CBSE Class 12 Chemistry Board Paper 2026 (56/1/1): Solutions

Four graphs are shown; pick the one that fits a first order reaction.

Uses a figure printed in CBSE's paper — what the figure shows: (A) t½ vs [R]₀: straight line rising from the origin. (B) t½ vs [R]₀: horizontal line. (C) rate vs concentration: horizontal line. (D) rate vs concentration: curve falling steeply then levelling off.

1. For a first order reaction, the half-life is given by t½ = 0.693/k, which is independent of the initial concentration [R]₀.
2. Since t½ does not change with [R]₀, the graph of t½ vs [R]₀ must be a horizontal line, corresponding to option (B).

Answer: For a first order reaction the half-life does not depend on initial concentration, so the t½ vs [R]₀ graph is a horizontal line — option (B).

Among 0.1 M aqueous solutions of glucose, CaCl₂, KCl and urea, which freezes at the lowest temperature?

1. Freezing point depression is a colligative property proportional to the total number of solute particles in solution.
2. Glucose and urea are non-electrolytes with van't Hoff factor i = 1, while KCl dissociates into 2 ions (i = 2).
3. CaCl₂ dissociates into 3 ions (Ca²⁺ and 2Cl⁻), giving the highest i ≈ 3, which causes the maximum depression and thus the lowest freezing temperature.

Answer: 0.1 M CaCl₂, because it gives 3 ions (i ≈ 3), the largest number of particles — option (B).

Which one is not a transition metal: Sc, Ag, Hg or Cu?

1. A transition element has a partly filled d subshell in the atom or in one of its common ions.
2. Hg is [Xe] 4f¹⁴ 5d¹⁰ 6s². Its common ions Hg²⁺ (5d¹⁰) and Hg₂²⁺ also keep a full 5d¹⁰ set, so it never has a partly filled d subshell.
3. Sc (3d¹), Cu²⁺ (3d⁹) and Ag²⁺ (4d⁹) do have partly filled d orbitals, so Hg is the one that is not a transition metal — option (C).

Answer: Hg, as it has a completely filled 5d¹⁰ configuration in the ground state and in its common Hg²⁺ ion — option (C).

Which expression gives the fraction of molecules having energy at least Ea?

1. The fraction of molecules with energy greater than or equal to the activation energy Ea is given by the Boltzmann factor.
2. This fraction is expressed as e^(−Ea/RT), which corresponds to option (B).

Answer: e^(−Ea/RT) (the Boltzmann factor) — option (B).

Aqueous CuCl₂ is electrolysed with platinum electrodes. What happens?

1. At the cathode, Cu²⁺ (E° = +0.34 V) is reduced far more easily than H⁺ or water, so copper metal deposits: Cu²⁺ + 2e⁻ → Cu.
2. At the anode, oxidation of water to O₂ has the lower E°, but O₂ needs a large overpotential, so Cl⁻ is oxidised instead: 2Cl⁻ → Cl₂ + 2e⁻.
3. So Cu deposits at the cathode and Cl₂ is given off at the anode — option (D).

Answer: Cu deposits at the cathode and Cl₂ gas is given off at the anode — option (D).

Acetylating which compound gives aspirin?

1. Aspirin is synthesized by the acetylation of salicylic acid, which is chemically known as 2-hydroxybenzoic acid.
2. Therefore, the correct compound is 2-hydroxybenzoic acid (salicylic acid), corresponding to option (C).

Answer: 2-Hydroxybenzoic acid (salicylic acid) — option (C).

What is the secondary valency of cobalt in [Co(NH₃)₅(NO₂)]²⁺?

1. Identify the ligands in the complex [Co(NH₃)₅(NO₂)]²⁺ as five NH₃ molecules and one NO₂⁻ ion.
2. Count the total number of monodentate ligands, which is 5 + 1 = 6.
3. Conclude that the secondary valency, or coordination number, is 6.

Answer: 6 (five NH₃ and one NO₂⁻ ligands, coordination number 6) — option (D).

What does phenol give with dilute HNO₃ at low temperature?

1. Dilute nitric acid acts as a mild nitrating agent, and the hydroxyl group in phenol is an ortho/para-directing group.
2. At low temperatures, electrophilic substitution occurs at the ortho and para positions, yielding a mixture of ortho-nitrophenol and para-nitrophenol.

Answer: A mixture of ortho- and para-nitrophenol — option (D).

Which statement about enantiomers is false: same chemical reactivity, same specific rotation, same melting/boiling point, or same refractive index?

1. Enantiomers are non-superimposable mirror images that share identical physical properties like melting point, boiling point, and refractive index.
2. However, they rotate plane-polarized light by equal magnitudes but in opposite directions, so the statement that they have the same specific rotation is false.

Answer: Same specific rotation is false — enantiomers rotate plane-polarised light by equal amounts in opposite directions — option (B).

What product distribution does direct nitration of aniline give?

1. Direct nitration of aniline is carried out in a strongly acidic medium (conc. HNO3 and conc. H2SO4).
2. In this acidic medium, aniline is protonated to form the anilinium ion, which is a meta-directing group due to the electron-withdrawing nature of the -NH3+ group.
3. Consequently, the reaction yields a mixture of about 51% para, 47% meta, and 2% ortho nitro derivatives, making option (C) the correct answer.

Answer: About 51% para, 47% meta and 2% ortho nitro derivatives (meta forms because the anilinium ion is produced in acid) — option (C).

Which has the lowest pKb: C₆H₅N(CH₃)₂, C₆H₅NHCH₃, C₆H₅NH₂ or p-nitroaniline (O₂N–C₆H₄–NH₂)?

1. pKb is lowest for the strongest base, the one whose nitrogen lone pair is most available.
2. In all four, the lone pair is partly delocalised into the benzene ring, which is why aryl amines are weak bases.
3. Methyl groups push electrons onto N (+I effect): two CH₃ groups help more than one, and aniline has none; the –NO₂ group in p-nitroaniline pulls electrons away and weakens it most.
4. So C₆H₅N(CH₃)₂ is the strongest base here (pKb about 8.92) and has the lowest pKb — option (A).

Answer: N,N-Dimethylaniline, C₆H₅N(CH₃)₂ (pKb about 8.92, the strongest base of the four) — option (A).

Lack of which vitamin makes red blood cells more fragile and causes muscle weakness?

1. Vitamin E acts as an antioxidant that protects cell membranes, including those of red blood cells, from oxidative damage.
2. A deficiency in Vitamin E leads to hemolytic anemia (fragile red blood cells) and muscle weakness, making option (C) the correct answer.

Answer: Vitamin E — option (C).

Assertion: An azeotrope's components cannot be separated by fractional distillation. Reason: In an azeotrope, liquid and vapour have identical composition and it boils at a fixed temperature.

1. An azeotrope is a mixture that boils at a constant temperature and has the same composition in both the liquid and vapour phases.
2. Since the vapour composition equals the liquid composition, fractional distillation cannot separate the components, making the Reason the correct explanation for the Assertion.

Answer: Both are true and the Reason correctly explains the Assertion — option (A).

Assertion: The –OH group of phenol sends incoming groups to the meta position. Reason: –OH activates the benzene ring for electrophilic substitution.

1. The assertion is false because the –OH group is an ortho/para directing group, not meta directing.
2. The reason is true because the –OH group activates the benzene ring towards electrophilic substitution via resonance.
3. Therefore, the correct option is (D) where the assertion is false but the reason is true.

Answer: Assertion is false (–OH is ortho/para directing) but the Reason is true — option (D).

Assertion: Actinoids show many different oxidation states. Reason: Actinoids are radioactive.

1. Both the assertion and the reason are factually correct statements.
2. However, the range of oxidation states in actinoids is due to the close energy levels of 5f, 6d, and 7s orbitals, not radioactivity.
3. Therefore, the reason is not the correct explanation for the assertion, making option (B) the correct choice.

Answer: Both are true but radioactivity does not explain the range of oxidation states (that comes from close 5f, 6d and 7s energies) — option (B).

Assertion: Glucose pentaacetate gives no reaction with hydroxylamine (H₂N–OH). Reason: This shows that glucose has a free –CHO group.

1. Hydroxylamine forms an oxime only with a free –CHO group.
2. In glucose pentaacetate the –CHO carbon is tied up in the cyclic (ring) form and is not free, so no oxime forms: the Assertion is true.
3. This result shows glucose has no free –CHO group, which supports its cyclic structure; the Reason says the opposite, so it is false — option (C).

Answer: Assertion is true but the Reason is false — it shows glucose has no free –CHO group (cyclic structure) — option (C).

A phenol–aniline mixture: which deviation from Raoult's law does it show and why? How does its boiling point change on mixing?

1. The phenol–aniline mixture shows a negative deviation from Raoult’s law.
2. This occurs because strong hydrogen bonds form between the –OH group of phenol and the nitrogen atom of aniline, which are stronger than the intermolecular forces in the pure components.
3. Due to these stronger attractions, the vapour pressure of the mixture decreases, causing the boiling point to rise.

Answer: Negative deviation. Hydrogen bonding between the phenolic –OH hydrogen and the nitrogen lone pair of aniline is stronger than the attractions within pure phenol or pure aniline. So vapour pressure falls and the boiling point of the mixture rises.

Give two reasons why haloarenes react poorly in nucleophilic substitution.

1. Reason 1: Resonance in the haloarene gives the carbon-halogen bond partial double bond character, making it shorter and stronger, which makes it difficult to break.
2. Reason 2: The carbon atom of the aryl group is sp2 hybridized and more electronegative than the sp3 hybridized alkyl carbon, so it holds the bonding electrons more tightly, resisting nucleophilic attack.

Answer: Any two: (1) resonance gives the C–X bond partial double-bond character, so it is harder to break; (2) the sp² carbon is more electronegative and holds the bond electrons more tightly; (3) the electron-rich ring repels the incoming nucleophile; (4) the phenyl cation that would form is unstable.

Give IUPAC names: (i) [Ag(NH₃)₂][Ag(CN)₂] (ii) K₃[Fe(C₂O₄)₃].

1. For [Ag(NH₃)₂][Ag(CN)₂], identify the cation [Ag(NH₃)₂]⁺ as diamminesilver(I) and the anion [Ag(CN)₂]⁻ as dicyanidoargentate(I), giving the full name diamminesilver(I) dicyanidoargentate(I).
2. For K₃[Fe(C₂O₄)₃], identify the cation as potassium and the anion [Fe(C₂O₄)₃]³⁻ as trioxalatoferrate(III), giving the full name potassium trioxalatoferrate(III).

Answer: (i) Diamminesilver(I) dicyanidoargentate(I). (ii) Potassium trioxalatoferrate(III).

(i) Suggest a chemical test proving [Co(NH₃)₅SO₄]Cl and [Co(NH₃)₅Cl]SO₄ are ionisation isomers. (ii) Explain the chelate effect with one example.

1. (i) Add AgNO₃ solution to both complexes: [Co(NH₃)₅SO₄]Cl gives a white precipitate of AgCl, while [Co(NH₃)₅Cl]SO₄ does not, proving they are ionisation isomers.
2. (ii) The chelate effect states that complexes formed by di- or polydentate ligands are more stable than those with monodentate ligands, e.g., [Co(en)₃]³⁺ is more stable than [Co(NH₃)₆]³⁺.

Answer: (i) With AgNO₃ solution, [Co(NH₃)₅SO₄]Cl gives a white precipitate of AgCl but [Co(NH₃)₅Cl]SO₄ does not; or with BaCl₂, [Co(NH₃)₅Cl]SO₄ gives white BaSO₄ but the other does not. (ii) Chelate effect: complexes formed by a di- or polydentate ligand binding one metal ion are extra stable, e.g. [Co(en)₃]³⁺.

Distinguish: (i) peptide linkage vs glycosidic linkage (ii) essential vs non-essential amino acids.

1. (i) A peptide linkage is an amide bond (–CO–NH–) formed between the –COOH group of one amino acid and the –NH₂ group of another.
2. (i) A glycosidic linkage is an ether bond that joins two monosaccharide units through an oxygen atom.
3. (ii) Essential amino acids cannot be synthesized by the body and must be obtained from food.
4. (ii) Non-essential amino acids can be synthesized by the body.

Answer: (i) A peptide linkage is the amide bond –CO–NH– formed between the –COOH of one amino acid and the –NH₂ of another; a glycosidic linkage joins two monosaccharide units through an oxygen atom. (ii) Essential amino acids cannot be made by the body and must come from food; non-essential ones are synthesised in the body.

The elementary (one-step) reaction 2A + B → 2C: how does the rate change when the vessel volume is cut to one-third? Does the order change?

1. For the elementary reaction 2A + B → 2C, the rate law is Rate = k[A]²[B].
2. When the volume is cut to one-third, the concentration of each reactant triples, so [A] becomes 3[A] and [B] becomes 3[B].
3. Substituting these into the rate law gives New Rate = k(3[A])²(3[B]) = k(9[A]²)(3[B]) = 27k[A]²[B].
4. Thus, the rate becomes 27 times the original rate, while the order of the reaction remains 3.

Answer: Rate = k[A]²[B]. One-third volume triples each concentration, so the new rate = k(3[A])²(3[B]), i.e. the rate becomes 27 times. The order (3) stays the same.

First order gas reaction C₂H₅Cl(g) → C₂H₄(g) + HCl(g): total pressure is 0.30 atm at t = 0 s and 0.50 atm at t = 30 s. Find k. (log 3 = 0.48)

1. Let the initial pressure of C₂H₅Cl be P₀ = 0.30 atm.
2. At time t, let x atm of C₂H₅Cl have decomposed. The total pressure Pt = P₀ + x.
3. Substitute the given values: 0.50 = 0.30 + x, so x = 0.20 atm.
4. The pressure of the reactant remaining at t = 30 s is P = P₀ - x = 0.30 - 0.20 = 0.10 atm.
5. Use the first-order rate constant formula: k = (2.303/t) log(P₀/P).
6. Substitute values: k = (2.303/30) log(0.30/0.10) = (2.303/30) log(3).
7. Calculate k: k = (2.303/30) × 0.48 ≈ 0.0368 s⁻¹, which rounds to 0.037 s⁻¹.

Answer: If x atm reacts, Pt = Pi + x, so x = 0.20 atm and the reactant left = 2Pi − Pt = 0.10 atm. k = (2.303/30) log(0.30/0.10) = (2.303/30) × 0.48 ≈ 0.0368, i.e. k ≈ 0.037 s⁻¹.

Note: CBSE's marking scheme prints the unit of k as atm⁻¹; that is a slip — a first order rate constant has unit s⁻¹ (time⁻¹).

(i) Why is Kc linked to E°cell rather than Ecell? (ii) Metals A and B have E° = −0.24 V and +0.80 V. Which one releases H₂ from dilute H₂SO₄? (iii) Give the cell reaction of a lead storage battery while it is being charged.

1. At equilibrium, the cell potential Ecell becomes zero, so the Nernst equation simplifies to E°cell = (0.059/n) log Kc, linking Kc to E°cell.
2. Metal A has a negative standard electrode potential (−0.24 V), which is lower than the hydrogen electrode (0 V), so it displaces hydrogen from dilute H₂SO₄.
3. During charging, the discharge reaction is reversed: 2PbSO₄(s) + 2H₂O(l) → Pb(s) + PbO₂(s) + 2H₂SO₄(aq).

Answer: (i) At equilibrium Ecell = 0, so the Nernst equation relates Kc to E°cell only. (ii) Metal A (negative E°, it can reduce H⁺). (iii) 2PbSO₄(s) + 2H₂O(l) → Pb(s) + PbO₂(s) + 2H₂SO₄(aq).

What kind of battery is the mercury cell, what advantage does it have over a dry cell, and what is its overall reaction?

1. The mercury cell is a primary cell, meaning it is a non-rechargeable battery.
2. Its main advantage over a dry cell is that it provides a constant voltage of about 1.35 V throughout its entire life.
3. The overall reaction for the mercury cell is Zn(Hg) + HgO(s) → ZnO(s) + Hg(l).

Answer: It is a primary cell. Its voltage stays constant (about 1.35 V) through its whole life. Overall: Zn(Hg) + HgO(s) → ZnO(s) + Hg(l).

0.61 g benzoic acid (M = 122 g mol⁻¹) in 5 g CS₂ dimerises 88%. CS₂ boils at 46.2 °C, Kb = 2.3 K kg mol⁻¹. Find the solution's boiling point.

1. Calculate the molality of the solution: m = (0.61 / 122) × (1000 / 5) = 1.0 mol kg⁻¹.
2. Determine the van't Hoff factor for dimerisation: i = 1 − α/2 = 1 − 0.44 = 0.56.
3. Calculate the boiling point elevation: ΔTb = i Kb m = 0.56 × 2.3 × 1.0 = 1.288 K.
4. Find the boiling point of the solution: T = 46.2 + 1.288 = 47.488 °C (about 320.64 K).

Answer: For dimerisation i = 1 − α/2 = 1 − 0.44 = 0.56. ΔTb = i Kb m = 0.56 × 2.3 × (0.61/122) × (1000/5) = 1.288 K. Boiling point = 46.2 + 1.288 = 47.488 °C (about 320.64 K).

Give the reaction of D-glucose with (a) HI (b) bromine water (c) concentrated HNO₃.

1. (a) Prolonged heating with HI reduces glucose completely to n-hexane, which shows its six carbons form a straight chain: OHC(CHOH)₄CH₂OH → CH₃CH₂CH₂CH₂CH₂CH₃.
2. (b) Bromine water is a mild oxidant; it oxidises only the –CHO group to –COOH, giving gluconic acid: OHC(CHOH)₄CH₂OH → HOOC(CHOH)₄CH₂OH.
3. (c) Concentrated HNO₃ is a stronger oxidant; it oxidises both the –CHO group and the terminal –CH₂OH group to –COOH, giving saccharic acid: OHC(CHOH)₄CH₂OH → HOOC(CHOH)₄COOH.

Answer: (a) Prolonged heating with HI: OHC(CHOH)₄CH₂OH → CH₃CH₂CH₂CH₂CH₂CH₃ (n-hexane). (b) Br₂ water: OHC(CHOH)₄CH₂OH → HOOC(CHOH)₄CH₂OH (gluconic acid). (c) Conc. HNO₃: OHC(CHOH)₄CH₂OH → HOOC(CHOH)₄COOH (saccharic acid).

Explain: (a) carboxylic acids boil higher than alcohols of similar molar mass; (b) α-hydrogens of aldehydes and ketones are acidic; (c) ammonia and its derivatives do not add to C=O in strongly acidic medium.

1. Carboxylic acids form stable dimers via strong intermolecular hydrogen bonding between the carbonyl oxygen and the hydroxyl hydrogen of adjacent molecules.
2. This dimerization increases the effective molecular size and intermolecular forces, requiring more energy to separate molecules, thus raising the boiling point compared to alcohols of similar molar mass.
3. The carbonyl group in aldehydes and ketones is electron-withdrawing, which stabilizes the negative charge on the alpha-carbon in the conjugate base.
4. The resulting enolate ion is resonance-stabilized, with the negative charge delocalized between the alpha-carbon and the oxygen atom, making the alpha-hydrogens acidic.
5. In a strongly acidic medium, ammonia and its derivatives are protonated to form ammonium ions, which lack a lone pair of electrons.
6. Without a lone pair, these species cannot act as nucleophiles to attack the electrophilic carbonyl carbon, preventing the addition reaction.

Answer: (a) Carboxylic acids associate strongly through intermolecular hydrogen bonds (they form dimers). (b) The carbonyl group strongly withdraws electrons and the conjugate base (enolate) is resonance-stabilised. (c) In strong acid the nucleophile (NH₃ or its derivative) gets protonated and loses its lone pair.

Give the equations for Kolbe's reaction, the Reimer–Tiemann reaction (both on phenol) and Friedel–Crafts acylation of anisole.

1. Kolbe's reaction: C₆H₅OH + NaOH → C₆H₅O⁻Na⁺ + H₂O; sodium phenoxide with CO₂ (then H⁺) → o-HOC₆H₄COOH (salicylic acid).
2. Reimer–Tiemann reaction: C₆H₅OH + CHCl₃ + aq. NaOH, then H⁺ → o-HOC₆H₄CHO (salicylaldehyde, 2-hydroxybenzaldehyde).
3. Friedel–Crafts acylation: C₆H₅OCH₃ + CH₃COCl, with anhydrous AlCl₃ → CH₃OC₆H₄COCH₃ + HCl, a mixture of 2-methoxyacetophenone (minor) and 4-methoxyacetophenone (major).

Answer: (a) Reimer–Tiemann: phenol + CHCl₃ + aq. NaOH, then H⁺ → salicylaldehyde (2-hydroxybenzaldehyde, o-HOC₆H₄CHO). (b) Kolbe: phenol + NaOH → sodium phenoxide; with CO₂ then H⁺ → salicylic acid (o-HOC₆H₄COOH). (c) C₆H₅OCH₃ + CH₃COCl with anhydrous AlCl₃ → 2-methoxyacetophenone (minor) + 4-methoxyacetophenone (major), i.e. CH₃OC₆H₄COCH₃.

X (C₄H₉Br) gives an alcohol with aqueous KOH at a rate depending only on [X]. Y, an optically active isomer of X, reacts with aqueous KOH at a rate depending on both [Y] and [KOH]. (a) Structures of X and Y. (b) Which racemises and why? (c) Which gives inversion of configuration and why?

1. Rate depends only on [X], so X reacts by SN1; that is fastest for a tertiary halide: X = 2-bromo-2-methylpropane, (CH₃)₃CBr.
2. Rate depends on [Y] and [KOH], so Y reacts by SN2; Y is optically active, so it needs a chiral carbon: Y = 2-bromobutane, CH₃CH(Br)CH₂CH₃.
3. (b) The SN1 path (X) racemises: the planar carbocation can be attacked from either face equally, so a chiral centre would give both forms in equal amounts.
4. (c) The SN2 path (Y) gives inversion: OH⁻ attacks from the side opposite to Br, so the configuration flips.

Answer: (a) X = 2-bromo-2-methylpropane, (CH₃)₃CBr; Y = 2-bromobutane, CH₃CH(Br)CH₂CH₃. (b) X — it reacts by SN1 through a planar carbocation that can be attacked from either face. (c) Y — it reacts by SN2, with the nucleophile attacking from the back side, so configuration is inverted.

Note: Part (b) is loosely set: X is achiral, so racemisation only shows up for a chiral substrate reacting by SN1. CBSE's marking scheme gives X and also gives full marks for Y.

In water, why is CH₃NH₂ a stronger base than (CH₃)₃N?

1. Basicity in water depends on the combined effects of the +I effect, solvation of the ammonium ion, and steric hindrance.
2. Trimethylamine is a weaker base because its conjugate acid, (CH₃)₃NH⁺, is crowded and poorly solvated due to steric hindrance.

Answer: Basicity in water is a combined result of the +I effect, solvation (hydrogen bonding) of the ammonium ion and steric hindrance; (CH₃)₃NH⁺ is poorly solvated and crowded, so trimethylamine is weaker.

Identify A and B: CH₃CONH₂ with NaOBr gives A; A with C₆H₅COCl in presence of base gives B.

1. Step 1: CH₃CONH₂ reacts with NaOBr via Hoffmann bromamide degradation to lose one carbon atom, forming A as CH₃NH₂ (methanamine).
2. Step 2: A (CH₃NH₂) reacts with C₆H₅COCl in the presence of a base via nucleophilic acyl substitution to form B as CH₃NHCOC₆H₅ (N-methylbenzamide).

Answer: A = CH₃NH₂ (methanamine, by Hoffmann bromamide degradation); B = CH₃NHCOC₆H₅ (N-methylbenzamide).

Amine X, C₃H₉N, gives a Hinsberg-reagent product that does not dissolve in alkali. Identify X.

1. Hinsberg's reagent (C₆H₅SO₂Cl) gives a primary amine a sulphonamide that dissolves in alkali (it still has an acidic N–H), gives a secondary amine a sulphonamide with no N–H that does not dissolve in alkali, and does not react with a tertiary amine.
2. So X is a secondary amine. The only secondary amine with formula C₃H₉N is CH₃CH₂NHCH₃ (N-methylethanamine).

Answer: X is a secondary amine: CH₃CH₂NHCH₃ (N-methylethanamine).

How is aniline changed into benzonitrile?

1. Aniline is first treated with sodium nitrite and hydrochloric acid at 0–5 °C to form benzene diazonium chloride.
2. The diazonium salt is then heated with cuprous cyanide (CuCN) to replace the diazonium group with a cyano group, yielding benzonitrile.

Answer: Diazotise: C₆H₅NH₂ + NaNO₂ + HCl at 0–5 °C → C₆H₅N₂⁺Cl⁻; then treat with CuCN → C₆H₅CN.

Why is aniline's –NH₂ acetylated before nitration?

1. Acetylation converts the strongly activating -NH2 group into a less activating -NHCOCH3 group, which prevents the formation of the meta-directing anilinium ion in the acidic medium.
2. This controlled reactivity ensures that nitration occurs predominantly at the para position, yielding p-nitroaniline as the main product.

Answer: Acetylation controls the reactivity so p-nitroaniline becomes the main product, and it stops formation of the meta-directing anilinium ion.

In an octahedral crystal field, which d-orbitals rise in energy as ligands approach, and why?

1. In an octahedral complex, six ligands approach the central metal ion along the x, y, and z axes.
2. The d(x²−y²) and d(z²) orbitals (the eg set) have their lobes pointing directly along these axes.
3. Because these orbitals point toward the incoming ligands, they experience greater electron-electron repulsion compared to the d(xy), d(xz), and d(yz) orbitals.
4. This increased repulsion causes the energy of the eg orbitals to rise relative to the t2g set.

Answer: dx²−y² and dz² (the eg set). Ligands approach along the axes, where these orbitals point, so they are repelled more.

By crystal field theory, give the d-electron configuration of the metal ion in (i) [CoF₆]³⁻ (ii) [Co(NH₃)₆]³⁺. (Co = 27)

1. Oxidation state of Co in both complexes is +3, giving a 3d⁶ configuration.
2. In [CoF₆]³⁻, F⁻ is a weak field ligand, so electrons remain unpaired: t2g⁴ eg².
3. In [Co(NH₃)₆]³⁺, NH₃ acts as a strong field ligand, causing pairing: t2g⁶ eg⁰.

Answer: Both contain Co³⁺ (3d⁶). (i) F⁻ is weak field: t2g⁴ eg². (ii) NH₃ is strong field here: t2g⁶ eg⁰.

Both tetrahedral, yet [NiCl₄]²⁻ is paramagnetic and [Ni(CO)₄] diamagnetic. Explain. (Ni = 28)

1. Ni in [NiCl₄]²⁻ is in +2 oxidation state with configuration 3d⁸; Cl⁻ is a weak field ligand so electrons remain unpaired, making it paramagnetic.
2. Ni in [Ni(CO)₄] is in 0 oxidation state with configuration 3d¹⁰; CO is a strong field ligand causing all electrons to pair up, making it diamagnetic.

Answer: Cl⁻ is a weak field ligand and does not pair the 3d electrons of Ni²⁺ (3d⁸, two unpaired), so it is paramagnetic. CO is a strong field ligand; in Ni(0) the electrons pair up (3d¹⁰), so it is diamagnetic.

Give the hybridisation and magnetic nature of [Fe(CN)₆]³⁻. (Fe = 26)

1. Oxidation state of Fe is +3, so electronic configuration is 3d⁵.
2. CN⁻ is a strong field ligand causing pairing of electrons, resulting in d²sp³ hybridisation.
3. One unpaired electron remains, making the complex paramagnetic.

Answer: Fe³⁺ (3d⁵) with strong field CN⁻: d²sp³ hybridisation, one unpaired electron, so paramagnetic.

(i) Compound X is C₅H₁₀O. Give a structure of X if: (I) Tollens' negative, iodoform positive; (II) Tollens' and iodoform negative but it shows aldol condensation; (III) it shows Cannizzaro reaction. (3 marks) (ii) Convert to benzoic acid: (I) acetophenone (II) ethylbenzene. (2 marks)

1. Part (i)(I): The compound is a methyl ketone (CH3CO-R) because it is Tollens' negative (not an aldehyde) and iodoform positive. The structure is Pentan-2-one, CH3COCH2CH2CH3.
2. Part (i)(II): The compound is a ketone that is Tollens' and iodoform negative but shows aldol condensation, meaning it has alpha-hydrogens but no CH3CO- group. The structure is Pentan-3-one, CH3CH2COCH2CH3.
3. Part (i)(III): The compound shows the Cannizzaro reaction, indicating it is an aldehyde with no alpha-hydrogens. The structure is 2,2-Dimethylpropanal, (CH3)3CCHO.
4. Part (ii)(I): Acetophenone is converted to benzoic acid via the haloform reaction using NaOI, followed by acidification with H+ to yield C6H5COOH.
5. Part (ii)(II): Ethylbenzene is oxidized using alkaline KMnO4 (or KMnO4/KOH) under heat, followed by acidification with H+ to yield benzoic acid, C6H5COOH.

Answer: (i) (I) Pentan-2-one, CH₃COCH₂CH₂CH₃ (a methyl ketone). (II) Pentan-3-one, CH₃CH₂COCH₂CH₃ (has α-H, no CH₃CO– group). (III) 2,2-Dimethylpropanal, (CH₃)₃CCHO (aldehyde with no α-H). (ii) (I) C₆H₅COCH₃ + NaOI (iodoform reaction), then H⁺ → C₆H₅COOH; or KMnO₄/KOH, heat, then H₃O⁺. (II) C₆H₅CH₂CH₃ with alkaline KMnO₄, then H⁺ → C₆H₅COOH.

Note: In part (i), the Tollens-negative, iodoform-positive compound can also be 3-methylbutan-2-one, CH₃COCH(CH₃)₂, another methyl ketone.

(i) Structure of benzaldehyde 2,4-dinitrophenylhydrazone. (ii) Order di-tert-butyl ketone, acetaldehyde, acetone by increasing reactivity with HCN. (iii) A simple test to tell benzoic acid from ethyl benzoate. (iv) Reagent for ethanenitrile → ethanal. (v) Product X when cyclohexanol is treated with CrO₃.

1. (i) Draw the structure of benzaldehyde 2,4-dinitrophenylhydrazone as C6H5CH=N-NH-C6H3(NO2)2, with the nitro groups at positions 2 and 4 on the second benzene ring.
2. (ii) The order of increasing reactivity with HCN is di-tert-butyl ketone < acetone < acetaldehyde, because steric hindrance decreases from left to right.
3. (iii) Add NaHCO3 solution to the samples; benzoic acid produces brisk effervescence of CO2, while ethyl benzoate shows no reaction.
4. (iv) The reagent for converting ethanenitrile to ethanal is DIBAL-H followed by H3O+ (or SnCl2/HCl followed by H3O+).
5. (v) The product X formed when cyclohexanol is treated with CrO3 is cyclohexanone (C6H10O), which contains a ring C=O group.

Answer: (i) C₆H₅CH=N–NH–C₆H₃(NO₂)₂, the NO₂ groups at positions 2 and 4 of the second ring. (ii) Di-tert-butyl ketone < acetone < acetaldehyde. (iii) Add NaHCO₃ solution: benzoic acid gives brisk effervescence of CO₂, ethyl benzoate does not. (iv) DIBAL-H followed by H₃O⁺ (or SnCl₂/HCl then H₃O⁺, Stephen reduction). (v) X = cyclohexanone (C₆H₁₀O, ring C=O).

(i) E°(M²⁺/M) in V: V −1.18, Cr −0.91, Mn −1.18, Fe −0.44, Co −0.28, Ni −0.25, Cu +0.34. (I) Why is the trend irregular? (II) Why is Cu's value unusually positive? (III) Why is Mn's value so negative? (3 marks) (ii) Ionic equations for KMnO₄ oxidising I⁻ in acidic and in alkaline solution. (2 marks)

1. (i) (I) The irregular trend in E° values arises because the ionisation enthalpies (ΔiH₁ + ΔiH₂) and sublimation enthalpies of the metals vary irregularly across the series.
2. (i) (II) The value for Cu is unusually positive because it has a high enthalpy of atomisation, and the hydration enthalpy of Cu²⁺ is too low to compensate for this energy cost.
3. (i) (III) The value for Mn is very negative because the Mn²⁺ ion is exceptionally stable due to its half-filled 3d⁵ electronic configuration.
4. (ii) In acidic solution, the balanced ionic equation is: 2MnO₄⁻ + 10I⁻ + 16H⁺ → 2Mn²⁺ + 5I₂ + 8H₂O.
5. (ii) In alkaline (or neutral) solution, the balanced ionic equation is: 2MnO₄⁻ + I⁻ + H₂O → 2MnO₂ + IO₃⁻ + 2OH⁻.

Answer: (i) (I) Ionisation enthalpies (ΔiH₁ + ΔiH₂) and sublimation enthalpies vary irregularly. (II) Cu has a high enthalpy of atomisation and a hydration enthalpy of Cu²⁺ too low to compensate. (III) Mn²⁺ is very stable because of its half-filled 3d⁵ configuration. (ii) Acidic: 2MnO₄⁻ + 10I⁻ + 16H⁺ → 2Mn²⁺ + 5I₂ + 8H₂O. Alkaline (or neutral): 2MnO₄⁻ + I⁻ + H₂O → 2MnO₂ + IO₃⁻ + 2OH⁻.

(i) Name a lanthanoid that shows (I) +4 and (II) +2 oxidation state. (1 mark) (ii) Why are transition metals good catalysts? (1) (iii) Why does Cr melt higher than Mn? (1) (iv) What happens to acidified KMnO₄ solution left standing for some time? Equation and type of reaction. (2)

1. For part (i), name Cerium for the +4 state and Europium or Ytterbium for the +2 state.
2. For part (ii), state that transition metals are good catalysts due to variable oxidation states, complex formation, and large surface area.
3. For part (iii), explain that Cr has a higher melting point because it uses more d and s electrons in metallic bonding.
4. Note that Mn has a stable 3d5 configuration, so its d-electrons participate less in bonding, lowering its melting point.
5. For part (iv), state that acidified KMnO4 slowly decomposes to form MnO2 and O2.
6. Write the equation: 4MnO4- + 4H+ → 4MnO2 + 3O2 + 2H2O.
7. Identify the reaction as a redox process where MnO4- is reduced and oxygen is oxidized.

Answer: (i) (I) Cerium (Ce⁴⁺). (II) Europium or ytterbium (Eu²⁺/Yb²⁺). (ii) They show variable oxidation states, form complexes and offer large surface area. (iii) Cr uses more (n−1)d and ns electrons in metallic bonding (Mn's stable 3d⁵ takes part less). (iv) It slowly decomposes, giving MnO₂ and O₂: 4MnO₄⁻ + 4H⁺ → 4MnO₂ + 3O₂ + 2H₂O. It is a redox reaction (MnO₄⁻ is reduced, oxygen is oxidised).

Note: CBSE's marking scheme also accepts the half-reaction MnO₄⁻ + 4H⁺ + 3e⁻ → MnO₂ + 2H₂O, called a reduction, for part (iv).

A cell at 298 K has a Mg anode dipping in 0.01 M Mg²⁺ and an Ag cathode dipping in 0.001 M Ag⁺. Standard potentials: Ag⁺/Ag +0.80 V, Mg²⁺/Mg −2.37 V. Take F = 96500 C mol⁻¹ and log 10 = 1. Find emf and ΔG.

1. Write the balanced cell reaction: Mg(s) + 2Ag⁺(aq) → Mg²⁺(aq) + 2Ag(s), so n = 2.
2. Calculate standard cell potential: E°cell = E°cathode − E°anode = 0.80 − (−2.37) = 3.17 V.
3. Write the Nernst equation: Ecell = E°cell − (0.059/n) log([Mg²⁺]/[Ag⁺]²).
4. Substitute concentrations: [Mg²⁺] = 0.01 M and [Ag⁺] = 0.001 M, so [Ag⁺]² = (0.001)² = 10⁻⁶.
5. Calculate the reaction quotient term: log(0.01 / 10⁻⁶) = log(10⁴) = 4.
6. Calculate Ecell: Ecell = 3.17 − (0.059/2) × 4 = 3.17 − 0.0295 × 4 = 3.17 − 0.118 = 3.052 V.
7. Calculate Gibbs free energy: ΔG = −nFEcell = −2 × 96500 × 3.052 = −589036 J mol⁻¹ ≈ −589 kJ mol⁻¹.

Answer: Mg + 2Ag⁺ → Mg²⁺ + 2Ag, n = 2. E°cell = 0.80 − (−2.37) = 3.17 V. Ecell = 3.17 − (0.059/2) log[0.01/(0.001)²] = 3.17 − 0.0295 × 4 = 3.052 V and ΔG = -589036 J mol⁻¹ (ΔG = −nFEcell = −2 × 96500 × 3.052, about −589 kJ mol⁻¹).

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

Cell reaction: 2AgCl(s) + H₂ (at 0.4 atm) gives 2Ag(s) with H⁺ at 0.1 M and Cl⁻ at 0.2 M (2 mol of each ion). ΔG° = −43500 J mol⁻¹. Find the emf at 25 °C, taking F = 96500 C mol⁻¹ and log 10 = 1.

1. Identify the number of electrons transferred (n) from the balanced reaction: 2AgCl + H₂ → 2Ag + 2H⁺ + 2Cl⁻, so n = 2.
2. Calculate the standard cell potential using E° = −ΔG°/nF: E° = 43500 / (2 × 96500) ≈ 0.225 V.
3. Write the expression for the reaction quotient Q: Q = [H⁺]²[Cl⁻]² / p(H₂).
4. Substitute the given concentrations and pressure into Q: Q = (0.1)²(0.2)² / 0.4 = 0.0004 / 0.4 = 10⁻³.
5. Apply the Nernst equation at 25 °C: E = E° − (0.059/n) log Q.
6. Substitute values into the Nernst equation: E = 0.225 − (0.059/2) log(10⁻³).
7. Calculate the final emf: E = 0.225 − (0.0295)(−3) = 0.225 + 0.0885 = 0.3135 V.

Answer: n = 2. E°cell = −ΔG°/nF = 43500/(2 × 96500) ≈ 0.225 V. Q = [H⁺]²[Cl⁻]²/p(H₂) = (0.1)²(0.2)²/0.4 = 10⁻³. Ecell = 0.225 − (0.059/2) log 10⁻³ = 0.225 + 0.0885 = 0.3135 V.

Note: Uses 2.303RT/F = 0.059 V, which the paper does not print. E° is rounded to 0.225 V as in CBSE's marking scheme; unrounded 0.2254 V gives 0.3139 V.