Solid State****Q.No.1  Which of the following is not a characteristic property of Solids?(a) Intermolecular distances are short(b) Intermolecular forces are weak(c) Constituent particles have fixed positions(d) Solids oscillate about their mean positionsAns.(b) Intermolecular forces are strong in solids.
Q.No.2 Most crystals show good cleavage because their atoms, ions or molecules are(a) weakly bonded together(b) strongly bonded together(c) spherically symmetrical(d) arranged in planesAns. (d) crystals show good cleavage because their constituent particles are arranged in planes.
Q.NO.3 "crystalline solids are anisotropic in nature.  What is the meaning of anisotropic in the given statement?(a) a regular pattern of arrangement of particles which repeats itself periodically over the entire crystal(b) Different values of some of physical properties are shown when measured along different directions in the same crystals(c) An irregular arrangement of particles over the entire crystal(d) Same values of some of physical properties are shown when measured along different directions in the same crystals.Ans.(b) Different values of some of physical properties are shown when measured along different directions in the same crystals.
Q.No.4 A crystalline solid (a) changes abruptly from solid to liquid when heated(b) has no definite melting point(c) undergoes deformation of its geometry easily(d) has irregular 3D arrangementsAns. (a) changes abruptly from solid to liquid when heated
Q.No.5 which of the following is not a characteristic of a crystalline solid?(a) definite and characteristic heat of fusion(b) Isotropic nature(c) A regular periodically repeated pattern of arrangement of constituent particles in the entire crystal(d) A true solidAns. (b) isotropic nature
Q.No.6 Which of the following is not a crystalline solid?(a) KCl(b) CsCl(c) Glass(d) Rhombic SAns.(c) Glass
Q.No.7 Which of the following statements about amorphous solids is incorrect?(a) They melt over a range of temperature(b) They are anisotropic(c) There is no orderly arrangement of particles(d) They are rigid and incompressibleAns. (b) Amorphous solids are isotropic, because these substances show same properties in all directions.
Q.No.9 Which of the following is an amorphous solid?(a) Graphite (b) Quartz glass (c) Chrome alum(d) Silicon CarbideAns.(b) Quartz glass
Q.NO.10 Which of the following statement is not true about amorphous solids?(a) On heating they may become crystalline at certain temperature(b) They may become crystalline on keeping for long time(c) Amorphous solids can be moulded by heating(d) They are anisotropic in natureAns.(d) They are anisotropic in nature.
Q.No.11 The sharp melting point of crystalline solids is due to ____(a) a regular arrangement of constituent particles observed over a short distance in the crystal lattice(b) a regular arrangement of constituent particles observed over a long distance in the crystal lattice(c) same arrangement of constituent particles in different directions (d) different arrangement of constituent particles in different directions.Ans. (b) a regular arrangement of constituent particles observed over a long distance in the crystal lattice.
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Q1 The electrical conductivity of semiconductor is a. 107 ohm-1 cm-1b. 10-15 ohm-1 cm-1c. in the range of 10-6 to 104 ohm-1 cm-1d. noneAns. c. in the range of 10-6 to 104 ohm-1 cm-1
Q2. Pure silicon and germanium at 0 K (-273.15 'C) area. conductorsb. insulatorsc. semiconductorsd. may be or may not be aboveAns. b.insulators
Q.3 A solid has a structure in which W atoms are located at the corners of a cubic lattice, O atoms at the centre of edges and Na atom at centre of the cube.  The formula for the compound isa. NaWO2b. NaWO3c.Na2WO3d. NaWO4 Ans. b. NaWO3
Q4 The number of atoms in 100g of a fcc crystal with density = 10.0g/cm3 and cell edge equal to 200 pm is equal to a. 5x 1024 b. 5x1025c. 6x1023d. 2x1025Ans.a. 5x1024
Q Schottky defect in crystals is observed whena. unequal no. of cations and anions are missing from the latticeb. equal no. of cations and anions are missing from the latticec. an ion leaves its normal site and occupies an interstitial sited. density of the crystal is increasedAns. b. equal no. of cations and anions are missing from the lattice.
Q5 Due to Frenkel defect, the density of the ionic solids a. increasesb. decreasesc. does not changed. changesAns.c. does not change
Q6 In a tetragonal crystala. a=b=c, α = β = 90 ≠  γb. α = β = γ = 90, a = b ≠ cc. α = β = γ = 90 , a ≠ b ≠ cd. α = β = 90, γ = 120, a=b=cAns. b. α = β = γ = 90, a = b ≠ c
Q7 The ratio of Fe+3  and Fe+2  ions in Fe0.9 S1.0 a.0.28b.0.5c.2d. 4Ans.a.0.28
Q8 The material used in the solar cells containsa. Csb. Sic. Snd. TiAns.b. Si
Q9 In the calcium fluoride structure, the coordination no. of the cations and the anions are respectively (V Imp)a. 6 and 6b. 8 and 4c. 4 and 4d. 4 and 8Ans.b.8 and 4
Q10 A metallic crystal crystallizes into a lattice containing a sequence of layers AB AB AB ... Any packing of spheres leaves out voids in the lattice.  What percentage by volume of this lattice is empty space?a.74%b. 26% c.50%d.NoneAns.b.26%
Q11 The intermetallic compound LiAg crystallizes in cubic lattice in which both lithium and silver have coordination number of eight.  The crystal class isa. Simple cubicb. Body centred cubicc. Face-centred cubicd. NoneAns.b.Body centred cubic
Q12 The edge length of face centred cubic unit cell is 508pm.  If the radius of the cation is 110pm, the radius of the anion isa.144pmb. 288pmc. 618 pmd. 398pmAns.a.144pm
Q13. In the crystals of which of the following ionic compounds would you expect maximum distance between the centres of the cations and anions?a. LiFb. CsFc. CsId. LiIAns. c.CsI
Q14 Schottky defect in crystals is observed whena. unequal no. of cations and anions are missing from the latticeb. equal no. of cations and anions are missing from the latticec. an ion leaves its normal site and occupies an interstitial sited. density of the crystal is increasedAns. b. equal no. of cations and anions are missing from the lattice.
Q15 How many kinds of space lattice are possible in a crystal?a. 23b. 7c. 230d. 14Ans. d. 14
Q16. Potassium crystallizes with a a. Face-centred cubic latticeb. Body-centred cubic latticec. simple cubic latticed. orthorhombic latticeAns. b. Body-centred cubic lattice.
Q17 The coordination no. of a metal crystallising in a hexagonal close packed structure isa. 12b. 4c. 8d. 6Ans. a.12
Q18 A compound formed by element A and B crystalizes in the cubic structure where A atoms are at the corners of a cube and B atoms are at the face-centres.  The formula of the compound is a. AB3b. ABc. A3Bd. A2B2Ans. a. AB3
Q19 The number of unit cells in 58.5g of NaCl is nearlya. 6 x 1020b. 3 x 1022c. 1.5 x 1023d. 0.5 x 1024Ans. c. 1.5 x 1023
Q20 The number of octahedral sites per sphere in fcc structure is a. 8b. 4c. 2d. 1Ans. d.1
Q21 the packing fraction for a body-centred cube is a. 0.42b. 0.53c. 0.68d. 0.82Ans. c. 0.68
Q22 Which of the following has Frenkel defect?a. NaClb. Graphitec. Silver Bromided. DiamondAns. c. Silver Bromide
Q23 In NaCl, the chloride ions occupy the space in a fashion ofa. fccb. bccc. bothd. noneAns. a.fcc
Q24 To get n-type doped semiconductor, impurity to be added to silicon should have the following number of valence electrons?a. 2b. 5c. 3d. 1Ans. b.5
Q25 The range of radius ratio (cationic to anionic) for an octahedral arrangement of ions in an ionic solid is a. 0 - 0.155b. 0.155 - 0.225c. 0.225 - 0.414d. 0.414 - 0.732e. 0.732 - 1.000Ans. d. 0.414 - 0.732
Q26 When molten zinc is cooled to solid state, it assumes HCP structure.  Then the number of nearest neighbours of zinc atom will be a. 4b. 6c. 8 d. 12Ans. d. 12
Q27 In a solid 'AB' having the NaCl structure, 'A' atoms occupy the corners of the cubic unit cell.  If all the face-centred atoms along one of the axes are removed, then the resultant stoichiometry of the solid isa. AB2b. A2Bc. A4B3d. A3B4Ans. d. A3B4
Q28 A substance AxBy crystallizes in a face centred cubic (FCC) lattice in which atoms 'A' occupy each corner of the cube and atoms 'B' occupy the centres of each face of the cube. Identify the correct composition of the substance AxBy  a. AB3 b. A4B3 c. A3Bd. Compound cannot be specified. Ans. a. AB3 
Q.29 Superconductors are derived from the compounds a. p-block elementsb. lanthanidesc. actinidesd. transition elementse. scandiumAns. a. p block elements
Q30 A semiconductor of Ge can be made p type by addinga. trivalent impurityb. tetravalent impurityc. pentavalent impurityd. divalent impurityAns. a.trivalent impurity
Q31 the interionic distance for cesium chloride crystal will be a. ab. a/2c. √3 a/2d. 2a/ √3 Ans. c. √3 a/2--------------------------------------------------------------------------------------------------------Q1 1. The pure crystalline substance on being heated gradually first forms a turbid liquid at constant temperature and still at higher temperature turbidity completely disappears. The behaviour is a characteristic of substance forming(a) allotropic crystals (b) liquid crystals(c) isomeric crystals (d) isomorphous crystals.Ans. 
Q2 Glass is a (a) liquid(b) solid(c) supercooled liquid(d) transparent organic polymer. Ans.
Q3. Most crystals show good cleavage because their atoms, ions or molecules are(a) weakly bonded together(b) strongly bonded together(c) spherically symmetrical(d) arranged in planes. Ans.
Q4. The ability of a substance to assume two or more crystalline structures is called(a) isomerism (b) polymorphism(c)  isomorphism (d) amorphism.Ans.
Q5. Cation and anion combines in a crystal to form following type of compound(a) ionic (b) metallic(c) covalent (d) dipole-dipole.Ans.
Q6. For two ionic solids CaO and KI, identify the wrong statement among the following.(a) CaO has high melting point.(b) Lattice energy of CaO is much larger than that of KI.(c) KI has high melting point.(d) KI is soluble in benzene.Ans.
Q7. For orthorhombic system axial ratios are a ≠ b ≠ c and the axial angles are(a) α = β = γ ≠ 90º (b)  α = β = γ = 90º(c) α = γ = 90º, β ≠ 90º (d)   α ≠ β ≠ γ ≠ 90ºAns. 
Q8 The number of carbon atoms per unit cell of diamond unit cell isa. 6b. 1c. 4d. 8Ans. 
Q9. In a face-centred cubic lattice, a unit cell is shared equally by how many unit cells?(a) 2 (b) 4(c) 6 (d) 8 Ans.
Q10. When Zn converts from melted state to its solid state, it has hcp structure, then find the number of nearest atoms.(a) 6 (b) 8(c) 12 (d) 4 Ans.
Q11. The fcc crystal contains how many atoms in each unit cell?(a) 6 (b) 8(c) 4 (d) 5 Ans. 
Q12. The number of atoms contained in a fcc unit cell of a monatomic substance is(a) 1 (b) 2(c) 4 (d) 6Ans.
Q13. A compound is formed by cation C and anion A. The anions form hexagonal close packed (hcp) lattice and the cations occupy 75% of octahedral voids. The formula of the compound is(a) C4A3 (b) C2A3(c) C3A2 (d) C3A4Ans.
Q14. In calcium fluoride, having the fluorite structure, the coordination numbers for calcium ion (Ca2+) and fluoride ion (F–) are(a)  4 and 2 (b) 6 and 6(c)  8 and 4 (d) 4 and 8 Ans.
Q15. The ionic radii of A+ and B– ions are 0.98 × 10–10 m and 1.81 × 10–10 m. The coordination number of each ion in AB is(a) 8 (b) 2(c) 6 (d) 4Ans. 
Q16. The number of octahedral void(s) per atom present in a cubic close-packed structure is(a) 1 (b) 3(c) 2(d) 4Ans.
Q17. Structure of a mixed oxide is cubic close packed (ccp). The cubic unit cell of mixed oxide is composed of oxide ions. One fourth of the tetrahedral voids are occupied by divalent metal A and the octahedral voids are occupied by a monovalent metal B. The formula of the oxide isa. ABO2b. A2BO2c. A2B3O4d. AB2O2Ans. Explanation: 
Q18. A solid compound XY has NaCl structure. If the radius of the cation is 100 pm, the radius of the anion (Y–) will be (a) 275.1 pm (b) 322.5 pm(c) 241.5 pm (d) 165.7 pmAns.
19. A compound formed by elements X and Y crystallises in a cubic structure in which the X atoms are at the corners of a cube and the Y atoms are at the face- centres. The formula of the compound is(a) XY3 (b) X3Y(c) XY (d) XY2Ans.
20. In cube of any crystal A-atom placed at every corners and B-atom placed at every centre of face. The formula of compound is(a) AB (b) AB3(c) A2B2 (d) A2B3Ans.
Q21. In crystals of which one of the following ionic compounds would you expect maximum distance between centres of cations and anions?(a)CsI(b) CsF(c)LiF(d) LiIAns.
Q22. The second order Bragg diffraction of X-rays with λ = 1.00 Å from a set of parallel planes in a metal occurs at an angle 60°. The distance between the scattering planes in the crystal is(a) 2.00 Å (b) 1.00 Å(c) 0.575 Å(d) 1.15 ÅAns.
Q23. The intermetallic compound LiAg crystallizes in cubic lattice in which both lithium and silver have coordination number of eight. The crystal class is(a) face-centred cube (b) simple cube(c) body-centred cube (d) none of these. Ans.
Q24. In the fluorite structure, the coordination number of Ca2+ ionis(a) 4 (b) 6(c) 8 (d) 3Ans.
Q25. An element has a body centered cubic (bcc) structure with a cell edge of 288 pm. The atomic radius isa. √3/4 x 288pmb. √2/4 x 288pmc. 4/√3 x 288pmd. 4/√2 x 288pm Ans. 
Q26. The vacant space in bcc lattice unit cell is (a) 48% (b) 23%(c) 32% (d) 26% Ans.
Q27. If a is the length of the side of a cube, the distance between the body-centred atom and one corner atom in the cube will bea. 2/ √3 a b. 4/√3 ac. √3/4 ad. √3/2 aAns. 
28. A metal crystallises with a face-centred cubic lattice. The edge of the unit cell is 408 pm. The diameter of the metal atom is(a) 288 pm (b) 408 pm(c) 144 pm (d) 204 pm Ans.
29. AB crystallizes in a body-centred  cubic  lattice with edge length ‘a’ equal to 387 pm. The distance between two oppositely charged ions in the lattice is(a) 335 pm (b) 250 pm(c) 200 pm (d) 300 pm Ans.
Q30. Lithium metal crystallises in a body-centred cubic crystal. If the length of the side of the unit cell of lithium is 351 pm, the atomic radius of lithium will be(a) 151.8 pm (b) 75.5 pm(c) 300.5 pm (d) 240.8 pm Ans.
Q31. Copper crystallises in a face-centred cubic lattice with a unit cell length of 361 pm. What is the radius of copper atom in pm?(a) 157 (b) 181(c) 108 (d) 128Ans.d.Explanation: Since Cu crystallises in a fcc so, r = a/2√2a = edge length = 361 pm r = 361/ 2√2 = 128 pm 
Q32. Which of the following statements is not correct?(a) The number of carbon atoms in a unit cell of diamond is 8.(b) The number of Bravais lattices in which a crystal can be categorized is 14.(c) The fraction of the total volume occupied by the atoms in a primitive cell is 0.48.(d) Molecular solids are generally volatileAns. c. The fraction of the total volume occupied by the atoms in a primitive cell is 0.48.
Q33 If a stands for the edge length of the cubic systems: simple cubic, body-centred cubic and face-centred cubic, then the ratio of radii of the spheres in these systems will be respectivelya. 1a/2: √3a/2 : √2a/2b. 1a: √3a:√2ac. 1a/2: √3a/4: 1a/2√2d. 1a/2: √3a:1a/√2Ans.  c. c. 1a/2: √3a/4: 1a/2√2Explanation: For simple cubic: r = a/2For body centred: r = a√3/4For face centred : r = a/2√2a = edge length and r = radiusRatio of radii of the three will be a/2: a √3/4: a/2√2
Q34 The fraction of total volume occupied by the atoms present in a simple cube isa. P/ 3√2b. P/4√2c. P/4d. P/6Ans. d. P/6
Q35 The pyknometric density of sodium chloride crystal is 2.165 ×103 kg m–3 while its X-ray density is 2.178 × 103 kg m–3. The fraction of unoccupied sites in sodium chloride crystal is(a) 5.96(b) 5.96 × 10–2(c)  5.96 × 10–1 (d)  5.96 × 10–3Ans. d. 5.96x10-3 Explanation: Molar volume from pyknometric density = M/ 2.165 x 103 Molar volume from X-ray density = M / 2.178 x 103 m3  Volume occupied = M/ 103 (1/2.165 - 1/2.178) m3Fraction unoccupied = (0.013 M x 10-3 / 2.165x2.178)/(Mx10-3/2.165) = 5.96 x 10-3
Q36. The edge length of face-centred unit cubic cells is 508 pm. If the radius of the cation is 110 pm, the radius of the anion is(a) 144 pm (b) 398 pm(c) 288 pm (d) 618 pmAns. a. 144 pm Explanation: In the face–centred cubic lattice, the edge length of the unit cell, a = r + 2R + rwhere r = Radius of cation, R = Radius of anion 508 = 2 × 110 + 2R  R = 144 pm
Q37 Iron exhibits bcc structure at room temperature. Above 900°C, it transforms to fcc structure. The ratio of density of iron at room temperature to that at 900°C (assuming molar mass and atomic radii of iron remains constant with temperature) isa. √3/√2b. 4√3/ 3√2c. 3√3/4√2d. 1/2Ans. c. 3√3/4√2Explanation: for bcc, Z = 2 and a = 4r/√3for fcc Z = 4 and a = 2√2 r dRT/d900oC = ZM/ a3xNA /ZM/ a3xNAmolar mass and atomic radii are constant=  3√3/4√2
Q38 Lithium  has  a  bcc   structure.   Its   density   is 530 kg m–3 and its atomic mass is 6.94 g mol–1. Calculate the edge length of a unit cell of lithium metal. (NA=6.022 x 1023 mol-1 )(a) 527 pm (b)  264 pm(c) 154 pm (d)  352 pmAns. d. 352 pm Explanation: as bcc Z = 2density = 530 kg m-3at mass of Li = 6.94 g / mol NA = 6.022 x 1023 ρ = ZM/ a3xNA   = 2 x 6.94/a3 x 6.022 x 1023 a = 352 pm 
Q39 A metal has a fcc lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g cm–3. The molar mass of the metal is (NA Avogadro’s constant = 6.02 × 1023  mol–1)(a) 27 g mol–1 (b) 20 g mol–1(c)  40 g mol–1 (d) 30 g mol–1Ans. a. 27 g mol-1
Q40. CsBr crystallises in a body-centred cubic lattice. The unit cell length is 436.6 pm. Given that the atomic mass of Cs = 133 and that of Br = 80 amu and Avogadro number being 6.02 × 1023  mol–1, the density of CsBr is(a) 4.25 g/cm3 (b) 42.5 g/cm3(c) 0.425 g/cm3 (d) 8.25 g/cm3Ans. a. 4.25 g/cm3
Q41. An element (atomic mass = 100 g/mol) having bcc structure has unit cell edge 400 pm. The density of element is(a) 7.289 g/cm3(b) 2.144 g/cm3(c) 10.376 g/cm3(d) 5.188 g/cm3Ans. d. 5.188 g/cm3Explanation: cell edge = 400 pmAs it is bcc, so Z = 2at. mass = 100 g/ mlmass of each element = 100 / 6.022 x 1023 = 16.6 x 10 -23 gdensity = mass of unit cell/ volume of unit celldensity = 33.2 x 10 -23/ 64 x 10 -24 = 5.188 g/cm3
Q42.Formula of nickel oxide with metal deficiency defect in its crystal is Ni0.98O. The crystal contains Ni2+ and Ni3+ ions. The fraction of nickel existing as Ni2+ ions in the crystal is(a) 0.96 (b) 0.04(c) 0.50 (d) 0.3Ans. a. 0.96Explanation: Let the fraction of metal which exists as Ni2+ ion be x. Then the fraction of metal as Ni3+ = 0.98 – x2x + 3(0.98 – x) = 22x + 2.94 – 3x = 2 x = 0.94
Q43. The correct statement regarding defects in crystalline solids is(a) Frenkel    defects   decrease   the density of crystalline solids(b) Frenkel defect is a dislocation defect(c) Frenkel defect is found in halides of alkaline metals(d) Schottky defects have no effect on the density of crystalline solids.Ans. b. Frenkel defect is a dislocation defectExplanation: (b) : Frenkel defect is a dislocation defect as smaller ions (usually cations) are dislocated from normal sites to interstitial sites. Frenkel defect is shown by compounds having large difference in the size of cations and anions hence, alkali metal halides do not show Frenkel defect. Also, Schottky defect decreases the density of crystal while Frenkel defect has no effect on the density of crystal.
Q44. The appearance of colour in solid alkali metal halides is generally due to(a) interstitial positions(b) F-centres(c) Schottky defect(d) Frenkel defect.Ans. b. F-centresExplanation: F-centres are the sites where anions are missing and instead electrons are present. They are responsible for colours.
Q45. Schottky defect in crystals is observed when(a) density of the crystal is increased(b) unequal number of cations and anions are missing from the lattice(c) an ion leaves its normal site and occupies an interstitial site(d) equal number of cations and anions are missing from the lattice.Ans. d) equal no. of cations and anions are missing from the lattice.Explanation: In Schottky defect, equal no. of cations and anions are missing from the lattice. So, the crystal remains neutral. Such defect is more common in highly ionic compounds of similar cationic and anionic size, i.e. NaCl.
Q46. Ionic solids, with Schottky defects, contain in their structure(a) cation vacancies only(b) cation vacancies and interstitial cations(c) equal number of cation and anion vacancies(d) anion vacancies and interstitial anions.Ans. (c)equal number of cation and anion vacancies.Explanation: When an atom is missing from its normal lattice site, a lattice vacancy is created. Such a defect, which involves equal number of cation and anion vacancies in the crystal lattice is called a Schottky defect.
Q47. Which is the incorrect statement?(a) Density decreases in case of crystals with Schottky defect.(b) NaCl(s) is insulator, silicon is semiconductor, silver is conductor, quartz is piezoelectric crystal.(c) Frenkel defect is favoured in those ionic compounds in which sizes of cation and anions are almost equal.(d) FeO0.98 has non-stoichiometric metal deficiency defect.Ans. (c, d) Explanation: Frenkel defect is favoured in those ionic compounds in which there is large difference in the size of cations and anions.Non-stoichiometric defects due to metal deficiency is shown by FexO where x = 0.93 to 0.96.
Q48. With which one  of  the  following  elements silicon should be doped so as to give p-type of semiconductor?(a)  Selenium(b) Boron(c)  Germanium(d) ArsenicAns. (b) BoronExplanation: If silicon is doped with any of the elements of group 13 (B, Al, Ga, In, Tl) of the periodic table p-type semiconductor will be obtained. 
Q49. If  NaCl  is  doped  with  10-4   mol  %  of  SrCl2,  the concentration of cation vacancies will be(NA = 6.02 × 1023  mol–1)(a)  6.02 × 1016  mol–1 (b)  6.02 × 1017  mol–1(c)  6.02 × 1014 mol–1 (d)  6.02 × 1015 mol–1Ans. (b)  6.02 × 1017  mol–1Explanation: As each Sr2+ ion introduces one cation vacancy, therefore, concentration of cation vacancies = mole % of SrCl2 added.Therefore, Concentration of cation vacancies = 10–4 mole%1015x 6.023 x 1023/ 100 = 6.023 x 1017 
Q50. If we mix a pentavalent impurity in a crystal lattice of germanium, what type of semiconductor formation will occur?(a) n-type semiconductor(b) p-type semiconductor(c) Both (a) and (b)(d) None of these Ans. (a) n-type semiconductorExplanation: When an impurity atom with 5 valence electrons (as arsenic) is introduced in a germanium crystal, it replaces one of the germanium atoms. Four of the five valence electrons of the impurity atom form covalent bonds with each valence electron of four germanium atoms and fifth valence electron becomes free to move in the crystal structure. This free electron acts as a charge carrier. Such as an impure germanium crystal is called n-type semiconductor because in it charge carriers are negative (free electrons).
Q51. On doping Ge metal with a little of In or Ga, one gets(a) p-type semiconductor(b) n-type semiconductor(c) insulator(d) rectifier.Ans. a. p-type semiconductorExplanation: (a)due to metal deficiency defects (b) by adding impurity containing less  electrons  (i.e.  atoms  of  group   13). Ge belongs to Group 14 and In or Ga to Group 13. Hence on doping p-type semiconductor is obtained. This doping of Ge with In increase the electrical conductivity of the Ge crystal.

Solutions******
Q.No.1 The importance of many pure substance in life depends on their composition.  Which of the following statement justify the above fact?(a) 1ppm of fluoride ions in water prevents tooth decay(b) 1.5 ppm of fluoride ions causes tooth decay(c) Concentration above 1.5ppm can be poisonous(d) AllAns. 
Q.NO.2 Which of the following fluoride is used as rat poison?(a) CaF2(b) KF(c) NaF(d) MgF2Ans.
Q.No.3 Most of the processes in our body occur in (a) solid solution(b) liquid solution (c) gaseous solution(d) colloidal solutionAns.
Q.No.4 Homogenous mixtures signifies(a) its composition is uniform throughout the mixture(b) its properties are uniform throughout the mixture(c) both composition and properties are uniform throughout the mixture(d) neither composition nor properties are uniform throughout the mixtureAns. Q.No.5 Which of the following mixture is called solution.(i) water + ammonia (ii) water + acetone (iii) acetone + alcohol (iv) hexane + water(a) i, ii, iii(b) i, iii, iv(c) i, iv(d) ii, iiiAns. Q.No.6 Which of the following is a quantitative description of the solution(a) Dilute(b) Concentrated(c) Saturated(d) MolarAns.Q.No.7 When a solute is present in trace quantities the following expression is used(a) gram per million (b) milligram percent (c) microgram percent (d) Parts per millionAns.Q.No.8 Molality of liquid HCl will be, if density of solution is 1.17gm/cc(a) 36.5(b) 32.05(c) 18.25(d) 42.10Ans.Q.No.9 1M, 2.5 L of NaOH solution is mixed with another 0.5M, 3 L NaOH solution. Then find out the molarity of resultant solution.(a) 0.80M(b) 1.0M(c) 0.73 M(d) 0.50MAns.Q.No.10 An X molal solution of a compound in benzene has mole fraction of solute equal to 0.2. The value of X is (a) 14(b) 3.2(c) 1.4(d) 2Q11. Benzoic acid undergoes dimerisation in benzene solution. The van't Hoff factor (i) is related to the degree of association x of the acid asa. i = (1-x)b. i = (1+x)c. i = (1-x/2)d. i = (1+x/2)Ans.  c. i=(1-x/2)Q12. Normality of 0.3M phosphorous acid H3PO3 is a. 0.5b. 0.6c. 0.9d. 0.1Ans. b. 0.6Q13. A liquid is in equilibrium with its vapour at its boiling point on the average, the molecules in the two phases have equala. intermolecular forcesb. potential energyc. total energyd. kinetic energyAns. c. total energyQ14 In which mode of expression, the concentration of solution remains independent of temperature?a. Molarityb. Normalityc. Formalityd. MolalityAns. d. MolalityExplanation: - Molality is the moles of solute dissolved in mass of solvent in Kg.Q15. The molal depression constant depends upon a. nature of the soluteb. nature of the solventc. heat of solution of the solute in the solventd. vapour pressure of the solutionAns. b. nature of the solventQ16. Increasing the temp of an aqueous solution will causea. decrease in molalityb. decrease in molarityc. decrease in mole fractiond. decrease in % w/w .Ans. b. decrease in molarity.Q17. which one is unitlessa. molarityb. molalityc. mole fractiond. all do not have unitsAns. c. mole fractionQ18 Which of the following colligative properties can provide molar mass of proteins (or polymers or colloids) with greatest precision.a. Depression in Freezing pointb. Osmotic pressurec. relative lowering of vapour pressured. elevation in boiling pointAns. b. Osmotic pressureQ19. The vapour pressure of a solvent A is 0.80 atm. When a non-volatile substance B is added to this solvent, its vapour pressure drops to 0.60 atm. The mole fraction of B in this solution is a. 0.25b. 0.50c. 0.75d. 0.90Ans. 0.25Q20 The aqueous solution that has the lower vapour pressure at a given temp is a. 0.1 molal sodium phosphateb. 0.1 molal barium chloridec. 0.1 molal sodium chlorided. 0.1 molal glucosee. 0.1 molal acetic acidAns. a. 0.1 molal sodium phosphateQ21. Molarity of the liquid HCl if density of the solution is 1.17 g/ cca. 36.5b. 18.25c. 32.05d. 42.10Ans. c. 32.05 M.Q22. Which among the following will have the highest boiling point at 1atm pressure?a. 0.1 M NaClb. 0.1 M Sucrosec. 0.1 M BaCl2d. 0.1 M GlucoseAns. c. 0.1 M BaCl2
Q1. Which of the following is dependent on temperature?a. Molarityb. Mole fractionc. weight percentaged. molality Ans. a. molarityExplanation: molarity is a function of temp as volume depends on temp. Q2. what is the mole fraction of the solute in a 1.00m aqueous solution?a. 1.770 b. 0.0354c. 0.0177d. 0.177Ans. c. 0.0177Explanation: 1.00 m means 1 mole dissolved in 1 kg of water.moles of 1kg water = 55.55 molesXsolute = moles of solute/ moles of solute + moles of water X solute = 1/1+55.55 = 0.0177Q3. How many grams of conc. HNO3 solution should be used to prepare 250mL of 2.0M HNO3? The concentrated acid is 70% HNO3.a. 70.0 g conc. HNO3b. 54.0 g conc. HNO3c. 45.0 g conc. HNO3d. 90.0 g conc. HNO3Ans. c. 45 g conc. HNO3Explanation: 2M = 2 moles / 1 L of solution2 M = 0.5 moles / 250 mL of solutionmolar mass of HNO3 = 63weight in 0.5 moles of HNO3 = 0.5 x 63 but as question saya 70% of HNO3 then mass of acid x 70/100 = 0.5 x 63mass of acid = 45 gQ4. Which of the following compounds can be used as antifreeze in automobile radiators?a. Methyl alcoholb. Glycolc. Nitrophenold. EthylalcoholAns. b. glycolExplanation: a 35% v/v solution of ethylene glycol is used as an antifreeze in cars for cooling the engine. At this conc. the antifreeze lowers the freezing point of water to 255.4 K (-17.6 *C)Q5. Conc. aq. H2SO4 is 98% H2SO4 by mass and has a density of 1.80 g /mL. Volume of acid required to make one litre of 0.1M H2SO4 solution isa. 16.65 mLb. 22.20 mL c. 5.55 mLd. 11.10 mLAns. c. 5.55mL Explanation: 98% means in 100 g of solution only 98 g of H2SO4 is present.moles of H2SO4 in 98 g = 1 mol = given mass / molar mass = 98/ 98 = 1 moledensity = mass / volumevolume = mass / densityvolume of solution = 100g/ 1.80g = 55.55 mL = 0.0555 L Molarity of solution = 1 mol/0.0555 mL = 18.02 MLet  V mL of this H2SO4 is used to prepare 1 L of 0.1 M H2SO4M1V1 = M2V2Vx18.02 = 1000x0.1V = 5.55 mL Q6. The mole fraction of the solute in one molal aq solution isa. 0.009b. 0.018c. 0.027d. 0.036Ans. b. 0.018Explanation: 1 molal means 1 mole in 1 kg of water moles of1 kg of water = 55.55 molesX solute = 1/1+55.55 = 0.018 Q7 2.5 L of 1M NaOH solution is mixed with another 3L of 0.5M NaOH solution. Then find out molarity of resultant solution. a. 0.80Mb. 1.0M c. 0.73M d. 0.50M Ans. c. 0.73 M Explanation: Molar mass of NaOH = 40 g / mol2.5 L of 1 M NaOH contain = 40 g /mol x 1 mol / L x 2.5 L = 40 x 2.5 g of NaOH = 100 g3 L of 0.5 M NaOH solution contain = 40 g / mol x 0.5 mol / L x 3 L = 60 g if two solution of 3 L and 2.5 L are mixed then total volume = 5.5 L and this 5.5 L of NaOH comprises 160 g of NaOH then the resultant molarity will be Molarity = moles/ volume of solution in L = given mass/ molar mass/ volumeMolarity = 160/40/5.5 = 0.73MQ8. How many g of dibasic acid (mol. weight 200) should be present in 100mL of the aq solution to give strength of 0.1 N?a. 10g b. 2gc. 1g d. 20gAns. c. 1gExplanation: dibasic means acidity is 2. Equivalent weight = molecular weight/2 = 200/2 = 100W/E = VXN/1000W= 100x0.1x100/1000 = 1gQ9. What is the molarity of H2SO4 solution, that has a density 1.84 g/ cc of 35*C and contains 98% by weight?a. 18.4 M b. 18 M c. 4.18 M d. 8.14 M Ans. a. 18.4MExplanation; 98% means 98 g in 100 g.Moles of 98g of H2SO4 = 1 molevolume in 1.84 g /cc = 100/1.84 = 54.35 mL = 0.05435L molarity = 1 mole/ 0.054 L = 18.4 M Q10. The concentration unit, independent of temperature, would be a. normality b. weight volume percentc. molality d. molarity Ans. c. molalityQ11. How many grams of CH3OH should be added to water to prepare 150mL solution of 2M CH3OH?a. 9.6x103b. 2.4 x 103c. 9.6d. 2.4Ans. a. 9.6Explanation: Molarity = mol/ Volume in LMolarity = given mass/ M Mass/ Volume in L given mass = Molarity x molar mass x volume in L given mass = 2M x 32 x 0.15 = 9.6gQ12 In water saturated air, the mole fraction of water vapour is 0.02. If the total pressure of the saturated air is 1.2 atm, the partial pressure of dry air is a. 1.18 atmb. 1.76 atmc. 1.176 atmd. 0.98 atmAns. c. 1.176 atmExplanation: p water vapour = X water vapour x P total p total = p water vapour + p dry air1.2 = 0.024 + p dry airp dry air = 1.2 - 0.024 = 1.176 atmPartial vapour pressure is directly proportional to mole fraction;  p ∝ xQ13. pA and pS are the vapour pressure of pure liquid components of A and B respectively of an ideal binary solution. If xA represents the mole fraction of component A, the total pressure of the solution will be a. pA + xA (pB-pA)b. pA + xA (pA-pB)c. pB + xA (pB-pA)d. pB + xA (pA-pB)Ans. d. pB+xA (pA-pB)Explanation: according to Raoult's law P = xApA+xBpBfir binary solutionxA+xB=1 or we can say xB = 1-xAP=xApA + (1-xA)pB= xApA + pB - xApBP=pB+xA(pA-pB)Q14. vapour pressure of chloroform (CHCl3) and dichloromethane (CH2Cl2) at 25*C are 200 mm Hg and 41.5 mm Hg respectively.  Vapour pressure of the solution obtained by mixing 25.5 g of CHCl3 and 40g of CH2Cl2 at the same temp will be a. 173.9 mm Hgb. 615.0 mm Hgc. 347.9 mm Hgd. 285.5 mm HgAns. p*CHCl3 = 200 mm Hg and p*CH2Cl2 = 41.5 mm HgMoles of CHCl3 = given mass / molar mass = 25.5/ 119.5 = 0.213 molesMoles of CH2Cl2 = given mass / molar mass = 40/85 = 0.470 molesxCHCl3 = 0.213/0.213+0.470 = 0.31xCH2Cl2= 0.470/0.213+0.470= 0.69Ptotal=xCHCl3p*CHCl3+xCH2Cl2p*CH2Cl2Ptotal=200x0.31+41.5x0.69=62+28.63=90.63mm HgQ15. A solution has a 1:4 mole ratio of pentane to hexane.  The vapour pressure of the pure hydrocarbons at 20*C are 440 mm Hg for pentane and 120 mm Hg for hexane.  The mole fraction of pentane in the vapour phase would bea. 0.200b. 0.549c. 0.786d. 0.478Ans. d. 0.478explanation: - mole of C5H12: mole of C6H14 = 1:4xC5H12=1/5xC6H14=4/5p*C5H12=440 mm Hg  and p*C6H14= 120 mm HgPtotal=xC5H12p*C5H12 + xC6H14p*C6H14Ptotal=440x1/5 + 120x4/5=88+96= 184mm of HgAccording to Raoult's LawpC5H12=xC5H12p*C5H12 = 440 x 1/5 = 88 mm of HgxC5H12 = mole fraction of pentane in solutionBy dalton's law pC5H12 = x'C5H12Ptotal88 =x'C5H12 184 = 0.478Q16. the vapour pressure of two liquids P and Q are 80 and 60 torr, respectively.  The total vapour pressure of solution obtained by mixing 3 mole of P and 2 mol of Q would bea. 72 torrb. 140 torrc. 68 torrd. 20 torrAns. a. 72torrExplanation: Pt=xPpP+xQpQwhere pP=80 torr and pQ=60torrxP=3/5 and xQ=2/5P=80x3/5 +60x2/5 = 48+24 = 72 torrQ17. The mixture which shows +ve deviation from Raoult's law is a. ethanol + acetoneb. benzene + toluene c. acetone + chloroformd. chloroethane + bromoethaneAns. a.ethanol+acetoneExplanation: mixture of ethanol and acetone shows positive deviation from Raoult's lawIn pure ethanol, molecules are hydrogen bonded. On adding acetone, its molecules get in between the host molecules and break some of the hydrogen bonds between them.  Due to weakening of interactions, the solution shows positive deviation from Raoult's law. Q18 for an ideal solution the correct option is a. ΔmixG = O at constant T and P b. ΔmixS = 0 at constant T and P c. ΔmixV ≠  at constant T and P d. ΔmixH = 0 at constant T and P Ans. d.  ΔmixH = 0 at constant T and Pexplanation: for an ideal solution  ΔmixH = 0 at constant T and PQ19 The mixture that forms maximum boiling azeotrope is a. heptane + octaneb. water + nitric acidc. ethanol + waterd. acetone + carbon disulphideAns. b. water + nitric acidexplanation: maximum boiling azeotropes are formed by those solutions which show negative deviations from Raoult's. H2O and HNO3, mixture shows negative deviations. Q20 which of the following statements is correct regarding a solution of two components A and B exhibiting +ve deviation from ideal behaviour?a. Intermolecular attractive forces between A-A and B-B are stronger than those between A-Bb. ΔmixH=0 at constant T and Pc. ΔmixV= 0 at constant T and P d. Intermolecular attractive forces between A-A and B-B are equal to those between A-BAns. a.Intermolecular attractive forces between A-A and B-B are stronger than those between A-BExplanation: In case of +ve deviations from Raoult's Law, A-B interactions are weaker than those between A-A or B-B i.e. in this case the intermolecular attractive forces between the solute-solvent molecules are weaker than those between solute-solute and solvent-solvent molecules.  This means that in such solution, molecules of A (or B) will find it easier to escape than in pure state.  This will increase the vapour pressure and result in positive deviation.  Q21. Which one of the following is incorrect for ideal solution?a. Δmix H = 0b. ΔmixU = 0c. ΔP = Pobs - P calculated by Raoult's law = 0d. Δmix G = 0Ans. d. Δmix G = 0Explanation: for an ideal solutionΔmixH=0, ΔmixV=0, now ΔmixU=ΔmixH - PΔmixVΔmixU = 0also for an ideal solutionpA=xp*ApB=xp*BΔP=Pobs-Pcalculated by Raoult's law = 0Δmix G = ΔmixH-TΔmixSfor an ideal solution ΔmixS≠0ΔmixG≠0Q22. Which of the following statements about the composition of the vapour over an ideal 1:1 molar mixture of benzene and toluene is correct? Assume that the temperature is constant at 25^C (given vapour pressure data at 25*C, benzene = 12.8kPa toluene=3.85kPa)a. the vapour will contain equal amounts of benzene and toluene b. not enough information is given to make a predictionc. the vapour will contain a higher percentage of benzened. the vapour will contain a higher percentage of tolueneAns. c. the vapour will contain a higher percentage of benzeneExplanation: - pBenzene = xBenzenep*BenzenepToluene = xToluenep*Tolouenefor an ideal 1:1 molar mixture of benzene and toluene, xBenzene=1/2 and xToluene=1/2pBenzene = 1/2 p*Benzene = 1/2 x 12.8kPa = 6.4 kPapToluene = 1/2 p*Toluene = 1/2 x 3.85 kPa = 1,925 kPaThus, the vapour will contain a high percentage of benzene as the partial vapour pressure of benzene is higher as compared to that of tolueneQ23 Which condition is not satisfied by an ideal solution?a. ΔmixV=0b. ΔmixS = 0c. Obeyance to Raoult's lawd. ΔmixH = 0Ans. b. ΔmixS = 0Explanation: - for an ideal solutionvolume change ΔV on mixing should be zeroheat change ΔH on mixing should be zeroObeys Raoult's law at every range of concentrationEntropy Change ΔS on mixing ≠ 0Q24. A solution of acetone in ethanola. obeys raoult's lawb. shows -ve deviation from Raoult's lawc. shows a +ve deviation from Raoult's lawd. behaves like a near ideal solutionAns. c. shows a +ve deviation from Raoult's lawExplanation: both the components escape easily showing higher vapour pressure than the expected value. This is due to breaking of some hydrogen bonds between ethanol molecules.Q25 All form ideal solution except a. C6H6 and C6H5CH3b. C2H6 and C2H5Ic. C6H5Cl and C6H5Brd. C2H5I and C2H5OhAns. c. C6H5Cl and C6H5BrExplanation: Raoult's law is valid for ideal solutions only. The element of non-ideality enters into the picture when the molecules of the solute and solvent affect each others intermolecular forces.  A solution containing components of A and B behaves as ideal solution when A-B attraction force remains same as A-A and B-B attraction forces.Q26. All form ideal solution excepta. C6H6 and C6H9CH3b. C2H6 and C2H5Ic. C6H5Cl and C6H5Brd. C2H5I and C2H5OHAns. d. C2H5I and C2H5OHExplanation: Because C2H5I and C2H5OH are dissimilar liquids.Q27 An ideal solution is formed when its componentsa. have no volume change on mixing b. have no enthalpy change on mixingc. have both the above characteristics d. have high solubilityAns. c. have both the above characteristicsExplanation: for ideal solution  ΔmixV=0 and  ΔmixH=0Q28. The freezing point depression constant Kf of benzene is 5.12 K kg / mol.  The freezing point depression for the solution of molality 0.078 m containing a non-electrolyte solute in benzene is (rounded off up to two decimal places)a. 0.20 K b. 0.80 K c. 0.40 K d. 0.60 K Ans. c. 0.40Kexplanation: Given Kf=5.12K kg / mol m= 0.078 molal ΔTf=Kfxm=5.12x0.078=0.39936=0.40KQ29. If molality of the dilute solution is obtained, the value of molal depression constant Kf will be a. halvedb. tripledc. unchangedd. doubledAns. c.unchangedExplanation: The value of molal depression constant, Kf is constant for a particular solvent, thus, it will be unchanged when molality of the dilute solution is doubled. Q30. At 100*C the vapour pressure of a solution of 6.5 g of a solute in 100 g water is 732 mm.  If Kb = 0.52, the boiling point of this solution will be a. 102 *Cb. 103 *Cc. 101 *Cd. 100 *CAns. c.101*CExplanation: - wB = 6.5 g and wA=100 g ps=732mm and Kb= 0.52, T*b=100*C and p*=760mmp*-ps/p* = n2/n1 760-732/760 = n2/100/18n2= 28x100 / 760x18  = 0.2046 mol ΔTb=KbxmTb-T*b=Kbxn2x1000/wATb-100*C = 0.52x0.2046x1000/100 = 1.06Tb=100+1-06=101.06*CQ31. 200 mL of an aq solution of a protein contains its 1.26 g. The osmotic pressure of this solution at 300K is found to be 2.57x10-3 bar. The molar mass of protein will be (R=0.083 L bar / mol K)a. 51022 g /molb. 122044 g / molc. 31011 g/ mold. 61038 g/ molAns. d. 61038 g/ molwe know that pV=nRT where n=w/M𝜋V=wRT/MM=wRT/𝜋V = 1.26x0.083x300/ 2.57x10-3x200/1000M=1.26x0.083x300/ 2.57x10-3x0.2 = 61038 g / molQ32. A solution of sucrose (molar mass = 342 g / mol) has been prepared by dissolving by 68.5 g of sucrose in 1000g of water.  The freezing point of the solution obtained will be (Kf for water = 1.86 K kg / mol)a. -0.372*Cb. -0.520*Cc. +0.372 *C d. -0.570 *CAns. a.-0.372*CExplanation: we know that ΔTf = Kf x mm = w 1000/ MB wA = 68.5x1000/ 342x1000 = 68.5 /342ΔTf = 1.86 x 68.5/342 = 0.372*CTf=0-0.372*C= -0.372*CQ33. During osmosis, flow of water through a semipermeable membrane is a. from solution having lower concentration onlyb. from solution having higher concentration onlyc. from both sides of semipermeable membrane with equal flow ratesd. from both sides of semipermeable membrane with unequal flow ratesAns. a. from solution having lower concentration onlyQ34 1.00 g of non-electrolyte solute (molar mass 250 g/ mol) was dissolved in 51.2 g of benzene.  If the freezing point constant, Kf of benzene is 5.12 K kg /mol, the freezing point of benzene will be lowered by a. 0.2 Kb. 0.4 Kc. 0.3 K d. 0.5 K Ans. b. 0.4 KExplanation: MB = 1000x Kf x wB/ wA x ΔTf or 250 = 1000 x 5.12 x 1 / 51.2 x ΔTf ΔTf = 1000x 5.12 x 1/51.2 x 250 = 0.4KQ35 A solution containing 10g per dm3 of urea (molecular mass = 60 g / mol ) is isotonic with a 5% solution of a non-volatile solute.  The molecular mass of this non-volatile solute is a. 200 g/molb. 250 g/molc. 300 g/ mold. 350 g/ molAns. c. 300 g/ mol explanation: Molar concentration of urea = 10/60 dm-3Molar concentration of non-volatile solution = 50/MB L-1 = 50/MB dm-3for isotonic solution, 10/60 = 50/MBMB = 300 g mol–1Q36. A solution of urea (mol. mass 56 g/mol) boils at 100.18 *C at the atmospheric pressure.  If K4 and Kb for water are 1.86 and 0.512 K kg / mol respectively, the above solution will freeze at a. 0.654 °C b. -0.654 °Cc. 6.54 °C d. -6.54 °CAns. b.-0.654°C Explanation: ΔTf = Kf mΔTb = Kb mΔTb / ΔTb = Kf/Kb ΔTf  is depression in freezing pointΔTb is elevation in boiling pointKf = 1.86 K kg mol–1Kb = 0.512 K kg mol–1, ΔTb = 100.18 – 100 = 0.18therefore, 1.86/0.18 = 0.512Tf = 0.654 = T°f – Tf = 0 – Tf  Tf = – 0.654°C Freezing point of urea in water = –0.654°CQ37. A solution contains non-volatile solute of molecular mass M2. Which of the following can be used to calculate the molecular mass of solute in terms of osmotic pressure?a. M2=m2VRT/𝜋 b. M2=m2RT/V𝜋 c. M2=m2𝜋  RT/Vd. M2=m2𝜋 /VRTAns.b. M2=m2RT/V𝜋 Q38. Pure water can be obtained from sea water bya. centrifugationb. plasmolysisc. reverse osmosisd. sedimentationAns. c. reverse osmosisQ39. From the colligative properties the solution, which one is the best methods for the determination of molecular weight of proteins and polymers?a. osmotic pressureb. lowering in vapour pressurec. lowering in freezing pointd. elevation in boiling pointAns. a. osmotic pressureQ40. The vapour pressure of benzene at a certain temp is 640 mm of Hg. A non-volatile and non-electrolyte solid, weighing 2.175 g is added to 39.08 of benzene.  The vapour pressure of the solution is 600 mm of  Hg. What is the molecular weight of solid substance?a. 69.5b. 59.6c. 49.50d. 79.8Ans. a. 69.5Q41. If 0.15 g of a solute, dissolved in 15 g of solvent, is boiled at a temperature higher by 0.216°C, than that of the pure solvent. The molecular weight of the substance (Molal elevation constant for the solvent is 2.16°C) is(a) 10.1 (b) 100(c) 1.01 (d) 1000Ans.(b) 100Q42. A  5%   solution  of  cane  sugar   (mol.   wt.  =   342) is isotonic with 1% solution of a substance X. The molecular weight of X is(a) 68.4(b) 171.2(c) 34.2(d) 136.8Ans.(a) 68.4Q43. The vapour pressure of a solvent decreased by 10 mm of mercury when a non-volatile solute was added to the solvent. The mole fraction of the solute in the solution is 0.2. What should be the mole fraction of the solvent if the decrease in the vapour pressure is to be 20 mm of mercury?(a) 0.4 (b) 0.6(c) 0.8 (d) 0.2Ans.(b) 0.6Q44. The vapour pressure of CCl4 at 25°C is 143 mm Hg. If 0.5 g of a non-volatile solute (mol. weight = 65) is dissolved in 100 g CCl4, the vapour pressure of the solution will be(a) 199.34 mm Hg (b) 143.99 mm Hg(c) 141.43 mm Hg (d) 94.39 mm Hg.Ans.(c) 141.43 mm Hg Q45. The relationship between osmotic pressure at 273 K when 10 g glucose (p1), 10 g urea (p2), and 10 g sucrose (p3) are dissolved in 250 mL of water is(a)  p2 > p1 > p3 (b) p2 > p3 > p1(c)  p1 > p2 > p3 (d) p3  > p1 > p2 Ans.(a)  p2 > p1 > p3Q46. According to Raoult’s law, the relative lowering of vapour pressure for a solution is equal to(a) mole fraction of solute(b) mole fraction of solvent(c) moles of solute (d) moles of solvent.Ans.(a) mole fraction of soluteQ47. If 0.1 M solution of glucose and 0.1 M solution of urea are placed on two sides of the semipermeable membrane to equal heights, then it will be correct to say that(a) there will be no net movement across the membrane(b) glucose will flow towards glucose solution(c) urea will flow towards glucose solution(d) water will flow from urea solution to glucose.Ans. (a) there will be no net movement across the membraneQ48. Which one is a colligative property?(a)Boiling point (b) Vapour pressure(c) Osmotic pressure (d) Freezing point Ans.(c) Osmotic pressure Q49. Blood cells retain their normal shape in solution which are(a) hypotonic to blood (b) isotonic to blood(c) hypertonic to blood (d) equinormal to blood.Ans.(b) isotonic to bloodQ50. The relative lowering of the vapour pressure is equal to the ratio between the number of(a) solute molecules to the solvent molecules(b) solute molecules to the total molecules in the solution(c) solvent molecules to the total molecules in the solution(d) solvent molecules to the total number of ions of the solute.Ans.(b) solute molecules to the total molecules in the solutionQ51. 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) 3Ans.(d) 3Q52. The boiling point of 0.2 mol kg–1 solution of X in water is greater than equimolal solution of Y in water. Which one of the following statements is truein this case?(a) Molecular mass of X is less than the molecular mass of Y.(b) Y is undergoing dissociation in water while X undergoes no change.(c)  X is undergoing dissociation in water.(d) Molecular mass of X is greater than the molecular mass of Y.Ans.(c) X is undergoing dissociation in water.Q53. Of the following 0.10 m aqueous solutions, which one will exhibit the largest freezing point depression?(a) KCl (b) C6H12O6 (c) Al2(SO4)3 (d) K2SO4 Ans.(c) Al2(SO4)3Q54. The van’t Hoff factor i for a compound which undergoes dissociation in one solvent and association in other solvent is respectively(a) less than one and greater than one(b) less than one and less than one(c) greater than one and less than one(d) greater than one and greater than one. Ans. (c) greater than one and less than oneQ55. The freezing point depression constant for water is –1.86 °C m–1. If 5.00 g Na2SO4 is dissolved in 45.0 g H2O, the freezing point is changed by –3.82 °C. Calculate the van’t Hoff factor for Na2SO4.(a) 2.05 (b) 2.63(c) 3.11 (d) 0.381 Ans.(b) 2.63Q56. A  0.1  molal  aqueous   solution   of   a   weak  acid is 30% ionized. If Kf for water is 1.86°C/m, the freezing point of the solution will be(a)  – 0.18 °C (b) – 0.54 °C(c)  – 0.36 °C (d) – 0.24 °CAns. (d) – 0.24 °CQ57. An aqueous solution is 1.00 molal in KI. Which change will cause the vapour pressure of the solution to increase?(a)Addition of NaCl (b) Addition of Na2SO4(c)Addition of 1.00 molal KI(d)Addition of waterAns. (d)Addition of water Q58. A 0.0020 m aqueous solution of an ionic compound [Co(NH3)5(NO2)]Cl freezes at –0.00732 °C. Number of moles of ions which 1 mol of ionic compound produces on being dissolved in water will be (Kf = –1.86 °C/m)(a) 3 (b) 4 (c)  1 (d) 2 Ans.(d) 2Explanation: The number of moles of ions produced by 1 mol of ionic compound = i Applying  ΔTf = i x Kf x m0.00732 = i x 1.86 x 0.002i = 0.00732 / 1.86 x 0.002 = 1.96 = 2Q59. 0.5 molal aqueous solution of a weak acid (HX) is 20% ionised. If Kf for water is 1.86 K kg mol–1, the lowering in freezing point of the solution is(a) 0.56 K (b) 1.12 K(c) –0.56 K (d) –1.12 KAns. (b) 1.12 KExplanation: HX → H+ + X- α .........0.......01-α ......α ......α Total = 1 + α therefore, i = 1 + α = 1 + 0.2 = 1.2 ΔTf = i x Kf x m = 1.2 x 1.86 x 0.5 = 1.1116 K = 1.12KQ60. Which of the following 0.10 m aqueous solution will have the lowest freezing point?(a)KI (b)C12H22O11(c)Al2(SO4)3 (d)C5H10O5 Ans. (c)Al2(SO4)3Explanation: Since Al2(SO4)3 gives maximum number of ions on dissociation, therefore it will have the lowest freezing point.ΔTf = i x Kf x mQ61. Which of the following salts has the same value of van’t Hoff factor (i) as that of K3[Fe(CN)6]? (a) Na2SO4 (b) Al(NO3)3 (c) Al2(SO4)3 (d) NaCl Ans.(b) Al(NO3)3Explanation: K3[Fe(CN)6] → 3K+ + [Fe(CN)6]3- and Al(NO3)3 → Al3+ + 3NO3-Since both Al(NO3)3 and K3[Fe(CN)6] give the same number of ions, therefore they have the same van’t Hoff factor.Q62. At 25°C, the highest osmotic pressure is exhibited by 0.1 M solution of(a)glucose (b) urea(c) CaCl2(d) KCl.Ans. (c) CaCl2Explanation: In solution, CaCl2  gives three ions, KCl gives two ions while glucose and urea are covalent molecules so they do not undergo ionisation. Since osmotic pressure is a colligative property and it depends upon the number of solute particles (ions), therefore, 0.1 M solution of CaCl2  exhibits the highest osmotic pressure.Q63. Which of the following aqueous solution has minimum freezing point?(a) 0.01 m NaCl(b) 0.005 m C2H5OH (c) 0.005 m MgI2(d) 0.005 m MgSO4Ans. (a) 0.01 m NaClExplanation: Here ΔTf = i x Kf x mvan't Hoff factor, i = 2 for NaCl, so concentration = 0.02 which is maximum in the present case.Hence, ΔTf is maximum or freezing point is minimum in 0.01 in NaCl.

Biomolecules*******Q1 the protein responsible for blood clotting isa. Albuminsb. Globulinsc. Fibroind. FibrinogenAns. d. FibrinogenQ2 The disease diabetes mellitus is caused by deficiency of a. Iodineb. Insulinc. Phenylalanine hydroxylased. LysineAns. b. InsulinQ3 The carbohydrate which cannot be hydrolysed by the human digestive systema. starchb. glycogenc. cellulosed. AllAns. c. CelluloseQ4 Genetic material of the cella. Nucleic acidsb. proteinsc. Lipids d. CarbohydratesAns. a. Nucleic AcidsQ5 Double Helix structure DNA. Are the base pairs? a. part of the backbone structureb. inside the helixc. outside the helixd. noneAns. b. inside the helixQ6 Mutation in DNA occurs due to changes in the sequence of a. Nitrogeneous basesb. ribose unitsc. phosphate unitsd. noneAns. a. Nitrogeneous basesQ7 Which is a branched chain structurea. Amylopectin b. Amylosec. Cellulosed. NylonAns. a. AmylopectinExplanation: - Starch is made of two components: amylose, amylopectin.Q8 Antibodies area. carbohydratesb. proteinsc. lipdsd. enzymesAns. b. proteinQ9 The main structural feature of protein isa. ether linkageb. ester linkagec. peptide linkaged. allAns. c. peptide linkageQ9a. The main structural feature of sugars/ carbohydrates isa. ether linkageb. ester linkagec. peptide linkaged. allAns. a. ether linkageQ10 Enzymes belongs to which class of the compoundsa. Polysaccharidesb. Polypeptidesc. Polynitro heterocyclic compoundsd. HydrocarbonsAns. b. polypeptideExplanation: - if we take Amino Acid as AA.  AA-AA-AA... = polypeptide = proteinQ11. Glucose reacts with acetic anhydride to forma. Monoacetateb. Tetra-acetatec. Penta-acetated. Hexa-acetateAns. c. Penta-acetateQ12 Oxygen atom of which position is missing from ribose sugar so as to make it deoxyribosea. 2'b. 3'c. 4'd. 5'Ans. a. 2'Q13 The function of enzymes in the living system is to a. transport oxygen b. provide immunityc. catalyse biochemical reactiond. provide energy.Ans. c. catalyse biochemical reactionQ14. In DNA, the complementary bases are: a. Uracil and Adenine: Cytosine and Guanineb. Adenine and Thymine: Guanine and Cytosinec. Adenine and Thymine: Guanine and Uracild. Adenine and Guanine: Thymine and CytosineAns. b.Adenine and Thymine: Guanine and CytosineQ14a. Purines are a. adenine & guanineb. cytosine & thyminec. adenine & thymined. guanine & cytosineAns. a. adenine & guanineQ14a. Pyrimidines are a. adenine & guanineb. cytosine & thyminec. adenine & thymined. guanine & cytosineAns. b. cytosine & thymineQ15. Glucose molecule reacts with X no. of molecules of phenylhydrazine to yield osazone.  The value of X is a. Threeb. Twoc. Oned. FourAns. a. ThreeExplanation: - Phenylhydrazine = (C6H8N2) = C6H5NHNH2


*****1. Sucrose on hydrolysis gives(a) β-D-glucose + α-D-fructose(b) α-D-glucose + β-D-glucose(c) α-D-glucose + β-D-fructose(d) α-D-fructose + β-D-fructose.Ans. c2. The difference between amylose and amylopectin is(a) amylopectin  have 1→4 α-linkage and 1→6 α-linkage(b) amylose have 1→4 α-linkage and 1→6 β-linkagec. amylopectin have 1→4 α-linkage and 1→6 β-linkaged. amylose is made up of glucose and galactoseAns. aQ3.Q4. Which one given below is a non-reducing sugar?(a) Glucose (b) Sucrose(c) Maltose (d) Lactose Ans.b5. D(+)-glucose reacts with hydroxyl amine and yields an oxime. The structure of the oxime would beQ6. Which one of the following sets of monosaccharides forms sucrose?(a) α-D-galactopyranose and β-D-glucopyranose(b) α-D-glucopyranose and β-D-fructofuranose(c) β-D-glucopyranose and α-D-fructofuranose(d) α-D-glucopyranose and β-D-fructopyranoseAns. b. Q7 Which one of the following statements is not true regarding (+)–lactose?(a) On hydrolysis (+)–lactose gives equal amount of D(+)–glucose and D(+)–galactose.(b) (+)–Lactose is a β-glucoside formed by theunion of a molecule of D(+)–glucose and a molecule of D(+)–galactose.(c) (+)–Lactose is a reducing sugar and does not exhibit mutarotation.(d) (+)–Lactose, C12H22O11 contains 8 –OH groups.Ans. cQ8. Which one of the following does not exhibit the phenomenon of mutarotation?(a) (+)–Sucrose (b) (+)–Lactose(c) (+)–Maltose (d) (–)–FructoseAns. a9. Fructose reduces Tollens’ reagent due to(a) asymmetric carbons(b) primary alcoholic group(c) secondary alcoholic group(d) enolisation of fructose followed by conversion to aldehyde by base.Ans. dQ10. Number of chiral carbons in -D-(+) glucose is(a) five (b) six(c) three (d) four.Ans.dQ11. Glycolysis is(a) oxidation of glucose to glutamate(b) conversion of pyruvate tocitrate(c) oxidation of glucose to pyruvate(d) conversion of glucose to haem.Ans.cQ12. Cellulose is polymer of(a) glucose (b) fructose(c)  ribose (d) sucroseAns.a.Q13. Which of the following gives positive Fehling solution test?(a) Sucrose (b) Glucose(c) Fats (d) Protein Ans.b.Q14. α-D-glucose and β-D-glucose are(a) epimers (b) anomers(c)  enantiomers (d) diastereomers.Ans. bQ15. Which of the following is the sweetest sugar?(a) Fructose (b) Glucose(c) Sucrose(d) MaltoseAns.a.Q16. Glucose molecule reacts with X number of molecules of phenyl hydrazine to yield osazone. The value of X is(a) two (b) one(c) four (d) three.Ans.dQ17. The oxidation of glucose is one of the most important reactions in a living cell. What is the number of ATP molecules generated in cells from one molecule of glucose?(a) 28 (b) 38(c) 12 (d) 18Ans.b. Q18. The α-D-glucose and β-D-glucose differ from each other due to difference in carbon atom with respect to its(a) number of OH groups(b) size of hemiacetal ring(c) conformation(d) configuration. Ans. d.Q19. Chemically considering digestion is basically(a) anabolism(b) hydrogenation(c) hydrolysis(d) dehydrogenation. Ans.c.20. On hydrolysis of starch, we finally get(a) glucose (b) fructose(c)  both (a) and (b) (d) sucrose. Ans.a21. Which of the following is a basic amino acid?(a) Serine (b) Alanine(c)  Tyrosine (d) Lysine Ans. d22. The non-essential amino acid among the following is(a) lysine (b) valine(c) leucine (d) alanine.Ans. d23. Which structure(s) of proteins remain(s) intact during denaturation process?(a) Both secondary and tertiary structures(b) Primary structure only(c) Secondary structure only(d) Tertiary structure only Ans. bQ24. Which of the following compounds can form a zwitter ion?(a) Aniline(b) Acetanilide(c) Benzoic acid(d) GlycineAns. dQ25. In a protein molecule various amino acids are linked together by(a) peptide bond(b) dative bond(c) α-glycosidic bond(d) β-glycosidic bond.Ans. aQ26. Which of the statements about “Denaturation” given below are correct?(1) Denaturation of proteins causes loss of secondary and tertiary structures of the protein.(2) Denaturation leads to the conversion of double strand of DNA into single strand.(3) Denaturation affects primary structure which gets distorted.(a)  (2) and (3) (b) (1) and (3)(c)  (1) and (2) (d) (1), (2) and(3)Ans. cQ27. Which functional group participates in disulphide bond formation in proteins?(a) Thioester (b) Thioether(c) Thiol (d) Thiolactone Ans. c28. Which of the following structures represents the peptide chain?a. -NH-CO-NH-C-NH-CO=NH-b. -NH-CO-C-C-C-C-NH-C-C-C-c. -NH-C-CO-NH-C-CO-NH-C-CO-d. -NH-C-C-CO-NH-C-C-NH-CO-C-C-Ans. cQ29. The correct statement in respect of protein haemoglobin is that it a. functions as a catalyst for biological reactionsb. maintains blood sugar levelc. acts as an oxygen carrier in the bloodd. forms antibodies and offers resistance to diseases.Ans. cQ30. The helical structure of protein is stabilised by(a) dipeptide bonds (b) hydrogen bonds(c)  ether bonds (d) peptide bondsAns. bQ31. Which is not true statement?a. alpha C of alpha amino acid is asymmetricb. all proteins are found in L-formc. human body can synthesise all proteins they need.d. at pH=7 both amino and carboxylic groups exist in ionised formAns. bQ32 -CO-NH peptide bondwhich statement is incorrect about peptide bond?a. C-N bond length in proteins is longer than usual bond length of N-C bondb. Spectroscopic analysis shows planar structurec. C-N bond length in proteins is smaller than usual bond length of C-N bondd. None of the above.Ans. aQ33. Which is the correct statement?(a) Starch is a polymer of α-glucose.(b) Amylose is a component of cellulose.(c) Proteins are composed of only one type of amino acid.(d) In cyclic structure of fructose, there are four carbons and one oxygen atom.Ans. aQ34. Haemoglobin is(a) a vitamin (b) a carbohydrate(c) an enzyme(d) a globular protein.Ans. d. Q35 The secondary structure of a protein refers to(a) regular folding patterns of continuous portions of the polypeptide chain(b) three-dimensional structure, specially the bond between amino acid residues that are distant from each other in the polypeptide chain(c) mainly denatured proteins and structures of prosthetic groups(d) linear sequence of amino acid residues in the polypeptide chain.Ans. aQ36 36. During the process of digestion, the proteins present in food materials are hydrolysed to amino acids. The two enzymes involved in the process protein + enzyme A --> polypeptides + enzyme B----> amino acidsa. invertase and zymaseb. amylase and maltasec. diastase and lipased. pepsin and trypsinAns. dQ37. Enzymes are made up of(a) edible proteins(b) proteins with specific structure(c) nitrogen  containing carbohydrates(d) carbohydrates.Ans. b38. Which of the following is correct?(a) Cycloheptane is an aromatic compound.(b) Diastase is an enzyme.(c) Acetophenone is an ether.(d) All of these.Ans. b39. The function of enzymes in the living system is to(a) catalyse biochemical reactions(b) provide energy(c) transport oxygend. provide immunityAns. aQ40 Which of the following statements about enzymes is true?(a) Enzymes catalyse chemical reactions by increasing the activation energy.(b) Enzymes are highly specific both in binding chiral substrates and in catalysing their reactions.(c) Enzymes lack in nucleophilic groups.(d) Pepsin is proteolytic enzyme.Ans. b.Q 41. Enzymes take part in a reaction and(a) decrease the rate of a chemical reaction(b) increase the rate of a chemical reaction(c) both (a) and (b)(d) none of these.Ans. b.42. Deficiency of vitamin B1 causes the disease(a) convulsions (b) beri-beri(c) cheilosis (d) sterilityAns.b.43. Which of the following is not a fat soluble vitamin?(a) Vitamin B complex(b) Vitamin D(c) Vitamin E(d) Vitamin A Ans. a44. Which of the following vitamins is water soluble?(a) Vitamin E (b) Vitamin K (c) Vitamin A (d) Vitamin B Ans. d. 45. The human body does not produce(a) enzymes (b) DNA (c) vitamins (d) hormones. Ans. c. 46. Vitamin B12 contains (a) Fe (II) (b) Co (III) (c) Zn (II) (d) Ca (II)Ans. b47. The central dogma of molecular genetics states that the genetic information flows from(a) Amino acids→ Proteins→ DNA(b) DNA →Carbohydrates→Proteins(c) DNA →RNA →Proteins(d) DNA →RNA →CarbohydratesAns.c48. The correct statement regarding RNA and DNA, respectively is(a) the sugar component in RNA is a arabinose and the sugar component in DNA is ribose(b) the sugar component in RNA is 2-deoxyribose and the sugar component in DNA is arabinose(c) the sugar component in RNA is arabinose and the sugar component in DNA is 2-deoxyribose(d) the sugar component in RNA is ribose and the sugar  component  in  DNA  is  2-deoxyribose.Ans. d49. In DNA, the linkages between different nitrogenous bases are(a) phosphate linkage(b) H-bonding(c) glycosidic linkage(d) peptide linkage. Ans. b50. The segment of DNA which acts as the instrumental manual for the synthesis of the protein is(a) ribose (b) gene(c) nucleoside (d) nucleotide. Ans. b51. In DNA, the complimentary bases are(a) adenine and guanine; thymine and cytosine(b) uracil and adenine; cytosine and guanine(c) adenine and thymine; guanine and cytosine(d) adenine and thymine; guanine and uracil.Ans. c52. RNA and DNA are chiral molecules, their chirality is due to(a) chiral bases(b) chiral phosphate ester units(c) D-sugar component(d) L-sugar componentAns. c53. A sequence of how many nucleotides in messenger RNA makes a codon for an amin oacid?(a) Three (b) Four(c)  One(d) TwoAns. a. three54. Chargaff’s rule states that in an organism(a) amount of adenine (A) is equal to that of thymine (T) and the amount of guanine (G) is equal to that of cytosine (C)(b) amount of adenine (A) is equal to that of guanine (G) and the amount of thymine (T) is equal to that of cytosine (C)(c) amount of adenine (A) is equal to that of cytosine (C) and the amount of thymine (T) is equal to that of guanine (G)(d) amounts of all bases areequal. Ans. a55. Which of the following is correct about H-bonding in nucleotide?(a) A – T,  G – C (b) A – G , T – C(c)  G – T,  A – C (d) A – A, T – TAns. a56. An example of biopolymer is(a) teflon (b) neoprene(c) nylon-6, 6 (d) DNA. Ans. d57. The couplings between base units of DNA is through(a) hydrogen bonding(b) electrostatic bonding(c) covalent bonding(d) van der Waals’ forcesAns. aQ58. Which of the following statements is not correct?(a) Ovalbumin is a simple food reserve in egg- white.(b) Blood proteins thrombin and fibrinogen are involved in blood clotting.(c) Denaturation makes the proteins more active.(d) Insulin maintains sugar level in the blood of a human bodyAns. c59. Which of the following hormones is produced under the conditions of stress which stimulate glycogenolysis in the liver of human beings?(a) Thyroxin (b) Insulin(c) Adrenaline (d) EstradiolAns. c. Adrenaline60. Which of the following hormones contains iodine?(a) Testosterone (b) Adrenaline (c) Thyroxine (d) InsulinAns. C61. Which of the following is an amine hormone?(a) Insulin (b) Progesterone(c) Thyroxine (d) OxypurinAns. C62. Which one of the following is a peptide hormone?(a) Adrenaline (b) Glucagon(c)  Testosterone (d) Thyroxine Ans. C. testosterone63. The hormone that helps in the conversion of glucose to glycogen is(a) cortisone (b) bile acids(c) adrenaline (d) insulin.Ans. d. insulinQ64. Which one is responsible for production of energy in biochemical reaction?(a) Thyroxine (b) Adrenaline(c) Oestrogen (d) Progesterone Ans. a65. The cell membranes are mainly composed of(a) fats (b) proteins(c) phospholipids(d) carbohydrates.Ans. c66. Phospholipids are esters of glycerol with(a) three carboxylic acid residues(b) two carboxylic acid residues and one phosphate group(c) one carboxylic acid residue and two phosphate groups(d) three phosphate groups.Ans. b67. The number of molecules of ATP produced in the lipid metabolism of a molecule of palmitic acid is (a) 56 (b) 36(c) 130 (d) 86Ans. c. 130

P Block Elements************
Q1. In which of the following compounds, nitrogen exhibits highest oxidation state?a. N2H4b. NH3c. N3Hd. NH2OHAns. c. N2H4Explanation: highest oxidation state is -1/3.Q2. Nitrogen forms N2, but phosphorous does not form P2, however, it converts P4, reason is a. triple bond present between phosphorous atomb. pπ-pπ bonding is weakc. pπ-pπ bonding is strongd. multiple bonds form easily.Ans. b. pπ-pπ bonding is weakExplanation: for strong π bonding, pπ-pπ bonding should be strong. In case of P, due to larger size as compared to N atom, pπ-pπ bonding is not so strong. Q3 Which of the following oxides is most acidic?a. As2O5b. P2O5c. N2O5d. Sb2O5Ans. c. N2O5Explanation: Among N, P, As and Sb, the former has highest electronegativity so its oxide is most acidic.  As the electronegativity value of element increases, the acidic character of the oxide also increases. Q4. Which of the following has the highest dipole moment?a. SbH3b. AsH3c. NH3d. PH3Ans. c. NH3Explanation: Due to greater electronegativity of nitrogen, dipole moment for NH3 is greater. Q5 The basic character of hydrides of the V group elements decreases in the ordera. NH3>PH3>AsH3>SbH3b. SbH3>AsH3>PH3>NH3c. SbH3>PH3>AsH3>NH3d. NH3>SbH3>PH3>AsH3Ans. a. NH3>PH3>AsH3>SbH3Explanation: All the hydrides of group V elements have one lone pair of electrons on their central atom. Therefore, they can act as Lewis bases.  The basic character of these hydrides decreases down the group.Q6 Among the following oxides, the lowest acidic is a. As4O6b. As4O10c. P4O6d. P4O10Ans. a. As4O6Explanation: the acidic character of the oxides decreases with the decrease in the oxidation state and also decreases down the group. Q7. Which of the following fluorides does not exist?a. NF5b. PF5c. AsF5d. SbF5Ans.a. NF5Explanation: Nitrogen cannot form pentahalides because it cannot expand its octet due to non-availability of d-orbitals.Q8. Which one has the lowest boiling point?a. NH3b. PH3c. AsH3d. SbH3Ans. b. PH3Explanation: boiling point of hydrides increase with increase in atomic no. but ammonia has exceptionally high boiling point due to hydrogen bonding. Thus, the correct order of boiling point is BiH3>SbH3>NH3>AsH3>PH3Q9. Number of electrons shared in the formation of nitrogen molecule is a. 6b. 10c. 2d. 8Ans. a. 6Explanation: Nitrogen molecule is diatomic containing a triple bond between two N atoms, N≡N therefore, nitrogen molecule is formed by sharing six electrons. Q10. Nitrogen is relatively inactive element becausea. its atom has a stable electronic configurationb. it has low atomic radiusc. its electronegativity is fairly highd. dissociation energy of its molecule is fairly high.Ans. d. dissociation energy of its molecule is fairly high.Explanation: N2 molecule contains triple bond between N atoms having very high dissociation energy (946 kJ / mol) due to which it is relatively inactive. Ans. d. dissociation energy of its molecule is fairly high.Q11. Pure nitrogen is prepared in the laboratory by heating a mixture of a. NH4OH + NaClb. NH4NO3 + NaClc. NH4Cl + NaOHd. NH4Cl + NaNO2Ans. d. NH4Cl + NaNO2Explanation:NH4Cl+NaNO2→NH4NO2 which on heating gives NH4NO2→N2+H2OQ12. which of the following statement is not correct for nitrogen?a. its electronegativity is very highb. d-orbitals are available for bonding.c. it is a typical non-metald. its molecular size is smallAns. b. d-orbitals are available for bondingQ13. Urea reacts with water to form A which will decompose to form B. B when passed through Cu2+, deep blue colour solution C is formed. What is the formula of C from the following?a. CuSO4b. [Cu(NH3)4]2+c. Cu(OH)2d. CuCO3.Cu(OH)2Ans. b Cu(OH)2Explanation: [Cu(NH3)4]2+Urea reacts with water to give ammonium carbamate
NH2CONH2+H2O→NH2COO−NH4+                                                 (A)
Ammonium carbamate on decomposition gives ammonia.NH2COO−NH4+Δ2NH3+CO2                                          (B)
When NH3 is passsed through Cu2+ solution it forms a deep blue coloured complex. The complex  formed is [Cu(NH3)4]2+Q14. Aqueous solution of ammonia consists of a. H+b. OH-c. NH+d. NH and OH-Ans. dExplanation: aq solution of ammonia contains NH4+ and OH- ions Q15 which of the following oxides of nitrogen is paramagnetica. NO2b. N2O3c. N2Od. N2O5Ans. a. NO2explanation: NO2 is paramagnetic due to the presence of one unpaired electronQ16. which of the following is a nitric acid anhydride?a. NOb. NO2c. N2O5d. N2O3Ans. c. N2O5Explanation: when two molecules of nitric acid undergoes heating, loose a water molecule to form anhydride. HONO2             → N2O5 + H2OHONO2 Q17. when copper is heated with conc. HNO3 it produces a. Cu(NO3)2, NO and NO2b. Cu(NO3)2 and N2)c. Cu(NO3)2 and NO2d. Cu(NO3)2 and NOAns. c. Cu(NO3)2 and NO2 Explanation: Cu+4HNO3(conc.) → Cu(NO3)2 + 2NO2 + 2H2OQ18. Zn gives H2 gas with H2SO4 and HCl but not with HNO3 becausea. Zn act as oxidising agent when react with HNO3b. HNO3 is weaker acid than H2SO4 and HClc. In electrochemical series Zn is above hydrogen d. NO3- is reduced in preference to hydronium ion. Ans. d. NO3- is reduced in preference to hydronium ion. Zn+4HNO3 (conc) →Zn(NO3)2 + 2NO2 + 2H2OZinc is on the top position of hydrogen in electrochemical series. So, Zn displaces H2 from dilute H2SO4 and HCl with liberation of H2.  On the other hand, HNO3 is one oxidising agent. Hydrogen obtained in the reaction is converted into H2O.Q19. Sugarcane on reaction with nitric acid givesa. CO2 and SO2b. (COOH)2c. 2HCOOH (two moles)d. no reactionAns. b. (COOH)2C12H22O11 + 18[O] → 6 (COOH)2 + 5H2OQ20. Which of the following phosphorous is the most reactive?a. Scarletb. whitec. redd. violetAns. b. whiteexplanation: white phosphorous has low ignition temp so it is most reactive among all the allotropes.Q21. Each of the following is true for white and red phosphorous except that they?a. are both soluble in CS2b. can be oxidised by heating in airc. consist of the same kind of atomsd. can be converted into one anotherAns. a. are both soluble in CS2explanation: red phosphorous is insoluble in CS2 and only white P is soluble in CS2. Q22 A compound X upon reaction with H2O produces a colourless gas Y with rotten fish smell.  Gas Y is absorbed in a solution of CuSO4 to give Cu3P2 as one of the products.  Predict the compound Xa. Ca3P2b. NH4Clc. As2O3d. Ca3(PO4)2Ans. a. Ca3P2Explanation: (X) Ca3P2 + 6H2O  → 3Ca (OH)2 + 2PH3 (Y)3CuSO4 + 2PH3 (Y)  → Cu3P2 + H2SO4Q23. PH4I + NaOH formsa. PH3b. NH3c. P4O6d. P4O10Ans. a. PH3explanation: PH4I + NaOH  → NaI + PH3 + H2OQ24. Identify the incorrect statement related to PCl from the following: a. PCl5 molecule is non-reactiveb. three equatorial P-Cl bonds make an angle of 120* with each other.c. Two axial P-Cl bonds make an angle of 180* with each otherd. Axial P-Cl bonds are longer than equatorial P-Cl bondsAns. a. PCl5 molecule is non-reactiveexplanation: it is a reactive gas as it easily provides Cl2 gas.Q25. PCl3 reacts with water to form(a) PH3 (b) H3PO3, HCl(c) POCl3 (d) H3PO4Ans.(b) H3PO3, HClQ26. Which of the following oxoacids of phosphorus has strongest reducing property?(a) H4P2O7 (b) H3PO3(c)  H3PO2 (d) H3PO4Ans.(c)  H3PO2 Q27. Which is the correct statement for the given acids?(a) Phosphinic acid is a monoprotic acid while phosphonic acid is a diprotic acid.(b) Phosphinic acid is a diprotic acid while phosphonic acid is a monoprotic acid.(c) Both are diprotic acids.(d) Both are triprotic acids. Ans. (a) Phosphinic acid is a monoprotic acid while phosphonic acid is a diprotic acid.Q28. Strong reducing behaviour of H3PO2 is due to(a) high electron gain enthalpy of phosphorus(b) high oxidation state of phosphorus(c) presence of two —OH groups and one P—H bond(d) presence of one —OH group and two P—H bonds.Ans.(d) presence of one —OH group and two P—H bonds.Q29. Which of the following statements is not valid for oxoacids of phosphorus?(a) Orthophosphoric acid is used in the manufacture of triple superphosphate.(b) Hypophosphorous acid is a diprotic acid.(c) All oxoacids contain tetrahedral four coordinated phosphorus. (d) All oxoacids contain at least one P=O unit and one P-OH group.Ans. (b) Hypophosphorous acid is a diprotic acid.Q30. Oxidation states of P in H4P2O5, H4P2O6, H4P2O7 are respectively(a)  +3, +5, +4 (b) +5, +3, +4(c)  +5, +4, +3 (d)  +3, +4, +5Ans.(d)  +3, +4, +5Q31. How many bridging oxygen atoms are present in P4O10?(a) 6 (b) 4(c) 2 (d) 5Ans.a.6Q32. The structural formula of hypophosphorous acid isa. H-P=O(OH)(OH)b. HO-P=O(OH)(OH)c. H-P=O(H)(OH)d. none of these Ans. c. H-P=O(H)(OH)Q33. H3PO2 is the molecular formula of an acid of phosphorus. Its name and basicity respectively are(a) phosphorous acid and two(b) hypophosphorous acid and two(c) hypophosphorous acid and one(d) hypophosphoric acid and two. Ans. (c) hypophosphorous acid and oneQ34. Which one of the following substance is used in the laboratory for fast drying of neutral gases?(a) Phosphorus pentoxide (b) Active charcoal(c) Anhydrous calcium chloride(d) Na3PO4 Ans.(a) Phosphorus pentoxide Q35. P2O5 is heated with water to give(a) hypophosphorous acid(b) phosphorous acid (c) hypophosphoric acid(d) orthophosphoric acid. Ans. (d) orthophosphoric acid. Q36. Basicity of orthophosphoric acid is(a) 2 (b) 3(c) 4 (d) 5 Ans. (b) 3Q37. When orthophosphoric acid is heated to 600°C, the product formed is(a) PH3 (b) P2O5(c) H3PO3 (d) HPO3Ans.(d) HPO3Q38. Which is the correct thermal stability order for H2E (E = O, S, Se, Te and Po) ?(a) H2Se < H2Te < H2Po < H2O < H2S(b) H2S < H2O < H2Se < H2Te < H2Po(c) H2O < H2S < H2Se < H2Te < H2Po(d) H2Po < H2Te < H2Se < H2S < H2O Ans.(d) H2Po < H2Te < H2Se < H2S < H2OQ39. Acidity of diprotic acids in aqueous solutions increases in the order(a) H2S < H2Se < H2Te (b) H2Se < H2S < H2Te(c) H2Te < H2S < H2Se (d) H2Se < H2Te < H2SAns.(a) H2S < H2Se < H2Te Q40. Which of the following bonds has the highest energy?(a)S–S (b) O–O(c) Se–Se (d) Te–TeAns.(a)S–S Q41. Which of the following does not give oxygen on heating?(a) K2Cr2O7 (b) (NH4)2Cr2O7 (c) KClO3 (d) Zn(ClO3)2Ans.(b) (NH4)2Cr2O7 Q42. Which would quickly absorb oxygen?(a) Alkaline solution of pyrogallol(b)Conc. H2SO4(c) Lime water(d) Alkaline solution of CuSO4Ans. (a) Alkaline solution of pyrogallolQ43. Oxygen will directly react with each of the following elements except(a) P (b) Cl(c) Na (d) S Ans. b. ClQ44. It is possible to obtain oxygen from air by fractional distillation because(a) oxygen is in a different group of the periodic table from nitrogen(b) oxygen is more reactive than nitrogen(c) oxygen has higher b.pt. than nitrogen(d) oxygen has a lower density than nitrogen. Ans.(c) oxygen has higher b.pt. than nitrogenQ45. Match the following: .........Oxide..............NatureA...........CO...............(i) BasicB..........BaO...............(ii)  NeutralC..........Al2O3............(iii) AcidicD......... Cl2O7............(iv) Amphotericwhich of the following is correct option?a. Ai Bii Ciii Divb. Aii Bi Civ  Diiic. Aiii Biv Ci Diid. Aiv Biii Cii DiAns.b. Aii Bi Civ  DiiiQ46. The angular shape of ozone molecule (O3) consists of(a) 1σ  and 1𝜋  bond (b) 2σ and 1𝜋 bond(c) 1σ and 2𝜋 bonds (d) 2σ and 2𝜋 bonds.Ans. (b) 2σ and 1𝜋 bondQ47. The gases respectively absorbed by alkaline pyrogallol and oil of cinnamon are(a) O3, CH4 (b) O2, O3(c) SO2, CH4 (d) N2O, O3 Ans. (b) O2, O3Q48. Nitrogen dioxide and sulphur dioxide have some properties in common. Which property is shown by one of these compounds, but not by the other?(a) Is soluble in water.(b) Is used as a food preservative.(c) Forms ‘acid-rain’.(d) Is a reducing agent. Ans. (b) Is used as a food preservative.Q49. Sulphur trioxide can be obtained by which of the following reaction?(a) CaSO4 + C   → (b) Fe2(SO4)3  →   (c) S + H2SO4 →    (d) H2SO4 + PCl5 → Ans. (b) Fe2(SO4)3  →Q50. Which of the following oxoacid of sulphur has  -O-O-linkage(a) H2SO3, sulphurous acid(b) H2SO4, sulphuric acid(c) H2S2O8, peroxodisulphuric acid(d) H2S2O7, pyrosulphuric acidAns.(c) H2S2O8, peroxodisulphuric acidQ51. Identify the correct formula of oleum from the following :(a) H2S2O7 (b) H2SO3(c) H2SO4 (d) H2S2O8Ans.(a) H2S2O7Q52. In which pair of ions both the species contain S-S bond?a. S4O62-       S2O32- b. S2O72-      S2O82-c. S4O62-    S2O72-   d. S2O32- S2O72- Ans.a. S4O62-       S2O32- Q53. Oleum is(a) castor oil(b) oil of vitriol(c) fuming H2SO4 (d) none of these.  Ans.(c) fuming H2SO4Q54. Match List I (substances) with List II (processes) employed in the manufacture of the substances and select the correct option..............List I...................List II.......(Substances)...........(Processes)(A)...Sulphuric acid.........(i) Haber’s process(B)..........Steel.................(ii) Bessemer’s process(C).....Sodium hydroxide...(iii) Leblanc Process(D).....Ammonia................(iv) Contact Process(a) A - (i), B - (iv), C - (ii), D - (iii)(b) A - (i), B - (ii), C - (iii), D - (iv)(c) A - (iv), B - (iii), C - (ii), D - (i)(d) A - (iv), B - (ii), C - (iii), D - (i)Ans. d. (d) A - (iv), B - (ii), C - (iii), D - (i)Q55. Which of the following statements is not true for halogens?(a) All form monobasic oxyacids.(b) All are oxidizing agents.(c) All but fluorine show positive oxidation states.(d) Chlorine has the highest electron-gain enthalpy. Ans.Q56. Which one of the following orders is correct for the bond dissociation enthalpy of halogen molecules?(a) Br2 > I2 > F2 > Cl2 (b) F2 > Cl2 > Br2 > I2(c) I2  > Br2 > Cl2 > F2 (d) Cl2 > Br2 > F2 > I2Ans.Q57. The variation of the boiling points of the hydrogen halides is in the order HF > HI > HBr > HCl. What explains the higher boiling point of hydrogen fluoride?(a) There is strong hydrogen bonding between HF molecules.(b) The bond energy of HF molecules is greater than in other hydrogen halides.(c) The effect of nuclear shielding is much reduced in fluorine which polarises the HF molecule.(d) The electronegativity of fluorine is much higher than for other elements in the group.Ans.(a) There is strong hydrogen bonding between HF molecules.Q58. Among the following which is the strongest oxidising agent?a. Brb. Ic.Cl2d. I2Ans.d. I2Q59. Which one of the following arrangements  does  not give the correct picture of the trends indicated against it?(a) F2 > Cl2 > Br2 > I2 : Bond dissociation energy(b) F2 > Cl2 > Br2 > I2 : Electronegativity(c) F2 > Cl2 > Br2 > I2 : Oxidizing power(d) F2 > Cl2 > Br2 > I2 : Electron gain enthalpyAns. a and d(a) F2 > Cl2 > Br2 > I2 : Bond dissociation energy(d) F2 > Cl2 > Br2 > I2 : Electron gain enthalpyQ60. Which one of the following orders is not in accordance with the property stated against it?(a) F2 > Cl2 > Br2 > I2 : Bond dissociation energy(b) F2 > Cl2 > Br2 > I2 : Oxidising power(c) HI > HBr > HCl > HF : Acidic property in water(d) F2 > Cl2 > Br2 > I2 : ElectronegativityAns. (a) F2 > Cl2 > Br2 > I2 : Bond dissociation energyQ61. Which statement is wrong?(a) Bond energy of F2 > Cl2(b) Electronegativity of F > Cl(c) F is more oxidising than Cl(d) Electron affinity of Cl > F Ans. (a) Bond energy of F2 > Cl2Q62. Which of the following has the greatest electron affinity?(a) I (b) Br(c) F (d) Cl Ans. d. ClQ63. Which of the following displaces Br2 from an aqueous solution containing bromide ions?(a) I2 (b) I –(c) Cl2 (d) Cl Ans. c. Cl2Q64. Which of the following species has four lone pairs of electrons?(a) I (b) O(c) Cl–(d) HeAns. c. Cl-Q65. Match the followingA. pure nitrogen.............i. chlorineB. Haber Process............ii. Sulphuric acidC. Contact Process..........iii. AmmoniaD. Deacon's Process........iv. Sodium azide or Barium azideWhich of the following is the correct option?.......A.....B........C.........Da......iv....iii........ii.........ib......i......ii........iii.........ivc......ii.....iv.........i.........iiid......iii....iv........ii.........iAns. a.......A.....B........C.........Da......iv....iii........ii.........iQ66. When Cl2 gas reacts with hot and concentrated sodium hydroxide solution, the oxidation number of chlorine changes from a. zero to +1 and zero to -5b. zero to -1 and zero to +5(c) zero to –1 and zero to +3d. zero to +1 and zero to –3Ans. b. zero to -1 and zero to +5Q67 Which of the following is used in the preparation of chlorine?(a) Both MnO2 and KMnO4(b) Only KMnO4(c) Only MnO2(d) Either MnO2 or KMnO4 Ans. (a) Both MnO2 and KMnO4Q68. Which of the following elements is extracted commercially by the electrolysis of an aqueous solution of its compound?(a) Cl(b) Br(c) Al (d) Na Ans. a. ClQ69. When chlorine is passed over dry slaked lime at room temperature, the main reaction product is(a) Ca(ClO2)2 (b) CaCl2(c) CaOCl2 (d) Ca(OCl)2 Ans. c. CaOCl2Q70. In the manufacture of bromine from sea water, the mother liquor containing bromides is treated with(a) carbon dioxide (b) chlorine(c) iodine (d) sulphur dioxide.Ans. b. ChlorineQ71. The bleaching action of chlorine is due to(a) reduction (b) hydrogenation(c)  chlorination (d) oxidation. Ans. d. OxidationQ72. Bleaching powder reacts with a few drops of conc. HCl to give(a) chlorine (b) hypochlorous acid(c) calcium oxide (d) oxygen.Ans. a. chlorineQ73. Among the following, the correct order of acidity is(a) HClO2 < HClO < HClO3 < HClO4(b) HClO4 < HClO2 < HClO < HClO3(c) HClO3 < HClO4 < HClO2 < HClO(d) HClO < HClO2 < HClO3 < HClO4Ans.(d) HClO < HClO2 < HClO3 < HClO4Q74. Which of the statements given below is incorrect?(a) O3  molecule is bent. (b) ONF is isoelectronic with O2N(c) OF2 is an oxide of fluorine(d) Cl2O7  is an anhydride of perchloric acidAns.(c) OF2 is an oxide of fluorineQ75. The correct order of increasing bond angles in the following species is –(a)Cl2O < ClO2  < ClO2- (b)ClO2  <Cl2O < ClO-(c) Cl2O < ClO2- < ClO2 (d) ClO2-<Cl2O < ClO2Ans. (d) ClO2-<Cl2O < ClO2Q76. Which one of the following oxides is expected to exhibit paramagnetic behaviour?(a)CO2(b) SiO2 (c)SO2(d)ClO2Ans.(d)ClO2Q77. Match the interhalogen compounds of column-I with the geometry in column-II and assign the correct code..........Column I...........Column II(A) ........XX.................(i)  T-shape(B).........XX3..............(ii) Pentagonal bipyramidal(C).........XX5..............(iii)  Linear(D).........XX7..............(iv) Square pyramidal...................................(v) TetrahedralCode : A........B........C........D(a)......(iii)....... (i)......(iv)......(ii)(b)......(v)....... (iv)......(iii)......(ii)(c)......(iv)....... (iii)......(ii)......(i)(d)......(iii)....... (iv)......(i).......(ii)Ans. (a)Code : A........B........C........D(a)......(iii)....... (i)......(iv)......(ii)
Q78. Match the Xenon compounds in Column-I with its structure in Column-II and assign the correct code........Column I ............Column IIA. XeF4.....................i. pyramidalB. XeF6.....................ii. Square planarC. XeOF4...................iii. distorted octahedralD. XeO3.....................iv. square pyramidal
....A.......B.......C......Da. iii.......iv.......i.......iib. i.........ii.......iii......ivc. ii........iii.......iv......id. ii........iii.......i.......ivAns.Q79. Identify the incorrect statement, regarding the molecule XeO4(a) XeO4 molecule is square planar.(b) There are four p - d bonds.(c) There are four sp3 - p,  σ bonds.(d) XeO4 molecule is tetrahedralAns.(a) XeO4 molecule is square planar.Q80. Which compound has planar structure?(a) XeF4 (b) XeOF2(c) XeO2F2 (d) XeO4Ans.(a) XeF4

HaloalkarenesAssertion /Reasona. both are correct and reason is the correct explanationb. both are correct but reason is not the correct explanationc. a is true but r is falsed. both are false
Q1. Assertion: Alkyl halides can be prepared on mixing alcohols with halogen acidsReason: 3* alcohols favours the reaction as 3* carbocation is more stable than 1* carbocationAns. a. Alkyl halides can be prepared on mixing alcohols with halogen acids in the presence of lewis acids. Carbocation was formed as the intermediate.  3* Carbocations are more stable than 2* which in turn more stable than 1*.
Q2. Assertion: Alkyl halides can be prepared on mixing alcohols with halogen acidsReason: Hydrogen iodide favours the reaction as the bond strength of Hydrogen and Iodine is weak.Ans. a. 
Q3. Assertion: Alkyl iodide can be prepared by treating alkyl chloride / bromide with NaI in acetone.Reason: NaCl/ NaBr are soluble in acetone while NaI is notAns.c.Explanation: R-Cl + NaI  (acetone) = R-INaCl and NaBr are polar compounds and acetone is non polar. Hence they are not soluble in acetone.
Q4. Assertion: Addition of HBr to 2-butene gives two isomeric productsReason: Addition of HBr to 2-butene follows Markonikov's ruleAns. d.
Q5. Assertion: Diastereomers have different physical propertiesReason: They are non-superimposible mirror images. Ans. cExplanation: Diasteromers are not mirror images.
Q6 Assertion: the presence of nitro group facilitates nucleophilic substitution reaction in aryl halidesReason: The intermediate carbanion is stabilized due to the presence of the nitro group. Ans. aExplanation: 
Q7. Assertion: Rate of hydrolysis of methyl chloride to methanol is higher in DMF than in waterReason; Hydrolysis of methyl chloride follows second order kineticsAns. aExplanation: DMF is a polar aprotic solvent i.e. does not contain any OH group while H2O is a polar protic solvent. Therefore, nucleophile i.e. OH- ion is solvated in H2O but not in DMF.  As a result, the OH- ion is more nucleophilic in DMF than in H2O and hence the rate of the reaction is increased. DMF = Dimethylformamide (CH2)2N-CHO
Q8. Assertion: 2-Bromobutane on reaction with sodium ethoxide in ethanol gives 1-butene as a major productReason: 1-Butene is more stable than 2-butene.Ans. dExplanation: 2-Bromobutane on reaction with sodium ethoxide eliminates Br and forms 2-butene as a major product. It happens as per the Saytzeff Rule.
Q9. Assertion: Hydroxy ketones are not directly used in Grignard Reagents.Reason: Grignard reagents react with Hydroxyl Group. Ans. a



MCQQ3. When chlorine is passed through propene at 400°C, which of the following is formed?(a) PVC(b) Allyl chloride(c) Propyl chloride(d) 1, 2-DichloroethaneAns. b. Explanation: At 400°C temperature, substitution occurs instead of addition.
Q4. Elimination reaction of 2-bromopentane to form pent-2-ene is(A) -Elimination reaction(B) Follows Zaitsev rule(C) Dehydrohalogenation reaction(D) Dehydration reaction(a)  (A), (B), (C) (b)  (A), (C), (D)(c)  (B), (C), (D) (d)  (A), (B),(D)Ans. (a)  (A), (B), (C)
Q6. The compound A on treatment with Na gives B, and with PCl5 gives C. B and C react together to give diethyl ether. A, B and C are in the order(a) C2H5OH, C2H6, C2H5Cl(b) C2H5OH, C2H5Cl, C2H5ONa(c) C2H5Cl, C2H6, C2H5OH(d) C2H5OH, C2H5ONa, C2H5Cl Ans. dExplanation: compound A is C2H5OH which on reaction with Na gives C2H5ONa which is compound B. Compound A on reaction with PCl5 gives C2H5Cl which is C.  Then, compound B and C together follows SN2 and forms C2H5OC2H5.
7. The compound C7H8 undergoes the following reactions :C7H8 +  3Cl  →  A + Br2 /Fe →  B + Zn/HCl →  CThe product C is(a) m-bromotoluene(b) o-bromotoluene(c) 3-bromo-2,4,6-trichlorotoluene(d) p-bromotoluene.Ans. (a) m-bromotoluene
Q9 Consider the reactionCH3CH2CH2Br + NaCN → CH3CH2CH2CN + NaBrThis reaction will be the fastest in(a) ethanol(b) methanol(c) N,N-dimethylformamide (DMF)(d) water.Ans. c. 
12. Two possible stereo-structures of CH3CH(OH)COOH, which are optically active, are called(a) atropisomers (b) enantiomers(c) mesomers (d) diastereomers. Ans. (b). enantiomers
13. In an SN1 reaction on chiral centres, there is(a) inversion more than retention leading to partial racemisation(b) 100% retention(c) 100% inversion(d) 100% racemisation.Ans. a.Explanation: In case of optically active alkyl halides, SN1 reaction is accompanied by racemisation. The carbocation formed in the slow step being sp2 hybridised is planar and attack of nucleophile may take place from either side resulting in a mixture of products, one having the same configuration and other having inverted configuration.  The isomer corresponding to inversion is present in slight excess because SN1 also depends upon the degree of shielding of the front side of the reacting carbon.
16. Which of the following acids does not exhibit optical isomerism?(a) Maleic acid (b) -Amino acids(c) Lactic acid (d) Tartaric acidAns. a. Maleic acid Explanation: Maleic acid shows geometrical isomerism and not optical isomerism
19. Which one is most reactive towards SN1 reaction?(a) C6H5CH(C6H5)Br(b) C6H5CH(CH3)Br(c) C6H5C(CH3)(C6H5)Br(d) C6H5CH2BrAns. c. Explanation: In SN1 mechanism only one nucleophile is attached with the optically active carbon atom. Bulky the group that are attached to the optically active carbon chain, more difficult it is for the nucleophile to attack. Hence, carbocation formation takes place in case bulky groups are attached to the chiral carbon/ optically active carbon atom. First carbocation is formed with the release of nucleophile. Then substituent nucleophile attacks on that carbocation. Bulky the groups SN1 mechanism prevails. 
Q21 in the following reactionC6H5CHBr + Mg (Ether) + H3O+    -----> Xthe product X is the product ‘X’ is(a) C6H5CH2OCH2C6H5(b) C6H5CH2OH(c) C6H5CH3(d) C6H5CH2CH2C6H5Ans. c. Q22. Which of the following reactions is an example of nucleophilic substitution reaction?(a) 2RX + 2Na ----> R – R + 2NaX(b) RX + H2 -----> RH + HX(c) RX + Mg -----> RMgX(d) RX + KOH ---> ROH + KXAns. dQ23. How many stereoisomers does this molecule have?CH3CHCHCH2CHBrCH3(a) 8 (b) 2(c) 4 (d) 6Ans. c. Explanation: Both geometrical isomerism (cis-trans form) and optical isomerism is possible in the given compound.No. of optical isomer = 2n = 21 = 2 (where n = no. of asymmetric carbon)Hence total no. of stereoisomers = 2 + 2 = 4Q24 In a SN2 substitution reaction of the type R-Br + Cl (DMF) ------> R-Cl + Brwhich one of the following has the highest relative rate?a. (CH3)3CCH2Brb. CH3CH2Brc. CH3CH2CH2Brd. (CH3)2CHCH2BrAns. b.Explanation: SN2 mechanism is followed in case of primary and secondary alkyl halides i.e. SN2 reaction is favoured by small groups on the carbon atoms attached to halogen so, CH3 – X > R – CH2 – X > R2CH – X > R3C – X. Primary is more reactive than secondary and tertiary alkyl halides.
25. If there is no rotation of plane polarised light by a compound in a specific solvent, though to be chiral, it may mean that(a) the compound is certainly meso(b) there is no compound in the solvent(c) the compound may be a racemic mixture(d) the compound is certainly a chiralAns.  a Explanation: Meso compound does not rotate plane polarised light. Compound which contains tetrahedral atoms with four different groups but the whole molecule is achiral, is known as meso compound. It possesses a plane of symmetry and is optically inactive. One of the asymmetric carbon atoms turns the plane of polarised light to the right and other to the left and to the same extent so that the rotation due to upper half is compensated by the lower half, i.e., internally compensated, and finally there is no rotation of plane polarised lightQ27 which of the following is not chiral?a. 2-Hydroxypropanoic acidb. 2-Butanolc. 2,3-Dibromopentaned. 3-BromopentaneAns. d28. Which of the following undergoes nucleophilic substitution exclusively by SN1 mechanism?(a) Ethyl chloride(b) Isopropyl chloride(c) Chlorobenzene(d) Benzyl chlorideAns. d. Explanation: SN1 reaction is favoured by heavy (bulky) groups on the carbon atom attached to halogens and nature of carbonium ion in substrate is Benzyl > Allyl > Tertiary > Secondary > Primary > Methyl halides.30. Which of the following is least reactive in a nucleophilic substitution reaction?(a) (CH3)3C – Cl(b) CH2=CHCl(c) CH3CH2Cl(d) CH2=CHCH2ClAns. b. Q32 Reactivity order of halides for dehydrohalogenation is(a) R – F > R – Cl > R – Br > R – I(b) R – I > R – Br > R – Cl > R – F(c) R – I > R – Cl > R – Br > R – F(d) R – F > R – I > R – Br > R – ClAns. bExplanation: I > Br > Cl > F atomic radiiF, Cl, Br, I belong to the same group orderly. Atomic radii go on increasing as the nuclear charge increases in preceding downwards in a group. The decreasing order of bond lengthQ33 CH3CH2Cl + NaCN --------> X + Ni/H2 -------> Y + acetic anhydride --------> ZZ in the above reaction sequence is(a) CH3CH2CH2NHCOCH3(b) CH3CH2CH2NH2(c) CH3CH2CH2CONHCH3(d) CH3CH2CH2CONHCOCH3Ans. a.Q34 CH3CH2CH(Cl)CH3 obtained by chlorination of n-butane will be a. meso formb. racemic mixturec. d-formd. l-formAns. bQ35 An organic compound A(C4H9Cl) on reaction with Na/diethyl ether gives a hydrocarbon which on monochlorination gives only one chloro derivative then, A is(a) t-butyl chloride (b) s-butyl chloride(c) iso-butyl chloride (d) n-butyl chloride.Ans. aQ 36. A compound of molecular formula C7H16 shows optical isomerism, compound will be(a) 2,3-dimethylpentane(b) 2,2-dimethylbutane(c) 2-methylhexane(d) none of these. Ans. a 37. Which of the following compounds is not chiral?(a) CH3CHDCH2Cl (b) CH3CH2CHDCl(c) DCH2CH2CH2Cl (d) CH3CHClCH2DAns. c. Explanation: The above compound has no chiral ‘C’-atom. All the ‘C’ atoms are attached to two identical ‘H’ atoms, so they are not asymmetrical
38. Replacement of Cl of chlorobenzene to  give phenol requires  drastic conditions.  But  chlorine of 2,4-dinitrochlorobenzene is readily replaced because(a) NO2 donates e– at meta position(b) NO2 withdraws e– from ortho/para positions(c) NO2 makes ring electron rich at ortho and para(d) NO2 withdraws e– from meta position.Ans. b. NO2 withdraws e- from ortho / para positionQ 39. The alkyl halide is converted into an alcoholby(a) elimination(b) dehydrohalogenation(c) additiond. substitution Ans. d. substitutionQ41. Reaction of t-butyl bromide with sodium methoxide produces(a) sodium t-butoxide(b) t-butyl methyl ether(c) isobutane(d) isobutylene.Ans. d. isobutylene42. Grignard reagent is prepared by the reaction between(a) magnesium and alkane(b) magnesium and aromatic hydrocarbon(c) zinc and alkyl halide(d) magnesium and alkyl halide.Ans. d. 43. Chlorobenzene reacts with Mg in dry ether to give a compound (A) which further reacts with ethanol to yield(a) phenol (b) benzene(c) ethyl benzene (d) phenyl ether.Ans. b. benzene44. Benzene reacts with n-propyl chloride in the presence of anhydrous AlCl3 to give(a) 3-propyl-1-chlorobenzene(b) n-propylbenzene(c) no reaction(d) isopropylbenzeneAns. d. 46. Which of the following is an optically active compound?(a) 1-Butanol (b) 1-Propanol(c)  2-Chlorobutane (d) 4-HydroxyheptaneAns. c. 2-Chlorobutane48. Industrial preparation of chloroform employs acetone and(a) phosgene(b) calcium hypochlorite(c) chlorine gas(d) sodium chloride. Ans. b49. Phosgene is a common name for(a) phosphoryl chloride(b) thionyl chloride(c) carbon dioxide and phosphine(d) carbonyl chloride.Ans. d. 

AlcoholsQ1 Assertion: Benezenediazonium chloride on boiling with water gives phenolReason: C-N bond is polar.Ans. cQ2. Assertion: The presence of nitro group facilitates nucleophilic substitution reaction in aryl halidesReason: The intermediate carbanion is stabilized due to the presence of the nitro group. Ans. aQ3. Assertion: t-butyl methyl ether is not prepared by the reaction of t-butyl bromide with Sodium methoxideReason: Sodium methoxide is a strong nucleophile. Ans. b. Explanation: In presence of strong base, i.e. sodium methoxide, tert-butyl bromide undergoes dehydrohalogenation to form isobutylene.Q4. Assertion: the major products formed by heating C6H5CH2-O-CH3 with HI are C6H5CH2I and CH3OHReason: Benzyl cation is more stable than methyl cationAns. a.


Ordinary MCQ type: Q1. The general molecular formula, which representsthe homologous series of alkanols is(a) CnH2n+2O (b) CnH2nO2(c)  CnH2nO (d)  CnH2n + 1OAns. a2. Reaction between acetone and methyl magnesium chloride followed by hydrolysis will give(a) iso-propyl alcohol (b) sec-butyl alcohol(c) tert-butyl alcohol (d) iso-butyl alcohol.Ans. cQ3. The structure of intermediate A in the following reaction is C6H5-CH-(CH3)2 + O2 ---> A + H3O+ ---> C6H5OH + CH3COCH3a. C6H5-CH(CH3)-CH2-O-O-Hb. C6H5-O-CH-(CH3)2c. C6H5-C(CH3)2-O-O-Hd. C6H5-O-O-CH(CH3)2Ans.c4. When vapours of a secondary alcohol is passed over heated copper at 573 K, the product formed is(a) a carboxylic acid (b) an aldehyde(c) a ketone (d) an alkene.Ans. cQ5 In the reactionC6H5-OH + CHCl3 + NaOH ---> C6H5-ONa  (o-CHO)the electrophile involved isa. dichloromethyl cation (CHCl2+)b. formyl cation (CHO+)c. dichloromethyl anion (CHCl2-)d. dichlorocarbene (:CCl2)Ans. d6. Compound A, C8H10O, is found to react with NaOI (produced by reacting Y with NaOH) and yields a yellow precipitate with characteristic smell. A and Y are respectivelya. CH3-C6H4-CH2-OH and I2b. C6H5-CH2-CH2-OH and I2c. C6H5-CHOH-CH3 and I2d. CH3C6H3(o-CH3)OH(p) and I2Ans. cQ7 Identify the major products P, Q and R in the following sequence of reactions: C6H6 + CH3CH2CH2Cl + Anhy AlCl3 ---> P + (i) O2 + (ii) H3O+ --> Q+Ra. P = C6H5-CH2-CH2-CH3 Q= C6H5-CHO R= CH3CH2-OHb. P = C6H5-CH2-CH2-CH3 Q= C6H5-CHO R= C6H5-COOHc. P=C6H5-CH(CH3)2 Q=C6H5-OH R= CH3CH(OH)CH3d. P = C6H5-CH(CH3)2 Q =  C6H5-OH R = CH3-O-CH3Ans. dQ8 most acidic compounda. phenolb. p-nitrophenolc. 2,4,6-trinitrophenold. p-CresolAns. c9. Reaction of phenol with chloroform in presence of dilute sodium hydroxide finally introduces which one of the following functional group?(a) –COOH (b) –CHCl2(c) –CHO(d) –CH2ClAns. c10. Which of the following reaction(s) can be used for the preparation of alkyl halides?(I) CH3CH2OH + HCl Anh.ZnCl2(II) CH3CH2OH + HCl(III) (CH3)3COH + HCl(IV) (CH3)2CHOH + HClAnh.ZnCl2(a) (I) and (II) only (b) (IV) only(c) (III) and (IV) only(d) (I), (III) and (IV) onlyAns. d11. Which of the following will not be soluble in sodium hydrogen carbonate?(a) 2,4,6 - Trinitrophenol(b) Benzoic acid(c) o-Nitrophenol(d) Benzenesulphonic acidAns. cQ12 Number of isomeric alcohols of molecular formula C6H14O give positive iodoform testsa. threeb. fourc. fived. twoAns. bQ13 in the following sequence of reactionsCH3-Br + KCN ---> A + H3O+ ---> B + LiAlH4 (ether) ---> Cthe end product C is a. acetoneb. methanec. acetaldehyded. ethyl alcoholAns. dQ14. In the following reactionsi. CH3-CH-(CH3)-CH(OH)-CH3 + H+ (heat) --> A (major) + B (Minor)ii. A + HBr, dark (in absence of peroxide) ---> C (major) + D (Minor) the major products A and C respectively area. CH2=C(CH3)2-CH2-CH3 and Br-CH2-CH(CH3)-CH2-CH3b. CH3-C(CH3)2=CH-CH3 and CH3-C(CH3)Br-CH2-CH3c. CH2=C(CH3)-CH2-CH3 and CH3-C(CH3)Br-CH2-CH3d. CH3-C(CH3)=CH-CH3 and CH3-CH(CH3)-CH(Br)-CH3Ans.  b15. Given are cyclohexanol (I), acetic acid (II), 2,4,6- trinitrophenol (III) and phenol (IV). In these the order of decreasing acidic character will be(a)  III > II > IV > I (b) II > III > I > IV(c)  II > III > IV > I (d) III > IV > II > IAns.aQ16. which of the following compounds has the most acidic naturea. C6H5-CH2-OHb. C6H5-OHc. C6H11-OHd. C6H11-CH(OH)-C6H11Ans. b17. Among the following four compounds(i) Phenol (ii) Methyl phenol(iii) Meta-nitrophenol (iv) Para-nitrophenol The acidity order is(a) (iv) > (iii) > (i) > (ii) (b) (iii) > (iv) > (i) > (ii)(c) (i) > (iv) > (iii) > (ii) (d) (ii) > (i) > (iii) > (iv)Ans.a18. When glycerol is treated with excess of HI, it produces(a) 2-iodopropane (b) allyl iodide(c)  propene (d) glycerol triiodide.Ans.aQ19 Consider the following reacction: Ethanol +PBr3 ------> X + alc. KOH ----> Y + (i) H2SO4 + (ii) H2O ------> Zthen the product Z is a. CH3CH2-O-CH2-CH3b. CH3-CH2-O-SO3Hc. CH3CH2-OHd. CH2=CH2Ans.cQ20. HOCH2CH2OH on heating with periodic acid givesa. 2HCOOHb. CHO-CHOc. 2 HCHOd. 2CO2Ans.cQ21. Consider the following reaction: Phenol + Zn dust ---> X + CH3Cl (anhy AlCl3) ----> Y + alk. KMNO4 ----> Zthe product Z is a. benzaldehydeb. benzoic acidc. benzened. tolueneAns. b22. Ethylene oxide when treated with Grignard reagent yields(a) primary alcohol (b) secondary alcohol(c) tertiary alcohol (d) cyclopropyl alcoholAns. aQ23. which one of the following compounds is most acidica. Cl-CH2-CH2-OHb. C6H5-OHc. o-Nitrophenold. o-CresolAns. csQ24. Which one of the following will not form a yellow precipitate on heating with an alkaline solution of iodine?(a)  CH3CH(OH)CH3 (b)  CH3CH2CH(OH)CH3(c)  CH3OH (d) CH3CH2OHAns. c25. n-Propyl alcohol and isopropyl alcohol can be chemically distinguished by which reagent?(a)  PCl5 (b) Reduction(c) Oxidation with potassium dichromate(d) OzonolysisAns. c26. When phenol is treated with CHCl3 and NaOH, the product formed is(a) benzaldehyde (b) salicylaldehyde(c) salicylic acid (d) benzoic acidAns. b27. Which of the following is correct?(a) On reduction, any aldehyde gives secondary alcohol.(b) Reaction of vegetable oil with H2SO4 gives glycerine.(c) Alcoholic iodine with NaOH gives iodoform.(d) Sucrose on reaction with NaCl gives invert sugar.Ans. cQ28 The correct acidic order of the following is I. C6H5-OH II. CH3-C6H4-OH III. O2N-C6H4-OHa. I>II>IIIb. III>I>IIc. II>III>Id. I>III>IIAns. b29. Reaction of oxirane with RMgX leads to the formation of(a)  RCH2CH2OH (b) RCHOHCH3(c) R-CH-OH-R (d)  (R)2-CHCH2OHAns.a30. When 3,3-dimethyl-2-butanol is heated with H2SO4, the major product obtained is(a) 2,3-dimethyl-2-butene(b) cis and trans isomers of2,3-dimethyl-2-butene(c) 2,3-dimethyl-1-butene(d) 3,3-dimethyl-1-butene.Ans. aQ31. The alkene R-CH=CH2 reacts readily with B2H6 and the product on oxidation with alkaline hydrogen peroxides producesa. R-(CH3)-C=Ob. R-CHOH-CH2-OHc. R-CH2-CHOd. R-CH2-CH2-OHAns. d32. On heating glycerol with conc. H2SO4, a compound is obtained which has bad odour. The compound is(a) acrolein (b) formic acid(c) allyl alcohol (d) glycerol sulphate.Ans. a33. Ethanol and dimethyl ether form a pair of functional isomers. The boiling point of ethanol is higher than that of dimethyl ether, due to the presence of(a) H-bonding in ethanol(b) H-bonding in dimethyl ether(c) CH3 group in ethanol(d) CH3  group in dimethyl ether.Ans.a34. Increasing order of acid strength among p- methoxyphenol, p-methylphenol and p-nitrophenol is(a) p-nitrophenol, p-methoxyphenol, p-methylphenol(b) p-methylphenol, p-methoxyphenol, p-nitrophenol(c) p-nitrophenol, p-methylphenol, p-methoxyphenol(d) p-methoxyphenol, p-methylphenol,p-nitrophenol.Ans. d35. Which one of the following on oxidation gives a ketone?(a)  Primary alcohol (b) Secondary alcoholc. Tertiary alcohold. AllAns.bQ36. what is formed when a primary alcohol undergoes catalytic dehydrogenation?(a) Aldehyde (b) Ketone(c) Alkene (d) AcidAns.a37. How many isomers of C5H11OH will be primary alcohols?(a) 5 (b) 4 (c)  2(d) 3Ans.bQ38 HBr reacts fastest with a. 2-methylpropan-1-olb. methylpropan-2-olc. propan-2-old. propan-1-olAns.bQ39. When phenol is treated with excess bromine water. It gives(a) m-bromophenol(b) o- and p-bromophenols(c) 2,4-dibromophenol(d) 2,4,6-tribromophenolAns.dQ40. The compound which reacts fastest with Lucas reagent at room temperature is(a) butan-1-ol (b) butan-2-ol(c)2-methylpropan-1-ol(d)2-methylpropan-2-ol.Ans. d41.Which one of the following compounds will be most readily attacked by anelectrophile?(a)Chlorobenzene (b) Benzene(c) Phenol (d) TolueneAns. c42. Propene, CH3CH CH2 can be converted into 1- propanol by oxidation. Indicate which set of reagents amongst the following is ideal for the above conversion?(a) KMnO4 (alkaline)(b) Osmium tetroxide (OsO4/CH2Cl2)(c) B2H6  and alk. H2O2   (d) O3/Zn Ans. c43. Phenol is heated with CHCl3 and aqueous KOH when salicylaldehyde is produced. This reaction is known as(a) Rosenmund’s reaction(b) Reimer-Tiemann reaction(c) Friedel-Crafts reaction(d) Sommelet reaction. Ans. b44. Lucas reagent is(a) conc. HCl and anhydrous ZnCl2(b) conc. HNO3  and hydrous ZnCl2(c) conc. HCl and hydrous ZnCl2(d) conc. HNO3  and anhydrous ZnCl2.Ans.a.Q45. Methanol is industrially prepared by a. oxidation of CH4 by steam at 900*Cb. reduction of HCHO using LiAlH4c. conc. HCl hydrous ZnCl2d. conc. HCl and anhydrous ZnCl2Ans. dQ46. Anisole on cleavage with HI givesa. C6H5-OH +CH3Ib. C6H5-I + CH3OHc. C6H5-OH + C2H5Id. C6H5-I + C2H5OHAns. aQ47. the compound that is most difficult to protonate is a. Ph-O-Hb. H-O-Hc. H3C-O-H d. H3C-O-CH3Ans. a48. The major products C and D formed in the following reactions respectively areH3C-CH2-CH2-O-C(CH3)3+ excess HI--------->C + D(a) H3C-CH2-CH2-I and I-C(CH3)3(b) H3C-CH2-CH2-OH  and IC-(CH3)3(c) H3C-CH2-CH2I and HO-C(CH3)3(d) H3C-CH2-CH2-OH and HO-C(CH3)3Ans. a49. The heating of phenyl methyl ether with HI produces(a) iodobenzene (b) phenol(c) benzene (d) ethyl chlorideAns. bQ50. The reaction ⬠-OH + NaH → ⬠-ONa + MeI  →  ⬠-OMe is calleda. dehydration reactionb. williamson alcohol synthesis reactionc. williamson ether synthesis reactiond. alcohol formation reactionAns. cQ51. The reaction(CH3)3-C-ONa+CH3CH2Cl ------> (CH3)3-C-O-CH2-CH3 is calleda. Etard reactionb. Gattermann Koch reactionc. Williamson synthesisd. Williamson continuous etherification process.Ans. c52. Among the following sets of reactants which one produces anisole?(a) CH3CHO ; RMgX(b) C6H5OH ; NaOH ; CH3I(c) C6H5OH ; neutral FeCl3(d) C6H5CH3 ; CH3COCl ; AlCl3Ans. b.Q53. Identify Z in the sequence of reactions: CH3-CH2-CH=CH2 + HBr/H2O2-----------> Y + C2H5ONa-----------> Za. CH3-CH2-CH2-CH2-O-CH2CH3b. (CH3)2CH-O-CH2CH3c. CH3(CH2)4-O-CH3d. CH3CH2-CH(CH3)-O-CH2CH3Ans. a.Q54. Among the following ethers, which one will produce methyl alcohol on treatment with hot concentrated HI?a. (CH3)3-C-O-CH3b. (CH3)2-CH-CH2-O-CH3c. CH3-CH2-CH2-O-CH3d. CH3-CH2-CH(CH3)-O-CH3Ans. aQ55. in the reaction: CH3CH(CH3) CH2-O-CH2-CH3 + HI ------------->  which of the following compounds will be formed?a. CH3-CH(CH3)2 + CH3CH2OH b. (CH3)2-CH-CH2OH + CH3CH3c. (CH3)2-CH-CH2OH + CH3CH2Id. (CH3)2CH-CH2-I + CH3CH2OHAns. cQ56. the major organic product in the reaction is CH3-O-CH(CH3)2 + HI ----------> productsa. CH3I + (CH3)2CHOHb. CH3OH + (CH3)2CHIc. ICH2oCH(CH3)2d. CH2OC(CH3)2Ans.  a57. Ethyl chloride is converted into diethyl ether by(a) Perkins reaction (b) Grignard reaction(c) Wurtz synthesis(d) Williamson’s synthesis.Ans. d. 58. Which one of the following compounds is resistant to nucleophilic attack by hydroxyl ions?(a) Diethyl ether (b) Acetonitrile(c) Acetamide (d) Methyl acetateAns. a.59.The compound which does not react with sodiumis(a)CH3COOH (b)CH3CHOHCH3 (c) C2H5OH (d)CH3OCH3 Ans. d. 60. Which one is formed when sodium phenoxide isheated with ethyl iodide?a. Phenetole(b) Ethyl phenylalcoholc. Phenol(d) None of these Ans. a.