ISC Class 12 Chemistry Sample Paper 2024 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]
[ decreases, CN^- ion, activation energy, catalyst, two, Fe²⁺ ion, carbon, lattice energy, enzyme, five, double, halogen, triple, increases]
(i) In the Haber process, iron changes the _______ of reaction while molybdenum increases the efficiency of the _________. [1 Mark]

Answer: activation energy, catalyst

Teacher's Note:
a) Iron acts as a catalyst which lowers the activation energy of the reaction.
b) Molybdenum acts as a promoter (or catalyst activator) that increases the efficiency of the catalyst.

(ii) The number of ions that will be produced when potassium ferrocyanide, K₄[Fe(CN)₆], dissolves in water is _________. This shows that __________ is the ligand in the coordination compound. [1 Mark]

Answer: five, CN^- ion

Teacher's Note:
a) K₄[Fe(CN)₆] ionises to give 4 K⁺ ions and 1 [Fe(CN)₆]^4- ion, totalling 5 ions.
b) Cyanide ion (CN^-) coordinates with the central metal atom/ion as a ligand.

(iii) Haloalkenes undergo both nucleophilic and electrophilic reactions due to the presence of __________bond and the __________ atom. [1 Mark]

Answer: double, halogen

Teacher's Note:
a) The presence of a carbon-carbon double bond allows electrophilic addition reactions.
b) The presence of a halogen atom attached to carbon allows nucleophilic substitution reactions.

(iv) In case of alcohols, as the carbon chain length increases, the boiling point ____________ and the solubility in water____________. [1 Mark]

Answer: increases, decreases

Teacher's Note:
a) Boiling point increases due to an increase in molecular mass and Van der Waals forces.
b) Solubility in water decreases due to the increasing hydrophobic hydrocarbon part offsetting the hydrophilic hydroxyl group.

(B) Select and write the correct alternative from the choices given below: [7×1]
(i) A potassium iodide (KI) solution containing starch turns blue on the addition of chlorine. Which one of the following statements explain this?
(P) The reduction potential of Cl₂ is more than that of I₂.
(Q) The oxidation potential of Cl₂ is more than that of I₂.
(R) The product formed when Cl₂ combines with starch is blue.
(S) The product formed when I₂ combines with starch is blue. [1 Mark]
(a) Only P and R
(b) Only Q and R
(c) Only Q and S
(d) Only P and S

Answer: (d) Only P and S

Chlorine displaces iodine from potassium iodide because chlorine has a higher reduction potential than iodine (statement P). The liberated iodine reacts with starch to give a characteristic blue-black complex (statement S).

Teacher's Note:
a) Higher standard reduction potential means greater oxidising power, so Cl₂ oxidises I^- to I₂.
b) The starch-iodine inclusion complex is intensely blue in colour.

(ii) Crystal field splitting energy (CFSE) for high spin d⁴ octahedral complex is: [1 Mark]
(a) \(-1.6 \Delta_o\)
(b) \(-1.2 \Delta_o\)
(c) \(-0.8 \Delta_o\)
(d) \(-0.6 \Delta_o\)

Answer: (c) \(-0.8 \Delta_o\)

For a high spin \(d^4\) octahedral complex, three electrons enter the \(t_{2g}\) orbitals and one electron enters the \(e_g\) orbital. CFSE = \(3 \times (-0.4 \Delta_o) + 1 \times (+0.6 \Delta_o) = -1.2 \Delta_o + 0.6 \Delta_o = -0.6 \Delta_o\).

Teacher's Note:
a) The official key shows \(-0.6 \Delta_o\); the correct calculation is \(3(-0.4) + 1(+0.6) = -0.6 \Delta_o\).
b) Students must remember the energy levels: \(t_{2g}\) is \(-0.4 \Delta_o\) per electron and \(e_g\) is \(+0.6 \Delta_o\) per electron.

(iii) Acidified K₂Cr₂O₇ solution turns green when Na₂SO₃ is added to it. This is due to formation of: [1 Mark]
(a) CrO₄^2-
(b) Cr₂(SO₃)₃
(c) Cr₂O₃
(d) Cr₂(SO₄)₃

Answer: (d) Cr₂(SO₄)₃

Acidified potassium dichromate oxidises sodium sulphite to sodium sulphate, while dichromate itself is reduced to chromium(III) sulphate, which imparts a green colour to the solution.

Teacher's Note:
a) Orange dichromate (\(Cr^{6+}\)) is reduced to green chromium(III) ions (\(Cr^{3+}\)).
b) Sulphite ions are oxidized to sulphate ions in an acidic medium.

(iv) Which of the following product is formed when benzene diazonium chloride is reduced by hypophosphorous acid (H₃PO₂) in the presence of cuprous ion as catalyst? [1 Mark]
(a) Phenol
(b) Aniline
(c) Benzene
(d) Benzene cyanide

Answer: (c) Benzene

Reduction of benzene diazonium chloride with H₃PO₂ replaces the diazonium group with hydrogen, forming benzene.

Teacher's Note:
a) This is a deamination reaction (reduction of diazonium salt to arene).
b) H₃PO₂ gets oxidized to phosphorous acid (H₃PO₃).

(v) Which of the following aqueous solution has lowest vapour pressure? [1 Mark]
(a) 1M NaCl
(b) 1M K₂SO₄
(c) 1M Glucose
(d) 1M Sucrose

Answer: (b) 1M K₂SO₄

Lowering of vapour pressure is a colligative property directly proportional to the total number of particles (van 't Hoff factor \(i \times M\)). K₂SO₄ dissociates into 3 ions (\(2K^+ + SO_4^{2-}\)), giving the highest particle concentration (effective molality = \(3 \times 1 = 3\)).

Teacher's Note:
a) Higher particle concentration leads to a greater lowering of vapour pressure and hence the lowest vapour pressure.
b) NaCl gives 2 ions, whereas glucose and sucrose do not dissociate (\(i = 1\)).

(vi) Assertion: Adding water to two beakers ‘A’ and ‘B’ containing NaOH and CH₃COOH solutions respectively will increase the molar conductance (\(\Lambda_m\)) of the solutions sharply in beaker ‘A’ and slowly in beaker ‘B’.
Reason: Molar conductance (\(\Lambda_m\)) increases with a decrease in concentration or upon dilution. [1 Mark]
(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not the correct explanation for Assertion.
(c) Assertion is true but Reason is false.
(d) Assertion is false but Reason is true.

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

NaOH is a strong electrolyte whose molar conductance increases only slightly on dilution. CH₃COOH is a weak electrolyte whose degree of dissociation increases sharply on dilution, leading to a sharp increase in molar conductance in beaker B, not beaker A. Thus the assertion is reversed.

Teacher's Note:
a) Strong electrolytes show a gradual, linear increase in molar conductance with dilution due to reduced interionic forces.
b) Weak electrolytes show a sharp increase in molar conductance upon dilution due to a rapid increase in degree of dissociation.

(vii) Assertion: Aniline is soluble in HCl while it is only slightly soluble in water.
Reason: Aniline cannot make hydrogen bonds with water but gets protonated easily by acids. [1 Mark]
(a) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not the correct explanation for Assertion.
(c) Assertion is true but Reason is false.
(d) Assertion is false but Reason is true.

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

Aniline can form hydrogen bonds with water molecules due to its amino group, but its low solubility in water is due to the large hydrophobic benzene ring. It dissolves in HCl because it forms a soluble salt (anilinium chloride) via protonation. Thus the reason statement claiming it cannot make hydrogen bonds with water is false.

Teacher's Note:
a) Primary aromatic amines form intermolecular hydrogen bonds with water, but solubility is limited by the aromatic ring.
b) Reaction with mineral acids forms water-soluble amine hydrochloride salts.

(C) Read the passage given below carefully and answer the questions that follow. [3×1]
During the winter season in a particular year, Kashmir experienced heavy snowfall. It was an unexpected snowfall. Thousands of visitors were stranded because it was dangerous to travel on snowy roads and vehicles could not move as water froze in the car radiators. In such conditions officials decided to sprinkle rock salt or CaCl₂ on roads.
(i) Why was it decided to sprinkle rock salt or CaCl₂ on the roads? [1 Mark]

Answer:
To depress the freezing point of water and melt the ice/snow on the roads.

Teacher's Note:
a) Addition of ionic solutes causes depression in freezing point of water.
b) This prevents ice formation and melts existing snow at sub-zero temperatures.

(ii) A mixture of ethylene glycol and water is used as coolant in car radiators. Why? [1 Mark]

Answer:
It acts as an antifreeze, lowering the freezing point of water in winter and raising the boiling point in summer.

Teacher's Note:
a) Ethylene glycol forms hydrogen bonds with water, altering vapour pressure and freezing/boiling points.
b) It prevents the coolant liquid from freezing during winter or boiling over during summer.

(iii) How many grams of ethylene glycol (mol. wt. = 62 g mol^-1) should be added to 10 kg of water so that the solution freezes at -10^oC? (K_f for water = 1.86 K kg mol^-1) [1 Mark]

Answer:
\(w_2 = \frac{\Delta T_f \times M_2 \times w_1}{K_f \times 1000} = \frac{10 \times 62 \times 10000}{1.86 \times 1000} = 3333.33 \text{ g}\).

Teacher's Note:
a) Use the depression in freezing point formula: \(\Delta T_f = K_f \times m\).
b) Substitute \(\Delta T_f = 10 \text{ K}\), \(K_f = 1.86\), and solvent mass \(w_1 = 10 \text{ kg} = 10000 \text{ g}\).

SECTION B - 20 MARKS

Question 2 [2]
(i) Arrange the following alcohols in order of decreasing activity towards Lucas reagent.
2-butanol, 2-methyl-2-propanol and 1-butanol [1 Mark]

Answer:
2-methyl-2-propanol > 2-butanol > 1-butanol.

Teacher's Note:
a) Lucas reagent reactivity depends on carbocation stability formed during the reaction (\(3^{\circ} \gt 2^{\circ} \gt 1^{\circ}\)).
b) 2-methyl-2-propanol is tertiary, 2-butanol is secondary, and 1-butanol is primary.

(ii) Ethanol has a higher boiling point than methoxymethane. Justify the statement. [1 Mark]

Answer:
Ethanol molecules associate through strong intermolecular hydrogen bonding, whereas methoxymethane (an ether) has only weak dipole-dipole interactions.

Teacher's Note:
a) Hydrogen bonds require extra thermal energy to break compared to ether dipole interactions.
b) Consequently, ethanol has a significantly higher boiling point than methoxymethane of comparable molecular mass.

Question 3 [2]
Give a reason for each of the following:
(i) The size of the trivalent cations in Lanthanoid series decreases steadily as the atomic number increases. [1 Mark]

Answer:
Due to lanthanoid contraction caused by the poor shielding effect of 4f electrons.

Teacher's Note:
a) As atomic number increases, nuclear charge increases by one unit at each step.
b) The 4f electrons shield the outer electrons very poorly, resulting in a greater effective nuclear charge pulling the electron cloud inward.

(ii) The third ionization energy of manganese (Z = 25) is unexpectedly high. [1 Mark]

Answer:
Removal of the third electron involves breaking the exceptionally stable half-filled \(3d^5\) electronic configuration of \(Mn^{2+}\).

Teacher's Note:
a) Electronic configuration of Mn is \([Ar] 3d^5 4s^2\) and of \(Mn^{2+}\) is \([Ar] 3d^5\).
b) Half-filled orbitals have extra exchange energy and spherical symmetry, making them very stable.

Question 4 [2]
Give balanced chemical equations to convert the following:
(i) Benzene to biphenyl [1 Mark]

Answer:
\(2C_6H_6 + 2Na \xrightarrow{Dry ether} C_6H_5-C_6H_5 + 2NaH\) (or via Fittig reaction using chlorobenzene: \(2C_6H_5Cl + 2Na \xrightarrow{Dry ether} C_6H_5-C_6H_5 + 2NaCl\)).

Teacher's Note:
a) This is an example of the Wurtz-Fittig or Fittig reaction coupling aryl halides with sodium.
b) Ensure conditions like dry ether are mentioned in the equation.

(ii) Propene to propane-1-ol [1 Mark]

Answer:
\(CH_3-CH=CH_2 + HBr \xrightarrow{Peroxide} CH_3-CH_2-CH_2Br\)
\(CH_3-CH_2-CH_2Br + KOH(aq) \rightarrow CH_3-CH_2-CH_2OH + KBr\)

Teacher's Note:
a) Anti-Markovnikov addition of HBr in the presence of peroxide gives 1-bromopropane.
b) Nucleophilic substitution with aqueous KOH converts 1-bromopropane to propane-1-ol.

Question 5 [2]
Account for the following:
(i) Salts of cuprous (Cu⁺) ion are colourless whereas the salts of cupric (Cu²⁺) ion are coloured. [1 Mark]

Answer:
\(Cu^+\) has a completely filled \(3d^{10}\) configuration with no unpaired d-electrons, so d-d transition is not possible. \(Cu^{2+}\) has a \(3d^9\) configuration with one unpaired electron, permitting d-d transitions.

Teacher's Note:
a) Colour in transition metal ions is due to d-d electronic transitions in the visible region.
b) Diamagnetic \(Cu^+\) is colourless, whereas paramagnetic \(Cu^{2+}\) salts are blue or green.

(ii) Zinc is not regarded as a transition element. (at. no. of Zn = 30) [1 Mark]

Answer:
Zinc has a completely filled \(3d^{10}\) electronic configuration in its ground state as well as in its common oxidation state (\(Zn^{2+}\)).

Teacher's Note:
a) By definition, a transition element must have partially filled d-orbitals in its ground state or in stable oxidation states.
b) Hence Zn, Cd, and Hg are block d-elements but not transition elements.

Question 6 [2]
Two compounds, D-2-chlorobutane and L-2-chlorobutane, are enantiomers of each other.
Name one physical property that is:
(i) same for D-2-chlorobutane and L-2-chlorobutane. [1 Mark]

Answer:
Boiling point (or melting point, density, refractive index, solubility in achiral solvents).

Teacher's Note:
a) Enantiomers have identical scalar physical properties.
b) Only interaction with plane-polarised light or chiral reagents differs.

(ii) different for D-2-chlorobutane and L-2-chlorobutane. [1 Mark]

Answer:
Direction of optical rotation (specific rotation).

Teacher's Note:
a) D-isomer rotates plane-polarised light to the right (dextrorotatory), while L-isomer rotates it to the left (laevorotatory).
b) The magnitude of optical rotation is identical, only the sign differs.

Question 7 [2]
(i) A rusted piece of iron undergoes electrochemical reactions. Write the chemical reaction taking place at:
(a) the electrode that behaves as an anode. [1 Mark]

Answer:
\(Fe_{(s)} \rightarrow Fe^{2+}_{(aq)} + 2e^-\)

Teacher's Note:
a) Oxidation occurs at the anodic region of the iron surface.
b) Iron loses electrons to form ferrous ions.

(b) the electrode that behaves as a cathode. [1 Mark]

Answer:
\(O_{2(g)} + 4H^+_{(aq)} + 4e^- \rightarrow 2H_2O_{(l)}\)

Teacher's Note:
a) Reduction occurs at the cathodic region using dissolved oxygen and hydrogen ions.
b) Electrons released at the anode are consumed here.

(ii) Given that the standard reduction potential for Al³⁺/Al = -1.66 V and \(\frac{1}{2} I_2/I^- = 0.54V\), what will be the standard potential of the cell made by using Al³⁺ and I^-? [1 Mark]

Answer:
\(E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.54 - (-1.66) = 2.20 \text{ V}\).

Teacher's Note:
a) Aluminium has a lower reduction potential, so it acts as the anode.
b) Iodine/iodide couple has a higher reduction potential, so it acts as the cathode.

Question 8 [2]
(i) What happens when (write chemical reactions only)
(a) Diethyl ether is treated with phosphorous pentachloride. [1 Mark]

Answer:
\(C_2H_5-O-C_2H_5 + PCl_5 \rightarrow 2C_2H_5Cl + POCl_3\)

Teacher's Note:
a) Ethers react with PCl₅ to form alkyl chlorides and phosphorus oxychloride.
b) Two moles of ethyl chloride are produced from one mole of diethyl ether.

(b) Ethyl alcohol is treated with methyl magnesium bromide. [1 Mark]

Answer:
\(C_2H_5OH + CH_3MgBr \rightarrow CH_4 + CH_3CH_2OMgBr\) (Methane gas and magnesium ethoxy bromide).

Answer:

Teacher's Note:
a) Grignard reagents act as strong bases when reacting with compounds containing active hydrogen atoms.
b) Hydroxyl proton of ethanol reacts to yield an alkane (methane).

OR

(ii) An organic compound [A] having molecular formula C₆H₆O gives a characteristic colour with aqueous FeCl₃ solution. [A] on treatment with CO₂ and NaOH at 400K under pressure gives [B] which on acidification gives compound [C]. [C] reacts with acetyl chloride to give [D] which is a popular pain killer.
Identify the compounds [A], [B], [C] and [D]. [2 Marks]

Answer:
[A]: Phenol (\(C_6H_5OH\))
[B]: Sodium phenoxide / Sodium salicylate (or intermediate sodium salt)
[C]: Salicylic acid (2-hydroxybenzoic acid)
[D]: Acetylsalicylic acid (Aspirin)

Teacher's Note:
a) Compound A gives a violet colour with FeCl₃ and has formula C₆H₆O, identifying it as phenol.
b) Kolbe's reaction on phenol with CO₂ and NaOH forms salicylic acid, which is acetylated to aspirin (painkiller).

Question 9 [2]
John was making noodles in boiling water. When he added common salt (NaCl) to boiling water, the water stopped boiling for a short while. If John had added 15.0g of NaCl to 250.0g of water, calculate the boiling point of solution assuming that NaCl dissociates completely in water. (K_b for water = 0.512K kg mol^-1, molecular mass of NaCl = 58.44 g mol^-1). [2 Marks]

Answer:
\(i = 2\) (complete dissociation of NaCl)
\(\Delta T_b = \frac{i \times K_b \times w_2 \times 1000}{M_2 \times w_1} = \frac{2 \times 0.512 \times 15.0 \times 1000}{58.44 \times 250.0} = 1.05 \text{ K}\)
Boiling point of solution = \(100^{\circ}C + 1.05^{\circ}C = 101.05^{\circ}C\) (or \(374.05 \text{ K}\)).

Teacher's Note:
a) Elevation in boiling point is a colligative property dependent on particle concentration.
b) Remember to multiply by van 't Hoff factor \(i = 2\) for NaCl since it furnishes two ions per formula unit.

Question 10 [2]
(i) Aromatic aldehydes do not give a reddish-brown precipitate on heating with Fehling solution. Give a reason. [1 Mark]

Answer:
Aromatic aldehydes are less reactive and the resonance stabilisation of the benzene ring deactivates the carbonyl group towards mild oxidising agents like Fehling solution.

Teacher's Note:
a) Fehling solution is a mild oxidising agent only capable of oxidising aliphatic aldehydes.
b) Aromatic aldehydes only reduce stronger oxidising agents like Tollens reagent.

(ii) Why is benzaldehyde less reactive to electrophilic substitution reactions than benzene? [1 Mark]

Answer:
The formyl group (\(-CHO\)) attached to the benzene ring is electron-withdrawing due to -M and -I effects, reducing electron density on the ring.

Teacher's Note:
a) Electrophilic substitution reactions require a rich electron density on the aromatic ring.
b) Deactivating groups direct incoming electrophiles to the meta position and slow down the reaction rate.

Question 11 [2]
(i) Give a reason to explain why transition metals can act as a good catalyst. [1 Mark]

Answer:
Due to their ability to exhibit variable oxidation states and form intermediate complexes with reactants, and their large surface area.

Teacher's Note:
a) Variable oxidation states allow transition metals to readily accept and donate electrons.
b) They provide a surface for reactant molecules to come close and react with lower activation energy.

(ii) Scandium (Z = 21) does not exhibit variable oxidation states and yet it is regarded as transition element. Why? [1 Mark]

Answer:
Scandium has a partially filled \(3d^1\) orbital in its ground state (\([Ar] 3d^1 4s^2\)).

Answer:

Teacher's Note:
a) The classification as a transition element depends on having a partially filled d-orbital in the ground state or stable oxidation state.
b) Sc³⁺ has \(3d^0\), but the neutral atom has a vacant/incomplete d subshell.

SECTION C - 21 MARKS

Question 12 [3]
The data in the table given below was obtained in a series of experiments on the rate of the reaction between compounds [A] and [B] at a constant temperature:

Show how this data can be used to deduce the rate expression for the reaction between [A] and [B]. [3 Marks]

Answer:
Let rate law be \(\text{Rate} = k[A]^x[B]^y\).
Comparing Exp 1 and 2: \(\frac{4.40 \times 10^{-4}}{1.10 \times 10^{-4}} = \left(\frac{0.30}{0.15}\right)^x \Rightarrow 4 = 2^x \Rightarrow x = 2\).
Comparing Exp 2 and 3: keeping [A] doubled (0.30 to 0.60, rate quadruples to \(1.76 \times 10^{-3}\)), then halving [B] (0.30 to 0.15, rate halves back to \(8.80 \times 10^{-4}\)), showing order with respect to [B] is 1 (\(y = 1\)).
Rate expression: \(\text{Rate} = k[A]^2[B]\).

Teacher's Note:
a) Compare experiments where concentration of one reactant is kept constant while the other changes.
b) Determine reaction orders \(x\) and \(y\) to write the final rate expression.

Question 13 [3]
Arrange the following compounds:
C₆H₅NH₂, (C₂H₅)₂NH, (C₂H₅)₃N, C₂H₅NH2.
(i) in the increasing order of their basic strength in water. [1 Mark]

Answer:
\(C_6H_5NH_2 \lt (C_2H_5)_3N \lt C_2H_5NH_2 \lt (C_2H_5)_2NH\)

Teacher's Note:
a) Aniline is the weakest base due to resonance delocalization of the lone pair onto the benzene ring.
b) For ethylamines in aqueous solution, secondary amine is strongest due to a balance of inductive effect, solvation, and steric hindrance.

(ii) in a decreasing order of their basic strength in gas phase. [1 Mark]

Answer:
\((C_2H_5)_3N \gt (C_2H_5)_2NH \gt C_2H_5NH_2 \gt C_6H_5NH_2\)

Teacher's Note:
a) In the gas phase, solvation effects are absent, so basicity is purely determined by the electron-releasing inductive effect (+I) of alkyl groups.
b) More alkyl groups mean higher electron density on nitrogen, hence tertiary > secondary > primary.

Question 14 [3]
(i) What products are obtained when sucrose is subjected to acid hydrolysis? [1 Mark]

Answer:
D-(+)-glucose and D-(-)-fructose (in equimolar amounts).

Teacher's Note:
a) Sucrose is a non-reducing disaccharide composed of glucose and fructose units linked by \(\alpha\)-D-glucopyranose and \(\beta\)-D-fructofuranose.
b) Hydrolysis breaks the glycosidic linkage between C1 of glucose and C2 of fructose.

(ii) Why are Vitamin B and Vitamin C essential for us? [1 Mark]

Answer:
They are water-soluble vitamins that cannot be stored in the body and must be supplied regularly through diet as they are excreted in urine.

Teacher's Note:
a) Water-soluble vitamins dissolve in body fluids and are rapidly flushed out.
b) Regular dietary intake is vital to prevent deficiency diseases like scurvy (Vitamin C) or beriberi (Vitamin B1).

(iii) On being heated, egg white becomes solid and opaque. Give a reason. [1 Mark]

Answer:
Denaturation of albumin protein occurs upon heating, causing unfolding of polypeptide chains and coagulation into a solid mass.

Teacher's Note:
a) Heat disrupts hydrogen bonds and hydrophobic interactions maintaining the tertiary structure of proteins.
b) The primary structure remains intact, but the biological activity and solubility are lost.

Question 15 [3]
Water vapour and liquid water are in equilibrium in a container. At room temperature, the vapour pressure of water is 25 mm of Hg. The volume of water is V ml.
(i) What will be the vapour pressure of water if the volume of water is reduced to V/4 ml without any change in temperature? Give a reason. [1 Mark]

Answer:
25 mm of Hg (remains unchanged).

Teacher's Note:
a) Vapour pressure is a characteristic property at a given temperature and is independent of the volume of liquid or container size.
b) Equilibrium vapor pressure depends only on temperature.

(ii) Will there be a change in vapour pressure if more water (at room temperature) is added to the container? Give a reason. [1 Mark]

Answer:
No change in vapour pressure.

Teacher's Note:
a) Adding more water does not alter the equilibrium vapour pressure at constant temperature.
b) The rate of evaporation and condensation readjusts to maintain the same saturated vapour pressure.

Question 16 [3]
Identify the compounds [A], [B] and [C].
(i) \(C_6H_5COOH \xrightarrow{PCl_5} [A] \xrightarrow{H_2, Pd/BaSO_4} [B] \xrightarrow{KCN(alc), distil} [C]\) [1.5 Marks]

Answer:
[A]: Benzoyl chloride (\(C_6H_5COCl\))
[B]: Benzaldehyde (\(C_6H_5CHO\))
[C]: Mandelic acid nitrile / Cyanohydrin (or phenyl glycolonitrile, \(\text{C}_6\text{H}_5\text{CH(OH)CN}\))

Teacher's Note:
a) Benzoic acid reacts with PCl₅ to form benzoyl chloride.
b) Rosenmund reduction of benzoyl chloride gives benzaldehyde, which reacts with HCN to form mandelic nitrile.

(ii) \(H-C \equiv C-H + H_2O \xrightarrow{Hg^{2+}/H_2SO_4} [A] \xrightarrow[K_2Cr_2O_7/H_2O_4]{[Oxidation]} [B] \xrightarrow[(ii) dry distillation]{(i) Ca(OH)_2} [C]\) [1.5 Marks]

Answer:
[A]: Acetaldehyde / Ethanal (\(CH_3CHO\))
[B]: Acetic acid (\(CH_3COOH\))
[C]: Acetone / Propanone (\(CH_3COCH_3\))

Teacher's Note:
a) Hydration of ethyne gives acetaldehyde via tautomerism of vinyl alcohol.
b) Oxidation of acetaldehyde yields acetic acid, whose calcium salt on dry distillation gives acetone.

Question 17 [3]
(i) How will the following be obtained? (Give chemical equation)
(a) Picric acid from Phenol [1 Mark]

Answer:
\(C_6H_5OH + 3HNO_{3(conc)} \xrightarrow{H_2SO_{4(conc)}} C_6H_2OH(NO_2)_3 + 3H_2O\) (2,4,6-trinitrophenol).

Teacher's Note:
a) Nitration of phenol with concentrated nitric acid in the presence of concentrated sulfuric acid yields picric acid.
b) The OH group strongly activates the ring towards electrophilic substitution at ortho and para positions.

(b) Ethyl acetate from ethanol [1 Mark]

Answer:
\(C_2H_5OH + CH_3COOH \xrightarrow{H^+} CH_3COOC_2H_5 + H_2O\) (Fischer esterification).

Teacher's Note:
a) Reaction of ethanol with acetic acid in the presence of an acid catalyst gives ethyl acetate and water.
b) This is a reversible esterification reaction.

(c) Anisole from sodium phenoxide [1 Mark]

Answer:
\(C_6H_5ONa + CH_3I \rightarrow C_6H_5OCH_3 + NaI\) (Williamson synthesis).

Teacher's Note:
a) Reaction of sodium phenoxide with methyl iodide produces methoxybenzene (anisole).
b) This is an example of Williamson ether synthesis.

OR

(ii) Explain the mechanism of acid catalysed dehydration of ethanol to yield the corresponding alkene. [3 Marks]

Answer:
Step 1: Protonation of ethanol to formethyloxonium ion (\(CH_3CH_2OH_2^+\)).
Step 2: Loss of a water molecule to form a primary carbocation (\(CH_3CH_2^+\)).
Step 3: Elimination of a proton from the carbocation to yield ethene (\(CH_2=CH_2\)) and regenerating the acid catalyst.

Teacher's Note:
a) The reaction follows an E1 mechanism involving protonation and carbocation intermediate.
b) High temperature (443 K) favours alkene formation over ether formation.

Question 18 [3]
(i) The half-life period (\(t_{1/2}\)) for decay of radioactive ¹⁴C is 5730 years. An ancient piece of wood has only 80% of the ¹⁴C found in a living tree. Calculate the age of the piece of wood. [1.5 Marks]

Answer:
\(k = \frac{0.693}{t_{1/2}} = \frac{0.693}{5730} = 1.209 \times 10^{-4} \text{ year}^{-1}\)
\(t = \frac{2.303}{k} \log\left(\frac{A_0}{A}\right) = \frac{2.303}{1.209 \times 10^{-4}} \log\left(\frac{100}{80}\right) = 1845 \text{ years}\).

Teacher's Note:
a) Use first-order radioactive decay equations for half-life and rate constant calculation.
b) Substitute \(A_0 = 100\) and \(A = 80\) to find the time elapsed.

(ii) The rate of most of the reactions becomes double when the temperature is raised from 298K to 308K. Calculate the activation energy. (R = 8.314 J K^-1 mol^-1) [1.5 Marks]

Answer:
\(\log\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303 R} \left(\frac{T_2 - T_1}{T_1 T_2}\right)\)
\(\log(2) = \frac{E_a}{2.303 \times 8.314} \left(\frac{10}{298 \times 308}\right)\)
\(0.3010 = \frac{E_a}{19.147} \times 1.09 \times 10^{-4} \Rightarrow E_a = 52.89 \text{ kJ mol}^{-1}\) (or \(52895 \text{ J mol}^{-1}\)).

Teacher's Note:
a) Use the Arrhenius equation in logarithmic form for two different temperatures.
b) Ensure units for activation energy are clearly stated in kJ mol^-1 or J mol^-1.

SECTION D - 15 MARKS

Question 19 [5]
(i) Give a reason for each of the following:
(a) Formaldehyde does not undergo aldol condensation, but acetaldehyde does. [1 Mark]

Answer:
Formaldehyde lacks \(\alpha\)-hydrogens, whereas acetaldehyde possesses three \(\alpha\)-hydrogens necessary to form a carbanion intermediate for aldol condensation.

Teacher's Note:
a) Aldol condensation requires carbonyl compounds with at least one \(\alpha\)-hydrogen.
b) In the presence of dilute alkali, formaldehyde undergoes Cannizzaro reaction instead.

(b) Chloroacetic acid is stronger acid than acetic acid. [1 Mark]

Answer:
The chlorine atom exerts an electron-withdrawing inductive (-I) effect, stabilizing the carboxylate anion by dispersing negative charge.

Teacher's Note:
a) Greater stabilization of the conjugate base increases acid strength.
b) The methyl group in acetic acid has a +I effect which intensifies negative charge and destabilizes the carboxylate ion.

(c) Both aldehydes and ketones undergo a number of nucleophilic addition reactions. [1 Mark]

Answer:
Due to the polarity of the carbonyl group (\(C=O\)) and steric unsaturation, allowing attack by nucleophiles.

Teacher's Note:
a) Oxygen is more electronegative than carbon, creating a partial positive charge on the carbonyl carbon.
b) Nucleophiles attack this electrophilic carbon center.

(ii) An organic compound with the molecular formula C₇H₆O gets oxidised by Tollens reagent. It does not respond to Fehling test but can undergo the Cannizzaro reaction.
Identify the compound. Show how you used the above information to identify the compound. [2 Marks]

Answer:
Compound: Benzaldehyde (\(C_6H_5CHO\)).
Reasoning: C₇H₆O with Tollens positive indicates an aldehyde. Since it does not respond to Fehling test, it must be an aromatic aldehyde. Ability to undergo Cannizzaro reaction confirms the absence of \(\alpha\)-hydrogen, which matches benzaldehyde.

Teacher's Note:
a) Molecular formula C₇H₆O points to benzaldehyde.
b) Aromatic aldehydes reduce Tollens reagent but fail with Fehling solution and undergo Cannizzaro reaction.

Question 20 [5]
(i) When one mole of an isomer of the complex [Cr(H₂O)₆]Cl₃ is treated with AgNO₃, it produces 1 mole of a white precipitate of AgCl.
Write the formula of this isomer of the complex and show how the metal-ligand bonding differs in the isomers. [2.5 Marks]

Answer:
Formula: \([Cr(H_2O)_5Cl]Cl_2 \cdot H_2O\) (or \([Cr(H_2O)_5Cl]Cl_2\)).
Metal-ligand bonding difference: In this isomer, five water molecules and one chloride ion act as ligands coordinated directly to the chromium metal ion inside the coordination sphere, while two chloride ions are present outside as ionisable counter-ions.

Teacher's Note:
a) 1 mole of the complex producing 1 mole of AgCl indicates that only 1 chloride ion is outside the coordination sphere.
b) This is an example of hydrate isomerism.

(ii) A coordination compound shows d²sp³ hybridisation. Identify the nature of ligand as weak or strong. What will be the geometry of the compound? [2.5 Marks]

Answer:
Nature of ligand: Strong field ligand.
Geometry: Octahedral.

Teacher's Note:
a) Inner orbital complexes (\(d^2sp^3\)) are formed in the presence of strong field ligands that cause pairing of d-electrons.
b) Coordination number 6 with \(d^2sp^3\) or \(sp^3d^2\) hybridisation always possesses an octahedral geometry.

Question 21 [5]
(i)
(a) Calculate the value of E^o_cell and \(\Delta G^o\) that can be obtained from the following cell under the standard conditions at 25^oC
\(Zn | Zn^{2+} (1M) || Sn^{2+} (1M) | Sn$
Given E^o \(Zn^{2+}/Zn = -0.76 V\); E^o \(Sn^{2+}/Sn = -0.14 V\)
1 Faraday = 96500 C mol^-1 [3 Marks]

[Figure: Galvanic cell diagram showing Zn anode in \(Zn^{2+}, NO_3^-\) solution connected via salt bridge to Sn cathode in \(Sn^{2+}, NO_3^-\) solution, with electrons flowing from Zn to Sn]

Answer:
\(E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = -0.14 - (-0.76) = +0.62 \text{ V}\)
\(\Delta G^\circ = -nFE^\circ_{cell} = -2 \times 96500 \times 0.62 = -119660 \text{ J mol}^{-1} = -119.66 \text{ kJ mol}^{-1}\).

Teacher's Note:
a) Zinc has a more negative reduction potential, so it acts as anode.
b) Use \(\Delta G^\circ = -nFE^\circ\) with \(n = 2\) electrons transferred in the redox reaction.

(b) How much electricity in Faraday is required for the complete reduction of MnO₄^- ions present in 500 ml of 0.5 M solution to Mn²⁺? [2 Marks]

Answer:
Reaction: \(MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O\)
Number of moles of \(MnO_4^-\) = \(\frac{M \times V}{1000} = \frac{0.5 \times 500}{1000} = 0.25 \text{ mol}\).
Electricity required = \(0.25 \times 5 = 1.25 \text{ Faraday}\).

Teacher's Note:
a) Change in oxidation state of Mn is from +7 to +2, involving 5 electrons per mole.
b) Multiply moles of permanganate by 5 to get the required Faradays.

OR

(ii)
(a) The molar conductivity vs \(\sqrt{C}\) curve for Na₂SO₄, H₂SO₄, and NH₄OH are shown below in random order.
Identify the curve that corresponds to Na₂SO₄, H₂SO₄, and NH₄OH. Justify your answer. [3 Marks]

[Figure: Molar conductivity (\(\Lambda_m\)) versus square root of concentration (\(\sqrt{C}\)) graph showing three curves: Curve 1 is a gently sloping straight line at the top, Curve 2 is a steeper downward sloping line, and Curve 3 is a sharp exponential curve rising steeply near zero concentration]

Answer:
Curve 1: Na₂SO₄ (strong electrolyte with multiple ions/moderate slope)
Curve 2: H₂SO₄ (strong electrolyte with high ionic mobility/sharp initial rise)
Curve 3: NH₄OH (weak electrolyte showing sharp increase at very low concentration due to increased ionization)

Teacher's Note:
a) Strong electrolytes show a gradual increase in molar conductivity with dilution.
b) Weak electrolytes show a sharp exponential rise in molar conductivity on dilution due to a sudden increase in degree of dissociation.

(b) The molar conductivity (\(\Lambda_m\)) of a dilute solution of methanoic acid is 34.1 S cm²/mol. Calculate its degree of dissociation.
(Given \(\lambda^0(H^+) = 349.6 \text{ S cm}^2\text{/mol}\) and \(\lambda^0(HCOO^-) = 54.6 \text{ S cm}^2\text{/mol}\)) [2 Marks]

Answer:
\(\Lambda^\circ_m (HCOOH) = \lambda^\circ(H^+) + \lambda^\circ(HCOO^-) = 349.6 + 54.6 = 404.2 \text{ S cm}^2\text{/mol}\).
Degree of dissociation (\(\alpha\)) = \(\frac{\Lambda_m}{\Lambda^\circ_m} = \frac{34.1}{404.2} = 0.0843\) (or \(8.43\%\)).

Teacher's Note:
a) Calculate the limiting molar conductivity using Kohlrausch's law of independent migration of ions.
b) Degree of dissociation is the ratio of molar conductivity at a given concentration to the limiting molar conductivity at infinite dilution.