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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.Write a short note on electrochemical principles of metallurgy.[1]
2.Write the reason for the anomalous behaviour of Nitrogen.[1]
3.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]
4.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]
5.3,3‑Dimethylbutan‑2‑ol on treatment with conc. H2SO4 gives tetramethyl ethylene as a major product. Suggest a suitable mechanism.[1]
6.Write a note on Frenkel defect.[1]
7.Deduce the oxidation number of oxygen in hypofluorous acid - HOF.[1]
8.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]
9.Is the following sugar, D - sugar or L - sugar? CHO H OH H OH H OH CH2 - OH[1]
10.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]
11.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]
12.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]
13.Why do transition metals show high melting points?[1]
14.Write a note on vulcanization of rubber[1]
15.Explain why fluorine always exhibit an oxidation state of -1?[1]
Part II — Short Answer Questions 14 × 2 = 28

Answer briefly. (Answer all questions.)

16.Ksp of Ag2CrO4 is 1.1×10^-12. What is solubility of Ag2CrO4 in 0.1 M K2CrO4?[2]
17.Write the structure of the major product of the aldol condensation of benzaldehyde with acetone.[2]
18.How will you convert diethylamine into i) N,N‑diethylacetamide ii) N‑nitrosodiethylamine?[2]
19.What type of linkages hold together monomers of DNA?[2]
20.What is the role of quick lime in the extraction of Iron from its oxide Fe2O3?[2]
21.What are reducing and non - reducing sugars?[2]
22.What is the major product obtained when two moles of ethyl magnesium bromide is treated with methyl benzoate followed by acid hydrolysis.[2]
23.Give an example of coordination compound used in medicine and two examples of biologically important coordination compounds.[2]
24.Give any three characteristics of ionic crystals.[2]
25.What are Lewis acids and bases? Give two examples for each.[2]
26.What is the coordination entity formed when excess of liquid ammonia is added to an aqueous solution of copper sulphate?[2]
27.Identify A, B and C: benzoic acid --PCl5--> A; Benzene + Anhydrous AlCl3 + A --> B; H+ + C2H5OH on B gives C; C6H5MgBr ...[2]
28.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]
29.Aluminium crystallizes in a cubic close packed structure. Its metallic radius is 125 pm. calculate the edge length of unit cell.[2]
Part III — Long Answer Questions 10 × 5 = 50

Answer in detail. (Answer all questions.)

30.State Kohlrausch Law. How is it useful to determine the molar conductivity of weak electrolyte at infinite dilution.[5]
31.Describe briefly allotropism in p- block elements with specific reference to carbon.[5]
32.Explain common ion effect with an example.[5]
33.Explain the principle of electrolytic refining with an example.[5]
34.Why is the first ionisation enthalpy of chromium lower than that of zinc?[5]
35.Account for the acidic nature of HClO4 in terms of Bronsted–Lowry theory. Identify its conjugate base.[5]
36.Write short note on metal excess and metal deficiency defect with an example.[5]
37.Explain the rate determining step with an example.[5]
38.Explain the electrometallurgy of aluminium.[5]
39.Complete the following reactions a b c . . . B OH +NH Na B O + H SO H O B H + 2NaOH ( ) →  + →  + 2H O B H + CH OH BF + 9 H O HCOOH + H SO →  →  →  d e f . . . SiCl + NH SiCl + C H OH B + NaOH →  →  →  →  g h i . . . j. H B O Red hot  → [5]
🔑 Show Answer Key — Set 1
  1. 1. Electrochemical metallurgy uses electrode potentials and cell EMFs to predict and carry out extraction/refining of metals by electrolysis or electrowinning.
  2. 2. Small size, high electronegativity, absence of d-orbitals, strong tendency to form pπ–pπ multiple bonds and N≡N triple bond.
  3. 3. 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).
  4. 4. 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.
  5. 5. 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).
  6. 6. Frenkel defect: cation vacancy–interstitial pair; stoichiometry unchanged.
  7. 7. 0
  8. 8. See solution
  9. 9. D-sugar
  10. 10. i) Acetic anhydride: Dehydrate acetic acid, e.g. 2 CH3COOH \xrightarrow{P2O5, heat} (CH3CO)2O + H2O. Or CH3COOH + CH3COCl → (CH3CO)2O + HCl.
  11. 11. k ≈ 1.0216 × 10^-2 min^-1. Time for 80% completion ≈ 158 minutes.
  12. 12. Equivalent conductance Λ°eq = 349 mho·cm2·equiv−1. Molar conductance Λ°m = 2094 mho·cm2·mol−1.
  13. 13. Because of strong metallic bonding from delocalised d and s electrons and many unpaired d electrons leading to high cohesive energy.
  14. 14. 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.
  15. 15. -1
  16. 16. 1.66×10^-6 M
  17. 17. Acetone (enolate donor) condenses with two equivalents of benzaldehyde to give dibenzalacetone (major product), structure: PhCH=CH–CO–CH=CHPh (trans,trans‑dibenzalacetone). (This is the crossed aldol condensation product where both α‑positions of acetone are condensed with benzaldehyde.)
  18. 18. 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).
  19. 19. Phosphodiester linkages (and N‑glycosidic bonds between sugar and base).
  20. 20. Quicklime (CaO) acts as a flux to form a removable slag with acidic impurities (silica).
  21. 21. Reducing sugars have a free aldehyde or free ketose (in open chain) that can reduce mild oxidants (e.g., glucose, fructose, maltose); non‑reducing sugars lack a free anomeric hydroxyl (e.g., sucrose).
  22. 22. Tertiary alcohol: phenyl(ethyl)2-carbinol i.e. Ph–C(OH)(Et)2 (a tertiary alcohol obtained by double addition).
  23. 23. Medicine: cisplatin, Pt(NH3)2Cl2. Biologically important: heme (iron porphyrin in haemoglobin) and chlorophyll (magnesium porphyrin); also vitamin B12 (cobalamin).
  24. 24. 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.
  25. 25. Lewis acid: electron-pair acceptor. Examples: BF3, AlCl3. Lewis base: electron-pair donor. Examples: NH3, OH−.
  26. 26. [Cu(NH3)4(H2O)2]^{2+} (tetraamminecopper(II) complex, deep blue).
  27. 27. 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.)
  28. 28. 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
  29. 29. a = 2√2 r = 353.6 pm
  30. 30. Kohlrausch's law of independent migration: at infinite dilution the molar conductivity Λm° of an electrolyte equals the sum of the ionic contributions: Λm° = ν+ λ+° + ν- λ-°, where λ±° are limiting ionic molar conductivities and ν± stoichiometric numbers. For a weak electrolyte, Λm measured at various dilutions can be extrapolated to infinite dilution, or Λm° obtained by adding λ° values of constituent ions determined from strong electrolytes; then degree of dissociation and K can be evaluated.
  31. 31. Allotropy in p-block elements — example of carbon
  32. 32. Addition of an ion common to an equilibrium suppresses the ionization or solubility of the weak electrolyte. Example: Adding NaCl to a saturated AgCl solution (which provides Cl−) decreases AgCl solubility because AgCl ⇌ Ag+ + Cl− shifts left.
  33. 33. Impure metal anode dissolves; pure metal plates at cathode; impurities either remain in solution or collect as anode sludge. Example: electrolytic refining of copper.
  34. 34. Because Cr has a single 4s electron that is weakly held and its removal gives the stable half‑filled 3d5 configuration; Zn has a filled 3d10 and higher effective nuclear charge.
  35. 35. HClO4 is a strong Bronsted–Lowry acid because it readily donates H+ to solvents (e.g. water), producing H3O+. Its conjugate base is ClO4− (perchlorate), which is highly stabilized by resonance.
  36. 36. Metal excess: occurs when there are extra metal atoms or electrons (e.g. Na excess in NaCl gives F-centres — electrons trapped in anion vacancies causing colour). Metal deficiency: occurs when metal sites are vacant or metals have mixed valency causing vacancies (e.g. FeO is metal deficient: some Fe2+ oxidises to Fe3+ and vacancies form).
  37. 37. Rate-determining step (RDS) is the slowest step in a reaction mechanism which controls the overall rate. Example: mechanism: (1) A + B → I (slow) (2) I + B → products (fast). Overall rate ≈ rate of step (1) → rate = k1 [A][B]. The RDS sets the observable rate law because faster steps equilibrate quickly relative to the slow step.
  38. 38. Hall–Héroult electrolytic process: Al2O3 dissolved in molten cryolite is electrolysed to produce Al.
  39. 39. Reconstructed common transformations (a–j) related to boron and silicon chemistry: (a) Na2B4O7·10H2O + H2SO4 + 5 H2O → 4 H3BO3 + Na2SO4 (b) H3BO3 + NaOH → Na[B(OH)4] (or on heating → NaBO2 + H2O) (c) BF3 + 3 H2O → B(OH)3 + 3 HF (d) H3BO3 + CH3OH ⇌ B(OCH3)3 + 3 H2O (esterification) (e) HCOOH —(conc. H2SO4)→ CO + H2O (dehydration) (f) B2H6 + 6 H2O → 2 B(OH)3 + 6 H2 (hydrolysis) (g) SiCl4 + 4 NH3 → Si3N4 + 12 HCl (formation of silicon nitride under forcing conditions) (h) SiCl4 + 4 C2H5OH → Si(OC2H5)4 + 4 HCl (formation of tetraalkoxysilane) (i) B + NaOH → (depends on conditions) e.g. B2O3 + 2 NaOH → 2 NaBO2 + H2O (j) 2 H3BO3 —(red hot)→ B2O3 + 3 H2O
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.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]
2.Identify A, B and C: Benzyl bromide --NaCN/THF--> (A) --H3O+--> (B). i) CO2 ii) H3O+, Mg/ether --> (C) ...[1]
3.CO is a reducing agent. Justify with an example.[1]
4.Which one of the following structures represents nylon 6,6 polymer? (intended answer: repeating unit of nylon-6,6)[1]
5.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]
6.What is linkage isomerism? Explain with an example.[1]
7.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]
8.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]
9.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]
10.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]
11.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]
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.Describe adsorption theory of catalysis.[1]
14.How is propanoic acid is prepared starting from (a) an alcohol (b) an alkylhalide (c) an alkene[1]
15.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]
Part II — Short Answer Questions 14 × 2 = 28

Answer briefly. (Answer all questions.)

16.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]
17.How are vitamins classified[2]
18.Give the uses of helium.[2]
19.Sodium metal crystallizes in bcc structure with the edge length of the unit cell 4 3 . × -cm. calculate the radius of sodium atom.[2]
20.Give the uses of argon.[2]
21.Give three uses of emulsions.[2]
22.Give one example for each of the following (i) icosagens (ii) tetragens (iii) pnictogens (iv) chalcogens[2]
23.Name the Vitamins whose deficiency cause i) rickets ii) scurvy[2]
24.The rate constant for a first order reaction is 1.54 × 10^-3 s^-1. Calculate its half life time.[2]
25.If NaCl is doped with 10-2 mol percentage of strontium chloride, what is the concentration of cation vacancy?[2]
26.50 mL of 0.05 M HNO3 is mixed with 50 mL of 0.025 M KOH. Calculate the pH of the resultant solution.[2]
27.Define unit cell.[2]
28.Ksp of AgCl is 1.8×10^-10. Calculate molar solubility in 1 M AgNO3[2]
29.Describe the graphical representation of first order reaction.[2]
Part III — Long Answer Questions 10 × 5 = 50

Answer in detail. (Answer all questions.)

30.Derive an expression for Ostwald's dilution law for a weak acid.[5]
31.Using the Ellingham diagram, (A) Predict the conditions under which (i) Aluminium might be expected to reduce magnesia. (ii) Magnesium could reduce alumina. (B) is it possible to reduce Fe2O3 by coke at a temperature around 1200K?[5]
32.Draw all possible geometrical isomers of the complex [Co(en)2Cl2]+ and identify the optically active isomer.[5]
33.Which of 0.1 M HCl and 0.1 M KCl do you expect to have greater Λm and why?[5]
34.Actinoid contraction is greater from element to element than the lanthanoid contraction, why?[5]
35.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]
36.Identify A, B and C: HOOC-(CH2)3-COOH --(SOCl2)--> A --(NH3)--> B --(LiAlH4)--> C.[5]
37.Is it possible to oxidise t - butyl alcohol using acidified dichromate to form a carbonyl compound.[5]
38.How will you identify borate radical?[5]
39.An atom crystallizes in fcc crystal lattice and has a density of 10 gcm - with unit cell edge length of 100pm. calculate the number of atoms present in 1 g of crystal.[5]
🔑 Show Answer Key — Set 2
  1. 1. 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…
  2. 2. See solution
  3. 3. CO reduces metal oxides to metals; e.g., Fe2O3 + 3 CO → 2 Fe + 3 CO2 (in blast furnace) or CuO + CO → Cu + CO2.
  4. 4. d
  5. 5. See solution
  6. 6. 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).
  7. 7. A–B3N3H6, B–B(OH)3, C–SiO2, D–Na2[B4O5(OH)4]·8H2O
  8. 8. See solution
  9. 9. pH = 4.00 (precipitation occurs when pH > 4.00)
  10. 10. A = 1,1-dibromobutane; B = butanoic acid (butyric acid), CH3CH2CH2COOH; C = propylamine, CH3CH2CH2NH2; D = propanoic acid (propionic acid), CH3CH2COOH.
  11. 11. (i) 4 times (ii) 4 times (iii) 1/2 times (iv) unchanged (1 times).
  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. 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.
  14. 14. See solution
  15. 15. A = LiH, B = BCl3 (or BCl3 source), C = LiBH4 (lithium borohydride). Example reaction: 4 LiH + BCl3 → LiBH4 + 3 LiCl.
  16. 16. k = 1.155 × 10^{-2} s^{-1}
  17. 17. Vitamins are classified as fat‑soluble (A, D, E, K) and water‑soluble (vitamin C and B‑complex).
  18. 18. Common uses of helium include: Filling balloons and airships (non-flammable, lighter-than-air gas). Breathing gas mixtures for deep-sea diving to avoid nitrogen narcosis. Liquid helium for cryogenics and cooling superconducting magnets (e.g., MRI machines). Inert shielding gas in welding and in semiconductor production. Leak detection and as an inert atmosphere for manufacture of reactive metals.
  19. 19. r ≈ 1.86×10^{-8} cm = 186 pm
  20. 20. Uses of argon include: Inert shielding gas in arc welding (prevents oxidation of the weld). Filling incandescent and fluorescent lamps to prevent filament oxidation and prolong life. Providing an inert atmosphere in the production of reactive metals and in semiconductor manufacturing. Used in gas-discharge tubes and as an inert blanket in laboratory and industrial processes.
  21. 21. Uses: food products (milk, mayonnaise), pharmaceuticals/cosmetics (creams, lotions), and paints/coatings or insecticide formulations.
  22. 22. (i) Icosagen: Boron (B) (ii) Tetragen: Carbon (C) (iii) Pnictogen: Nitrogen (N) (iv) Chalcogen: Oxygen (O)
  23. 23. i) Vitamin D ii) Vitamin C (ascorbic acid)
  24. 24. t1/2 = 450 s (approximately).
  25. 25. 1.0×10^-4 (fraction) = 0.01% vacancies
  26. 26. pH ≈ 1.90
  27. 27. A unit cell is the smallest repeating structural unit of a crystal lattice that, by translation in three dimensions, reproduces the entire crystal. It is defined by the edge lengths a, b, c and interaxial angles α, β, γ and contains the arrangement of atoms for the crystal.
  28. 28. 1.8×10^-10 M
  29. 29. 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.
  30. 30. For a weak acid HA at concentration c with degree of dissociation α: Ka = \dfrac{α^2 c}{1−α}.
  31. 31. (A)(i) Aluminium cannot reduce MgO under normal conditions; (ii) Magnesium cannot reduce Al2O3. (B) Yes — Fe2O3 can be reduced by coke (via CO/CO2) near 1200 K.
  32. 32. For [Co(en)2Cl2]+ (octahedral, en = bidentate) there are two geometric types: cis and trans. The trans isomer is achiral. The cis isomer exists as a pair of enantiomers (Δ and Λ) and is optically active.
  33. 33. 0.1 M HCl has greater molar conductivity. Although both are strong electrolytes at same concentration, H+ has very high ionic mobility (very large λ°) compared to K+, so Λm(HCl) > Λm(KCl).
  34. 34. Actinoid contraction per element is greater because 5f electrons shield the increasing nuclear charge even less effectively than 4f electrons, and relativistic effects and stronger penetration of 5f orbitals enhance the contraction.
  35. 35. The data fit first order: ln(V∞ − Vt) vs t is linear. Estimated k ≈ 8.0 × 10^-2 min^-1.
  36. 36. A = glutaryl chloride, ClCO-(CH2)3-COCl; B = glutaramide (pentane-1,5-diamide), NH2CO-(CH2)3-CONH2; C = pentane-1,5-diamine, H2N-(CH2)5-NH2. HOOC-(CH2)3-COOH --(SOCl2)--> ClCO-(CH2)3-COCl --(NH3)--> NH2CO-(CH2)3-CONH2 --(LiAlH4)--> H2N-(CH2)5-NH2.
  37. 37. No. Tertiary alcohols (like t‑butyl alcohol) lack a hydrogen on the carbon bearing the OH and cannot be oxidized to a carbonyl by acidified dichromate under normal conditions.
  38. 38. Convert the borate to trimethyl borate by adding methanol and conc. H2SO4; the ester on ignition burns with a characteristic green flame (boron test).
  39. 39. 4.0×10^{23} atoms
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.Write a short note on electrochemical principles of metallurgy.[1]
2.Write the reason for the anomalous behaviour of Nitrogen.[1]
3.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]
4.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]
5.3,3‑Dimethylbutan‑2‑ol on treatment with conc. H2SO4 gives tetramethyl ethylene as a major product. Suggest a suitable mechanism.[1]
6.Write a note on Frenkel defect.[1]
7.Deduce the oxidation number of oxygen in hypofluorous acid - HOF.[1]
8.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]
9.Is the following sugar, D - sugar or L - sugar? CHO H OH H OH H OH CH2 - OH[1]
10.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]
11.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]
12.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]
13.Why do transition metals show high melting points?[1]
14.Write a note on vulcanization of rubber[1]
15.Explain why fluorine always exhibit an oxidation state of -1?[1]
Part II — Short Answer Questions 14 × 2 = 28

Answer briefly. (Answer all questions.)

16.Which metal in the 3d series exhibits +1 oxidation state most frequently and why?[2]
17.Define ionic product of water. Give its value at room temperature.[2]
18.Define enzymes[2]
19.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]
20.Draw the major product formed when 1-ethoxyprop-1-ene is heated with one equivalent of HI[2]
21.Classify the following into monosaccharides, oligosaccharides and polysaccharides. i) Starch ii) fructose iii) sucrose iv) lactose iv) maltose[2]
22.What are antibiotics?[2]
23.Write the expression for the solubility product of Hg2Cl2.[2]
24.In an octahedral crystal field, show the splitting of the five d orbitals.[2]
25.A saturated solution, prepared by dissolving CaF2(s) in water, has [Ca2+] = 3.3×10^-4 M. What is the Ksp of CaF2 ?[2]
26.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]
27.Write the rate law for the following reactions. (a) A reaction that is 3/2 order in x and zero order in y. (b) A reaction that is second order in NO and first order in Br2.[2]
28.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]
29.A gas phase reaction has energy of activation 200 kJ mol-1. If the frequency factor of the reaction is 1.6 × 10^13 s^-1. Calculate the rate constant at 600 K. (Use R = 8.314 J mol^-1 K^-1.)[2]
Part III — Long Answer Questions 10 × 5 = 50

Answer in detail. (Answer all questions.)

30.What are the functions of lipids in living organism.[5]
31.What are interstitial compounds?[5]
32.A solution of [Ni(H2O)6]2+ is green, whereas a solution of [Ni(CN)4]2− is colourless. Explain.[5]
33.Write the mechanism of acid catalysed dehydration of ethanol to give ethene.[5]
34.How will you convert benzaldehyde into the following compounds? (i) benzophenone (ii) benzoic acid (iii)α-hydroxyphenylacetic acid.[5]
35.Describe the variable oxidation state of 3d series elements.[5]
36.Discuss the Lowry–Bronsted concept of acids and bases.[5]
37.Identify A, B and C in the sequence: CH 3 NO 2 --(LiAlH 4 )--> A --(2CH 3 CH 2 Br)--> B --(H 2 SO 4 )--> C.[5]
38.Describe the structure of diborane.[5]
39.What are different types of RNA which are found in cell[5]
🔑 Show Answer Key — Set 3
  1. 1. Electrochemical metallurgy uses electrode potentials and cell EMFs to predict and carry out extraction/refining of metals by electrolysis or electrowinning.
  2. 2. Small size, high electronegativity, absence of d-orbitals, strong tendency to form pπ–pπ multiple bonds and N≡N triple bond.
  3. 3. 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).
  4. 4. 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.
  5. 5. 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).
  6. 6. Frenkel defect: cation vacancy–interstitial pair; stoichiometry unchanged.
  7. 7. 0
  8. 8. See solution
  9. 9. D-sugar
  10. 10. i) Acetic anhydride: Dehydrate acetic acid, e.g. 2 CH3COOH \xrightarrow{P2O5, heat} (CH3CO)2O + H2O. Or CH3COOH + CH3COCl → (CH3CO)2O + HCl.
  11. 11. k ≈ 1.0216 × 10^-2 min^-1. Time for 80% completion ≈ 158 minutes.
  12. 12. Equivalent conductance Λ°eq = 349 mho·cm2·equiv−1. Molar conductance Λ°m = 2094 mho·cm2·mol−1.
  13. 13. Because of strong metallic bonding from delocalised d and s electrons and many unpaired d electrons leading to high cohesive energy.
  14. 14. 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.
  15. 15. -1
  16. 16. Copper has electronic configuration 3d 10 4s¹. It can easily lose 4s¹ electron to give the stable 3d 10 configuration. Hence, it shows +1 oxidation state.
  17. 17. Ionic product of water Kw = [H+][OH−]; at 25 °C Kw = 1.0 × 10^-14.
  18. 18. Enzymes are biological catalysts (mostly proteins) that accelerate biochemical reactions without being consumed, by lowering activation energy and showing substrate specificity.
  19. 19. k = 0.03 L mol^{-1} s^{-1}
  20. 20. Propanal (CH3CH2CHO) and ethyl iodide (C2H5I) — major carbonyl product: propanal.
  21. 21. i) Starch — polysaccharide; ii) Fructose — monosaccharide; iii) Sucrose — disaccharide (oligosaccharide); iv) Lactose — disaccharide (oligosaccharide); v) Maltose — disaccharide (oligosaccharide).
  22. 22. Antibiotics are substances (natural or synthetic) that kill or inhibit the growth of microorganisms, especially bacteria, at low concentrations. They can be bactericidal or bacteriostatic; examples include penicillin and streptomycin.
  23. 23. Ksp = [Hg2^{2+}][Cl^-]^2
  24. 24. In an octahedral field the five d orbitals split into two sets: the lower‑energy t2g set (dxy, dxz, dyz) and the higher‑energy eg set (dz2, dx2−y2). The t2g orbitals are stabilized by −0.4Δo each and the eg orbitals are destabilized by +0.6Δo each; the energy gap between the sets is Δo.
  25. 25. 1.44×10^-10
  26. 26. A ≈ 9.1 × 10^10 s^-1.
  27. 27. (a) rate = k [x]^{3/2} (since zero order in y means [y]^0 = 1). (b) rate = k [NO]^2 [Br2]^1 = k [NO]^2 [Br2].
  28. 28. NH3 is the stronger Lewis base.
  29. 29. k = 6.2 × 10^-5 s^-1 (approximately).
  30. 30. Lipids function as energy stores, membrane components, thermal/electrical insulators, protective padding, signalling molecules (hormones), and sources of fat-soluble vitamins and essential fatty acids.
  31. 31. Interstitial compounds are non-stoichiometric compounds formed when small atoms (H, C, N, B, etc.) occupy interstitial sites in a metal lattice without disrupting the metal framework.
  32. 32. [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.
  33. 33. Stepwise acid‑catalysed mechanism: (1) Protonation of ethanol to give ethyl oxonium ion. (2) Loss of water to form ethyl carbocation. (3) Deprotonation of the carbocation to give ethene and H^+.
  34. 34. (i) benzaldehyde → benzophenone: Oxidize benzaldehyde to benzoic acid, convert to benzoyl chloride, then Friedel–Crafts acylation with benzene: \(PhCHO \xrightarrow{[O]} PhCOOH \xrightarrow{SOCl2} PhCOCl;\; PhCOCl + PhH \xrightarrow{AlCl3} PhCOPh\).
  35. 35. 3d elements show multiple oxidation states due to similar energies of 3d and 4s electrons; highest states up to group number; trend peaks at Mn.
  36. 36. Bronsted–Lowry acid: proton (H+) donor. Bronsted–Lowry base: proton acceptor. Reactions occur as acid + base ⇌ conjugate base + conjugate acid.
  37. 37. LiAlH 4 reduces nitromethane to methylamine, so A = CH 3 NH 2 . Alkylation with two moles of ethyl bromide gives the tertiary amine B = (C 2 H 5 ) 2 NCH 3 , N,N-diethylmethylamine. Treatment with sulphuric acid gives the ammonium salt C = [(C 2 H 5 ) 2 N(H)CH 3 ] + HSO 4 - .
  38. 38. 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.
  39. 39. Major types: messenger RNA (mRNA), transfer RNA (tRNA), ribosomal RNA (rRNA); other types: snRNA, snoRNA, miRNA, siRNA, piRNA.

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