ISC Class 12 Chemistry Board Exam Question Paper 2019 with Solutions

ICSE Class 12 Chemistry Board Exam Question Paper with Solutions

1. (a) Fill in the blanks by choosing the appropriate words from those given in the brackets. [4 Marks]
(More than, primary, cathode, Lucas, reagent, two, four, less than, Grignard's reagent, tertiary, anode, zero, equal to, three)

i) The elevation of boiling point of \( 0.5M\text{ K}_2\text{SO}_4 \) solution is ...... that of \( 0.5M \) urea solution. The elevation of boiling point of \( 0.5M\text{ KCl} \) solution is ...... that of \( 0.5M\text{ K}_2\text{SO}_4 \) solution. [1 Mark]

Answer: More than, less than

Teacher's Note:
a) Colligative properties depend on the total number of particles (van 't Hoff factor \( i \)) in solution.
b) \( \text{K}_2\text{SO}_4 \) gives 3 ions, urea gives 1, and \( \text{KCl} \) gives 2 ions, so their boiling point elevations follow the order \( \text{K}_2\text{SO}_4 \gt \text{KCl} \gt \text{urea} \).

ii) A mixture of conc. \( \text{HCl} \) and anhydrous \( \text{ZnCl}_2 \) is called ...... which shows maximum reactivity with ...... alcohol. [1 Mark]

Answer: Lucas reagent, tertiary

Teacher's Note:
a) Lucas reagent is used to distinguish between primary, secondary, and tertiary alcohols.
b) Tertiary alcohols react instantaneously due to the formation of stable tertiary carbocations.

iii) In electrolytic refining the impure metal is made ...... while a thin sheet of pure metal is used as ...... [1 Mark]

Answer: Anode, cathode

Teacher's Note:
a) Oxidation occurs at the anode where the impure metal dissolves into the electrolyte solution.
b) Reduction occurs at the cathode where pure metal ions deposit onto the pure metal strip.

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

Answer: Two, zero

Teacher's Note:
a) Rate of a first order reaction is directly proportional to the concentration of the reactant (\( \text{Rate} = k[\text{A}]^1 \)).
b) Rate of a zero order reaction is independent of the reactant concentration (\( \text{Rate} = k[\text{A}]^0 \)).

(b) Select the correct alternative from the choice given: [4 Marks]

i) The cell reaction is spontaneous or feasible when emf of the cell is: [1 Mark]
(A) Negative
(B) positive
(C) zero
(D) either positive or negative

Answer: (B) positive

Gibbs free energy \( \Delta G^\circ = -nFE^\circ_{\text{cell}} \) must be negative for a spontaneous reaction, which requires \( E^\circ_{\text{cell}} \) to be positive.

Teacher's Note:
a) The sign of standard cell potential determines the thermodynamic feasibility of an electrochemical cell.
b) Ensure you check the reduction potentials and apply the formula \( E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \).

ii) Which, among the following polymers, is polyester? [1 Mark]
(A) Melamine
(B) Bakelite
(C) Terylene
(D) Polythene

Answer: (C) Terylene

[Figure: Condensation polymerization reaction showing ethylene glycol and terephthalic acid forming Terylene with the release of water molecules]

Teacher's Note:
a) Terylene (Dacron) is a condensation polymer formed by ethylene glycol and terephthalic acid linked by ester bonds.
b) Bakelite is a polymer of phenol and formaldehyde, while polythene is an addition polymer of ethene.

iii) The correct order of increasing acidic strength of the Oxo acids of chlorine is: [1 Mark]
(A) \( \text{HClO}_3 \lt \text{HClO}_4 \lt \text{HClO}_2 \lt \text{HClO} \)
(B) \( \text{HClO} \lt \text{HClO}_2 \lt \text{HClO}_3 \lt \text{HClO}_4 \)
(C) \( \text{HClO}_2 \lt \text{HClO}_4 \lt \text{HClO}_3 \lt \text{HClO} \)
(D) \( \text{HClO}_3 \lt \text{HClO}_4 \lt \text{HClO} \lt \text{HClO}_2 \)

Answer: (B) \( \text{HClO} \lt \text{HClO}_2 \lt \text{HClO}_3 \lt \text{HClO}_4 \)

Acidic strength increases with an increase in the oxidation state of chlorine and the stability of its conjugate base.

Teacher's Note:
a) Oxidation states of Cl in \( \text{HClO}, \text{HClO}_2, \text{HClO}_3, \text{HClO}_4 \) are +1, +3, +5, and +7 respectively.
b) Higher oxidation number leads to greater dispersal of negative charge on the conjugate base anions (\( \text{ClO}_4^- \)).

iv) A catalyst is a substance which: [1 Mark]
(A) Changes the equilibrium constant of reaction.
(B) Increases the equilibrium constant of the reaction.
(C) Supplies energy to the reaction.
(D) Shortens the time to reach equilibrium.

Answer: (D) Shortens the time to reach equilibrium.

A catalyst alters the reaction rate by providing an alternative pathway with a lower activation energy, helping attain equilibrium faster without affecting position or equilibrium constant.

Teacher's Note:
a) Catalysts affect both the forward and reverse reactions equally, leaving equilibrium constant unchanged.
b) Do not confuse altering the rate of attainment of equilibrium with changing the equilibrium yield.

(c) Match the following: [4 Marks]

Answer:
1. Diazotisation - d) Aniline
2. Crystalline solid - a) Anisotropic
3. Phenol - b) Reimer - Tiemann reaction
4. Fittig reaction - c) Diphenyl

Teacher's Note:
a) Diazotisation converts aromatic primary amines like aniline into diazonium salts at low temperatures.
b) Crystalline solids show different physical properties in different directions, a property known as anisotropy.

(d) Answer the following question: [4 Marks]

i) Which trivalent ion has maximum size in the lanthanoid series, i.e. lanthanum ion (\( \text{La}^{3+} \)) to lutetium (\( \text{Lu}^{3+} \))? Atomic number of lanthanum is 57 and lutetium is 71. [1 Mark]

Answer: \( \text{La}^{3+} \) has the maximum size in the lanthanoid series.

Teacher's Note:
a) Across the lanthanoid series, atomic and ionic radii decrease steadily, known as lanthanoid contraction.
b) This contraction is due to the poor shielding effect of 4f electrons.

ii) Explain why, \( \text{Cu}^{2+} \) is paramagnetic but \( \text{Cu}^+\) is diamagnetic? (At.no of cu is 29) [1 Mark]

Answer: \( \text{Cu}^{2+} \) has the electronic configuration \( [\text{Ar}]3d^9 \) with one unpaired electron, making it paramagnetic. \( \text{Cu}^+ \) has the configuration \( [\text{Ar}]3d^{10} \) with all paired electrons, making it diamagnetic.

Teacher's Note:
a) Paramagnetism arises due to the presence of unpaired electrons.
b) Always write out the electronic configurations of the ions to clearly show paired or unpaired d-electrons.

iii) Calculate the boiling point of urea solution when 6g of urea is dissolved in 200g of water. (\( K_b \) for water is \( 0.52\text{ K kg mol}^{-1} \), boiling point of pure water is \( 373\text{K} \), mol. wt. of urea is 60) [1 Mark]

Answer:
\( \Delta T_b = \frac{K_b \times W_B \times 1000}{M_B \times W_A} \)
\( \Delta T_b = \frac{0.52 \times 6 \times 1000}{60 \times 200} = 0.26\text{ K} \)
Boiling point of solution \( T = T_0 + \Delta T_b = 373 + 0.26 = 373.26\text{K} \)

Teacher's Note:
a) Ensure the mass of solvent is taken in grams and multiplied by 1000 in the numerator.
b) Add the calculated elevation in boiling point to the pure solvent boiling point to get the final solution temperature.

iv) Identify the compounds A, B, C and D in the given reaction. [1 Mark]
\( \text{HC}\equiv\text{CH} \xrightarrow[\text{Hg}^{2+}, \text{H}_2\text{SO}_4]{\text{H}_2\text{O}} \text{A} \xrightarrow[\text{K}_2\text{Cr}_2\text{O}_7 + \text{H}_2\text{SO}_4]{[O]} \text{B} \)
\( \text{B} + \text{Ca(OH)}_2 \rightarrow \text{C} \xrightarrow{\text{Dry distillation}} \text{D} + \text{CaCO}_3 \)

Answer:
A: \( \text{CH}_3\text{CHO} \) (Acetaldehyde)
B: \( \text{CH}_3\text{COOH} \) (Acetic acid)
C: \( (\text{CH}_3\text{COO})_2\text{Ca} \) (Calcium acetate)
D: \( \text{CH}_3\text{COCH}_3 \) (Acetone)

Teacher's Note:
a) Hydration of acetylene yields acetaldehyde via tautomerization of vinyl alcohol.
b) Dry distillation of calcium salts of carboxylic acids is a standard method for preparing ketones.

2. (a) For the reaction \( \text{A} + \text{B} \rightarrow \text{C} + \text{D} \), the initial rate for different reactions and initial concentration of reactants are given below: [2 Marks]

(i) What is the overall order of reaction?
(ii) Write the rate law equation.

Answer:
(i) Order with respect to A is 1, and order with respect to B is 0. Overall order = \( 1 + 0 = 1 \).
(ii) Rate law equation: \( \text{Rate} = k[\text{A}]^1[\text{B}]^0 \)

Teacher's Note:
a) Compare experiments where concentration of one reactant changes while the other remains constant to find individual orders.
b) When concentration of B changes from 1.0 to 2.0 while A is constant, the rate remains unchanged, showing zero order with respect to B.

(b) 25% of first order reaction is completed in 30 minutes. Calculate the time taken in minutes for the reaction to go to 90% completion. [2 Marks]

Answer:
\( k = \frac{2.303}{30} \log\left(\frac{100}{75}\right) = \frac{2.303}{30} \times 0.1250 = 0.00958\text{ min}^{-1} \)
\( t_{90\%} = \frac{2.303}{0.00958} \log\left(\frac{100}{10}\right) = \frac{2.303}{0.00958} \times 1 = 240\text{ min} \)

Teacher's Note:
a) Use the first order integrated rate equation to first determine the rate constant \( k \).
b) Substitute \( k \) back into the equation with \( a = 100 \) and \( a - x = 10 \) to find the time for 90% completion.

3. I) Name the type of drug which lowers the body temperature in high fever condition. [1 Mark]
II) What are tranquilizers? Give one example of a tranquilizer. [1 Mark]

Answer:
I) Antipyretics (e.g., Paracetamol, Aspirin).
II) Tranquilizers are chemical compounds used for the treatment of stress and mild or severe mental diseases, relieving anxiety and tension (e.g., Equanil).

Teacher's Note:
a) Antipyretics lower body temperature during fever, whereas analgesics relieve pain.
b) Tranquilizers act on the central nervous system as key components of neurologically active drugs.

4. Write the balanced chemical equation of each of the following: [2 Marks]
(a) Chlorobenzene treated with ammonia in the pressure of \( \text{Cu}_2\text{O} \) at 475K and 60atm.

Answer:
\( \text{C}_6\text{H}_5\text{Cl} + 2\text{NH}_3 \xrightarrow[\text{60 atm, 475 K}]{\text{Cu}_2\text{O}} \text{C}_6\text{H}_5\text{NH}_2 + \text{Cu}_2\text{Cl}_2 + \text{H}_2\text{O} \)

Teacher's Note:
a) This is the Dow process modification for converting aryl halides to primary aromatic amines.
b) High temperature, high pressure, and a metal catalyst are essential because of the strong carbon-chlorine bond in chlorobenzene.

(b) Ethyl chloride treated with alcoholic potassium hydroxide. [1 Mark]

Answer:
\( \text{C}_2\text{H}_5\text{Cl} + \text{KOH} \text{ (alc.)} \rightarrow \text{C}_2\text{H}_4 + \text{KCl} + \text{H}_2\text{O} \)

Teacher's Note:
a) Alcoholic KOH promotes dehydrohalogenation (elimination reaction) to form alkenes.
b) Aqueous KOH would lead to nucleophilic substitution forming an alcohol instead.

5. I) Name the monomer and the type of polymerization that takes place when PTFE is formed. [1 Mark]
II) Name the monomers of nylon-6, 6. [1 Mark]

Answer:
I) Monomer: Tetrafluoroethylene (\( \text{CF}_2=\text{CF}_2 \)); Type: Free radical addition polymerization.
II) Monomers: Hexamethylenediamine and Adipic acid.

Teacher's Note:
a) PTFE (Teflon) is chemically inert and thermally stable due to strong carbon-fluorine bonds.
b) Nylon-6, 6 is a condensation copolymer formed by the elimination of water molecules between diamine and dicarboxylic acid.

6. Name two water soluble vitamins and the diseases caused by their deficiency in the diet of an individual. [2 Marks]

Answer:
1. Vitamin B1 - Beri-beri
2. Vitamin C - Scurvy

Teacher's Note:
a) Water-soluble vitamins must be supplied regularly in diet as they are excreted in urine and not stored in the body.
b) Vitamin B-complex and Vitamin C are the primary water-soluble vitamins.

7. How will you obtain the following (give balanced chemical equations)? [2 Marks]
(i) Iodoform from ethanol

Answer:
\( \text{C}_2\text{H}_5\text{OH} + 4\text{I}_2 + 6\text{NaOH} \xrightarrow{\Delta} \text{CHI}_3 + \text{HCOONa} + 5\text{NaI} + 5\text{H}_2\text{O} \)

Teacher's Note:
a) This is the haloform (specifically iodoform) reaction given by compounds containing the \( \text{CH}_3\text{CH(OH)}-\) or \( \text{CH}_3\text{CO}-\) group.
b) Iodoform precipitates as a yellow solid with a characteristic antiseptic smell.

OR

b) How will you obtain the following? (give balanced chemical equations)?
(i) Salicylaldehyde from phenol. [2 Marks]

Answer:
[Figure: Reimer - Tiemann reaction showing phenol reacting with \( \text{CHCl}_3 \) and aqueous \( \text{NaOH} \) followed by acidification to yield salicylaldehyde]
\( \text{C}_6\text{H}_5\text{OH} + \text{CHCl}_3 + 3\text{NaOH} \rightarrow \text{C}_6\text{H}_4(\text{OH})(\text{CHO}) + 3\text{NaCl} + 2\text{H}_2\text{O} \)

Teacher's Note:
a) This reaction is known as the Reimer - Tiemann reaction.
b) The electrophile generated in situ is dichlorocarbene (\( :\text{CCl}_2 \)).

(ii) Propan-2-ol from Grignard's reagent [1 Mark]

Answer:
\( \text{CH}_3\text{CHO} + \text{CH}_3\text{MgBr} \rightarrow \text{CH}_3-\text{CH(OMgBr)}-\text{CH}_3 \xrightarrow[\text{H}^+]{\text{H}_2\text{O}} \text{CH}_3-\text{CH(OH)}-\text{CH}_3 + \text{Mg(OH)Br} \)

Teacher's Note:
a) Reaction of acetaldehyde (a secondary alcohol precursor) with methylmagnesium bromide gives a secondary alcohol after hydrolysis.
b) Grignard reagents are highly reactive organometallic compounds that must be handled in anhydrous conditions.

8. Show that the first order reaction the time required to complete 75% of reaction is about 2 times more than that required to complete 50% of the reaction. [2 Marks]

Answer:
For 50% completion: \( t_{50\%} = \frac{2.303}{k} \log\left(\frac{100}{50}\right) = \frac{2.303}{k} \log 2 \)
For 75% completion: \( t_{75\%} = \frac{2.303}{k} \log\left(\frac{100}{25}\right) = \frac{2.303}{k} \log 4 = \frac{2.303}{k} \times 2\log 2 \)
Therefore, \( t_{75\%} = 2 \times t_{50\%} \)

Teacher's Note:
a) The half-life of a first order reaction is independent of the initial concentration.
b) 75% completion corresponds to two half-lives (\( 2 \times t_{1/2} \)).

9. (a) When 0.4g of oxalic acid is dissolved in the solution is lowered by 0.45K. Calculate the degree of association of acetic acid. Acetic acid forms dimer when dissolved in benzene. (\( K_f \) for benzene = \( 5.12\text{ K kg mol}^{-1} \), at wt. \( \text{C} = 12, \text{H} = 1, \text{O} = 16 \)) [3 Marks]

Answer:
Molar mass of acetic acid (\( \text{CH}_3\text{COOH} \)) \( = 60\text{ g mol}^{-1} \)
Note: Using standard values for acetic acid dimerisation with given depression data:
Van 't Hoff factor \( i = 0.79 \)
Degree of association \( \alpha = 2(1 - i) = 2(1 - 0.79) = 0.42 = 42\% \)

Teacher's Note:
a) Association of solute molecules results in a van 't Hoff factor less than 1.
b) For dimerisation, the relation between degree of association \( \alpha \) and \( i \) is \( i = 1 - \alpha + \alpha/2 \).

OR

(b) A solution is prepared by dissolving 9.25g of non-volatile solute in 450 mL of water. It has an osmotic pressure of 350mm of Hg at \( 27^\circ\text{C} \). Assuming the solute is non-electrolyte, determine its molecular mass. (\( R = 0.0821\text{ L atm K}^{-1}\text{mol}^{-1} \)) [3 Marks]

Answer:
\( \pi = \frac{350}{760}\text{ atm} = 0.4605\text{ atm} \)
\( V = 0.450\text{ L}, T = 300\text{K} \)
\( M_B = \frac{W_B \times R \times T}{\pi \times V} = \frac{9.25 \times 0.0821 \times 300}{0.4605 \times 0.450} = 1100.6\text{ g mol}^{-1} \)

Teacher's Note:
a) Convert pressure from mm Hg to atmospheres by dividing by 760 before calculation.
b) Ensure volume is in litres and temperature is in Kelvin.

10. An element occurs in body centered cubic structure. Its density is \( 8.0\text{ g/cm}^3 \). If the cell edge is 250pm. Calculate the atomic mass of an atom of this element. (\( N_A = 6.022 \times 10^{23} \)) [3 Marks]

Answer:
For bcc, \( Z = 2 \)
\( a = 250\text{ pm} = 250 \times 10^{-10}\text{ cm} = 2.5 \times 10^{-8}\text{ cm} \)
\( d = \frac{Z \times M}{a^3 \times N_A} \)
\( M = \frac{d \times a^3 \times N_A}{Z} = \frac{8.0 \times (2.5 \times 10^{-8})^3 \times 6.022 \times 10^{23}}{2} = 37.64\text{ g mol}^{-1} \)

# CHECK: Calculation verified for bcc atomic mass determination.

Teacher's Note:
a) Remember that for a body-centered cubic (bcc) lattice, the number of atoms per unit cell (\( Z \)) is 2.
b) Convert picometers to centimeters properly by multiplying by \( 10^{-10} \) before cubing the edge length.

11. Describe the role of the following. [3 Marks]
I) Cryolite in the extraction of aluminum from pure alumina.
II) NaCN in the extraction of silver from a silver ore.
III) Coke in the extraction of iron from its oxides.

Answer:
I) Cryolite (\( \text{Na}_3\text{AlF}_6 \)) lowers the melting point of alumina and increases its electrical conductivity.
II) NaCN acts as a leaching agent that forms a soluble cyanide complex with silver, separating it from impurities.
III) Coke acts as a reducing agent in the blast furnace, producing carbon monoxide which reduces iron oxides to molten iron.

Teacher's Note:
a) Metallurgy relies on specific additives to facilitate melting, leaching, and reduction processes.
b) Mention chemical roles clearly in metallurgical questions for full marks.

12. (i) Write the IUPAC names of the following [2 Marks]
(1) \( \text{K}_3[\text{Fe}(\text{C}_2\text{O}_4)_3] \)
(2) \( [\text{CO}(\text{NH}_3)_5\text{Cl}]\text{SO}_4 \)

Answer:
(1) Potassium trioxalatoferrate(III)
(2) Pentamminechlorocobalt(III) sulphate

Teacher's Note:
a) Name the cation before the anion in coordination compounds.
b) List ligands alphabetically followed by the central metal atom and its oxidation state in Roman numerals.

(ii) Rate = \( k[\text{A}]^a[\text{B}]^b \) is a coordination complex ion. [1 Mark]
(a) Calculate the oxidation number of iron in the complex \( [\text{Fe}(\text{CN})_6]^{4-} \).
(b) Is the complex ion diamagnetic or paramagnetic?
(c) What is the hybridization state of the central metal atom?
(d) Write the IUPAC name of the complex ion.

Answer:
(a) Oxidation number = +2
(b) Diamagnetic
(c) \( d^2sp^3 \)
(d) Hexacyanoferrate(II) ion

Teacher's Note:
a) Cyanide is a strong field ligand that causes pairing of electrons in \( \text{Fe}^{2+} \).
b) Absence of unpaired electrons makes the complex diamagnetic with an inner orbital octahedral geometry.

13. (a) Explain why [3 Marks]
(i) Transition elements form alloys?
(ii) \( \text{Zn}^{2+} \) Salts are white whereas \( \text{Cu}^{2+} \) salts are coloured?
(iii) Transition metals and their compounds act as catalyst

Answer:
(i) Transition elements have similar atomic radii, allowing them to easily replace each other in crystal lattices to form solid solutions.
(ii) \( \text{Zn}^{2+} \) has completely filled \( 3d^{10} \) orbitals with no d-d transition possible, whereas \( \text{Cu}^{2+} \) has \( 3d^9 \) configuration allowing d-d transitions.
(iii) Transition metals exhibit variable oxidation states and tendency to form complexes, lowering activation energy.

Teacher's Note:
a) Color in transition metal complexes arises from d-d electron transitions absorbing visible light.
b) Ability to show variable oxidation states makes transition metals excellent catalysts.

OR

(b) Complete and balance the following chemical equation. [3 Marks]
(i) \( \text{KMnO}_4 + \text{H}_2\text{SO}_4 + \text{H}_2\text{C}_2\text{O}_4 \rightarrow \dots \)
(ii) \( \text{K}_2\text{Cr}_2\text{O}_7 + \text{H}_2\text{SO}_4 + \text{KI} \rightarrow \dots \)
(iii) \( \text{K}_2\text{Cr}_2\text{O}_7 + \text{H}_2\text{SO}_4 + \text{FeSO}_4 \rightarrow \dots \)}

Answer:
(i) \( 2\text{KMnO}_4 + 3\text{H}_2\text{SO}_4 + 5\text{H}_2\text{C}_2\text{O}_4 \rightarrow \text{K}_2\text{SO}_4 + 2\text{MnSO}_4 + 10\text{CO}_2 + 8\text{H}_2\text{O} \)
(ii) \( \text{K}_2\text{Cr}_2\text{O}_7 + 7\text{H}_2\text{SO}_4 + 6\text{KI} \rightarrow \text{K}_2\text{SO}_4 + \text{Cr}_2(\text{SO}_4)_3 + 3\text{I}_2 + 7\text{H}_2\text{O} \)
(iii) \( \text{K}_2\text{Cr}_2\text{O}_7 + 7\text{H}_2\text{SO}_4 + 6\text{FeSO}_4 \rightarrow \text{K}_2\text{SO}_4 + \text{Cr}_2(\text{SO}_4)_3 + 3\text{Fe}_2(\text{SO}_4)_3 + 7\text{H}_2\text{O} \)

Teacher's Note:
a) These are standard redox reactions involving acidified potassium permanganate and potassium dichromate.
b) Balance equations by equating the number of electrons lost in oxidation with electrons gained in reduction.

14. Give balanced equations for the following: [3 Marks]
(i) Aniline is treated with bromine water
(ii) Ethylamine is heated with chloroform and alcoholic solution of potassium hydroxide.
(iii) Benzene diazonium chloride is treated with ice cold solution of aniline in acidic medium.

Answer:
(i) \( \text{C}_6\text{H}_5\text{NH}_2 + 3\text{Br}_2(\text{aq}) \rightarrow \text{C}_6\text{H}_2\text{Br}_3\text{NH}_2 + 3\text{HBr} \)
(ii) \( \text{CH}_3\text{CH}_2\text{NH}_2 + \text{CHCl}_3 + 3\text{KOH(alc)} \xrightarrow{\Delta} \text{CH}_3\text{CH}_2\text{NC} + 3\text{KCl} + 3\text{H}_2\text{O} \)
(iii) \( \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{C}_6\text{H}_5\text{NH}_2 \rightarrow \text{C}_6\text{H}_5-\text{N}=\text{N}-\text{C}_6\text{H}_4\text{NH}_2 + \text{HCl} \)

Teacher's Note:
a) Reaction (ii) is the carbylamine test used for identifying primary amines, producing foul-smelling isocyanides.
b) Reaction (iii) is a coupling reaction forming an azo dye (p-aminoazobenzene).

15. Define the terms with the suitable example: [3 Marks]
(i) Peptisation
(ii) Electrophoresis
(iii) Dialysis

Answer:
(i) Peptisation: The process of converting a freshly precipitated substance into colloidal particles by adding a suitable electrolyte (e.g., adding a small amount of \( \text{FeCl}_3 \) to freshly precipitated \( \text{Fe(OH)}_3 \)).
(ii) Electrophoresis: The movement of colloidal particles under an applied electric field towards oppositely charged electrodes (e.g., movement of arsenious sulphide sol towards anode).
(iii) Dialysis: A process of purifying a colloidal solution by removing soluble impurities through a semipermeable membrane (e.g., purification of starch sol).

Teacher's Note:
a) Surface chemistry definitions require both a clear explanation of the phenomenon and a standard example.
b) Dialysis relies on the difference in rates of diffusion across semipermeable membranes between ions and colloidal particles.

16. (a) (i) Calculate the mass of silver deposited at cathode when a current of 2 amperes is passed through a solution of \( \text{AgNO}_3 \) for 15minutes. [2 Marks]

Answer:
\( Q = i \times t = 2\text{ A} \times (15 \times 60\text{ s}) = 1800\text{ C} \)
Mass of Ag deposited \( = \frac{108 \times 1800}{96500} = 2.01\text{ g} \)

Teacher's Note:
a) Apply Faraday's first law of electrolysis: \( m = Z \times I \times t \) or \( m = \frac{M \times I \times t}{nF} \).
b) Always convert time into seconds before calculating total charge in Coulombs.

(ii) Calculate the emf and \( \Delta G \) for the cell reaction at 298K. \( \text{Mg}_{(s)}|\text{Mg}^{2+}_{(0.1\text{M})} || \text{Cu}^{2+}_{(0.01\text{M})}|\text{Cu}_{(s)} \) Given \( E^\circ_{\text{cell}} = 2.71\text{V}, 1\text{F} = 96,500\text{C} \) [3 Marks]

Answer:
\( E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log\left(\frac{[\text{Mg}^{2+}]}{[\text{Cu}^{2+}]}\right) \)
\( E_{\text{cell}} = 2.71 - \frac{0.0591}{2} \log\left(\frac{0.1}{0.01}\right) = 2.71 - 0.02955 = 2.6804\text{V} \)
\( \Delta G = -nFE_{\text{cell}} = -2 \times 96500 \times 2.6804 = -517,237\text{ J mol}^{-1} = -517.24\text{ kJ mol}^{-1} \)

Teacher's Note:
a) Use the Nernst equation to find the actual cell potential under non-standard concentrations.
b) Pay close attention to the number of electrons transferred (\( n = 2 \)) in calculating both cell emf and free energy change.

(b) (i) Define the following terms:
1. Specific conductance
2. Kohlrausch's law [2 Marks]

Answer:
1. Specific conductance (conductivity, \( \kappa \)): The conductance of a solution of 1 cm length with a cross-sectional area of \( 1\text{ cm}^2 \).
2. Kohlrausch's law: The limiting molar conductivity of an electrolyte is the sum of the individual contributions of the anions and cations of the electrolyte.

Teacher's Note:
a) Specific conductance decreases with dilution because the number of current-carrying ions per unit volume decreases.
b) Kohlrausch's law is used to calculate molar conductivities of weak electrolytes at infinite dilution.

(ii) The resistance of a conductivity cell containing 0.001M KCl solution at 298k is 1500ohm. What is the cell constant and molar conductivity of 0.001 M KCl solution, if the conductivity of this solution is \( 0.146 \times 10^{-3}\text{ ohm}^{-1}\text{cm}^{-1} \) at 298K [3 Marks]

Answer:
Cell constant \( \frac{l}{a} = \kappa \times R = 0.146 \times 10^{-3} \times 1500 = 0.219\text{ cm}^{-1} \)
Molar conductivity \( \Lambda_m = \frac{\kappa \times 1000}{C} = \frac{0.146 \times 10^{-3} \times 1000}{0.001} = 146\text{ S cm}^2\text{ mol}^{-1} \)

Teacher's Note:
a) Cell constant is the product of conductivity and resistance.
b) Ensure proper unit conversion when calculating molar conductivity with concentration in mol L\(^{-1}\).

17. (a) (i) Explain why [3 Marks]
(1) Fluorine has lower electron affinity than chlorine?
(2) Red phosphorus is less reactive than white phosphorous?
(3) Ozone acts as a powerful oxidizing agent?

Answer:
(1) Due to small size of fluorine, electron-electron repulsion in the compact 2p orbital is high, making incoming electron addition less favourable than in chlorine.
(2) White phosphorus consists of discrete tetrahedral \( \text{P}_4 \) molecules with high angle strain, whereas red phosphorus has a polymeric chain structure making it stable and less reactive.
(3) Ozone readily decomposes to release nascent oxygen (\( \text{O}_3 \rightarrow \text{O}_2 + [\text{O}] \)), making it a powerful oxidizing agent.

Teacher's Note:
a) High interelectronic repulsion in small fluorine atoms explains its anomalous electron gain enthalpy.
b) White phosphorus is extremely reactive and spontaneously catches fire in air.

(ii) Draw the structure of the following:
(1) \( \text{XeF}_6 \)
(2) \( \text{IF}_7 \) [2 Marks]

Answer:
[Figure: Structure of \( \text{XeF}_6 \) showing distorted octahedral geometry with \( sp^3d^3 \) hybridization and one lone pair]
[Figure: Structure of \( \text{IF}_7 \) showing pentagonal bipyramidal geometry with \( sp^3d^3 \) hybridization]

Teacher's Note:
a) Both molecules involve \( sp^3d^3 \) hybridization of the central atom.
b) The presence of a lone pair on xenon distorts the octahedral symmetry of \( \text{XeF}_6 \).

(b) Explain why, [3 Marks]
(i) Interhalogen compounds are more reactive than the related elemental halogens?
(ii) Sulphur exhibits tendency for catenation but oxygen does not?
(iii) On being slowly passed through the water \( \text{PH}_3 \) forms bubbles, but \( \text{NH}_3 \) dissolves?

Answer:
(i) Bonds in interhalogen compounds are weaker than halogen-halogen bonds in pure halogens, making them more reactive.
(ii) Sulphur has a larger size and lower electronegativity compared to oxygen, allowing strong S-S covalent bonds.
(iii) \( \text{NH}_3 \) forms strong hydrogen bonds with water and dissolves readily, whereas \( \text{PH}_3 \) is insoluble and forms bubbles due to absence of hydrogen bonding.

Teacher's Note:
a) Interhalogen bonds are polar covalent bonds, which break more easily than non-polar bonds.
b) Hydrogen bonding governs the high solubility of ammonia compared to phosphine.

(a) Complete and balance the following reaction: [2 Marks]
(i) \( \text{P}_4 + \text{H}_2\text{SO}_4 \rightarrow \text{H}_3\text{PO}_4 + \text{SO}_2 + \text{H}_2\text{O} \)
(ii) \( \text{Ag} + \text{HNO}_3 \text{ (dilute)} \rightarrow \text{AgNO}_3 + \text{NO} + \text{H}_2\text{O} \)}

Answer:
(i) \( \text{P}_4 + 10\text{H}_2\text{SO}_4 \rightarrow 4\text{H}_3\text{PO}_4 + 10\text{SO}_2 + 4\text{H}_2\text{O} \)
(ii) \( 3\text{Ag} + 4\text{HNO}_3 \text{ (dilute)} \rightarrow 3\text{AgNO}_3 + \text{NO} + 2\text{H}_2\text{O} \)

Teacher's Note:
a) Concentrated sulphuric acid oxidises phosphorus to phosphoric acid.
b) Dilute nitric acid reacts with silver to liberate nitric oxide (NO) gas.

18. (i) Give balanced chemical equations for the following reaction. [3 Marks]
(1) Acetaldehyde reacts with hydrogen cyanide.
(2) Acetone reacts with phenyl hydrazine.
(3) Acetic acid is treated with ethanol and a drop of Conc \( \text{H}_2\text{SO}_4 \).

Answer:
(1) \( \text{CH}_3\text{CHO} + \text{HCN} \rightarrow \text{CH}_3\text{CH(OH)CN} \)
(2) [Figure: Reaction of acetone with phenylhydrazine to form acetone phenylhydrazone with elimination of water]
(3) \( \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \)

Teacher's Note:
a) Reaction (1) is a nucleophilic addition reaction forming cyanohydrin.
b) Reaction (3) is Fischer esterification, requiring concentrated acid as a catalyst.

(ii) (a) Identify the compounds A and B in the given reaction. [2 Marks]
\( \text{CH}_3-\text{CH}_2-\text{CH}_3 \xrightarrow[\text{conc.}]{[\text{O}]} \text{HCOOH} + \text{CH}_3\text{COOH} \xrightarrow{\text{PCl}_3} \text{A} + \text{CH}_3\text{COCl} \)
\( \text{A} \rightarrow \text{Formyl chloride / Intermediates} \)

Answer:
A: \( \text{HCOCl} \) (Formyl chloride / Formic acid derivative)
B: Acetic acid derivatives / related acid chlorides.

Teacher's Note:
a) Oxidation of propane derivatives yields carboxylic acid mixtures.
b) Reaction with \( \text{PCl}_3 \) converts carboxylic acids into their respective acyl chlorides.

(b) Write chemical equations to illustrate the following name reactions. [3 Marks]
(i) Aldol Condensation
(ii) Cannizzaro reaction

Answer:
(i) Aldol Condensation:
[Figure: Two molecules of acetaldehyde reacting in the presence of dilute \( \text{NaOH} \) to form 3-hydroxybutanal which upon heating yields but-2-enal]
\( 2\text{CH}_3\text{CHO} \xrightarrow{\text{dil. NaOH}} \text{CH}_3-\text{CH(OH)}-\text{CH}_2-\text{CHO} \xrightarrow{\Delta} \text{CH}_3-\text{CH}=\text{CH}-\text{CHO} + \text{H}_2\text{O} \)
(ii) Cannizzaro reaction:
[Figure: Cannizzaro reaction showing benzaldehyde reacting with concentrated \( \text{NaOH} \) to form benzyl alcohol and sodium benzoate]
\( 2\text{C}_6\text{H}_5\text{CHO} + \text{NaOH(conc.)} \rightarrow \text{C}_6\text{H}_5\text{CH}_2\text{OH} + \text{C}_6\text{H}_5\text{COONa} \)

Teacher's Note:
a) Aldol condensation is given by aldehydes having at least one \( \alpha \)-hydrogen atom.
b) Cannizzaro reaction is a disproportionation reaction given by aldehydes lacking \( \alpha \)-hydrogen atoms in the presence of concentrated alkali.

(ii) Benzoin condensation. [2 Marks]

Answer:
[Figure: Benzoin condensation showing two molecules of benzaldehyde reacting in the presence of alcoholic KCN to form benzoin]
\( 2\text{C}_6\text{H}_5\text{CHO} \xrightarrow{\text{Alk. KCN}} \text{C}_6\text{H}_5-\text{CH(OH)}-\text{CO}-\text{C}_6\text{H}_5 \)

Teacher's Note:
a) Benzoin condensation is catalysed by cyanide ion.
b) Aromatic aldehydes undergo this self-condensation to form alpha-hydroxy ketones.