CBSE Class 12 Chemistry Question Paper 2026 Solved Code 56-1-3

SECTION A

1. In the ring structure of fructose, the anomeric carbon is [1 Mark]
(A) C - 1
(B) C - 2
(C) C - 3
(D) C - 5

Answer: (B) C - 2

Teacher's Note:
a) Fructose is a ketose; its keto group is at C-2, so the ring closes at C-2.
b) In glucose the anomeric carbon is C-1, so do not mix up the two sugars.

2. Which of the following is the correct expression of K_f which depends upon the nature of solvent ? [1 Mark]
(A) \( K_f = \dfrac{M_1 \times T_f^2}{R \times 1000 \times \Delta_{fus}H} \)
(B) \( K_f = \dfrac{R \times M_1 \times \Delta_{fus}H}{1000 \times T_f^2} \)
(C) \( K_f = \dfrac{R \times T_f^2 \times \Delta_{fus}H}{1000 \times M_1} \)
(D) \( K_f = \dfrac{R \times M_1 \times T_f^2}{1000 \times \Delta_{fus}H} \)

Answer: (D) \( K_f = \dfrac{R \times M_1 \times T_f^2}{1000 \times \Delta_{fus}H} \)

Teacher's Note:
a) K_f depends on molar mass, freezing point and enthalpy of fusion of the solvent only.
b) Remember that \( \Delta_{fus}H \) is in the denominator: a larger enthalpy of fusion gives a smaller K_f.

3. Which of the following transition metals has lowest enthalpy of atomisation ? [1 Mark]
(A) Cr
(B) V
(C) Mn
(D) Fe

Answer: (C) Mn

Teacher's Note:
a) Mn has a stable half-filled 3d⁵ configuration, so its d-electrons take little part in metallic bonding.
b) Weaker metallic bonding means lower enthalpy of atomisation.

4. Which of the following curve represents the first order reaction ? [1 Mark]

[Figure: Four graphs. (A) t_1/2 on the y-axis against [R]₀ on the x-axis: a straight line rising from the origin. (B) t_1/2 against [R]₀: a horizontal straight line. (C) Rate against Concentration: a horizontal straight line. (D) Rate against Concentration: a curve falling steeply and then levelling off.]

(A) t_1/2 vs. [R]₀: straight line through the origin
(B) t_1/2 vs. [R]₀: horizontal line
(C) Rate vs. Concentration: horizontal line
(D) Rate vs. Concentration: falling curve

Answer: (B) \( t_{1/2} \) vs. \( [R]_0 \) graph that is a horizontal line

Teacher's Note:
a) For a first order reaction, \( t_{1/2} = \dfrac{0.693}{k} \), which does not depend on initial concentration.
b) Graph (A) is for zero order, where \( t_{1/2} \) is proportional to [R]₀.

5. Aniline on direct nitration yields [1 Mark]
(A) 51%-ortho, 47%-para, 2%-meta derivatives
(B) 51%-meta, 47%-ortho, 2%-para derivatives
(C) 51%-para, 47%-meta, 2%-ortho derivatives
(D) 51%-ortho, 47%-meta, 2%-para derivatives

Answer: (C) 51%-para, 47%-meta, 2%-ortho derivatives

Teacher's Note:
a) In strongly acidic nitrating mixture, aniline forms the anilinium ion, which is meta directing.
b) This is why a large amount (47%) of meta product forms along with the para product.

6. Which of the following is 'not' true about enantiomers ? [1 Mark]
(A) They have the same chemical reactivity.
(B) They have the same specific rotation.
(C) They have the same melting or boiling point.
(D) They have the same refractive index.

Answer: (B) They have the same specific rotation.

Teacher's Note:
a) Enantiomers rotate plane polarised light by equal amounts but in opposite directions.
b) All their other physical properties are identical, so only (B) is false.

7. The secondary valency of Co in [Co(en)₃]³⁺ is [1 Mark]
(A) 4
(B) 3
(C) 5
(D) 6

Answer: (D) 6

Teacher's Note:
a) Secondary valency equals the coordination number of the metal.
b) Ethane-1,2-diamine (en) is bidentate, so 3 en ligands give 3 × 2 = 6 donor atoms.

8. At low temperature, phenol on reaction with Br₂ in CS₂ gives [1 Mark]
(A) 2, 4, 6-Tribromophenol
(B) a mixture of ortho-and para-bromophenol
(C) ortho-bromophenol only
(D) para-bromophenol only

Answer: (B) a mixture of ortho-and para-bromophenol

Teacher's Note:
a) CS₂ is a solvent of low polarity, so bromination is limited to mono-substitution.
b) With bromine water, 2,4,6-tribromophenol (white precipitate) is formed instead.

9. How the conductivity varies on decreasing concentration for both weak and strong electrolytes ? [1 Mark]
(A) It increases for weak electrolyte and decreases for strong electrolyte.
(B) It decreases for weak electrolyte and increases for strong electrolyte.
(C) It increases for both weak and strong electrolytes.
(D) It decreases for both weak and strong electrolytes.

Answer: (D) It decreases for both weak and strong electrolytes.

Teacher's Note:
a) Conductivity depends on the number of ions per unit volume, which falls on dilution.
b) Do not confuse conductivity with molar conductivity, which increases on dilution.

10. CH₃ - NH₂ on reaction with (CH₃CO)₂O gives [1 Mark]
(A) CH₃CONH₂
(B) CH₃COONHCH₃
(C) CH₃ - NH - CO - CH₃
(D) CH₃ - CO - CH₂ - NH₂

[Figure: Options (C) and (D) are drawn as structures with the C=O shown as a C atom with a double-bonded O below it: (C) CH₃-NH-C(=O)-CH₃, (D) CH₃-C(=O)-CH₂-NH₂.]

Answer: (C) \( CH_3 - NH - CO - CH_3 \) (N-methylethanamide)

Teacher's Note:
a) This is acetylation: the H of -NH₂ is replaced by an acetyl (CH₃CO-) group.
b) The by-product is CH₃COOH.

11. The activation energy of a reaction can be determined from the slope of which of the following curves ? [1 Mark]
(A) \( \ln k \) vs. \( T \)
(B) \( \ln k \) vs. \( \dfrac{1}{T} \)
(C) \( \ln \dfrac{k}{T} \) vs. \( T \)
(D) \( \dfrac{T}{\ln k} \) vs. \( \dfrac{1}{T} \)

Answer: (B) \( \ln k \) vs. \( \dfrac{1}{T} \)

Teacher's Note:
a) From the Arrhenius equation, \( \ln k = \ln A - \dfrac{E_a}{RT} \).
b) So the plot of \( \ln k \) against \( \dfrac{1}{T} \) is a straight line with slope \( -\dfrac{E_a}{R} \).

12. Identify 'X' in the following reaction :
CH₃ - CO - CH₂ - CO - O - CH₃ → (NaBH₄) 'X'
'X' is [1 Mark]
(A) CH₃ - CH(OH) - CH₂ - CO - O - CH₃
(B) CH₃ - CO - CH₂ - CH₂ - OH
(C) CH₃ - CH(OH) - CH₂ - CH₂ - OH
(D) CH₃ - CH(OH) - CH₂ - COOH

[Figure: Reactant CH₃-C(=O)-CH₂-C(=O)-O-CH₃ with NaBH₄ written over the arrow giving 'X'. In the options, OH groups are drawn below the carbon atoms and C=O groups are drawn with O below the carbon.]

Answer: (A) \( CH_3 - CH(OH) - CH_2 - CO - O - CH_3 \) (methyl 3-hydroxybutanoate)

Teacher's Note:
a) NaBH₄ is a mild reducing agent that reduces the ketone group to a secondary alcohol.
b) It does not reduce the ester group, so -COOCH₃ stays unchanged.

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) : The presence of -OH group in phenols directs the incoming group at ortho- and para- positions.
Reason (R) : -OH group in phenols deactivates the aromatic ring. [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) The -OH group activates the ring by resonance (+R effect), so R is false.
b) Electron density increases at ortho and para positions, which explains the orientation.

14. Assertion (A) : Actinoids show irregularities in their electronic configurations.
Reason (R) : Due to varying stabilities of f⁰, f⁷ and f¹⁴ occupancies of the 5f orbitals. [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) Empty, half-filled and fully filled 5f levels are extra stable.
b) This extra stability causes electrons to shift between 5f and 6d orbitals, giving irregular configurations.

15. Assertion (A) : The pentaacetate of glucose does not react with H₂N - OH.
Reason (R) : It indicates the absence of free -CHO group in glucose. [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) Glucose pentaacetate has no free -CHO group, so it cannot form an oxime with hydroxylamine.
b) This fact supports the cyclic (ring) structure of glucose.

16. Assertion (A) : It is not possible to separate the components of an azeotrope by fractional distillation.
Reason (R) : Minimum boiling azeotrope is formed by the solutions showing large positive deviation from Raoult's law. [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) An azeotrope boils at constant temperature and its vapour has the same composition as the liquid, so it cannot be separated.
b) R is a correct fact about azeotropes but does not explain why separation is impossible.

SECTION B

17. What type of deviation is shown by a mixture of chloroform and acetone from Raoult's law ? Give reason. What will happen to the boiling point of the solution on mixing chloroform and acetone ? [2 Marks]

Answer:
1. The mixture shows negative deviation from Raoult's law.
2. Reason: chloroform and acetone form hydrogen bonds with each other, which are stronger than the chloroform-chloroform and acetone-acetone interactions.
3. The boiling point of the solution increases.

Teacher's Note:
a) Stronger A-B attraction lowers the vapour pressure, so the boiling point rises.
b) Mention hydrogen bonding clearly; it carries the reason mark.

18. Following reaction takes place in one step :
2A + B → 2C
How will the rate of above reaction change if the volume of the reaction vessel is decreased to one third of its original volume ? Will there be any change in the order of reaction with the reduced volume ? [2 Marks]

Answer:
1. Since the reaction is elementary (one step), Rate = k[A]²[B].
2. When the volume is reduced to one third, each concentration becomes 3 times.
3. New rate = k[3A]²[3B] = 27 k[A]²[B], so the rate increases 27 times.
4. The order of reaction remains the same (3).

Teacher's Note:
a) For an elementary reaction, the order equals the stoichiometric coefficients.
b) Order depends on the rate law, not on the volume of the vessel.

19. (a) Write IUPAC names of the following coordination compounds :
(i) [Ag(NH₃)₂] [Ag(CN)₂]
(ii) K₃[Fe(C₂O₄)₃] [2 Marks]

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

Teacher's Note:
a) In an anionic complex, the metal name ends in "-ate" and Latin names are used (argentate, ferrate).
b) Name the cation first, then the anion, and show oxidation states in Roman numerals.

OR

(b) (i) Give a chemical test to show that [Co(NH₃)₅SO₄]Cl and [Co(NH₃)₅Cl]SO₄ are ionisation isomers.
(ii) What is meant by the 'Chelate effect' ? Give an example. [2 Marks]

Answer:
(i) [Co(NH₃)₅SO₄]Cl gives a white precipitate of AgCl with AgNO₃ solution, while [Co(NH₃)₅Cl]SO₄ does not. (Or: [Co(NH₃)₅Cl]SO₄ gives a white precipitate of BaSO₄ with BaCl₂, while the other does not.)
(ii) Chelate effect: the formation of more stable complexes when a di- or polydentate ligand binds to a single metal atom or ion. Example: [Co(en)₃]³⁺.

Teacher's Note:
a) Only the ion outside the square bracket is free to give a precipitate test.
b) Always give an example with a bidentate ligand such as en or oxalate for the chelate effect.

20. Differentiate between the following :
(i) Fibrous protein and Globular protein
(ii) Peptide linkage and Phosphodiester linkage [2 Marks]

Answer:
(i) In fibrous proteins, polypeptide chains run parallel and are held together by hydrogen and disulphide bonds, giving a fibre-like structure; they are insoluble in water. In globular proteins, the chains coil around to give a spherical shape; they are soluble in water.
(ii) A peptide linkage is an amide linkage (-CONH-) formed between the -COOH group of one amino acid and the -NH₂ group of another. A phosphodiester linkage joins two nucleotides through a phosphate group.

Teacher's Note:
a) Give examples to strengthen the answer: keratin (fibrous) and insulin (globular).
b) Peptide links are found in proteins; phosphodiester links are found in nucleic acids.

21. Why are haloarenes less reactive towards nucleophilic substitution reaction ? Give two reasons. [2 Marks]

Answer:
1. Due to resonance, the C-X bond gets partial double bond character, so it is difficult to break.
2. The carbon of the C-X bond is sp² hybridised; it is more electronegative and holds the bond electron pair more tightly, making the bond shorter and stronger.

Teacher's Note:
a) Other accepted reasons: instability of the phenyl cation, and repulsion between the nucleophile and the electron-rich ring.
b) Give exactly two reasons as asked.

SECTION C

22. Calculate the boiling point of a solution containing 0.61 g of benzoic acid (Molar mass = 122 g mol^-1) in 5 g of CS₂ in which it dimerises to the extent of 88%. The boiling point and K_b of CS₂ are 46.2 °C and 2.3 K kg mol^-1 respectively. [3 Marks]

Answer:
Given: \( W_B = 0.61 \, g \), \( M_B = 122 \, g \, mol^{-1} \), \( W_A = 5 \, g \), \( K_b = 2.3 \, K \, kg \, mol^{-1} \), \( T_b^{\circ} = 46.2^{\circ}C \), \( \alpha = 88\% = 0.88 \)
For dimerisation \( 2C_6H_5COOH \rightarrow (C_6H_5COOH)_2 \): \( \alpha = \dfrac{1 - i}{1 - \dfrac{1}{2}} \)
\( 0.88 = \dfrac{1 - i}{0.5} \Rightarrow i = 1 - 0.44 = 0.56 \)
Formula: \( \Delta T_b = i \, K_b \, \dfrac{W_B}{M_B} \times \dfrac{1000}{W_A} \)
Substitution: \( \Delta T_b = 0.56 \times 2.3 \times \dfrac{0.61}{122} \times \dfrac{1000}{5} = 0.56 \times 2.3 \times 1 = 1.288 \, K \)
\( T_b = 46.2 + 1.288 = 47.488^{\circ}C \) (or 320.638 K)
Boiling point of the solution = 47.488 °C

Teacher's Note:
a) For association into dimers, \( i = 1 - \dfrac{\alpha}{2} \), so i is less than 1.
b) A rise of 1.288 K equals a rise of 1.288 °C; add it to the boiling point of pure CS₂.

23. Write the reaction involved in the following :
(a) Reimer-Tiemann reaction
(b) Kolbe's reaction
(c) Friedal-Crafts acylation of anisole [3 Marks]

Answer:
(a) Reimer-Tiemann reaction: Phenol (C₆H₅OH) + CHCl₃ + aq. NaOH, then H⁺ → Salicylaldehyde (2-hydroxybenzaldehyde, -CHO group at the ortho position to -OH).
(b) Kolbe's reaction: C₆H₅OH + aq. NaOH → C₆H₅O^-Na⁺ (sodium phenoxide); C₆H₅O^-Na⁺ + CO₂, then H⁺ → Salicylic acid (2-hydroxybenzoic acid, -COOH at the ortho position to -OH).
(c) Friedel-Crafts acylation of anisole: C₆H₅OCH₃ + CH₃COCl (anhydrous AlCl₃) → 2-Methoxyacetophenone (minor) + 4-Methoxyacetophenone (major).

Teacher's Note:
a) Always write the reagents and conditions (aq. NaOH, CO₂, H⁺, anhydrous AlCl₃) over the arrow.
b) In (c), the -OCH₃ group is ortho/para directing, so show both products.

24. Give reasons for the following :
(a) Carboxylic acids have higher boiling point than alcohols of comparable molecular masses.
(b) Alpha (α) hydrogens of aldehydes and ketones are acidic in nature.
(c) Nucleophilic addition of ammonia and its derivatives does not occur with carbonyl group in strongly acidic medium. [3 Marks]

Answer:
(a) Carboxylic acid molecules are extensively associated through intermolecular hydrogen bonding and form dimers, so more energy is needed to separate them.
(b) The carbonyl group has a strong electron withdrawing effect, and the conjugate base (carbanion/enolate) formed after losing an α-hydrogen is stabilised by resonance.
(c) In strongly acidic medium, the nucleophile (ammonia or its derivative) gets protonated, loses its lone pair and can no longer attack the carbonyl carbon.

Teacher's Note:
a) The keywords "dimer", "resonance stabilisation" and "protonation of nucleophile" carry the marks.
b) Such reactions are carried out in weakly acidic medium, where the carbonyl oxygen is protonated but the nucleophile is free.

25. For the first order thermal decomposition reaction, following data was obtained :
C₂H₅Cl(g) → C₂H₄(g) + HCl(g)
S. No. | Time(s) | Total Pressure (atm)
1 | 0 | 0.30
2 | 30 | 0.50
Calculate rate constant. [Given : log 3 = 0.48] [3 Marks]

Answer:
C₂H₅Cl(g) → C₂H₄(g) + HCl(g)
At t = 0: \( P_i \), 0, 0
At time t: \( P_i - x \), \( x \), \( x \)
Total pressure \( P_t = P_i - x + x + x = P_i + x \), so \( x = P_t - P_i \) and \( P_i - x = 2P_i - P_t \)
Formula: \( k = \dfrac{2.303}{t} \log \dfrac{P_i}{2P_i - P_t} \)
Substitution: \( k = \dfrac{2.303}{30} \log \dfrac{0.30}{2 \times 0.30 - 0.50} = \dfrac{2.303}{30} \log \dfrac{0.30}{0.10} \)
\( k = \dfrac{2.303}{30} \times 0.48 \)
\( k = 0.037 \, s^{-1} \) (approximately \( 3.7 \times 10^{-2} \, s^{-1} \))

Teacher's Note:
a) The key step is expressing the pressure of the reactant as \( 2P_i - P_t \).
b) The unit of a first order rate constant is time^-1, here s^-1.

26. How do you explain the presence of following in open structure of glucose ?
(a) all the six carbon atoms are in a straight chain.
(b) five -OH groups which are attached to different carbon atoms.
(c) an aldehyde group. [3 Marks]

Answer:
(a) On prolonged heating with HI, glucose forms n-hexane: CHO-(CHOH)₄-CH₂OH → (HI, heat) CH₃-(CH₂)₄-CH₃. This shows all six carbon atoms are in a straight chain.
(b) On acetylation with acetic anhydride, glucose forms a stable pentaacetate: CHO-(CHOH)₄-CH₂OH → ((CH₃CO)₂O) CHO-(CH-OCOCH₃)₄-CH₂-OCOCH₃. This confirms five -OH groups on different carbon atoms.
(c) On oxidation with bromine water, glucose gives gluconic acid: CHO-(CHOH)₄-CH₂OH → (Br₂ water) COOH-(CHOH)₄-CH₂OH. This confirms the presence of an aldehyde group.

Teacher's Note:
a) Write the reagent and the product for each part; the reaction carries the mark.
b) The pentaacetate is stable, which is why the -OH groups must be on different carbon atoms.

27. Compound 'X' with molecular formula C₄H₉Br reacts with aqueous KOH to give an alcohol. The rate of this reaction depends only on the concentration of the compound 'X'. When an optically active isomer 'Y' of the compound 'X' was treated with aqueous KOH solution, the rate of reaction was found to be dependent on concentration of compound 'Y' and aqueous KOH both.
(a) Write down the structural formula of both 'X' and 'Y'.
(b) Out of 'X' and 'Y', which one will undergo racemisation and why ?
(c) Out of 'X' and 'Y', which one will form product with inversion of configuration and why ? [3 Marks]

Answer:
(a) X = (CH₃)₃C-Br (2-bromo-2-methylpropane); Y = CH₃-CH(Br)-CH₂-CH₃ (2-bromobutane).
(b) X, because its reaction follows the S_N1 mechanism, in which a planar carbocation intermediate is formed and can be attacked from either side. (The marking scheme also gives full marks for Y, since Y is the optically active compound.)
(c) Y, because its reaction follows the S_N2 mechanism, in which the nucleophile attacks from the rear side opposite to the leaving group.

Teacher's Note:
a) Rate depending on one reactant means S_N1; rate depending on both reactants means S_N2.
b) Tertiary halides prefer S_N1 and secondary chiral halides can undergo S_N2 with inversion.

28. (a) Answer the following :
(i) Why is the Equilibrium Constant (K_c) related to \( E^{\circ}_{cell} \) and not to E_cell ?
(ii) Two metals 'A' and 'B' have standard electrode potential values of -0.24 V and +0.80 V respectively. Which of these will liberate hydrogen gas from dil. H₂SO₄ ?
(iii) Write the cell reaction which occurs in lead storage battery when it is in charging. [3 Marks]

Answer:
(i) At equilibrium, E_cell = 0, so the Nernst equation gives \( E^{\circ}_{cell} = \dfrac{2.303RT}{nF} \log K_c \). Hence K_c is related to \( E^{\circ}_{cell} \).
(ii) Metal 'A' (E° = -0.24 V), because its reduction potential is lower than that of hydrogen (0.00 V).
(iii) 2PbSO₄(s) + 2H₂O(l) → PbO₂(s) + Pb(s) + 2H₂SO₄(aq)

Teacher's Note:
a) Metals with negative E° values lie above hydrogen and can displace it from dilute acids.
b) Charging is the reverse of discharging; PbSO₄ is converted back to Pb and PbO₂.

OR

(b) What type of battery is Mercury cell ? Why it is more advantageous than dry cell ? Write overall reaction taking place in Mercury cell. [3 Marks]

Answer:
1. Mercury cell is a primary cell (battery).
2. It is more advantageous because it provides a constant potential (about 1.35 V) throughout its life, as the overall reaction involves no ions in solution.
3. Overall reaction: Zn(Hg) + HgO(s) → ZnO(s) + Hg(l)

Teacher's Note:
a) A primary cell cannot be recharged; its reaction occurs only once.
b) Write the physical states (s) and (l) in the overall reaction.

SECTION D

29. The reaction of amines with mineral acids to form ammonium salts shows that these are basic in nature. Aliphatic amines are stronger bases than ammonia whereas aromatic amines are weaker bases than ammonia. Aliphatic and aromatic primary and secondary amines react with acid chlorides, anhydrides and esters by nucleophilic substitution reaction. The main problem encountered during electrophilic substitution reactions of aromatic amines is that of their high reactivity. Substitution tends to occur at ortho-and para-positions. Hinsberg reagent is used for the identification and distinction between primary, secondary and tertiary amines. Aryldiazonium salts, usually obtained from arylamines, undergo replacement of the diazonium group with a variety of nucleophiles to provide advantageous methods for producing aryl halides, cyanides, phenols and arenes.
Answer the following questions :

(a) (i) Why CH₃ - NH₂ is a stronger base than (CH₃)₃N in aqueous solution ?
(ii) Write structural formulae of the compound A and B :
CH₃CONH₂ → (NaOBr) A → (C₆H₅COCl, Base) B [2 Marks]

Answer:
(i) In aqueous solution, basic strength depends on the combined effect of the inductive effect, solvation (hydrogen bonding) effect and steric hindrance of the alkyl groups. In (CH₃)₃N, the three methyl groups cause steric hindrance and it has no N-H to stabilise its cation by hydrogen bonding with water, so CH₃NH₂ is the stronger base.
(ii) A = CH₃NH₂ (methanamine, by Hoffmann bromamide degradation); B = CH₃-NH-CO-C₆H₅ (N-methylbenzamide, by benzoylation).

Teacher's Note:
a) Name all three factors (inductive, solvation, steric) to get the full mark in (i).
b) Hoffmann bromamide degradation gives an amine with one carbon less than the amide.

(b) A compound 'X' with molecular formula C₃H₉N reacts with Hinsberg reagent to give a product insoluble in alkali. Identify 'X'. [1 Mark]

Answer: X is a secondary amine: CH₃-CH₂-NH-CH₃ (N-methylethanamine).

Teacher's Note:
a) Secondary amines give a sulphonamide with no H on nitrogen, so it is insoluble in alkali.
b) A primary amine gives an alkali-soluble product; a tertiary amine does not react.

OR

(b) How can you convert aniline to benzonitrile ? [1 Mark]

Answer:
C₆H₅NH₂ + NaNO₂ + HCl (0-5 °C) → C₆H₅N₂⁺Cl^- (benzenediazonium chloride)
C₆H₅N₂⁺Cl^- + CuCN → C₆H₅CN (benzonitrile) + N₂

Teacher's Note:
a) Diazotisation must be done at 0-5 °C, as the diazonium salt is unstable at higher temperature.
b) Replacement of the diazonium group by -CN using CuCN is a Sandmeyer reaction.

(c) Why is -NH₂ group of aniline acetylated before carrying out nitration ? [1 Mark]

Answer: Acetylation reduces the activating effect of -NH₂, so the nitration can be controlled and the p-nitro derivative is obtained as the major product. It also prevents the formation of the anilinium ion, which is meta directing.

Teacher's Note:
a) Direct nitration of aniline also causes oxidation and gives tarry products.
b) After nitration, the acetyl group is removed by hydrolysis to get p-nitroaniline.

30. The Valence Bond Theory (VBT) explains the formation, magnetic behaviour and geometry of coordination compounds. The Crystal Field Theory (CFT) of coordination compounds is based on the effect of different crystal fields (provided by the ligands taken as point charges), on the degeneracy of d-orbital energies of the central metal atom/ion. The splitting of the d-orbitals provides different electronic arrangements in strong and weak crystal fields.
Answer the following questions :

(a) In octahedral crystal field, energies of which d-orbitals will be raised when ligands approach the central metal atom/ion ? Give reason in support of your answer. [2 Marks]

Answer:
1. The energies of the \( d_{x^2 - y^2} \) and \( d_{z^2} \) orbitals (the e_g set) are raised.
2. Reason: in an octahedral field, the ligands approach the metal atom/ion along the axes. The \( d_{x^2 - y^2} \) and \( d_{z^2} \) orbitals point along the axes, so they experience more repulsion from the ligands.

Teacher's Note:
a) The t_2g orbitals (d_xy, d_yz, d_xz) lie between the axes and are lowered in energy.
b) Mention "along the axes" clearly; it is the key point of the reason.

(b) Using crystal field theory, write the electronic configuration of central metal atom/ion of the following :
(i) [CoF₆]^3-
(ii) [Co(NH₃)₆]³⁺ [At. No. : Co = 27] [1 Mark]

Answer:
(i) [CoF₆]^3-: Co³⁺ = 3d⁶; F^- is a weak field ligand, so the configuration is t_2g⁴ e_g².
(ii) [Co(NH₃)₆]³⁺: Co³⁺ = 3d⁶; NH₃ is a strong field ligand, so the configuration is t_2g⁶ e_g⁰.

Teacher's Note:
a) Weak field ligands give high spin complexes because \( \Delta_o \) is less than the pairing energy.
b) Strong field ligands force pairing in t_2g before filling e_g.

(c) [NiCl₄]^2- is paramagnetic while [Ni(CO)₄] is diamagnetic though both are tetrahedral. Why ? [Atomic No. : Ni = 28] [1 Mark]

Answer:
1. In [NiCl₄]^2-, Ni²⁺ is 3d⁸ and Cl^- is a weak field ligand that cannot pair up the two unpaired 3d electrons, so it is paramagnetic.
2. In [Ni(CO)₄], Ni is in the zero oxidation state (3d⁸4s²); CO is a strong field ligand that pairs up all the electrons (3d¹⁰), so it is diamagnetic.

Teacher's Note:
a) First find the oxidation state of Ni in each complex.
b) Unpaired electrons make a complex paramagnetic; no unpaired electrons make it diamagnetic.

OR

(c) Write hybridization and magnetic behaviour of the complex [Fe(CN)₆]^3-. [Atomic No. : Fe = 26] [1 Mark]

Answer:
Hybridisation: d²sp³
Magnetic behaviour: Paramagnetic (Fe³⁺ is 3d⁵; with strong field CN^- one electron stays unpaired, t_2g⁵).

Teacher's Note:
a) CN^- is a strong field ligand, so inner d orbitals (3d) are used, giving d²sp³.
b) Even after pairing, one unpaired electron remains, so the complex is paramagnetic.

SECTION E

31. (a) (i) From the given data of E° values, answer the following questions :
\( E^{\circ}_{M^{2+}/M} \) : V | Cr | Mn | Fe | Co | Ni | Cu
Value (V) : -1.18 | -0.91 | -1.18 | -0.44 | -0.28 | -0.25 | +0.34
(I) Why \( E^{\circ}_{M^{2+}/M} \) show irregular trend in the above values ?
(II) Why is \( E^{\circ}_{Cu^{2+}/Cu} \) value exceptionally positive ?
(III) Why \( E^{\circ}_{Mn^{2+}/Mn} \) value is highly negative ?
(ii) Write the ionic equations for the oxidising action of potassium permanganate for its reaction with I^- in both acidic and alkaline solutions. [5 Marks]

Answer:
(i) (I) The irregular trend is due to irregular variation of ionisation enthalpies (\( \Delta_i H_1 + \Delta_i H_2 \)) and sublimation (atomisation) enthalpies of these metals.
(II) Cu has a high enthalpy of atomisation (\( \Delta_a H^{\circ} \)) and a low hydration enthalpy (\( \Delta_{hyd} H^{\circ} \)), which is not enough to compensate for the energy needed to form Cu²⁺(aq). So its E° value is positive.
(III) Mn²⁺ is highly stable due to its half-filled 3d⁵ configuration, so the value is highly negative.
(ii) Acidic solution: 2MnO₄^- + 10I^- + 16H⁺ → 2Mn²⁺ + 8H₂O + 5I₂
Alkaline (neutral or faintly alkaline) solution: 2MnO₄^- + I^- + H₂O → 2MnO₂ + IO₃^- + 2OH^-

Teacher's Note:
a) In acidic medium I^- is oxidised to I₂, but in alkaline medium it is oxidised to iodate (IO₃^-).
b) Balance charges on both sides; unbalanced ionic equations lose marks.
c) Link each E° anomaly to a stable configuration or an enthalpy term.

OR

(b) Answer the following questions :
(i) Name a member of the lanthanoid series
(I) which exhibits +4 oxidation state
(II) which exhibits +2 oxidation state.
(ii) Why transition metals act as good catalyst ?
(iii) Why Cr has higher melting point than Mn ?
(iv) What happens when acidic solution of potassium permanganate is allowed to stand for sometime ? Give the equation involved. What is this type of reaction called ? [5 Marks]

Answer:
(i) (I) Cerium (Ce) shows +4 oxidation state.
(II) Europium (Eu) (or Ytterbium) shows +2 oxidation state.
(ii) Transition metals act as good catalysts because of their ability to show multiple oxidation states and to form complexes; they also provide a large surface area for the reaction.
(iii) In Cr, a greater number of (n-1)d and ns electrons take part in interatomic metallic bonding, making the bonding stronger. In Mn, the stable 3d⁵ configuration makes d electrons less available, so its melting point is lower.
(iv) On standing, acidic KMnO₄ slowly decomposes and brown MnO₂ is deposited with evolution of oxygen:
4MnO₄^- + 4H⁺ → 4MnO₂ + 3O₂ + 2H₂O
This is a redox reaction (the marking scheme also accepts the reduction half reaction MnO₄^- + 4H⁺ + 3e^- → MnO₂ + 2H₂O, called a reduction reaction).

Teacher's Note:
a) Ce⁴⁺ (4f⁰) and Eu²⁺ (4f⁷) are stable because of empty and half-filled f subshells.
b) Part (iv) carries 2 marks: 1 for the equation and 1 for naming the type of reaction.

32. (a) Calculate emf and ΔG for the following cell at 298 K :
Mg(s) / Mg²⁺(0.01 M) // Ag⁺(0.001 M) / Ag(s)
Given : \( E^{\circ}_{Mg^{2+}/Mg} = -2.37 \, V \) \( E^{\circ}_{Ag^{+}/Ag} = +0.80 \, V \)
[1 F = 96500 C mol^-1, log 10 = 1] [5 Marks]

Answer:
Cell reaction: Mg(s) + 2Ag⁺(aq) → Mg²⁺(aq) + 2Ag(s); n = 2
\( E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.80 - (-2.37) = 3.17 \, V \)
Formula: \( E_{cell} = E^{\circ}_{cell} - \dfrac{0.059}{2} \log \dfrac{[Mg^{2+}]}{[Ag^{+}]^2} \)
Substitution: \( E_{cell} = 3.17 - \dfrac{0.059}{2} \log \dfrac{0.01}{(0.001)^2} = 3.17 - \dfrac{0.059}{2} \log 10^4 \)
\( E_{cell} = 3.17 - \dfrac{0.059}{2} \times 4 = 3.17 - 0.118 = 3.052 \, V \)
Formula: \( \Delta G = -nFE_{cell} \)
Substitution: \( \Delta G = -2 \times 96500 \times 3.052 \)
\( \Delta G = -589036 \, J \, mol^{-1} \) or \( -589.036 \, kJ \, mol^{-1} \)

Teacher's Note:
a) Square the Ag⁺ concentration because 2 Ag⁺ ions take part in the balanced reaction.
b) A negative ΔG confirms that the cell reaction is spontaneous.
c) Always write units: V for emf and J mol^-1 for ΔG.

OR

(b) For the reaction :
2AgCl(s) + H₂(g) (0.4 atm) → 2Ag(s) + 2H⁺(0.1 M) + 2Cl^-(0.2 M)
Calculate emf of the cell at 25 °C.
Given : ΔG° = - 43500 J mol^-1
[log 10 = 1, 1 F = 96500 C mol^-1] [5 Marks]

Answer:
Given: \( \Delta G^{\circ} = -43500 \, J \, mol^{-1} \), n = 2, \( P_{H_2} = 0.4 \, atm \), [H⁺] = 0.1 M, [Cl^-] = 0.2 M
Formula: \( \Delta G^{\circ} = -nFE^{\circ}_{cell} \)
\( -43500 = -2 \times 96500 \times E^{\circ}_{cell} \)
\( E^{\circ}_{cell} = 0.225 \, V \)
Formula: \( E_{cell} = E^{\circ}_{cell} - \dfrac{0.059}{2} \log \dfrac{[H^{+}]^2 [Cl^{-}]^2}{P_{H_2}} \)
Substitution: \( E_{cell} = 0.225 - \dfrac{0.059}{2} \log \dfrac{(0.1)^2 (0.2)^2}{0.4} = 0.225 - \dfrac{0.059}{2} \log 10^{-3} \)
\( E_{cell} = 0.225 + \dfrac{0.059}{2} \times 3 = 0.225 + 0.0885 \)
\( E_{cell} = 0.3135 \, V \)

Teacher's Note:
a) Solids (AgCl and Ag) are not written in the reaction quotient; gases are written as partial pressures.
b) \( \dfrac{0.01 \times 0.04}{0.4} = 10^{-3} \); a negative log makes the second term positive.
c) First find \( E^{\circ}_{cell} \) from ΔG°, then apply the Nernst equation.

33. (a) (i) An organic compound (X) has the molecular formula C₅H₁₀O. Draw structures for (X) if it :
(I) does not give Tollen's test but gives a positive iodoform test.
(II) does not give Tollen's test and iodoform test but undergoes Aldol condensation.
(III) undergoes Cannizzaro's reaction.
(ii) Show how each of the following compounds can be converted to benzoic acid :
(I) Acetophenone (II) Ethyl benzene [5 Marks]

Answer:
(i) (I) Pentan-2-one: CH₃-CO-CH₂-CH₂-CH₃ (a methyl ketone, so it gives the iodoform test).
(II) Pentan-3-one: CH₃-CH₂-CO-CH₂-CH₃ (a ketone with α-hydrogens but no CH₃CO- group).
(III) 2,2-Dimethylpropanal: (CH₃)₃C-CHO (an aldehyde with no α-hydrogen).
(ii) (I) C₆H₅COCH₃ → (NaOI, then H⁺) C₆H₅COOH + CHI₃
(or C₆H₅COCH₃ → (KMnO₄-KOH, heat) C₆H₅COOK → (H₃O⁺) C₆H₅COOH)
(II) C₆H₅CH₂CH₃ → ((i) KMnO₄-KOH, heat (ii) H⁺) C₆H₅COOH

Teacher's Note:
a) No Tollen's test means the compound is a ketone; iodoform test needs a CH₃CO- group.
b) Cannizzaro reaction is given only by aldehydes without α-hydrogen.
c) On strong oxidation, the whole side chain of an alkylbenzene becomes -COOH.

OR

(b) Answer the following questions :
(i) Draw structure of the 2, 4-dinitrophenyl hyzdrazone derivative of benzaldehyde.
(ii) Arrange the following in increasing order of their reactivity towards HCN :
Di-tert. butyl ketone, Acetaldehyde, Acetone
(iii) Give a simple chemical test to distinguish between benzoic acid and ethyl benzoate.
(iv) Write the name of the reagent to convert Ethanenitrile to Ethanal.
(v) Draw the structure of 'X' in the following reaction : [5 Marks]

[Figure: In (v), the reaction shows cyclohexanol (a cyclohexane ring with an -OH group attached) with CrO₃ written over the arrow, giving 'X'.]

Answer:
(i) C₆H₅-CH=N-NH-C₆H₃(NO₂)₂: the benzylidene carbon is double bonded to N, which is bonded to NH attached to a benzene ring carrying -NO₂ groups at positions 2 and 4.
(ii) Di-tert-butyl ketone \( \lt \) Acetone \( \lt \) Acetaldehyde
(iii) Add NaHCO₃ solution to both. Benzoic acid gives brisk effervescence of CO₂, while ethyl benzoate does not.
(iv) (i) DIBAL-H (ii) H₃O⁺ (or SnCl₂ + HCl followed by H₃O⁺, the Stephen reaction).
(v) X = Cyclohexanone: a six-membered carbon ring with a C=O group in place of the CH-OH group.

Teacher's Note:
a) Reactivity towards nucleophiles falls as bulky alkyl groups increase steric hindrance and electron donation.
b) CrO₃ oxidises a secondary alcohol to a ketone.
c) In the 2,4-DNP derivative, the C=O of the aldehyde is replaced by C=N.

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