ISC Class 12 Chemistry Board Exam Question Paper 2018 with Solutions

ISC Class 12 Chemistry Board Exam Question Paper with Solutions

Question 1

(a) Fill in the blanks by choosing the appropriate word/words from those given in the brackets: [4×1]
(square pyramidal, electrical, 74; 26, sp³d², sp³d, chemical, 68, 32, tetrahedral, yellow, white, iodoform, Lucas)
(i) A Galvanic cell converts _______ energy into _______ energy. [1 Mark]

Answer: A Galvanic cell converts chemical energy into electrical energy.

Teacher's Note:
a) A Galvanic cell is an electrochemical device that generates electricity from spontaneous redox reactions.
b) Ensure you do not confuse it with an electrolytic cell, which performs the exact reverse conversion.

(ii) The percentage of unoccupied spaces in bcc and fcc arrangements are _______ and _______ respectively. [1 Mark]

Answer: The percentage of unoccupied spaces in bcc and fcc arrangements are 32 and 26 respectively.

Teacher's Note:
a) Packing efficiency in bcc is 68% (leaving 32% empty space), while in fcc/hcp it is 74% (leaving 26% empty space).
b) Read the question carefully to check whether occupied space or unoccupied (void) space is asked.

(iii) Propan-2-ol on reaction with iodine and sodium hydroxide gives _______ precipitate and the reaction is called _______ test. [1 Mark]

Answer: Propan-2-ol on reaction with iodine and sodium hydroxide gives yellow precipitate and the reaction is called iodoform test.

Teacher's Note:
a) Alcohols containing the CH₃-CH(OH)- group give a positive iodoform test.
b) The yellow precipitate formed is triiodomethane (iodoform, CHI₃).

(iv) The geometry of XeOF₄ molecule is _______ and the hybridisation of Xenon atom in the molecule is _______. [1 Mark]

Answer: The geometry of XeOF₄ molecule is pyramidal and the hybridisation of Xenon atom in the molecule is sp³d².

Teacher's Note:
a) Xenon in XeOF₄ has 6 electron pairs (5 sigma bonds + 1 lone pair), giving octahedral electronic geometry and square pyramidal molecular shape (often referred to simply as pyramidal based on atom positions).
b) Count lone pairs and bond pairs accurately using VSEPR theory.

(b) Complete the following statements by selecting the correct alternative from the choices given: [4×1]
(i) During the course of an S_N1 reaction, the intermediate species formed is: [1 Mark]
(1) a carbocation
(2) a free radical
(3) a carbanion
(4) an intermediate complex

Answer: (1) a carbocation

S_N1 reactions proceed via a two-step mechanism where the rate-determining step involves the formation of a planar carbocation intermediate.

Teacher's Note:
a) S_N1 stands for Substitution Nucleophilic Unimolecular.
b) Tertiary alkyl halides favour S_N1 due to greater stability of the tertiary carbocation.

(ii) Purification of aluminium by electrolytic refining is called: [1 Mark]
(1) Serpeck's process
(2) Hoope's process
(3) Hall's process
(4) Baeyer's process

Answer: (2) Hoope's process

Hoope's process is used for the electrolytic refining of impure aluminium to obtain 99.99% pure aluminium.

Teacher's Note:
a) Hoope's cell uses three liquid layers of different densities.
b) Do not confuse with Hall-Heroult process, which is the electrolytic reduction of alumina.

(iii) An aqueous solution of urea freezes at - 0.186°C, K_f for water = 1.86 K kg mol^-1, K_b for water = 0.512 K kg mol^-1. The boiling point of urea solution will be: [1 Mark]
(1) 373.065 K
(2) 373.186 K
(3) 373.512 K
(4) 373.0512 K

Answer: (4) 373.0512 K

\(\Delta T_f = K_f \times m \implies 0 - (-0.186) = 1.86 \times m \implies m = 0.1 \, \text{mol kg}^{-1}\).
\(\Delta T_b = K_b \times m = 0.512 \times 0.1 = 0.0512 \, \text{K}\).
Boiling point \(T = 373.15 + 0.0512 = 373.2012 \, \text{K}\) (Using standard normal boiling point \(373.15\) K or standard approximation \(373 + 0.0512 = 373.0512\) K as per official solution key).

Teacher's Note:
a) Molality remains constant for a dilute solution of the same solute concentration.
b) Use the exact numerical values provided in the question.

(iv) In the dehydration of alcohols to alkenes by heating with concentrated sulphuric acid, the initiation step is: [1 Mark]
(1) formation of carbocation
(2) formation of an ester
(3) protonation of alcohol molecule
(4) elimination of water

Answer: (3) protonation of alcohol molecule

The mechanism begins with the protonation of the oxygen atom of the alcohol by an acid catalyst to form an oxonium ion.

Teacher's Note:
a) Protonation makes the leaving group (\(-OH_2^+\)) a good leaving group.
b) The second step is the loss of a water molecule to form a carbocation.

(c) Match the following: [4×1]
(i) Rate constant                                 (a) Dialysis
(ii) Biodegradable polymer                 (b) Glycine
(iii) Zwitter ion                                 (c) Arrhenius equation
(iv) Purification of colloids                 (d) PHBV

Answer:
(i) Rate constant - Arrhenius equation
(ii) Biodegradable polymer - PHBV
(iii) Zwitter ion - Glycine
(iv) Purification of colloids - Dialysis

Teacher's Note:
a) Match each term directly with its correct definition or example.
b) Glycine exhibits dipolar (zwitterionic) character due to the presence of both acidic carboxyl and basic amino groups.

(d) Answer the following questions: [4×2]
(i) (1) Why does the density of transition elements increase from Titanium to Copper? (at. no. Ti = 22, Cu = 29) [1 Mark]

Answer: Across the period from Ti to Cu, as atomic size decreases due to effective nuclear charge, the atomic mass increases significantly, resulting in an increase in density.

Teacher's Note:
a) Density depends upon mass and volume (\(d = m/V\)).
b) Across a transition series, atomic volume decreases slightly while atomic mass increases noticeably.

(2) Why is zinc not regarded as a transition element? (at. no. Zn = 30) [1 Mark]

Answer: The electronic configuration of Zn (atomic number 30) is \([Ar]_{18} 4s^2 3d^{10}\). Since zinc has completely filled \(3d\) orbitals in its ground state as well as in its stable oxidation state (\(Zn^{2+}\)), it does not have vacant or partially filled \(d\)-orbitals. Therefore, it is not regarded as a transition element.

Teacher's Note:
a) Transition elements are defined as elements that have incompletely filled \(d\)-orbitals either in ground state or in stable oxidation state.
b) Zn, Cd, and Hg belong to group 12 and are classic examples of non-transition metals.

(ii) Identify the compounds A, B, C and D.
CH₃CN \(\xrightarrow{H_2O / H^+} A \xrightarrow{NH_3} B \xrightarrow{\text{heat}} C \xrightarrow{Br_2 / KOH} D\) [2 Marks]

Answer:
A = CH₃COOH (Acetic acid)
B = CH₃COONH₄ (Ammonium acetate)
C = CH₃CONH₂ (Acetamide)
D = CH₃NH₂ (Methylamine)

Teacher's Note:
a) Hydrolysis of cyanides yields carboxylic acids.
b) Hofmann bromamide degradation reaction (\(Br_2 / KOH\)) converts amides into primary amines with one carbon atom less.

(iii) Calculate the osmotic pressure of a solution prepared by dissolving 0.025 g of K₂SO₄ in 2.0 litres of water at 25°C assuming that K₂SO₄ is completely dissociated. (mol. wt. of K₂SO₄ = 174 g mol^-1) [2 Marks]

Answer:
Mass of K₂SO₄ = \(0.025\) g
Molar mass of K₂SO₄ = \(174\) g/mol
Volume \(V = 2\) L
Temperature \(T = 25^{\circ}\text{C} = 298\) K
Van't Hoff factor \(i\) for complete dissociation of K₂SO₄ (\(\text{K}_2\text{SO}_4 \rightarrow 2\text{K}^+ + \text{SO}_4^{2-}\)) is \(3\).
\(\pi = i \cdot \frac{w}{M \cdot V} \cdot R \cdot T\)
\(\pi = 3 \times \frac{0.025}{174 \times 2} \times 0.8314 \times 298 = 0.173\) atm

Teacher's Note:
a) Always include the van't Hoff factor \(i\) for ionic solutes undergoing dissociation.
b) Double-check units of volume (litres) and gas constant (\(R = 0.0821\) L atm K^-1 mol^-1 or use corresponding units).

(iv) What type of isomerism is shown by the following coordination compounds: \([PtCl_2(NH_3)_4]Br_2\) and \([PtBr_2(NH_3)_4]Cl_2\). Write their IUPAC names. [2 Marks]

Answer:
These coordination compounds show ionization isomerism.
\([PtCl_2(NH_3)_4]Br_2\): Tetraamminedichloroplatinum(IV) bromide
\([PtBr_2(NH_3)_4]Cl_2\): Tetraamminedibromoplatinum(IV) chloride

Teacher's Note:
a) Ionization isomerism arises when compounds yield different ions in solution.
b) Ligands are named in alphabetical order in IUPAC nomenclature.

Question 2

(a) (i) Write the rate law expression for the reaction \(A + B + C \rightarrow D + E\), if the order of reaction is first, second and zero with respect to A, B and C, respectively. [1 Mark]

Answer:
Rate = \(K[A]^1[B]^2[C]^0\)

Teacher's Note:
a) Rate law expresses the reaction rate in terms of molar concentrations of reactants with each term raised to its individual order.
b) Any concentration raised to power zero equals one.

(ii) How many times the rate of reaction will increase if the concentration of A, B and C are doubled in the equation given in (i) above? [1 Mark]

Answer:
If concentration of A is doubled, rate increases by \(2\) times. If concentration of B is doubled, rate increases by \(4\) ($2^2$) times. If concentration of C is doubled, there is no change in rate (\(2^0 = 1\)). Overall, the rate increases by \(2 \times 4 \times 1 = 8\) times.

Teacher's Note:
a) Substitute doubled concentration terms into the rate law expression.
b) The overall rate multiplication factor is the product of individual increases.

OR

(b) The rate of reaction becomes four times when the temperature changes from 293 K to 313 K. Calculate the energy of activation (\(E_a\)) of the reaction assuming that it does not change with temperature. (\(R = 8.314\) J K^-1 mol^-1) [2 Marks]

Answer:
\(T_1 = 293\) K, \(T_2 = 313\) K
\(K_1 = K\), \(K_2 = 4K\)
\(\log \left(\frac{K_2}{K_1}\right) = \frac{E_a}{2.303 R} \left(\frac{T_2 - T_1}{T_1 T_2}\right)\)
\(\log(4) = \frac{E_a}{2.303 \times 8.314} \left(\frac{313 - 293}{293 \times 313}\right)\)
\(0.6020 = \frac{E_a}{2.303 \times 8.314} \left(\frac{20}{293 \times 313}\right)\)
\(E_a = \frac{0.6020 \times 2.303 \times 8.314 \times 293 \times 313}{20} = 52854\) J mol^-1 = \(52.854\) kJ mol^-1

Teacher's Note:
a) Use the Arrhenius equation in logarithmic form for two different temperatures.
b) Ensure units are consistent (convert joules to kilojoules if required).

Question 3

(a) How do antiseptics differ from disinfectants? [2 Marks]

Answer:
1. Antiseptics are applied safely to living tissues such as wounds, cuts, and diseased skin to kill or prevent the growth of microorganisms.
2. Disinfectants are applied to non-living objects such as floors, drainage systems, and instruments because they are harmful and toxic to living tissues.

Teacher's Note:
a) The key distinction lies in the surface of application (living vs. non-living objects).
b) Certain chemical substances can act as both depending on their concentration (e.g., phenol).

(b) State the role of the following chemicals in the food industry: [2 Marks]
(i) Sodium benzoate
(ii) Aspartame

Answer:
(i) Sodium benzoate: Acts as a chemical food preservative to prevent spoilage by microbial growth.
(ii) Aspartame: Acts as an artificial sweetener used in low-calorie and diabetic foods.

Teacher's Note:
a) Preservatives inhibit the growth of bacteria, yeasts, and molds in food items.
b) Aspartame is unstable at cooking temperatures and is therefore restricted to cold foods and soft drinks.

Question 4

An aromatic organic compound [A] on heating with NH₃ and Cu₂O at high pressure gives [B]. The compound [B] on treatment with ice cold solution of NaNO₂ and HCl gives [C], which on heating with Cu/HCl gives compound [A] again. Identify the compounds [A], [B] and [C]. Write the name of the reaction for the conversion of [B] to [C]. [3 Marks]

Answer:
[A] = Chlorobenzene
[B] = Aniline
[C] = Benzenediazonium chloride
Reaction name for conversion of [B] to [C]: Diazotisation reaction.

Teacher's Note:
a) Chlorobenzene reacts with aqueous ammonia in presence of \(\text{Cu}_2\text{O}\) at high temperature and pressure to form aniline.
b) Aniline reacts with nitrous acid (\(\text{NaNO}_2 + \text{HCl}\)) at \(0 - 5^{\circ}\text{C}\) to form benzenediazonium chloride.

Question 5

Write the names of the monomers for each of the following polymers: [2 Marks]
(a) Bakelite
(b) Nylon-2-nylon-6

Answer:
(a) Monomers of Bakelite: Phenol and formaldehyde.
(b) Monomers of Nylon-2-nylon-6: Glycine (\(\text{H}_2\text{NCH}_2\text{COOH}\)) and aminocaproic acid (\(\text{H}_2\text{N(CH}_2)_5\text{COOH}\)).

Teacher's Note:
a) Bakelite is a condensation copolymer formed via novolac intermediate.
b) Nylon-2-nylon-6 is a polyamide copolymer containing amino acid monomers of different chain lengths.

Question 6

Name the purine bases and pyrimidine bases present in RNA and DNA. [2 Marks]

Answer:
Purine bases (common to both DNA and RNA): Adenine and Guanine.
Pyrimidine bases in DNA: Cytosine and Thymine.
Pyrimidine bases in RNA: Cytosine and Uracil.

Teacher's Note:
a) Purines have a double-ring structure, whereas pyrimidines have a single-ring structure.
b) Thymine is present in DNA, while uracil replaces it in RNA.

Question 7

(a) How will you obtain the following? (Give balanced equation.) [2 Marks]
(i) Picric acid from phenol
(ii) Ethyl chloride from diethyl ether.

Answer:
(i) Phenol reacts with concentrated nitric acid in the presence of concentrated sulfuric acid to yield 2,4,6-trinitrophenol (picric acid).
\(\text{C}_6\text{H}_5\text{OH} + 3\text{HNO}_3 \xrightarrow{\text{conc. H}_2\text{SO}_4} \text{C}_6\text{H}_2\text{OH(NO}_2)_3 + 3\text{H}_2\text{O}\)
(ii) Diethyl ether reacts with phosphorus pentachloride upon heating to form ethyl chloride.
\(\text{CH}_3\text{CH}_2-\text{O}-\text{CH}_2\text{CH}_3 + \text{PCl}_5 \xrightarrow{\Delta} 2\text{CH}_3\text{CH}_2\text{Cl} + \text{POCl}_3\)

Teacher's Note:
a) Nitration of phenol with conc. \(\text{HNO}_3\) directly gives picric acid though with low yield due to oxidation.
b) Ethers undergo cleavage upon heating with phosphorus pentachloride.

OR

(b) How will you obtain the following? (Give balanced equation.) [2 Marks]
(i) Anisole from phenol
(ii) Ethyl acetate from ethanol.

Answer:
(i) Phenol is first treated with sodium hydroxide to form sodium phenoxide, which on methylation with methyl bromide gives anisole.
\(\text{C}_6\text{H}_5\text{OH} + \text{NaOH} \rightarrow \text{C}_6\text{H}_5\text{ONa} + \text{H}_2\text{O}\)
\(\text{C}_6\text{H}_5\text{ONa} + \text{CH}_3\text{Br} \rightarrow \text{C}_6\text{H}_5\text{OCH}_3 + \text{NaBr}\)
(ii) Ethanol undergoes esterification with acetic acid in the presence of concentrated sulfuric acid to form ethyl acetate.
\(\text{CH}_3\text{COOH} + \text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{conc. H}_2\text{SO}_4} \text{CH}_3\text{COOCH}_2\text{CH}_3 + \text{H}_2\text{O}\)

Teacher's Note:
a) Williamson synthesis is used for the preparation of ethers like anisole.
b) Esterification is a reversible reaction catalyzed by mineral acids.

Question 8

40% of a first order reaction is completed in 50 minutes. How much time will it take for the completion of 80% of this reaction? [3 Marks]

Answer:
For a first-order reaction:
\(K = \frac{2.303}{t} \log\left(\frac{[A]_0}{[A]_t}\right)\)
When 40% is completed, \([A]_t = 100 - 40 = 60\), \(t = 50\) min.
\(K = \frac{2.303}{50} \log\left(\frac{100}{60}\right) = \frac{2.303}{50} \times 0.2218 = 0.0102\) min^-1
Now, for 80% completion, remaining concentration \([A]_t = 100 - 80 = 20\).
\(t = \frac{2.303}{0.0102} \log\left(\frac{100}{20}\right) = \frac{2.303}{0.0102} \times \log(5) = \frac{2.303}{0.0102} \times 0.6990 = 157.81\) minutes.

Teacher's Note:
a) Always substitute the remaining concentration of reactant for \([A]_t\) in the integrated rate expression.
b) Calculate the rate constant first before applying it to find the unknown time.

Question 9

(a) The freezing point of a solution containing 5.85 g of NaCl in 100 g of water is -3.348°C. Calculate van't Hoff factor for this solution. What will be the experimental molecular weight of NaCl? (\(K_f\) for water = 1.86 K kg mol^-1, at wt, Na = 23, Cl = 35.5) [3 Marks]

Answer:
Molality \(m = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} = \frac{5.85 / 58.5}{0.1} = \frac{0.1}{0.1} = 1\) mol kg^-1
\(\Delta T_f = i \cdot K_f \cdot m\)
\(3.348 = i \times 1.86 \times 1 \implies i = 1.8\)
\(i = \frac{\text{Theoretical molar mass}}{\text{Experimental molar mass}}\)
\(1.8 = \frac{58.5}{\text{Experimental molar mass}}\)
Experimental molar mass = \(\frac{58.5}{1.8} = 32.5\) g mol^-1

Teacher's Note:
a) Van't Hoff factor relates abnormal colligative properties to normal calculated values.
b) The experimental molar mass of an electrolyte undergoing dissociation is always less than its theoretical formula mass.

OR

(b) An aqueous solution containing 12.48 g of barium chloride (\(\text{BaCl}_2\)) in 1000 g of water, boils at 100.0832°C. Calculate the degree of dissociation of barium chloride. (\(K_b\) for water = 0.52 K kg mol^-1, at. wt. Ba = 137, Cl = 35.5) [3 Marks]

Answer:
Molar mass of \(\text{BaCl}_2 = 137 + 35.5 \times 2 = 208\) g mol^-1.
Molality \(m = \frac{12.48 / 208}{1 \text{ kg}} = 0.06\) mol kg^-1.
\(\Delta T_b = 100.0832 - 100 = 0.0832^{\circ}\text{C}\).
\(\Delta T_b = i \cdot K_b \cdot m \implies 0.0832 = i \times 0.52 \times 0.06 \implies i = 2.66\).
For \(\text{BaCl}_2 \rightarrow \text{Ba}^{2+} + 2\text{Cl}^{-}\), number of ions \(n = 3\).
Degree of dissociation \(\alpha = \frac{i - 1}{n - 1} = \frac{2.66 - 1}{3 - 1} = \frac{1.66}{2} = 0.83\) (or 83%).

Teacher's Note:
a) Use the dissociation formula \(\alpha = \frac{i - 1}{n - 1}\) for electrolytes.
b) Ensure proper calculation of molecular masses and temperature increments.

Question 10

Examine the defective crystal given below and answer the question that follows:
[Figure: Ionic crystal lattice showing missing cation and anion pairs, e.g., \(A^+ B^- A^+ B^-\) with vacant lattice sites for both positive and ions]
State if the above defect is stoichiometric or non-stoichiometric. How does this defect affect the density of the crystal? Also, write the term used for this type of defect. [2 Marks]

Answer:
1. The above defect is stoichiometric.
2. Density of the crystal decreases because equal number of cations and anions are missing from their lattice sites.
3. This defect is known as Schottky defect.

Teacher's Note:
a) Schottky defect is a vacancy defect shown by ionic compounds with high coordination numbers where cation and anion sizes are similar.
b) It differs from Frenkel defect, where ions occupy interstitial sites without changing density.

Question 11

Give reason for each of the following: [3 Marks]
(a) For ferric hydroxide sol the coagulating power of phosphate ion is more than chloride ion.
(b) Medicines are more effective in their colloidal form.
(c) Gelatin is added to ice creams.

Answer:
(a) According to the Schulze-Hardy rule, greater the valency of the flocculating ion, higher is its coagulating power. Ferric hydroxide sol is positively charged, so anions cause coagulation. Phosphate ion (\(\text{PO}_4^{3-}\)) carries three negative charges, whereas chloride ion (\(\text{Cl}^{-}\)) carries one negative charge; hence phosphate is far more effective.
(b) Colloidal particles possess a large surface area, which leads to enhanced adsorption and better assimilation/absorption by living tissues.
(c) Gelatin acts as a protective colloid/stabilizer in ice creams to prevent the growth of ice crystals and maintain a smooth, creamy texture.

Teacher's Note:
a) State the Schulze-Hardy rule clearly for coagulation-related reasoning.
b) Colloidal medicines are easily absorbed due to increased surface area-to-volume ratio.

Question 12

(a) For the complex ion \([Fe(CN)_6]^{3-}\), state: [3 Marks]
(i) the type of hybridisation
(ii) the magnetic behaviour
(iii) the oxidation number of the central metal atom

Answer:
(i) Hybridisation: \(d^2sp^3\) (inner orbital complex).
(ii) Magnetic behaviour: Paramagnetic (since \(\text{CN}^{-}\) is a strong field ligand, Fe(III) with \(d^5\) configuration has 1 unpaired electron).
(iii) Oxidation number of Fe: +3.

Teacher's Note:
a) Strong field ligands cause electron pairing in the \(d\)-orbitals.
b) Magnetic moment can be calculated using the spin-only formula \(\mu = \sqrt{n(n+2)}\).

(b) Write the IUPAC name of \([Co(en)_2Cl_2]^+\) ion and draw the structures of its geometrical isomers. [3 Marks]

Answer:
IUPAC name: Dichlorobis(ethane-1,2-diamine)cobalt(III) ion.
[Figure: Two stereoisomers showing cis-form (identical chlorine ligands adjacent to each other) and trans-form (identical chlorine ligands opposite to each other) in an octahedral coordination geometry around the central cobalt atom with two chelating ethylenediamine groups.]

Teacher's Note:
a) Ethylenediamine (\(en\)) is a neutral bidentate ligand.
b) The cis-isomer of this complex is optically active and exists in dextro and laevo forms.

Question 13

(a) Explain why: [3 Marks]
(i) \text{Mn}^{2+} is more stable than \text{Fe}^{2+} towards oxidation to +3 state. (At. no. of Mn = 25, Fe = 26)
(ii) Transition elements usually form coloured ions
(iii) Zr and Hf exhibit similar properties (At. no. of Zr = 40, Hf = 72)

Answer:
(i) \(\text{Mn}^{2+}\) has a stable half-filled \(3d^5\) electronic configuration, making it difficult to oxidize to \(\text{Mn}^{3+}\). On the other hand, \(\text{Fe}^{2+}\) has \(3d^6\) configuration, and losing one electron gives it a stable half-filled \(3d^5\) configuration (\(\text{Fe}^{3+}\)), so it easily undergoes oxidation.
(ii) Transition metal ions possess partially filled \(d\)-orbitals, allowing electrons to undergo \(d-d\) transition by absorbing specific wavelengths of light from the visible region, resulting in complementary colours.
(iii) Zirconium and Hafnium exhibit similar properties due to lanthanoid contraction, which results in almost identical atomic and ionic radii (chemical twins).

Teacher's Note:
a) Half-filled and completely filled subshells impart extra stability.
b) Lanthanoid contraction is caused by the poor shielding effect of \(4f\) electrons.

OR

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

Answer:
(i) \(2\text{KMnO}_4 + 10\text{KI} + 8\text{H}_2\text{SO}_4 \rightarrow 6\text{K}_2\text{SO}_4 + 2\text{MnSO}_4 + 5\text{I}_2 + 8\text{H}_2\text{O}\)
(ii) \(\text{K}_2\text{Cr}_2\text{O}_7 + 4\text{H}_2\text{SO}_4 + 3\text{H}_2\text{S} \rightarrow \text{K}_2\text{SO}_4 + \text{Cr}_2(\text{SO}_4)_3 + 3\text{S} + 7\text{H}_2\text{O}\)
(iii) \(2\text{KMnO}_4 + 8\text{H}_2\text{SO}_4 + 10\text{FeSO}_4 \rightarrow 5\text{Fe}_2(\text{SO}_4)_3 + \text{K}_2\text{SO}_4 + 2\text{MnSO}_4 + 8\text{H}_2\text{O}\)

Teacher's Note:
a) Acidified potassium permanganate and potassium dichromate are powerful oxidizing agents.
b) Balance redox equations by equating the number of electrons lost and gained.

Question 14

(a) Arrange the following in the increasing order of their basic strength: \(\text{C}_2\text{H}_5\text{NH}_2, \text{C}_6\text{H}_5\text{NH}_2, (\text{C}_2\text{H}_5)_2\text{NH}\) [1 Mark]

Answer:
\(\text{C}_6\text{H}_5\text{NH}_2 < \text{C}_2\text{H}_5\text{NH}_2 < (\text{C}_2\text{H}_5)_2\text{NH}\)

Teacher's Note:
a) Aliphatic amines are more basic than aromatic amines due to resonance stabilization of the lone pair in aniline.
b) Secondary aliphatic amines are generally more basic than primary amines due to cumulative inductive effects (+I effect) and solvation effects.

(b) Give a balanced chemical equation to convert methyl cyanide to ethyl alcohol. [2 Marks]

Answer:
\(\text{CH}_3\text{CN} + 4[\text{H}] \xrightarrow{\text{Na} / \text{C}_2\text{H}_5\text{OH}} \text{CH}_3\text{CH}_2\text{NH}_2 \xrightarrow{\text{NaNO}_2 / \text{HCl}} [\text{CH}_3\text{CH}_2\text{N}_2^+\text{Cl}^-] \xrightarrow{\text{H}_2\text{O}} \text{CH}_3\text{CH}_2\text{OH} + \text{N}_2 + \text{HCl}\)

Teacher's Note:
a) Reduction of alkyl cyanides yields primary amines (Mendius reduction).
b) Aliphatic primary amines react with nitrous acid to form unstable diazonium salts, which decompose to yield alcohols.

(c) What happens when benzene diazonium chloride reacts with phenol in weak alkaline medium? (Give balanced equation.) [2 Marks]

Answer:
Benzene diazonium chloride couples with phenol in a weak alkaline medium to form p-hydroxyazobenzene, which is an orange-coloured azo dye.
\(\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{C}_6\text{H}_5\text{OH} \xrightarrow{\text{OH}^-} \text{C}_6\text{H}_5-\text{N}=\text{N}-\text{C}_6\text{H}_4-\text{OH} + \text{HCl}\)

Answer: It gives an orange-coloured dye.

Teacher's Note:
a) This is an example of an electrophilic aromatic substitution (coupling) reaction.
b) Coupling reactions occur typically at the para position to the hydroxyl or amino group.

Question 15

Name the sulphide ore of Copper. Describe how pure copper is extracted from this ore. [5 Marks]

Answer:
Sulphide ore of copper: Copper pyrites (\(\text{CuFeS}_2\)).
Extraction steps:
1. Concentration: Powdered ore is concentrated by froth floatation process.
2. Roasting: The concentrated ore is roasted in a reverberatory furnace in a limited supply of air to remove volatile impurities and form a mixture of FeS and \(\text{Cu}_2\text{S}\) (matte).
\(2\text{CuFeS}_2 + \text{O}_2 \rightarrow \text{Cu}_2\text{S} + 2\text{FeS} + \text{SO}_2\)
3. Smelting: Roasted ore is mixed with silica and coke and heated in a blast furnace. Iron oxide slag is removed.
4. Bessemerization: Molten matte is transferred to a Bessemer converter and air is blown. \(\text{Cu}_2\text{S}\) is partially oxidized to \(\text{Cu}_2\text{O}\), which reacts with remaining \(\text{Cu}_2\text{S}\) to produce blister copper (auto-reduction).
\(2\text{Cu}_2\text{S} + 3\text{O}_2 \rightarrow 2\text{Cu}_2\text{O} + 2\text{SO}_2\)
\(2\text{Cu}_2\text{O} + \text{Cu}_2\text{S} \rightarrow 6\text{Cu} + \text{SO}_2\)
5. Refining: Blister copper is purified by electrolytic refining using impure copper as anode, pure copper strip as cathode, and acidified copper sulfate solution as electrolyte.

Teacher's Note:
a) Blister copper gets its name from the blisters formed due to the evolution of \(\text{SO}_2\) gas during solidification.
b) Electrolytic refining yields 99.99% pure copper.

Question 16

(a) (i) Calculate the emf and \(\Delta^{\circ} G\) for the cell reaction at 25°C:
\(\text{Zn}(s) | \text{Zn}^{2+}_{(aq)} (0.1M) || \text{Cd}^{2+}_{(aq)} (0.01M) | \text{Cd}(s)\)
Given \(E^{\circ}_{\text{Zn}^{2+}/\text{Zn}} = -0.763\) V and \(E^{\circ}_{\text{Cd}^{2+}/\text{Cd}} = -0.403\) V. [3 Marks]

Answer:
\(E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = -0.403 - (-0.763) = +0.360\) V.
Using Nernst equation:
\(E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{n} \log\left(\frac{[\text{Zn}^{2+}]}{[\text{Cd}^{2+}]}\right)\)
\(E_{\text{cell}} = 0.36 - \frac{0.0591}{2} \log\left(\frac{0.1}{0.01}\right) = 0.36 - \frac{0.0591}{2} \log(10) = 0.36 - 0.0295 = 0.3305\) V (\(0.331\) V).
\(\Delta^{\circ} G = -nFE^{\circ} = -2 \times 96500 \times 0.36 = -69480\) J = \(-69.48\) kJ.

Teacher's Note:
a) Ensure the number of electrons transferred (\(n = 2\)) is correctly applied.
b) \(\Delta G\) should be converted from Joules to kilo-Joules as standard practice.

(ii) Define the following terms: [2 Marks]
(1) Equivalent conductivity
(2) Corrosion of metals

Answer:
(1) Equivalent conductivity: It is defined as the conducting power of all the ions produced by dissolving one gram equivalent of an electrolyte in a given volume of solution. (\(\Lambda_e = \frac{\kappa \times 1000}{C}\))
(2) Corrosion of metals: The slow deterioration and destruction of metals due to chemical or electrochemical attack by atmospheric oxygen, moisture, and gases.

Teacher's Note:
a) Equivalent conductivity units are \(\text{S cm}^2 \text{eq}^{-1}\).
b) Rusting of iron is a classic electrochemical example of metal corrosion.

(b) (i) The specific conductivity of a solution containing 5 g of anhydrous \(\text{BaCl}_2\) (mol. wt = 208) in \(1000 \text{ cm}^3\) of a solution is found to be \(0.0058 \text{ ohm}^{-1} \text{cm}^{-1}\). Calculate the molar and equivalent conductivity of the solution. [3 Marks]

Answer:
Molarity \(M = \frac{5}{208} \times \frac{1000}{1000} = 0.024\) mol L^-1.
Molar conductivity \(\Lambda_m = \frac{\kappa \times 1000}{M} = \frac{0.0058 \times 1000}{0.024} = 241.66 \text{ S cm}^2 \text{ mol}^{-1}\).
Normality \(N = \frac{5}{104} \times \frac{1000}{1000} = 0.048\) g eq L^-1 (since n-factor for \(\text{BaCl}_2 = 2\)).
Equivalent conductivity \(\Lambda_{eq} = \frac{\kappa \times 1000}{N} = \frac{0.0058 \times 1000}{0.048} = 120.83 \text{ S cm}^2 \text{ eq}^{-1}\).

Teacher's Note:
a) Note the difference between molarity and normality for salts like \(\text{BaCl}_2\).
b) Apply correct multiplier factors for \(1000 \text{ cm}^3\) volume.

(ii) What is an electrochemical series? How is it useful in predicting whether a metal can liberate hydrogen from acid or not? [2 Marks]

Answer:
1. Electrochemical series: The arrangement of elements in order of increasing standard reduction potential values.
2. Prediction: Only metals with negative standard reduction potential values (placed above hydrogen in the series) can displace hydrogen from acids because they act as stronger reducing agents than hydrogen.

Teacher's Note:
a) Hydrogen has a standard reduction potential of \(0.00\) V.
b) Metals below hydrogen (like copper) cannot liberate hydrogen gas from dilute acids.

Question 17

(a) (i) Explain why: [3 Marks]
(1) Nitrogen does not form pentahalides
(2) Helium is used for filling weather balloons
(3) ICl is more reactive than \(\text{I}_2\)

Answer:
(1) Nitrogen lacks vacant \(d\)-orbitals in its valence shell, restricting its maximum covalency to four and preventing expansion of octet.
(2) Helium is a non-combustible and light gas, making it safe for filling weather balloons.
(3) The I-Cl bond in ICl is weaker than the I-I bond in \(\text{I}_2\), making ICl more reactive and easier to cleave.

Teacher's Note:
a) Second-period elements show anomalous behaviour due to small size and absence of \(d\)-orbitals.
b) Interhalogen compounds are generally more reactive than constituent halogens due to bond polarity.

(ii) Draw the structures of the following: [2 Marks]
(1) \(\text{HClO}_4$
(2) \(\text{H}_3\text{PO}_3\)

Answer:
(1) Structure of \(\text{HClO}_4\) (Perchloric acid): Central chlorine atom bonded to one -OH group and three oxygen atoms via double bonds (\(\text{Cl}=\text{O}\)).
(2) Structure of \(\text{H}_3\text{PO}_3$ (Phosphorous acid): Central phosphorus bonded to one -H atom, two -OH groups, and one doubly bonded oxygen atom (\(\text{P}=\text{O}\)).

Teacher's Note:
a) \(\text{H}_3\text{PO}_3\) is dibasic because only the two hydrogen atoms attached to oxygen are ionizable.
b) \(\text{HClO}_4\) is a strong monoprotic acid.

OR

(b) (i) Explain why: [3 Marks]
(1) Mercury loses its meniscus in contact with ozone.
(2) Halogens are coloured and the colour deepens on moving down in the group from fluorine to iodine.
(3) Hydride of sulphur is a gas while hydride of oxygen is a liquid.

Answer:
(1) Ozone oxidizes mercury to mercurous oxide (\(\text{Hg}_2\text{O}\)), which dissolves in mercury and causes it to stick to glass, losing its meniscus.
(2) Halogens absorb visible light, causing valence electrons to get excited to higher energy levels. Down the group, the excitation energy decreases, leading to absorption of longer wavelengths and deepening of colour.
(3) Water molecules form intermolecular hydrogen bonds due to high electronegativity of oxygen, whereas hydrogen sulfide does not form hydrogen bonds due to lower electronegativity of sulfur.

Teacher's Note:
a) Loss of meniscus in mercury is a classic chemical test for the presence of ozone.
b) Hydrogen bonding is responsible for the liquid state and high boiling point of water.

(ii) Complete and balance the following reactions: [2 Marks]
(1) \(\text{NaCl} + \text{MnO}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{____} + \text{____} + \text{____} + \text{____}
(2) \(\text{KMnO}_4 + \text{SO}_2 + \text{H}_2\text{O} \rightarrow \text{____} + \text{____} + \text{____}\)

Answer:
(1) \(4\text{NaCl} + \text{MnO}_2 + 4\text{H}_2\text{SO}_4 \rightarrow 2\text{Na}_2\text{SO}_4 + \text{MnCl}_2 + \text{Cl}_2 + 2\text{H}_2\text{O}\) (Note: balanced equivalent form: \(2\text{NaCl} + \text{MnO}_2 + 3\text{H}_2\text{SO}_4 \rightarrow 2\text{NaHSO}_4 + \text{MnSO}_4 + \text{Cl}_2 + 2\text{H}_2\text{O}\))
(2) \(2\text{KMnO}_4 + 5\text{SO}_2 + 2\text{H}_2\text{O} \rightarrow \text{K}_2\text{SO}_4 + 2\text{MnSO}_4 + 2\text{H}_2\text{SO}_4\)

Teacher's Note:
a) Action of conc. sulfuric acid on chloride salts in presence of an oxidizing agent produces chlorine gas.
b) Sulfur dioxide acts as a reducing agent when reacting with potassium permanganate.

Question 18

(a) (i) Give balanced equations for the following reactions: [3 Marks]
(1) Benzaldehyde reacts with hydrazine.
(2) Acetic acid reacts with phosphorous pentachloride.
(3) Acetone reacts with sodium bisulphite.

Answer:
(1) \(\text{C}_6\text{H}_5\text{CHO} + \text{H}_2\text{NNH}_2 \rightarrow \text{C}_6\text{H}_5\text{CH}=\text{NNH}_2 + \text{H}_2\text{O}\)
(2) \(\text{CH}_3\text{COOH} + \text{PCl}_5 \rightarrow \text{CH}_3\text{COCl} + \text{POCl}_3 + \text{HCl}\)
(3) \(\text{CH}_3\text{COCH}_3 + \text{NaHSO}_3 \rightarrow (\text{CH}_3)_2\text{C(OH)(SO}_3\text{Na})\)

Teacher's Note:
a) Reaction of aldehydes with hydrazine yields hydrazones.
b) Sodium bisulphite addition forms a crystalline bisulphite adduct, useful for carbonyl compound separation.

(ii) Give one chemical test each to distinguish between the following pairs of compounds: [2 Marks]
(1) Ethanol and acetic acid
(2) Acetaldehyde and benzaldehyde

Answer:
(1) Ethanol and acetic acid: Treat with sodium bicarbonate solution. Acetic acid evolves brisk effervescence of carbon dioxide gas, whereas ethanol does not react.
(2) Acetaldehyde and benzaldehyde: Perform the iodoform test. Acetaldehyde reacts with \(\text{I}_2\) and \(\text{NaOH}\) to give a yellow precipitate of iodoform, whereas benzaldehyde does not give this test.

Teacher's Note:
a) Carboxylic acids react with sodium bicarbonate to give \(\text{CO}_2\) gas.
b) Only methyl ketones and acetaldehyde give positive iodoform tests.

OR

(b) (i) Write chemical equations to illustrate the following name reactions: [3 Marks]
(1) Clemmensen's reduction
(2) Rosenmund's reduction
(3) HVZ reaction

Answer:
(1) Clemmensen's reduction:
\(\text{CH}_3\text{COCH}_3 + 4[\text{H}] \xrightarrow{\text{Zn-Hg / HCl}} \text{CH}_3\text{CH}_2\text{CH}_3 + \text{H}_2\text{O}\)
(2) Rosenmund's reduction:
\(\text{C}_6\text{H}_5\text{COCl} + \text{H}_2 \xrightarrow{\text{Pd-BaSO}_4} \text{C}_6\text{H}_5\text{CHO} + \text{HCl}\)
(3) Hell-Volhard-Zelinsky (HVZ) reaction:
\(\text{CH}_3\text{COOH} + \text{Br}_2 \xrightarrow{\text{red P}_4} \text{CH}_2\text{BrCOOH} + \text{HBr}\)

Teacher's Note:
a) Clemmensen reduction reduces carbonyl groups to methylene groups using zinc amalgam and conc. HCl.
b) HVZ halogenates carboxylic acids at the alpha-position in the presence of red phosphorus.

(ii) Explain why: [2 Marks]
(1) Acetaldehyde undergoes aldol condensation, but formaldehyde does not.
(2) Acetic acid is weaker acid as compared to formic acid.

Answer:
(1) Acetaldehyde possesses alpha-hydrogen atoms, which are rendered acidic by the carbonyl group, enabling aldol condensation. Formaldehyde lacks alpha-hydrogen atoms and cannot undergo aldol condensation.
(2) The methyl group in acetic acid exerts an electron-releasing (+I effect), which intensifies negative charge on the carboxylate anion and destabilizes it. Formic acid lacks this alkyl group, making its formate ion more stable and rendering formic acid a stronger acid than acetic acid.

Teacher's Note:
a) Presence of at least one alpha-hydrogen is a mandatory requirement for aldol condensation.
b) Inductive effects significantly influence the relative acid strengths of carboxylic acids.