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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.An isotope of hydrogen (A) reacts with diatomic molecule of element which occupies group number 16 and period number 2 to give compound (B) is used as a moderatorin nuclear reaction. (A) adds on to a compound (C), which has the molecular formula C 3 H 6 to give (D). Identify A, B, C and D.[1]
2.The value of the universal gas constant depends upon(a) Temperature of the gas(b) Volume of the gas(c) Number of moles of the gas(d) units of pressure and volume[1]
3.Freon – 12 manufactured from tetrachloro methane by a) Wurtz reaction b) Swarts reaction c) Haloform reaction d) Gattermann reaction[1]
4.The Van’t Hoff factor (i) for a dilute aqueous solution of the strong electrolyte barium hydroxide is a) 0 b) 1 c) 2 d) 3[1]
5.Which one of the following statements is incorrect with regard to ortho and para dihydrogen ?(a) They are nuclear spin isomers(b) Ortho isomer has zero nuclear spin whereas the para isomer has one nuclear spin(c) The para isomer is favoured at low temperatures(d) The thermal conductivity of the para isomer is 50% greater than that of the ortho isomer.[1]
6.The C – H bond and C – C bond in ethane are formed by which of the following types of overlap a) sp 3 – s and sp 3 – sp 3 b) sp 2 – s and sp 2 – sp 2 c) sp – sp and sp – sp d) p – s and p – p[1]
7.The molecules having same hybridisation, shape and number of lone pairs of electrons are a) SeF 4, XeO 2 F 2 b) SF 4, XeF 2 c) XeOF 4, TeF 4 d) SeCl 4, XeF 4[1]
8.The general formula for cyclo alkanes a) C n H n b) C n H 2n C) C n H 2n-2 d) C n H 2n + 2[1]
9.Which of the following carbocation will be most stable? a) Ph 3 + C – b) CH 3 – + CH 2 – c) (CH 3 ) 2 – + CH d) CH 2 = CH – + CH 2[1]
10.The ratio of number of sigma (σ) bond and pi (π) bonds in 2 – butynal is a) 8/3 b) 5/3 c) 8/2 d) 9/2[1]
11.Which of the following statement is false?(a) Ca 2+ ions are not important in maintaining the regular beating of the heart(b) Mg 2+ ions are important in the green parts of the plants(c) Mg 2+ ions form a complex with ATP(d) Ca 2+ ions are important in blood clotting[1]
12.Which of the following species does not exert a resonance effect? a) C 6 H 5 OH b) C 6 H 5 Cl c) C 6 H 5 NH 2 d) C 6 H 5 NH 3[1]
13.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]
14.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]
15.The empirical formula of a non – electrolyte (X) is CH 2 O. A solution containing six grams of X exerts the same osmotic pressure as that of 0.025 M glucose solution at the same temperature. The molecular formula of X is a) C 2 H 4 O 2 b) C 8 H 16 O 8 c) C 4 H 8 O 4 d) CH 2 O[1]
Part II — Short Answer Questions 14 × 2 = 28

Answer briefly. (Answer all questions.)

16.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]
17.CO 2 and H 2 O both are triatomic molecule but their dipole moment values are different. Why?[2]
18.Which is the correct sequence of solubility of carbonates of alkaline earth metals ? (a) BaCO 3 > SrCO 3 > CaCO 3 > MgCO 3 (b) MgCO 3 > CaCO 3 > SrCO 3 > BaCO 3 (c) CaCO 3 > BaCO 3 > SrCO 3 > MgCO 3 (d) BaCO 3 > CaCO 3 > SrCO 3 > MgCO 3[2]
19.P 1 and P 2 are the vapour pressures of pure liquid components, 1 and 2 respectively of an ideal binary solution If x 1 represents the mole fraction of component 1, the total pressure of the solution formed by 1 and 2 will be a) P 1 + x 1 (P 2 – P 1 ) b) P 2 – x 1 (P 2 + P 1 ) c) P 1 – x 2 (P 1 – P 2 ) d) P 1 + x 2 (P 1 – P 2 )[2]
20.Define the term ‘isotonic solution’.[2]
21.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]
22.Find the wrong statement (a) sodium metal is used in organic qualitative analysis (b) sodium carbonate is soluble in water and it is used in inorganic qualitative analysis (c) potassium carbonate can be prepared by solvay process (d) potassium bicarbonate is acidic salt[2]
23.The major products obtained when chlorobenzene is nitrated with HNO 3 and con H 2 SO 4 a) 1 – chloro – 4 – nitrobenzene b) 1 – chloro – 2 – nitrobenzene c) 1 – chloro – 3 – nitrobenzene d) 1 – chloro – 1 – nitrobenzene[2]
24.Lassaigne’s test for the detection of nitrogen fails in a) H 2 N – CO – NH.NH 2.HCl b) NH 2 – NH 2.HCl c) C 6 H 5 – NH – NH 2.HCl d) C 6 H 5 CONH 2[2]
25.Choose the disproportionation reaction among the following redox reactions. (a) 3Mg (s) + N 2(g) → Mg 3 N 2(s) (b) P 4(s) + 3NaOH + 3H 2 O → PH 3(g) + 3NaH 2 PO 2(aq) (c) Cl 2(g) + 2KI (aq) → 2KCl (aq) + I 2(s) (d) Cr 2 O 3(s) + 2Al (s) → Al 2 O 3(s) + 2Cr (s)[2]
26.Hybridisation of central atom in PCl 5 involves the mixing of orbitals. a) s, P x, P y, d x 2, d x 2 – y 2 b) s, p x, p y, p xy, d x 2 – y 2 c) s, p x, p y, p z, d x 2 – y 2 d) s, p x, P y, d xy, d x 2 – y 2[2]
27.The IUPAC name of the following compound is a) trans – 2- chloro – 3- iodo – 2- pentane b) cis – 3- iodo – 4 chloro – 3 – pentane c) trans – 3 – iodo – 4 – chloro – 3 – pentene d) cis – 2 – chloro – 3 iodo – 2 – pentene[2]
28.Give the general electronic configuration of lanthanides and actinides?[2]
29.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]
Part III — Long Answer Questions 10 × 5 = 50

Answer in detail. (Answer all questions.)

30.Henry’s law constant for solubility of methane in benzene is 4.2 × 10 -5 mm Hg at a particular constant temperature. At this temperature calculate the solubility of methane at i) 750 mm Hg ii) 840 mm Hg.[5]
31.Pick out the incorrect statement from the following: a) sp 3 hybrid orbitals are equivalent and are at an angle of 109°28’ with each other b) dsp 2 hybrid orbitals are equivalent and bond angle between any two of them is 90° c) All five sp 3 d hybrid orbitals are not equivalent out of these five sp 3 d hybrid orbitals, three are at an angle of 120° remaining two are perpendicular to the plane containing the other three d) none of these[5]
32.What are isotopes? Write the names of isotopes of hydrogen.[5]
33.A sample of gas at 15°C at 1 atm. has a volume of 2.58 dm 3. When the temperature is raised to 38°C at 1 atm does the volume of the gas Increase? If so, calculate the final volume.[5]
34.What do you understand by the Linear combination of atomic orbitals in MO theory?[5]
35.Consider the following reactions, a) H 2 (g) + I 2 (g) ⇌ 2HI(g) b) CaCO 3 (s) ⇌ CaO (s) + CO 2 (g) c) S(s) + 3 F 2 (g) ⇌ SF 6 (g) In each of the above reactions find out whether you have to increase (or) decrease the volume to increase the yield of the product.[5]
36.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]
37.Explain how does greenhouse effect cause global warming.[5]
38.Write short notes on the following. i) Raschig process ii) Dows process iii) Darzen’s process[5]
39.Classify the following compounds in the form of alkyl, allylic, vinyl, benzylic halides. i) CH 3 – CH = CH – Cl ii) C 6 H 5 CH 2 I iii) iv) CH 2 = CH – Cl[5]
🔑 Show Answer Key — Set 1
  1. 1. An isotope of hydrogen (A) reacts with diatomic molecule of element which occupies group number 16 and period number 2 to give compound (B) is used as a moderator in nuclear reaction. 2D 2 + O 2 → 2D 2 O The isotope of hydrogen is deuterium(A), diatomic element is oxygen and the compound (B) is heavy water. (A) adds on to a compound (C), which has the molecular formula C 3 H 6 to give (D). CH 3 – CH = CH 2 + D 2 → CH 3 – CHD – CH 2 D (C) (D) The compound (C) is propene and (D) is 1, 2 deutropropane. A – Deuterium (D 2 ) B – Heavy water (D 2 O) C – Propene (CH 3 – CH = CH 2 ) D – 1, 2 deutero propene (CH 3 – CHD – CH 2 D)
  2. 2. (d) units of pressure and volume
  3. 3. b) Swarts reaction
  4. 4. d) 3
  5. 5. (b) Ortho isomer has zero nuclear spin whereas the para isomer has one nuclear spin
  6. 6. a) sp 3 – s and sp 3 – sp 3
  7. 7. a) SeF 4, XeO 2 F 2
  8. 8. b) C n H 2n
  9. 9. d) CH 2 = CH – + CH 2
  10. 10. a) 8/3
  11. 11. (a) Ca 2+ ions are not important in maintaining the regular beating of the heart
  12. 12. d) C 6 H 5 NH 3
  13. 13. c) CO 2
  14. 14. (d) -632.68 kJ
  15. 15. b) C 8 H 16 O 8
  16. 16. (c) Minute solid particles in air are known as particulate pollutants
  17. 17. Sum of the dipole moment are cancelled. Water is ‘v’ shape sum of the dipole moments are not equal to zero.
  18. 18. (b) MgCO 3 > CaCO 3 > SrCO 3 > BaCO 3
  19. 19. c) P 1 – x 2 (P 1 – P 2 )
  20. 20. Two solutions having same osmotic pressure at a given temperature are called isotonic solutions.
  21. 21. K p >K c n p > n R
  22. 22. (c) potassium carbonate can be prepared by solvay process
  23. 23. a) 1 – chloro – 4 – nitrobenzene
  24. 24. c) C 6 H 5 – NH – NH 2.HCl
  25. 25. b) P 4(s) + 3NaOH + 3H 2 O → PH 3(g) + 3NaH 2 PO 2(aq)
  26. 26. c) s, p x, p y, p z, d x 2 – y 2
  27. 27. a) trans – 2- chloro – 3- iodo – 2- pentane
  28. 28. 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.
  29. 29. (a) Amongst the isoelectronic species, smaller the positive charge on cation, smaller is the ionic radius
  30. 30. (K H ) benzene = 4.2 × 10 -5 mm Solubility of methane =? P = 750 mm Hg P = 840 mm Hg According to Henrys Law, P = K H X in solution 750 mm Hg = 4.2 × 10 -5 mm Hg. X in solution ⇒ X in solution = \(\frac{750}{4.2 \times 10^{-5}}\) i. e solubility = 178. 5 × 10 5 similarly at P = 840 mm Hg solubility = \(\frac{840}{4.2 \times 10^{-5}}\) = 200 × 10 -5
  31. 31. c) All five sp 3 d hybrid orbitals are not equivalent out of these five sp 3 d hybrid orbitals, three are at an angle of 120° remaining two are perpendicular to the plane containing the other three
  32. 32. Atoms of the same element having same atomic number and different mass number are called isotopes. Hydrogen has three naturally occurring isotopes, viz., protium ( 1 H 1 or H), deuterium ( 1 H 2 or D) and tritium ( 1 H 3 or T). Protium 1 H 1 is the predominant form (99.985 %) and it is the only isotope that does not contain a neutron. Deuterium, also known as heavy hydrogen, constitutes about 0.015 %. The third isotope, tritium is a radioactive isotope of hydrogen which occurs only in traces (~1 atom per 1018 hydrogen atoms). Due to the existence of these isotopes naturally occurring hydrogen exists as H 2, HD, D 2, HT, T 2, and DT.
  33. 33. T 1 = 15°C + 273; T 2 = 38 + 273 T 1 = 228; T 2 = 311K V 1 = 2.58dm 3; V 2 = ? (P = 1 atom constant) \(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\) V 2 = \(\left[\frac{V_{1}}{T_{1}}\right]\) × T 2 = \(\frac{2.58 d m^{3}}{288 K}\) × 311K V 2 = 2.78 dm 3 i.e, volume increased from 2.58 dm 3 to 2.78 dm 3
  34. 34. The wave functions for the molecular orbitals can be obtained by solving the Schrodinger wave equation for the molecule. Since solving the Schrodinger equation is too complex, approximation methods are used to obtain the wave function for molecular orbitals. The most common method is the linear combination of atomic orbitals (LCAO). We know that the atomic orbitals are represented by the wave function ψ. Let us consider two atomic orbitals represented by the wave function ψ A and ψ B with comparable energy, combines to form two molecular orbitals. One is bonding molecular orbital(ψ bonding ) and the other is antibonding molecular orbital (ψ antibonding ) The wave functions for these two molecular orbitals can be obtained by the linear combination of the atomic orbitals ψ A and ψ B as below, ψ bonding = ψ A + ψ B; ψ antibonding = ψ A – ψ B The formation of bonding molecular orbital can be considered as the result of constructive interference of the atomic orbitals and the formation of anti-bonding molecular orbital can be the result of the destructive interference of the atomic orbitals. The formation of the two molecular orbitals from two is orbitals is shown below. Constructive interaction: The two 1s orbitals are in phase and have the same sign, Destructive interaction: The two orbitals are out phase
  35. 35. a) H 2 (g) + I 2 (g) ⇌ 2HI(g) In this reaction, there is no effect on changing the volume. ∆ng = 0 b) CaCO 3 (s) ⇌ CaO (s) + CO 2 (g) In this reaction, increases in volume favours forward reaction. c) S(s) + 3 F 2 (g) ⇌ SF 6 (g) In this reaction, decreases in volume favours forward reaction.
  36. 36. (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 –
  37. 37. Greenhouse effect may be defined as the heating up of the earth surface due to trapping of infrared radiations reflected by earth’s surface by CO 2 layer in the atmosphere”. The heating up of earth through the greenhouse effect is called global warming. Without the heating caused by the greenhouse effect, Earth’s average surface temperature would be only about -18 °C (CPF). Although the greenhouse effect is a naturally occurring phenomenon, it is intensified by the continuous emission of greenhouse gases into the atmosphere. During the past 100 years, the amount of carbon dioxide in the atmosphere increased by roughly 30 percent and the amount of methane more than doubled. If these trends continue, the average global temperature will increase which can lead to melting of polar ice caps and flooding of low lying areas. This will increase incidence of infectious diseases like dengue, malaria etc.
  38. 38. i) Raschig process: Chloro benzene is commercially prepared by passing a mixture of benzene vapour, air and HCl overheated cupric chloride, this reaction is called the Raschig process, ii) Dows Process: C 6 H 5 Cl + NaOH C 6 H 5 OH + NaCl This reaction is known as Dows process. iii) Darzen’s process: CH 3 CH 2 OH + SOCl CH 3 CH 2 Cl + SO 2 ↑ + HCl↑ Ethanol Chloro ethane This reaction is known as Darzen’s process.
  39. 39. i) CH 3 – CH = CH – Cl = Allylic halide ii) C 6 H 5 CH 2 I = Benzylic halide iii) = Alkyl halide iv) CH 2 = CH – Cl = Vinyl halide
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.An organic Compound weighing 0.15 g gave on carius estimation, 0.12 g of silver bromide. The percentage of bromine in the Compound will be close to a) 46 % b) 34 % c) 3.4 % d) 4.6 %[1]
2.According to the Bohr Theory, which of the following transitions in the hydrogen atom will give rise to least energetic photon?(a) n = 6 to n = 1(b) n = 5 to n = 4(c) n = 5 to n = 3(d) n = 6 to n = 5[1]
3.A commercial sample of hydrogen peroxide marked as 100 volume H 2 O 2, it means that(a) 1 ml of H 2 O 2 will give 100 ml O 2 at STP(b) 1 L of H 2 O 2 will give 100 ml O 2 at STP(c) 1 L of H 2 O 2 will give 22.4 L O 2(d) 1 ml of H 2 O2 will give 1 mole of O 2 at STP[1]
4.Which of the following molecule contain no π bond? a) SO 2 b) NO 2 c) CO 2 d) H 2 O[1]
5.The First ionisation potential of Na, Mg and Si are 496, 737 and 786 kJ mol -1 respectively. The ionisation potential of Al will be closer to(a) 760 kJ mol -1(b) 575 kJ mol -1(c) 801 kJ mol -1(d) 419 kJ mol -1[1]
6.Which one of the following represents 180g of water? (a) 5 Moles of water (b) 90 moles of water (c) \(\frac{6.022 \times 10^{23}}{180}\) (d) \(\frac{6.022 \times 10^{23}}{1.7}\)[1]
7.The variation of volume V, with temperature T, keeping pressure constant is called the coefficient of thermal expansion i.e., α = 1\(\left[\frac{\partial V}{\delta T}\right]\)Vp. For an ideal gas, α is equal to (a) T (b) 1/T (c) P (d) none of these[1]
8.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]
9.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]
10.Which of the following species is not electrophilic in nature? a) Cl + b) BH 3 c) H 3 O + d) + NO 2[1]
11.How many electrons in an atom with atomic number 105 can have (n + l) = 8?(a) 30(b) 17(c) 15(d) unpredictable[1]
12.The IUPAC name of the compound is a) 2, 3 – Dimethylheptane b) 3 – methyl – 4 – ethyloctane c) 5 – ethyl – 6- methyloctane d) 4 – Ethyl – 3 methyloctane[1]
13.For a solution, the plot of osmotic pressure (π) versus the concentration (c in mol L -1 ) gives a straight line with slope 310 R where ‘R’ is the gas constant. The temperature at which osmotic pressure measured is a) 310 × 0.082 K b) 310° C c) 37°C d) \(\frac{310}{0.082}\) K[1]
14.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]
15.A bottle of ammonia and a bottle of HCl connected through a long tube are opened simultaneously at both ends. The white ammonium chloride ring first formed will be(a) At the center of the tube(b) Near the hydrogen chloride bottle(c) Near the ammonia bottle(d) Throughout the length of the tube[1]
Part II — Short Answer Questions 14 × 2 = 28

Answer briefly. (Answer all questions.)

16.A person was using water supplied by corporation. Due to shortage of water he started using underground water. He felt laxative effect. What could be the cause?[2]
17.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]
18.∆S is expected to be maximum for the reaction (a) Ca(S) + 1/2 O 2 (g) → CaO(S) (b) C(S) + O 2 (g) → CO 2 (g) (c) N 2 (g) + O 2 (g) → 2NO(g) (d) CaCO 3 (S) → CaO(S) + CO 2 (g)[2]
19.Give Kelvin a statement of the second law of thermodynamics.[2]
20.The equilibrium constants of the following reactions are: N 2 + 3H 2 ⇌ 2NH 3; K 1 N 2 + O 2 ⇌ 2NO; K 2 H 2 + 1/2O 2 ⇌ H 2 O; K 3 The equilibrium constant (K) for the reaction; 2NH 3 + 5/2 O 2 ⇌ 2NO + 3H 2 O, will be a) K 2 3 K 3 /K 1 b) K 1 K 3 3 /K 2 c) K 2 K 3 3 /K 1 d) K 2 K 3 /K 1[2]
21.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]
22.In what period and group will an element with Z = 118 will be present?[2]
23.The correct order of O – O bond length in hydrogen peroxide, ozone and oxygen is a) H 2 O 2 > O 3 > O 2 b) O 2 > O 3 > H 2 O 2 c) O 2 > H 2 O 2 > O 3 d) O 3 > O 2 > H 2 O 2[2]
24.What is osmosis?[2]
25.Acetone X, X is a) 2 – propanol b) 2 – methyl – 2 – propanol c) 1 – propanol d) acetonol[2]
26.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]
27.Suggest a simple chemical test to distinguish propane and propene.[2]
28.Which of these represents the correct order of their increasing bond order. a) C 2 + < C 2 2- < O 2 2- < O 2 b) C 2 2- < C 2 + < O 2 < O 2 2- c) O 2 2- < O 2 < C 2 2- < C 2 + d) O 2 2- < C 2 + < O 2 < C 2 2-[2]
29.Match the List I with List-II and select the correct answer using the code given below the lists: List I List II A. Depletion of the ozone layer 1. CO 2 B. Acid rain 2. NO C. Photochemical smog 3. SO 2 D. Greenhouse effect 4. CFC Code:[2]
Part III — Long Answer Questions 10 × 5 = 50

Answer in detail. (Answer all questions.)

30.A sealed container was filled with 0.3 mol H 2 (g), 0.4 mol I 2 (g) and 0.2 mol HI(g) at 800 K and total pressure 1.00 bar. Calculate the amounts of the components in the mixture at equilibrium given that K = 870 for the reaction, H 2 (g) + I 2 (g) ⇌ 2 HI (g).[5]
31.Give the principle involved in the estimation of halogen in an organic compound by Carius method.[5]
32.The partial pressure of carbon dioxide in the reaction CaCO 3 (s) ⇌ CaO(s) + CO 2 (g) is 1.017 × 10 -3 atm at 500°C. Calculate K p at 600°C for the reaction. H for the reaction is 181 KJ mol -1 and does not change in the given range of temperature.[5]
33.Even though the use of pesticides increases crop production, they adversely affect the living organisms. Explain the function and the adverse effects of the pesticides.[5]
34.Why do halogens act as oxidizing agents?[5]
35.How will distinguish 1 – butyne and 2 – butyne?[5]
36.Oxidation of nitrogen monoxide was studied at 200 with initial pressures of 1 atm NO and 1 atm of O 2. At equilibrium partial pressure of oxygen is found to be 0. 5 atm calculate K p value.[5]
37.Arrange NH 3, H 2 O, and HF in the order of increasing magnitude of hydrogen bonding and explain the basis for your arrangement.[5]
38.Explain the Pauling method for the determination of ionic radius.[5]
39.When 22.4 litres of H 2 (g) is mixed with 11.2 litres of Cl 2 (g), each at 273 K at 1 atm the moles of HCl (g), formed is equal to (a) 2 moles of HCl (g) (b) 0.5 moles of HCl (g) (c) 1.5 moles of HCl (g) (d) 1 moles of HCl (g)[5]
🔑 Show Answer Key — Set 2
  1. 1. b) 34 %
  2. 2. (d) n = 6 to n = 5
  3. 3. (a) 1 ml of H 2 O 2 will give 100 ml O 2 at STP
  4. 4. d) H 2 O
  5. 5. (b) 575 kJ mol -1
  6. 6. (d) \(\frac{6.022 \times 10^{23}}{1.7}\) No. of moles of water present in 180 g = Mass of water / Molar mass of water = 180 g/18 gmol -1 = 10 moles One mole of water contains = 6.022 × 10 23 water molecules 10 mole of water contains = 6.022 × 10 23 × 10 × 6.022 × 10 24 water molecules
  7. 7. (a) T
  8. 8. b) x + 0.5
  9. 9. a) 25, 25, 33.3, 50
  10. 10. c) H 3 O +
  11. 11. (b) 17
  12. 12. d) 4 – Ethyl – 3 methyloctane
  13. 13. c) 37°C
  14. 14. (c) Hydrogen has three isotopes of which tritium is the most common.
  15. 15. (b) Near the hydrogen chloride bottle
  16. 16. Drinking water containing moderate level of sulphatcs is harmless. But excessive concentration (>500 ppm) of suiphates in drinking water causes laxative effect.
  17. 17. a) sp, sp, sp 3, sp 2, sp 3
  18. 18. (d) CaCO 3 (S) → CaO(S) + CO 2 (g)
  19. 19. Kelvin-Planck statement: It is impossible to take heat from a hotter reservoir and convert a cyclic process heat to a cooler reservoir.
  20. 20. c) K 2 K 3 3 /K 1
  21. 21. d) sp 3 d, sp 3 and sp hybridised respectively
  22. 22. The element Ununoctium (Oganesson, Z – 118) present in the 7 th period and 18 th group of the periodic table.
  23. 23. b) O 2 > O 3 > H 2 O 2
  24. 24. Osmosis is a spontaneous process by which the solvent molecules pass through a semipermeable membrane from a solution of lower concentration to the solution of higher concentration.
  25. 25. b) 2 – methyl – 2 – propanol
  26. 26. d) C 6 H 5 NH 2 and ClCH 2 – CHO
  27. 27. Propene decolourises Br 2 /H 2 O it forms dibromo compound but propane does not react with Br 2 / H 2 O.
  28. 28. d) O 2 2- < C 2 + < O 2 < C 2 2-
  29. 29. (a)
  30. 30. A 2 (g) + B 2 (g) ⇌ 2 AB (g) Given that, K p = 1; \(\frac{4 x^{2}}{(1-x)^{2}}\) = 1 ⇒ 4x 2 = (1 – x) 2 = 1 ⇒ 4x 2 = 1 + x 2 – 2x 3x 2 + 2x – 1 = 0 x = 0.33 – 1(not possible) ∴ [A 2 ] eq = 1 – x = 1 – 0.33 = 0.67 [B 2 ] eq = 1 – x = 1 – 0.33 = 0.67 [AB 2 ] eq = 2x × 0.33 = 0.66
  31. 31. Estimation of halogens (Carius method): A known mass of the organic compound is heated with fuming HNO 3 along with AgNO 3. C, H & S gets oxidized to CO 2, H 2 O, SO 2 and halogen combines with AgNO 3 to form a precipitate of silver halide. The ppt of AgX is filtered, washed, dried and weighed. From the mass of AgX and the mass of the organic compound taken, percentage of halogens are calculated. A known mass of the substance is taken along with fuming HNO 3 and AgNO 3 is taken in a clean carius tube. The open end of the Carius tube is sealed and placed in a iron tube for 5 hours in the range at 530 – 540 K Then the tube is allowed to cool and a small hole is made in the tube to allow gases produced to escape. The tube is broken and the ppt is filtered, washed, dried and weighed. From the mass of AgX obtained, calculations are made. Calculation: Weight of the organic compound: w g Weight of AgCl precipitate = a g 143. 5 g of AgCl contains 35.5 g of Cl ∴ a g of AgCl contains \(\frac{35.5}{143.5}\) × a w g Organic compound gives a g AgCl Percentage of Cl in w g organic compound = \(\left(\frac{35.5}{143.5} \times \frac{\mathrm{a}}{\mathrm{w}} \times 100\right) \%\) Let Weight of silver Bromide be ‘b’ g 188 g of AgBr contains 80 g of Br ∴ b g of AgBr contains \(\frac{80}{180} \times \frac{b}{w}\) of Br w g Organic compound gives b g AgBr Percentage of Br in w g organic compound = \(\left(\frac{80}{180} \times \frac{b}{w} \times 100\right) \%\) Let Weight of silver Iodide be ‘c’ g 235 g of AgI contains 127 g of I ∴ C g of AgI contains \(\left(\frac{127}{235} \times \frac{c}{w}\right)\) of I w g Organic compound gives c g AgI percentage of I in w g organic compound = \(\left(\frac{80}{180} \times \frac{b}{w} \times 100\right) \%\)
  32. 32. P CO 2 = 1.017 × 10 -3 atm T = 500°C; K p = P CO 2 ∴ K p 1 = 1.017 × 10 -3; T = 500 + 273 = 773 K K p 2 = ? T = 600 + 273 = 873 K ∆H° = 181 KJ mol -1
  33. 33. Pesticides are the chemicals that are used to kill or stop the growth of unwanted organisms. But these pesticides can affect the health of human beings. Pesticides are classified as (a) insecticides, (b) Fungicides and (c) Herbicides. (a) Insecticides: Insecticides like DDT, BHC, Aidrin can stay in soil for a long period of time and are absorbed by soil. They contaminate root crops like carrot, radish. (b) Fungicides: Organomercury compounds dissociate in soil to produce mercury which is highly toxic. (c) Herbicides: They are used to control unwanted plants and are also known as weed killers. Eg, Sodium chlorate, sodium nitrate. They are toxic to mammals.
  34. 34. Halogens are having the general electronic configuration of ns 2, np 5 and readily accept an electron to get the stable noble gas electronic configuration. Therefore, halogens have high electron affinity. Hence, halogens act as oxidizing agents.
  35. 35. 1 – butyne reacts with ammoniacal AgNO 3 solution it forms white precipitate of silver acetylide but, 2 – butyne doesnot reacts with ammoniacal AgNO 3 solution.
  36. 36. 2 NO(g) + O 2 (g) ⇌ 2NO 2 (g) K p = \(\frac{\left(P_{N O_{2}}\right)^{2}}{\left(P_{N O}\right)^{2}\left(P_{O}\right)}\) = \(\frac{0.96 \times 0.96}{0.04 \times 0.04 \times 0.52}\) K p = 1.017 × 10 3.
  37. 37. The increasing magnitude of hydrogen bonding among NH 3, H 2 O, and HF is HF > H 2 O > NH 3 The extent of hydrogen bonding depends upon electronegativity and the number of hydrogen atoms available for bonding. Among N, F, and O the increasing order of their electronegativities are N < O, H 2 O > NH 3.
  38. 38. Ionic radius is defined as the distance from the centre of the nucleus of the ion upto which it exerts its influence on the electron cloud of the ion. The ionic radius of a uni-univalent crystal can be calculated using Pauling’s method from the interionic distance between the nuclei of the cation and anion. Pauling assumed that ions present in a crystal lattice are perfect spheres, and they are in contact with each other and therefore, d = r c+ + r A- …………..(1) where ‘d’ is the distance between the centre of the nucleus of the cation C + and A –. r c+ and r A- are the radius of the cation and anion respectively. Pauling also assumed that the radius of the ion having noble gas electronic configuration (Na + and Cl – having 1s 2, 2s 2, 2p 6 configuration) is inversely proportional to the effective nuclear charge felt at the periphery of the ion. i.e., r c+ ∝ \(\frac{1}{\left(Z_{e f f}\right)^{C+}}\) ………(2) and r A- ∝ \(\frac{1}{\left(Z_{e f f}\right)^{A-}}\) ………….(3) where Z eff is the effective nuclear charge. Z eff = Z – S. Dividing the equation (2) by (3) \(\frac{r_{c^{+}}}{r_{A^{-}}}=\frac{\left(Z_{e f f}\right)^{A-}}{\left(Z_{e f f}\right)^{C+}}\) On solving the equations (1) and (4), the ionic radius of cation and anion are calculated.
  39. 39. (d) 1 moles of HCl (g) H 2(g) + Cl 2(g) → 2HCl (g) Content CH 4 0 2 CO 2 Stoichiometric coefficient 1 1 2 No. of moles of reactants allowed to react at 273 K and 1 atm pressure 22.4 L (1 mol) 11.2 L (0.5 mol) _ No. of moles of a reactant reacted and product formed 0.5 0.5 _ Amount of HCl formed = 1 mol
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.According to Valence bond theory, a bond between two atoms is formed when a) fully filled atomic orbitals overlap b) half filled atomic orbitals overlap c) non – bonding atomic orbitals overlap d) empty atomic orbitals overlap[1]
2.The reaction H 3 PO 2 + D 2 O → H 2 DPO 2 + HDO indicates that hypo-phosphorus acid is(a) tribasic acid(b) dibasic acid(c) monobasic acid(d) none of these[1]
3.The values of ∆H and ∆S for a reaction are respectively 30 kJ mol -1 and 100 JK -1 mol -1. Then the temperature above which the reaction will become spontaneous is(a) 300 K(b) 30 K(c) 100 K(d) 20°C[1]
4.Which of the following belongs to secondary air pollutant?(a) Hydrocarbon(b) Peroxy acetyl nitrate(c) Carbon monoxide(d) Nitric oxide[1]
5.The cause of permanent hardness of water is due to(a) Ca(HCO 3 ) 2(b) Mg(HCO 3 ) 2(c) CaCl 2(d) MgCO 3[1]
6.The freezing point depression constant for water is 1.86° K Kg mol -1. If 5 g Na 2 SO 4 is dissolved in 45 g water, the depression in freezing point is 3.64°C. The Vant Hoff factor for Na 2 SO 4 is a) 2.57 b) 2.63 c) 3.64 d) 5.50[1]
7.40 ml of methane is completely burnt using 80 ml of oxygen at room temperature. The volume of gas left after cooling to room temperature(a) 40 ml CO 2(b) 40 ml CO 2 gas and 80 ml H 2 o gas(c) 60 ml CO 2 gas and 60 ml H 2 o gas(d) 120 ml CO 2 gas[1]
8.Which of the following elements will have the highest electronegativity?(a) Chlorine(b) Nitrogen(c) Cesium(d) Fluorine[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.The element with positive electron gain enthalpy is(a) Hydrogen(b) Sodium(c) Argon(d) Fluorine[1]
11.Four gases P, Q, R, and S have almost the same values of ‘b’ but their a’ values (a. h are Vander Waals Constants) are in the order Q < R < S < p. At a particular temperature, among the four gases, the most easily liquelìable one is(a) P(b) Q(c) R(d) S[1]
12.How does electron affinity change when we move from left to right in a period in the periodic table?(a) Generally increases(b) Generally decreases(c) Remains unchanged(d) First increases and then decreases[1]
13.Which of the following concentration terms is / are independent of temperature a) molality b) molarity c) mole fraction d) a and b[1]
14.Which of the following represent a set of nucleophiles? a) BF 3, H 2 O, NH 2- b) AlCl 3, BF 3, NH 3 c) CN, RCH 2 –, ROH d) H +, RNH 3 +,:CCl 2[1]
15.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]
Part II — Short Answer Questions 14 × 2 = 28

Answer briefly. (Answer all questions.)

16.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]
17.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]
18.When one s and three p orbitals hybridise, a) four equivalent orbitals at 90° to each other will be formed b) four equivalent orbitals at 109°28’ to each other will be formed c) four equivalent orbitals, that are lying the same plane will be formed d) none of these[2]
19.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]
20.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]
21.Which of the group has highest + I effect? a) CH 3 – b) CH 3 – CH 2 – c) (CH 3 ) 2 – CH- d) (CH 3 ) 3 – C –[2]
22.The equilibrium for the dissociation of XY 2 is given as, 2 XY 2 (g) ⇌ 2 XY(g) + Y 2 (g) if the degree of dissociation x is so small compared to one. Show that 2 K p = PX 3 where P is the total pressure and K p is the dissociation equilibrium constant of XY 2.[2]
23.Match the flame colours of the alkali and alkaline earth metal salts in the bunsen burner (p) Sodium (1) Brick red (q) Calcium (2) Yellow (r) Barium (3) Violet (s) Strontium (4) Apple green (t) Cesium (5) Crimson red (u) Potassium (6) Blue (a) p – 2, q – 1, r – 4, s – 5, t – 6, u – 3 (b) p – 1, q – 2, r – 4, s – 5, t – 6, u – 3 (c) p – 4, q – 1, r – 2, s – 3, t – 5, u – 6 (d) p – 6, q – 5, r – 4, s – 3, t – 1, u – 2[2]
24.How will you convert ethyl chloride into (i) ethane (ii) n – butane[2]
25.What type of hybridization is possible in the following geometries? * octahedral * tetrahedral * square planar[2]
26.Mention the standards prescribed by BIS for quality of drinking water.[2]
27.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]
28.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]
29.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]
Part III — Long Answer Questions 10 × 5 = 50

Answer in detail. (Answer all questions.)

30.An alkali metal (x) forms a hydrated sulphate, X 2 SO 4.10H 2 O. Is the metal more likely to be sodium (or) potassium?[5]
31.Give reasons for the polarity of C – X bond in haloalkane.[5]
32.State the first law of thermodynamics.[5]
33.How is plaster of paris prepared?[5]
34.Differentiate the following: (i) BOD and COD (ii) Viable and non-viable particulate pollutants.[5]
35.Define smog.[5]
36.Substantiate lithium fluoride has the lowest solubility among group one metal fluorides.[5]
37.In a constant-volume calorimeter, 3.5 g of gas with molecular weight 28 was burnt in excess oxygen at 298 K. The temperature of the calorimeter was found to increase from 298 K to 298.45 K due to the combustion process. Given that the calorimeter constant is 2.5kJ K -1. Calculate the enthalpy of combustion of the gas in kJ mol -1.[5]
38.How will you prepare n propyl iodide from n – propyl bromide?[5]
39.Discuss the aromatic nucleophilic substitutions reaction of chlorobenzene.[5]
🔑 Show Answer Key — Set 3
  1. 1. b) half filled atomic orbitals overlap
  2. 2. (c) monobasic acid Hypophosphorous acid on reaction with D 2 O, only one hydrogen is replaced by deuterium and hence it is monobasic.
  3. 3. (a) 300 K II. Write brief answers to the following questions:
  4. 4. (a) Hydrocarbon
  5. 5. (c) CaCl 2
  6. 6. a) 2.57
  7. 7. (a) 40 ml CO 2 CH 4 (g) + 2O 2 → CO 2 (g) + 2H 2 O (1) Content CH 4 O 2 CO 2 Stoichiometric coefficient 1 2 1 Volume of reactants allowed to react 40 mL 80 mL – Volume of reactant reacted and product formed 40 mL 80 mL 40 mL Volume of gas after cooling to the room temperature – – – Since the product was cooled to room temperature, water exists mostly as liquid. Hence, option (a) is correct.
  8. 8. (d) Fluorine
  9. 9. (b) (ii), (iv) and (v)
  10. 10. (c) Argon
  11. 11. (c) R
  12. 12. (a) Generally increases
  13. 13. d) a and b
  14. 14. c) CN, RCH 2 –, ROH
  15. 15. (d) Be and Al II. Write brief answer to the following questions:
  16. 16. Average atomic mass = \(\frac{(78.9923.99)(1024.99)(11.0125.98)}{100}\) = \(\frac{2430.9}{100}\) = 24.31 u
  17. 17. When Q > K C the reaction will proceed in the reverse direction, i.e, formation of reactants.
  18. 18. b) four equivalent orbitals at 109°28’ to each other will be formed
  19. 19. (a) ns → (n – 2)f → (n – 1)d → np
  20. 20. b) Pent – 3- en – 1- yne
  21. 21. d) (CH 3 ) 3 – C –
  22. 22. 2 XY 2 (g) ⇌ 2 XY(g) + Y 2 (g)
  23. 23. (a) p – 2, q – 1, r – 4, s – 5, t – 6, u – 3
  24. 24. (i) ethane (ii) n – butane
  25. 25. octahedral: sp 3 d 2 tetrahedral: sp 3 square planar: dsp 2
  26. 26. Standard characteristics prescribed for deciding the quality of drinking water by BIS, in 1991 are shown in Table.
  27. 27. A gas’s partial pressure formula is the gas pressure exerted if that gas were alone.
  28. 28. 1) 3 – chloro nitrobenzene 2) 4 – chlorotoluene 3) Bromo benzene 4) m – dinitro benzene
  29. 29. 1) 2) 3) 4) ethyl isopropyl acetylene 5)
  30. 30. X forms X 2 SO 2. 10H 2 O. The metal is more likely to be sodium. So X is Na 2 SO 4. 10H 2 O. It is otherwise called as Glauber’s salt.
  31. 31. Carbon halogen bond is a polar bond as halogens are more electronegative than carbon. The carbon atom exhibits a partial positive charge (δ + ) and halogen atom a partial negative charge (δ – ) The C -X bond is formed by overlap of sp 3 orbital of a carbon atom with half-filled p- orbital of the halogen atom. The atomic size of halogen increases from fluorine to iodine, which increases the C – X bond length. Larger the size, greater is the bond length, and the weaker is the bond formed. The bond strength of C – X decreases from C – F to C – I in CH 3 X.
  32. 32. The first law of thermodynamics states that “the total energy of an isolated system remains constant though it may change from one form to another” (or) Energy can neither be created nor destroyed but may be converted from one form to another.
  33. 33. It is a hemihydrate of calcium sulphate. It is obtained when gypsum, CaSO 4.2H 2 O is heated to 393 K. 2CaSO 4.2H 2 O(s) → 2CaSO 4.H 2 O + 3H 2 O Above 393 K, no water of crystallisation is left and anhydrous calcium sulphate, CaSO 4 is formed. This is known as ‘dead burnt plaster’.
  34. 34. (i) BOD and COD Biochemical oxygen demand (BOD): The total amount of oxygen in milligrams consumed by microorganisms in decomposing the waste in one litre of water at 200°C for a period of 5 days is called biochemical oxygen demand (BOD) and its value is expressed in ppm. BOD is used as a measure of degree of water pollution. Clean water would have BOD value less than 5 ppm whereas highly polluted water has BOD value of 17 ppm or more. Chemical Oxygen Demand (COD): BOD measurement takes 5 days so another parameter called the Chemical Oxygen Demand (COD) is measured. Chemical oxygen demand (COD) is defined as the amount of oxygen required by the organic matter in a sample of water for its oxidation by a strong oxidising agent like K 2 Cr 2 O 7 in acid medium for a period of 2 hrs. (ii) Viable and non – viable particulate pollutants: Viable particulates: The viable particulates are the small size living organisms such as bacteria, fungi, moulds, algae, etc. which are dispersed in air. Some of the fungi cause allergy in human beings and diseases in plants. Non-viable particulates: The non- viable particulates are small solid particles and liquid droplets suspended in air. They help in the transportation of viable particles. There are four types of non-viable particulates in the atmosphere. Example: Smoke, Dust, Mists, Fumes.
  35. 35. Smog is a combination of smoke and fog which form droplets that remain suspended in the air. Smog is a chemical mixture of gases that forms a brownish-yellow haze. It mainly consists of ground-level ozone, oxides of nitrogen, volatile organic compounds, SO 2, acidic aerosols and some other gases.
  36. 36. Lithium fluoride has high lattice enthalpy due to the small size of Li + and F –. So, due to the high lattice enthalpy, LiF is less soluble in water.
  37. 37. Given, T i = 298 K T f = 298.45 K k = 2.5 kJ K m = 3.5g M m = 28 heat evolved = k∆T ∆H C = k (T f – T i ) ∆H C = 2.5 kJ K’ (298.45 – 298) K -1 ∆H C = 1.125 kJ ∆H C = \(\frac {1.125}{3.5}\) x 28 kJ mol -1 ∆H C = 9 kJ mol -1
  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. The halogen of haloarenes can be substituted by OH –, NH 2 – or CN – with appropriate nucleophilic reagents at high temperature and pressure. Example: (i) Chlorobenzene reacts with ammonium at 250 and at 50 atm to give aniline. C 6 H 5 Cl + 2NH 3 C 6 H 5 NH 2 + NH 4 Cl Chlorobenzene Aniline (ii) Chlorobenzcne reacts with CuCN in presence of pyridine at 250 to give phenyl cyanide. C 6 H 5 Cl + CuCN C 6 H 5 CN + CuCl Chlorobenzene Phenyl cyanide (iii) Dows process: C 6 H 5 Cl + NaOH C 6 H 5 OH + NaCl Chlorobenzene Phenol This reaction is known as Dow’s process.

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