CBSE Class 12 Chemistry Board Paper 2023 (56/1/1): Solutions

Which colligative property is preferred for finding molar masses of polymers and proteins?

1. Osmotic pressure is the preferred colligative property for determining the molar masses of polymers and proteins.
2. This is because it is measurable at room temperature and yields a large value even for very dilute solutions, unlike other properties which are too small to measure accurately for large molecules.

Answer: Osmotic pressure, as it is measurable at room temperature and gives a large value even for very dilute solutions — option (a).

Why do people at high altitudes have a low oxygen level in blood and tissues?

1. At high altitudes, the atmospheric pressure is lower, which results in a lower partial pressure of oxygen in the air.
2. According to Henry's law, the solubility of a gas in a liquid is directly proportional to its partial pressure, so less oxygen dissolves in the blood.

Answer: Because the atmospheric pressure (and so the partial pressure of oxygen) is low there, less oxygen dissolves (Henry's law) — option (c).

Which cell notation correctly represents Zn + 2Ag⁺ → Zn²⁺ + 2Ag?

1. Identify the oxidation half-reaction: Zn loses electrons to form Zn²⁺, so Zn acts as the anode and is written on the left side.
2. Identify the reduction half-reaction: Ag⁺ gains electrons to form Ag, so Ag acts as the cathode and is written on the right side.
3. Combine the half-cells using the standard notation format (Anode | Anode ion || Cathode ion | Cathode) to get Zn | Zn²⁺ || Ag⁺ | Ag, which corresponds to option (d).

Answer: Zn is the anode (left) and Ag the cathode (right): Zn | Zn²⁺ || Ag⁺ | Ag — option (d).

For a spontaneous cell reaction, what are the signs of ΔG and E°cell?

1. For a spontaneous reaction, the Gibbs free energy change (ΔG) must be negative.
2. Using the relation ΔG = −nFE°cell, a negative ΔG implies that E°cell must be positive.
3. Thus, ΔG is negative and E°cell is positive, which corresponds to option (c).

Answer: ΔG = −nFE°cell, so ΔG is negative and E°cell is positive — option (c).

Among ΔH, ΔG, Ea and ΔS, which quantity does a catalyst change?

1. A catalyst provides an alternative reaction pathway with a lower energy barrier, thereby decreasing the activation energy (Ea).
2. It does not alter the thermodynamic state functions such as enthalpy (ΔH), Gibbs free energy (ΔG), or entropy (ΔS), so only Ea is changed.

Answer: Only the activation energy Ea (a catalyst gives a path of lower Ea) — option (c).

What is the order of the photochemical reaction H₂(g) + Cl₂(g) → 2HCl(g) (in light, hν)?

1. NCERT gives the photochemical reaction H₂(g) + Cl₂(g) → 2HCl(g) (in light, over water) as an example of a zero order reaction.
2. For it, rate = k: the rate does not change with [H₂] or [Cl₂], so the order is zero — option (c).

Answer: Zero order — option (c).

Note: CBSE's marking scheme keys (c) but gave full marks for any option chosen.

Which oxidation state is the most common and stable one for lanthanoids?

1. The principal (most common) oxidation state of all lanthanoids is +3.
2. +2 and +4 appear only for a few elements, mostly when they give an empty, half-filled or full 4f shell (f⁰, f⁷, f¹⁴), so the answer is +3 — option (b).

Answer: +3 — option (b).

What type of isomerism is shown by the pair [Co(SO₄)(NH₃)₅]Br and [Co(Br)(NH₃)₅]SO₄?

1. Identify that the two complexes have the same composition but differ in which anion is coordinated to the cobalt center versus which is present as a counter-ion.
2. Recognize that [Co(SO₄)(NH₃)₅]Br releases Br⁻ ions in solution, while [Co(Br)(NH₃)₅]SO₄ releases SO₄²⁻ ions, which is the defining characteristic of ionisation isomerism.

Answer: Ionisation isomerism — they swap the ion inside and outside the coordination sphere, so they give different ions in solution — option (c).

Which method is best for making alkyl fluorides?

1. The Swarts reaction involves heating an alkyl chloride or bromide with a metallic fluoride like AgF, Hg₂F₂, CoF₂, or SbF₃ to form alkyl fluorides.
2. This method is preferred because direct fluorination is too violent and other halogen exchange reactions do not yield fluorides efficiently.

Answer: Heating an alkyl chloride/bromide with a metallic fluoride such as AgF, Hg₂F₂, CoF₂ or SbF₃ (Swarts reaction) — option (b).

Order of reactivity of 1°, 2° and 3° alcohols in R–OH + HCl (ZnCl₂) → RCl + H₂O.

1. With HCl and ZnCl₂, the –OH is replaced by –Cl; tertiary (and secondary) alcohols react through a carbocation formed in the slow step.
2. Carbocation stability is 3° > 2° > 1° because alkyl groups release electrons (inductive effect and hyperconjugation).
3. So 3° alcohols react fastest and 1° slowest: 1° < 2° < 3° — option (a).

Answer: 1° < 2° < 3° (tertiary react fastest via the most stable carbocation) — option (a).

Product when CH₃CONH₂ is treated with Br₂ and NaOH (alcoholic).

1. The reaction of an amide with bromine and alcoholic sodium hydroxide is the Hoffmann bromamide degradation reaction.
2. This reaction converts the amide into a primary amine with one less carbon atom, so CH₃CONH₂ yields CH₃NH₂ (methanamine).

Answer: Methanamine, CH₃NH₂ (one carbon less than the amide) — option (b).

Least basic among (CH₃)₂NH, NH₃, C₆H₅NH₂ and (CH₃)₃N.

1. Identify that in aniline (C₆H₅NH₂), the lone pair on the nitrogen atom is delocalised into the benzene ring through resonance.
2. Conclude that this delocalisation reduces the availability of the lone pair for protonation, making aniline the least basic among the given amines.

Answer: Aniline, C₆H₅NH₂, because the N lone pair is delocalised into the benzene ring — option (c).

Which glycosidic linkage joins the glucose units in the amylose part of starch?

1. Amylose is a linear polymer of glucose where the monomers are connected via alpha-glycosidic bonds.
2. The specific linkage joining the glucose units in amylose is the C1–C4 α-glycosidic linkage, corresponding to option (c).

Answer: C1–C4 α-glycosidic linkage — option (c).

Note: The paper prints 'amylase' (an enzyme), but it means amylose. CBSE gave full marks to everyone who attempted this question.

The α-helix is a structural feature of which class of substances?

1. The α-helix is a secondary structure formed by hydrogen bonding between peptide bonds in the backbone of a polypeptide chain.
2. Therefore, it is a structural feature of polypeptides (proteins), making option (c) the correct answer.

Answer: Polypeptides (proteins) — option (c).

Assertion: –NH₂ directs electrophiles to ortho/para positions. Reason: aniline does not give Friedel-Crafts reaction.

1. Assertion is true because the -NH2 group donates electron density via resonance (+R effect), activating the ortho and para positions.
2. Reason is true because aniline forms a salt with the Lewis acid AlCl3, which deactivates the ring and prevents the Friedel-Crafts reaction.
3. The reason does not explain the assertion because the o/p direction is caused by the +R effect, not by the salt formation that inhibits the reaction.

Answer: Both are true, but the reason (salt formation with AlCl₃) does not explain o/p direction (which is due to +R effect) — option (b).

Assertion: acetylated aniline gives a monosubstituted product. Reason: –NHCOCH₃ activates the ring more than –NH₂.

1. In acetanilide, the lone pair on N is also delocalised towards the C=O oxygen, so less of it is given to the benzene ring.
2. So –NHCOCH₃ activates the ring less than –NH₂; substitution stops at one group (mainly para), and the assertion is true.
3. The reason says –NHCOCH₃ activates more than –NH₂, which is false — option (c).

Answer: Assertion true; reason false — the acetamido group activates less than –NH₂, which is why substitution stops at one — option (c).

Assertion: H₂ + Br₂ → 2HBr seems to have molecularity 2. Reason: in this elementary step two reactant molecules take part.

1. Molecularity is the number of reacting species that collide at the same time in an elementary step.
2. If H₂ + Br₂ → 2HBr is treated as one elementary step, one H₂ and one Br₂ collide, so its molecularity would be 2; the reason states exactly this.
3. On that reading both statements are true and the reason explains the assertion, which is CBSE's key — option (a).

Answer: Both true and the reason explains the assertion (as per the marking scheme) — option (a).

Note: In reality H₂ + Br₂ → 2HBr is a complex reaction (NCERT gives rate = k[H₂][Br₂]^(1/2), order 3/2), and molecularity has no meaning for a complex reaction. CBSE's marking scheme still keys (a).

Assertion: low-spin tetrahedral complexes are rare. Reason: for tetrahedral complexes Δt is smaller than the pairing energy.

1. Both the assertion and the reason are true statements.
2. Since Δt is smaller than the pairing energy (P), electrons prefer to occupy higher energy orbitals singly rather than pairing up.
3. This results in high-spin complexes, making low-spin tetrahedral complexes rare, so the reason correctly explains the assertion.

Answer: Both true and the reason explains it: since Δt < P, electrons do not pair, so complexes are high spin — option (a).

State Henry's law and give one use of it.

1. State Henry's law: The partial pressure of a gas over a solution is directly proportional to its mole fraction in the solution, expressed as p = K_H·x.
2. Give one use: Soft-drink bottles are sealed under high CO₂ pressure to increase the solubility of CO₂ in the liquid.

Answer: The partial pressure of a gas over a solution is proportional to its mole fraction in the solution: p = K_H·x. Use: soft-drink bottles are sealed under high CO₂ pressure to raise CO₂ solubility (or: divers' tanks use O₂ diluted with helium to avoid bends).

On dilution, Λm of electrolyte A rises 25 times and that of B rises 1.5 times. Which is the strong electrolyte? Support with a graph.

1. Electrolyte B is the strong electrolyte because its molar conductivity (Λm) increases only slightly (1.5 times) upon dilution, which is characteristic of strong electrolytes that are already highly dissociated.
2. Electrolyte A is the weak electrolyte because its Λm increases significantly (25 times) upon dilution, indicating a large increase in the degree of dissociation as concentration decreases.
3. Plot Λm on the y-axis and √c on the x-axis; the curve for B (strong) will be nearly linear with a small slope, while the curve for A (weak) will show a steep rise at low concentrations.

Answer: B is the strong electrolyte. Plot Λm against √c: for a strong electrolyte Λm rises only slightly and almost linearly on dilution; for a weak electrolyte (A) Λm rises very steeply at low concentration because its degree of dissociation increases.

NaOH solution (0.05 mol L⁻¹) fills a tube 50 cm long and 1 cm in diameter; its resistance is 5.55 × 10³ Ω. Find the conductivity κ.

1. Calculate the cross-sectional area of the tube: A = πr² = 3.14 × (0.5)² = 0.785 cm².
2. Substitute values into the conductivity formula: κ = l/(R·A) = 50/(5550 × 0.785) ≈ 1.15 × 10⁻² S cm⁻¹ (0.01148 S cm⁻¹).

Answer: A = πr² = 3.14 × (0.5)² = 0.785 cm²; κ = l/(R·A) = 50/(5550 × 0.785) ≈ 1.15 × 10⁻² S cm⁻¹ (0.01148 S cm⁻¹).

Note: CBSE's marking scheme writes 11.47 × 10⁻³ S cm⁻¹; that is the same value, which rounds to 1.15 × 10⁻² S cm⁻¹.

Complete: (a) 2MnO₄⁻ + 5NO₂⁻ + 6H⁺ → ; (b) Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → .

1. For part (a), identify that permanganate ion (MnO₄⁻) is reduced to Mn²⁺ and nitrite ion (NO₂⁻) is oxidized to nitrate ion (NO₃⁻) in acidic medium.
2. Balance the atoms and charges to complete the equation: 2MnO₄⁻ + 5NO₂⁻ + 6H⁺ → 2Mn²⁺ + 5NO₃⁻ + 3H₂O.
3. For part (b), identify that dichromate ion (Cr₂O₇²⁻) is reduced to Cr³⁺ in acidic medium by gaining electrons.
4. Balance the atoms and charges to complete the half-reaction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O.

Answer: (a) 2MnO₄⁻ + 5NO₂⁻ + 6H⁺ → 2Mn²⁺ + 5NO₃⁻ + 3H₂O. (b) Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O.

Identify A and B: (a) CH₃CH(OH)CH₃ with PCl₅ gives A, which with AgCN gives B; (b) CH₃CH₂CH₂Cl with KOH in ethanol gives A, which with HBr gives B.

1. (a) CH₃CH(OH)CH₃ reacts with PCl₅ to replace the -OH group with -Cl, forming A = CH₃CHClCH₃ (2-chloropropane).
2. (a) A reacts with AgCN to form B = (CH₃)₂CH–NC (2-propyl isocyanide), as AgCN primarily yields isocyanides.
3. (b) CH₃CH₂CH₂Cl undergoes dehydrohalogenation with KOH in ethanol to form A = CH₃CH=CH₂ (propene).
4. (b) A reacts with HBr via Markovnikov addition to form B = CH₃CHBrCH₃ (2-bromopropane).

Answer: (a) A = CH₃CHClCH₃ (2-chloropropane); B = (CH₃)₂CH–NC (2-propyl isocyanide), since AgCN gives isocyanides. (b) A = CH₃CH=CH₂ (propene); B = CH₃CHBrCH₃ (2-bromopropane, Markovnikov addition).

Explain: (i) phenol is more acidic than alcohols; (ii) more branching in the alkyl chain lowers an alcohol's boiling point.

1. Phenol is more acidic than alcohols because the phenoxide ion formed after deprotonation is stabilised by resonance, which delocalises the negative charge over the aromatic ring.
2. In contrast, the alkoxide ion formed from alcohols lacks this resonance stabilisation, making the O-H bond in alcohols harder to break and thus less acidic.
3. More branching in an alkyl chain reduces the surface area of the alcohol molecule, which weakens the intermolecular van der Waals forces.
4. Weaker van der Waals forces require less energy to overcome, resulting in a lower boiling point for branched alcohols compared to their straight-chain isomers.

Answer: (i) The phenoxide ion is stabilised by resonance (charge spread over the ring), unlike the alkoxide ion. (ii) Branching reduces surface area, so van der Waals forces become weaker and the boiling point falls.

(i) Mechanism of CH₃CH₂OH → CH₂=CH₂ + H₂O with H⁺ at 443 K. (ii) Equation of the Reimer-Tiemann reaction.

1. (i) Step 1: ethanol is protonated by the acid: CH₃CH₂OH + H⁺ → CH₃CH₂–OH₂⁺.
2. Step 2 (slow): the protonated alcohol loses water to give a carbocation: CH₃CH₂–OH₂⁺ → CH₃CH₂⁺ + H₂O.
3. Step 3: the carbocation loses H⁺ from the next carbon to give ethene: CH₃CH₂⁺ → CH₂=CH₂ + H⁺ (the acid is regenerated).
4. (ii) Reimer-Tiemann: phenol heated with CHCl₃ and aqueous NaOH (about 340 K) gives a –CHO group at the ortho position, as the sodium salt of salicylaldehyde.
5. Acidifying (H⁺) this salt gives salicylaldehyde, o-HOC₆H₄CHO.

Answer: (i) Step 1: CH₃CH₂OH + H⁺ → CH₃CH₂–OH₂⁺ (protonated ethanol). Step 2: CH₃CH₂–OH₂⁺ → CH₃CH₂⁺ + H₂O (slow, carbocation). Step 3: CH₃CH₂⁺ → CH₂=CH₂ + H⁺. (ii) C₆H₅OH + CHCl₃ + aq NaOH (heat) → sodium salt of salicylaldehyde; acidifying (H⁺) gives o-HOC₆H₄CHO (salicylaldehyde).

Briefly explain (a) the carbylamine reaction and (b) the Gabriel phthalimide synthesis.

1. For part (a), state that primary amines react with chloroform and ethanolic KOH to form isocyanides, a reaction known as the carbylamine reaction.
2. Write the balanced equation: R–NH₂ + CHCl₃ + 3KOH → R–NC + 3KCl + 3H₂O, noting that the product has a foul smell and serves as a test for primary amines.
3. For part (b), explain that phthalimide reacts with ethanolic KOH to form potassium phthalimide, which then reacts with an alkyl halide upon heating to yield N-alkylphthalimide.
4. Conclude that alkaline hydrolysis of N-alkylphthalimide releases the primary amine R–NH₂, completing the Gabriel phthalimide synthesis.

Answer: (a) A primary amine heated with CHCl₃ and ethanolic KOH gives a foul-smelling isocyanide: R–NH₂ + CHCl₃ + 3KOH → R–NC + 3KCl + 3H₂O (test for 1° amines). (b) Phthalimide with ethanolic KOH gives potassium phthalimide; heating with an alkyl halide gives N-alkylphthalimide, and alkaline hydrolysis releases the primary amine R–NH₂.

(a) Give a reaction proving open-chain D-glucose has a straight (unbranched) carbon chain. (b) Which linkage forms proteins?

1. Prolonged heating of D-glucose with HI yields n-hexane, CH₃(CH₂)₄CH₃.
2. The formation of a straight-chain alkane proves that all six carbon atoms in glucose are arranged in an unbranched chain.
3. Proteins are formed by peptide linkages, represented as –CO–NH–.

Answer: (a) Prolonged heating of glucose with HI gives n-hexane, CH₃(CH₂)₄CH₃, so all six carbons are in a straight chain. (b) Peptide linkage (–CO–NH–).

(a) Give a difference between ideal and non-ideal solutions. (b) Urea (30 g) is dissolved in water (846 g). Pure water has vapour pressure 23.8 mm Hg at 298 K; find the vapour pressure of water over this solution.

1. (a) An ideal solution obeys Raoult's law at all concentrations with ΔmixH = 0 and ΔmixV = 0; a non-ideal one does not obey Raoult's law and ΔmixH, ΔmixV ≠ 0.
2. (b) Moles of urea = 30/60 = 0.5 mol (M of urea, NH₂CONH₂ = 60 g mol⁻¹); moles of water = 846/18 = 47 mol.
3. By Raoult's law, p = p° × x(water), with x(water) = 47/(47 + 0.5) = 47/47.5.
4. p = 23.8 × 47/47.5 = 23.55 mm Hg (≈ 23.5 mm Hg).

Answer: (a) An ideal solution obeys Raoult's law at all concentrations with ΔmixH = 0 and ΔmixV = 0; a non-ideal one does not obey Raoult's law and ΔmixH, ΔmixV ≠ 0. (b) n(urea) = 30/60 = 0.5 mol, n(water) = 846/18 = 47 mol; p = 23.8 × 47/47.5 = 23.55 mm Hg (≈ 23.5 mm Hg).

Note: The paper does not give the molar mass of urea; it is 60 g mol⁻¹ (NH₂CONH₂).

Main product: (a) CH₃Cl with NaI in acetone; (b) hydrolysis of 2,4,6-trinitrochlorobenzene; (c) n-butyl chloride with alcoholic KOH.

1. (a) CH₃Cl reacts with NaI in acetone via an SN2 mechanism; the iodide ion displaces the chloride to form CH₃I (iodomethane) and NaCl.
2. (b) Hydrolysis of 2,4,6-trinitrochlorobenzene involves nucleophilic aromatic substitution where the -OH group replaces the -Cl group, yielding 2,4,6-trinitrophenol (picric acid).
3. (c) n-butyl chloride undergoes dehydrohalogenation with alcoholic KOH, eliminating HCl to form the major product CH₃CH₂CH=CH₂ (but-1-ene).

Answer: (a) CH₃I (iodomethane). (b) 2,4,6-trinitrophenol (picric acid). (c) CH₃CH₂CH=CH₂ (but-1-ene).

Do any three conversions: (a) phenol → picric acid; (b) propanone → 2-methylpropan-2-ol; (c) phenol → anisole; (d) propene → propan-1-ol.

1. For conversion (a), treat phenol with concentrated H₂SO₄ followed by concentrated HNO₃ to form 2,4,6-trinitrophenol (picric acid).
2. For conversion (b), react propanone with CH₃MgBr and then hydrolyze with H₃O⁺ to obtain 2-methylpropan-2-ol.
3. For conversion (c), react phenol with NaOH to form sodium phenoxide, then react with CH₃I via Williamson synthesis to yield anisole.
4. For conversion (d), treat propene with B₂H₆ followed by H₂O₂/OH⁻ to achieve anti-Markovnikov addition and form propan-1-ol.

Answer: (a) Treat phenol with conc. H₂SO₄ (sulphonation) and then conc. HNO₃, giving 2,4,6-trinitrophenol. (b) CH₃COCH₃ + CH₃MgBr, then hydrolysis (H₃O⁺) → (CH₃)₃C–OH. (c) Phenol + NaOH → C₆H₅ONa; then + CH₃I → C₆H₅OCH₃ (Williamson synthesis). (d) CH₃CH=CH₂ + B₂H₆ (hydroboration), then H₂O₂/OH⁻ → CH₃CH₂CH₂OH (anti-Markovnikov).

Note: Only three of the four conversions are needed in the exam.

Why: (i) –COOH in benzoic acid directs to meta position; (ii) NaHSO₃ helps purify aldehydes/ketones; (iii) –COOH does not show the usual reactions of C=O?

1. (i) The –COOH group is electron-withdrawing due to the –R effect, which deactivates the ortho and para positions more than the meta position, directing substitution to meta.
2. (ii) Aldehydes and ketones form crystalline bisulphite addition products with NaHSO₃, which can be separated from impurities and then hydrolysed to recover the pure carbonyl compound.
3. (iii) Resonance in the –COOH group delocalizes the pi electrons, reducing the electrophilic character of the carbonyl carbon and preventing usual C=O reactions.

Answer: (i) –COOH withdraws electrons (–R effect), deactivating ortho/para positions more, so substitution goes to meta. (ii) Aldehydes/ketones form crystalline bisulphite addition products that can be separated and hydrolysed back to the pure carbonyl compound. (iii) Resonance in –COOH makes the carboxyl carbon less electrophilic.

Compound A (C₃H₈O) over Cu at 573 K gives B. B does not reduce Fehling's solution but gives a yellow precipitate C with I₂/NaOH. Find A, B, C.

1. The molecular formula C₃H₈O corresponds to a saturated alcohol, and heating over Cu at 573 K causes dehydrogenation to form carbonyl compound B.
2. Since B does not reduce Fehling's solution, it is a ketone rather than an aldehyde, which identifies B as propanone (CH₃COCH₃).
3. The yellow precipitate C with I₂/NaOH is the iodoform test, confirming B has a CH₃CO- group, so C is iodoform (CHI₃).
4. Therefore, A is the secondary alcohol propan-2-ol (CH₃CH(OH)CH₃), which dehydrogenates to propanone.

Answer: A = CH₃CH(OH)CH₃ (propan-2-ol); B = CH₃COCH₃ (propanone, a ketone, so no Fehling's test); C = CHI₃ (iodoform).

(a) Hydrolysis products of lactose and of maltose. (b) Basic structural difference between starch and cellulose.

1. Lactose is a disaccharide that hydrolyzes into one molecule of β-D-galactose and one molecule of β-D-glucose.
2. Maltose is a disaccharide that hydrolyzes into two molecules of α-D-glucose.
3. Starch is a polymer of α-D-glucose units, existing as a mixture of amylose and amylopectin.
4. Cellulose is a straight-chain polymer of β-D-glucose units linked by C1–C4 β-glycosidic bonds.

Answer: (a) Lactose → β-D-galactose + β-D-glucose; maltose → two α-D-glucose units. (b) Starch is a polymer of α-D-glucose (amylose + amylopectin), while cellulose is a straight-chain polymer of β-D-glucose (C1–C4 β links).

Define the average rate of a reaction.

1. The average rate of a reaction is defined as the change in concentration of a reactant or product divided by the time interval taken.
2. For a reactant, this is expressed as −Δ[R]/Δt, where the negative sign indicates the decrease in concentration over time.

Answer: The change in concentration of a reactant or product divided by the time interval taken, e.g. −Δ[R]/Δt.

Name two factors on which reaction rate depends.

1. The rate of a chemical reaction depends on the concentration of the reactants, as higher concentration increases the frequency of collisions.
2. It also depends on temperature, since increasing temperature provides more energy to overcome the activation barrier.

Answer: Any two: concentration of reactants, temperature, catalyst, surface area.

For a zero order reaction: (1) how does the rate depend on concentration? (2) what is the unit of k?

1. For a zero order reaction, the rate is independent of the reactant concentration, so the rate remains constant and equals k.
2. The unit of the rate constant k is mol L⁻¹ s⁻¹.

Answer: (1) Rate does not depend on reactant concentration; it stays constant (rate = k). (2) mol L⁻¹ s⁻¹.

(1) Reaction P + 2Q → products follows rate = k[P]^(1/2)[Q]. Give the overall order. (2) Define a pseudo first order reaction with an example.

1. Identify the powers of the reactant concentrations in the rate law: [P] is raised to the power 1/2 and [Q] is raised to the power 1.
2. Calculate the overall order by summing the individual orders: 1/2 + 1 = 3/2 (or 1.5).
3. Define a pseudo first order reaction as a reaction that is actually higher order but exhibits first-order kinetics because one reactant is present in large excess.
4. Provide the example of acid hydrolysis of ethyl acetate where water is in excess: CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH.

Answer: (1) Order = 1/2 + 1 = 1.5 (i.e. 3/2). (2) A reaction that looks higher order but follows first-order kinetics because one reactant is in large excess, e.g. acid hydrolysis of ethyl acetate in excess water: CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH.

PtCl₂·2NH₃ gives no precipitate with AgNO₃. Write its formula.

1. No precipitate with AgNO₃ means no free Cl⁻ ions are present in the solution.
2. Thus, both chlorides are inside the coordination sphere, giving the formula [Pt(NH₃)₂Cl₂].

Answer: [Pt(NH₃)₂Cl₂] — both chlorides are inside the coordination sphere, so no free Cl⁻.

Secondary valency of cobalt in [Co(en)₃]³⁺?

1. Each ethylenediamine (en) ligand is bidentate, meaning it forms two coordinate bonds with the central metal ion.
2. Since there are three en ligands in the complex, the total number of coordinate bonds is 3 × 2 = 6, so the secondary valency is 6.

Answer: 6 (three bidentate en ligands).

(1) Formula of iron(III) hexacyanidoferrate(II). (2) IUPAC name of [Co(NH₃)₅Cl]Cl₂.

1. Identify the ions in iron(III) hexacyanidoferrate(II) as Fe³⁺ and the complex ion [Fe(CN)₆]⁴⁻.
2. Balance the charges by combining four Fe³⁺ ions with three [Fe(CN)₆]⁴⁻ ions to form the neutral salt Fe₄[Fe(CN)₆]₃.
3. Determine the oxidation state of cobalt in [Co(NH₃)₅Cl]Cl₂ by noting that the complex cation is [Co(NH₃)₅Cl]²⁺ and NH₃ is neutral, so Co is +3.
4. Name the complex by listing ligands alphabetically (pentaamminechlorido), the metal with its oxidation state (cobalt(III)), and the counter ion (chloride) to get Pentaamminechloridocobalt(III) chloride.

Answer: (1) Fe₄[Fe(CN)₆]₃. (2) Pentaamminechloridocobalt(III) chloride.

Hybridisation and magnetic nature of [Ni(CN)₄]²⁻ (Ni: Z = 28).

1. Ni (Z = 28) is 3d⁸4s², so Ni²⁺ is 3d⁸.
2. CN⁻ is a strong field ligand, so the eight 3d electrons pair up in four 3d orbitals, leaving one 3d orbital empty.
3. That empty 3d orbital, the 4s and two 4p orbitals give dsp² hybridisation: square planar, holding the four CN⁻ lone pairs.
4. No unpaired electrons are left, so [Ni(CN)₄]²⁻ is diamagnetic.

Answer: Ni²⁺ is 3d⁸; strong-field CN⁻ pairs the electrons, so dsp² hybridisation (square planar) with no unpaired electron — diamagnetic.

(i) State Kohlrausch's law and write Λ°m for acetic acid using it. (ii) For Ni(s) + 2Ag⁺(aq) ⇌ Ni²⁺(aq) + 2Ag(s) at 298 K, find the maximum work and log Kc, using standard potentials −0.25 V for Ni²⁺/Ni and +0.80 V for Ag⁺/Ag (F = 96500 C mol⁻¹).

1. Kohlrausch's law states that the limiting molar conductivity of an electrolyte is the sum of the limiting ionic conductivities of its constituent ions.
2. For acetic acid, the limiting molar conductivity is written as Λ°m(CH₃COOH) = λ°(H⁺) + λ°(CH₃COO⁻).
3. The standard cell potential is calculated as E°cell = E°cathode − E°anode = 0.80 − (−0.25) = 1.05 V.
4. The number of electrons transferred in the reaction Ni(s) + 2Ag⁺(aq) ⇌ Ni²⁺(aq) + 2Ag(s) is n = 2.
5. The maximum work is equal to the negative of the standard Gibbs free energy change, calculated as −ΔG° = nFE°cell = 2 × 96500 × 1.05 = 202650 J mol⁻¹.
6. Using the relation log Kc = nE°cell / 0.059, the value is calculated as log Kc = 2 × 1.05 / 0.059 = 35.6.

Answer: (i) The limiting molar conductivity of an electrolyte is the sum of the limiting ionic conductivities of its cation and anion: Λ°m(CH₃COOH) = λ°(H⁺) + λ°(CH₃COO⁻). (ii) E°cell = 0.80 − (−0.25) = 1.05 V, n = 2. Maximum work = −ΔG° = nFE° = 2 × 96500 × 1.05 = 202650 J mol⁻¹ (202.65 kJ mol⁻¹). log Kc = nE°/0.059 = 2 × 1.05/0.059 = 35.6.

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

(i) State Faraday's first law of electrolysis; how many faradays reduce 1 mol Cu²⁺ to Cu? (ii) Find the emf at 298 K of Mg(s) | Mg²⁺(0.1 M) || Cu²⁺(0.01 M) | Cu(s); standard emf is +2.71 V (F = 96500 C mol⁻¹; take log 10 = 1).

1. State Faraday's first law: The mass of substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed (m = ZIt).
2. For the reduction Cu²⁺ + 2e⁻ → Cu, 2 moles of electrons are required per mole of copper, so 2 Faradays are needed.
3. Identify the number of electrons transferred in the cell reaction as n = 2.
4. Write the Nernst equation for the cell at 298 K: Ecell = E°cell − (0.0591/n) log([Mg²⁺]/[Cu²⁺]).
5. Substitute the given values: Ecell = 2.71 − (0.0591/2) log(0.1/0.01).
6. Calculate the log term: log(0.1/0.01) = log 10 = 1, so the correction is (0.0591/2) × 1 = 0.02955.
7. Compute the final emf: Ecell = 2.71 − 0.02955 = 2.68 V.

Answer: (i) The mass of substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed (m = Z·I·t). Cu²⁺ + 2e⁻ → Cu needs 2 F. (ii) n = 2; Ecell = 2.71 − (0.0591/2) log([Mg²⁺]/[Cu²⁺]) = 2.71 − 0.02955 × log 10 = 2.68 V.

Give reasons: (i) Mn reaches +7, the highest oxidation state in the 3d series; (ii) transition metals and their compounds are good catalysts; (iii) Cr²⁺ (d⁴) is a reductant but Mn³⁺ (d⁴) is an oxidant; (iv) Zn has the lowest enthalpy of atomisation; (v) Cu⁺ is unstable in water.

1. Mn has the configuration 3d⁵4s², allowing it to lose all 7 valence electrons to achieve the +7 oxidation state.
2. Transition metals act as catalysts because they exhibit variable oxidation states, provide a surface for adsorption, and form unstable intermediate complexes.
3. Cr²⁺ (d⁴) acts as a reductant because it oxidizes to Cr³⁺ (d³), which has a stable half-filled t₂g orbital configuration.
4. Mn³⁺ (d⁴) acts as an oxidant because it reduces to Mn²⁺ (d⁵), which has a stable half-filled d orbital configuration.
5. Zn has the lowest enthalpy of atomisation because its 3d¹⁰4s² configuration has no unpaired d electrons, resulting in weak metallic bonding.
6. Cu⁺ is unstable in water because it disproportionates into Cu²⁺ and Cu(s), driven by the much more negative hydration enthalpy of Cu²⁺.

Answer: (i) Mn (3d⁵4s²) can use all its 3d and 4s electrons (7) in bonding. (ii) They show variable oxidation states, provide surfaces and form intermediate complexes. (iii) Cr²⁺ → Cr³⁺ gives stable half-filled t₂g³ (d³), while Mn³⁺ → Mn²⁺ gives stable half-filled d⁵. (iv) Zn (3d¹⁰4s²) has no unpaired d electrons, so metallic bonding is weak. (v) Cu⁺ disproportionates: 2Cu⁺(aq) → Cu²⁺(aq) + Cu(s), because Cu²⁺(aq) has a much more negative hydration enthalpy.

(i) Convert ethanal into but-2-en-1-al, and convert propanoic acid into 2-chloropropanoic acid. (ii) 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. (i)(1) Aldol: 2CH₃CHO with dil. NaOH → CH₃CH(OH)CH₂CHO (3-hydroxybutanal).
2. On heating, the aldol loses water: CH₃CH(OH)CH₂CHO → CH₃CH=CHCHO + H₂O (but-2-en-1-al).
3. (i)(2) Hell-Volhard-Zelinsky: CH₃CH₂COOH with Cl₂ and red phosphorus, then H₂O → CH₃CHClCOOH (Cl enters the α-carbon).
4. (ii) B gives Fehling's test (an aldehyde) and the iodoform test (has CH₃CO–), so B = CH₃CHO (ethanal).
5. C gives iodoform but not Fehling's test, so it is a methyl ketone: C = CH₃COCH₃ (propanone).
6. Joining the two carbonyl carbons by a double bond gives A = (CH₃)₂C=CHCH₃ (2-methylbut-2-ene), C₅H₁₀.

Answer: (i)(1) 2CH₃CHO with dil. NaOH → CH₃CH(OH)CH₂CHO (aldol); heating removes H₂O → CH₃CH=CHCHO. (2) CH₃CH₂COOH + Cl₂ with red P, then H₂O → CH₃CHClCOOH (Hell-Volhard-Zelinsky). (ii) B = CH₃CHO (ethanal), C = CH₃COCH₃ (propanone), so A = (CH₃)₂C=CHCH₃ (2-methylbut-2-ene).

(i) Chemical test to tell apart (1) CH₃COCH₂CH₃ and CH₃CH₂CH₂CHO, (2) ethanal and ethanoic acid. (ii) Structure of acetone oxime. (iii) CH₃COOH with PCl₅ gives A; A with H₂/Pd–BaSO₄ gives B; B with CH₃MgBr then H₃O⁺ gives C; B with LiAlH₄ gives D. Identify A–D.

1. (i)(1) Use the iodoform test with I₂/NaOH: butanone gives a yellow precipitate of CHI₃, while butanal does not.
2. (i)(2) Add NaHCO₃: ethanoic acid gives brisk CO₂ effervescence, while ethanal shows no reaction.
3. (ii) The structure of acetone oxime is (CH₃)₂C=N–OH.
4. (iii) CH₃COOH reacts with PCl₅ to form A, which is CH₃COCl.
5. A (CH₃COCl) is reduced by H₂/Pd–BaSO₄ to form B, which is CH₃CHO.
6. B (CH₃CHO) reacts with CH₃MgBr followed by H₃O⁺ to form C, which is (CH₃)₂CHOH.
7. B (CH₃CHO) is reduced by LiAlH₄ to form D, which is CH₃CH₂OH.

Answer: (i)(1) Iodoform test (I₂/NaOH, warm): butanone gives yellow CHI₃, butanal does not (or Tollens' test: only butanal gives a silver mirror). (2) NaHCO₃: ethanoic acid gives brisk CO₂ effervescence, ethanal does not. (ii) (CH₃)₂C=N–OH. (iii) A = CH₃COCl, B = CH₃CHO, C = (CH₃)₂CHOH (propan-2-ol), D = CH₃CH₂OH.