CBSE Class 12 Chemistry Question Paper 2026 Solved Code 56-1-2

SECTION A

1. Which of the following is most basic ? [1 Mark]
(A) Mn₂O₇
(B) MnO₂
(C) MnO
(D) Mn₂O₃

Answer: (C) MnO

Teacher's Note:
a) Basic character of an oxide decreases as the oxidation state of the metal increases.
b) Mn is in +2 state in MnO (lowest), so it is most basic; Mn₂O₇ (+7) is acidic.

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

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

(A) Graph of t_1/2 against [R]₀ - straight line rising from the origin
(B) Graph of t_1/2 against [R]₀ - horizontal straight line
(C) Graph of Rate against Concentration - horizontal straight line
(D) Graph of Rate against Concentration - falling curve

Answer: (B) Graph of \( t_{1/2} \) against \( [R]_0 \) - horizontal straight 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) Graph (A) is for zero order (\( t_{1/2} \) proportional to [R]₀) and graph (C) shows rate independent of concentration, which is also zero order.

3. Which of the following solutions will have the highest osmotic pressure ? [1 Mark]
(A) 0.1 M KCl
(B) 0.1 M CaCl₂
(C) 0.1 M Glucose
(D) 0.1 M Urea

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

Teacher's Note:
a) Osmotic pressure \( \pi = iCRT \), so at equal concentration the solute with the largest van't Hoff factor wins.
b) CaCl₂ gives 3 ions (i = 3), KCl gives 2 ions, while glucose and urea do not ionise (i = 1).

4. For a reaction : N₂ + 3H₂ → 2NH₃, the rate of reaction with respect to NH₃ is [1 Mark]
(A) \( +\frac{1}{3}\frac{\Delta [NH_3]}{\Delta t} \)
(B) \( -\frac{1}{2}\frac{\Delta [NH_3]}{\Delta t} \)
(C) \( +\frac{1}{4}\frac{\Delta [NH_3]}{\Delta t} \)
(D) \( +\frac{1}{2}\frac{\Delta [NH_3]}{\Delta t} \)

Answer: (D) \( +\frac{1}{2}\frac{\Delta [NH_3]}{\Delta t} \)

Teacher's Note:
a) Divide the rate of change of concentration by the stoichiometric coefficient (2 for NH₃).
b) NH₃ is a product, so its concentration increases and the sign is positive.

5. How many Faradays are required to reduce 1 mol of Cr₂O₇^2- to Cr³⁺ in acidic medium ? [1 Mark]
(A) 2
(B) 3
(C) 6
(D) 4

Answer: (C) 6

Teacher's Note:
a) Cr changes from +6 to +3, gaining 3 electrons per Cr atom.
b) One Cr₂O₇^2- has 2 Cr atoms, so 6 mol electrons = 6 F are needed.

6. Which of the following Grignard Reagent will be used to prepare C₆H₁₁CH₂OH when treated with methanal ? [1 Mark]

[Figure: The compound to be prepared is a cyclohexane ring with a CH₂OH group attached (cyclohexylmethanol). Options (B), (C) and (D) are cyclohexane rings carrying CH₂MgBr, MgBr and CH₂CH₂MgBr groups respectively.]

(A) CH₃MgBr
(B) Cyclohexane ring with a CH₂MgBr group
(C) Cyclohexane ring with a MgBr group
(D) Cyclohexane ring with a CH₂CH₂MgBr group

Answer: (C) Cyclohexane ring with a MgBr group (cyclohexylmagnesium bromide)

Teacher's Note:
a) Methanal (HCHO) adds one carbon as CH₂OH to the R group of RMgX, giving a primary alcohol RCH₂OH.
b) Here R must be the cyclohexyl group, so the reagent is C₆H₁₁MgBr.

7. The secondary valency of Pt in [Pt(en)₂Cl₂]²⁺ is [1 Mark]
(A) 6
(B) 5
(C) 4
(D) 2

Answer: (A) 6

Teacher's Note:
a) Secondary valency equals the coordination number of the metal.
b) Each 'en' is bidentate (2 x 2 = 4) and two Cl^- add 2, giving 6.

8. (CH₃)₃C - OC₂H₅ on reaction with HI gives [1 Mark]
(A) (CH₃)₃ C - I and C₂H₅ - I
(B) (CH₃)₃ C - OH and C₂H₅ - I
(C) (CH₃)₃ C - I and C₂H₅ - OH
(D) (CH₃)₃ C - OH and C₂H₅ - OH

Answer: (C) \( (CH_3)_3C-I \) and \( C_2H_5-OH \)

Teacher's Note:
a) With a tertiary alkyl group the cleavage follows S_N1, because the stable tertiary carbocation forms.
b) So the iodide goes to the tertiary carbon and the ethyl part becomes ethanol.

9. 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 is protonated to the anilinium ion, which is meta directing.
b) This is why a large amount (47%) of meta product forms along with the para product.

10. 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 other physical properties such as melting point, boiling point and refractive index are identical.

11. An example of non-reducing sugar is [1 Mark]
(A) Glucose
(B) Sucrose
(C) Lactose
(D) Maltose

Answer: (B) Sucrose

Teacher's Note:
a) In sucrose, C1 of glucose and C2 of fructose are both used in the glycosidic bond.
b) No free aldehyde or ketone group is left, so it does not reduce Tollens' or Fehling's reagent.

12. Benzene diazonium chloride on reaction with phenol in weakly basic medium gives [1 Mark]
(A) Azobenzene
(B) Benzene
(C) Chlorobenzene
(D) p-hydroxyazobenzene

Answer: (D) p-hydroxyazobenzene

Teacher's Note:
a) This is a coupling reaction; the diazonium ion attacks the para position of phenol.
b) The product is an orange dye; remember "phenol + diazonium salt in mild alkali = coupling".

13. Assertion (A) : The presence of -OH group in phenols directs the incoming group to ortho and para positions.
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: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

Teacher's Note:
a) The -OH group donates electrons to the ring by resonance.
b) This raises electron density mainly at ortho and para positions, so the ring is activated and o/p directing.

14. Assertion (A) : Actinoids show irregularities in their electronic configurations.
Reason (R) : In actinoids 5f, 6d and 7s orbitals are of comparable energies. [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) Both statements are true: 5f, 6d and 7s orbitals are close in energy.
b) The irregularities are explained by the extra stability of f⁰, f⁷ and f¹⁴ configurations, so R is not the correct explanation.

15. Assertion (A) : Components of azeotropes are easily separated by fractional distillation.
Reason (R) : Components of an azeotrope have same composition in liquid and vapour phase. [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) Azeotropes boil at constant temperature with the same composition in liquid and vapour.
b) Hence they cannot be separated by fractional distillation, so the Assertion is false.

16. Assertion (A) : The two strands of DNA are complementary to each other.
Reason (R) : The hydrogen bonds are formed between specific pairs of bases. [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) Adenine pairs only with thymine and guanine pairs only with cytosine.
b) This specific base pairing makes the two strands complementary.

SECTION B

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

Answer:
1. The mixture shows positive deviation from Raoult's law.
2. Reason: ethanol-ethanol interactions (hydrogen bonding) are stronger than ethanol-acetone interactions; acetone breaks some hydrogen bonds, so molecules escape more easily.
3. The boiling point of the solution will decrease.

Teacher's Note:
a) Positive deviation means higher vapour pressure than expected, and so a lower boiling point.
b) Mention "A-B interactions weaker than A-A interactions" as the key reason.

18. (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) - the cation is [Ag(NH₃)₂]⁺ and the anion is [Ag(CN)₂]^-.
(ii) Potassium trioxalatoferrate(III) - three oxalate ligands around Fe in the +3 state.

Teacher's Note:
a) Name the cation first; an anionic complex takes the "-ate" ending (argentate, ferrate).
b) Oxalate is C₂O₄^2-, so in (ii) Fe is +3: 3(+1) + x + 3(-2) = 0 gives x = +3.

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, whereas [Co(NH₃)₅Cl]SO₄ does not. (Or: with BaCl₂, [Co(NH₃)₅Cl]SO₄ gives a white precipitate of BaSO₄, whereas the other does not.)
(ii) Chelate effect: a di- or polydentate ligand bound to a single metal atom or ion forms a more stable complex than similar monodentate ligands. Example: [Co(en)₃]³⁺.

Teacher's Note:
a) Ionisation isomers give different ions in solution, so test the ion outside the square bracket.
b) For the chelate effect, the word "stable" and a correct example each carry marks.

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

Answer:
1. Resonance effect: the lone pair on halogen is delocalised into the ring, so the C-X bond gets partial double bond character and is hard to break.
2. The carbon of C-X 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 only two reasons as asked, each in one clear sentence.

20. 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 it is a one-step (elementary) reaction: Rate = \( k[A]^2[B] \).
2. When volume becomes 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) Only for an elementary reaction can the rate law be written from the coefficients.
b) Order depends on the rate law, not on volume or concentration, so it does not change.

21. Differentiate between the following :
(i) Acidic amino acids and basic amino acids
(ii) Nucleotide and Nucleoside [2 Marks]

Answer:
(i) Acidic amino acids have more -COOH groups than -NH₂ groups (e.g. aspartic acid), whereas basic amino acids have more -NH₂ groups than -COOH groups (e.g. lysine).
(ii) Nucleotide = base + sugar + phosphate group, whereas nucleoside = base + sugar only.

Teacher's Note:
a) Compare the number of carboxyl and amino groups for (i).
b) Remember: nucleoTide has the exTra phosphate.

SECTION C

22. 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, a tertiary bromide). Y = CH₃-CH(Br)-CH₂-CH₃ (2-bromobutane, which is chiral).
(b) X, because it reacts by the S_N1 mechanism through a planar carbocation intermediate, which the nucleophile can attack from either side.
(c) Y, because it reacts by the S_N2 mechanism; the nucleophile attacks from the rear side, opposite to the leaving group, which inverts the configuration.

Teacher's Note:
a) Rate depending only on [X] means first order, so S_N1; rate depending on [Y] and [OH^-] means S_N2.
b) The marking scheme also gives full marks in (b) if Y is written, since only Y is optically active.
c) Key words: "planar carbocation" for racemisation and "backside attack" for inversion.

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) Phenol (C₆H₅OH) + CHCl₃ + aq. NaOH, then H⁺ → salicylaldehyde (2-hydroxybenzaldehyde, o-HO-C₆H₄-CHO).
(b) Phenol + aq. NaOH → sodium phenoxide (C₆H₅O^-Na⁺); sodium phenoxide + CO₂, then H⁺ → salicylic acid (2-hydroxybenzoic acid, o-HO-C₆H₄-COOH).
(c) Anisole (C₆H₅OCH₃) + CH₃COCl in presence of anhydrous AlCl₃ → 2-methoxyacetophenone (minor) + 4-methoxyacetophenone (major) + HCl.

Teacher's Note:
a) Always write the reagents and conditions on the arrow; they carry marks.
b) Reimer-Tiemann puts -CHO and Kolbe's puts -COOH at the ortho position of phenol.
c) -OCH₃ is ortho/para directing, so show both products in the acylation of anisole.

24. Give reasons for the following :
(a) Carboxylic acids have higher boiling point than alcohols of comparable molecular masses.
(b) Alpha (\( \alpha \)) 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 more extensively associated through intermolecular hydrogen bonding; they exist as dimers even in the vapour phase.
(b) The carbonyl group has a strong electron withdrawing effect, and the conjugate base (enolate ion) formed after losing an \( \alpha \)-hydrogen is stabilised by resonance.
(c) In strongly acidic medium, ammonia and its derivatives (nucleophiles) get protonated, so they lose their lone pair and can no longer act as nucleophiles.

Teacher's Note:
a) Use the keyword "dimer" for (a) and "resonance stabilised conjugate base" for (b).
b) In (c), the reaction needs a weakly acidic medium (pH about 3.5) to activate the carbonyl without protonating the nucleophile.

25. Define the following terms :
(a) Anomers
(b) Invert sugar
(c) Glycosidic linkage [3 Marks]

Answer:
(a) Anomers: isomers of a monosaccharide which differ only in the configuration of the -OH group at C-1 (aldoses) or C-2 (ketoses), for example \( \alpha \)-D-glucose and \( \beta \)-D-glucose.
(b) Invert sugar: sucrose is dextrorotatory; on hydrolysis it gives dextrorotatory glucose and laevorotatory fructose. The laevorotation of fructose is more than the dextrorotation of glucose, so the mixture is laevorotatory. This product is called invert sugar.
(c) Glycosidic linkage: the oxide linkage through an oxygen atom by which two or more monosaccharide units are joined together.

Teacher's Note:
a) Mention C-1 or C-2 in the definition of anomers; this is the key point.
b) For invert sugar, stress the change of sign of rotation from (+) to (-).

26. 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: \( k = \frac{2.303}{30}\log\frac{0.30}{2 \times 0.30 - 0.50} = \frac{2.303}{30}\log\frac{0.30}{0.10} \)
\( k = \frac{2.303}{30} \times 0.48 \)
Answer: \( k = 0.037\ \text{s}^{-1} \) (approximately)

Teacher's Note:
a) Deriving \( 2P_i - P_t \) for the reactant pressure is the key step that carries marks.
b) A first order rate constant always has the unit s^-1 (time^-1).

27. (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 \( E^{\circ}_{cell} = \frac{0.059}{n}\log K_c \); hence K_c is related to \( E^{\circ}_{cell} \) only.
(ii) Metal 'A' (E^o = -0.24 V), because it has a negative reduction potential and lies above hydrogen, so it can reduce H⁺ to H₂.
(iii) On charging: 2PbSO₄(s) + 2H₂O(l) → PbO₂(s) + Pb(s) + 2H₂SO₄(aq)

Teacher's Note:
a) A metal with negative E^o (below 0 V of SHE) displaces hydrogen from dilute acids.
b) Charging is the reverse of discharging; PbSO₄ is on the reactant side.

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 better than a dry cell because it gives a constant potential (about 1.35 V) throughout its life, as no ions are involved in the overall reaction.
3. Overall reaction: Zn(Hg) + HgO(s) → ZnO(s) + Hg(l)

Teacher's Note:
a) Primary cells cannot be recharged; the reaction occurs only once.
b) Write the physical states in the overall reaction; Zn is used as zinc-mercury amalgam.

28. 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 \( 46.2\,^{\circ}\text{C} \) 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, \( T_b^{\circ} = 46.2\,^{\circ}\text{C} \), \( \alpha = 88\% = 0.88 \).
For dimerisation (n = 2): \( \alpha = \frac{1 - i}{1 - \frac{1}{2}} \), so \( 0.88 = \frac{1 - i}{0.5} \) and \( i = 0.56 \).
Formula: \( \Delta T_b = i K_b \frac{W_B \times 1000}{M_B \times W_A} \)
Substitution: \( \Delta T_b = 0.56 \times 2.3 \times \frac{0.61 \times 1000}{122 \times 5} = 0.56 \times 2.3 \times 1 \)
\( \Delta T_b = 1.288 \) K (or \( ^{\circ}\text{C} \))
Answer: \( T_b = 46.2 + 1.288 = 47.488\,^{\circ}\text{C} \) (320.638 K)

Teacher's Note:
a) For association into dimers, i is less than 1: \( i = 1 - \frac{\alpha}{2} \).
b) Add \( \Delta T_b \) to the boiling point of the pure solvent; a rise of 1 K equals a rise of \( 1\,^{\circ}\text{C} \).

SECTION D

29. 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, so these axial orbitals, which point towards the ligands, experience more repulsion.

Teacher's Note:
a) The t_2g orbitals (d_xy, d_yz, d_xz) lie between the axes, so they are lowered in energy.
b) Mention "along the axes" and "greater repulsion" to get the reason mark.

(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) In [CoF₆]^3-, Co³⁺ = 3d⁶; F^- is a weak field ligand, so the configuration is \( t_{2g}^{4}\,e_g^{2} \).
(ii) In [Co(NH₃)₆]³⁺, Co³⁺ = 3d⁶; NH₃ is a strong field ligand, so the configuration is \( t_{2g}^{6}\,e_g^{0} \).

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 electrons pair up in t_2g first.

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

Answer: In [NiCl₄]^2-, Cl^- is a weak field ligand and cannot pair up the two unpaired 3d electrons of Ni²⁺, so it is paramagnetic. In [Ni(CO)₄], CO is a strong field ligand and pairs up the electrons of Ni(0), so it is diamagnetic.

Teacher's Note:
a) Ni is +2 (3d⁸) in [NiCl₄]^2- but 0 (3d⁸4s²) in [Ni(CO)₄].
b) Both are sp³ hybridised; the difference is only in unpaired electrons.

OR

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

Answer: Hybridisation: d²sp³ (inner orbital octahedral complex). Magnetic behaviour: paramagnetic (Fe³⁺, 3d⁵, one unpaired electron after pairing by strong field CN^-).

Teacher's Note:
a) CN^- pairs the five 3d electrons into three orbitals, leaving one unpaired electron.
b) Two 3d orbitals stay empty, so inner d orbitals are used: d²sp³.

30. 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. (CH₃)₃N has more steric hindrance and its cation is poorly solvated (no N-H for hydrogen bonding), so CH₃NH₂ is the stronger base.
(ii) A = CH₃NH₂ (methanamine, by Hoffmann bromamide degradation); B = CH₃-NHCOC₆H₅ (N-methylbenzamide, by benzoylation).

Teacher's Note:
a) In aqueous phase, solvation of the ammonium ion by hydrogen bonding is the deciding factor, not the inductive effect alone.
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) A secondary amine gives a sulphonamide with no H on nitrogen, so it is insoluble in alkali.
b) A primary amine gives an alkali-soluble product, and a tertiary amine does not react.

OR

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

Answer: C₆H₅NH₂ → (NaNO₂ + HCl, 273-278 K) C₆H₅N₂⁺Cl^- → (CuCN) C₆H₅CN. Aniline is first diazotised to benzene diazonium chloride, which on treatment with CuCN gives benzonitrile.

Teacher's Note:
a) Diazotisation must be done at 0-5 degree Celsius, since the diazonium salt decomposes at higher temperature.
b) Replacement of -N₂⁺ 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 meta directing anilinium ion.

Teacher's Note:
a) Direct nitration of aniline gives oxidation products and a large amount of meta product.
b) The acetyl group is removed by hydrolysis after nitration to get p-nitroaniline.

SECTION E

31. (a) Calculate emf and \( \Delta G \) 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: \( E_{cell} = 3.17 - \frac{0.059}{2}\log\frac{0.01}{(0.001)^2} = 3.17 - \frac{0.059}{2}\log 10^{4} \)
\( E_{cell} = 3.17 - \frac{0.059}{2} \times 4 = 3.17 - 0.118 \)
\( E_{cell} = 3.052 \) V
Formula: \( \Delta G = -nFE_{cell} \)
Substitution: \( \Delta G = -2 \times 96500 \times 3.052 \)
Answer: \( \Delta G = -589036\ \text{J mol}^{-1} \) or \( -589.036\ \text{kJ mol}^{-1} \)

Teacher's Note:
a) Square [Ag⁺] in the Nernst equation because 2 Ag⁺ ions appear in the balanced reaction.
b) A negative \( \Delta G \) confirms that the cell reaction is spontaneous.
c) Always write the unit J mol^-1 (or kJ mol^-1) with \( \Delta G \).

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 \( 25\,^{\circ}\text{C} \).
Given : \( \Delta G^{\circ} = -43500\ \text{J mol}^{-1} \)
[log 10 = 1, 1 F = 96500 C mol^-1] [5 Marks]

Answer:
Given: n = 2, \( \Delta G^{\circ} = -43500 \) J mol^-1, \( P_{H_2} = 0.4 \) atm, [H⁺] = 0.1 M, [Cl^-] = 0.2 M.
Formula: \( \Delta G^{\circ} = -nFE^{\circ}_{cell} \)
Substitution: \( -43500 = -2 \times 96500 \times E^{\circ}_{cell} \)
\( E^{\circ}_{cell} = 0.225 \) 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) Solids (AgCl, Ag) are not written in the reaction quotient; the gas enters as its partial pressure.
b) \( \log 10^{-3} = -3 \), so the correction term becomes positive; watch this sign.
c) First find \( E^{\circ}_{cell} \) from \( \Delta G^{\circ} \), then apply the Nernst equation.

32. (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. [3 Marks]

Answer:
(I) CH₃-CO-CH₂-CH₂-CH₃ (pentan-2-one): a methyl ketone, so it gives the iodoform test but not Tollens' test.
(II) CH₃-CH₂-CO-CH₂-CH₃ (pentan-3-one): a ketone without a CH₃CO- group, but it has \( \alpha \)-hydrogens, so it undergoes aldol condensation.
(III) (CH₃)₃C-CHO (2,2-dimethylpropanal): an aldehyde with no \( \alpha \)-hydrogen, so it undergoes Cannizzaro's reaction.

Teacher's Note:
a) Iodoform test needs a CH₃CO- group; aldol needs at least one \( \alpha \)-hydrogen.
b) Cannizzaro reaction needs an aldehyde with no \( \alpha \)-hydrogen.

(ii) Show how each of the following compounds can be converted to benzoic acid :
(I) Acetophenone (II) Ethyl benzene [2 Marks]

Answer:
(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) Alkaline KMnO₄ oxidises the whole side chain to -COOH, whatever its length.
b) Write the acidification step, because the first product is the potassium salt.

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: Cyclohexanol (a cyclohexane ring carrying an -OH group) reacting with CrO₃ written over the arrow to give 'X'.]

Answer:
(i) C₆H₅-CH=N-NH-C₆H₃(NO₂)₂: the benzylidene 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 less than Acetone less than Acetaldehyde.
(iii) Add NaHCO₃ solution to both. Benzoic acid gives brisk effervescence of CO₂, while ethyl benzoate does not.
(iv) (i) DIBAL-H (diisobutylaluminium hydride), (ii) H₃O⁺. (Or: (i) SnCl₂ + HCl, (ii) H₃O⁺ - Stephen reduction.)
(v) X is cyclohexanone: a cyclohexane ring with a C=O group in place of CH-OH.

Teacher's Note:
a) Reactivity towards HCN falls as bulky alkyl groups increase steric hindrance and electron donation at the carbonyl carbon.
b) CrO₃ oxidises a secondary alcohol to a ketone.
c) Only carboxylic acids react with NaHCO₃; esters and phenols do not.

33. (a) (i) From the given data of \( E^{\circ} \) 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 ? [3 Marks]

Answer:
(I) Because of the irregular variation of ionisation enthalpies (\( \Delta_i H_1 + \Delta_i H_2 \)) and sublimation (atomisation) enthalpies across the series.
(II) Because Cu has a high enthalpy of atomisation (\( \Delta_a H^{\circ} \)) and a low hydration enthalpy (\( \Delta_{hyd} H^{\circ} \)) of Cu²⁺, which do not balance the energy needed.
(III) Because Mn²⁺ is highly stable due to its half-filled 3d⁵ configuration.

Teacher's Note:
a) \( E^{\circ} \) depends on three energy terms: atomisation, ionisation and hydration enthalpies.
b) Half-filled (d⁵) and full (d¹⁰) configurations explain most exceptions in the 3d series.

(ii) Write the ionic equations for the oxidising action of potassium permanganate for its reaction with I^- in both acidic and alkaline solutions. [2 Marks]

Answer:
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 acid, iodide is oxidised to iodine; in alkaline medium it is oxidised to iodate.
b) Check that both atoms and charges are balanced in each equation.

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 (Ce⁴⁺). (II) Europium (Eu²⁺) or Ytterbium (Yb²⁺).
(ii) Transition metals can show multiple (variable) oxidation states and can form complexes; they also provide a large surface area for the reaction.
(iii) In Cr, a greater number of electrons from (n-1)d and ns orbitals take part in interatomic metallic bonding, so its metallic bonds are stronger. Mn (3d⁵4s²) holds its stable half-filled d electrons more tightly.
(iv) Acidified KMnO₄ slowly decomposes to give MnO₂ and oxygen:
4MnO₄^- + 4H⁺ → 4MnO₂ + 3O₂ + 2H₂O
This is a redox reaction (manganese is reduced and oxygen is oxidised).

Teacher's Note:
a) Ce⁴⁺ (f⁰) and Eu²⁺, Yb²⁺ (f⁷, f¹⁴) are stable because of empty, half-filled and full f subshells.
b) The marking scheme also accepts the reduction half reaction MnO₄^- + 4H⁺ + 3e^- → MnO₂ + 2H₂O with "reduction reaction" for (iv).
c) Store KMnO₄ solution in dark bottles, since light speeds up this decomposition.

Please click the link below to download pdf file of CBSE Class 12 Chemistry Question Paper 2026 Solved Code 56-1-2