ISC Class 12 Chemistry Board Exam Question Paper 2013 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: [5 Marks]
(hydrolysis, reduction, oxidation, vacant, osmotic, above, benzoic acid, phenol, aniline, below, can, decreases, increases, cannot, crystal, ionization, rate, rate constant.)
(i) A catalyst cannot start a reaction but it can increase the rate of the reaction.
(ii) Electrons trapped in the vacant sites of the crystal lattice are called F-centres.
(iii) An aqueous solution of sugar boils above 100\( ^{\circ} \)C and freezes below 0\( ^{\circ} \)C.
(iv) Toluene on oxidation with alkaline potassium permanganate gives benzoic acid.
(v) The degree of ionization of ammonium hydroxide decreases on addition of ammonium chloride.

Answer:
(i) cannot, rate
(ii) vacant, crystal
(iii) above, below
(iv) oxidation, benzoic acid
(v) ionization, decreases

Teacher's Note:
a) Remember that a catalyst only alters the activation energy to change the rate of a thermodynamically feasible reaction; it cannot initiate a non-spontaneous reaction.
b) The suppression of ionization of a weak electrolyte by the addition of a strong electrolyte with a common ion is an application of Le Chatelier's principle and the common-ion effect.

(b) Complete the following statements by selecting the correct alternative from the choices given: [5 Marks]
(i) For reaction \( 2\text{N}_2\text{O}_5 \rightarrow 2\text{NO}_2 + \text{O}_2 \), the rate and rate constants are \( 1.02 \times 10^{-4} \) mole litre\( ^{-1} \) sec\( ^{-1} \) and \( 3.4 \times 10^{-5} \) sec\( ^{-1} \) respectively. The concentration of \( \text{N}_2\text{O}_5 \) at that time will be:
(1) \( 1.732 \) mol lit\( ^{-1} \)
(2) \( 3 \) mol lit\( ^{-1} \)
(3) \( 1.02 \times 10^{-4} \) mol lit\( ^{-1} \)
(4) \( 3.2 \times 10^5 \) mol lit\( ^{-1} \)

Answer: (2) \( 3 \) mol lit\( ^{-1} \)

Rate \( = k[\text{N}_2\text{O}_5] \Rightarrow [\text{N}_2\text{O}_5] = \frac{\text{Rate}}{k} = \frac{1.02 \times 10^{-4}}{3.4 \times 10^{-5}} = 3\text{ mol L}^{-1} \).

Teacher's Note:
a) Use the rate law expression for a first-order reaction where rate is directly proportional to the concentration of the reactant.
b) Pay careful attention to the units of rate (\( \text{mol L}^{-1}\text{s}^{-1} \)) and rate constant (\( \text{s}^{-1} \)) to confirm the reaction order.

(ii) Ethanoic acid dimerises in solution. Its molecular mass determined from its depression of freezing point of the solution will be:
(1) Same as the theoretical value
(2) Half its theoretical value
(3) Double its theoretical value
(4) One third of its theoretical value.

Answer: (3) Double its theoretical value

Due to association (dimerisation), the number of particles in solution decreases, hence the observed colligative property decreases, resulting in a molecular mass that is double the normal value.

Teacher's Note:
a) Association of solute molecules results in abnormal molar mass which is higher than the theoretical value by a factor equal to the association factor.
b) Van't Hoff factor \( i \) is less than 1 for associating solutes like carboxylic acids in non-polar solvents.

(iii) Magnesium displaces hydrogen from dilute acid solution because:
(1) The oxidation potential of magnesium is less than that of hydrogen.
(2) The reduction potential of magnesium is less than that of hydrogen.
(3) Both magnesium and hydrogen have same oxidation potential.
(4) Both magnesium and hydrogen have same reduction potential.

Answer: (2) The reduction potential of magnesium is less than that of hydrogen.

Magnesium has a standard reduction potential of \( -2.37\text{ V} \), which is much lower (more negative) than that of hydrogen (\( 0.00\text{ V} \)), making it a stronger reducing agent.

Teacher's Note:
a) A metal with a more negative reduction potential can displace metals or hydrogen with higher reduction potentials from their solutions.
b) Metals lying above hydrogen in the electrochemical series liberate hydrogen gas on reacting with dilute acids.

(iv) In the series of reactions \( \text{CH}_3\text{COOH} \xrightarrow{\text{NH}_3} \text{A} \xrightarrow{\text{heat}} \text{B} \xrightarrow{\text{P}_2\text{O}_5} \text{C} \), the product C is:
(1) Acetyl chloride
(2) Ammonium acetate
(3) Acetic anhydride
(4) Methyl cyanide.

Answer: (4) Methyl cyanide.

\( \text{CH}_3\text{COOH} + \text{NH}_3 \rightarrow \text{CH}_3\text{COONH}_4 \) (A - Ammonium acetate) \( \xrightarrow{\text{heat}} \text{CH}_3\text{CONH}_2 + \text{H}_2\text{O} \) (B - Acetamide) \( \xrightarrow{\text{P}_2\text{O}_5} \text{CH}_3\text{CN} + \text{H}_2\text{O} \) (C - Methyl cyanide / Acetonitrile).

Teacher's Note:
a) Phosphorus pentoxide (\( \text{P}_2\text{O}_5 \)) is a powerful dehydrating agent that converts primary amides into alkyl cyanides.
b) Trace each step carefully: neutralization gives salt, thermal dehydration gives amide, and further dehydration gives nitrile.

(v) In the reaction \( \text{PCl}_3(\text{g}) + \text{Cl}_2(\text{g}) \rightarrow \text{PCl}_5(\text{g}) \), the equilibrium will shift in the opposite direction, if:
(1) Chlorine is added.
(2) \( \text{PCl}_3 \) is added
(3) Pressure is increased
(4) Pressure is reduced.

Answer: (4) Pressure is reduced.

According to Le Chatelier's principle, decreasing the pressure shifts the equilibrium towards the side with a greater number of moles of gas (reactants side).

Teacher's Note:
a) A decrease in pressure favors the forward direction if number of moles decreases, and the backward direction if number of moles increases.
b) Addition of reactants shifts equilibrium in the forward direction, not the opposite direction.

(c) Answer the following questions: [5 Marks]
(i) Among equimolal aqueous solutions of \( \text{MgCl}_2 \), \( \text{NaCl} \), \( \text{FeCl}_3 \) and \( \text{C}_{12}\text{H}_{22}\text{O}_{11} \), which will show minimum osmotic pressure? Why?

Answer:
\( \text{C}_{12}\text{H}_{22}\text{O}_{11} \) (sucrose) will show the minimum osmotic pressure. Osmotic pressure is a colligative property directly proportional to the van't Hoff factor (\( i \)). Since sucrose is a non-electrolyte, \( i = 1 \), whereas \( \text{NaCl} \) gives 2 ions, \( \text{MgCl}_2 \) gives 3 ions, and \( \text{FeCl}_3 \) gives 4 ions, resulting in higher osmotic pressures for the ionic compounds.

Teacher's Note:
a) Use the formula \( \pi = iCRT \) to explain that osmotic pressure depends on the total number of particles in solution.
b) Non-electrolytes do not dissociate, hence they have the lowest value of van't Hoff factor among equimolal solutions of electrolytes.

(ii) If \( K_c \) for the reaction \( \text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3 \) is \( 1.5 \times 10^{-5} \) \( (\text{mol / lit})^{-2} \), write the value of \( K_{c'} \) for the reaction \( \frac{1}{2}\text{N}_2 + \frac{3}{2}\text{H}_2 \rightleftharpoons \text{NH}_3 \).

Answer:
\( K_{c'} = \sqrt{K_c} = \sqrt{1.5 \times 10^{-5}} = 3.87 \times 10^{-3} \) \( (\text{mol / lit})^{-1} \).

Teacher's Note:
a) When a chemical equation is multiplied by a factor \( n \), the equilibrium constant is raised to the power \( n \).
b) Ensure that units are appropriately adjusted according to the stoichiometric coefficients of the new equation.

(iii) The pH of acetic acid decreases on dilution. State the Law governing this statement.

Answer:
Ostwald's Dilution Law. (Note: As dilution increases, degree of dissociation increases, leading to an increase in \( [\text{H}^+] \) concentration and consequently a decrease in pH).

Teacher's Note:
a) Ostwald's dilution law relates the degree of dissociation of a weak electrolyte to its ionization constant and concentration.
b) Note that while total hydrogen ion concentration increases due to enhanced dissociation, concentration per unit volume decreases, but overall pH drops as measured.

(iv) Xenon gives a series of fluorides, but Helium and Neon do not. Why? (At. No: Xe = 54, Ne = 10, He = 2)

Answer:
Xenon has a large atomic size and low ionization energy, and possesses vacant d-orbitals which allow it to expand its octet and form compounds with highly electronegative fluorine. Helium and Neon have very small atomic sizes, extremely high ionization energies, and lack d-orbitals in their valence shell, making them chemically inert.

Teacher's Note:
a) Inert gas reactivity increases down the group due to a decrease in ionization energy.
b) The presence of vacant d-orbitals is essential for the promotion of electrons and formation of multiple covalent bonds in xenon fluorides.

(v) Calculate the number of coulombs required to deposite \( 20.25\text{ g} \) of aluminium (at. mass = 27) from a solution containing \( \text{Al}^{3+} \).

Answer:
Reaction: \( \text{Al}^{3+} + 3\text{e}^- \rightarrow \text{Al} \)
Mass of 1 mole of Al \( = 27\text{ g} \)
Charge required to deposit \( 27\text{ g} \) of Al \( = 3 \times 96500\text{ C} \)
Charge required to deposit \( 20.25\text{ g} \) of Al \( = \frac{3 \times 96500 \times 20.25}{27} = 2,17,125\text{ C} \).

Teacher's Note:
a) One mole of any metal ion requires \( n \times F \) coulombs of electricity for deposition, where \( n \) is the valency of the metal ion.
b) Always write the balanced half-cell reduction reaction to determine the exact number of Faradays required per mole.

(d) Match the following: [5 Marks]
(i) \( \text{CHCl}_3 + \text{NaOH} \)                  (a) Fluorine
(ii) Proteins                                      (b) Starch
(iii) Carbohydrate                            (c) Ammonia
(iv) Lewis base                                (d) Peptide linkage
(v) \( \text{KHF}_2 \)                                     (e) Isocyanide test

Answer:
(i) - (e)
(ii) - (d)
(iii) - (b)
(iv) - (c)
(v) - (a)

Teacher's Note:
a) Chloroform with alcoholic KOH and primary amine gives carbylamine reaction (isocyanide test).
b) Proteins are linked by peptide bonds, carbohydrates include starch, ammonia acts as a Lewis base due to a lone pair, and \( \text{KHF}_2 \) contains fluorine.

PART II (50 Marks)

SECTION A

Question 2

(a) (i) A certain aqueous solution boils at \( 100.303^{\circ}\text{C} \). What is its freezing point? \( K_b \) for water = \( 0.5\text{ K mol}^{-1} \) and \( K_f = 1.87\text{ K mol}^{-1} \). [2 Marks]

Answer:
Elevation in boiling point \( \Delta T_b = 100.303 - 100 = 0.303^{\circ}\text{C} \)
Molality \( m = \frac{\Delta T_b}{K_b} = \frac{0.303}{0.5} = 0.606\text{ mol kg}^{-1} \)
Depression in freezing point \( \Delta T_f = K_f \times m = 1.87 \times 0.606 = 1.133^{\circ}\text{C} \)
Freezing point of the solution \( = 0 - 1.133 = -1.133^{\circ}\text{C} \).

Teacher's Note:
a) Molality of the solution remains constant for a dilute solution regardless of whether it is boiling or freezing.
b) Always calculate molality as the intermediate bridge between boiling point elevation and freezing point depression.

(ii) A solution containing \( 1\text{ g} \) of sodium chloride in \( 100\text{ g} \) of water freezes at \( -0.604^{\circ}\text{C} \). Calculate the degree of dissociation of sodium chloride. (\(\text{Na} = 23\), \(\text{Cl} = 35.5\), \(K_f\) for water = \(1.87\text{ K kg mol}^{-1}\)) [4 Marks]

Answer:
Molar mass of \( \text{NaCl} = 23 + 35.5 = 58.5\text{ g mol}^{-1} \)
Observed molality \( m = \frac{1 / 58.5}{100 / 1000} = \frac{10}{58.5} = 0.1709\text{ mol kg}^{-1} \)
Observed \( \Delta T_f = 0 - (-0.604) = 0.604^{\circ}\text{C} \)
Experimental \( K_f \) (or calculated \( \Delta T_f \) theoretically):
Theoretical \( \Delta T_f = K_f \times m = 1.87 \times 0.1709 = 0.3196^{\circ}\text{C} \)
Van't Hoff factor \( i = \frac{\text{Observed } \Delta T_f}{\text{Calculated } \Delta T_f} = \frac{0.604}{0.3196} = 1.89 \)
For dissociation of \( \text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^- \), number of ions \( n = 2 \).
Degree of dissociation \( \alpha = \frac{i - 1}{n - 1} = \frac{1.89 - 1}{2 - 1} = 0.89 \) or \( 89\% \).

Teacher's Note:
a) Van't Hoff factor \( i \) for electrolytes is greater than 1 due to dissociation into ions.
b) Ensure proper substitution in the formula \( \alpha = \frac{i - 1}{n - 1} \) using the correct number of ions \( n \).

(b) (i) Explain graphically how the rate of a reaction changes with every \( 10^{\circ}\text{C} \) rise in temperature. [2 Marks]

Answer:
The rate of a reaction becomes approximately double or triple for every \( 10^{\circ}\text{C} \) rise in temperature. Graphically, the Maxwell-Boltzmann distribution curve shifts to the right and broadens at a higher temperature, increasing the fraction of molecules possessing energy equal to or greater than the threshold activation energy.

[Figure: Maxwell-Boltzmann distribution curves for kinetic energy at two temperatures T and T + 10\( ^{\circ} \)C showing higher area under the curve beyond activation energy \( E_a \)]

Teacher's Note:
a) The temperature coefficient of a reaction is defined as the ratio of rate constants at temperatures separated by \( 10^{\circ}\text{C} \).
b) Emphasize that the increase in rate is primarily due to an exponential increase in the number of effective collisions, not merely a rise in average kinetic energy.

(ii) How is the activation energy of a reaction related to its rate constant? [1 Mark]

Answer:
According to the Arrhenius equation, \( k = A \cdot e^{-E_a / RT} \), where rate constant \( k \) is inversely related to activation energy \( E_a \); higher activation energy results in a smaller rate constant and a slower reaction.

Teacher's Note:
a) State the logarithmic form: \( \log \left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303 R} \left(\frac{T_2 - T_1}{T_1 T_2}\right) \).
b) Clearly mention the negative exponential relationship between activation energy and rate constant.

(iii) The half life period for the decomposition of a substance is \( 2.5\text{ hours} \). If the initial weight of the substance is \( 160\text{ g} \), how much of the substance will be left after \( 10\text{ hours} \)? [1 Mark]

Answer:
Number of half-lives \( n = \frac{\text{Total time}}{\text{Half-life}} = \frac{10}{2.5} = 4 \)
Amount left \( W = \frac{W_0}{2^n} = \frac{160}{2^4} = \frac{160}{16} = 10\text{ g} \).

Teacher's Note:
a) Radioactive decay and first-order reactions follow the same half-life decay formula.
b) Always determine the number of half-life periods elapsed before calculating the remaining mass.

Question 3

(a) (i) Define Frenkel defects of an ionic crystal. [1 Mark]

Answer:
Frenkel defect is a point defect in ionic solids where an ion (usually the smaller cation) is missing from its normal lattice site and occupies an interstitial site, maintaining electrical neutrality without changing the density of the solid.

Teacher's Note:
a) This defect is typically shown by ionic compounds with a large difference in size between cations and anions.
b) Mention that density remains unchanged because atoms only shift positions internally.

(ii) Iron has an edge length \( 288\text{ pm} \). Its density is \( 7.86\text{ g cm}^{-3} \). Find the type of cubic lattice to which the crystal belongs. (at. mass of iron = 56) [3 Marks]

Answer:
Edge length \( a = 288\text{ pm} = 2.88 \times 10^{-8}\text{ cm} \)
Density \( d = 7.86\text{ g cm}^{-3} \)
Molar mass \( M = 56\text{ g mol}^{-1} \)
Using density formula \( d = \frac{Z \cdot M}{a^3 \cdot N_A} \):
\( Z = \frac{d \cdot a^3 \cdot N_A}{M} = \frac{7.86 \times (2.88 \times 10^{-8})^3 \times 6.023 \times 10^{23}}{56} \)
\( Z = \frac{7.86 \times 23.887 \times 10^{-24} \times 6.023 \times 10^{23}}{56} = \frac{113.04}{56} \approx 2.01 \approx 2 \)
Since \( Z = 2 \), the crystal belongs to a Body-Centred Cubic (BCC) lattice.

Teacher's Note:
a) Carefully convert picometres to centimetres (\( 1\text{ pm} = 10^{-10}\text{ cm} \)) before calculating volume \( a^3 \).
b) Round off the calculated value of \( Z \) to the nearest integer to identify the unit cell type (\( Z = 1 \) for simple cubic, \( 2 \) for BCC, \( 4 \) for FCC).

(b) Explain giving reasons why:
(i) \( \text{Mg(OH)}_2 \) is sparingly soluble in water but highly soluble in ammonium chloride solution. [2 Marks]

Answer:
\( \text{Mg(OH)}_2 \) is a sparingly soluble weak base. On adding \( \text{NH}_4\text{Cl} \), the ammonium ions combine with hydroxyl ions from \( \text{Mg(OH)}_2 \) to form weakly ionized ammonium hydroxide (\( \text{NH}_4\text{OH} \)), reducing \( [\text{OH}^-] \). According to Le Chatelier's principle and common-ion effect, the equilibrium shifts forward, dissolving more \( \text{Mg(OH)}_2 \).

Teacher's Note:
a) The addition of ammonium chloride suppresses the ionization of ammonium hydroxide, lowering free hydroxyl ion concentration.
b) This keeps the ionic product of \( \text{Mg(OH)}_2 \) below its solubility product, dissolving the precipitate.

(ii) When \( \text{H}_2\text{S} \) is passed through acidified zinc sulphate solution, white precipitate of zinc sulphide is not formed. [2 Marks]

Answer:
In the presence of a mineral acid like \( \text{H}_2\text{SO}_4 \), the ionization of \( \text{H}_2\text{S} \) is heavily suppressed due to the common-ion effect of \( \text{H}^+ \) ions. As a result, the concentration of sulfide ions \( [\text{S}^{2-}] \) becomes very low, so the ionic product of \( \text{ZnS} \) does not exceed its solubility product, and precipitation does not occur.

Teacher's Note:
a) Sulfides of Group II are precipitated in acidic medium because they have very low solubility products.
b) Group IV sulfides like ZnS require a neutral or alkaline medium to ensure a high enough sulfide ion concentration.

(c) The equilibrium constant for the reaction \( \text{H}_2(\text{g}) + \text{I}_2(\text{g}) \rightleftharpoons 2\text{HI}(\text{g}) \) is \( 49.5 \) at \( 440^{\circ}\text{C} \). If \( 0.2\text{ mole} \) of \( \text{H}_2 \) and \( 0.2\text{ mole} \) of \( \text{I}_2 \) are allowed to react in a \( 10\text{ litre} \) flask at this temperature, calculate the concentration of each at equilibrium. [2 Marks]

Answer:
Initial concentration of \( \text{H}_2 = \frac{0.2}{10} = 0.02\text{ M} \); Initial concentration of \( \text{I}_2 = \frac{0.2}{10} = 0.02\text{ M} \).
Let \( x \) be the amount reacted at equilibrium.
\( [\text{H}_2]_{\text{eq}} = 0.02 - x \); \( [\text{I}_2]_{\text{eq}} = 0.02 - x \); \( [\text{HI}]_{\text{eq}} = 2x \)
\( K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} \Rightarrow 49.5 = \frac{(2x)^2}{(0.02 - x)(0.02 - x)} \)
Taking square root on both sides:
\( \sqrt{49.5} = \frac{2x}{0.02 - x} \Rightarrow 7.035 = \frac{2x}{0.02 - x} \)
\( 0.1407 - 7.035x = 2x \Rightarrow 9.035x = 0.1407 \Rightarrow x = 0.01556\text{ M} \)
Equilibrium concentrations:
\( [\text{H}_2] = 0.02 - 0.01556 = 0.00444\text{ M} \)
\( [\text{I}_2] = 0.02 - 0.01556 = 0.00444\text{ M} \)
\( [\text{HI}] = 2 \times 0.01556 = 0.03112\text{ M} \).

Teacher's Note:
a) Since the number of moles of reactants and products are equal, volume terms cancel out in the equilibrium expression.
b) Taking square roots of both sides simplifies quadratic equations considerably in such equilibrium problems.

Question 4

(a) (i) What is specific conductance of a solution and what is its unit? How is it related to the equivalent conductance of the solution? [2 Marks]

Answer:
Specific conductance (\( \kappa \)) is the conductance of a solution of an electrolyte enclosed between two electrodes of unit cross-sectional area and unit distance apart. Its unit is \( \text{ohm}^{-1}\text{ cm}^{-1} \) or \( \text{S m}^{-1} \).
Relation with equivalent conductance (\( \Lambda_eq \)): \( \Lambda_eq = \frac{\kappa \times 1000}{\text{Normality}} \).

Teacher's Note:
a) Specific conductance decreases with dilution because the number of current-carrying ions per unit volume decreases.
b) Always include proper units in your final expressions.

(ii) \( 2.5\text{ amperes} \) of current is passed through copper sulphate solution for \( 30\text{ minutes} \). Calculate the number of copper atoms deposited at the cathode (\(\text{Cu} = 63.54\)). [2 Marks]

Answer:
Current \( I = 2.5\text{ A} \); Time \( t = 30 \times 60 = 1800\text{ s} \)
Charge \( Q = I \times t = 2.5 \times 1800 = 4500\text{ C} \)
Reaction: \( \text{Cu}^{2+} + 2\text{e}^- \rightarrow \text{Cu} \)
Charge required to deposit 1 mole of Cu \( = 2 \times 96500\text{ C} = 1,93,000\text{ C} \)
Molecules/atoms deposited by \( 4500\text{ C} = \frac{6.023 \times 10^{23} \times 4500}{193000} = 1.404 \times 10^{22}\text{ atoms} \).

Teacher's Note:
a) Convert time into seconds before calculating total charge in Coulombs.
b) Use Avogadro's number directly when the question asks for the number of atoms rather than mass.

(iii) Four metals W, X, Y and Z have the following values of \( E^{\circ}_{\text{red}} \):
\( W = -0.140\text{ V} \)
\( X = -2.93\text{ V} \)
\( Y = +0.80\text{ V} \)
\( Z = +1.50\text{ V} \)
Arrange them in the increasing order of reducing power. [2 Marks]

Answer:
Reducing power is inversely proportional to standard reduction potential (\( E^{\circ}_{\text{red}} \)).
Order of \( E^{\circ}_{\text{red}} \): \( Z (+1.50\text{ V}) \gt Y (+0.80\text{ V}) \gt W (-0.140\text{ V}) \gt X (-2.93\text{ V}) \)
Increasing order of reducing power: \( \text{Z} \lt \text{Y} \lt \text{W} \lt \text{X} \).

Answer: (Z < Y < W < X)

Teacher's Note:
a) A substance with a high positive reduction potential is a strong oxidizing agent and a weak reducing agent.
b) The most negative reduction potential indicates the strongest reducing agent.

(b) (i) On adding sodium acetate to aqueous solution of acetic acid, what happens to the pH of the solution? Give a reason for your answer. [2 Marks]

Answer:
The pH of the solution increases. Reason: The addition of sodium acetate introduces acetate ions, which suppresses the dissociation of acetic acid due to the common-ion effect, lowering the hydrogen ion concentration \( [\text{H}^+] \) and making the solution less acidic (higher pH).

Teacher's Note:
a) This forms an acidic buffer solution consisting of a weak acid and its salt with a strong base.
b) A decrease in hydrogen ion concentration corresponds directly to an increase in pH values.

(ii) Calculate the pH of an aqueous solution of ammonium formate assuming complete dissociation. \( pK_a \) for formic acid = \( 3.8 \) and \( pK_b \) of ammonia = \( 4.8 \). [1 Mark]

Answer:
Ammonium formate is a salt of a weak acid and a weak base.
Formula for pH: \( \text{pH} = 7 + \frac{1}{2}(pK_a - pK_b) \)
\( \text{pH} = 7 + \frac{1}{2}(3.8 - 4.8) = 7 + \frac{1}{2}(-1.0) = 7 - 0.5 = 6.5 \).

Teacher's Note:
a) The hydrolysis of a salt of weak acid and weak base depends on the relative strengths of the acid and base components.
b) Since \( pK_a \) is less than \( pK_b \), the acid is slightly stronger than the base, making the solution mildly acidic (\(\text{pH} = 6.5\)).

(c) Explain auto catalysis with one example. [1 Mark]

Answer:
Auto catalysis is a phenomenon where one of the products formed during a chemical reaction acts as a catalyst for the same reaction. Example: The hydrolysis of an ester (ethyl acetate) in the presence of an acid, where acetic acid produced during the reaction catalyzes the further hydrolysis.
Equation: \( \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \).

Teacher's Note:
a) Initially, the reaction proceeds slowly, but as the catalytic product accumulates, the reaction rate increases sharply.
b) Another classic example is the oxidation of oxalic acid by acidified potassium permanganate where \(\text{Mn}^{2+}\) ions act as autocatalysts.

SECTION B

Question 5

(a) (i) State the geometry and magnetic property of tetracarbonyl nickel according to the valence bond theory. [1 Mark]

Answer:
Geometry: Tetrahedral.
Magnetic property: Diamagnetic.

Teacher's Note:
a) Nickel in \( \text{[Ni(CO)}_4\text{]} \) has an oxidation state of zero, with electronic configuration \( \text{3d}^8\text{4s}^2 \).
b) Strong field ligand CO forces electron pairing, leading to \( \text{sp}^3 \) hybridization and zero unpaired electrons.

(ii) What type of structural isomers are \( \text{[Pt(OH)}_2\text{(NH}_3\text{)}_4\text{]SO}_4 \) and \( \text{[Pt SO}_4\text{(NH}_3\text{)}_4\text{](OH)}_2 \)? How will you identify the isomers with a chemical test? [2 Marks]

Answer:
Type of isomerism: Ionisation isomerism.
Chemical test: Dissolve both isomers in water. Add barium chloride (\(\text{BaCl}_2\)) solution to the first isomer solution; a white precipitate of barium sulphate indicates the presence of outside \( \text{SO}_4^{2-} \) ions. The second isomer will not give a white precipitate with \( \text{BaCl}_2 \), but will give a basic test with phenolphthalein due to free \( \text{OH}^- \) ions.

Teacher's Note:
a) Ionisation isomers give different ions in solution despite having the same molecular formula.
b) Always use a reagent that selectively tests for the counter-ion present outside the coordination sphere.

(b) Name the co-ordination compound used for the following: [2 Marks]
(i) Treatment of cancer.
(ii) Treatment of lead poisoning.

Answer:
(i) Cisplatin (\( \text{cis-}\text{[Pt(NH}_3\text{)}_2\text{Cl}_2\text{]} \))
(ii) Calcium disodium EDTA (\( \text{CaNa}_2\text{EDTA} \))

Teacher's Note:
a) Cisplatin binds to DNA and inhibits replication in cancer cells.
b) EDTA forms a stable chelate complex with toxic lead ions, facilitating their excretion through urine.

Question 6

(a) Explain giving reasons why:
(i) The halogens are coloured and the colour deepens from fluorine to iodine. [2 Marks]

Answer:
Halogens are coloured because their molecules absorb visible light, causing excitation of electrons from the highest occupied molecular orbital (HOMO) to the lowest unoccupied molecular orbital (LUMO). As we move from fluorine to iodine, the energy gap between HOMO and LUMO decreases, requiring lower energy (longer wavelengths) for excitation, so the color deepens (\( \text{F}_2 \) - yellow, \( \text{Cl}_2 \) - greenish-yellow, \( \text{Br}_2 \) - red, \( \text{I}_2 \) - violet).

Teacher's Note:
a) The absorption of specific wavelengths of visible light imparts complementary colors to the halogens.
b) Larger size and lower electronegativity down the group reduce excitation energy requirements.

(ii) In a given transition series, the atomic radius does not change very much with increasing atomic number. [2 Marks]

Answer:
Along a transition series, nuclear charge increases by one unit at each step, adding electrons to inner d-orbitals. The d-electrons provide a poor shielding or screening effect for the outer s-electrons. This increased effective nuclear charge balances out the added electron repulsion, keeping atomic radii nearly constant across the period.

Teacher's Note:
a) The balance between increased nuclear charge and poor d-orbital shielding is the primary reason for atomic radius constancy.
b) Mention both screening effect and effective nuclear charge to earn full credit.

(b) Draw the resonating structures of ozone molecule. [1 Mark]

Answer:
Ozone has two major contributing resonance structures:
\( \text{O}=\text{-O}^{+}-\text{-O}^- \longleftrightarrow \text{-O}^-\text{-O}^{+}=\text{O} \).

Teacher's Note:
a) Show formal charges on oxygen atoms and double/single bond alternation in resonance hybrids.
b) The actual bond length in ozone is intermediate between a single and double bond due to resonance delocalization.

Question 7

(a) (i) Give equations to show the use of aqua regia in dissolving platinum. [1 Mark]

Answer:
Aqua regia is a \( 3:1 \) mixture of concentrated \( \text{HCl} \) and concentrated \( \text{HNO}_3 \).
\( \text{HNO}_3 + 3\text{HCl} \rightarrow \text{NOCl} + 2\text{H}_2\text{O} + 2\text{[Cl]} \)
\( \text{Pt} + 4\text{[Cl]} \rightarrow \text{PtCl}_4 \)
\( \text{PtCl}_4 + 2\text{HCl} \rightarrow \text{H}_2\text{[PtCl}_6\text{]} \) (Chloroplatinic acid).

Teacher's Note:
a) Nascent chlorine generated in situ by aqua regia is a powerful oxidizing agent capable of dissolving noble metals like gold and platinum.
b) Write balanced chemical equations showing the generation of nascent chlorine and subsequent complex formation.

(ii) Draw the structure of Xenon hexafluoride molecule and state the hybridisation of the central atom and the structure of the molecule. [2 Marks]

Answer:
Hybridisation of central atom: \( \text{sp}^3\text{d}^3 \).
Structure: Distorted octahedral (capped octahedron due to the presence of a lone pair).

[Figure: Distorted octahedral geometry of \( \text{XeF}_6 \) with six fluorine atoms around a central Xenon atom and a lone pair occupying an octahedral face]

Teacher's Note:
a) Count total valence electron pairs around xenon: \( 8 + 6 = 14 \Rightarrow 7 \) pairs (\( 6 \) bond pairs + \( 1 \) lone pair).
b) State that VSEPR theory predicts a distorted geometry due to the lone pair's repulsion.

(b) Write balanced equations for the following reactions: [2 Marks]
(i) Ozone and alkaline potassium iodide.
(ii) Sodium sulphite and acidified potassium permanganate.

Answer:
(i) \( 2\text{KI} + \text{H}_2\text{O} + \text{O}_3 \rightarrow 2\text{KOH} + \text{I}_2 + \text{O}_2 \)
(ii) \( 2\text{KMnO}_4 + 3\text{H}_2\text{SO}_4 + 5\text{Na}_2\text{SO}_3 \rightarrow \text{K}_2\text{SO}_4 + 2\text{MnSO}_4 + 5\text{Na}_2\text{SO}_4 + 3\text{H}_2\text{O} \)

Teacher's Note:
a) Ozone acts as a strong oxidizing agent, oxidizing potassium iodide to liberate iodine gas.
b) Ensure all ionic charges and stoichiometric coefficients balance correctly in redox reactions.

SECTION C

Question 8

(a) Write equations for the following reactions and name the reactions: [3 Marks]
(i) Benzene diazonium chloride is treated with copper and hydrochloric acid.
(ii) Formaldehyde is treated with 50% caustic soda solution.

Answer:
(i) \( \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{HCl} \xrightarrow{\text{Cu}} \text{C}_6\text{H}_5\text{Cl} + \text{N}_2 + \text{HCl} \). Name: Gattermann reaction.
(ii) \( 2\text{HCHO} + \text{NaOH (50\%)} \rightarrow \text{HCOONa} + \text{CH}_3\text{OH} \). Name: Cannizzaro reaction.

Teacher's Note:
a) Gattermann reaction uses copper powder and halogen acid to replace the diazonium group with a halogen.
b) Cannizzaro reaction is a disproportionation reaction given by aldehydes lacking an alpha-hydrogen.

(b) (i) Write the structures of the isomers of 3 phenyl prop-2-enoic acid. [1 Mark]

Answer:
3-phenylprop-2-enoic acid (Cinnamic acid) exhibits geometrical isomerism (cis-trans isomerism):
1. Trans-isomer (Cinnamic acid): Phenyl group and carboxylic acid group are on opposite sides of the double bond.
2. Cis-isomer (Allocinnamic acid): Phenyl group and carboxylic acid group are on the same side of the double bond.

Teacher's Note:
a) Restricted rotation about the carbon-carbon double bond gives rise to geometrical isomers.
b) Clearly draw cis and trans configurations to earn full marks.

(ii) What type of isomerism is exhibited by the following pairs of compounds:
(1) \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} \) and \( (\text{C}_2\text{H}_5)_2\text{CHOH} \)
(2) [Figure: structural pair shown in paper] [2 Marks]

Answer:
(1) Chain isomerism.
(2) Position isomerism (or Functional / Metamerism based on precise structure).

Teacher's Note:
a) Chain isomerism arises due to difference in carbon chain branching while keeping the same functional group.
b) Position isomerism involves the same carbon skeleton with different positions for the functional group.

(c) Give one good chemical test to distinguish between the following pairs of compounds: [3 Marks]
(i) Urea and acetamide
(ii) 1-propanol and 2 methyl 2-propanol.

Answer:
(i) Biuret test: Heat urea gently until it melts and forms biuret; on adding sodium hydroxide and a drop of dilute copper sulphate solution, a violet-pink color appears. Acetamide does not give this test.
(ii) Lucas test: Treat with anhydrous \( \text{ZnCl}_2 \) and concentrated \( \text{HCl} \). 2-methyl-2-propanol (tertiary alcohol) gives immediate turbidity at room temperature, whereas 1-propanol (primary alcohol) shows no turbidity at room temperature.

Teacher's Note:
a) The biuret test is a characteristic test for compounds containing peptide-like linkages or -CONH- groups bonded together.
b) Lucas test relies on the stability of carbocations to distinguish between primary, secondary, and tertiary alcohols.

(d) Name the monomeric units of Nylon 66. [1 Mark]

Answer:
Hexamethylenediamine and Adipic acid.

Teacher's Note:
a) Nylon 66 is a condensation polymer formed by the elimination of water molecules.
b) The numbers 6, 6 represent the number of carbon atoms present in each of the two monomer molecules.

Question 9

(a) Identify the compounds A, B, C, D, E and F. [3 Marks]
\( \text{C}_6\text{H}_6 \xrightarrow{\text{A}} \text{C}_6\text{H}_5\text{CH}_3 \xrightarrow{\text{B}} \text{C}_6\text{H}_5\text{CHO} \xrightarrow{\text{C}} \text{C}_6\text{H}_5\text{CH}_2\text{OH} + \text{D} \)
\( \text{alc. KCN} \rightarrow \text{E} \)
\( \text{F} \rightarrow \text{C}_6\text{H}_5\text{COOH} \)}

Answer:
A: \( \text{CH}_3\text{Cl} \) / Anhydrous \( \text{AlCl}_3 \) (Friedel-Crafts methylation)
B: \( \text{CrO}_2\text{Cl}_2 \) / \( \text{CS}_2 \) followed by hydrolysis (Etard reaction)
C: Conc. \( \text{NaOH} \) (Cannizzaro reaction)
D: Sodium benzoate (\( \text{C}_6\text{H}_5\text{COONa} \))
E: Benzyl cyanide (\( \text{C}_6\text{H}_5\text{CH}_2\text{CN} \))
F: Benzyl chloride (\( \text{C}_6\text{H}_5\text{CH}_2\text{Cl} \))

Teacher's Note:
a) Follow the sequence of reagents carefully: alkylation gives toluene, Etard oxidation gives benzaldehyde, and Cannizzaro disproportionation gives alcohol and salt.
b) Verify conversion of benzyl chloride with alcoholic KCN to form benzyl cyanide (E).

(b) How can the following conversions be brought about?
(i) Ethanoic acid to ethylamine. [3 Marks]
(ii) Aniline to benzoic acid. [3 Marks]

Answer:
(i) Ethanoic acid to ethylamine:
1. \( \text{CH}_3\text{COOH} + \text{NH}_3 \xrightarrow{\Delta} \text{CH}_3\text{CONH}_2 \) (Acetamide)
2. \( \text{CH}_3\text{CONH}_2 + \text{Br}_2 + 4\text{KOH} \rightarrow \text{CH}_3\text{NH}_2 + \text{K}_2\text{CO}_3 + 2\text{KBr} + 2\text{H}_2\text{O} \) (Methylamine - wait, question asks for ethylamine, so convert via Grignard or reduction of acetonitrile from acetic acid).
Correction pathway for ethanoic acid to ethylamine (2 carbon chain preserved):
1. \( \text{CH}_3\text{COOH} + \text{LiAlH}_4 \rightarrow \text{CH}_3\text{CH}_2\text{OH} \) (Ethanol)
2. \( \text{CH}_3\text{CH}_2\text{OH} + \text{PBr}_3 \rightarrow \text{CH}_3\text{CH}_2\text{Br} \) (Bromoethane)
3. \( \text{CH}_3\text{CH}_2\text{Br} + \text{NH}_3 \rightarrow \text{CH}_3\text{CH}_2\text{NH}_2 \) (Ethylamine).
(ii) Aniline to benzoic acid:
1. \( \text{C}_6\text{H}_5\text{NH}_2 + \text{HNO}_2 (\text{NaNO}_2 + \text{HCl}) \xrightarrow{0-5^{\circ}\text{C}} \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \) (Benzene diazonium chloride)
2. \( \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{CuCN} + \text{KCN} \rightarrow \text{C}_6\text{H}_5\text{CN} \) (Benzonitrile)
3. \( \text{C}_6\text{H}_5\text{CN} + 2\text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{C}_6\text{H}_5\text{COOH} \) (Benzoic acid).

Teacher's Note:
a) Step-down or step-up conversion strategies require careful selection of reagents like LiAlH\( _4 \) or diazonium intermediates.
b) Hydrolysis of nitriles yields corresponding carboxylic acids.

(c) What is a zwitter ion? Represent the zwitter ion of glycine. [1 Mark]

Answer:
A zwitter ion is a dipolar ion formed by the internal transfer of a proton from the carboxylic acid group to the amino group within an amino acid molecule, resulting in a net zero charge.
Representation for glycine: \( ^+\text{H}_3\text{NCH}_2\text{COO}^- \).

Teacher's Note:
a) Amino acids exist predominantly as zwitter ions in aqueous solution near their isoelectric point.
b) Show both positive and negative charges clearly within the same molecule.

Question 10

(a) An organic compound A on treatment with ethanol gives a carboxylic acid B and a compound C. Hydrolysis of C under acidic condition gives B and D. Oxidation of D with acidified potassium permanganate also gives B. B on heating with calcium hydroxide gives E with molecular formula \( \text{C}_3\text{H}_6\text{O} \). E does not give Tollen's test but reacts with iodine and caustic potash to give a yellow precipitate. [4 Marks]
(i) Identify A, B, C, D and E.
(ii) Write balanced equation of E with iodine and caustic potash and name the reaction.

Answer:
(i) Identification:
E: Acetone (\( \text{CH}_3\text{COCH}_3 \) - formula \( \text{C}_3\text{H}_6\text{O} \), gives iodoform test, negative Tollen's test).
B: Acetic acid (\( \text{CH}_3\text{COOH} \), obtained by dry distillation of calcium acetate E).
D: Ethanol (\( \text{CH}_3\text{CH}_2\text{OH} \), oxidation gives B).
C: Ethyl acetate (\( \text{CH}_3\text{COOC}_2\text{H}_5 \), ester formed from A and ethanol).
A: Acetic anhydride or Acetyl chloride (\( (\text{CH}_3\text{CO})_2\text{O} \) or \( \text{CH}_3\text{COCl} \)).
(ii) Reaction of E with iodine and caustic potash (Iodoform test):
\( \text{CH}_3\text{COCH}_3 + 3\text{I}_2 + 4\text{KOH} \rightarrow \text{CHI}_3\downarrow + \text{CH}_3\text{COOK} + 3\text{KI} + 3\text{H}_2\text{O} \).
Name of reaction: Haloform (Iodoform) reaction.

Teacher's Note:
a) Work backwards from the iodoform-positive ketone E (\(\text{acetone}\)) to deduce all precursor compounds step by step.
b) Calcium salt of acetic acid on dry distillation gives acetone.

(b) (i) Name the functional groups that distinguish glucose and fructose. How will you distinguish between the two compounds? [2 Marks]

Answer:
Functional groups: Glucose contains an aldehyde group (\( -\text{CHO} \)), while fructose contains a keto group (\( >\text{C}=\text{O} \)).
Distinguishing test: Seliwanoff's test. When heated with resorcinol and concentrated \( \text{HCl} \), fructose gives a rapid cherry-red colored complex, whereas glucose reacts very slowly or gives a faint pink color after a long time.

Teacher's Note:
a) Both are monosaccharides with the same molecular formula (\(\text{C}_6\text{H}_{12}\text{O}_6\)) but different functional groups (aldose vs ketose).
b) Seliwanoff's test specifically differentiates ketoses from aldoses based on dehydration rate in acid.

(ii) What are polyesters? Give one example of polyester and the monomers. [2 Marks]

Answer:
Polyesters are synthetic polymers containing multiple ester linkages formed by condensation polymerization between dicarboxylic acids and diols.
Example: Dacron (Terylene).
Monomers: Ethylene glycol (\( \text{HO}-\text{CH}_2-\text{CH}_2-\text{OH} \)) and Terephthalic acid (\( \text{HOOC}-\text{C}_6\text{H}_4-\text{COOH} \)).

Teacher's Note:
a) Condensation polymers release small molecules like water during chain growth.
b) Dacron is widely used in textile industries and for making magnetic recording tapes.

(c) Give balanced equations for the following reactions: [2 Marks]
(i) Aniline and benzoyl chloride.
(ii) Diethyl ether and hydroiodic acid (cold).

Answer:
(i) \( \text{C}_6\text{H}_5\text{NH}_2 + \text{C}_6\text{H}_5\text{COCl} \rightarrow \text{C}_6\text{H}_5\text{NHCOC}_6\text{H}_5 + \text{HCl} \) (Benzoylation / Schotten-Baumann reaction)
(ii) \( \text{C}_2\text{H}_5\text{OC}_2\text{H}_5 + \text{HI (cold)} \rightarrow \text{C}_2\text{H}_5\text{I} + \text{C}_2\text{H}_5\text{OH} \)

Teacher's Note:
a) Benzoylation of primary amines introduces a benzoyl group to form secondary amides in the presence of a base.
b) Cold hydroiodic acid cleaves ethers to form an alkyl halide and an alcohol.