CBSE Class 12 – Chemistry Question Paper 2023

SECTION A

Questions no. 1 to 18 are Multiple Choice (MCQ) type Questions, carrying 1 mark each.

Colligative Property for Molar Mass Determination
The colligative property used for the determination of molar mass of polymers and proteins is:
(a) Osmotic pressure
(b) Depression in freezing point
(c) Relative lowering in vapour pressure
(d) Elevation in boiling point
Answer: (a) Osmotic pressure
Low Oxygen Concentration at High Altitude
Low concentration of oxygen in the blood and tissues of people living at high altitude is due to:
(a) High atmospheric pressure
(b) Low temperature
(c) Low atmospheric pressure
(d) Both low temperature and high atmospheric pressure
Answer: (c) Low atmospheric pressure
Correct Cell Representation of Reaction
The correct cell to represent the following reaction is: Zn + 2Ag+ → Zn2+ + 2Ag
(a) 2Ag | Ag+ | Zn | Zn2+
(b) Ag+ | Ag | Zn2+ | Zn
(c) Ag | Ag+ | Zn | Zn2+
(d) Zn | Zn2+ | Ag+ | Ag
Answer: (d) Zn | Zn2+ | Ag+ | Ag
G and E for Spontaneous Reaction
G and E for a spontaneous reaction will be:
(a) Positive, negative
(b) Negative, negative
(c) Negative, positive
(d) Positive, positive
Answer: (c) Negative, positive
Effect of Catalyst
Which of the following is affected by a catalyst?
(a) H
(b) G
(c) Ea
(d) S
Answer: (c) Ea
Order of Reaction
The order of the reaction H2(g) + Cl2(g) → 2HCl(g) is:
(a) 2
(b) 1
(c) 0
(d) 3
Answer: (b) 1
Common Oxidation State of Lanthanoids
The most common and stable oxidation state of a Lanthanoid is:
(a) +2
(b) +3
(c) +4
(d) +6
Answer: (b) +3
Ionisation Isomerism
The compounds [Co(SO4)(NH3)5]Br and [Co(Br)(NH3)5]SO4 represent:
(a) Optical isomerism
(b) Linkage isomerism
(c) Ionisation isomerism
(d) Coordination isomerism
Answer: (c) Ionisation isomerism
Synthesis of Alkyl Fluoride
The synthesis of alkyl fluoride is best obtained from:
(a) Free radicals
(b) Swartz reaction
(c) Sandmeyer reaction
(d) Finkelstein reaction
Answer: (b) Swartz reaction
Reactivity Order of Alcohols
In the reaction R-OH + HCl (ZnCl2) → RCl + H2O, what is the correct order of reactivity of alcohol?
(a) 1 < 2 < 3
(b) 1 > 3 > 2
(c) 1 > 2 > 3
(d) 3 > 1 > 2
Answer: (c) 1 > 2 > 3
Reaction of CH3CONH2 with NaOH and Br2
CH3CONH2 on reaction with NaOH and Br2 in alcoholic medium gives:
(a) CH3COONa
(b) CH3NH2
(c) CH3CH2Br
(d) CH3CH2NH2
Answer: (b) CH3NH2
Least Basic Compound
Which of the following is least basic?
(a) (CH3)2NH
(b) NH3
(c) NH2
(d) (CH3)3N
Answer: (b) NH3
Glycosidic Linkage in Starch
The glycosidic linkage involved in linking the glucose units in the amylase part of starch is:
(a) C1-C6 linkage
(b) C1-C6 linkage
(c) C1-C4 linkage
(d) C1-C4 linkage
Answer: (c) C1-C4 linkage
Structural Feature of α-Helix
An α-helix is a structural feature of:
(a) Sucrose
(b) Starch
(c) Polypeptides
(d) Nucleotides
Answer: (c) Polypeptides

For Questions number 15 to 18, 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.

15.Assertion (A): NH2 group is o- and p-directing in electrophilic substitution reactions.
Reason (R): Aniline cannot undergo Friedel-Crafts reaction.
Answer: (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

16.Assertion (A): Acetylation of aniline gives a monosubstituted product.
Reason (R): Activating effect of NHCOCH3 group is more than that of amino group.
Answer: (c) Assertion (A) is true, but Reason (R) is false.

17.Assertion (A): The molecularity of the reaction H2 + Br2 → 2HBr appears to be 2.
Reason (R): Two molecules of the reactants are involved in the given elementary reaction.
Answer: (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

18.Assertion (A): Low spin tetrahedral complexes are rarely observed.
Reason (R): Crystal field splitting energy is less than pairing energy for tetrahedral complexes.
Answer: (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

SECTION B

What is Henry’s law? Give one application of it.

Answer:
Henry’s law states that “the amount of a gas that dissolves in a liquid is directly proportional to the partial pressure of the gas above the liquid.”
Mathematically, it can be expressed as:
$C = k_H P$

$C = k_{H} P$
where $C$

$C$ is the concentration of the gas in the liquid, $k_H$

$k_{H}$ is Henry’s law constant, and $P$

$P$ is the partial pressure of the gas.

Application:
One application of Henry’s law is in the carbonation of soft drinks. The dissolved carbon dioxide (CO2) in the drink depends on the pressure of CO2 gas applied during the manufacturing process.

(a) On diluting two electrolytes ‘A’ and ‘B’, the $A_m$
$A_{m}$ of ‘A’ increases 25 times while that of ‘B’ increases by 1.5 times. Which of the two electrolytes is strong? Justify your answer graphically.

Answer:
The increase in the molar conductivity ( $A_m$

$A_{m}$ ) with dilution indicates the strength of the electrolyte. A higher increase in $A_m$

$A_{m}$ suggests a stronger electrolyte.
Since the $A_m$

$A_{m}$ of ‘A’ increases 25 times, while that of ‘B’ increases by only 1.5 times, electrolyte ‘A’ is stronger.
Graphically, for strong electrolytes, the molar conductivity increases rapidly with dilution.

(b) The electrical resistance of a column of 0.05 mol L⁻¹ NaOH solution of diameter 1 cm and length 50 cm is $5.55 \times 10^8 \, \Omega$

$5.55 \times 1 0^{8} Ω$ . Calculate the conductivity.

Answer:
Given:

Resistance, $R = 5.55 \times 10^8 \, \Omega$
Concentration, $C = 0.05 \, \text{mol L}^{-1}$
Diameter of column, $d = 1 \, \text{cm} = 0.01 \, \text{m}$
Length of column, $L = 50 \, \text{cm} = 0.5 \, \text{m}$

First, calculate the cross-sectional area of the column: $\text{Area} = \pi \left( \frac{d}{2} \right)^2 = \pi \left( \frac{0.01}{2} \right)^2 = 7.854 \times 10^{-5} \, \text{m}^2$

$Area = π (\frac{d}{2})^{2} = π (\frac{0.01}{2})^{2} = 7.854 \times 1 0^{-5} m^{2}$

Now, calculate the conductivity ( $\kappa$

$κ$ ) using the formula: $\kappa = \frac{1}{\rho} = \frac{L}{R A}$

$κ = \frac{1}{ρ} = \frac{L}{R A}$ where $A$

$A$ is the cross-sectional area and $\rho$

$ρ$ is the resistivity.

Substitute the values: $\kappa = \frac{0.5}{(5.55 \times 10^8) \times (7.854 \times 10^{-5})} = 1.13 \times 10^{-3} \, \text{S m}^{-1}$

$κ = \frac{0.5}{( 5.55 \times 10 ^{8} ) \times ( 7.854 \times 10 ^{-5} )} = 1.13 \times 1 0^{-3} S m^{-1}$

So, the conductivity is $1.13 \times 10^{-3} \, \text{S m}^{-1}$

$1.13 \times 1 0^{-3} S m^{-1}$ .

21. Complete the following equations:

(a) 2MnO + 5NO + 6H →
Answer:
$2MnO + 5NO + 6H^+ → Mn_2O_3 + 5/2 N_2 + 3H_2O$

$2 M n O + 5 NO + 6 H^{+} \to M n_{2} O_{3} + 5/2 N_{2} + 3 H_{2} O$

(b) Cr₂O₇²⁻ + 14H⁺ + 6e⁻ →
Answer:
$Cr_2O_7^{2-} + 14H^+ + 6e^- → 2Cr^{3+} + 7H_2O$

$C r_{2} O_{7 2-} + 14 H^{+} + 6 e^{-} \to 2 C r^{3+} + 7 H_{2} O$

22.(a) CH₃-CH-CH₃ + PCl₅ → ‘A’ → AgCN → ‘B’
Answer:

CH₃-CH-CH₃ + PCl₅ → CH₃-CH-CH₃Cl
CH₃-CH-CH₃Cl + AgCN → CH₃-CH-CH₃CN

(b) CH₃CH₂CH₂Cl + KOH (ethanol) → ‘A’ → HBr → ‘B’
Answer:

CH₃CH₂CH₂Cl + KOH (ethanol) → CH₃CH=CH₂
CH₃CH=CH₂ + HBr → CH₃CH₂CH₂Br

23. (a) Account for the following: (1+1=2)

(i) Phenol is a stronger acid than an alcohol.
Answer:
Phenol is a stronger acid than alcohol because in phenol, the negative charge after deprotonation is stabilized by resonance with the aromatic ring. In alcohol, there is no such resonance stabilization, so the negative charge on the oxygen atom is less stabilized, making it a weaker acid than phenol.

(ii) The boiling point of alcohols decreases with an increase in branching of the alkyl chain.
Answer:
The boiling point of alcohols decreases with increasing branching because branching reduces the surface area for intermolecular forces (like Van der Waals forces) to act. Less surface area means weaker intermolecular forces, resulting in a lower boiling point.

(b) (i) Write the mechanism of the following reaction:

$\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{443K} \text{CH}_2 = \text{CH}_2 + \text{H}_2\text{O}$

$CH_{3} CH_{2} OH 443 K CH_{2} = CH_{2} + H_{2} O$
Answer:
This is an example of an elimination (dehydration) reaction where ethanol is converted to ethene. The mechanism proceeds as follows:

Protonation: The alcohol (ethanol) is protonated to form an oxonium ion.
Elimination of water: The protonated alcohol loses a molecule of water, forming a carbocation at the adjacent carbon.
Formation of the double bond: A proton is abstracted from the adjacent carbon, and a double bond forms between the carbons, resulting in ethene.

(ii) Write the equation involved in the Reimer-Tiemann reaction.
Answer:
The Reimer-Tiemann reaction is the formylation of phenol using chloroform (CHCl₃) and a base (usually NaOH).

Equation: $\text{Phenol} + \text{CHCl}_3 \xrightarrow{NaOH} \text{Salicylaldehyde}$

$Phenol + CHCl_{3} N a O H Salicylaldehyde$

24. Explain briefly: (1+1=2)

(a) Carbylamine reaction
Answer:
The carbylamine reaction is the reaction between a primary amine and chloroform (CHCl₃) in the presence of a strong base (like KOH) to form an isocyanide (carbylamine). The reaction is:

$\text{RNH}_2 + \text{CHCl}_3 + \text{KOH} → \text{RNC} + \text{KCl} + 2\text{H}_2\text{O}$

$RNH_{2} + CHCl_{3} + KOH \to RNC + KCl + 2 H_{2} O$

(b) Gabriel phthalimide synthesis
Answer:
Gabriel phthalimide synthesis is a method to prepare primary amines. It involves the reaction of phthalimide with an alkyl halide in the presence of potassium hydroxide to form a phthalimide derivative, which is then hydrolyzed to give the primary amine.

Reaction: $\text{Phthalimide} + \text{RCl} + \text{KOH} → \text{R-phthalimide}$

$Phthalimide + RCl + KOH \to R-phthalimide$ Hydrolysis of the intermediate gives the primary amine: $\text{R-phthalimide} + \text{H}_2\text{O} \rightarrow \text{RNH}_2 + \text{Phthalic acid}$

$R-phthalimide + H_{2} O \to RNH_{2} + Phthalic acid$

25. (a) Write a chemical reaction to show that the open structure of D-glucose contains the straight chain.

Answer:
The open-chain structure of D-glucose can be demonstrated by its reaction with an oxidizing agent like Benedict’s reagent or Tollens’ reagent. This shows the presence of an aldehyde group (-CHO) in the open-chain form.

Reaction: $\text{D-glucose} + \text{Tollens’ reagent} → \text{Glucitol (sorbitol)} + \text{Silver mirror}$

$D-glucose + Tollens’ reagent \to Glucitol (sorbitol) + Silver mirror$

(b) What type of linkage is responsible for the formation of protein?
Answer:
The formation of proteins is due to peptide linkages (also called amide bonds) between the amino group (-NH₂) of one amino acid and the carboxyl group (-COOH) of another amino acid.

SECTION C

26. (a) Differentiate between Ideal solution and Non-ideal solution.

Answer:

Ideal solution: In an ideal solution, the intermolecular forces between different molecules are equal to the intermolecular forces between like molecules. The vapor pressure of each component follows Raoult’s Law at all concentrations. Examples include benzene-toluene mixtures.
Non-ideal solution: In a non-ideal solution, the intermolecular forces between different molecules are not equal to the intermolecular forces between like molecules. These solutions deviate from Raoult’s Law and exhibit either positive or negative deviations. Examples include acetic acid-water mixtures.

(b) 30 g of urea is dissolved in 846 g of water. Calculate the vapor pressure of water for this solution if the vapor pressure of pure water at 298 K is 23.8 mm Hg.

Answer: To calculate the vapor pressure of the solution, we will use Raoult’s Law: $P_{\text{solution}} = P_{\text{pure solvent}} \times X_{\text{solvent}}$

$P_{solution} = P_{pure solvent} \times X_{solvent}$

Where:

$P_{\text{solution}}$
$P_{\text{pure solvent}}$
$X_{\text{solvent}}$

First, calculate the mole fraction of water.

Moles of urea (solute): Molar mass of urea = 60 g/mol
Moles of urea = $\frac{30 \text{ g}}{60 \text{ g/mol}} = 0.5 \text{ mol}$

$\frac{30 g}{60 g/mol} = 0.5 mol$
Moles of water (solvent): Molar mass of water = 18 g/mol
Moles of water = $\frac{846 \text{ g}}{18 \text{ g/mol}} = 47 \text{ mol}$

$\frac{846 g}{18 g/mol} = 47 mol$
Mole fraction of water (X_water): $X_{\text{water}} = \frac{\text{moles of water}}{\text{moles of water} + \text{moles of urea}} = \frac{47}{47 + 0.5} = 0.9905$

$X_{water} = \frac{moles of water}{moles of water + moles of urea} = \frac{47}{47 + 0.5} = 0.9905$
Vapor pressure of the solution: $P_{\text{solution}} = P_{\text{pure water}} \times X_{\text{water}}$

$P_{solution} = P_{pure water} \times X_{water}$ $P_{\text{solution}} = 23.8 \, \text{mm Hg} \times 0.9905 = 23.6 \, \text{mm Hg}$

$P_{solution} = 23.8 mm Hg \times 0.9905 = 23.6 mm Hg$

27. Write the main product formed when:

(a) Methyl chloride is treated with NaI/Acetone.

Answer: The reaction involves a nucleophilic substitution (SN2) mechanism. Methyl chloride (CH₃Cl) reacts with NaI in acetone, where iodide ion (I⁻) displaces chloride (Cl⁻) to form methyl iodide (CH₃I).

Reaction: $\text{CH₃Cl} + \text{NaI} → \text{CH₃I} + \text{NaCl}$

$CH₃Cl + NaI \to CH₃I + NaCl$

(b) 2,4,6-trinitrochlorobenzene is subjected to hydrolysis.

Answer: Hydrolysis of 2,4,6-trinitrochlorobenzene will replace the chlorine atom with a hydroxyl group, forming 2,4,6-trinitrophenol (also known as picric acid).

Reaction: $\text{2,4,6-trinitrochlorobenzene} + \text{H}_2\text{O} → \text{2,4,6-trinitrophenol} + \text{HCl}$

$2,4,6-trinitrochlorobenzene + H_{2} O \to 2,4,6-trinitrophenol + HCl$

Answer: The reaction involves elimination (E2) of HCl, forming but-2-ene as the main product.

Reaction: $\text{C₄H₉Cl} + \text{KOH (alcoholic)} → \text{C₄H₈ (but-2-ene)} + \text{KCl} + \text{H₂O}$

$C₄H₉Cl + KOH (alcoholic) \to C₄H₈ (but-2-ene) + KCl + H₂O$

28. How do you convert the following: (Any three)

(a) Phenol to picric acid

Answer: Phenol can be converted to picric acid (2,4,6-trinitrophenol) by nitrating phenol with a mixture of concentrated nitric acid and sulfuric acid.

Reaction: $\text{Phenol} + 3\text{HNO}_3 \xrightarrow{\text{H}_2\text{SO}_4} \text{Picric acid (2,4,6-trinitrophenol)}$

$Phenol + 3 HNO_{3} H^{2} SO^{4} Picric acid (2,4,6-trinitrophenol)$

(b) Propanone to 2-Methylpropan-2-ol

Answer: Propanone (acetone) can be converted to 2-methylpropan-2-ol (tert-butyl alcohol) through a reaction with isobutylmagnesium chloride (Grignard reagent) followed by hydrolysis.

Reaction: $\text{CH₃COCH₃} + \text{CH₃C(Cl)CH₃} \xrightarrow{\text{Mg, Ether}} \text{2-Methylpropan-2-ol}$

$CH₃COCH₃ + CH₃C(Cl)CH₃ Mg, Ether 2-Methylpropan-2-ol$

Answer: Phenol can be converted to anisole (methyl phenyl ether) by methylation with methyl iodide (CH₃I) in the presence of a base (such as KOH).

Reaction: $\text{Phenol} + \text{CH₃I} \xrightarrow{\text{KOH}} \text{Anisole (C₆H₅OCH₃)}$

$Phenol + CH₃I KOH Anisole (C₆H₅OCH₃)$

(d) Propene to Propan-1-ol

Answer: Propene can be converted to propan-1-ol by hydroboration-oxidation. The reaction involves the addition of borane (BH₃) followed by oxidation with hydrogen peroxide (H₂O₂).

Reaction: $\text{CH₃CH=CH₂} + \text{BH₃} \xrightarrow{\text{H₂O₂}} \text{Propan-1-ol (CH₃CH₂CH₂OH)}$

$CH₃CH=CH₂ + BH₃ H₂O₂ Propan-1-ol (CH₃CH₂CH₂OH)$

29. (a) Explain why: (1+1+1=3)

(i) Carboxyl group in benzoic acid is meta directing.

Answer:
The carboxyl group (-COOH) is an electron-withdrawing group, and it deactivates the benzene ring. Due to the delocalization of the electron pair from the oxygen in the carboxyl group, it tends to pull electron density away from the ring, especially at the ortho and para positions. This makes the meta position more reactive towards electrophilic substitution reactions, which is why the carboxyl group is meta directing.

(ii) Sodium bisulfite is used for the purification of aldehydes and ketones.

Answer:
Sodium bisulfite (NaHSO₃) forms a soluble adduct with aldehydes and certain ketones, making it a useful reagent for purifying these compounds. The aldehyde or ketone reacts with sodium bisulfite to form a crystalline addition product, which can be easily separated from other impurities. Upon acid hydrolysis, the pure aldehyde or ketone is recovered.

(iii) Carboxylic acids do not give characteristic reactions of the carbonyl group.

Answer:
Carboxylic acids contain a carbonyl group, but the presence of the -OH group attached to the carbonyl carbon makes the carboxyl group behave differently compared to other carbonyl-containing compounds like aldehydes or ketones. Carboxylic acids do not react in the typical nucleophilic addition reactions (such as with nucleophiles like alcohols or amines) that are characteristic of aldehydes and ketones because the -OH group in carboxylic acids is already involved in hydrogen bonding and stabilizes the molecule, reducing the reactivity of the carbonyl group.

29. (b) An organic compound ‘A’, having the molecular formula C₃H₈O on treatment with Cu at 573 K, gives ‘B’. ‘B’ does not reduce Fehling’s solution but gives a yellow precipitate of the compound ‘C’ with I₂/NaOH. Deduce the structures of A, B, and C.

Answer:

Molecular formula of A: C₃H₈O suggests the presence of a single hydroxyl group in the compound.
Treatment of A with Cu at 573 K: When an alcohol (A) is heated with copper at high temperature (573 K), it undergoes dehydration to form an alkene. This suggests that A is propan-2-ol (isopropanol), which will undergo dehydration to form propene (B).

Structure of A: Propan-2-ol (isopropanol)
Chemical formula: C₃H₈O
Reaction:
$\text{CH₃CH(OH)CH₃} \xrightarrow{\text{Cu at 573 K}} \text{CH₂=CHCH₃}$

$CH₃CH(OH)CH₃ Cu at 573 K CH₂=CHCH₃$
Structure of B: Propene (C₃H₆)
B does not reduce Fehling’s solution:
Propene (B) is an alkene and does not have the aldehyde group necessary to reduce Fehling’s solution, confirming that B is not an aldehyde.
Yellow precipitate of C with I₂/NaOH:
The reaction of propene (B) with iodine in the presence of sodium hydroxide forms a yellow precipitate of iodoform (CHI₃), which is characteristic of methyl ketones or secondary alcohols with a methyl group adjacent to the carbonyl group. Since propene is a product of propan-2-ol, this reaction indicates the presence of a methyl ketone group.

Structure of C: Iodoform (CHI₃)

Thus, the structures are:

A: Propan-2-ol (Isopropanol)
B: Propene (C₃H₆)
C: Iodoform (CHI₃)

30. (a) What are the hydrolysis products of (i) Lactose, (ii) Maltose?

(i) Lactose:
Lactose is a disaccharide made up of glucose and galactose units. When lactose undergoes hydrolysis (in the presence of an acid or enzyme lactase), it breaks down into glucose and galactose.

Hydrolysis products of lactose: $\text{Lactose} \xrightarrow{\text{H}_2\text{O}} \text{Glucose} + \text{Galactose}$

$Lactose H^{2} O Glucose + Galactose$

(ii) Maltose:
Maltose is a disaccharide composed of two glucose units. Upon hydrolysis, maltose breaks down into two glucose molecules.

Hydrolysis products of maltose: $\text{Maltose} \xrightarrow{\text{H}_2\text{O}} \text{Glucose} + \text{Glucose}$

$Maltose H^{2} O Glucose + Glucose$

30. (b) Give the basic structural difference between starch and cellulose.

Answer:
The basic structural difference between starch and cellulose lies in the type of glycosidic bond connecting the glucose units:

Starch: Starch is made up of glucose units connected by α-1,4-glycosidic bonds. This allows starch to form a helical structure, which is flexible and easily digestible by humans.
Cellulose: Cellulose is also made up of glucose units, but they are connected by β-1,4-glycosidic bonds. This creates long, straight chains that are rigid and form strong, fibrous structures. Cellulose cannot be digested by humans due to the β-linkage.

SECTION D

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

31. The rate of reaction is concerned with decrease in concentration of
reactants or increase in the concentration of products per unit time. It can
be expressed as instantaneous rate at a particular instant of time and
average rate over a large interval of time. Mathematical representation of
rate of reaction is given by rate law. Rate constant and order of a reaction
can be determined from rate law or its integrated rate equation.

(i) What is the average rate of reaction? (1)

Answer:
The average rate of a reaction is the change in the concentration of reactants or products divided by the time interval during which the change occurs.

$\text{Average rate} = \frac{\Delta[\text{Concentration}]}{\Delta t}$

$Average rate = \frac{Δ [ Concentration ]}{Δ t}$

(ii) Write two factors that affect the rate of reaction. (1)

Answer:

Concentration of reactants: Higher concentration leads to a faster reaction.
Temperature: Higher temperature increases the reaction rate.

(iii) (1) What happens to the rate of reaction for zero order reaction? (2)

Answer:
For a zero-order reaction, the rate is independent of the concentration of reactants. The rate remains constant throughout the reaction.

(iii) (2) What is the unit of $k$
$k$ for zero order reaction? (1)

Answer:
The unit of the rate constant $k$

$k$ for a zero-order reaction is M/s (molar per second).

OR

(iii) (1) For a reaction $P + 2Q \rightarrow \text{Products}$
$Rate = k [P]^{1/2} [Q]^{1}$ . What is the order of the reaction? (2)

Answer:
The order of the reaction is the sum of the exponents in the rate law:

$\frac{1}{2} + 1 = \frac{3}{2} \text{ or } 1.5$

$\frac{1}{2} + 1 = \frac{3}{2} or 1.5$

(iii) (2) Define pseudo first order reaction with an example. (1)

Answer:
A pseudo-first-order reaction occurs when a higher-order reaction behaves like a first-order reaction due to one reactant being in large excess.
Example: The reaction between ethanol and HCl, where HCl is in excess, behaves as a pseudo-first-order reaction with respect to ethanol.

32. In coordination compounds, metals show two types of linkages, primary
and secondary. Primary valencies are ionisable and are satisfied by
negatively charged ions. Secondary valencies are non-ionisable and are
satisfied by neutral or negative ions having lone pair of electrons. Primary
valencies are non-directional while secondary valencies decide the shape
of the complexes.

(i) If PtCl₂. 2NH₃ does not react with AgNO₃, what will be its formula? (1)

Answer:
If PtCl₂. 2NH₃ does not react with AgNO₃, it indicates that the chloride ions are not ionizable (they are bound in a non-ionizable way). This suggests the formula is:

$\text{[Pt(NH₃)₂Cl₂]}$

$[Pt(NH₃)₂Cl₂]$

Here, the chloride ions are part of the coordination sphere and do not dissociate in solution.

(ii) What is the secondary valency of [Co(en)₃]³⁺? (1)

Answer:
In the complex [Co(en)₃]³⁺, en (ethylenediamine) is a bidentate ligand, meaning it can form two bonds with the central metal ion. Since there are three en ligands, each contributing two bonds, the secondary valency of Co is:

$\text{Secondary valency} = 3 \times 2 = 6$

$Secondary valency = 3 \times 2 = 6$

Thus, the secondary valency of [Co(en)₃]³⁺ is 6.

(iii) (1) Write the formula of Iron(III)hexacyanidoferrate(II). (1)

Answer:
The formula for Iron(III)hexacyanidoferrate(II) is:

$\text{[Fe(CN)₆]³⁻[Fe(CN)₆]⁴⁻}$

$[Fe(CN)₆]³⁻[Fe(CN)₆]⁴⁻$

Here, the two iron ions are in different oxidation states: Fe(III) and Fe(II).

(iii) (2) Write the IUPAC name of [Co(NH₃)₅Cl]Cl₂. (1)

Answer:
The IUPAC name of [Co(NH₃)₅Cl]Cl₂ is:

Pentaamminechloridocobalt(III) chloride

OR

(iii) Write the hybridization and magnetic behavior of [Ni(CN)₄]²⁻. (2)

Answer:

Hybridization of [Ni(CN)₄]²⁻: The central metal ion Ni²⁺ has an electronic configuration of 3d⁸. With four CN⁻ ligands, the hybridization is sp³ (since it forms four sigma bonds with the ligands).
Magnetic behavior of [Ni(CN)₄]²⁻: Since CN⁻ is a strong field ligand and causes pairing of electrons in Ni²⁺ (which has 2 unpaired electrons in the d orbitals), [Ni(CN)₄]²⁻ is low-spin and diamagnetic (having no unpaired electrons).

SECTION E

33. (a)

(i) State Kohlrausch’s law of independent migration of ions. Write an expression for the limiting molar conductivity of acetic acid according to Kohlrausch’s law. (2)

Answer:
Kohlrausch’s Law of Independent Migration of Ions states that at infinite dilution, the molar conductivity of an electrolyte is the sum of the contributions from the individual ions, and these contributions are independent of the type of electrolyte and only depend on the ions themselves.

Mathematically:

$\Lambda_m^\circ = \lambda_+^\circ + \lambda_-^\circ$

$Λ_{m \circ} = λ_{+ \circ} + λ_{- \circ}$

Where:

$\Lambda_m^\circ$
$\lambda_+^\circ$

For acetic acid ( $CH_3COOH$

$C H_{3} COO H$ ):

$\Lambda_m^\circ(\text{CH}_3\text{COOH}) = \lambda_{\text{H}^+}^\circ + \lambda_{\text{CH}_3\text{COO}^-}^\circ$

$Λ_{m \circ} (CH_{3} COOH) = λ_{H + \circ} + λ_{CH 3 COO - \circ}$

(ii) Calculate the maximum work and log Kc for the given reaction at 298 K:

$\text{Ni (s)} + 2\text{Ag}^+ (\text{aq}) \rightleftharpoons \text{Ni}^{2+} (\text{aq}) + 2\text{Ag (s)}$

$Ni (s) + 2 Ag^{+} (aq) ⇌ Ni^{2+} (aq) + 2 Ag (s)$

Given:
$E_{\text{Ni}^{2+}/\text{Ni}} = 0.25 \, \text{V}, \, E_{\text{Ag}^+/ \text{Ag}} = +0.80 \, \text{V}$

$E_{Ni 2+ / Ni} = 0.25 V, E_{Ag + / Ag} = + 0.80 V$
$F = 96500 \, \text{C/mol}$

$F = 96500 C/mol$

Answer:
The cell potential for the reaction can be calculated using the Nernst equation:

$E_{\text{cell}} = E_{\text{cathode}} – E_{\text{anode}}$

$E_{cell} = E_{cathode} - E_{anode}$

Here, Ag $^+$

$^{+}$ /Ag is the cathode and Ni $^2+$

$^{2} +$ /Ni is the anode.

$E_{\text{cell}} = 0.80 \, \text{V} – 0.25 \, \text{V} = 0.55 \, \text{V}$

$E_{cell} = 0.80 V - 0.25 V = 0.55 V$

Next, calculate the maximum work done ( $W_{\text{max}}$

$W_{max}$ ):

$W_{\text{max}} = -nFE_{\text{cell}}$

$W_{max} = - n F E_{cell}$

Where:

$n = 2$
$F = 96500 \, \text{C/mol}$
$E_{\text{cell}} = 0.55 \, \text{V}$

Thus,

$W_{\text{max}} = -2 \times 96500 \, \text{C/mol} \times 0.55 \, \text{V} = -106150 \, \text{J/mol} = -106.15 \, \text{kJ/mol}$

$W_{max} = - 2 \times 96500 C/mol \times 0.55 V = - 106150 J/mol = - 106.15 kJ/mol$

Now, calculate $\log K_c$

$lo g K_{c}$ using the relation:

$\Delta G^\circ = -nFE_{\text{cell}}$

$Δ G^{\circ} = - n F E_{cell}$ $\Delta G^\circ = -RT \ln K_c$

$Δ G^{\circ} = - RT ln K_{c}$

At 298 K, $R = 8.314 \, \text{J/mol·K}$

$R=8.314J/mol\cdotpK$ , and $\Delta G^\circ = W_{\text{max}}$

$Δ G^{\circ} = W_{max}$ , so:

$-106150 = -2 \times 96500 \times \ln K_c$

$- 106150 = - 2 \times 96500 \times ln K_{c}$ $\ln K_c = \frac{106150}{2 \times 96500} = 0.549$

$ln K_{c} = \frac{106150}{2 \times 96500} = 0.549$ $K_c = 10^{0.549} = 3.55$

$K_{c} = 1 0^{0.549} = 3.55$

Thus,
$\log K_c = 0.549$

$lo g K_{c} = 0.549$ .

OR

(b)

(i) State Faraday’s law of electrolysis. How much charge, in terms of Faraday, is required for the reduction of 1 mol Cu²⁺ to Cu? (2)

Answer:
Faraday’s Law of Electrolysis states that the amount of substance deposited or liberated at an electrode during electrolysis is directly proportional to the quantity of electric charge passed through the electrolyte.

$\text{Mass of substance} \propto \text{Charge} \quad \text{or} \quad m = \frac{M}{n} \times Q$

$Mass of substance \propto Charge or m = \frac{M}{n} \times Q$

Where:

For the reduction of 1 mole of $\text{Cu}^{2+}$

$Cu^{2+}$ to Cu, the number of electrons involved is $n = 2$

$n = 2$ .
The charge required for the reduction of 1 mole of $\text{Cu}^{2+}$

$Cu^{2+}$ is equal to 2 Faradays (since 1 Faraday = 96500 C/mol).

Thus, the charge required is 2 Faradays.

(ii) Calculate the emf of the following cell at 298 K:

$\text{Mg (s)} | \text{Mg}^{2+} (0.1 M) || \text{Cu}^{2+} (0.1 M) | \text{Cu (s)}$

$Mg (s) ∣ Mg^{2+} (0.1 M) ∣∣ Cu^{2+} (0.1 M) ∣ Cu (s)$

Given:
$E_{\text{cell}} = +2.71 \, \text{V}, \, 1 \, F = 96500 \, \text{C/mol}, \, \log_{10} = 1$

$E_{cell} = + 2.71 V, 1 F = 96500 C/mol, lo g_{10} = 1$

Answer:
The cell potential can be calculated using the Nernst equation:

$E_{\text{cell}} = E^\circ_{\text{cell}} – \frac{0.0592}{n} \log Q$

$E_{cell} = E_{cell \circ} - \frac{0.0592}{n} lo g Q$

For the given reaction, $n = 2$

$n = 2$ (since 2 moles of electrons are involved) and $Q = \frac{[\text{Mg}^{2+}]^2}{[\text{Cu}^{2+}]^2} = 1$

$Q = \frac{[ Mg ^{2+} ] ^{2}}{[ Cu ^{2+} ] ^{2}} = 1$ , since both concentrations are 0.1 M. Therefore, the equation becomes:

$E_{\text{cell}} = 2.71 – \frac{0.0592}{2} \log(1)$

$E_{cell} = 2.71 - \frac{0.0592}{2} lo g (1)$

Since $\log(1) = 0$

$lo g (1) = 0$ , the emf remains:

$E_{\text{cell}} = 2.71 \, \text{V}$

$E_{cell} = 2.71 V$

Thus, the emf of the cell is 2.71 V.

34. Assign reason for each of the following:

(i) Manganese exhibits the highest oxidation state of +7 among the 3d series of transition elements. (1)

Answer:
Manganese exhibits the highest oxidation state of +7 because it has a half-filled d-orbital configuration ( $3d^5$

$3 d^{5}$ ) when it reaches the +7 oxidation state. This allows for a stable electronic configuration. Manganese is also relatively small in size, and its high ionization energy allows it to lose 7 electrons easily, giving it the +7 oxidation state.

(ii) Transition metals and their compounds are generally found to be good catalysts in chemical reactions. (1)

Answer:
Transition metals are good catalysts because they have multiple oxidation states and can easily donate and accept electrons. This allows them to form temporary bonds with reactants, lowering the activation energy of the reaction. Additionally, their d-orbitals can participate in bonding and stabilize the transition state.

(iii) Cr²⁺ is reducing in nature while Mn³⁺ is an oxidizing agent with the same d-orbital configuration (d⁴). (1)

Answer:
Although both Cr²⁺ and Mn³⁺ have the same d⁴ configuration, the difference in their behavior arises from their position in the periodic table. Cr²⁺ is a reducing agent because it can easily lose an electron to reach the stable Cr³⁺ state. On the other hand, Mn³⁺ is an oxidizing agent because it is more stable in the +3 oxidation state and can readily accept an electron to convert to Mn²⁺.

(iv) Zn has the lowest enthalpy of atomization. (1)

Answer:
Zinc has the lowest enthalpy of atomization among the transition metals because it has a completely filled d-orbital (3d¹⁰), resulting in fewer metallic bonds. This makes it easier for zinc atoms to be separated into individual atoms, reducing the energy required for atomization.

(v) Cu⁺ is unstable in an aqueous solution. (1)

Answer:
Cu⁺ is unstable in aqueous solution because it is easily oxidized to Cu²⁺ due to the lack of a stable electronic configuration in the Cu⁺ state. The higher stability of Cu²⁺ compared to Cu⁺ leads to the disproportionation of Cu⁺ into Cu²⁺ and Cu (metal), making Cu⁺ unstable in solution.

CBSE Class 12 – Chemistry Question Paper 2023

SECTION A

21. Complete the following equations:

23. (a) Account for the following: (1+1=2)

24. Explain briefly: (1+1=2)

25. (a) Write a chemical reaction to show that the open structure of D-glucose contains the straight chain.

SECTION C

26. (a) Differentiate between Ideal solution and Non-ideal solution.

27. Write the main product formed when:

28. How do you convert the following: (Any three)

29. (a) Explain why: (1+1+1=3)

29. (b) An organic compound ‘A’, having the molecular formula C₃H₈O on treatment with Cu at 573 K, gives ‘B’. ‘B’ does not reduce Fehling’s solution but gives a yellow precipitate of the compound ‘C’ with I₂/NaOH. Deduce the structures of A, B, and C.

30. (a) What are the hydrolysis products of (i) Lactose, (ii) Maltose?

30. (b) Give the basic structural difference between starch and cellulose.

(i) What is the average rate of reaction? (1)

(ii) Write two factors that affect the rate of reaction. (1)

(iii) (1) What happens to the rate of reaction for zero order reaction? (2)

(iii) (2) What is the unit of kk k for zero order reaction? (1)

OR

(iii) (1) For a reaction P+2Q→ProductsP + 2Q \rightarrow \text{Products} P+2Q→Products, the rate law is given by Rate=k[P]1/2[Q]1\text{Rate} = k[P]^{1/2}[Q]^1 Rate=k[P]1/2[Q]1. What is the order of the reaction? (2)

(iii) (2) Define pseudo first order reaction with an example. (1)

(i) If PtCl₂. 2NH₃ does not react with AgNO₃, what will be its formula? (1)

(ii) What is the secondary valency of [Co(en)₃]³⁺? (1)

(iii) (1) Write the formula of Iron(III)hexacyanidoferrate(II). (1)

(iii) (2) Write the IUPAC name of [Co(NH₃)₅Cl]Cl₂. (1)

OR

(iii) Write the hybridization and magnetic behavior of [Ni(CN)₄]²⁻. (2)

SECTION E

33. (a)

(i) State Kohlrausch’s law of independent migration of ions. Write an expression for the limiting molar conductivity of acetic acid according to Kohlrausch’s law. (2)

(ii) Calculate the maximum work and log Kc for the given reaction at 298 K:

OR

(b)

(i) State Faraday’s law of electrolysis. How much charge, in terms of Faraday, is required for the reduction of 1 mol Cu²⁺ to Cu? (2)

(ii) Calculate the emf of the following cell at 298 K:

34. Assign reason for each of the following:

(i) Manganese exhibits the highest oxidation state of +7 among the 3d series of transition elements. (1)

(ii) Transition metals and their compounds are generally found to be good catalysts in chemical reactions. (1)

(iii) Cr²⁺ is reducing in nature while Mn³⁺ is an oxidizing agent with the same d-orbital configuration (d⁴). (1)

(iv) Zn has the lowest enthalpy of atomization. (1)

(v) Cu⁺ is unstable in an aqueous solution. (1)

(iii) (2) What is the unit of $k$
$k$ for zero order reaction? (1)

(iii) (1) For a reaction $P + 2Q \rightarrow \text{Products}$
$Rate = k [P]^{1/2} [Q]^{1}$ . What is the order of the reaction? (2)