ISC Class 12 Chemistry Board Exam Question Paper 2025 with Solutions

ISC Class 12 Chemistry Board Exam Question Paper with Solutions

SECTION - A

Question 1

(A) Fill in the blanks by choosing the appropriate word(s) from those given in the brackets: [4×1]
[+2, ethane, tetrahedral, square planar, zero dry cell, nickel-cadmium cell, propane, Wolff-Kishner, Stephen, completely filled, incompletely filled, paramagnetic, diamagnetic]

(i) _________ is an example of a primary cell but _________ is an example of a secondary cell [1 Mark]

Answer: Dry cell, nickel-cadmium cell

Teacher's Note:
a) Primary cells cannot be recharged easily whereas secondary cells are rechargeable.
b) Remember that Leclanche cell is a standard dry cell.

(ii) The complex compound [Ni(CO)₄] is _________ in shape and nickel is in _________ oxidation state in this complex compound. [1 Mark]

Answer: tetrahedral, zero

Teacher's Note:
a) CO is a strong field ligand, leading to \( sp^3 \) hybridisation for tetrahedral geometry.
b) Metal carbonyls generally exhibit zero oxidation state for the central metal atom.

(iii) When acetaldehyde is treated with hydrazine and KOH in a high boiling solvent glycol, _________ is formed and the reaction is known as _________ reduction. [1 Mark]

Answer: ethane, Wolff-Kishner

Teacher's Note:
a) Wolff-Kishner reduction reduces aldehydes and ketones to hydrocarbons using hydrazine and a strong base.
b) Acetaldehyde (\( CH_3CHO \)) gives ethane upon complete reduction of the carbonyl group.

(iv) The transition metal ions having _________ d-orbitals are colourless and _________ in nature. [1 Mark]

Answer: completely filled, diamagnetic

Teacher's Note:
a) Completely filled d-orbitals (\( d^{10} \)) prevent d-d transitions, making ions colourless.
b) Absence of unpaired electrons results in diamagnetic properties.

(B) Select and write the correct alternative from the choices given below. [7×1]

(i) Which one of the following can produce the foul smelling compound methyl isocyanide in presence of alcoholic KOH? [1 Mark]
(a) Chloroform and aniline
(b) Chloroform and methanol
(c) Chloroform and dimethyl amine
(d) Chloroform and methyl amine

Answer: (d) Chloroform and methyl amine

Primary aliphatic or aromatic amines react with chloroform and alcoholic KOH to form foul-smelling isocyanides (carbylamine test).

Teacher's Note:
a) Methyl amine is a primary amine which gives the positive carbylamine test.
b) Aniline gives phenyl isocyanide, whereas methyl amine gives methyl isocyanide.

(ii) The osmotic pressure of a solution: [1 Mark]
(P) increases with an increase in number of moles of solute.
(Q) decreases with an increase in number of moles of solute.
(R) increases at a higher temperature.
(S) is dependent on the nature of solute.
Which one of the following is correct?
(a) Only (P) and (Q) are correct
(b) Only (P) and (R) are correct
(c) Only (P) and (S) are correct
(d) Only (Q) and (S) are correct

Answer: (b) Only (P) and (R) are correct

Osmotic pressure formula is \( \pi = CRT = \frac{n}{V} RT \), showing direct proportionality to moles (\( n \)) and temperature (\( T \)).

Teacher's Note:
a) Osmotic pressure is a colligative property depending on the number of particles, not their chemical nature.
b) Higher temperature increases kinetic energy and thereby osmotic pressure.

(iii) Which one of the following alcohols is the strongest acid? [1 Mark]
(a) Phenol
(b) Methanol
(c) Ethanol
(d) t-butyl alcohol

Answer: (a) Phenol

Phenol is a stronger acid than aliphatic alcohols due to resonance stabilization of the phenoxide ion.

Teacher's Note:
a) Aliphatic alcohols have electron-releasing alkyl groups that destabilize the alkoxide ion.
b) Phenoxide ion charge is delocalized over the benzene ring via resonance.

(iv) Which one of the following does NOT form a silver mirror on heating with Tollen's reagent? [1 Mark]
(a) Glucose
(b) Fructose
(c) Sucrose
(d) Lactose

Answer: (c) Sucrose

Sucrose is a non-reducing disaccharide as its reducing groups (glycosidic hydroxyls of glucose and fructose) are involved in bond formation.

Teacher's Note:
a) Reducing sugars like glucose, fructose, and lactose reduce Tollen's reagent.
b) Sucrose lacks free aldehyde or ketone groups in solution.

(v) The coordination number and the oxidation state of the central metal 'D' in the complex [D(en)₂(H₂O)₂]Cl₃ are: [1 Mark]
(a) 6 and 2, respectively
(b) 4 and 2, respectively
(c) 3 and 6, respectively
(d) 6 and 3, respectively

Answer: (d) 6 and 3, respectively

Ethylenediamine (en) is bidentate (contributes 2 coords each, total 4) and \( H_2O \) is monodentate (2 coords), yielding coordination number 6. Oxidation state is +3 balanced by three chloride ions.

Teacher's Note:
a) Coordination number = \( (2 \times 2) + (2 \times 1) = 6 \).
b) Let metal oxidation state be \( x \); \( x + 2(0) + 2(0) = +3 \), so \( x = +3 \).

(vi) Given below are two statements marked Assertion and Reason. Read the two statements carefully and select the correct option.
Assertion: The process of halogenation of benzene takes place in the presence of anhydrous \( FeCl_3 \).
Reason: Anhydrous \( FeCl_3 \) prepares nucleophile to attack the benzene ring. [1 Mark]
(a) Both Assertion and Reason are true and Reason is the correct explanation for Assertion.
(b) Both Assertion and Reason are true but Reason is not the correct explanation for Assertion.
(c) Assertion is true and Reason is false.
(d) Both Assertion and Reason are false.

Answer: (c) Assertion is true and Reason is false.

Anhydrous \( FeCl_3 \) acts as a Lewis acid to generate an electrophile (\( Cl^+ \)), not a nucleophile.

Teacher's Note:
a) Electrophilic aromatic substitution requires an electrophile generated by a Lewis acid catalyst.
b) Reason states it prepares a nucleophile, which is chemically incorrect.

(vii) Given below are two statements marked Assertion and Reason. Read the two statements carefully and select the correct option.
Assertion: The Zr-Hf pair of elements has the same value of atomic radii though Zr and Hf are placed in different periods in the periodic table.
Reason: The lanthanoid contraction prevents the expected increase in atomic radii of Hf. [1 Mark]
(a) Both Assertion and Reason are true and Reason is the correct explanation for Assertion.
(b) Both Assertion and Reason are true but Reason is not the correct explanation for Assertion.
(c) Assertion is true and Reason is false.
(d) Both Assertion and Reason are false.

Answer: (a) Both Assertion and Reason are true and Reason is the correct explanation for Assertion.

Due to the filling of 4f orbitals prior to 5d transition series, imperfect shielding causes lanthanoid contraction, making Zr and Hf radii nearly identical.

Teacher's Note:
a) Lanthanoid contraction counteracts the normal size increase down a group.
b) This results in similar chemical and physical properties for second and third row transition metal pairs.

(C) Read the passage carefully and answer the questions that follow. [3×1]

The rate of a reaction depends on the concentration of reactants. The rate law for a hypothetical reaction \( aA + bB \rightarrow cC + dD \) is rate = \( k[A]^x[B]^y \) where \( x \) and \( y \) are calculated experimentally and known as order of reaction. In most cases, the mechanism of a reaction is not straight forward but broken down in simple elementary steps. These steps represent the progress of overall reaction at the molecular level.

(i) What will be the order of reaction if the unit of \( k \) is \( mol^{-2}\ L^2s^{-1} \)? [1 Mark]

Answer: Third order reaction

Teacher's Note:
a) General unit for rate constant is \( mol^{1-n}\ L^{n-1}\ s^{-1} \).
b) Equating \( 1-n = -2 \) gives \( n = 3 \).

(ii) State any one difference between order of reaction and molecularity of reaction. [1 Mark]

Answer:

Teacher's Note:
a) Order is an experimental quantity.
b) Molecularity is a theoretical concept applicable to elementary steps.

(iii) For the reaction \( 2A \rightarrow B + C \) the rate law is rate = \( k[A]^{3/2} \). What is the order and molecularity of the reaction? [1 Mark]

Answer: Order = \( \frac{3}{2} \), Molecularity = 2

Teacher's Note:
a) Order is the sum of powers of concentration terms in the rate law equation.
b) Molecularity refers to the number of reacting species colliding simultaneously in an elementary step (assumed here as a bimolecular elementary reaction).

SECTION - B

Question 2 [2 Marks]

Calculate the number of coulombs required to electroplate \( 4.75\ g \) of aluminium when electrode reaction is \( Al^{3+} + 3e^- \rightarrow Al \)
(Given: Atomic weight of \( Al = 27\ g\ mol^{-1} \), 1 Faraday = \( 96500\ coulombs \))

Answer:
1. According to the reaction, 1 mole of \( Al \) (27 g) requires 3 moles of electrons or \( 3 \times 96500\ C \).
2. Moles of \( Al \) in \( 4.75\ g = \frac{4.75}{27} = 0.1759\ mol \).
3. Total charge required = \( 0.1759 \times 3 \times 96500 = 50923.05\ C \).

Teacher's Note:
a) Use Faraday's first law of electrolysis: \( Q = n \times F \times z \).
b) Ensure proper substitution of equivalent mass or stoichiometric electron moles.

Question 3 [2 Marks]

An organic compound [A] has molecular formula \( C_7H_6O_2 \). When compound [A] is treated with \( SOCl_2 \), it yields compound [B]. On heating with \( NH_3 \), compound [B] forms compound [C]. Compound [C] forms compound [D] on reaction with \( Br_2/KOH \). Compound [D] responds to carbylamine test. Identify compounds [A], [B], [C] and [D].

Answer:
[A] = Benzoic acid (\( C_6H_5COOH \))
[B] = Benzoyl chloride (\( C_6H_5COCl \))
[C] = Benzamide (\( C_6H_5CONH_2 \))
[D] = Aniline (\( C_6H_5NH_2 \))

Teacher's Note:
a) Benzoic acid reacts with \( SOCl_2 \) to substitute -OH with -Cl.
b) Hofmann bromamide degradation converts an amide into a primary amine with one carbon less, which gives a positive carbylamine test.

Question 4 [2 Marks]

In the reaction \( 2NO + O_2 \rightarrow 2NO_2 \), the rate law is rate = \( k[NO][O_2]^2 \).
(i) How will the rate of reaction change if [NO] concentration is doubled and [O_2] concentration is halved at the same time?
(ii) Write the order of reaction if [NO] concentration is in large excess.

Answer:
(i) Initial rate \( r_0 = k[NO][O_2]^2 \).
New rate \( r_n = k \times (2[NO]) \times \left(\frac{1}{2}[O_2]\right)^2 = k \times 2[NO] \times \frac{1}{4}[O_2]^2 = \frac{1}{2} r_0 \).
The rate of the reaction will become half of the initial rate.
(ii) When [NO] is in large excess, its concentration does not change significantly during the reaction. Therefore, it can be considered constant, making the reaction a pseudo second order reaction with respect to \( O_2 \) (overall order = 2).

Teacher's Note:
a) Substitute concentration multipliers directly into the rate expression.
b) Excess reactant concentration is absorbed into the pseudo rate constant.

Question 5 [2 Marks]

When an organic compound [A] having molecular formula \( C_4H_9Br \) is treated with aqueous KOH, the rate of reaction depends on concentration of compound [A] only. But when compound [B], with the same molecular formula, reacts with aqueous KOH, the rate of reaction depends on the concentration of compound [B] as well as of KOH. Compound [B] is a structural isomer of [A].
Identify compounds [A] and [B].

Answer:
[A] = tert-butyl bromide (2-bromo-2-methylpropane), as it undergoes \( S_N1 \) mechanism.
[B] = n-butyl bromide (1-bromobutane), as it undergoes \( S_N2 \) mechanism.

Teacher's Note:
a) Unimolecular nucleophilic substitution (\( S_N1 \)) rate depends only on substrate concentration.
b) Bimolecular nucleophilic substitution (\( S_N2 \)) rate depends on both substrate and nucleophile concentrations.

Question 6 [2 Marks]

Write a balanced chemical equation for each of the following:
(i) Ethyl cyanide is reduced with \( LiAlH_4 \).
(ii) Aniline is treated with bromine water.

Answer:
(i) \( CH_3CH_2CN + 4[H] \xrightarrow{LiAlH_4} CH_3CH_2CH_2NH_2 \)
(ii) \( C_6H_5NH_2 + 3Br_2 \rightarrow C_6H_2Br_3NH_2 + 3HBr \)

Teacher's Note:
a) Reduction of cyanides with \( LiAlH_4 \) yields primary amines.
b) Aniline undergoes rapid tribromination at 2, 4, 6 positions with bromine water to give a white precipitate.

Question 7 [2 Marks]

While conducting an experiment on coordination compounds, Shirin observed a white precipitate when \( AgNO_3 \) solution was added to the aqueous solution of the complex [Co(NH_3)_5(EN)SO_4]Cl.
[Note: The formula is printed as [Co(NH_3)_5(EN)SO_4]Cl, where EN stands for ethylenediamine; treating it as a typo in original paper text or interpreting standard coordination formula, let's follow the standard key structure].
How can the ionisation isomer of this complex be detected by a chemical test? Write the structure and the IUPAC name of the ionisation isomer.

Answer:
The ionisation isomer is \( [Co(NH_3)_5(en)Cl]SO_4 \).
Detection: This ionisation isomer generates \( SO_4^{2-} \) ions in solution, which can be tested by adding Barium chloride solution to give a thick white precipitate of \( BaSO_4 \).

Teacher's Note:
a) Ionisation isomers yield different ions in solution.
b) IUPAC nomenclature: Pentaammine(ethylenediamine)chlorocobalt(III) sulphate.

Question 8 [2 Marks]

(i) Arrange the following in the increasing order of their Acidic strength:
\( CH_3CH_2OH, CF_3CH_2OH, CCl_3CH_2OH \)
(ii) Arrange the following in the increasing order of their Boiling points:
\( CH_3CH_2OH, CH_3OH, CH_3CH_2Cl, CH_3OH \)
[Note: The paper repeats \( CH_3OH \); let's arrange the distinct given chemical species: \( CH_3CH_2OH, CH_3OH, CH_3CH_2Cl \)]

Answer:
(i) \( CH_3CH_2OH \lt CCl_3CH_2OH \lt CF_3CH_2OH \)
(ii) \( CH_3CH_2Cl \lt CH_3OH \lt CH_3CH_2OH \)

Teacher's Note:
a) Strong -I effect groups like \( -CF_3 \) and \( -CCl_3 \) disperse negative charge on conjugate base, enhancing acidity.
b) Alcohols have stronger intermolecular hydrogen bonding than chloroalkanes, raising boiling points.

Question 9 [2 Marks]

(i) Identify the compounds [A], [B], [C] and [D] in the following reaction:
\( CH_3COOH + PCl_5 \rightarrow [A] \xrightarrow[Boiling\ xylene]{H_2,\ Pd/BaSO_4} [B] \xrightarrow[|O|]{K_2Cr_2O_7 + H_2SO_4} [C] \xrightarrow[Ca(OH)_2,\ heat]{} [D] $
OR
(ii) Write chemical equations to convert the following:
(a) Benzaldehyde to benzene
(b) Benzoic acid to benzaldehyde

Answer:
(i) [A] = Acetyl chloride (\( CH_3COCl \))
[B] = Acetaldehyde (\( CH_3CHO \))
[C] = Acetic acid (\( CH_3COOH \))
[D] = Acetone (\( CH_3COCH_3 \))
OR
(ii) (a) \( C_6H_5CHO \xrightarrow{K_2Cr_2O_7/H^+} C_6H_5COOH \xrightarrow{NaOH} C_6H_5COONa \xrightarrow{CaO/NaOH,\ heat} C_6H_6 \)
(b) \( C_6H_5COOH + PCl_5 \rightarrow C_6H_5COCl \xrightarrow[Boiling\ xylene]{H_2,\ Pd/BaSO_4} C_6H_5CHO \)

Teacher's Note:
a) Rosenmund reduction converts acid chlorides to aldehydes.
b) Decarboxylation of sodium benzoate with soda-lime gives benzene.

Question 10 [2 Marks]

Calculate the \( E^0_{cell} \) of the following if the cell potential (\( E_{cell} \)) is 0.59 V.
(Given: \( Ni/Ni^{2+}\ (0.1M)\ ||\ Cu^{2+}\ (0.01\ M)/Cu \))

Answer:
Cell reaction: \( Ni + Cu^{2+} \rightarrow Ni^{2+} + Cu \)
Using Nernst equation:
\( E_{cell} = E^0_{cell} - \frac{0.0591}{n} \log \frac{[Ni^{2+}]}{[Cu^{2+}]} \)
\( 0.59 = E^0_{cell} - \frac{0.0591}{2} \log \frac{0.1}{0.01} \)
\( 0.59 = E^0_{cell} - 0.02955 \log(10) \)
\( E^0_{cell} = 0.59 + 0.02955 = 0.6195\ V \)

Teacher's Note:
a) Number of electrons transferred \( n = 2 \).
b) Substitute concentrations correctly into the logarithmic reaction quotient term.

Question 11 [2 Marks]

Write the chemical test to distinguish between each of the following pairs of compounds.
(i) Ethanol and propan-1-al
(ii) Phenol and benzoic acid

Answer:
(i) Propan-1-al gives a silver mirror with Tollen's reagent, whereas ethanol does not respond to this test.

(ii) Benzoic acid gives brisk effervescence with the evolution of \( CO_2 \) gas on reaction with sodium bicarbonate, while phenol does not respond to this test.

Teacher's Note:
a) Aldehydes can be distinguished from alcohols using Tollen's or Fehling's reagents.
b) Carboxylic acids are stronger acids than phenols and liberate carbon dioxide from sodium bicarbonate.

SECTION - C

Question 12 [3 Marks]

Identify the compounds [A], [B] and [C] in each of the following reactions:
(i) \( C_2H_5Br \xrightarrow{KCN} [A] \xrightarrow[|4H|]{LiAlH_4} [B] \xrightarrow[0^{\circ}-5^{\circ}C]{HNO_2} [C] \)
(ii) \( C_2H_5CONH_2 \xrightarrow{Br_2/KOH} [A] \xrightarrow{CHCl_3/KOH} [B] \xrightarrow[Na/C_2H_5OH]{} [C] \)

Answer:
(i) [A] = \( C_2H_5CN \)
[B] = \( C_2H_5CH_2NH_2 \) (or \( C_3H_7NH_2 \))
[C] = \( C_3H_7CH_2OH \) (or \( C_3H_7OH \)-propyl alcohol)
(ii) [A] = \( C_2H_5NH_2 \)
[B] = \( C_2H_5NC \)
[C] = \( C_2H_5NHCH_3 \)

Teacher's Note:
a) Reduction of alkyl cyanides with \( LiAlH_4 \) yields primary amines.
b) Carbylamine reaction converts ethylamine into ethyl isocyanide, which upon reduction yields secondary methyl-ethylamine.

Question 13 [3 Marks]

The scientist van't Hoff introduced a factor (\( i \)) to account for the extent of association or dissociation of solutes. It is mathematically expressed as:
\( i = \frac{\text{normal molecular mass}}{\text{experimental molecular mass}} \)
In case of association, \( i \lt 1 \) and in case of dissociation \( i \gt 1 \).
(i) In the calculation of molecular mass of \( K_4[Fe(CN)_6] \) by using a colligative property, what will be the value of van't Hoff factor if the solute is \( 25\% \) dissociated?
(ii) Find the value of van't Hoff factor for a dilute aqueous solution of benzoic acid in water when it is completely associated to form a dimer.

Answer:
(i) \( K_4[Fe(CN)_6] \rightarrow 4K^+ + [Fe(CN)_6]^{4-} \). Total ions \( n = 5 \).
Degree of dissociation \( \alpha = 0.25 \).
\( i = 1 + \alpha(n - 1) = 1 + 0.25(5 - 1) = 1 + 0.25(4) = 1 + 1 = 2 \).
(ii) For dimer association (\( n = 2 \)), complete association (\( \alpha = 1 \)):
\( i = 1 + \alpha\left(\frac{1}{n} - 1\right) = 1 + 1\left(\frac{1}{2} - 1\right) = 1 - 0.5 = 0.5 \).

Teacher's Note:
a) Use the standard formula \( i = 1 + \alpha(n - 1) \) for dissociation.
b) For association into dimers, \( i = 1 - \frac{\alpha}{2} \), which gives 0.5 for complete association.

Question 14 [3 Marks]

Write chemical equations to illustrate the following name reactions:
(i) Finkelstein reaction
(ii) Williamson's synthesis
(iii) Reimer-Tiemann reaction

Answer:
(i) \( CH_3CH_2Cl + NaI \xrightarrow{acetone} CH_3CH_2I + NaCl \)
(ii) \( RCH_2ONa + R'X \rightarrow RCH_2 - O - R' + NaX \)
(iii) Phenol reacts with chloroform and aqueous NaOH followed by acidification to form salicylaldehyde:
\( C_6H_5OH + CHCl_3 + 3NaOH \rightarrow C_6H_4(OH)CHO + 3NaCl + 2H_2O \)

Teacher's Note:
a) Finkelstein reaction is a halogen exchange method to prepare alkyl iodides.
b) Reimer-Tiemann introduces an aldehyde group at the ortho-position of phenol.

Question 15 [3 Marks]

According to Crystal-Field Theory, the electronic configuration of complex compound [A] is \( t_{2g}^4 e_g^2 \) and that of complex compound [B] is \( t_{2g}^6 e_g^0 \).
(i) Which of the two complex compounds, [A] or [B], is a low spin complex?
(ii) Write the number of unpaired electrons in complex compounds [A] and [B].
(iii) Does complex [A] have strong field ligands or weak field ligands? Give a reason.

Answer:
(i) Complex [B] is a low spin complex.
(ii) Complex [A] has 4 unpaired electrons; Complex [B] has 0 unpaired electrons.
(iii) Complex [A] has weak field ligands because the energy difference \( \Delta \) between \( t_{2g} \) and \( e_g \) is small, allowing electrons to jump to higher orbitals rather than pairing up (high spin complex).

Teacher's Note:
a) Low spin complexes form when pairing energy is greater than crystal field splitting energy (\( \Delta_o \)).
b) Count unpaired electrons by looking at individual orbital occupations.

Question 16 [3 Marks]

(i) Answer the following questions.
(a) By referring to electrochemical series, how can anode and cathode half cells be identified in a galvanic cell?
(b) What is the role of salt bridge in a galvanic cell?
(c) Write an advantage and a disadvantage of a fuel cell.
OR
(ii) Answer the following questions.
(a) Specific conductance of a solution decreases upon dilution. Why?
(b) The emf of a cell should be positive for a spontaneous reaction. Give a reason.
(c) Name the type of cell in which reaction occurs only in one direction and cannot be reversed by an external energy source. Write any one disadvantage of this type of cell.

Answer:
(i) (a) The electrode with lower standard reduction potential acts as anode, and the one with higher standard reduction potential acts as cathode.
(b) Salt bridge maintains electrical neutrality and completes the internal circuit.
(c) Advantage: High efficiency and continuous energy supply; Disadvantage: Expensive to manufacture and catalyst degradation.
OR
(ii) (a) Conductivity depends on the number of ions per unit volume. Upon dilution, the number of ions per unit volume decreases.
(b) Free energy change \( \Delta G = -nFE_{cell} \). For spontaneity, \( \Delta G \) must be negative, making \( E_{cell} \) positive.
(c) Primary cell. Disadvantage: It cannot be recharged once depleted.

Teacher's Note:
a) Standard reduction potentials dictate electrode polarities.
b) Primary batteries are discarded after single use.

Question 17 [3 Marks]

(i) The structure of amino acid exists in the following two forms:
[Figure: zwitterionic equilibrium structure showing \( R - CH(NH_2) - COOH \) and \( R - CH(NH_3^+) - COO^- \)]
The above structure is an example of _________. If the side chain R is replaced by hydrogen, the amino acid is known as _________.
(ii) Janice notices that her gums bleed while brushing and eating food. Name the water soluble vitamin which she should consume to prevent bleeding of gums.
(iii) Which linkage holds two units of monosaccharides in a disaccharide?

Answer:
(i) Zwitter ion, Glycine.
(ii) Vitamin C (Ascorbic acid).
(iii) Glycosidic linkage.

Teacher's Note:
a) Zwitter ions contain both positive and negative charges internally.
b) Bleeding gums are a classic symptom of scurvy, caused by Vitamin C deficiency.

Question 18 [3 Marks]

The data given below is for the reaction between [NO] and [Cl_2] to form \( NOCl \) at \( 25^{\circ}C \)

Answer:
(i) Rate law: rate = \( k[NO]^x[Cl_2]^y \)
From experiments 1 and 2, when [NO] is constant, [Cl_2] triples, rate triples, so \( y = 1 \).
From experiments 1 and 3, when [Cl_2] is constant, [NO] triples, rate increases 9 times (\( 3^2 \)), so \( x = 2 \).
Order w.r.t NO = 2, w.r.t \( Cl_2 \) = 1.
(ii) Overall order = \( 2 + 1 = 3 \).
(iii) Rate constant \( k = \frac{\text{rate}}{[NO]^2[Cl_2]} = \frac{2.0 \times 10^{-3}}{(2.0)^2 \times 2.0} = \frac{2.0 \times 10^{-3}}{8} = 0.25 \times 10^{-3}\ L^2\ mol^{-2}\ s^{-1} \).

Teacher's Note:
a) Compare initial rates while keeping one reactant concentration constant.
b) Include proper units for third-order rate constants.

SECTION - D

Question 19 [5 Marks]

Phenol is an aromatic alcohol that is used to prepare many important compounds such as picric acid. Phenol is widely used in household and industrial settings as a cleaner and disinfectant. It is also used as a primary chemical to make plastics. Phenol is less soluble in water as compared to aliphatic alcohol. Some aliphatic alcohols are toxic and can be addictive.
(i) How is the acid mentioned above prepared from phenol? Write the chemical reaction involved in this preparation.
(ii) 'Phenol is less soluble in water as compared to aliphatic alcohol'. Explain.
(iii) Write the chemical equation for the preparation of phenol from chlorobenzene.
(iv) An organic compound [A] having molecular formula \( C_4H_{10}O \) gives positive Lucas test within five minutes at room temperature. Compound [A] upon oxidation with \( K_2Cr_2O_7/H_2SO_4 \) forms compound [B] which does not respond to Tollen's test. Identify compounds [A] and [B].
(v) In the above reaction, compound [B] gets reduced with Zn/Hg and HCl and forms compound (C). Identify compound [C] and write the balanced reaction for the conversion of compound [B] to compound [C].

Answer:
(i) Nitration of phenol with concentrated \( HNO_3 \) yields 2,4,6-trinitrophenol (picric acid):
\( C_6H_5OH + 3HNO_{3(conc)} \xrightarrow{H_2SO_4} C_6H_2OH(NO_2)_3 + 3H_2O \)
(ii) Aliphatic alcohols form intermolecular hydrogen bonds with water effectively and dissolve. Phenol features a bulky hydrophobic benzene ring, making it less soluble.
(iii) Chlorobenzene to phenol (Dow's process):
\( C_6H_5Cl + NaOH \xrightarrow{623K,\ 300\ atm} C_6H_5ONa \xrightarrow{HCl} C_6H_5OH \)
(iv) [A] = Butan-2-ol, [B] = Butan-2-one (ketone).
(v) [C] = n-butane.
Reaction: \( CH_3COCH_2CH_3 + 4[H] \xrightarrow{Zn-Hg/HCl} CH_3CH_2CH_2CH_3 + H_2O \)

Teacher's Note:
a) Lucas test distinguishes secondary alcohols by cloudiness within 5 minutes.
b) Clemmensen reduction converts ketones into corresponding alkanes using amalgamated zinc and hydrochloric acid.

Question 20 [5 Marks]

(i) Give a reason for each of the following:
(a) \( Ti^{3+} \) salts are coloured whereas \( Ti^{4+} \) salts are colourless.
(Given: Atomic number of Ti = 22)
(b) Transition elements form alloys.
(c) The pink coloured \( KMnO_4 \) solution turns colourless when reacted with Mohr's salt (\( Fe^{2+} \)) in acidic medium.
(ii) Complete and balance the following reactions:
(a) \( KMnO_4 + H_2SO_4 + H_2C_2O_4 \rightarrow \text{____} + \text{____} + \text{____} + \text{____} \)
(b) \( K_2Cr_2O_7 + H_2SO_4 + H_2S \rightarrow \text{____} + \text{____} + \text{____} + \text{____} + \text{____} \)

Answer:
(i) (a) \( Ti^{3+} \) has one unpaired electron in d-orbital allowing d-d transition absorbing visible light, whereas \( Ti^{4+} \) has no unpaired electrons.
(b) Transition metals have similar atomic sizes, allowing atoms of one metal to easily replace another in the crystal lattice.
(c) Mohr's salt acts as a reducing agent, reducing \( Mn^{7+} \) to colourless \( Mn^{2+} \).
(ii) (a) \( 2KMnO_4 + 3H_2SO_4 + 5H_2C_2O_4 \rightarrow K_2SO_4 + 2MnSO_4 + 8H_2O + 10CO_2 \)
(b) \( K_2Cr_2O_7 + 3H_2S + 4H_2SO_4 \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + 7H_2O + 3S \)

Teacher's Note:
a) Colour in transition metal complexes arises primarily from d-d electron transitions.
b) Balance redox equations by equating oxidation number increases and decreases.

Question 21 [5 Marks]

Osmotic pressure is the external pressure which should be applied to stop the flow of solvent into the solution when the two are separated by a semipermeable membrane. The osmotic pressure is a colligative property. Two solutions having the same osmotic pressure are called isotonic. If there are two solutions and one of them is of lower osmotic pressure, it is called hypotonic while the other is called hypertonic.
Answer the questions given below.
(a) What will happen if red blood corpuscles are placed in a 5% NaCl solution which is a hypertonic solution?
(b) Show that osmotic pressure (\( \pi \)) is a colligative property.
(c) Calculate the amount of pressure required to stop osmosis of a solution when \( 40\ g \) of \( Na_2SO_4 \) is added to 1 L of water at \( 298\ K \).
(Given: \( Na = 23, O = 16, S = 32 \) and \( R = 0.0821\ L\ atm\ K^{-1}\ mol^{-1} \))
(d) Briefly discuss the process of reverse osmosis followed to desalinate sea water and convert it into drinking water.
OR
(ii) (a) An aqueous solution is made by dissolving \( 10\ g \) of glucose (\( C_6H_{12}O_6 \)) in \( 90\ g \) of water at \( 300\ K \). If the vapour pressure of pure water at \( 300\ K \) is \( 32.8\ mm\ Hg \), what would be the vapour pressure of the solution?
(b) A solution containing \( 12.5\ g \) of a non-electrolyte solute in \( 175\ g \) of water gave boiling point of \( 100.70^{\circ}C \). Calculate the molecular mass of the solute.
(Given: \( K_b \) for water = \( 0.52\ K\ kg\ mol^{-1} \))
(c) Why are soda water bottles sealed under high pressure?

Answer:
(a) Water will flow out of the red blood corpuscles due to exosmosis, causing them to shrink (crenation).
(b) Osmotic pressure \( \pi = iCRT = i \left(\frac{n}{V}\right)RT \). Since \( \pi \) depends on the number of moles of solute (\( n \)) and not on their chemical identity, it is a colligative property.
(c) Molar mass of \( Na_2SO_4 = (2 \times 23) + 32 + (4 \times 16) = 142\ g\ mol^{-1} \).
Moles \( n = \frac{40}{142} = 0.281\ mol \). Volume \( V = 1\ L \).
Van't Hoff factor \( i = 3 \) (since \( Na_2SO_4 \rightarrow 2Na^+ + SO_4^{2-} \)).
\( \pi = iCRT = 3 \times 0.281 \times 0.0821 \times 298 = 20.62\ atm \).
(d) When a pressure higher than the osmotic pressure is applied on the solution side, pure solvent flows out of the solution through a semipermeable membrane. This process is called reverse osmosis.
OR
(ii) (a) Moles of glucose = \( \frac{10}{180} = 0.0555\ mol \). Moles of water = \( \frac{90}{18} = 5\ mol \).
\( \frac{P_0 - P_s}{P_0} = \frac{n_2}{n_1 + n_2} \implies \frac{32.8 - P_s}{32.8} = \frac{0.0555}{5 + 0.0555} = 0.0109 \).
\( 32.8 - P_s = 0.357 \implies P_s = 32.44\ mm\ Hg \).
(b) \( \Delta T_b = K_b \times m \implies 0.70 = 0.52 \times \left(\frac{12.5}{M_2} \times \frac{1000}{175}\right) \).
\( M_2 = \frac{0.52 \times 12.5 \times 1000}{0.70 \times 175} = 53.06\ g\ mol^{-1} \).
(c) High pressure increases the solubility of \( CO_2 \) gas in soft drinks according to Henry's law.

Teacher's Note:
a) Hypertonic solutions cause cells to shrink due to outward osmosis.
b) Always account for ionization via the van't Hoff factor when calculating colligative properties for ionic solutes.