CBSE_Class_12_chemistry_Model_Paper (2024)

SECTION A: Multiple Choice Questions

1. 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 can be calculated using the equation:Ecell=Ecathode−EanodeE_{\text{cell}} = E_{\text{cathode}} – E_{\text{anode}}Ecell=Ecathode−Eanode

The cathode (reduction half-reaction) is the half-reaction with the more positive electrode potential.
The anode (oxidation half-reaction) is the half-reaction with the more negative electrode potential.

Given:

Sn4+/Sn2+ has a standard electrode potential of +0.15 V (this will be reduced, so it is the cathode).
Cr3+/Cr has a standard electrode potential of –0.73 V (this will be oxidized, so it is the anode).

The cell potential will therefore be:Ecell=(+0.15)−(−0.73)=+0.15+0.73=+0.88 VE_{\text{cell}} = (+0.15) – (-0.73) = +0.15 + 0.73 = +0.88 \, \text{V}Ecell=(+0.15)−(−0.73)=+0.15+0.73=+0.88V

Since the cell potential is positive, the reaction is spontaneous.

2. The most stable complex among the following is:

(A) [Pt(NH3)2Cl2]
(B) [Ag(NH3)2]Cl
(C) [Pt(en)2Cl2]2+
(D) K4 [Fe(CN)6]

Answer: (D) K4 [Fe(CN)6]

Explanation: The stability of metal-ligand complexes depends on several factors, including the nature of the metal, the ligand, and the geometry of the complex.

[Pt(NH3)2Cl2]: This is a relatively stable complex due to platinum being a transition metal, but ammonia and chloride are not the strongest ligands.
[Ag(NH3)2]Cl: Silver forms relatively less stable complexes, especially with ammonia, because silver tends to have lower coordination numbers and doesn’t form very stable bonds with NH3.
[Pt(en)2Cl2]2+: Ethylenediamine (en) is a bidentate ligand, meaning it can form two bonds with the metal ion. This chelation increases the stability of the complex, but it still isn’t as stable as the cyanide complex.
K4 [Fe(CN)6]: Cyanide (CN-) is a very strong field ligand. It forms very stable complexes with transition metals, especially with iron(III). The cyanide ion stabilizes the iron center significantly, making this the most stable complex among the options.

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

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

Answer: (A) Tetrahedral

Explanation:
The geometry of the complex [Ni(CN)4]²– is tetrahedral. Nickel in the +2 oxidation state has an electron configuration that leads to a tetrahedral arrangement when bonded to four cyanide ions. Additionally, the cyanide ion is a strong field ligand, leading to the pairing of electrons, which makes the complex diamagnetic (no unpaired electrons), and this is characteristic of tetrahedral geometry in such complexes.

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

(A) Fe²+
(B) Co²+
(C) Cr³+
(D) Ni²+

Answer: (A) Fe²+

Explanation:
The magnetic moment of a metal ion depends on the number of unpaired electrons. To calculate the magnetic moment (μ\muμ), we use the formula:μ=n(n+2)\mu = \sqrt{n(n+2)}μ=n(n+2)

where nnn is the number of unpaired electrons.

Fe²+ (d6): 4 unpaired electrons, so it has the highest magnetic moment.
Co²+ (d7): 3 unpaired electrons.
Cr³+ (d3): 3 unpaired electrons.
Ni²+ (d8): 2 unpaired electrons.

Thus, Fe²+ has the highest magnetic moment.

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

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

Answer: (D) NH₃ < CH₃NH₂ < (CH₃)₂NH < (CH₃)₃N

Explanation:
The basic strength of amines in the gaseous phase increases with the availability of the lone pair on nitrogen for protonation. In general:

Ammonia (NH₃) has the least basic strength as it has no alkyl groups to donate electron density to the nitrogen.
Methylamines (CH₃NH₂) show higher basicity because the methyl group donates electron density to nitrogen, increasing its availability for protonation.
Dimethylamine ((CH₃)₂NH) and trimethylamine ((CH₃)₃N) are progressively more basic as the number of methyl groups increases, further donating electron density to nitrogen and making the lone pair more available for protonation.

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

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

Answer: (D) C₆H₅CHO

Explanation:
Aldol condensation involves the reaction of an aldehyde or ketone with an enolate ion to form a β-hydroxy ketone or aldehyde, which can then undergo dehydration to form an α,β-unsaturated carbonyl compound.

CH₃CHO (Acetaldehyde), CH₃COCH₃ (Acetone), and CH₃CH₂CHO (Propionaldehyde) all undergo Aldol condensation because they contain a hydrogen atom on the α-carbon next to the carbonyl group, allowing enolate formation.
C₆H₅CHO (Benzaldehyde) does not undergo Aldol condensation under normal conditions because the aromatic ring’s electron-withdrawing nature makes it less reactive, and it does not easily form an enolate.

7. 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: (A) 2,2-Dimethyl-2-bromopropane

Explanation:
In this compound, the main chain consists of a 3-carbon alkane (propane), with a bromine (Br) and a methyl group (-CH₃) attached. The position of the bromine and the methyl groups on the carbon chain should be indicated with the correct numbering, giving the name as 2,2-Dimethyl-2-bromopropane. The two methyl groups are on the second carbon, and the bromine is also on the second carbon, making this the correct IUPAC name.

8. Considering the strength of the ligand, the highest 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:
In coordination chemistry, the ligand field strength directly affects the splitting of the metal’s d-orbitals and, consequently, the excitation energy. Cyanide (CN⁻) is a very strong field ligand, meaning it causes a larger splitting of the d-orbitals and hence requires more energy to promote an electron from the lower energy orbitals to the higher ones. Therefore, [Co(CN)₆]³⁻ will have the highest excitation energy.

9. 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 of a reaction can be expressed as rate = k[A]^n, where n is the order of the reaction with respect to A. In this case, the rate of reaction doubles when the concentration of A is increased by a factor of 4. From the rate law, we know that the rate should be proportional to the concentration raised to the power of n. So, if the concentration of A is increased by a factor of 4, and the rate doubles, we can use the following equation:

(2) = (4)^n
Solving for n gives n = 2. Therefore, the order of the reaction is 2.

10. 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: (A) Pentaamminenitrito-O-cobalt(III) chloride

Explanation:
In this complex, the ligand nitrite (NO₂) can coordinate to the metal center either through the nitrogen atom (nitro) or the oxygen atom (nitrito). The nitrite ion is coordinated through the oxygen atom in this case, which is denoted as “nitrito-O”. The complex contains five ammonia molecules (amminen), and the metal is cobalt in the +3 oxidation state. Hence, the correct IUPAC name is “Pentaamminenitrito-O-cobalt(III) chloride”.

11. 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 of ether involves the reaction of an alkoxide ion with an alkyl halide. The mechanism of this reaction is bimolecular nucleophilic substitution (SN2), where the nucleophile (alkoxide) attacks the electrophilic carbon of the alkyl halide from the opposite side, displacing the leaving group. This leads to the formation of the ether. Hence, this process follows the SN2 mechanism.

12. 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 is used to distinguish between primary, secondary, and tertiary amines. Ethanamine (ethylamine) is a primary amine, and aniline (C₆H₅NH₂) is also a primary amine, but they behave differently in the Hinsberg test. When treated with benzenesulfonyl chloride (Hinsberg reagent), primary amines like ethanamine form a soluble product, whereas aniline does not. Secondary and tertiary amines react differently as well. Hence, the Hinsberg test can be used to distinguish between different types of amines, including ethanamine and aniline.

For Questions number 13 to 16, two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C), and (D) as given below:

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

13. Assertion (A): 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.

Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

Explanation:

Assertion (A): Maltose is indeed a reducing sugar because it contains a free aldehyde group (–CHO) on one of its glucose units, which can reduce other substances, making it a reducing sugar.
Reason (R): Maltose is composed of two glucose molecules, and the linkage between them is formed between the anomeric carbon (C-1) of one glucose molecule and the hydroxyl group on C-4 of the other glucose molecule. This linkage is a glycosidic bond. The key point here is that in maltose, one of the glucose units retains a free anomeric carbon, which contributes to its reducing nature. Hence, Reason (R) correctly explains why maltose is a reducing sugar.

14. Assertion (A): The boiling point of 1-butanol is higher than that of 1-butene.
Reason (R): 1-Butanol is a polar compound and forms hydrogen bonds, while 1-butene is a nonpolar compound.

Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

Explanation:

Assertion (A): The boiling point of 1-butanol is indeed higher than that of 1-butene. This is because 1-butanol is an alcohol that can form hydrogen bonds due to the –OH group, which results in stronger intermolecular forces and thus a higher boiling point.
Reason (R): 1-Butanol is polar because of the hydroxyl (–OH) group, and it can form hydrogen bonds with other 1-butanol molecules. On the other hand, 1-butene is a nonpolar molecule, as it consists of only carbon and hydrogen atoms with no groups capable of hydrogen bonding. The presence of hydrogen bonds in 1-butanol is the reason it has a higher boiling point compared to the nonpolar 1-butene.

15. Assertion (A): Chlorine water reacts with excess ammonia to form ammonium chloride.
Reason (R): Chlorine acts as an oxidizing agent and oxidizes ammonia to nitrogen gas.

Answer: (D) Assertion (A) is false, but Reason (R) is true.

Explanation:

Assertion (A): Chlorine water reacts with ammonia to form ammonium chloride, but this reaction requires only a small amount of chlorine, not excess. When excess ammonia is added, it reacts to form a mixture of ammonium chloride and ammonium hydroxide, not just ammonium chloride.
Reason (R): Chlorine is indeed an oxidizing agent, and it can oxidize ammonia, leading to the formation of nitrogen gas (N₂) under certain conditions. This is true in reactions such as the reaction of chlorine with ammonia in excess. Chlorine oxidizes ammonia to nitrogen gas and forms ammonium chloride as a byproduct.

16. Assertion (A): Alkenes show geometric isomerism.
Reason (R): Geometric isomerism occurs in compounds that have restricted rotation around a double bond, and alkenes possess a double bond.

Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

Explanation:

Assertion (A): Alkenes do show geometric (cis-trans) isomerism, which occurs due to the restricted rotation around the carbon-carbon double bond.
Reason (R): Geometric isomerism occurs in compounds where there is restricted rotation around a bond. In the case of alkenes, the presence of a double bond between carbon atoms restricts the free rotation, leading to the possibility of cis-trans isomerism. This is the reason why alkenes exhibit geometric isomerism.

15. 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.

Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

Explanation:

The rate constant of a chemical reaction is related to the temperature through the Arrhenius equation:k=Ae−Ea/RTk = A e^{-E_a / RT}k=Ae−Ea/RT

where kkk is the rate constant, AAA is the pre-exponential factor, EaE_aEa is the activation energy, RRR is the gas constant, and TTT is the temperature in Kelvin. As the temperature increases, the exponential factor increases, which leads to an increase in the rate constant.

Reason (R) explains that the increase in temperature results in a higher fraction of molecules possessing energy greater than the activation energy. This means more molecules can effectively collide and react, leading to an increase in the rate constant. Therefore, both the assertion and reason are true, and the reason provides the correct explanation for the assertion.

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

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

Answer: (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

Explanation:

Copper (II), or Cu²⁺, is a strong oxidizing agent. When Cu²⁺ reacts with iodide ions (I⁻), it readily oxidizes I⁻ to iodine (I₂). This oxidation process leads to the reduction of Cu²⁺ to Cu⁺. As a result, Cu²⁺ does not form a stable iodide compound, CuI₂, because the iodide ions are oxidized before the compound can be formed. Therefore, Cu²⁺ iodide is not known in nature.

Reason (R) explains that the strong oxidizing tendency of Cu²⁺ toward iodide ions is the reason why Cu²⁺ iodide does not exist. Thus, both the assertion and reason are true, and the reason correctly explains the assertion.

SECTION B

17. Write the reaction involved in the following:

(a) Reimer-Tiemann reaction

Answer:

The Reimer-Tiemann reaction is a method for the formylation of phenols to give ortho- and para-hydroxybenzaldehydes. The reaction is typically carried out in the presence of chloroform (CHCl₃) and a strong base such as sodium hydroxide (NaOH).

Reaction:C6H5OH+CHCl3→NaOHC6H4(CHO)OH\text{C}_6\text{H}_5\text{OH} + \text{CHCl}_3 \xrightarrow{\text{NaOH}} \text{C}_6\text{H}_4(\text{CHO})\text{OH}C6H5OH+CHCl3NaOHC6H4(CHO)OH

In the above reaction, phenol (C₆H₅OH) reacts with chloroform (CHCl₃) in the presence of sodium hydroxide (NaOH) to form salicylaldehyde (C₆H₄(CHO)OH).

The reaction results in the formation of hydroxybenzaldehyde, primarily in the ortho- and para-positions relative to the hydroxyl group.

(b) Kolbe’s reaction

Answer:

The Kolbe’s reaction (also called Kolbe’s electrolysis) is a reaction where the sodium salts of carboxylic acids undergo electrolysis to form an alkyl group at the benzylic position.

Reaction:C6H5COONa→electrolysisC6H4COOH\text{C}_6\text{H}_5\text{COONa} \xrightarrow{\text{electrolysis}} \text{C}_6\text{H}_4\text{COOH}C6H5COONaelectrolysisC6H4COOH

In this reaction, sodium phenoxide (C₆H₅COONa) is heated under electrolysis conditions, which results in the formation of the carboxylated product, typically benzoic acid (C₆H₅COOH).

This reaction is important for the preparation of carboxylic acids from sodium salts.

Question 19:

The thermal decomposition of an acid is a first-order reaction with a rate constant of k=2.3×10−3 s−1k = 2.3 \times 10^{-3} \, \text{s}^{-1}k=2.3×10−3s−1 at a certain temperature. Calculate how long it will take for three-fourths of the initial quantity of acid to decompose.
Given:

Rate constant, k=2.3×10−3 s−1k = 2.3 \times 10^{-3} \, \text{s}^{-1}k=2.3×10−3s−1
Fraction decomposed = 34\frac{3}{4}43
log⁡4=0.6021\log 4 = 0.6021log4=0.6021, log⁡2=0.301\log 2 = 0.301log2=0.301

Solution:

For a first-order reaction, the integrated rate law is given by:ln⁡([A]0[A])=k⋅t\ln \left( \frac{[A]_0}{[A]} \right) = k \cdot tln([A][A]0)=k⋅t

Where:

[A]0[A]_0[A]0 is the initial concentration of the reactant
[A][A][A] is the concentration at time ttt
kkk is the rate constant
ttt is the time

Now, when three-fourths of the acid has decomposed, 14\frac{1}{4}41 of the initial quantity remains, so:[A][A]0=14\frac{[A]}{[A]_0} = \frac{1}{4}[A]0[A]=41

Substituting into the integrated rate law:ln⁡(14)=k⋅t\ln \left( \frac{1}{4} \right) = k \cdot tln(41)=k⋅t

We know that ln⁡(14)=ln⁡(4−1)=−ln⁡(4)\ln \left( \frac{1}{4} \right) = \ln(4^{-1}) = -\ln(4)ln(41)=ln(4−1)=−ln(4). From the given data, log⁡4=0.6021\log 4 = 0.6021log4=0.6021, so:ln⁡(4)=2log⁡2=2×0.301=0.602\ln(4) = 2 \log 2 = 2 \times 0.301 = 0.602ln(4)=2log2=2×0.301=0.602

Thus:ln⁡(14)=−0.602\ln \left( \frac{1}{4} \right) = -0.602ln(41)=−0.602

Now substitute the values:−0.602=(2.3×10−3)⋅t-0.602 = (2.3 \times 10^{-3}) \cdot t−0.602=(2.3×10−3)⋅t

Solving for ttt:t=−0.602−2.3×10−3=0.6022.3×10−3=261 secondst = \frac{-0.602}{-2.3 \times 10^{-3}} = \frac{0.602}{2.3 \times 10^{-3}} = 261 \, \text{seconds}t=−2.3×10−3−0.602=2.3×10−30.602=261seconds

Thus, it will take 261 seconds for three-fourths of the initial quantity of acid to decompose.

Question 20:

(a) Account for the following:

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

Explanation:

CH₃CHO (acetaldehyde) is more reactive than CH₃COCH₃ (acetone) in reactions with HCN due to the electronic effect of the substituents attached to the carbonyl group.
In CH₃CHO, the carbonyl group (C=O) is attached to a hydrogen atom, which is less electron-donating than the methyl group in CH₃COCH₃ (acetone).
The methyl group in acetone donates electron density through inductive effects, making the carbonyl carbon less electrophilic and therefore less reactive towards nucleophiles like HCN.
In contrast, the hydrogen in acetaldehyde does not donate electron density, making the carbonyl carbon more electrophilic and more susceptible to attack by the cyanide ion.

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

Explanation:

Carboxylic acids have higher boiling points compared to aldehydes and ketones due to the ability to form strong intermolecular hydrogen bonds.
The –OH group of carboxylic acids can form hydrogen bonds with the carbonyl group of adjacent molecules, leading to the formation of dimers. This increases the boiling point.
Aldehydes and ketones, on the other hand, cannot form such strong hydrogen bonds due to the absence of an –OH group, so their boiling points are lower compared to carboxylic acids.

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

(i) Propanal and Propanone

Test:

Tollens’ Test (Silver Mirror Test):
- Propanal (an aldehyde) will react with Tollens’ reagent (ammoniacal silver nitrate) to form a silver mirror.
- Propanone (a ketone) will not react with Tollens’ reagent.
So, Tollens’ test can distinguish between propanal and propanone, as only propanal will give a positive result (silver mirror).

(ii) Benzaldehyde and Benzoic Acid

Test:

Fehling’s Test:
- Benzaldehyde (an aldehyde) will reduce Fehling’s solution, resulting in a red precipitate of copper(I) oxide.
- Benzoic Acid (a carboxylic acid) will not react with Fehling’s solution and will not produce a precipitate.
Fehling’s test can distinguish between benzaldehyde (which gives a positive result) and benzoic acid (which does not react).

Question 21: Write the reaction of glucose with:

(a) HI (Hydriodic Acid)

Answer: When glucose reacts with HI (Hydriodic acid), it undergoes a reduction reaction where the aldehyde group of glucose is reduced and broken down to form ethyl alcohol (ethanol) and iodoform (CHI₃).

Reaction:C6H12O6→HICH3CH2OH+CHI3C_6H_{12}O_6 \xrightarrow{\text{HI}} CH_3CH_2OH + CHI_3C6H12O6HICH3CH2OH+CHI3

Glucose (C₆H₁₂O₆) reacts with HI to form ethyl alcohol (CH₃CH₂OH) and iodoform (CHI₃).

(b) Br₂ Water (Bromine Water)

Answer: When glucose reacts with Br₂ water (bromine water), it undergoes oxidation, where the aldehyde group of glucose is oxidized to a carboxyl group, forming gluconic acid.

Reaction:C6H12O6→Br2/H2OC6H12O7C_6H_{12}O_6 \xrightarrow{\text{Br}_2/\text{H}_2O} C_6H_{12}O_7C6H12O6Br2/H2OC6H12O7

Glucose (C₆H₁₂O₆) reacts with Br₂ water to form gluconic acid (C₆H₁₂O₇).

SECTION C

Question 22:

(a) Draw the geometrical isomers of the complex [Pt(NH₃)₂Cl₂].

Answer: The complex [Pt(NH₃)₂Cl₂] can have two geometrical isomers:

Cis-Isomer: Both Cl ligands are adjacent to each other (on the same side of the platinum center).
- Cis-[Pt(NH₃)₂Cl₂]
In this isomer, the two NH₃ ligands are opposite each other, while the two Cl ligands are next to each other.
Trans-Isomer: The Cl ligands are opposite each other (across from each other).
- Trans-[Pt(NH₃)₂Cl₂]
In this isomer, the NH₃ ligands are adjacent to each other, while the two Cl ligands are opposite each other.

(b) Give the electronic configuration of the d4d^4d4 ion when Δo>P\Delta_o > PΔo>P.

Answer: For a d4d^4d4 ion, when the ligand field splitting energy Δo\Delta_oΔo is greater than the pairing energy PPP (i.e., in the case of strong field ligands), the electrons will pair up in the lower energy t2gt_{2g}t2g orbitals.

The electronic configuration for a d4d^4d4 ion in a strong field would be:

t2g4eg0t_{2g}^4 e_g^0t2g4eg0

In this case, all four electrons will occupy the lower energy t2gt_{2g}t2g orbitals and pair up, resulting in a low-spin configuration.

(c) Solution of [Ni(H₂O)₆]²⁺ is green in colour whereas [Ni(CN)₄]²⁻ is colourless. Give reason.

Answer: The difference in colour between [Ni(H₂O)₆]²⁺ and [Ni(CN)₄]²⁻ is due to the nature of the ligands and their effect on the ddd-orbital splitting:

[Ni(H₂O)₆]²⁺ is green because water (H₂O) is a weak field ligand. This results in partial splitting of the ddd-orbitals of the nickel ion, which allows for some absorption of visible light, giving the solution its green colour.
[Ni(CN)₄]²⁻ is colourless because cyanide (CN⁻) is a strong field ligand, which causes a large splitting of the ddd-orbitals. This large splitting leads to a low-spin configuration, where all the ddd-electrons are paired. Since no unpaired electrons are present, no visible light is absorbed, and the solution appears colourless.

Question 23:

The electrical resistance of a column of 0.05 M NaOH solution of cell constant 50 cm⁻¹ is 4.5 × 10³ ohms. Calculate its resistivity, conductivity, and molar conductivity.

Answer:

Given:

Resistance R=4.5×103 ΩR = 4.5 \times 10^3 \, \OmegaR=4.5×103Ω
Concentration of NaOH C=0.05 MC = 0.05 \, \text{M}C=0.05M
Cell constant k=50 cm−1k = 50 \, \text{cm}^{-1}k=50cm−1

We need to calculate:

Resistivity ( ρ\rhoρ )
Conductivity ( κ\kappaκ )
Molar conductivity ( Λm\Lambda_mΛm )

Step 1: Conductivity ( κ\kappaκ )

Conductivity is given by:κ=1R×k\kappa = \frac{1}{R} \times kκ=R1×k

Substituting the values:κ=14.5×103×50=504.5×103=1.11×10−2 S/cm\kappa = \frac{1}{4.5 \times 10^3} \times 50 = \frac{50}{4.5 \times 10^3} = 1.11 \times 10^{-2} \, \text{S/cm}κ=4.5×1031×50=4.5×10350=1.11×10−2S/cm

Step 2: Resistivity ( ρ\rhoρ )

Resistivity is the reciprocal of conductivity:ρ=1κ=11.11×10−2=90.09 Ω⋅cm\rho = \frac{1}{\kappa} = \frac{1}{1.11 \times 10^{-2}} = 90.09 \, \Omega \cdot \text{cm}ρ=κ1=1.11×10−21=90.09Ω⋅cm

Step 3: Molar Conductivity ( Λm\Lambda_mΛm )

Molar conductivity is given by:Λm=κ×M\Lambda_m = \kappa \times MΛm=κ×M

Where MMM is the molarity of NaOH, i.e., 0.05 M:Λm=1.11×10−2×0.05=5.55×10−4 S\cdotpcm2/mol\Lambda_m = 1.11 \times 10^{-2} \times 0.05 = 5.55 \times 10^{-4} \, \text{S·cm}^2/\text{mol}Λm=1.11×10−2×0.05=5.55×10−4S\cdotpcm2/mol

Question 24:

Calculate the elevation of the boiling point of the solution when 4 g of MgSO₄ (molar mass = 120 g/mol) was dissolved in 100 g of water, assuming MgSO₄ undergoes complete ionization. (Kb for water = 0.52 K·kg/mol)

Answer:

Given:

Mass of MgSO₄ = 4 g
Molar mass of MgSO₄ = 120 g/mol
Mass of water = 100 g
KbK_bKb for water = 0.52 K·kg/mol
MgSO₄ ionizes completely into 2 ions: Mg²⁺ and SO₄²⁻

Step 1: Calculate the number of moles of MgSO₄

Moles of MgSO₄=MassMolar mass=4120=0.03333 mol\text{Moles of MgSO₄} = \frac{\text{Mass}}{\text{Molar mass}} = \frac{4}{120} = 0.03333 \, \text{mol}Moles of MgSO₄=Molar massMass=1204=0.03333mol

Step 2: Calculate the molality of the solution

Molality mmm is given by:m=moles of solutemass of solvent in kg=0.033330.1=0.3333 mol/kgm = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.03333}{0.1} = 0.3333 \, \text{mol/kg}m=mass of solvent in kgmoles of solute=0.10.03333=0.3333mol/kg

Step 3: Calculate the elevation in boiling point

The elevation in boiling point ΔTb\Delta T_bΔTb is given by:ΔTb=Kb×m×i\Delta T_b = K_b \times m \times iΔTb=Kb×m×i

Where iii is the van’t Hoff factor (number of ions produced by the solute). Since MgSO₄ ionizes completely into 2 ions (Mg²⁺ and SO₄²⁻), i=2i = 2i=2.ΔTb=0.52×0.3333×2=0.3467 K\Delta T_b = 0.52 \times 0.3333 \times 2 = 0.3467 \, \text{K}ΔTb=0.52×0.3333×2=0.3467K

Thus, the elevation in boiling point is 0.347 K.

Question 25: Account for the following:

(a) The dipole moment of chlorobenzene is lower than that of cyclohexylchloride.

Answer: The dipole moment of chlorobenzene is lower than that of cyclohexylchloride due to the following reasons:

Chlorobenzene has a chlorine atom attached to a benzene ring. The benzene ring is planar and delocalizes the electron density over the ring, which reduces the overall polarity of the C–Cl bond. In addition, the electron-withdrawing effect of chlorine on the ring is countered by the resonance effect of the aromatic ring, which weakens the dipole.
On the other hand, cyclohexylchloride has a chlorine atom attached to a cyclohexane ring, which does not have the resonance delocalization effect like the benzene ring. As a result, the dipole moment of the C–Cl bond in cyclohexylchloride is more pronounced and the overall dipole moment of the molecule is higher.

(b) Alkyl halides are immiscible in water.

Answer: Alkyl halides are generally immiscible in water due to the following reasons:

Alkyl halides have a nonpolar or slightly polar nature, especially in the alkyl group, which does not interact strongly with the polar water molecules. While the halogen (such as Cl or Br) can be polar, it is not enough to overcome the strong hydrogen bonding between water molecules.
Water is a highly polar solvent, and for two liquids to mix, they must have similar polarity (like dissolves like). Since alkyl halides are mostly nonpolar, they do not dissolve easily in water, making them immiscible.

(c) t-Butyl bromide has a lower boiling point than n-butyl bromide.

Answer: t-Butyl bromide has a lower boiling point than n-butyl bromide due to the following reasons:

t-Butyl bromide is a branched compound, and the branching reduces the surface area available for intermolecular interactions. This leads to weaker van der Waals forces between the molecules, which means less energy is required to overcome these forces and vaporize the liquid.
n-Butyl bromide, being unbranched, has a larger surface area, allowing for stronger van der Waals forces (more surface area for interactions), which results in a higher boiling point.

In summary, the branching in t-butyl bromide reduces the boiling point compared to the more linear n-butyl bromide.

Question 27: Give plausible explanation for the following:

(a) Diazonium salts of aromatic amines are stable.

Answer: The diazonium salts of aromatic amines are relatively stable due to the resonance stabilization provided by the aromatic ring. The positive charge on the nitrogen in the diazonium group can be delocalized onto the ring through resonance, making the diazonium ion less reactive and thus more stable. Additionally, substituents on the aromatic ring (like alkyl groups) can further stabilize the diazonium ion through inductive effects. However, these salts are still reactive and can undergo nucleophilic substitution reactions under suitable conditions.

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

Answer: Aniline (C₆H₅NH₂) does not undergo a Friedel-Crafts reaction (either alkylation or acylation) because the amine group (–NH₂) is a strong electron-donating group through both resonance and inductive effects. It increases the electron density on the benzene ring, which makes the ring more reactive towards electrophiles. However, the Friedel-Crafts reaction requires a Lewis acid (such as AlCl₃) as a catalyst, and the amine group reacts with the Lewis acid, deactivating it. As a result, the electrophilic aromatic substitution (which is key to Friedel-Crafts reactions) is inhibited.

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

Answer: In nitration of aniline, the amine group (–NH₂) is an activating group that increases the electron density on the aromatic ring, making it more reactive toward electrophiles like the nitronium ion (NO₂⁺). However, the amine group directs the incoming nitro group to the ortho and para positions. Despite this, due to the strong electron-donating nature of the amine group, the meta position also becomes significantly reactive. Moreover, in the presence of excess nitration conditions (highly electrophilic), the formation of meta products becomes more substantial. This is due to the steric hindrance and electronic effects that influence the distribution of products.

Question 28: Hydrolysis of sucrose takes place by the chemical reaction:

C12H22O11+H2O→H2C6H12O6C_{12}H_{22}O_{11} + H_2O \xrightarrow{\text{H}} 2 C_6H_{12}O_6C12H22O11+H2OH2C6H12O6

Based on the above reaction, answer the following:

(a) Rate law equation

Answer: The hydrolysis of sucrose in the presence of an acid is a second-order reaction (first-order with respect to sucrose and first-order with respect to water). This is due to the fact that both sucrose and water molecules are involved in the rate-determining step of the reaction. Therefore, the rate law can be written as:Rate=k⋅[sucrose]⋅[water]\text{Rate} = k \cdot [\text{sucrose}] \cdot [\text{water}]Rate=k⋅[sucrose]⋅[water]

Where:

[sucrose][\text{sucrose}][sucrose] is the concentration of sucrose.
[water][\text{water}][water] is the concentration of water.
kkk is the rate constant.

(b) Molecularity and order of reaction

Answer:

Molecularity: The reaction involves the collision of two reactant molecules (sucrose and water), so it is bimolecular in terms of molecularity.
Order: The overall order of the reaction is second-order because the rate depends on the concentration of sucrose and water, both of which are involved in the rate-determining step.

(c) What do you call such reactions?

Answer: This reaction is called a hydrolysis reaction, specifically a biomolecular acid-catalyzed hydrolysis of sucrose. The process involves the splitting of a disaccharide (sucrose) into two monosaccharides (glucose and fructose) by the action of water, in the presence of an acidic catalyst.

Question 29: Case-based question

The case describes the structure and functions of nucleic acids, namely DNA and RNA, which are composed of pentose sugar, phosphoric acid, and nitrogen-containing heterocyclic compounds. They have various functions like genetic information storage, protein synthesis, and energy generation.

(a) Write two functions of DNA.

Answer:

Genetic Information Storage: DNA stores the hereditary information necessary for the growth, development, and functioning of all living organisms.
Protein Synthesis: DNA provides the instructions for protein synthesis through the process of transcription and translation, where it guides the synthesis of proteins from amino acids.

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

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

Adenine (a nitrogenous base)
Deoxyribose (a pentose sugar)
Phosphoric acid (the phosphate group)

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

Answer:

Nucleic acids are large biomolecules composed of long chains of nucleotides. They are the carriers of genetic information in cells and are involved in various cellular processes like protein synthesis, cell division, and energy transfer. DNA and RNA are the two main types of nucleic acids.
Difference between nucleotide and nucleoside:
- A nucleotide consists of three components: a nitrogenous base, a pentose sugar (ribose or deoxyribose), and a phosphate group.
- A nucleoside consists only of two components: a nitrogenous base and a pentose sugar. It lacks the phosphate group.

SECTION D

The following questions are case-based questions. Read the case carefully and answer the questions that follow.

Passage:

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 nucleic acids are mainly of two types: DNA (Deoxyribonucleic Acid) and RNA (Ribonucleic Acid). Nucleic acids, on hydrolysis, yield a pentose sugar, phosphoric acid, and a nitrogen-containing heterocyclic compound (a nitrogenous base). 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 for 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 carries the genetic blueprint of an organism. It stores the information needed for the development, functioning, and reproduction of all living organisms.
Protein Synthesis: DNA is involved in protein synthesis by encoding the instructions necessary to produce proteins through transcription and translation processes.

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

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

Adenine (a nitrogenous base)
Deoxyribose (a pentose sugar)
Phosphoric acid (a phosphate group)

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

Answer:

Nucleic acids are large biomolecules composed of long chains of monomers called nucleotides. Nucleic acids, such as DNA and RNA, are responsible for storing genetic information and facilitating various cellular functions, including protein synthesis and energy transfer.
Difference between nucleotide and nucleoside:
- A nucleotide is composed of three components: a nitrogenous base (adenine, thymine, cytosine, or guanine), a pentose sugar (deoxyribose for DNA or ribose for RNA), and one or more phosphate groups.
- A nucleoside consists of only two components: a nitrogenous base and a pentose sugar. It lacks the phosphate group that is present in nucleotides.

OR

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

Answer:

Similarity: Both DNA and RNA are nucleic acids made up of similar monomer units called nucleotides, which consist of a nitrogenous base, a pentose sugar, and a phosphate group.
Difference: The sugar in DNA is deoxyribose (lacking one oxygen atom), while the sugar in RNA is ribose (with a hydroxyl group on the 2′ carbon). Additionally, DNA is usually double-stranded, while RNA is typically single-stranded.

Question 30:

The cause for deviation from Raoult’s law in the colligative properties of non-ideal solutions lies 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 ethanol–water mixture. Comment.

Answer: A mixture of ethanol and water forms an azeotrope, which behaves as if it were a single substance. This means that the ethanol-water mixture has the same composition in both the liquid and vapor phases and boils at a constant temperature. Due to this, fractional distillation cannot separate the components to achieve pure ethanol because both the ethanol and water will vaporize together in the same proportion, making it impossible to isolate pure ethanol.

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

Answer: A mixture of chloroform and acetone shows deviation from ideal behavior because of strong intermolecular interactions between the molecules of the two substances. These interactions are different from the interactions between the molecules of each individual substance. Specifically, hydrogen bonding between chloroform and acetone molecules causes the mixture to deviate from the ideal solution behavior, resulting in a change in the vapor pressures that is not predicted by Raoult’s law.

(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:

Given:

Vapor pressure of pure benzene (P₀) = 1.25 atm
Vapor pressure of the solution (P) = 1.237 atm
Mass of solute (m) = 1.2 g
Mass of benzene (M_benzene) = 60 g
Molar mass of benzene = 78 g/mol

We can use Raoult’s law for this calculation. Raoult’s law states that the relative decrease in vapor pressure is proportional to the mole fraction of the solute.ΔP=P0−P=vapor pressure lowering\Delta P = P_0 – P = \text{vapor pressure lowering}ΔP=P0−P=vapor pressure lowering ΔP=1.25 atm−1.237 atm=0.013 atm\Delta P = 1.25 \, \text{atm} – 1.237 \, \text{atm} = 0.013 \, \text{atm}ΔP=1.25atm−1.237atm=0.013atm

Using the relation for the vapor pressure lowering:ΔPP0=nsolutensolvent+nsolute\frac{\Delta P}{P_0} = \frac{n_{\text{solute}}}{n_{\text{solvent}} + n_{\text{solute}}}P0ΔP=nsolvent+nsolutensolute

Since the amount of solute is very small compared to the solvent, we can approximate nsolvent+nsolute≈nsolventn_{\text{solvent}} + n_{\text{solute}} \approx n_{\text{solvent}}nsolvent+nsolute≈nsolvent, so:ΔPP0=nsolutensolvent\frac{\Delta P}{P_0} = \frac{n_{\text{solute}}}{n_{\text{solvent}}}P0ΔP=nsolventnsolute

Now, the number of moles of solvent (benzene):nsolvent=mass of solventmolar mass of solvent=60 g78 g/mol=0.769 moln_{\text{solvent}} = \frac{\text{mass of solvent}}{\text{molar mass of solvent}} = \frac{60 \, \text{g}}{78 \, \text{g/mol}} = 0.769 \, \text{mol}nsolvent=molar mass of solventmass of solvent=78g/mol60g=0.769mol

The number of moles of solute (nsoluten_{\text{solute}}nsolute) can be found from the equation:0.0131.25=nsolute0.769\frac{0.013}{1.25} = \frac{n_{\text{solute}}}{0.769}1.250.013=0.769nsolute nsolute=0.013×0.7691.25=0.00769 moln_{\text{solute}} = \frac{0.013 \times 0.769}{1.25} = 0.00769 \, \text{mol}nsolute=1.250.013×0.769=0.00769mol

Now, to find the molar mass of the solute:Molar mass of solute=mass of solutensolute=1.2 g0.00769 mol=156.25 g/mol\text{Molar mass of solute} = \frac{\text{mass of solute}}{n_{\text{solute}}} = \frac{1.2 \, \text{g}}{0.00769 \, \text{mol}} = 156.25 \, \text{g/mol}Molar mass of solute=nsolutemass of solute=0.00769mol1.2g=156.25g/mol

Thus, the molar mass of the non-volatile solute is approximately 156.25 g/mol.

OR

(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⁻¹.

Answer:

To calculate the molar mass of the solute, we can use the boiling point elevation formula:ΔTb=Kb⋅m\Delta T_b = K_b \cdot mΔTb=Kb⋅m

Where:

ΔTb\Delta T_bΔTb is the elevation in boiling point,
KbK_bKb is the ebullioscopic constant (boiling point elevation constant) for benzene,
mmm is the molality of the solution.

Step 1: Calculate the boiling point elevation (ΔTb\Delta T_bΔTb).

The initial boiling point of pure benzene is given as 353.23 K, and the final boiling point of the solution is 354.11 K.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

Step 2: Use the boiling point elevation formula to calculate molality.

Rearrange the formula to solve for molality (mmm):m=ΔTbKbm = \frac{\Delta T_b}{K_b}m=KbΔTb

Substitute the given values:m=0.88 K2.53 K kg/mol=0.3486 mol/kgm = \frac{0.88 \, \text{K}}{2.53 \, \text{K kg/mol}} = 0.3486 \, \text{mol/kg}m=2.53K kg/mol0.88K=0.3486mol/kg

Step 3: Calculate the number of moles of solute.

Molality is defined as the number of moles of solute per kilogram of solvent. We are given that the mass of the solvent (benzene) is 90 g (which is 0.090 kg). Using the molality (mmm):moles of solute=m×mass of solvent (in kg)\text{moles of solute} = m \times \text{mass of solvent (in kg)}moles of solute=m×mass of solvent (in kg) moles of solute=0.3486 mol/kg×0.090 kg=0.031374 mol\text{moles of solute} = 0.3486 \, \text{mol/kg} \times 0.090 \, \text{kg} = 0.031374 \, \text{mol}moles of solute=0.3486mol/kg×0.090kg=0.031374mol

Step 4: Calculate the molar mass of the solute.

Now, we can calculate the molar mass of the solute using the formula:Molar mass=mass of solutemoles of solute\text{Molar mass} = \frac{\text{mass of solute}}{\text{moles of solute}}Molar mass=moles of solutemass of solute

We are given that the mass of the solute is 1.80 g.Molar mass=1.80 g0.031374 mol=57.4 g/mol\text{Molar mass} = \frac{1.80 \, \text{g}}{0.031374 \, \text{mol}} = 57.4 \, \text{g/mol}Molar mass=0.031374mol1.80g=57.4g/mol

Thus, the molar mass of the solute is approximately 57.4 g/mol.

SECTION E

31. Attempt any five of the following: (5 × 1 = 5)

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

Answer: Cu⁺ (copper(I)) is not stable in aqueous solution because it is easily oxidized to Cu²⁺ (copper(II)) in the presence of oxygen or water. Cu⁺ has a high tendency to lose an electron, and in aqueous solutions, the stability of Cu⁺ is compromised by the formation of Cu²⁺. The disproportionation reaction is as follows:2 Cu+→Cu2++Cu2 \, \text{Cu}^+ \rightarrow \text{Cu}^2+ + \text{Cu}2Cu+→Cu2++Cu

This instability of Cu⁺ is a result of the lower stability of the 3d⁹ configuration compared to the more stable 3d⁸ configuration of Cu²⁺.

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

Answer: Cr²⁺ (Chromium(II)) is a stronger reducing agent than Fe²⁺ (Iron(II)). This is because Cr²⁺ has a greater tendency to lose electrons and get oxidized to Cr³⁺. The reduction potential for Cr²⁺/Cr³⁺ is more negative than that for Fe²⁺/Fe³⁺, indicating that Cr²⁺ is more easily oxidized and hence a stronger reducing agent.

The standard reduction potential for the half-reaction:Cr3++e−→Cr2+E∘=−0.41 V\text{Cr}^{3+} + e^- \rightarrow \text{Cr}^{2+} \quad E^\circ = -0.41 \, \text{V}Cr3++e−→Cr2+E∘=−0.41V

is more negative than for Fe²⁺/Fe³⁺, indicating Cr²⁺ is a stronger reducing agent.

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

Answer: The actinoid contraction is greater than the lanthanoid contraction because the 5f orbitals in actinoids are less effective at shielding the nuclear charge compared to the 4f orbitals in lanthanoids. As a result, the atomic size decreases more rapidly across the actinoid series due to poorer shielding of the increasing nuclear charge by the 5f electrons. This leads to a greater contraction in the atomic radii of actinoid elements from element to element than in the lanthanoid series.

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

Answer: In acidic medium, KMnO₄ (potassium permanganate) acts as a strong oxidizing agent. The ionic equation for the reduction of MnO₄⁻ (permanganate ion) to Mn²⁺ in acidic medium is:MnO4−+8 H++5 e−→Mn2++4 H2O\text{MnO}_4^- + 8 \, \text{H}^+ + 5 \, e^- \rightarrow \text{Mn}^{2+} + 4 \, \text{H}_2OMnO4−+8H++5e−→Mn2++4H2O

This reaction shows that MnO₄⁻ is reduced to Mn²⁺ while donating 5 electrons, and H⁺ ions from the acid medium are involved in the reaction.

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

Answer: The metal in the first transition series that exhibits the +1 oxidation state most frequently is Copper (Cu). Copper predominantly exists in the +1 oxidation state as Cu⁺ in many of its compounds, such as Cu₂O (copper(I) oxide). This is due to the relatively stable d¹⁰ configuration that Cu⁺ achieves.

(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 exist in multiple oxidation states, which allows them to participate in a variety of redox reactions. Their d-orbitals facilitate the adsorption of reactants and provide a surface for reaction. Additionally, their ability to form intermediate complexes with reactants lowers the activation energy for reactions. The presence of partially filled d-orbitals allows them to effectively interact with substrates and provide an alternative reaction pathway with lower energy.

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

Answer: Scandium (Sc) is regarded as a transition element because it has a partially filled d-orbital in its ground state or common oxidation states. However, Sc³⁺ (the most common oxidation state of Sc) has no d-electrons because it has lost all three electrons from the 3d orbitals, making it colorless. Despite this, Sc is classified as a transition metal because it has the capacity to form compounds with partially filled d-orbitals in its lower oxidation states (Sc²⁺), and thus it fits the criteria for transition elements.

32. (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 rechargeable battery commonly used in vehicles. It is also known as the lead-acid battery.

Anode reaction (during discharge):Pb(s)+HSO₄−(aq)→PbSO₄(s)+H+(aq)+2e−\text{Pb(s)} + \text{HSO₄}⁻(aq) \rightarrow \text{PbSO₄(s)} + \text{H}⁺(aq) + 2e⁻Pb(s)+HSO₄−(aq)→PbSO₄(s)+H+(aq)+2e−(Lead is oxidized to lead sulfate, releasing electrons)
Cathode reaction (during discharge):PbO₂(s)+H+(aq)+HSO₄−(aq)+2e−→PbSO₄(s)+H₂O(l)\text{PbO₂(s)} + \text{H}⁺(aq) + \text{HSO₄}⁻(aq) + 2e⁻ \rightarrow \text{PbSO₄(s)} + \text{H₂O}(l)PbO₂(s)+H+(aq)+HSO₄−(aq)+2e−→PbSO₄(s)+H₂O(l)(Lead dioxide is reduced to lead sulfate)
Overall reaction (during discharge):Pb(s)+PbO₂(s)+2HSO₄−(aq)→2PbSO₄(s)+H₂O(l)\text{Pb(s)} + \text{PbO₂(s)} + 2\text{HSO₄}⁻(aq) \rightarrow 2\text{PbSO₄(s)} + \text{H₂O}(l)Pb(s)+PbO₂(s)+2HSO₄−(aq)→2PbSO₄(s)+H₂O(l)

The lead storage battery works by converting electrical energy into chemical energy during charging and converting it back into electrical energy during discharging.

(ii) Calculate the time to deposit 1.5 g of silver at cathode when a current of 1.5 A was passed through the solution of AgNO₃.

Given:

Molar mass of Ag = 108 g/mol
Current (I) = 1.5 A
Faraday constant (F) = 96500 C/mol

Step 1: Calculate the number of moles of silver to be deposited.Moles of Ag=Mass of AgMolar Mass of Ag=1.5 g108 g/mol=0.01389 mol\text{Moles of Ag} = \frac{\text{Mass of Ag}}{\text{Molar Mass of Ag}} = \frac{1.5 \, \text{g}}{108 \, \text{g/mol}} = 0.01389 \, \text{mol}Moles of Ag=Molar Mass of AgMass of Ag=108g/mol1.5g=0.01389mol

Step 2: Use Faraday’s law of electrolysis to calculate the charge required.

The reduction reaction for Ag⁺ at the cathode is:Ag++e−→Ag\text{Ag}^+ + e^- \rightarrow \text{Ag}Ag++e−→Ag

This means 1 mole of Ag requires 1 mole of electrons.

Therefore, the charge required for depositing 0.01389 mol of Ag is:Q=Moles of Ag×F=0.01389 mol×96500 C/mol=1344.9 CQ = \text{Moles of Ag} \times F = 0.01389 \, \text{mol} \times 96500 \, \text{C/mol} = 1344.9 \, \text{C}Q=Moles of Ag×F=0.01389mol×96500C/mol=1344.9C

Step 3: Calculate the time (t) required to deposit 1.5 g of silver.

Using the formula Q=I×tQ = I \times tQ=I×t, where III is the current and ttt is the time, we can rearrange to solve for time:t=QI=1344.9 C1.5 A=896.6 secondst = \frac{Q}{I} = \frac{1344.9 \, \text{C}}{1.5 \, \text{A}} = 896.6 \, \text{seconds}t=IQ=1.5A1344.9C=896.6seconds

So, the time required to deposit 1.5 g of silver is approximately 897 seconds or 14.95 minutes.

OR

(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.

Answer: Kohlrausch’s law of independent migration of ions states that the molar conductivity at infinite dilution of an electrolyte can be considered as the sum of the contributions of the individual ions, each of which migrates independently in the solution. This means:Λm∞=λcation∞+λanion∞\Lambda_m^\infty = \lambda_{\text{cation}}^\infty + \lambda_{\text{anion}}^\inftyΛm∞=λcation∞+λanion∞

Where Λm∞\Lambda_m^\inftyΛm∞ is the molar conductivity at infinite dilution, and λcation∞\lambda_{\text{cation}}^\inftyλcation∞ and λanion∞\lambda_{\text{anion}}^\inftyλanion∞ are the individual conductivities of the cation and anion at infinite dilution.

To calculate the molar conductivity of NH₄OH, we can use Kohlrausch’s law:Λm∞(NH₄OH)=Λm∞(NH₄Cl)+Λm∞(OH⁻)−Λm∞(Cl⁻)\Lambda_m^\infty (\text{NH₄OH}) = \Lambda_m^\infty (\text{NH₄Cl}) + \Lambda_m^\infty (\text{OH⁻}) – \Lambda_m^\infty (\text{Cl⁻})Λm∞(NH₄OH)=Λm∞(NH₄Cl)+Λm∞(OH⁻)−Λm∞(Cl⁻)

Using the given values:

Λm∞(NH₄Cl)=110 S cm² mol⁻¹\Lambda_m^\infty (\text{NH₄Cl}) = 110 \, \text{S cm² mol⁻¹}Λm∞(NH₄Cl)=110S cm² mol⁻¹
Λm∞(Cl⁻)=76.3 S cm² mol⁻¹\Lambda_m^\infty (\text{Cl⁻}) = 76.3 \, \text{S cm² mol⁻¹}Λm∞(Cl⁻)=76.3S cm² mol⁻¹ (from NaCl, since Cl⁻ is the same for NaCl and NH₄Cl)
Λm∞(OH⁻)=198 S cm² mol⁻¹\Lambda_m^\infty (\text{OH⁻}) = 198 \, \text{S cm² mol⁻¹}Λm∞(OH⁻)=198S cm² mol⁻¹

Λm∞(NH₄OH)=110+198−76.3=231.7 S cm² mol⁻¹\Lambda_m^\infty (\text{NH₄OH}) = 110 + 198 – 76.3 = 231.7 \, \text{S cm² mol⁻¹}Λm∞(NH₄OH)=110+198−76.3=231.7S cm² mol⁻¹

Thus, the molar conductivity of NH₄OH is 231.7 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+}(aq) || \text{Cu}^{2+}(aq) | \text{Cu(s)}Zn(s)∣Zn2+(aq)∣∣Cu2+(aq)∣Cu(s)

Given:

EZn²⁺/Zn∘=−0.76 VE^\circ_{\text{Zn²⁺/Zn}} = -0.76 \, \text{V}EZn²⁺/Zn∘=−0.76V
ECu²⁺/Cu∘=+0.34 VE^\circ_{\text{Cu²⁺/Cu}} = +0.34 \, \text{V}ECu²⁺/Cu∘=+0.34V
F=96500 C/molF = 96500 \, \text{C/mol}F=96500C/mol
Temperature = 25°C = 298 K

Step 1: Calculate the cell potential (Ecell∘E^\circ_{\text{cell}}Ecell∘).

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

For this cell:

The cathode is Cu (reduction half-reaction: Cu²⁺ + 2e⁻ → Cu)
The anode is Zn (oxidation half-reaction: Zn → Zn²⁺ + 2e⁻)

Ecell∘=0.34 V−(−0.76 V)=1.10 VE^\circ_{\text{cell}} = 0.34 \, \text{V} – (-0.76 \, \text{V}) = 1.10 \, \text{V}Ecell∘=0.34V−(−0.76V)=1.10V

Step 2: Calculate ΔG° using the equation:ΔG∘=−nFEcell∘\Delta G^\circ = -nFE^\circ_{\text{cell}}ΔG∘=−nFEcell∘

Where n=2n = 2n=2 (since 2 electrons are involved in the half-reactions), and F=96500 C/molF = 96500 \, \text{C/mol}F=96500C/mol.ΔG∘=−2×96500 C/mol×1.10 V\Delta G^\circ = -2 \times 96500 \, \text{C/mol} \times 1.10 \, \text{V}ΔG∘=−2×96500C/mol×1.10V ΔG∘=−2×96500×1.10=−212300 J/mol=−212.3 kJ/mol\Delta G^\circ = -2 \times 96500 \times 1.10 = -212300 \, \text{J/mol} = -212.3 \, \text{kJ/mol}ΔG∘=−2×96500×1.10=−212300J/mol=−212.3kJ/mol

Thus, ΔG∘=−212.3 kJ/mol\Delta G^\circ = -212.3 \, \text{kJ/mol}ΔG∘=−212.3kJ/mol.

33. (a) (i) Explain with the help of chemical reaction when:

(I) Acetone is treated with semicarbazide.

Answer: When acetone (a ketone) reacts with semicarbazide, it forms a semicarbazone derivative. This is an example of a condensation reaction, where the carbonyl group of acetone reacts with the amino group of semicarbazide.

The reaction is:CH₃COCH₃+H₂N-NH-C(O)-NH₂→CH₃C=NNH-C(O)-NH₂+H₂O\text{CH₃COCH₃} + \text{H₂N-NH-C(O)-NH₂} \rightarrow \text{CH₃C=NNH-C(O)-NH₂} + \text{H₂O}CH₃COCH₃+H₂N-NH-C(O)-NH₂→CH₃C=NNH-C(O)-NH₂+H₂O

(II) Two molecules of benzaldehyde are treated with conc. NaOH.

Answer: When benzaldehyde reacts with concentrated NaOH, the reaction undergoes aldol condensation. In this reaction, two molecules of benzaldehyde condense in the presence of a strong base (NaOH) to form 2-phenyl-2-propenal (also called cinnamaldehyde).

The reaction is:2 C₆H₅CHO→NaOHC₆H₅CH=CHCHO2 \, \text{C₆H₅CHO} \xrightarrow{\text{NaOH}} \text{C₆H₅CH=CHCHO}2C₆H₅CHONaOHC₆H₅CH=CHCHO

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

Answer: When butan-2-one is treated with Zn/Hg and concentrated HCl, it undergoes a Clemmensen reduction, which reduces the ketone to an alkane by removing the oxygen atom.

The reaction is:CH₃COCH₂CH₃→Zn/Hg, HClCH₃CH₂CH₃\text{CH₃COCH₂CH₃} \xrightarrow{\text{Zn/Hg, HCl}} \text{CH₃CH₂CH₃}CH₃COCH₂CH₃Zn/Hg, HClCH₃CH₂CH₃

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

(I) CH₃CH₂CH₂COOH, BrCH₂CH₂CH₂COOH, CH₃CHBrCH₂COOH, CH₃CH₂CHBrCOOH

Answer: The presence of electronegative substituents such as Br increases the acidity by stabilizing the conjugate base (the carboxylate anion) through inductive effects.

The increasing order of acidity is:CH₃CH₂CH₂COOH<CH₃CH₂CHBrCOOH<CH₃CHBrCH₂COOH<BrCH₂CH₂CH₂COOH\text{CH₃CH₂CH₂COOH} < \text{CH₃CH₂CHBrCOOH} < \text{CH₃CHBrCH₂COOH} < \text{BrCH₂CH₂CH₂COOH}CH₃CH₂CH₂COOH<CH₃CH₂CHBrCOOH<CH₃CHBrCH₂COOH<BrCH₂CH₂CH₂COOH

(II) Benzoic acid, 4-Methoxybenzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitrobenzoic acid

Answer: The nitro group (NO2\text{NO}_2NO2) is an electron-withdrawing group, which increases acidity by stabilizing the conjugate base. The methoxy group (OCH3\text{OCH}_3OCH3) is an electron-donating group, which decreases acidity.

The increasing order of acidity is:4-Methoxybenzoic acid<Benzoic acid<4-Nitrobenzoic acid<3,4-Dinitrobenzoic acid\text{4-Methoxybenzoic acid} < \text{Benzoic acid} < \text{4-Nitrobenzoic acid} < \text{3,4-Dinitrobenzoic acid}4-Methoxybenzoic acid<Benzoic acid<4-Nitrobenzoic acid<3,4-Dinitrobenzoic acid

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

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

Answer:

Step 1: CH₃CHO [O][O][O] → A
The oxidation of acetaldehyde (CH₃CHO) with an oxidizing agent leads to the formation of acetic acid (CH₃COOH).
A = CH₃COOH (Acetic acid)
Step 2: A →PCl₅\xrightarrow{\text{PCl₅}}PCl₅ → B
The reaction of acetic acid with PCl₅ (phosphorus pentachloride) converts the carboxyl group into an acyl chloride.
B = CH₃COCl (Acetyl chloride)
Step 3: B →(CH₃)₂Cd\xrightarrow{\text{(CH₃)₂Cd}}(CH₃)₂Cd → D
The reaction of acetyl chloride (CH₃COCl) with dimethylcadmium ((CH₃)₂Cd) results in the formation of acetone ((CH₃)₂CO) due to the nucleophilic attack of the methyl group from dimethylcadmium.
D = (CH₃)₂CO (Acetone)
Step 4: D →Zn/Hg, conc. HCl\xrightarrow{\text{Zn/Hg, conc. HCl}}Zn/Hg, conc. HCl → C
The reduction of acetone with Zn/Hg and conc. HCl (Clemmensen reduction) leads to the formation of propane (CH₃CH₂CH₃).
C = CH₃CH₂CH₃ (Propane)

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

(I) Propanone to Propene

Answer: To convert propanone (CH₃COCH₃) to propene (CH₃CH=CH₂), you can use a reduction reaction known as Wolff-Löffler reduction or clemmensen reduction.

Reaction: CH₃COCH₃→Zn/Hg, HClCH₃CH=CH₂\text{CH₃COCH₃} \xrightarrow{\text{Zn/Hg, HCl}} \text{CH₃CH=CH₂}CH₃COCH₃Zn/Hg, HClCH₃CH=CH₂ This is a reductive dehalogenation that removes the carbonyl group and forms a double bond, resulting in propene.

(II) Benzoic acid to Benzaldehyde

Answer: To convert benzoic acid (C₆H₅COOH) to benzaldehyde (C₆H₅CHO), you can use the reduction of the carboxyl group while keeping the benzene ring intact. One of the common methods is partial reduction using diborane (B₂H₆) or LiAlH₄ in a controlled environment.

Reaction: C₆H₅COOH→Diborane (B₂H₆)C₆H₅CHO\text{C₆H₅COOH} \xrightarrow{\text{Diborane (B₂H₆)}} \text{C₆H₅CHO}C₆H₅COOHDiborane (B₂H₆)C₆H₅CHO

Alternatively, the Rosenmund reduction (reduction of acyl chlorides using hydrogen and palladium on barium sulfate) can be employed.

(III) Ethanal to But-2-enal

Answer: To convert ethanal (CH₃CHO) to but-2-enal (CH₃CH=CHCHO), you can use an aldol condensation reaction followed by dehydration.

Step 1: Aldol condensation of ethanal with itself
In an aldol condensation, ethanal undergoes self-condensation in the presence of a base to form but-3-en-2-ol.CH₃CHO+CH₃CHO→BaseCH₃CH=CHOHCHO\text{CH₃CHO} + \text{CH₃CHO} \xrightarrow{\text{Base}} \text{CH₃CH=CHOHCHO}CH₃CHO+CH₃CHOBaseCH₃CH=CHOHCHO
Step 2: Dehydration
The aldol product undergoes dehydration (removal of water) to form but-2-enal (CH₃CH=CHCHO).CH₃CH=CHOHCHO→DehydrationCH₃CH=CHCHO\text{CH₃CH=CHOHCHO} \xrightarrow{\text{Dehydration}} \text{CH₃CH=CHCHO}CH₃CH=CHOHCHODehydrationCH₃CH=CHCHO

CBSE_Class_12_chemistry_Model_Paper (2024)

SECTION A: Multiple Choice Questions

For Questions number 13 to 16, two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C), and (D) as given below:

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

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

17. Write the reaction involved in the following:

(a) Reimer-Tiemann reaction

(b) Kolbe’s reaction

Question 19:

Solution:

Question 20:

(a) Account for the following:

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

Question 21: Write the reaction of glucose with:

(a) HI (Hydriodic Acid)

(b) Br₂ Water (Bromine Water)

Question 22:

(a) Draw the geometrical isomers of the complex [Pt(NH₃)₂Cl₂].

(b) Give the electronic configuration of the d4d^4d4 ion when Δo>P\Delta_o > PΔo​>P.

(c) Solution of [Ni(H₂O)₆]²⁺ is green in colour whereas [Ni(CN)₄]²⁻ is colourless. Give reason.

Question 23:

Step 1: Conductivity ( κ\kappaκ )

Step 2: Resistivity ( ρ\rhoρ )

Step 3: Molar Conductivity ( Λm\Lambda_mΛm​ )

Question 24:

Step 1: Calculate the number of moles of MgSO₄

Step 2: Calculate the molality of the solution

Step 3: Calculate the elevation in boiling point

Question 25: Account for the following:

(a) The dipole moment of chlorobenzene is lower than that of cyclohexylchloride.

(b) Alkyl halides are immiscible in water.

(c) t-Butyl bromide has a lower boiling point than n-butyl bromide.

Question 27: Give plausible explanation for the following:

(a) Diazonium salts of aromatic amines are stable.

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

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

Question 28: Hydrolysis of sucrose takes place by the chemical reaction:

(a) Rate law equation

(b) Molecularity and order of reaction

(c) What do you call such reactions?

Question 29: Case-based question

(a) Write two functions of DNA.

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

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

SECTION D

Passage:

Based on the above information, answer the following questions:

(a) Write two functions of DNA.

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

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

OR

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

Question 30:

Based on the above passage, answer the following questions:

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

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

(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.

OR

(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⁻¹.

Step 1: Calculate the boiling point elevation (ΔTb\Delta T_bΔTb​).

Step 2: Use the boiling point elevation formula to calculate molality.

Step 3: Calculate the number of moles of solute.

Step 4: Calculate the molar mass of the solute.

SECTION E

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

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

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

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

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

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

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

32. (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.

(ii) Calculate the time to deposit 1.5 g of silver at cathode when a current of 1.5 A was passed through the solution of AgNO₃.

OR

(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:

33. (a) (i) Explain with the help of chemical reaction when:

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

(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?

(b) Give the electronic configuration of the d4d^4d4 ion when Δo>P\Delta_o > PΔo>P.

Step 3: Molar Conductivity ( Λm\Lambda_mΛm )

Step 1: Calculate the boiling point elevation (ΔTb\Delta T_bΔTb).