CBSE • Class 12 • Theory

रसायन विज्ञान (सैद्धांतिक)

CHEMISTRY (Theory)

निर्धारित समय : 3 घंटे

Time allowed : 3 hours

अधिकतम अंक : 70

Maximum Marks : 70

CBSE Class 12 Chemistry Solved Paper 2026 (New)

Showing all questions

(1) Questions No. 1 to 16 are Multiple Choice type questions carrying 1 mark each.

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

(A) t_1/2 vs [R]₀ (straight line increasing)
(B) t_1/2 vs [R]₀ (horizontal straight line)
(C) Rate vs Concentration (horizontal straight line)
(D) Rate vs Concentration (decreasing curve)

Answer ✅

Correct option: (B)

Explanation (Step by Step) 👉

In a first-order reaction, the half-life does not depend on the initial concentration [R]₀.

The half-life formula is:
t_1/2 = 0.693 / k

Since [R]₀ does not appear in the formula, t_1/2 remains constant even when [R]₀ changes.

So, the graph of t_1/2 vs [R]₀ is a horizontal straight line, which is Option (B).

Did You Know? 🤔📌

✅ For a zero-order reaction, half-life depends on initial concentration:
t_1/2 = [R]₀ / (2k)

So t_1/2 increases linearly with [R]₀, which matches Option (A).

2. Which of the following solutions will have the lowest freezing point in water?

(A) 0.1 M Glucose
(B) 0.1 M CaCl₂
(C) 0.1 M KCl
(D) 0.1 M Urea

Answer ✅

Correct option: (B) 0.1 M CaCl₂ ✅

Explanation (Step by Step) 👉

Freezing point depression depends on the number of particles in solution:

ΔT_f = i K_f m

Where i = van’t Hoff factor (number of ions/particles formed)

For 0.1 M solutions (same concentration), the one with highest i will have the largest ΔT_f, so the lowest freezing point.

Glucose (non-electrolyte): i = 1

Urea (non-electrolyte): i = 1

KCl → K⁺ + Cl⁻ : i ≈ 2

CaCl₂ → Ca²⁺ + 2Cl⁻ : i ≈ 3 ✅

Since CaCl₂ produces the most particles, it causes the maximum freezing point depression, hence lowest freezing point.

Did You Know? 🤔📌

✅ Colligative properties (like freezing point depression) depend on number of solute particles, not their nature.

✅ In real solutions, i can be slightly less than ideal due to ion pairing, but CaCl₂ still gives the greatest effect among these options.

3. Which of the following is not a transition metal?

(A) Sc
(B) Ag
(C) Hg
(D) Cu

Answer ✅

Correct option: (C) Hg ✅

Explanation (Step by Step) 👉

A transition metal is an element that forms at least one ion with a partially filled d-subshell.

Let’s check each:

Sc (Z = 21): Sc → Sc³⁺ is 3d⁰ (but Sc is generally included in d-block and treated as transition in many exam contexts; however its ions have d⁰).

Cu (Z = 29): Cu²⁺ is 3d⁹ ✅ (partially filled d)

Ag (Z = 47): Ag²⁺ can be 4d⁹ ✅ (partially filled d)

Hg (Z = 80): Hg is [Xe] 4f¹⁴ 5d¹⁰ 6s² and Hg²⁺ is 5d¹⁰ (d-subshell remains completely filled) ❌

So Hg does not form ions with incomplete d-orbitals, hence not a transition metal.

Did You Know? 🤔📌

✅ Zn, Cd, Hg are classic examples of d-block elements that are not transition metals because their common ions have d¹⁰ configuration (completely filled d-orbitals).

4. Which of the following represents the fraction of molecules with energies equal to or greater than E_a?

(A) + E_a/RT
(B) e^−E_a/RT
(C) − E_a/RT
(D) e^+E_a/RT

Answer ✅

Correct option: (B) e^−E_a/RT ✅

Explanation (Step by Step) 👉

According to the Maxwell–Boltzmann distribution, the fraction of molecules having energy ≥ E_a (activation energy) is given by the Boltzmann factor:

✅ Fraction = e^−E_a/RT

If E_a increases, the fraction decreases (harder to cross the barrier).

If T increases, the fraction increases (more molecules can reach ≥ E_a).

So the correct expression is e^−E_a/RT , which is Option (B).

Did You Know? 🤔📌

✅ This same term appears in the Arrhenius equation:
k = A e^−E_a/RT

That’s why reaction rate constant k increases rapidly with temperature.

5. What will happen during the electrolysis of aqueous solution of CuCl₂ by using platinum electrodes?

(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 ✅

Correct option: (D) Cl₂ gas will be released at anode ✅

Explanation (Step by Step) 👉

At Cathode (reduction):
In aqueous CuCl₂, possible reductions are:

Cu²⁺ + 2e⁻ → Cu(s) ✅ (easy)

2H₂O + 2e⁻ → H₂ + 2OH⁻ (less preferred here)

Since Cu²⁺ has higher tendency to get reduced than water,
✅ Copper deposits at the cathode (not at anode).

At Anode (oxidation):
Possible oxidations are:

2Cl⁻ → Cl₂ + 2e⁻ ✅

2H₂O → O₂ + 4H⁺ + 4e⁻

With Pt (inert) electrodes and chloride ions present, chloride ions are preferentially oxidized, so:
✅ Cl₂ gas is released at the anode.

So, option (D) is correct.

Did You Know? 🤔📌

✅ In electrolysis of aqueous CuCl₂, you often observe:

Reddish-brown Cu coating on the cathode

Greenish-yellow Cl₂ gas bubbles at the anode

✅ Overall reaction: CuCl₂(aq) → Cu(s) + Cl₂(g)

6. Aspirin is obtained by acetylation of:

(A) Phenol
(B) Salicylaldehyde
(C) 2-Hydroxybenzoic acid
(D) Benzoic acid

Answer ✅

Correct option: (C) 2-Hydroxybenzoic acid ✅

Explanation (Step by Step) 👉

Aspirin is acetylsalicylic acid. It is prepared by acetylating salicylic acid, which is the same as 2-hydroxybenzoic acid.

✅ 2-Hydroxybenzoic acid (Salicylic acid) has an –OH group on the benzene ring.

During acetylation, this –OH group gets acetylated:

2-Hydroxybenzoic acid + (CH₃CO)₂O → Aspirin (acetylsalicylic acid) + CH₃COOH

So, the correct starting compound is 2-hydroxybenzoic acid .

Did You Know? 🤔📌

✅ Aspirin is an ester (acetylated –OH group) and that’s why it is less acidic and less irritating to the stomach than salicylic acid.

7. The secondary valency of Co in the complex [Co(NH₃)₅(NO₂)]²⁺ is:

(A) 5
(B) 1
(C) 4
(D) 6

Answer ✅

Correct option: (D) 6 ✅

Explanation (Step by Step) 👉

In Werner’s theory:

Primary valency = oxidation state (ionisable)

Secondary valency = coordination number (non-ionisable) = number of ligands directly attached to the metal

In [Co(NH₃)₅(NO₂)]²⁺:

NH₃ ligands = 5

NO₂ ligand = 1

Total ligands attached to Co = 5 + 1 = 6

✅ So, secondary valency (coordination number) = 6

Did You Know? 🤔📌

✅ Complexes with coordination number 6 usually show octahedral geometry (like this one).

✅ NO₂⁻ is an ambidentate ligand: it can bind through N (nitro) or O (nitrito), giving linkage isomerism.

8. At low temperature, phenol reacts with dil. HNO₃ to yield:

(A) 2, 4, 6-Trinitrophenol
(B) o-Nitrophenol only
(C) p-Nitrophenol only
(D) ortho- and para-nitrophenol

Answer ✅

Correct option: (D) ortho- and para-nitrophenol ✅

Explanation (Step by Step) 👉

Phenol has the –OH group, which shows a strong +M (resonance donating) effect.

✅ This increases electron density at the ortho and para positions, making phenol highly activated towards electrophilic substitution.

With dilute HNO₃ at low temperature, nitration is mild, so phenol forms a mixture of:

o-nitrophenol
p-nitrophenol

So the correct product is ortho + para nitrophenol, i.e., Option (D).

Did You Know? 🤔📌

✅ With concentrated HNO₃, phenol undergoes multiple nitrations to form 2,4,6-trinitrophenol (picric acid), which is Option (A) but only under stronger conditions.

9. Which of the following is ‘not’ true about enantiomers?

(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 ✅

Correct option: (B) They have the same specific rotation. ✅

Explanation (Step by Step) 👉

Enantiomers are non-superimposable mirror images of each other. In an achiral environment, they usually have the same physical properties, such as:

melting point / boiling point ✅

refractive index ✅

density and many other properties ✅

They also generally show the same chemical reactivity with achiral reagents ✅.

But they differ in optical rotation:

One enantiomer rotates plane-polarized light clockwise (+)

The other rotates it anticlockwise (−)

And the magnitude is equal, but the sign is opposite.

So they do not have the same specific rotation.

Did You Know? 🤔📌

✅ Enantiomers show different behavior only in chiral environments, like with chiral reagents, enzymes, or chiral drugs receptors.

That’s why in biology, one enantiomer of a drug can be useful while the other can be less effective or even harmful.

10. Aniline on direct nitration yields

(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 ✅

Correct option: (C) 51% para, 47% meta, 2% ortho derivatives ✅

Explanation (Step by Step) 👉

In the nitrating mixture (conc. HNO₃/H₂SO₄), aniline (–NH₂) gets protonated to form anilinium ion (–NH₃⁺).

–NH₃⁺ is deactivating and meta-directing, so a significant amount of meta product forms.

Still, some unprotonated aniline gives para product, and ortho is least due to stronger deactivation/steric effects in acidic medium.

Hence the observed distribution is approximately:

✅ p-nitroaniline ~51%

✅ m-nitroaniline ~47%

✅ o-nitroaniline ~2%

Did You Know? 🤔📌

✅ To get mainly para-nitroaniline, chemists first protect –NH₂ by acetylation (make acetanilide), then do nitration, and finally hydrolyse back to aniline.

11. Which of the following amines has lowest pK_b value?

(A) C₆H₅ – N(CH₃)₂
(B) C₆H₅ – NH(CH₃)
(C) C₆H₅ – NH₂
(D) O₂N—(benzene ring)—NH₂

Answer ✅

Correct option: (A) C₆H₅ – N(CH₃)₂ ✅

Explanation (Step by Step) 👉

Lower pK_b = stronger base (because K_b is larger).

Basic strength depends on how available the lone pair on N is to accept H⁺.

Compare the options:

(C) Aniline (C₆H₅–NH₂): lone pair is partly delocalized into benzene ring by resonance → less basic.

(B) N-methylaniline and (A) N,N-dimethylaniline: CH₃ groups donate electron density (+I effect), making the N lone pair more available → more basic than aniline.

More alkyl groups = stronger +I effect → (A) > (B) > (C) in basicity.

(D) Nitroaniline: –NO₂ is strongly electron-withdrawing (−I and −M), pulls electron density away and reduces basicity a lot → weakest base (highest pK_b).

So the strongest base here is (A), hence it has the lowest pK_b.

Did You Know? 🤔📌

✅ In aromatic amines, anything that withdraws electrons (like –NO₂, –CN, –COOH) raises pK_b (makes the base weaker).

✅ Anything that donates electrons (like alkyl groups) usually lowers pK_b (makes the base stronger).

12. The deficiency of which of the following vitamins causes increased fragility of red blood cells and muscular weakness?

(A) Vitamin A
(B) Vitamin K
(C) Vitamin E
(D) Vitamin D

Answer ✅

Correct option: (C) Vitamin E ✅

Explanation (Step by Step) 👉

Vitamin E (tocopherol) is a major antioxidant in the body. It protects cell membranes, especially the membranes of red blood cells (RBCs), from oxidative damage.

When Vitamin E is deficient:

RBC membranes get damaged easily → RBC fragility increases → hemolysis can occur

Nerve and muscle tissues are affected → muscular weakness (and sometimes neurological issues)

So, the symptoms given match Vitamin E deficiency.

Did You Know? 🤔📌

✅ Vitamin K deficiency mainly causes excess bleeding due to poor blood clotting.

✅ Vitamin D deficiency causes rickets/osteomalacia (weak bones).

✅ Vitamin A deficiency is famous for night blindness.

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 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 Assertion (A).

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

Answer ✅

✅ Both A and R are true, and R is the correct explanation of A.

Explanation (Step by Step) 👉

In fractional distillation, separation happens because vapour phase becomes richer in the more volatile component than the liquid phase.

But in an azeotrope, the vapour and liquid have the same composition at the azeotropic point.

Also, an azeotrope boils at a constant temperature like a pure substance.

So, during distillation, the vapour that rises has no composition difference from the boiling liquid. That means no enrichment, so fractional distillation cannot separate the components at the azeotropic composition.

That’s exactly why the Reason correctly explains the Assertion.

Did You Know? 🤔📌

✅ Ethanol–water forms a minimum boiling azeotrope (around 95.6% ethanol by volume) , so you can’t get absolute ethanol by simple fractional distillation.

✅ To break azeotropes, methods like azeotropic distillation, pressure-swing distillation, or using drying agents (like CaO for ethanol) are used.

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.

Answer ✅

✅ Assertion (A) is false, but Reason (R) is true.

Explanation (Step by Step) 👉

In phenol, the –OH group is an activating group because oxygen donates electron density into the benzene ring by +M (resonance) effect.

This resonance donation increases electron density mainly at the ortho and para positions.

✅ इसलिए phenol ortho/para directing होता है, meta directing नहीं.

So:

Assertion (A) is false (phenol does not direct to meta).

Reason (R) is true (phenol strongly activates the ring).

Did You Know? 🤔📌

✅ Meta-directing groups are usually electron-withdrawing groups like –NO₂, –CHO, –COOH, –CN, –SO₃H (they show −M/−I effects).

✅ Phenol reacts so fast in EAS that even bromine water can give 2,4,6-tribromophenol without a catalyst.

15. Assertion (A): Actinoids show wide range of oxidation states.

Reason (R): Actinoids are radioactive in nature.

Answer ✅

✅ Both A and R are true, but R is NOT the correct explanation of A.

Explanation (Step by Step) 👉

Assertion (A) is true: Actinoids show variable oxidation states (commonly +3, and also +4, +5, +6; a few even +7).

Reason (R) is also true: Actinoids are generally radioactive.

But radioactivity is not the reason for variable oxidation states.

The real reason is:
✅ 5f, 6d and 7s orbitals have comparable energies , so different numbers of electrons can take part in bonding, giving multiple oxidation states.

Did You Know? 🤔📌

✅ Early actinoids (Th, U, Np, Pu) show more oxidation states because their 5f electrons participate more in bonding.

✅ Later actinoids mostly show +3 state as 5f electrons become more “inner” and less available.

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.

Answer ✅

✅ Assertion (A) is true, but Reason (R) is false.

Explanation (Step by Step) 👉

Hydroxylamine (H₂N–OH) reacts only if a free carbonyl group (–CHO / >C=O) is available (it forms an oxime).

In glucose pentaacetate, the anomeric carbon (C-1) is acetylated too, and glucose gets “locked” in a cyclic acetal-like form, so no free –CHO group is available.

✅ इसलिए pentaacetate H₂N–OH से react नहीं करता.

The given Reason says it indicates presence of free –CHO in glucose. That is opposite of what this observation tells us.

❌ Actually, non-reaction of pentaacetate indicates absence of free –CHO (glucose exists mainly in cyclic form).

Did You Know? 🤔📌

✅ Glucose itself can react with hydroxylamine because in water it is in equilibrium with a small open-chain aldehyde form.

✅ But after forming pentaacetate, the structure can’t open easily, so aldehyde-type reactions stop.

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?

Answer ✅

A phenol + aniline mixture shows negative deviation from Raoult’s law, and its boiling point increases (often forming a maximum-boiling azeotrope).

Explanation (Step by Step) 👉

1) Why negative deviation occurs

Phenol (C₆H₅OH) can form hydrogen bonds.

Aniline (C₆H₅NH₂) can also form hydrogen bonds.

When mixed, they form strong intermolecular H-bonding between unlike molecules (phenol–aniline):

O–H···N hydrogen bonding is quite strong.

So, A–B interactions (phenol–aniline) become stronger than A–A and B–B interactions.

✅ Stronger attraction → molecules escape less into vapour → vapour pressure decreases compared to ideal solution.

➡️ Hence negative deviation from Raoult’s law.

2) Effect on boiling point

Lower vapour pressure means the solution needs higher temperature to reach atmospheric pressure.

✅ Therefore, the boiling point increases (and such systems may show maximum boiling azeotrope).

Did You Know? 🤔📌

✅ Negative deviation is usually due to strong A–B interactions (like hydrogen bonding).

✅ Positive deviation happens when A–B interactions are weaker than A–A and B–B, leading to higher vapour pressure and a lower boiling point (minimum-boiling azeotrope).

18. Why are haloarenes less reactive towards nucleophilic substitution reaction? Give two reasons.

Answer ✅

Haloarenes are less reactive towards nucleophilic substitution because the C–X bond is stronger and the usual substitution mechanisms (SN1/SN2) are not easy on an aromatic carbon.

Explanation (Two Reasons) 👉

1) Resonance gives partial double-bond character to C–X bond

In haloarenes (like chlorobenzene), the halogen lone pair overlaps with the benzene π-system:
✅ resonance: C–X bond gets partial double-bond character

➡️ So the bond becomes shorter and stronger, and harder to break than in haloalkanes.

2) Aromatic carbon is sp² hybridized, so SN2 is difficult

The carbon attached to halogen in haloarenes is sp² (planar).

✅ Nucleophile cannot do easy backside attack (required for SN2) on an sp² carbon

➡️ So SN2 is strongly hindered.

(Also, SN1 is not favored because forming a phenyl carbocation would destroy aromatic stability and is highly unstable.)

Did You Know? 🤔📌

✅ Haloarenes can undergo nucleophilic substitution more easily if there are strong electron-withdrawing groups (like –NO₂) at ortho/para positions, because they stabilize the intermediate (SNAr mechanism).

19. (a) Write IUPAC names of the following coordination compounds:

(i) [Ag(NH₃)₂][Ag(CN)₂]

(ii) K₃[Fe(C₂O₄)₃]

Answer ✅

(i) Diamminesilver(I) dicyanoargentate(I)

(ii) Potassium tris(oxalato)ferrate(III)

Explanation (Step by Step) 👉

(i) [Ag(NH₃)₂][Ag(CN)₂]

It is a salt containing:

Cation: [Ag(NH₃)₂]⁺

Anion: [Ag(CN)₂]⁻

Cation naming:

NH₃ = ammine (neutral ligand)

2 NH₃ = diammine

Ag oxidation state in cation = +1
✅ Diamminesilver(I)

Anion naming:

CN⁻ = cyano

2 CN = dicyano

For anionic complex, metal name ends with -ate → silver → argentate

Ag oxidation state here also = +1
✅ Dicyanoargentate(I)

So full name: Diamminesilver(I) dicyanoargentate(I)

(ii) K₃[Fe(C₂O₄)₃]

Complex ion is: [Fe(C₂O₄)₃]³⁻

Oxalate ligand (C₂O₄)²⁻ = oxalato (bidentate)

Oxidation state of Fe:
Let Fe = x
3(oxalato) = 3 × (−2) = −6
Overall charge = −3
So, x − 6 = −3 ⇒ x = +3
✅ Fe(III)

Name:

3 oxalato ligands → tris(oxalato) (use tris because ligand name is complex)

Anionic complex → ferrate

✅ Potassium tris(oxalato)ferrate(III)

Did You Know? 🤔📌

✅ For anionic complexes, metal names often change:

silver → argentate, iron → ferrate, copper → cuprate, gold → aurate.

✅ For ligands like oxalato, CN, NO₂, etc., the ligand name changes slightly in coordination nomenclature.

OR

19 (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.

Answer ✅

(i) Chemical test for ionisation isomerism

Use AgNO₃ solution and BaCl₂ solution (separately).

For [Co(NH₃)₅SO₄]Cl
Cl⁻ is outside the coordination sphere (free ion).
✅ On adding AgNO₃, it gives white precipitate of AgCl.

For [Co(NH₃)₅Cl]SO₄
SO₄²⁻ is outside the coordination sphere (free ion).
✅ On adding BaCl₂, it gives white precipitate of BaSO₄.

These different ionic precipitate tests show they are ionisation isomers.

(ii) Chelate effect (meaning + example)

Chelate effect: Complexes formed by polydentate (chelating) ligands are more stable than those formed by similar monodentate ligands.

Example:
✅ [Cu(en)₂]²⁺ (en = ethylenediamine, bidentate) is more stable than
[Cu(NH₃)₄]²⁺ (NH₃ is monodentate).

Explanation (Step by Step) 👉

(i) Why this works

Ionisation isomers have the same overall formula, but the anion inside the coordination sphere and the counter ion outside get interchanged.

So each complex releases different ions in solution, giving different precipitates.

(ii) Why chelate complexes are more stable

Chelating ligands form ring structures with the metal ion. Ring formation increases stability mainly due to higher entropy (more particles released/greater randomness) compared to monodentate ligands.

Did You Know? 🤔📌

✅ EDTA is a famous chelating ligand used in:

water softening

heavy metal poisoning treatment

complexometric titrations (like hardness of water).

20. Differentiate between the following:

(i) Peptide linkage and Glycosidic linkage

(ii) Essential amino acids and Non-essential amino acids

Answer ✅

(i) Peptide linkage vs Glycosidic linkage

Peptide linkage (–CO–NH–):

Formed between: Two amino acids

How it forms: Condensation between –COOH of one amino acid and –NH₂ of another (loss of H₂O)

Bond type: Amide bond (–CO–NH–)

Found in: Proteins and peptides

Example: Gly–Ala (a dipeptide) contains –CO–NH– bond

Glycosidic linkage (–O–):

Formed between: Two monosaccharides (sugars)

How it forms: Condensation between –OH groups of sugars (loss of H₂O)

Bond type: Acetal/ether-like C–O–C bond

Found in: Disaccharides and polysaccharides

Example: Maltose has an α(1→4) glycosidic bond

(ii) Essential amino acids vs Non-essential amino acids

Essential amino acids:

Meaning: Cannot be synthesized (or not enough) in the human body

Must be obtained from: Diet

Examples: Lysine, Valine, Leucine, Isoleucine, Methionine, Threonine, Phenylalanine, Tryptophan (Histidine is essential especially in growing children)

Non-essential amino acids:

Meaning: Can be synthesized in the human body

Diet requirement: Not compulsory (though still useful)

Examples: Glycine, Alanine, Serine, Aspartic acid, Glutamic acid, Proline

Explanation (Step by Step) 👉

Peptide bond joins amino acids to build proteins.

Glycosidic bond joins sugars to build carbohydrates like starch and cellulose.

Essential vs non-essential classification depends on whether the body can make that amino acid or must take it from food.

Did You Know? 🤔📌

✅ A protein’s nutritional quality depends a lot on the presence of essential amino acids. That’s why foods like egg, milk, soy, and pulses + cereals combination are considered good protein sources.

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?

Answer ✅

The rate becomes 27 times the original rate.

No, the order of reaction does not change.

Explanation (Step by Step) 👉

1) Write the rate law (one-step / elementary reaction)

For an elementary reaction:
✅ Rate ∝ [A]²[B]

So, Rate = k [A]²[B]

Overall order = 2 + 1 = 3

2) Effect of decreasing volume to one third

If volume becomes V/3, concentrations become 3 times (because concentration ∝ 1/Volume):

[A] → 3[A]
[B] → 3[B]

New rate:

Rate′ = k (3[A])²(3[B])
Rate′ = k × 9[A]² × 3[B]
Rate′ = 27 × k[A]²[B]

✅ Rate′ = 27 × Rate

3) Does the order change?

No. Order depends on the mechanism (rate law), not on concentration/volume. So the reaction remains third order.

Did You Know? 🤔📌

✅ Volume change affects rate by changing concentrations, but it does not change the order (unless the mechanism itself changes, which is not assumed here).

22. For the first order thermal decomposition reaction:

C₂H₅Cl(g) ⟶ C₂H₄(g) + HCl(g)

Given data:

Time (s)	Total Pressure (atm)
0	0.30
30	0.50

Given: log 3 = 0.48

Calculate rate constant (k).

Answer ✅

k = 3.68 × 10⁻² s⁻¹ (≈ 0.0368 s⁻¹) ✅

Explanation (Step by Step) 👉

Step 1: Find degree of dissociation (α)

For A ⟶ B + C (starting with only A):

P_t = P₀(1 + α)

α = (P_t/P₀) − 1

α = (0.50/0.30) − 1
α = 1.6667 − 1 = 0.6667 = 2/3

Step 2: Find partial pressure of reactant at 30 s

P_A = P₀(1 − α)

P_A = 0.30 × (1 − 2/3)
P_A = 0.30 × (1/3) = 0.10 atm

Step 3: Apply first-order formula

k = (2.303/t) log(P₀/P_A)

k = (2.303/30) log(0.30/0.10)
k = (2.303/30) log(3)

Given log 3 = 0.48:

k = (2.303 × 0.48)/30
k = 1.10544/30

✅ k = 0.03685 s⁻¹ ≈ 3.68 × 10⁻² s⁻¹

Did You Know? 🤔📌

✅ In gas-phase reactions like A ⟶ B + C, total pressure increases because total moles increase. That’s why we convert total pressure → reactant partial pressure before using the first-order equation.

23. (a) Answer the following:

(i) Why is the Equilibrium Constant (K_c) related to E°_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.

Answer ✅

(i) Because K_c is defined for equilibrium under standard conditions, and E°_cell is the cell potential under standard conditions.

(ii) ✅ Metal A (E° = −0.24 V) will liberate H₂ gas from dilute H₂SO₄.

(iii) During charging, the reaction is the reverse of discharge:

✅ 2PbSO₄(s) + 2H₂O(l) ⟶ Pb(s) + PbO₂(s) + 2H₂SO₄(aq)

Explanation (Step by Step) 👉

(i)

E_cell changes with concentration/pressure (non-standard conditions).

E°_cell is measured at standard state (1 M, 1 atm, 298 K).

The relation is:

ΔG° = −nF E°_cell and ΔG° = −RT ln K

Combining:

E°_cell = (RT/nF) ln K

So K is linked to E°_cell, not to E_cell.

(ii)

Hydrogen electrode has E° = 0.00 V.

A metal will displace H₂ from acids if it is more reactive than hydrogen, i.e., it has more negative E° than 0.

Metal A: −0.24 V (negative) → can reduce H⁺ to H₂ ✅

Metal B: +0.80 V (positive) → less reactive than H, will not liberate H₂ ❌

(iii)

On charging, PbSO₄ on both plates converts back to:

Pb (negative plate)

PbO₂ (positive plate)

Sulfuric acid concentration increases again.

Did You Know? 🤔📌

✅ At 298 K: E°_cell = (0.0591/n) log K

✅ In a lead-acid battery, the acid becomes more dilute during discharge and becomes more concentrated during charging (that’s why specific gravity of acid indicates battery charge).

OR

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

Answer ✅

A mercury cell is a primary (non-rechargeable) battery. It is more advantageous than a dry cell because it gives a nearly constant voltage and has a longer life.

✅ Overall reaction: Zn(Hg) + HgO(s) ⟶ ZnO(s) + Hg(l)

Explanation (Step by Step) 👉

1) Type of battery

✅ Mercury cell = Primary cell (used in button cells like watches, hearing aids, calculators)

2) Why it is more advantageous than dry cell

Mercury cell advantages:

✅ Constant cell potential (~1.35 V) during use (very stable voltage)

✅ Long shelf life and more reliable output

✅ Compact size (high energy per volume)

✅ Less leakage compared to many dry cells

3) Overall reaction in mercury cell

Anode (oxidation):

Zn(Hg) + 2OH⁻ ⟶ ZnO(s) + H₂O + 2e⁻

Cathode (reduction):

HgO(s) + H₂O + 2e⁻ ⟶ Hg(l) + 2OH⁻

✅ Overall reaction:

Zn(Hg) + HgO(s) ⟶ ZnO(s) + Hg(l)

Did You Know? 🤔📌

✅ Mercury cells were popular because of their steady voltage, but their use has been restricted in many places due to mercury toxicity, so silver-oxide cells are common replacements.

24. Calculate the boiling point of a solution containing 0.61 g of benzoic acid (M = 122 g mol⁻¹) in 5 g of CS₂ in which it dimerises to the extent of 88%.

Boiling point and K_b of CS₂ are 46.2 °C and 2.3 K kg mol⁻¹ respectively.

Answer ✅

Boiling point of solution = 47.49 °C (approx.) ✅

Explanation (Step by Step) 👉

Step 1: Moles of benzoic acid

Moles = 0.61 / 122 = 0.005 mol

Step 2: Molality (m)

Mass of CS₂ = 5 g = 0.005 kg

m = moles / kg solvent = 0.005 / 0.005 = 1 m

Step 3: Van’t Hoff factor (i) for dimerisation

For dimerisation: 2A ⟶ A₂

If degree of association = α, then:
i = 1 − α/2

Given α = 0.88
i = 1 − 0.88/2 = 1 − 0.44 = 0.56

Step 4: Boiling point elevation

ΔT_b = i K_b m

ΔT_b = 0.56 × 2.3 × 1 = 1.288 K

Step 5: Boiling point of solution

T_b(solution) = 46.2 + 1.288 = 47.488 °C ≈ 47.49 °C

Did You Know? 🤔📌

✅ Association (like dimerisation) decreases the number of solute particles, so i < 1, and colligative effects (ΔT_b, ΔT_f) become smaller than expected.

25. Write the reactions of D-Glucose with the following:

(a) HI
(b) Br₂ water
(c) Conc. HNO₃

Answer ✅

(a) D-Glucose + HI

On heating D-glucose with excess HI (often with red phosphorus), glucose is reduced to n-hexane.

C₆H₁₂O₆ + (excess) HI ⟶ C₆H₁₄ + I₂ + H₂O
(Product: n-Hexane)

(b) D-Glucose + Br₂ water

Br₂/water is a mild oxidizing agent. It oxidizes the –CHO group of glucose to –COOH, forming gluconic acid.

C₆H₁₂O₆ + Br₂ + H₂O ⟶ C₆H₁₂O₇ + 2HBr
(Product: Gluconic acid)

(c) D-Glucose + Conc. HNO₃

Conc. HNO₃ strongly oxidizes both ends:

–CHO (aldehyde) → –COOH

–CH₂OH (primary alcohol) → –COOH

forming a dicarboxylic acid called saccharic acid (glucaric acid).

C₆H₁₂O₆ + (conc.) HNO₃ ⟶ HOOC–(CHOH)₄–COOH
(Product: Saccharic acid / Glucaric acid)

Explanation (Step by Step) 👉

(a) HI

On heating D-glucose with excess HI (often with red phosphorus), all oxygen-containing groups are reduced, giving n-hexane.

(b) Br₂ water

Br₂/water is mild, so it oxidizes only the aldehyde (–CHO) to carboxylic acid (–COOH), producing gluconic acid.

(c) Conc. HNO₃

Conc. HNO₃ is strong, so it oxidizes both –CHO and –CH₂OH ends to –COOH, giving the diacid saccharic acid (glucaric acid).

Did You Know? 🤔📌

✅ Br₂ water oxidizes only the aldehyde group (mild), but conc. HNO₃ oxidizes both aldehyde and primary alcohol, giving a diacid.

✅ The HI reduction to n-hexane is an important proof that glucose has a straight 6-carbon chain.

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.

Answer ✅

(a) Carboxylic acids have higher boiling point than alcohols (same molar mass)

(b) α-Hydrogens of aldehydes and ketones are acidic

(c) NH₃ and its derivatives do not add to carbonyl in strongly acidic medium

Explanation (Step by Step) 👉

(a) Carboxylic acids have higher boiling point than alcohols

Carboxylic acids form strong intermolecular hydrogen bonding.

They commonly exist as dimers in vapour as well as liquid phase due to two H-bonds:

R–COOH ··· HOOC–R (dimer)

✅ Dimer formation increases effective molecular mass and intermolecular attraction, so more heat is needed to separate molecules.
➡️ Hence higher boiling point than alcohols.

(b) α-Hydrogens of aldehydes and ketones are acidic

α-H is the hydrogen on the carbon next to the carbonyl group (C=O).

When α-H is removed (by a base), an enolate ion is formed.

This enolate is stabilized by resonance:

✅ negative charge delocalizes between α-carbon and oxygen (more stable conjugate base).

➡️ Greater stability of conjugate base = higher acidity of α-hydrogen.

(c) NH₃ and its derivatives do not add to carbonyl in strongly acidic medium

Ammonia and amine derivatives act as nucleophiles using the lone pair on nitrogen.

In a strongly acidic medium, they get protonated:

NH₃ + H⁺ → NH₄⁺

RNH₂ + H⁺ → RNH₃⁺

✅ After protonation, the lone pair is not available, so nucleophilicity drops sharply.
➡️ Hence nucleophilic addition does not occur (or becomes very slow) in strongly acidic medium.

Did You Know? 🤔📌

✅ Alcohols also hydrogen bond, but carboxylic acids form two H-bonds per pair (dimer), making association much stronger.

✅ This is the reason aldehydes/ketones undergo aldol condensation (it starts via enolate formation).

✅ Mildly acidic conditions can help because they activate the carbonyl (make it more electrophilic), but too much acid protonates the nucleophile, stopping the reaction.

27. Write the reaction involved in the following:

(a) Reimer–Tiemann reaction

(b) Kolbe’s reaction

(c) Friedel–Crafts acylation of anisole

Answer ✅

(a) Reimer–Tiemann reaction

Phenol on treatment with CHCl₃ + aq. NaOH (heat) gives mainly salicylaldehyde (o-hydroxybenzaldehyde).

C₆H₅OH + CHCl₃ + 3NaOH ⟶ o-HO–C₆H₄–CHO + 3NaCl + 2H₂O
(major product: o-hydroxybenzaldehyde; p-isomer minor)

(b) Kolbe’s reaction (Kolbe–Schmitt reaction)

Sodium phenoxide reacts with CO₂ (pressure, heat) to form sodium salicylate, which on acidification gives salicylic acid.

C₆H₅ONa + CO₂ (pressure, heat) ⟶ o-HO–C₆H₄–COONa

o-HO–C₆H₄–COONa + H⁺ ⟶ o-HO–C₆H₄–COOH
(product: salicylic acid)

(c) Friedel–Crafts acylation of anisole

Anisole (methoxybenzene) reacts with an acyl chloride in presence of anhyd. AlCl₃ to give mainly p-acyl anisole (o-product minor).

Example (acetylation):

C₆H₅OCH₃ + CH₃COCl (AlCl₃) ⟶ p-CH₃CO–C₆H₄–OCH₃ + o-isomer (minor)
(major product: p-methoxyacetophenone)

Explanation (Step by Step) 👉

Reimer–Tiemann: introduces –CHO group on phenol (mainly at ortho).

Kolbe–Schmitt: introduces –COOH group on phenol (mainly at ortho).

Friedel–Crafts acylation: introduces –COR group; anisole directs ortho/para, para major due to less steric hindrance.

Did You Know? 🤔📌

✅ Reimer–Tiemann uses :CCl₂ (dichlorocarbene) as the key electrophile.

✅ Kolbe–Schmitt is an industrial route to salicylic acid, which is used to make aspirin.

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?

Answer ✅

(a) Structural formula of X and Y

X (shows first-order rate, depends only on [X] → S_N1):
✅ tert-Butyl bromide (2-bromo-2-methylpropane)
Structural formula: (CH₃)₃C–Br

Y (optically active and rate depends on [Y] and [KOH] → S_N2):
✅ (R)/(S)-2-bromobutane (sec-butyl bromide)
Structural formula: CH₃–CH(Br)–CH₂–CH₃ (chiral at C-2)

(b) Which undergoes racemisation and why? ✅

✅ X undergoes racemisation (in general for S_N1) because it forms a planar carbocation intermediate, allowing nucleophile attack from both sides.

Note: tert-butyl bromide itself is not chiral, so you won’t “observe” racemisation here, but the S_N1 pathway is the racemisation-type mechanism (attack from both sides).

(c) Which gives inversion of configuration and why? ✅

✅ Y gives inversion of configuration because S_N2 occurs by backside attack of OH⁻, producing Walden inversion at the chiral carbon.

Explanation (Step by Step) 👉

Why X is S_N1 (rate depends only on [X])

Rate law given: Rate = k[X] → unimolecular substitution → S_N1

Aqueous KOH (polar protic) supports carbocation formation.

Among C₄H₉Br isomers, tert-butyl bromide forms a very stable 3° carbocation, so it reacts by S_N1.

Why Y is S_N2 (rate depends on [Y] and [KOH])

Rate law: Rate = k[Y][OH⁻] → bimolecular → S_N2

“Optically active” requires a chiral center.

2-bromobutane is chiral (exists as R/S enantiomers) and can undergo S_N2 with OH⁻.

Did You Know? 🤔📌

✅ S_N1 → carbocation → racemisation tendency (often partial due to ion-pair effects).

✅ S_N2 → one-step backside attack → 100% inversion at the stereocenter.

Section - D

The following questions are case-based questions. Each question has an internal choice and carries 4(2 + 1 + 1) marks each. Read the passage carefully and answer the questions that follow.

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.

(a) Answer the following questions :

(i) Why CH₃–NH₂ is a stronger base than (CH₃)₃N in aqueous solution?

(ii) Write structural formulae of compound A and B :
CH₃CONH₂ —NaOBr→ A —C₆H₅COCl / Base→ B

(b) A compound ‘X’ with molecular formula C₃H₉N reacts with Hinsberg reagent to give a product insoluble in alkali. Identify ‘X’.

Answer ✅

(a)(i) Reason

In aqueous solution, CH₃NH₂ is a stronger base than (CH₃)₃N because its conjugate acid is better solvated (more hydrogen bonding with water).

(a)(ii) Compounds A and B

✅ A = CH₃NH₂ (methylamine)

✅ B = C₆H₅CONHCH₃ (N-methylbenzamide)

(b) Identify X

✅ X = CH₃NHCH₂CH₃ (ethylmethylamine) ✅

Explanation (Step by Step) 👉

(a)(i) Why CH₃NH₂ > (CH₃)₃N as a base in water

Basicity in water depends a lot on stability of the conjugate acid after protonation.

CH₃NH₂ + H⁺ → CH₃NH₃⁺

(CH₃)₃N + H⁺ → (CH₃)₃NH⁺

✅ CH₃NH₃⁺ gets strongly solvated by water (forms more effective H-bonds).
❌ (CH₃)₃NH⁺ is bulky, so solvation is hindered (steric effect), making it less stable in water.

So in aqueous medium, methylamine behaves as a stronger base.

(a)(ii) Identify A and B

Step 1: CH₃CONH₂ —NaOBr→ A

This is Hofmann bromamide reaction (amide → amine with one carbon less).
So: CH₃CONH₂ ⟶ CH₃NH₂ (A)

Step 2: A —C₆H₅COCl / Base→ B

Methylamine reacts with benzoyl chloride to give an amide:

CH₃NH₂ + C₆H₅COCl ⟶ C₆H₅CONHCH₃ (B) + HCl
(Base neutralizes HCl)

(b) Hinsberg test logic

Hinsberg reagent = benzenesulfonyl chloride (C₆H₅SO₂Cl)

Primary amine → sulfonamide with N–H, which becomes soluble in alkali (forms salt).

Secondary amine → sulfonamide with no N–H, so it does NOT dissolve in alkali. ✅

Given: product insoluble in alkali → indicates a secondary amine.

For formula C₃H₉N, the secondary amine is:
✅ CH₃NHCH₂CH₃ (ethylmethylamine)

Did You Know? 🤔📌

✅ In water, basic strength of amines depends not only on +I effect, but also on solvation of the conjugate acid. That’s why the order can change between gas phase and aqueous solution.

OR

29. (b) How can you convert aniline to benzonitrile?

(c) Why is –NH₂ group of aniline acetylated before carrying out nitration?

Answer ✅

(b) Aniline → Benzonitrile

Convert aniline to benzene diazonium chloride, then replace –N₂⁺ by –CN using CuCN (Sandmeyer reaction).

Diazotisation (0–5 °C):

C₆H₅NH₂ + NaNO₂ + 2HCl ⟶ C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O

Sandmeyer (CuCN):

C₆H₅N₂⁺Cl⁻ + CuCN ⟶ C₆H₅CN + CuCl + N₂↑

✅ Product: Benzonitrile (C₆H₅CN)

(c) Why acetylate –NH₂ before nitration?

Because in nitrating mixture (HNO₃/H₂SO₄), aniline gets protonated to anilinium ion (–NH₃⁺), which is deactivating and meta-directing, and also nitration becomes uncontrolled (oxidation/tar formation).

Acetylation converts aniline to acetanilide (–NHCOCH₃), which:

✅ is less basic, so it is not easily protonated in acid

✅ is still ortho/para directing, but moderately activating, so nitration is controlled

✅ gives mainly para-nitroacetanilide (clean major product)

Explanation (Step by Step) 👉

(b) First make diazonium salt (cold conditions), then do Sandmeyer to introduce –CN. This is a clean way to convert –NH₂ into other groups on benzene ring.

(c) Acetylation acts like a protecting group step: it reduces excessive activation of the ring and avoids protonation problems, giving a more controlled o/p nitration.

Did You Know? 🤔📌

✅ This “protecting group” strategy (convert –NH₂ to –NHCOCH₃) is a classic example of controlling orientation and avoiding side reactions in aromatic substitutions.

30. 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.

(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)

(c) [NiCl₄]²⁻ is paramagnetic while [Ni(CO)₄] is diamagnetic though both are tetrahedral. Why? (At. No.: Ni = 28)

Answer ✅

(a) Orbitals raised in octahedral field

✅ d_x²−y² and d_z² (the e_g set) are raised in energy.

(b) Electronic configuration (CFT)

(i) [CoF₆]³⁻
✅ Co is +3 → d⁶
F⁻ is weak field → high spin
Configuration: t_2g⁴ e_g² (4 unpaired)

(ii) [Co(NH₃)₆]³⁺
✅ Co is +3 → d⁶
NH₃ is stronger field than F⁻ → low spin (for Co³⁺)
Configuration: t_2g⁶ e_g⁰ (0 unpaired)

(c) Paramagnetic vs diamagnetic (both tetrahedral)

✅ [NiCl₄]²⁻ is paramagnetic because Cl⁻ is weak field, so electrons remain unpaired.

✅ [Ni(CO)₄] is diamagnetic because CO is very strong field, causing complete pairing (no unpaired electrons).

Explanation (Step by Step) 👉

(a) Why e_g orbitals are raised

In an octahedral complex, ligands approach along the x, y, z axes.

d_x²−y² and d_z² point directly along the axes, so they face maximum repulsion from ligands.
➡️ Their energy increases (raised).

The other three orbitals (d_xy, d_xz, d_yz) lie between axes, so they face less repulsion and are lower (t_2g).

(b) Details (Co³⁺ : d⁶)

Co (Z=27): [Ar] 3d⁷4s²
Co³⁺ loses 2 from 4s and 1 from 3d → 3d⁶

With F⁻ (weak field) → Δ_o small → high spin: t_2g⁴ e_g²

With NH₃ (stronger field) and Co³⁺ (high charge) → Δ_o large → low spin: t_2g⁶ e_g⁰

(c) Why both tetrahedral but different magnetism

[NiCl₄]²⁻: Ni is +2 → d⁸; weak ligand field → unpaired electrons remain → paramagnetic.

[Ni(CO)₄]: Ni is 0 → d¹⁰; all electrons paired → diamagnetic.

Did You Know? 🤔📌

✅ For Co³⁺ (d⁶), many complexes become low spin because higher oxidation state increases Δ_o.

✅ CO is a strong ligand due to π-back bonding, which stabilizes paired-electron arrangements.

OR

Answer ✅

Hybridization: d²sp³ (inner orbital complex, octahedral)

Magnetic behaviour: Paramagnetic (1 unpaired electron)

Explanation (Step by Step) 👉

Step 1: Oxidation state of Fe

In [Fe(CN)₆]³⁻:

Let oxidation state of Fe = x
x + 6(−1) = −3
x − 6 = −3 ⇒ x = +3

So Fe³⁺ configuration:
Fe (26) = [Ar] 3d⁶ 4s²
Fe³⁺ = [Ar] 3d⁵ (d⁵)

Step 2: Nature of ligand

CN⁻ is a strong field ligand → causes pairing (low spin).

So in octahedral field (low spin d⁵):

t_2g⁵ e_g⁰

→ 1 unpaired electron

Step 3: Hybridization (VBT view)

Because electrons pair up, two 3d orbitals become available for bonding:
✅ d²sp³ hybridization → octahedral, inner orbital complex.

Did You Know? 🤔📌

✅ [Fe(CN)₆]⁴⁻ (ferrocyanide, Fe²⁺ = d⁶) is diamagnetic, while [Fe(CN)₆]³⁻ (ferricyanide, Fe³⁺ = d⁵) is paramagnetic due to that one unpaired electron.

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

Answer ✅

(i) Structures of X (C₅H₁₀O)

(I) No Tollen’s, but positive iodoform

✅ X = Pentan-2-one
CH₃–CO–CH₂–CH₂–CH₃

(II) No Tollen’s, no iodoform, but Aldol condensation

✅ X = Pentan-3-one
CH₃–CH₂–CO–CH₂–CH₃

(III) Undergoes Cannizzaro

✅ X = 2,2-Dimethylpropanal (pivaldehyde)
(CH₃)₃C–CHO

(ii) Conversions to benzoic acid

(I) Acetophenone → Benzoic acid

C₆H₅COCH₃ —(hot KMnO₄ / alkaline, then acidify)→ C₆H₅COOH

(II) Ethyl benzene → Benzoic acid

C₆H₅CH₂CH₃ —(hot KMnO₄ / alkaline, then acidify)→ C₆H₅COOH

Explanation (Step by Step) 👉

For (i)

Iodoform + no Tollen’s → methyl ketone (–COCH₃) → pentan-2-one

Aldol + no Tollen’s + no iodoform → ketone with α-H but not methyl ketone → pentan-3-one

Cannizzaro → aldehyde with no α-H → 2,2-dimethylpropanal

For (ii)

Any aromatic compound having a side chain with at least one benzylic hydrogen on strong oxidation gives C₆H₅COOH.

Acetophenone also oxidizes under strong conditions to benzoic acid.

Did You Know? 🤔📌

✅ Hot KMnO₄ side-chain oxidation converts many alkyl benzenes (toluene, ethyl benzene, propyl benzene, etc.) into benzoic acid, regardless of chain length, as long as benzylic H is present.

OR

31. (b) Answer the following questions :

(i) Draw structure of the 2,4-dinitrophenylhydrazone 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:
(Cyclohexanol) —CrO₃→ ‘X’

Answer ✅

(i) 2,4-dinitrophenylhydrazone derivative of benzaldehyde

C₆H₅–CH=N–NH–C₆H₃(NO₂)₂
(NO₂ at 2- and 4-positions on the second ring)

(ii) Increasing reactivity towards HCN

✅ Di-tert. butyl ketone < Acetone < Acetaldehyde

(iii) Test: Benzoic acid vs Ethyl benzoate

✅ Sodium bicarbonate test (NaHCO₃)

Benzoic acid: brisk effervescence of CO₂

C₆H₅COOH + NaHCO₃ → C₆H₅COONa + CO₂↑ + H₂O

Ethyl benzoate: no effervescence (ester does not react)

(iv) Reagent: ethanenitrile → ethanal

✅ Stephen reduction: SnCl₂/HCl, followed by hydrolysis.

(v) Structure of X

✅ X = Cyclohexanone
(six-membered ring with a C=O group)

Explanation (Step by Step) 👉

(ii) HCN adds by nucleophilic addition to C=O. Reactivity increases with:

less steric hindrance and more electrophilic carbonyl carbon.

Aldehydes > ketones (less +I effect and less hindrance). Di-tert-butyl ketone is most hindered → least reactive.

(v) CrO₃ oxidizes secondary alcohols → ketones
cyclohexanol → cyclohexanone .

Did You Know? 🤔📌

✅ 2,4-DNP test is used to identify aldehydes and ketones by forming orange/yellow crystalline hydrazones.

✅ CrO₃ oxidizes secondary alcohols → ketones.

32. (a) (i) From the given data of E° values, answer the following questions:

E°_M²⁺/M (V)	Cr	Mn	Fe	Co	Ni	Cu
Values	−0.91	−1.18	−0.44	−0.28	−0.25	+0.34

(Cr = −0.91, Mn = −1.18, Fe = −0.44, Co = −0.28, Ni = −0.25, Cu = +0.34)

(I) Why E°_M²⁺/M show irregular trend in the above values?

(II) Why is E°_Cu²⁺/Cu value exceptionally positive?

(III) Why E°_Mn²⁺/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.

Answer ✅

(i)

(I) Irregular trend is due to multiple competing factors like atomisation enthalpy, ionisation enthalpy, hydration enthalpy and special stability of some d-electron configurations.

(II) E°_Cu²⁺/Cu is exceptionally positive because Cu does not oxidise easily to Cu²⁺; Cu²⁺ prefers to get reduced to Cu.

(III) E°_Mn²⁺/Mn is highly negative because Mn²⁺ (3d⁵) is especially stable in aqueous solution, so Mn²⁺ is difficult to reduce to Mn.

(ii)

Acidic medium:

2MnO₄⁻ + 10I⁻ + 16H⁺ → 2Mn²⁺ + 5I₂ + 8H₂O

Alkaline medium:

2MnO₄⁻ + I⁻ + H₂O → 2MnO₂(s) + IO₃⁻ + 2OH⁻

Explanation (Step by Step) 👉

(i)(I) Why irregular E° trend?

E°_M²⁺/M depends on overall energy change when:

M(s) → M(g) (atomisation)

M(g) → M²⁺(g) (I₁ + I₂)

M²⁺(g) → M²⁺(aq) (hydration)

These terms vary irregularly across Cr to Cu. Also, extra stability of d⁵ and d¹⁰ configurations affects values.

(i)(II) Why Cu²⁺/Cu is positive?

For Cu, the energy needed to convert Cu(s) → Cu²⁺ (high atomisation + high ionisation) is large. So oxidation of Cu is not favoured, and Cu²⁺ more readily accepts electrons and becomes Cu. Hence E° is positive (+0.34 V).

(i)(III) Why Mn²⁺/Mn is very negative?

Mn²⁺ has 3d⁵ (half-filled), which is extra stable in aqueous solution. Therefore Mn²⁺ does not easily get reduced to Mn(s), making E° very negative (−1.18 V).

(ii) KMnO₄ oxidising I⁻

In acidic medium, MnO₄⁻ is reduced to Mn²⁺ (strong oxidising action), and I⁻ is oxidised to I₂.

In alkaline medium, MnO₄⁻ is reduced to MnO₂ (brown ppt), and I⁻ is oxidised to IO₃⁻.

Did You Know? 🤔📌

✅ KMnO₄ is a stronger oxidising agent in acidic medium because it forms Mn²⁺. In alkaline medium it often forms MnO₂, so its oxidising power is lower.

✅ Extra stability of half-filled (d⁵) and fully filled (d¹⁰) orbitals is a key reason for “odd” behaviour in transition metal trends.

OR

32. (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 KMnO₄ is allowed to stand for sometime? Give the equation involved. What is this type of reaction called?

Answer ✅

(i) Lanthanoids

(I) Ce (Cerium) shows +4 oxidation state. ✅

(II) Eu (Europium) shows +2 oxidation state. ✅

(ii) Transition metals are good catalysts because of variable oxidation states and ability to form intermediate complexes / adsorb reactants. ✅

(iii) Cr has higher melting point because it has stronger metallic bonding (higher enthalpy of atomisation) than Mn. ✅

(iv) Acidic KMnO₄ on standing

It slowly decomposes, giving MnO₂ and O₂. ✅

4MnO₄⁻ + 4H⁺ → 4MnO₂(s) + 3O₂(g) + 2H₂O

Type of reaction: Redox (self-redox / decomposition) ✅

Explanation (Step by Step) 👉

(i)

Ce⁴⁺ is common because Ce can lose one more electron beyond +3 to reach a relatively stable state.

Eu²⁺ is common because Eu²⁺ has 4f⁷ (half-filled) configuration, which is extra stable.

(ii)

Transition metals help reactions by:

providing an alternative pathway with lower activation energy via intermediate complexes, and

easily switching between oxidation states to assist electron transfer steps.

(iii)

Cr has stronger metal–metal bonding than Mn (reflected in higher atomisation enthalpy), so more energy is needed to break the lattice → higher melting point.

(iv)

In acidic medium, permanganate can undergo slow self-redox decomposition on standing, forming brown MnO₂ and releasing O₂ gas. The balanced equation above matches atoms and charge.

Did You Know? 🤔📌

✅ Other examples: Tb can also show +4, and Yb/Sm can show +2 oxidation state.

✅ The brown solid formed in many KMnO₄ decompositions is MnO₂.

33. 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°_Mg²⁺/Mg = −2.37 V, E°_Ag⁺/Ag = +0.80 V
1 F = 96500 C mol⁻¹, log 10 = 1

Answer ✅

E_cell (298 K): 3.05 V (approx.)

ΔG: −5.89 × 10⁵ J mol⁻¹ = −589 kJ mol⁻¹ (approx.)

Explanation (Step by Step) 👉

1) Write the cell reaction and find n

Anode (oxidation):
Mg → Mg²⁺ + 2e⁻

Cathode (reduction):
Ag⁺ + e⁻ → Ag

Overall:
Mg + 2Ag⁺ → Mg²⁺ + 2Ag

So, n = 2

2) Standard cell potential

E°_cell = E°_cathode − E°_anode

= (+0.80) − (−2.37)

= 3.17 V

3) Reaction quotient (Q)

Q = [Mg²⁺] / [Ag⁺]²

= 0.01 / (0.001)²

= 0.01 / 10⁻⁶

= 10⁴

So, log Q = 4

4) Nernst equation at 298 K

E_cell = E°_cell − (0.0591/n) log Q

= 3.17 − (0.0591/2)×4

= 3.17 − 0.1182

= 3.0518 V ≈ 3.05 V

5) Gibbs free energy change

ΔG = −n F E

= −(2)(96500)(3.0518)

= −5.88997 × 10⁵ J mol⁻¹

≈ −589 kJ mol⁻¹

Did You Know? 🤔📌

✅ Negative ΔG means the cell reaction is spontaneous.

✅ Because Q = 10⁴ is large, the cell emf becomes a bit less than E°.

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

Answer ✅

✅ E_cell (25 °C) ≈ 0.314 V

Explanation (Step by Step) 👉

1) Find n (electrons transferred)

H₂ ⟶ 2H⁺ + 2e⁻

So, n = 2

2) Calculate E° from ΔG°

ΔG° = −nFE°

E° = −ΔG°/(nF)

E° = −(−43500) / (2 × 96500)

E° = 43500 / 193000

E° ≈ 0.225 V

3) Write reaction quotient Q

Solids (AgCl, Ag) are not included in Q.

Q = ([H⁺]² [Cl⁻]²) / P(H₂)

= (0.1² × 0.2²) / 0.4

= (0.01 × 0.04) / 0.4

= 0.0004 / 0.4

= 0.001 = 10⁻³

So, log Q = −3

4) Apply Nernst equation (298 K)

E = E° − (0.0591/n) log Q

E = 0.225 − (0.0591/2)(−3)

E = 0.225 + 0.08865

E ≈ 0.31365 V ≈ 0.314 V

Did You Know? 🤔📌

✅ Here Q < 1, so log Q is negative, which makes E > E° (cell emf increases compared to standard conditions).