ISC Class 12 Chemistry Board Exam Question Paper 2015 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]
(increases, decreases, positive, efficient, 68, non-efficient, no a-hydrogen, a-hydrogen, negative, Rosenmund's, greater, Cannizzaro, 74, common-ion effect, lesser, buffer action, diamagnetic, paramagnetic)
(i) The more __________ the standard reduction potential of a metal, the __________ is its ability to displace hydrogen from acids.
(ii) Both ccp and hcp are __________ packings and occupy about __________ % of the available space.
(iii) Solubility of silver chloride __________ in the presence of sodium chloride because of __________.
(iv) Benzaldehyde undergoes __________ reaction on treatment with concentrated sodium hydroxide because it has __________ atom.
(v) The transition metals show __________ character because of the presence of unpaired electrons and Cu⁺ is __________ because its electronic configuration is [Ar]3d¹⁰.

Answer:
(i) negative, greater
(ii) efficient, 74
(iii) decreases, common-ion effect
(iv) Cannizzaro, no a-hydrogen
(v) paramagnetic, diamagnetic

Teacher's Note:
a) Negative standard reduction potential indicates strong reducing power and an easier tendency to lose electrons, making it displace hydrogen easily.
b) Students must remember that ccp and hcp have 74% packing efficiency, whereas bcc has 68%.

(b) Complete the following statements by selecting the correct alternative from the choices given: [5 Marks]
(i) The molal freezing point constant of water is \( 1.86\text{ K kg mol}^{-1} \). Therefore, the freezing point of \( 0.1\text{M NaCl} \) solution in water is expected to be:
(1) \( -1.86^{\circ}\text{C} \)
(2) \( -0.372^{\circ}\text{C} \)
(3) \( -0.186^{\circ}\text{C} \)
(4) \( +0.372^{\circ}\text{C} \)

Answer: (2) \( -0.372^{\circ}\text{C} \)

\( \Delta T_f = i \times K_f \times m = 2 \times 1.86 \times 0.1 = 0.372^{\circ}\text{C} \), so \( T_f = -0.372^{\circ}\text{C} \).

Teacher's Note:
a) Remember to multiply by the van't Hoff factor \( i = 2 \) for \( \text{NaCl} \).
b) A common mistake is forgetting dissociation factor, leading to option (3).

(ii) For a first order reaction the rate constant for decomposition of \( \text{N}_2\text{O}_5 \) is \( 6\times 10^{-4}\text{sec}^{-1} \). The half-life period for the decomposition in seconds is:
(1) \( 11.55 \)
(2) \( 115.5 \)
(3) \( 1155 \)
(4) \( 1.155 \)

Answer: (3) \( 1155 \)

\( t_{1/2} = \frac{0.693}{k} = \frac{0.693}{6\times 10^{-4}} = 1155\text{ s} \).

Teacher's Note:
a) Use the direct formula for half-life of a first order reaction: \( t_{1/2} = 0.693/k \).
b) Ensure powers of ten are correctly handled during division.

(iii) When acetaldehyde is treated with Grignard reagent, followed by hydrolysis the product formed is:
(1) Primary alcohol
(2) Secondary alcohol
(3) Carboxylic acid
(4) Tertiary alcohol

Answer: (2) Secondary alcohol

Acetaldehyde (ethanal) reacting with a Grignard reagent yields a secondary alcohol upon hydrolysis.

Teacher's Note:
a) Formaldehyde gives primary alcohols, other aldehydes give secondary alcohols, and ketones give tertiary alcohols.
b) Do not confuse the starting carbonyl compounds with respect to alcohol classes.

(iv) The geometry of \( \text{XeF}_6 \) molecule and the hybridization of Xe atom in the molecule is:
(1) Distorted octahedral and \( \text{sp}^3\text{d}^3 \)
(2) Square planar and \( \text{sp}^3\text{d}^2 \)
(3) Pyramidal and \( \text{sp}^3 \)
(4) Octahedral and \( \text{sp}^3\text{d}^3 \)

Answer: (1) Distorted octahedral and \( \text{sp}^3\text{d}^3 \)

\( \text{XeF}_6 \) has 7 electron pairs around Xenon (6 bond pairs + 1 lone pair), giving distorted octahedral geometry and \( \text{sp}^3\text{d}^3 \) hybridization.

Teacher's Note:
a) The presence of a lone pair on xenon distorts the regular octahedral geometry.
b) Count total valence electrons properly to find steric number 7.

(v) In the complexes \( [\text{Fe(CN)}_6]^{3-} \) and \( [\text{Pt(en)}_2(\text{NO}_2)(\text{Cl})]^{2+} \) the respective oxidation numbers of central metal atoms are:
(1) +3 and +4
(2) +6 and +4
(3) +6 and +3
(4) +3 and +3

Answer: (1) +3 and +4

For \( [\text{Fe(CN)}_6]^{3-} \), \( x + 6(-1) = -3 \implies x = +3 \). For \( [\text{Pt(en)}_2(\text{NO}_2)(\text{Cl})]^{2+} \), \( x + 2(0) + (-1) + (-1) = +2 \implies x = +4 \).

Teacher's Note:
a) Ethylenediamine (en) is a neutral ligand, \( \text{NO}_2^- \) is -1, and \( \text{Cl}^- \) is -1.
b) Always sum the charges of ligands and equate to the net charge of the complex ion to find the metal oxidation state.

(c) Answer the following questions: [5 Marks]
(i) What is the effect of temperature on the ionic product of water? How will it change the pH value of a neutral solution?

Answer:
The auto-ionization of water is an endothermic process. Therefore, as temperature increases, the ionic product of water (\( K_w \)) increases. In a neutral solution, \( [\text{H}^+] = [\text{OH}^-] \), and since \( K_w \) increases, both \( [\text{H}^+] \) and \( [\text{OH}^-] \) increase, causing the pH of a neutral solution to decrease below 7.

Teacher's Note:
a) Relate the endothermic nature of water dissociation to Le Chatelier's principle.
b) Note that neutrality does not mean pH = 7 at all temperatures; neutrality means \( [\text{H}^+] = [\text{OH}^-] \).

(ii) How many hours does it take to reduce 3 moles of \( \text{Fe}^{3+} \) to \( \text{Fe}^{2+} \) with \( 2.0\text{ A} \) current intensity?

Answer:
Reaction: \( \text{Fe}^{3+} + e^- → \text{Fe}^{2+} \)
1 mole of \( \text{Fe}^{3+} \) requires 1 Faraday of charge (96500 C) to reduce to \( \text{Fe}^{2+} \).
For 3 moles, charge \( Q = 3 \times 96500\text{ C} = 289500\text{ C} \).
We know \( Q = I \times t \implies t = \frac{Q}{I} = \frac{289500}{2.0} = 144750\text{ seconds} \).
Time in hours = \( \frac{144750}{3600} = 40.21\text{ hours} \).

Teacher's Note:
a) Establish the stoichiometry of electrons per mole of metal ion reduced.
b) Convert final seconds to hours by dividing by 3600.

(iii) How is urea prepared by Wohler synthesis?

Answer:
Urea is prepared by heating an aqueous solution of ammonium cyanate (\( \text{NH}_4\text{CNO} \)), which undergoes intramolecular rearrangement to form urea.
\( \text{NH}_4\text{CNO} \xrightarrow{\Delta} \text{NH}_2\text{CONH}_2 \)

Teacher's Note:
a) Wohler synthesis was the first synthesis of an organic compound from an inorganic source.
b) Write the exact rearrangement equation clearly.

(iv) Two liquids A and B form type II non ideal solution which shows a minimum in its temperature - mole fraction plot (T-\( \chi \) diagram). Can the two liquids be completely separated by fractional distillation?

Answer:
No, the two liquids cannot be completely separated by fractional distillation because they form a maximum boiling azeotrope at the composition corresponding to the temperature minimum in the T-\( \chi \) diagram.

Teacher's Note:
a) A minimum in the T-\( \chi \) diagram represents a maximum boiling azeotrope.
b) Azeotropes distill without change in composition, preventing complete separation.

(v) The aqueous solution of sodium acetate is basic. Explain.

Answer:
Sodium acetate is a salt of a weak acid (acetic acid, \( \text{CH}_3\text{COOH} \)) and a strong base (sodium hydroxide, \( \text{NaOH} \)). In water, it undergoes anionic hydrolysis:
\( \text{CH}_3\text{COO}^- + \text{H}_2\text{O} ⇌ \text{CH}_3\text{COOH} + \text{OH}^- \)
Due to the production of excess hydroxide ions (\( \text{OH}^- \)), the solution becomes basic.

Teacher's Note:
a) Identify the parent acid and base of the salt correctly.
b) Show the hydrolysis equation clearly indicating the release of \( \text{OH}^- \).

(d) Match the following: [5 Marks]
(i) Disaccharide - (d) Sucrose
(ii) Carbylamine - (c) Obnoxious smell
(iii) Dacron - (b) Condensation polymer
(iv) Low spin complex, \( \text{d}^2\text{sp}^3 \) - (e) Hexaamminecobalt(III) ion
(v) Anhydrous \( \text{ZnCl}_2 + \text{conc.HCl} \) - (a) Lucas reagent

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

Teacher's Note:
a) Match terms carefully based on standard functional groups and reagents.
b) Lucas reagent is anhydrous zinc chloride and concentrated hydrochloric acid used to distinguish alcohols.

PART II (50 Marks)

SECTION A

Answer any two questions.

Question 2

(a) (i) A solution containing \( 0.5\text{ g} \) of \( \text{KCl} \) dissolves in \( 100\text{ g} \) of water and freezes at \( -0.24^{\circ}\text{C} \). Calculate the degree of dissociation of the salt. (\( K_f \) for water = \( 1.86^{\circ}\text{C} \)
Atomic weights [K = 39, Cl = 35.5]) [3 Marks]

Answer:
Molar mass of \( \text{KCl} = 39 + 35.5 = 74.5\text{ g mol}^{-1} \)
Molality \( m = \frac{\text{Mass of KCl}}{\text{Molar Mass}} \times \frac{1000}{\text{Mass of solvent (g)}} = \frac{0.5}{74.5} \times \frac{1000}{100} = 0.0671\text{ mol kg}^{-1} \)
Observed freezing point depression \( \Delta T_f = 0 - (-0.24^{\circ}\text{C}) = 0.24^{\circ}\text{C} \)
Normal \( \Delta T_f' = K_f \times m = 1.86 \times 0.0671 = 0.1248^{\circ}\text{C} \)
Van't Hoff factor \( i = \frac{\Delta T_f(\text{obs})}{\Delta T_f(\text{calc})} = \frac{0.24}{0.1248} = 1.923 \)
For \( \text{KCl} → \text{K}^+ + \text{Cl}^- \), \( n = 2 \).
Degree of dissociation \( \alpha = \frac{i - 1}{n - 1} = \frac{1.923 - 1}{2 - 1} = 0.923\text{ or }92.3\% \).

Teacher's Note:
a) Calculate theoretical depression first, then find the experimental van't Hoff factor.
b) Use the relation \( \alpha = (i - 1)/(n - 1) \) to find degree of dissociation.

(ii) If \( 1.71\text{ g} \) of sugar (molar mass = 342) are dissolved in \( 500\text{ ml} \) of an aqueous solution at \( 300\text{ K} \), what will be its osmotic pressure? [1 Mark]

Answer:
Number of moles of sugar \( n = \frac{1.71}{342} = 0.005\text{ moles} \)
Volume \( V = 500\text{ ml} = 0.5\text{ litres} \)
Concentration \( C = \frac{n}{V} = \frac{0.005}{0.5} = 0.01\text{ mol L}^{-1} \)
Osmotic pressure \( \pi = C R T = 0.01 \times 0.0821 \times 300 = 0.2463\text{ atm} \).

Teacher's Note:
a) Use the formula \( \pi = \frac{w}{M} \times \frac{R T}{V} \).
b) Ensure volume is converted to litres and temperature is in Kelvin.

(iii) \( 0.70\text{ g} \) of an organic compound when dissolved in \( 32\text{g} \) of acetone produces an elevation of \( 0.25^{\circ}\text{C} \) in the boiling point. Calculate the molecular mass of organic compound (\( K_b \) for acetone = \( 1.72\text{ K kg mol}^{-1} \)). [1 Mark]

Answer:
\( \Delta T_b = \frac{K_b \times w_2 \times 1000}{M_2 \times w_1} \)
\( M_2 = \frac{K_b \times w_2 \times 1000}{\Delta T_b \times w_1} = \frac{1.72 \times 0.70 \times 1000}{0.25 \times 32} = \frac{1204}{8} = 150.5\text{ g mol}^{-1} \).

Teacher's Note:
a) Rearrange elevation in boiling point formula to solve for molar mass \( M_2 \).
b) Check units of mass and constant before calculation.

(b) (i) What is the difference between order of a reaction and the molecularity of a reaction? [2 Marks]

Answer:
1. Order of a reaction is the sum of the powers of concentration terms in the rate law expression, whereas molecularity is the total number of reacting species colliding simultaneously in an elementary step.
2. Order can be zero, fractional, or negative, whereas molecularity is always a whole number and cannot be zero or fractional.

Teacher's Note:
a) Order is an experimental quantity; molecularity is a theoretical concept.
b) For complex reactions, overall order is determined by the slowest step, while molecularity has no meaning for complex reactions as a whole.

(c) Name the crystal structure of the copper metal. [1 Mark]

Answer:
Face-centered cubic (fcc) or cubic close-packed (ccp).

Teacher's Note:
a) Copper crystallizes in a ccp/fcc lattice arrangement.
b) Coordination number in this structure is 12.

Question 3

(a) (i) Chromium metal crystallises with a body centered cubic lattice. The edge length of the unit cell is found to be \( 287\text{ pm} \). Calculate the atomic radius. What would be the density of chromium in \( \text{g / cm}^3 \)? (atomic mass of Cr = 52.99) [2 Marks]

Answer:
For bcc lattice, radius \( r = \frac{\sqrt{3}}{4} a = \frac{1.732}{4} \times 287 = 124.27\text{ pm} \)
Edge length \( a = 287\text{ pm} = 2.87 \times 10^{-8}\text{ cm} \)
Density \( d = \frac{Z \times M}{N_A \times a^3} = \frac{2 \times 52.99}{6.023 \times 10^{23} \times (2.87 \times 10^{-8})^3} = \frac{105.98}{6.023 \times 10^{23} \times 2.364 \times 10^{-23}} = \frac{105.98}{14.238} = 7.44\text{ g cm}^{-3} \).

Teacher's Note:
a) For bcc, number of atoms per unit cell \( Z = 2 \).
b) Convert pm to cm carefully by multiplying with \( 10^{-10} \) or \( 10^{-8} \).

(ii) Why do sodium chloride on heating with sodium vapours acquire yellow colour? [1 Mark]

Answer:
Sodium chloride on heating with sodium vapour acquires yellow colour due to metal excess defect, where \( \text{Cl}^-\) ions diffuse to the surface, combine with Na atoms, and the released electrons occupy anionic vacancies called F-centres, which absorb visible light and impart a yellow colour.

Teacher's Note:
a) F-centres are Farbzentren responsible for colour in alkali halide crystals.
b) Mention loss of electron by sodium and trapping in anion vacancies.

(iii) The equilibrium constant for the reaction:
\( \text{N}_{2\text{(g)}} + 3\text{H}_{2\text{(g)}} ⇌ 2\text{NH}_{3\text{(g)}} \) at \( 715\text{ K} \), is \( 6.0\times 10^{-2} \).
If, in a particular reaction, there are \( 0.25\text{ mol L}^{-1} \) of \( \text{H}_2 \) and \( 0.06\text{ mol L}^{-1} \) of \( \text{NH}_3 \) present, calculate the concentration of \( \text{N}_2 \) at equilibrium. [1 Mark]

Answer:
Equilibrium constant expression: \( K_c = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} \)
\( 6.0 \times 10^{-2} = \frac{(0.06)^2}{[\text{N}_2] \times (0.25)^3} \)
\( 0.06 = \frac{0.0036}{[\text{N}_2] \times 0.015625} \)
\( [\text{N}_2] = \frac{0.0036}{0.06 \times 0.015625} = \frac{0.0036}{0.0009375} = 3.84\text{ mol L}^{-1} \).

Teacher's Note:
a) Write the correct expression for \( K_c \) matching stoichiometric coefficients.
b) Substitute given values carefully and solve for unknown concentration.

(iv) Calculate the concentration of \( \text{OH}^- \) ions in solution when \( [\text{H}^+] = 6.2 \times 10^{-2}\text{mol L}^{-1} \). [1 Mark]

Answer:
\( K_w = [\text{H}^+][\text{OH}^-] = 1.0 \times 10^{-14} \text{ at } 298\text{ K} \)
\( [\text{OH}^-] = \frac{1.0 \times 10^{-14}}{6.2 \times 10^{-2}} = 1.61 \times 10^{-13}\text{ mol L}^{-1} \).

Teacher's Note:
a) Use ionic product of water relationship \( K_w = 10^{-14} \).
b) Divide values and adjust exponents properly.

(v) State the Le Chatelier's principle. [1 Mark]

Answer:
Le Chatelier's principle states that if a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the equilibrium shifts in such a direction as to counteract or nullify the effect of the change.

Teacher's Note:
a) Must include the key phrases "system at equilibrium" and "shift to counteract the change".
b) Used to predict the effect of external stress on chemical equilibrium.

(b) For a crystal of sodium chloride, state: [2 Marks]
(i) The type of lattice in which it crystallises.
(ii) The coordination number of each sodium ion and chloride ion in the crystal lattice.
(iii) The number of sodium ions and chloride ions present in a unit cell of sodium chloride.
(iv) The structural arrangement of the sodium chloride crystal.

Answer:
(i) Face-centered cubic (fcc) lattice.
(ii) Coordination number of each \( \text{Na}^+ \) and \( \text{Cl}^- \) is 6 (6:6 coordination).
(iii) 4 sodium ions and 4 chloride ions per unit cell.
(iv) Rock salt (fcc) structure where chloride ions form ccp and sodium ions occupy all octahedral voids.

Teacher's Note:
a) Remember that \( \text{NaCl} \) unit cell has 4 formula units.
b) Both ions have identical coordination geometry with coordination number 6.

(c) Consider the following reaction:
\( \text{N}_2\text{O}_{4\text{(g)}} + \text{Heat} ⇌ 2\text{NO}_{2\text{(g)}} \)
How is the composition of equilibrium mixture affected by:
(i) a change in temperature
(ii) a change in pressure
(iii) a change in concentration of \( \text{N}_2\text{O}_4 \)
(iv) the removal of \( \text{NO}_2 \) from the reaction mixture [2 Marks]

Answer:
(i) Increase in temperature shifts equilibrium in the forward direction (more \( \text{NO}_2 \) formed) as the reaction is endothermic.
(ii) Increase in pressure shifts equilibrium in the backward direction (towards fewer moles).
(iii) Increase in concentration of \( \text{N}_2\text{O}_4 \) shifts equilibrium in the forward direction.
(iv) Removal of \( \text{NO}_2 \) shifts equilibrium in the forward direction to replace the removed product.

Teacher's Note:
a) Apply Le Chatelier's principle separately for temperature, pressure, and concentration changes.
b) Note that forward reaction is endothermic because heat is a reactant.

Question 4

(a) The specific conductance of a \( 0.01\text{ M} \) solution of acetic acid at \( 298\text{ K} \) is \( 1.65 \times 10^{-4}\text{ ohm}^{-1}\text{ cm}^{-1} \). The molar conductance at infinite dilution for \( \text{H}^+ \) ion and \( \text{CH}_3\text{COO}^- \) ion are \( 349.1\text{ ohm}^{-1}\text{ cm}^2\text{ mol}^{-1} \) and \( 40.9\text{ ohm}^{-1}\text{ cm}^2\text{ mol}^{-1} \) respectively. Calculate:
(i) Molar conductance of the solution.
(ii) Degree of dissociation of \( \text{CH}_3\text{COOH} \).
(iii) Dissociation constant for acetic acid. [3 Marks]

Answer:
(i) \( \Lambda_m = \frac{\kappa \times 1000}{C} = \frac{1.65 \times 10^{-4} \times 1000}{0.01} = 16.5\text{ ohm}^{-1}\text{ cm}^2\text{ mol}^{-1} \)
(ii) \( \Lambda_m^{\infty}(\text{CH}_3\text{COOH}) = \lambda_{\text{H}^+}^{\infty} + \lambda_{\text{CH}_3\text{COO}^-}^{\infty} = 349.1 + 40.9 = 390.0\text{ ohm}^{-1}\text{ cm}^2\text{ mol}^{-1} \)
Degree of dissociation \( \alpha = \frac{\Lambda_m}{\Lambda_m^{\infty}} = \frac{16.5}{390.0} = 0.0423 \)
(iii) Dissociation constant \( K_a = \frac{C\alpha^2}{1 - \alpha} = \frac{0.01 \times (0.0423)^2}{1 - 0.0423} = \frac{0.01 \times 0.001789}{0.9577} = 1.87 \times 10^{-5}\text{ mol L}^{-1} \).

Teacher's Note:
a) Apply Kohlrausch's law to find limiting molar conductivity of weak acid.
b) Use Ostwald's dilution law expression \( K_a = C\alpha^2/(1-\alpha) \) to calculate dissociation constant.

(b) (i) Calculate the e.m.f. of the following cell reaction at \( 298\text{ K} \):
\( \text{Mg}_{\text{(s)}} + \text{Cu}^{2+}_{\text{(0.0001 M)}} → \text{Mg}^{2+}_{\text{(0.001 M)}} + \text{Cu}_{\text{(s)}} \)
The standard potential (\( E^0 \)) of the cell is \( 2.71\text{ V} \). [2 Marks]

Answer:
Number of electrons transferred \( n = 2 \)
Using Nernst equation:
\( E_{\text{cell}} = E^0_{\text{cell}} - \frac{0.0591}{n} \log\frac{[\text{Mg}^{2+}]}{[\text{Cu}^{2+}]} \)
\( E_{\text{cell}} = 2.71 - \frac{0.0591}{2} \log\frac{0.001}{0.0001} \)
\( E_{\text{cell}} = 2.71 - 0.02955 \log(10) \)
\( E_{\text{cell}} = 2.71 - 0.02955 = 2.68045\text{ V} \).

Teacher's Note:
a) Substitute concentration values correctly into the logarithmic term of Nernst equation.
b) Note that \( \log(10) = 1 \).

(ii) The solubility product (\( K_{sp} \)) of \( \text{BaSO}_4 \) is \( 1.5 \times 10^{-9} \). Calculate the solubility of barium sulphate in pure water and in \( 0.1\text{ M BaCl}_2 \). [2 Marks]

Answer:
1. In pure water: Let solubility be \( s \).
\( K_{sp} = [Ba^{2+}][SO_4^{2-}] = s^2 \)
\( s = \sqrt{1.5 \times 10^{-9}} = \sqrt{15 \times 10^{-10}} = 3.87 \times 10^{-5}\text{ mol L}^{-1} \)
2. In \( 0.1\text{ M BaCl}_2 \): Let solubility be \( s' \).
\( [Ba^{2+}] = 0.1 + s' \approx 0.1\text{ M} \)
\( K_{sp} = [Ba^{2+}][SO_4^{2-}] \implies 1.5 \times 10^{-9} = 0.1 \times s' \)
\( s' = \frac{1.5 \times 10^{-9}}{0.1} = 1.5 \times 10^{-8}\text{ mol L}^{-1} \).

Teacher's Note:
a) In the presence of a common ion (\( \text{Ba}^{2+} \) from \( \text{BaCl}_2 \)), solubility decreases drastically due to the common-ion effect.
b) Neglect \( s' \) alongside 0.1 to simplify calculation.

(c) Explain the following: [2 Marks]
(i) When \( \text{NH}_4\text{Cl} \) and \( \text{NH}_4\text{OH} \) are added to a solution containing both, \( \text{Fe}^{3+} \) and \( \text{Ca}^{2+} \) ions, which ion is precipitated first and why?
(ii) Dissociation of \( \text{H}_2\text{S} \) is suppressed in acidic medium.

Answer:
(i) \( \text{Fe}^{3+} \) is precipitated first as ferric hydroxide because the solubility product (\( K_{sp} \)) of \( \text{Fe(OH)}_3 \) is very low compared to that of calcium hydroxide, and the addition of \( \text{NH}_4\text{Cl} \) and \( \text{NH}_4\text{OH} \) provides sufficient hydroxide ion concentration to exceed the \( K_{sp} \) of \( \text{Fe(OH)}_3 \) first.
(ii) The dissociation of \( \text{H}_2\text{S} \) produces sulfide ions in equilibrium with hydrogen ions. In an acidic medium, excess \( \text{H}^+ \) ions are present, which shifts the equilibrium backward due to the common-ion effect, thereby suppressing its dissociation.

Teacher's Note:
a) Precipitation occurs when ionic product exceeds solubility product.
b) Common-ion effect is responsible for suppressing weak electrolyte dissociation.

SECTION B

Answer any two questions.

Question 5

(a) Write the IUPAC names of the following coordination compounds: [1 Mark]
(i) \( [\text{Cr(NH}_3)_4(\text{H}_2\text{O})_2]\text{Cl}_3 \)
(ii) \( [\text{PtCl}_2(\text{NH}_3)_4][\text{PtCl}_4] \)

Answer:
(i) Tetraamminediaquachromium(III) chloride
(ii) Tetraamminedichloroplatinum(IV) tetrachloroplatinate(II)

Teacher's Note:
a) Ligands are named alphabetically (ammine before aqua before chloro).
b) Determine oxidation state of the central metal properly from counter ions or complex charges.

(b) State the hybridization and magnetic property of \( [\text{Fe(CN)}_6]^{3-} \) ion according to the valence bond theory. [1 Mark]

Answer:
Hybridization: \( \text{d}^2\text{sp}^3 \) (inner orbital complex)
Magnetic property: Paramagnetic (due to 1 unpaired electron)

Teacher's Note:
a) Cyanide is a strong field ligand, causing pairing of electrons in Fe(III) (\( 3\text{d}^5 \)).
b) One unpaired electron remains in the 3d orbitals, resulting in paramagnetism.

(c) (i) What type of isomers are \( [\text{Co(NH}_3)_5\text{Br}]\text{SO}_4 \) and \( [\text{Co(NH}_3)_5\text{SO}_4]\text{Br} \)? Give a chemical test to distinguish between them. [2 Marks]

Answer:
They are ionization isomers.
Chemical test: Dissolve both complexes in water and add barium chloride solution. \( [\text{Co(NH}_3)_5\text{Br}]\text{SO}_4 \) gives a white precipitate of barium sulfate (\( \text{BaSO}_4 \)), whereas \( [\text{Co(NH}_3)_5\text{SO}_4]\text{Br} \) does not give a white precipitate with \( \text{BaCl}_2 \).

Teacher's Note:
a) Ionization isomers yield different ions in solution.
b) Sulfate ions are detected using barium chloride solution.

(ii) Write the structures of optical isomers of the complex ion \( [\text{Co(en)}_2\text{Cl}_2]^+ \). [1 Mark]

Answer:
The cis-isomer of \( [\text{Co(en)}_2\text{Cl}_2]^+ \) shows optical isomerism and exists in two optically active forms (dextro and laevo mirror images), where two chlorine atoms are adjacent to each other. The trans-isomer has a plane of symmetry and is optically inactive.

Teacher's Note:
a) Only the cis-form lacks a plane of symmetry and is optically active.
b) Clearly mention the cis-configuration for optical isomerism.

Question 6

(a) Give balanced chemical equations for the following reactions: [3 Marks]
(i) Fluorine is passed through cold, dilute \( \text{NaOH} \) solution.
(ii) Hydrogen peroxide is treated with acidified \( \text{KMnO}_4 \) solution.
(iii) Sulphuric acid is treated with hydrogen sulphide.

Answer:
(i) \( 2\text{F}_{2\text{(g)}} + 2\text{NaOH}_{\text{(aq)}} (\text{cold, dilute}) → 2\text{NaF}_{\text{(aq)}} + \text{OF}_{2\text{(g)}} + \text{H}_2\text{O}_{\text{(l)}} \)
(ii) \( 2\text{KMnO}_{4\text{(aq)}} + 3\text{H}_2\text{SO}_{4\text{(aq)}} + 5\text{H}_2\text{O}_{2\text{(aq)}} → K_2\text{SO}_4 + 2\text{MnSO}_4 + 8\text{H}_2\text{O} + 5\text{O}_{2\text{(g)}} \)
(iii) \( \text{H}_2\text{SO}_{4\text{(aq)}} + \text{H}_2\text{S}_{\text{(g)}} → \text{SO}_{2\text{(g)}} + 2\text{H}_2\text{O}_{\text{(l)}} + S_{\text{(s)}} \)

Teacher's Note:
a) Fluorine reacts vigorously with NaOH; with cold dilute NaOH, oxygen difluoride is formed.
b) Hydrogen peroxide acts as a reducing agent towards acidified potassium permanganate.

(b) Draw the structure of xenon tetrafluoride molecule and state the hybridization of the central atom and the geometry of the molecule. [2 Marks]

Answer:
Hybridization: \( \text{sp}^3\text{d}^2 \)
Geometry: Square planar (Octahedral with two trans lone pairs)
Structure: Xenon atom at the centre surrounded by four fluorine atoms in a square planar arrangement and two lone pairs perpendicular to the plane.

Teacher's Note:
a) Steric number of \( \text{XeF}_4 \) is 6 (4 bond pairs + 2 lone pairs).
b) Lone pairs occupy axial positions to minimize repulsion, yielding a square planar geometry.

Question 7

(a) Name the important ore of silver. Write all the steps and reactions involved in the Cyanide process for the extraction of silver from its ore. [3 Marks]

Answer:
Important ore of silver: Argentite or Silver Glance (\( \text{Ag}_2\text{S} \)).
Cyanide process steps:
1. Leaching: Finely powdered ore is treated with a dilute solution of sodium cyanide in the presence of air or oxygen to form soluble sodium dicyanoargentate(I).
\( \text{Ag}_2\text{S} + 4\text{NaCN} + \text{O}_{2\text{(air)}} ⇌ 2\text{Na}[\text{Ag(CN)}_2] + \text{Na}_2\text{SO}_4 \)
2. Displacement (Reduction): The soluble complex is treated with electropositive zinc dust to precipitate silver.
\( 2\text{Na}[\text{Ag(CN)}_2] + \text{Zn}_{\text{(s)}} → \text{Na}_2[\text{Zn(CN)}_4] + 2\text{Ag}_{\text{(s)}} \)

Teacher's Note:
a) Mention argentite as the chief ore.
b) Write both leaching and zinc reduction equations correctly with balancing.

(b) Explain the following: [2 Marks]
(i) Why do transition metal ions possess a great tendency to form complexes?
(ii) The paramagnetic character in 3d-transition series elements increases upto Mn and then decreases.

Answer:
(i) Transition metal ions have small ionic sizes, high nuclear charges, and vacant d-orbitals of suitable energy to accept electron pairs from ligands.
(ii) Across the 3d-series from Sc to Mn, the number of unpaired d-electrons increases from 1 to 5, increasing paramagnetism. After Mn, pairing of electrons begins, leading to a decrease in the number of unpaired electrons and hence decreasing paramagnetism.

Teacher's Note:
a) High charge-to-size ratio favors complex formation.
b) Paramagnetic moment depends directly on the number of unpaired electrons.

SECTION C

Answer any two questions.

Question 8

(a) How can the following conversions be brought about: [1 Mark each for (i), (ii), (iii); 2 Marks for (iv)]
(i) Glycerol to formic acid
(ii) Chlorobenzene to phenol
(iii) Diethyl ether to ethanol
(iv) Phenol to aniline.

Answer:
(i) Glycerol is heated with oxalic acid at \( 110^{\circ}\text{C} \) to form glycerol monoxalate, which decomposes to give formin and then formic acid on further heating with water.
(ii) Chlorobenzene is heated with aqueous NaOH at \( 623\text{ K} \) and \( 300\text{ atm} \) (Dow's process) followed by acidification to yield phenol.
(iii) Diethyl ether is treated with hot concentrated hydroiodic acid (HI) to form ethyl iodide, which on hydrolysis with aqueous KOH yields ethanol.
(iv) Phenol is heated with zinc dust to form benzene. Benzene is nitrated with nitrating mixture to give nitrobenzene, which is then reduced with \( \text{Sn} + \text{HCl} \) followed by alkali treatment to give aniline.

Teacher's Note:
a) Mention exact reagents and conditions for each step.
b) Multi-step conversions like phenol to aniline require passing through benzene and nitrobenzene intermediates.

(b) (i) How is iodoform prepared from ethanol? Give balanced equation. [1 Mark]

Answer:
Ethanol is heated with iodine and sodium hydroxide solution (haloform reaction).
\( \text{CH}_3\text{CH}_2\text{OH} + 4\text{I}_2 + 6\text{NaOH} → \text{CHI}_{3\text{(s)}} + \text{HCOONa} + 5\text{NaI} + 5\text{H}_2\text{O} \)

Teacher's Note:
a) Ethanol contains the \( \text{CH}_3\text{CH(OH)}- \) group, so it responds positively to the iodoform test.
b) Iodoform precipitates as a yellow solid.

(ii) What will be the product formed when chlorobenzene is heated with sodium metal in the presence of dry ether? [1 Mark]

Answer:
Diphenyl (or Biphenyl) is formed along with sodium chloride (Fittig reaction).
\( 2\text{C}_6\text{H}_5\text{Cl} + 2\text{Na} \xrightarrow{\text{dry ether}} \text{C}_6\text{H}_5-\text{C}_6\text{H}_5 + 2\text{NaCl} \)

Teacher's Note:
a) This is the Fittig reaction for coupling aryl halides.
b) Do not confuse with Wurtz reaction which uses alkyl halides.

(c) Identify the compounds A, B, C, D, E and F: [3 Marks]
\( \text{CH}_3\text{COCH}_3 \xrightarrow{\text{Conc. }\text{HNO}_3} \text{A} \xrightarrow{\text{SOCl}_2} \text{B} \xrightarrow{\text{NH}_3} \text{C} \xrightarrow{\text{LiAlH}_4} \text{D} \)
\( \text{A} \xrightarrow{\text{HNO}_2} \text{E} \xrightarrow{\text{CH}_3\text{COCl}} \text{F} \)
*(Note: As per standard reaction sequence of acetone oxidation / related pathways or specific carbon acids)*

Answer:
A: Acetic acid (\( \text{CH}_3\text{COOH} \))
B: Acetyl chloride (\( \text{CH}_3\text{COCl} \))
C: Acetamide (\( \text{CH}_3\text{CONH}_2 \))
D: Ethylamine (\( \text{CH}_3\text{CH}_2\text{NH}_2 \))
E: Nitrous acid reaction intermediate / remains acetic acid or similar derivative (or as per standard sequence, let's trace: A is \( \text{CH}_3\text{COOH} \), E is \( \text{CH}_3\text{COOH} \) with \( \text{HNO}_2 \) or similar)
F: Acetic anhydride (\( (\text{CH}_3\text{CO})_2\text{O} \)) or acetyl derivative.

Teacher's Note:
a) Oxidation of acetone with conc. HNO3 yields acetic acid as compound A.
b) Follow each reagent step systematically to derive functional group changes.

Question 9

(a) Give balanced equations for the following name reactions: [3 Marks]
(i) Reimer-Tiemann reaction.
(ii) Rosenmund reduction.
(iii) Hoffmann's degradation reaction

Answer:
(i) Reimer-Tiemann reaction:
\( \text{C}_6\text{H}_5\text{OH} + \text{CHCl}_3 + 3\text{NaOH} → \text{C}_6\text{H}_4(\text{OH})(\text{CHO}) + 3\text{NaCl} + 2\text{H}_2\text{O} \)
(ii) Rosenmund reduction:
\( \text{CH}_3\text{COCl} + \text{H}_2 \xrightarrow{\text{Pd/BaSO}_4} \text{CH}_3\text{CHO} + \text{HCl} \)
(iii) Hoffmann's degradation reaction:
\( \text{CH}_3\text{CONH}_2 + \text{Br}_2 + 4\text{NaOH} → \text{CH}_3\text{NH}_2 + \text{Na}_2\text{CO}_3 + 2\text{NaBr} + 2\text{H}_2\text{O} \)

Teacher's Note:
a) Reimer-Tiemann introduces a formyl group into phenol using chloroform and alkali.
b) Rosenmund reduction converts acid chlorides to aldehydes using poisoned palladium catalyst.

(b) Give one chemical test to distinguish between the following pairs of compounds: [3 Marks]
(i) Ethylamine and diethylamine.
(ii) Acetaldehyde and benzaldehyde

Answer:
(i) Carbylamine test: Ethylamine (primary amine) when warmed with chloroform and ethanolic KOH gives a foul-smelling isocyanide, whereas diethylamine (secondary amine) does not give this test.
(ii) Iodoform test: Acetaldehyde has a \( \text{CH}_3\text{CO}- \) group and gives a yellow precipitate of iodoform with \( \text{I}_2/\text{NaOH} \), whereas benzaldehyde does not give this test.

Teacher's Note:
a) Primary amines give carbylamine test; secondary and tertiary amines do not.
b) Methyl ketones and acetaldehyde give positive iodoform tests.

(c) (i) Arrange the following compounds in the ascending order of their basic strength and give reasons for your answer:
Methylamine, Aniline, Ethylamine, Diethyl ether
(ii) Name the monomers and the type of polymerization in each of the following polymers:
(a) Polyester
(b) Bakelite [2 Marks each]

Answer:
(i) Ascending order of basic strength: Diethyl ether < Aniline < Methylamine < Ethylamine.
Reason: Diethyl ether is neutral/weakly basic oxygen compound; Aniline is less basic because the lone pair on nitrogen is delocalized into the benzene ring; Ethylamine is more basic than methylamine due to greater +I effect of the ethyl group enhancing electron density on nitrogen.
(ii) (a) Polyester: Monomers are Ethylene glycol and Terephthalic acid; Type of polymerization: Condensation polymerization (step-growth).
(b) Bakelite: Monomers are Phenol and Formaldehyde; Type of polymerization: Condensation polymerization.

Teacher's Note:
a) Aliphatic amines are more basic than aromatic amines due to resonance stabilization in aniline.
b) Identify condensation polymers by the elimination of small molecules like water during formation.

Question 10

(a) An organic compound A with molecular formula \( \text{C}_2\text{H}_7\text{N} \) on reaction with nitrous acid gives a compound B. B on controlled oxidation gives compound C. C reduces Tollen's reagent to give silver mirror and D. B reacts with D in the presence of concentrated sulphuric acid to give sweet smelling compound E. Identify A, B, C, D and E. Give the reaction of C with ammonia. [3 Marks]

Answer:
A: Ethylamine (\( \text{CH}_3\text{CH}_2\text{NH}_2 \))
B: Ethanol (\( \text{CH}_3\text{CH}_2\text{OH} \))
C: Acetaldehyde (\( \text{CH}_3\text{CHO} \))
D: Acetic acid (\( \text{CH}_3\text{COOH} \))
E: Ethyl acetate (\( \text{CH}_3\text{COOC}_2\text{H}_5 \))
Reaction of C with ammonia: Acetaldehyde reacts with ammonia to form acetaldehyde ammonia addition product.
\( \text{CH}_3\text{CHO} + \text{NH}_3 → \text{CH}_3\text{CH(OH)NH}_2 \)

Teacher's Note:
a) Trace reactions step-by-step from formula \( \text{C}_2\text{H}_7\text{N} \) representing ethylamine.
b) Esterification of ethanol (B) and acetic acid (D) yields ethyl acetate (E).

(b) Give balanced equations for the following reactions: [4 Marks]
(i) How will you convert ethyl amine to methyl amine?
(ii) What is the effect of denaturation on the structure of proteins?
(iii) Name the nitrogen base residues present in DNA.

Answer:
(i) Conversion of ethylamine to methylamine:
Ethylamine + \( \text{HNO}_2 → \text{Ethanol} \xrightarrow{\text{Oxidation}} \text{Acetic acid} \xrightarrow{\text{NaOH/CaO, }\Delta} \text{Methane} \xrightarrow{\text{Cl}_2/\text{U.V.}} \text{Chloromethane} \xrightarrow{\text{NH}_3} \text{Methylamine} \)
(ii) Denaturation disrupts the secondary and tertiary structures of proteins by unfolding the polypeptide chain and breaking hydrogen bonds, while the primary structure remains intact, causing loss of biological activity.
(iii) Nitrogen bases in DNA: Adenine (A), Guanine (G), Cytosine (C), and Thymine (T).

Answer:
(i) \( \text{C}_6\text{H}_5\text{NH}_2 + \text{HNO}_2 + \text{HCl} \xrightarrow{0-5^{\circ}\text{C}} \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + 2\text{H}_2\text{O} \)
(ii) \( \text{CH}_3\text{COCl} + \text{C}_2\text{H}_5\text{OH} → \text{CH}_3\text{COOC}_2\text{H}_5 + \text{HCl} \)
(iii) \( 6\text{HCHO} + 4\text{NH}_3 → (\text{CH}_2)_6\text{N}_4 \text{ (Urotropine)} + 6\text{H}_2\text{O} \)

Teacher's Note:
a) Diazotization of aniline occurs strictly between \( 0^{\circ}\text{C} \) and \( 5^{\circ}\text{C} \).
b) Formaldehyde reacts with ammonia to form hexamethylenetetramine (Urotropine).