ISC Class 12 Chemistry Sample Paper 2023 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]
[two, Williamson's synthesis, three, anisole, toluene, Friedel-Crafts alkylation, iodoform, sec^-1, mol^-1L sec^-1, Lewis base, acetone, Lewis acid, chloroform, formaldehyde]

(i) Sodium phenoxide reacts with methyl chloride to give anisole. The reaction is known as Williamson's synthesis. [1 Mark]

Answer: anisole, Williamson's synthesis

Teacher's Note:
a) Williamson's synthesis involves the reaction of sodium alkoxide or sodium phenoxide with an alkyl halide to form ethers.
b) Ensure correct spelling of scientific names like Williamson's synthesis.

(ii) When the concentration of a reactant of first order reaction is tripled, the rate of reaction becomes three times. The unit of rate constant (k) for the first order reaction is sec^-1. [1 Mark]

Answer: three, sec^-1

Teacher's Note:
a) For a first order reaction, Rate = \( k[A] \), so rate is directly proportional to concentration.
b) The unit of rate constant depends on the overall order of the reaction.

(iii) In coordination complexes, the central metal atom or ion behaves as Lewis acid and the ligands behave as Lewis base. [1 Mark]

Answer: Lewis acid, Lewis base

Teacher's Note:
a) Central metal accepts electron pairs (Lewis acid), and ligands donate electron pairs (Lewis base).
b) Do not confuse Lewis acids and bases with Bronsted-Lowry definitions in this context.

(iv) Calcium acetate on dry distillation gives acetone which gives iodoform on heating with iodine and alkali. [1 Mark]

Answer: acetone, iodoform

Teacher's Note:
a) Dry distillation of calcium salts of fatty acids yields ketones or aldehydes.
b) Methyl ketones undergo haloform (iodoform) reaction with \( I_2 \) and \( NaOH \).

(B) Select and write the correct alternative from the choices given below: [4×1]

(i) An alkyl isocyanide on complete reduction gives : [1 Mark]
(A) Primary amine.
(B) Secondary amine.
(C) Tertiary amine.
(D) Carboxylic acid.

Answer: (B) Secondary amine.

\( R-NC + 4[H] \rightarrow R-NH-CH_3 \) (Secondary amine)

Teacher's Note:
a) Alkyl cyanides on reduction give primary amines, whereas alkyl isocyanides give secondary amines.
b) Remember that reduction of isocyanides yields N-substituted methylamines.

(ii) For a spontaneous reaction E^o_cell and \(\Delta G^o\) will be respectively: [1 Mark]
(A) -ve and -ve
(B) -ve and +ve
(C) +ve and -ve
(D) +ve and +ve

Answer: (C) +ve and -ve

\(\Delta G^o = -nFE^o_{\text{cell}}\). For spontaneity, \(\Delta G^o\) must be negative, which requires \(E^o_{\text{cell}}\) to be positive.

Teacher's Note:
a) A positive standard cell potential indicates a feasible and spontaneous redox reaction.
b) Gibbs free energy change must be negative for any spontaneous process at constant temperature and pressure.

(iii) Which of the following pairs of transition elements have exceptional electronic configuration? [1 Mark]
(A) Sc and Cu
(B) Fe and Ni
(C) Cr and Cu
(D) Mn and Zn

Answer: (C) Cr and Cu

Cr (\(3d^5 4s^1\)) and Cu (\(3d^{10} 4s^1\)) show stability due to half-filled and completely filled d-orbitals.

Teacher's Note:
a) Chromium and copper deviate from the Aufbau principle to achieve extra exchange energy stability.
b) Always write exceptional electronic configurations correctly in exams.

(iv) For a first order reaction, when 100g of the reactant is taken, 75g of the reactant reacts in 8 minutes. If 200g of the same reactant is taken, in how much time 150g of the reactant will react? [1 Mark]
(A) 8 minutes
(B) 16 minutes
(C) 20 minutes
(D) 24 minutes.

Answer: (A) 8 minutes

The time taken for a given fraction of a first-order reaction to react is independent of the initial concentration.

Teacher's Note:
a) In case 1, 75g out of 100g reacted, which is 75% reacted in 8 minutes.
b) In case 2, 150g out of 200g is also 75% reacted, hence the time taken remains identical (8 minutes).

(C) Match the following: [4×1]

Answer: (i) -> (c), (ii) -> (d), (iii) -> (a), (iv) -> (b)

Teacher's Note:
a) Phenol gives a violet coloration with neutral \(FeCl_3\) solution.
b) Ethylenediamine (en) donates two pairs of electrons, acting as a bidentate ligand.

(D) [2×1]

(i) Assertion: Specific conductance of all electrolytes decreases on dilution.
Reason: On dilution, number of ions per unit volume decreases. [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: (A) Both assertion and reason are true and reason is the correct explanation of assertion.

Specific conductance is the conductance of ions present in unit volume of solution. On dilution, volume increases, so the number of ions per unit volume decreases, lowering specific conductance.

Teacher's Note:
a) Specific conductivity decreases on dilution for both strong and weak electrolytes.
b) Molar conductivity, however, increases on dilution due to increased degree of dissociation or interionic distance.

(ii) Assertion: Nitration of chlorobenzene leads to the formation of m-nitro chlorobenzene.
Reason: Nitro (-NO₂) group is a m-directing group. [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.

Halogen atom (-Cl) in chlorobenzene is ortho-para directing due to +R effect, so nitration yields ortho and para nitrochlorobenzenes, not meta.

Teacher's Note:
a) The reason is a true chemical fact, but the assertion is false because chloro group directs incoming electrophiles to ortho and para positions.
b) Always consider the directing nature of the group already present on the benzene ring.

SECTION B - 20 MARKS

Question 2 [2 Marks]
The osmotic pressure of 20g haemoglobin in 500ml of solution is 0·016atm at 25^oC. Calculate the molecular mass of haemoglobin.

Answer:
Given: \( w_2 = 20\text{ g} \), \( V = 500\text{ ml} = 0.5\text{ L} \), \(\pi = 0.016\text{ atm}\), \( T = 25 + 273 = 298\text{ K}\), \( R = 0.0821\text{ L atm K}^{-1}\text{mol}^{-1} \).
Formula: \(\pi V = \frac{w_2}{M_2} R T\)
\( M_2 = \frac{w_2 R T}{\pi V} = \frac{20 \times 0.0821 \times 298}{0.016 \times 0.5} \)
\( M_2 = \frac{489.316}{0.008} = 61164.5\text{ g mol}^{-1} \)

Teacher's Note:
a) Ensure all units are consistent (Volume in litres, pressure in atmospheres, temperature in Kelvin).
b) High molecular mass is characteristic of macromolecular proteins like haemoglobin.

Question 3 [2 Marks]
Give reason for the following:
(i) Transition metals form large number of complex compounds.
(ii) Transition elements show variable oxidation states.

Answer:
(i) Due to small ionic size, high nuclear charge, and availability of vacant d-orbitals of suitable energy to accept electron pairs from ligands.
(ii) Due to the very small energy difference between (n-1)d and ns orbitals, electrons from both can participate in bond formation.

Teacher's Note:
a) Mentioning vacant d-orbitals and high charge density is essential for complex formation.
b) Variable oxidation states arise because ns and (n-1)d electrons have comparable ionization energies.

Question 4 [2 Marks]
Identify compounds [A] and [B] in the following reactions.
(i) \( CH_3Br + KCN_{(alc)} \rightarrow [A] \xrightarrow[ \text{(complete hydrolysis)} ]{ +HOH/H^{+} } [B] \)
(ii) \( C_6H_5NH_2 + HNO_2 + HCl \xrightarrow{ 0-5^{o}C } [A] \xrightarrow{ Cu_2Cl_2/HCl } [B] \)

Answer:
(i) [A] is \( CH_3CN \) (Acetonitrile / Methyl cyanide); [B] is \( CH_3COOH \) (Acetic acid).
(ii) [A] is \( C_6H_5N_2^{+}Cl^{-} \) (Benzenediazonium chloride); [B] is \( C_6H_5Cl \) (Chlorobenzene - Sandmeyer reaction).

Teacher's Note:
a) Alcoholic KCN undergoes nucleophilic substitution to form cyanides, which hydrolyze to carboxylic acids.
b) Primary aromatic amines react with nitrous acid at 0-5 degrees Celsius to form diazonium salts, which react with Cu₂Cl₂/HCl to form aryl halides.

Question 5 [2 Marks]
State reasons for the following:
(i) Ethylamine is soluble in water whereas aniline is not soluble in water.
(ii) Aliphatic amines are stronger bases than aromatic amines.

Answer:
(i) Ethylamine forms intermolecular hydrogen bonds with water molecules due to its small alkyl group, whereas aniline has a large hydrophobic benzene ring that disrupts hydrogen bonding.
(ii) In aliphatic amines, the alkyl group has a +I effect, increasing electron density on nitrogen, whereas in aromatic amines, the lone pair on nitrogen is delocalized into the benzene ring due to resonance, making it less available for protonation.

Teacher's Note:
a) Hydrogen bonding explains the solubility difference between lower aliphatic amines and aromatic amines.
b) Resonance stabilization of aniline and the +I effect of alkyl groups are key concepts for basicity comparisons.

Question 6 [2 Marks]
Calculate the standard free energy change (\(\Delta G^o\)) for the following chemical reaction:
(i) \( Cd_{(s)} + 2Ag^{+} \rightarrow Cd^{2+} + 2Ag_{(s)} \)
(ii) \( E^o_{Cd^{2+}/Cd} = -0\cdot40\text{V}, E^o_{Ag^{+}/Ag} = +0\cdot80\text{V} \)

Answer:
\( E^o_{\text{cell}} = E^o_{\text{cathode}} - E^o_{\text{anode}} = 0.80 - (-0.40) = 1.20\text{ V} \)
Number of electrons transferred \( n = 2 \).
\(\Delta G^o = -n F E^o_{\text{cell}}\)
\(\Delta G^o = -2 \times 96500\text{ C mol}^{-1} \times 1.20\text{ V} = -231600\text{ J mol}^{-1} = -231.6\text{ kJ mol}^{-1} \)

Teacher's Note:
a) Always calculate standard cell potential first by subtracting anode potential from cathode potential.
b) Convert Joules to kilojoules if required and state proper units for Gibbs free energy.

Question 7 [2 Marks]
Complete and balance the following chemical equations:
(i) \( KMnO_4 + H_2SO_4 + FeSO_4 \rightarrow \text{____} + \text{____} + \text{____} + \text{____} \)
(ii) \( K_2Cr_2O_7 + KI + H_2SO_4 \rightarrow \text{____} + \text{____} + \text{____} + \text{____} \)

Answer:
(i) \( 2KMnO_4 + 8H_2SO_4 + 10FeSO_4 \rightarrow K_2SO_4 + 2MnSO_4 + 5Fe_2(SO_4)_3 + 8H_2O \)
(ii) \( K_2Cr_2O_7 + 6KI + 7H_2SO_4 \rightarrow Cr_2(SO_4)_3 + 3I_2 + 4K_2SO_4 + 7H_2O \)

Teacher's Note:
a) These are standard redox titration equations that must be memorized with correct stoichiometric coefficients.
b) Check mass and charge balance on both sides of the equation.

Question 8 [2 Marks]
(i) How will the following be obtained? (Give chemical equation)
(a) Picric acid from phenol
(b) Ethanol from formaldehyde

OR

(ii) Write the chemical equations for the dehydration of ethanol with conc. \(H_2SO_4\) at 140^oC and 170^oC. [2 Marks]

Answer:
(i)(a) Treatment of phenol with concentrated nitric acid in the presence of concentrated sulfuric acid:
\( C_6H_5OH + 3HNO_{3(conc)} \xrightarrow{ conc. H_2SO_4 } 2,4,6\text{-trinitrophenol (Picric acid)} + 3H_2O \)
(i)(b) Reaction of formaldehyde with Grignard reagent followed by hydrolysis:
\( HCHO + CH_3MgBr \rightarrow CH_3-CH_2-OMgBr \xrightarrow{ H_2O / H^{+} } CH_3CH_2OH + Mg(OH)Br \)

OR Answer:
At 140^oC (Diethyl ether formation):
\( 2C_2H_5OH \xrightarrow{ conc. H_2SO_4, 140^{o}C } C_2H_5-O-C_2H_5 + H_2O \)
At 170^oC (Ethene formation):
\( C_2H_5OH \xrightarrow{ conc. H_2SO_4, 170^{o}C } CH_2=CH_2 + H_2O \)

Teacher's Note:
a) Picric acid synthesis requires nitrating mixture (conc. \(HNO_3\) and \(H_2SO_4\)).
b) Temperature control is critical in alcohol dehydration reactions to yield either ethers or alkenes.

Question 9 [2 Marks]
A solution of urea in water has boiling point 100·128^oC. Calculate the freezing point of the same solution. Molal constants for water are \(K_b = 0\cdot512\text{ K kg mol}^{-1}\) and \(K_f = 1\cdot86\text{ K kg mol}^{-1}\) respectively.

Answer:
\(\Delta T_b = T_b - T_b^o = 100.128 - 100 = 0.128^{\circ}\text{C}\)
\( \Delta T_b = K_b \times m \implies m = \frac{\Delta T_b}{K_b} = \frac{0.128}{0.512} = 0.25\text{ mol kg}^{-1} \)
Now, calculate depression in freezing point (\(\Delta T_f\)):
\( \Delta T_f = K_f \times m = 1.86 \times 0.25 = 0.465^{\circ}\text{C} \)
Freezing point of solution \( T_f = 0 - 0.465 = -0.465^{\circ}\text{C} \)

Teacher's Note:
a) Molality of the solution remains the same whether calculating elevation in boiling point or depression in freezing point.
b) Subtract the depression value from the normal freezing point of water (0 degrees Celsius).

Question 10 [2 Marks]
Give one chemical test for each to distinguish between the following pair of compounds.
(i) Formaldehyde and acetic acid
(ii) Acetaldehyde and acetone

Answer:
(i) Sodium bicarbonate test: Acetic acid reacts with \(NaHCO_3\) to give brisk effervescence of \(CO_2\), whereas formaldehyde does not show this reaction.
(ii) Tollen's test: Acetaldehyde gives a silver mirror with Tollen's reagent, whereas acetone does not.

Teacher's Note:
a) Carboxylic acids react with sodium bicarbonate liberating carbon dioxide gas.
b) Aldehydes reduce Tollen's reagent to form a silver mirror, while ketones generally do not respond.

Question 11 [2 Marks]
Why are Zn, Cd and Hg not regarded as transition elements?

Answer:
Because they have completely filled d-orbitals in their ground state as well as in their common oxidation states (\(d^{10}\)), and do not show typical transition metal properties such as variable oxidation states or colored ion formation.

Teacher's Note:
a) The formal definition of a transition element requires a partially filled d-subshell in ground state or oxidation state.
b) Zinc, cadmium, and mercury belong to group 12 and behave more like main group metals.

SECTION C - 21 MARKS

Question 12 [3 Marks]
The rate constant for a first order reaction becomes six times when the temperature is increased from 350 K to 410 K. Calculate activation energy (\(E_a\)) for the reaction.

Answer:
Given: \( T_1 = 350\text{ K} \), \( T_2 = 410\text{ K} \), \(\frac{k_2}{k_1} = 6\), \( R = 8.314\text{ J K}^{-1}\text{mol}^{-1} \).
Using Arrhenius equation:
\( \log\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303 R} \left(\frac{T_2 - T_1}{T_1 T_2}\right) \)
\( \log(6) = \frac{E_a}{2.303 \times 8.314} \left(\frac{410 - 350}{350 \times 410}\right) \)
\( 0.7782 = \frac{E_a}{19.147} \left(\frac{60}{143500}\right) \)
\( E_a = \frac{0.7782 \times 19.147 \times 143500}{60} = 35649.5\text{ J mol}^{-1} = 35.65\text{ kJ mol}^{-1} \)

Teacher's Note:
a) Use logarithm base 10 values correctly (\(\log 6 = 0.7782\)).
b) Express the final activation energy in kJ/mol with proper rounding.

Question 13 [3 Marks]
An organic compound 'A' on treatment with aq.KCN produces compound 'B'. Compound 'B' on reduction with \(Na/C_2H_5OH\) gives compound 'C' with molecular formula \(C_2H_7N\). Compound 'C' reacts with \(NaNO_2\) and HCl to form compound 'D'. Compound 'D' on treatment with acetic acid in presence of conc. \(H_2SO_4\) produces a sweet smelling compound 'E'.
(i) Identify the compounds 'A' to 'E'.
(ii) Name the reaction for the formation of compound 'E' from compound 'D'.

Answer:
(i) Identification:
- 'C' is \(CH_3CH_2NH_2\) (Ethylamine / Ethanamine, formula \(C_2H_7N\)).
- 'B' is \(CH_3CN\) (Acetonitrile, formed by reduction of which gives ethylamine).
- 'A' is \(CH_3X\) (Methyl halide, e.g., Methyl chloride, \(CH_3Cl\)).
- 'D' is \(CH_3CH_2OH\) (Ethanol, formed from ethylamine and \(NaNO_2\)/HCl).
- 'E' is \(CH_3COOCH_2CH_3\) (Ethyl acetate, sweet smelling ester).
(ii) Esterification reaction.

Teacher's Note:
a) Work backwards from the molecular formula \(C_2H_7N\) and sweet smelling compound (ester) to trace all intermediates.
b) Clearly write the chemical name and structure for each identified compound.

Question 14 [3 Marks]
(i) Name the four bases present in DNA. Which one of these is not present in RNA?
(ii) Deficiency of which vitamin causes the following diseases.
(a) Scurvy
(b) Night blindness

Answer:
(i) The four bases in DNA are Adenine, Guanine, Cytosine, and Thymine. Thymine is not present in RNA (replaced by Uracil).
(ii) (a) Scurvy is caused by the deficiency of Vitamin C.
(ii) (b) Night blindness is caused by the deficiency of Vitamin A.

Teacher's Note:
a) Distinguish clearly between DNA and RNA nucleic acid bases.
b) Vitamin deficiencies are standard factual questions in biomolecules.

Question 15 [3 Marks]
An aqueous solution containing 12·48g of barium chloride in 1000g of water boils at 373·0832K. Calculate the degree of dissociation (\(\alpha\)) of barium chloride. \(K_b\) for \(H_2O = 0\cdot52\text{ K kg mol}^{-1}\), molecular mass of \(BaCl_2 = 208\cdot34\text{ g mol}^{-1}\)

Answer:
Observed \(\Delta T_b = 373.0832 - 373 = 0.0832\text{ K}\).
Molality (\(m\)) = \(\frac{w_2 \times 1000}{M_2 \times w_1} = \frac{12.48 \times 1000}{208.34 \times 1000} = 0.0599\text{ mol kg}^{-1}\)
Calculated \(\Delta T_b = K_b \times m = 0.52 \times 0.0599 = 0.03115\text{ K}\).
Van't Hoff factor (\(i\)) = \(\frac{\text{Observed } \Delta T_b}{\text{Calculated } \Delta T_b} = \frac{0.0832}{0.03115} = 2.67\)
For \(BaCl_2 \rightarrow Ba^{2+} + 2Cl^{-}\), total ions \(n = 3\).
\(i = 1 + (n - 1)\alpha \implies 2.67 = 1 + (3 - 1)\alpha \implies 2\alpha = 1.67 \implies \alpha = 0.835\text{ (or } 83.5\% \text{)}\)

Teacher's Note:
a) Calculate Van't Hoff factor using experimental and theoretical boiling point elevation.
b) Use the dissociation formula \(i = 1 + (n-1)\alpha\) to find the degree of dissociation.

Question 16 [3 Marks]
Write the chemical equation for the following named organic reactions.
(i) Haloform reaction
(ii) Reimer - Tiemann reaction
(iii) Kolbe - Schmidt reaction or Kolbe reaction

Answer:
(i) Haloform reaction:
\( CH_3COCH_3 + 3NaOI \rightarrow CHI_3\downarrow + CH_3COONa + 2NaOH \)
(ii) Reimer - Tiemann reaction:
\( C_6H_5OH + CHCl_3 + 3NaOH \rightarrow salicylaldehyde + 3NaCl + 2H_2O \)
(iii) Kolbe - Schmidt reaction:
\( C_6H_5ONa + CO_2 \xrightarrow{ 400K, 4-7 atm } sodium\ salicylate \xrightarrow{ H^{+} } salicylic\ acid \)

Teacher's Note:
a) Ensure all reagents and reaction conditions are mentioned alongside equations.
b) Named reactions carry direct marks in board examinations.

Question 17 [3 Marks]
(i) Identify the compounds A, B and C in the following reactions:
(a) \( C_6H_5NO_2 \xrightarrow{ Sn + HCl } A \xrightarrow[ 273K - 278K ]{ NaNO_2 + HCl } B \xrightarrow{ H_2O } C \)
(b) \( CH_3CN \xrightarrow{ H_2O / H^{+} } A \xrightarrow{ NH_3 / heat } B \xrightarrow{ Br_2 + KOH } C \)

OR

(ii) How will the following be converted? (Give chemical equations) [3 Marks]
(a) Benzenediazonium chloride to Benzene
(b) Ethylamine to ethyl alcohol
(c) Methylamine to methyl isocyanide

Answer:
(i)(a) A is \(C_6H_5NH_2\) (Aniline); B is \(C_6H_5N_2^{+}Cl^{-}\) (Benzenediazonium chloride); C is \(C_6H_5OH\) (Phenol).
(i)(b) A is \(CH_3COOH\) (Acetic acid); B is \(CH_3COONH_4 \rightarrow CH_3CONH_2\) (Acetamide); C is \(CH_3NH_2\) (Methylamine - Hofmann bromamide degradation).

OR Answer:
(ii)(a) \( C_6H_5N_2^{+}Cl^{-} + H_3PO_2 + H_2O \rightarrow C_6H_6 + N_2 + H_3PO_3 + HCl \)
(ii)(b) \( C_2H_5NH_2 + HNO_2 \rightarrow C_2H_5OH + N_2 + H_2O \)
(ii)(c) \( CH_3NH_2 + CHCl_3 + 3KOH_{(alc)} \xrightarrow{ \Delta } CH_3NC + 3KCl + 3H_2O \) (Carbylamine reaction)

Teacher's Note:
a) Sequence conversions step-by-step keeping functional group transformations in mind.
b) Mentioning reagents clearly above the reaction arrows is mandatory.

Question 18 [3 Marks]
Suppose 50 bacteria are placed in a flask containing nutrients, so that they can multiply. A study at 35^oC gave the following results:

Answer the following questions:
(i) This multiplication of bacteria follows:
(a) Zero order reaction
(b) First order reaction
(c) Second order reaction
(d) Third order reaction
(ii) The rate constant for the reaction is:
(a) \(0\cdot0462\text{ min}^{-1}\)
(b) \(0\cdot462\text{ min}^{-1}\)
(c) \(4\cdot62\text{ min}^{-1}\)
(d) \(46\cdot2\text{ min}^{-1}\)
(iii) The half life period (\(t_{1/2}\)) of the reaction is:
(a) 1500 minutes
(b) 150 minutes
(c) 15 minutes
(d) 1·5 minutes

Answer:
(i) (b) First order reaction
(ii) (a) \(0\cdot0462\text{ min}^{-1}\)
(iii) (c) 15 minutes

Teacher's Note:
a) The time required for bacterial count to double is constant (15 minutes), which defines a first-order half-life.
b) Rate constant \(k = \frac{0.693}{t_{1/2}} = \frac{0.693}{15} = 0.0462\text{ min}^{-1}\).

SECTION D - 15 MARKS

Question 19 [5 Marks]
(i) Starting with methyl magnesium bromide, how will the following compounds be synthesised?
(a) Acetaldehyde
(b) Acetone
(c) Acetic acid
(ii) Explain the following:
(a) Chloroacetic acid is stronger acid than acetic acid.
(b) Formic acid reduces Tollen's reagent but acetic acid does not.

Answer:
(i)(a) Acetaldehyde: Reaction of methyl magnesium bromide with ethyl formate or hydrogen cyanide followed by hydrolysis (Note: reaction with HCN gives acetaldehyde after hydrolysis of imine intermediate).
(i)(b) Acetone: \( CH_3MgBr + CH_3CN \rightarrow \text{addition product} \xrightarrow{ H_3O^{+} } CH_3COCH_3 + NH_3 + Mg(OH)Br \)
(i)(c) Acetic acid: \( CH_3MgBr + CO_2 \rightarrow CH_3COOMgBr \xrightarrow{ H_3O^{+} } CH_3COOH + Mg(OH)Br \)
(ii)(a) Chlorine is electronegative and exhibits a -I effect, which stabilizes the carboxylate anion by dispersing negative charge, whereas methyl group in acetic acid has +I effect which intensifies negative charge and destabilizes it.

(ii)(b) Formic acid (\(HCOOH\)) contains a formyl group (\(-CHO\)) in addition to the carboxyl group, enabling it to act as a reducing agent and reduce Tollen's reagent, whereas acetic acid lacks this aldehyde hydrogen.

Teacher's Note:
a) Grignard reagents are versatile nucleophiles used for chain extension to form aldehydes, ketones, and acids.
b) Inductive effect (-I and +I) explains acid strength differences in substituted carboxylic acids.

Question 20 [5 Marks]
(i) Name the type of isomerism shown by the following pairs of coordination compounds.
(a) \([Pt(H_2O)_4Cl_2]Cl_2\cdot H_2O\) and \([Pt(H_2O)_3Cl_3]Cl\cdot 2H_2O\)
(b) \([Co(NH_3)_4Cl_2]Br_2\) and \([Co(NH_3)_4Br_2]Cl_2\)
(c) \([Cr(H_2O)_5(SCN)]Cl_2\) and \([Cr(H_2O)_5(NCS)]Cl_2$
(ii) Consider the complex ion \([Co(CN)_6]^{3-}\) and answer the following questions:
(atomic number of Co = 27)
(a) Type of hybridisation of central metal atom
(b) Magnetic nature
(c) Geometry of the complex ion
(d) Low spin complex or high spin complex

Answer:
(i)(a) Hydrate isomerism (or Solvate isomerism)
(i)(b) Ionisation isomerism
(i)(c) Linkage isomerism
(ii) For \([Co(CN)_6]^{3-}\):
- Cobalt oxidation state is +3. Electronic configuration of \(Co^{3+}\) is \(3d^6\).
- Cyanide (\(CN^{-}\)) is a strong field ligand, causing electron pairing.
(a) Hybridisation: \(d^2sp^3$
(b) Magnetic nature: Diamagnetic (all electrons paired)
(c) Geometry: Octahedral
(d) Low spin complex

Teacher's Note:
a) Isomerism types depend on whether water molecules, anions inside/outside sphere, or ambidentate ligands differ.
b) Strong field ligands force inner d-orbital pairing, leading to inner orbital / low spin complexes (\(d^2sp^3\)).

Question 21 [5 Marks]
(i) A 0·06 molar \(CH_3COOH\) solution offers a resistance of 55 ohms to a conductivity cell at 25^oC. If the cell constant is \(0\cdot45\text{ cm}^{-1}\) and the molar conductance of \(CH_3COOH\) at infinite dilution is \(398\cdot5\text{ ohm}^{-1}\text{cm}^2\text{mol}^{-1}\). Calculate:
(a) Specific conductance
(b) Molar conductance
(c) Degree of dissociation
(ii) Calculate the number of coulombs of charge required to deposit 24·35g of aluminium from a solution containing \(Al^{3+}\) ions. (Atomic weight of Al = 27)

OR

(i) Write the Nernst equation for the cell reaction given below and calculate the emf of the cell at 298K.
\( 2Cr_{(s)} + 3Fe^{2+}_{(0.1M)} \rightarrow 2Cr^{3+}_{(0.01M)} + 3Fe_{(s)} \)
Given \(E^o_{(Cr^{3+}/Cr)} = -0\cdot74\text{V}, E^o_{(Fe^{2+}/Fe)} = -0\cdot44\text{V}\)
(ii) Calculate the molar conductance at infinite dilution (\(\Lambda_m^{\infty}\)) for \(NH_4OH\). Given that \(\Lambda_m^{\infty}\) for \(Ba(OH)_2\), \(BaCl_2\) and \(NH_4Cl\) are \(457\text{ ohm}^{-1}\text{cm}^2\text{mol}^{-1}\), \(240\text{ ohm}^{-1}\text{cm}^2\text{mol}^{-1}\) and \(129\text{ ohm}^{-1}\text{cm}^2\text{mol}^{-1}\) respectively. [5 Marks]

Answer:
(i)(a) Specific conductance (\(\kappa\)) = \(\frac{\text{Cell constant}}{\text{Resistance}} = \frac{0.45}{55} = 0.00818\text{ ohm}^{-1}\text{cm}^{-1}\)
(i)(b) Molar conductance (\(\Lambda_m\)) = \(\frac{\kappa \times 1000}{M} = \frac{0.00818 \times 1000}{0.06} = 136.33\text{ ohm}^{-1}\text{cm}^2\text{mol}^{-1}\)
(i)(c) Degree of dissociation (\(\alpha\)) = \(\frac{\Lambda_m}{\Lambda_m^{\infty}} = \frac{136.33}{398.5} = 0.342\)
(ii) For reduction of aluminium: \(Al^{3+} + 3e^{-} \rightarrow Al\)
Charge required for 27g of Al = \(3 \times 96500\text{ C}\)
Charge required for 24.35g of Al = \(\frac{3 \times 96500 \times 24.35}{27} = 261159.26\text{ C}\)

OR Answer:
(i) Nernst equation:
\( E_{\text{cell}} = E^o_{\text{cell}} - \frac{0.0591}{6} \log\left(\frac{[Cr^{3+}]^2}{[Fe^{2+}]^3}\right) \)
\( E^o_{\text{cell}} = -0.44 - (-0.74) = +0.30\text{ V} \)
\( E_{\text{cell}} = 0.30 - \frac{0.0591}{6} \log\left(\frac{(0.01)^2}{(0.1)^3}\right) = 0.30 - \frac{0.0591}{6} \log\left(\frac{10^{-4}}{10^{-3}}\right) = 0.30 - \frac{0.0591}{6}(-1) = 0.30 + 0.00985 = 0.30985\text{ V} \)
(ii) Applying Kohlrausch's Law:
\( \Lambda_m^{\infty}(NH_4OH) = \frac{1}{2}\Lambda_m^{\infty}(Ba(OH)_2) + \Lambda_m^{\infty}(NH_4Cl) - \frac{1}{2}\Lambda_m^{\infty}(BaCl_2) \)
\( \Lambda_m^{\infty}(NH_4OH) = \frac{1}{2}(457) + 129 - \frac{1}{2}(240) = 228.5 + 129 - 120 = 237.5\text{ ohm}^{-1}\text{cm}^2\text{mol}^{-1} \)

Teacher's Note:
a) Correct formula application for specific conductance, molar conductance, and degree of dissociation is vital.
b) For Kohlrausch's law calculations, balance the constituent ions properly using stoichiometric coefficients.