CBSE Class 12 Chemistry Question Paper 2025 Solved Code 56-1-1

SECTION A

1. In case of association, abnormal molar mass of solute will [1 Mark]
(A) increase
(B) decrease
(C) remain same
(D) first increase and then decrease

Answer: (A) increase

Teacher's Note:
a) Association reduces the number of solute particles, so colligative properties become smaller than expected.
b) Molar mass is inversely related to the colligative property, so the calculated (abnormal) molar mass increases.

2. Standard electrode potential for Sn⁴⁺/Sn²⁺ couple is +0.15 V and that for the Cr³⁺/Cr couple is -0.74 V. The two couples in their standard states are connected to make a cell. The cell potential will be [1 Mark]
(A) +1.19 V
(B) +0.89 V
(C) +0.18 V
(D) +1.83 V

Answer: (B) +0.89 V

Teacher's Note:
a) The couple with lower reduction potential (Cr³⁺/Cr) acts as the anode.
b) \( E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.15 - (-0.74) = 0.89 \) V.

3. The magnetic moment is associated with its spin angular momentum and orbital angular momentum. Spin only magnetic moment value of Cr³⁺ ion (Atomic no. : Cr = 24) is ________. [1 Mark]
(A) 2.87 B.M.
(B) 3.87 B.M.
(C) 3.47 B.M.
(D) 3.57 B.M.

Answer: (B) 3.87 B.M.

Teacher's Note:
a) Cr³⁺ has the configuration 3d³, so it has 3 unpaired electrons.
b) \( \mu = \sqrt{n(n+2)} = \sqrt{3 \times 5} = \sqrt{15} \approx 3.87 \) B.M.

4. Acidified KMnO₄ oxidises sulphite to [1 Mark]
(A) S₂O₃^2-
(B) S₂O₈^2-
(C) SO₂(g)
(D) SO₄^2-

Answer: (D) \( SO_4^{2-} \)

Teacher's Note:
a) In acidic medium: 5SO₃^2- + 2MnO₄^- + 6H⁺ → 2Mn²⁺ + 5SO₄^2- + 3H₂O.
b) Sulphur is oxidised from +4 (sulphite) to +6 (sulphate).

5. The correct IUPAC name of [Pt(NH₃)₂Cl₂]²⁺ is [1 Mark]
(A) Diamminedichloridoplatinum (II)
(B) Diamminedichloridoplatinum (IV)
(C) Diamminedichloridoplatinum (O)
(D) Diamminedichloridoplatinate (IV)

Answer: (B) Diamminedichloridoplatinum (IV)

Teacher's Note:
a) Oxidation state of Pt: x + 2(0) + 2(-1) = +2, so x = +4.
b) The complex is a cation, so the metal name stays "platinum"; the ending "-ate" is used only for anionic complexes.

6. Arrange the following compounds in increasing order of their boiling points :
(i) (CH₃)₂CH - CH₂Br (ii) CH₃CH₂CH₂CH₂Br (iii) (CH₃)₃C - Br
The correct order is [1 Mark]
(A) \( (ii) \lt (i) \lt (iii) \)
(B) \( (i) \lt (ii) \lt (iii) \)
(C) \( (iii) \lt (i) \lt (ii) \)
(D) \( (iii) \lt (ii) \lt (i) \)

[Figure: Structures of the three compounds: (i) a CH carbon bonded to two CH₃ groups and a CH₂Br group; (ii) straight chain CH₃CH₂CH₂CH₂Br; (iii) H₃C - C - CH₃ with a CH₃ group above and Br below the central carbon.]

Answer: (C) \( (iii) \lt (i) \lt (ii) \)

Teacher's Note:
a) All three are isomers (C₄H₉Br), so boiling point depends on branching.
b) More branching gives a more compact shape, less surface area and weaker van der Waals forces, so a lower boiling point.

7. Alkyl halides undergoing nucleophilic bimolecular substitution reaction involve [1 Mark]
(A) retention of configuration
(B) formation of racemic mixture
(C) inversion of configuration
(D) formation of carbocation

Answer: (C) inversion of configuration

Teacher's Note:
a) In S_N2, the nucleophile attacks from the side opposite to the leaving group (backside attack).
b) Racemisation and carbocation formation are features of S_N1, not S_N2.

8. Which is the correct order of acid strength from the following ? [1 Mark]
(A) \( C_6H_5OH \gt H_2O \gt ROH \)
(B) \( C_6H_5OH \gt ROH \gt H_2O \)
(C) \( ROH \gt C_6H_5OH \gt H_2O \)
(D) \( H_2O \gt C_6H_5OH \gt ROH \)

Answer: (A) \( C_6H_5OH \gt H_2O \gt ROH \)

Teacher's Note:
a) Phenoxide ion is stabilised by resonance, so phenol is the strongest acid here.
b) The +I effect of the alkyl group makes alcohols weaker acids than water.

9. The acid formed when propyl magnesium bromide is treated with CO₂ followed by acid hydrolysis is : [1 Mark]
(A) C₃H₇COOH
(B) C₂H₅COOH
(C) CH₃COOH
(D) C₃H₇OH

Answer: (A) \( C_3H_7COOH \)

Teacher's Note:
a) C₃H₇MgBr + CO₂ → C₃H₇COOMgBr, which on hydrolysis (H₃O⁺) gives C₃H₇COOH (butanoic acid).
b) Grignard reagent with CO₂ adds one extra carbon atom to the chain.

10. The best reagent for converting propanamide into propanamine is ________. [1 Mark]
(A) excess H₂
(B) Br₂ in aqueous NaOH
(C) iodine in the presence of red phosphorus
(D) LiAlH₄ in ether

Answer: (D) \( LiAlH_4 \) in ether

Teacher's Note:
a) LiAlH₄ reduces the amide to an amine with the same number of carbon atoms: CH₃CH₂CONH₂ → CH₃CH₂CH₂NH₂.
b) Br₂/NaOH (Hoffmann bromamide degradation) would give ethanamine, which has one carbon less.

11. Which of the following statements is not true about glucose ? [1 Mark]
(A) It is an aldohexose.
(B) On heating with HI it forms n-hexane.
(C) It exists in furanose form.
(D) It does not give Schiff's test.

Answer: (C) It exists in furanose form.

Teacher's Note:
a) The cyclic form of glucose is a six-membered ring, called the pyranose form.
b) Glucose not giving Schiff's test is a true fact, which shows the free -CHO group is absent in its cyclic form.

12. An unripe mango placed in a concentrated salt solution to prepare pickle, shrivels because ________. [1 Mark]
(A) it gains water due to osmosis
(B) it loses water due to reverse osmosis
(C) it gains water due to reverse osmosis
(D) it loses water due to osmosis

Answer: (D) it loses water due to osmosis

Teacher's Note:
a) The salt solution is hypertonic, so water moves out of the mango cells into the solution.
b) Reverse osmosis needs external pressure, which is not applied here.

For questions number 13 to 16, two statements are given - one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below :
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

13. Assertion (A) : [Cr(H₂O)₆]Cl₂ and [Fe(H₂O)₆]Cl₂ are examples of homoleptic complexes.
Reason (R) : All the ligands attached to the metal are the same. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

Teacher's Note:
a) In both complexes the metal is bonded only to H₂O ligands, so they are homoleptic.
b) Complexes with more than one kind of ligand are called heteroleptic.

14. Assertion (A) : The boiling points of alkyl halides decrease in the order : \( RI \gt RBr \gt RCl \gt RF \).
Reason (R) : The boiling points of alkyl chlorides, bromides and iodides are considerably higher than that of the hydrocarbon of comparable molecular mass. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

Teacher's Note:
a) The order in A is due to increasing size and mass of the halogen, which increases van der Waals forces.
b) R compares alkyl halides with hydrocarbons, so it does not explain the order among the halides.

15. Assertion (A) : The boiling point of ethanol is higher than that of methoxymethane.
Reason (R) : There is intramolecular hydrogen bonding in ethanol. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (C) Assertion (A) is true, but Reason (R) is false.

Teacher's Note:
a) Ethanol has intermolecular (not intramolecular) hydrogen bonding, which raises its boiling point.
b) Watch for the words "intra" and "inter" in such reasons; this is a common trap.

16. Assertion (A) : All naturally occurring \( \alpha \)-amino acids except glycine are optically active.
Reason (R) : Most naturally occurring amino acids have L-configuration. [1 Mark]
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(C) Assertion (A) is true, but Reason (R) is false.
(D) Assertion (A) is false, but Reason (R) is true.

Answer: (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

Teacher's Note:
a) Optical activity is due to a chiral \( \alpha \)-carbon; glycine (R = H) has no chiral carbon.
b) L-configuration describes the arrangement of groups, it does not explain why the molecules are optically active.

SECTION B

17. (A) Give reasons :
(a) Cooking is faster in pressure cooker than in an open pan.
(b) On mixing liquid X and liquid Y, volume of the resulting solution decreases. What type of deviation from Raoult's law is shown by the resulting solution ? What change in temperature would you observe after mixing liquids X and Y ? [2 Marks]

Answer:
(a) Inside a pressure cooker the pressure is high, so the boiling point of water is raised. Food cooks at a higher temperature, so cooking is faster.
(b) The solution shows negative deviation from Raoult's law. The temperature increases on mixing (heat is released).

Teacher's Note:
a) Link higher pressure directly to higher boiling point; this is the value point in (a).
b) For negative deviation remember: \( \Delta V_{mix} \lt 0 \) and \( \Delta H_{mix} \lt 0 \), so the mixture warms up.

OR

(B) Define Azeotrope. What type of Azeotrope is formed by negative deviation from Raoult's law ? Give an example. [2 Marks]

Answer:
1. Azeotropes are binary mixtures which have the same composition in the liquid and vapour phase and boil at a constant temperature.
2. Negative deviation forms a maximum boiling azeotrope. Example: a mixture of 68% HNO₃ and 32% H₂O by mass.

Teacher's Note:
a) The definition must mention both "same composition in liquid and vapour" and "constant boiling temperature".
b) Positive deviation gives minimum boiling azeotropes (e.g. about 95% ethanol and water); do not mix them up.

18. Complete and balance the following chemical equations : [2 Marks]
(a) 2MnO₄^-(aq) + 10I^-(aq) + 16H⁺(aq) →
(b) Cr₂O₇^2-(aq) + 6Fe²⁺(aq) + 14H⁺(aq) →

Answer:
(a) 2MnO₄^-(aq) + 10I^-(aq) + 16H⁺(aq) → 2Mn²⁺(aq) + 5I₂(s) + 8H₂O(l)
(b) Cr₂O₇^2-(aq) + 6Fe²⁺(aq) + 14H⁺(aq) → 2Cr³⁺(aq) + 6Fe³⁺(aq) + 7H₂O(l)

Teacher's Note:
a) In acidic medium, MnO₄^- is reduced to Mn²⁺ and Cr₂O₇^2- is reduced to Cr³⁺.
b) Check both atoms and charges on each side; for example in (b) the charge is +24 on both sides.

19. Would you expect benzaldehyde to be more reactive or less reactive in nucleophilic addition reactions than propanal ? Justify your answer. [2 Marks]

Answer:
1. Benzaldehyde is less reactive than propanal in nucleophilic addition reactions.
2. In benzaldehyde, the polarity of the carbonyl group is reduced due to resonance with the benzene ring. So its carbonyl carbon is less electrophilic than the carbonyl carbon of propanal.

Teacher's Note:
a) The first mark is only for the correct choice "less reactive".
b) The key word for the second mark is "less electrophilic" carbonyl carbon due to resonance.

20. Identify A and B in each of the following reaction sequence : [2 Marks]
(a) CH₃CH₂Cl (NaCN) → A (H₂/Ni) → B
(b) C₆H₅NH₂ (NaNO₂/HCl, 0 - 5 °C) → A (C₆H₅NH₂, H⁺) → B

Answer:
(a) A = CH₃CH₂CN (propanenitrile); B = CH₃CH₂CH₂NH₂ (propan-1-amine)
(b) A = C₆H₅N₂⁺Cl^- (benzenediazonium chloride); B = p-aminoazobenzene, C₆H₅-N=N-C₆H₄-NH₂ (the -NH₂ group is at the para position)

Teacher's Note:
a) Each of A and B carries half a mark, so name or write all four clearly.
b) Diazonium salts couple with aniline at the para position in mildly acidic medium to give a yellow azo dye.

21. Write the reactions involved when D-glucose is treated with the following reagents : [2 Marks]
(a) HCN
(b) Br₂ water

Answer:
(a) CHO-(CHOH)₄-CH₂OH + HCN → CH(OH)(CN)-(CHOH)₄-CH₂OH (glucose cyanohydrin)
(b) CHO-(CHOH)₄-CH₂OH (Br₂ water) → COOH-(CHOH)₄-CH₂OH (gluconic acid)

Teacher's Note:
a) Reaction with HCN proves the presence of a carbonyl (-CHO) group in glucose.
b) Bromine water is a mild oxidising agent; it oxidises only the -CHO group to -COOH.

SECTION C

22. A solution of glucose (molar mass = 180 g mol^-1) in water has a boiling point of 100.20 °C. Calculate the freezing point of the same solution. Molal constants for water K_f and K_b are 1.86 K kg mol^-1 and 0.512 K kg mol^-1 respectively. [3 Marks]

Answer:
Given: \( T_b = 100.20^{\circ}C \), \( T_b^{\circ} = 100^{\circ}C \), \( K_b = 0.512 \) K kg mol^-1, \( K_f = 1.86 \) K kg mol^-1
Step 1: \( \Delta T_b = T_b - T_b^{\circ} = 100.20 - 100 = 0.20^{\circ}C \) or 0.20 K
Step 2: \( \Delta T_b = K_b \times m \)
\( m = \dfrac{\Delta T_b}{K_b} = \dfrac{0.20}{0.512} = 0.390 \) mol kg^-1
Step 3: \( \Delta T_f = K_f \times m = 1.86 \times 0.390 = 0.725 \) K
Step 4: Freezing point of solution \( = 273.15 - 0.725 = 272.425 \) K (or \( -0.725^{\circ}C \))

Teacher's Note:
a) The molality is the link between the two colligative properties; the molar mass of glucose is not needed.
b) Subtract \( \Delta T_f \) from the freezing point of pure water (273.15 K), and always write the unit.

23. (a) State the following :
(i) Kohlrausch law of independent migration of ions and
(ii) Faraday's first law of electrolysis.
(b) Using \( E^{\circ}_{values} \) of X and Y given below, predict which is better for coating the surface of iron to prevent corrosion and why ?
Given \( E^{\circ}_{X^{2+}/X} = -2.36 \) V,
\( E^{\circ}_{Y^{2+}/Y} = -0.14 \) V,
\( E^{\circ}_{Fe^{2+}/Fe} = -0.44 \) V [3 Marks]

Answer:
(a) (i) Kohlrausch law: The limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and the cation of the electrolyte.
(ii) Faraday's first law: The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte.
(b) X is better for coating iron, because X has a more negative electrode potential than Fe (-2.36 V compared to -0.44 V). So X is oxidised in preference to iron and protects it.

Teacher's Note:
a) Faraday's first law can also be written as \( w = Z \times I \times t \).
b) A metal with more negative E° (higher oxidation potential) acts as a sacrificial coating; Y, with less negative E°, cannot protect iron.

24. A certain reaction is 50% complete in 20 minutes at 300 K and the same reaction is 50% complete in 5 minutes at 350 K. Calculate the activation energy if it is a first order reaction.
[R = 8.314 J K^-1 mol^-1; log 4 = 0.602] [3 Marks]

Answer:
Given: \( t_{1/2} = 20 \) min at \( T_1 = 300 \) K; \( t_{1/2} = 5 \) min at \( T_2 = 350 \) K
Step 1: For a first order reaction, \( t_{1/2} = \dfrac{0.693}{k} \)
\( k_1 = \dfrac{0.693}{20} = 0.03465 \) min^-1
\( k_2 = \dfrac{0.693}{5} = 0.1386 \) min^-1
Step 2: \( \log \dfrac{k_2}{k_1} = \dfrac{E_a}{2.303R} \left[ \dfrac{T_2 - T_1}{T_1 T_2} \right] \)
Step 3: \( \log \dfrac{0.1386}{0.03465} = \dfrac{E_a}{2.303 \times 8.314} \left[ \dfrac{350 - 300}{350 \times 300} \right] \)
\( \log 4 = \dfrac{E_a}{19.15} \times \dfrac{50}{105000} \)
\( E_a = \dfrac{0.602 \times 19.15 \times 105000}{50} \)
Step 4: \( E_a = 24209 \) J mol^-1 or 24.209 kJ mol^-1 (about 24.2 kJ mol^-1)

Teacher's Note:
a) Since k is inversely proportional to \( t_{1/2} \), the ratio \( \dfrac{k_2}{k_1} = \dfrac{20}{5} = 4 \) directly.
b) Half a mark is deducted for a missing or wrong unit of \( E_a \).

25. The elements of 3d transition series are given as :
Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn
Answer the following :
(a) Copper has exceptionally positive \( E^{\circ}_{M^{2+}/M} \) value, why ?
(b) Which element is a strong reducing agent in +2 oxidation state and why ?
(c) Zn²⁺ salts are colourless. Why ? [3 Marks]

Answer:
(a) Copper has a high enthalpy of atomisation \( (\Delta_a H^{\circ}) \) and a low enthalpy of hydration \( (\Delta_{hyd} H^{\circ}) \), which do not balance its ionisation enthalpy. So its \( E^{\circ}_{Cu^{2+}/Cu} \) is positive.
(b) Chromium (Cr). Cr²⁺ easily changes to Cr³⁺ (d⁴ to d³), which has a stable half-filled t_2g level.
(c) Zn²⁺ has fully filled d-orbitals (3d¹⁰) and no unpaired electron, so no d-d transition is possible and its salts are colourless.

Teacher's Note:
a) In (b), half a mark is for naming Cr and half for the reason.
b) Colour of transition metal ions needs unpaired d-electrons for d-d transitions; d⁰ and d¹⁰ ions are colourless.

26. (A) Draw the structure of the major monohalo product for each of the following reaction : [3 Marks]
(a) Cl-C₆H₄-CH₂ - CH₃ (Br₂, Heat) → ?
(b) + HBr → ?
(c) HO - H₂C-C₆H₄-OH (HCl, Heat) → ?

[Figure: (a) A benzene ring with Cl at one position and a CH₂ - CH₃ group at the para position, reacting with Br₂, Heat. (b) A cyclohexene ring with a CH₃ group on one carbon of the double bond, reacting with HBr. (c) A benzene ring with an OH group and, at the para position, a HO - H₂C group, reacting with HCl, Heat.]

Answer:
(a) 1-(1-Bromoethyl)-4-chlorobenzene: Cl-C₆H₄-CH(Br)-CH₃ (bromine replaces an H on the benzylic carbon).
(b) 1-Bromo-1-methylcyclohexane: a cyclohexane ring with Br and CH₃ on the same carbon (Markovnikov addition).
(c) 4-(Chloromethyl)phenol: HO-C₆H₄-CH₂Cl (only the benzylic -OH is replaced by Cl; the phenolic -OH stays).

Teacher's Note:
a) Free radical bromination with heat occurs at the benzylic position because the benzylic radical is resonance stabilised.
b) The C-O bond of phenol has partial double bond character, so the phenolic -OH is not replaced by HCl.

OR

(B) How do you convert :
(a) Chlorobenzene to biphenyl
(b) Propene to 1-Iodopropane
(c) 2-bromobutane to but-2-ene. [3 Marks]

Answer:
(a) Fittig reaction: 2C₆H₅Cl + 2Na (dry ether) → C₆H₅-C₆H₅ + 2NaCl
(b) CH₃CH=CH₂ + HBr (peroxide) → CH₃CH₂CH₂Br; then CH₃CH₂CH₂Br + NaI (acetone) → CH₃CH₂CH₂I + NaBr
(c) CH₃-CH(Br)-CH₂-CH₃ (ethanolic KOH, heat) → CH₃-CH=CH-CH₃ + HBr

Teacher's Note:
a) In (b), peroxide gives anti-Markovnikov addition, and the Finkelstein reaction (NaI in acetone) then gives the iodide.
b) In (c), Saytzeff rule makes the more substituted alkene, but-2-ene, the major product.

27. (a) Arrange the following compounds in increasing order of their boiling point :
(CH₃)₂NH, CH₃CH₂NH₂, CH₃CH₂OH.
(b) Give plausible explanation for each of the following :
(i) Aromatic primary amines cannot be prepared by Gabriel Phthalimide synthesis.
(ii) Amides are less basic than amines. [3 Marks]

Answer:
(a) \( (CH_3)_2NH \lt CH_3CH_2NH_2 \lt CH_3CH_2OH \)
(b) (i) Aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide. So aromatic primary amines cannot be made this way.
(ii) In amides, the lone pair on nitrogen is delocalised onto the C=O group by resonance (O=C-NH₂ ↔ ^-O-C=NH₂⁺). So the lone pair is less available for donation and amides are less basic than amines.

Teacher's Note:
a) Alcohols form stronger hydrogen bonds than amines since O is more electronegative than N; primary amines have more H-bonding than secondary amines.
b) For (ii), writing the resonance structures of the amide earns the mark quickly.

28. (a) What is the difference between native protein and denatured protein ?
(b) Which one of the following is a disaccharide ?
Glucose, Lactose, Amylose, Fructose
(c) Which vitamin is responsible for the coagulation of blood ? [3 Marks]

Answer:
(a) A native protein has its three-dimensional structure intact and is biologically active. In a denatured protein the three-dimensional structure is destroyed and it is biologically inactive.
(b) Lactose
(c) Vitamin K

Teacher's Note:
a) Denaturation destroys the secondary and tertiary structures, but the primary structure stays the same.
b) Glucose and fructose are monosaccharides and amylose is a polysaccharide, so lactose is the only disaccharide.

SECTION D

The following questions are case based questions. Read the passage carefully and answer the questions that follow.

29. The rate of a chemical reaction is expressed either in terms of decrease in the concentration of reactants or increase in the concentration of a product per unit time. Rate of the reaction depends upon the nature of reactants, concentration of reactants, temperature, presence of catalyst, surface area of the reactants and presence of light. Rate of reaction is directly related to the concentration of reactant. Rate law states that the rate of reaction depends upon the concentration terms on which the rate of reaction actually depends, as observed experimentally. The sum of powers of the concentration of the reactants in the Rate law expression is called order of reaction while the number of reacting species taking part in an elementary reaction which must collide simultaneously in order to bring about a chemical reaction is called molecularity of the reaction.
Answer the following questions :

(a) (i) What is a rate determining step ?
(ii) Define complex reaction. [2 Marks]

Answer:
(i) The slowest step in the mechanism of a reaction is called the rate determining step.
(ii) A complex reaction is one that takes place through a series of elementary reactions, that is, it involves two or more steps.

Teacher's Note:
a) The key word for (i) is "slowest step".
b) The overall rate of a complex reaction is controlled by its rate determining step.

(b) What is the effect of temperature on the rate constant of a reaction ? [1 Mark]

Answer: The rate constant increases with increase in temperature.

Teacher's Note:
a) For many reactions, the rate constant nearly doubles for a 10 K rise in temperature.
b) This is given by the Arrhenius equation, \( k = A e^{-E_a/RT} \).

OR

(b) Why is molecularity applicable only for elementary reactions whereas order is applicable for elementary as well as complex reactions ? [1 Mark]

Answer: Molecularity is defined only for an elementary reaction, as each step of a complex reaction may have a different molecularity, which has no meaning for the overall reaction. Order is determined experimentally (by the slowest step), so it applies to both elementary and complex reactions.

Teacher's Note:
a) Molecularity is a theoretical concept; order is an experimental quantity.
b) Mention "experimentally determined" for order to earn the mark.

(c) The conversion of molecule X to Y follows second order kinetics. If concentration of X is increased 3 times, how will it affect the rate of formation of Y ? [1 Mark]

Answer: Rate \( = k[X]^2 \). New rate \( = k(3[X])^2 = 9k[X]^2 \). So the rate of formation of Y increases 9 times.

Teacher's Note:
a) For order n, increasing the concentration a times increases the rate \( a^n \) times.
b) A common mistake is to answer "3 times", which is true only for a first order reaction.

30. Phenols undergo electrophilic substitution reactions readily due to the strong activating effect of OH group attached to the benzene ring. Since, the OH group increases the electron density more to o- and p- positions therefore OH group is ortho, para-directing. Reimer-Tiemann reaction is one of the examples of aldehyde group being introduced on the aromatic ring of phenol, ortho to the hydroxyl group. This is a general method used for the ortho-formylation of phenols.
Answer the following questions :

(a) What happens when phenol reacts with
(i) Br₂/CS₂
(ii) Conc. HNO₃ [2 Marks]

Answer:
(i) With bromine in CS₂ at low temperature (273 K), phenol gives a mixture of 2-bromophenol and 4-bromophenol (monobromophenols).
C₆H₅OH + Br₂ (CS₂, 273 K) → o-BrC₆H₄OH + p-BrC₆H₄OH
(ii) With concentrated HNO₃, phenol gives 2,4,6-trinitrophenol (picric acid).
C₆H₅OH (Conc. HNO₃) → 2,4,6-(NO₂)₃C₆H₂OH

Teacher's Note:
a) A solvent of low polarity (CS₂) gives only monobromination; bromine water gives 2,4,6-tribromophenol.
b) Name the products clearly; the name alone is accepted for each mark.

(b) Why phenol does not undergo protonation readily ? [1 Mark]

Answer: Due to resonance, the lone pair of electrons on the oxygen of phenol is delocalised into the benzene ring, so it is not easily available for protonation.

Teacher's Note:
a) The key word is "resonance" (delocalisation of the lone pair on oxygen).
b) The same reason makes the C-O bond in phenol stronger than in alcohols.

(c) Which is a stronger acid - phenol or cresol ? Give reason. [1 Mark]

Answer: Phenol is the stronger acid. In cresol, the methyl group has an electron releasing (+I) effect, so the phenoxide ion formed from cresol is less stable.

Teacher's Note:
a) Electron releasing groups decrease acidity; electron withdrawing groups (like -NO₂) increase acidity of phenols.
b) Half a mark is for "phenol" and half for the reason.

OR

(c) Write the IUPAC name of the product formed in the Reimer-Tiemann reaction. [1 Mark]

Answer: 2-Hydroxybenzaldehyde (salicylaldehyde), also named 2-hydroxybenzenecarbaldehyde.

Teacher's Note:
a) In the Reimer-Tiemann reaction, phenol is treated with CHCl₃ and aqueous NaOH, followed by acid hydrolysis.
b) The -CHO group enters the ortho position to -OH, as stated in the passage.

SECTION E

31. (A) (a) Write the cell reaction and calculate the e.m.f. of the following cell at 298 K :
Sn(s) | Sn²⁺ (0.004 M) || H⁺ (0.02 M) | H₂(g) (1 Bar) | Pt (s)
(Given : \( E^{\circ}_{Sn^{2+}/Sn} = -0.14 \) V, \( E^{\circ}_{H^{+}|H_2(g), Pt} = 0.00 \) V)
(b) Account for the following ;
(i) On the basis of E° values, O₂ gas should be liberated at anode but it is Cl₂ gas which is liberated in the electrolysis of aqueous NaCl.
(ii) Conductivity of CH₃COOH decreases on dilution. [5 Marks]

Answer:
(a) Cell reaction: Sn(s) + 2H⁺(aq) → Sn²⁺(aq) + H₂(g)
Given: [Sn²⁺] = 0.004 M, [H⁺] = 0.02 M, \( p_{H_2} = 1 \) bar, n = 2
Formula: \( E_{cell} = (E^{\circ}_{c} - E^{\circ}_{a}) - \dfrac{0.059}{n} \log \dfrac{[Sn^{2+}]}{[H^{+}]^2} \)
Substitution: \( E_{cell} = [0 - (-0.14)] - \dfrac{0.059}{2} \log \dfrac{0.004}{(0.02)^2} \)
\( E_{cell} = 0.14 - 0.0295 \log \dfrac{0.004}{0.0004} = 0.14 - 0.0295 \log 10 \)
\( E_{cell} = 0.14 - 0.0295 = 0.1105 \) V
(b) (i) Oxygen has a high overpotential (overvoltage) at the anode, so the oxidation of Cl^- to Cl₂ takes place in preference.
(ii) On dilution, the number of ions carrying current per unit volume decreases, so the conductivity of CH₃COOH decreases.

Teacher's Note:
a) Square the H⁺ concentration in the Nernst equation because 2 H⁺ ions appear in the balanced cell reaction.
b) Do not confuse conductivity (decreases on dilution) with molar conductivity (increases on dilution).
c) Write the unit V with the final e.m.f.; the cell reaction itself carries 1 mark.

OR

(B) (a) Write the anode and cathode reactions and the overall cell reaction occurring in a lead storage battery during its use.
(b) Calculate the potential for half-cell containing 0.01 M K₂Cr₂O₇(aq), 0.01 M Cr³⁺ (aq) and 1.0 × 10^-4 M H⁺(aq).
The half cell reaction is
Cr₂O₇^2-(aq) + 14H⁺(aq) + 6e^- → 2Cr³⁺(aq) + 7H₂O(l)
and the standard electrode potential is given as E° = 1.33 V.
[Given : log 10 = 1] [5 Marks]

Answer:
(a) At anode: Pb(s) + SO₄^2-(aq) → PbSO₄(s) + 2e^-
At cathode: PbO₂(s) + SO₄^2-(aq) + 4H⁺(aq) + 2e^- → PbSO₄(s) + 2H₂O(l)
Overall reaction: Pb(s) + PbO₂(s) + 2SO₄^2-(aq) + 4H⁺(aq) → 2PbSO₄(s) + 2H₂O(l)
(b) Given: [Cr₂O₇^2-] = 0.01 M = 10^-2 M, [Cr³⁺] = 10^-2 M, [H⁺] = 10^-4 M, n = 6, E° = 1.33 V
Formula: \( E = E^{\circ} - \dfrac{0.059}{n} \log \dfrac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}][H^{+}]^{14}} \)
Substitution: \( E = 1.33 - \dfrac{0.059}{6} \log \dfrac{(10^{-2})^2}{(10^{-2})(10^{-4})^{14}} \)
\( \dfrac{(10^{-2})^2}{(10^{-2})(10^{-4})^{14}} = \dfrac{10^{-4}}{10^{-58}} = 10^{54} \)
\( E = 1.33 - \dfrac{0.059}{6} \times 54 \log 10 = 1.33 - 0.059 \times 9 \)
\( E = 1.33 - 0.531 = 0.799 \) V

Teacher's Note:
a) In the lead storage battery, PbSO₄ forms at both electrodes during discharge and H₂SO₄ is used up.
b) Water is a pure liquid, so it is not written in the Nernst equation expression.
c) Keep track of powers carefully: \( (10^{-4})^{14} = 10^{-56} \); most errors happen here.

32. (A) Answer the following :
(a) Low spin tetrahedral complexes are not known.
(b) Co²⁺ is easily oxidised to Co³⁺ in the presence of a strong ligand [At. No. of Co = 27]
(c) What type of isomerism is shown by the complex [Co(NH₃)₆] [Cr(CN)₆] ?
(d) Why a solution of [Ni(H₂O)₆]²⁺ is green while a solution of [Ni(CN)₄]^2- is colourless. (At. No. of Ni = 28)
(e) Write the IUPAC name of the following complex : [Co(NH₃)₅(CO₃)]Cl [5 Marks]

Answer:
(a) In tetrahedral complexes the crystal field splitting energy \( (\Delta_t) \) is small and is not large enough to force pairing of electrons. So low spin tetrahedral complexes are not formed.
(b) Co²⁺ has 3d⁷ configuration. In the presence of a strong field ligand, the d⁷ ion changes to the more stable d⁶ (Co³⁺, t_2g⁶) configuration, so the strong field stabilises the higher oxidation state.
(c) Coordination isomerism.
(d) [Ni(H₂O)₆]²⁺ has unpaired electrons (H₂O is a weak ligand), so d-d transitions give it a green colour. [Ni(CN)₄]^2- has no unpaired electron (CN^- is a strong ligand), so it is colourless.
(e) Pentaamminecarbonatocobalt(III) chloride

Teacher's Note:
a) Remember \( \Delta_t = \dfrac{4}{9} \Delta_o \); this is why tetrahedral complexes are always high spin.
b) Coordination isomerism needs both cation and anion to be complex ions, as in (c).
c) In (e), list ligands alphabetically (ammine before carbonato) and give the oxidation state of Co in Roman numerals.

OR

(B) (a) What is meant by 'Chelate effect' ? Give an example.
(b) Write the hybridization and magnetic behaviour of [Fe(CN)₆]^4-.
(Atomic number : Fe = 26)
(c) If PtCl₂ · 2NH₃ does not react with AgNO₃, what will be its formula ? [5 Marks]

Answer:
(a) Chelate effect is the higher stability of complexes containing chelating (polydentate) ligands compared to complexes having similar non-chelating ligands. Example: [Co(en)₃]³⁺.
(b) In [Fe(CN)₆]^4-, Fe is in the +2 state (3d⁶). CN^- is a strong field ligand, so all six d-electrons pair up in the 3d orbitals. Hybridisation is d²sp³ and the complex is diamagnetic (no unpaired electron).
(c) [Pt(NH₃)₂Cl₂]

Teacher's Note:
a) In (b), hybridisation and magnetic behaviour carry one mark each.
b) No reaction with AgNO₃ means no free Cl^- ions; both Cl atoms are inside the coordination sphere.

33. (A) (a) Carry out the following conversions :
(i) Ethanal to But-2-enal
(ii) Propanoic acid to ethane
(b) An alkene A with molecular formula C₅H₁₀ on ozonolysis gives a mixture of two compounds B and C. Compound B gives positive Fehling test and also reacts with iodine and NaOH solution. Compound C does not give Fehling solution test but forms iodoform. Identify the compounds A, B and C. [5 Marks]

Answer:
(a) (i) Aldol condensation: 2CH₃CHO (dil. NaOH) → CH₃-CH(OH)-CH₂-CHO (heat) → CH₃-CH=CH-CHO + H₂O
(ii) Decarboxylation: CH₃CH₂COOH + NaOH → CH₃CH₂COONa + H₂O; then CH₃CH₂COONa (NaOH + CaO, heat) → CH₃-CH₃ + Na₂CO₃
(b) A = (CH₃)₂C=CHCH₃ (2-Methylbut-2-ene)
B = CH₃CHO (Ethanal)
C = CH₃COCH₃ (Propanone / Acetone)
Ozonolysis: (CH₃)₂C=CHCH₃ (O₃, then Zn/H₂O) → CH₃COCH₃ + CH₃CHO

Teacher's Note:
a) B is an aldehyde (Fehling positive) with a CH₃CO- group (iodoform positive), so it must be ethanal.
b) C is a methyl ketone (iodoform positive, Fehling negative), so it is acetone; joining the two carbonyl carbons by C=C gives A.
c) Soda-lime decarboxylation removes one carbon, so propanoic acid gives ethane.

OR

(B) An organic compound (A) (molecular formula C₈H₁₆O₂) was hydrolysed with dilute sulphuric acid to get a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives But-l-ene. Identify (A), (B) and (C) and write chemical equations for the reactions involved. [5 Marks]

Answer:
1. A = C₃H₇COOC₄H₉, that is CH₃CH₂CH₂COOCH₂CH₂CH₂CH₃ (Butyl butanoate)
2. B = C₃H₇COOH, that is CH₃CH₂CH₂COOH (Butanoic acid); C = C₄H₉OH, that is CH₃CH₂CH₂CH₂OH (Butan-1-ol)
3. Hydrolysis: C₃H₇COOC₄H₉ + H₂O (dil. H₂SO₄) → C₃H₇COOH + C₄H₉OH
4. Dehydration: CH₃CH₂CH₂CH₂OH (conc. H₂SO₄, heat) → CH₃CH₂CH=CH₂ + H₂O
5. Oxidation: CH₃CH₂CH₂CH₂OH (CrO₃/CH₃COOH, chromic acid) → CH₃CH₂CH₂COOH

Teacher's Note:
a) Since oxidation of C gives B, both must have the same carbon skeleton: 4 carbons each, which fits C₈H₁₆O₂.
b) Dehydration giving but-1-ene shows that C is a straight chain primary alcohol, butan-1-ol.
c) Identification of A carries 1 mark, B and C half a mark each, and each equation 1 mark.

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