CBSE Class 12 Chemistry Board Paper 2024 (56/1/1): Solutions

Among Fe (Z = 26), Mn (Z = 25), Cr (Z = 24) and V (Z = 23), which should have the highest third ionisation enthalpy?

1. Write the electronic configuration of Mn (Z = 25) as [Ar] 3d⁵ 4s².
2. Remove two electrons to form Mn²⁺, resulting in the configuration [Ar] 3d⁵.
3. Recognize that the 3d⁵ configuration is half-filled and exceptionally stable due to exchange energy and symmetry.
4. Conclude that removing a third electron from this stable half-filled shell requires the highest energy, so Mn has the highest third ionisation enthalpy.

Answer: Manganese: Mn²⁺ is 3d⁵ (half-filled, extra stable), so removing a third electron is hardest — option (B).

Which alcohol gives a ketone when oxidised by CrO₃: (CH₃)₂CHCH₂OH, CH₃CH₂CH₂OH, (CH₃)₃COH or CH₃CH₂CH(OH)CH₃?

1. CrO₃ oxidises a primary alcohol to an aldehyde (or further to an acid), a secondary alcohol to a ketone, and leaves a tertiary alcohol unchanged under normal conditions.
2. (CH₃)₂CHCH₂OH and CH₃CH₂CH₂OH are primary, and (CH₃)₃COH is tertiary.
3. In CH₃CH₂CH(OH)CH₃ the –OH carbon is bonded to two other carbons, so it is secondary (butan-2-ol).
4. Butan-2-ol is oxidised to the ketone butan-2-one, CH₃COCH₂CH₃ — option (D).

Answer: CH₃CH₂CH(OH)CH₃ (butan-2-ol), a secondary alcohol, which is oxidised to butan-2-one — option (D).

DNA is made of β-D-2-deoxyribose, a heterocyclic base and one more component. Name the third component.

1. A DNA nucleotide consists of three components: a pentose sugar, a nitrogenous base, and a phosphate group.
2. The third component is phosphoric acid (or phosphate), which forms the backbone of the DNA strand.

Answer: Phosphoric acid (phosphate) — option (B).

What can be said about the van't Hoff factor i of an electrolyte that associates in a solvent?

1. The van't Hoff factor i is defined as the ratio of the observed colligative property to the normal colligative property, which corresponds to the ratio of the actual number of particles to the initial number of particles.
2. When an electrolyte associates in a solvent, multiple molecules combine to form a single larger particle, thereby reducing the total number of solute particles in the solution.
3. Since the number of particles decreases upon association, the ratio of actual to initial particles becomes less than 1, meaning i is always less than 1.

Answer: Association lowers the number of particles, so i is always less than 1 — option (D).

Write the differential rate expression linking d[Y]/dt and d[P]/dt for X + 2Y → P.

1. For the reaction X + 2Y → P, the stoichiometric coefficient of Y is 2 and that of P is 1.
2. The differential rate expression is Rate = −(1/2) d[Y]/dt = +d[P]/dt, which matches option (A).

Answer: Rate = −(1/2) d[Y]/dt = +d[P]/dt — option (A).

Which of the four bromides drawn in the paper reacts fastest by the SN1 path?

Uses a figure printed in CBSE's paper — what the figure shows: Options drawn as rings: (A) cyclohexene ring with Br on the carbon next to the double bond (allylic bromide); (B) cyclohexyl-CH₂Br; (C) bromocyclohexane; (D) bromobenzene.

1. SN1 reactions are fastest when the carbocation intermediate is most stable.
2. Option (A) forms an allylic carbocation which is resonance-stabilized, making it the most stable among the options.
3. Therefore, 3-bromocyclohexene (Option A) reacts fastest by the SN1 path.

Answer: 3-Bromocyclohexene (Br on the carbon next to the ring C=C), because it forms a resonance-stabilised allylic carbocation — option (A).

What does CH₃COOH give with PCl₅?

1. Acetic acid (CH₃COOH) reacts with phosphorus pentachloride (PCl₅) to replace the hydroxyl group with a chlorine atom.
2. The reaction produces acetyl chloride (CH₃COCl) along with phosphorus oxychloride (POCl₃) and hydrogen chloride (HCl).

Answer: Acetyl chloride, CH₃COCl (CH₃COOH + PCl₅ → CH₃COCl + POCl₃ + HCl) — option (C).

An aldehyde reacting with HCN to give a cyanohydrin is which type of reaction?

1. Identify the reagent HCN as a source of the cyanide ion (CN⁻), which acts as a nucleophile.
2. Recognize that the carbonyl carbon in an aldehyde is electrophilic, so the reaction involves the nucleophile attacking the carbon atom.
3. Conclude that this mechanism is a nucleophilic addition reaction, corresponding to option (A).

Answer: Nucleophilic addition (CN⁻ attacks the carbonyl carbon) — option (A).

For a plot of log k versus 1/T (Arrhenius equation), state the slope and the intercept.

1. Write the Arrhenius equation in logarithmic form: log k = log A − Ea/(2.303 RT).
2. Compare this with the straight-line equation y = mx + c, where y = log k and x = 1/T, to identify the slope as −Ea/(2.303 R) and the intercept as log A.

Answer: log k = log A − Ea/(2.303 RT), so slope = −Ea/(2.303 R) and intercept = log A — option (D).

Name the reaction in which an alkyl halide and a sodium alkoxide give an ether.

1. The reaction between an alkyl halide and a sodium alkoxide to form an ether is known as the Williamson synthesis.
2. Therefore, the correct option is (C).

Answer: Williamson synthesis — option (C).

Order (CH₃)₃C–OH, (CH₃)₂CH–OH and CH₃CH₂–OH by ease of dehydration with conc. H₂SO₄.

1. Dehydration of alcohols with conc. H₂SO₄ proceeds via an E1 mechanism where the rate-determining step is the formation of a carbocation.
2. The ease of dehydration depends on the stability of the carbocation formed, following the order: tertiary > secondary > primary.
3. Therefore, the order of ease of dehydration is (CH₃)₃C–OH > (CH₃)₂CH–OH > CH₃CH₂–OH.

Answer: Tertiary > secondary > primary (carbocation stability): (CH₃)₃C–OH > (CH₃)₂CH–OH > CH₃CH₂–OH — option (A).

In glucose, which two groups react to give the cyclic hemiacetal (pyranose ring)?

1. The aldehyde group at C-1 reacts with the hydroxyl group at C-5 to form the six-membered pyranose ring.
2. Option (B) is correct because it identifies the C-1 aldehyde and C-5 hydroxyl as the reacting groups.

Answer: The –CHO group (C-1) and the –OH group on C-5 — option (B).

Assertion: adding NaCl to water lowers its freezing point. Reason: NaCl dissociates in water. Choose the A-R code.

1. Both the assertion and the reason are factually true statements.
2. The freezing point depression is caused by the presence of a non-volatile solute, not specifically by dissociation.
3. Dissociation of NaCl increases the number of particles, which makes the depression larger, but it is not the fundamental reason for the phenomenon.
4. Therefore, the correct code is (B), where both are true but the reason is not the correct explanation.

Answer: Both statements are true, but the freezing point falls because a non-volatile solute is present; dissociation only makes the fall larger, so R is not the correct explanation — option (B).

Note: This pair is debatable: some teachers take dissociation as the explanation and pick (A), but CBSE's marking scheme gives (B).

Assertion: Zr and Hf are hard to separate. Reason: lanthanoid contraction makes their radii almost equal. Choose the A-R code.

1. Both the assertion and the reason are factually correct.
2. The reason correctly explains the assertion because the nearly equal atomic radii of Zr and Hf result in similar chemical properties, making them difficult to separate.

Answer: Both are true and R explains A (nearly equal size gives near-identical chemistry) — option (A).

Assertion: ethanoic acid has a lower pKa than ClCH₂COOH. Reason: the −I effect of Cl makes ClCH₂COOH more acidic. Choose the A-R code.

1. Assertion is false because the −I effect of the chlorine atom in ClCH₂COOH withdraws electron density, stabilizing the conjugate base and making it a stronger acid than ethanoic acid, which results in a lower pKa.
2. Reason is true because the electron-withdrawing inductive effect (−I) of the chlorine atom increases the acidity of the carboxylic acid.
3. Since the Assertion is false and the Reason is true, the correct code is (D).

Answer: A is false (ClCH₂COOH is the stronger acid, so it has the lower pKa); R is true — option (D).

Assertion: aniline is more basic than NH₃. Reason: resonance makes the N lone pair of aniline less available for a proton. Choose the A-R code.

1. In aniline, the nitrogen lone pair delocalizes into the benzene ring via resonance, reducing its availability for protonation.
2. This resonance stabilization makes aniline a weaker base than ammonia, where the lone pair is fully available.
3. Thus, Assertion is false and Reason is true, and the Reason correctly explains why aniline is less basic than NH₃, so option (D) is correct.

Answer: A is false (aniline is weaker than ammonia); R is true and is the reason aniline is weaker — option (D).

Find the electrode potential of Fe²⁺(0.01 M)/Fe at 298 K. E°(Fe²⁺/Fe) = −0.45 V, log 10 = 1.

1. Write the reduction half-reaction: Fe²⁺ + 2e⁻ → Fe, so n = 2.
2. Apply the Nernst equation: E = E° − (0.059/n) log(1/[Fe²⁺]).
3. Substitute values: E = −0.45 − (0.059/2) log(1/0.01).
4. Simplify the log term: log(100) = 2, so (0.059/2) × 2 = 0.059.
5. Calculate final potential: E = −0.45 − 0.059 = −0.509 V.

Answer: Fe²⁺ + 2e⁻ → Fe, n = 2. E = E° − (0.059/2) log(1/[Fe²⁺]) = −0.45 − 0.0295 × log 100 = −0.45 − 0.059 = −0.509 V.

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

What is molecularity? Give one condition under which a bimolecular reaction behaves as first order.

1. Molecularity is defined as the number of reacting species (atoms, ions, or molecules) that must collide simultaneously in an elementary step.
2. A bimolecular reaction behaves as first order when one reactant is present in large excess, such as water in the hydrolysis of an ester.

Answer: Molecularity is the number of reacting species (atoms, ions or molecules) that must collide at the same time in an elementary step. A bimolecular reaction becomes (pseudo) first order when one reactant is taken in large excess, e.g. water in the hydrolysis of an ester.

What is formed when D-glucose reacts with (a) HI (prolonged heating) and (b) conc. HNO₃?

1. (a) On long heating with HI, glucose is fully reduced: every oxygen is removed and n-hexane, CH₃CH₂CH₂CH₂CH₂CH₃, is formed.
2. This shows that the six carbon atoms of glucose lie in one straight chain.
3. (b) Conc. HNO₃ oxidises both ends: the –CHO at C-1 and the –CH₂OH at C-6 become –COOH, giving saccharic (glucaric) acid, HOOC–(CHOH)₄–COOH.

Answer: (a) n-Hexane, CH₃CH₂CH₂CH₂CH₂CH₃ (shows the six carbons are in a straight chain). (b) Saccharic (glucaric) acid, HOOC–(CHOH)₄–COOH, as both end groups are oxidised.

Give the main monohalo product: (i) C₆H₅CH=CH₂ + HBr; (ii) cyclohexene + Br₂ in UV light.

1. For C₆H₅CH=CH₂ + HBr, the reaction follows Markovnikov's rule where the H⁺ adds to the terminal carbon to form the more stable benzylic carbocation at the benzylic position.
2. The bromide ion then attacks this carbocation to yield the main product C₆H₅CH(Br)CH₃, which is 1-bromo-1-phenylethane.
3. For cyclohexene + Br₂ in UV light, the reaction proceeds via a free-radical mechanism involving allylic substitution rather than addition.
4. A bromine radical abstracts an allylic hydrogen from the ring, and the resulting allylic radical reacts with Br₂ to form 3-bromocyclohexene as the main monohalo product.

Answer: (i) C₆H₅CH(Br)CH₃, 1-bromo-1-phenylethane (Markovnikov addition via the benzylic carbocation). (ii) 3-Bromocyclohexene (free-radical substitution at the allylic carbon).

Explain: (i) a Grignard reagent must be made in dry conditions; (ii) aqueous KOH turns alkyl halides into alcohols but alcoholic KOH gives alkenes.

1. Grignard reagents are strong bases that react with water to form alkanes (RMgX + H₂O → RH + Mg(OH)X), so dry conditions are required to prevent destruction.
2. In aqueous solution, OH⁻ acts as a nucleophile and substitutes the halogen to form alcohols.
3. In alcoholic solution, the alkoxide ion (RO⁻) is a stronger base than OH⁻, promoting β-elimination (dehydrohalogenation) to form alkenes.

Answer: (i) Grignard reagents react with water (RMgX + H₂O → RH + Mg(OH)X), so moisture would destroy them to give the alkane. (ii) In water OH⁻ is mainly a nucleophile and substitutes the halogen; in alcohol the alkoxide (RO⁻) present is a stronger base, so β-elimination (dehydrohalogenation) takes place.

Write the equation and name the reaction: (a) butan-2-one with zinc amalgam and conc. HCl; (b) two moles of benzaldehyde heated with conc. NaOH.

1. (a) Butan-2-one undergoes Clemmensen reduction with Zn-Hg and conc. HCl, where the carbonyl group is reduced to a methylene group, yielding butane: CH₃COCH₂CH₃ + 4[H] → CH₃CH₂CH₂CH₃ + H₂O.
2. (b) Benzaldehyde, lacking alpha-hydrogens, undergoes a Cannizzaro reaction with conc. NaOH, where one molecule is oxidized to sodium benzoate and the other is reduced to benzyl alcohol: 2C₆H₅CHO + NaOH → C₆H₅CH₂OH + C₆H₅COONa.

Answer: (a) Clemmensen reduction: CH₃COCH₂CH₃ + 4[H] → CH₃CH₂CH₂CH₃ (butane) + H₂O. (b) Cannizzaro reaction: 2C₆H₅CHO + NaOH (conc.) → C₆H₅CH₂OH + C₆H₅COONa.

A cell shows 100 Ω with 0.05 M KCl (κ = 1.35 × 10⁻² Ω⁻¹ cm⁻¹) and 90 Ω with 0.02 M AgNO₃ at 25 °C. Find κ and Λm of the AgNO₃ solution.

1. Calculate the cell constant using the KCl data: G* = κ × R = 1.35 × 10⁻² Ω⁻¹ cm⁻¹ × 100 Ω = 1.35 cm⁻¹.
2. Determine the conductivity of the AgNO₃ solution: κ = G* / R = 1.35 cm⁻¹ / 90 Ω = 0.015 S cm⁻¹.
3. Calculate the molar conductivity: Λm = (κ × 1000) / C = (0.015 S cm⁻¹ × 1000) / 0.02 mol L⁻¹ = 750 S cm² mol⁻¹.

Answer: Cell constant G* = κR = 1.35 × 10⁻² × 100 = 1.35 cm⁻¹. For AgNO₃: κ = G*/R = 1.35/90 = 0.015 S cm⁻¹. Λm = κ × 1000/C = 0.015 × 1000/0.02 = 750 S cm² mol⁻¹.

For 2NO + Br₂ → 2NOBr: [NO], [Br₂] (mol/L) and initial rate (mol L⁻¹ s⁻¹) are 0.05, 0.05, 1.0 × 10⁻³; 0.05, 0.15, 3.0 × 10⁻³; 0.15, 0.05, 9.0 × 10⁻³. Find (a) the orders in NO and Br₂, (b) k, (c) the rate at [NO] = 0.4 M, [Br₂] = 0.2 M.

1. Compare experiments 1 and 2 where [NO] is constant at 0.05 M. Tripling [Br2] from 0.05 to 0.15 M triples the rate from 1.0 × 10⁻³ to 3.0 × 10⁻³, so the order with respect to Br2 is 1.
2. Compare experiments 1 and 3 where [Br2] is constant at 0.05 M. Tripling [NO] from 0.05 to 0.15 M increases the rate by a factor of 9 (from 1.0 × 10⁻³ to 9.0 × 10⁻³), so the order with respect to NO is 2.
3. The rate law is Rate = k[NO]²[Br2]. Using data from experiment 1, calculate k as 1.0 × 10⁻³ / (0.05² × 0.05) = 8 L² mol⁻² s⁻¹.
4. Substitute the given concentrations into the rate law to find the new rate: Rate = 8 × (0.4)² × 0.2 = 0.256 mol L⁻¹ s⁻¹.

Answer: (a) Tripling [Br₂] triples the rate, so order in Br₂ = 1; tripling [NO] makes it 9 times, so order in NO = 2. Rate = k[NO]²[Br₂]. (b) k = 1.0 × 10⁻³/(0.05² × 0.05) = 8 L² mol⁻² s⁻¹. (c) Rate = 8 × (0.4)² × 0.2 = 0.256 mol L⁻¹ s⁻¹.

Note: CBSE's marking scheme writes the rate as 2.56 × 10⁻¹ mol L⁻¹; the correct unit is mol L⁻¹ s⁻¹.

(a) Formula of potassium tetrahydroxidozincate(II). (b) Put [Cr(NH₃)₅Cl]Cl₂, [Cr(NH₃)₃Cl₃], [Cr(NH₃)₆]Cl₃ in increasing order of solution conductivity. (c) Name the isomerism shown by (i) [Co(NH₃)₅NO₂]²⁺ and (ii) [Co(en)₃]Cl₃.

1. (a) The complex ion is [Zn(OH)₄]²⁻, where Zn is +2 and four OH⁻ ligands give -4, so the charge is -2; balancing with K⁺ gives K₂[Zn(OH)₄].
2. (b) Conductivity depends on the number of ions produced in solution: [Cr(NH₃)₃Cl₃] is neutral (0 ions), [Cr(NH₃)₅Cl]Cl₂ gives 3 ions, and [Cr(NH₃)₆]Cl₃ gives 4 ions, so the increasing order is [Cr(NH₃)₃Cl₃] < [Cr(NH₃)₅Cl]Cl₂ < [Cr(NH₃)₆]Cl₃.
3. (c)(i) [Co(NH₃)₅NO₂]²⁺ shows linkage isomerism because the ambidentate NO₂⁻ ligand can coordinate through N (nitro) or O (nitrito).
4. (c)(ii) [Co(en)₃]Cl₃ shows optical isomerism because the three bidentate en ligands create a chiral complex with non-superimposable mirror images (d and l forms).

Answer: (a) K₂[Zn(OH)₄]. (b) [Cr(NH₃)₃Cl₃] (no ions) < [Cr(NH₃)₅Cl]Cl₂ (3 ions) < [Cr(NH₃)₆]Cl₃ (4 ions). (c)(i) Linkage isomerism (NO₂⁻ can bind through N or O). (ii) Optical isomerism (d and l forms).

(a) Which is allylic: CH₃–CH=CH–Br or CH₂=CH–CH(Br)–CH₃? (b) Which is more reactive to nucleophilic substitution, chlorobenzene or 2,4,6-trinitrochlorobenzene, and why? (c) Which C₄H₉Cl isomer boils lowest?

1. (a) In CH₂=CH–CH(Br)–CH₃, the bromine is attached to the carbon adjacent to the double bond, making it allylic; in CH₃–CH=CH–Br, bromine is on the double-bonded carbon, making it vinylic.
2. (b) 2,4,6-Trinitrochlorobenzene is more reactive because the three electron-withdrawing nitro groups at ortho and para positions stabilize the carbanion intermediate formed during nucleophilic substitution.
3. (c) (CH₃)₃CCl (tert-butyl chloride) has the lowest boiling point because it is the most branched isomer, resulting in the smallest surface area and weakest van der Waals forces.

Answer: (a) CH₂=CH–CH(Br)–CH₃ (Br on the carbon next to C=C); the other is vinylic. (b) 2,4,6-Trinitrochlorobenzene: the three electron-withdrawing –NO₂ groups at o/p positions stabilise the carbanion intermediate. (c) (CH₃)₃CCl, tert-butyl chloride (most branched, smallest surface area).

(a) Give the mechanism of CH₃CH₂OH → CH₂=CH₂ with H⁺ at 443 K. (b) Main products: (i) CH₃CH₂CH=CH₂ treated with B₂H₆ then H₂O₂/OH⁻; (ii) phenol treated with aq. NaOH, then CO₂ and H⁺.

1. (a) Step 1 (fast): the –OH oxygen takes up a proton, giving the protonated alcohol: CH₃CH₂OH + H⁺ ⇌ CH₃CH₂OH₂⁺.
2. Step 2 (slow, rate-determining): the C–O bond breaks, water leaves and the ethyl carbocation forms: CH₃CH₂OH₂⁺ → CH₃CH₂⁺ + H₂O.
3. Step 3: a proton is lost from the β-carbon, forming the double bond: CH₃CH₂⁺ → CH₂=CH₂ + H⁺.
4. (b)(i) Hydroboration–oxidation of but-1-ene puts –OH on the end carbon (anti-Markovnikov product): butan-1-ol, CH₃CH₂CH₂CH₂OH.
5. (b)(ii) NaOH turns phenol into sodium phenoxide; CO₂ attacks the ortho position (Kolbe reaction) and acidification gives salicylic acid, o-HOC₆H₄COOH.

Answer: (a) Step 1 (fast): CH₃CH₂OH + H⁺ ⇌ CH₃CH₂OH₂⁺ (protonated alcohol). Step 2 (slow, rate-determining): CH₃CH₂OH₂⁺ → CH₃CH₂⁺ + H₂O. Step 3: CH₃CH₂⁺ → CH₂=CH₂ + H⁺ (loss of a β-proton). (b)(i) Butan-1-ol, CH₃CH₂CH₂CH₂OH (anti-Markovnikov hydration). (ii) Salicylic acid, 2-hydroxybenzoic acid, o-HOC₆H₄COOH (Kolbe reaction).

Answer any three: (a) define a peptide linkage; (b) bonds holding the two DNA strands together; (c) the polysaccharide among sucrose, glucose, starch, fructose; (d) one water-soluble and one fat-soluble vitamin.

1. Define a peptide linkage as the amide bond –CO–NH– formed between the –COOH of one amino acid and the –NH₂ of another with the loss of water.
2. Identify that the two DNA strands are held together by hydrogen bonds between their complementary nitrogenous bases.
3. Select starch as the polysaccharide from the given list, distinguishing it from the monosaccharides glucose and fructose and the disaccharide sucrose.
4. List Vitamin C as a water-soluble vitamin and Vitamin A as a fat-soluble vitamin to satisfy the requirement for one of each type.

Answer: (a) The amide link –CO–NH– formed between the –COOH of one amino acid and the –NH₂ of another (with loss of water). (b) Hydrogen bonds between complementary bases. (c) Starch. (d) Water-soluble: vitamin C (or B group); fat-soluble: vitamin A (or D, E, K).

Ester A (C₆H₁₂O₂) gives B and C with LiAlH₄. B + PCC gives D; D with dil. NaOH then heat gives E; E + H₂ (catalyst) gives C. D oxidises to a monobasic acid F of molar mass 60. Identify A to F.

1. F is a monobasic acid of molar mass 60: CH₃COOH (24 + 4 + 32 = 60).
2. So D, which oxidises to F, is ethanal CH₃CHO, and B, which PCC oxidises to D, is ethanol CH₃CH₂OH.
3. With dilute NaOH and heat, ethanal undergoes aldol condensation to E = CH₃CH=CHCHO (but-2-enal).
4. Hydrogenating E reduces both C=C and C=O, giving C = CH₃CH₂CH₂CH₂OH (butan-1-ol).
5. LiAlH₄ reduces ester A to two alcohols, ethanol and butan-1-ol, so A is CH₃COOCH₂CH₂CH₂CH₃ (butyl ethanoate) or CH₃CH₂CH₂COOCH₂CH₃ (ethyl butanoate); both are C₆H₁₂O₂.

Answer: F (monobasic, M = 60) is CH₃COOH, so D = CH₃CHO and B = CH₃CH₂OH. Aldol condensation of D gives E = CH₃CH=CHCHO (but-2-enal), which on hydrogenation gives C = CH₃CH₂CH₂CH₂OH (butan-1-ol). A gives ethanol + butan-1-ol, so A = CH₃COOCH₂CH₂CH₂CH₃ (butyl ethanoate) or CH₃CH₂CH₂COOCH₂CH₃ (ethyl butanoate); both are C₆H₁₂O₂.

Note: CBSE's marking scheme accepts either butyl ethanoate or ethyl butanoate as A.

Case on batteries and fuel cells: how is a primary battery different from a secondary one?

1. A primary battery is a non-rechargeable cell where the electrochemical reaction occurs only once and cannot be reversed.
2. A secondary battery is a rechargeable cell where the reaction can be reversed by passing an electric current, allowing it to be reused.

Answer: In a primary battery the reaction occurs only once and it cannot be recharged after it is dead; a secondary battery can be recharged and reused.

Why does the mercury cell keep a steady potential of 1.35 V throughout its life?

1. The overall reaction of the mercury cell is Zn(Hg) + HgO(s) → ZnO(s) + Hg(l).
2. No ion in solution appears in this overall reaction, so no concentration term changes as the cell is used.
3. Hence the cell potential stays steady at about 1.35 V throughout its life.

Answer: Its overall reaction has no ion in solution whose concentration changes during use, so the cell potential stays constant.

Give the reactions taking place when a lead storage battery is recharged.

1. During recharging, the lead storage battery functions as an electrolytic cell, and the overall discharge reaction is reversed.
2. The overall reaction is: 2PbSO₄(s) + 2H₂O(l) → Pb(s) + PbO₂(s) + 2H₂SO₄(aq).
3. At the cathode, lead(II) sulfate is reduced to lead metal: PbSO₄ + 2e⁻ → Pb + SO₄²⁻.
4. At the anode, lead(II) sulfate is oxidized to lead(IV) oxide: PbSO₄ + 2H₂O → PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻.

Answer: On charging the cell works as an electrolytic cell and the discharge reaction is reversed: 2PbSO₄(s) + 2H₂O(l) → Pb(s) + PbO₂(s) + 2H₂SO₄(aq). (Cathode: PbSO₄ + 2e⁻ → Pb + SO₄²⁻; anode: PbSO₄ + 2H₂O → PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻.)

State two advantages that fuel cells have over other galvanic cells.

1. Advantage 1: Fuel cells have a higher efficiency, approximately 70%, in converting fuel energy into electricity compared to other galvanic cells.
2. Advantage 2: They are pollution-free because the hydrogen-oxygen fuel cell produces only water as a by-product.

Answer: (i) Higher efficiency (about 70%) in converting fuel energy to electricity. (ii) Pollution-free, as the H₂–O₂ cell gives only water.

Case on VBT and crystal field theory: define crystal field splitting energy.

1. Crystal field splitting energy is defined as the energy difference between the two sets of d orbitals created when ligands split the degenerate d orbitals.
2. For an octahedral field, this energy difference is denoted by Δo.

Answer: The energy difference between the two sets of d orbitals (e.g. t₂g and e_g) produced when the ligand field splits the degenerate d orbitals; written Δo for an octahedral field.

Using crystal field theory, explain why [Ti(H₂O)₆]³⁺ is violet.

1. Ti³⁺ has a 3d¹ configuration, placing the single electron in the lower-energy t₂g orbital.
2. Absorption of visible light promotes this electron from t₂g to e_g via a d–d transition.
3. The complementary colour of the absorbed light is observed, making the complex appear violet.

Answer: Ti³⁺ is 3d¹ (t₂g¹). It absorbs visible light to promote the electron from t₂g to e_g (d–d transition); the complementary colour seen is violet.

Why is [Cr(NH₃)₆]³⁺ paramagnetic but [Ni(CN)₄]²⁻ diamagnetic? (Cr = 24, Ni = 28)

1. Cr³⁺ is 3d³; in [Cr(NH₃)₆]³⁺ the three electrons stay unpaired in t₂g and two empty 3d orbitals are used for d²sp³ bonding.
2. With three unpaired electrons, [Cr(NH₃)₆]³⁺ is paramagnetic.
3. Ni²⁺ is 3d⁸; the strong-field ligand CN⁻ pairs these electrons into four 3d orbitals, leaving one 3d orbital empty for dsp² (square planar) bonding.
4. With no unpaired electron, [Ni(CN)₄]²⁻ is diamagnetic.

Answer: Cr³⁺ is 3d³: the three electrons stay unpaired in t₂g (d²sp³), so the complex is paramagnetic. Ni²⁺ is 3d⁸: strong-field CN⁻ pairs up the electrons, leaving one 3d orbital empty for dsp² (square planar) bonding, so no unpaired electron remains and the complex is diamagnetic.

Why is [Fe(CN)₆]³⁻ an inner orbital complex while [Fe(H₂O)₆]³⁺ is outer orbital? (Fe = 26)

1. Fe (Z=26) is 3d⁶ 4s², so Fe³⁺ has a 3d⁵ configuration with five unpaired electrons.
2. CN⁻ is a strong field ligand that forces electron pairing in the 3d orbitals, leaving two 3d orbitals vacant.
3. Since inner 3d orbitals are available, [Fe(CN)₆]³⁻ uses d²sp³ hybridization and is an inner orbital complex.
4. H₂O is a weak field ligand that does not cause pairing, so the 3d orbitals remain occupied and no inner d orbital is free.
5. Thus, [Fe(H₂O)₆]³⁺ uses outer 4d orbitals for sp³d² hybridization and is an outer orbital complex.

Answer: Fe³⁺ is 3d⁵. Strong-field CN⁻ forces pairing, freeing two 3d orbitals, so (n−1)d orbitals are used: d²sp³ (inner orbital). Weak-field H₂O causes no pairing, so the outer 4d orbitals are used: sp³d² (outer orbital).

(i) CO₂ dissolves in water more than O₂ at one temperature; which has the larger KH and why? (ii) What happens to blood cells in aqueous NaCl stronger than 0.9% (mass/volume)? (iii) A 1 molal aqueous solution of A₂B₃ is 60% ionised; find its boiling point. Kb(water) = 0.52 K kg mol⁻¹.

1. (i) By Henry's law p = KH·x, so at the same pressure a larger KH means a smaller mole fraction dissolved. O₂ is less soluble, so O₂ has the larger KH.
2. (ii) NaCl solution stronger than 0.9% is hypertonic to blood cells; water moves out of the cells by osmosis and they shrink.
3. (iii) A₂B₃ → 2A³⁺ + 3B²⁻, so one formula unit gives n = 5 ions.
4. i = 1 + α(n − 1) = 1 + 0.6 × 4 = 3.4.
5. ΔTb = i × Kb × m = 3.4 × 0.52 × 1 = 1.768 K.
6. Tb = 373 + 1.768 = 374.768 K.

Answer: (i) O₂, because a larger KH means lower solubility (p = KH·x). (ii) The solution is hypertonic, water leaves the cells and they shrink. (iii) A₂B₃ → 2A³⁺ + 3B²⁻, n = 5; i = 1 + α(n − 1) = 1 + 0.6 × 4 = 3.4. ΔTb = iKb·m = 3.4 × 0.52 × 1 = 1.768 K, so Tb = 373 + 1.768 = 374.768 K (374.918 K if water boils at 373.15 K).

Note: CBSE's marking scheme takes the boiling point of water as 373 K; with 373.15 K the answer is 374.918 K.

(i) At 25 °C p°A = 75 mm Hg and p°B = 25 mm Hg; liquid mixture has xA = 0.4. Find the mole fraction of B in the vapour. (ii) Define a colligative property; which one is preferred for molar masses of macromolecules? (iii) Why are equimolar NaCl and glucose solutions not isotonic?

1. Given xA = 0.4, so xB = 1 - 0.4 = 0.6.
2. Calculate partial pressures: pA = 75 × 0.4 = 30 mm Hg and pB = 25 × 0.6 = 15 mm Hg.
3. Total pressure P = 30 + 15 = 45 mm Hg, so mole fraction of B in vapour yB = 15/45 = 1/3 ≈ 0.33.
4. Colligative properties depend only on the number of solute particles, not their nature.
5. Osmotic pressure is preferred for determining molar masses of macromolecules.
6. NaCl dissociates into Na⁺ and Cl⁻ (i ≈ 2) while glucose does not (i = 1).
7. Since osmotic pressure π = iCRT, the different van't Hoff factors mean equimolar solutions are not isotonic.

Answer: (i) pA = 75 × 0.4 = 30, pB = 25 × 0.6 = 15, total = 45 mm Hg; yB = 15/45 = 1/3 ≈ 0.33. (ii) A property that depends only on the number of solute particles, not on their nature; osmotic pressure is preferred. (iii) NaCl dissociates into Na⁺ and Cl⁻ (i ≈ 2) while glucose does not (i = 1), so their osmotic pressures (π = iCRT) differ.

Note: CBSE's marking scheme writes yB = 0.33 mm Hg, but a mole fraction has no unit.

Answer any five: (a) why N,N-diethylbenzenesulphonamide does not dissolve in alkali; (b) why aniline fails in Friedel-Crafts reactions; (c) a test to tell methylamine from aniline; (d) the Gabriel phthalimide reaction; (e) aniline to p-bromoaniline; (f) product of C₆H₅N₂⁺Cl⁻ with (i) HBF₄ then (ii) NaNO₂/Cu, heat; (g) A and B in C₆H₅COOH –(NH₃, heat)→ A –(Br₂ + NaOH)→ B.

1. (a) The sulphonamide nitrogen carries no hydrogen (both are replaced by ethyl groups), so it has no acidic H and cannot form a salt with alkali.
2. (b) Aniline is a base and forms a salt with the Lewis acid AlCl₃; the nitrogen becomes positive and strongly deactivates the ring, so Friedel-Crafts reaction fails.
3. (c) Azo dye test: with HNO₂ at 0–5 °C aniline gives benzenediazonium chloride, which couples with alkaline phenol to an orange dye; methylamine gives N₂ gas and no dye.
4. (d) Gabriel synthesis: phthalimide + KOH → potassium phthalimide; heating with R–X gives N-alkylphthalimide; hydrolysis with aqueous NaOH gives the primary amine R–NH₂ and sodium phthalate.
5. (e) Acetylate aniline with (CH₃CO)₂O/pyridine to acetanilide, brominate with Br₂ in CH₃COOH to p-bromoacetanilide, then hydrolyse (H⁺ or OH⁻) to p-bromoaniline.
6. (f) With HBF₄ the diazonium salt gives benzenediazonium fluoroborate, C₆H₅N₂⁺BF₄⁻; heating this with aqueous NaNO₂ and copper replaces the diazo group by –NO₂, giving nitrobenzene, C₆H₅NO₂.
7. (g) C₆H₅COOH + NH₃ on heating gives A = benzamide, C₆H₅CONH₂; Hofmann bromamide degradation with Br₂ + NaOH gives B = aniline, C₆H₅NH₂.

Answer: (a) It has no H on the nitrogen, so it is not acidic and cannot form a salt with alkali. (b) Aniline (a base) forms a salt with the Lewis acid AlCl₃ catalyst; the N becomes positive and deactivates the ring. (c) Azo dye test: with HNO₂ at 0–5 °C aniline gives benzenediazonium chloride, which couples with alkaline phenol to an orange dye; methylamine gives no dye (gives N₂ gas). (d) Phthalimide + KOH → potassium phthalimide; + R–X → N-alkylphthalimide; + NaOH(aq) → R–NH₂ (primary amine) + sodium phthalate. (e) Acetylate with (CH₃CO)₂O/pyridine to acetanilide, brominate with Br₂/CH₃COOH to p-bromoacetanilide, then hydrolyse (H⁺ or OH⁻) to p-bromoaniline. (f) C₆H₅N₂⁺BF₄⁻ forms first, then nitrobenzene, C₆H₅NO₂. (g) A = benzamide, C₆H₅CONH₂; B = aniline, C₆H₅NH₂ (Hofmann bromamide degradation).

(i) Explain: (1) Zn, Cd, Hg have low melting and boiling points; (2) among d⁴ ions Cr²⁺ is a strong reductant but Mn³⁺ a strong oxidant; (3) E°(Cu²⁺/Cu) is positive (+0.34 V). (ii) Finish and balance: (1) KMnO₄ on heating; (2) Cr₂O₇²⁻ + 6I⁻ + 14H⁺ →.

1. (i)(1) Zn, Cd, and Hg have completely filled d-orbitals, resulting in no unpaired d-electrons.
2. This lack of unpaired electrons leads to weak metallic bonding, causing low melting and boiling points.
3. (i)(2) Cr²⁺ is a strong reductant because it loses an electron to form Cr³⁺, which has a stable d³ configuration.
4. Mn³⁺ is a strong oxidant because it gains an electron to form Mn²⁺, which has a stable half-filled d⁵ configuration.
5. (i)(3) The positive E° value for Cu²⁺/Cu arises because the high enthalpy of atomisation and ionisation energy are not compensated by the hydration enthalpy.
6. (ii)(1) On heating, potassium permanganate decomposes into potassium manganate, manganese dioxide, and oxygen: 2KMnO₄ → K₂MnO₄ + MnO₂ + O₂.
7. (ii)(2) The dichromate ion oxidizes iodide ions to iodine while being reduced to chromium(III) ions: Cr₂O₇²⁻ + 6I⁻ + 14H⁺ → 2Cr³⁺ + 3I₂ + 7H₂O.

Answer: (i)(1) Their d orbitals are full (no unpaired d electrons), so metallic bonding is weak. (2) Cr²⁺ → Cr³⁺ gives the stable d³ (t₂g³) state, so Cr²⁺ reduces; Mn³⁺ → Mn²⁺ gives the stable half-filled d⁵ state, so Mn³⁺ oxidises. (3) Copper's high enthalpy of atomisation (plus ionisation) is not balanced by its hydration enthalpy, so E° is positive. (ii)(1) 2KMnO₄ → K₂MnO₄ + MnO₂ + O₂. (2) Cr₂O₇²⁻ + 6I⁻ + 14H⁺ → 2Cr³⁺ + 3I₂ + 7H₂O.

(i) Which is more stable in water, Cu₂Cl₂ or CuCl₂, and why? (ii) Outer configuration common to all f-block elements. (iii) Which of Sc³⁺, Fe³⁺, Zn²⁺ (Z = 21, 26, 30) is coloured in water, and why? (iv) How is K₂Cr₂O₇ made from sodium chromate? (v) Why are transition metals and their compounds good catalysts?

1. (i) CuCl₂ is more stable: Cu²⁺(aq) has a much more negative hydration enthalpy than Cu⁺(aq), and Cu⁺ disproportionates in water (2Cu⁺ → Cu²⁺ + Cu).
2. (ii) The common outer configuration of f-block elements is (n−2)f¹⁻¹⁴ (n−1)d⁰⁻¹ ns².
3. (iii) Fe³⁺ (3d⁵) is coloured because its unpaired d electrons allow d–d transitions; Sc³⁺ (3d⁰) and Zn²⁺ (3d¹⁰) cannot, so they are colourless.
4. (iv) Acidify sodium chromate: 2Na₂CrO₄ + 2H⁺ → Na₂Cr₂O₇ + 2Na⁺ + H₂O; then add KCl: Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl.
5. (v) Transition metals show variable oxidation states and form unstable intermediate compounds, and they provide a large surface for reactants to adsorb, so they are good catalysts.

Answer: (i) CuCl₂: Cu²⁺(aq) has a much more negative hydration enthalpy, and Cu⁺ disproportionates in water (2Cu⁺ → Cu²⁺ + Cu). (ii) (n−2)f¹⁻¹⁴ (n−1)d⁰⁻¹ ns². (iii) Fe³⁺ (3d⁵, unpaired electrons allow d–d transitions); Sc³⁺ (3d⁰) and Zn²⁺ (3d¹⁰) are colourless. (iv) Acidify: 2Na₂CrO₄ + 2H⁺ → Na₂Cr₂O₇ + 2Na⁺ + H₂O; then Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl. (v) They show variable oxidation states and form complexes/intermediates, and provide a large surface area for adsorption.