ISC Class 12 Chemistry Sample Paper 2025 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]
[four, three, remains same, propane, ns^1-2, increases, six, propan-2-ol, (n-1)d^1-10, two, decreases, Clemmensen's, Wolff-Kishner]

(i) The molar conductance of a solution _________ on dilution but the specific conductance of a solution _________ on dilution. [1 Mark]

Answer: increases, decreases

Teacher's Note:
a) Molar conductance increases due to the total volume containing 1 mole of electrolyte increasing significantly upon dilution.
b) Specific conductance decreases because the number of current-carrying ions per unit volume of the solution decreases on dilution.

(ii) In the complex ion [CoCl(en)₂ONO]⁺, the coordination number and the oxidation number of the central metal ion are ______ and ________. [1 Mark]

Answer: six, three

Teacher's Note:
a) Coordination number is six because ethylenediamine (en) is a bidentate ligand (contributing 4) while chloride and nitrite are monodentate ligands (contributing 2).
b) Oxidation number is +3, calculated as \( x + (-1) + 2(0) + (-1) = +1 \), giving \( x = +3 \).

(iii) Propanone on reaction with zinc amalgam in the presence of concentrated HCl gives ______ and the reaction is called _______ reduction. [1 Mark]

Answer: propane, Clemmensen's

Teacher's Note:
a) Clemmensen reduction specifically converts aldehydes and ketones into their corresponding alkanes using zinc-amalgam and concentrated hydrochloric acid.
b) Propanone contains 3 carbon atoms, hence it yields propane as the hydrocarbon product.

(iv) The variation and magnitude of oxidation state in coordination complexes is due to the participation of inner ________ electrons in addition to outer __________ electrons. [1 Mark]

Answer: (n-1)d^1-10, ns^1-2

Teacher's Note:
a) Transition elements use both (n-1)d and ns electrons for bonding due to very small energy differences between them.
b) This dual participation allows transition metals to exhibit multiple variable oxidation states in their complexes.

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

(i) The correct order of the increasing basic nature of Ammonia, Methylamine and Aniline is: [1 Mark]
(a) Methylamine < Aniline < Ammonia
(b) Aniline < Methylamine < Ammonia
(c) Ammonia < Aniline < Methylamine
(d) Aniline < Ammonia < Methylamine

Answer: (d) Aniline < Ammonia < Methylamine

Methylamine is strongly basic due to the +I effect of the methyl group; aniline is weakly basic due to resonance delocalization of the lone pair into the benzene ring.

Teacher's Note:
a) Electron-donating alkyl groups increase basicity by increasing electron density on nitrogen.
b) Electron-withdrawing aryl groups decrease basicity by delocalizing the lone pair of electrons.

(ii) Colligative properties of a solution are: [1 Mark]
(P) Independent of the nature of solute
(Q) Proportional to molecular mass of solute
(R) Proportional to concentration of solute
(S) Independent of the amount of solvent
(a) Only P and Q are correct
(b) Only P and R are correct
(c) Only P and S are correct
(d) Only Q and S are correct

Answer: (b) Only P and R are correct

Colligative properties depend only on the number of solute particles (concentration) and not on their chemical nature.

Teacher's Note:
a) They are inversely proportional to molecular mass, not directly proportional.
b) They depend on the amount of solvent (mass of solvent in kg for molality).

(iii) According to Markownikoff's rule, "The negative part of the halogen acid attaches itself with that C atom of an unsymmetrical alkene which has least number of H-atom."
Anti-Markownikoff's addition of HBr is NOT observed in: [1 Mark]
(a) Propene
(b) But-2-ene
(c) But-1-ene
(d) Pent-2-ene

Answer: (b) But-2-ene

But-2-ene is a symmetrical alkene, so addition of HBr yields the same product regardless of the rule followed.

Teacher's Note:
a) Anti-Markownikoff's addition (peroxide effect) is strictly applicable only to unsymmetrical alkenes.
b) Symmetrical alkenes do not produce positional isomers upon addition.

(iv) Most of the naturally occurring alpha-amino acids are optically active and have L-configuration. Which of the following amino acids is NOT asymmetric? [1 Mark]
(a) Valine
(b) Alanine
(c) Glycine
(d) Leucine

Answer: (c) Glycine

Glycine has two hydrogen atoms attached to the alpha-carbon, making it achiral (symmetric).

Teacher's Note:
a) An asymmetric carbon requires four different groups attached to it.
b) Glycine is the only naturally occurring standard amino acid that is optically inactive.

(v) Which one of the following complex ions has geometrical isomers? [1 Mark]
(a) [Ni(NH₃)₅Br]⁺
(b) [Cr(NH₃)₄en]³⁺
(c) [Co(NH₃)₂(en)₂]³⁺
(d) [Co(NH₃)₆]³⁺

Answer: (c) [Co(NH₃)₂(en)₂]³⁺

Octahedral complexes of the type [MA₂B₄] or [MA₂(en)₂] exhibit cis-trans geometrical isomerism.

Teacher's Note:
a) Complexes of formula [MA₆] or [MA₅B] do not show geometrical isomerism.
b) The cis and trans forms are possible due to the relative spatial positions of the two monodentate ammonia ligands.

(vi) Given below are two statements marked Assertion and Reason. Read the two statements carefully and select the correct option.
Assertion: Tertiary butyl bromide undergoes Wurtz reaction to give 2,2,3,3 tetramethyl butane.
Reason: In Wurtz reaction, when same alkyl halides react with sodium in dry ether to give hydrocarbon; the hydrocarbon formed contains double the number of carbon atoms present in the alkyl halide. [1 Mark]
(a) Both Assertion and Reason are true and Reason is the correct explanation of 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) Both Assertion and Reason are false.

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

Coupling of two tert-butyl radicals yields 2,2,3,3-tetramethylbutane, doubling the carbon framework as stated by the general Wurtz principle.

Teacher's Note:
a) Although tertiary alkyl halides predominantly undergo elimination, coupling can occur under specific conditions.
b) The doubling of carbon atoms is the fundamental characteristic of standard Wurtz reactions.

(vii) Given below are two statements marked Assertion and Reason. Read the two statements carefully and select the correct option.
Assertion: There is a continuous increase in size among Lanthanoids with an increase in atomic number.
Reason: Lanthanoids do not show Lanthanoid contraction. [1 Mark]
(a) Both Assertion and Reason are true and Reason is the correct explanation of 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) Both Assertion and Reason are false.

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

Atomic and ionic radii steadily decrease (lanthanoid contraction) across the series due to poor shielding by 4f electrons.

Teacher's Note:
a) Both statements contradict well-established periodic properties of f-block elements.
b) Poor shielding of nuclear charge by 4f orbitals causes a net contraction in atomic size.

(C) Read the passage carefully and answer the questions that follow. [3×1]

A plant or any other living being maintains a reasonable balance of C-14 (radioactive carbon) in its tissue during its lifetime. C-14 is used to determine the age of fossils. In the upper atmosphere, neutrons present in the cosmic rays are captured to produce the following nuclear reaction.
₇N¹⁴ + ₀n¹ → ₆C¹⁴ + ₁H¹
C-14 isotope is circulated in the atmosphere and gets absorbed by living organism during photosynthesis. The ratio of C-14 to C-12 in living being is 1:10¹². Once the living being dies, the level of C-14 in the dead being decreases due to the following reaction.
₆C¹⁴ → ₇N¹⁴ + _-1e⁰ + γ rays
The death of the plant brings an end to its tendency to take up C-14. The half-life period of C-14 is 5770 years. By knowing the concentration of C-14 in the living plant and the piece of dead material at a particular time, the age of the material (fossil) can be determined.

(i) Write the relation between the decay constant and half-life period. [1 Mark]

Answer: \( k = \frac{0.693}{t_{1/2}} \)

Teacher's Note:
a) This standard relationship applies universally to all first-order radioactive decay processes.
b) Ensure proper units are used for time when calculating decay constants.

(ii) In a piece of dead wood, the activity or concentration of C-14 is found to be one third of its initial activity. Calculate the age of the old wood. [1 Mark]

Answer: 9146.5 years

Using first-order rate equation: \( t = \frac{2.303}{k} \log\left(\frac{[A]_0}{[A]}\right) = \frac{2.303 \times 5770}{0.693} \log(3) = 9146.5 \) years.

Teacher's Note:
a) Substitute the half-life into the rate constant formula first to find \( k \).
b) Use base-10 logarithm values correctly (\( \log(3) = 0.4771 \)).

(iii) The half-life period (t_1/2) for a first order reaction is 30 minutes. Calculate the time taken to complete 87.5% of the reaction. [1 Mark]

Answer: 90.03 minutes

For 87.5% completion, remaining concentration is 12.5% or \( \frac{1}{8} \), which corresponds to 3 half-lives (\( 3 \times 30 = 90 \) minutes).

Teacher's Note:
a) Recognizing fraction multiples of half-life simplifies calculations for first-order reactions.
b) Alternatively, apply the integrated rate law directly to verify the result.

SECTION B - 20 MARKS

Question 2 [2 Marks]

The solution of two electrolytes A and B are diluted. Λ_m of B increases 1.5 times while that of A increases 25 times. Which of the two is a strong electrolyte? Give a reason.

Answer: B is a strong electrolyte because on dilution, molar conductance increases only to a small extent (1.5 times), whereas for a weak electrolyte (A), the degree of dissociation increases sharply, leading to a large increase in molar conductance (25 times).

Teacher's Note:
a) Strong electrolytes are already completely dissociated in solution prior to dilution.
b) Weak electrolytes undergo further ionization upon dilution, drastically increasing the number of free ions.

Question 3 [2 Marks]

An organic compound [A] which has characteristic odour, reacts with conc. NaOH to give two compounds [B] and [C]. Compound [B] has molecular formula C₇H₈O which upon oxidation gives back compound [A]. Compound [C] is the sodium salt of the acid and upon treatment with sodalime yields an aromatic hydrocarbon [D]. Identify the compounds [A], [B], [C] and [D].

Answer:
[A] = C₆H₅CHO (Benzaldehyde)
[B] = C₆H₅CH₂OH (Benzyl alcohol)
[C] = C₆H₅COONa (Sodium benzoate)
[D] = C₆H₆ (Benzene)

Teacher's Note:
a) Benzaldehyde undergoes Cannizzaro reaction in the presence of concentrated NaOH since it lacks alpha-hydrogens.
b) Decarboxylation of sodium benzoate with sodalime produces benzene.

Question 4 [2 Marks]

Devise a scheme for the synthesis of n-butane using CH₃I as the only source of carbon. Is it possible to obtain propane in pure state by applying this scheme? Give a reason for your answer.

Answer:
1. Carry out Wurtz reaction on methyl iodide using sodium in dry ether: \( 2\text{CH}_3\text{I} + 2\text{Na} \rightarrow \text{CH}_3-\text{CH}_3 + 2\text{NaI} \).
2. Halogenate ethane to form ethyl iodide, then subject ethyl iodide to Wurtz reaction to obtain n-butane.
No, propane cannot be prepared in a pure state using Wurtz reaction because coupling of a mixture of methyl and ethyl halides yields a statistical mixture of propane along with side products ethane and butane.

Teacher's Note:
a) Wurtz reaction is best suited for symmetric alkanes with an even number of carbon atoms.
b) Mixed Wurtz reactions lead to inseparable mixtures of symmetrical and unsymmetrical alkanes.

Question 5 [2 Marks]

It is generally observed that the rate of a chemical reaction becomes double with every 10°C rise in temperature. If the generalization holds true for a reaction in the temperature range of 298K to 308K, what would be the value of activation energy (E_a) for the reaction?

Answer:
Using Arrhenius equation:
\( \log\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303 R} \left(\frac{T_2 - T_1}{T_1 T_2}\right) \)
Here, \( \frac{k_2}{k_1} = 2 \), \( T_1 = 298\text{ K} \), \( T_2 = 308\text{ K} \), \( R = 8.314\text{ J K}^{-1}\text{mol}^{-1} \).
\( \log(2) = \frac{E_a}{2.303 \times 8.314} \left(\frac{308 - 298}{298 \times 308}\right) \)
\( 0.3010 = \frac{E_a}{19.147} \left(\frac{10}{91784}\right) \)
\( E_a = \frac{0.3010 \times 19.147 \times 91784}{10} = 52897.78\text{ J mol}^{-1} \).

Teacher's Note:
a) Ensure correct temperature values in Kelvin are used in the denominator product.
b) Express the final activation energy in joules per mole or kilojoules per mole clearly.

Question 6 [2 Marks]

An organic compound [X] of molecular formula C₃H₅N, in one reaction produced a primary amine with LiAlH₄. In another case, the organic compound [Y] of same molecular formula C₃H₅N produced a secondary amine with LiAlH₄. Identify both the compounds [X] and [Y]. Why do these compounds form two different products?

Answer:
[X] = C₂H₅CN (Propanenitrile / Ethyl cyanide)
[Y] = C₂H₅NC (Ethyl isocyanide)
They form different products because reduction of cyanides (-CN) yields primary amines (since carbon is attached to carbon), whereas reduction of isocyanides (-NC) yields secondary methyl alkylamines (since nitrogen is attached to the main alkyl chain).

Teacher's Note:
a) Cyanides contain a carbon-nitrogen triple bond where reduction adds hydrogens across both atoms.
b) Isocyanides possess a coordinate covalent bond to nitrogen, dictating the formation of secondary amine derivatives.

Question 7 [2 Marks]

During chemistry class, a teacher wrote [Ni(CN)₄]^2- as a coordination complex ion on the board. The students were asked to find out the magnetic behaviour and shape of the complex. Pari, a student, wrote the answer paramagnetic and tetrahedral whereas another student Suhail wrote diamagnetic and square planar. Evaluate Pari's and Suhail's responses.

Answer:
Suhail's response was correct. According to Valence Bond Theory (VBT), Nickel in [Ni(CN)₄]^2- is in the +2 oxidation state with a d⁸ configuration. Cyanide is a strong field ligand which forces electron pairing, resulting in dsp² hybridization, a square planar geometry, and diamagnetic behavior.

Teacher's Note:
a) Weak field ligands typically form tetrahedral complexes with d⁸ metal ions, leading to paramagnetism.
b) Strong field ligands cause inner orbital pairing, favoring square planar geometries.

Question 8 [2 Marks]

An organic compound 'X' with molecular formula C₄H₁₀O is found to be soluble in conc. H₂SO₄ and does not react with sodium metal or KMnO₄. Compound 'X' when heated with excess of HI gives a single alkyl halide. Deduce the structure of the compound 'X'. Explain all the reactions involved.

Answer:
[X] = C₂H₅OC₂H₅ (Diethyl ether)
Reaction: \( \text{C}_2\text{H}_5\text{OC}_2\text{H}_5 + 2\text{HI} \xrightarrow{\Delta} 2\text{C}_2\text{H}_5\text{I} + \text{H}_2\text{O} \)

Teacher's Note:
a) Solubility in conc. H₂SO₄ indicates the presence of an oxygen-containing functional group (ether).
b) Lack of reaction with sodium and KMnO₄ rules out alcohols and active functional groups, confirming it is a symmetrical ether.

Question 9 [2 Marks]

(i) Identify the compounds [A] and [B] in the following reactions.
(a) C₆H₅CHO &xrightarrow[(2)\text{ SOCl}_2]{(1)\text{ K}_2\text{Cr}_2\text{O}_7/\text{H}_2\text{SO}_4} [\text{A}] \xrightarrow{\text{C}_6\text{H}_6(\text{anhy. AlCl}_3)} [\text{B}]
(b) CH₃-C≡CH &xrightarrow[(2)\text{ vigorous oxidation, conc. HNO}_3]{(1)\text{ H}_2\text{SO}_4/\text{HgSO}_4} [\text{A}] \xrightarrow{\text{C}_2\text{H}_5\text{OH}[\text{conc. H}_2\text{SO}_4]} [\text{B}]

Answer:
(a) [A] = C₆H₅COCl, [B] = C₆H₅COC₆H₅
(b) [A] = CH₃COOH, [B] = CH₃COOC₂H₅

Teacher's Note:
a) Benzoic acid treated with thionyl chloride yields benzoyl chloride, which on Friedel-Crafts acylation gives benzophenone.
b) Hydration of propyne yields acetone initially, but vigorous oxidation yields acetic acid, which esterifies with ethanol to give ethyl acetate.

OR

(ii) Three organic compounds A, B and C are non cyclic functional isomers of carbonyl compounds with molecular formula C₄H₈O. Isomers A and C give positive Tollen's test while compound B does not give positive Tollen's test but gives positive iodoform test. Compounds A and B on reduction with Zn amalgam and conc. HCl give the same product.
(a) Write the structures of the compounds A, B and C.
(b) Out of the compounds A, B and C, which one will be the least reactive towards addition of HCN.

Answer:
(a) [A] = CH₃CH₂CH₂CHO, [B] = CH₃COCH₂CH₃, [C] = CH₃CH(CH₃)CHO
(b) [B] (Butanone) will be the least reactive towards addition of HCN.

Teacher's Note:
a) Aldehydes (A and C) give positive Tollen's tests, while methyl ketones (B) give positive iodoform tests.
b) Ketones are less reactive than aldehydes toward nucleophilic addition due to steric and electronic factors.

Question 10 [2 Marks]

Calculate the value of E⁰_cell, E_cell and ΔG that can be obtained from the following cell at 298K.
Al / Al³⁺ (0.01M) // Sn²⁺ (0.015M) / Sn
Given E⁰_Al3+/Al = -1.66V; E⁰_Sn2+/Sn = -0.14V

Answer:
Cell reaction: \( 2\text{Al} + 3\text{Sn}^{2+} \rightarrow 2\text{Al}^{3+} + 3\text{Sn} \)
\( E^0_{\text{cell}} = E^0_{\text{cathode}} - E^0_{\text{anode}} = -0.14 - (-1.66) = +1.52\text{ V} \)
Using Nernst equation:
\( E_{\text{cell}} = E^0_{\text{cell}} - \frac{0.0591}{n} \log\left(\frac{[\text{Al}^{3+}]^2}{[\text{Sn}^{2+}]^3}\right) \)
\( E_{\text{cell}} = 1.52 - \frac{0.0591}{6} \log\left(\frac{(0.01)^2}{(0.015)^3}\right) = 1.5056\text{ V} \)
\( \Delta G = -nFE_{\text{cell}} = -6 \times 96500 \times 1.5056 = -871742\text{ J} \)

Teacher's Note:
a) Balance electrons carefully by multiplying oxidation and reduction half-reactions before applying Nernst equation.
b) Pay close attention to stoichiometric coefficients when setting up concentration quotient terms.

Question 11 [2 Marks]

Write the chemical test to distinguish between the following pairs of compounds.
(i) Propan-1-ol and Propan-2-ol
(ii) Phenol and ethanol

Answer:
(i) Lucas test: Propan-2-ol reacts rapidly with Lucas reagent (anhydrous ZnCl₂ + conc. HCl) to form turbidity within 5 minutes, whereas propan-1-ol shows turbidity only upon heating.
(ii) Ferric chloride test: Phenol reacts with neutral FeCl₃ to give a violet/purple coloured complex, whereas ethanol does not react.

Teacher's Note:
a) Lucas test relies on carbocation stability differences between primary, secondary, and tertiary alcohols.
b) Phenols exhibit weakly acidic phenolic hydroxyl groups, producing characteristic colored complexes with iron(III).

SECTION C - 21 MARKS

Question 12 [3 Marks]

Identify the compounds [A], [B] and [C] in the following reactions.
(i) C₆H₅NO₂ &xrightarrow[Sn/HCl]{6[\text{H}]} [\text{A}] &xrightarrow[273\text{K}-278\text{K}]{\text{HNO}_2 + \text{HCl}} [\text{B}] \xrightarrow{\text{C}_6\text{H}_5\text{NH}_2} [\text{C}]
(ii) CH₃CH₂NH₂ &xrightarrow[(2)\text{ P/Cl}_2}{(1)\text{ HNO}_2} [\text{A}] &xrightarrow[(2)\text{ H}_2\text{O}/\text{H}^+}{(1)\text{ KCN}} [\text{B}] &xrightarrow[(2)\text{ Br}_2/\text{KOH}}{(1)\text{ NH}_3/\text{heat}} [\text{C}]

Answer:
(i) [A] = C₆H₅NH2, [B] = C₆H₅N₂Cl, [C] = C₆H₅-N=N-C₆H₄NH₂
(ii) [A] = CH₃CH₂Cl, [B] = CH₃CH₂COOH, [C] = CH₃CH₂NH₂

Teacher's Note:
a) Reduction of nitrobenzene gives aniline, diazotization gives benzene diazonium chloride, and coupling yields an azo dye.
b) Conversion steps involving halogenation, cyanation, hydrolysis, and Hoffmann bromamide degradation alter carbon chain lengths accordingly.

Question 13 [3 Marks]

The molar mass calculated through colligative properties is sometimes different from that of experimentally determined molecular mass and is known as abnormal molecular mass. This difference is due to the solute particles that undergo association or dissociation.
Consider the two cases given below and answer the questions that follow.
(A) The freezing point of a solution containing 5.85g of NaCl in 100g of water is -3.348°C. (K_f of water = 1.86 K kg mol^-1, molecular mass of NaCl = 58.5)
(B) The freezing point of benzene solution decreases by 0.45°C when 0.2g of acetic acid is added to 20g of benzene. (K_f of benzene = 5.12 K kg mol^-1, molecular mass of acetic acid = 60)
Answer the following questions.
(i) Calculate the abnormal molecular mass of solute in both the cases.
(ii) Inferring the value of van't Hoff factor in both the cases, find out which solution undergoes association and which solution undergoes dissociation.
(iii) Given below is the increasing order of depression in freezing point of water observed for equimolar concentrations of the compounds, acetic acid < trichloro acetic acid < trifluoro acetic acid. Provide a reason to explain if the above ordered arrangement is correct or not.

Answer:
(i) For Case (A): \( M_2 = \frac{K_f \times w_2 \times 1000}{\Delta T_f \times w_1} = \frac{1.86 \times 5.85 \times 1000}{3.348 \times 100} = 32.5 \)
For Case (B): \( M_2 = \frac{5.12 \times 0.2 \times 1000}{0.45 \times 20} = 113.79 \)
(ii) Case (A) undergoes dissociation (i > 1), while Case (B) undergoes association (i < 1).
(iii) The order is correct because depression in freezing point (ΔT_f) is a colligative property depending on particle count. Stronger acid substituents (trifluoro > trichloro > acetic) increase ionization, yielding more ions and greater freezing point depression.

Teacher's Note:
a) Abnormal molecular mass calculations reflect effective particle counts resulting from solute association/dissociation.
b) Acid strength correlates with degree of ionization, directly scaling colligative property magnitudes.

Question 14 [3 Marks]

Write chemical equations to illustrate the following name reactions.
(i) Rosenmund's reaction
(ii) Benzoin condensation
(iii) Perkin's reaction

Answer:
(i) Rosenmund's reaction:
\( \text{RCOCl} + \text{H}_2 \xrightarrow{\text{Pd/BaSO}_4, \text{heat}} \text{RCHO} + \text{HCl} \)
(ii) Benzoin condensation:
\( 2\text{C}_6\text{H}_5\text{CHO} \xrightarrow{\text{KCN (alc)}} \text{C}_6\text{H}_5\text{CH(OH)COC}_6\text{H}_5 \)
(iii) Perkin's reaction:
\( \text{C}_6\text{H}_5\text{CHO} + (\text{CH}_3\text{CO})_2\text{O} \xrightarrow[\text{Heat}]{\text{CH}_3\text{COONa}} \text{C}_6\text{H}_5\text{CH}=\text{CHCOOH} + \text{CH}_3\text{COOH} \)

Teacher's Note:
a) Poisoned catalysts like Pd/BaSO₄ prevent further reduction of aldehydes to alcohols in Rosenmund's reaction.
b) Perkin's reaction involves condensation of aromatic aldehydes with acid anhydrides in the presence of a salt of the acid.

Question 15 [3 Marks]

(i) Assuming that cyanide is a strong ligand, write the formula of two coordination complex ions with cyanide and iron having coordination number six. Name the complex ions.
(ii) On the basis of Crystal Field Theory, write the electronic configuration of d⁴ ion if:
(a) Δ_o > P
(b) Δ_o < P

Answer:
(i) [Fe(CN)₆]^4- : hexacyanidoferrate(II) ion
[Fe(CN)₆]^3- : hexacyanidoferrate(III) ion
(ii) (a) For Δ_o > P (Strong field): t_2g⁴ e_g⁰ (Low spin)
(b) For Δ_o < P (Weak field): t_2g³ e_g¹ (High spin)

Teacher's Note:
a) Naming coordination complexes requires listing ligands alphabetically followed by the metal and its oxidation state in Roman numerals.
b) Crystal field splitting energy (Δ_o) versus pairing energy (P) determines electron distribution in d-orbitals.

Question 16 [3 Marks]

(i) Dry cells are commonly used in clocks, torches, calculators etc. They are the most familiar type of commercial cells.
(a) What acts as anode and cathode in dry cell?
(b) Which chemical compounds are filled between anode and cathode of this cell?
(c) Why do dry cells not have a long life?

Answer:
(a) Anode is the zinc container, and cathode is a graphite rod surrounded by powdered manganese dioxide.
(b) A moist paste of ammonium chloride (NH₄Cl) and zinc chloride (ZnCl₂).
(c) The acidic nature of ammonium chloride continuously corrodes the zinc container even when the cell is not in use.

Teacher's Note:
a) Commercial primary cells cannot be recharged because chemical reactions are irreversible.
b) Zinc corrosion accounts for self-discharge and limited shelf-life.

OR

(ii) Answer the following:
(a) What happens to the voltage when the salt bridge is removed from the two half cells and why?
(b) Copper sulphate solution cannot be stored in a zinc pot. Why?
(c) Why does the density of H₂SO₄ in a lead storage battery decrease as it is discharged?

Answer:
(a) Voltage drops to zero because electrical neutrality is lost, stopping current flow.
(b) Zinc is more reactive than copper (placed above copper in the electrochemical series) and displaces copper from CuSO₄ solution.
(c) Sulfuric acid is consumed during the discharge process of the battery, lowering the concentration and density of the solution.

Teacher's Note:
a) Salt bridges complete the internal electrical circuit and maintain charge balance across half-cells.
b) Electrochemical series reactivity predicts spontaneous metal displacement reactions.

Question 17 [3 Marks]

(i) The structures of glycine and alanine are given below. Show the peptide bond linkage in glycylalanine.
[Figure: Glycine: H₂N-CH₂-COOH, Alanine: H₂N-CH(CH₃)-COOH]
(ii) What products would be formed when a nucleotide from DNA containing thymine is hydrolysed?
(iii) What are reducing sugars? Give an example.

Answer:
(i) Peptide linkage: H₂NCH₂-CONH-CH(CH₃)COOH + H₂O
(ii) Hydrolysis yields β-D-2-deoxyribose sugar, thymine base, and phosphoric acid.
(iii) Carbohydrates that contain free aldehydic or ketonic groups and are capable of reducing Tollen's reagent or Fehling's solution. Example: Glucose.

Teacher's Note:
a) Peptide bonds form via condensation between the amino group of one amino acid and the carboxyl group of another.
b) DNA nucleotides consist of a deoxyribose sugar, a nitrogenous base, and a phosphate group.

Question 18 [3 Marks]

Consider the data given below for the reaction A + B → Product
[Table: S.No. | conc. of [A] mol L^-1 | conc. of [B] mol L^-1 | Rate; mol L^-1 sec^-1
1. | 0.1 | 0.1 | 4.0 × 10^-4
2. | 0.2 | 0.2 | 1.6 × 10^-3
3. | 0.5 | 0.1 | 1.0 × 10^-2
4. | 0.5 | 0.5 | 1.0 × 10^-2]
Answer the following questions.
(i) What is the order of reaction with respect to A and B?
(ii) Calculate the rate constant.
(iii) Determine the reaction rate when the concentration of A and B are 0.2 mol L^-1 and 0.35 mol L^-1 respectively.

Answer:
(i) Order with respect to [A] is 2 and with respect to [B] is 0.
(ii) Rate constant \( k = \frac{\text{Rate}}{[\text{A}]^2} = \frac{4.0 \times 10^{-4}}{(0.1)^2} = 4 \times 10^{-2}\text{ mol}^{-1}\text{L sec}^{-1} \).
(iii) Rate = \( k[\text{A}]^2[\text{B}]^0 = 4 \times 10^{-2} \times (0.2)^2 = 1.6 \times 10^{-3}\text{ mol L}^{-1}\text{sec}^{-1} \).

Teacher's Note:
a) Compare experimental runs where one reactant concentration changes while the other remains constant to determine individual reaction orders.
b) Zero-order dependence on B means its concentration changes do not affect the overall reaction rate.

SECTION D - 15 MARKS

Question 19 [5 Marks]

Phenol, one of the deadliest acids, contains a hydroxy group directly attached to the aromatic ring. A hydroxy group attached to an aromatic ring is also referred to as a phenolic group. Due to the presence of -OH group, phenols show many reactions similar to those of alcohols. The direct attachment of -OH group to the aromatic ring makes its behaviour very different from that of alcohols. That is why phenol behaves differently from alcohols in many reactions. The chemical reactions of phenol can be categorised as follows:
• Reactions involving -OH group
• Reactions involving benzene ring
(i) Write the chemical equation for the preparation of phenol from benzene using H₂SO₄ and NaOH.
(ii) What happens when phenol reacts with Br₂/CS₂ at low temperature?
(iii) What is observed when phenol is treated with neutral ferric chloride solution?
(iv) Write the reaction when phenol is treated with conc. HNO₃ in the presence of conc. H₂SO₄.
(v) Write the reaction when phenol is treated with chloroform in the presence of aqueous NaOH at 340K.

Answer:
(i) Benzene → Benzene sulphonic acid → Sodium benzene sulphonate → Sodium phenoxide → Phenol.
(ii) o-Bromophenol and p-bromophenol are formed (with para isomer as major product).
(iii) A violet/purple colored solution/complex is obtained.
(iv) 2,4,6-Trinitrophenol (picric acid) is formed.
(v) Salicylaldehyde is formed (Reimer-Tiemann reaction).

Teacher's Note:
a) Phenol synthesis via sulfonation and alkali fusion is an industrial preparation method.
b) Electrophilic aromatic substitution on phenol occurs readily due to strong activating effects of the hydroxyl group.

Question 20 [5 Marks]

(i) Give a reason for each of the following.
(a) The purple colour of KMnO₄ disappears in the presence of acidified solution of oxalic acid.
(b) Some transition metals and their compounds get attracted towards the magnetic field.
(c) Why are Mn²⁺ compounds more stable than Fe²⁺ towards oxidation to the +3 state? (atomic no. of Mn = 25 and Fe = 26)
(ii) Complete and balance the following reactions.
(a) KMnO₄ + H₂SO₄ + FeSO₄ → ______ + ______ + ______ + ______
(b) K₂Cr₂O₇ + H₂SO₄ + KI → ______ + ______ + ______ + ______

Answer:
(i) (a) Oxidation state of Mn changes from +7 (purple) to +2 (colorless) as MnO₄^- is reduced by oxalic acid.
(b) Due to the presence of unpaired d-electrons.
(c) Mn²⁺ has a stable half-filled d⁵ electronic configuration, whereas Fe²⁺ has a d⁶ configuration and readily loses an electron to achieve a stable d⁵ state.
(ii) (a) 2KMnO₄ + 8H₂SO₄ + 10FeSO₄ → K₂SO₄ + 2MnSO₄ + 5Fe₂(SO₄)₃ + 8H₂O
(b) K₂Cr₂O₇ + 7H₂SO₄ + 6KI → 4K₂SO₄ + Cr₂(SO₄)₃ + 3I₂ + 7H₂O

Teacher's Note:
a) Half-filled and fully-filled d-orbitals confer extra stability due to symmetrical charge distribution and exchange energy.
b) Balance redox equations carefully using ion-electron or oxidation number methods.

Question 21 [5 Marks]

(i) Osmosis and osmotic pressure play a very significant role in biological processes. The osmotic pressure of human blood is 8.21 atm at 27°C. It is interesting to know that a 0.91% (mass/volume) solution of NaCl, known as saline water is isotonic with fluids inside the human red blood cells. In this solution, the blood corpuscles neither swell nor shrink. Therefore, medicines are mixed with saline water before being injected into the veins.
Answer the following questions.
(a) What is an isotonic solution?
(b) How much glucose should be used per 100ml (aqueous solution) for intravenous injection that is isotonic with blood at 27°C?
(c) What would be the osmotic pressure of blood if the temperature is 37°C?
(d) A pure NaCl solution with salt concentration less than 0.91% (mass/volume) is hypertonic solution to human blood. Is this statement true or false? Give a reason.
(e) Name the disease caused by taking a lot of salt or salty food, which results in water retention in tissue cells causing puffiness or swelling.

Answer:
(a) Solutions having the same osmotic pressure at a given temperature are called isotonic solutions.
(b) w = 6 grams of glucose.
(c) π = 8.48 atm.
(d) False, it is hypotonic (lower concentration causes water to enter RBCs, making them swell).
(e) Edema.

Teacher's Note:
a) Isotonic solutions have identical molar concentrations and exert equal osmotic pressures.
b) Osmotic pressure calculations follow the gas-law analogy equation πV = nRT.

OR

(ii) (a) Calculate the boiling point of urea solution when 8.0g of urea is dissolved in 250g of water. Boiling point of pure water is 373K. (K_b of water = 0.52 K kg mol^-1, molecular mass of urea = 60 g mol^-1)
(b) Ethylene glycol is used as an antifreeze agent. Calculate the amount of ethylene glycol to be added to four kilogram of water to prevent it from freezing at -6°C. (K_f of water = 1.86 K kg mol^-1) Assume that ethylene glycol (CH₂OH.CH₂OH) does not dissociate or associate in aqueous solution.
(c) What will be the effect on the value of molality and molarity of a solution with change in temperature?

Answer:
(a) Boiling point of urea solution = 373.277 K
(b) Amount of ethylene glycol = 800 grams
(c) Molality is unaffected by temperature changes because it is based on mass of solvent. Molarity decreases as temperature increases due to thermal expansion and increased volume.

Teacher's Note:
a) Boiling point elevation (ΔT_b = K_b × m) depends directly on solution molality.
b) Molality uses mass units which are temperature-independent, whereas molarity depends on volume, which varies with temperature.