ISC Class 12 Chemistry Board Exam Question Paper 2023 with Solutions

ISC Class 12 Chemistry Board Exam Question Paper with Solutions

SECTION - A

Question 1

(A) Fill in the blanks by choosing the appropriate word(s) from those given in the brackets: [stable, low, aldehyde, unstable, 6, 4, ethane, Clemmensen's, 2, 3, carboxylic acid, high, propane, Rosenmund's]

(i) The primary alcohols are easily oxidized first into ______ and then into ________. [1 Mark]

Answer: aldehyde, carboxylic acid

Teacher's Note:
a) Primary alcohols undergo oxidation to yield aldehydes as intermediate products, which upon further oxidation produce carboxylic acids.
b) Ensure correct spelling of functional groups while writing chemical conversions.

(ii) The intermediate activated complex in a chemical reaction is highly _______ due to _______ energy. [1 Mark]

Answer: unstable, high

Teacher's Note:
a) The transition state or activated complex possesses maximum potential energy along the reaction coordinate, making it highly unstable.
b) Remember that stability is inversely proportional to energy in chemical systems.

(iii) The coordination number and oxidation state of the complex K₄[Fe(CN)₆] are ______ and ______ respectively. [1 Mark]

Answer: 6, 2

Teacher's Note:
a) Coordination number is determined by the number of coordinate bonds formed by ligands with the central metal atom (here, six CN^- ligands).
b) Oxidation state of Fe is calculated as x + 6(-1) = -4, giving x = +2.

(iv) Propanone on reaction with zinc-amalgam in presence of conc. HCl gives _______ and the reaction is known as ________ reduction. [1 Mark]

Answer: propane, Clemmensen's

Teacher's Note:
a) Clemmensen reduction is used to convert carbonyl groups (aldehydes and ketones) into corresponding alkanes using zinc amalgam and concentrated hydrochloric acid.
b) Do not confuse Clemmensen reduction with Wolff-Kishner reduction, which uses hydrazine and a base.

(B) Select and write the correct alternative from the choices given below:

(i) The reaction of a primary amine with chloroform and ethanolic KOH is called: [1 Mark]
(a) Carbylamine reaction
(b) Kolbe's reaction
(c) Reimer-Tiemann reaction
(d) Wurtz-Fitting reaction

Answer: (a) Carbylamine reaction

Aliphatic or aromatic primary amines on heating with chloroform give foul smelling products called isocyanides or carbylamines and the reaction is known as carbylamine reaction.

Teacher's Note:
a) This reaction is a specific chemical test used exclusively for the detection of primary amines.
b) Secondary and tertiary amines do not give this test.

(ii) Which one of the following statements is TRUE for the galvanic cell? [1 Mark]
(a) Electrons flow from copper electrode to zinc electrode.
(b) Current flows from zinc electrode to copper electrode
(c) Cations move towards copper electrode.
(d) Cations move towards zinc electrode.

Answer: (c) Cations move towards copper electrode.

In Galvanic cell, following reaction occurs:
\( Zn + Cu^{2+} \to Zn^{2+} + Cu \)
At anode: \( Zn \to Zn^{2+} + 2e^{-} \)
At cathode: \( Cu^{2+} + 2e^{-} \to Cu \)
In this cell, electrons flow from anode to cathode i.e. zinc electrode to copper electrode whereas current flows from copper electrode to zinc electrode in an external circuit. And cations i.e., \( Cu^{2+} \) ions move towards the copper electrode.

Teacher's Note:
a) In any galvanic cell, oxidation takes place at the anode (negative terminal) and reduction takes place at the cathode (positive terminal).
b) Salt bridge maintains electrical neutrality by allowing cations to migrate towards the cathode compartment.

(iii) Which one of the following compounds is diamagnetic and colourless? [1 Mark]
(a) K₂Cr₂O₇
(b) ZnSO₄
(c) KMnO₄
(d) Cr₂(SO₄)₃

Answer: (b) ZnSO₄

\( ZnSO_{4} \) is a diamagnetic and colourless compound, because here \( Zn \) is in +2 oxidation state i.e., its electronic configuration will be: [Ar]\( 3d^{10} \). Thus, \( Zn^{2+} \) has no unpaired electrons and it is colourless.

Teacher's Note:
a) Presence of unpaired electrons causes d-d transitions and paramagnetism; fully filled d-orbitals result in colourless and diamagnetic species.
b) Always check the oxidation state and d-electron count of the transition or inner transition metal ion.

(iv) For a first order reaction, the half-life period (\( t_{1/2} \)) is: [1 Mark]
(a) Proportional to the initial concentration.
(b) Inversely proportional to the initial concentration.
(c) Proportional to the square root of the initial concentration.
(d) Independent of the initial concentration.

Answer: (d) Independent of the initial concentration.

For a first order reaction, half-life is independent of the initial concentration of the reactant.
\( t_{1/2} = \frac{0.693}{k} \)
Hence, option (D) is the correct answer.

Teacher's Note:
a) The half-life of a zero-order reaction is directly proportional to the initial concentration.
b) For a first-order reaction, half-life remains constant throughout the reaction course.

(C) Match the following: [4 Marks]

Answer:
(i) Phenol -> (c) Azo dye
(ii) EDTA -> (d) Hexadentate ligand
(iii) Ideal solution -> (a) Hexane + heptane
(iv) Insulin -> (b) Globular protein

Teacher's Note:
a) EDTA forms six coordinate bonds with a central metal ion, making it a classic hexadentate ligand.
b) Solutions of structurally similar liquids like hexane and heptane form nearly ideal solutions.

(D)

(i) Assertion: If a solution contains both H⁺ and Na⁺ ions, the H⁺ ions are reduced first at cathode.
Reason: Cations with higher E⁰ value are reduced first at cathode.
(a) Both Assertion and Reason are true, and Reason is the correct explanation for Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation for Assertion.
(c) Assertion is true but Reason is false.
(d) Assertion is false but Reason is true. [1 Mark]

Answer: (a) Both Assertion and Reason are true, and Reason is the correct explanation for Assertion.

Reactions at cathode:
\( H^{+} + e^{-} \to \frac{1}{2} H_{2}; E^{0} = 0.0\text{ V} \)
\( Na^{+} + e^{-} \to Na; E^{0} = -2.71\text{ V} \)
The cations having higher positive reduction potential will be reduced first at the cathode, that's why H⁺ reduced first at the cathode.

Teacher's Note:
a) A species with a more positive standard reduction potential acts as a better oxidizing agent and gets reduced more easily.
b) Hydrogen ion discharge potential is favourable compared to alkali metal cations in aqueous electrolytic cells.

(ii) Assertion: Addition of bromine water to 1-butene gives two optical isomers.
Reason: The product formed contains two asymmetric carbon atoms.
(a) Both Assertion and Reason are true, and Reason is the correct explanation for Assertion.
(b) Both Assertion and Reason are true, but Reason is not correct explanation for Assertion.
(c) Assertion is true but Reason is false.
(d) Assertion is false but Reason is true. [1 Mark]

Answer: (c) Assertion is true but Reason is false.

Addition of bromine water to 1-butene gives 1,2-dibromobutane which contains only one chiral (asymmetric) carbon atom, leading to a pair of enantiomers (two optical isomers).
Hence, the assertion is true, but reason is false.

Teacher's Note:
a) 1,2-dibromobutane formed from 1-butene has only carbon-2 as chiral: \( CH_{3}-CH_{2}-CH(Br)-CH_{2}Br \).
b) Compounds with one chiral center always exist as optically active d and l isomers.

SECTION - B

Question 2

Calculate the mass of ascorbic acid (molecular mass = 176 g/mol) that should be dissolved in 155 g of acetic acid to cause a depression of freezing point by 1.15 K. Assume that ascorbic acid does not dissociate or associate in the solution.
(K_f for acetic acid = 3.9 K kg/mol) [2 Marks]

Answer:
Depression in freezing point \( \Delta T_{f} = 1.15\text{ K} \)
As we know, \( \Delta T_{f} = i K_{f} m \)
\( \Delta T_{f} = 1 \times 3.9 \times \frac{x}{176 \times 0.155} \)
\( 1.15 = 3.9 \times \frac{x}{27.28} \)
\( x = \frac{1.15 \times 176 \times 0.155}{3.9} = 8.044\text{ g} \)
Thus, mass of ascorbic acid will be 8.044 g.

Teacher's Note:
a) Ensure mass of solvent is converted to kilograms when applying the molality formula.
b) The van't Hoff factor (i) is taken as 1 since the solute neither associates nor dissociates.

Question 3

Given a reason for the following:
(i) Cu²⁺ salts are paramagnetic while Cu⁺ salts are diamagnetic.
(ii) Mn²⁺ compounds are more stable than Fe²⁺ compounds. [2 Marks]

Answer:
(i) \( Cu^{2+} \) has one unpaired electron (\( [Ar]3d^{9} \)). Therefore, it is paramagnetic. \( Cu^{+} \) has no unpaired electrons (\( [Ar]3d^{10} \)). Therefore, it is diamagnetic.
(ii) \( Mn^{2+} \) configuration: \( 1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3p^{6}, 3d^{5} \). \( Fe^{2+} \) configuration: \( 1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3p^{6}, 3d^{6} \). \( Mn^{2+} \) compounds are more stable due to half-filled d-orbitals. \( Fe^{2+} \) compounds are comparatively less stable as they have six electrons in their 3d-orbitals. So, they tend to lose one electron (form \( Fe^{3+} \)) and get stable \( 3d^{5} \) configuration.

Teacher's Note:
a) Half-filled and completely filled electronic configurations impart extra stability due to symmetrical distribution of electrons and high exchange energy.
b) Magnetic moment formula \( \mu = \sqrt{n(n+2)} \) Bohr Magnetons is useful for verifying paramagnetism where n is the number of unpaired electrons.

Question 4

Given chemical equations for each of the following:
(i) Ethyl chloride is treated with aqueous KOH solution.
(ii) Chlorobenzene is treated with ammonia at 573 K and high pressure. [2 Marks]

Answer:
(i) Ethyl chloride undergoes a nucleophilic substitution reaction when treated with aqueous KOH solution and gives ethanol as a major product.
\( C_{2}H_{5}Cl + \text{aq. KOH} \to C_{2}H_{5}OH + KCl \)
(ii) Chlorobenzene on reaction with ammonia at 573 K and high pressure gives aniline as a major product.
\( C_{6}H_{5}Cl + NH_{3} \to C_{6}H_{5}NH_{2} + HCl \)

Teacher's Note:
a) Aqueous KOH provides OH^- ions which act as strong nucleophiles for aliphatic nucleophilic substitution.
b) Chlorobenzene requires drastic conditions (high temperature and pressure) with ammonia in the presence of catalysts like Cu₂O to undergo nucleophilic substitution due to partial double bond character of the C-Cl bond.

Question 5

State one reason for each of the following:
(i) Alkylamine is soluble in water whereas arylamine is insoluble in water.
(ii) Methylamine is a stronger base than methyl alcohol. [2 Marks]

Answer:
(i) Lower aliphatic alkylamine is soluble in water owing to their potential to form intermolecular hydrogen bonds with water. On the other hand, arylamine does not undergo hydrogen bonding because of the presence of the benzene ring which is highly hydrophobic. Therefore, arylamine is insoluble in water.
(ii) Methylamine is a stronger base than methyl alcohol. This is because nitrogen in methylamine is less electronegative than oxygen in methanol. So, methylamine can easily donate a lone pair of electrons to a proton of an acid.

Teacher's Note:
a) Basicity depends on the ease of donation of the lone pair of electrons; nitrogen has lower electronegativity than oxygen, making its lone pair more readily available.
b) Extent of hydration and hydrophobic bulk of the aryl group determine amine solubility in water.

Question 6

Calculate the emf of the following cell at 298 K.
Cu / Cu²⁺ (0.025 M) // Ag⁺ (0.005 M) / Ag
Given \( E^{0}_{Cu^{2+}/Cu} = 0.34\text{ V}, E^{0}_{Ag^{+}/Ag} = 0.80\text{ V} \), 1 Faraday = 96500 C mol^-1 [2 Marks]

Answer:
Cell reaction: \( Cu + 2Ag^{+} \to 2Ag + Cu^{2+} \)
\( E^{0}_{cell} = E^{0}_{cathode} - E^{0}_{anode} = 0.80\text{ V} - 0.34\text{ V} = 0.46\text{ V} \)
According to Nernst equation:
\( E_{cell} = E^{0}_{cell} - \frac{0.0591}{n} \log \frac{[Cu^{2+}]}{[Ag^{+}]^{2}} \)
\( E_{cell} = 0.46 - \frac{0.0591}{2} \log \frac{0.025}{(0.005)^{2}} \)
\( E_{cell} = 0.46 - \frac{0.0591}{2} \log \frac{0.025}{0.000025} \)
\( E_{cell} = 0.46 - \frac{0.0591}{2} \log (1000) \)
\( E_{cell} = 0.46 - \frac{0.0591}{2} \times 3 \)
\( E_{cell} = 0.46 - 0.08865 = 0.371\text{ V} \)

Teacher's Note:
a) Ensure the number of electrons transferred (n) is correctly identified as 2 based on the balanced overall redox equation.
b) Pay close attention to stoichiometric coefficients when writing the concentration quotient for the Nernst equation.

Question 7

Complete and balance the following chemical equations:
(i) \( KMnO_{4} + H_{2}SO_{4} + KI \to \) _____ + _______ + _______ + _______
(ii) \( K_{2}Cr_{2}O_{7} + H_{2}SO_{4} + H_{2}S \to \) _______ + _______ + _______ + ________ [2 Marks]

Answer:
(i) \( 2KMnO_{4} + 8H_{2}SO_{4} + 10KI \to 2MnSO_{4} + 6K_{2}SO_{4} + 5I_{2} + 8H_{2}O \)
(ii) \( K_{2}Cr_{2}O_{7} + 4H_{2}SO_{4} + 3H_{2}S \to K_{2}SO_{4} + Cr_{2}(SO_{4})_{3} + 3S + 7H_{2}O \)

Teacher's Note:
a) Acidified potassium permanganate oxidizes potassium iodide to iodine gas and gets reduced to manganous sulfate.
b) Acidified potassium dichromate oxidizes hydrogen sulfide to free sulfur and gets reduced to green chromium(III) sulfate.

Question 8

(i) How will the following be obtained? (Give chemical equation)
a) Ethanol from Grignard's reagent.
b) Diethyl ether from sodium ethoxide.

OR

(ii) An organic compound [A] \( C_{2}H_{6}O \), on heating with conc. \( H_{2}SO_{4} \) at 413 K gives a neutral compound [B] \( C_{4}H_{10}O \). Compound [B] on treatment with \( PCl_{5} \) gives a product, which on subsequent treatment with KCN yields compound [C] \( C_{3}H_{5}N \). [C] on hydrolysis gives an acid [D] \( C_{3}H_{6}O_{2} \). Identify the compounds [A], [B], [C] and [D]. [2 Marks]

Answer:
(i) a) Reaction of formaldehyde (methanal) with methyl magnesium bromide followed by acid hydrolysis gives ethanol:
\( HCHO + CH_{3}MgBr \to CH_{3}-CH_{2}-O-MgBr \)
\( CH_{3}-CH_{2}-O-MgBr + H_{2}O \to CH_{3}CH_{2}OH + Mg(OH)Br \)
b) Preparation of Diethyl ether by Williamson's ether synthesis:
\( CH_{3}-CH_{2}-O^{-}Na^{+} + Br-CH_{2}-CH_{3} \xrightarrow{\Delta} CH_{3}-CH_{2}-O-CH_{2}-CH_{3} + NaBr \)

OR

(ii) Identification of compounds:
[A] = Ethanol (\( C_{2}H_{5}OH \))
[B] = Diethyl ether (\( C_{2}H_{5}-O-C_{2}H_{5} \))
[C] = Propanenitrile / Ethyl cyanide (\( CH_{3}CH_{2}CN \))
[D] = Propanoic acid (\( CH_{3}CH_{2}COOH \))

Teacher's Note:
a) Williamson synthesis involves nucleophilic substitution of an alkyl halide by an alkoxide ion.
b) Hydrolysis of cyanides (nitriles) yields corresponding carboxylic acids with the same number of carbon atoms plus one carbon from the cyanide group.

Question 9

The osmotic pressure of blood at \( 37^{\circ}C \) is 8.21 atm. How much glucose in grams should be used per litre of aqueous solution for an intravenous injection so that it is isotonic with blood? (Molecular weight of glucose = 180 g/mol) [2 Marks]

Answer:
\( \pi = 8.21\text{ atm}, T = 273 + 37 = 310\text{ K} \)
Isotonic solutions are solutions having equal osmotic pressure.
We know, \( \pi = i C R T \)
\( 8.21 = \frac{m}{M \times 1} \times 0.082 \times 310 \)
\( 8.21 = \frac{m}{180 \times 1} \times 0.082 \times 310 \)
\( m = \frac{8.21 \times 180}{0.082 \times 310} = 58.13\text{ g} \)

Teacher's Note:
a) Temperature must always be converted from Celsius to Kelvin by adding 273.
b) The gas constant R value of \( 0.0821\text{ L atm K}^{-1}\text{mol}^{-1} \) is appropriate when osmotic pressure is given in atmospheres.

Question 10

An aromatic carboxylic acid [A] which readily sublimes on heating, produces compound [B] on treatment with \( PCl_{5} \). Compound [B], when reduced in the presence of Pd catalyst over \( BaSO_{4} \) poisoned by sulphur in xylene solution gives compound [C]. When compound [C] is condensed in the presence of alcoholic KCN, it gives compound [D]. (Molecular formula of compound [D] is \( C_{14}H_{12}O_{2} \))
Identify the compounds [A], [B], [C] and [D]. [2 Marks]

Answer:
[A] = Benzoic acid (\( C_{6}H_{5}COOH \))
[B] = Benzoyl chloride (\( C_{6}H_{5}COCl \))
[C] = Benzaldehyde (\( C_{6}H_{5}CHO \))
[D] = Benzoin (\( C_{6}H_{5}-CH(OH)-CO-C_{6}H_{5} \))

Teacher's Note:
a) Rosenmund reduction converts an acid chloride into corresponding aldehyde using poisoned palladium catalyst.
b) Benzoin condensation occurs when benzaldehyde is treated with alcoholic potassium cyanide.

Question 11

State a reason for each of the following:
(i) \( La(OH)_{3} \) is more basic than \( Lu(OH)_{3} \).
(ii) Transition elements and their compounds act as catalyst. [2 Marks]

Answer:
(i) The basic strength of hydroxides is \( La(OH)_{3} > Lu(OH)_{3} \). Due to lanthanide contraction, size of \( M^{3+} \) ions decrease from La to Lu. Thus, there is an increase in the covalent character of \( Lu - OH \) bond according to Fajan's rules (smaller cation has higher polarizing power, increasing covalency).
(ii) Transition metals and their compounds function as catalysts because of their ability to show variable oxidation states and to form complexes.

Teacher's Note:
a) Lanthanide contraction causes a steady decrease in atomic and ionic radii with increasing atomic number across the lanthanide series.
b) Variable oxidation states enable transition metals to easily form intermediate complexes and provide an alternative reaction path with lower activation energy.

SECTION C - 21 MARKS

Question 12

20% of a first order reaction is completed in five minutes. How much time will the 60% reaction take to complete? Calculate the half-life period (\( t_{1/2} \)) for the above reaction. [3 Marks]

Answer:
In 5 minutes, the reaction is 20% complete.
\( [A]_{0} = 100 \) and \( [A] = 100 - 20 = 80 \)
\( k = \frac{2.303}{t} \log_{10} \frac{[A]_{0}}{[A]} \)
\( k = \frac{2.303}{5\text{ min}} \log_{10} \frac{100}{80} = \frac{2.303}{5} \times 0.0969 = 0.0446\text{ min}^{-1} \)
Now, the reaction is 60% complete.
\( [A]_{0} = 100 \) and \( [A] = 100 - 60 = 40 \)
\( t = \frac{2.303}{k} \log_{10} \frac{[A]_{0}}{[A]} \)
\( t = \frac{2.303}{0.0446\text{ min}^{-1}} \log_{10} \frac{100}{40} = \frac{2.303}{0.0446} \times 0.3979 = 20.5\text{ min} \)
And, \( t_{1/2} = \frac{0.693}{k} = \frac{0.693}{0.0446} = 15.538\text{ min} \)

Teacher's Note:
a) Always calculate the rate constant (k) first from the initial given data before solving for time or half-life in first-order kinetic problems.
b) Logarithmic values like \( \log 2 = 0.3010 \) and \( \log 2.5 = 0.3979 \) should be memorized for speed.

Question 13

Write the balanced chemical equations for the following name reactions:
(i) Sandmeyer's reaction
(ii) Wurtz reaction
(iii) Finkelstein reaction [3 Marks]

Answer:
(i) Sandmeyer reaction is a type of substitution reaction that is widely used in the production of aryl halides from aryl diazonium salts using cuprous halides:
\( C_{6}H_{5}N_{2}^{+}Cl^{-} \xrightarrow{CuX/HX} C_{6}H_{5}X + N_{2} \)
(ii) Wurtz's reaction is an organic chemical coupling reaction wherein sodium metal is reacted with two alkyl halides in the environment provided by a solution of dry ether in order to form a higher alkane:
\( 2R - X + 2Na \xrightarrow{\text{Dry Ether}} R - R + 2NaX \)
(iii) In the Finkelstein reaction, alkyl iodides are prepared by the reaction of alkyl chlorides or bromides with NaI in dry acetone:
\( 2R - X + 2NaI \xrightarrow{\text{Acetone}} R - I + 2NaX \) (where X = Cl, Br)

Teacher's Note:
a) Sandmeyer reaction proceeds via free radical mechanism facilitated by copper(I) salts.
b) Finkelstein reaction is driven forward by the precipitation of sodium chloride or sodium bromide in dry acetone.

Question 14

(i) Give an example each of reducing sugar and non-reducing sugar.
(ii) What is denaturation of proteins?
(iii) Give an example each of water soluble vitamin and fat soluble vitamin. [3 Marks]

Answer:
(i) Reducing sugar: Maltose
Non-reducing sugar: Sucrose
(ii) When a protein in its native form is subjected to physical change like change in temperature or chemical change like change in pH, the hydrogen bonds are disturbed. Due to this, globules unfold and the helix gets uncoiled and protein loses its biological activity. This is called denaturation of protein.
(iii) Water soluble vitamin: Vitamin C
Fat soluble vitamin: Vitamin A (or D, E, K)

Teacher's Note:
a) Reducing sugars contain free aldehydic or ketonic groups capable of reducing Tollens' or Fehling's reagent.
b) Denaturation disrupts secondary and tertiary structures of proteins while primary structure remains intact.

Question 15

When 2g of benzoic acid is dissolved in 25 g of benzene, it shows depression in freezing point equal to 1.62 K. Molal depression constant (\( K_{f} \)) of benzene is 4.9 K kg mol^-1 and molecular weight of benzoic acid = 122 g/mol. What will be the percentage association of the benzoic acid?
(Benzoic acid forms dimer when dissolved in benzene) [3 Marks]

Answer:
Given:
\( W_{B} = 2\text{ g}, K_{f} = 4.9\text{ K kg mol}^{-1}, W_{A} = 25\text{ g}, \Delta T_{f} = 1.62\text{ K}, M_{B} = 122\text{ g mol}^{-1} \)
Now, \( \Delta T_{f} = i \times K_{f} \times \frac{W_{B}}{M_{B}} \times \frac{1000}{W_{A}} \)
\( 1.62 = i \times 4.9 \times \frac{2}{122} \times \frac{1000}{25} \)
\( i = \frac{1.62 \times 122 \times 25}{4.9 \times 2 \times 1000} = 0.504 \)
As, \( i = 1 + (\frac{1}{n} - 1)\alpha \) (Here, n = 2, since benzoic acid undergoes dimerization in benzene)
\( 0.504 = 1 + (\frac{1}{2} - 1)\alpha \)
\( 0.504 = 1 - 0.5\alpha \)
\( 0.5\alpha = 1 - 0.504 = 0.496 \)
\( \alpha = \frac{0.496}{0.5} = 0.992 = 99.2\% \)
\( \therefore \) Degree of association of benzoic acid = 99.2%

Teacher's Note:
a) Association of solute molecules results in a van't Hoff factor (i) less than 1.
b) For dimerization, the number of particles coalescing into one aggregate is 2, so n = 2 in the association formula.

Question 16

Account for the following:
(i) Phenol is a stronger acid than aliphatic alcohols.
(ii) Ethanol gives iodoform reaction whereas methanol does not give iodoform reaction.
(iii) Ethers should not be distilled to dryness. [3 Marks]

Answer:
(i) When a molecule of phenol loses a proton, it forms a phenoxide ion which is stabilized by resonance as the negative charge is delocalized over the aromatic nucleus. No such resonance is present when an alcohol loses a proton to form an alkoxide ion. Hence, phenols are more acidic than alcohols.
(ii) Iodoform (haloform) reaction is given by ketones having a methyl group attached to the carbonyl carbon or compounds that can be oxidized to a methyl ketone. Ethyl alcohol (\( CH_{3}-CH_{2}-OH \)) can be oxidized to acetaldehyde which contains a methyl keto group. Hence, ethyl alcohol gives the iodoform reaction. But methyl alcohol (\( CH_{3}-OH \)) does not give iodoform as it has only one carbon atom and cannot be oxidized to a methyl ketone.
(iii) Heating ethers can cause the formation of explosive peroxides, especially towards the end of a distillation when a large amount of heat is being passed through a decreasing amount of liquid. For this reason, it is a standing rule in chemistry labs that ethers should never be distilled to dryness.

Teacher's Note:
a) Resonance stabilization of the conjugate base increases acidity significantly.
b) Peroxide formation in ethers poses a severe explosion hazard during distillation.

Question 17

(i) Identify the compounds [A], [B] and [C] in the following reactions:
(a) \( CH_{3}COOH \xrightarrow{NH_{3}/\Delta} [A] \xrightarrow{Br_{2}+KOH} [B] \xrightarrow{CHCl_{3}+NaOH(alc)} [C] \)
(b) \( CH_{3}Br \xrightarrow{KCN} [A] \xrightarrow{LiAlH_{4}} [B] \xrightarrow{HNO_{2}/273K} [C] \)
OR
(ii) How will the following be converted? (Give chemical equation)
(a) Ethyl bromide to ethyl isocyanide.
(b) Aniline to benzene diazonium chloride.
(c) Benzene diazonium chloride to phenol. [3 Marks]

Answer:
(i)
(a) [A] = Acetamide (\( CH_{3}CONH_{2} \)), [B] = Methylamine (\( CH_{3}NH_{2} \)), [C] = Methyl isocyanide / Methyl carbylamine (\( CH_{3}NC \))
(b) [A] = Acetonitrile (\( CH_{3}CN \)), [B] = Ethylamine (\( CH_{3}CH_{2}NH_{2} \)), [C] = Ethanol (\( CH_{3}CH_{2}OH \))

OR

(ii)
(a) Ethyl bromide will react with alcoholic AgCN to form ethyl isocyanide:
\( C_{2}H_{5}Br + AgCN \to C_{2}H_{5}NC + AgBr \)
(b) Diazotization of aniline with \( NaNO_{2} \) and \( HCl \) at 273-278 K gives benzene diazonium chloride:
\( C_{6}H_{5}NH_{2} + NaNO_{2} + 2HCl \to C_{6}H_{5}N_{2}^{+}Cl^{-} + NaCl + 2H_{2}O \)
(c) The diazonium salt on heating in aqueous acidic medium liberates nitrogen and forms phenol as a product:
\( C_{6}H_{5}N_{2}^{+}Cl^{-} + H_{2}O \xrightarrow{\Delta} C_{6}H_{5}OH + HCl + N_{2} \)

Teacher's Note:
a) Hofmann bromamide degradation shortens the carbon chain by one carbon atom, converting an amide into a primary amine.
b) Silver cyanide is a covalent compound, hence yielding isocyanides predominantly due to attack by the carbon-free lone pair on nitrogen, unlike KCN which yields cyanides.

Question 18

A first order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction. [3 Marks]

Answer:
When the first order reaction is 50% completed, the time is equal to the half-life period.
At \( T_{1} = 300\text{ K}, t_{1/2} = 40\text{ minutes} \)
At \( T_{2} = 320\text{ K}, t'_{1/2} = 20\text{ minutes} \)
Since \( t_{1/2} = \frac{0.693}{k} \), rate constants are:
\( k_{1} = \frac{0.693}{40}\text{ min}^{-1} \)
\( k_{2} = \frac{0.693}{20}\text{ min}^{-1} \)
The Arrhenius equation for temperature variation is:
\( \log \left(\frac{k_{2}}{k_{1}}\right) = \frac{E_{a}}{2.303 R} \left[\frac{T_{2} - T_{1}}{T_{1} T_{2}}\right] \)
\( \log \left(\frac{0.693 / 20}{0.693 / 40}\right) = \frac{E_{a}}{2.303 \times 8.314} \left[\frac{320 - 300}{300 \times 320}\right] \)
\( \log(2) = \frac{E_{a}}{19.147} \left[\frac{20}{96000}\right] \)
\( 0.3010 = \frac{E_{a}}{19.147} \times 0.0002083 \)
\( E_{a} = \frac{0.3010 \times 19.147}{0.0002083} = 27668\text{ J mol}^{-1} = 27.67\text{ kJ mol}^{-1} \)

Teacher's Note:
a) Half-life is inversely proportional to the rate constant for a first-order reaction, so \( \frac{k_{2}}{k_{1}} = \frac{t_{1/2}(1)}{t_{1/2}(2)} \).
b) Activation energy units should be stated clearly in joules per mole or kilojoules per mole.

Question 19

(i) Write the chemical equations to illustrate the following name reactions:
(a) Cannizzaro's reaction
(b) HVZ reaction
(c) Aldol condensation
(ii) How will the following be converted? (Give chemical equation)
(a) Acetaldehyde to acetone
(b) Formaldehyde to urotropine [3 Marks]

Answer:
(i)
(a) Cannizzaro's reaction: Aldehydes which do not contain alpha hydrogen when treated with a concentrated solution of an alkali undergo self oxidation-reduction (disproportionation). One molecule is reduced to alcohol and another is oxidized to carboxylic acid salt:
\( 2HCHO + NaOH \to HCOONa + CH_{3}OH \)
(b) HVZ reaction: Carboxylic acids having an alpha-hydrogen are halogenated at the alpha-position on treatment with chlorine or bromine in the presence of small amount of red phosphorus to give alpha-halocarboxylic acids:
\( R-CH_{2}-COOH \xrightarrow[\text{(ii) } H_{2}O]{\text{(i) } X_{2}/\text{Red } P} R-CH(X)-COOH \) (where X = Cl, Br)
(c) Aldol condensation: Aldehydes and ketones having at least one alpha-hydrogen undergo reaction in the presence of dilute alkali as catalyst to form beta-hydroxy aldehydes (aldol) or beta-hydroxy ketones:
\( 2CH_{3}CHO \xrightarrow{\text{dil. NaOH}} CH_{3}-CH(OH)-CH_{2}-CHO \xrightarrow{\Delta} CH_{3}-CH=CH-CHO + H_{2}O \)
(ii)
(a) Acetaldehyde to acetone:
Oxidation of acetaldehyde gives acetic acid, treatment with calcium hydroxide yields calcium acetate, which on dry distillation gives acetone:
\( CH_{3}CHO \xrightarrow{[O]} CH_{3}COOH \xrightarrow{Ca(OH)_{2}} (CH_{3}COO)_{2}Ca \xrightarrow{\Delta} CH_{3}COCH_{3} + CaCO_{3} \)
(b) Formaldehyde to urotropine:
When formaldehyde reacts with ammonia, hexamethylenetetramine (urotropine) is obtained:
\( 6HCHO + 4NH_{3} \to (CH_{2})_{6}N_{4} + 6H_{2}O \)

Teacher's Note:
a) Cannizzaro reaction is characteristic of aldehydes lacking alpha-hydrogens, such as formaldehyde and benzaldehyde.
b) Urotropine is used medicinally as a urinary antiseptic.

Question 20

(i) Name the type of isomerism exhibited by the following pairs of compounds.
(a) [Co(NH₃)₅(ONO)]Cl₂ and [Co(NH₃)₅(NO2)]Cl₂
(b) [Cr(H₂O)₅Cl]Cl₂.H₂O and [Cr(H₂O)₄Cl₂]Cl.2H₂O
(c) [Pt(NH₃)₄Cl₂]Br₂ and [Pt(NH₃)₄Br₂]Cl₂
(ii) Write the IUPAC names of the following complexes:
(a) [Co(NH₃)₄(H₂O)₂]Cl₃
(b) K₂[Ni(CN)₄] [3 Marks]

Answer:
(i)
(a) Linkage Isomers
(b) Hydrate Isomers
(c) Ionization Isomers
(ii) IUPAC Name:
(a) [Co(NH₃)₄(H₂O)₂]Cl₃ : Tetraamminediaquacobalt(III) chloride
(b) K₂[Ni(CN)₄] : Potassium tetracyanonickelate(II)

Teacher's Note:
a) Linkage isomerism arises when an ambidentate ligand (like \( NO_{2}^{-} \) or \( SCN^{-} \) coordinates through different donor atoms.
b) In naming coordination compounds, ligands are listed in alphabetical order before the central metal name, followed by its oxidation number in Roman numerals.

Question 21

(i) The specific conductance of \( 2.5 \times 10^{-4} \) M formic acid is \( 5.25 \times 10^{-5}\text{ ohm}^{-1}\text{ cm}^{-1} \). Calculate its molar conductivity and degree of dissociation.
Given \( \lambda^{0}_{(H)^{+}} = 349.5\text{ ohm}^{-1}\text{ cm}^{2}\text{ mol}^{-1} \) and \( \lambda^{0}_{(HCOO)^{-}} = 50.5\text{ ohm}^{-1}\text{ cm}^{2}\text{ mol}^{-1} \)
(ii) Calculate the time taken to deposit 1.27 g of copper at cathode when a current of 2 amp. is passed through the solution of CuSO₄. (Atomic weight of Cu = 63.5 g mol^-1)

OR

(i) The resistance of a conductivity cell with 0.1 M KCl solution is 200 ohm. When the same cell is filled with 0.02 M NaCl solution, the resistance is 1100 ohm. If the conductivity of 0.1 M KCl solution is \( 0.0129\text{ ohm}^{-1}\text{ cm}^{-1} \). Calculate the cell constant and molar conductivity of 0.02 M NaCl solution.
(ii) The emf (\( E^{0}_{cell} \)) of the following reaction is 0.89 V.
\( 3Sn^{4+} + 2Cr \to 3Sn^{2+} + 2Cr^{3+} \)
Calculate the value of \( \Delta G^{0} \) for the reaction. Predict whether the above reaction will be spontaneous or not. [5 Marks]

Answer:
(i)
We know molar conductivity, \( \lambda_{m} = \frac{1000 \times \text{conductivity}(k)}{\text{concentration}(c)} \)
\( \lambda_{m} = \frac{1000 \times 5.25 \times 10^{-5}}{2.5 \times 10^{-4}} = 210\text{ S cm}^{2}\text{ mol}^{-1} \)
\( \lambda^{0}_{HCOOH} = \lambda^{0}_{H^{+}} + \lambda^{0}_{HCOO^{-}} = 349.5 + 50.5 = 400\text{ S cm}^{2}\text{ mol}^{-1} \)
\( \therefore \alpha = \frac{\lambda_{m}}{\lambda^{0}_{m}} = \frac{210}{400} = 0.525 \) or \( 52.5\% \)
(ii)
Given: Mass of Cu deposited (W) = 1.27 g, Current (I) = 2 A, Molar mass of Cu = \( 63.5\text{ g mol}^{-1} \), 1F = \( 96500\text{ C mol}^{-1} \)
Reaction: \( Cu^{2+} + 2e^{-} \to Cu \)
Electrochemical equivalent \( z = \frac{\text{Atomic weight}}{\text{n-factor} \times F} = \frac{63.5}{2 \times 96500} \)
By Faraday's first law, \( W = z I t \)
\( 1.27 = \frac{63.5 \times 2 \times t}{2 \times 96500} \)
\( t = \frac{1.27 \times 2 \times 96500}{63.5 \times 2} = 1930\text{ seconds} \)

OR

(i)
For 0.1 M KCl: Resistance = \( 200\text{ }\Omega \), \( k = 0.0129\text{ }\Omega^{-1}\text{cm}^{-1} \)
As, \( k = G \times \frac{l}{A} \Rightarrow \frac{1}{R} \times x^* \)
Cell constant \( x^* = k \times R = 0.0129 \times 200 = 2.58\text{ cm}^{-1} \)
For 0.02 M NaCl: R = \( 1100\text{ }\Omega \)
Conductivity \( k = \frac{x^*}{R} = \frac{2.58}{1100}\text{ }\Omega^{-1}\text{cm}^{-1} \)
Molar conductivity \( \lambda_{m} = \frac{k}{c} \times 1000 = \frac{2.58}{1100 \times 0.02} \times 1000 = 117.27\text{ }\Omega^{-1}\text{ cm}^{2}\text{ mol}^{-1} \)
(ii)
\( \Delta G^{0} = -n F E^{0}_{cell} \)
Given \( E^{0} = 0.89\text{ V} \)
Reaction: \( 3Sn^{4+} + 2Cr \to 3Sn^{2+} + 2Cr^{3+} \)
Here n = 6 electrons transferred.
\( \Delta G^{0} = -6 \times 96500 \times 0.89 = -515310\text{ J} = -515.3\text{ kJ} \)
Hence, as the value of standard Gibbs free energy is negative, the reaction will be spontaneous in nature.

Teacher's Note:
a) Cell constant depends only on the cell geometry (distance between electrodes and cross-sectional area) and remains constant for a given conductivity cell.
b) A negative value of standard Gibbs free energy (\( \Delta G^{0} \)) confirms the thermodynamic spontaneity of a redox reaction.