I. Multiple Choice Questions (Type-I) 1. On addition of conc. H2SO4 to a chloride salt, colourless fumes are evolved but in case of iodide salt, violet fumes come out. This is because (i) H2SO4 reduces HI to I2 (ii) HI is of violet colour (iii) HI gets oxidised to I2 (iv) HI changes to HIO3 2. In qualitative analysis when H2S is passed through an aqueous solution of salt acidified with dil. HCl, a black precipitate is obtained. On boiling the precipitate with dil. HNO3, it forms a solution of blue colour. Addition of excess of aqueous solution of ammonia to this solution gives _________. (i) deep blue precipitate of Cu (OH)2 ]2+(ii) deep blue solution of [Cu (NH3)4(iii) deep blue solution of Cu(NO3)2 (iv) deep blue solution of Cu(OH)2.Cu(NO3)2 3. In a cyclotrimetaphosphoric acid molecule, how many single and double bonds are present? (i) 3 double bonds; 9 single bonds (ii) 6 double bonds; 6 single bonds (iii) 3 double bonds; 12 single bonds (iv) Zero double bonds; 12 single bonds 4. Which of the following elements can be involved in pπ–dπ bonding? (i) Carbon (ii) Nitrogen (iii) Phosphorus (iv) Boron 5. Which of the following pairs of ions are isoelectronic and isostructural? (i) CO32–, NO3– (ii) ClO3– , CO32– (iii) SO32–, NO3– (iv) ClO3– , SO32– 6. Affinity for hydrogen decreases in the group from fluorine to iodine. Which of the halogen acids should have highest bond dissociation enthalpy? (i) HF (ii) HCl (iii) HBr (iv) HI 7. Bond dissociation enthalpy of E—H (E = element) bonds is given below. Which of the compounds will act as strongest reducing agent? Compound NHPHAsH SbH33 33 Δdiss (E—H)/kJ mol–1 389 322 297 255 (i) NH3 (ii) PH3 (iii) AsH3 (iv) SbH3 8. On heating with concentrated NaOH solution in an inert atmosphere of CO2, white phosphorus gives a gas. Which of the following statement is incorrect about the gas? (i) It is highly poisonous and has smell like rotten fish. (ii) It’s solution in water decomposes in the presence of light. (iii) It is more basic than NH3. (iv) It is less basic than NH3. 9. Which of the following acids forms three series of salts? (i) H3PO2 (ii) H3BO3 (iii) H3PO4 (iv) H3PO3 10. Strong reducing behaviour of H3PO2 is due to (i) Low oxidation state of phosphorus (ii) Presence of two –OH groups and one P–H bond 91 p-Block Elements (iii) Presence of one –OH group and two P–H bonds (iv) High electron gain enthalpy of phosphorus 11. On heating lead nitrate forms oxides of nitrogen and lead. The oxides formed are ______. (i) N2O, PbO (ii) NO2, PbO (iii) NO, PbO (iv) NO, PbO2 12. Which of the following elements does not show allotropy? (i) Nitrogen (ii) Bismuth (iii) Antimony (iv) Arsenic 13. Maximum covalency of nitrogen is ______________. (i) 3 (ii) 5 (iii) 4 (iv) 6 14. Which of the following statements is wrong? (i) Single N–N bond is stronger than the single P–P bond. (ii) PH3 can act as a ligand in the formation of coordination compound with transition elements. (iii) NO2 is paramagnetic in nature. (iv) Covalency of nitrogen in N2O5 is four. 15. A brown ring is formed in the ring test for NO3– ion. It is due to the formation of (i) [Fe(H2O)5 (NO)]2+ (ii) FeSO4.NO2 ]2+(iii) [Fe(H2O)4(NO)2(iv) FeSO4.HNO3 16. Elements of group-15 form compounds in +5 oxidation state. However, bismuth forms only one well characterised compound in +5 oxidation state. The compound is (i) Bi2O5 (ii) BiF5 (iii) BiCl5 (iv) Bi2S5 17. On heating ammonium dichromate and barium azide separately we get (i) N2 in both cases (ii) N2 with ammonium dichromate and NO with barium azide (iii) N2O with ammonium dichromate and N2 with barium azide (iv) N2O with ammonium dichromate and NO2 with barium azide 18. In the preparation of HNO3, we get NO gas by catalytic oxidation of ammonia. The moles of NO produced by the oxidation of two moles of NH3 will be ______. (i) 2 (ii) 3 (iii) 4 (iv) 6 19. The oxidation state of central atom in the anion of compound NaH2PO2 will be ______. (i) +3 (ii) +5 (iii) +1 (iv) –3 20. Which of the following is not tetrahedral in shape? (i) NH4+ (ii) SiCl4 (iii) SF4 (iv) SO42– 21. Which of the following are peroxoacids of sulphur? (i) HSO and HSO(ii) HSO and HSO25228 25227 (iii) HSO and HSO227228 (iv) HSO and HSO226227 22. Hot conc. H2SO4 acts as moderately strong oxidising agent. It oxidises both metals and nonmetals. Which of the following element is oxidised by conc. H2SO4 into two gaseous products? (i) Cu (ii) S (iii) C (iv) Zn 23. A black compound of manganese reacts with a halogen acid to give greenish yellow gas. When excess of this gas reacts with NH3 an unstable trihalide is formed. In this process the oxidation state of nitrogen changes from _________. (i) – 3 to +3 (ii) – 3 to 0 (iii) – 3 to +5 (iv) 0 to – 3 93 p-Block Elements 24. In the preparation of compounds of Xe, Bartlett had taken O2+ Pt F6– as a base compound. This is because (i) both O2 and Xe have same size. (ii) both O2 and Xe have same electron gain enthalpy. (iii) both O2 and Xe have almost same ionisation enthalpy. (iv) both Xe and O2 are gases. 25. In solid state PCl5 is a _________. (i) covalent solid (ii) octahedral structure (iii) ionic solid with [PCl6]+ octahedral and [PCl4]– tetrahedra ]– (iv) ionic solid with [PCl4]+tetrahedral and [PCl6octahedra 26. Reduction potentials of some ions are given below. Arrange them in decreasing order of oxidising power. –– –Ion ClOIOBrO44 4 Reduction EV=1.19VE V=1.65VE V=1.74V potential EV/V (i) ClO– > IO– > BrO– 4 4 4 (ii) IO– > BrO– > ClO– 4 44 (iii) BrO– > IO– > ClO– 4 4 4 (iv) BrO– > ClO– > IO– 4 44 27. Which of the following is isoelectronic pair? (i) ICl2, ClO2 (ii) BrO2– , BrF2+ (iii) ClO2, BrF (iv) CN –, O3 II. Multiple Choice Questions (Type-II) Note : In the following questions two or more options may be correct. 28. If chlorine gas is passed through hot NaOH solution, two changes are observed in the oxidation number of chlorine during the reaction. These are ________ and _________. (i) 0 to +5 (ii) 0 to +3 (iii) 0 to –1 (iv) 0 to +1 29. Which of the following options are not in accordance with the property mentioned against them? (i) F2 > Cl2 > Br2 > I2 Oxidising power. (ii) MI > MBr > MCl > MF Ionic character of metal halide. (iii) F2 > Cl2 > Br2 > I2 Bond dissociation enthalpy. (iv) HI < HBr < HCl < HF Hydrogen-halogen bond strength. 30. Which of the following is correct for P4 molecule of white phosphorus? (i) It has 6 lone pairs of electrons. (ii) It has six P–P single bonds. (iii) It has three P–P single bonds. (iv) It has four lone pairs of electrons. 31. Which of the following statements are correct? (i) Among halogens, radius ratio between iodine and fluorine is maximum. (ii) Leaving F—F bond, all halogens have weaker X—X bond than X—X' bond in interhalogens. (iii) Among interhalogen compounds maximum number of atoms are present in iodine fluoride. (iv) Interhalogen compounds are more reactive than halogen compounds. 32. Which of the following statements are correct for SO2 gas? (i) It acts as bleaching agent in moist conditions. (ii) It’s molecule has linear geometry. (iii) It’s dilute solution is used as disinfectant. (iv) It can be prepared by the reaction of dilute H2SO4 with metal sulphide. 33. Which of the following statements are correct? (i) All the three N—O bond lengths in HNO3 are equal. (ii) All P—Cl bond lengths in PCl5 molecule in gaseous state are equal. (iii) P4 molecule in white phohsphorus have angular strain therefore white phosphorus is very reactive. (iv) PCl is ionic in solid state in which cation is tetrahedral and anion is octahedral. 34. Which of the following orders are correct as per the properties mentioned against each? (i) AsO < SiO < PO < SOAcid strength.232232 (ii) AsH3 < PH3 < NH3 Enthalpy of vapourisation. (iii) S < O < Cl < F More negative electron gain enthalpy. (iv) H2O > H2S > H2Se > H2Te Thermal stability. 35. Which of the following statements are correct? (i) S–S bond is present in H2S2O6. (ii) In peroxosulphuric acid (H2SO5) sulphur is in +6 oxidation state. (iii) Iron powder along with Al2O3 and K2O is used as a catalyst in the preparation of NH3 by Haber’s process. (iv) Change in enthalpy is positive for the preparation of SO3 by catalytic oxidation of SO2. 36. In which of the following reactions conc. H2SO4 is used as an oxidising reagent? (i) CaF + HSO⎯→ CaSO + 2HF2 24 4(ii) 2HI + HSO⎯→ I + SO + 2HO24 222(iii) Cu + 2HSO⎯→ CuSO + SO + 2HO24 422(iv) NaCl + H2SO4 ⎯→ NaHSO4 + HCl 37. Which of the following statements are true? (i) Only type of interactions between particles of noble gases are due to weak dispersion forces. (ii) Ionisation enthalpy of molecular oxygen is very close to that of xenon. (iii) Hydrolysis of XeF6 is a redox reaction. (iv) Xenon fluorides are not reactive. III. Short Answer Type 38. In the preparation of H2SO4 by Contact Process, why is SO3 not absorbed directly in water to form H2SO4? 39. Write a balanced chemical equation for the reaction showing catalytic oxidation of NH3 by atmospheric oxygen. 40. Write the structure of pyrophosphoric acid. 41. PH3 forms bubbles when passed slowly in water but NH3 dissolves. Explain why? 42. In PCl5, phosphorus is in sp3d hybridised state but all its five bonds are not equivalent. Justify your answer with reason. 43. Why is nitric oxide paramagnetic in gaseous state but the solid obtained on cooling it is diamagnetic? 44. Give reason to explain why ClF3 exists but FCl3 does not exist. 45. Out of H2O and H2S, which one has higher bond angle and why? 46. SF6 is known but SCl6 is not. Why? 47. On reaction with Cl2, phosphorus forms two types of halides ‘A’ and ‘B’. Halide A is yellowish-white powder but halide ‘B’ is colourless oily liquid. Identify A and B and write the formulas of their hydrolysis products. –48. In the ring test of NO3 ion, Fe2+ ion reduces nitrate ion to nitric oxide, which combines with Fe2+ (aq) ion to form brown complex. Write the reactions involved in the formation of brown ring. 49. Explain why the stability of oxoacids of chlorine increases in the order given below: HClO < HClO2 < HClO3 < HClO4 50. Explain why ozone is thermodynamically less stable than oxygen. 51. PO reacts with water according to equation PO + 6HO ⎯→ 4HPO.46 46233Calculate the volume of 0.1 M NaOH solution required to neutralise the acid formed by dissolving 1.1 g of P4O6 in H2O. 52. White phosphorus reacts with chlorine and the product hydrolyses in the presence of water. Calculate the mass of HCl obtained by the hydrolysis of the product formed by the reaction of 62 g of white phosphorus with chlorine in the presence of water. 53. Name three oxoacids of nitrogen. Write the disproportionation reaction of that oxoacid of nitrogen in which nitrogen is in +3 oxidation state. 54. Nitric acid forms an oxide of nitrogen on reaction with P4O10. Write the reaction involved. Also write the resonating structures of the oxide of nitrogen formed. 55. Phosphorus has three allotropic forms — (i) white phosphorus (ii) red phosphorus and (iii) black phosphorus. Write the difference between white and red phosphorus on the basis of their structure and reactivity. 56. Give an example to show the effect of concentration of nitric acid on the formation of oxidation product. 57. PCl5 reacts with finely divided silver on heating and a white silver salt is obtained, which dissolves on adding excess aqueous NH3 solution. Write the reactions involved to explain what happens. 58. Phosphorus forms a number of oxoacids. Out of these oxoacids phosphinic acid has strong reducing property. Write its structure and also write a reaction showing its reducing behaviour. IV. Matching Type Note : Match the items of Column I and Column II in the following questions. 59. Match the compounds given in Column I with the hybridisation and shape given in Column II and mark the correct option. Column I Column II (A) Xe F6 (1) sp3d3 – distorted octahedral (B) Xe O3 (2) sp3d2 - square planar (C) Xe OF4 (3) sp3 - pyramidal (D) Xe F4 (4) sp3 d2 - square pyramidal Code : (i) A(1) B (3) C (4) D(2) (ii) A(1) B (2) C (4) D(3) (iii) A(4) B (3) C (1) D(2) (iv) A(4) B (1) C (2) D(3) 60. Match the formulas of oxides given in Column I with the type of oxide given in Column II and mark the correct option. 61. Match the items of Columns I and II and mark the correct option. Column I Column II (A) Pb3O4 (1) Neutral oxide (B) N2O (2) Acidic oxide (C) Mn2O7 (3) Basic oxide (D) Bi2O3 (4) Mixed oxide Code : (i) A (1) B (2) C (3) D (4) (ii) A (4) B (1) C (2) D (3) (iii) A (3) B (2) C (4) D (1) (iv) A (4) B (3) C (1) D (2) Column I Column II (A) H2SO4 (1) Highest electron gain enthalpy (B) CCl3NO2 (2) Chalcogen (C) Cl2 (3) Tear gas (D) Sulphur (4) Storage batteries Code : (i) A(4) B (3) C (1) D(2) (ii) A(3) B (4) C (1) D(2) (iii) A(4) B (1) C (2) D(3) (iv) A(2) B (1) C (3) D(4) 62. Match the species given in Column I with the shape given in Column II and mark the correct option. Column I Column II (A) SF4 (1) Tetrahedral (B) BrF3 (2) Pyramidal –(C) BrO3 (3) Sea-saw shaped (D) NH+4 (4) Bent T-shaped Code : (i) A(3) B (2) C (1) D(4) (ii) A(3) B (4) C (2) D(1) (iii) A(1) B (2) C (3) D(4) (iv) A(1) B (4) C (3) D(2) 63. Match the items of Columns I and II and mark the correct option. Column I Column II (A) Its partial hydrolysis does not (1) He change oxidation state of central atom (B) It is used in modern diving apparatus (2) XeF6 (C) It is used to provide inert atmosphere (3) XeF4 for filling electrical bulbs (D) Its central atom is in sp 3d 2 hybridisation (4) Ar Code : (i) A(1) B (4) C (2) D(3) (ii) A(1) B (2) C (3) D(4) (iii) A(2) B (1) C (4) D(3) (iv) A(1) B (3) C (2) D(4) V. Assertion and Reason Type Note : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (i) Both assertion and reason are correct statements, and reason is the correct explanation of the assertion. (ii) Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion. (iii) Assertion is correct, but reason is wrong statement. (iv) Assertion is wrong but reason is correct statement. (v) Both assertion and reason are wrong statements. 64. Assertion : N2 is less reactive than P4. Reason : Nitrogen has more electron gain enthalpy than phosphorus. 65. Assertion : HNO3 makes iron passive. Reason : HNO3 forms a protective layer of ferric nitrate on the surface of iron. 66. Assertion : HI cannot be prepared by the reaction of KI with concentrated H2SO4 Reason : HI has lowest H–X bond strength among halogen acids. 67. Assertion : Both rhombic and monoclinic sulphur exist as S8 but oxygen exists as O2. Reason : Oxygen forms pπ – pπ multiple bond due to small size and small bond length but pπ – pπ bonding is not possible in sulphur. 68. Assertion : NaCl reacts with concentrated H2SO4 to give colourless fumes with pungent smell. But on adding MnO2 the fumes become greenish yellow. Reason : MnO2 oxidises HCl to chlorine gas which is greenish yellow. 69. Assertion : SF6 cannot be hydrolysed but SF4 can be. Reason : Six F atoms in SF6 prevent the attack of H2O on sulphur atom of SF6. VI. Long Answer Type 70. An amorphous solid “A” burns in air to form a gas “B” which turns lime water milky. The gas is also produced as a by-product during roasting of sulphide ore. This gas decolourises acidified aqueous KMnO4 solution and reduces Fe3+ to Fe2+. Identify the solid “A” and the gas “B” and write the reactions involved. 71. On heating lead (II) nitrate gives a brown gas “A”. The gas “A” on cooling changes to colourless solid “B”. Solid “B” on heating with NO changes to a blue solid ‘C’. Identify ‘A’, ‘B’ and ‘C’ and also write reactions involved and draw the structures of ‘B’ and ‘C’. 72. On heating compound (A) gives a gas (B) which is a constituent of air. This gas when treated with 3 mol of hydrogen (H2) in the presence of a catalyst gives another gas (C) which is basic in nature. Gas C on further oxidation in moist condition gives a compound (D) which is a part of acid rain. Identify compounds (A) to (D) and also give necessary equations of all the steps involved. ANSWERS I. Multiple Choice Questions (Type-I) 1. (iii) 2. (ii) 3. (i) 4. (iii) 5. (i) 6. (i) 7. (iv) 8. (iii) 9. (iii) 10. (iii) 11. (ii) 12. (i) 13. (iii) 14. (i) 15. (i) 16. (ii) 17. (i) 18. (i) 19. (iii) 20. (iii) 21. (i) 22. (iii) 23. (i) 24. (iii) 25. (iv) 26. (iii) 27. (ii) II. Multiple Choice Questions (Type-II) 28. (i), (iii) 29. (ii), (iii) 30. (ii), (iv) 31. (i), (iii), (iv) 32. (i), (iii) 33. (iii), (iv) 34. (i), (iv) 35. (i), (ii) 36. (ii), (iii) 37. (i), (ii) III. Short Answer Type 38. Acid fog is formed, which is difficult to condense. Pt/Rh gauge catalyst→39. 4NH + 5O⎯⎯⎯⎯⎯⎯⎯⎯⎯ 4NO + 6HO32 500K, 9 bar 2 (From air) 40. Pyrophosphoric acid 41. NH3 forms hydrogen bonds with water therefore it is soluble in it but PH3 cannot form hydrogen bond with water so it escapes as gas. 42. [Hint : It has trigonal bipyramidal geometry] 43. In gaseous state NO2 exists as monomer which has one unpaired electron but in solid state it dimerises to N2O4 so no unpaired electron is left hence solid form is diamagnetic. 44. Because fluorine is more electronegative as compared to chlorine. 45. Bond angle of H2O is larger, because oxygen is more electronegative than sulphur therefore bond pair electron of O–H bond will be closer to oxygen and there will be more bond-pair bond-pair repulsion between bond pairs of two O–H bonds. 46. Due to small size of fluorine six F – ion can be accomodated around sulphur whereas chloride ion is comparatively larger in size, therefore, there will be interionic repulsion. 47. 48. 49. 50. 51. 52. A is PCl5 (It is yellowish white powder) P4 + 10Cl2 ⎯→ 4PCl5 B is PCl3 (It is a colourless oily liquid) P4 + 6Cl2 ⎯→ 4PCl3 Hydrolysis products are formed as follows : PCl3 + 3H2O ⎯→ H3PO3+3HCl PCl5 + 4H2O ⎯→ H3PO4 + 5HCl –NO3 + 3Fe2+ + 4H+ ⎯→ NO + 3Fe3+ + 2H2O [Fe(HO)]2+ + NO ⎯→ [Fe(HO)(NO)]2+ + HO26252 (brown complex) Oxygen is more electronegative than chlorine, therefore dispersal of negative charge present on chlorine increases from ClO– to ClO4– ion because number of oxygen atoms attached to chlorine is increasing. Therefore, stability of ions will increase in the order given below : ClO– < ClO– < ClO– < ClO– 234 Thus due to increase in stability of conjugate base, acidic strength of corresponding acid increases in the following order HClO < HClO2 < HClO3 < HClO4 See the NCERT textbook for Class XII, page 186. PO + 6HO ⎯→ 4HPO46233 HPO + 2NaOH ⎯→ Na HPO + 2HO] × 4 (Neutralisation reaction)33232PO + 8NaOH ⎯→ 4Na HPO + 2HO462421 mol 8 mol Product formed by 1 mol of P4O6 is neutralised by 8 mols of NaOH 1.1 1.1 ∴ Product formed by mol of P4O6 will be neutralised by ×8 mol220220 of NaOH Molarity of NaOH solution is 0.1M ⇒ 0.1 mol NaOH is present in 1 L solution 1.1 1.1 × 8 884 ∴ ×8 mol NaOH is present in L = L = L = 0.4 L =220 220 ×0.1 22010400 mL of NaOH solution. P4 + 6Cl2 ⎯→ 4PCl3 PCl3 + 3H2O ⎯→ H3PO3 + 3HCl] × 4 P + 6Cl + 12HO ⎯→ 4HPO + 12HCl422331 mol of white phosphorus produces 12 mol of HCl 62 1 62g of white phosphorus has been taken which is equivalent to = mol.124 2 Therefore 6 mol HCl will be formed. Mass of 6 mol HCl = 6 × 36.5 = 219.0 g HCl 53. Three oxoacids of nitrogen are (i) HNO2, Nitrous acid (ii) HNO3 , Nitric acid (iii) Hyponitrous acid, H2N2O2 3HNO2 ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯→Disproportionation HNO3 + H2O + 2NO 54. 4HNO3 + P4O10 ⎯→ 4HPO3 + 2N2O5 55. (a) • Structures (See NCERT textbook for Class XII) • White phosphorus is discrete tetrahedral molecule. Thus it has tetrahedral structure with six P–P bonds. • Red phosphorus has polymeric structure in which P4 tetrahedra are linked together through P—P bonds to form chain. (b) Reactivity White phosphorus is much more reactive than red phosphorus. This is because in white phosphorus there is angular strain in P4 molecules because the bond angles are only of 60°. 56. Dilute and concentrated nitric acid give different oxidation products on reaction with copper metal. 3Cu + 8HNO3 (dil.) ⎯→ 3Cu(NO3)2 + 2NO + 4H2O Cu + 4HNO3 (Conc.) ⎯→ 3Cu(NO3)2 + 2NO + 2H2O 57. PCl5 + 2Ag ⎯→ 2AgCl + PCl3 AgCl + 2NH3(aq) ⎯→ [Ag(NH3)2]+Cl– (soluble complex) 58. Structure of phosphinic acid (Hypophosphorous acid) is as follows : Reducing behaviour of phosphinic acid is observable in the reaction with silver nitrate given below : 4AgNO + 2HO + HPO⎯→ 4Ag + 4HNO + HPO3 232 334 IV. Matching Type 59. (i) 60. (ii) 61. (i) 62. (ii) 63. (iii) V. Assertion and Reason Type 64. (iii) 65. (iii) 66. (ii) 67. (i) 68. (i) 69. (i) VI. Long Answer Type 70. ‘A’ is S8 ‘B’ is SO2 gas S + 8O→ 8SO82 ⎯⎯⎯Δ 2 2MnO– + 5SO + 2HO ⎯→ 5 SO2– + 4H+ + 2Mn2+ 4224 (violet) (colourless) 2Fe3+ + SO2 + 2H2O ⎯→ 2Fe2+ + SO42– + 4H+ Δ 71. Pb(NO)2PbO + 4NO32 673K 2 (A) (Brown colour) On cooling2NO� NO2 Heating24 (B) (Colourless solid) ⎯⎯⎯⎯⎯⎯Δ 250 K2NO + NO→ 2 NO24 23 (C) (Blue solid) (Structure of N2O4) (Structure of N2O3) 72. A = NH NOB = NC = NHD = HNO4 223 3 (i) NH NO→ N + 2HO4 222(ii) N2 + 3H2 → 2NH3 (iii) 4NH3 + 5O2 → 4NO + 6H2O 4NO + O2 → 2NO2 3NO2 + H2O → 2HNO3 + NO

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