ISC Class 12 Chemistry Board Exam Question Paper 2026 with Solutions

ISC Class 12 Chemistry Board Exam Question Paper with Solutions

SECTION A - 14 MARKS

Question 1

(A) Fill in the blanks by choosing the appropriate word(s) from those given in the brackets: [4×1]
[first, alc. AgCN, primary, decrease, third, increase, zero, Lucas, alc. KCN, KOH, no change, two, Tollen's, AgOH, tertiary]

(i) A mixture of anhydrous ZnCl₂ and conc. HCl is called _________ reagent which shows maximum reactivity with _________ alcohol. [1 Mark]

Answer: Lucas, tertiary.

Teacher's Note:
a) Lucas reagent consists of anhydrous zinc chloride and concentrated hydrochloric acid.
b) Tertiary alcohols react instantaneously with Lucas reagent due to the formation of stable tertiary carbocations.

(ii) Methyl chloride on treatment with _________ forms methyl cyanide, whereas on treatment with _________ it forms methyl isocyanide. [1 Mark]

Answer: alc. KCN, alc. AgCN.

Teacher's Note:
a) Potassium cyanide (KCN) is predominantly ionic and furnishes cyanide ions, leading to the formation of alkyl cyanides.
b) Silver cyanide (AgCN) is more covalent, making the carbon atom less available, so nitrogen attacks to form isocyanides.

(iii) The hydrolysis of ethyl acetate in acidic medium is a pseudo order reaction. However, the molecularity of the reaction is _________. [1 Mark]

Answer: two

Teacher's Note:
a) Molecularity is the number of reacting species colliding simultaneously in an elementary step, which is two for this ester hydrolysis.
b) It is a pseudo-first-order reaction because water is present in large excess and its concentration remains practically constant.

(iv) Dissociation of solute particles leads to a / an _________ in the magnitude of a colligative property and a / an _________ in the value of the observed molecular mass. [1 Mark]

Answer: increase, decrease

Teacher's Note:
a) Colligative properties depend on the total number of solute particles in solution, so dissociation increases their values.
b) Since observed molecular mass is inversely proportional to colligative properties, it decreases when particles dissociate.

(B) Select and write the correct alternative from the choices given below. [7×1]

(i) Which one of the following is a tetrahedral complex as well as diamagnetic in nature? [1 Mark]
(A) [Ni(CN)₄]^2-
(B) [NiCl₄]^2-
(C) [Cu(NH₃)₄]²⁺
(D) [Ni(CO)₄]

Answer: (D) [Ni(CO)₄]

In [Ni(CO)₄], nickel is in oxidation state zero with configuration \( 3d^8 4s^2 \). CO is a strong field ligand, causing pairing of electrons to give \( sp^3 \) hybridization (tetrahedral) and zero unpaired electrons (diamagnetic).

Teacher's Note:
a) [NiCl₄]^2- is tetrahedral but paramagnetic because Cl^- is a weak field ligand.
b) [Ni(CN)₄]^2- is diamagnetic but square planar (\( dsp^2 \) hybridization).

(ii) When H₂S gas is passed through acidified K₂Cr₂O₇ solution, the colour of the solution changes to: [1 Mark]
(A) green.
(B) orange.
(C) black.
(D) red.

Answer: (A) green.

Acidified potassium dichromate oxidizes H₂S to colloidal sulphur (yellow/turbidity) while dichromate ions (\( \text{Cr}_2\text{O}_7^{2-} \), orange) are reduced to chromic ions (\( \text{Cr}^{3+} \), green).

Teacher's Note:
a) The green colour is due to the formation of chromium(III) sulfate.
b) This reaction is commonly used as a confirmatory test for reducing agents.

(iii) Identify the end product [C] in the following sequence of chemical reaction:
\( \text{CH}\equiv\text{CH} \xrightarrow{\text{dil. H}_2\text{SO}_4 / \text{HgSO}_4} [\text{A}] \xrightarrow{[\text{O}]} [\text{B}] \xrightarrow{\text{PCl}_5} [\text{C}] \) [1 Mark]
(A) Acetic anhydride
(B) Acetic acid
(C) Acetyl chloride
(D) Ethyl chloride

Answer: (C) Acetyl chloride

Acetylene hydration gives acetaldehyde ([A] = \( \text{CH}_3\text{CHO} \)), oxidation gives acetic acid ([B] = \( \text{CH}_3\text{COOH} \)), and treatment with PCl₅ gives acetyl chloride ([C] = \( \text{CH}_3\text{COCl} \)).

Teacher's Note:
a) Follow each step of functional group transformation carefully.
b) Carboxylic acids react with PCl₅ to form acyl chlorides, POCl₃, and HCl.

(iv) Rohit is studying the iodoform reaction by using the following compounds:
(P) \( \text{CH}_3\text{OH} \)
(Q) \( \text{CH}_3\text{CHO} \)
(R) \( \text{CH}_3\text{COCH}_3 \)
(S) \( \text{CH}_3\text{CH}_2-\text{CO}-\text{CH}_2-\text{CH}_3 \)
Which of the following combinations will give an iodoform test when the compounds are heated separately with iodine and aqueous NaOH? [1 Mark]
(A) Only (P) and (Q)
(B) Only (Q) and (R)
(C) Only (P) and (S)
(D) Only (Q) and (S)

Answer: (B) Only (Q) and (R)

Compounds containing a \( \text{CH}_3-\text{CO}- \) group or a \( \text{CH}_3-\text{CH(OH)}- \) group give the iodoform test. Acetaldehyde (Q) and acetone (R) both contain the \( \text{CH}_3-\text{CO}- \) group.

Teacher's Note:
a) Methanol (P) and diethyl ketone (S) do not have the required methyl carbonyl group.
b) The iodoform test results in a yellow precipitate of CHI₃ with methyl ketones and acetaldehyde.

(v) Which one of the following nitrogen bases is absent in RNA? [1 Mark]
(A) Adenine
(B) Guanine
(C) Uracil
(D) Thymine

Answer: (D) Thymine

RNA contains adenine, guanine, cytosine, and uracil. Thymine is present in DNA in place of uracil.

Teacher's Note:
a) Thymine is chemically 5-methyluracil.
b) Remember the base pairing rules: A pairs with U in RNA and T in DNA.

(vi) Given below are two statements marked Assertion and Reason. Read the two statements carefully and select the correct option.
Assertion: Phenols are more acidic than aliphatic alcohols.
Reason: Phenoxide ion is stabilised by resonance but alkoxide ion is not stabilised by resonance. [1 Mark]
(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 and Reason is false.
(D) Both Assertion and Reason are false.

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

Phenols dissociate to give phenoxide ions which are stabilized by delocalization of the negative charge into the benzene ring (resonance), whereas alkoxide ions lack such resonance stabilization.

Teacher's Note:
a) Greater stability of the conjugate base shifts equilibrium towards the right, increasing acidity.
b) Aliphatic alcohols form alkoxides where the negative charge is localized on oxygen.

(vii) Given below are two statements marked Assertion and Reason. Read the two statements carefully and select the correct option.
Assertion: The depression of freezing point of 0.1 molal aqueous solution of MgCl₂ is less than that of 0.1 molal aqueous solution of NaCl.
Reason: The number of particles furnished by 0.1 molal MgCl₂ is less than that of 0.1 molal NaCl. [1 Mark]
(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 and Reason is false.
(D) Both Assertion and Reason are false.

Answer: (D) Both Assertion and Reason are false.

MgCl₂ gives 3 particles (\( \text{Mg}^{2+} + 2\text{Cl}^- \)) while NaCl gives 2 particles (\( \text{Na}^+ + \text{Cl}^- \)). Hence, depression of freezing point for MgCl₂ is greater, making both assertion and reason incorrect statements.

Teacher's Note:
a) Colligative properties depend directly on the Van't Hoff factor (i).
b) For MgCl₂, \( i \approx 3 \) (assuming complete dissociation), which is greater than \( i \approx 2 \) for NaCl.

(C) Read the information given below and answer the questions that follow. [3×1]

Kohlrausch observed an interesting pattern between the values of molar conductance at infinite dilution (\( \Lambda^\infty_m \)) for strong electrolytes. It was observed that different pairs of electrolytes having a common cation or anion had almost same difference of \( \Lambda^\infty_m \). On the basis of his observation, he postulated a law known as Kohlrausch's Law of Independent Migration of ions.
The values of molar conductivities at infinite dilution for some cations and anions are as follows:

(i) State the Kohlrausch's Law of Independent Migration of ions. [1 Mark]

Answer: Kohlrausch's Law states that the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anions and cations of the electrolyte.

Teacher's Note:
a) Mathematically, \( \Lambda^\infty_m = \nu_+ \lambda^\infty_+ + \nu_- \lambda^\infty_- \).
b) Each ion contributes to the total molar conductivity independent of the nature of the other ion present.

(ii) Calculate the molar conductance at infinite dilution \( \Lambda^\infty_m \) for BaCl₂. [1 Mark]

Answer:
\( \Lambda^\infty_m(\text{BaCl}_2) = \lambda^\infty_{\text{Ba}^{2+}} + 2\lambda^\infty_{\text{Cl}^-} \)
\( = 127.2 + 2(76.3) = 127.2 + 152.6 = 279.8 \text{ S cm}^2\text{mol}^{-1} \)

Teacher's Note:
a) Multiply the anion molar conductivity by 2 because one molecule of BaCl₂ yields two chloride ions.
b) Units of molar conductivity must be included in the final answer.

(iii) Arrange the values of \( \Lambda^\infty_m \) for CaSO₄ and BaCl₂ in increasing order. [1 Mark]

Answer:
\( \Lambda^\infty_m(\text{CaSO}_4) \) (\( 119.0 + 160.0 = 279.0 \text{ S cm}^2\text{mol}^{-1} \)) < \( \Lambda^\infty_m(\text{BaCl}_2) \) (\( 279.8 \text{ S cm}^2\text{mol}^{-1} \)).
Increasing order: \( \text{CaSO}_4 < \text{BaCl}_2 \)

Teacher's Note:
a) Calculate \( \Lambda^\infty_m \) for CaSO₄ as \( 119.0 + 160.0 = 279.0 \text{ S cm}^2\text{mol}^{-1} \).
b) Compare the calculated values to determine the correct order.

SECTION B - 20 MARKS

Question 2 [2 Marks]

A reaction is of first order with respect to reactant [A] and second order with regard to reactant [B]. What is the effect on rate of reaction when:
(i) concentration of only [B] is increased three times?
(ii) concentration of both [A] and [B] is doubled?

Answer:
Rate law is \( \text{Rate} = k[\text{A}][\text{B}]^2 \).
(i) When [B] is increased 3 times: \( \text{Rate}' = k[\text{A}](3[\text{B}])^2 = 9 \times k[\text{A}][\text{B}]^2 \). The rate increases by 9 times.
(ii) When both [A] and [B] are doubled: \( \text{Rate}' = k(2[\text{A}])(2[\text{B}])^2 = 8 \times k[\text{A}][\text{B}]^2 \). The rate increases by 8 times.

Teacher's Note:
a) Rate depends directly on the concentration terms raised to their stoichiometric coefficients in the rate law.
b) Ensure powers corresponding to the order of each reactant are applied correctly.

Question 3 [2 Marks]

(i) 0.680 g of a compound is dissolved in 15.0 g of benzene and the freezing point of solution is lowered by 1.44°C.
Calculate the experimental molecular mass of the compound.
(\( K_f \) for benzene = \( 5.12 \text{ K kg mol}^{-1} \))
(ii) If the theoretical molecular mass of the compound referred to above is \( 80.5 \text{ g mol}^{-1} \), suggest whether it is undergoing association or dissociation.

Answer:
(i) \( \Delta T_f = \frac{K_f \times W_2 \times 1000}{M_2 \times W_1} \)
\( 1.44 = \frac{5.12 \times 0.680 \times 1000}{M_2 \times 15.0} \)
\( M_2 = \frac{5.12 \times 680}{1.44 \times 15} = \frac{3481.6}{21.6} = 161.18 \text{ g mol}^{-1} \).
(ii) Since the experimental molecular mass (\( 161.18 \text{ g mol}^{-1} \)) is greater than the theoretical molecular mass (\( 80.5 \text{ g mol}^{-1} \)), the compound is undergoing association.

Teacher's Note:
a) Association of molecules decreases the total number of particles, thereby increasing the observed molecular mass.
b) Use proper unit conversions between grams and kilograms for molality calculations.

Question 4 [2 Marks]

Write the IUPAC names of the following complex compounds:
(i) [Pt(NH₃)₄Cl₂]SO₄
(ii) K₂[Ni(CN)₄]

Answer:
(i) Tetraamminedichlorideplatinum(IV) sulfate
(ii) Potassium tetracyanonickelate(II)

Teacher's Note:
a) Ligands are named in alphabetical order before the central metal atom.
b) Oxidation state of the metal is calculated and written in Roman numerals inside parentheses.

Question 5 [2 Marks]

Write chemical equations to convert each of the following:
(i) Methyl magnesium bromide to propan-2-ol
(ii) Phenol to benzene

Answer:
(i) \( \text{CH}_3\text{MgBr} + \text{CH}_3\text{CHO} \rightarrow \text{CH}_3-\text{CH(OMgBr)}-\text{CH}_3 \xrightarrow{\text{H}_2\text{O} / \text{H}^+} \text{CH}_3-\text{CH(OH)}-\text{CH}_3 + \text{Mg(OH)Br} \)
(ii) \( \text{C}_6\text{H}_5\text{OH} + \text{Zn} \xrightarrow{\Delta} \text{C}_6\text{H}_6 + \text{ZnO} \)

Teacher's Note:
a) Grignard reagent reacts with acetaldehyde to give a secondary alcohol after hydrolysis.
b) Reduction of phenol with zinc dust removes the phenolic oxygen to yield benzene.

Question 6 [2 Marks]

Calculate the value of \( \Delta G^\circ \) for the following cell at \( 25^\circ\text{C} \).
\( \text{Zn(s)} \mid \text{Zn}^{2+}(1\text{M}) \parallel \text{Sn}^{2+}(1\text{M}) \mid \text{Sn(s)} \)
Given \( E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76\text{ V}, E^\circ_{\text{Sn}^{2+}/\text{Sn}} = -0.14\text{ V} \)
1 faraday = \( 96,500 \text{ coulombs} \)

Answer:
\( E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = -0.14 - (-0.76) = +0.62\text{ V} \)
\( \Delta G^\circ = -nFE^\circ_{\text{cell}} \)
\( n = 2 \)
\( \Delta G^\circ = -2 \times 96500 \times 0.62 = -119660 \text{ J mol}^{-1} = -119.66 \text{ kJ mol}^{-1} \)

Teacher's Note:
a) Anode is the electrode with lower reduction potential (Zn), and cathode is Sn.
b) Convert the final Gibbs free energy from Joules to kiloJoules by dividing by 1000.

Question 7 [2 Marks]

Write chemical reactions for the following named organic reactions:
(i) Diazotisation reaction
(ii) Sandmeyer's reaction

Answer:
(i) \( \text{C}_6\text{H}_5\text{NH}_2 + \text{NaNO}_2 + 2\text{HCl} \xrightarrow{0-5^\circ\text{C}} \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{NaCl} + 2\text{H}_2\text{O} \)
(ii) \( \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{CuCl} / \text{HCl} \rightarrow \text{C}_6\text{H}_5\text{Cl} + \text{N}_2 + \text{CuCl} \)

Teacher's Note:
a) Temperature must be maintained between 0 to 5°C during diazotisation to prevent decomposition of the diazonium salt.
b) Sandmeyer reaction replaces the diazonium group with chlorine or bromine using cuprous halide.

Question 8 [2 Marks]

A first order reaction is 40% complete in 20 minutes. Calculate the time required for 90% completion of the same reaction.

Answer:
For a first order reaction, \( k = \frac{2.303}{t} \log\left(\frac{[\text{A}]_0}{[\text{A}]}\right) \)
When \( t = 20 \text{ min} \), \( [\text{A}] = 100 - 40 = 60 \):
\( k = \frac{2.303}{20} \log\left(\frac{100}{60}\right) = \frac{2.303}{20} \times 0.2218 = 0.0255 \text{ min}^{-1} \)
For 90% completion, \( [\text{A}] = 100 - 90 = 10 \):
\( t_{90\%} = \frac{2.303}{0.0255} \log\left(\frac{100}{10}\right) = \frac{2.303}{0.0255} \times 1 = 90.3 \text{ minutes} \)

Teacher's Note:
a) First calculate the rate constant \( k \) using the given 40% completion data.
b) Use the calculated \( k \) to find the time required for 90% completion.

Question 9 [2 Marks]

(i) Write chemical tests to distinguish between the following pairs of compounds:
(a) Acetaldehyde and acetone
(b) Acetic acid and benzaldehyde
OR
(ii) An organic compound [A] having molecular formula C₃H₈O gives turbidity with Lucas reagent within five minutes. On heating compound [A] with Cu at 573K, compound [B] is formed. Compound [B] does not reduce Fehling's solution but when heated with iodine and sodium hydroxide, it gives a yellow precipitate of compound [C]. Identify the compounds [A], [B] and [C].
Write the chemical reaction for the conversion of compound [B] to compound [C].

Answer:
(i) (a) Acetaldehyde gives Tollen's test (silver mirror) or Fehling's test (red precipitate), whereas acetone does not.
(b) Acetic acid reacts with NaHCO₃ to evolve brisk effervescence of CO₂ gas, whereas benzaldehyde does not.
OR
(ii) [A] = Propan-2-ol (\( \text{CH}_3-\text{CH(OH)}-\text{CH}_3 \)), [B] = Acetone (\( \text{CH}_3\text{COCH}_3 \)), [C] = Iodoform (\( \text{CHI}_3 \)).
Conversion reaction:
\( \text{CH}_3\text{COCH}_3 + 3\text{I}_2 + 4\text{NaOH} \rightarrow \text{CHI}_3\downarrow + \text{CH}_3\text{COONa} + 3\text{NaI} + 3\text{H}_2\text{O} \)

Teacher's Note:
a) Secondary alcohols give turbidity with Lucas reagent in 5 minutes.
b) Dehydrogenation of secondary alcohol gives a ketone, which gives iodoform test.

Question 10 [2 Marks]

(i) Why is the first ionisation enthalpy of chromium less than that of zinc?
(Atomic Number of Cr = 24 and Zn = 30)
(ii) Why does pink colour of aqueous KMnO₄ solution disappear when warmed with oxalic acid solution in acidic medium?

Answer:
(i) Chromium has a configuration \( 3d^5 4s^1 \), where the 4s electron is easily removed due to stable half-filled d-subshell configuration. Zinc has a fully-filled stable \( 3d^{10} 4s^2 \) configuration with higher nuclear charge, making removal of an electron difficult.
(ii) Permanganate ions (\( \text{MnO}_4^- \), pink) are reduced by oxalic acid to colourless manganese(II) ions (\( \text{Mn}^{2+} \)).

Teacher's Note:
a) Stability of half-filled and fully-filled subshells plays a vital role in ionisation enthalpies.
b) This is a classic redox titration reaction in acidic medium.

Question 11 [2 Marks]

Write chemical equations to convert each of the following:
(i) Chlorobenzene to aniline
(ii) Iodoform to acetylene

Answer:
(i) \( \text{C}_6\text{H}_5\text{Cl} + 2\text{NH}_3 \xrightarrow{\text{Cu}_2\text{O}, 473\text{K}, 60\text{ atm}} \text{C}_6\text{H}_5\text{NH}_2 + \text{NH}_4\text{Cl} \)
(ii) \( 2\text{CHI}_3 + 6\text{Ag} \xrightarrow{\Delta} \text{CH}\equiv\text{CH} + 6\text{AgI} \)

Teacher's Note:
a) Chlorobenzene requires high temperature and pressure with ammonia in presence of cuprous oxide to form aniline.
b) Heating iodoform with silver powder deiodinates it to form acetylene.

SECTION C - 21 MARKS

Question 12 [3 Marks]

Write chemical equations to convert each of the following:
(i) Methanol to ethanol
(ii) Benzene diazonium chloride to phenol
(iii) Sodium ethoxide to ethoxyethane

Answer:
(i) \( \text{CH}_3\text{OH} + \text{PCl}_3 \rightarrow \text{CH}_3\text{Cl} \xrightarrow{\text{KCN}} \text{CH}_3\text{CN} \xrightarrow{\text{H}_2/\text{Ni}} \text{CH}_3\text{CH}_2\text{NH}_2 \xrightarrow{\text{HNO}_2} \text{CH}_3\text{CH}_2\text{OH} \)
(ii) \( \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{H}_2\text{O} \xrightarrow{\text{Warm}} \text{C}_6\text{H}_5\text{OH} + \text{N}_2 + \text{HCl} \)
(iii) \( \text{C}_2\text{H}_5\text{ONa} + \text{C}_2\text{H}_5\text{Cl} \rightarrow \text{C}_2\text{H}_5-\text{O}-\text{C}_2\text{H}_5 + \text{NaCl} \)

Teacher's Note:
a) Step-up conversion involves increasing carbon chain length via cyanide formation.
b) Williamson synthesis is used for preparing ethers from sodium alkoxide and alkyl halide.

Question 13 [3 Marks]

(i) Complete and balance the following reactions:
(a) \( \text{K}_2\text{Cr}_2\text{O}_7 + \text{H}_2\text{SO}_4 + \text{KI} \rightarrow \) _________ + _________ + _________ + _________
(b) \( \text{KMnO}_4 + \text{H}_2\text{SO}_4 + \text{FeSO}_4 \rightarrow \) _________ + _________ + _________ + _________
(ii) Which ion, \( \text{Co}^{2+} \) or \( \text{Zn}^{2+} \), is attracted to a magnetic field? Give reasons.
(Atomic Number of Co = 27 and Zn = 30)

Answer:
(i) (a) \( \text{K}_2\text{Cr}_2\text{O}_7 + 4\text{H}_2\text{SO}_4 + 6\text{KI} \rightarrow \text{Cr}_2(\text{SO}_4)_3 + 3\text{I}_2 + 4\text{K}_2\text{SO}_4 + 7\text{H}_2\text{O} \)
(b) \( 2\text{KMnO}_4 + 8\text{H}_2\text{SO}_4 + 10\text{FeSO}_4 \rightarrow \text{K}_2\text{SO}_4 + 2\text{MnSO}_4 + 5\text{Fe}_2(\text{SO}_4)_3 + 8\text{H}_2\text{O} \)
(ii) \( \text{Co}^{2+} \) is attracted to a magnetic field because it has unpaired electrons (\( 3d^7 \)), making it paramagnetic. \( \text{Zn}^{2+} \) has a fully filled d-subshell (\( 3d^{10} \)) with no unpaired electrons, making it diamagnetic.

Teacher's Note:
a) Balance redox reactions using ion-electron method or oxidation number method.
b) Paramagnetic species are attracted to magnetic fields due to presence of unpaired electrons.

Question 14 [3 Marks]

(i) Give one example each of fibrous protein and globular protein.
(ii) The structures of two amino acids, glycine and alanine, are given below.
Draw the peptide linkage between these two amino acids.
[Figure: Glycine (\( \text{H}_2\text{N}-\text{CH}_2-\text{COOH} \)) and Alanine (\( \text{H}_2\text{N}-\text{CH(CH}_3)-\text{COOH} \))]
(iii) Justify that glucose is an aldose form of sugar, with the help of a chemical reaction.

Answer:
(i) Fibrous protein: Keratin (or Myosin); Globular protein: Insulin (or Albumin).
(ii) Peptide linkage structure:
\( \text{H}_2\text{N}-\text{CH}_2-\text{CO}-\text{NH}-\text{CH(CH}_3)-\text{COOH} \)
(iii) Glucose reacts with hydroxylamine to form an oxime and with hydrogen cyanide to form a cyanohydrin, which confirms the presence of a carbonyl group (\(-\text{CHO}\)), proving it is an aldose:
\( \text{CHO}(\text{CHOH})_4\text{CH}_2\text{OH} + \text{H}_2\text{NOH} \rightarrow \text{CH=NOH}(\text{CHOH})_4\text{CH}_2\text{OH} + \text{H}_2\text{O} \)

Teacher's Note:
a) Fibrous proteins are water-insoluble thread-like structures, whereas globular proteins are soluble folded chains.
b) A peptide bond is formed by the elimination of a water molecule between the carboxyl group of one amino acid and the amino group of another.

Question 15 [3 Marks]

Write the chemical reactions when:
(i) Chlorobenzene is heated with conc. HNO₃ in the presence of conc. H₂SO₄.
(ii) Ethyl alcohol is treated with SOCl₂ in the presence of pyridine.
(iii) Chloroform is slowly oxidized with oxygen in the presence of sunlight.

Answer:
(i) \( \text{C}_6\text{H}_5\text{Cl} + \text{HNO}_3 \xrightarrow{\text{conc. H}_2\text{SO}_4} \text{o-nitrochlorobenzene} + \text{p-nitrochlorobenzene} + \text{H}_2\text{O} \)
(ii) \( \text{C}_2\text{H}_5\text{OH} + \text{SOCl}_2 \xrightarrow{\text{pyridine}} \text{C}_2\text{H}_5\text{Cl} + \text{SO}_2\uparrow + \text{HCl}\uparrow \)
(iii) \( 2\text{CHCl}_3 + \text{O}_2 \xrightarrow{\text{sunlight}} 2\text{COCl}_2\uparrow + 2\text{HCl} \)

Teacher's Note:
a) Nitration of chlorobenzene yields a mixture of ortho and para nitro isomers.
b) Oxidation of chloroform by air and sunlight produces phosgene (carbonyl chloride), a poisonous gas.

Question 16 [3 Marks]

(i) The rate constant of a first order reaction increases five times when the temperature is raised from 350K to 500K. Calculate the activation energy of the reaction. (\( R = 8.314 \text{ J K}^{-1}\text{mol}^{-1} \))
(ii) The unit of rate constant of a reaction is same as that of its rate of reaction. What is the order of this reaction?

Answer:
(i) Using Arrhenius equation:
\( \log\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303 R} \left(\frac{T_2 - T_1}{T_1 T_2}\right) \)
\( \log(5) = \frac{E_a}{2.303 \times 8.314} \left(\frac{500 - 350}{350 \times 500}\right) \)
\( 0.6990 = \frac{E_a}{19.147} \left(\frac{150}{175000}\right) \)
\( E_a = \frac{0.6990 \times 19.147 \times 175000}{150} = 15609 \text{ J mol}^{-1} = 15.61 \text{ kJ mol}^{-1} \)
(ii) Zero order reaction.

Teacher's Note:
a) Apply the logarithmic form of the Arrhenius equation for two different temperatures.
b) For a zero-order reaction, rate = \( k[\text{A}]^0 = k \), so the units of rate constant and rate are identical.

Question 17 [3 Marks]

(i) (a) The molar conductivity at infinite dilution for \( \lambda^\infty_{\text{H}^+} = 348.65 \text{ S cm}^2\text{mol}^{-1} \) and for \( \lambda^\infty_{\text{CH}_3\text{COO}^-} = 41.4 \text{ S cm}^2\text{mol}^{-1} \) respectively.
Calculate the degree of dissociation (\( \alpha \)) of acetic acid if its molar conductivity (\( \Lambda_m \)) is \( 40.65 \text{ S cm}^2\text{mol}^{-1} \).
(b) Compounds [A] and [B] are two electrolytes. Upon dilution, the molar conductivity of compound [A] increases 3 times while that of [B] increases 30 times. Which one of the two is a weak electrolyte? Why?
(c) Can copper sulphate solution be stored in a zinc pot? Give a reason for your answer by referring to the values given below.
\( E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76\text{ V}, E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34\text{ V} \)
OR
(ii) (a) When a current of 0.75 ampere is passed through a \( \text{CuSO}_4 \) solution for 25 minutes, 0.370 g of copper is deposited. Calculate the atomic weight of copper by using the given information.
(b) Two metals A and B have standard reduction potential values, -0.76V and +0.34V respectively.
Which of these metals will liberate H₂ gas from dil. H₂SO₄? Why?
(c) Why does the specific conductivity (\( \kappa \)) of a solution decrease on dilution?

Answer:
(i) (a) \( \Lambda^\infty_m(\text{CH}_3\text{COOH}) = \lambda^\infty_{\text{H}^+} + \lambda^\infty_{\text{CH}_3\text{COO}^-} = 348.65 + 41.4 = 390.05 \text{ S cm}^2\text{mol}^{-1} \).
\( \alpha = \frac{\Lambda_m}{\Lambda^\infty_m} = \frac{40.65}{390.05} = 0.104 \)
(b) Compound [B] is a weak electrolyte because weak electrolytes show a sharp increase in molar conductivity upon dilution due to a large increase in degree of dissociation.
(c) No, copper sulphate cannot be stored in a zinc pot because zinc has a lower reduction potential (\( -0.76\text{ V} \)) than copper (\( +0.34\text{ V} \)), so zinc will displace copper and corrode the pot.
OR
(ii) (a) \( W = \frac{Z \times I \times t}{96500} \Rightarrow 0.370 = \frac{M \times 0.75 \times 25 \times 60}{2 \times 96500} \)
\( M = \frac{0.370 \times 2 \times 96500}{0.75 \times 1500} = 63.38 \text{ g mol}^{-1} \)
(b) Metal A (with \( E^\circ = -0.76\text{ V} \)) will liberate H₂ gas because it has a negative reduction potential, meaning it is a stronger reducing agent than hydrogen.
(c) Specific conductivity decreases on dilution because the number of current-carrying ions per unit volume of the solution decreases.

Teacher's Note:
a) Degree of dissociation is the ratio of molar conductivity at a given concentration to that at infinite dilution.
b) Metals with negative standard reduction potentials can displace hydrogen from dilute acids.

Question 18 [3 Marks]

Identify the compounds [A], [B] and [C] in the following reactions:
(i) \( \text{CH}_3\text{CH}_2\text{Br} \xrightarrow{\text{KCN}} [\text{A}] \xrightarrow{\text{LiAlH}_4} [\text{B}] \xrightarrow{\text{HNO}_2} [\text{C}] \)
(ii) \( \text{CH}_3\text{CH}_2\text{COOH} \xrightarrow{\text{NH}_3} [\text{A}] \xrightarrow{\Delta} [\text{B}] \xrightarrow{\text{Br}_2 / \text{KOH}} [\text{C}] \)

Answer:
(i) [A] = Propanenitrile (\( \text{CH}_3\text{CH}_2\text{CN} \)), [B] = Propan-1-amine (\( \text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2 \)), [C] = Propan-1-ol (\( \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \)).
(ii) [A] = Ammonium propanoate (\( \text{CH}_3\text{CH}_2\text{COONH}_4 \)), [B] = Propanamide (\( \text{CH}_3\text{CH}_2\text{CONH}_2 \)), [C] = Ethanamine (\( \text{CH}_3\text{CH}_2\text{NH}_2 \)).

Teacher's Note:
a) Reduction of cyanides with LiAlH₄ yields primary amines.
b) Hofmann bromamide degradation reaction converts an amide into a primary amine with one carbon atom less.

SECTION D - 15 MARKS

Question 19 [5 Marks]

(i) Calculate the mass of ascorbic acid (molecular mass = \( 176 \text{ g mol}^{-1} \)) to be dissolved in 75 g of acetic acid to lower its freezing point by \( 1.5^\circ\text{C} \). Assume that the solute neither associates nor dissociates in solution.
(\( K_f \) for acetic acid = \( 3.9 \text{ K kg mol}^{-1} \))
(ii) A solution of an organic compound was prepared by dissolving 6.8g in 0.1 litre of water. Calculate the osmotic pressure of this solution at \( 25^\circ\text{C} \).
(Molecular mass of organic compound = \( 321.45 \text{ g mol}^{-1} \))
(iii) Why are aquatic species more comfortable in cold water than warm water?

Answer:
(i) \( \Delta T_f = \frac{K_f \times W_2 \times 1000}{M_2 \times W_1} \)
\( 1.5 = \frac{3.9 \times W_2 \times 1000}{176 \times 75} \)
\( W_2 = \frac{1.5 \times 176 \times 75}{3.9 \times 1000} = \frac{19800}{3900} = 5.076 \text{ g} \)
(ii) \( \pi = \frac{n}{V} RT = \frac{W_2}{M_2 \times V} RT \)
\( \pi = \frac{6.8}{321.45 \times 0.1} \times 0.0821 \times 298 = 5.17 \text{ atm} \)
(iii) Oxygen gas is more soluble in cold water than in warm water. Therefore, higher dissolved oxygen availability in cold water makes aquatic species more comfortable.

Teacher's Note:
a) Ensure all temperatures are converted to Kelvin before applying gas equation formulas.
b) Solubility of gases in liquids decreases with increase in temperature according to Henry's Law.

Question 20 [5 Marks]

(i) What happens when (write chemical equations):
(a) Calcium acetate is subjected to dry distillation.
(b) Acetaldehyde is reduced with hydrogen in the presence of Zn / Hg and conc. HCl.
(c) Acetic acid is heated with ethyl alcohol in the presence of conc. H₂SO₄.
(ii) An organic compound [X] with molecular formula C₂H₄O forms compound [Y] on oxidation. Compound [X] undergoes haloform reaction. On treatment with HCN, compound [X] forms a product [Z] which on hydrolysis gives 2-hydroxy propanoic acid.
(a) Write down the structures of compounds [X] and [Y].
(b) Name the product formed when [X] reacts with dil. NaOH.

Answer:
(i) (a) \( (\text{CH}_3\text{COO})_2\text{Ca} \xrightarrow{\text{Dry distillation}} \text{CH}_3\text{COCH}_3 + \text{CaCO}_3 \)
(b) \( \text{CH}_3\text{CHO} + 4[H] \xrightarrow{\text{Zn-Hg} / \text{conc. HCl}} \text{CH}_3\text{CH}_3 + \text{H}_2\text{O} \)
(c) \( \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH} \rightleftharpoons \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \)
(ii) [X] = Acetaldehyde (\( \text{CH}_3\text{CHO} \)), [Y] = Acetic acid (\( \text{CH}_3\text{COOH} \)).
(a) Structure of [X]: \( \text{CH}_3-\text{CHO} \); Structure of [Y]: \( \text{CH}_3-\text{COOH} \).
(b) Aldol condensation product: 3-hydroxybutanal (\( \text{CH}_3-\text{CH(OH)}-\text{CH}_2-\text{CHO} \)).

Teacher's Note:
a) Dry distillation of calcium salts of carboxylic acids yields ketones.
b) Clemmensen reduction reduces aldehydes and ketones to corresponding alkanes using zinc-amalgam and conc. HCl.

Question 21 [5 Marks]

(i) (a) With reference to Valence Bond Theory (VBT), answer the following questions regarding the complex ion [Cr(NH₃)₆]³⁺.
(1) What is the oxidation number of chromium in the complex?
(2) State the magnetic behaviour of the complex.
(3) How many unpaired electrons are there in the complex?
(4) State the type of hybridisation of the central metal atom.
(b) The coordination complex \( \text{CoNO}_2\text{Cl}\cdot 5\text{NH}_3 \) exists in two isomeric forms 'P' and 'Q'. Isomer 'P' reacts with AgNO₃ solution to give white precipitate whereas 'Q' does not give any precipitate with AgNO₃ solution.
(1) Write the structural formula of isomers 'P' and 'Q'.
(2) Name the type of isomerism involved.
(c) Explain why an aqueous solution of potassium hexacyanidoferrate(II) does not give the test of ferrous ion.
OR
(ii) (a) Write the electronic configuration of the following:
(1) \( d^4 \) (high spin octahedral)
(2) \( d^6 \) (low spin octahedral)
(b) Based on the above configuration, calculate the value of Crystal Field Splitting Energy (CFSE). (Ignore pairing energy)
(c) Name the type of isomerism exhibited by the following pairs of coordination compounds.
(1) [Co(NH₃)₅(ONO)]Cl₂ and [Co(NH₃)₅NO₂]Cl₂
(2) [PtCl₂(NH₃)₄]Br₂ and [PtBr₂(NH₃)₄]Cl₂
(d) Which of the following coordination complexes is an outer orbital complex? Explain.
[Fe(H₂O)₆]²⁺ or [Fe(CN)₆]⁴⁺

Answer:
(i) (a) (1) +3
(2) Paramagnetic
(3) 3 unpaired electrons
(4) \( d^2sp^3 \)
(b) (1) 'P': [Co(NH₃)₅(NO₂)]ClCl, 'Q': [Co(NH₃)₅(NO₂)Cl] (or ionization isomers [Co(NH₃)₅(NO₂)]Cl and [Co(NH₃)₅Cl]NO₂)
(2) Ionization isomerism
(c) Potassium hexacyanidoferrate(II) is a coordination compound that does not dissociate to give free \( \text{Fe}^{2+} \) ions in aqueous solution; instead, it remains as the stable complex ion \( [\text{Fe}(\text{CN})_6]^{4-} \).
OR
(ii) (a) (1) \( t_{2g}^3 e_g^1 \)
(2) \( t_{2g}^6 e_g^0 \)
(b) For high spin \( d^4 \): \( \text{CFSE} = (-0.4 \times 3 + 0.6 \times 1)\Delta_o = -0.6\Delta_o \); For low spin \( d^6 \): \( \text{CFSE} = (-0.4 \times 6)\Delta_o = -2.4\Delta_o \).
(c) (1) Linkage isomerism
(2) Ionization isomerism
(d) [Fe(H₂O)₆]²⁺ is an outer orbital complex because H₂O is a weak field ligand, resulting in \( sp^3d^2 \) hybridization using outer d-orbitals.

Teacher's Note:
a) Coordination number and ligand field strength determine whether a complex is inner or outer orbital.
b) Ionization isomers yield different ions in solution upon reacting with reagents like silver nitrate.