ISC Class 12 Chemistry Board Exam Question Paper 2024 with Solutions

ISC (Science) Class 12 Chemistry Board Exam Question Paper with Solutions

SECTION A - 14 MARKS

Q1.

1. (a) Fill in the blanks by choosing the appropriate word(s) from those given below in the options: [4 Marks]
1. lead poisoning
2. zero
3. phosgene
4. dependent
5. cancer
6. independent
7. diethyl ether
8. first
9. ethyl carbonate
10. ethene

1. (a) (i) For a particular reaction, the value of the rate constant is 0.05 \( \text{sec}^{-1} \). The reaction is of _______ order and will be _______ of the initial concentration. [1 Mark]

Answer: first, dependent

Teacher's Note:
a) The unit of rate constant is \( \text{sec}^{-1} \), which corresponds to a first-order reaction.
b) For a first-order reaction, the rate depends on the initial concentration of reactants.

1. (a) (ii) EDTA is used in the treatment of _______ while Cis platin is used in the treatment of _______. [1 Mark]

Answer: lead poisoning, cancer

Teacher's Note:
a) EDTA forms a stable complex with lead ions to help excrete them from the body.
b) Cis-platin is a well-known coordination complex used in chemotherapy for cancer treatment.

1. (a) (iii) The addition of small quantity of ethanol to chloroform prevents the formation of _______ and converts it into the harmless compound _______. [1 Mark]

Answer: phosgene, ethyl carbonate

Teacher's Note:
a) Ethanol acts as a negative catalyst and destroys poisonous phosgene gas formed by air oxidation of chloroform.
b) It reacts with phosgene to form harmless diethyl carbonate (or ethyl carbonate).

1. (a) (iv) The dehydration of ethyl alcohol with conc. \( \text{H}_2\text{SO}_4 \) at \( 140 \, ^{\circ}\text{C} \) mainly yields _______ while at \( 170 \, ^{\circ}\text{C} \) the main product formed is _______. [1 Mark]

Answer: diethyl ether, ethene

Teacher's Note:
a) Bimolecular dehydration at lower temperature (\( 140 \, ^{\circ}\text{C} \)) yields ether.
b) Intramolecular dehydration at higher temperature (\( 170 \, ^{\circ}\text{C} \)) yields alkene (ethene).

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

1. (b) (i) Which one of the following statements is correct regarding the dry cell? [1 Mark]
(P) Zinc container acts as an anode in dry cell.
(Q) Zinc container touches the paste of \( \text{MnO}_2 \) and carbon.
(R) Dry cell can be charged easily.
(S) Graphite rod acts as a cathode in dry cell.
1. Only (P) and (R)
2. Only (Q) and (R)
3. Only (P) and (S)
4. Only (Q) and (S)

Answer: (3) Only (P) and (S)

The zinc container acts as the anode, and the graphite rod surrounded by \( \text{MnO}_2 \) and carbon acts as the cathode.

Teacher's Note:
a) Primary cells like dry cells cannot be recharged easily.
b) Zinc container acts as the negative terminal (anode).

1. (b) (ii) The metal complex ion that is paramagnetic is _______.
(Atomic number of \( \text{Fe} = 26, \text{Cu} = 29, \text{Co} = 27 \) and \( \text{Ni} = 28 \)) [1 Mark]
1. \( [\text{Fe}(\text{CN})_4]^{2-} \)
2. \( [\text{Co}(\text{NH}_3)_6]^{3+} \)
3. \( [\text{Ni}(\text{CN})_4]^{2-} \)
4. \( [\text{Cu}(\text{NH}_3)_4]^{2+} \)

Answer: (4) \( [\text{Cu}(\text{NH}_3)_4]^{2+} \)

\( \text{Cu}^{2+} \) has \( 3\text{d}^9 \) configuration with one unpaired electron, making it paramagnetic.

Teacher's Note:
a) Paramagnetic species contain unpaired electrons.
b) Check the oxidation state and d-electron configuration to find unpaired electrons.

1. (b) (iii) When \( \text{KMnO}_4 \) is heated with acidified oxalic acid, gas bubbles are evolved. These gas bubbles are evolved due to the formation of _______. [1 Mark]
1. \( \text{SO}_2 \)
2. \( \text{CO}_2 \)
3. \( \text{SO}_3 \)
4. \( \text{O}_2 \)

Answer: (2) \( \text{CO}_2 \)

Oxalic acid gets oxidized to carbon dioxide gas during the titration with \( \text{KMnO}_4 \).

Teacher's Note:
a) Permanganate oxidation of organic acids containing carbon yields carbon dioxide.
b) Ensure balanced redox reactions are written for confirmation.

1. (b) (iv) The reaction of ethanamide with alcoholic sodium hydroxide and bromine gives _______. [1 Mark]
1. Ethylamine
2. Methylamine
3. Propylamine
4. Aniline

Answer: (2) Methylamine

This is Hofmann bromamide degradation reaction where one carbon atom is lost, converting ethanamide to methylamine.

Teacher's Note:
a) Hofmann degradation reduces the chain length by one carbon atom.
b) Ethanamide has two carbons, so the amine formed has one carbon (methylamine).

1. (b) (v) An equimolar solution of non-volatile solutes A and B shows a depression in freezing point in the ratio of 2:1. If A remains in its normal state in the solution, the state of B in the solution will be _______. [1 Mark]
1. Normal
2. Hydrolysed
3. Associated
4. Dissociated

Answer: (4) Dissociated

Higher depression in freezing point for equimolar solutions indicates a higher van 't Hoff factor due to dissociation.

Teacher's Note:
a) Depression in freezing point is directly proportional to the van 't Hoff factor (\( i \)).
b) \( i \gt 1 \) implies dissociation of the solute.

1. (b) (vi) Assertion: Specific conductivity of all electrolytes decreases on dilution.
Reason: On dilution, the number of ions per unit volume decreases. [1 Mark]
1. Both Assertion and Reason are true and Reason is the correct explanation for Assertion.
2. Both Assertion and Reason are true but Reason is not the correct explanation for Assertion.
3. Assertion is true but Reason is false.
4. Assertion is false but Reason is true.

Answer: (1) Both Assertion and Reason are true and Reason is the correct explanation for Assertion.

Conductivity depends on the number of current-carrying ions present per unit volume of solution.

Teacher's Note:
a) Specific conductivity is defined for unit volume of solution.
b) On dilution, volume increases, so the number of ions per unit volume decreases, lowering the conductivity.

1. (b) (vii) Assertion: Ammonolysis of alkyl halides involves the reaction between alkyl halides and alcoholic ammonia.
Reason: Ammonolysis of alkyl halides produces secondary amines only. [1 Mark]
1. Both Assertion and Reason are true and Reason is the correct explanation for Assertion.
2. Both Assertion and Reason are true but Reason is not the correct explanation for Assertion.
3. Assertion is true but Reason is false.
4. Assertion is false but Reason is true.

Answer: (3) Assertion is true but Reason is false.

Ammonolysis yields a mixture of primary, secondary, and tertiary amines along with quaternary ammonium salts.

Teacher's Note:
a) The reaction does not stop at secondary amines because the newly formed amine also acts as a nucleophile.
b) Reason is false since multiple products are formed.

1. (c) Read the passage given below and answer the questions that follow. [3 Marks]
When two solutions are separated by a semi-permeable membrane, the solvent molecules move from a solution of lower molar concentration to a solution of higher molar concentration through osmosis.

1. (c) (i) Samar removed the outer hard shell of two different eggs while cooking at home. He then placed one egg in pure water and the other egg in a saturated solution of sucrose. What change is he likely to observe in the eggs after a few hours? [1 Mark]

Answer: The egg placed in pure water will swell due to endosmosis, whereas the egg placed in a saturated sucrose solution will shrink due to exosmosis.

Teacher's Note:
a) Water moves from a region of lower solute concentration to higher solute concentration through the semi-permeable membrane of the egg.
b) Pure water is hypotonic to the egg contents, while concentrated sucrose is hypertonic.

1. (c) (ii) Which solution, hypertonic or hypotonic, has a higher amount of solute in same quantity of solution? [1 Mark]

Answer: Hypertonic solution has a higher amount of solute in the same quantity of solution.

Teacher's Note:
a) A hypertonic solution has a higher osmotic pressure and solute concentration compared to another solution.
b) Use correct terminology while comparing concentrations.

1. (c) (iii) A \( 5\% \) aqueous solution of glucose (molar mass = \( 180\text{ g mol}^{-1} \)) is isotonic with \( 1.66\% \) aqueous solution of urea. Calculate the molar mass of urea. [1 Mark]

Answer: Molar mass of urea = \( 59.76\text{ g mol}^{-1} \)

Teacher's Note:
a) Isotonic solutions have equal concentrations (\( C_1 = C_2 \)).
b) Apply the formula \( \frac{W_1}{M_1} = \frac{W_2}{M_2} \) for equal volumes of percentage solutions.

SECTION B - 20 MARKS

Q2.

2. (a) Write a chemical test to distinguish between ethanol and phenol. [2 Marks]

Answer: Phenol reacts with neutral \( \text{FeCl}_3 \) solution to give a violet or purple coloration, whereas ethanol does not react with neutral \( \text{FeCl}_3 \).

Teacher's Note:
a) Phenols show acidic character due to resonance stabilization of the phenoxide ion.
b) Neutral ferric chloride test is a classic identification test for phenolic groups.

2. (b) Give a chemical reaction to convert acetaldehyde into secondary propyl alcohol. [2 Marks]

Answer:
Step 1: Reaction of acetaldehyde with methylmagnesium iodide (Grignard reagent) followed by hydrolysis.
\( \text{CH}_3\text{CHO} + \text{CH}_3\text{MgI} \rightarrow \text{CH}_3-\text{CH}(\text{OMgI})-\text{CH}_3 \
\text{CH}_3-\text{CH}(\text{OMgI})-\text{CH}_3 + \text{H}_2\text{O}/\text{H}^+ \rightarrow \text{CH}_3-\text{CH}(\text{OH})-\text{CH}_3 + \text{Mg}(\text{OH})\text{I} \)

Teacher's Note:
a) Grignard reagents add across the carbonyl double bond to form an addition product.
b) Acid hydrolysis of the addition product yields a secondary alcohol.

Q3.

3. (a) Zinc, cadmium and mercury are considered as d-block elements but not regarded as transition elements. [2 Marks]

Answer: Zinc, cadmium, and mercury have completely filled d-orbitals (\( (n-1)\text{d}^{10} \)) in their ground state as well as in their common oxidation states. Therefore, they do not exhibit typical properties of transition elements such as variable oxidation states or colored ion formation.

Teacher's Note:
a) Transition elements are defined as those having incompletely filled d-orbitals.
b) Mention the complete \( \text{d}^{10} \) configuration to justify the statement.

3. (b) Transition metals possess a great tendency to form complex compounds. [2 Marks]

Answer: Transition metals form complex compounds due to: (1) their small ionic size, (2) high nuclear charge, and (3) availability of vacant d-orbitals of suitable energy to accept lone pairs of electrons from ligands.

Teacher's Note:
a) High charge density enables them to attract electron pairs strongly.
b) Vacant d-orbitals allow coordinate bonding with various ligands.

Q4.

4. (a) Convert the following by giving chemical equations for each: Ethyl bromide to diethyl ether. [2 Marks]

Answer:
Ethyl bromide is heated with sodium ethoxide (Williamson synthesis).
\( \text{C}_2\text{H}_5\text{Br} + \text{C}_2\text{H}_5\text{ONa} \rightarrow \text{C}_2\text{H}_5-\text{O}-\text{C}_2\text{H}_5 + \text{NaBr} \)

Teacher's Note:
a) Williamson synthesis is a standard laboratory method for the preparation of symmetrical and unsymmetrical ethers.
b) The reaction proceeds via an \( \text{S}_\text{N}2 \) mechanism.

4. (b) Phenol to salicylaldehyde. [2 Marks]

Answer:
Phenol is treated with chloroform and aqueous sodium hydroxide at \( 340\text{ K} \) followed by acidification (Reimer-Tiemann reaction).
\( \text{C}_6\text{H}_5\text{OH} + \text{CHCl}_3 + 3\text{NaOH} \rightarrow \text{C}_6\text{H}_4(\text{OH})(\text{CHO}) + 3\text{NaCl} + 2\text{H}_2\text{O} \)

Teacher's Note:
a) This reaction is known as the Reimer-Tiemann reaction.
b) An ised formyl group (\( -\text{CHO} \)) is introduced at the ortho-position of the phenol ring.

Q5.

5. (a) Account for each of the following: Zirconium (Zr) and Hafnium (Hf) are difficult to separate. [2 Marks]

Answer: Zirconium and hafnium belong to different periods of the same group, but due to lanthanide contraction, they have almost identical atomic and ionic radii. Consequently, their chemical properties are very similar, making them difficult to separate.

Teacher's Note:
a) Lanthanide contraction leads to a regular decrease in atomic size, offsetting the expected increase down the group.
b) Mention identical radii as the primary reason for chemical similarity.

5. (b) Salts of Cupric (\( \text{Cu}^{2+} \)) ion are coloured whereas salts of the Cuprous (\( \text{Cu}^+\)) ion are colourless. [2 Marks]

Answer: \( \text{Cu}^{2+} \) has a \( 3\text{d}^9 \) configuration with one unpaired electron, permitting d-d transitions and imparting color. In contrast, \( \text{Cu}^+ \) has a fully filled \( 3\text{d}^{10} \) configuration with no unpaired electrons, making d-d transitions impossible, so its salts are colorless.

Teacher's Note:
a) Color in transition metal ions is typically due to d-d electron transitions.
b) Fully filled or empty d-orbitals result in colorless compounds.

Q6.

6. (a) How will you bring the following conversion? Benzene to biphenyl. [2 Marks]

Answer:
Benzene is first chlorinated in the presence of anhydrous \( \text{AlCl}_3 \) to form chlorobenzene. Chlorobenzene is then treated with sodium in dry ether (Fittig reaction) to form biphenyl.
\( \text{C}_6\text{H}_6 + \text{Cl}_2 \xrightarrow{\text{Anhy. AlCl}_3} \text{C}_6\text{H}_5\text{Cl} + \text{HCl} \
2\text{C}_6\text{H}_5\text{Cl} + 2\text{Na} \xrightarrow{\text{Dry Ether}} \text{C}_6\text{H}_5-\text{C}_6\text{H}_5 + 2\text{NaCl} \)

Teacher's Note:
a) Stepwise conversion requires halogenation followed by a coupling reaction.
b) Mention Fittig reaction clearly for coupling two aryl groups.

6. (b) Iodoform to acetylene. [2 Marks]

Answer:
Heating iodoform with powdered silver metal drives off iodine, producing acetylene gas.
\( \text{CHI}_3 + 6\text{Ag} + \text{CHI}_3 \xrightarrow{\Delta} \text{CH}\equiv\text{CH} + 6\text{AgI} \)

Teacher's Note:
a) Silver powder abstracts iodine atoms from iodoform to form silver iodide precipitate.
b) Two molecules of iodoform are required to supply two carbon atoms for acetylene.

Q7. Calculate the maximum possible electrical work that can be obtained from a galvanic cell under standard conditions at \( 298\text{ K} \).
\( \text{Zn} | \text{Zn}^{2+}_{(aq)} || \text{Ag}^+_{(aq)} | \text{Ag} \)
Given \( \text{E}^{\circ}_{(\text{Zn}^{2+}/\text{Zn})} = -0.76\text{ V}; \text{E}^{\circ}_{(\text{Ag}^+/\text{Ag})} = +0.80\text{ V} \) [2 Marks]

Answer:
\( \text{E}^{\circ}_{\text{cell}} = \text{E}^{\circ}_{\text{cathode}} - \text{E}^{\circ}_{\text{anode}} = 0.80 - (-0.76) = +1.56\text{ V} \
W_{\max} = -\Delta G^{\circ} = nFE^{\circ}_{\text{cell}} = 2 \times 96500\text{ C mol}^{-1} \times 1.56\text{ V} = 301080\text{ J mol}^{-1} = 301.08\text{ kJ mol}^{-1} \)

Teacher's Note:
a) Maximum electrical work equals the standard Gibbs free energy change with a negative sign.
b) Ensure proper substitution of \( n = 2 \) electrons transferred in the redox reaction.

Q8.

8. (a) Give a reason for each of the following: Ethoxy ethane does not react with sodium, but ethanol does. [2 Marks]

Answer: Ethanol contains a reactive, polar hydroxyl hydrogen atom which can be easily replaced by sodium metal to form sodium ethoxide and hydrogen gas. Ethoxy ethane (ether) lacks an active hydrogen atom attached to an electronegative oxygen atom, hence it does not react with sodium.

Teacher's Note:
a) Active hydrogen attached to oxygen confers acidic character to alcohols.
b) Ethers are relatively inert towards active metals due to the absence of \( -\text{OH} \) groups.

8. (a) (ii) Ethoxy ethane with conc. \( \text{HI} \) at \( 373\text{ K} \) gives \( \text{C}_2\text{H}_5\text{OH} \) and \( \text{CH}_3\text{I} \) but not \( \text{CH}_3\text{OH} \) and \( \text{C}_2\text{H}_5\text{I} \). [2 Marks]

Answer: The cleavage of asymmetrical ethers with cold concentrated \( \text{HI} \) proceeds via an \( \text{S}_\text{N}2 \) mechanism where the iodide ion attacks the smaller alkyl group to avoid steric hindrance, forming methyl iodide and ethanol.

Teacher's Note:
a) Nucleophilic attack occurs preferably at the less sterically hindered carbon.
b) Mention \( \text{S}_\text{N}2 \) pathway clearly.

OR

8. (b) An organic compound [A] having the molecular formula \( \text{C}_4\text{H}_{10}\text{O} \) forms a compound [B] with the molecular formula \( \text{C}_4\text{H}_8\text{O} \) on oxidation. Compound [B] gives a positive iodoform test. The reaction of compound [B] with \( \text{CH}_3\text{MgBr} \) followed by hydrolysis, gives compound [C] with the molecular formula \( \text{C}_5\text{H}_{12}\text{O} \). Identify the compounds [A], [B] and [C]. Write the reaction for the conversion of compound [A] to compound [B]. [2 Marks]

Answer:
Compound [A]: Butan-2-ol (\( \text{CH}_3-\text{CH}(\text{OH})-\text{CH}_2-\text{CH}_3 \))
Compound [B]: Butanone (\( \text{CH}_3-\text{CO}-\text{CH}_2-\text{CH}_3 \))
Compound [C]: 2-Methylbutan-2-ol
Conversion equation: Oxidation of butan-2-ol with PCC gives butanone.
\( \text{CH}_3-\text{CH}(\text{OH})-\text{CH}_2-\text{CH}_3 \xrightarrow{\text{PCC}} \text{CH}_3-\text{CO}-\text{CH}_2-\text{CH}_3 \)

Teacher's Note:
a) Positive iodoform test for [B] indicates a methyl ketone structure.
b) Grignard addition to a ketone yields a tertiary alcohol.

Q9. If \( 200\text{ cm}^3 \) of an aqueous solution of a protein contains \( 1.26\text{ g} \) of protein, the osmotic pressure of the solution at \( 300\text{ K} \) is found to be \( 2.57 \times 10^{-3}\text{ atm} \). Calculate the molar mass of protein. (\( \text{R} = 0.0821\text{ L atm K}^{-1}\text{ mol}^{-1} \)) [2 Marks]

Answer:
\( \pi = \frac{W}{M} \times \frac{R T}{V} \
M = \frac{W \times R \times T}{\pi \times V} \
M = \frac{1.26\text{ g} \times 0.0821\text{ L atm K}^{-1}\text{ mol}^{-1} \times 300\text{ K}}{2.57 \times 10^{-3}\text{ atm} \times 0.200\text{ L}} = 60377\text{ g mol}^{-1} \)

Teacher's Note:
a) Convert volume from \( \text{cm}^3 \) to liters (\( 0.2\text{ L} \)) before applying the formula.
b) Osmotic pressure formula \( \pi V = nRT \) is ideal for high molar mass polymers like proteins.

Q10.

10. (a) Benzaldehyde is less reactive than propionaldehyde. Why? [2 Marks]

Answer: In benzaldehyde, the carbonyl carbon is involved in resonance with the benzene ring, which reduces the positive charge on the carbonyl carbon and makes it less susceptible to nucleophilic attack. In propionaldehyde, only the electron-donating inductive effect of the alkyl group operates, making its carbonyl carbon more electrophilic and reactive.

Teacher's Note:
a) Resonance stabilization deactivates the carbonyl group in aromatic aldehydes.
b) Aliphatic aldehydes experience greater electrophilicity at the carbonyl carbon.

10. (b) In the preparation of ethanal by the oxidation of ethanol, ethanal should be removed immediately as it is formed. Why? [2 Marks]

Answer: Ethanal is a volatile aldehyde which, if not removed immediately, undergoes further oxidation in the presence of oxidizing agents to form ethanoic acid.

Teacher's Note:
a) Aldehydes are easily oxidized to carboxylic acids.
b) Rapid distillation prevents over-oxidation during preparation.

Q11.

11. (a) Why is \( \text{Mn}^{2+} \) ion more stable than \( \text{Fe}^{2+} \) ion? (Atomic numbers of \( \text{Mn} = 25 \) and \( \text{Fe} = 26 \)) [2 Marks]

Answer: The outer electronic configuration of \( \text{Mn}^{2+} \) is \( 3\text{d}^5 \), which represents a half-filled d-subshell with extra exchange energy and high stability. In contrast, \( \text{Fe}^{2+} \) has a \( 3\text{d}^6 \) configuration, which lacks this symmetrical half-filled stability.

Teacher's Note:
a) Half-filled and fully filled subshells possess extra stability due to symmetrical distribution of electrons.
b) State the exact d-orbital configurations for clarity.

11. (b) Trivalent lanthanoid ions such as \( \text{La}^{3+} \) (\( \text{Z} = 57 \)) and \( \text{Lu}^{3+} \) (\( \text{Z} = 71 \)) do not show any colour in their solution. Give a reason. [2 Marks]

Answer: \( \text{La}^{3+} \) has an empty f-orbital (\( 4\text{f}^0 \)) and \( \text{Lu}^{3+} \) has a completely filled f-orbital (\( 4\text{f}^{14} \)). Due to the absence of unpaired f-electrons, f-f transitions are not possible, rendering their solutions colorless.

Teacher's Note:
a) Color in lanthanoids arises due to f-f transitions.
b) \( 4\text{f}^0 \) and \( 4\text{f}^{14} \) configurations do not exhibit f-f transitions.

SECTION C - 21 MARKS

Q12. For the reaction \( \text{A} + \text{B} \rightarrow \text{Product} \), the following data was obtained: [3 Marks]

Calculate the following:
i. The overall order of the reaction
ii. The rate law equation
iii. The value of rate constant

Answer:
i. Overall order = 3
ii. Rate law = \( k[\text{A}]^2[\text{B}]^1 \)
iii. Rate constant \( k = 28.44\text{ mol}^{-2}\text{ L}^2\text{ min}^{-1} \)

Teacher's Note:
a) Compare rates between experiments where one reactant concentration is kept constant to find individual orders.
b) Check the units of the rate constant according to the overall third order.

Q13.

13. (a) Illustrate the following reaction by giving one suitable example: Coupling reaction. [1½ Marks]

Answer: Benzenediazonium chloride reacts with phenol in a mildly alkaline medium to form an orange-red azo dye (p-hydroxyazobenzene).
\( \text{C}_6\text{H}_5\text{N}_2\text{Cl} + \text{C}_6\text{H}_5\text{OH} \xrightarrow{\text{NaOH}} \text{C}_6\text{H}_5-\text{N}=\text{N}-\text{C}_6\text{H}_4-\text{OH} + \text{HCl} \)

Teacher's Note:
a) Coupling reactions are electrophilic substitution reactions involving diazonium salts.
b) They are used for the synthesis of colored azo dyes.

13. (a) (ii) Acetylation of ethylamine. [1½ Marks]

Answer: Ethylamine reacts with acetyl chloride or acetic anhydride in the presence of a base to form N-ethylacetamide.
\( \text{C}_2\text{H}_5\text{NH}_2 + (\text{CH}_3\text{CO})_2\text{O} \rightarrow \text{C}_2\text{H}_5\text{NHCOCH}_3 + \text{CH}_3\text{COOH} \)

Teacher's Note:
a) Acetylation replaces a hydrogen atom of the amino group with an acyl group.
b) The reaction is typically carried out in the presence of pyridine as a base.

13. (b) Aniline does not give Friedel-Crafts reaction. Give a reason. [3 Marks]

Answer: Aniline is a strong Lewis base due to the lone pair on the nitrogen atom. It reacts with the Lewis acid catalyst (\( \text{AlCl}_3 \)) used in Friedel-Crafts reactions to form a stable salt complex. This deactivates the benzene ring and prevents further electrophilic substitution.

Teacher's Note:
a) Catalyst poisoning via salt formation is the primary cause.
b) Protection of the amino group via acetylation is required prior to carrying out substitution.

Q14.

14. (a) Aradhana visits a physician as she is suffering from rickets and joint pain. Which fat-soluble vitamin should the physician prescribe to her? [1 Mark]

Answer: Vitamin D.

Teacher's Note:
a) Rickets is caused due to the deficiency of Vitamin D in children and adults.
b) Vitamin D regulates calcium and phosphorus metabolism for bone health.

14. (b) Somesh put a few drops of vinegar in the milk. What change do you think he observed in the milk after some time? What is this phenomenon known as? [1 Mark]

Answer: The milk curdles and forms white solid bits (curd). This phenomenon is known as the coagulation or curdling of milk proteins.

Teacher's Note:
a) Addition of acid alters the pH and neutralizes the charge on casein micelles.
b) This leads to precipitation of milk proteins.

14. (c) Name the product of hydrolysis of sucrose. Is it a reducing sugar or a non-reducing sugar? [1 Mark]

Answer: Hydrolysis of sucrose yields an equimolar mixture of D-(+)-glucose and D-(-)-fructose. Sucrose itself is a non-reducing sugar.

Teacher's Note:
a) Glycosidic linkage between C-1 of glucose and C-2 of fructose involves reducing centers.
b) Hence, free aldehyde or ketone groups are absent in sucrose.

Q15. An aqueous solution containing \( 12.50\text{ g} \) of barium chloride in \( 1000\text{ g} \) of water boils at \( 373.0834\text{ K} \). Calculate the degree of dissociation of barium chloride.
Given \( \text{K}_\text{b} \) for \( \text{H}_2\text{O} = 0.52\text{ K kg mol}^{-1} \); molecular mass of \( \text{BaCl}_2 = 208.34\text{ g mol}^{-1} \). [3 Marks]

Answer:
\( \Delta T_\text{b} = 373.0834 - 373 = 0.0834\text{ K} \
\text{Molality } (m) = \frac{12.50 / 208.34}{1\text{ kg}} = 0.06\text{ m} \
i = \frac{\Delta T_\text{b}}{K_\text{b} \times m} = \frac{0.0834}{0.52 \times 0.06} = 2.67 \
\text{Degree of dissociation } (\alpha) = \frac{i - 1}{n - 1} = \frac{2.67 - 1}{3 - 1} = 0.835 \text{ or } 83.5\% \)

Teacher's Note:
a) Calculate the van 't Hoff factor first using experimental elevation in boiling point.
b) Use \( n = 3 \) since \( \text{BaCl}_2 \) dissociates into three ions (\( \text{Ba}^{2+} + 2\text{Cl}^- \)).

Q16. An organic compound \( \text{C}_2\text{H}_4\text{O} \) gives a red precipitate when heated with Fehling solution. It also undergoes aldol condensation in the presence of dilute \( \text{NaOH} \).
i. Identify the organic compound and write its IUPAC name.
ii. Which compound will be formed when this organic compound reacts with hydroxylamine?
iii. What is observed when the compound, referred to in subpart (i), is heated with ammonical silver nitrate? [3 Marks]

Answer:
i. Acetaldehyde (Ethanal, \( \text{CH}_3\text{CHO} \))
ii. Ethanal oxime (\( \text{CH}_3\text{CH}=\text{NOH} \))
iii. A shining silver mirror is formed on the inner walls of the test tube (Tollen's test).

Teacher's Note:
a) Positive Fehling and Tollen's tests confirm the presence of an aldehyde group.
b) Presence of alpha-hydrogens allows aldol condensation.

Q17.

17. (a) (i) Identify the compound A, B and C:
\( \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{PCl}_5} \text{A} \xrightarrow{\text{KCN}} \text{B} \xrightarrow{\text{H}_3\text{O}^+} \text{C}_2\text{H}_5\text{COOH} \xrightarrow{\text{NH}_3/\Delta} \text{C} \) [1½ Marks]

Answer:
Compound [A]: Ethyl chloride (\( \text{C}_2\text{H}_5\text{Cl} \))
Compound [B]: Ethyl cyanide or Propanenitrile (\( \text{C}_2\text{H}_5\text{CN} \))
Compound [C]: Propanamide (\( \text{C}_2\text{H}_5\text{CONH}_2 \))

Teacher's Note:
a) Halogenation of alcohol gives alkyl halide.
b) Hydrolysis of cyanide yields carboxylic acid, which on ammonolysis forms an amide.

17. (a) (ii) Identify the compound [A], [B] and [C] in the following reaction:
\( \text{C}_6\text{H}_5\text{OH} \xrightarrow{\text{Zn/dust}} \text{A} \xrightarrow[\text{Anhy. AlCl}_3]{\text{CH}_3\text{Cl}} \text{B} \xrightarrow[\text{K}_2\text{Cr}_2\text{O}_7+\text{H}_2\text{SO}_4]{[\text{O}]} \text{C} \) [1½ Marks]

Answer:
Compound [A]: Benzene (\( \text{C}_6\text{H}_6 \))
Compound [B]: Toluene (\( \text{C}_6\text{H}_5\text{CH}_3 \))
Compound [C]: Benzoic acid (\( \text{C}_6\text{H}_5\text{COOH} \))

Teacher's Note:
a) Reduction of phenol with zinc dust gives benzene.
b) Friedel-Crafts alkylation of benzene gives toluene, which oxidizes to benzoic acid.

OR

17. (b) Give a chemical test to distinguish between the following pair of compound: Ethanol and methanol. [3 Marks]

Answer: Iodoform test: Ethanol reacts with iodine and sodium hydroxide to give a yellow precipitate of iodoform, whereas methanol does not give this test.

Teacher's Note:
a) Compounds containing the \( \text{CH}_3-\text{CH}(\text{OH})- \) group or \( \text{CH}_3-\text{CO}- \) group give a positive iodoform test.
b) Methanol lacks this structural unit.

17. (b) (ii) Ethanol and Ethanal. [3 Marks]

Answer: Fehling solution test: Ethanal reduces Fehling solution to a red precipitate of cuprous oxide upon heating, whereas ethanol does not react with Fehling solution.

Teacher's Note:
a) Aldehydes act as reducing agents and reduce metal cations.
b) Alcohols do not reduce Fehling solution under normal conditions.

17. (b) (iii) Propan-2-ol and 2-methyl propan-2-ol. [3 Marks]

Answer: Lucas test: 2-methylpropan-2-ol (tertiary alcohol) reacts immediately with Lucas reagent (anhydrous \( \text{ZnCl}_2 \) + conc. \( \text{HCl} \)) to form turbidity, whereas propan-2-ol (secondary alcohol) produces turbidity only after 3 to 5 minutes.

Teacher's Note:
a) Reactivity towards Lucas reagent follows the order: Tertiary \( \gt \) Secondary \( \gt \) Primary.
b) Carbocation stability dictates the rate of the reaction.

Q18.

18. (a) The rate constant of a reaction at \( 500\text{ K} \) and \( 700\text{ K} \) are \( 0.02\text{ sec}^{-1} \) and \( 0.07\text{ sec}^{-1} \) respectively. Calculate the value of \( \text{E}_\text{a} \). (activation energy) [3 Marks]

Answer:
\( \log\left(\frac{k_2}{k_1}\right) = \frac{\text{E}_\text{a}}{2.303 \times R} \left(\frac{T_2 - T_1}{T_1 T_2}\right) \
\log\left(\frac{0.07}{0.02}\right) = \frac{\text{E}_\text{a}}{2.303 \times 8.314} \left(\frac{700 - 500}{500 \times 700}\right) \
\text{E}_\text{a} = 18.228\text{ kJ mol}^{-1} \)

Teacher's Note:
a) Use the Arrhenius equation in logarithmic form for two different temperatures.
b) Ensure units for gas constant and activation energy are consistent.

18. (b) A radioactive substance which emits alpha particle follows a first-order reaction. The half-life period of this radioactive substance is 30 hours. Calculate the fraction in percent (%) of the radioactive substance which remains after 90 hours. [3 Marks]

Answer:
Number of half-lives \( n = \frac{t}{t_{1/2}} = \frac{90}{30} = 3 \) half-lives.
Remaining fraction \( = \left(\frac{1}{2}\right)^n = \left(\frac{1}{2}\right)^3 = \frac{1}{8} = 0.125 \) or \( 12.5\% \).
Percentage remaining = \( 12.5\% \).

Teacher's Note:
a) Radioactive decay strictly follows first-order kinetics.
b) The half-life method provides a quick alternative to calculate remaining amounts when time is an exact multiple of half-life.

SECTION D - 15 MARKS

Q19.

19. (a) An organic compound [A], having a specific smell forms two compounds [B] and [C] by reacting with conc. sodium hydroxide. The molecular formula of compound [B] is \( \text{C}_7\text{H}_8\text{O} \), which forms compound [A] again on oxidation. Compound [C] forms benzene on heating with soda lime.
Write the structures of compounds [A], [B] and [C]. Also, write the reactions involved. [5 Marks]

Answer:
Compound [A]: Benzaldehyde (\( \text{C}_6\text{H}_5\text{CHO} \))
Compound [B]: Benzyl alcohol (\( \text{C}_6\text{H}_5\text{CH}_2\text{OH} \))
Compound [C]: Sodium benzoate (\( \text{C}_6\text{H}_5\text{COONa} \))
Reaction: Cannizzaro reaction of benzaldehyde with conc. \( \text{NaOH} \) yields benzyl alcohol and sodium benzoate.

Teacher's Note:
a) Aldehydes without alpha-hydrogens undergo Cannizzaro disproportionation reaction in the presence of concentrated alkali.
b) Decarboxylation of sodium benzoate with soda lime yields benzene.

19. (b) (i) Identify the compound [A] and [B] in the reaction given below:
\( \text{C}_6\text{H}_6 \xrightarrow[\text{AlCl}_3 (\text{anhy.})]{\text{CH}_3\text{Cl}} [\text{A}] \xrightarrow[\text{K}_2\text{Cr}_2\text{O}_7+\text{H}_2\text{SO}_4]{[\text{O}]} [\text{B}] \) [2½ Marks]

Answer:
Compound [A]: Toluene (\( \text{C}_6\text{H}_5\text{CH}_3 \))
Compound [B]: Benzoic acid (\( \text{C}_6\text{H}_5\text{COOH} \))

Teacher's Note:
a) Friedel-Crafts methylation of benzene produces toluene.
b) Oxidation of side chain in toluene yields benzoic acid.

19. (b) (ii) Identify the compound [A] and [B] in the reaction given below:
\( \text{CH}_3-\text{CHOH}-\text{CH}_3 \xrightarrow[\text{K}_2\text{Cr}_2\text{O}_7+\text{H}_2\text{SO}_4]{[\text{O}]} [\text{A}] \xrightarrow{\text{NH}_2\text{OH}} [\text{B}] \) [2½ Marks]

Answer:
Compound [A]: Acetone (\( \text{CH}_3\text{COCH}_3 \))
Compound [B]: Acetone oxime (\( \text{CH}_3\text{C}(=\text{NOH})\text{CH}_3 \))

Teacher's Note:
a) Oxidation of secondary alcohol gives a ketone.
b) Reaction of a ketone with hydroxylamine produces an oxime.

Q20.

20. (a) A coordination compound has the formula \( \text{CoCl}_3 \cdot 4\text{NH}_3 \). It precipitates silver ions as \( \text{AgCl} \) and its molar conductance corresponds to a total of two ions.
Based on this information, answer the following question:
a. Deduce the structural formula of the complex compound.
b. Write the IUPAC name of the complex compound.
c. Draw the geometrical isomers of the complex compound. [5 Marks]

Answer:
a. Structural formula: \( [\text{Co}(\text{NH}_3)_4\text{Cl}_2]\text{Cl} \)
b. IUPAC name: Tetraamminedichloridocobalt(III) chloride
c. Geometrical isomers: Cis and trans isomers of \( [\text{Co}(\text{NH}_3)_4\text{Cl}_2]^+ \).

Teacher's Note:
a) Precipitation of one mole of \( \text{AgCl} \) per mole of compound indicates one chloride ion outside the coordination sphere.
b) Cis-trans isomerism is exhibited by octahedral complexes of type \( \text{MA}_4\text{B}_2 \).

20. (b) Give a chemical test to show that \( [\text{Co}(\text{NH}_3)_5\text{Cl}]\text{SO}_4 \) and \( [\text{Co}(\text{NH}_3)_5\text{SO}_4]\text{Cl} \) are ionisation isomers. [5 Marks]

Answer: Aqueous solution of \( [\text{Co}(\text{NH}_3)_5\text{Cl}]\text{SO}_4 \) gives a white precipitate of barium sulfate with \( \text{BaCl}_2 \) solution because sulfate ions are outside the coordination sphere. On the other hand, \( [\text{Co}(\text{NH}_3)_5\text{SO}_4]\text{Cl} \) gives a white precipitate of silver chloride with \( \text{AgNO}_3 \) solution because chloride ions are outside the sphere.

Teacher's Note:
a) Ionization isomers yield different ions in solution.
b) Use group reagents like \( \text{BaCl}_2 \) for sulfate and \( \text{AgNO}_3 \) for chloride to distinguish them.

Q21.

21. (a) (i) Study the diagram given below that represents Cu-Ag electrochemical cell and answer the questions that follow.
[Figure: Electrochemical cell with copper electrode in copper nitrate solution and silver electrode in silver nitrate solution connected via salt bridge and voltmeter]
1. Write the cell reaction for the above cell.
2. Calculate the standard emf of the cell.
3. If the concentration of \( [\text{Cu}^{2+}] \) is \( 0.1\text{ M} \) and \( \text{E}_{\text{cell}} \) is \( 0.422\text{ V} \), at \( 25^{\circ}\text{C} \), calculate the concentration of \( [\text{Ag}^+] \).
4. Calculate \( \Delta G \) for the cell. [5 Marks]

Answer:
1. Cell reaction: \( \text{Cu} + 2\text{Ag}^+ \rightarrow \text{Cu}^{2+} + 2\text{Ag} \)
2. Standard EMF = \( 0.462\text{ V} \)
3. Concentration of \( [\text{Ag}^+] = 7.5 \times 10^{-2}\text{ M} \)
4. \( \Delta G = -81.446\text{ kJ mol}^{-1} \)

Teacher's Note:
a) Apply Nernst equation for calculating concentrations of ions under non-standard conditions.
b) Gibbs free energy change relates to cell potential via \( \Delta G = -nFE_{\text{cell}} \).

21. (a) (ii) Calculate \( \Lambda^\circ_{\text{m}} \) for \( \text{BaCl}_2 \) and \( \text{Al}_2(\text{SO}_4)_3 \) from the following data.
Given \( \Lambda^\circ_{\text{m}} (\text{Ba}^{2+}) = 127.2\text{ S cm}^2\text{ mol}^{-1} \), \( \Lambda^\circ_{\text{m}} (\text{Al}^{3+}) = 189\text{ S cm}^2\text{ mol}^{-1} \), \( \Lambda^\circ_{\text{m}} (\text{Cl}^-) = 76.3\text{ S cm}^2\text{ mol}^{-1} \), \( \Lambda^\circ_{\text{m}} (\text{SO}_4^{2-}) = 160\text{ S cm}^2\text{ mol}^{-1} \) [5 Marks]

Answer:
\( \Lambda^\circ_{\text{m}}(\text{BaCl}_2) = \Lambda^\circ(\text{Ba}^{2+}) + 2\Lambda^\circ(\text{Cl}^-) = 127.2 + 2(76.3) = 279.8\text{ S cm}^2\text{ mol}^{-1} \)
\( \Lambda^\circ_{\text{m}}(\text{Al}_2(\text{SO}_4)_3) = 2\Lambda^\circ(\text{Al}^{3+}) + 3\Lambda^\circ(\text{SO}_4^{2-}) = 2(189) + 3(160) = 858\text{ S cm}^2\text{ mol}^{-1} \)

Teacher's Note:
a) Apply Kohlrausch law of independent migration of ions.
b) Multiply ionic conductivities by their stoichiometric coefficients from the dissociation formula.

OR

21. (b) (i) A \( 0.05\text{ M NH}_4\text{OH} \) solution offers the resistance of \( 30.8\text{ ohms} \) to a conductivity cell at \( 298\text{K} \). If the cell constant is \( 0.343\text{ cm}^{-1} \) and the molar conductance of \( \text{NH}_4\text{OH} \) at infinite dilution is \( 471.4\text{ S cm}^2\text{ mol}^{-1} \), calculate the following:
1. Specific conductance
2. Molar conductance
3. Degree of dissociation [5 Marks]

Answer:
1. Specific conductance \( \kappa = 0.011\text{ \(\Omega\)^{-1} cm}^{-1} \)
2. Molar conductance \( \Lambda_\text{m} = 220\text{ \(\Omega\)^{-1} cm}^2\text{ mol}^{-1} \)
3. Degree of dissociation \( \alpha = 0.47 \) (or \( 47\% \))

Teacher's Note:
a) Use formulas connecting resistance, cell constant, and specific conductivity.
b) Degree of dissociation is the ratio of molar conductance at a given concentration to that at infinite dilution.

21. (b) (ii) In the diagram of the electrolytic cell given below, A, B and C are connected in series having electrolytes of \( \text{ZnSO}_4 \), \( \text{AgNO}_3 \) and \( \text{CuSO}_4 \), respectively.
[Figure: Three electrolytic cells in series labeled A, B, and C containing zinc sulfate, silver nitrate, and copper sulfate solutions connected to a DC source]
A steady current of \( 1.5\text{ A} \) was passed until \( 1.45\text{ g} \) of Ag was deposited at the cathode of cell B.
(Atomic mass of \( \text{Ag} = 108, \text{Cu} = 63.5, \text{Zn} = 65.3 \))
Answer the following questions:
1. How long did the current flow?
2. What weight of Cu and Zn was deposited at the cathode? [5 Marks]

Answer:
1. Time \( t = 863.7\text{ seconds} \)
2. Weight of Zn deposited = \( 0.438\text{ g} \); Weight of Cu deposited = \( 0.426\text{ g} \)

Teacher's Note:
a) Apply Faraday's laws of electrolysis for cells connected in series.
b) Masses of substances deposited are proportional to their chemical equivalents when the same charge passes through them.