CBSE Class 12 Chemistry Board Paper 2025 (56/1/1): Solutions

When solute particles associate in solution, how does the abnormal (observed) molar mass change?

1. Association of solute particles reduces the total number of particles in the solution, which decreases the magnitude of the colligative property.
2. Since the observed molar mass is inversely proportional to the colligative property, a decrease in the property results in an increase in the observed molar mass.

Answer: It increases, because association lowers the number of particles and so the colligative property, giving a larger apparent molar mass — option (A).

E° of Sn⁴⁺/Sn²⁺ is +0.15 V and of Cr³⁺/Cr is −0.74 V. Find E°cell when the two standard couples form a cell.

1. Identify the cathode as the couple with the higher reduction potential, which is Sn⁴⁺/Sn²⁺ at +0.15 V.
2. Identify the anode as the couple with the lower reduction potential, which is Cr³⁺/Cr at −0.74 V.
3. Calculate the cell potential using the formula E°cell = E°cathode − E°anode = 0.15 − (−0.74) = +0.89 V.

Answer: E°cell = E°cathode − E°anode = 0.15 − (−0.74) = +0.89 V — option (B).

Spin-only magnetic moment of Cr³⁺ (Cr: Z = 24)?

1. Cr (Z=24) is [Ar] 3d⁵ 4s¹, so Cr³⁺ loses 3 electrons to become [Ar] 3d³.
2. With n = 3 unpaired electrons, μ = √(3(3+2)) = √15 ≈ 3.87 B.M.

Answer: Cr³⁺ is 3d³ with n = 3 unpaired electrons; μ = √(n(n+2)) = √15 ≈ 3.87 B.M. — option (B).

To what does acidified KMnO₄ oxidise sulphite ion?

1. Acidified KMnO₄ is a strong oxidising agent; it oxidises sulphite ion (SO₃²⁻) to sulphate ion (SO₄²⁻) and is itself reduced to Mn²⁺.
2. So the product is sulphate ion — option (D).

Answer: Sulphate ion, SO₄²⁻ — option (D).

Give the IUPAC name of [Pt(NH₃)₂Cl₂]²⁺.

1. Determine the oxidation state of Pt: x + 2(0) + 2(-1) = +2, so x = +4.
2. Name the ligands alphabetically: two ammine (NH3) and two chlorido (Cl), giving diamminedichlorido.
3. Combine with the metal name and oxidation state in Roman numerals: diamminedichloridoplatinum(IV).

Answer: Pt is +4 here (x + 0 − 2 = +2), and the ion is a cation, so: diamminedichloridoplatinum(IV) — option (B).

Order by increasing boiling point: (i) (CH₃)₂CHCH₂Br, (ii) CH₃CH₂CH₂CH₂Br, (iii) (CH₃)₃CBr.

1. All three compounds are alkyl bromides with the same molecular formula, so their boiling points depend on surface area and intermolecular forces.
2. More branching reduces the surface area of the molecule, which weakens van der Waals forces and lowers the boiling point.
3. Therefore, the order of increasing boiling point is (iii) < (i) < (ii).

Answer: More branching gives a smaller surface area and lower boiling point: (iii) < (i) < (ii) — option (C).

What happens to configuration when an alkyl halide reacts by the SN2 path?

1. In an SN2 reaction, the nucleophile attacks the electrophilic carbon from the side opposite to the leaving group.
2. This backside attack causes the tetrahedral geometry to flip, resulting in an inversion of configuration at the chiral center.

Answer: The nucleophile attacks from the back side, so configuration is inverted — option (C).

Correct order of acid strength among C₆H₅OH, H₂O and ROH?

1. In phenol the –OH is on an sp² carbon, and the phenoxide ion formed on losing H⁺ is stabilised by resonance (the negative charge spreads over the ring), so phenol is the strongest acid here.
2. In ROH the alkyl group pushes electrons (+I effect) onto oxygen, destabilising the alkoxide ion, so ROH is weaker than water.
3. Order of acid strength: C₆H₅OH > H₂O > ROH — option (A).

Answer: C₆H₅OH > H₂O > ROH — option (A).

Which acid forms when propylmagnesium bromide reacts with CO₂ and the product is hydrolysed with acid?

1. Grignard reagents react with carbon dioxide to form a carboxylate salt, adding one carbon atom to the alkyl chain.
2. Acid hydrolysis of the salt yields butanoic acid (C₃H₇COOH), making option (A) correct.

Answer: Butanoic acid, C₃H₇COOH (one carbon added) — option (A).

Best reagent to change propanamide into propanamine (same number of carbons)?

1. Identify that converting propanamide to propanamine requires reducing the carbonyl group of the amide to a methylene group while retaining the nitrogen.
2. Select LiAlH₄ in ether as the reagent because it reduces the –CONH₂ group to –CH₂NH₂ without losing any carbon atoms, unlike Hofmann bromamide degradation which removes one carbon.

Answer: LiAlH₄ in ether reduces –CONH₂ to –CH₂NH₂ without losing a carbon — option (D).

Pick the false statement about glucose: aldohexose; gives n-hexane with HI; exists in furanose form; no Schiff's test.

1. Glucose is an aldohexose, reacts with HI to form n-hexane, and does not give Schiff's test due to its cyclic structure, so options A, B, and D are true.
2. Glucose primarily exists in the six-membered pyranose form, not the five-membered furanose form, making option C the false statement.

Answer: Glucose exists in the six-membered pyranose form, not furanose — option (C).

Why does a raw mango kept in strong salt solution for pickling shrivel?

1. Strong salt solution acts as a hypertonic medium compared to the cell sap inside the raw mango.
2. Water moves out of the mango cells into the brine through osmosis, causing the fruit to lose water and shrivel.

Answer: Water moves out of the mango into the more concentrated brine, i.e. it loses water by osmosis — option (D).

A: [Cr(H₂O)₆]Cl₂ and [Fe(H₂O)₆]Cl₂ are homoleptic. R: every ligand on the metal is the same.

1. Identify that in [Cr(H₂O)₆]Cl₂ and [Fe(H₂O)₆]Cl₂, the only ligands coordinated to the central metal ions are water molecules, making them homoleptic complexes, so Assertion A is true.
2. Recognize that the definition of a homoleptic complex is one where all ligands attached to the metal are identical, so Reason R is true.
3. Conclude that Reason R provides the direct definition and explanation for why the complexes in Assertion A are classified as homoleptic, making option (A) correct.

Answer: Both true, and R explains A (only H₂O ligands) — option (A).

A: boiling points fall as RI > RBr > RCl > RF. R: alkyl chlorides, bromides and iodides boil much higher than hydrocarbons of similar molar mass.

1. Assertion A is true because the boiling points of alkyl halides increase with the size of the halogen atom, following the order RI > RBr > RCl > RF.
2. Reason R is true because alkyl halides have higher boiling points than hydrocarbons of similar molar mass due to stronger intermolecular forces.
3. Reason R does not explain Assertion A because it compares alkyl halides to hydrocarbons, whereas the trend in A is explained by the increasing van der Waals forces due to the increasing size and polarizability of the halogen atoms.

Answer: Both true, but R (comparison with hydrocarbons) does not explain the trend in A, which comes from size/van der Waals forces of the halogen — option (B).

A: ethanol boils higher than methoxymethane. R: ethanol shows intramolecular H-bonding.

1. Ethanol has a higher boiling point than methoxymethane because ethanol molecules form intermolecular hydrogen bonds, which require more energy to break.
2. Methoxymethane lacks an O-H bond and cannot form hydrogen bonds, so it has weaker intermolecular forces and a lower boiling point.
3. The reason is false because ethanol shows intermolecular hydrogen bonding, not intramolecular hydrogen bonding.
4. Therefore, Assertion A is true but Reason R is false, making option (C) the correct answer.

Answer: A is true but R is false: ethanol has intermolecular hydrogen bonding — option (C).

A: apart from glycine, all natural α-amino acids are optically active. R: most natural amino acids have L-configuration.

1. Statement A is true because all natural α-amino acids except glycine have a chiral α-carbon, making them optically active.
2. Statement R is true because most natural amino acids possess the L-configuration, but this stereochemical label does not explain the cause of optical activity.
3. Since both statements are true but R is not the correct explanation for A, the correct option is (B).

Answer: Both true, but R does not explain A (optical activity is due to the chiral α-carbon) — option (B).

(a) Why does food cook faster in a pressure cooker than in an open pan? (b) Liquids X and Y on mixing give a smaller total volume. Which deviation from Raoult's law is this, and how does temperature change on mixing?

1. (a) In a pressure cooker, the internal pressure is higher than atmospheric pressure, which increases the boiling point of water.
2. Since the water boils at a higher temperature, the food cooks faster due to the increased thermal energy.
3. (b) The decrease in total volume upon mixing indicates a negative deviation from Raoult's law.
4. Negative deviation implies stronger intermolecular forces, releasing heat and causing the temperature to rise.

Answer: (a) Pressure inside the cooker is high, so water boils at a higher temperature and food cooks faster. (b) Negative deviation (ΔVmix < 0); heat is released, so the temperature rises.

What is an azeotrope? Which kind arises from negative deviation from Raoult's law? Give one example.

1. An azeotrope is a liquid mixture that has the same composition in the liquid and vapour phases and boils at a constant temperature.
2. Negative deviation from Raoult's law results in a maximum-boiling azeotrope.
3. An example of a maximum-boiling azeotrope is a mixture of approximately 68% HNO₃ and 32% water.

Answer: An azeotrope is a liquid mixture that has the same composition in liquid and vapour and boils at a constant temperature. Negative deviation gives a maximum-boiling azeotrope, e.g. about 68% HNO₃ + 32% water.

Complete and balance: (a) 2MnO₄⁻ + 10I⁻ + 16H⁺ → ? (b) Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → ?

1. (a) In acidic medium, permanganate ions (MnO₄⁻) are reduced to Mn²⁺ while iodide ions (I⁻) are oxidized to I₂. The balanced equation is 2MnO₄⁻ + 10I⁻ + 16H⁺ → 2Mn²⁺ + 5I₂ + 8H₂O.
2. (b) Dichromate ions (Cr₂O₇²⁻) oxidize iron(II) ions (Fe²⁺) to iron(III) ions (Fe³⁺) while being reduced to Cr³⁺. The balanced equation is Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O.

Answer: (a) 2MnO₄⁻ + 10I⁻ + 16H⁺ → 2Mn²⁺ + 5I₂ + 8H₂O. (b) Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O.

Is benzaldehyde more or less reactive than propanal towards nucleophilic addition? Why?

1. Benzaldehyde is less reactive than propanal towards nucleophilic addition.
2. In benzaldehyde, the carbonyl group is conjugated with the benzene ring, allowing resonance delocalization of the pi electrons.
3. This resonance reduces the polarity of the C=O bond, making the carbonyl carbon less electrophilic compared to propanal.

Answer: Less reactive. Resonance of the C=O group with the benzene ring lowers the polarity of the carbonyl, so its carbon is less electrophilic than in propanal.

Name A and B: (a) CH₃CH₂Cl + NaCN → A; A + H₂/Ni → B. (b) C₆H₅NH₂ + NaNO₂/HCl at 0–5 °C → A; A + C₆H₅NH₂ in acid (H⁺) → B.

1. (a) Ethyl chloride reacts with sodium cyanide to form propanenitrile, so A is CH₃CH₂CN.
2. (a) Propanenitrile undergoes reduction with hydrogen and nickel to form propan-1-amine, so B is CH₃CH₂CH₂NH₂.
3. (b) Aniline reacts with sodium nitrite and hydrochloric acid at 0–5 °C to form benzenediazonium chloride, so A is C₆H₅N₂⁺Cl⁻.
4. (b) Benzenediazonium chloride couples with aniline in acidic medium to form para-aminoazobenzene, so B is C₆H₅–N=N–C₆H₄–NH₂.

Answer: (a) A = CH₃CH₂CN (propanenitrile), B = CH₃CH₂CH₂NH₂ (propan-1-amine). (b) A = C₆H₅N₂⁺Cl⁻ (benzenediazonium chloride), B = C₆H₅–N=N–C₆H₄–NH₂ (para-aminoazobenzene, yellow dye from coupling at the para position).

Write the reactions of D-glucose with (a) HCN and (b) bromine water.

1. For part (a), write the reaction where D-glucose reacts with HCN to form glucose cyanohydrin: HOCH₂(CHOH)₄CHO + HCN → HOCH₂(CHOH)₄CH(OH)CN.
2. For part (b), write the reaction where D-glucose is oxidized by bromine water to form gluconic acid: HOCH₂(CHOH)₄CHO + [O] (Br₂ water) → HOCH₂(CHOH)₄COOH.

Answer: (a) HOCH₂(CHOH)₄CHO + HCN → HOCH₂(CHOH)₄CH(OH)CN (glucose cyanohydrin). (b) HOCH₂(CHOH)₄CHO + [O] (Br₂ water) → HOCH₂(CHOH)₄COOH (gluconic acid).

An aqueous glucose solution (M = 180 g/mol) boils at 100.20 °C. Find its freezing point. Kf = 1.86 and Kb = 0.512 K kg mol⁻¹.

1. Calculate the boiling point elevation: ΔTb = 100.20 °C − 100.00 °C = 0.20 K.
2. Determine the molality of the solution using the boiling point constant: m = ΔTb / Kb = 0.20 / 0.512 = 0.391 mol/kg.
3. Calculate the freezing point depression using the molality and freezing point constant: ΔTf = Kf × m = 1.86 × 0.391 = 0.73 K.
4. Find the freezing point of the solution: T_f = 0.00 °C − 0.73 K = −0.73 °C.

Answer: ΔTb = 0.20 K, so m = 0.20/0.512 = 0.391 mol/kg. ΔTf = Kf·m = 1.86 × 0.391 = 0.73 K, so freezing point ≈ −0.73 °C.

Note: CBSE's marking scheme rounds m to 0.390 and gets −0.725 °C; unrounded working gives −0.727 °C. Both round to −0.73 °C.

(a) State (i) Kohlrausch's law of independent migration of ions, (ii) Faraday's first law of electrolysis. (b) E°(X²⁺/X) = −2.36 V, E°(Y²⁺/Y) = −0.14 V, E°(Fe²⁺/Fe) = −0.44 V. Which metal, X or Y, is better for coating iron against corrosion, and why?

1. Kohlrausch's law states that the limiting molar conductivity of an electrolyte is the sum of the separate limiting contributions of its cation and anion.
2. Faraday's first law of electrolysis states that the mass of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity (charge) passed through the electrolyte.
3. Compare the standard reduction potentials: E°(X²⁺/X) = −2.36 V is more negative than E°(Fe²⁺/Fe) = −0.44 V, while E°(Y²⁺/Y) = −0.14 V is less negative than iron.
4. Since X has a more negative potential than iron, it acts as a sacrificial anode, oxidizing in preference to iron and thus protecting it from corrosion.
5. Metal Y is less reactive than iron (less negative potential), so it cannot protect iron sacrificially; therefore, X is the better choice for coating.

Answer: (a)(i) Limiting molar conductivity of an electrolyte is the sum of the separate limiting contributions of its cation and anion. (ii) Mass of substance reacting at an electrode is proportional to the charge passed (m = Z·Q). (b) X, because its E° is more negative than iron's, so X is oxidised in preference and protects iron sacrificially; Y (−0.14 V) is less reactive than iron.

A first order reaction is half done in 20 min at 300 K and in 5 min at 350 K. Find Ea. R = 8.314 J K⁻¹ mol⁻¹, log 4 = 0.602.

1. For a first order reaction, the rate constant is given by k = 0.693 / t½.
2. Calculate k₁ at 300 K: k₁ = 0.693 / 20 = 0.03465 min⁻¹.
3. Calculate k₂ at 350 K: k₂ = 0.693 / 5 = 0.1386 min⁻¹.
4. Determine the ratio of rate constants: k₂ / k₁ = 0.1386 / 0.03465 = 4.
5. Use the Arrhenius equation in logarithmic form: log(k₂/k₁) = Ea(T₂ − T₁) / (2.303 R T₁ T₂).
6. Substitute the values: log 4 = Ea(350 − 300) / (2.303 × 8.314 × 300 × 350).
7. Solve for Ea: Ea = (2.303 × 8.314 × 0.602 × 300 × 350) / 50 ≈ 24206 J mol⁻¹ ≈ 24.2 kJ mol⁻¹.

Answer: k = 0.693/t½: k₁ = 0.03465 min⁻¹, k₂ = 0.1386 min⁻¹, k₂/k₁ = 4. log 4 = Ea(T₂ − T₁)/(2.303 R T₁T₂) gives Ea = 2.303 × 8.314 × 0.602 × 300 × 350/50 ≈ 24206 J/mol ≈ 24.2 kJ mol⁻¹.

Note: CBSE's marking scheme prints 24209 J/mol because it rounds 2.303 × 8.314 to 19.15; either value is about 24.2 kJ/mol.

For the 3d series Sc to Zn: (a) why is E°(Cu²⁺/Cu) unusually positive? (b) which element is a strong reducing agent in the +2 state, and why? (c) why are Zn²⁺ salts colourless?

1. (a) The standard reduction potential depends on the sum of atomisation, ionisation, and hydration enthalpies.
2. Copper has high atomisation and ionisation enthalpies which are not compensated by its hydration enthalpy, making E°(Cu²⁺/Cu) unusually positive.
3. (b) Chromium is the strong reducing agent in the +2 state because Cr²⁺ has a d⁴ configuration.
4. Cr²⁺ easily loses an electron to form Cr³⁺, which has a stable d³ configuration with a half-filled t₂g set.
5. (c) The electronic configuration of Zn²⁺ is 3d¹⁰, meaning all d-orbitals are completely filled.
6. Since there are no unpaired electrons, d–d transitions are not possible, so Zn²⁺ salts are colourless.

Answer: (a) Copper's high enthalpy of atomisation and ionisation is not balanced by its hydration enthalpy. (b) Cr: Cr²⁺ (d⁴) easily loses an electron to become Cr³⁺ (d³), which has a stable half-filled t₂g set. (c) Zn²⁺ is 3d¹⁰ with no unpaired electrons, so no d–d transition is possible.

Give the main monohalogen product: (a) 4-chloroethylbenzene (Cl–C₆H₄–CH₂CH₃, para) with Br₂ and heat; (b) 1-methylcyclohexene + HBr; (c) 4-(hydroxymethyl)phenol (HOCH₂–C₆H₄–OH, para) with HCl and heat.

Uses a figure printed in CBSE's paper — what the figure shows: (a) benzene ring with Cl and CH₂–CH₃ para to each other, reagent Br₂, heat. (b) cyclohexene ring with CH₃ on one carbon of the C=C, reagent HBr. (c) benzene ring with OH and CH₂OH para to each other, reagent HCl, heat.

1. (a) Under heat, Br2 reacts with the benzylic position of the ethyl group via a free radical mechanism. The hydrogen on the carbon attached to the ring is replaced by bromine, yielding 1-(1-bromoethyl)-4-chlorobenzene.
2. (b) The addition of HBr to 1-methylcyclohexene follows Markovnikov's rule. The bromine atom attaches to the tertiary carbon (the one bearing the methyl group), forming 1-bromo-1-methylcyclohexane.
3. (c) With HCl and heat, the benzylic alcohol group (-CH2OH) undergoes nucleophilic substitution to form -CH2Cl. The phenolic -OH group remains unchanged, resulting in 4-(chloromethyl)phenol.

Answer: (a) 1-(1-bromoethyl)-4-chlorobenzene, p-Cl–C₆H₄–CH(Br)CH₃ (benzylic bromination). (b) 1-bromo-1-methylcyclohexane (Markovnikov: Br on the ring carbon carrying CH₃). (c) 4-(chloromethyl)phenol, p-HO–C₆H₄–CH₂Cl (only the benzylic –OH is replaced; phenolic –OH stays).

Convert: (a) chlorobenzene → biphenyl; (b) propene → 1-iodopropane; (c) 2-bromobutane → but-2-ene.

1. (a) Chlorobenzene undergoes the Fittig reaction with sodium metal in dry ether to form biphenyl: 2C₆H₅Cl + 2Na → C₆H₅–C₆H₅ + 2NaCl.
2. (b) Propene reacts with HBr in the presence of peroxides to give 1-bromopropane via anti-Markovnikov addition: CH₃CH=CH₂ + HBr → CH₃CH₂CH₂Br.
3. (b) 1-Bromopropane is converted to 1-iodopropane using NaI in dry acetone (Finkelstein reaction): CH₃CH₂CH₂Br + NaI → CH₃CH₂CH₂I + NaBr.
4. (c) 2-Bromobutane undergoes dehydrohalogenation with alcoholic KOH and heat to form but-2-ene as the major product: CH₃CHBrCH₂CH₃ + KOH → CH₃CH=CHCH₃ + KBr + H₂O.

Answer: (a) Fittig reaction: 2C₆H₅Cl + 2Na (dry ether) → C₆H₅–C₆H₅ + 2NaCl. (b) CH₃CH=CH₂ + HBr (peroxide) → CH₃CH₂CH₂Br; then NaI in dry acetone (Finkelstein) → CH₃CH₂CH₂I + NaBr. (c) CH₃CHBrCH₂CH₃ + alcoholic KOH, heat → CH₃CH=CHCH₃ (major, Saytzeff) + KBr + H₂O.

(a) Put (CH₃)₂NH, CH₃CH₂NH₂ and CH₃CH₂OH in increasing boiling point. (b) Explain: (i) Gabriel phthalimide synthesis cannot make aromatic primary amines; (ii) amides are weaker bases than amines.

1. (a) Ethanol forms the strongest hydrogen bonds because O is more electronegative than N, so it boils highest.
2. A primary amine (CH₃CH₂NH₂) has two H atoms on N for hydrogen bonding, while the secondary amine (CH₃)₂NH has only one, so the secondary amine boils lowest. Order: (CH₃)₂NH < CH₃CH₂NH₂ < CH₃CH₂OH.
3. (b)(i) Gabriel synthesis needs the phthalimide anion to displace a halide by nucleophilic substitution; aryl halides do not undergo this, so aromatic primary amines cannot be made.
4. (b)(ii) In an amide the lone pair on N is delocalised onto the C=O group by resonance, so it is less available to accept a proton; amides are weaker bases than amines.

Answer: (a) (CH₃)₂NH < CH₃CH₂NH₂ < CH₃CH₂OH. (b)(i) Aryl halides do not undergo nucleophilic substitution with the phthalimide anion. (ii) In amides the nitrogen lone pair is delocalised onto the C=O group by resonance, so it is less available to accept a proton.

(a) How does a native protein differ from a denatured one? (b) Which of glucose, lactose, amylose, fructose is a disaccharide? (c) Which vitamin is needed for blood clotting?

1. (a) A native protein retains its specific three-dimensional secondary and tertiary structures, which are essential for it to be biologically active.
2. (a) In contrast, a denatured protein has lost this specific three-dimensional structure due to external stress, resulting in the loss of its biological activity.
3. (b) Among the given options, lactose is the disaccharide because it is composed of two monosaccharide units, glucose and galactose, linked by a glycosidic bond.
4. (c) Vitamin K is the specific vitamin required for the synthesis of clotting factors in the liver, which is necessary for blood clotting.

Answer: (a) A native protein keeps its 3-D (secondary/tertiary) structure and is biologically active; a denatured protein has lost that structure and its activity. (b) Lactose. (c) Vitamin K.

Case on reaction rate, order and molecularity. (i) What is the rate determining step? (ii) What is a complex reaction?

1. (i) The rate determining step is the slowest step in the reaction mechanism, which fixes the overall rate of the reaction.
2. (ii) A complex reaction is defined as a reaction that takes place through a sequence of two or more elementary steps.

Answer: (i) The slowest step of a reaction mechanism, which fixes the overall rate. (ii) A reaction that takes place through a sequence of two or more elementary steps.

How does temperature affect a reaction's rate constant?

1. According to the Arrhenius equation, the rate constant k is given by k = A e^(−Ea/RT), where A is the frequency factor, Ea is activation energy, R is the gas constant, and T is temperature.
2. Since T appears in the denominator of the negative exponent, an increase in temperature decreases the magnitude of the negative term, causing the exponential value and thus the rate constant k to increase.

Answer: The rate constant increases as temperature rises (k = A e^(−Ea/RT)).

Why does molecularity apply only to elementary reactions while order applies to both elementary and complex ones?

1. Molecularity is defined as the number of species colliding in a single elementary step, so it is meaningless for a multi-step reaction.
2. Order is determined experimentally from the rate law, which depends on the slowest step, so it applies to both elementary and complex reactions.

Answer: Molecularity counts species colliding in a single step, so it has no meaning for a multi-step reaction; order is found experimentally from the rate law (set by the slowest step), so it can be given for any reaction.

X → Y is second order. How does the rate of forming Y change if [X] is made 3 times?

1. For a second-order reaction, the rate is proportional to the square of the reactant concentration, so Rate = k[X]².
2. If the concentration of X is tripled, the new rate becomes k(3[X])² = 9k[X]², which is 9 times the original rate.

Answer: Rate ∝ [X]², so it becomes 9 times the original rate.

Case on electrophilic substitution of phenols and the Reimer-Tiemann reaction. What forms when phenol reacts with (i) Br₂ in CS₂, (ii) concentrated HNO₃?

1. Phenol reacts with Br2 in CS2 to form a mixture of o-bromophenol and p-bromophenol, with the para isomer being the major product.
2. The low polarity of the CS2 solvent restricts the reaction to monobromination, preventing further substitution.
3. Phenol reacts with concentrated HNO3 to undergo extensive nitration, forming 2,4,6-trinitrophenol, also known as picric acid.

Answer: (i) A mixture of o-bromophenol and p-bromophenol (para major), since the low-polarity solvent allows only monobromination. (ii) 2,4,6-trinitrophenol (picric acid).

Why is phenol not easily protonated?

1. The lone pair of electrons on the oxygen atom in phenol is delocalised into the benzene ring through resonance.
2. Because this lone pair is involved in resonance, it is not readily available to accept a proton, making phenol difficult to protonate.

Answer: The lone pair on the oxygen is delocalised into the benzene ring by resonance, so it is not readily available to accept a proton.

Which is the stronger acid, phenol or cresol? Why?

1. Phenol is the stronger acid because the electron-releasing (+I) CH3 group in cresol destabilises the phenoxide ion, making cresol a weaker acid.

Answer: Phenol. The electron-releasing (+I) CH₃ group in cresol destabilises the phenoxide ion, making cresol a weaker acid.

Give the IUPAC name of the product of the Reimer-Tiemann reaction of phenol.

1. The Reimer-Tiemann reaction involves the formylation of phenol using chloroform and alkali, which introduces an aldehyde group (-CHO) at the ortho position relative to the hydroxyl group.
2. The resulting product is 2-hydroxybenzaldehyde, commonly known as salicylaldehyde.

Answer: 2-Hydroxybenzaldehyde (salicylaldehyde).

(a) For Sn | Sn²⁺ (0.004 M) || H⁺ (0.02 M) | H₂ (1 bar) | Pt at 298 K, write the cell reaction and find the emf. E°(Sn²⁺/Sn) = −0.14 V, E°(H⁺/H₂) = 0.00 V. (b) Explain: (i) electrolysis of aqueous NaCl gives Cl₂, not O₂, at the anode although E° values favour O₂; (ii) conductivity of CH₃COOH falls on dilution.

1. Anode: Sn(s) → Sn²⁺(aq) + 2e⁻; Cathode: 2H⁺(aq) + 2e⁻ → H₂(g); Overall: Sn(s) + 2H⁺(aq) → Sn²⁺(aq) + H₂(g) with n = 2.
2. Calculate E°cell = E°cathode − E°anode = 0.00 − (−0.14) = 0.14 V.
3. Apply Nernst equation: E = E°cell − (0.059/2) log([Sn²⁺]/[H⁺]²).
4. Substitute values: E = 0.14 − 0.0295 log(0.004/0.02²) = 0.14 − 0.0295 log(10).
5. Calculate final emf: E = 0.14 − 0.0295(1) = 0.1105 V.
6. For (b)(i), state that oxygen evolution has a high overpotential, so Cl₂ is liberated instead of O₂.
7. For (b)(ii), explain that conductivity falls on dilution because the number of ions per unit volume decreases.

Answer: (a) Sn(s) + 2H⁺(aq) → Sn²⁺(aq) + H₂(g), n = 2. E = [0 − (−0.14)] − (0.059/2) log(0.004/0.02²) = 0.14 − 0.0295 log 10 = 0.1105 V. (b)(i) Oxygen evolution has a high overpotential, so Cl₂ is liberated instead. (ii) Conductivity depends on ions per unit volume, which decrease on dilution.

Note: Uses 2.303RT/F = 0.059 V at 298 K, which the paper does not print.

(a) Give the anode, cathode and overall reactions of a lead storage battery while discharging. (b) Half-cell: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O, E° = 1.33 V. Find its potential with 0.01 M K₂Cr₂O₇, 0.01 M Cr³⁺ and 1.0 × 10⁻⁴ M H⁺. log 10 = 1.

1. Write the anode reaction for the lead storage battery during discharge: Pb + SO₄²⁻ → PbSO₄ + 2e⁻.
2. Write the cathode reaction for the lead storage battery during discharge: PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄ + 2H₂O.
3. Combine the half-reactions to write the overall cell reaction: Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O.
4. Apply the Nernst equation for the given half-cell: E = E° − (0.059/n) log([Cr³⁺]² / ([Cr₂O₇²⁻][H⁺]¹⁴)), where n = 6.
5. Substitute the given concentrations into the log term: log((0.01)² / (0.01 × (10⁻⁴)¹⁴)) = log(10⁻⁴ / (10⁻² × 10⁻⁵⁶)) = log(10⁵⁴) = 54.
6. Calculate the final potential: E = 1.33 − (0.059/6) × 54 = 1.33 − 0.531 = 0.799 V.

Answer: (a) Anode: Pb + SO₄²⁻ → PbSO₄ + 2e⁻. Cathode: PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄ + 2H₂O. Overall: Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O. (b) E = 1.33 − (0.059/6) log([Cr³⁺]²/([Cr₂O₇²⁻][H⁺]¹⁴)) = 1.33 − (0.059/6) log(10⁻⁴/(10⁻² × 10⁻⁵⁶)) = 1.33 − (0.059/6) × 54 = 1.33 − 0.531 = 0.799 V.

Note: Uses 2.303RT/F = 0.059 V at 298 K, which the paper does not print.

(a) Why are low-spin tetrahedral complexes not known? (b) Why is Co²⁺ (Z = 27) readily oxidised to Co³⁺ with a strong ligand? (c) Which isomerism does [Co(NH₃)₆][Cr(CN)₆] show? (d) Why is [Ni(H₂O)₆]²⁺ green but [Ni(CN)₄]²⁻ colourless (Ni: Z = 28)? (e) IUPAC name of [Co(NH₃)₅(CO₃)]Cl.

1. Tetrahedral splitting Δt is approximately 4/9 of octahedral splitting Δo, which is too small to overcome the electron pairing energy.
2. Therefore, electrons remain unpaired in tetrahedral complexes, making low-spin configurations energetically unfavorable and unknown.
3. Co²⁺ is d⁷; in a strong field, one electron occupies the high-energy eg level and is easily lost to form the stable low-spin d⁶ (t₂g⁶) Co³⁺.
4. The complex [Co(NH₃)₆][Cr(CN)₆] shows coordination isomerism because ligands can be exchanged between the cationic and anionic metal centers.
5. [Ni(H₂O)₆]²⁺ has two unpaired electrons due to weak-field H₂O, allowing d–d transitions that give it a green color.
6. [Ni(CN)₄]²⁻ is square planar dsp² with all electrons paired due to strong-field CN⁻, so no d–d transitions occur and it is colorless.
7. The IUPAC name of [Co(NH₃)₅(CO₃)]Cl is Pentaamminecarbonatocobalt(III) chloride.

Answer: (a) Tetrahedral splitting Δt (≈ 4/9 Δo) is too small to force electron pairing. (b) Co²⁺ is d⁷; in a strong field one electron sits in the high-energy eg level and is easily lost, giving the more stable low-spin d⁶ (t₂g⁶) Co³⁺. (c) Coordination isomerism. (d) [Ni(H₂O)₆]²⁺ has two unpaired electrons (weak-field H₂O), so d–d transitions give colour; [Ni(CN)₄]²⁻ is square planar dsp² with all electrons paired, so no d–d transition and it is colourless. (e) Pentaamminecarbonatocobalt(III) chloride.

(a) What is the chelate effect? Give an example. (b) Hybridisation and magnetic nature of [Fe(CN)₆]⁴⁻ (Fe: Z = 26)? (c) PtCl₂·2NH₃ gives no reaction with AgNO₃. Write its formula.

1. The chelate effect states that complexes formed with polydentate ligands are more stable than those with monodentate ligands.
2. An example of this effect is the formation of the stable complex [Co(en)₃]³⁺ from Co³⁺ and ethylenediamine.
3. For [Fe(CN)₆]⁴⁻, the oxidation state of Fe is +2, giving the electronic configuration 3d⁶.
4. Since CN⁻ is a strong field ligand, it causes pairing of electrons in the d-orbitals, resulting in d²sp³ hybridisation.
5. With all electrons paired, the complex [Fe(CN)₆]⁴⁻ is diamagnetic in nature.
6. PtCl₂·2NH₃ does not react with AgNO₃, indicating that no chloride ions are present outside the coordination sphere.
7. Therefore, both chlorine atoms are coordinated to platinum, and the formula is written as [Pt(NH₃)₂Cl₂].

Answer: (a) Complexes with chelating (polydentate) ligands are more stable than similar complexes with monodentate ligands, e.g. [Co(en)₃]³⁺. (b) Fe²⁺ is 3d⁶; strong-field CN⁻ pairs all electrons: d²sp³ hybridisation, diamagnetic. (c) [Pt(NH₃)₂Cl₂] — both Cl are inside the coordination sphere, so no free Cl⁻ precipitates with AgNO₃.

(a) Convert (i) ethanal to but-2-enal, (ii) propanoic acid to ethane. (b) Alkene A (C₅H₁₀) on ozonolysis gives B and C. B gives Fehling's test and the iodoform test; C gives iodoform but not Fehling's test. Identify A, B, C.

1. (a)(i) Two molecules of ethanal undergo aldol condensation with dilute NaOH: 2CH₃CHO → CH₃CH(OH)CH₂CHO (3-hydroxybutanal).
2. On heating, the aldol loses water: CH₃CH(OH)CH₂CHO → CH₃CH=CHCHO (but-2-enal) + H₂O.
3. (a)(ii) Propanoic acid with NaOH gives sodium propanoate: CH₃CH₂COOH + NaOH → CH₃CH₂COONa + H₂O.
4. Heating the salt with soda lime (NaOH + CaO) removes CO₂: CH₃CH₂COONa + NaOH → CH₃CH₃ + Na₂CO₃.
5. (b) B gives Fehling's test, so it is an aldehyde, and the iodoform test, so it has a CH₃CO– group: B is ethanal, CH₃CHO.
6. C gives iodoform but not Fehling's test, so it is a methyl ketone: C is propanone, CH₃COCH₃.
7. Joining the carbonyl carbons of B and C by a double bond gives A = (CH₃)₂C=CHCH₃, 2-methylbut-2-ene (C₅H₁₀).

Answer: (a)(i) 2CH₃CHO + dil. NaOH → CH₃CH(OH)CH₂CHO (aldol); heat → CH₃CH=CHCHO + H₂O. (ii) CH₃CH₂COOH + NaOH → CH₃CH₂COONa; heat with soda lime (NaOH + CaO) → CH₃CH₃ + Na₂CO₃. (b) A = (CH₃)₂C=CHCH₃ (2-methylbut-2-ene); B = CH₃CHO (ethanal); C = CH₃COCH₃ (propanone).

Note: CBSE's marking scheme prints A as (CH₃)₂CH=CHCH₃; the correct structure for 2-methylbut-2-ene is (CH₃)₂C=CHCH₃.

Compound A (C₈H₁₆O₂) is hydrolysed by dilute H₂SO₄ into acid B and alcohol C. Chromic acid oxidises C to B, and C on dehydration gives but-1-ene. Identify A, B, C and give the equations.

1. Chromic acid oxidises C to B, so the alcohol and the acid have the same carbon chain; A (C₈H₁₆O₂) is their ester, so each part has 4 carbons.
2. C is oxidised to an acid, so it is a primary alcohol, and its dehydration gives but-1-ene: C = CH₃CH₂CH₂CH₂OH (butan-1-ol), B = CH₃CH₂CH₂COOH (butanoic acid).
3. A = CH₃CH₂CH₂COOCH₂CH₂CH₂CH₃ (butyl butanoate), which is C₈H₁₆O₂.
4. Hydrolysis: CH₃CH₂CH₂COOC₄H₉ + H₂O (dil. H₂SO₄) → CH₃CH₂CH₂COOH + C₄H₉OH.
5. Oxidation: CH₃CH₂CH₂CH₂OH + 2[O] (H₂CrO₄) → CH₃CH₂CH₂COOH + H₂O.
6. Dehydration: CH₃CH₂CH₂CH₂OH (conc. H₂SO₄, heat) → CH₃CH₂CH=CH₂ + H₂O.

Answer: A = CH₃CH₂CH₂COOCH₂CH₂CH₂CH₃ (butyl butanoate); B = CH₃CH₂CH₂COOH (butanoic acid); C = CH₃CH₂CH₂CH₂OH (butan-1-ol). Equations: C₃H₇COOC₄H₉ + H₂O (dil. H₂SO₄) → C₃H₇COOH + C₄H₉OH; CH₃CH₂CH₂CH₂OH + 2[O] (H₂CrO₄) → CH₃CH₂CH₂COOH + H₂O; CH₃CH₂CH₂CH₂OH (conc. H₂SO₄, heat) → CH₃CH₂CH=CH₂ + H₂O.