ISC Class 12 Chemistry Board Exam Question Paper 2016 with Solutions

ISC Class 12 Chemistry Board Exam Question Paper with Solutions

PART I (20 Marks)

Question 1

(a) Fill in the blanks by choosing the appropriate word/words from those given in the brackets:
(Henry’s, aldol condensation, absence, do not, ohm, Raoult’s, increases, common ion effect, easily, three, solubility product, ohm^-1, two, four, ohm^-1cm², cannizzaro, ohm^-1cm^-1, zero, decreases, presence) [5 Marks]
(i) Ideal solutions obey _________ law and they _______ form azeotropic mixtures.
(ii) Benzaldehyde undergoes _________ reaction due to _________ of \(\alpha\)-hydrogen atom.
(iii) The solubility of silver chloride _________ in the presence of sodium chloride because of _________.
(iv) The unit of conductance is _________ and that of specific conductance is _________.
(v) When the concentration of a reactant of first order reaction is doubled, the rate becomes _________ times, but for _________ order reaction, the rate remains same.

Answer:
(i) Raoult's, do not
(ii) Cannizzaro, absence
(iii) decreases, common ion effect
(iv) ohm^-1, ohm^-1cm^-1
(v) two, zero

Teacher's Note:
a) Ideal solutions follow Raoult's law across all concentrations and do not form azeotropic mixtures as they show no deviation.
b) Students must carefully distinguish between terms like conductance units (ohm^-1 or Siemens) and specific conductance units (ohm^-1cm^-1).

(b) Complete the following statements by selecting the correct alternative from the choices given: [5 Marks]
(i) Electrochemical equivalent is the amount of substance which gets deposited from its solution on passing electrical charge equal to:
(1) 96,500 Coulombs
(2) 1 Coulomb
(3) 60 Coulombs
(4) 965 Coulombs
(ii) The complex ion \([Ni(CN)_4]^{2-}\) is:
(1) Square planar and diamagnetic
(2) Tetrahedral and paramagnetic
(3) Square planar and paramagnetic
(4) Tetrahedral and diamagnetic
(iii) Wohler’s synthesis is used for the preparation of:
(1) Glycine
(2) Amino acids
(3) Urea
(4) Proteins
(iv) When \(\text{SO}_2\) gas is passed through acidified \(\text{K}_2\text{Cr}_2\text{O}_7\) solution, the colour of the solution changes to:
(1) Red
(2) Black
(3) Orange
(4) Green
(v) In the equation \(\text{CH}_3\text{COOH} + \text{Cl}_2 \xrightarrow[\text{-HCl}]{\text{Red P}} \text{A}\), the compound A is:
(1) \(\text{CH}_3\text{CH}_2\text{Cl}\)
(2) \(\text{ClCH}_2\text{COOH}\)
(3) \(\text{CH}_3\text{Cl}\)
(4) \(\text{CH}_3\text{COCl}\)

Answer:
(i) (2) 1 Coulomb
(ii) (1) Square planar and diamagnetic
(iii) (3) Urea
(iv) (4) Green
(v) (2) \(\text{ClCH}_2\text{COOH}\)

Teacher's Note:
a) Electrochemical equivalent (Z) is defined as mass deposited per coulomb of charge (\(m = Z \cdot Q\); when \(Q = 1\) C, \(m = Z\)).
b) For \(\text{[Ni(CN)}_4]^{2-}\), nickel is in \(+2\) oxidation state with \(3d^8\) configuration; strong field \(\text{CN}^-$

Teacher's Note (continued):
causes \(dsp^2\) hybridization leading to a square planar geometry with all electrons paired (diamagnetic).

(c) Answer the following questions: [5 Marks]
(i) What is the order of reaction whose rate constant has the same unit as the rate of reaction?
(ii) What is the pH value of a solution whose hydroxyl ion concentration is \(1 \times 10^{-2}\text{ M}\)?
(iii) Calculate the number of coulombs required to deposit \(5\cdot4\text{ g}\) of \(\text{Al}\) when the electrode reaction is: \(\text{Al}^{3+} + 3e^- \rightarrow \text{Al}\) [Atomic Weight of \(\text{Al} = 27\text{ g/mol}\)].
(iv) Write the reaction to prepare acetaldehyde from hydrogen gas and an acid chloride.
(v) The edge length of unit cell of a body centered cubic (bcc) crystal is \(352\text{ pm}\). Calculate the radius of the atom.

Answer:
(i) Zero order reaction, because \(\text{rate} = k[\text{A}]^0 = k\).
(ii) \(\text{pOH} = -\log_{10}[\text{OH}^-] = -\log_{10}(1 \times 10^{-2}) = 2\).
Therefore, \(\text{pH} = 14 - 2 = 12$.
(iii) For \(\text{Al}^{3+} + 3e^- \rightarrow \text{Al}\), \(1\text{ mole}\) (\(27\text{ g}\)) of \(\text{Al}\) requires \(3\text{ Faraday}\) ($3 \times 96,500\text{ C}$).
Coulombs required for \(5\cdot4\text{ g}\) of \(\text{Al}\) = \(\frac{3 \times 5\cdot4}{27} \times 96,500 = 0\cdot6 \times 96,500 = 57,900\text{ Coulombs}\).
(iv) \(\text{CH}_3\text{COCl} + \text{H}_2 \xrightarrow{\text{Pd/BaSO}_4} \text{CH}_3\text{CHO} + \text{HCl}\).
(v) For bcc structure, radius \(r = \frac{\sqrt{3}a}{4}\).
Given \(a = 352\text{ pm}\).
\(r = \frac{\sqrt{3} \times 352}{4} = \frac{1\cdot732 \times 352}{4} = 152\cdot42\text{ pm}\).

Teacher's Note:
a) Always substitute units correctly in Faraday calculations and relate moles to stoichiometry.
b) Ensure proper geometric relations are used for different unit cells (bcc: \(r = \sqrt{3}a/4\), fcc: \(r = \sqrt{2}a/4\)).

(d) Match the following: [5 Marks]
(i) Weak electrolyte      (a) \(\text{pH}\) of a solution
(ii) Colour in crystals      (b) Iodoform
(iii) Acetone      (c) Tollen’s reagent
(iv) Sorensen      (d) Ostwald dilution law
(v) Ammonical silver nitrate      (e) F - centre

Answer:
(i) - (d) Ostwald dilution law
(ii) - (e) F - centre
(iii) - (b) Iodoform
(iv) - (a) \(\text{pH}\) of a solution
(v) - (c) Tollen’s reagent

Teacher's Note:
a) Ostwald dilution law governs the dissociation constants of weak electrolytes.
b) F-centres are anion vacancies in crystals that trap electrons, imparting colour.

PART II (50 Marks)
SECTION A
Answer any two questions.

Question 2

(a) (i) A \(10\%\) aqueous solution of cane sugar (mol. wt. 342) is isotonic with \(1\cdot754\%\) aqueous solution of urea. Find the molecular mass of urea. [2 Marks]
(ii) The molecular weight of an organic compound is \(58\text{ g mol}^{-1}\). What will be the boiling point of a solution containing \(48\text{ grams}\) of the solute in \(1200\text{ grams}\) of water? [\(\text{K}_b\) for water = \(0\cdot513^{\circ}\text{C kg mole}^{-1}\); Boiling point of water = \(100^{\circ}\text{C}\).] [2 Marks]
(iii) What will be the value of van’t Hoff factor \((i)\) of benzoic acid if it dimerises in aqueous solution? How will the experimental molecular weight vary as compared to the normal molecular weight? [1 Mark]

Answer:
(i) For cane sugar: \(\text{wt.} = 10\text{ g}\) in \(100\text{ mL}\), \(\text{mol. wt.} = 342\text{ g/mol}\).
Moles of cane sugar \((n_1) = \frac{10}{342} = 0\cdot0292\text{ mol}\).
For urea: \(\text{wt.} = 1\cdot754\text{ g}\) in \(100\text{ mL}\), let molecular mass be \(x\).
Moles of urea \((n_2) = \frac{1\cdot754}{x}\).
Since solutions are isotonic, \(\pi_1 = \pi_2 \implies \frac{n_1}{V}RT = \frac{n_2}{V}RT\).
\(0\cdot0292 = \frac{1\cdot754}{x} \implies x = \frac{1\cdot754}{0\cdot0292} = 60\cdot06\text{ g/mol}\).
(ii) Molality \((m) = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} = \frac{48 / 58}{1200 / 1000} = \frac{0\cdot8276}{1\cdot2} = 0\cdot6897\text{ mol/kg}\).
Elevation in boiling point (\(\Delta\text{T}_b\)) = \(\text{K}_b \times m = 0\cdot513 \times 0\cdot6897 = 0\cdot353^{\circ}\text{C}\).
Boiling point of solution = \(100 + 0\cdot353 = 100\cdot353^{\circ}\text{C}\).
(iii) Van’t Hoff factor \((i) = \frac{1}{2} = 0\cdot5\).
Experimental molecular weight will be twice the normal molecular weight due to association.

Teacher's Note:
a) Isotonic solutions have identical molar concentrations at the same temperature.
b) For associating solutes like benzoic acid in benzene, the van't Hoff factor is less than one.

(b) (i) Determine the pH value of \(0\cdot001\text{ M}\) acetic acid solution if it is \(2\%\) ionised at this concentration. How can the degree of dissociation of this acetic acid solution be increased? [2 Marks]
(ii) The solubility product of \(\text{PbCl}_2\) at \(298\text{ K}\) is \(1\cdot7 \times 10^{-5}\). Calculate the solubility of \(\text{PbCl}_2\) in \(\text{g/lit.}\) at \(298\text{ K}\). Atomic Weights: [\(\text{Pb} = 207\) and \(\text{Cl} = 35\cdot5\)] [2 Marks]

Answer:
(i) Concentration \(C = 0\cdot001\text{ M} = 10^{-3}\text{ M}\).
Degree of dissociation \(\alpha = \frac{2}{100} = 0\cdot02\).
Hydrogen ion concentration \([\text{H}^+] = C \cdot \alpha = 10^{-3} \times 0\cdot02 = 2 \times 10^{-5}\text{ M}\).
\(\text{pH} = -\log[\text{H}^+] = -\log(2 \times 10^{-5}) = 5 - \log 2 = 5 - 0\cdot3010 = 4\cdot69\).
(Note: Official marking scheme uses alternative rounding/formula route yielding \(\text{pH} = 4\cdot69\)).
The degree of dissociation can be increased by dilution.
(ii) For \(\text{PbCl}_2 \rightleftharpoons \text{Pb}^{2+} + 2\text{Cl}^-\), let solubility be \(S\text{ mol L}^{-1}\).
\(\text{K}_{sp} = [S][2S]^2 = 4S^3\).
\(4S^3 = 1\cdot7 \times 10^{-5} \implies S^3 = 4\cdot25 \times 10^{-6} \implies S = 0\cdot01619\text{ mol L}^{-1}\).
Molar mass of \(\text{PbCl}_2 = 207 + 2(35\cdot5) = 278\text{ g/mol}\).
Solubility in \(\text{g/L} = 0\cdot01619 \times 278 = 4\cdot50\text{ g/L}\).

Teacher's Note:
a) Ostwald's dilution law states that dilution increases the degree of dissociation of a weak electrolyte.
b) Always remember to multiply molar solubility by the molar mass to get solubility in grams per litre.

(c) Graphite is anisotropic with respect to conduction of electric current. Explain. [1 Mark]

Answer:
Graphite consists of layered hexagonal carbon structures where delocalized electrons are free to move within the layers. Consequently, electrical conductivity is much higher parallel to the layers than perpendicular to them, showing anisotropy.

Teacher's Note:
a) Anisotropy refers to directional dependence of physical properties.
b) Mentioning the layered structure and delocalized electrons is essential for full credit.

Question 3

(a) (i) In a body centred and face centred arrangement of atoms of an element, what will be the number of atoms present in respective unit cells? Justify your answer with calculation. [2 Marks]
(ii) A compound AB has a simple cubic structure and has molecular mass \(99\). Its density is \(3\cdot4\text{ g cm}^{-3}\). What will be the edge length of the unit cell? [2 Marks]

Answer:
(i) For BCC unit cell: Corner atoms = \(8 \times \frac{1}{8} = 1\); Body centred atom = \(1 \times 1 = 1\). Total = \(1 + 1 = 2\) atoms.
For FCC unit cell: Corner atoms = \(8 \times \frac{1}{8} = 1\); Face centred atoms = \(6 \times \frac{1}{2} = 3\). Total = \(1 + 3 = 4\) atoms.
(ii) Density formula: \(\rho = \frac{Z \cdot M}{a^3 \cdot \text{N}_A}\).
For simple cubic, \(Z = 1\). Given \(M = 99\), \(\rho = 3\cdot4\text{ g cm}^{-3}\), \(\text{N}_A = 6\cdot023 \times 10^{23}\).
\(a^3 = \frac{1 \times 99}{3\cdot4 \times 6\cdot023 \times 10^{23}} = 4\cdot834 \times 10^{-23}\text{ cm}^3 = 48\cdot34 \times 10^{-24}\text{ cm}^3\).
\(a = 3\cdot64 \times 10^{-8}\text{ cm}\).

Teacher's Note:
a) Clearly specify the contribution of atoms at corners, face-centres, and body-centre.
b) Pay close attention to powers of 10 when calculating cube roots for unit cell edge lengths.

(b) (i) For the reaction: \(2\text{NO}_{(g)} \rightleftharpoons \text{N}_{2(g)} + \text{O}_{2(g)}\); \(\Delta\text{H} = -\text{heat}\), \(\text{K}_c = 2\cdot5 \times 10^2\) at \(298\text{ K}\), what will happen to the concentration of \(\text{N}_2\) if: [2 Marks]
(1) Temperature is decreased to \(273\text{ K}\).
(2) Pressure is reduced.
(ii) In a first order reaction, \(10\%\) of the reactant is consumed in \(25\text{ minutes}\). Calculate: [2 Marks]
(1) The half-life period of the reaction.
(2) The time required for completing \(87\cdot5\%\) of the reaction.

Answer:
(i) (1) Since the reaction is exothermic (\(\Delta\text{H} = -\text{heat}\)), decrease in temperature favours the forward reaction. Therefore, the concentration of \(\text{N}_2\) will increase.
(2) Since total moles of gaseous reactants equal total moles of gaseous products (\(\Delta n = 0\)), change in pressure has no effect on the equilibrium.
(ii) (1) Rate constant \(k = \frac{2\cdot303}{t} \log\left(\frac{a}{a - x}\right) = \frac{2\cdot303}{25} \log\left(\frac{100}{90}\right) = 0\cdot0042\text{ min}^{-1}\).
\(t_{1/2} = \frac{0\cdot693}{k} = \frac{0\cdot693}{0\cdot0042} = 165\text{ minutes}\).
(2) For \(87\cdot5\%\) completion, remaining amount = \(12\cdot5\%\).
\(t = \frac{2\cdot303}{0\cdot0042} \log\left(\frac{100}{12\cdot5}\right) = \frac{2\cdot303}{0\cdot0042} \log 8 = 495\cdot14\text{ minutes}\).

Teacher's Note:
a) Apply Le Chatelier's principle correctly based on enthalpy change and gaseous mole differences.
b) First-order half-life is independent of initial concentration; use logarithmic standard formulas for completion times.

(c) Water acts as Bronsted acid as well as a Bronsted base. Give one example each to illustrate this statement. [2 Marks]

Answer:
As a Bronsted acid (proton donor): \(\text{H}_2\text{O}_{(l)} + \text{NH}_{3}_{(l)} \rightleftharpoons \text{NH}_{4(aq)}^+ + \text{OH}^-_{(aq)}\).
As a Bronsted base (proton acceptor): \(\text{HCl}_{(aq)} + \text{H}_2\text{O}_{(l)} \rightleftharpoons \text{H}_3\text{O}^+_{(aq)} + \text{Cl}^-_{(aq)}\).

Teacher's Note:
a) Amphoteric species like water donate a proton in the presence of a stronger base and accept a proton in the presence of a stronger acid.
b) Clearly indicate conjugate acid-base pairs in equations.

Question 4

(a) (i) Consider the following cell reaction at \(298\text{ K}\):
\(2\text{Ag}^^+ + \text{Cd} \rightarrow 2\text{Ag} + \text{Cd}^{2+}\)
The standard reduction potentials (\(\text{E}^{\circ}\)) for \(\text{Ag}^+/\text{Ag}\) and \(\text{Cd}^{2+}/\text{Cd}\) are \(0\cdot80\text{ V}\) and \(-0\cdot40\text{ V}\) respectively: [3 Marks]
(1) Write the cell representation.
(2) What will be the emf of the cell if the concentration of \(\text{Cd}^{2+}\) is \(0\cdot1\text{ M}\) and that of \(\text{Ag}^+\) is \(0\cdot2\text{ M}\)?
(3) Will the cell work spontaneously for the condition given in (2) above?
(ii) What is a buffer solution? How is it prepared? Explain the buffer action of a basic buffer with a suitable example. [2 Marks]

Answer:
(i) (1) Cell representation: \(\text{Cd}_{(s)} / \text{Cd}^{2+}_{(aq)} // \text{Ag}^+_{(aq)} / \text{Ag}_{(s)}\).
(2) \(\text{E}^{\circ}_{\text{cell}} = \text{E}^{\circ}_{\text{cathode}} - \text{E}^{\circ}_{\text{anode}} = 0\cdot80 - (-0\cdot40) = 1\cdot2\text{ V}\).
Using Nernst equation: \(\text{E}_{\text{cell}} = \text{E}^{\circ}_{\text{cell}} - \frac{0\cdot0591}{n} \log\left(\frac{[\text{Cd}^{2+}]}{[\text{Ag}^+]^2}\right)\).
\(\text{E}_{\text{cell}} = 1\cdot2 - \frac{0\cdot0591}{2} \log\left(\frac{0\cdot1}{(0\cdot2)^2}\right) = 1\cdot2 - 0\cdot02955 \log\left(\frac{0\cdot1}{0\cdot04}\right) = 1\cdot2 - 0\cdot02955 \log(2\cdot5) = 1\cdot188\text{ V}\) (or \(1\cdot18\text{ V}\)).
(3) Since \(\text{E}_{\text{cell}}\) is positive, \(\Delta\text{G}\) is negative (\(\Delta\text{G} = -n\text{FE}_{\text{cell}}\)), hence the cell will work spontaneously.
(ii) A buffer solution is one that resists changes in its \(\text{pH}\) when small amounts of acid or alkali are added.
Preparation: Prepared by mixing a weak base and its salt with a strong acid (e.g., \(\text{NH}_4\text{OH}\) and \(\text{NH}_4\text{Cl}\)).
Buffer action: When \(\text{OH}^-\) is added, it combines with \(\text{NH}_4^+\) ions from the salt to form unionized \(\text{NH}_4\text{OH}\). When \(\text{H}^+\) is added, it combines with \(\text{NH}_4\text{OH}\) to form \(\text{NH}_4^+\) and \(\text{H}_2\text{O}\).

Teacher's Note:
a) Proper Nernst equation setup with correct stoichiometry powers is vital.
b) Emphasize the relationship between cell potential and Gibbs free energy for spontaneity.

(b) Explain the following: [2 Marks]
(i) When \(\text{NaCl}\) is added to \(\text{AgNO}_3\) solution, a white precipitate is formed.
(ii) An aqueous solution of ammonium chloride is acidic in nature.

Answer:
(i) \(\text{NaCl}\) reacts with \(\text{AgNO}_3\) to form silver chloride (\(\text{AgCl}\)), which is sparingly soluble and precipitates as a white solid: \(\text{NaCl} + \text{AgNO}_3 \rightarrow \text{AgCl}\downarrow + \text{NaNO}_3\).
(ii) \(\text{NH}_4\text{Cl}\) is a salt of a strong acid (\(\text{HCl}\)) and a weak base (\(\text{NH}_4\text{OH}\)). Due to cationic hydrolysis, the resulting solution contains excess hydronium ions, making it acidic.

Teacher's Note:
a) Precipitation reactions depend on ionic product exceeding solubility product.
b) Highlight salt hydrolysis principles clearly for acidic or basic salt solutions.

(c) A \(0\cdot05\text{ M}\) \(\text{NH}_4\text{OH}\) solution offers the resistance of \(50\text{ ohms}\) to a conductivity cell at \(298\text{ K}\). If the cell constant is \(0\cdot50\text{ cm}^{-1}\) and molar conductance of \(\text{NH}_4\text{OH}\) at infinite dilution is \(471\cdot4\text{ ohm}^{-1}\text{ cm}^2\text{ mol}^{-1}\), calculate: [3 Marks]
(i) Specific conductance
(ii) Molar conductance
(iii) Degree of dissociation

Answer:
(i) Specific conductance (\(\kappa\)) = \(\frac{1}{R} \times \text{cell constant} = \frac{1}{50} \times 0\cdot50 = 0\cdot01\text{ ohm}^{-1}\text{ cm}^{-1}\).
(ii) Molar conductance (\(\Lambda_m\)) = \(\frac{1000 \times \kappa}{C} = \frac{1000 \times 0\cdot01}{0\cdot05} = 200\text{ ohm}^{-1}\text{ cm}^2\text{ mol}^{-1}\).
(iii) Degree of dissociation (\(\alpha\)) = \(\frac{\Lambda_m}{\Lambda_m^{\infty}} = \frac{200}{471\cdot4} = 0\cdot4242\).

Teacher's Note:
a) Ensure correct units are used throughout electrolytic conductance calculations.
b) Molar conductance formula requires concentration in moles per litre and specific conductance in \(\text{ohm}^{-1}\text{cm}^{-1}\).

SECTION B
Answer any two questions

Question 5

(a) Write the IUPAC names of the following: [2 Marks]
(i) \(\text{[Co(NH}_3)_4\text{SO}_4\text{]NO}_3\)
(ii) \(\text{K[Pt(NH}_3)\text{Cl}_3\text{]}\)

Answer:
(i) Tetraamminesulphatocobalt(III) nitrate
(ii) Potassium amminetrichloridoplatinate(II)

Teacher's Note:
a) Follow alphabetical ordering of ligands before naming the central metal atom.
b) Oxidation state of the metal must be correctly determined and specified in Roman numerals.

(b) What type of isomerism is exhibited by the following pairs of compounds: [1 Mark]
(i) \(\text{[PtCl}_2\text{(NH}_3)_4\text{]Br}_2\) and \(\text{[PtBr}_2\text{(NH}_3)_4\text{]Cl}_2$
(ii) \(\text{[Cr(SCN)(H}_2\text{O)}_5\text{]}^{2+}\) and \(\text{[Cr(NCS)(H}_2\text{O)}_5\text{]}^{2+}\)

Answer:
(i) Ionisation isomerism
(ii) Linkage isomerism

Teacher's Note:
a) Ionisation isomers yield different ions in solution.
b) Linkage isomerism arises due to the presence of ambidentate ligands like \(\text{SCN}^-\).

(c) How does \(\text{K}_2\text{[PtCl}_4\)] get ionised when dissolved in water? Will it form precipitate when \(\text{AgNO}_3\) solution is added to it? Give a reason for your answer. [2 Marks]

Answer:
Ionisation: \(\text{K}_2\text{[PtCl}_4\text{]} \rightarrow 2\text{K}^+ + \text{[PtCl}_4\text{]}^{2-}\).
It will not form a white precipitate with \(\text{AgNO}_3\) solution because chloride ions are coordinated inside the complex sphere and are not free in solution.

Teacher's Note:
a) Complex species do not dissociate into individual constituent ions inside the coordination sphere.
b) Only free chloride ions precipitate with silver nitrate.

Question 6

(a) Give balanced equations for the following reactions: [3 Marks]
(i) Silver nitrate is added to dilute solution of sodium thiosulphate.
(ii) Potassium dichromate is treated with acidified ferrous sulphate solution.
(iii) Phosphorous reacts with conc. sulphuric acid.

Answer:
(i) \(2\text{AgNO}_3 + \text{Na}_2\text{S}_2\text{O}_3 \rightarrow \text{Ag}_2\text{S}_2\text{O}_3\downarrow\text{ (white)} + 2\text{NaNO}_3\)
\(\text{Ag}_2\text{S}_2\text{O}_3 + \text{H}_2\text{O} \rightarrow \text{Ag}_2\text{S}\downarrow\text{ (black)} + \text{H}_2\text{SO}_4\)
(ii) \(\text{K}_2\text{Cr}_2\text{O}_7 + 7\text{H}_2\text{SO}_4 + 6\text{FeSO}_4 \rightarrow \text{K}_2\text{SO}_4 + \text{Cr}_2(\text{SO}_4)_3 + 3\text{Fe}_2(\text{SO}_4)_3 + 7\text{H}_2\text{O}\)
(iii) \(\text{P}_4 + 10\text{H}_2\text{SO}_4\text{ (conc.)} \rightarrow 4\text{H}_3\text{PO}_4 + 10\text{SO}_2 + 4\text{H}_2\text{O}\)

Teacher's Note:
a) Ensure all equations are fully balanced with proper state changes or intermediates noted.
b) The reaction of silver thiosulphate transitioning from white to black precipitate is a classic analytical observation.

(b) How will you obtain pure potassium permanganate (\(\text{KMnO}_4\)) crystals from its ore, pyrolusite? Give the steps involved and the reactions. [2 Marks]

Answer:
Step 1: Conversion of pyrolusite (\(\text{MnO}_2\)) to potassium manganate by fusing with \(\text{KOH}\) in the presence of air or an oxidizing agent like \(\text{KNO}_3\):
\(2\text{MnO}_2 + 4\text{KOH} + \text{O}_2 \xrightarrow{\Delta} 2\text{K}_2\text{MnO}_4 + 2\text{H}_2\text{O}\).
Step 2: Oxidation of potassium manganate to potassium permanganate by bubbling carbon dioxide or chlorine:
\(3\text{K}_2\text{MnO}_4 + 2\text{CO}_2 \rightarrow 2\text{KMnO}_4 + \text{MnO}_2 + 2\text{K}_2\text{CO}_3\).

Teacher's Note:
a) Emphasize both fusion and electrolytic/acidic or gaseous oxidation steps.
b) Potassium manganate is green, whereas potassium permanganate forms dark purple crystals.

Question 7

(a) (i) Sulphur dioxide acts as an oxidizing agent as well as a reducing agent. Give one reaction each to show its oxidizing nature and its reducing nature. [3 Marks]
(ii) Explain why an aqueous solution of potassium hexacyanoferrate (II) does not give the test for ferrous ion.

Answer:
(i) Oxidation state of \(\text{S}\) in \(\text{SO}_2\) is \(+4\), which is intermediate, allowing it to act as both oxidant and reductant.
Oxidizing nature: \(3\text{Fe} + \text{SO}_2 \rightarrow 2\text{FeO} + \text{FeS}\).
Reducing nature: \(\text{SO}_2 + \text{I}_2 + 2\text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4 + 2\text{HI}\).
(ii) Aqueous solution of \(\text{K}_4\text{[Fe(CN)}_6]\) ionizes to give \(\text{4K}^+\) and \(\text{[Fe(CN)}_6\text{]}^{4-}\). The iron remains inside the stable complex ion and is not present as free \(\text{Fe}^{2+}\) ions in solution.

Teacher's Note:
a) Intermediate oxidation states permit elements to undergo both oxidation and reduction.
b) Complex ions do not dissociate to release transition metal ions into free ionic form.

(b) What is meant by Lanthanide contraction? Write the general electronic configuration of inner transition elements. [2 Marks]

Answer:
The steady decrease in atomic and ionic radii of lanthanides from \(\text{La}^{3+}\) to \(\text{Lu}^{3+}\) due to imperfect shielding of nuclear charge by \(4f\) electrons is called Lanthanide contraction.
General electronic configuration: \(\text{ns}^2 (\text{n}-1)d^{0-1} (\text{n}-2)f^{1-14}\).

Teacher's Note:
a) Poor shielding by \(f\)-orbitals causes effective nuclear charge to pull electrons closer.
b) Precise shell indexing is required for electronic configurations.

SECTION C
Answer any two questions.

Question 8

(a) How can the following conversions be brought about:
(i) Acetaldehyde to acetaldehyde phenyl hydrazone. [1 Mark]
(ii) Benzoic acid to aniline. [1 Mark]
(iii) Methyl chloride to acetone. [2 Marks]
(iv) Benzene to benzene diazonium chloride. [1 Mark]

Answer:
(i) \(\text{CH}_3\text{CHO} + \text{H}_2\text{NNHC}_6\text{H}_5 \rightarrow \text{CH}_3\text{CH}=\text{NNHC}_6\text{H}_5 + \text{H}_2\text{O}\).
(ii) \(\text{C}_6\text{H}_5\text{COOH} \xrightarrow[\text{CaO}]{\text{NaOH, }\Delta} \text{C}_6\text{H}_6 \xrightarrow[\text{H}_2\text{SO}_4]{\text{HNO}_3} \text{C}_6\text{H}_5\text{NO}_2 \xrightarrow[\text{HCl}]{\text{Sn}} \text{C}_6\text{H}_5\text{NH}_2\).
(iii) \(\text{CH}_3\text{Cl} \xrightarrow{\text{KCN}} \text{CH}_3\text{CN} \xrightarrow[\text{H}^+]{\text{H}_2\text{O}} \text{CH}_3\text{COOH} \xrightarrow{\text{Ca(OH)}_2} (\text{CH}_3\text{COO})_2\text{Ca} \xrightarrow{\text{dry dist.}} \text{CH}_3\text{COCH}_3\).
(iv) \(\text{C}_6\text{H}_6 \xrightarrow[\text{H}_2\text{SO}_4]{\text{conc. HNO}_3} \text{C}_6\text{H}_5\text{NO}_2 \xrightarrow[\text{HCl}]{\text{Sn / }6[\text{H}]} \text{C}_6\text{H}_5\text{NH}_2 \xrightarrow[\text{0}^{\circ}\text{-5}^{\circ}\text{C}]{\text{NaNO}_2 + \text{HCl}} \text{C}_6\text{H}_5\text{N}_2\text{Cl}\).

Teacher's Note:
a) Multi-step organic conversions require exact reagent specifications and intermediate naming.
b) Diazotization reactions must strictly maintain low temperatures (\(0^{\circ}\text{C}\) to \(5^{\circ}\text{C}\)).

(b) (i) Glycerol (propane 1, 2, 3 triol) is more viscous than ethylene glycol (ethane 1, 2 diol). Explain. [1 Mark]
(ii) How can urea be detected by Biuret test? [1 Mark]

Answer:
(i) Glycerol contains three hydroxyl groups per molecule compared to two in ethylene glycol, resulting in a much greater extent of intermolecular hydrogen bonding, making it more viscous.
(ii) Biuret test: When urea is heated above \(132^{\circ}\text{C}\), biuret is formed. When an alkaline solution of biuret is treated with copper sulphate solution, a violet colour is produced.

Teacher's Note:
a) Viscosity in polyhydric alcohols increases directly with the number of hydroxyl groups available for hydrogen bonding.
b) Biuret test is a characteristic colorimetric test for compounds containing peptide-like linkages.

(c) Identify the compounds A, B and C: [3 Marks]
(i) \(\text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{PCl}_5} \text{A} \xrightarrow{\text{KCN}} \text{B} \xrightarrow[\text{H}_2\text{O}]{\text{H}^+} \text{C}_2\text{H}_5\text{COOH} \xrightarrow[\text{ }\Delta\text{ }](\text{NH}_4)_2\text{CO}_3\text{ or NH}_3 \text{C}\)
(ii) \(\text{C}_6\text{H}_5\text{COOH} \xrightarrow{\text{SOCl}_2} \text{A} \xrightarrow{\text{NH}_3} \text{B} \xrightarrow[\text{KOH}]{\text{Br}_2} \text{C}\)

Answer:
(i) \(\text{A} = \text{C}_2\text{H}_5\text{Cl}\); \(\text{B} = \text{C}_2\text{H}_5\text{CN}\); \(\text{C} = \text{C}_2\text{H}_5\text{CONH}_2\).
(ii) \(\text{A} = \text{C}_6\text{H}_5\text{COCl}\); \(\text{B} = \text{C}_6\text{H}_5\text{CONH}_2\); \(\text{C} = \text{C}_6\text{H}_5\text{NH}_2\).

Teacher's Note:
a) Trace each functional group transformation carefully along the reaction sequence.
b) Hoffmann bromamide degradation converts amides to primary amines with one carbon atom less.

Question 9

(a) Give balanced equations for the following name reactions: [3 Marks]
(i) Benzoin condensation
(ii) Wurtz-Fittig reaction
(iii) Carbylamine reaction

Answer:
(i) Benzoin condensation: \(2\text{C}_6\text{H}_5\text{CHO} \xrightarrow[\text{alc. KCN}]{\Delta} \text{C}_6\text{H}_5\text{CH(OH)COC}_6\text{H}_5\) (Benzoin).
(ii) Wurtz-Fittig reaction: \(\text{C}_6\text{H}_5\text{Cl} + 2\text{Na} + \text{ClCH}_3 \xrightarrow{\text{dry ether}} \text{C}_6\text{H}_5\text{-CH}_3 + 2\text{NaCl}\).
(iii) Carbylamine reaction: \(\text{RNH}_2 + \text{CHCl}_3 + 3\text{KOH}_{\text{(alc)}} \xrightarrow{\Delta} \text{R-N}\equiv\text{C} + 3\text{KCl} + 3\text{H}_2\text{O}\).

Teacher's Note:
a) Name reactions must include correct catalysts, solvents, and side products.
b) Carbylamine reaction is specific to primary amines and produces foul-smelling isocyanides.

(b) Give chemical test to distinguish: [3 Marks]
(i) Formaldehyde and acetaldehyde
(ii) Dimethyl ether and ethyl alcohol.

Answer:
(i) Acetaldehyde responds to the iodoform test (gives a yellow precipitate with \(\text{I}_2\) and \(\text{NaOH}\)), whereas formaldehyde does not.
(ii) Ethyl alcohol reacts with sodium metal to evolve hydrogen gas or gives iodoform test, whereas dimethyl ether does not give these tests.

Teacher's Note:
a) Iodoform test is positive for compounds containing the \(\text{CH}_3\text{CO-}\) group or secondary alcohols yielding it upon oxidation.
b) Active hydrogen tests (like sodium metal) distinguish alcohols from ethers.

(c) (i) Write the structure of three ethers with molecular formula \(\text{C}_4\text{H}_{10}\text{O}\). [4 Marks]
(ii) Starting with Grignard’s reagent, how will you prepare propanoic acid?

Answer:
(i) Structures of three ethers:
1. \(\text{CH}_3-\text{O}-\text{CH}_2-\text{CH}_2-\text{CH}_3\) (1-Methoxypropane)
2. \(\text{CH}_3\text{CH}_2-\text{O}-\text{CH}_2\text{CH}_3\) (Ethoxyethane)
3. \(\text{CH}_3-\text{O}-\text{CH}(\text{CH}_3)_2\) (2-Methoxypropane)
(ii) Preparation of propanoic acid from Grignard reagent:
\(\text{C}_2\text{H}_5\text{MgBr} + \text{O}=\text{C}=\text{O} \rightarrow \text{C}_2\text{H}_5\text{COOMgBr} \xrightarrow[\text{H}^+]{\text{H}_2\text{O}} \text{C}_2\text{H}_5\text{COOH} + \text{Mg(OH)Br}\).

Teacher's Note:
a) Isomers of ethers should cover both straight-chain and branched-chain arrangements.
b) Carbon dioxide insertion into Grignard reagents provides a reliable route for chain elongation into carboxylic acids.

Question 10

(a) An organic compound A has the molecular formula \(\text{C}_7\text{H}_6\text{O}\). When A is treated with \(\text{NaOH}\) followed by acid hydrolysis, it gives two products B and C. When B is oxidized, it gives A, when A and C are each treated separately with \(\text{PCl}_5\), they give two different products D and E. [3 Marks]
(i) Identify A, B, C, D and E.
(ii) Give the chemical reaction when A is treated with \(\text{NaOH}\) and name the reaction.

Answer:
(i) \(\text{A} = \text{C}_6\text{H}_5\text{CHO}\) (Benzaldehyde); \(\text{B} = \text{C}_6\text{H}_5\text{CH}_2\text{OH}\) (Benzyl alcohol); \(\text{C} = \text{C}_6\text{H}_5\text{COOH}\) (Benzoic acid); \(\text{D} = \text{C}_6\text{H}_5\text{CHCl}_2\) (Benzal chloride); \(\text{E} = \text{C}_6\text{H}_5\text{COCl}\) (Benzoyl chloride).
(ii) Reaction: \(2\text{C}_6\text{H}_5\text{CHO} + \text{NaOH} \rightarrow \text{C}_6\text{H}_5\text{COONa} + \text{C}_6\text{H}_5\text{CH}_2\text{OH}\).
Name of reaction: Cannizzaro reaction.

Teacher's Note:
a) Aldehydes lacking \(\alpha\)-hydrogens undergo self-oxidation and reduction in concentrated alkali.
b) Product identifications rely on stepwise functional group transformations.

(b) Answer the following: [4 Marks]
(i) What do you observe when glucose solution is heated with Tollen’s reagent?
(ii) Name the monomers and the type of polymerisation in each of the following polymers:
(1) Terylene
(2) Polyvinyl chloride

Answer:
(i) A shiny silver mirror is formed on the inner walls of the test tube due to the reduction of silver ions by glucose.
(ii) (1) Terylene: Monomers are ethylene glycol and terephthalic acid; Polymerisation type is condensation polymerisation.
(2) Polyvinyl chloride: Monomer is vinyl chloride; Polymerisation type is addition polymerisation.

Teacher's Note:
a) Tollen's test confirms reducing sugars through silver mirror deposition.
b) Distinguish clearly between condensation polymers (involving elimination of small molecules) and addition polymers.

(c) Give balanced equations for the following reactions: [3 Marks]
(i) Ethylamine with nitrous acid.
(ii) Diethyl ether with phosphorous pentachloride.
(iii) Aniline with acetyl chloride.

Answer:
(i) \(\text{C}_2\text{H}_5\text{NH}_2 + \text{HNO}_2 \rightarrow \text{C}_2\text{H}_5\text{OH} + \text{N}_2 + \text{H}_2\text{O}\).
(ii) \(\text{C}_2\text{H}_5-\text{O}-\text{C}_2\text{H}_5 + \text{PCl}_5 \rightarrow 2\text{C}_2\text{H}_5\text{Cl} + \text{POCl}_3\).
(iii) \(\text{C}_6\text{H}_5\text{NH}_2 + \text{CH}_3\text{COCl} \rightarrow \text{C}_6\text{H}_5\text{NHCOCH}_3\) (Acetanilide) \(+\text{HCl}\).

Teacher's Note:
a) Primary aliphatic amines react with nitrous acid to liberate nitrogen gas and form alcohols.
b) Acylation of aniline produces acetanilide, protecting the amino group during substitutions.