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Samacheer Kalvi Class 12 Chemistry Practice Question Papers

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Brain Grain · braingrain.in
Chemistry — Practice Paper · Set 1
Class: 12Samacheer KalviMax Marks: 93
Name: ____________________Reg No: ____________
Part I — Multiple Choice Questions 15 × 1 = 15

Choose the correct answer. (Answer all questions.)

1.An alkene (A) on ozonolysis gives propanone and aldehyde (B). When (B) is oxidised (C) is obtained. (C) is treated with Br2/P gives (D) which on hydrolysis gives (E). When propanone is treated with HCN followed by hydrolysis gives (E). Identify A, B, C, D and E.[1]
2.The rate law for a reaction of A, B and L has been found to be rate = k [A][B][L]. How would the rate of reaction change when (i) Concentration of [L] is quadrupled (ii) Concentration of both [A] and [B] are doubled (iii) Concentration of [A] is halved (iv) Concentration of [A] is reduced to 1/4 and concentration of [L] is quadrupled.[1]
3.Match items in column - I with the items of column - II and assign the correct code. Column-I: A Borazole B(OH)3 B Boric acid B3N3H6 C Quartz Na2[B4O5(OH)4]·8H2O D Borax SiO2[1]
4.Ksp of Al(OH)3 is 1.0×10^-33. At what pH does 1.0×10^-3 M Al3+ precipitate on the addition of buffer of NH4Cl and NH4OH solution?[1]
5.Write the reason for the anomalous behaviour of Nitrogen.[1]
6.Account for the following: i. Aniline does not undergo Friedel–Crafts reaction. ii. Diazonium salts of aromatic amines are more stable than those of aliphatic amines. iii. pKb of aniline is more than that of methylamine. iv. Gabriel phthalimide synthesis is preferred for synthesising primary amines. v. Ethylamine is soluble in water whereas aniline is not. vi. Amines are more basic than amides. vii. Although –NH2 is o‑ and p‑ directing, nitration of aniline gives a substantial amount of m‑nitroaniline.[1]
7.A Compound (A) with molecular formula C7H5N on acid hydrolysis gives (B) which reacts with thionyl chloride to give compound (C). Benzene reacts with compound (C) in presence of anhydrous AlCl3 to give compound (D). Compound (D) on reduction with Zn/Hg and Conc.HCl gives (E). Identify (A), (B), (C) and (D) and (E). Write the equations.[1]
8.Which one of the following structures represents nylon 6,6 polymer? (intended answer: repeating unit of nylon-6,6)[1]
9.Explain why fluorine always exhibit an oxidation state of -1?[1]
10.Describe adsorption theory of catalysis.[1]
11.Write a short note on electrochemical principles of metallurgy.[1]
12.A double salt which contains fourth period alkali metal (A) on heating at 500K gives (B). Aqueous solution of (B) gives white precipitate with BaCl2 and gives a red colour compound with alizarin. Identify A and B.[1]
13.A hydride of 2nd period alkali metal (A) on reaction with compound of Boron (B) to give a reducing agent (C). identify A, B and C.[1]
14.Identify A, B and C: Benzyl bromide --NaCN/THF--> (A) --H3O+--> (B). i) CO2 ii) H3O+, Mg/ether --> (C) ...[1]
15.Identify A to E in the following sequence of reactions. CH3Cl, AlCl3 → A; HNO3 / H2SO4 → B; Sn / HCl → (C); NaNO2 / HCl → D; CuCN → E; (then oxidation/hydrolysis) → (Major product)[1]
Part II — Short Answer Questions 14 × 2 = 28

Answer briefly. (Answer all questions.)

16.What type of linkages hold together monomers of DNA?[2]
17.Atoms X and Y form bcc crystalline structure. Atom X is present at the corners of the cube and Y is at the centre of the cube. What is the formula of the compound?[2]
18.The concentration of hydroxide ion in a water sample is found to be 2.5 × 10^-6 M. Identify the nature of the solution.[2]
19.Ksp of AgCl is 1.8×10^-10. Calculate molar solubility in 1 M AgNO3[2]
20.The half life of the homogeneous gaseous reaction SO2Cl2 → SO2 + Cl2 which obeys first order kinetics is 8.0 minutes. How long will it take for the concentration of SO2Cl2 to be reduced to 1% of the initial value?[2]
21.What happens when 1-phenyl ethanol is treated with acidified KMnO4.[2]
22.The rate of formation of a dimer in a second order reaction is 7.5 × 10^{-5} mol L^{-1} s^{-1} at 0.05 mol L^{-1} monomer concentration. Calculate the rate constant.[2]
23.The Ka value for HCN is 10^-9. What is the pH of 0.4 M HCN solution?[2]
24.Define pH.[2]
25.Why does bleeding stop by rubbing moist alum[2]
26.Give any three characteristics of ionic crystals.[2]
27.Explain Schottky defect.[2]
28.Differentiate crystalline solids and amorphous solids.[2]
29.Calculate the pH of 0.04 M HNO3 solution.[2]
Part III — Long Answer Questions 10 × 5 = 50

Answer in detail. (Answer all questions.)

30.How do nature of the reactant influence rate of reaction.[5]
31.What are inner transition elements?[5]
32.What is the difference between homogenous and hetrogenous catalysis?[5]
33.From the following data, show that the decomposition of hydrogen peroxide is a reaction of the first order: t (min) 0,10,20; V (ml) 46.1,29.8,19.3. (V is the volume of KMnO4 used.)[5]
34.Why tetrahedral complexes do not exhibit geometrical isomerism.[5]
35.For the reaction 2x + y → L find the rate law from the following data. [x] (M) [y] (M) rate (M s-1) 0.2 0.02 0.15 0.4 0.02 0.30 0.4 0.08 1.20[5]
36.Give two difference between Hormones and vitamins[5]
37.Comment on the statement: Colloid is not a substance but it is a state of substance.[5]
38.Explain why [Ti(H2O)6]3+ is coloured, while [Sc(H2O)6]3+ is colourless.[5]
39.Out of Lu(OH)3 and La(OH)3 which is more basic and why?[5]
🔑 Show Answer Key — Set 1
  1. 1. See solution
  2. 2. (i) 4 times (ii) 4 times (iii) 1/2 times (iv) unchanged (1 times).
  3. 3. A–B3N3H6, B–B(OH)3, C–SiO2, D–Na2[B4O5(OH)4]·8H2O
  4. 4. pH = 4.00 (precipitation occurs when pH > 4.00)
  5. 5. Small size, high electronegativity, absence of d-orbitals, strong tendency to form pπ–pπ multiple bonds and N≡N triple bond.
  6. 6. Short accounts: i) Aniline is strongly deactivated toward Friedel–Crafts because the –NH2 coordinates to Lewis acids (AlCl3), forming anilinium salt or complex; this removes the lone pair from resonance donation, preventing the electrophilic substitution and often destroying the catalyst. ii) Aromatic diazonium salts (Ar–N2+) are stabilized by resonance with the aromatic ring; aliphatic diazonium ions lack such resonance and readily decompose, so they are unstable. iii) pKb: aniline is less basic (higher pKb) than methylamine because the lone pair on N in aniline is delocalized into the benzene ring (resonance), reducing availability for protonation; in methylamine the lone pair is fully available and is electron‑donated by the methyl group (+I), increasing basicity. iv) Gabriel synthesis gives primary amines selectively because the phthalimide anion undergoes alkylation at nitrogen (SN2) and after hydrolysis liberates a primary amine without over‑alkylation (no formation of secondary/tertiary amines). v) Ethylamine is small and can hydrogen‑bond with water to give a soluble salt; aniline is less soluble because the aromatic ring is hydrophobic and the N‑lone pair is involved in resonance, reducing hydrogen bonding and solubility. vi) Amines are more basic than amides because in amides the lone pair on nitrogen is delocalized into the carbonyl (resonance), greatly reducing its availability to accept a proton; in amines the lone pair is localized and more basic. vii) In nitration of aniline under strongly acidic conditions the –NH2 is protonated (to –NH3+), which is a strong meta‑director; moreover to prevent oxidation and protonation aniline is often first acetylated to acetanilide before nitration. Hence direct nitration of aniline gives appreciable m‑product due to protonation under acidic nitrating conditions.
  7. 7. See solution
  8. 8. d
  9. 9. -1
  10. 10. Adsorption theory: catalysis occurs because reactant molecules are adsorbed on the catalyst surface, which brings them together, weakens bonds, provides active centres and proper orientation, thereby lowering activation energy and increasing reaction rate.
  11. 11. Electrochemical metallurgy uses electrode potentials and cell EMFs to predict and carry out extraction/refining of metals by electrolysis or electrowinning.
  12. 12. A = K (potassium); the double salt is potash alum KAl(SO4)2·12H2O. On heating it yields potassium sulfate (B = K2SO4).
  13. 13. A = LiH, B = BCl3 (or BCl3 source), C = LiBH4 (lithium borohydride). Example reaction: 4 LiH + BCl3 → LiBH4 + 3 LiCl.
  14. 14. See solution
  15. 15. A = toluene (C6H5CH3); B = p-nitrotoluene (major) (p-CH3C6H4NO2); C = p-toluidine (p-CH3C6H4NH2); D = p-toluenediazonium salt (p-CH3C6H4–N2+); E = p-cyanotoluene (p-CH3C6H4CN). Major product after hydrolysis/oxidation of the nitrile = p-methylbenzoic acid (p-toluic acid, p-CH3C6H4COOH).
  16. 16. Phosphodiester linkages (and N‑glycosidic bonds between sugar and base).
  17. 17. Y – body centre atom 1 × 1/1 = 1 ∴ The formula is X 1 Y 1 or XY.
  18. 18. Basic (pH ≈ 8.40).
  19. 19. 1.8×10^-10 M
  20. 20. We know that, k = 0.693/ t 1/2 k = 0.693/8.0 minutes = 0.087 minutes -1 For a first order reaction, k = \(\frac { 2.303 }{ k }\) log \(\left( \frac { \left[ { A }_{ 0 } \right] }{ \left[ A \right] } \right)\) t = \(\frac { 2.303 }{ 0.087{ min }^{ -1 } }\) log\(\frac { 100 }{ 1 }\) t = 52.93 mm
  21. 21. 1‑Phenylethanol (a secondary benzylic alcohol) is oxidized to acetophenone (phenyl methyl ketone, PhCOCH3).
  22. 22. k = 0.03 L mol^{-1} s^{-1}
  23. 23. K a =10 -9 c = O.4M pH = – log 10 [H 3 O + ] ∴ pH = – log(2 x 10 -5 ) = – log 2 – log (10 -5 ) = – 0.3010 + 5 pH = 4.699 Question17. Calculate the extent of hydrolysis and the pH of 0.1 M ammonium acetate Given that. K a = K b = 1.8 x 10 -5 K a = K b = 1.8 x 10 -5
  24. 24. pH = −log10[H+], where [H+] is the hydrogen ion concentration in mol L^-1.
  25. 25. Alum causes protein coagulation and vasoconstriction; the precipitated proteins form a plug sealing the wound and stopping bleeding.
  26. 26. 1. High melting and boiling points due to strong electrostatic forces. 2. Hard but brittle (cleave along planes). 3. Conduct electricity only in molten state or in solution (ions mobile); poor conductors as solids.
  27. 27. Schottky defect: paired vacancies of both cations and anions in an ionic solid such that stoichiometry is maintained (e.g., NaCl, KCl). It lowers the density of the solid and increases ionic mobility; its concentration depends on temperature and lattice energy.
  28. 28. Crystalline solids: long-range order, definite geometric shape, sharp melting points, anisotropic. Amorphous solids: no long-range order, no definite shape, show glass transition rather than sharp melting, isotropic.
  29. 29. Concentration of HNO 3 = 0.04M [H 3 O + ] = 0.04 mol dm -3 pH = – 1og[H 3 O + ] = – log (0.04) = – log(4 x 10 -2 ) = 2 – log4 = 2 – 0.6021 = 1.3979 = 1.40 Identify the degree and leading coefficient calculator of polynomial functions.
  30. 30. Nature affects rate via bond strengths, molecular structure, phase (gas/liquid/solid), and electronic factors: weaker bonds and more reactive functional groups react faster. Physical state and surface area (for solids) and presence of ionic vs covalent bonds also matter.
  31. 31. Inner transition elements are elements in which the differentiating (last) electron enters an f-orbital; they comprise the lanthanoids (4f) and actinoids (5f).
  32. 32. Homogeneous catalysis: catalyst and reactants in same phase; Heterogeneous catalysis: catalyst and reactants in different phases (usually solid catalyst with gaseous/liquid reactants).
  33. 33. The reaction is first order. k ≈ 4.35 × 10^-2 min^-1.
  34. 34. In a tetrahedral ML4 complex all four ligand positions are equivalent; swapping positions produces superposable arrangements, so cis/trans type geometrical isomerism (as in octahedral complexes) does not occur for simple tetrahedral complexes.
  35. 35. Rate law: rate = k [x]^1 [y]^1 = k[x][y], with k = 37.5 M^-1 s^-1.
  36. 36. 1) Origin/function: Hormones are endogenous chemical messengers produced by endocrine glands to regulate physiology; vitamins are dietary micronutrients required for normal metabolism. 2) Quantity and role: Hormones act at low concentrations as signaling molecules; vitamins act mainly as coenzymes/antioxidants and are required in small dietary amounts.
  37. 37. A colloid describes a dispersion state (particle size ~1–1000 nm) of a substance in a medium; the same chemical can exist in molecular, colloidal or coarse states depending on condition.
  38. 38. [Ti(H2O)6]3+ is coloured because Ti3+ is d1 (one d electron) — d–d transitions are possible. [Sc(H2O)6]3+ is colourless because Sc3+ is d0 (no d electrons) — no d–d transitions.
  39. 39. La(OH)3 is more basic than Lu(OH)3 because La3+ has a larger ionic radius and is less polarizing, so its hydroxide is less covalent and more basic.
Brain Grain · braingrain.in
Chemistry — Practice Paper · Set 2
Class: 12Samacheer KalviMax Marks: 93
Name: ____________________Reg No: ____________
Part I — Multiple Choice Questions 15 × 1 = 15

Choose the correct answer. (Answer all questions.)

1.How are the following conversions effected(a) propanal into butanone(b) Hex-3-yne into hexan-3-one.(c) phenylmethanal into benzoic acid(d) phenylmethanal into benzoin[1]
2.Deduce the oxidation number of oxygen in hypofluorous acid - HOF.[1]
3.3,3‑Dimethylbutan‑2‑ol on treatment with conc. H2SO4 gives tetramethyl ethylene as a major product. Suggest a suitable mechanism.[1]
4.A dibromo derivative (A) on treatment with KCN followed by acid hydrolysis and heating gives a monobasic acid (B) along with liberation of CO2. (B) on heating with liquid ammonia followed by treating with Br2/KOH gives (C) which on treating with NaNO2 and HCl at low temperature followed by oxidation gives a monobasic acid (D) having molecular mass 74. Identify A to D.[1]
5.Ionic conductances at infinite dilution of Al3+ and SO42− are 189 and 160 mho·cm2·equiv−1 respectively. Calculate the equivalent and molar conductance at infinite dilution of the electrolyte Al2(SO4)3.[1]
6.CO is a reducing agent. Justify with an example.[1]
7.How will you prepare i. Acetic anhydride from acetic acid ii. Ethyl acetate from methyl acetate iii. Acetamide from methylcyanide iv. Lactic acid from ethanal v. Acetophenone from acetylchloride vi. Ethane from sodium acetate vii. Benzoic acid from toluene viii. Malachitegreen from benzaldehyde ix. Cinnamic acid from benzaldehyde x. Acetaldehyde from ethyne[1]
8.How is propanoic acid is prepared starting from (a) an alcohol (b) an alkylhalide (c) an alkene[1]
9.A first order reaction is 40% complete in 50 minutes. Calculate the value of the rate constant. In what time will the reaction be 80% complete?[1]
10.What is linkage isomerism? Explain with an example.[1]
11.For the complex K3[Mn(CN)6], write the oxidation state, coordination number, nature of ligand, magnetic property and electronic configuration in an octahedral crystal field.[1]
12.Write a note on vulcanization of rubber[1]
13.Write a note on Frenkel defect.[1]
14.Why do transition metals show high melting points?[1]
15.Is the following sugar, D - sugar or L - sugar? CHO H OH H OH H OH CH2 - OH[1]
Part II — Short Answer Questions 14 × 2 = 28

Answer briefly. (Answer all questions.)

16.What is the two dimensional coordination number of a molecule in square close packed layer?[2]
17.What is the action of HCN on (i) propanone (ii) 2,4-dichlorobenzaldehyde (iii) ethanal[2]
18.Give one example for each of the following (i) icosagens (ii) tetragens (iii) pnictogens (iv) chalcogens[2]
19.A carbonyl compound A having molecular formula C5H10O forms crystalline precipitate with sodium bisulphite and gives positive iodoform test. A does not reduce Fehling solution. Identify A.[2]
20.A lab assistant prepared a solution by adding a calculated quantity of HCl gas at 25 °C to get a solution with [H3O+] = 4 × 10^-5 M. Is the solution neutral, acidic, or basic?[2]
21.How are vitamins classified[2]
22.The decomposition of Cl2O7 at 500 K in the gas phase to Cl2 and O2 is a first order reaction. After 1 minute at 500 K, the pressure of Cl2O7 falls from 0.08 to 0.04 atm. Calculate the rate constant in s^{-1}.[2]
23.Suggest a suitable reagent to prepare secondary alcohol with identical group using Grignard reagent.[2]
24.Aluminium crystallizes in a cubic close packed structure. Its metallic radius is 125 pm. calculate the edge length of unit cell.[2]
25.What happens when a colloidal sol of Fe(OH)3 and As2S3 are mixed?[2]
26.Name one substance which can act as both analgesic and antipyretic[2]
27.The activation energy of a reaction is 22.5 kcal mol-1 and the value of rate constant at 40°C is 1.8 × 10^-5 s^-1. Calculate the frequency factor, A.[2]
28.Write a note on denaturation of proteins[2]
29.Name the Vitamins whose deficiency cause i) rickets ii) scurvy[2]
Part III — Long Answer Questions 10 × 5 = 50

Answer in detail. (Answer all questions.)

30.Identify X and Y. \(CH_3COCH_2CH_2COOC_2H_5 \xrightarrow{CH_3MgBr} X \xrightarrow{H_3O^+} Y\)[5]
31.Given E°(Fe2+/Fe) = -0.44 V. Two metals M1 and M2 have standard reduction potentials E°(M1) and E°(M2). Which metal is better for coating (electroplating) iron?[5]
32.Describe the structure of diborane.[5]
33.A solution of [Ni(H2O)6]2+ is green, whereas a solution of [Ni(CN)4]2− is colourless. Explain.[5]
34.A zero order reaction is 20% complete in 20 minutes. Calculate the value of the rate constant. In what time will the reaction be 80% complete?[5]
35.Distinguish tetrahedral and octahedral voids.[5]
36.Describe the construction of Daniel cell. Write the cell reaction.[5]
37.Calculate the extent of hydrolysis and the pH of 0.1 M ammonium acetate. Given that Ka (acetic acid) = Kb (NH3) = 1.8×10^-5 and Kw = 1.0×10^-14.[5]
38.Give the uses of silicones.[5]
39.Benzene diazonium chloride in aqueous solution decomposes according to the equation C6H5N2Cl → C6H5Cl + N2. Starting with an initial concentration, the volume of N2 gas obtained at 50 °C at different intervals of time was found to be: Volumes (ml): 19.3, 32.6, 41.3, 46.5, 50.4, 58.3 (final). Show that the above reaction follows first order kinetics. What is the value of the rate constant?[5]
🔑 Show Answer Key — Set 2
  1. 1. See solution
  2. 2. 0
  3. 3. Mechanism: acid‑catalyzed dehydration (E1). Protonation of OH → loss of H2O to give a secondary carbocation; a methyl shift (1,2‑shift) from the gem‑dimethyl center produces a more stable tertiary carbocation; deprotonation of the tertiary carbocation gives the highly substituted alkene (tetramethyl ethylene, i.e. (CH3)2C=C(CH3)2).
  4. 4. A = 1,1-dibromobutane; B = butanoic acid (butyric acid), CH3CH2CH2COOH; C = propylamine, CH3CH2CH2NH2; D = propanoic acid (propionic acid), CH3CH2COOH.
  5. 5. Equivalent conductance Λ°eq = 349 mho·cm2·equiv−1. Molar conductance Λ°m = 2094 mho·cm2·mol−1.
  6. 6. CO reduces metal oxides to metals; e.g., Fe2O3 + 3 CO → 2 Fe + 3 CO2 (in blast furnace) or CuO + CO → Cu + CO2.
  7. 7. i) Acetic anhydride: Dehydrate acetic acid, e.g. 2 CH3COOH \xrightarrow{P2O5, heat} (CH3CO)2O + H2O. Or CH3COOH + CH3COCl → (CH3CO)2O + HCl.
  8. 8. See solution
  9. 9. k ≈ 1.0216 × 10^-2 min^-1. Time for 80% completion ≈ 158 minutes.
  10. 10. Linkage isomerism arises when an ambidentate ligand can bind through two different donor atoms. Example: nitrite ion NO2− binds via N (nitro) or via O (nitrito): [Co(NH3)5(NO2)]2+ (nitro, N‑bound) vs [Co(NH3)5(ONO)]2+ (nitrito, O‑bound).
  11. 11. Oxidation state of Mn: +3 (Mn3+, d4). Coordination number: 6. Nature of ligand: CN− (monodentate, strong‑field). In an octahedral crystal field CN− causes pairing → low‑spin d4 with electronic configuration t2g4 eg0. Magnetic property: paramagnetic with two unpaired electrons.
  12. 12. Vulcanization is the process of cross-linking natural rubber (polyisoprene) chains with sulfur to form S–S and C–S bridges, improving elasticity, strength and heat resistance; typically heating rubber with sulfur and accelerators.
  13. 13. Frenkel defect: cation vacancy–interstitial pair; stoichiometry unchanged.
  14. 14. Because of strong metallic bonding from delocalised d and s electrons and many unpaired d electrons leading to high cohesive energy.
  15. 15. D-sugar
  16. 16. Square close packing – When the spheres of the second row are placed exactly above those of the first row. This way the spheres are aligned horizontally as well as vertically. The arrangement is AAA type. Coordination number is 4.
  17. 17. HCN adds to carbonyl compounds to give cyanohydrins. (i) Propanone (acetone): \((CH3)2CO + HCN \to (CH3)2C(OH)CN\) (acetone cyanohydrin, 2‑hydroxy‑2‑methylpropanenitrile). (ii) 2,4‑Dichlorobenzaldehyde: \(Cl2C6H3CHO + HCN \to Cl2C6H3CH(OH)CN\) (the corresponding aromatic cyanohydrin). (iii) Ethanal (acetaldehyde): \(CH3CHO + HCN \to CH3CH(OH)CN\) (acetaldehyde cyanohydrin, 2‑hydroxypropanenitrile). Cyanohydrins can be further hydrolysed to α‑hydroxy acids on acidic hydrolysis.
  18. 18. (i) Icosagen: Boron (B) (ii) Tetragen: Carbon (C) (iii) Pnictogen: Nitrogen (N) (iv) Chalcogen: Oxygen (O)
  19. 19. Positive iodoform and no Fehling reduction identifies a methyl ketone (CH3CO–R). A C5H10O methyl ketone is pentan‑2‑one (CH3COCH2CH2CH3) or pentan‑2‑one (also written 2‑pentanone). 2‑pentanone forms a bisulphite adduct (bisulphite soluble) and gives iodoform. Therefore A = pentan‑2‑one (2‑pentanone).
  20. 20. Acidic.
  21. 21. Vitamins are classified as fat‑soluble (A, D, E, K) and water‑soluble (vitamin C and B‑complex).
  22. 22. k = 1.155 × 10^{-2} s^{-1}
  23. 23. React a Grignard reagent R–MgX with an aldehyde of the same R group, R–CHO, followed by acid workup: R–MgX + R–CHO → R–CH(OMgX)–R → (H^+/H2O) → R–CH(OH)–R.
  24. 24. a = 2√2 r = 353.6 pm
  25. 25. Mutual coagulation (precipitation) occurs due to neutralization of opposite surface charges.
  26. 26. Aspirin (acetylsalicylic acid) or paracetamol (acetaminophen).
  27. 27. A ≈ 9.1 × 10^10 s^-1.
  28. 28. Denaturation = loss of native 3D structure (secondary/tertiary/quaternary) caused by heat, pH, solvents, detergents or heavy metals, resulting in loss of biological activity; may be reversible or irreversible.
  29. 29. i) Vitamin D ii) Vitamin C (ascorbic acid)
  30. 30. Starting compound is a β‑keto ester (ethyl 4‑oxobutanoate, shown as CH3COCH2CH2COOEt). Addition of methyl Grignard (1 equiv.) attacks the ketone carbonyl to give the magnesium alkoxide intermediate X; acidic workup gives the corresponding tertiary/secondary alcohol Y.
  31. 31. The metal whose E° is more positive than −0.44 V (i.e. more noble than Fe) is better for coating iron.
  32. 32. Diborane B2H6 has a bridged structure with two terminal B–H bonds on each B and two bridging hydrogen atoms forming two three-center two-electron (3c–2e) B–H–B bonds; electron-deficient and C2h symmetry.
  33. 33. [Ni(H2O)6]2+ is an octahedral, weak‑field (water) complex of Ni2+ (d8). Weak‑field ligands give moderate d–d splitting so d–d transitions absorb visible light, producing a green colour. [Ni(CN)4]2− has CN−, a strong‑field ligand, and the complex adopts square‑planar geometry for d8 Ni2+; electrons are paired and d–d transitions are suppressed or shifted out of the visible region, so the complex appears colourless.
  34. 34. k = (0.2[A]0)/20 min = 0.01[A]0 min^-1. Time for 80% completion = 80 min.
  35. 35. Tetrahedral void: formed by four atoms surrounding a small sphere; coordination number 4; number of tetrahedral voids = 2 per atom in close packing; radius ratio r/R ≈ 0.225 for fit. Octahedral void: formed by six atoms; coordination number 6; number of octahedral voids = 1 per atom; radius ratio r/R ≈ 0.414.
  36. 36. Daniel cell: two half-cells: Zn(s) | Zn2+(aq) (usually ZnSO4) and Cu(s) | Cu2+(aq) (CuSO4). Salt bridge (porous pot or KNO3 bridge) completes circuit. Anode (Zn): Zn → Zn2+ + 2e- (oxidation). Cathode (Cu): Cu2+ + 2e- → Cu (reduction). Overall: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s).
  37. 37. Extent of hydrolysis h ≈ 7.45×10^-5; pH = 7.00 (neutral).
  38. 38. Silicones (polysiloxanes, e.g., PDMS) are used as sealants, lubricants, heat-resistant coatings, electrical insulators, moulding/adhesive materials, medical implants and devices, water-repellent treatments and in cosmetics.
  39. 39. The data fit first order: ln(V∞ − Vt) vs t is linear. Estimated k ≈ 8.0 × 10^-2 min^-1.
Brain Grain · braingrain.in
Chemistry — Practice Paper · Set 3
Class: 12Samacheer KalviMax Marks: 93
Name: ____________________Reg No: ____________
Part I — Multiple Choice Questions 15 × 1 = 15

Choose the correct answer. (Answer all questions.)

1.An alkene (A) on ozonolysis gives propanone and aldehyde (B). When (B) is oxidised (C) is obtained. (C) is treated with Br2/P gives (D) which on hydrolysis gives (E). When propanone is treated with HCN followed by hydrolysis gives (E). Identify A, B, C, D and E.[1]
2.The rate law for a reaction of A, B and L has been found to be rate = k [A][B][L]. How would the rate of reaction change when (i) Concentration of [L] is quadrupled (ii) Concentration of both [A] and [B] are doubled (iii) Concentration of [A] is halved (iv) Concentration of [A] is reduced to 1/4 and concentration of [L] is quadrupled.[1]
3.Match items in column - I with the items of column - II and assign the correct code. Column-I: A Borazole B(OH)3 B Boric acid B3N3H6 C Quartz Na2[B4O5(OH)4]·8H2O D Borax SiO2[1]
4.Ksp of Al(OH)3 is 1.0×10^-33. At what pH does 1.0×10^-3 M Al3+ precipitate on the addition of buffer of NH4Cl and NH4OH solution?[1]
5.Write the reason for the anomalous behaviour of Nitrogen.[1]
6.Account for the following: i. Aniline does not undergo Friedel–Crafts reaction. ii. Diazonium salts of aromatic amines are more stable than those of aliphatic amines. iii. pKb of aniline is more than that of methylamine. iv. Gabriel phthalimide synthesis is preferred for synthesising primary amines. v. Ethylamine is soluble in water whereas aniline is not. vi. Amines are more basic than amides. vii. Although –NH2 is o‑ and p‑ directing, nitration of aniline gives a substantial amount of m‑nitroaniline.[1]
7.A Compound (A) with molecular formula C7H5N on acid hydrolysis gives (B) which reacts with thionyl chloride to give compound (C). Benzene reacts with compound (C) in presence of anhydrous AlCl3 to give compound (D). Compound (D) on reduction with Zn/Hg and Conc.HCl gives (E). Identify (A), (B), (C) and (D) and (E). Write the equations.[1]
8.Which one of the following structures represents nylon 6,6 polymer? (intended answer: repeating unit of nylon-6,6)[1]
9.Explain why fluorine always exhibit an oxidation state of -1?[1]
10.Describe adsorption theory of catalysis.[1]
11.Write a short note on electrochemical principles of metallurgy.[1]
12.A double salt which contains fourth period alkali metal (A) on heating at 500K gives (B). Aqueous solution of (B) gives white precipitate with BaCl2 and gives a red colour compound with alizarin. Identify A and B.[1]
13.A hydride of 2nd period alkali metal (A) on reaction with compound of Boron (B) to give a reducing agent (C). identify A, B and C.[1]
14.Identify A, B and C: Benzyl bromide --NaCN/THF--> (A) --H3O+--> (B). i) CO2 ii) H3O+, Mg/ether --> (C) ...[1]
15.Identify A to E in the following sequence of reactions. CH3Cl, AlCl3 → A; HNO3 / H2SO4 → B; Sn / HCl → (C); NaNO2 / HCl → D; CuCN → E; (then oxidation/hydrolysis) → (Major product)[1]
Part II — Short Answer Questions 14 × 2 = 28

Answer briefly. (Answer all questions.)

16.Identify A, B and C: benzoic acid --PCl5--> A; Benzene + Anhydrous AlCl3 + A --> B; H+ + C2H5OH on B gives C; C6H5MgBr ...[2]
17.Why ionic crystals are hard and brittle?[2]
18.What is the role of quick lime in the extraction of Iron from its oxide Fe2O3?[2]
19.What are harmones? Give examples[2]
20.Sodium metal crystallizes in bcc structure with the edge length of the unit cell 4 3 . × -cm. calculate the radius of sodium atom.[2]
21.Describe the graphical representation of first order reaction.[2]
22.How will you convert diethylamine into i) N,N‑diethylacetamide ii) N‑nitrosodiethylamine?[2]
23.A saturated solution, prepared by dissolving CaF2(s) in water, has [Ca2+] = 3.3×10^-4 M. What is the Ksp of CaF2 ?[2]
24.If NaCl is doped with 10-2 mol percentage of strontium chloride, what is the concentration of cation vacancy?[2]
25.Give an example of coordination compound used in medicine and two examples of biologically important coordination compounds.[2]
26.Write the expression for the solubility product of Hg2Cl2.[2]
27.Which will be adsorbed more readily on the surface of charcoal and why? NH3 or O2 ?[2]
28.What are Lewis acids and bases? Give two examples for each.[2]
29.When aqueous ammonia is added to CuSO4 solution, the solution turns deep blue due to the formation of tetramminecopper(II) complex, [Cu(H2O)4]2+ + 4NH3 ⇌ [Cu(NH3)4]2+ + 4H2O. Between H2O and NH3 which is the stronger Lewis base?[2]
Part III — Long Answer Questions 10 × 5 = 50

Answer in detail. (Answer all questions.)

30.Give the uses of sulphuric acid.[5]
31.Write the postulates of Werner's theory.[5]
32.Explain the variation in E°(M3+/M2+) in 3d series.[5]
33.Explain Kolbe's reaction[5]
34.Calculate the standard emf of the cell Cd(s)|Cd2+||Cu2+|Cu(s) and determine the cell reaction. Given E°(Cu2+/Cu) = +0.34 V and E°(Cd2+/Cd) = −0.40 V. Predict feasibility.[5]
35.Complete the following reactions: a) MnO4^2- + H+ -> ?; b) C6H5CH3 --(acidified KMnO4)--> ?; c) MnO4^- + Fe2+ + H+ -> ?; d) KMnO4 --(red heat)--> ?; e) Cr2O7^2- + I^- + H+ -> ?; f) Na2Cr2O7 + KCl -> ?[5]
36.Why is anode in galvanic cell considered to be negative and cathode positive electrode?[5]
37.What are narcotic and non - narcotic drugs. Give examples[5]
38.Write the structure of α-D (+) glucophyranose[5]
39.Explain briefly how +2 states becomes more and more stable in the first half of the first row transition elements with increasing atomic number.[5]
🔑 Show Answer Key — Set 3
  1. 1. See solution
  2. 2. (i) 4 times (ii) 4 times (iii) 1/2 times (iv) unchanged (1 times).
  3. 3. A–B3N3H6, B–B(OH)3, C–SiO2, D–Na2[B4O5(OH)4]·8H2O
  4. 4. pH = 4.00 (precipitation occurs when pH > 4.00)
  5. 5. Small size, high electronegativity, absence of d-orbitals, strong tendency to form pπ–pπ multiple bonds and N≡N triple bond.
  6. 6. Short accounts: i) Aniline is strongly deactivated toward Friedel–Crafts because the –NH2 coordinates to Lewis acids (AlCl3), forming anilinium salt or complex; this removes the lone pair from resonance donation, preventing the electrophilic substitution and often destroying the catalyst. ii) Aromatic diazonium salts (Ar–N2+) are stabilized by resonance with the aromatic ring; aliphatic diazonium ions lack such resonance and readily decompose, so they are unstable. iii) pKb: aniline is less basic (higher pKb) than methylamine because the lone pair on N in aniline is delocalized into the benzene ring (resonance), reducing availability for protonation; in methylamine the lone pair is fully available and is electron‑donated by the methyl group (+I), increasing basicity. iv) Gabriel synthesis gives primary amines selectively because the phthalimide anion undergoes alkylation at nitrogen (SN2) and after hydrolysis liberates a primary amine without over‑alkylation (no formation of secondary/tertiary amines). v) Ethylamine is small and can hydrogen‑bond with water to give a soluble salt; aniline is less soluble because the aromatic ring is hydrophobic and the N‑lone pair is involved in resonance, reducing hydrogen bonding and solubility. vi) Amines are more basic than amides because in amides the lone pair on nitrogen is delocalized into the carbonyl (resonance), greatly reducing its availability to accept a proton; in amines the lone pair is localized and more basic. vii) In nitration of aniline under strongly acidic conditions the –NH2 is protonated (to –NH3+), which is a strong meta‑director; moreover to prevent oxidation and protonation aniline is often first acetylated to acetanilide before nitration. Hence direct nitration of aniline gives appreciable m‑product due to protonation under acidic nitrating conditions.
  7. 7. See solution
  8. 8. d
  9. 9. -1
  10. 10. Adsorption theory: catalysis occurs because reactant molecules are adsorbed on the catalyst surface, which brings them together, weakens bonds, provides active centres and proper orientation, thereby lowering activation energy and increasing reaction rate.
  11. 11. Electrochemical metallurgy uses electrode potentials and cell EMFs to predict and carry out extraction/refining of metals by electrolysis or electrowinning.
  12. 12. A = K (potassium); the double salt is potash alum KAl(SO4)2·12H2O. On heating it yields potassium sulfate (B = K2SO4).
  13. 13. A = LiH, B = BCl3 (or BCl3 source), C = LiBH4 (lithium borohydride). Example reaction: 4 LiH + BCl3 → LiBH4 + 3 LiCl.
  14. 14. See solution
  15. 15. A = toluene (C6H5CH3); B = p-nitrotoluene (major) (p-CH3C6H4NO2); C = p-toluidine (p-CH3C6H4NH2); D = p-toluenediazonium salt (p-CH3C6H4–N2+); E = p-cyanotoluene (p-CH3C6H4CN). Major product after hydrolysis/oxidation of the nitrile = p-methylbenzoic acid (p-toluic acid, p-CH3C6H4COOH).
  16. 16. benzoic acid --(PCl5)--> A = benzoyl chloride (C6H5COCl). Benzene + benzoyl chloride/AlCl3 --> B = benzophenone (C6H5COC6H5) (Friedel–Crafts acylation). B + H^+ / C2H5OH (acidic ethanol) converts the ketone to its acetal (diethyl acetal): C = benzophenone diethyl acetal, (C6H5)2C(OC2H5)2. (Alternatively, protonation/acetalization of the ketone with ethanol produces the corresponding ketal/acetal.)
  17. 17. Hardness: strong electrostatic (ionic) bonds hold ions rigidly in place. Brittleness: when a stress shifts ionic planes, like-charged ions can be brought adjacent causing strong repulsive forces and the crystal cleaves along planes.
  18. 18. Quicklime (CaO) acts as a flux to form a removable slag with acidic impurities (silica).
  19. 19. Hormones are chemical messengers secreted by endocrine glands into the blood to regulate target organs; examples: insulin (peptide), adrenaline (amine), estrogen/testosterone (steroids).
  20. 20. r ≈ 1.86×10^{-8} cm = 186 pm
  21. 21. For first order: [A] = [A]0 e^{-kt}. Plots: (i) ln[A] vs t is a straight line with slope −k and intercept ln[A]0. (ii) log10[A] vs t is straight with slope −k/2.303. (iii) [A] vs t is an exponential decay curve. From ln[A] vs t one obtains k from slope.
  22. 22. i) Diethylamine (Et2NH) + acetyl chloride (CH3COCl) or acetic anhydride → N,N‑diethylacetamide (Et2N–COCH3) via acylation (Schotten–Baumann conditions or pyridine). ii) N‑nitrosation: Diethylamine treated with NaNO2/HCl at 0–5 °C (or with nitrosyl chloride) gives N‑nitrosodiethylamine (Et2N–NO).
  23. 23. 1.44×10^-10
  24. 24. 1.0×10^-4 (fraction) = 0.01% vacancies
  25. 25. Medicine: cisplatin, Pt(NH3)2Cl2. Biologically important: heme (iron porphyrin in haemoglobin) and chlorophyll (magnesium porphyrin); also vitamin B12 (cobalamin).
  26. 26. Ksp = [Hg2^{2+}][Cl^-]^2
  27. 27. NH3 will be adsorbed more readily.
  28. 28. Lewis acid: electron-pair acceptor. Examples: BF3, AlCl3. Lewis base: electron-pair donor. Examples: NH3, OH−.
  29. 29. NH3 is the stronger Lewis base.
  30. 30. Fertiliser production, chemical manufacture, petroleum refining, dehydrating agent, lead–acid batteries, metal processing.
  31. 31. Werner's postulates (concise): 1) Metals have two kinds of valences — primary (ionisable, oxidation state) and secondary (non‑ionisable, coordination number). 2) Primary valences are satisfied by negative ions; secondary valences are directed in space and fixed in number giving definite geometry (e.g. 4 → tetrahedral/square planar, 6 → octahedral). 3) Secondary valences are responsible for stereochemistry (isomerism) of complexes. 4) In solution, complexes retain the arrangement of ligands (secondary valences) while primary valence ions may be ionisable.
  32. 32. E°(M3+/M2+) varies across the 3d series because of changes in third ionization energy, ionic radius (hence hydration energy), electronic configuration (stability of dN vs dN−1), and ligand/hydration stabilization; these combined factors cause a non-linear trend with peaks where d-electron configurations are especially stable.
  33. 33. Kolbe–Schmitt (Kolbe) reaction: carboxylation of sodium phenoxide with CO2 under pressure and heat to give salicylate; acidification yields salicylic acid (ortho‑hydroxybenzoic acid).
  34. 34. E°cell = 0.74 V. Cell reaction: Cd(s) + Cu2+ → Cd2+ + Cu(s). Reaction is spontaneous (feasible).
  35. 35. 3MnO4^2- + 4H+ -> 2MnO4^- + MnO2 + 2H2O C6H5CH3 --(acidified KMnO4)--> C6H5COOH 2MnO4^- + 10Fe2+ + 16H+ -> 2Mn2+ + 10Fe3+ + 8H2O 2KMnO4 --(red heat)--> K2MnO4 + MnO2 + O2 Cr2O7^2- + 6I^- + 14H+ -> 2Cr3+ + 3I2 + 7H2O Na2Cr2O7 + 2KCl -> K2Cr2O7 + 2NaCl
  36. 36. In a galvanic cell oxidation at the anode releases electrons into the electrode, making it electron-rich (negative). Electrons flow through the external circuit to the cathode where reduction consumes electrons, making the cathode electron-deficient relative to anode (positive).
  37. 37. Narcotics are addictive drugs that produce sleep and relieve pain (e.g., morphine, heroin, codeine). Non-narcotic drugs provide pain relief or other effects without narcotic-type addiction (e.g., aspirin, paracetamol, ibuprofen).
  38. 38. α-D-Glucopyranose: six-membered (pyranose) ring with the anomeric OH (C1) axial (down) and CH2OH at C5 up.
  39. 39. +2 states become increasingly stable across the first half of the 3d series because the third ionization energy (to form M3+) increases as nuclear charge and 3d electron binding increase; thus removing only two electrons (giving M2+) is relatively easier and more stable.

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