CBSE_Class_12_Chemistry_Model_Paper (2) (2024)

General Instructions:

This question paper contains 33 questions. All questions are compulsory.
The question paper is divided into five sections: Section A, B, C, D, and E.
- Section A – Questions number 1 to 16 are multiple-choice type questions. Each question carries 1 mark.
- Section B – Questions number 17 to 21 are very short answer type questions. Each question carries 2 marks.
- Section C – Questions number 22 to 28 are short answer type questions. Each question carries 3 marks.
- Section D – Questions number 29 and 30 are case-based questions. Each question carries 4 marks.
- Section E – Questions number 31 to 33 are long answer type questions. Each question carries 5 marks.
There is no overall choice in the question paper. However, internal choice has been provided in some questions in all sections except Section A.
Kindly note that there is a separate question paper for Visually Impaired candidates.
Use of calculators is not allowed.

Instructions for Answering:

Section A contains multiple choice questions. Choose the correct option for each question and write it down in your answer sheet.
Section B contains very short answer type questions. Write the answer in a concise manner (maximum 2-3 lines).
Section C contains short answer type questions. Provide detailed answers in approximately 3-4 lines.
Section D contains case-based questions. Answer the questions based on the given case and explain your reasoning.
Section E contains long answer type questions. Write detailed answers in 5-6 sentences.

Ensure that you answer all the questions, and write your answers clearly.

SECTION A (Multiple Choice Questions)

1. The most stable complex among the following is:

(A) Na₂[Na(CN)₅(NO)]
(B) K₃[Fe(CN)₆]
(C) [Cr(NH₃)₆]³⁺
(D) Na₂[Ni(EDTA)]

Answer: (B) K₃[Fe(CN)₆)

Explanation: The complex K₃[Fe(CN)₆] is stable due to the strong field ligand CN⁻ which causes significant splitting of the d-orbitals, leading to a more stable arrangement.

2. If amines are arranged in decreasing order of their basic strength in the gaseous phase, then the correct order will be:

(A) (CH₃)₃N > (CH₃)₂NH > CH₃NH₂ > NH₃
(B) NH₃ > CH₃NH₂ > (CH₃)₂NH > (CH₃)₃N
(C) CH₃NH₂ > (CH₃)₃N > (CH₃)₂NH > NH₃
(D) (CH₃)₂NH > (CH₃)₃N > CH₃NH₂ > NH₃

Answer: (A) (CH₃)₃N > (CH₃)₂NH > CH₃NH₂ > NH₃

Explanation: The basicity of amines in the gaseous phase decreases as the alkyl groups increase in size. Tertiary amines are more basic due to better electron-donating effects of alkyl groups compared to primary and secondary amines.

3. Considering the strength of the ligand, the lowest excitation energy will be observed in:

(A) [Co(H₂O)₆]³⁺
(B) [Co(NH₃)₆]³⁺
(C) [Co(CN)₆]³⁻
(D) [CoCl₆]³⁻

Answer: (C) [Co(CN)₆]³⁻

Explanation: The ligand strength follows the order: CN⁻ > NH₃ > H₂O > Cl⁻. CN⁻ is a strong field ligand, causing a large splitting of the d-orbitals in the metal complex, leading to a lower excitation energy.

4. The standard electrode potential for Sn⁴⁺/Sn²⁺ couple is + 0.15 V and for Cr³⁺/Cr couple is – 0.73 V. These two couples are connected to make an electrochemical cell. The redox reaction is spontaneous. The cell potential will be:

(A) + 0.88 V
(B) + 0.58 V
(C) – 0.88 V
(D) – 0.58 V

Answer: (A) + 0.88 V

Explanation: The cell potential EcellE_{\text{cell}}Ecell can be calculated using the formula:Ecell=Ecathode−EanodeE_{\text{cell}} = E_{\text{cathode}} – E_{\text{anode}}Ecell=Ecathode−Eanode

Since the redox reaction is spontaneous, the couple with a higher standard electrode potential will act as the cathode (reduction), and the couple with a lower potential will act as the anode (oxidation).

Ecathode=+0.15 VE_{\text{cathode}} = +0.15 \, \text{V}Ecathode=+0.15V (Sn⁴⁺/Sn²⁺)
Eanode=−0.73 VE_{\text{anode}} = -0.73 \, \text{V}Eanode=−0.73V (Cr³⁺/Cr)

Ecell=(+0.15)−(−0.73)=0.88 VE_{\text{cell}} = (+0.15) – (-0.73) = 0.88 \, \text{V}Ecell=(+0.15)−(−0.73)=0.88V

Thus, the cell potential is +0.88 V.

5. The chemical test which can be used to distinguish between ethanamine and aniline is:

(A) Haloform test
(B) Tollens’ test
(C) Azo dye test
(D) Hinsberg test

Answer: (D) Hinsberg test

Explanation: The Hinsberg test distinguishes between primary, secondary, and tertiary amines. Ethanamine (a primary amine) reacts with benzenesulfonyl chloride (Hinsberg reagent) to form a soluble product, whereas aniline (an aromatic amine) does not react in the same way.

6. The geometry of diamagnetic nickel complex [Ni(CN)₄]²⁻ is:

(A) Tetrahedral
(B) Octahedral
(C) Square planar
(D) Distorted octahedral

Answer: (C) Square planar

Explanation: The [Ni(CN)₄]²⁻ complex has a square planar geometry due to the presence of a d⁸ metal ion (Ni²⁺), which favors a square planar arrangement, particularly when paired with the strong field CN⁻ ligands that induce low-spin behavior.

7. Williamson’s synthesis of preparing dimethyl ether is a/an:

(A) electrophilic substitution
(B) SN1 reaction
(C) electrophilic addition
(D) SN2 reaction

Answer: (D) SN2 reaction

Explanation: Williamson’s synthesis involves the nucleophilic substitution (SN2 mechanism) where an alkoxide ion (RO⁻) attacks the electrophilic carbon in an alkyl halide (R-X), displacing the halide ion and forming an ether (R-O-R’).

8. Out of Fe²⁺, Co²⁺, Cr³⁺, Ni²⁺, the one which shows highest magnetic moment is:

(A) Fe²⁺
(B) Co²⁺
(C) Cr³⁺
(D) Ni²⁺
[Atomic number: Cr = 24, Fe = 26, Co = 27, Ni = 28]

Answer: (A) Fe²⁺

Explanation: The magnetic moment of a metal ion is related to the number of unpaired electrons. Fe²⁺ (with 4 unpaired electrons in its d-orbitals) will have the highest magnetic moment compared to the other ions.

For Fe²⁺:

Electronic configuration: [Ar] 3d⁶
Unpaired electrons = 4

Thus, Fe²⁺ will exhibit the highest magnetic moment.

9. The IUPAC name of the complex [Co(NH₃)₅(NO₂)]Cl₂ is:

(A) Pentaamminenitrito-O-cobalt(III) chloride
(B) Pentaamminenitrito-N-cobalt(III) chloride
(C) Pentaamminenitro-cobalt(III) chloride
(D) Pentaaminenitrito-cobalt(II) chloride

Answer: (B) Pentaamminenitrito-N-cobalt(III) chloride

Explanation: The complex ion is coordinated with five ammine (NH₃) ligands and one nitrite (NO₂⁻) ligand. The NO₂⁻ ligand is attached through the nitrogen atom, so it is named nitrito-N. The central metal is cobalt in the +3 oxidation state, so the complex is cobalt(III). The correct IUPAC name is Pentaamminenitrito-N-cobalt(III) chloride.

10. Which of the following does not undergo Aldol condensation?

(A) CH₃CHO
(B) CH₃COCH₃
(C) CH₃CH₂CHO
(D) C₆H₅CHO

Answer: (D) C₆H₅CHO

Explanation: Aldol condensation typically involves aldehydes or ketones that have at least one α-hydrogen (hydrogen atom attached to the carbon next to the carbonyl group). In the case of C₆H₅CHO (benzaldehyde), there is no α-hydrogen because the phenyl group (C₆H₅) is attached to the carbonyl carbon, making it incapable of undergoing the Aldol condensation reaction.

11. For a chemical reaction, A → B, it was observed that the rate of reaction doubles when the concentration of A is increased four times. The order of the reaction is:

(A) 2
(B) 1
(C) 1/2
(D) Zero

Answer: (A) 2

Explanation: The rate law for the reaction can be written as:Rate=k[A]n\text{Rate} = k[A]^nRate=k[A]n

Where k is the rate constant, [A] is the concentration of A, and n is the order of the reaction.

When the concentration of A is increased by a factor of 4, the rate doubles:Rate2Rate1=([A]2[A]1)n\frac{\text{Rate}_2}{\text{Rate}_1} = \left( \frac{[A]_2}{[A]_1} \right)^nRate1Rate2=([A]1[A]2)n

Given that the rate doubles (Rate2/Rate1=2\text{Rate}_2/\text{Rate}_1 = 2Rate2/Rate1=2) and the concentration of A increases by a factor of 4 ([A]2/[A]1=4[A]_2/[A]_1 = 4[A]2/[A]1=4):2=4n2 = 4^n2=4n

Taking the logarithm on both sides:log⁡2=nlog⁡4\log 2 = n \log 4log2=nlog4n=log⁡2log⁡4=12n = \frac{\log 2}{\log 4} = \frac{1}{2}n=log4log2=21

Thus, the order of the reaction is 2.

12. The correct IUPAC name of (CH₃)₃C–CH₂Br is:

(A) 2,2-Dimethyl-2-bromopropane
(B) 1-Bromo-2,2,2-trimethylethane
(C) 2-Bromo-1,1,1-trimethylethane
(D) 1-Bromo-2,2-dimethylpropane

Answer: (D) 1-Bromo-2,2-dimethylpropane

Explanation: The structure of the compound shows a bromine (Br) attached to the first carbon of the propane chain, with two methyl groups attached to the second carbon. The IUPAC name is 1-Bromo-2,2-dimethylpropane, where “1-Bromo” indicates the position of the bromine atom, and “2,2-dimethyl” refers to the two methyl groups attached to the second carbon.

For Questions number 13 to 16: Assertion and Reasoning

13. Assertion (A): Rate constant increases with increase in temperature.

Reason (R): Increasing the temperature of the substance increases the fraction of molecules, which collide with energies greater than activation energy.

(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).

Explanation: According to the Arrhenius equation, the rate constant k increases with temperature. The fraction of molecules that have energy greater than the activation energy increases as the temperature rises, which leads to more effective collisions and a higher rate constant. Hence, both the assertion and the reason are true, and the reason correctly explains the assertion.

14. Assertion (A): Cu²⁺ iodide is not known.

Reason (R): Cu²⁺ has a strong tendency to oxidize I⁻ to iodine.

(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).

Explanation: Cu²⁺ has a strong tendency to oxidize iodide ions (I⁻) to iodine (I₂), so Cu²⁺ cannot form a stable complex with iodide ions. Therefore, Cu²⁺ iodide (CuI₂) is not known. The reason given explains why the complex is not stable.

Let me know if you’d like further clarification or more questions!

You said:

give same 15. Assertion (A) : Maltose is a reducing sugar. Reason (R) : Maltose is composed of two glucose units in which C-1 of one glucose unit is linked to C-4 of another glucose unit. 16. Assertion (A) : Acetanilide is less basic than aniline. Reason (R) : Acetylation of aniline results in decrease of electron density on nitrogen.

ChatGPT said:

ChatGPT

For Questions 15 and 16: Assertion and Reasoning

15. Assertion (A): Maltose is a reducing sugar.

Reason (R): Maltose is composed of two glucose units in which C-1 of one glucose unit is linked to C-4 of another glucose unit.

(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).

Explanation: Maltose is a reducing sugar because it has a free aldehyde group (-CHO) in one of its glucose units, which is capable of reducing other substances. In maltose, two glucose molecules are linked by an α(1→4) glycosidic bond, and the second glucose unit has a free reducing group at the C-1 carbon, making maltose a reducing sugar. The reason given is also correct because the glycosidic linkage at C-1 of one glucose and C-4 of the other does not involve the aldehyde group, leaving it free to reduce other substances.

16. Assertion (A): Acetanilide is less basic than aniline.

Reason (R): Acetylation of aniline results in a decrease of electron density on nitrogen.

(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).

Explanation: Acetanilide is less basic than aniline because the acetyl group (-COCH₃) attached to the nitrogen in acetanilide withdraws electron density from the nitrogen atom, making the nitrogen less basic. In aniline, the amino group (-NH₂) donates electron density to the benzene ring, making the nitrogen more available for protonation. Therefore, acetanilide is less basic than aniline, and the reason correctly explains this by describing the electron-withdrawing effect of the acetyl group.

SECTION B (Very Short Answer Type Questions)

17. Write the reactions of glucose with:

(a) HCN

Answer: Glucose (C₆H₁₂O₆) reacts with hydrogen cyanide (HCN) to form a cyanohydrin. The reaction is as follows:C₆H₁₂O₆+HCN→C₆H₁₁(OH)(CN)\text{C₆H₁₂O₆} + \text{HCN} \rightarrow \text{C₆H₁₁(OH)(CN)}C₆H₁₂O₆+HCN→C₆H₁₁(OH)(CN)

This reaction forms a glucose cyanohydrin by adding a cyanide group (-CN) to the carbonyl group (aldehyde) of glucose.

(b) Conc. HNO₃

Answer: Glucose reacts with concentrated nitric acid (HNO₃) to undergo oxidation. The reaction results in the formation of glucaric acid (a dicarboxylic acid):C₆H₁₂O₆+2HNO₃→C₆H₈O₆(glucaric acid)\text{C₆H₁₂O₆} + 2 \text{HNO₃} \rightarrow \text{C₆H₈O₆} (\text{glucaric acid})C₆H₁₂O₆+2HNO₃→C₆H₈O₆(glucaric acid)

In this process, the aldehyde group and the alcohol groups of glucose are oxidized to carboxylic acid groups.

18. (a) Account for the following:

(i) CH₃CHO is more reactive than CH₃COCH₃ towards reaction with HCN.

Answer: CH₃CHO (acetaldehyde) is more reactive towards HCN than CH₃COCH₃ (acetone) because the aldehyde group (in CH₃CHO) is more electrophilic than the ketone group (in CH₃COCH₃). The carbonyl carbon in an aldehyde is less sterically hindered, making it more susceptible to nucleophilic attack by HCN. In contrast, the ketone group in acetone has two alkyl groups (CH₃) which donate electron density through inductive effects, reducing the electrophilicity of the carbonyl carbon and making it less reactive toward nucleophiles like HCN.

(ii) Carboxylic acids are higher boiling liquids than aldehydes and ketones.

Answer: Carboxylic acids have strong hydrogen bonding due to the -OH group, which leads to higher boiling points compared to aldehydes and ketones. In aldehydes and ketones, only dipole-dipole interactions exist, which are weaker than the hydrogen bonding in carboxylic acids. This makes carboxylic acids have higher boiling points than aldehydes and ketones.

(b) Give chemical tests to distinguish between the following pair of compounds:

(i) Propanal and Propanone

Answer:

Test:Tollens’ test
- Propanal (an aldehyde) will reduce Tollens’ reagent (AgNO₃ in ammonia) to give a silver mirror on the inner surface of the test tube.
- Propanone (a ketone) will not reduce Tollens’ reagent and will not give a silver mirror.
Test:Iodoform test
- Propanal does not react with iodine in the presence of sodium hydroxide to give the yellow precipitate of iodoform (CHI₃).
- Propanone will give a yellow precipitate of iodoform when treated with iodine and sodium hydroxide.

(ii) Benzaldehyde and Benzoic acid

Answer:

Test:Tollens’ test
- Benzaldehyde (an aldehyde) will reduce Tollens’ reagent to form a silver mirror.
- Benzoic acid (a carboxylic acid) will not reduce Tollens’ reagent.
Test:Sodium bicarbonate test
- Benzoic acid will effervesce when treated with sodium bicarbonate (NaHCO₃), forming CO₂ gas.
- Benzaldehyde will not react with sodium bicarbonate.

19. Write the reactions involved in the following:

(a) Reimer-Tiemann reaction

Answer: The Reimer-Tiemann reaction is used to synthesize salicylic acid (2-hydroxybenzoic acid) from phenol. The reaction is as follows:C₆H₆OH (phenol)+CHCl₃ (chloroform)+NaOH→C₆H₄(OH)COOH (salicylic acid)+CH₄\text{C₆H₆OH (phenol)} + \text{CHCl₃ (chloroform)} + \text{NaOH} \rightarrow \text{C₆H₄(OH)COOH (salicylic acid)} + \text{CH₄}C₆H₆OH (phenol)+CHCl₃ (chloroform)+NaOH→C₆H₄(OH)COOH (salicylic acid)+CH₄

In this reaction, phenol reacts with chloroform (CHCl₃) in the presence of sodium hydroxide (NaOH) to form salicylic acid (2-hydroxybenzoic acid) and methane (CH₄).

(b) Kolbe’s reaction

Answer: The Kolbe’s reaction is a method to form carboxylate salts by the electrolysis of sodium phenoxide. The reaction is as follows:C₆H₅O⁻ (phenoxide ion)+Na⁺→C₆H₄COO⁻ (salicylic acid)+CO₂\text{C₆H₅O⁻ (phenoxide ion)} + \text{Na⁺} \rightarrow \text{C₆H₄COO⁻ (salicylic acid)} + \text{CO₂}C₆H₅O⁻ (phenoxide ion)+Na⁺→C₆H₄COO⁻ (salicylic acid)+CO₂

In this process, phenoxide ions are subjected to electrolysis, and carbon dioxide (CO₂) is evolved at the anode, leading to the formation of sodium salicylate (sodium salt of salicylic acid).

20. The thermal decomposition of an acid is a first-order reaction with a rate constant of 2.3 × 10⁻³ s⁻¹ at a certain temperature. Calculate how long it will take for three-fourths of the initial quantity of acid to decompose.

Answer:

For a first-order reaction, the rate law is given by the equation:ln⁡([A]0[A])=kt\ln \left( \frac{[A]_0}{[A]} \right) = ktln([A][A]0)=kt

Where:

[A]0[A]_0[A]0 is the initial concentration of the acid,
[A][A][A] is the concentration of the acid at time ttt,
kkk is the rate constant, and
ttt is the time taken.

We are asked to find the time when three-fourths of the initial quantity of the acid decomposes. This means that one-fourth of the initial amount remains unreacted, so:[A][A]0=14\frac{[A]}{[A]_0} = \frac{1}{4}[A]0[A]=41

Substitute this into the first-order rate equation:ln⁡([A]0[A])=ln⁡([A]0[A]04)=ln⁡(4)\ln \left( \frac{[A]_0}{[A]} \right) = \ln \left( \frac{[A]_0}{\frac{[A]_0}{4}} \right) = \ln(4)ln([A][A]0)=ln(4[A]0[A]0)=ln(4)

We know that ln⁡(4)\ln(4)ln(4) can be written as:ln⁡(4)=log⁡(4)×2.303=0.6021×2.303=1.388\ln(4) = \log(4) \times 2.303 = 0.6021 \times 2.303 = 1.388ln(4)=log(4)×2.303=0.6021×2.303=1.388

Now, substitute the values into the rate equation:1.388=(2.3×10−3)×t1.388 = (2.3 \times 10^{-3}) \times t1.388=(2.3×10−3)×t

Solve for ttt:t=1.3882.3×10−3=1.3880.0023≈603.48 secondst = \frac{1.388}{2.3 \times 10^{-3}} = \frac{1.388}{0.0023} \approx 603.48 \, \text{seconds}t=2.3×10−31.388=0.00231.388≈603.48seconds

Thus, it will take approximately 603.48 seconds (or about 10.06 minutes) for three-fourths of the initial quantity of acid to decompose.

SECTION C (Short Answer Type Questions)

22. What happens when: (any three)

(a) CH₃MgBr is treated with cyclohexanone followed by hydrolysis

Answer: When methylmagnesium bromide (CH₃MgBr) is treated with cyclohexanone followed by hydrolysis, a Grignard reaction occurs. The Grignard reagent (CH₃MgBr) acts as a nucleophile and attacks the carbonyl carbon of the cyclohexanone. This leads to the formation of a tert-butyl alcohol after hydrolysis. The reaction proceeds as follows:

CH₃MgBr attacks the carbonyl group of cyclohexanone.
After hydrolysis, the final product is 2-methylcyclohexanol.

(b) Phenol is treated with Br₂ in the presence of CS₂

Answer: When phenol is treated with bromine (Br₂) in the presence of carbon disulfide (CS₂), the reaction leads to the formation of 2,4,6-tribromophenol. This is an example of electrophilic aromatic substitution, where bromine substitutes hydrogen atoms on the aromatic ring. In this case, bromine reacts at the positions ortho and para to the hydroxyl group (-OH) on the benzene ring, forming 2,4,6-tribromophenol.

The reaction proceeds as:

Phenol + Br₂ (CS₂) → 2,4,6-tribromophenol

(c) Benzene diazonium chloride is treated with H₂O

Answer: When benzene diazonium chloride (C₆H₅N₂Cl) is treated with water (H₂O), the diazonium group is replaced by a hydroxyl group (-OH), leading to the formation of phenol (C₆H₅OH). This is an example of a diazonium coupling reaction, where water acts as a nucleophile and replaces the diazonium group.

The reaction proceeds as:

C₆H₅N₂Cl + H₂O → C₆H₅OH (phenol) + N₂ (gas)

(d) Anisole is treated with HI

Answer: When anisole (C₆H₅OCH₃) is treated with hydroiodic acid (HI), the ether bond between the phenyl group (C₆H₅) and the methoxy group (OCH₃) is cleaved, and methyl iodide (CH₃I) is produced as one product, while phenol (C₆H₅OH) is the other. This is an example of ether cleavage using a strong acid.

The reaction proceeds as:

C₆H₅OCH₃ + HI → C₆H₅OH (phenol) + CH₃I (methyl iodide)

23. Hydrolysis of sucrose takes place by the chemical reaction:

C₁₂H₂₂O₁₁+H₂O→HC₆H₁₂O₆+C₆H₁₂O₆\text{C₁₂H₂₂O₁₁} + \text{H₂O} \xrightarrow{\text{H}} \text{C₆H₁₂O₆} + \text{C₆H₁₂O₆}C₁₂H₂₂O₁₁+H₂OHC₆H₁₂O₆+C₆H₁₂O₆

Based on the above reaction, write:

(a) Rate law equation

Answer: The rate law for the hydrolysis of sucrose can be written as:Rate=k[C₁₂H₂₂O₁₁]\text{Rate} = k[\text{C₁₂H₂₂O₁₁}]Rate=k[C₁₂H₂₂O₁₁]

Where:

k is the rate constant,
[C₁₂H₂₂O₁₁] is the concentration of sucrose.

Since the reaction is first-order with respect to sucrose, the rate law depends only on the concentration of sucrose.

(b) Molecularity and order of the reaction

Answer:

Order of the reaction: The order of the reaction is 1 (first-order) because the rate depends only on the concentration of sucrose.
Molecularity: The reaction involves the interaction of one molecule of sucrose with water molecules in the presence of an acid catalyst, making the molecularity of the reaction 1 (unimolecular).

(c) What do you call such reactions?

Answer: The reaction is called a hydrolysis reaction. Specifically, since it involves the breaking down of sucrose into two monosaccharides (glucose and fructose) by the addition of water, it is referred to as acid-catalyzed hydrolysis of sucrose.

24. Account for the following:

(a) n-butyl bromide has a higher boiling point than t-butyl bromide.

Answer: The higher boiling point of n-butyl bromide compared to t-butyl bromide is due to the shape and intermolecular forces. n-Butyl bromide is a straight-chain compound and has a larger surface area, leading to stronger van der Waals forces (induced dipole-induced dipole) between molecules. On the other hand, t-butyl bromide is a highly branched compound, which reduces its surface area and results in weaker intermolecular forces, thus a lower boiling point.

(b) Alkyl halides are insoluble in water.

Answer: Alkyl halides are generally insoluble in water because they cannot form hydrogen bonds with water molecules. While water is highly polar, alkyl halides are non-polar (or only slightly polar), and thus do not interact strongly with water molecules. The hydrogen bonding capability of water molecules is not sufficiently utilized in alkyl halides, leading to their poor solubility in water.

(c) Cyclohexyl chloride possesses a higher dipole moment than chlorobenzene.

Answer: Cyclohexyl chloride has a higher dipole moment than chlorobenzene because in cyclohexyl chloride, the C-Cl bond is more polar, and there is no resonance effect that would delocalize the electron density from the chlorine atom. In contrast, in chlorobenzene, the lone pair on chlorine interacts with the benzene ring through resonance, partially reducing the polarity of the C-Cl bond. This makes chlorobenzene’s dipole moment lower than cyclohexyl chloride’s.

25. Give plausible explanation for the following:

(a) Diazonium salts of aromatic amines are stable.

Answer: Diazonium salts of aromatic amines are stable due to resonance stabilization. The aromatic ring can delocalize the positive charge on the nitrogen atom through resonance, which stabilizes the diazonium ion. This delocalization is enhanced when the amine group (–NH₂) is in the para or ortho positions relative to the diazonium group, which makes these salts more stable.

(b) Aniline does not undergo Friedel-Crafts reaction.

Answer: Aniline does not undergo the Friedel-Crafts alkylation reaction because the amine group (–NH₂) is an electron-donating group that increases the electron density on the aromatic ring, making it more reactive towards electrophiles. However, the Lewis acid catalyst (such as AlCl₃) required for the Friedel-Crafts reaction coordinates with the nitrogen lone pair in aniline, which deactivates the aromatic ring towards electrophilic attack, preventing the reaction.

(c) Aniline on nitration gives a substantial amount of meta product.

Answer: Aniline undergoes nitration to give a substantial amount of meta product because the amine group (–NH₂), although an electron-donating group, makes the ring more reactive, but it also deactivates the meta positions. This deactivation of the meta position is due to steric and electronic effects in which the lone pair on nitrogen interacts with the nitronium ion (NO₂⁺) and directs the substitution predominantly to the meta position, rather than the ortho or para positions.

26. Calculate elevation of the boiling point of solution when 2 g of MgSO₄ (molar mass = 120 g/mol) was dissolved in 100 g of water, assuming MgSO₄ undergoes complete dissociation. [Kb for water = 0.52 K·kg·mol⁻¹]

Answer:

To calculate the elevation of the boiling point, we use the formula:ΔTb=Kb×m×i\Delta T_b = K_b \times m \times iΔTb=Kb×m×i

Where:

ΔTb\Delta T_bΔTb is the elevation in boiling point,
KbK_bKb is the ebullioscopic constant of water (0.52 K⋅kg/mol0.52 \, \text{K} \cdot \text{kg/mol}0.52K⋅kg/mol),
mmm is the molality of the solution, and
iii is the van’t Hoff factor, which is the number of ions produced upon dissociation.

Step 1: Calculate molality (m)
Molality is calculated as:m=moles of solutemass of solvent in kgm = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}m=mass of solvent in kgmoles of solute

Moles of MgSO₄ = 2 g120 g/mol=0.01667 mol\frac{2 \, \text{g}}{120 \, \text{g/mol}} = 0.01667 \, \text{mol}120g/mol2g=0.01667mol

Mass of solvent (water) = 100 g = 0.1 kgm=0.016670.1=0.1667 mol/kgm = \frac{0.01667}{0.1} = 0.1667 \, \text{mol/kg}m=0.10.01667=0.1667mol/kg

Step 2: Determine the van’t Hoff factor (i)
Since MgSO₄ dissociates into two ions (Mg²⁺ and SO₄²⁻), the van’t Hoff factor i=2i = 2i=2.

Step 3: Calculate the boiling point elevationΔTb=0.52 K\cdotpkg/mol×0.1667 mol/kg×2=0.1733 K\Delta T_b = 0.52 \, \text{K·kg/mol} \times 0.1667 \, \text{mol/kg} \times 2 = 0.1733 \, \text{K}ΔTb=0.52K\cdotpkg/mol×0.1667mol/kg×2=0.1733K

Thus, the elevation of the boiling point is approximately 0.17 K.

27. The electrical resistance of a column of 0.05 M KOH solution of cell constant 60 cm⁻¹ is 5.4 × 10³ ohm. Calculate its resistivity, conductivity, and molar conductivity.

Answer:

We can use the following formulas:

Resistivity (ρ):

R=ρ×lAR = \rho \times \frac{l}{A}R=ρ×Al

Where RRR is the resistance, lll is the length of the column, and AAA is the cross-sectional area.

Conductivity (κ):

κ=1ρ\kappa = \frac{1}{\rho}κ=ρ1

Molar conductivity (Λm):

Λm=κ×c\Lambda_m = \kappa \times cΛm=κ×c

Where ccc is the concentration of the solution.

Given:

R=5.4×103 ΩR = 5.4 \times 10^3 \, \OmegaR=5.4×103Ω,
Cell constant = 60 cm⁻¹,
Concentration = 0.05 M.

From the formula for cell constant, κ\kappaκ can be calculated as:κ=Cell constantR=605.4×103=1.11×10−2 S/cm\kappa = \frac{\text{Cell constant}}{R} = \frac{60}{5.4 \times 10^3} = 1.11 \times 10^{-2} \, \text{S/cm}κ=RCell constant=5.4×10360=1.11×10−2S/cm

Now, calculate molar conductivity:Λm=κ×c=1.11×10−2 S/cm×0.05 mol/L=5.55×10−4 S\cdotpcm²/mol\Lambda_m = \kappa \times c = 1.11 \times 10^{-2} \, \text{S/cm} \times 0.05 \, \text{mol/L} = 5.55 \times 10^{-4} \, \text{S·cm²/mol}Λm=κ×c=1.11×10−2S/cm×0.05mol/L=5.55×10−4S\cdotpcm²/mol

So, molar conductivity is 5.55×10−4 S\cdotpcm²/mol5.55 \times 10^{-4} \, \text{S·cm²/mol}5.55×10−4S\cdotpcm²/mol.

28. (a) Draw the geometrical isomers of the complex [Mn(Br)₂(Cl)₂]²⁻.

Answer: The complex [Mn(Br)₂(Cl)₂]²⁻ is an octahedral complex with four ligands. The two bromine (Br⁻) and two chlorine (Cl⁻) ligands can be arranged in two ways:

Cis isomer: Both Br⁻ ligands are adjacent to each other, and both Cl⁻ ligands are adjacent to each other.
Trans isomer: The Br⁻ ligands are opposite to each other, and the Cl⁻ ligands are opposite to each other.

(b) Give the electronic configuration of d⁴ ion when Δo < P.

Answer: When the splitting energy of the d-orbital (Δo\Delta_oΔo) is smaller than the pairing energy (PPP), the complex exhibits high-spin configuration. The electronic configuration for a d⁴ ion in an octahedral field is:

High-spin configuration: t2g2 eg2t_{2g}^2 \, e_g^2t2g2eg2

(c) Solution of [CoF₆]³⁻ is coloured whereas [Ni(CN)₄]²⁻ is colourless. Explain.

Answer: The color of a coordination compound depends on the d-d transitions between split d-orbitals. In [CoF₆]³⁻, Co³⁺ has a partially filled d-orbital, allowing for d-d transitio

SECTION D (Case-Based Questions)

29. The cause for deviation from Raoult’s law in the colligative properties of non-ideal solutions lie in the nature of interactions at the molecular level. These properties show deviations from Raoult’s law due to differences in interactions between solute-solvent, solute-solute, and solvent-solvent. Some liquids on mixing form azeotropes, which are binary mixtures having the same composition in the liquid and vapor phase and boil at a constant temperature. In such cases, it is not possible to separate the components by fractional distillation. There are two types of azeotropes: minimum boiling azeotrope and maximum boiling azeotrope.

Based on the above passage, answer the following questions:

(a) Pure ethanol cannot be prepared by fractional distillation of an ethanol-water mixture. Comment.

Answer: Pure ethanol cannot be prepared by fractional distillation of an ethanol-water mixture because ethanol and water form an azeotropic mixture at a certain composition (approximately 95.6% ethanol and 4.4% water by volume). This azeotrope has the same composition in both the liquid and vapor phases and boils at a constant temperature. As a result, fractional distillation will not be effective in separating the components beyond this azeotropic composition, and it will not be possible to obtain pure ethanol from the mixture.

(b) Why does a mixture of chloroform and acetone show deviation from ideal behavior?

Answer: A mixture of chloroform and acetone shows a deviation from ideal behavior because the intermolecular interactions between chloroform and acetone are different from the interactions between their own molecules. Specifically, there is hydrogen bonding between chloroform (CHCl₃) and acetone (CH₃COCH₃), which leads to stronger solute-solvent interactions compared to solute-solute and solvent-solvent interactions. This results in either a positive or negative deviation from Raoult’s law depending on the strength of these interactions, and the mixture may show behavior that is not predicted by the ideal solution model.

(c) (i) The vapour pressure of pure benzene at a certain temperature is 1.25 atm. When 1.2 g of non-volatile, non-electrolyte solute is added to 60 g of benzene (M = 78 g/mol), the vapour pressure of the solution becomes 1.237 atm. Calculate the molar mass of the non-volatile solute.

Answer:

To calculate the molar mass of the non-volatile solute, we will use Raoult’s law and the relation for relative lowering of vapour pressure.

The relative lowering of vapour pressure is given by:ΔPPinitial=nsolutensolvent\frac{\Delta P}{P_{\text{initial}}} = \frac{n_{\text{solute}}}{n_{\text{solvent}}}PinitialΔP=nsolventnsolute

Where:

Pinitial=1.25 atmP_{\text{initial}} = 1.25 \, \text{atm}Pinitial=1.25atm is the vapour pressure of pure benzene.
ΔP=Pinitial−Pfinal=1.25−1.237=0.013 atm\Delta P = P_{\text{initial}} – P_{\text{final}} = 1.25 – 1.237 = 0.013 \, \text{atm}ΔP=Pinitial−Pfinal=1.25−1.237=0.013atm is the change in vapour pressure.
nsoluten_{\text{solute}}nsolute is the number of moles of solute.
nsolventn_{\text{solvent}}nsolvent is the number of moles of solvent (benzene).

Step 1: Calculate the number of moles of benzene (solvent).nsolvent=mass of benzenemolar mass of benzene=60 g78 g/mol=0.769 moln_{\text{solvent}} = \frac{\text{mass of benzene}}{\text{molar mass of benzene}} = \frac{60 \, \text{g}}{78 \, \text{g/mol}} = 0.769 \, \text{mol}nsolvent=molar mass of benzenemass of benzene=78g/mol60g=0.769mol

Step 2: Use the relative lowering of vapour pressure equation to find the moles of solute.ΔPPinitial=nsolutensolvent\frac{\Delta P}{P_{\text{initial}}} = \frac{n_{\text{solute}}}{n_{\text{solvent}}}PinitialΔP=nsolventnsolute 0.0131.25=nsolute0.769\frac{0.013}{1.25} = \frac{n_{\text{solute}}}{0.769}1.250.013=0.769nsolute nsolute=0.0131.25×0.769=0.00757 moln_{\text{solute}} = \frac{0.013}{1.25} \times 0.769 = 0.00757 \, \text{mol}nsolute=1.250.013×0.769=0.00757mol

Step 3: Calculate the molar mass of the solute.Molar mass of solute=mass of solutemoles of solute=1.2 g0.00757 mol=158.52 g/mol\text{Molar mass of solute} = \frac{\text{mass of solute}}{\text{moles of solute}} = \frac{1.2 \, \text{g}}{0.00757 \, \text{mol}} = 158.52 \, \text{g/mol}Molar mass of solute=moles of solutemass of solute=0.00757mol1.2g=158.52g/mol

Thus, the molar mass of the non-volatile solute is approximately 158.5 g/mol.

(c) (ii) The boiling point of benzene is 353.23 K. When 1.80 g of a non-volatile solute was dissolved in 90 g of benzene, the boiling point is raised to 354.11 K. Calculate the molar mass of the solute. Kb for benzene is 2.53 K·kg/mol.

Answer:

To calculate the molar mass of the solute, we will use the formula for boiling point elevation:ΔTb=Kb×m\Delta T_b = K_b \times mΔTb=Kb×m

Where:

ΔTb=354.11 K−353.23 K=0.88 K\Delta T_b = 354.11 \, \text{K} – 353.23 \, \text{K} = 0.88 \, \text{K}ΔTb=354.11K−353.23K=0.88K is the elevation in boiling point,
Kb=2.53 K\cdotpkg/molK_b = 2.53 \, \text{K·kg/mol}Kb=2.53K\cdotpkg/mol is the ebullioscopic constant for benzene,
m=molality of the solution=moles of solutemass of solvent in kgm = \text{molality of the solution} = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}m=molality of the solution=mass of solvent in kgmoles of solute.

Step 1: Calculate the molality.m=ΔTbKb=0.88 K2.53 K\cdotpkg/mol=0.348 mol/kgm = \frac{\Delta T_b}{K_b} = \frac{0.88 \, \text{K}}{2.53 \, \text{K·kg/mol}} = 0.348 \, \text{mol/kg}m=KbΔTb=2.53K\cdotpkg/mol0.88K=0.348mol/kg

Step 2: Calculate the number of moles of solute.

Molality is given by:m=moles of solutemass of solvent in kgm = \frac{\text{moles of solute}}{\text{mass of solvent in kg}}m=mass of solvent in kgmoles of solute moles of solute=m×mass of solvent in kg=0.348 mol/kg×0.090 kg=0.03132 mol\text{moles of solute} = m \times \text{mass of solvent in kg} = 0.348 \, \text{mol/kg} \times 0.090 \, \text{kg} = 0.03132 \, \text{mol}moles of solute=m×mass of solvent in kg=0.348mol/kg×0.090kg=0.03132mol

Step 3: Calculate the molar mass of the solute.Molar mass of solute=mass of solutemoles of solute=1.80 g0.03132 mol=57.47 g/mol\text{Molar mass of solute} = \frac{\text{mass of solute}}{\text{moles of solute}} = \frac{1.80 \, \text{g}}{0.03132 \, \text{mol}} = 57.47 \, \text{g/mol}Molar mass of solute=moles of solutemass of solute=0.03132mol1.80g=57.47g/mol

Thus, the molar mass of the solute is approximately 57.5 g/mol.

30. The particles in the nucleus of the cell, responsible for heredity, are called chromosomes, which are made up of proteins and another type of biomolecules called nucleic acids. These are mainly of two types, DNA and RNA. Nucleic acids on hydrolysis yield a pentose sugar, phosphoric acid, and nitrogen-containing heterocyclic compound. Nucleic acids have a very diverse set of functions, such as cell creation, the storage and processing of genetic information, protein synthesis, and the generation of energy cells. Although their functions may differ, the structure of DNA and RNA are very similar, with only a few fundamental differences in their molecular make-up.

Based on the above information, answer the following questions:

(a) Write two functions of DNA.

Answer:

Storage of genetic information: DNA stores the genetic information required for the development, functioning, and reproduction of living organisms.
Protein synthesis: DNA serves as a template for the synthesis of RNA during transcription, which then guides the synthesis of proteins during translation.

(b) What products will be formed when a nucleotide from DNA containing Adenine is hydrolyzed?

Answer: When a nucleotide containing Adenine is hydrolyzed, the products formed will be:

A pentose sugar (such as deoxyribose in DNA),
Phosphoric acid (H₃PO₄),
A nitrogenous base – in this case, Adenine.

(c) (i) What are nucleic acids? What is the difference between a nucleotide and a nucleoside?

Answer:

Nucleic acids are large biomolecules made up of chains of nucleotides, which include DNA (Deoxyribonucleic acid) and RNA (Ribonucleic acid). These molecules are essential for storing, transmitting, and expressing genetic information in cells.
Difference between nucleotide and nucleoside:
- A nucleotide consists of a pentose sugar (ribose or deoxyribose), a phosphate group, and a nitrogenous base (adenine, thymine, cytosine, guanine, or uracil).
- A nucleoside consists only of a pentose sugar and a nitrogenous base, but lacks the phosphate group.

(c) (ii) Give one similarity and one difference between DNA and RNA.

Answer:

Similarity: Both DNA and RNA are made up of nucleotides, which contain a pentose sugar, a phosphate group, and a nitrogenous base. They are essential for storing and transmitting genetic information.
Difference:
- DNA is double-stranded and contains the sugar deoxyribose, whereas RNA is single-stranded and contains the sugar ribose.
- Additionally, DNA has thymine (T) as one of its nitrogenous bases, while RNA has uracil (U) instead.

SECTION E (Long Answer Questions)

31. (a) (i) Explain with the help of chemical reactions when:

Acetone is treated with semicarbazide.

Answer: When acetone is treated with semicarbazide, a semicarbazone is formed. The reaction is:CH₃COCH₃+H₂N-NH-C(O)-NH₂→CH₃C(NH-NH-C(O)-NH₂)CH₃\text{CH₃COCH₃} + \text{H₂N-NH-C(O)-NH₂} \rightarrow \text{CH₃C(NH-NH-C(O)-NH₂)CH₃}CH₃COCH₃+H₂N-NH-C(O)-NH₂→CH₃C(NH-NH-C(O)-NH₂)CH₃

This reaction is used to test for the presence of carbonyl compounds (like aldehydes and ketones) as semicarbazones are typically formed with these compounds.

Two molecules of benzaldehyde are treated with conc. NaOH.

Answer: When benzaldehyde reacts with concentrated NaOH, it undergoes Aldol condensation to form 4-Phenyl-3-buten-2-one (also known as α,β-unsaturated ketone). The reaction is:2 C₆H₅CHO+NaOH (conc.)→C₆H₅CH=CHC₆H₅+H₂O2 \, \text{C₆H₅CHO} + \text{NaOH (conc.)} \rightarrow \text{C₆H₅CH=CHC₆H₅} + \text{H₂O}2C₆H₅CHO+NaOH (conc.)→C₆H₅CH=CHC₆H₅+H₂O

This is an example of aldol condensation, where an enolate ion is formed from the aldehyde and then attacks another molecule of aldehyde, leading to the formation of a β-hydroxy ketone, which then undergoes dehydration.

Butan-2-one is treated with Zn/Hg and conc. HCl.

Answer: When Butan-2-one (a ketone) is treated with Zn/Hg and conc. HCl, it undergoes a Clemmensen reduction, where the carbonyl group is reduced to a methylene group (–CH₂–). The reaction is:CH₃COCH₂CH₃→Zn/Hg, HClCH₃CH₂CH₂CH₃\text{CH₃COCH₂CH₃} \xrightarrow{\text{Zn/Hg, HCl}} \text{CH₃CH₂CH₂CH₃}CH₃COCH₂CH₃Zn/Hg, HClCH₃CH₂CH₂CH₃

This reduction removes the oxygen from the carbonyl group, converting the ketone into an alkane.

(ii) Arrange the following in the increasing order of their acidic strength:

CH₃CH₂CH₂COOH, BrCH₂CH₂CH₂COOH, CH₃CHBrCH₂COOH, CH₃CH₂CHBrCOOH

Answer: The acidic strength increases with the presence of electron-withdrawing groups (like Br), which stabilize the conjugate base (the carboxylate anion). The order of increasing acidity is:

CH₃CH₂CH₂COOH < CH₃CH₂CHBrCOOH < CH₃CHBrCH₂COOH < BrCH₂CH₂CH₂COOH

Benzoic acid, 4-Methoxybenzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid

Answer: The order of increasing acidity is based on the electron-withdrawing or electron-donating effects of the substituents on the benzene ring. Nitro (–NO₂) is an electron-withdrawing group, and methoxy (–OCH₃) is an electron-donating group. The order is:

4-Methoxybenzoic acid < Benzoic acid < 4-Nitrobenzoic acid < 3,4-Dinitrobenzoic acid

31. (b) (i) Identify the products A, B, C, and D in the following sequence of reactions:

CH₃CHO[O]→(O)A→PCl₅B→(CH₃)₂CdD→Zn-Hg, Conc. HClC\text{CH₃CHO} [O] \xrightarrow{\text{(O)}} A \xrightarrow{\text{PCl₅}} B \xrightarrow{\text{(CH₃)₂Cd}} D \xrightarrow{\text{Zn-Hg, Conc. HCl}} CCH₃CHO[O](O)APCl₅B(CH₃)₂CdDZn-Hg, Conc. HClC

Answer:

CH₃CHO [O] → A: Oxidation of acetaldehyde (CH₃CHO) will form acetic acid (CH₃COOH).
- A = CH₃COOH
A (CH₃COOH) + PCl₅ → B: The reaction of acetic acid with phosphorus pentachloride (PCl₅) will result in the formation of acetyl chloride (CH₃COCl).
- B = CH₃COCl
B (CH₃COCl) + (CH₃)₂Cd → D: The reaction of acetyl chloride with dimethylcadmium (CH₃)₂Cd will undergo nucleophilic substitution to give dimethylacetic acid (CH₃COCH₃).
- D = CH₃COCH₃ (acetone)
D (CH₃COCH₃) + Zn-Hg, Conc. HCl → C: The reaction of acetone (CH₃COCH₃) with Zn/Hg and conc. HCl leads to the Clemmensen reduction, which reduces the ketone to an alkane, propane (CH₃CH₂CH₃).
- C = CH₃CH₂CH₃ (propane)

(ii) How will you bring about the following conversions?

Propanone to Propene:

Answer:

Propanone (CH₃COCH₃) can be converted to Propene (CH₃CH=CH₂) by dehydration using an acid catalyst like H₂SO₄ at high temperature:

CH₃COCH₃→H₂SO₄, heatCH₃CH=CH₂+H2O\text{CH₃COCH₃} \xrightarrow{\text{H₂SO₄, heat}} \text{CH₃CH=CH₂} + H₂OCH₃COCH₃H₂SO₄, heatCH₃CH=CH₂+H2O

Benzoic acid to Benzaldehyde:

Answer:

Benzoic acid (C₆H₅COOH) can be reduced to benzaldehyde (C₆H₅CHO) using LiAlH₄ (Lithium aluminium hydride), a selective reducing agent:

C₆H₅COOH→LiAlH₄C₆H₅CHO\text{C₆H₅COOH} \xrightarrow{\text{LiAlH₄}} \text{C₆H₅CHO}C₆H₅COOHLiAlH₄C₆H₅CHO

Ethanal to But-2-enal:

Answer:

Ethanal (CH₃CHO) can be converted to But-2-enal (CH₃CH=CHCHO) via Aldol condensation and subsequent dehydration:

2 CH₃CHO→NaOHCH₃CH=CHCHO+H2O2 \, \text{CH₃CHO} \xrightarrow{\text{NaOH}} \text{CH₃CH=CHCHO} + H₂O2CH₃CHONaOHCH₃CH=CHCHO+H2O

32. Attempt any five of the following:

(a) Cu⁺ is not stable in aqueous solution. Comment.

Answer:

Cu⁺ (Copper(I)) is unstable in aqueous solution because it easily undergoes oxidation to Cu²⁺ (Copper(II)) in the presence of oxygen. The stability of Cu⁺ is low due to its tendency to disproportionate (oxidize and reduce itself):2 Cu+→oxidationCu2++Cu2 \, \text{Cu}^+ \xrightarrow{\text{oxidation}} \text{Cu}^{2+} + \text{Cu}2Cu+oxidationCu2++Cu

(b) Out of Cr²⁺ and Fe²⁺, which one is a stronger reducing agent and why?

Answer:

Fe²⁺ is a stronger reducing agent than Cr²⁺ because it has a more negative reduction potential compared to Cr²⁺. The reduction potential of Fe²⁺/Fe³⁺ is -0.44 V, whereas for Cr²⁺/Cr³⁺ it is -0.91 V. The more negative the reduction potential, the stronger the reducing agent.

(c) Actinoid contraction is greater from element to element than lanthanoid contraction. Why?

Answer:

Actinoid contraction is greater than lanthanoid contraction because the 5f orbitals in actinoids are less effective in shielding the nuclear charge compared to the 4f orbitals in lanthanoids. This leads to a greater increase in effective nuclear charge across the actinoid series, resulting in a more significant contraction of the atomic radius.

(d) KMnO₄ acts as an oxidizing agent in acidic medium. Write the ionic equation to support this.

Answer:

In acidic medium, KMnO₄ (potassium permanganate) is reduced to Mn²⁺ and oxidizes other substances. The ionic equation for this reaction is:

MnO₄−+8H++5e−→Mn2++4H2O\text{MnO₄}^- + 8H⁺ + 5e⁻ \rightarrow \text{Mn}^{2+} + 4H₂OMnO₄−+8H++5e−→Mn2++4H2O

This shows KMnO₄ acting as an oxidizing agent, accepting electrons and being reduced.

(e) Name the metal in the first transition series which exhibits +1 oxidation state most frequently.

Answer:

Copper (Cu) is the metal in the first transition series that exhibits the +1 oxidation state most frequently. In compounds like Cu₂O, CuCl, and Cu₂SO₄, copper shows the +1 oxidation state.

(f) Transition metals and their compounds are good catalysts. Justify.

Answer:

Transition metals and their compounds are good catalysts because they have the ability to adopt multiple oxidation states, allowing them to facilitate the transfer of electrons during reactions. They also have empty or partially filled d-orbitals, which can form intermediate complexes with reactants, lowering the activation energy of the reaction.

(g) Scandium forms no coloured ions, yet it is regarded as a transition element. Why?

Answer:

Scandium (Sc) forms no coloured ions because its d-orbitals are empty in the Sc³⁺ state. However, it is still regarded as a transition element because it has an incomplete d-subshell in its Sc²⁺ state and can exhibit variable oxidation states.

33. (a) (i) What type of battery is the lead storage battery? Write the anode and the cathode reactions and the overall reaction occurring in a lead storage battery when current is drawn from it.

Answer:

The lead storage battery is a secondary (rechargeable) battery. During discharge, the anode and cathode reactions are as follows:

Anode Reaction:Pb (s)+SO₄²⁻→PbSO₄ (s)+2e−\text{Pb (s)} + \text{SO₄²⁻} \rightarrow \text{PbSO₄ (s)} + 2e⁻Pb (s)+SO₄²⁻→PbSO₄ (s)+2e−

Cathode Reaction:PbO₂ (s)+4H++SO42−+2e−→PbSO₄ (s)+2H2O\text{PbO₂ (s)} + 4H⁺ + SO₄²⁻ + 2e⁻ \rightarrow \text{PbSO₄ (s)} + 2H₂OPbO₂ (s)+4H++SO42−+2e−→PbSO₄ (s)+2H2O

Overall Reaction:Pb (s)+PbO₂ (s)+4H++2SO42−→2PbSO₄ (s)+2H2O\text{Pb (s)} + \text{PbO₂ (s)} + 4H⁺ + 2SO₄²⁻ \rightarrow 2\text{PbSO₄ (s)} + 2H₂OPb (s)+PbO₂ (s)+4H++2SO42−→2PbSO₄ (s)+2H2O

(ii) Calculate the time to deposit 1.5 g of silver at the cathode when a current of 1.5 A is passed through a solution of AgNO₃.
[Molar mass of Ag = 108 g mol⁻¹, 1 F = 96500 C mol⁻¹]

Answer: To calculate the time, we can use Faraday’s law of electrolysis:Mass=M⋅I⋅tn⋅F\text{Mass} = \frac{M \cdot I \cdot t}{n \cdot F}Mass=n⋅FM⋅I⋅t

Where:

M = molar mass of silver = 108 g/mol,
I = current = 1.5 A,
t = time in seconds,
n = number of electrons involved in the deposition of 1 mole of Ag = 1 (since Ag⁺ gains 1 electron to form Ag),
F = Faraday’s constant = 96500 C/mol.

We are given that the mass of silver deposited is 1.5 g, so:111

(b) (i) State Kohlrausch’s law of independent migration of ions. Molar conductivity at infinite dilution for NH₄Cl, NaOH, and NaCl solution at 298 K are 110, 100, and 105 S cm² mol⁻¹, respectively. Calculate the molar conductivity of NH₄OH solution.

Kohlrausch’s Law of Independent Migration of Ions:

Kohlrausch’s law states that at infinite dilution, the molar conductivity of an electrolyte can be expressed as the sum of the molar conductivities of the individual ions, each moving independently:λm∞=λ++λ−\lambda_m^\infty = \lambda^+ + \lambda^-λm∞=λ++λ−

Where:

λm∞\lambda_m^\inftyλm∞ is the molar conductivity at infinite dilution.
λ+\lambda^+λ+ is the molar conductivity of the cation.
λ−\lambda^-λ− is the molar conductivity of the anion.

To calculate the molar conductivity of NH₄OH solution, we can use Kohlrausch’s law, considering the individual ion contributions from NH₄Cl and NaOH, and then solving for NH₄OH.

Molar conductivity at infinite dilution for NH₄Cl is given as 110 S cm² mol⁻¹, which includes contributions from NH₄⁺ and Cl⁻.
Molar conductivity at infinite dilution for NaOH is given as 100 S cm² mol⁻¹, which includes contributions from Na⁺ and OH⁻.
Molar conductivity at infinite dilution for NaCl is given as 105 S cm² mol⁻¹, which includes contributions from Na⁺ and Cl⁻.

We can write the equation for the molar conductivity of NH₄OH solution at infinite dilution:λm∞(NH₄OH)=λNH₄⁺++λOH⁻−\lambda_m^\infty (\text{NH₄OH}) = \lambda^+_{\text{NH₄⁺}} + \lambda^-_{\text{OH⁻}}λm∞(NH₄OH)=λNH₄⁺++λOH⁻−

To find λm∞(NH₄OH)\lambda_m^\infty (\text{NH₄OH})λm∞(NH₄OH), we subtract the contribution of Na⁺ and Cl⁻ from NH₄Cl and NaOH:λNH₄⁺+=λm∞(NH₄Cl)−λm∞(NaCl)=110−105=5 S cm2mol−1\lambda^+_{\text{NH₄⁺}} = \lambda_m^\infty (\text{NH₄Cl}) – \lambda_m^\infty (\text{NaCl}) = 110 – 105 = 5 \, \text{S cm}^2 \text{mol}^{-1}λNH₄⁺+=λm∞(NH₄Cl)−λm∞(NaCl)=110−105=5S cm2mol−1 λOH⁻−=λm∞(NaOH)−λm∞(NaCl)=100−105=−5 S cm2mol−1\lambda^-_{\text{OH⁻}} = \lambda_m^\infty (\text{NaOH}) – \lambda_m^\infty (\text{NaCl}) = 100 – 105 = -5 \, \text{S cm}^2 \text{mol}^{-1}λOH⁻−=λm∞(NaOH)−λm∞(NaCl)=100−105=−5S cm2mol−1

Now, we can calculate the molar conductivity of NH₄OH:λm∞(NH₄OH)=5+(−5)=0 S cm2mol−1\lambda_m^\infty (\text{NH₄OH}) = 5 + (-5) = 0 \, \text{S cm}^2 \text{mol}^{-1}λm∞(NH₄OH)=5+(−5)=0S cm2mol−1

Thus, the molar conductivity of NH₄OH is 0 S cm² mol⁻¹.

(ii) Calculate ΔG° of the following cell at 25°C:

Zn (s)∣Zn2+(aq) ∣∣ Cu2+(aq)∣Cu (s)\text{Zn (s)} | \text{Zn}^{2+} (\text{aq}) \, || \, \text{Cu}^{2+} (\text{aq}) | \text{Cu (s)}Zn (s)∣Zn2+(aq)∣∣Cu2+(aq)∣Cu (s)

Given:

E°(Zn²⁺/Zn) = -0.76 V
E°(Cu²⁺/Cu) = +0.34 V
1 F = 96500 C mol⁻¹

The cell potential Ecell∘E^\circ_{\text{cell}}Ecell∘ is given by:Ecell∘=Ecathode∘−Eanode∘E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} – E^\circ_{\text{anode}}Ecell∘=Ecathode∘−Eanode∘

Where:

The cathode is the Cu²⁺/Cu half-reaction (reduction),
The anode is the Zn²⁺/Zn half-reaction (oxidation).

Substituting the given values:Ecell∘=+0.34 V−(−0.76 V)=0.34+0.76=1.10 VE^\circ_{\text{cell}} = +0.34 \, \text{V} – (-0.76 \, \text{V}) = 0.34 + 0.76 = 1.10 \, \text{V}Ecell∘=+0.34V−(−0.76V)=0.34+0.76=1.10V

Now, to calculate ΔG° (standard Gibbs free energy change), we use the relation:ΔG∘=−nFEcell∘\Delta G^\circ = -n F E^\circ_{\text{cell}}ΔG∘=−nFEcell∘

Where:

n = number of moles of electrons involved (for this reaction, n = 2 because two electrons are involved in the Zn²⁺ to Zn and Cu²⁺ to Cu half-reactions),
F = Faraday’s constant = 96500 C/mol,
E°_{\text{cell}} = 1.10 V.

Substituting the values:ΔG∘=−2×96500×1.10=−2×106150=−212300 J/mol\Delta G^\circ = -2 \times 96500 \times 1.10 = -2 \times 106150 = -212300 \, \text{J/mol}ΔG∘=−2×96500×1.10=−2×106150=−212300J/mol

Thus, the standard Gibbs free energy change for the reaction is:ΔG∘=−212.3 kJ/mol\Delta G^\circ = -212.3 \, \text{kJ/mol}ΔG∘=−212.3kJ/mol

This is the standard Gibbs free energy change for the reaction.

CBSE_Class_12_Chemistry_Model_Paper (2) (2024)

General Instructions:

Instructions for Answering:

SECTION A (Multiple Choice Questions)

1. The most stable complex among the following is:

2. If amines are arranged in decreasing order of their basic strength in the gaseous phase, then the correct order will be:

3. Considering the strength of the ligand, the lowest excitation energy will be observed in:

4. The standard electrode potential for Sn⁴⁺/Sn²⁺ couple is + 0.15 V and for Cr³⁺/Cr couple is – 0.73 V. These two couples are connected to make an electrochemical cell. The redox reaction is spontaneous. The cell potential will be:

5. The chemical test which can be used to distinguish between ethanamine and aniline is:

6. The geometry of diamagnetic nickel complex [Ni(CN)₄]²⁻ is:

7. Williamson’s synthesis of preparing dimethyl ether is a/an:

8. Out of Fe²⁺, Co²⁺, Cr³⁺, Ni²⁺, the one which shows highest magnetic moment is:

9. The IUPAC name of the complex [Co(NH₃)₅(NO₂)]Cl₂ is:

10. Which of the following does not undergo Aldol condensation?

11. For a chemical reaction, A → B, it was observed that the rate of reaction doubles when the concentration of A is increased four times. The order of the reaction is:

12. The correct IUPAC name of (CH₃)₃C–CH₂Br is:

For Questions number 13 to 16: Assertion and Reasoning

13. Assertion (A): Rate constant increases with increase in temperature.

14. Assertion (A): Cu²⁺ iodide is not known.

You said:

ChatGPT said:

For Questions 15 and 16: Assertion and Reasoning

15. Assertion (A): Maltose is a reducing sugar.

16. Assertion (A): Acetanilide is less basic than aniline.

SECTION B (Very Short Answer Type Questions)

17. Write the reactions of glucose with:

18. (a) Account for the following:

(b) Give chemical tests to distinguish between the following pair of compounds:

19. Write the reactions involved in the following:

20. The thermal decomposition of an acid is a first-order reaction with a rate constant of 2.3 × 10⁻³ s⁻¹ at a certain temperature. Calculate how long it will take for three-fourths of the initial quantity of acid to decompose.

SECTION C (Short Answer Type Questions)

22. What happens when: (any three)

23. Hydrolysis of sucrose takes place by the chemical reaction:

24. Account for the following:

25. Give plausible explanation for the following:

26. Calculate elevation of the boiling point of solution when 2 g of MgSO₄ (molar mass = 120 g/mol) was dissolved in 100 g of water, assuming MgSO₄ undergoes complete dissociation. [Kb for water = 0.52 K·kg·mol⁻¹]

27. The electrical resistance of a column of 0.05 M KOH solution of cell constant 60 cm⁻¹ is 5.4 × 10³ ohm. Calculate its resistivity, conductivity, and molar conductivity.

28. (a) Draw the geometrical isomers of the complex [Mn(Br)₂(Cl)₂]²⁻.

SECTION D (Case-Based Questions)

SECTION E (Long Answer Questions)

31. (a) (i) Explain with the help of chemical reactions when:

31. (b) (i) Identify the products A, B, C, and D in the following sequence of reactions:

(ii) How will you bring about the following conversions?

32. Attempt any five of the following:

33. (a) (i) What type of battery is the lead storage battery? Write the anode and the cathode reactions and the overall reaction occurring in a lead storage battery when current is drawn from it.

(b) (i) State Kohlrausch’s law of independent migration of ions. Molar conductivity at infinite dilution for NH₄Cl, NaOH, and NaCl solution at 298 K are 110, 100, and 105 S cm² mol⁻¹, respectively. Calculate the molar conductivity of NH₄OH solution.

(ii) Calculate ΔG° of the following cell at 25°C:

QUESTION BANK