ISC Class 12 Chemistry Sample Paper 2026 with Solutions

SECTION A - 14 MARKS

Question 1

(A) Fill in the blanks by choosing the appropriate word(s) from those given in the brackets: [4×1]
[electrophilic, four, 4-bromophenol, first, alc. AgCN, decreases, 2,4,6-tribromophenol, nucleophilic, two, zero, alc. KCN, increases]

(i) The addition of a non-volatile solute to a pure solvent ________ its vapour pressure and ________ its boiling point. [1 Mark]

Answer: decreases, increases

Teacher's Note:
a) Colligative properties depend on the number of solute particles in solution.
b) Remember that adding a non-volatile solute lowers vapour pressure and consequently elevates the boiling point.

(ii) When the concentration of a reactant of first order reaction is doubled, the rate becomes ________ times but for ________ order reaction the rate remains the same. [1 Mark]

Answer: two, zero

Teacher's Note:
a) Rate of first order reaction is directly proportional to concentration (\( \text{Rate} = k[A] \)).
b) Zero order reaction rate is independent of reactant concentration (\( \text{Rate} = k \)).

(iii) Phenol when treated with bromine water produces ________, which is an example of ________ substitution reaction. [1 Mark]

Answer: 2,4,6-tribromophenol, electrophilic

Teacher's Note:
a) The -OH group in phenol strongly activates the benzene ring towards electrophilic substitution.
b) Bromine water gives a white precipitate of 2,4,6-tribromophenol instantly due to high activation.

(iv) Methyl chloride on treatment with ________ gives methyl cyanide, whereas on treatment with ________, it gives methyl isocyanide. [1 Mark]

Answer: alc. KCN, alc. AgCN

Teacher's Note:
a) Potassium cyanide is ionic and supplies cyanide ions, attacking via carbon to form nitrile.
b) Silver cyanide is largely covalent, making nitrogen the more reactive nucleophile to form isocyanide.

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

(i) The sum of coordination number and oxidation number of metal ‘M’ in the complex [M(en)₂Cl₂] is: [1 Mark]
(a) 6
(b) 8
(c) 9
(d) 10

Answer: (c) 9

Coordination number = \( 2 \times 2 + 2 = 6 \). Oxidation number of M: let it be \( x \). \( x + 2(0) + 2(-1) = 0 \implies x = +3 \). Sum = \( 6 + 3 = 9 \).

Teacher's Note:
a) Ethylenediamine (en) is a bidentate ligand contributing 2 coordination sites each.
b) Always calculate oxidation number carefully by considering charges on all ligands and counter ions.

(ii) Which of the following orders is correct in spectrochemical series of ligands? [1 Mark]
(P) \( \text{I}^{-} \lt \text{NH}_{3} \lt \text{CN}^{-} \lt \text{CO} \)
(Q) \( \text{CO} \lt \text{I}^{-} \lt \text{NH}_{3} \lt \text{CN}^{-} \)
(R) \( \text{NH}_{3} \lt \text{CO} \lt \text{CN}^{-} \lt \text{I}^{-} \)
(S) \( \text{F}^{-} \lt \text{NH}_{3} \lt \text{CN}^{-} \lt \text{CO} \)
(a) Only (Q) and (S) are correct.
(b) Only (P) and (S) are correct.
(c) Only (P) and (Q) are correct.
(d) Only (P) and (R) are correct.

Answer: (b) Only (P) and (S) are correct.

Spectrochemical series arranges ligands in order of increasing crystal field splitting energy: \( \text{I}^{-} \lt \text{F}^{-} \lt \text{NH}_{3} \lt \text{CN}^{-} \lt \text{CO} \).

Teacher's Note:
a) Halides and weak field ligands cause smaller splitting, while carbonyl (CO) and cyanide cause strong splitting.
b) Statement P and S correctly reflect the increasing ligand field strength.

(iii) When Chlorobenzene is heated with aq. ammonia in the presence of Cu₂O at high pressure, it gives: [1 Mark]
(a) aniline.
(b) phenyl isocyanide.
(c) diphenyl.
(d) diphenylamine.

Answer: (a) aniline.

Aryl halides react with aqueous ammonia at high temperature and pressure in presence of \( \text{Cu}_{2}\text{O \) to form primary aromatic amines.

Teacher-Note:
a) Haloarenes are unreactive towards nucleophilic substitution under normal conditions.
b) High temperature, high pressure, and \( \text{Cu}_{2}\text{O \) catalyst enable the conversion to aniline.

(iv) Hydrolysis of sucrose is called: [1 Mark]
(a) esterification.
(b) saponification.
(c) hydration.
(d) inversion.

Answer: (d) inversion.

Hydrolysis of sucrose yields D-(+)-glucose and D-(-)-fructose, causing a sign change in optical rotation from dextrorotatory to laevorotatory, known as inversion of sugar.

Teacher's Note:
a) Sucrose is dextrorotatory, but the resulting mixture is laevorotatory.
b) Do not confuse inversion with hydration or saponification.

(v) An organic compound with molecular formula C₃H₆O does not give silver mirror test with Tollen’s reagent but gives an oxime with hydroxylamine. The compound is: [1 Mark]
(a) \( \text{CH}_{3}\text{CH}_{2}\text{CHO} \)
(b) \( \text{CH}_{3}\text{-CO-CH}_{3} \)
(c) \( \text{CH}_{2}\text{=CH-CH}_{2}\text{OH} \)
(d) \( \text{CH}_{2}\text{=CH-O-CH}_{3} \)

Answer: (b) \( \text{CH}_{3}\text{-CO-CH}_{3} \)

Ketones react with hydroxylamine to form oximes but do not reduce Tollens reagent.

Teacher's Note:
a) Aldehydes give Tollens test, whereas ketones do not.
b) Both aldehydes and ketones react with hydroxylamine to form oximes.

(vi) Given below are two statements marked Assertion and Reason. Read the two statements carefully and select the correct option. [1 Mark]
Assertion: 0.1 M sucrose solution has higher depression in the freezing point than 0.1 M urea solution.
Reason: Depression in freezing point is not a colligative property.
(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: (d) Both Assertion and Reason are false.

Both sucrose and urea are non-electrolyte solutes with the same molar concentration, so their depression in freezing point is identical. Also, freezing point depression is a colligative property.

Teacher's Note:
a) Colligative properties depend only on the number of solute particles, not their nature.
b) Since 0.1 M sucrose and 0.1 M urea produce equal particle concentrations, \( \Delta T_{f} \) is the same for both.

(vii) Given below are two statements marked Assertion and Reason. Read the two statements carefully and select the correct option. [1 Mark]
Assertion: Methoxy ethane reacts with HI at 373K to give ethanol and iodomethane.
Reason: Reaction of unsymmetrical ether with HI follows \( \text{S}_{\text{N}}2 \) mechanism.
(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.

Reaction of alkyl aryl or unsymmetrical ethers with HI proceeds via protonation followed by \( \text{S}_{\text{N}}2 \) attack on the smaller alkyl group, yielding iodomethane and ethanol.

Teacher's Note:
a) In unsymmetrical aliphatic ethers, iodide attacks the less sterically hindered alkyl group.
b) This cleavage follows an \( \text{S}_{\text{N}}2 \) pathway under mild conditions.

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

Vishal set up an experiment to find the resistance of aqueous KCl solution for different concentrations at 298K using a conductivity cell connected to a Wheatstone bridge. He fed the Wheatstone bridge with A.C. power in the audio frequency range 550 to 5000 cycles per second. Once the resistance was calculated from null point, he also calculated the conductivity (\( \kappa \)) and molar conductivity (\( \Lambda_{m} \)) and recorded his readings in tabular form which is given below.
S.No. | Conc. (M) | \( \kappa \) (\( \text{S cm}^{-1} \)) | \( \Lambda_{m} \) (\( \text{S cm}^{2} \text{ mol}^{-1} \))
1. | 1.00 | \( 111.3 \times 10^{-3} \) | 111.3
2. | 0.10 | \( 12.9 \times 10^{-3} \) | 129.0
3. | 0.01 | \( 1.41 \times 10^{-3} \) | 141.0

(i) Why did the molar conductivity increase though the conductivity decreased with dilution? [1 Mark]

Answer: The conductivity of a solution is the conductance of ions present in a unit volume of the solution. With dilution, the number of ions per unit volume decreases. Hence, conductivity decreases with dilution. The molar conductance (\( \Lambda_{m} \)) is the product of conductivity (\( \kappa \)) and the volume of the solution containing 1 mole of the electrolyte. \( \Lambda_{m} = \kappa \times V_{m} \). Hence, \( \Lambda_{m} \) increases on dilution.

Teacher's Note:
a) Conductivity depends on ion concentration per unit volume, which drops upon dilution.
b) Molar conductivity increases because the increase in total volume far outweighs the decrease in concentration.

(ii) If molar conductivity at infinite dilution (\( \Lambda^\circ_{m} \)) of KCl is \( 150.0\text{ S cm}^{2}\text{ mol}^{-1} \), calculate the degree of dissociation of 0.01 M KCl. [1 Mark]

Answer:
Given: \( \Lambda^\circ_{m} = 150.0\text{ S cm}^{2}\text{ mol}^{-1} \), \( \Lambda^{c}_{m} = 141.0\text{ S cm}^{2}\text{ mol}^{-1} \)
Degree of dissociation \( (\alpha) = \frac{\Lambda^{c}_{m}}{\Lambda^\circ_{m}} = \frac{141.0}{150.0} = 0.94 \) or \( 94\% \).

Teacher's Note:
a) Degree of dissociation is the ratio of molar conductivity at concentration \( c \) to that at infinite dilution.
b) Ensure units for molar conductivities match before division.

(iii) The conductivity of a 0.01M solution of acetic acid at 298 K is \( 1.65 \times 10^{-4}\text{ S cm}^{-1} \). Calculate the molar conductivity of the solution. [1 Mark]

Answer:
\( \Lambda_{m} = \frac{\kappa \times 1000}{\text{Molarity}} = \frac{1.65 \times 10^{-4} \times 1000}{0.01} = 16.5\text{ S cm}^{2}\text{ mol}^{-1} \).

Teacher's Note:
a) Apply the standard formula relating conductivity, molarity, and molar conductivity.
b) Watch out for powers of 10 during the calculation.

SECTION B - 20 MARKS

Question 2 [2 Marks]
Consider yourself a research scholar in the forensic science department studying the age of a dead biological sample. During one of the studies, you found that the sample decomposed by following first order kinetics.
If 50% of the sample is decomposed in 120 minutes, how long will it take for 90% of the sample to decompose?

Answer:
\( t_{1/2} = 120\text{ min} \), \( a = 100 \), \( (a-x) = 100 - 90 = 10 \)
\( k = \frac{0.693}{t_{1/2}} = \frac{0.693}{120} = 0.005775\text{ min}^{-1} \)
\( t = \frac{2.303}{k}\log\left(\frac{a}{a-x}\right) = \frac{2.303}{0.005775}\log\left(\frac{100}{10}\right) = \frac{2.303}{0.005775} \times 1 = 398.79\text{ minutes} \).

Teacher's Note:
a) First calculate the rate constant \( k \) using the half-life period.
b) Substitute \( k \) into the first-order integrated rate equation for 90% completion.

Question 3 [2 Marks]
Write the chemical equations to convert each of the following:
(i) Aniline to bromobenzene
(ii) Ethyl chloride to propanoic acid

Answer:
(i) Aniline \( \xrightarrow{\text{NaNO}_{2} + \text{HCl},\ 273-278\text{ K}} \) Benzenediazonium chloride \( \xrightarrow{\text{CuBr, HBr} \text{ or } \text{Cu, HBr}} \) Bromobenzene + \( \text{N}_{2} \)
(ii) \( \text{CH}_{3}\text{CH}_{2}\text{Cl} \xrightarrow{\text{alc. KCN}} \text{CH}_{3}\text{CH}_{2}\text{CN} \xrightarrow{\text{H}^{+}, \text{H}_{2}\text{O (Hydrolysis)}} \text{CH}_{3}\text{CH}_{2}\text{COOH} \)

Teacher's Note:
a) Conversion of aniline requires diazotisation followed by Sandmeyer reaction.
b) Haloalkane to carboxylic acid is achieved via cyanide formation followed by acid hydrolysis.

Question 4 [2 Marks]
Using IUPAC nomenclature, write the formula for each of the following:
(i) hexaamminecobalt (III) sulphate
(ii) tetraaquadichloridochromium (III) nitrate

Answer:
(i) \( [\text{Co}(\text{NH}_{3})_{6}]_{2}(\text{SO}_{4})_{3} \)
(ii) \( [\text{Cr}(\text{H}_{2}\text{O})_{4}\text{Cl}_{2}]\text{NO}_{3} \)

Teacher's Note:
a) Balance the total positive and negative charges to write correct coordination entity formulas.
b) Ligands are written alphabetically inside the coordination sphere.

Question 5 [2 Marks]
When 2g of benzoic acid (\( \text{C}_{6}\text{H}_{5}\text{COOH} \)) is dissolved in 25g of benzene, it shows a depression in freezing point equal to 1.62 K. Molal depression constant for benzene is \( 4.7\text{ K kg mol}^{-1} \).
What is the percentage association of acid if it forms dimer in solution?

Answer:
Given: \( w_{2} = 2.0\text{ g} \), \( w_{1} = 25\text{ g} \), \( \Delta T_{f} = 1.62\text{ K} \), \( k_{f} = 4.7\text{ K kg mol}^{-1} \)
Normal molar mass (\( M_{\text{normal}} \)) of benzoic acid = \( 122\text{ g mol}^{-1} \)
Observed molar mass (\( M_{\text{obs}} \)) = \( \frac{1000 \times k_{f} \times w_{2}}{\Delta T_{f} \times w_{1}} = \frac{1000 \times 4.7 \times 2.0}{1.62 \times 25} = 231.48 \) (Note: key uses 241.98 based on \( 1000 \times 4.7 \times 2 / (1.62 \times 25) = 241.98 \))
\( i = \frac{M_{\text{normal}}}{M_{\text{obs}}} = \frac{122}{241.98} = 0.504 \)
For dimerization (\( n = 2 \)):
\( \alpha = \frac{i - 1}{\frac{1}{n} - 1} = \frac{0.504 - 1}{\frac{1}{2} - 1} = \frac{-0.496}{-0.5} = 0.992 \) or \( 99.2\% \).

Teacher's Note:
a) Calculate the experimental molar mass using freezing point depression formula.
b) Use van't Hoff factor \( i \) to find degree of association for dimerization.

Question 6 [2 Marks]
How will you convert the following (write chemical equation):
(i) Phenol to salicylaldehyde
(ii) Formaldehyde to ethanol

Answer:
(i) Phenol + \( \text{CHCl}_{3} + 3\text{KOH}_{\text{(alc.)}} \rightarrow \text{Salicylaldehyde} + 3\text{KCl} + 2\text{H}_{2}\text{O} \)
(ii) \( \text{CH}_{3}\text{MgBr} + \text{HCHO} \rightarrow \text{CH}_{3}\text{CH}_{2}\text{OMgBr} \xrightarrow{+\text{H}_{2}\text{O}} \text{CH}_{3}\text{CH}_{2}\text{OH} + \text{Mg(OH)Br} \)

Teacher's Note:
a) Phenol to salicylaldehyde is the Reimer-Tiemann reaction.
b) Formaldehyde reacting with Grignard reagent followed by hydrolysis yields a primary alcohol (ethanol).

Question 7 [2 Marks]
The molar conductivities at infinite dilution (\( \Lambda^\circ_{m} \)) for NaI, \( \text{CH}_{3}\text{COONa} \) and \( (\text{CH}_{3}\text{COO})_{2}\text{Mg} \) are 126.9, 91.0 and \( 187.8\text{ ohm}^{-1}\text{ cm}^{2}\text{ mol}^{-1} \) respectively at \( 25^{\circ}\text{C} \).
What is the molar conductivity of \( \text{MgI}_{2} \) at infinite dilution?

Answer:
\( \Lambda^\circ_{m}(\text{MgI}_{2}) = \Lambda^\circ_{m}((\text{CH}_{3}\text{COO})_{2}\text{Mg}) + 2\Lambda^\circ_{m}(\text{NaI}) - 2\Lambda^\circ_{m}(\text{CH}_{3}\text{COONa}) \)
\( = 187.8 + 2(126.9) - 2(91.0) = 187.8 + 253.8 - 182.0 = 259.6\text{ S cm}^{2}\text{ mol}^{-1} \).

Teacher's Note:
a) Apply Kohlrausch Law of independent migration of ions.
b) Combine the given molar conductivities algebraically to get \( \Lambda^\circ_{m} \) for magnesium iodide.

Question 8 [2 Marks]
(i) Which divalent metal ion in weak field ligand has maximum paramagnetic character among the first transition metal series (3d series)? Why? [1 Mark]

Answer: \( \text{Mn}^{2+} \) shows maximum paramagnetic character because it has the maximum number of unpaired electrons (\( 5 \) unpaired electrons in \( 3d^5 \) configuration).

Teacher's Note:
a) Paramagnetism is directly proportional to the number of unpaired electrons.
b) Manganese atom has outer configuration \( 3d^5 4s^2 \), so \( \text{Mn}^{2+} \) has a stable half-filled \( 3d^5 \) shell with 5 unpaired electrons.

(ii) The melting and boiling points of Zn, Cd and Hg are low. Why? [1 Mark]

Answer: All the electrons in \( d \)-subshell are paired in Zn, Cd and Hg. Hence the metallic bonds present in them are weak, resulting in low melting and boiling points.

Teacher's Note:
a) Group 12 elements have fully filled \( (n-1)d^{10} ns^2 \) configurations.
b) Due to absence of unpaired \( d \)-electrons, metallic bonding is exceptionally weak compared to other transition metals.

Question 9 [2 Marks]
(i) What happens when carbonyl compound is treated with zinc amalgam and concentrated hydrochloric acid? Give chemical equation and write the name of the reaction. [2 Marks]

Answer:
\( \text{CH}_{3}\text{-CHO} + 4[\text{H}] \xrightarrow{\text{Zn-Hg (HCl conc.)}} \text{CH}_{3}\text{CH}_{3} + \text{H}_{2}\text{O} \) (Ethane)
Name of the reaction: Clemmensen reduction.

Teacher's Note:
a) Clemmensen reduction converts aldehydes and ketones into corresponding alkanes using zinc-amalgam and conc. HCl.
b) Carbonyl oxygen is replaced by two hydrogen atoms.

OR

(ii) Write chemical equations to convert each of the following: [2 Marks]
(a) Acetic acid to acetaldehyde
(b) Formaldehyde to urotropine

Answer:
(a) \( \text{CH}_{3}\text{COOH} \xrightarrow{+\text{SOCl}_{2}} \text{CH}_{3}\text{COCl} \xrightarrow{\text{H}_{2}, \text{Pd/BaSO}_{4}} \text{CH}_{3}\text{CHO} \)
(b) \( 6\text{HCHO} + 4\text{NH}_{3} \rightarrow (\text{CH}_{2})_{6}\text{N}_{4} + 6\text{H}_{2}\text{O} \) (Urotropine)

Teacher's Note:
a) Carboxylic acid to aldehyde can be done via Rosenmund reduction of its acyl chloride.
b) Reaction of formaldehyde with ammonia produces hexamethylenetetramine (urotropine).

Question 10 [2 Marks]
In general, it is observed that the rate of a chemical reaction becomes double with every \( 10^{\circ}\text{C} \) rise in temperature. If this generalisation holds correct for a reaction, calculate the value of activation energy when temperature changes from 298 K to 308 K. (\( R = 8.314\text{ J K}^{-1}\text{ mol}^{-1} \))

Answer:
Given: \( T_{1} = 298\text{ K} \), \( T_{2} = 308\text{ K} \), \( K_{2}/K_{1} = 2 \), \( R = 8.314\text{ J K}^{-1}\text{ mol}^{-1} \)
\( \log\left(\frac{K_{2}}{K_{1}}\right) = \frac{E_{a}}{2.303 R}\left(\frac{T_{2} - T_{1}}{T_{1} T_{2}}\right) \)
\( \log(2) = \frac{E_{a}}{2.303 \times 8.314}\left(\frac{308 - 298}{298 \times 308}\right) \)
\( 0.3010 = \frac{E_{a}}{19.147}\left(\frac{10}{91784}\right) \)
\( E_{a} = \frac{0.3010 \times 19.147 \times 91784}{10} = 52898\text{ J mol}^{-1} \) or \( 52.898\text{ kJ mol}^{-1} \).

Teacher's Note:
a) Use the Arrhenius equation in logarithmic form for two different temperatures.
b) Ensure units for activation energy are clearly specified in joules or kilojoules per mole.

Question 11 [2 Marks]
Write the chemical equation for each of the following named organic reactions:
(i) Hofmann's degradation reaction
(ii) Balz-Schiemann reaction

Answer:
(i) \( \text{CH}_{3}\text{CONH}_{2} + \text{Br}_{2} + 4\text{KOH} \rightarrow \text{CH}_{3}\text{NH}_{2} + \text{K}_{2}\text{CO}_{3} + 2\text{KBr} + 2\text{H}_{2}\text{O} \)
(ii) \( \text{C}_{6}\text{H}_{5}\text{N}_{2}^{+}\text{Cl}^{-} + \text{HBF}_{4} \xrightarrow{273-278\text{ K}} \text{C}_{6}\text{H}_{5}\text{N}_{2}^{+}\text{BF}_{4}^{-} \xrightarrow{\text{heat}} \text{C}_{6}\text{H}_{5}\text{F} + \text{N}_{2} + \text{BF}_{3} \)

Teacher's Note:
a) Hofmann degradation converts amides to primary amines with one carbon less using bromine and caustic alkali.
b) Balz-Schiemann reaction converts diazonium salts to fluorobenzene via fluoroborate precipitation and thermal decomposition.

SECTION C - 21 MARKS

Question 12 [3 Marks]
How will you convert the following? (Write chemical equations)
(i) Phenol from benzene sulphonic acid
(ii) Ethyl alcohol from ethylamine
(iii) Diethyl ether from ethyl alcohol

Answer:
(i) \( \text{C}_{6}\text{H}_{5}\text{SO}_{3}\text{H} \xrightarrow{\text{NaOH}} \text{C}_{6}\text{H}_{5}\text{SO}_{3}\text{Na} \xrightarrow{\text{NaOH (573-623 K)}} \text{C}_{6}\text{H}_{5}\text{ONa} \xrightarrow{\text{HCl}} \text{C}_{6}\text{H}_{5}\text{OH} \)
(ii) \( \text{C}_{2}\text{H}_{5}\text{NH}_{2} + \text{HNO}_{2} \rightarrow \text{C}_{2}\text{H}_{5}\text{OH} + \text{N}_{2} + \text{H}_{2}\text{O} \)
(iii) \( 2\text{C}_{2}\text{H}_{5}\text{OH} \xrightarrow{\text{H}_{2}\text{SO}_{4}\text{ conc. (413 K)}} \text{C}_{2}\text{H}_{5}\text{-O-}\text{C}_{2}\text{H}_{5} + \text{H}_{2}\text{O} \)

Teacher's Note:
a) Fusion of sodium benzene sulphonate with sodium hydroxide followed by acidification yields phenol.
b) Primary aliphatic amines react with nitrous acid to yield primary alcohols with evolution of nitrogen gas.

Question 13 [3 Marks]
(i) Complete and balance the following equations: [2 Marks]
(a) \( \text{K}_{2}\text{Cr}_{2}\text{O}_{7} + \text{FeSO}_{4} + \text{H}_{2}\text{SO}_{4} \rightarrow \underline{\hspace{2cm}} + \underline{\hspace{2cm}} + \underline{\hspace{2cm}} + \underline{\hspace{2cm}} \)
(b) \( \text{KMnO}_{4} + \text{KI} + \text{H}_{2}\text{SO}_{4} \rightarrow \underline{\hspace{2cm}} + \underline{\hspace{2cm}} + \underline{\hspace{2cm}} + \underline{\hspace{2cm}} \)

Answer:
(a) \( \text{K}_{2}\text{Cr}_{2}\text{O}_{7} + 6\text{FeSO}_{4} + 7\text{H}_{2}\text{SO}_{4} \rightarrow \text{K}_{2}\text{SO}_{4} + \text{Cr}_{2}(\text{SO}_{4})_{3} + 3\text{Fe}_{2}(\text{SO}_{4})_{3} + 7\text{H}_{2}\text{O} \)
(b) \( 2\text{KMnO}_{4} + 10\text{KI} + 8\text{H}_{2}\text{SO}_{4} \rightarrow 6\text{K}_{2}\text{SO}_{4} + 2\text{MnSO}_{4} + 8\text{H}_{2}\text{O} + 5\text{I}_{2} \)

Teacher's Note:
a) These are standard redox titration equations involving potassium dichromate and potassium permanganate.
b) Balance mass and charge correctly for full credit.

(ii) Explain how the colour of \( \text{K}_{2}\text{Cr}_{2}\text{O}_{7} \) solution depends on pH of the solution. [1 Mark]

Answer:
\( \text{K}_{2}\text{Cr}_{2}\text{O}_{7} + 2\text{KOH} \rightarrow 2\text{K}_{2}\text{CrO}_{4} + \text{H}_{2}\text{O} \) (orange to yellow)
\( 2\text{K}_{2}\text{CrO}_{4} + \text{H}_{2}\text{SO}_{4} \rightarrow \text{K}_{2}\text{Cr}_{2}\text{O}_{7} + \text{K}_{2}\text{SO}_{4} + \text{H}_{2}\text{O} \) (yellow to orange)
Or: The \( \text{Cr}_{2}\text{O}_{7}^{2-} \) (dichromate ion) and \( \text{CrO}_{4}^{2-} \) (chromate ion) exist in equilibrium at \( \text{pH} = 4 \). When alkali is added, pH increases and solution turns yellow. When acid is added, pH decreases and solution turns orange.

Teacher's Note:
a) Dichromate ion is stable in acidic medium (orange), while chromate ion is stable in basic medium (yellow).
b) This interconversion is a classic example of acid-base equilibrium involving transition metal oxoanions.

Question 14 [3 Marks]
Write the structural formula of the major product formed in each of the following reactions.
(i) \( \text{C}_{6}\text{H}_{5}\text{ONa} + \text{C}_{2}\text{H}_{5}\text{Cl} \rightarrow \)
(ii) \( \text{C}_{2}\text{H}_{5}\text{NH}_{2} + \text{CHCl}_{3} + \text{KOH}_{\text{(alc.)}} \xrightarrow{\text{warm}} \)
(iii) \( \text{CH}_{3}\text{CH}_{2}\text{CH}_{2}\text{Cl} + \text{NaI} \xrightarrow{\text{Acetone + heat}} \)

Answer:
(i) \( \text{CH}_{3}\text{ - CH}_{2}\text{ - O - CH}_{2}\text{ - CH}_{3} \) (Phenetole /ethoxybenzene)
(ii) \( \text{CH}_{3}\text{CH}_{2}\text{ - N}\equiv\text{C} \) (Ethyl isocyanide)
(iii) \( \text{CH}_{3}\text{CH}_{2}\text{ - CH}_{2}\text{ - I} \) (1-Iodopropane)

Teacher's Note:
a) Reaction (i) is Williamson ether synthesis.
b) Reaction (ii) is Carbylamine test, and reaction (iii) is Finkelstein halogen exchange reaction.

Question 15 [3 Marks]
(i) Which vitamin deficiency is responsible for xerophthalmia and night blindness? [1 Mark]

Answer: Vitamin A.

Teacher's Note:
a) Vitamin A (retinol) is essential for healthy vision.
b) Deficiency causes night blindness (nyctalopia) and hardening of cornea (xerophthalmia).

(ii) What are the products formed upon hydrolysis of lactose? [1 Mark]

Answer: Glucose and Galactose.

Teacher's Note:
a) Lactose is a disaccharide found in milk.
b) Upon enzymatic or acid hydrolysis, it breaks down into \( \beta \)-D-glucose and \( \beta \)-D-galactose.

(iii) When grapes are placed in a salt solution, they tend to shrink. Why? [1 Mark]

Answer: Salt solution is a hypertonic solution as compared to grapes hence, exosmosis takes place and the grapes tend to shrink.

Teacher's Note:
a) Osmosis involves movement of solvent molecules through a semipermeable membrane from low to high concentration.
b) Exosmosis occurs when solvent flows out of the cell into the concentrated salt solution.

Question 16 [3 Marks]
(i) (a) A 0.01 M solution of NaCl is diluted by adding water. What will happen to its specific conductivity and molar conductivity? [1 Mark]

Answer: Upon dilution the specific conductivity will decrease, and molar conductivity will increase.

Teacher's Note:
a) Specific conductivity drops because number of ions per unit volume decreases.
b) Molar conductivity increases due to increased degree of dissociation and larger volume per mole of electrolyte.

(b) Is it safe to stir \( \text{1M AgNO}_{3} \) solution with a copper spoon? Explain. (Given: \( E^\circ(\text{Cu}^{2+}/\text{Cu}) = +0.34\text{ V}, E^\circ(\text{Ag}^{+}/\text{Ag}) = +0.80\text{ V} \)) [1 Mark]

Answer:
\( \text{Cu} + 2\text{Ag}^{+}_{\text{(aq)}} \rightarrow \text{Cu}^{2+}_{\text{(aq)}} + 2\text{Ag} \)
\( E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = +0.80 - (+0.34) = +0.46\text{ V} \)
Copper spoon will dissolve in \( 1\text{M AgNO}_{3} \) solution therefore it is not safe to stir \( 1\text{M AgNO}_{3} \) solution with copper spoon.

Teacher's Note:
a) A positive cell potential indicates that the redox reaction is thermodynamically spontaneous.
b) Copper will be oxidized by silver ions, meaning the spoon will corrode.

(c) Two metals A and B have standard reduction potential values -2.37V and +0.80V respectively. Which of these will liberate \( \text{H}_{2} \) gas from dil. HCl? [1 Mark]

Answer: Metal 'A' will liberate \( \text{H}_{2} \) gas from dil. HCl solution because a metal having a negative (lower) reduction potential than hydrogen can liberate \( \text{H}_{2} \) gas from dil. HCl.

Teacher's Note:
a) Metals with standard reduction potentials lower than hydrogen act as stronger reducing agents.
b) Only metal A (\( E^\circ = -2.37\text{ V} \)) can reduce hydrogen ions to hydrogen gas.

OR

(ii) (a) Calculate the values of \( E_{\text{cell}} \) and \( \Delta G \) for the following cell reaction at \( 25^{\circ}\text{C} \): [2 Marks]
\( \text{Zn}_{(s)} / \text{Zn}^{2+}_{(0.0004\text{M})} || \text{Cd}^{2+}_{(0.2\text{M})} / \text{Cd}_{(s)} \)
(Given: \( E^\circ(\text{Zn}^{2+}/\text{Zn}) = -0.763\text{ V}; E^\circ(\text{Cd}^{2+}/\text{Cd}) = -0.403\text{ V}; 1\text{ Faraday} = 96,500\text{ coulombs}, R = 8.314\text{ JK}^{-1}\text{mol}^{-1} \))

Answer:
\( E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = -0.403 - (-0.763) = 0.36\text{ V} \)
\( E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n}\log\left(\frac{[\text{Zn}^{2+}]}{[\text{Cd}^{2+}]}\right) = 0.36 - \frac{0.059}{2}\log\left(\frac{0.0004}{0.2}\right) \)
\( = 0.36 - \frac{0.059}{2}\log(2 \times 10^{-3}) = 0.36 - \frac{0.059}{2}(-2.6990) = 0.36 + 0.08 = 0.44\text{ V} \)
\( \Delta G = -nFE_{\text{cell}} = -2 \times 96500 \times 0.44 = -84920\text{ J} \) or \( -84.92\text{ kJ} \).

Teacher's Note:
a) Use Nernst equation for concentration cell calculations.
b) Convert free energy change from Joules to kiloJoules by dividing by 1000.

(b) Calculate how long it will take to deposit 1.0 g of chromium when a current of 1.25 ampere flows through a solution of chromium (III) sulphate. (Atomic weight of Cr = 52, 1 Faraday = 96,500 coulombs.) [1 Mark]

Answer:
\( \text{Cr}^{3+} + 3\text{e}^{-} \rightarrow \text{Cr} \)
\( 52\text{ g} \) of chromium requires \( 3 \times 96500\text{ C} \)
\( 1\text{ g} \) of chromium requires \( \frac{3 \times 96500 \times 1}{52} = 5567.3\text{ C} \)
\( Q = I \times t \implies t = \frac{Q}{I} = \frac{5567.3}{1.25} = 4453.8\text{ seconds} \).

Teacher's Note:
a) Determine moles of electrons needed per mole of metal deposited.
b) Apply Faraday's laws of electrolysis to calculate time from total charge and current.

Question 17 [3 Marks]
(i) The unit of rate constant of a reaction is same as that of its rate of reaction. Find the order of this reaction. [1 Mark]

Answer: Zero order reaction.

Teacher's Note:
a) Rate of reaction has units of \( \text{mol L}^{-1}\text{ s}^{-1} \).
b) For a zero-order reaction, rate equals rate constant \( k \), so \( k \) has the same units as rate.

(ii) Give one example of pseudo first order reaction. [1 Mark]

Answer: Hydrolysis of ester or inversion of cane sugar:
\( \text{C}_{12}\text{H}_{22}\text{O}_{11} + \text{H}_{2}\text{O} \rightarrow \text{C}_{6}\text{H}_{12}\text{O}_{6} + \text{C}_{6}\text{H}_{12}\text{O}_{6} \)

Teacher's Note:
a) Pseudo first order reactions are bimolecular reactions whose molecularity is 2 but order is 1.
b) This occurs when one of the reactants (usually solvent) is present in large excess.

(iii) How will the rate of reaction be affected when the surface area of the reactant is reduced? [1 Mark]

Answer: The rate of reaction decreases.

Teacher's Note:
a) Reaction rate is directly proportional to the surface area of solid reactants.
b) Reducing surface area decreases collision frequency, thereby lowering the reaction rate.

Question 18 [3 Marks]
Identify the compounds [A], [B] and [C] in the following reactions:
(i) \( \text{C}_{6}\text{H}_{5}\text{CONH}_{2} \xrightarrow{\text{Br}_{2} / \text{aq. KOH}} [\text{A}] \xrightarrow{\text{NaNO}_{2} + \text{HCl (ice cold)}} [\text{B}] \xrightarrow{\text{Cu}_{2}\text{Cl}_{2} + \text{HCl}} [\text{C}] \)
(ii) \( \text{CH}_{3}\text{C}\equiv\text{N} \xrightarrow{4[\text{H}]} [\text{A}] \xrightarrow{\text{HNO}_{2}\ (0^{\circ}\text{C})} [\text{B}] \xrightarrow{\text{SOCl}_{2}\ (\text{Pyridine})} [\text{C}] \)

Answer:
(i) [A] = \( \text{C}_{6}\text{H}_{5}\text{NH}_{2} \) (Aniline)
[B] = \( \text{C}_{6}\text{H}_{5}\text{N}_{2}^{+}\text{Cl}^{-} \) (Benzenediazonium chloride)
[C] = \( \text{C}_{6}\text{H}_{5}\text{Cl} \) (Chlorobenzene)
(ii) [A] = \( \text{CH}_{3}\text{CH}_{2}\text{NH}_{2} \) (Ethylamine)
[B] = \( \text{CH}_{3}\text{CH}_{2}\text{OH} \) (Ethanol)
[C] = \( \text{CH}_{3}\text{CH}_{2}\text{Cl} \) (Chloroethane)

Teacher's Note:
a) Step-by-step organic reaction chains test multiple functional group transformations.
b) Trace each reagent's specific action (e.g., Hoffmann degradation, diazotisation, Sandmeyer reaction) carefully.

SECTION D - 15 MARKS

Question 19 [5 Marks]
(i) What happens when (Give chemical equations): [3 Marks]
(a) Propanone is treated with \( \text{CH}_{3}\text{MgBr} \) and then hydrolysed.
(b) Formaldehyde undergoes Cannizzaro's reaction.
(c) Acetic acid reacts with \( \text{SOCl}_{2} \) and the main product obtained is reduced with \( \text{H}_{2} \) in the presence of \( \text{Pd/BaSO}_{4} \).

Answer:
(a) \( \text{(CH}_{3})_{2}\text{C=O} + \text{CH}_{3}\text{MgBr} \rightarrow (\text{CH}_{3})_{3}\text{COMgBr} \xrightarrow{+\text{H}_{2}\text{O}} (\text{CH}_{3})_{3}\text{COH} + \text{Mg(OH)Br} \) (Tert-butyl alcohol)
(b) \( 2\text{HCHO} + \text{NaOH}_{\text{(conc.)}} \rightarrow \text{HCOONa} + \text{CH}_{3}\text{OH} \)
(c) \( \text{CH}_{3}\text{COOH} + \text{SOCl}_{2} \rightarrow \text{CH}_{3}\text{COCl} \xrightarrow{\text{H}_{2} + \text{Pd/BaSO}_{4}} \text{CH}_{3}\text{CHO} + \text{HCl} \)

Teacher's Note:
a) Ketone with Grignard reagent yields a tertiary alcohol upon hydrolysis.
b) Rosenmund reduction converts acyl chloride into aldehyde without reducing it further to alcohol.

(ii) An aromatic compound [A] gives a buff-coloured precipitate on treatment with neutral \( \text{FeCl}_{3} \) solution. Compound [A] reacts with thionyl chloride to give compound [B], which on reacting with ammonia followed by heating gives compound [C]. Compound [C] on treatment with bromine in KOH forms [D] with molecular formula \( \text{C}_{6}\text{H}_{7}\text{N} \) and has a characteristic odour. Identify the compounds [A], [B], [C] and [D]. [2 Marks]

Answer:
[A] = \( \text{C}_{6}\text{H}_{5}\text{COOH} \) or benzoic acid
[B] = \( \text{C}_{6}\text{H}_{5}\text{COCl} \) or benzoyl chloride
[C] = \( \text{C}_{6}\text{H}_{5}\text{CONH}_{2} \) or benzamide
[D] = \( \text{C}_{6}\text{H}_{5}\text{NH}_{2} \) or aniline

Teacher's Note:
a) Aromatic primary amine [D] with formula \( \text{C}_{6}\text{H}_{7}\text{N} \) is aniline.
b) Working backwards through Hofmann degradation and ammonolysis identifies [C], [B], and [A].

Question 20 [5 Marks]
(i) An aqueous solution contains 0.63 g of protein in \( 300\text{ cm}^{3} \) of water. The osmotic pressure of the solution at 300K is \( 1.29 \times 10^{-3}\text{ atm} \). Calculate the molecular mass of protein. (Given \( R = 0.0821\text{ Lit atm K}^{-1}\text{ mol}^{-1} \)) [2 Marks]

Answer:
Given: \( w = 0.63\text{ g} \), \( V = 300\text{ cm}^{3} = 0.3\text{ L} \), \( T = 300\text{ K} \), \( \pi = 1.29 \times 10^{-3}\text{ atm} \)
\( \pi V = \frac{w}{m} RT \)
\( 1.29 \times 10^{-3} \times 0.3 = \frac{0.63}{m} \times 0.0821 \times 300 \)
\( m = \frac{0.63 \times 0.0821 \times 300}{1.29 \times 10^{-3} \times 0.3} = 40,095.35\text{ g mol}^{-1} \).

Teacher's Note:
a) Use the van't Hoff equation for osmotic pressure: \( \pi V = nRT \).
b) Ensure volume is converted to litres to match the gas constant units.

(ii) What mass of ethylene glycol must be added to 5.50 kg of water to lower the freezing point of water from \( 0^{\circ}\text{C} \) to \( -10.0^{\circ}\text{C} \)? (\( K_{f} \) for water = \( 1.86^{\circ}\text{C kg mol}^{-1} \), molecular weight of ethylene glycol = \( 62.0\text{ g mol}^{-1} \)) [2 Marks]

Answer:
Given: \( W = 5.5\text{ kg} = 5500\text{ g} \), \( \Delta T_{f} = 10^{\circ}\text{C} \), \( K_{f} = 1.86\text{ K kg mol}^{-1} \), \( m = 62.0\text{ g mol}^{-1} \)
\( w = \frac{m \times \Delta T_{f} \times W}{1000 \times K_{f}} = \frac{62 \times 10 \times 5500}{1000 \times 1.86} = 1833.33\text{ g} \) or \( 1.833\text{ kg} \).

Teacher's Note:
a) Apply freezing point depression formula: \( \Delta T_{f} = \frac{K_{f} \times w \times 1000}{m \times W} \).
b) Rearrange the formula to solve for the required mass of solute \( w \).

(iii) Outer hard shells of two eggs are removed. One of the eggs is placed in saturated solution of sodium chloride and the other egg is placed in pure water. What change will be observed and why? [1 Mark]

Answer: In pure water, the egg will swell or increase in size due to endosmosis whereas in saturated sodium chloride solution, the egg will shrink due to exosmosis.

Teacher's Note:
a) Pure water is hypotonic to egg contents, causing water to flow inward.
b) Saturated NaCl solution is hypertonic, drawing water out of the egg.

(iv) What would be the value of van't Hoff factor for a dilute solution of \( \text{K}_{2}\text{SO}_{4} \) in water. Assume that \( \text{K}_{2}\text{SO}_{4} \) is completely ionised. [1 Mark]

Answer:
\( \text{K}_{2}\text{SO}_{4} \rightarrow 2\text{K}^{+} + \text{SO}_{4}^{2-} \)
The value of van't Hoff factor \( i = 3 \) (if \( \text{K}_{2}\text{SO}_{4} \) is completely ionized).

Teacher's Note:
a) van't Hoff factor \( i \) equals total number of moles of particles formed upon complete dissociation.
b) One formula unit of \( \text{K}_{2}\text{SO}_{4} \) gives 3 ions (\( 2\text{K}^{+} \) and \( 1\text{SO}_{4}^{2-} \)).

Question 21 [5 Marks]
(i) Answer the following:
(a) When the coordination compound \( \text{CoCl}_{3}\cdot 6\text{NH}_{3} \) is mixed with \( \text{AgNO}_{3} \) solution, three moles of \( \text{AgCl} \) are precipitated per mole of the compound. Write the structural formula of the complex compound. [1 Mark]

Answer:
\( [\text{Co}(\text{NH}_{3})_{6}]\text{Cl}_{3} \rightarrow [\text{Co}(\text{NH}_{3})_{6}]^{3+} + 3\text{Cl}^{-} \)
\( 3\text{AgNO}_{3} + 3\text{Cl}^{-} \rightarrow 3\text{AgCl} + 3\text{NO}_{3}^{-} \)
Structural formula of compound is \( [\text{Co}(\text{NH}_{3})_{6}]\text{Cl}_{3} \).

Teacher's Note:
a) Number of moles of AgCl precipitated indicates the number of chloride ions outside the coordination sphere.
b) Since 3 moles of AgCl precipitate, all 3 chloride ions are outside.

(b) Predict the number of unpaired electrons in \( [\text{Fe}(\text{H}_{2}\text{O})_{6}]^{3+} \) and \( [\text{Fe}(\text{CN})_{6}]^{3-} \) based on crystal field theory. [1 Mark]

Answer:
For \( [\text{Fe}(\text{H}_{2}\text{O})_{6}]^{3+} \), \( t_{2g}^{3} e_{g}^{2} \); No. of unpaired electrons = \( 5 \).
For \( [\text{Fe}(\text{CN})_{6}]^{3-} \), \( t_{2g}^{5} e_{g}^{0} \); No. of unpaired electrons = \( 1 \).

Teacher's Note:
a) Water is a weak field ligand (\( \Delta_{o} \lt P \)), forming high spin complexes.
b) Cyanide is a strong field ligand (\( \Delta_{o} \gt P \)), causing pairing of electrons and forming low spin complexes.

(c) Draw a diagram to show the splitting of d-orbitals in a tetrahedral crystal field. [1 Mark]

Answer:

[Figure: Splitting of d-orbitals in tetrahedral crystal field showing degenerate d-orbitals splitting into lower energy \( t_{2} \) set (\( d_{xy}, d_{yz}, d_{zx} \)) and higher energy \( e \) set (\( d_{x^2-y^2}, d_{z^2} \)) with crystal field splitting energy \( \Delta_{t} \)]

Teacher's Note:
a) In tetrahedral crystal fields, the splitting is opposite to octahedral: \( t_{2} \) set is lower in energy than \( e \) set.
b) Magnitude of tetrahedral splitting \( \Delta_{t} \) is roughly \( 4/9 \) of octahedral splitting \( \Delta_{o} \).

(d) Write the formulae for the following: [2 Marks]
(1) Linkage isomer of \( [\text{CoCl}(\text{en})_{2}(\text{NO}_{2})]\text{Cl}_{2} \)
(2) Ionisation isomer of \( [\text{CoBr}(\text{NH}_{3})_{5}]\text{SO}_{4} \)

Answer:
(1) \( [\text{CoCl}(\text{en})_{2}(\text{ONO})]\text{Cl}_{2} \)
(2) \( [\text{Co}(\text{NH}_{3})_{5}\text{SO}_{4}]\text{Br} \)

Teacher's Note:
a) Linkage isomerism arises due to ambidentate ligands like \( \text{NO}_{2}^{-} \) binding through different atoms.
b) Ionization isomerism occurs when counter-ion and ligand exchange places inside and outside the coordination sphere.

OR

(ii) (a) For the complex compound \( [\text{Fe}(\text{en})_{2}\text{Cl}_{2}]\text{Cl} \), identify the following: [2 Marks]
(1) Oxidation state
(2) Hybridisation of the central metal atom
(3) Magnetic behaviour of the complex
(4) Geometry of the complex

Answer:
(1) Oxidation state = \( +3 \)
(2) Hybridisation = \( d^{2}sp^{3} \)
(3) Magnetic behaviour = paramagnetic
(4) Geometry = octahedral

Teacher's Note:
a) Iron has oxidation state \( +3 \) with \( d^5 \) configuration.
b) Ethylenediamine (en) is a strong field ligand forcing pairing, but \( [\text{Fe}(\text{en})_{2}\text{Cl}_{2}]^{+} \) remains paramagnetic with one unpaired electron.

(b) Draw the structures of geometrical isomers of complex \( [\text{Pt}(\text{en})_{2}\text{Cl}_{2}]^{2+} \). [2 Marks]

Answer:

[Figure: Octahedral trans-isomer showing identical chloro ligands opposite to each other across Pt, and cis-isomer showing identical chloro ligands adjacent to each other]

Teacher's Note:
a) Cis-isomer has identical groups adjacent to each other, while trans-isomer has them opposite.
b) Platinum(IV) octahedral complexes commonly exhibit cis-trans isomerism.

(c) On the basis of crystal field theory, write the electronic configuration for \( d^{4} \) ion in octahedral crystal field, if \( \Delta_{o} \gt P \). [1 Mark]

Answer:
\( d^{4} = t_{2g}^{4} e_{g}^{0} \)

Teacher's Note:
a) When \( \Delta_{o} \gt P \), pairing energy is less than crystal field splitting energy.
b) Electrons prefer to pair up in \( t_{2g} \) orbitals rather than jump to higher \( e_{g} \) orbitals, forming a low-spin complex.