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

SECTION A

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

[Figure: Four graphs. (A) t_½ (y-axis) against [R]₀ (x-axis): a straight line rising from the origin. (B) t_½ against [R]₀: a horizontal straight line parallel to the x-axis. (C) Rate against Concentration: a horizontal straight line parallel to the x-axis. (D) Rate against Concentration: a curve falling steeply and then flattening out.]

(A) Graph of t_½ against [R]₀ - straight line through the origin
(B) Graph of t_½ against [R]₀ - horizontal line
(C) Graph of Rate against Concentration - horizontal line
(D) Graph of Rate against Concentration - falling curve

Answer: (B) Graph of \( t_{1/2} \) against \( [R]_0 \) - horizontal line

Teacher's Note:
a) For a first order reaction, \( t_{1/2} = \frac{0.693}{k} \), so half-life does not depend on initial concentration.
b) A straight line through the origin for t_½ against [R]₀ belongs to a zero order reaction.

2. Which of the following solutions will have the lowest freezing point in water ? [1 Mark]
(A) 0.1 M Glucose
(B) 0.1 M CaCl₂
(C) 0.1 M KCl
(D) 0.1 M Urea

Answer: (B) 0.1 M \( CaCl_2 \)

Teacher's Note:
a) CaCl₂ gives 3 ions (i = 3), so it causes the largest depression in freezing point.
b) At equal molarity, compare the van't Hoff factor i: glucose and urea have i = 1, KCl has i = 2.

3. Which of the following is not a transition metal ? [1 Mark]
(A) Sc
(B) Ag
(C) Hg
(D) Cu

Answer: (C) Hg

Teacher's Note:
a) Hg has a completely filled 5d¹⁰ configuration in the ground state and in its common oxidation state (Hg²⁺).
b) Zn, Cd and Hg are not regarded as transition metals for this reason.

4. Which of the following represents the fraction of molecules with energies equal to or greater than E_a ? [1 Mark]
(A) \( +\frac{E_a}{RT} \)
(B) \( e^{-E_a/RT} \)
(C) \( -\frac{E_a}{RT} \)
(D) \( e^{+E_a/RT} \)

Answer: (B) \( e^{-E_a/RT} \)

Teacher's Note:
a) This factor comes from the Arrhenius equation \( k = A e^{-E_a/RT} \).
b) A fraction must be less than 1, so the exponent must be negative.

5. What will happen during the electrolysis of aqueous solution of CuCl₂ by using platinum electrodes ? [1 Mark]
(A) Cu will deposit at Anode
(B) H₂ gas will be released at cathode
(C) O₂ gas will be released at anode
(D) Cl₂ gas will be released at anode

Answer: (D) \( Cl_2 \) gas will be released at anode

Teacher's Note:
a) At the cathode, Cu²⁺ is reduced to Cu because it is reduced more easily than water.
b) At the anode, Cl^- is oxidised to Cl₂ because of the overpotential of oxygen.

6. Aspirin is obtained by acetylation of [1 Mark]
(A) Phenol
(B) Salicylaldehyde
(C) 2-Hydroxybenzoic acid
(D) Benzoic acid

Answer: (C) 2-Hydroxybenzoic acid

Teacher's Note:
a) 2-Hydroxybenzoic acid is salicylic acid; its -OH group is acetylated by acetic anhydride.
b) Aspirin is acetylsalicylic acid (2-acetoxybenzoic acid).

7. The secondary valency of Co in the complex [Co(NH₃)₅(NO₂)]²⁺ is [1 Mark]
(A) 5
(B) 1
(C) 4
(D) 6

Answer: (D) 6

Teacher's Note:
a) Secondary valency equals the coordination number: 5 NH₃ + 1 NO₂^- = 6.
b) Do not confuse it with primary valency (oxidation state), which is +3 here.

8. At low temperature, phenol reacts with dil. HNO₃ to yield [1 Mark]
(A) 2, 4, 6-Trinitrophenol
(B) o-Nitrophenol only
(C) p-Nitrophenol only
(D) ortho-and para-nitrophenol

Answer: (D) ortho-and para-nitrophenol

Teacher's Note:
a) The -OH group is ortho and para directing, so a mixture of o- and p-nitrophenol forms.
b) 2, 4, 6-Trinitrophenol (picric acid) forms only with concentrated HNO₃.

9. 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.

10. 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 acidic medium, aniline is protonated to the anilinium ion, which is meta directing.
b) This is why a large amount (47%) of meta product forms.

11. Which of the following amines has lowest pK_b value ? [1 Mark]
(A) C₆H₅ - N(CH₃)₂
(B) C₆H₅ - NH(CH₃)
(C) C₆H₅ - NH₂
(D) O₂N-C₆H₄-NH₂ (p-nitroaniline)

[Figure: Option (D) is drawn as a benzene ring with an NO₂ group on one carbon and an NH₂ group on the opposite (para) carbon.]

Answer: (A) \( C_6H_5 - N(CH_3)_2 \)

Teacher's Note:
a) Lowest pK_b means the strongest base.
b) Two electron-releasing CH₃ groups increase electron density on N, while the -NO₂ group in p-nitroaniline decreases it.

12. The deficiency of which of the following vitamins causes increased fragility of red blood cells and muscular weakness : [1 Mark]
(A) Vitamin A
(B) Vitamin K
(C) Vitamin E
(D) Vitamin D

Answer: (C) Vitamin E

Teacher's Note:
a) Vitamin E deficiency causes increased fragility of RBCs and muscular weakness.
b) Remember: Vitamin K - blood clotting, Vitamin D - rickets, Vitamin A - night blindness.

13. Assertion (A) : It is not possible to separate the components of an azeotrope by fractional distillation.
Reason (R) : Components of an azeotrope have the same composition in liquid and vapour phase and boil at a constant temperature. [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) Fractional distillation works only when the vapour is richer in one component.
b) An azeotrope gives vapour of the same composition as the liquid, so no separation happens.

14. Assertion (A) : The presence of -OH group in phenols directs the incoming group to meta position in the ring.
Reason (R) : -OH group in phenols activates the aromatic ring towards electrophilic substitution reaction. [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: (D) Assertion (A) is false, but Reason (R) is true.

Teacher's Note:
a) The -OH group is ortho and para directing, not meta directing.
b) It activates the ring by increasing electron density through resonance.

15. Assertion (A) : Actinoids show wide range of oxidation states.
Reason (R) : Actinoids are radioactive in nature. [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 wide range of oxidation states is due to comparable energies of 5f, 6d and 7s orbitals.
b) Radioactivity is true but is not the reason for variable oxidation states.

16. Assertion (A) : The pentaacetate of glucose does not react with H₂N - OH.
Reason (R) : It indicates the presence 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: (C) Assertion (A) is true, but Reason (R) is false.

Teacher's Note:
a) No reaction with hydroxylamine shows the absence of a free -CHO group in the pentaacetate.
b) This supports the cyclic structure of glucose.

SECTION B

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

Answer:
1. The mixture shows negative deviation from Raoult's law.
2. Reason: intermolecular hydrogen bonding between the phenolic proton and the lone pair on the nitrogen atom of aniline is stronger than the hydrogen bonding between similar molecules. So solute-solvent interactions are stronger than solute-solute and solvent-solvent interactions.
3. The boiling point of the solution increases.

Teacher's Note:
a) Stronger A-B attractions lower the vapour pressure, which raises the boiling point.
b) Such mixtures form maximum boiling azeotropes.

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

Answer:
1. Resonance effect: the C-X bond acquires partial double bond character due to resonance, so it is difficult to break.
2. Hybridisation: the carbon of the C-X bond is sp² hybridised; it is more electronegative and holds the electron pair of the C-X bond more tightly.

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 with the key words "resonance" and "sp² carbon".

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; silver becomes argentate and iron becomes ferrate.
b) Name the cation first and write the oxidation state 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) Add AgNO₃ solution to both. [Co(NH₃)₅SO₄]Cl gives a white precipitate of AgCl, while [Co(NH₃)₅Cl]SO₄ does not. (Or: with BaCl₂ solution, [Co(NH₃)₅Cl]SO₄ gives a white precipitate of BaSO₄, while the other does not.)
(ii) Chelate effect: the formation of more stable complexes when a didentate or polydentate ligand binds to a single metal atom or ion. Example: [Co(en)₃]³⁺.

Teacher's Note:
a) Ionisation isomers give different ions in solution, so the test must detect the ion outside the bracket.
b) For the chelate effect, the key words are "polydentate ligand" and "more stable complex".

20. Differentiate between the following :
(i) Peptide linkage and Glycosidic linkage
(ii) Essential amino acids and Non-essential amino acids [2 Marks]

Answer:
(i) A peptide linkage is an amide linkage (-CONH-) formed between two amino acids, by reaction of the -COOH group of one with the -NH₂ group of the other. In a glycosidic linkage, two monosaccharide units are joined together through an oxygen atom.
(ii) Essential amino acids cannot be synthesised in the body and must be obtained through diet. Non-essential amino acids can be synthesised in the body.

Teacher's Note:
a) Peptide linkage is found in proteins; glycosidic linkage is found in carbohydrates.
b) Examples of essential amino acids: valine, leucine, lysine; non-essential: glycine, alanine.

21. 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, rate = k[A]²[B].
2. When the volume is decreased to one third, the concentration of each reactant 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 a one-step reaction, the stoichiometric coefficients give the order.
b) Order does not change with volume; only the rate changes.

SECTION C

22. 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:
Given: initial pressure \( P_i = 0.30 \) atm, total pressure \( P_t = 0.50 \) atm at t = 30 s.
At t = 0: C₂H₅Cl = \( P_i \), C₂H₄ = 0, HCl = 0
At time t: C₂H₅Cl = \( P_i - x \), C₂H₄ = \( x \), HCl = \( x \)
\( P_t = P_i - x + x + x = P_i + x \), so \( x = P_t - P_i \) and \( P_i - x = 2P_i - P_t \)
Formula: \( k = \frac{2.303}{t} \log \frac{P_i}{2P_i - P_t} \)
Substitution: \( k = \frac{2.303}{30} \log \frac{0.30}{2 \times 0.30 - 0.50} = \frac{2.303}{30} \log \frac{0.30}{0.10} \)
\( k = \frac{2.303}{30} \times \log 3 = \frac{2.303}{30} \times 0.48 \)
Answer: \( k = 0.037 \, s^{-1} \) (approximately)

Teacher's Note:
a) The key step is expressing the pressure of the reactant at time t as \( 2P_i - P_t \).
b) The unit of k for a first order reaction is s^-1 (time^-1), not atm^-1.

23. (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) Because at equilibrium, \( E_{cell} = 0 \); so K_c can only be related to \( E^{\circ}_{cell} \) through \( E^{\circ}_{cell} = \frac{0.059}{n} \log K_c \).
(ii) Metal 'A' (it has a negative reduction potential, so it can reduce H⁺ to H₂).
(iii) 2PbSO₄(s) + 2H₂O(l) → PbO₂(s) + Pb(s) + 2H₂SO₄(aq)

Teacher's Note:
a) Metals with E° less than 0 V (below hydrogen) displace H₂ 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.
3. Overall reaction: Zn(Hg) + HgO(s) → ZnO(s) + Hg(l)

Teacher's Note:
a) The overall reaction has no ions whose concentration changes, so the voltage stays constant.
b) Write Zn(Hg) to show zinc-mercury amalgam as the anode.

24. 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, \( \alpha = 88\% = 0.88 \)
For dimerisation, \( \alpha = \frac{1 - i}{1 - \frac{1}{2}} \), so \( 0.88 = \frac{1 - i}{0.5} \)
\( i = 1 - 0.44 = 0.56 \)
Formula: \( \Delta T_b = i K_b m = i K_b \times \frac{W_B}{M_B} \times \frac{1000}{W_A} \)
Substitution: \( \Delta T_b = 0.56 \times 2.3 \times \frac{0.61}{122} \times \frac{1000}{5} = 1.288 \) K (or °C)
\( T_b = T_b^{\circ} + \Delta T_b = 46.2 + 1.288 \)
Answer: \( T_b = 47.488 \) °C (320.638 K)

Teacher's Note:
a) In association, i is less than 1; for a dimer, \( i = 1 - \frac{\alpha}{2} \).
b) A rise of 1.288 K is the same as a rise of 1.288 °C, so add it directly to 46.2 °C.

25. Write the reactions of D-Glucose with the following :
(a) HI
(b) Br₂ water
(c) Conc. HNO₃ [3 Marks]

Answer:
(a) CHO-(CHOH)₄-CH₂OH + HI (heat) → CH₃-(CH₂)₄-CH₃ (n-hexane)
(b) CHO-(CHOH)₄-CH₂OH + Br₂ water → COOH-(CHOH)₄-CH₂OH (gluconic acid)
(c) CHO-(CHOH)₄-CH₂OH + Conc. HNO₃ → COOH-(CHOH)₄-COOH (saccharic acid)

Teacher's Note:
a) Formation of n-hexane shows that the six carbon atoms of glucose are in a straight chain.
b) Bromine water oxidises only the -CHO group, while nitric acid oxidises both the -CHO and the primary -CH₂OH groups.

26. 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 boil them.
(b) The carbonyl group has a strong electron withdrawing effect, and the conjugate base (enolate ion) formed after losing an α-hydrogen is stabilised by resonance.
(c) In strongly acidic medium, the nucleophile (ammonia or its derivative) gets protonated, so it loses its lone pair and can no longer act as a nucleophile.

Teacher's Note:
a) The key word for (a) is "dimer formation through hydrogen bonding".
b) For (c), remember that the reaction needs a weakly acidic medium (pH about 3.5).

27. 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) Phenol (C₆H₅OH) + CHCl₃ + aq. NaOH, then H⁺ → Salicylaldehyde (2-hydroxybenzaldehyde, HO-C₆H₄-CHO, with -CHO ortho to -OH)
(b) Phenol + aq. NaOH → Sodium phenoxide (C₆H₅ONa); then sodium phenoxide + CO₂, followed by H⁺ → Salicylic acid (2-hydroxybenzoic acid, HO-C₆H₄-COOH, with -COOH ortho to -OH)
(c) Anisole (C₆H₅OCH₃) + CH₃COCl, in presence of anhydrous AlCl₃ → 2-Methoxyacetophenone (minor) + 4-Methoxyacetophenone (major) (CH₃O-C₆H₄-COCH₃)

Teacher's Note:
a) Always write the reagents and conditions above the arrow; they carry marks.
b) The -OCH₃ group is ortho and para directing, so both products must be shown in (c).

28. 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. (The marking scheme also gives full marks if the student writes Y, since Y is the optically active one.)
(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 only on [X] means S_N1 (tertiary halide); rate depending on both means S_N2.
b) Y has a chiral carbon (C-2 attached to H, Br, CH₃ and C₂H₅), so it is optically active.

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 \( \xrightarrow{NaOBr} \) A \( \xrightarrow[Base]{C_6H_5COCl} \) B [2 Marks]

Answer:
(i) In aqueous solution, basic strength depends on the combined effect of the inductive effect, the solvation effect (hydrogen bonding with water) and the steric hindrance of the alkyl groups. The CH₃NH₃⁺ ion is better stabilised by solvation, while in (CH₃)₃N the bulky methyl groups cause steric hindrance and poor solvation.
(ii) A = CH₃NH₂ (methanamine, by Hoffmann bromamide degradation); B = CH₃-NHCOC₆H₅ (N-methylbenzamide, by benzoylation).

Teacher's Note:
a) Mention all three factors - inductive effect, solvation and steric hindrance - for full credit 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 CH₃-CH₂-NH-CH₃ (N-Methylethanamine), a secondary amine.

Teacher's Note:
a) A sulphonamide insoluble in alkali is formed only by a secondary amine, as it has no H on nitrogen.
b) Choose the secondary amine that fits C₃H₉N.

OR

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

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

Teacher's Note:
a) Diazotisation must be done at 0 - 5 °C because the diazonium salt is unstable at higher temperature.
b) Replacement of -N₂⁺ by -CN using CuCN is the Sandmeyer reaction.

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

Answer: Acetylation controls the reaction, so that the p-nitro derivative can be obtained as the major product; it also prevents the formation of the anilinium ion, which is meta directing.

Teacher's Note:
a) Acetylation reduces the activating effect of -NH₂ and protects it from oxidation by HNO₃.
b) The acetyl group is removed by hydrolysis after nitration to give 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 (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.

Teacher's Note:
a) The t_2g orbitals (d_xy, d_yz, d_xz) lie between the axes and are lowered in energy.
b) The key word in the reason is "ligands approach along the axes".

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

Teacher's Note:
a) Weak field ligands give \( \Delta_o \lt P \), so electrons enter e_g before pairing.
b) Strong field ligands give \( \Delta_o \gt P \), so all 6 electrons pair up in t_2g.

(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, so it does not pair up the two unpaired electrons in the 3d orbitals; hence it is paramagnetic.
2. In [Ni(CO)₄], Ni is in the 0 state (3d⁸ 4s²) and CO is a strong field ligand, which pairs up all the electrons (3d¹⁰); hence it is diamagnetic.

Teacher's Note:
a) Both are sp³ hybridised; the difference lies only in the presence of unpaired electrons.
b) Mention the nature of the ligand (weak field or strong field) for each complex.

OR

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

Answer:
1. Hybridisation: d²sp³
2. Magnetic behaviour: Paramagnetic

Teacher's Note:
a) Fe³⁺ is 3d⁵; strong field CN^- pairs the electrons, leaving one unpaired electron.
b) Two inner 3d orbitals are used, so it is an inner orbital complex.

SECTION E

31. (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) CH₃-CO-CH₂-CH₂-CH₃ (Pentan-2-one)
(II) CH₃-CH₂-CO-CH₂-CH₃ (Pentan-3-one)
(III) (CH₃)₃C-CHO (2,2-Dimethylpropanal)
(ii) (I) Acetophenone: C₆H₅COCH₃ + NaOI (I₂/NaOH), then H⁺ → C₆H₅COOH + CHI₃
(or C₆H₅COCH₃ + KMnO₄-KOH, heat → C₆H₅COOK; then C₆H₅COOK + H₃O⁺ → C₆H₅COOH)
(II) Ethyl benzene: C₆H₅CH₂CH₃ + (i) KMnO₄-KOH, heat (ii) H⁺ → C₆H₅COOH

Teacher's Note:
a) No Tollen's test means a ketone; positive iodoform test means a CH₃CO- group; Cannizzaro's reaction needs an aldehyde with no α-hydrogen.
b) Pentan-3-one has α-hydrogens (so it gives aldol condensation) but no CH₃CO- group.
c) Alkaline KMnO₄ oxidises the whole side chain of an alkylbenzene to -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 reactant is drawn as a cyclohexane ring with an -OH group attached to one carbon; an arrow labelled CrO₃ points to 'X'.]

Answer:
(i) C₆H₅-CH=N-NH-C₆H₃(NO₂)₂: the benzaldehyde carbon is joined by a double bond to N, which is joined 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 (i) SnCl₂ + HCl, (ii) H₃O⁺ - Stephen reaction)
(v) X is cyclohexanone: a cyclohexane ring with a C=O group in place of the CH-OH group (C₆H₁₀O).

Teacher's Note:
a) Reactivity towards HCN falls as bulky alkyl groups increase steric hindrance and reduce the positive charge on the carbonyl carbon.
b) CrO₃ oxidises a secondary alcohol to a ketone.
c) In the 2,4-DNP derivative, the C=O oxygen of benzaldehyde is replaced by =N-NH-.

32. (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 the irregular variation of ionisation enthalpies (\( \Delta_i H_1 + \Delta_i H_2 \)) and sublimation enthalpies of these metals.
(II) Cu has a low hydration enthalpy of Cu²⁺ (\( \Delta_{hyd} H^{\circ} \)) and a high enthalpy of atomisation (\( \Delta_a H^{\circ} \)), which do not balance its high ionisation enthalpy.
(III) Mn²⁺ is highly stable due to its half-filled 3d⁵ configuration.
(ii) Acidic solution: 2MnO₄^- + 10I^- + 16H⁺ → 2Mn²⁺ + 8H₂O + 5I₂
Alkaline solution: 2MnO₄^- + I^- + H₂O → 2MnO₂ + IO₃^- + 2OH^-

Teacher's Note:
a) In acidic medium, I^- is oxidised to I₂; in neutral or alkaline medium, it is oxidised to iodate (IO₃^-).
b) Always check that both atoms and charges are balanced in ionic equations.
c) Link Mn²⁺ stability to the half-filled d⁵ configuration - this is the key word.

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)
(II) Europium (Eu) or Ytterbium (Yb)
(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.
(iii) Cr has more unpaired electrons, so a greater number of (n-1)d and ns electrons take part in interatomic metallic bonding. Mn has a stable half-filled 3d⁵ configuration, so its metallic bonding is weaker.
(iv) Acidified KMnO₄ slowly decomposes to give MnO₂ and oxygen:
4MnO₄^- + 4H⁺ → 4MnO₂ + 3O₂ + 2H₂O
This is a redox reaction. (The marking scheme also accepts the reduction reaction MnO₄^- + 4H⁺ + 3e^- → MnO₂ + 2H₂O, called a reduction reaction.)

Teacher's Note:
a) Ce⁴⁺ gains the stable f⁰ configuration; Eu²⁺ (f⁷) and Yb²⁺ (f¹⁴) are also stable.
b) Part (iv) carries 2 marks: 1 for the equation and 1 for naming the reaction type.
c) Melting point depends on the strength of metallic bonding, which depends on the number of unpaired electrons.

33. (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: \( E_{cell} = E^{\circ}_{cell} - \frac{0.059}{2} \log \frac{[Mg^{2+}]}{[Ag^{+}]^2} \)
Substitution: \( E_{cell} = 3.17 - \frac{0.059}{2} \log \frac{0.01}{(0.001)^2} = 3.17 - \frac{0.059}{2} \log 10^4 \)
\( E_{cell} = 3.17 - \frac{0.059}{2} \times 4 = 3.17 - 0.118 \)
Answer: \( E_{cell} = 3.052 \) V
Formula: \( \Delta G = -nFE_{cell} \)
Substitution: \( \Delta G = -2 \times 96500 \times 3.052 \)
Answer: \( \Delta G = -589036 \) J mol^-1 = -589.036 kJ mol^-1

Teacher's Note:
a) Square the Ag⁺ concentration because 2 Ag⁺ ions take part in the balanced cell reaction.
b) Writing the Nernst equation correctly carries 1 mark even before substitution.
c) A negative ΔG confirms the cell reaction is spontaneous.

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: ΔG° = -43500 J mol^-1, n = 2, \( P_{H_2} = 0.4 \) atm, [H⁺] = 0.1 M, [Cl^-] = 0.2 M
Formula: \( \Delta G^{\circ} = -nFE^{\circ}_{cell} \)
Substitution: \( -43500 = -2 \times 96500 \times E^{\circ}_{cell} \)
\( E^{\circ}_{cell} = \frac{43500}{193000} = 0.225 \) V
Formula: \( E_{cell} = E^{\circ}_{cell} - \frac{0.059}{2} \log \frac{[H^{+}]^2 [Cl^{-}]^2}{P_{H_2}} \)
Substitution: \( E_{cell} = 0.225 - \frac{0.059}{2} \log \frac{(0.1)^2 (0.2)^2}{0.4} = 0.225 - \frac{0.059}{2} \log 10^{-3} \)
\( E_{cell} = 0.225 + \frac{0.059}{2} \times 3 = 0.225 + 0.0885 \)
Answer: \( E_{cell} = 0.3135 \) V

Teacher's Note:
a) First find E° from ΔG°, then apply the Nernst equation.
b) Solids (AgCl, Ag) are not included in the reaction quotient; the gas is taken as its partial pressure.
c) Since log 10^-3 = -3, the correction term is added to E°.

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