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Samacheer Kalvi Class 11 Chemistry Practice Question Papers

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Brain Grain · braingrain.in
Chemistry — Practice Paper · Set 1
Class: 11Samacheer KalviMax Marks: 93
Name: ____________________Reg No: ____________
Part I — Multiple Choice Questions 15 × 1 = 15

Choose the correct answer. (Answer all questions.)

1.The isomer of ethanol is a) acetaldehyde b) dimethyl ether c) acetone d) methyl carbinol[1]
2.Which one of the following is true about metallic character when we move from left to right in a period and top to bottom in a group? (a) Decreases in a period and increases along the group (b) Increases in a period and decreases in a group (c) Increases both in the period and the group (d) Decreases both in the period and in the group[1]
3.For d-electrons, the orbit angular momentum is (a) \(\frac{\sqrt{2} h}{2 \pi}\) (b) \(\frac{\sqrt{2 h}}{2 \pi}\) (c) \(\frac{\sqrt{2 \times 4} \mathrm{~h}}{2 \pi}\) (d) \(\frac{\sqrt{6} \mathrm{~h}}{2 \pi}\)[1]
4.The Henry’s law constants for two gases A and B are x and y respectively. The ratio of mole fractions of A to B 0.2. The ratio of mole fraction of B and A dissolved in water will be a) \(\frac{2 x}{y}\) b) \(\frac{y}{0.2 x}\) c) \(\frac{0.2 x}{y}\) d) \(\frac{5 x}{y}\)[1]
5.When 15.68 litres of a gas mixture of methane and propane are fully combusted at 0° C and 1 atmosphere, 32 litres of oxygen at the same temperature and pressure are consumed. The amount of heat released from this combustion in kJ is (∆H C(CH 4 ) ) = – 890 kJ mol and ∆H C(C 3 H 8 ) = -2220 kJ mol -1 )(a) -889 kJ(b) -1390 kJ(c) -3180 kJ(d) -632.68 kJ[1]
6.Molar heat of vapourisation of a liquid is 4.8 kJ mol -1 If the entropy change is 16 J mol -1 K -1. the boiling point of the liquid is(a) 323 K(b) 27°C(c) 164 K(d) 0.3 K[1]
7.C 2 H 5 Br + 2Na C 4 H 10 + 2NaBr. The above reaction is an example of which of the following a) Reimer Tiemann reaction b) Wurtz reaction c) Aldol condensation d) Hoffmann reaction[1]
8.Photo chemical smog formed in congested metropolitan cities mainly consists of(a) Ozone, SO 2 and hydrocarbons(b) Ozone, PAN and NO 2(c) PAN, smoke and SO 2(d) Hydrocarbons, SO 2 and CO 2[1]
9.Consider the following sets of quantum numbers: Which of the following sets of quantum numbers is not possible? (a) (i), (ii) and (iv) (b) (ii), (iv) and (v) (c) (i) and (iii) (d) (ii), (iii) and (iv)[1]
10.Which one of the following shows functional isomerism? a) ethylene b) Propane c) ethanol d) CH 2 Cl 2[1]
11.Which of the following statements about hydrogen is incorrect?(a) Hydrogen ion, H 3 O + exists freely in solution.(b) Dihydrogen acts as a reducing agent.(c) Hydrogen has three isotopes of which tritium is the most common.(d) Hydrogen never acts as cation in ionic salts.[1]
12.For decolourisation of 1mole of acidified KMnO 4, the moles of H 2 O 2 required is (a) \(\frac{1}{2}\) (b) \(\frac{3}{2}\) (c) \(\frac{5}{2}\) (d) \(\frac{7}{2}\)[1]
13.25g of each of the following gases are taken at 27°C and 600 mm Hg pressure. Which of these will have the least volume?(a) HBr(b) HCl(c) HF(d) HI[1]
14.If x is the fraction of PCl 5 dissociated at equilibrium in the reaction PCl 5 ⇌ PCl 3 + Cl 2 then starting with 0.5 mole of PCl 5, the total number of moles of reactants and products at equilibrium is a) 0.5 – x b) x + 0.5 c) 2x + 0.5 d) x + 1[1]
15.Which one of the following gases has the lowest value of Henry’s law constant? a) N 2 b) He c) CO 2 d) H 2[1]
Part II — Short Answer Questions 14 × 2 = 28

Answer briefly. (Answer all questions.)

16.State Le – Chateller principle.[2]
17.If n = 6, the sequence for filling electrons will be, (a) ns → (n – 2)f → (n – 1)d → np (b) ns → (n – 1 )d → (n – 2)f → np (c) ns → {n – 2)f → np → (n – 1 )d (d) none of these are correct[2]
18.Arrange the following alkyl halide in increasing order of bond enthalpy of RX. CH 3 Br, CH 3 F, CH 3 Cl, CH 3 I[2]
19.Silverpropionate when refluxed with Bromine in carbon tetrachloride gives a) propionic acid b) chloroethane c) Bromo ethane d) chloro propane[2]
20.The geometrical shape of carbocation is a) Linear b) tetrahedral c) Planar d) Pyramidal[2]
21.The electronic configuration of Eu (Atomic no, 63), Gd (Atomic no. 64), and Tb (Atomic no. 65) are (a) [Xe] 4f 6 5d 1 6s 2, [Xe] 4f 7 5d 1 6s 2 and [Xe] 4f 8 5d 1 6s 2 (b) [Xe] 4f 7 6s 2, [Xe] 4f 1 5d 1 6s 2 and [Xe] 4f 9 6s 2 (c) [Xe] 4f 7 6s 2, [Xe] 4f 8 6s 2 and [Xe] 4f 8 5d 1 6s 2 (d) [Xe] 4f 6 5d 1 6s 2, [Xe] 4f 1 5d 1 6s 2 and [Xe] 4f 9 6s 2[2]
22.Write the balanced chemical equation for an equilibrium reaction for which the equilibrium constant is given by expression.[2]
23.Two liquids X and Y on mixing gives a warm solution. The solution is a) ideal b) non-ideal and shows positive deviation from Raoult’s law c) ideal and shows negative deviation from Raoult’s Law d) non-ideal and shows negative deviation from Raoult’s Law[2]
24.Give Kelvin a statement of the second law of thermodynamics.[2]
25.Identify the wrong statement. (a) Amongst the isoelectronic species, smaller the positive charge on cation, smaller is the ionic radius (b) Amongst isoelectric species greater the negative charge on the anion, larger is the ionic radius (c) Atomic radius of the elements increases as one moves down the first group of the periodic table (d) Atomic radius of the elements decreases as one moves across from left to right in the 2nd period of the periodic table.[2]
26.How will you convert ethyl chloride into (i) ethane (ii) n – butane[2]
27.The IUPAC name of is a) 2 – Bromo – 3- methyl butanoic acid b) 2 – methyl – 3 bromo butanoic acid c) 3 – Bromo – 2 – methylbutanoic acid d) 3 – Bromo – 2, 3 – dimethyl propanoic acid[2]
28.Explain intensive properties with two examples.[2]
29.Match List I with List-II and select the correct answer using the code given below the lists. List I List II A. Stone leprosy 1. CO B. Biological magnification 2. Greenhouse gases C. Global warming 3. Acid rain D. Combination with hemoglobin 4. DDT Code:[2]
Part III — Long Answer Questions 10 × 5 = 50

Answer in detail. (Answer all questions.)

30.For the reaction at 298 K: 2A + B → C ∆H = 400 Jmol -1; ∆S = 0.2 JK∆ mol -1 Determine the temperature at which the reaction would be spontaneous.[5]
31.Write short notes on a) Resonance b) Hyper Conjugation[5]
32.Give a brief description of the principles of 1. Fractional distillation 2. Column Chromatography[5]
33.What is meant by a functional group? Identify the functional group in the following compounds. a) acetaldehyde b) oxalic acid e) dimethyl ether d) methylamine[5]
34.A sample of 12 M Concentrated hydrochloric acid has a density 1.2 M gL -1 calculate the molality.[5]
35.How does classical smog differ from photochemical smog?[5]
36.Suppose there is a tiny sticky area on the wall of a container of gas. Molecules hitting this area stick there permanently. Is the pressure greater or less than on the ordinary area of walls?[5]
37.Calculate the work done when 2 moles of an ideal gas expands reversibly and isothermally from a volume of 500 ml to a volume of 2 L at 25°C and normal pressure.[5]
38.How will you prepare n propyl iodide from n – propyl bromide?[5]
39.Explain whether a gas approaches ideal behavior or deviates from ideal behaviour if (a) it is compressed to a smaller volume at f constant pressure.[5]
🔑 Show Answer Key — Set 1
  1. 1. b) dimethyl ether
  2. 2. (b) Increases in a period and decreases in a group
  3. 3. (d) \(\frac{\sqrt{6} \mathrm{~h}}{2 \pi}\)
  4. 4. d) \(\frac{5 x}{y}\)
  5. 5. (d) -632.68 kJ
  6. 6. (b) 27°C
  7. 7. b) Wurtz reaction
  8. 8. (d) Hydrocarbons, SO 2 and CO 2
  9. 9. (b) (ii), (iv) and (v)
  10. 10. c) ethanol
  11. 11. (c) Hydrogen has three isotopes of which tritium is the most common.
  12. 12. (c) \(\frac{5}{2}\)
  13. 13. (d) HI II. Answer these questions briefly:
  14. 14. b) x + 0.5
  15. 15. c) CO 2
  16. 16. It states that If a system at equilibrium is disturbed, then the system shifts itself in a direction that nullifies the effect of that disturbance.
  17. 17. (a) ns → (n – 2)f → (n – 1)d → np
  18. 18. The order is: CH 3 I < CH 3 Br < CH 3 Cl < CH 3 F.
  19. 19. c) bromo ethane II. Write brief answer to the following questions:
  20. 20. c) Planar II. Write brief answer to the following questions:
  21. 21. (b) [Xe] 4f 7 6s 2, [Xe] 4f 1 5d 1 6s 2 and [Xe] 4f 9 6s 2
  22. 22. K c = \(\frac{\left[\mathrm{NH}_{3}\right]^{4}\left[\mathrm{O}_{2}\right]^{7}}{\left.[\mathrm{NO}]^{4} \mid \mathrm{H}_{2} \mathrm{O}\right]^{6}}\) Balanced equation is: 4NO + 6H 2 O ⇌ 4NH 3 + 7O 2
  23. 23. d) non-ideal and shows negative deviation from Raoult’s Law
  24. 24. Kelvin-Planck statement: It is impossible to take heat from a hotter reservoir and convert a cyclic process heat to a cooler reservoir.
  25. 25. (a) Amongst the isoelectronic species, smaller the positive charge on cation, smaller is the ionic radius
  26. 26. (i) ethane (ii) n – butane
  27. 27. c) 3 – Bromo – 2 – methylbutanoic acid
  28. 28. The property that is independent of the mass or size of the system is called an intensive property. e.g., Refractive index and surface tension.
  29. 29. (d)
  30. 30. Given, T = 298 ∆T ∆H = 400 J mol -1 = 400 J mol -1 ∆S = 0.2 JK -1 mol -1 ∆G = ∆H – T∆S if T = 2000 K ∆G = 400 – (0.2 × 2000) = 0 if T > 2000 K ∆G will be negative. The reaction would be spontaneous only beyond 2000 K
  31. 31. a) Resonance (or) Mesomeric effect: The resonance is a chemical phenomenon which is observed in certain organic compounds possessing double bonds at a suitable position. Certain organic compounds can be represented by more than one structure and they differ only in the position of bonding and lone pair of electrons. Such structures are called resonance structures (canonical structures) and this phenomenon is called resonance. This phenomenon is also called mesomerism or mesomeric effect. For example, the structure of aromatic compounds such as benzene and conjugated systems like 1,3 – butadiene cannot be represented by a single structure, and their, observed properties can be explained on…
  32. 32. 1. Fractional distillation: This method is used to purify and separate liquids present in the mixture having their boiling point close to each other. The process of separation of the components in liquid mixture at their respective boiling points in the form of vapours and the subsequent condensation of those vapours is called fractional distillation. This method is applied in distillation of petroleum. coal tar and crude oil. 2. Column chromatography: (a) The principle behind chromatography is selective distribution of mixture of organic substances between two phases a stationary phase and a moving phase. The stationary phase can be a solid or liquid, while the moving phase is a liquid o…
  33. 33. Functional group: An atom or group of atoms within a molecule that shows characteristics set of physical and chemical properties. a) acetaldehyde → – CHO b) oxalic acid → – COOH c) di methyl ether → – O – d) methyiamine → – NH 2
  34. 34. Given: Molarity = 12 M HCl density of solution = 1.2 g L -1 In 12 M HCl solution, there are 12 moles of HCl in 1 litre of the solution. Molality = \(\frac{\text { no of moles of solute }}{\text { mass of solvent (in } \mathrm{kg} \text { ) }}\) Calculate mass of water(solvent) mass of 1 litre HCl solution = density × volume = 1.2 × gmL -1 × 1000 mL = 1200 g mass of HCl = no. of moles of HCl × molar mass of HCl = 12 mol × 36.5 g mol -1 = 438 g. mass of water = mass of HCl solution – mass of HCl mass of water = 1200 – 438 = 762 g molality(m) = \(\frac{12}{0.762}\) = 15.75 m
  35. 35. Classical smog: * Classical smog is caused by coal smoke and fog. * It occurs in cold humid climate. * The chemical composition is the mixture of SO 2, SO 3 gases and humidity. * Chemically it is reducing in nature because of high concentration of SO, and so it is also called reducing smog. * It is primarily responsible for acid rain. * It also causes bronchial irritation. Photochemical smog: * Photochemical smog is cause by photochemical oxidants. * It occurs in warm, dry and sunny climate. * The chemical composition is the mixture of NO 2 and O 3 gases. * Chemically it is oxidising in nature because of high concentration of oxidising agents such as NO 2 and O 3 * and so it is also calle…
  36. 36. Gaseous pressure is developed by the continuous bombardment of the molecules of the gas among themselves and also with the walls of the container. When the molecules hit the sticky area of the container, the number of molecules decreases and hence, the pressure decreases. Therefore, pressure is less than the ordinary area of walls.
  37. 37. Given n = 2 moles V i = 500 ml = 0.5 lit V f = 2 lit T = 25°C = 298 K w = -2303 nRTIog (\(\frac{V_{f}}{V_{i}}\)) w = -2.303 × 2 × 8.314 × 298 × log (4) w = -2.303 × 2 × 8.314 × 298 × 0.6021 w = -6871 J w = -6.871 kJ.
  38. 38. Finkelstein reaction, nCH 3 – CH 2 – CH 2 – Br + NaI n – CH 3 – CH 2 – CH 2 – I + NaBr n – propyl iodide n- propyl bromide
  39. 39. When the gas is compressed to a smaller volume, the compressibility factor (Z) decreases. Hence, the gas deviates from ideal behavior (b) the temperature is raised while keeping the volume constant. When the temperature is increased, the compressibility factor approaches unity. Hence, the gas behaves ideally. (c) More gas is introduced into the same volume and at the same temperature. When more gas is introduced into a container of the same volume and at the same temperature, the compressibility factor tend to unity. Hence, the gas behaves ideally.
Brain Grain · braingrain.in
Chemistry — Practice Paper · Set 2
Class: 11Samacheer KalviMax Marks: 93
Name: ____________________Reg No: ____________
Part I — Multiple Choice Questions 15 × 1 = 15

Choose the correct answer. (Answer all questions.)

1.If one mole of ammonia and one mole of hydrogen chloride are mixed in a closed container to form ammonium chloride gas, then(a) ∆H > ∆U(b) ∆H – ∆U = 0(c) ∆H + ∆U = 0(d) ∆H < ∆U[1]
2.The correct equation for the degree of an associating solute, ‘n’ molecules of which undergoes association in solution, is a) α = \(\frac{\mathrm{n}(\mathrm{i}-1)}{\mathrm{n}-1}\) b) α 2 = \(\frac{n(1-i)}{(n-1)}\) c) α = \(\frac{\mathrm{n}(\mathrm{i}-1)}{1-\mathrm{n}}\) d) α = \(\frac{\mathrm{n}(1-\mathrm{i})}{\mathrm{n}(1-\mathrm{i})}\)[1]
3.The product obtained as a result of a reaction of nitrogen with CaC2 is(a) Ca(CN) 3(b) CaN 2(c) Ca(CN) 2(d) Ca 3 N 2[1]
4.Which of the following pairs of elements exhibit diagonal relationship?(a) Be and Mg(b) Li and Mg(c) Be and B(d) Be and Al[1]
5.In a given shell the order of screening effect is(a) s > p > d > f(b) s > p > f > d(c) f > d > p > s(d) f > p > s > d[1]
6.Heat of combustion is always(a) positive(b) negative(c) zero(d) either positive or negative[1]
7.Release of oxides of nitrogen and hydrocarbons into the atmosphere by motor vehicles is prevented by using(a) grit chamber(b) scrubbers(c) trickling filters(d) catalytic converters[1]
8.At a given temperature and pressure, the equilibrium constant values for the equilibria 3A 2 + B 2 + 2Cr 2A 3 BC and A 3 BC \(\frac{3}{2}\) [A 2 ] + \(\frac{1}{2}\) B 2 +C The relation between K 1 and K 2 is a) K 1 = \(\frac{1}{\sqrt{\mathrm{K}_{2}}}\) b)K 2 = \(\mathrm{K}_{1}^{-1 / 2}\) C) K 1 2 = 2K 2 d) \(\frac{\mathrm{K}_{1}}{2}\) = K 2[1]
9.Heterolytic fission of C – Br bond results in the formation of a) free radical b) Carbanion c) Carbocation d) Carbanion and Carbocation[1]
10.– I effect is shown by a) – Cl b) – Br c) both (a) and (b) d) – CH 3[1]
11.Which one of the following is diamagnetic? a) O 2 b) O 2 2- c) O 2 + d) None of these[1]
12.Stomach acid, a dilute solution of HCl can be neutralized by reaction with aluminium hydroxide Al(OH) 3 + 3HCl (aq) -> AlCl 3 + 3H 2 O. How many milliliters of 0.1 M Al(OH) 3 solution is needed to neutralize 21 ml of 0.1 M HCl? a) 14 mL b) 7 mL c) 21 mL d) none of these[1]
13.In a chemical equilibrium, the rate constant for the forward reaction is 2.5 × 10 -2, and the equilibrium constant is 50. The rate constant for the reverse reaction is, a) 11.5 b) 50 c) 2 × 10 2 d) 2 × 10 -3[1]
14.With respect to the position of – Cl in the compound CH3 – CH = CH – CH 2 – Cl, it is classified as a) Vinyl b) Allyl c) Secondary d) Aralkyl[1]
15.The percentage of s-character of the hybrid orbitals in methane, ethane, ethene, and ethyne are respectively a) 25, 25, 33.3, 50 b) 50, 50, 33.3, 25 c) 50, 25, 33.3, 50 d) 50, 25, 25, 50[1]
Part II — Short Answer Questions 14 × 2 = 28

Answer briefly. (Answer all questions.)

16.The IUPAC name of the compound is a) 3 – Ethyl – 2 – hexene b) 3 – Propyl – 3 – hexene c) 4 – Ethyl – 4 – hexene d) 3 – Propyl – 2 – hexene[2]
17.The IUPAC name of is a) 2,4,4 – Trimethylpent – 2 – ene b) 2,4,4 – Trimethylpent – 3 – ene c) 2,2,4 – Trimethylpent – 3 – ene d) 2,2,4 – Trirnethylpent – 2 – ene[2]
18.For a gaseous homogeneous reaction at equilibrium, number of moles of products are greater than the number of moles of reactants. Is K c is larger or smaller than K p.[2]
19.In the hydrocarbon the state of hybridization of carbon 1, 2, 3, 4 and 7 are in the following sequence. a) sp, sp, sp 3, sp 2, sp 3 b) sp 2, sp, sp 3, sp 2, sp 3 c) sp, sp, sp 2, sp, sp 3 d) none of these[2]
20.Which among the following alkenes on reductive ozonolysis produces only propanone? a) 2 – Methyl propene b) 2 – Methyl but – 2 – ene c) 2, 3 – Dimethyl but – 1- ene d) 2, 3 – Dimethyl but – 2 – ene[2]
21.When the numerical value of the reaction quotient (Q) is greater than the equilibrium constant, in which direction does the reaction proceed to reach equilibrium?[2]
22.Give the general electronic configuration of lanthanides and actinides?[2]
23.The IUPAC name of the compound CH 3 – CH = CH – C ≡ CH is a) Pent – 4- yn – 2 – ene b) Pent – 3- en – 1- yne c) Pent – 2 – en – 4 – yne d) Pent – 1 yn – 3 – ene[2]
24.Identify the wrong statement in the following. (a) The clean water would have a BOD value of less than 5 ppm (b) Greenhouse effect is also called Global warming (c) Minute solid particles in air are known as particulate pollutants (d) Biosphere is the protective blanket of gases surrounding the earth[2]
25.Give IUPAC names for the following compounds 1) CH 3 – CH = CH – CH = CH – C ≡ C – CH 3 2) 3) (CH 3 ) 3 C – C ≡ C – CH(CH 3 ) 2 4) ethyl isopropyl acetylene 5) CH ≡ C – C ≡ C – C ≡ CH[2]
26.Major product of the below mentioned reaction is, (CH 3 ) 2 C = CH 2 a) 2 – chloro – 1 – iode – 2 – methyl propane b) 1 – chloro – 2 – iodo – 2 – methyl propane c) 1, 2 – dichloro – 2 – methyl propane d) 1, 2 – diiodo – 2 – methyl propane[2]
27.Write the chemical equations for combustion of propane.[2]
28.In ClF 3, NF 3 and BF 3 molecules the chlorine, nitrogen and boron atoms are a) sp 3 hybridised b) sp 3, sp 3 and sp 2 respectively c) sp 3 hybridised d) sp 3 d, sp 3 and sp hybridised respectively[2]
29.What happens when ethylene is passed ‘ through cold dilute alkaline potassium permanganate.[2]
Part III — Long Answer Questions 10 × 5 = 50

Answer in detail. (Answer all questions.)

30.What is the relation between K p and K c ? Given one example for which K p is equal to K c.[5]
31.Linear form of carbondioxide molecule has two polar bonds. Yet the molecule has Zero dipole moment. Why?[5]
32.Hydrogen peroxide can function as an oxidizing agent as well as a reducing agent. Substantiate this statement with suitable examples.[5]
33.The stabilization of a half-filled d – orbital is more pronounced than that of the p-orbital. Why?[5]
34.Calculate the molar mass of the following compounds.[5]
35.How is acid rain formed? Explain its effect.[5]
36.Explain briefly the time-independent Schrodinger wave equation.[5]
37.Give the IUPAC names of the following compounds. i) (CH 3 ) 2 CH – CH 2 – CH(CH 3 ) – CH(CH 3 ) 2 ii) iii) CH 3 – O – CH 3 iv) v) CH 2 = CH – CH – CH 2 vi) vii) viii) ix) x) xi) xii) xiii) xiv) xv) xvi)[5]
38.The energy of an electron in the ground state of the hydrogen atom is -2.8 × 10 -8 J. Calculate the ionization enthalpy of atomic hydrogen in terms of kJ mol -1[5]
39.Show the heterolysis of covalent bond by using curved arrow notation and complete the following equations. Identify the nucelophile in each case. i) CH 3 – Br + KOH → ii) CH 3 – O – CH 3 + HI →[5]
🔑 Show Answer Key — Set 2
  1. 1. (d) ∆H < ∆U
  2. 2. c) α = \(\frac{\mathrm{n}(\mathrm{i}-1)}{1-\mathrm{n}}\)
  3. 3. (c) Ca(CN) 2
  4. 4. (d) Be and Al II. Write brief answer to the following questions:
  5. 5. (a) s > p > d > f
  6. 6. (b) negative
  7. 7. (b) scrubbers
  8. 8. b)K 2 = \(\mathrm{K}_{1}^{-1 / 2}\)
  9. 9. d) Carbanion and Carbocation
  10. 10. c) both (a) and (b)
  11. 11. b) O 2 2-
  12. 12. b) 7 mL
  13. 13. b) 50
  14. 14. b) Allyl
  15. 15. a) 25, 25, 33.3, 50
  16. 16. a) 3 – Ethyl – 2 – hexene
  17. 17. c) 2,2,4 – Trimethylpent – 3 – ene
  18. 18. K p >K c n p > n R
  19. 19. a) sp, sp, sp 3, sp 2, sp 3
  20. 20. d) 2, 3 – Dimethyl but – 2 – ene
  21. 21. When Q > K C the reaction will proceed in the reverse direction, i.e, formation of reactants.
  22. 22. The electronic configuration of lanthanides is 4f 1-14 5d 0-1 6s 2. The electronic configuration of actinides is 5f 1-14 6d 0-1 7s 2.
  23. 23. b) Pent – 3- en – 1- yne
  24. 24. (c) Minute solid particles in air are known as particulate pollutants
  25. 25. 1) 2) 3) 4) ethyl isopropyl acetylene 5)
  26. 26. a) 2 – chloro – 1 – iode – 2 – methyl propane
  27. 27. C 3 H 8 (g) + 5O 2 (g) ) → 3CO 2 (g) + 4H 2 O(l) propane
  28. 28. d) sp 3 d, sp 3 and sp hybridised respectively
  29. 29. CH 2 = CH + H 2 O + (O) CH 2 OH – CH 2 OH Ethylene Ethylene glycol
  30. 30. The relation between K p and K c is K p = K C (RT) ∆n g K p = equilibrium constant is terms of partial pressure. K c = equilibrium constant is terms of concentration. R = gas constant T = Temperature. ∆n g = Difference between the sum of the number of moles of products and the sum of number of moles of reactants in the gas phase. When ∆n g = 0 K p = K C (RT) 0 = K C i.e., K p = K C Example: H 2 (g) + I 2 (g) \(\rightleftharpoons\) 2HI(g) ∆n g = 2 – 2 = 0 ∴ K p = K C for the synthesis of HI.
  31. 31. The linear form of carbon dioxide has zero dipole moment, even though it has two polar bonds. In CO 2, the dipole moments of two polar bonds (CO) are equal in magnitude but have opposite direction. Hence, the net dipole moment of the CO 2 is, μ = μ 1 + μ 2 = μ 1 + (-μ 1 ) = 0
  32. 32. Hydrogen peroxide can act both as an oxidizing agent and a reducing agent. Oxidation is usually performed in an acidic medium while the reduction reactions are performed in a basic medium. In acidic conditions: H 2 O 2 + 2H + + 2e – → 2H 2 O (E° = +1.77 V) For example 2FeSO 4 + H 2 SO 4 + H 2 O 2 → Fe 2 (SO 4 ) 3 + 2H 2 O In basic conditions: HO 2 – + OH – → O 2 + H 2 O + 2e – (E° = + 0.08V) For Example, 2 KMnO 4(aq) + 3H 2 O 2(aq) → 2MnO 2 + 2KOH + 2H 2 O + 3O 2 (g)
  33. 33. The exactly half-filled orbitals have greater stability. The reason for their stability are – * symmetry * exchange energy. (1) Symmetry: The half-filled orbitals are more symmetrical than partially filled orbitals and this symmetry leads to greater stability. (2) Exchange energy: The electrons with the same spin in the different orbitals of the same sub-shell can exchange their position. Each such exchange releases energy and this is known as exchange energy. Greater the number of exchanges, the greater the exchange energy, and hence greater the stability. In d-orbital, 10 exchanges are possible but in p-orbital 6 exchanges are possible. So, d – orbital with 5 unpaired electrons (10 exch…
  34. 34. (i) urea [CO(NH 2 ) 2 ]: Molar mass of urea = (4 × Atomic mass of hydrogen) + ( 1 × Atomic mass of carbon) + ( 2 × Atomic mass of nitrogen) + ( 1 × Atomic mass of oxygen) = (4 × 1.008) +(1 × 12) + (2 × 14)+ ( 1 × 16) = 4.032 + 12 + 28 + 16 = 60.032 g mol -1 (ii) acetone[CH 3 COCH 3 ] Molar mass of Acetone = (6 × Atomic mass of hydrogen) + ( 3 × Atomic mass of carbon) + (1 × Atomic mass of oxygen) = (6 × 1.008) + (3 × 12) + ( 1 × 16) = 6.048 + 36 + 16 = 52.024 g mol -1. (iii) boric acid [H 3 BO 3 ]: Molar mass of Boric acid = (3 × Atomic mass of hydrogen) + ( 1 × Atomic mass of boron) + ( 3 × Atomic mass of oxygen) = (3 × 1.008) + (3 × 11) + ( 1 × 16) = 3.024 + 33 + 16 = 52.024 g mol -1. (…
  35. 35. 1. Rainwater has a pli of 5.6 due to the dissolution of CO., into it. Oxides of sulphur and nitrogen in the atmosphere may be absorbed by droplets of water that make up the clouds and get chemically converted into sulphuric acid and nitric acid. Due to this, the pH of rainwater drops below the level of 5.6. Hence it is called acid rain. 2. Acid rain is a by-product of sulphur and Nitrogen oxides in the atmosphere. Burning of fossil fuels in power stations, furnaces and petrol, diesel in motor engines produce SO 2 and NO 2 gases. They are converted into H 2 SO 4 and HNO 3 by the reaction with oxygen and water. 3. 2SO 2 + O 3 + 2H 2 O → 2H 2 SO 4 4NO 2 + O 2 + 2H 2 O → 4HNO 3 Harmful effect…
  36. 36. Erwin Schrodinger expressed the wave nature of electrons in terms of a differential equation. This equation determines the change of wave function in space depending on the field of force in which the electron moves. The time-independent Schrodinger equation can be expressed as Hψ = Eψ, where H is called Hamiltonian operator, ψ is the wave function and is a function of position coordinates of the particle and is denoted as ψ(x, y, z), E is the energy of the system. The above Schrodinger wave equation does not contain time as a variable and is referred to as time-independent Schrodinger wave equation. This equation can be solved only for certain values of E, the total energy, i.e., the ene…
  37. 37. (i) 2, 3, 5 – trimethyihexane (ii) 2 – bromo – 3 – methylbutane (iii) methoxymethane (iv) 2 – hydroxybutanal (y) buta – 1,3 – diene (vi) 4 – chloropent – 2 – yne (vii) 1 – bromobut – 2 – ene (viii) 5 – oxohexanoic acid (ix) 3 – ethyl – 4 – ethenylheptane (x) 2, 4, 4 – trimethylpent – 2 – ene (xi) 2 – methyl -1 – phenyipropan – 1 -amine (xii) 2, 2 – dimethyl – 4oxopentanenitrile (xiii) 2 – ethoxypropane (xiv) 1 – fluoro – 4 – methyl – 2 -nitrobenzene (xv) 3 – bromo – 2 – methylpentanal (xvi) Acetophenone
  38. 38. Energy of an electron in the ground state of the hydrogen atom is – 2.8 × 10 -8. The ionization energy of atomic hydrogen is 2.8 × 10 -18 × 6.023 × 10 23 J/mol = 16.86 × 10 5 J/mol = 1686 kJ/mol.
  39. 39. (i) CH 3 – Br + KOH → CH 3 – Br + KQH → CH 3 OH + KBr Nucleophile is: OH – ii) CH 3 – O – CH 3 + HI → H I CH 3 – O – CH 3 + HI → CH 3 OH + CH 3 I Nucleophile is: I –
Brain Grain · braingrain.in
Chemistry — Practice Paper · Set 3
Class: 11Samacheer KalviMax Marks: 93
Name: ____________________Reg No: ____________
Part I — Multiple Choice Questions 15 × 1 = 15

Choose the correct answer. (Answer all questions.)

1.The K H for the solution of oxygen dissolved in water is 4 × 10 4 atm at a given temperature. If the partial pressure of oxygen in air is 0.4 atm, the mole fraction of oxygen in solution is a) 4.6 × 10 3 b) 1.6 × 10 4 c) 1 × 10 -5 d) 1 × 10 5[1]
2.An ideal gas expands from the volume of 1 × 10 -3 m 3 to 1 × 10 -2 m 3 at 300 K against a constant pressure at 1 × 10 5 Nm -2. The work done is(a) -900 J(b) 900 kJ(c) 270 kJ(d) – 900 kJ[1]
3.What is the mass of precipitate formed when 50 ml of 8.5 % solution of AgNO 3 is mixed with 100 ml of 1.865 % potassium chloride solution?(a) 3.59 g(b) 7 g(c) 14 g(d) 28 g[1]
4.Among the following the least thermally stable is(a) K 2 CO 3(b) Na 2 CO 3(c) BaCO 3(d) Li 2 CO 3[1]
5.All the naturally occurring processes proceed spontaneously in a direction which leads to(a) decrease in entropy(b) increase in enthalpy(c) increase in free energy(d) decrease in free energy[1]
6.At identical temperature and pressure, the rate of diffusion of hydrogen gas is 3 times that of a hydrocarbon having molecular formula C n H 2n – 2. What is the value of n?(a) 8(b) 4(c) 3(d) 1[1]
7.The element with positive electron gain enthalpy is(a) Hydrogen(b) Sodium(c) Argon(d) Fluorine[1]
8.Which one of the following arrangements represent the correct order of least negative to most negative electron gain enthalpy(a) Al < O < C < Ca < F(b) Al < Ca < O < C < F(c) C < F < O < Al < Ca(d) Ca < Al < C < O < F[1]
9.Gases deviate from ideal behavior at high pressure. Which of the following statement(s) is correct for non-ideality? (a) at high pressure the collision between the gas molecule become enormous (b) at high pressure the gas molecules move only in one direction (c) at high pressure, the volume of gas become insignificant (d) at high pressure the intermolecular interactions become significant[1]
10.A colourless solid substance (A) on heating evolved CO 2 and also gave a white residue, soluble in water. Residue also gave CO 2 when treated with dilute HCl. (a) Na 2 CO 3 (b) NaHCO 3 (c) CaCO 3 (d) Ca(HCO 3 ) 2[1]
11.Maximum deviation from ideal gas is expected (a) CH 4 (g) (b) NH 3 (g) (c) H 2 (g) (d) N 2 (g)[1]
12.Which of the following is electron deficient? a) PH 3 b) (CH 3 ) 2 c) BH 3 d) NH 3[1]
13.Which of the following statements is incorrect?(a) Li + has minimum degree of hydration among alkali metal cations(b) The oxidation state of K in KO 2 is +1(c) Sodium is used to make Na / Pb alloy(d) MgSO 4 is readily soluble in water[1]
14.For the formation of Two moles of SO 3 (g) from SO 2 and O 2, the equilibrium constant is K1. The equilibrium constant for the dissociation of one mole of SO 3 into SO 2 and O 2 is a) \(1 / \mathrm{K}_{1}\) b) K 1 2 c) \(\left(\frac{1}{\mathrm{~K}_{1}}\right)^{1 / 2}\) d) \(\frac{\mathrm{K}_{1}}{2}\)[1]
15.The ratio of de Brogue wavelengths of a deuterium atom to that of an α – particle, when the velocity of the former is five times greater than that of later, is(a) 4(b) 0.2(c) 2.5(d) 0.4[1]
Part II — Short Answer Questions 14 × 2 = 28

Answer briefly. (Answer all questions.)

16.Consider the following electronic arrangements for the d 5 configuration. (i) Which of these represents the ground state? (ii) Which configuration has the maximum exchange energy?[2]
17.It takes 192 sec for an unknown gas to diffuse through a porous wall and 84 sec for N2 gas to effuse at the same temperature and pressure. What Is the molar mass of the unknown gas?[2]
18.You are given normal boiling points and standard enthalpies of vapourisation. Calculate the entropy of vapourisation of liquids listed below.[2]
19.Select the molecule which has only one π bond. a) CH 3 – CH = CH – CH 3 b) CH 3 – CH = CH – CHO c) CH 3 – CH = CH – COOH d) All of these[2]
20.According to VSEPR theory, the repulsion between different parts of electrons obey the order a) 1. p – 1. p > b. p – b. p > 1. p – b. p b) b. p – b. p > b. p – 1. p > 1. p – b. p c) 1. p – 1. p > b. p – 1. p > b. p – b. p d) b. p – b. p > 1. p – 1. p > b. p – 1. p[2]
21.Which quantum number reveals information about the shape, energy, orientation, and size of orbitals?[2]
22.Connect pair of compounds which give blue colouration / precipitate and white precipitate respectively, when their Lassaigne’s test is separately done. a) NH 2 NH 2 HCl and ClCH 2 – CHO b) NH 2 CS NH 2 and CH 3 – CH 2 Cl c) NH 2 CH 2 COOH and NH 2 CONH 2 d) C 6 H 5 NH 2 and ClCH 2 – CHO[2]
23.Suggest a simple chemical test to distinguish propane and propene.[2]
24.The types of hybridisation on the five-carbon atom from right to left in the, 2,3 pentadiene. a) sp 3, sp 2, sp, sp 2, sp 3 b) sp 3, sp, sp, sp, sp 3 c) sp 2, sp, sp 2, sp 2, sp 3 d) sp 3, sp 3, sp 2, sp 3, sp 3[2]
25.The IUPAC name of is a) 2 – Bromo pent – 3 – ene b) 4 – Bromo pent – 2 – ene c) 2 – Bromo pent – 4 – ene d) 4 – Bromo pent – 1 – ene[2]
26.Decreasing order of nucleophilicity is a) OH – > NH 2 – > -OCH 3 > RNH 2 b) NH 2 – > OH – > -OCH3 > RNH 2 c) NH 2 – > CH 3 O – > OH – > RNH 2 d) CH 3 O – > NH 2 – > OH – > RNH 2[2]
27.The compound formed at anode in the electrolysis of an aqueous solution of potassium acetate are a) CH 4 and H 2 b) CH 4 and CO 2 c) C 2 H 6 and CO 2 d) C 2 H 4 and Cl 2[2]
28.Calculate the average atomic mass of naturally occurring magnesium using the following data: Isotope Isotopic atomic mass Abundance (%) Mg 24 23.99 78.99 Mg 26 24.99 10.00 Mg 25 25.98 11.01[2]
29.Suggest the route for the preparation of the following from benzene. 1) 3 – chloro nitrobenzene 2) 4 – chlorotoluene 3) Bromo benzene 4) m – dinitro benzene[2]
Part III — Long Answer Questions 10 × 5 = 50

Answer in detail. (Answer all questions.)

30.Mention any three methods of preparation of haloalkanes from alcohols.[5]
31.Complete the following: i) 2 – butyne ii) CH 2 = CH 2 iii) iv) CaC 2[5]
32.Sulphur hexafluoride Is a colourless, odourless gas; calculate the pressure exerted by 1.82 moles of the gas In a steel vessel of volume 5.43 dm3 at 69.5°C, assumIng Ideal gas behaviour.[5]
33.At 33K, N 2 O 4 is fifty percent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere.[5]
34.Discuss the three types of Covalent hydrides.[5]
35.Argon is an Inert gas used In light bulbs to retard the vaporization of the tungsten filament. A certain light bulb containing argon at 1.2 atm and 18°C Is heated to 85°C at constant volume. Calculate its final pressure in atm.[5]
36.Write the structures of following alkanes. 1) 2, 3 – Dimethyl – 6 – (2 – methyl propyl) decane 2) 5 – (2 – Ethyl butyl ) – 3, 3, – dimethyldecane 3) 5 – (1, 2 – Dimethyl propyl) – 2 – methylnonane[5]
37.0.32 g of an organic compound, after heating with fuming nitric acid and barium nitrate crystals is a sealed tube gave 0. 466 g of barium sulphate. Determine the percentage of sulphur in the compound.[5]
38.Give the general characteristics of organic compounds.[5]
39.Explain the preparation of the following compounds. i) DDT ii) Chloroform iii) Biphrnyl iv) Chloropicrin v) Freon – 12[5]
🔑 Show Answer Key — Set 3
  1. 1. c) 1 × 10 -5
  2. 2. (a) -900 J
  3. 3. (a) 3.59 g AgNO 3 + KCl → KNO 3 + AgCl 50 mL of 8.5 % solution contains 4.25 g of AgNO 3 No. of moles of AgNO 3 present in 50 mL of 8.5 % AgNO 3 solution = Mass / Molar mass = 4.25 / 170 = 0.025 moles Similarly, No of moles of KCl present in 100 mL of 1.865 % KCl solution = 1.865 / 74.5 = 0.025 moles So total amount of AgCl formed is 0.025 moles (based on the stoichiometry) Amount of AgCl present in 0.025 moles of AgCl = No. of moles × molar mass = 0.025 × 143.5 = 3.59 g
  4. 4. (d) Li 2 CO 3 II. Write brief answer to the following questions:
  5. 5. (d) decrease in free energy
  6. 6. (d) 1
  7. 7. (c) Argon
  8. 8. (d) Ca < Al < C < O < F
  9. 9. (d) at high pressure the intermolecular interactions become significant
  10. 10. (b) NaHCO 3
  11. 11. (b) NH 3 (g)
  12. 12. c) BH 3
  13. 13. (a) Li + has minimum degree of hydration among alkali metal cations
  14. 14. c) \(\left(\frac{1}{\mathrm{~K}_{1}}\right)^{1 / 2}\)
  15. 15. (d) 0.4
  16. 16. (i) The ground state electronic configuration is (ii) The configuration has the maximum exchange energy is
  17. 17. A gas’s partial pressure formula is the gas pressure exerted if that gas were alone.
  18. 18. For ethanol Given:
  19. 19. a) CH 3 – CH = CH – CH 3
  20. 20. c) 1. p – 1. p > b. p – 1. p > b. p – b. p
  21. 21. Magnetic quantum numbers reveal information about the shape, energy, orientation, and size of orbitals.
  22. 22. d) C 6 H 5 NH 2 and ClCH 2 – CHO
  23. 23. Propene decolourises Br 2 /H 2 O it forms dibromo compound but propane does not react with Br 2 / H 2 O.
  24. 24. a) sp 3, sp 2, sp, sp 2, sp 3
  25. 25. b) 4 – Bromo pent – 2 – ene
  26. 26. b) NH 2 – > OH – > -OCH3 > RNH 2
  27. 27. c) C 2 H 6 and CO 2
  28. 28. Average atomic mass = \(\frac{(78.9923.99)(1024.99)(11.0125.98)}{100}\) = \(\frac{2430.9}{100}\) = 24.31 u
  29. 29. 1) 3 – chloro nitrobenzene 2) 4 – chlorotoluene 3) Bromo benzene 4) m – dinitro benzene
  30. 30. Haloalkanes are prepared by the following methods. From alcohols: Alcohol can be converted into halo alkenes by reacting it with any one of the following reagents. * Hydrogen halide * Phosphorous halides * Thionyl chloride. a) Reaction with hydrogen halide: Mixture of con. HCl and anhydrous ZnCl 2 is called Lucas Reagent. The order of reactivity of halo acids with alcohol is in the order HI > HBr > HCl. The order of reactivity of alcohols with halo acid is tertiary > secondary > primary. b) Reaction with phosphorous halides: Alcohols react with PX 5 or PX 3 to form haloalkanes. Example: CH 3 CH 2 OH + PCl 5 → CH 3 CH 2 Cl + POCl 3 + HCl Ethane Chloro ethane 3CH 3 CH 2 OH + PCl 3 → 3 CH 3…
  31. 31. i) 2 – butyne ii) CH 2 = CH 2 iii) iv) CaC 2 CaC 2 + 2H 2 O → Ca(OH) 2 + C 2 H 2 ↑ Calcium Carbide Acetylene
  32. 32. n = 1.82 mole V = 5.43 dm 3 T = 69.5 + 273 = 342.5 P =? PV = nRT P = nRT/V P = P = 94.25 atm
  33. 33. Given T= 33 K N 2 O 4 ⇌ 2NO 2 Initial concentration: 100% 0 Concentration dissociated 50% – Concentration remaining at equilibrium 50% 100% K eq = \(\frac{100}{50}\) = 2 ∆G° = -2.303 RT log K eq ∆G°= -2.303 × 8.314 × 33 × log 2 ∆G° = -190.18 Jmol -1
  34. 34. They are the compounds in which hydrogen is attached to another element by sharing electrons. The most common examples of covalent hydrides are methane, ammonia, water, and hydrogen chloride. Molecular hydrides of hydrogen are further classified into three categories, Electron precise (CH 4, C 2 H 6, SiH 4, GeH 4 ) Electron-deficient (B 2 H 6 ) and Electron-rich hydrides (NH 3, H 2 O) Since most of the covalent hydrides consist of discrete, small molecules that have relatively weak intermolecular forces, they are generally gases or volatile liquids.
  35. 35. P 1 = 1.2 atm T 1 = 180°C + 273 = 291 K T 2 = 850°C + 273 = 358 K P 2 =? \(\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}\) P 2 = \(\left[\frac{P_{1}}{T_{1}}\right] \times T_{2}\) P 2 = \(\frac{1.2 a t m}{291 K}\) × 358 K P 2 = 1.48 atm
  36. 36. 1) 2, 3 – Dimethyl – 6 – (2 – methyl propyl) decane 2) 5 – (2 – Ethyl butyl ) – 3, 3, – dimethyldecane 3) 5 – (1, 2 – Dimethyl propyl) – 2 – methylnonane
  37. 37. Mass of the substance taken = 0.32 g Mass of BaSO 4 formed = 0.466 g Molecular mass of BaSO 4 = 137 + 32 + 64 = 233 Then, mass of S in 0.466 g of BaSO 4 = \(\frac{0.466 \times 32}{233}\) Percentage of S in compound= \(\frac{0.466 \times 32 \times 100}{233 \times 0.32}\) = 20 %
  38. 38. All organic compounds have the following characteristic properties. * They are covalent compounds of carbon and generally insoluble in water and readily soluble in organic solvent such as benzene, toluene, ether, chloroform etc… * Many of the organic compounds are inflammable (except CCl 4 ). They possess low boiling and melting points due to their covalent nature. * Organic compounds are characterized by functional groups. A functional group is an atom or a specific combination of bonded atoms that react in a characteristic way, irrespective of the organic molecule in which it is present. In almost all the cases, the reaction of an organic compound takes place at the functional group. Th…
  39. 39. i) DDT: DDT can be prepared by heating a mixture of chlorobenzene with chloral (Trichloro acetaldehyde) in the presence of con.H 2 SO 4. ii) Chloroform: Preparation: Chloroform is prepared in the laboratory by the reaction between ethyl alcohol with bleaching powder followed by the distillation of the product chloroform. Bleaching powder act as a source of chlorine and calcium hydroxide. This reaction is called the haloform reaction. The reaction proceeds in three steps as shown below. Step – 1: Oxidation CH 3 CH 2 OH + Cl 2 → CH 3 CHO + 2HCl Ethyl alcohol Ethanal (Acetaldehyde) Step – 2: Chlorination CH 3 CHO + 3Cl 2 → CCl 3 CHO + 3HCl Acetaldehyde Trichloro acetaldehyde Step – 3: Hydrol…

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