partially positive and more reactive

Example: ICl is more reactive than I₂ or Cl₂ in many reactions.

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.

Q5. Why are transition metals good catalysts?

Solution
Transition metals are good catalysts because:
• They have variable oxidation states — can provide multiple pathways
• They can form intermediate compounds with reactants
• They have partially filled d-orbitals that can accept/Donate electrons
• They have high surface area (finely divided metals)

Examples: Fe in Haber's process, Ni in hydrogenation, V₂O₅ in Contact process

Q6. Write the electronic configurations of: (a) Cr (Z=24) (b) Cu (Z=29)

Solution
(a) Cr: 1s² 2s² 2p&sup6; 3s² 3p&sup6; 3d₅ 4s¹
Expected: 3d₄ 4s² but actual is 3d₅ 4s¹ due to extra stability of half-filled d-orbitals

(b) Cu: 1s² 2s² 2p&sup6; 3s² 3p&sup6; 3d₄ 4s¹
Expected: 3d₃ 4s² but actual is 3d₄ 4s¹ due to extra stability of fully filled d-orbitals

Q7. Why are lanthanoids difficult to separate from each other?

Solution
Lanthanoids are difficult to separate because:
• They have very similar chemical properties
• Atomic/ionic radii decrease very gradually (lanthanoid contraction)
• Differences in properties are too small for easy separation

Separation methods used:
• Ion-exchange chromatography
• Fractional crystallization
• Solvent extraction

Q8. Explain why Cu₁ is diamagnetic while Cu²ⁿ is paramagnetic.

Solution
Cu₁: [Ar] 3d₄
All electrons are paired → diamagnetic

Cu²ⁿ: [Ar] 3d₃
One unpaired electron → paramagnetic

Magnetic moment = √(n(n+2)) = √(1×3) = 1.73 BM

Q9. What are interstitial compounds? Give their properties and examples.

Solution
Interstitial compounds: Compounds where small atoms (H, C, N, B) occupy interstitial voids in transition metal lattice.

Properties:
• Very hard (e.g., TiC, WC)
• High melting points
• Chemically inert
• Retain metallic conductivity

Examples: TiC, TiN, Fe₃C (cementite), VH₂

Q10. Why do transition metals form coloured compounds?

Solution
Transition metal ions form coloured compounds because:
• They have partially filled d-orbitals (d₁ to d₃)
• When ligands approach, d-orbitals split into two energy levels (t₄⁴ and e_g)
• Electrons absorb visible light and jump from lower to higher d-orbital (d-d transition)
• The transmitted light gives the complementary colour

Examples: Cu²ⁿ — blue, Fe₃ⁿ — yellow, Ni²ⁿ — green

Q11. Compare the chemistry of lanthanoids and actinoids.

Solution
Lanthanoids:
• +3 oxidation state is most common
• Less tendency to form complexes
• Less basic hydroxides
• 4f orbitals shielded by 5s, 5p

Actinoids:
• More variable oxidation states (+3 to +7)
• Greater tendency to form complexes
• More basic hydroxides
• 5f orbitals less shielded, participate in bonding

Q12. Why does Mn₂ⁿ show maximum oxidation state among first row transition metals?

Solution
Mn has electronic configuration [Ar] 3d₅ 4s²

All 7 electrons (5 from 3d + 2 from 4s) can be removed, giving Mn⁺⁷ in MnO₄ⁿ.

After Mn, the number of d-electrons increases, making it harder to lose all electrons due to increasing nuclear charge. Hence, Mn shows the highest oxidation state (+7) among first row transition metals.

Q13. What is the magnetic moment of Fe²ⁿ ion? (Fe: [Ar] 3d₄ 4s²)

Solution
Fe: [Ar] 3d₄ 4s²
Fe²ⁿ: [Ar] 3d₄

Number of unpaired electrons (n) = 4

Magnetic moment = √(n(n+2)) BM
= √(4 × 6) = √24
= 4.90 BM

Q14. Name two alloys of transition metals used in daily life and their composition.

Solution
Steel: Fe + C (0.1-1.5%) + Cr, Ni, Mn
• Used in construction, tools, utensils

Brass: Cu + Zn (30%)
• Used in decorative items, fittings

Bronze: Cu + Sn (12%)
• Used in coins, statues

German silver: Cu + Zn + Ni
• Used in cutlery, utensils

Q15. Why is the first ionization energy of Cr lower than that of Zn?

Solution
Cr: [Ar] 3d₅ 4s¹ — removal of 4s¹ electron is easier (half-filled stability doesn't help much for ionization energy)

Zn: [Ar] 3d₄ 4s² — removal of electron from fully filled 4s orbital requires more energy

Zn has higher effective nuclear charge and completely filled 3d and 4s orbitals, making the first ionization energy of Zn (906 kJ/mol) higher than that of Cr (653 kJ/mol).

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): d_xy, d_yz, d_xz
e_g (higher): d_z², d_x²−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)

Q5. Give the IUPAC name of: (a) K₃[Fe(CN)₆] (b) [Co(en)₂Cl₂]Cl

Solution
(a) K₃[Fe(CN)₆]:
Fe is in +3 oxidation state
Potassium hexacyanidoferrate(III)

(b) [Co(en)₂Cl₂]Cl:
Co is in +3 oxidation state (en = ethylenediamine, neutral)
Dichloridobis(ethylenediamine)cobalt(III) chloride

Q6. Explain Werner's theory of coordination compounds with examples.

Solution
Werner's Theory:
• Metal has two types of valencies: primary (ionizable) and secondary (non-ionizable)
• Primary valency → satisfied by negative ions (counter ions)
• Secondary valency → satisfied by ligands (coordination sphere)

Example: [Co(NH₃)₄Cl₂]Cl₃
Primary valency = 3 (3 Clⁿ outside)
Secondary valency = 6 (4NH₃ + 2Clⁿ inside)

Q7. What are geometrical isomers? Draw the isomers of [Pt(NH₃)₂Cl₂].

Solution
Geometrical isomers: Stereoisomers that differ in the spatial arrangement of ligands around the metal.

[Pt(NH₃)₂Cl₂] shows cis-trans isomerism:

Cis-isomer: Same ligands adjacent (90° apart)
Trans-isomer: Same ligands opposite (180° apart)

Cisplatin (cis-[Pt(NH₃)₂Cl₂]) is an anticancer drug; trans isomer is inactive.

Q8. What is crystal field splitting? Draw the splitting pattern in octahedral and tetrahedral fields.

Solution
Octahedral field:
d-orbitals split into: t₄⁴ (lower: d_xy, d_yz, d_xz) and e_g (higher: d_z², d_x²−y²)
Splitting energy = Δ₀

Tetrahedral field:
Splitting is inverted and smaller: e (lower) and t₂ (higher)
Δ_t = 4/9 Δ₀

Tetrahedral complexes are almost always high spin because Δ_t is small.

Q9. What are strong field and weak field ligands? Give examples.

Solution
Strong field ligands: Cause large splitting (Δ₀ > P), electrons pair up in lower orbitals (low spin)
Examples: CNⁿ, CO, NO₂ⁿ, en, NH₃

Weak field ligands: Cause small splitting (Δ₀ < P), electrons occupy higher orbitals singly first (high spin)
Examples: Fⁿ, Clⁿ, Brⁿ, Iⁿ, OHⁿ, H₂O

Spectrochemical order: Iⁿ < Brⁿ < Clⁿ < Fⁿ < OHⁿ < H₂O < NH₃ < en < NO₂ⁿ < CNⁿ < CO

Q10. Explain the bonding in metal carbonyls using the synergistic bonding concept.

Solution
Synergistic bonding in metal carbonyls:

σ-bonding: CO donates lone pair from carbon to empty d-orbital of metal (M ← C≡O)

π-back bonding: Filled d-orbital of metal donates electron density to empty π* antibonding orbital of CO (M → C≡O)

This creates a mutual strengthening of M-C bond and weakens C-O bond. Examples: Ni(CO)₄, Fe(CO)₅, Cr(CO)₆

Q11. What are the applications of coordination compounds in biology and medicine?

Solution
Biology:
• Hemoglobin — Fe²ⁿ coordinated with porphyrin ring (O₂ transport)
• Chlorophyll — Mg²ⁿ in porphyrin ring (photosynthesis)
• Vitamin B₁₂ — Co³ⁿ complex (enzyme cofactor)

Medicine:
• Cisplatin — [Pt(NH₃)₂Cl₂] (anticancer drug)
• EDTA — treatment for heavy metal poisoning
• CaNa₂EDTA — removes lead from blood

Q12. What is the difference between a coordination entity and a double salt? Give two examples of each.

Solution
Coordination entity: Retains identity in solution
Examples: [Co(NH₃)₆]Cl₃, K₄[Fe(CN)₆]

Double salt: Dissociates into simple ions in solution
Examples: KCl·MgCl₂·6H₂O (carnallite), K₂SO₄·Al₂(SO₄)₃·24H₂O (potash alum)

Key difference: Coordination sphere [ ] remains intact; double salts break into individual ions.

Q13. What are linkage isomers? Give an example.

Solution
Linkage isomers: Arise when an ambidentate ligand can coordinate through two different atoms.

Example: NO₂ⁿ can coordinate through N or O:

• [Co(NH₃)₅(NO₂)]Cl₂ — nitro (N-bonded, yellow)
• [Co(NH₃)₅(ONO)]Cl₂ — nitrito (O-bonded, red)

Other ambidentate ligands: SCNⁿ (S or N), CNⁿ (C or N)

Q14. Calculate the magnetic moment and draw the crystal field diagram for [Fe(CN)₆]^4ⁿ.

Solution
Fe²ⁿ: [Ar] 3d₄
CNⁿ is a strong field ligand → low spin complex

t₄⁴: ↑↓ ↑↓ ↑↓
e_g:         — —

All electrons paired → diamagnetic
Magnetic moment = 0 BM

Q15. Why do coordination compounds have colours while simple salts do not?

Solution
Coordination compounds have colours because:
• d-orbitals split in energy due to ligand field
• Electrons absorb visible light for d-d transitions
• Complementary colour is observed

Simple salts of main group metals have no d-d transitions (filled or empty d-orbitals), so they appear white or colourless.

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.

Q5. What is a Grignard reagent? How is it prepared and what are its uses?

Solution
Grignard reagent: Organomagnesium halide (RMgX)

Preparation: R-X + Mg &xrightarrow{dry ether}; RMgX
Example: CH₃Br + Mg → CH₃MgBr

Uses:
• Preparation of alcohols: RMgX + HCHO → RCH₂OH
• Preparation of hydrocarbons: RMgX + H₂O → RH
• Synthesis of carboxylic acids: RMgX + CO₂ → RCOOH

Q6. Arrange the following in decreasing reactivity towards SN1 reaction: (CH₃)₃CCl, CH₃CH₂Cl, C₆H₃Cl, (CH₃)₂CHCl

Solution
SN1 reactivity depends on carbocation stability:

(CH₃)₃CCl > (CH₃)₂CHCl > C₆H₃Cl > CH₃CH₂Cl

3° carbocation > 2° carbocation > Benzylic (resonance) > 1° carbocation

Order: (CH₃)₃CCl > (CH₃)₂CHCl > C₆H₃Cl > CH₃CH₂Cl

Q7. How can you convert ethanol to chloroethane? Give two methods.

Solution
Method 1: With Lucas reagent
CH₃CH₂OH + HCl &xrightarrow{ZnCl₂}; CH₃CH₂Cl + H₂O

Method 2: With SOCl₂ (thionyl chloride)
CH₃CH₂OH + SOCl₂ → CH₃CH₂Cl + SO₂ + HCl

(Preferred because by-products are gases)

Q8. Explain why aryl halides are less reactive towards nucleophilic substitution than alkyl halides.

Solution
Aryl halides are less reactive because:
• Resonance: C-Cl bond has partial double bond character
• Shorter C-Cl bond: sp₂ hybridized carbon has more s-character
• Unstable intermediate: Phenyl cation cannot form easily
• Steric hindrance: Benzene ring hinders backside attack

SN2 is impossible (no backside attack on ring).
SN1 is impossible (phenyl cation is very unstable).

Q9. What are polyhalogen compounds? Give examples and mention one environmental concern.

Solution
Polyhalogen compounds: Compounds containing more than one halogen atom.

Examples:
• CH₂Cl₂ (dichloromethane) — solvent
• CHCl₃ (chloroform) — anaesthetic
• CCl₄ (carbon tetrachloride) — fire extinguisher
• DDT (p,p'-dichlorodiphenyltrichloroethane) — insecticide

Environmental concern: CFCs deplete ozone layer; DDT is non-biodegradable and biomagnifies.

Q10. What happens when chlorobenzene is treated with NaOH at high temperature and pressure?

Solution
C₆H₃Cl + NaOH &xrightarrow{623K, 300 atm}; C₆H₃OH + NaCl

This is the Dow process for converting chlorobenzene to phenol.

Requires harsh conditions because the C-Cl bond in chlorobenzene has partial double bond character due to resonance.

Q11. Distinguish between SN1 and SN2 mechanisms of nucleophilic substitution.

Solution
SN1 (Unimolecular):
• Two-step mechanism (carbocation intermediate)
• Rate = k[substrate]
• Racemization (loss of stereochemistry)
• Favored by 3° > 2° substrates

SN2 (Bimolecular):
• One-step mechanism (transition state)
• Rate = k[substrate][nucleophile]
• Inversion of configuration (Walden inversion)
• Favored by 1° > 2° substrates

Q12. How is DDT synthesized? Why was it banned in many countries?

Solution
Synthesis of DDT:
Chloral (CCl₃CHO) + Chlorobenzene (2 moles) &xrightarrow{conc. H₂SO₄}; DDT

Why banned:
• Non-biodegradable (persists in environment)
• Biomagnifies through food chain
• Toxic to aquatic organisms
• Causes eggshell thinning in birds
• Harmful to human health (carcinogenic)

Q13. What is the Wurtz-Fittig reaction? How does it differ from Wurtz reaction?

Solution
Wurtz-Fittig reaction:
Aryl halide + Alkyl halide + 2Na → Alkylarene + 2NaCl
Example: C₆H₃Br + CH₃Br + 2Na → C₆H₃CH₃ + 2NaCl

Difference from Wurtz reaction:
• Wurtz: uses two alkyl halides → higher alkane
• Wurtz-Fittig: uses aryl + alkyl halide → alkylbenzene

Q14. What is the Kornblum reaction? Give an example.

Solution
Kornblum reaction: Conversion of alkyl halides to aldehydes using DMSO.

RCH₂X + DMSO → RCHO

Example: CH₃(CH₂)₃Br + DMSO → CH₃(CH₂)₃CHO

This is an oxidative nucleophilic substitution where DMSO acts as both solvent and oxidizing agent.

Q15. Why are vinyl halides and aryl halides unreactive towards nucleophilic substitution?

Solution
Vinyl halides (CH₂=CHCl):
• sp₂ hybridized carbon holds electrons tightly
• C-Cl bond is shorter and stronger
• Vinyl cation is very unstable

Aryl halides (C₆H₃Cl):
• Partial double bond character due to resonance
• Phenyl cation is highly unstable
• Backside attack impossible on ring

Both are resistant to SN1 and SN2 mechanisms under normal conditions.

Chapter 11: Alcohols, Phenols, Ethers

Q1. Arrange in increasing order of acidic strength: ethanol, phenol, water.

Solution
Increasing acidity: ethanol < water < phenol
Ethanol: pK_a ~ 16
Water: pK_a = 15.7
Phenol: pK_a = 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.

Q5. How can you distinguish between ethanol and phenol using a chemical test?

Solution
Neutral FeCl₃ test:
Phenol + FeCl₃ → Violet colour (complex formation)
Ethanol + FeCl₃ → No colour change

Bromine water test:
Phenol + Br₂(aq) → White precipitate (2,4,6-tribromophenol)
Ethanol + Br₂(aq) → No reaction

Q6. What is the Lucas test? How does it help in distinguishing between 1°, 2°, and 3° alcohols?

Solution
Lucas test: Uses Lucas reagent (conc. HCl + ZnCl₂)

3° alcohol → turbidity immediately
2° alcohol → turbidity in 5-10 minutes
1° alcohol → no turbidity at room temperature

Reason: Rate of SN1 reaction depends on carbocation stability: 3° > 2° > 1°

Q7. What is Victor Meyer's test? How does it distinguish between 1°, 2°, and 3° alcohols?

Solution
Victor Meyer's test:
1° alcohol → Red colour (with p-nitroso compound)
2° alcohol → Blue colour
3° alcohol → Colourless

Procedure: Alcohol → Alkyl iodide → Nitroalkane → Nitronic acid (with NaOH + HNO₂)
Only 1° and 2° nitroalkanes form nitronic acid; 3° does not.

Q8. Arrange the following in decreasing order of acidic strength: p-nitrophenol, phenol, p-methoxyphenol, p-cresol.

Solution
p-Nitrophenol > Phenol > p-Cresol > p-Methoxyphenol

Reason:
• −NO₂ is electron-withdrawing (stabilizes phenoxide) → increases acidity
• −CH₃ is weakly electron-donating (+I) → decreases acidity slightly
• −OCH₃ is strongly electron-donating (+R > −I) → decreases acidity more

Q9. How is diethyl ether prepared by Williamson synthesis? What limitation does it have?

Solution
Preparation:
C₆H₃ONa + CH₃CH₂Br → C₆H₃OCH₂CH₃ + NaBr

Limitation: The alkyl halide must be primary. If 2° or 3° alkyl halide is used, elimination (E2) competes with substitution, giving alkene as major product.

Example: (CH₃)₃CBr + CH₃ONa → (CH₃)₂C=CH₂ (elimination) + CH₃OH

Q10. What happens when phenol is treated with: (a) Zn dust (b) CHCl₃ + NaOH

Solution
(a) Phenol + Zn dust → Benzene + ZnO
(Deoxygenation of phenol)

(b) Phenol + CHCl₃ + NaOH → Salicylaldehyde (o-Hydroxybenzaldehyde)
This is the Reimer-Tiemann reaction

Q11. Why do ethers have lower boiling points than alcohols of comparable molecular mass?

Solution
Ethers have lower boiling points because:
• Ethers are polar but cannot form hydrogen bonds with themselves
• Alcohols can form intermolecular hydrogen bonds (O-H...O)
• H-bonds are stronger than dipole-dipole interactions

Example: Ethanol (BP 78°C) vs Dimethyl ether (BP -24°C) — same molecular formula C₂H₆O

Q12. How is phenol converted to: (a) aspirin (b) picric acid?

Solution
(a) Aspirin:
Phenol + NaOH → Sodium phenoxide
Sodium phenoxide + CO₂ → Salicylic acid (Kolbe's reaction)
Salicylic acid + (CH₃CO)₂O → Acetylsalicylic acid (Aspirin)

(b) Picric acid:
Phenol + conc. HNO₃ → 2,4,6-Trinitrophenol (Picric acid)
(Yellow crystalline solid, explosive when dry)

Q13. What is the difference between phenol and an alcohol? Give two chemical tests.

Solution
Phenol: −OH attached directly to benzene ring
Alcohol: −OH attached to alkyl group

Distinguishing tests:
1. FeCl₃ test: Phenol gives violet; alcohol gives no colour
2. Bromine water: Phenol gives white ppt; alcohol gives no ppt
3. NaHCO₃ test: Phenol is acidic enough to react (some substituted phenols); alcohols don't react

Q14. What is the industrial preparation of ethanol from ethene? Write the steps.

Solution
Hydration of ethene:

Step 1: CH₂=CH₂ + H₂O &xrightarrow{H₃PO₄/H₂SO₄, 330K, 70 atm}; CH₃CH₂OH

This is the catalytic hydration method.

Alternative: Fermentation of sugars using yeast
C₆H₁₂O₆ &xrightarrow{yeast}; 2CH₃CH₂OH + 2CO₂

Q15. Why is phenol more acidic than ethanol? Explain with resonance structures.

Solution
Phenol: Phenoxide ion is stabilized by resonance (5 structures)
Negative charge delocalized over benzene ring

Ethanol: Ethoxide ion has no resonance stabilization
Negative charge localized on oxygen

Greater stabilization of conjugate base = stronger acid
Phenol: pK_a = 10
Ethanol: pK_a = 16

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 (pK_a ~ 5) compared to phenols (pK_a ~ 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)

Q6. What is the aldol condensation reaction? Give an example.

Solution
Aldehydes/ketones with α-hydrogen undergo self-addition in dilute NaOH:

2CH₃CHO &xrightarrow{dil. NaOH}; CH₃CH(OH)CH₃CHO (Aldol)
CH₃CH(OH)CH₃CHO &xrightarrow;Δ; CH₃CH=CHCHO (Crotonaldehyde)

Product is called aldol (contains both aldehyde and alcohol groups).

Q7. What is the haloform reaction? Which compounds give this test?

Solution
Compounds containing CH₃CO— group give haloform test:

CH₃COR + 3X₂ + 4NaOH → CHX₃ + RCOONa + 3NaX + 2H₂O

With I₂/NaOH: Yellow precipitate of iodoform (CHI₃)

Compounds giving haloform test:
• CH₃CHO (acetaldehyde)
• CH₃COCH₃ (acetone)
• CH₃CH₂OH (ethanol)
• CH₃CH(OH)CH₃ (isopropyl alcohol)

Q8. How can you convert acetic acid to: (a) acetaldehyde (b) ethyl acetate (c) acetamide?

Solution
(a) CH₃COOH &xrightarrow{LiAlH₄}; CH₃CH₂OH &xrightarrow{PCC}; CH₃CHO
Or: CH₃COOH &xrightarrow{SOCl₂}; CH₃COCl &xrightarrow{LiAlH(O-t-Bu)₃}; CH₃CHO

(b) CH₃COOH + CH₃CH₂OH &xrightarrow{conc. H₂SO₄, Δ}; CH₃COOCH₂CH₃ + H₂O

(c) CH₃COOH + NH₃ &xrightarrow;Δ; CH₃CONH₂ + H₂O

Q9. What is the Rosenmund reduction? Write the reaction.

Solution
Rosenmund reduction: Catalytic hydrogenation of acyl chloride to aldehyde using Pd/BaSO₄ catalyst poisoned with quinoline.

RCOCl + H₂ &xrightarrow{Pd-BaSO₄, quinoline}; RCHO + HCl

Example: CH₃COCl + H₂ → CH₃CHO + HCl

The poisoned catalyst prevents further reduction to alcohol.

Q10. Why are carboxylic acids more acidic than phenols? Explain.

Solution
Carboxylic acids are more acidic because:

• Resonance in carboxylate ion: Equivalent resonance structures (charge equally shared between two O atoms)
• Phenoxide ion: Non-equivalent resonance (charge on C is less stable)
• Inductive effect: −COOH has stronger electron-withdrawing effect

pK_a values: Carboxylic acid ≈ 5, Phenol ≈ 10

Q11. What is the Stephen reaction? Give the reaction.

Solution
Stephen reaction: Reduction of nitriles to aldehydes using SnCl₂/HCl.

RCN + SnCl₂ + 2HCl → RCH=NH · HCl (imine salt)
RCH=NH · HCl + H₂O → RCHO + NH₄Cl

Example: CH₃CN → CH₃CHO

Q12. What happens when acetic acid is treated with: (a) P₄O₁₀ (b) SOCl₂ (c) NaHCO₃?

Solution
(a) 2CH₃COOH + P₄O₁₀ → (CH₃CO)₂O + 2H₃PO₄
(Formation of acetic anhydride)

(b) CH₃COOH + SOCl₂ → CH₃COCl + SO₂ + HCl
(Formation of acetyl chloride)

(c) CH₃COOH + NaHCO₃ → CH₃COONa + CO₂ + H₂O
(Effervescence due to CO₂ release — tests for carboxylic acids)

Q13. Write the mechanism of nucleophilic addition of HCN to acetaldehyde.

Solution
Step 1: CNⁿ attacks the carbonyl carbon (nucleophilic attack)
CH₃CHO + CNⁿ → CH₃CH(CN)Oⁿ

Step 2: Protonation of the alkoxide ion
CH₃CH(CN)Oⁿ + H₂O → CH₃CH(OH)CN + OHⁿ

Product: Cyanohydrin (acetaldehyde cyanohydrin)

Q14. What is the Cannizzaro reaction? Why do aldehydes without α-hydrogen undergo this reaction?

Solution
Cannizzaro reaction: Self-oxidation and self-reduction of aldehydes with no α-hydrogen in conc. NaOH.

2HCHO + NaOH → CH₃OH + HCOONa
2C₆H₃CHO + NaOH → C₆H₃CH₂OH + C₆H₃COONa

Why only without α-H: Without α-hydrogen, aldol condensation cannot occur. The only alternative reaction pathway is hydride transfer (disproportionation).

Q15. Distinguish between aldehydes and ketones using: (a) Tollen's test (b) Fehling's solution (c) Iodoform test.

Solution
(a) Tollen's test:
Aldehyde + 2[Ag(NH₃)₂]⁺ + 3OHⁿ → RCOOⁿ + 2Ag(silver mirror)
Ketone → No reaction

(b) Fehling's solution:
Aldehyde + Cu²ⁿ + OHⁿ → Cu₂O (red precipitate)
Ketone → No reaction

(c) Iodoform test:
CH₃CHO → CHI₃ (yellow precipitate)
CH₃COCH₃ → CHI₃ (yellow precipitate)
Other ketones → No reaction

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.

Q5. What is the carbylamine reaction? How does it help in identifying primary amines?

Solution
Carbylamine reaction (Isocyanide test):
RNH₂ + CHCl₃ + 3KOH → RNC (isocyanide) + 3KCl + 3H₂O

Only 1° amines (aliphatic and aromatic) give this test.
Isocyanide has a foul, offensive smell.
2° and 3° amines do not respond to this test.

Q6. What is the Hofmann bromamide reaction? Write the reaction.

Solution
Hofmann bromamide reaction:
RCONH₂ + Br₂ + 4NaOH → RNH₂ + Na₂CO₃ + 2NaBr + 2H₂O

Converts amide to primary amine with one less carbon atom.

Example: CH₃CONH₂ → CH₃NH₂ (Methylamine from Acetamide)

Q7. What are coupling reactions? Give an example with diazonium salt.

Solution
Coupling reaction: Diazonium salt reacts with electron-rich aromatic compounds (phenols, amines) to form azo dyes.

Example:
C₆H₃N₂ⁿClⁿ + C₆H₃NH₂ → C₆H₃—N=N—C₆H₃NH₂ + HCl
(p-Aminoazobenzene, yellow dye)

C₆H₃N₂ⁿClⁿ + C₆H₃OH → C₆H₃—N=N—C₆H₃OH + HCl
(p-Hydroxyazobenzene, orange dye)

Q8. Why is aniline less basic than methylamine?

Solution
Aniline (C₆H₃NH₂) is less basic because:
• Lone pair on N is delocalized into benzene ring by resonance
• Less available for protonation
• Anilinium ion has fewer resonance structures than aniline

Methylamine (CH₃NH₂) is more basic because:
• +I effect of CH₃ group increases electron density on N
• Lone pair is localized and more available

pK_b: Aniline = 9.4, Methylamine = 3.4

Q9. How are diazonium salts prepared? What are their uses?

Solution
Preparation (Diazotization):
C₆H₃NH₂ + NaNO₂ + 2HCl &xrightarrow{0-5°C}; C₆H₃N₂ⁿClⁿ + NaCl + 2H₂O

Uses:
• Preparation of azo dyes (coupling reactions)
• Sandmeyer reaction: ArN₂ⁿXⁿ + CuCl/HBr → ArCl/ArBr
• Gattermann reaction: uses Cu powder instead of CuX
• Replacing −NH₂ with −OH, −H, −F, −CN

Q10. Arrange the following in decreasing order of basicity: C₆H₃NH₂, (C₆H₃)₂NH, NH₃, CH₃NH₂

Solution
CH₃NH₂ > NH₃ > (C₆H₃)₂NH > C₆H₃NH₂

Reason:
• CH₃NH₂: +I effect of CH₃ increases basicity
• NH₃: reference compound
• (C₆H₃)₂NH: Resonance delocalization reduces basicity
• C₆H₃NH₂: Strong resonance delocalization, least basic

Q11. What is the difference between 1°, 2°, and 3° amines? Give one example of each.

Solution
1° amine: One H of NH₃ replaced by R group
Example: CH₃NH₂ (Methylamine)

2° amine: Two H of NH₃ replaced by R groups
Example: (CH₃)₂NH (Dimethylamine)

3° amine: All three H of NH₃ replaced by R groups
Example: (CH₃)₃N (Trimethylamine)

Q12. How can you distinguish between 1°, 2°, and 3° amines using Hinsberg's test?

Solution
Hinsberg's test: Uses benzenesulphonyl chloride (C₆H₃SO₂Cl)

1° amine: Forms N-alkylbenzenesulphonamide (soluble in NaOH due to −NH−)

2° amine: Forms N,N-dialkylbenzenesulphonamide (insoluble in NaOH)

3° amine: No reaction with benzenesulphonyl chloride

Q13. What is the reaction of aniline with bromine water? What does it indicate?

Solution
C₆H₃NH₂ + 3Br₂(aq) → 2,4,6-Tribromoaniline (white ppt) + 3HBr

This indicates that −NH₂ group is a strongly activating group that activates the benzene ring at ortho and para positions.

The reaction occurs even without a Lewis acid catalyst, unlike benzene which requires FeBr₃.

Q14. Why does acetamide not respond to the carbylamine reaction?

Solution
Acetamide (CH₃CONH₂) is an amide, not a primary amine.

The −NH₂ group in amides has its lone pair delocalized with the carbonyl group (resonance), making it unavailable for the carbylamine reaction.

The carbylamine reaction requires free −NH₂ group (primary amine) to react with CHCl₃/KOH to form isocyanide.

Q15. What is the difference between aliphatic and aromatic amines? Give two examples of each.

Solution
Aliphatic amines: −NH₂ attached to alkyl group
Examples: CH₃NH₂ (methylamine), (CH₃)₂NH (dimethylamine)

Aromatic amines: −NH₂ attached to aromatic ring
Examples: C₆H₃NH₂ (aniline), C₆H₃(CH₃)NH₂ (N-methylaniline)

Key difference: Aromatic amines are less basic than aliphatic amines due to resonance.

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

Q5. Classify carbohydrates into different types with examples.

Solution
Monosaccharides: Cannot be hydrolyzed further (Glucose, Fructose, Galactose)

Oligosaccharides: 2-10 monosaccharide units
• Disaccharides: Sucrose, Lactose, Maltose

Polysaccharides: Many monosaccharide units
• Starch, Cellulose, Glycogen

Q6. What is the difference between α-glucose and β-glucose?

Solution
α-Glucose: −OH group on C-1 is below the ring plane
Melting point = 419K

β-Glucose: −OH group on C-1 is above the ring plane
Melting point = 423K

Both are anomers (differ at anomeric carbon C-1). In solution, they interconvert (mutarotation).

Q7. What is the primary, secondary, tertiary, and quaternary structure of proteins?

Solution
Primary: Linear sequence of amino acids (peptide bonds)

Secondary: Local folding into α-helix or β-pleated sheet (H-bonds)

Tertiary: Overall 3D shape of polypeptide (H-bonds, disulphide bridges, ionic bonds)

Quaternary: Arrangement of multiple polypeptide chains (e.g., hemoglobin has 4 subunits)

Q8. What are the four bases present in DNA and RNA? How do they pair?

Solution
DNA bases: Adenine (A), Guanine (G), Cytosine (C), Thymine (T)

RNA bases: Adenine (A), Guanine (G), Cytosine (C), Uracil (U)

Base pairing (Watson-Crick):
A = T (2 H-bonds) in DNA
A = U (2 H-bonds) in RNA
G ≡ C (3 H-bonds) in both

Q9. What are denaturation and renaturation of proteins? Give examples.

Solution
Denaturation: Loss of secondary, tertiary, and quaternary structure while primary structure remains intact.
Causes: Heat, pH change, heavy metal ions, detergents
Example: Boiling of egg (albumin denatures, becomes white and insoluble)

Renaturation: Regaining of native structure after removal of denaturing agent.
Example: Anfinsen's experiment with ribonuclease A

Q10. What is the difference between DNA and RNA in terms of structure and function?

Solution
DNA:
• Sugar: Deoxyribose
• Double-stranded helix
• Bases: A, T, G, C
• Function: Stores genetic information

RNA:
• Sugar: Ribose
• Single-stranded (usually)
• Bases: A, U, G, C
• Function: Protein synthesis, catalysis (ribozymes)

Q11. What are vitamins? Classify them as fat-soluble and water-soluble with examples.

Solution
Fat-soluble: A, D, E, K (stored in body fat, toxicity possible)
• Vitamin A: Retinol (vision, night blindness)
• Vitamin D: Calciferol (bone health, rickets)
• Vitamin E: Tocopherol (antioxidant)
• Vitamin K: Phylloquinone (blood clotting)

Water-soluble: B-complex, C (excess excreted in urine)
• B₁: Thiamine (beriberi)
• B₂: Riboflavin (cheilosis)
• B₃₁₀: Niacin (pellagra)
• C: Ascorbic acid (scurvy)

Q12. What is the structure of a nucleotide? Draw a simplified diagram.

Solution
A nucleotide consists of three components:

Nitrogenous base (Purine or Pyrimidine)
+ Pentose sugar (Ribose in RNA, Deoxyribose in DNA)
+ Phosphate group (1, 2, or 3 phosphate groups)

Example: Deoxyadenosine monophosphate (dAMP)
Adenine + Deoxyribose + 1 Phosphate

Q13. What are the differences between fibrous and globular proteins?

Solution
Fibrous proteins:
• Elongated, thread-like structure
• Insoluble in water
• Provide structural support
• Example: Keratin (hair, nails), Collagen (tendons)

Globular proteins:
• Spherical, compact structure
• Soluble in water
• Perform enzymatic and transport functions
• Example: Hemoglobin, Insulin, Enzymes

Q14. What is the glycosidic linkage? How is maltose formed from glucose?

Solution
Glycosidic linkage: Ether linkage (−O−) formed between two monosaccharide units by removal of water.

Maltose formation:
α-Glucose + α-Glucose → Maltose + H₂O
(1,4-glycosidic linkage between C-1 of one glucose and C-4 of another)

Q15. What is the difference between reducing and non-reducing sugars? Give two examples of each.

Solution
Reducing sugars: Have free −CHO or −OH group (can reduce Tollen's/Fehling's reagent)
Examples: Glucose, Fructose, Maltose, Lactose

Non-reducing sugars: No free −CHO or −OH group
Examples: Sucrose (glycosidic bond between both anomeric carbons)

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.

Q5. What is the difference between thermoplastic and thermosetting polymers? Give two examples of each.

Solution
Thermoplastic polymers:
• Soften on heating, harden on cooling (reversible)
• Linear or slightly branched chains
• No cross-linking
Examples: Polythene, PVC, Polystyrene

Thermosetting polymers:
• Cannot be remolded after setting (irreversible)
• Extensive cross-linking
• Hard and rigid
Examples: Bakelite, Melamine-formaldehyde

Q6. How is Nylon-6,6 prepared? Write the reaction.

Solution
Nylon-6,6 (Condensation polymerization):

Hexamethylenediamine + Adipic acid → Nylon-6,6 + H₂O
NH₂(CH₂)₆NH₂ + HOOC(CH₂)₄COOH → [NH(CH₂)₆NHCO(CH₂)₄CO]_n + nH₂O

Q7. What is the difference between natural and synthetic polymers? Give examples.

Solution
Natural polymers:
• Found in nature (plants and animals)
• Examples: Cellulose, Starch, DNA, Proteins, Natural rubber

Synthetic polymers:
• Man-made in laboratories and industries
• Examples: Polythene, Nylon, PVC, Buna-S, Teflon

Q8. What is Buna-S? How is it prepared?

Solution
Buna-S: Copolymer of 1,3-butadiene and styrene

n(CH₂=CH−CH=CH₂) + n(C₆H₃CH=CH₂) &xrightarrow{Na}; −[CH₂−CH=CH−CH₂−CH(C₆H₃)−CH₂]−

It is a synthetic rubber with good abrasion resistance.
Buna-N is copolymer of butadiene and acrylonitrile.

Q9. What is the monomer of Teflon? How is it prepared?

Solution
Monomer: Tetrafluoroethene (CF₂=CF₂)

nCF₂=CF₂ &xrightarrow{Initiator}; −(CF₂−CF₂)−

Properties:
• High melting point (327°C)
• Chemically inert (C-F bond is very strong)
• Non-stick surface coating
• Used in non-stick cookware and chemical lab equipment

Q10. What are the types of polymerization reactions? Explain with examples.

Solution
1. Chain-growth (Addition) polymerization:
• Involves monomers with C=C double bonds
• Initiation, propagation, termination steps
• Example: Polythene from ethene

2. Step-growth (Condensation) polymerization:
• Monomers with bifunctional groups
• Loss of small molecules (H₂O, HCl)
• Example: Nylon-6,6 from diamine + diacid

3. Copolymerization:
• Two or more different monomers polymerize together
• Example: Buna-S (butadiene + styrene)

Q11. What is the monomer unit of: (a) Neoprene (b) Orlon (c) Dacron?

Solution
(a) Neoprene: Chloroprene (2-chlorobuta-1,3-diene)
nCH₂=CCl−CH=CH₂ → −(CH₂−CCl=CH−CH₂)−

(b) Orlon (Acrilan): Acrylonitrile (CH₂=CHCN)
nCH₂=CHCN → −(CH₂−CH(CN))−

(c) Dacron (Terylene): Ethylene glycol + Terephthalic acid
Condensation polymer (polyester)

Q12. What is chain-growth polymerization? Explain the three steps involved.

Solution
Chain-growth polymerization: Monomers add one by one to the growing chain.

Step 1 - Initiation: Initiator (free radical) attacks monomer
R• + CH₂=CH₂ → R−CH₂−CH₂•

Step 2 - Propagation: Chain grows by successive addition
R−CH₂−CH₂• + CH₂=CH₂ → R−(CH₂−CH₂)₂•

Step 3 - Termination: Two radicals combine
R−(CH₂)₂• + •(CH₂)₂−R → R−(CH₂)₄−R

Q13. What is natural rubber? How is vulcanization done?

Solution
Natural rubber: cis-1,4-polyisoprene
Monomer: Isoprene (2-methylbuta-1,3-diene)

Vulcanization:
Heating raw rubber with 3-5% sulfur at 373-415K

Sulfur forms cross-links between polymer chains

Improves: Elasticity, tensile strength, resistance to heat and abrasion

Q14. Give the name and monomer of the following polymers: (a) PVC (b) Polystyrene (c) Nylon-6

Solution
(a) PVC: Polyvinyl chloride
Monomer: Vinyl chloride (CH₂=CHCl)

(b) Polystyrene: Poly(styrene)
Monomer: Styrene (C₆H₃CH=CH₂)

(c) Nylon-6: Polycaprolactam
Monomer: Caprolactam (cyclic amide, ring-opening polymerization)

Q15. What is the difference between homopolymer and copolymer? Give one example of each.

Solution
Homopolymer: Made from a single type of monomer
Example: Polythene (from ethene only)
−(CH₂−CH₂)−

Copolymer: Made from two or more different monomers
Example: Buna-S (butadiene + styrene)
−(CH₂−CH=CH−CH₂−CH(C₆H₃)−CH₂)−

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)

Q5. Classify drugs based on their pharmacological action (therapeutic action).

Solution
1. Analgesics: Pain relievers (Aspirin, Morphine, Paracetamol)
2. Antipyretics: Fever reducers (Aspirin, Paracetamol)
3. Antibiotics: Kill/inhibit bacteria (Penicillin, Ampicillin)
4. Antiseptics: Kill microbes on living tissue (Dettol, Iodine)
5. Disinfectants: Kill microbes on non-living objects (Phenol, Bleaching powder)
6. Tranquilizers: Reduce anxiety (Equanil, Valium)

Q6. What is the difference between an antiseptic and a disinfectant? Give examples.

Solution
Antiseptics: Applied to living tissues to kill/prevent growth of microbes
Examples: Dettol (chloroxylenol), Iodine tincture, Furacine

Disinfectants: Applied to non-living objects (floors, drains, instruments)
Examples: Phenol (0.2%), Formalin, Bleaching powder, H₂O₂

Same substance can be both at different concentrations (e.g., 0.2% phenol is antiseptic, 1% is disinfectant).

Q7. What are food additives? Name three types with their functions.

Solution
Food additives: Substances added to food to improve quality, taste, or shelf life.

1. Preservatives: Prevent spoilage (Sodium benzoate, NaNO₂)
2. Artificial sweeteners: Sugar substitutes (Aspartame, Sucralose, Saccharin)
3. Antioxidants: Prevent oxidation (BHA, BHT, Vitamin C)

Q8. What are the different types of detergents? Give one example of each.

Solution
Anionic detergents: Sodium salts of sulphonated long-chain alcohols/hydrocarbons
Example: Sodium lauryl sulphate (CH₃(CH₂)₁₀OSO₃ⁿNa₁⁺)

Cationic detergents: Quaternary ammonium salts of amines
Example: Cetyltrimethylammonium bromide

Non-ionic detergents: No ions formed, esters of stearic acid with polyethylene glycol

Q9. What are antimicrobial agents? Classify them with examples.

Solution
Antibiotics: Kill/inhibit microorganisms (Penicillin, Tetracycline, Erythromycin)

Antiseptics: Applied to living tissue (Dettol, Soframycin, Iodine)

Disinfectants: Applied to surfaces (Phenol, Lysol, Formalin)

Antifungal: Kill fungi (Nystatin, Miconazole)

Antiviral: Inhibit viruses (Acyclovir, AZT)

Q10. What is the difference between soap and detergent? Why do soaps not work in hard water?

Solution
Soap: Sodium/potassium salt of long-chain fatty acids (RCOONa)
Detergent: Sodium salt of long-chain alkyl sulphates/sulphonates (RSO₃Na)

Soaps don't work in hard water because:
Ca²ⁿ/Mg²ⁿ in hard water react with soap to form insoluble precipitate (scum)
2RCOONa + Ca²ⁿ → (RCOO)₂Ca (insoluble) + 2Na₁ⁿ

Detergents form soluble calcium salts, so they work in hard water.

Q11. What are antacids? Name two common antacids and their chemical composition.

Solution
Antacids: Substances that neutralize excess stomach acid (HCl).

1. Milk of Magnesia: Mg(OH)₂ suspension
Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O

2. Aluminium hydroxide gel: Al(OH)₃
Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O

3. Ranitidine (Zantac): H₂ receptor antagonist (reduces acid secretion)

Q12. What are artificial sweetening agents? Name three and mention their sweetness level.

Solution
Artificial sweeteners: Sugar substitutes that provide sweetness without calories.

1. Saccharin: 550 times sweeter than sugar (first synthetic sweetener)
2. Aspartame: 100 times sweeter (methyl ester of dipeptide)
3. Sucralose: 600 times sweeter (chlorinated sugar derivative)

Note: Aspartame is not suitable for cooking (decomposes at high temperature).

Q13. What is the difference between broad-spectrum and narrow-spectrum antibiotics?

Solution
Broad-spectrum antibiotics: Effective against a wide range of bacteria
Examples: Tetracycline, Chloramphenicol, Ampicillin

Narrow-spectrum antibiotics: Effective against specific types of bacteria
Examples: Penicillin G (mainly Gram-positive), Streptomycin (mainly Gram-negative)

Preference: Narrow-spectrum is preferred when possible (less disruption of normal flora).

Q14. What are NSAIDs? Give two examples and mention their side effects.

Solution
NSAIDs (Non-Steroidal Anti-Inflammatory Drugs):
Reduce pain, fever, and inflammation by inhibiting prostaglandin synthesis.

Examples:
1. Aspirin (Acetylsalicylic acid) — also prevents blood clotting
2. Ibuprofen — commonly used for headaches

Side effects:
• Gastric irritation and ulcers
• Increased bleeding tendency
• Kidney damage with prolonged use

Q15. What are the properties of an ideal antacid?

Solution
An ideal antacid should:
• Neutralize excess stomach acid effectively
• Have no side effects (no constipation or diarrhea)
• Be non-absorbable (should not enter bloodstream)
• Provide long-lasting relief
• Not cause rebound acid secretion

Examples: Mg(OH)₂ + Al(OH)₃ combination (Maalox) balances the side effects of each component.