Chapter 16 p-Block Elements

76 JEE MAIN CHAPTERWISE SOLUTIONS CHAPTER p-Block Elements 1. 2. 3. The compound that does not produce nitrogen gas by the thermal decomposition is (a) Ba(N3)2 (b) (NH4)2Cr2O7 (c) NH4NO2 (d) (NH4)2SO 4 When metal ‘M’ is treated with NaOH, a white gelatinous precipitate ‘X’ is obtained, which is soluble in excess of NaOH. Compound ‘X’ when heated strongly gives an oxide which is used in chromatography as an adsorbent. Then metal ‘M’ is (a) Zn (b) Ca (c) Al (d) Fe Xenon hexafluoride on partial hydrolysis produces compounds ‘X’ and ‘Y’. Compounds ‘X’ and ‘Y’ and the oxidation state of Xe are respectively (a) XeO2F2 (+6) and XeO2 (+4) (b) XeOF4 (+6) and XeO2F2 (+6) (c) XeOF4 (+6) and XeO3 (+6) (d) XeO2 (+4) and XeO3 (+6). 4. In graphite and diamond, the percentage of p-characters of the hybrid orbitals in hybridisation are respectively (a) 33 and 75 (b) 50 and 75 (c) 33 and 25 (d) 67 and 75 5. The number of P — O bonds in P4O6 is (a) 18 (b) 12 (c) 9 (d) 6 6. In XeO3F2, the number of bond pair(s), -bond(s) and lone pair(s) on Xe atom respectively are (a) 4, 2, 2 (b) 4, 4, 0 (c) 5, 2, 0 (d) 5, 3, 0 7. Lithium aluminium hydride reacts with silicon tetrachloride to form (a) LiCl, AlCl3 and SiH4 (b) LiCl, AlH3 and SiH4 (c) LiH, AlCl3 and SiCl2 (d) LiH, AlH3 and SiH4 (a) N2O3 and N2O5 (c) N2O3 and N2O4 (b) N2O4 and N2O5 (d) only N 2O 5 10. A group 13 element ‘X’ reacts with chlorine gas to produce a compound XCl 3. XCl 3 is electron deficient and easily reacts with NH3 to form Cl3X  NH3 adduct; however, XCl 3 does not dimerize. X is (a) Ga (c) In (b) Al (d) B 11. The products obtained when chlorine gas reacts with cold and dilute aqueous NaOH are – (a) Cl– and ClO– (b) Cl– and ClO2 – – – – (c) ClO and ClO3 (d) ClO2 and ClO3 12. Consider the following standard electrode potentials (E° in volts) in aqueous solution, Element M 3+ / M M+ / M Al 1.66 0.55 Tl 1.26 0.34 Based on these data, which of the following statements is correct? (a) Tl+ is more stable than Al+. (b) Tl3+ is more stable than Al3+. (c) Al+ is more stable than Al3+. (d) Tl + is more stable than Al3+. 13. A metal ‘M’ reacts with nitrogen gas to afford ‘M3N’. ‘M3N’ on heating at high temperature gives back ‘M’ and on reaction with water produces a gas ‘B’. Gas ‘B’ reacts with aqueous solution of CuSO4 to form deep blue compound. ‘M’ and ‘B’ respectively are (a) Li and NH3 (b) Al and N2 (c) Ba and N2 (d) Na and NH3 8. The correct order of electron affinity is (a) Cl > F > O (b) F > O > Cl (c) F > Cl > O (d) O > F > Cl 14. The number of S O and S — OH bonds present in peroxodisulphuric acid and pyrosulphuric acid respectively are (a) (4 and 2) and (4 and 2) (b) (2 and 4) and (2 and 4) (c) (4 and 2) and (2 and 4) (d) (2 and 2) and (2 and 2). 9. Among the oxides of nitrogen : N2O3, N2O4 and N2O5; the molecule(s) having nitrogen-nitrogen bond is/are 15. Which one of the following is an oxide? (a) BaO2 (b) SiO2 (c) KO2 (d) CsO 2 16. The correct sequence of decreasing number of -bonds in the structures of H2SO3, H2SO4 and H2S2O7 is (a) H2SO3 > H2SO4 > H2S2O7 (b) H2SO4 > H2S2O7 > H2SO3 (c) H2S2O7 > H2SO3 > H2SO4 (d) H2S2O7 > H2SO4 > H2SO3 17. XeF6 on partial hydrolysis with water produces a compound ‘X’. The same compound ‘X’ is formed when XeF6 reacts with silica. The compound ‘X’ is (a) XeO3 (b) XeF4 (c) XeF2 (d) XeOF4 18. The number of P — OH bonds and the oxidation state of phosphorus atom in pyrophosphoric acid (H 4 P 2 O 7 ) respectively are (a) five and four (b) four and five (c) four and four (d) five and five. 19. The reaction of zinc with dilute and concentrated nitric acid, respectively produces (a) N2O and NO2 (b) NO2 and NO (c) NO and N2O (d) NO2 and N2O 20. The pair in which phosphorus atoms have a formal oxidation state of +3 is (a) orthophosphorous and pyrophosphorous acids (b) pyrophosphorous and hypophosphoric acids (c) orthophosphorous and hypophosphoric acids (d) pyrophosphorous and pyrophosphoric acids. 21. The non-metal that does not exhibit positive oxidation state is (a) chlorine (b) iodine (c) fluorine (d) oxygen 22. Match the items in column I with its main use listed in column II. Column I Column II (A) Silica gel (i) Transistor (B) Silicon (ii) Ion-exchanger (C) Silicone (iii) Drying agent (D) Silicate (iv) Sealant (a) (A) – (iii), (B) – (i), (C) – (iv), (D) – (ii) (b) (A) – (iv), (B) – (i), (C) – (ii), (D) – (iii) (c) (A) – (ii), (B) – (i), (C) – (iv), (D) – (iii) (d) (A) – (ii), (B) – (iv), (C) – (i), (D) – (iii) 23. Identify the incorrect statement. (a) The S – S –S bond angles in the S8 and S6 rings are the same. (b) Rhombic and monoclinic sulphur have S8 molecules. (c) S2 is paramagnetic like oxygen. (d) S8 ring has a crown shape. 24. Assertion : Among the carbon allotropes, diamond is an insulator, whereas, graphite is a good conductor of electricity. Reason : Hybridization of carbon in diamond and graphite are sp3 and sp2, respectively. (a) Both assertion and reason are correct, but the reason is not the correct explanation for the assertion. (b) Both assertion and reason are correct, and the reason is the correct explanation for the assertion. (c) Both assertion and reason are incorrect. (d) Assertion is incorrect statement, but the reason is correct. 25. In the context of the Hall–Heroult process for the extraction of Al, which of the following statements is false? (a) Al3+ is reduced at the cathode to form Al. (b) Na 3AlF6 serves as the electrolyte. (c) CO and CO2 are produced in this process. (d) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity. 26. Which among the following is the most reactive? (a) I2 (b) ICl (c) Cl2 (d) Br2 27 . Assertion : Nitrogen and oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen. Reason : The reaction between nitrogen and oxygen requires high temperature. (a) The assertion is incorrect, but the reason is correct. (b) Both the assertion and reason are incorrect. (c) Both assertion and reason are correct and the reason is the correct explanation for the assertion. (d) Both assertion and reason are correct, but the reason is not the correct explanation for the assertion. 28. The least number of oxyacids are formed by (a) nitrogen (b) sulphur (c) fluorine (d) chlorine. 29. Choose the incorrect formula out of the four compounds for an element X given below. (a) X2Cl3 (b) X2O3 (c) X2(SO4)3 (d) XPO4 30. Which of the following compounds has a P – P bond? (a) H4P2O5 (b) H4P2O6 (c) H4P2O7 (d) (HPO3)3 31. Chlorine water on standing loses its colour and forms (a) HCl only (b) HOCl and HOCl2 (c) HCl and HOCl (d) HCl and HClO2 32. Among the following oxoacids, the correct decreasing order of acid strength is (a) HClO2 > HClO4 > HClO3 > HOCl (b) HOCl > HClO2 > HClO3 > HClO4 (c) HClO4 > HOCl > HClO2 > HClO3 (d) HClO4 > HClO3 > HClO2 > HOCl 33. Which of the following exists as covalent crystals in the solid state? (a) Phosphorus (b) Iodine (c) Silicon (d) Sulphur 34. Which of the following is the wrong statement? (a) Ozone is diamagnetic gas. (b) ONCl and ONO– are not isoelectronic. (c) O3 molecule is bent. (d) Ozone is violet-black in solid state. 35. Boron cannot form which one of the following anions? 3– – (a) BF6 (b) BH4 – – (c) B(OH)4 (d) BO2 36. Which of the following statement is wrong? (a) The stability of hydrides increases from NH3 to BiH3 in group 15 of the periodic table. (b) Nitrogen cannot form d-p bond. (c) Single N — N bond is weaker than the single P – P bond. (d) N2O4 has two resonance structure. 37. Which of the following statements regarding sulphur is incorrect? (a) S2 molecule is paramagnetic. (b) The vapour at 200°C consists mostly of S 8 rings. (c) At 600°C the gas mainly consists of S2 molecules. (d) The oxidation state of sulphur is never less than +4 in its compounds. 38. The bond dissociation energy of B – F in BF3 is 646 kJ mol–1 whereas that of C – F in CF4 is 515 kJ mol –1. The correct reason for higher B – F bond dissociation energy as compared to that of C – F is (a) smaller size of B-atom as compared to that of C-atom (b) stronger  bond between B and F in BF3 as compared to that between C and F in CF4 (c) significant p-p interaction between B and F in BF3 whereas there is no possibility of such interaction between C and F in CF4. (d) lower degree of p-p interaction between B and F in BF3 than that between C and F in CF4. 39. Which one of the following reactions of xenon compounds is not feasible? (a) XeO3 + 6HF  XeF6 + 3H2O (b) 3XeF4 + 6H2O  2Xe + XeO3 + 12HF + 1.5O2 (c) 2XeF2 + 2H2O  2Xe + 4HF + O2 (d) XeF6  RbF  Rb[XeF7] 40. In which of the following arrangements, the sequence is not strictly according to the property written against it? (a) CO2 < SiO2 < SnO2 < PbO2 : increasing oxidising power (b) HF < HCl < HBr < HI : increasing acid strength (c) NH3 < PH3 < AsH3 < SbH3 :increasing basic strength (d) B < C < O < N : increasing first ionization enthalpy 41. Among the following substituted silanes the one which will give rise to cross linked silicone polymer on hydrolysis is (a) R3SiCl (b) R4Si (c) RSiCl3 (d) R2SiCl2 42. Which one of the following is the correct statement? (a) B2H6·2NH3 is known as ‘inorganic benzene’. (b) Boric acid is a protonic acid. (c) Beryllium exhibits coordination number of six. (d) Chlorides of both beryllium and aluminium have bridged chloride structures in solid phase. 43. The stability of dihalides of Si, Ge, Sn and Pb increases steadily in the sequence (a) PbX2  SnX2  GeX2  SiX2 (b) GeX2  SiX2  SnX2  PbX2 (c) SiX2  GeX2  PbX2  SnX2 (d) SiX2  GeX2  SnX2  PbX2. 44. Identify the incorrect statement among the following. (a) Br2 reacts with hot and strong NaOH solution to give NaBr and H2O. (b) Ozone reacts with SO2 to give SO3. (c) Silicon reacts with NaOH(aq) in the presence of air to give Na2SiO3 and H2O. (d) Cl2 reacts with excess of NH3 to give N2 and HCl. 45. Regular use of the following fertilizers increases the acidity of soil? (a) Ammonium sulphate (b) Potassium nitrate (c) Urea (d) Superphosphate of lime 46. A metal, M forms chlorides in +2 and +4 oxidation states. Which of the following statements about these chlorides is correct? (a) MCl2 is more volatile than. MCl4. (b) MCl2 is more soluble in anhydrous ethanol than MCl4. (c) MCl2 is more ionic than MCl4. (d) MCl2 is more easily hydrolysed than MCl4. 47. What products are expected from the disproportionation reaction of hypochlorous acid? (a) HClO3 and Cl2O (b) HClO2 and HClO4 (c) HCl and Cl2O (d) HCl and HClO3 48. Which of the following statements is true? (a) H3PO3 is a stronger acid than H2SO3. (b) In aqueous medium HF is a stronger acid than HCl. (c) HClO4 is a weaker acid than HClO3. (d) HNO3 is a stronger acid than HNO2. 49. Heating an aqueous solution of aluminium chloride to dryness will give (a) AlCl3 (b) Al2Cl6 (c) Al2O3 (d) Al(OH)Cl2 50. The number and type of bonds between two carbon atoms in calcium carbide are (a) one sigma, one pi (b) one sigma, two pi (c) two sigma, one pi (d) two sigma, two pi. (a) (b) (c) (d) 51. The structure of diborane (B2H6) contains (a) four 2c-2e bonds and two 3c-2e bonds (b) two 2c-2e bonds and four 3c-2e bonds (c) two 2c-2e bonds and two 3c-3e bonds (d) four 2c-2e bonds and four 3c-2e bonds. 52. The molecular shapes of SF4, CF4 and XeF4 are (a) the same with 2, 0 and 1 lone pairs of electrons central atom respectively (b) the same with 1, 1 and 1 lone pair of electrons central atoms respectively (c) different with 0, 1 and 2 lone pairs of electrons central atom respectively (d) different with 1, 0 and 2 lone pairs of electrons central atom respectively. on the on the on the on the 53. The number of hydrogen atom(s) attached to phosphorus atom in hypophosphorous acid is (a) zero (b) two (c) one (d) three. 54. The correct order of the thermal stability of hydrogen halides (H – X) is (a) HI > HBr > HCl > HF (b) HF > HCl > HBr > HI (c) HCl < HF > HBr < HI (d) HI > HCl < HF > HBr 55. In silicon dioxide (a) each silicon atom is surrounded by four oxygen atoms and each oxygen atom is bonded to two silicon atoms (b) each silicon atom is surrounded by two oxygen atoms and each oxygen atom is bonded to two silicon atoms (c) silicon atom is bonded to two oxygen atoms (d) there are double bonds between silicon and oxygen atoms. 56. Which of the following oxides is amphoteric in character? (a) CaO (b) CO2 (c) SiO2 (d) SnO2 57. The soldiers of Napolean army while at Alps during freezing winter suffered a serious problem as regards to the tin buttons of their uniforms. White metallic tin buttons got converted to grey powder. This transformation is related to (a) an interaction with nitrogen of the air at very low temperatures (b) a change in the crystalline structure of tin (c) a change in the partial pressure of oxygen in the air (d) an interaction with water vapour contained in the humid air. 58. Aluminium chloride exists as dimer, Al2Cl6 in solid state as well as in solution of non-polar solvents such as benzene. When dissolved in water, it gives (a) Al3+ + 3Cl– (b) [Al(H2O)6]3+ + 3Cl– 3– (c) [Al(OH)6] + 3HCl (d) Al2O3 + 6HCl. 59. Among Al2O3, SiO2, P2O3 and SO2 the correct order of acid strength is SO2 < P2O3 < SiO2 < Al2O3 SiO2 < SO2 < Al2O3 < P2O3 Al2O3 < SiO2 < SO2 < P2O3 Al2O3 < SiO2 < P2O3 < SO2. 60. The states of hybridisation of boron and oxygen atoms in boric acid (H3BO3) are respectively (a) sp2 and sp2 (b) sp2 and sp3 3 2 (c) sp and sp (d) sp3 and sp3. 61. Which one of the following statements regarding helium is incorrect? (a) It is used to fill gas in balloons instead of hydrogen because it is lighter and non-inflammable. (b) It is used as a cryogenic agent for carrying out experiments at low temperatures. (c) It is used to produce and sustain powerful superconducting magnets. (d) It is used in gas-cooled nuclear reactors. 62. Glass is a (a) micro-crystalline solid (b) super-cooled liquid (c) gel (d) polymeric mixture. 63. Graphite is a soft solid lubricant extremely difficult to melt. The reason for this anomalous behaviour is that graphite (a) is a non-crystalline substance (b) is an allotropic form of diamond (c) has molecules of variable molecular masses like polymers (d) has carbon atoms arranged in large plates of rings of strongly bound carbon atoms with weak interplate bonds. 64. Which one of the following pairs of molecules will have permanent dipole moments for both members? (a) SiF4 and NO2 (b) NO2 and CO2 (c) NO2 and O3 (d) SiF4 and CO2 65. Which one of the following substances has the highest proton affinity? (a) H2O (b) H2S (c) NH3 (d) PH3 66. Which one of the following is an amphoteric oxide? (a) ZnO (b) Na2O (c) SO2 (d) B2O3 67. Concentrated hydrochloric acid when kept in open air sometimes produces a cloud of white fumes. The explanation for it is that (a) concentrated hydrochloric acid emits strongly smelling HCl gas all the time (b) oxygen in air reacts with the emitted HCl gas to form a cloud of chlorine gas (c) strong affinity of HCl gas for moisture in air results in forming of droplets of liquid solution which appears like a cloudy smoke (d) due to strong affinity for water, concentrated hydrochloric acid pulls moisture of air towards itself. This moisture forms droplets of water and hence the cloud. 68. What may be expected to happen when phosphine gas is mixed with chlorine gas? (a) The mixture only cools down (b) PCl3 and HCl are formed and the mixture warms up (c) PCl5 and HCl are formed and the mixture cools down (d) PH3·Cl2 is formed with warming up. 69. Which one of the following statements is correct? (a) Manganese salts give a violet borax test in the reducing flame. (b) From a mixed precipitate of AgCl and AgI, ammonia solution dissolves only AgCl. (c) Ferric ions give a deep green precipitate on adding potassium ferrocyanide solution. (d) On boiling a solution having K+, Ca2+ and HCO3– ions we get a precipitate of K2Ca(CO3)2. 70. Alum helps in purifying water by (a) forming Si complex with clay particles (b) sulphate part which combines with the dirt and removes it (c) coagulating the mud particles (d) making mud water soluble. 71. In case of nitrogen, NCl3 is possible but not NCl5 while in case of phosphorus, PCl3 as well as PCl5 are possible. It is due to (a) availability of vacant d orbitals in P but not in N (b) lower electronegativity of P than N (c) lower tendency of H-bond formation in P than N (d) occurrence of P in solid while N in gaseous state at room temperature. 72. In XeF2, XeF4, XeF6 the number of lone pairs on Xe are respectively (a) 2, 3, 1 (b) 1, 2, 3 (c) 4, 1, 2 (d) 3, 2, 1. 73. Which of the following statements is true? (a) HF is less polar than HBr. (b) Absolutely pure water does not contain any ions. (c) Chemical bond formation takes place when forces of attraction overcome the forces of repulsion. (d) In covalency transference of electron takes place. 74. When H2S is passed through Hg2S we get (a) HgS (b) HgS + Hg2S (c) Hg2S + Hg (d) Hg2S. ANSWER KEY 1. 13. 25. 37. 49. 61. 73. (d) (a) (b) (d) (b) (a) (c) 2. 14. 26. 38. 50. 62. 74. (c) (a) (b) (c) (b) (b) (c) 3. 15. 27. 39. 51. 63. (b) (b) (c) (a) (a) (d) 4. 16. 28. 40. 52. 64. (d) (d) (c) (c) (d) (c) 5. 17. 29. 41. 53. 65. (b) (d) (a) (c) (b) (c) 6. 18. 30. 42. 54. 66. (d) (b) (b) (d) (b) (a) 7 19. 31. 43. 55. 67. (a) (a) (c) (c) (a) (c) 8. 20. 32. 44. 56. 68. (a) (a) (d) (d) (d) (c) 9. 21. 33. 45. 57. 69. (c) (c) (c) (a) (b) (b) 10. 22. 34. 46. 58. 70. (d) (a) (None) (c) (b) (c) 11. 23. 35. 47. 59. 71. (a) (a) (a) (d) (d) (a) 12. 24. 36. 48. 60. 72. (a) (a) (a) (d) (b) (d) NH4NO2 (NH4)2SO4 2NH3 + H2SO4 NaOH (excess) (X) White ppt. Na[Al(OH)4] Sodium tetrahydroxoaluminate (III) (Soluble) 3. (b) : Partial hydrolysis of XeF6 gives XeOF4(X) and XeO2F2(Y) : XeF6 + H2O  XeOF4 + 2HF O.. .. O .. (N2O 4) O .. – .. O .. .. N O .. + .. – O .. .. O– .. .. .. .. + N—N .. (N 2O3) .. + + N O .. .. .. N2 + 2H2O 2. (c) : Metal M could be Al as Al(OH)3 is soluble in excess sodium hydroxide to form hydroxy aluminate ions. Al2O3 (oxide of metal M) is used as adsorbent in chromatography. (M) N—N .. O .. .. –O .. – O + .. N2 + 4H2O + Cr2O3 Al + NaOH  Al(OH)3 + 3Na+ .. .O . .. (NH4)2Cr2O7 .. Ba + 3N2 .. 1. (d) : Ba(N3)2 (N2O5) Thus, N2O3 and N2O4 contain one N — N bond. 10. (d) : Boron (X ) reacts with chlorine gas to produce boron trichloride, BCl 3. 2B + 3Cl2  2BCl 3 (X ) (XCl 3) BCl3 is electron deficient, i.e., it is a Lewis acid. Thus, it accepts an electron pair from ammonia to form an adduct. + BCl3  [H3N  BCl3] H3N: Donor (Lewis base) Acceptor (Lewis acid) (Adduct) BCl 3 does not dimerise due to p-p back bonding. 11. (a) : When Cl2 gas reacts with cold and dilute aqueous NaOH, chloride (Cl –) and hypochlorite (ClO–) ions are formed. (X ) XeF6 + 2H2O  XeO2F2 + 4HF (Y ) Let, oxidation states of Xe in XeOF4 and XeO2F2 be a and b respectively. XeOF4 : a + (–2) + 4 × (–1) = 0  a = + 6 XeO2F2 : b + 2 × (–2) + 2 × (–1) = 0  b = + 6 4. (d) : The hybridisation of carbon in graphite is sp2. 2  100 = 67% 3 The hybridisation of carbon in diamond is sp3. % p-character 12. (a) : Al3+ is more stable than Tl3+ because of negative E°Al3+/Al value. Tl+ is more stable than Al+ because of negative E° Tl+/Tl value. 13. (a) : 6Li + N2 6LiOH + 2NH3 O O O P P O Total P—O bonds = 12 F 6. (d) : Structure of XeO3F2 is O ''M [Cu(NH3)4]SO4 (deep blue coloured complex) O P 6Li + N2 6H2O CuSO4(aq) P 5. (b) : Structure of P4O6 is  ‘ B’ 3  % of p-character  100 = 75% 4 O 2Li3N ''M Xe F 14. (a) : Peroxodisulphuric acid (H2S2O8) : O O HO — S — O — O — S — OH; O O No. of S O bonds = 4, No. of S — OH bonds = 2 O O O No. of bond pairs = 5, No. of -bonds = 3, No. of lone pairs = 0 7. (a) : Lithium aluminium hydride reacts with silicon tetrachloride to form silane. LiAlH4 + SiCl4  LiCl + AlCl3 + SiH4 Silane 8. (a) : Electron affinity order : Cl > F > O Due to strong electron-electron repulsions, the electron affinity of fluorine is less than chlorine. 9. (c) : Structures of oxides of nitrogen are : O Pyrosulphuric acid (H2S2O7) : HO — S — O — S — OH; O O No. of S O bonds = 4, No. of S — OH bonds = 2 15. (b) : SiO2 is an oxide while KO2 , CsO2 are superoxides and BaO2 is a peroxide. O O 16. (d) : H2S2O7: HO — S — O — S — OH O O O .. H2SO4: HO — S — OH; H2SO3: HO — S — OH O O 17. (d) : XeF6 + H2O 2XeF6 + SiO2 O 33. (c) 34. (None) : All the statements are correct. 35. (a) : Due to non-availability of d-orbitals, boron is unable to expand its octet. Therefore, the maximum covalency of boron cannot exceed 4. 36. (a) : Thermal stability decreases gradually from NH3 to BiH3. So the stability also decreases. NH3 PH3 AsH 3 SbH 3 BiH3 XeOF4 + 2HF (X) 2XeOF4 + SiF4 (X) O P—O —P 18. (b) : HO OH OH OH (H 4P 2O 7) Oxidation state of P : 2x + 4 (+1) + 7 (–2) = 0  x = + 5 In this structure, four P—OH groups are present. 19. (a) : 4Zn + 10HNO3(dil.)  4Zn(NO3)2 + N2O + 5H2O Zn + 4HNO3(conc.)  Zn(NO3)2 + 2NO2 + 2H2O 20. (a) : Name Orthophosphorous acid Pyrophosphorous acid Hypophosphoric acid Pyrophosphoric acid Formula H3PO3 H4P2O5 H4P2O6 H4P2O7 Oxidation state +3 +3 +4 +5 21. (c) : Fluorine is the most electronegative element and it shows only –1 oxidation state. 22. (a) : Silica gel absorbs moisture thus, acts as drying agent. Silicon being a semiconductor is used in transistors. Silicone is a sealant while silicates including Zeolites are ion-exchangers. 23. (a) : Sulphur have puckered S8 rings with crown conformation and Engel’s sulphur contains S 6 rings arranged in a chair conformation. Decomposition 1300°C temperature 280°C 150°C Cl R OH R – Si – OH OH R R (ii) HO – Si – OH + H O – Si – OH + H O – Si – OH OH OH OH R –H2O R R R HO – Si – O – Si – O – Si – OH R OH R OH OH (iii) HO – Si – O – Si – O – Si – OH OH HO OH HO OH HO –3H2O HO – Si – O – Si – O – Si – OH R R R R 31. (c) : Cl 2 + H2O R R – O – Si – O – Si – O – Si – O – O 30. (b) : H4P2O6 : room temp. The size of the central atom increases from N to Bi therefore, the tendency to form a stable covalent bond with small atom like hydrogen decreases and therefore, stability decreases. 37. (d) : Sulphur exhibits –2, +2, +4, +6 oxidation states but +4 and +6 are more common. 38. (c) : In BF3, B is sp2 hybridised and has a vacant 2p-orbital which overlaps laterally with a filled 2p-orbital of F forming strong p-p bond. However in CF4, C does not have any vacant p-orbitals to undergo -bonding. Thus B.E.B – F > B.E.C – F. 39. (a) : The reaction is not feasible because XeF6 formed will further produce XeO 3 by getting hydrolysed. XeF6 + 3H2O XeO3 + 3H2F2. 40. (c) : In group15 hydrides, the basic character decreases on going down the group due to decrease in the availability of the lone pair of electrons because of the increase in size of elements from N to Bi. Thus, correct order of basicity is NH3 > PH3 > AsH3 > SbH3. 41. (c) : RSiCl3 on hydrolysis gives a cross linked silicone. The formation can be explained in three steps : Cl 3H2O (i) R – Si – Cl –3HCl 24. (a) : Diamond is a bad conductor of electricity because of the non-availability of free electrons while graphite is a good conductor of electricity due to the fourth valence electron of each carbon which is free to move. 25. (b) : In Hall – Heroult process for the extraction of Al, electrolyte is Al2O3 dissolved in Na3AlF6 containing a little of CaF2. 26. (b) : Interhalogen compounds are generally more reactive than the halogens (except F2). 27. (c) : Dinitrogen and dioxygen combine to form nitric oxide when the mixture is heated to 2273 – 3273 K in an electric arc. 28. (c) : Fluorine is the most electronegative element and has least tendency to form double bonds. 29. (a) : X2O3, X2(SO4)3 and XPO4 suggest the +3 oxidation state of X. Hence, the formula of its chloride is XCl 3. 440°C O O – O – Si – O – Si – O – Si – O – HCl Hydrochloric acid + HOCl Hypochlorous acid 32. (d) : No. of O-atoms (i.e. oxidation state)  Acidic strength Hence, the decreasing order of acidic strength will be HClO4 > HClO3 > HClO2 > HOCl R R R Cross linked silicone 42. (d) : Boric acid is a weak monobasic acid (Ka = 1.0 × 10–9). It is a notable part that boric acid does not act as a protonic acid (i.e., proton donor) but behaves as a Lewis acid by accepting a pair of electrons from OH– ions. B(OH)3 + 2H2O  [B(OH)4]– + H3O+ BeCl2 like Al2Cl 6 has a bridged polymeric structure in solid phase generally as shown below. Cl Cl – Be Be – Cl Cl Cl Polymeric structure of BeCl2 Cl Cl Al Al Cl Cl Cl Polymeric structure of Al2Cl6 Beryllium exhibits coordination number of four as it has only our available orbitals in its valency shell. Also, 3B2H6 + 6NH 3 3[BH2(NH3)2]+BH4– or B2H6 · 2NH3 or B2H6 · 2NH3 Heat, 450 K 2B3N3H6 + 12H2 Borazine Borazine has structure similar to benzene and therefore, it is called inorganic benzene. Hence option (d) is correct. 43. (c) : Due to the inert pair effect (the reluctance of ns2 electrons of outermost shell to participate in bonding) the stability of M2+ ions (of group IV elements) increases as we go down the group. 44. (d) : 3Br2 + 6NaOH  5NaBr + NaBrO3 + 3H2O O3 + SO2  O2 + SO3 Si + 2NaOH + O2  Na2SiO3 + H2O Cl2 reacts with excess of ammonia to produce ammonium chloride and nitrogen. 2NH3 + 3Cl2 N2 + 6HCl 6NH3 + 6HCl 6NH4Cl 8NH3 + 3Cl2 N2 + 6NH4Cl 45. (a) : (NH4)2SO4 + 2H2O 48. (d) : Higher is the oxidation state of the central atom, greater is the acidity. Hence, HClO4 is a stronger acid than HClO3. HNO3 is a stronger acid than HNO2. Now, greater is the electronegativity and higher is the oxidation state of the central atom, greater is the acidity. Hence H2SO3 is a stronger acid than H3PO3. Due to higher dissociation energy of H – F bond and molecular association due to hydrogen bonding in HF, HF is a weaker acid than HCl. 49 . (b) : Aluminium chloride in aqueous solution exists as ion pair. 2AlCl3 + aq.  [AlCl2(H2O)4]+(aq) + [AlCl4(H2O)2]–(aq) The crystallisation of AlCl3 from aqueous solution, therefore, yields an ionic solid of composition [AlCl2(H2O)4]+ [AlCl4(H2O)2]– · xH2O. This compound decomposes at about 190°C to give the nonionic dimer Al2Cl6. heat [AlCl2(H2O)4] + [AlCl 4(H2O)2]– · xH2O 190°C Al2Cl6 + H2O 50 . (b) : Calcium carbide is ionic carbide having [ .. C . Ca2+ [ . C 1 2 C .. ] C .. ] 2– . 2– 51. (a) : According to molecular orbital theory, each of the two boron atoms is in sp3 hybrid state. Of the four hybrid orbitals, three have one electron each while the fourth is empty. Two of the four orbitals of each of the boron atom overlap with two terminal hydrogen atoms forming two normal B – H -bonds. One of the remaining hybrid orbital (either filled or empty) of one of the boron atoms, 1s orbital of hydrogen atoms (bridge atom) and one of hybrid orbitals of the other boron atom overlap to form a delocalised orbital covering the three nuclei with a pair of electrons. Such a bond is known as three centre two electron (3c – 2e) bonds. . H H. H. . . B . . B .H . . H H. .. H B H .H . . H B H H Structure of diborane (2H+ + SO42–) + 2NH4OH Strong acid Weak base (NH4) 2SO 4 on hydrolysis produces strong acid H 2SO 4, which increases the acidity of the soil. 46 . (c) : The elements of group 14 show an oxidation state of +4 and +2. The compounds showing an oxidation state of +4 are covalent compound and have tetrahedral structures. e.g. SnCl4, PbCl4, SiCl4, etc. whereas those which show +2 oxidation state are ionic in nature and behave as reducing agent. e.g. SnCl2, PbCl2, etc. Further as we move down the group, the tendency of the element to form covalent compound decreases but the tendency to form ionic compound increases. 47. (d) : 3HClO4(aq)  HClO3(aq) + 2HCl(aq) It is a disproportionation reaction of hypochlorous acid where the oxidation number of Cl changes from +1 (in ClO–) to +5 (in ClO3–) and –1 (in Cl–). (sp 3d, 52. (d) : SF4 trigonal bipyramidal with one equatorial position occupied by 1 lone pair), CF4 (sp3, tetrahedral, no lone pair), XeF4 (sp3d2, square planar, two lone pairs). F F F F (SF4) F F C S F F F Xe F F (CF4) F (XeF4) 53. (b) : Hypophosphorous acid (H3PO2) : O P HO H H Number of hydrogen atom(s) attached to phosphorus atom = 2. 54. (b) : As the size of the halogen atom increases from F to I, H – X bond length in HX molecules also increases from H – F to H – I (H – F < H – Cl < H – Br < H – I). The increase in H – X bond length decreases the strength of H – X bond from H – F to H – I (H – F > H – Cl > H – Br > H – I). The decrease in the strength of H – X bond is evident from the fact that H – X bond dissociation energies decrease from H – F to H – I. Due to successive decrease in the strength of H – X bond from H – F to H – I, thermal stability of HX molecules also decreases from HF to HI (HF > HCl > HBr > HI). 55. (a) : Silicon dioxide exhibits polymorphism. It is a network solid in which each Si atom is surrounded tetrahedrally by four oxygen atoms. O 63. (d) : Graphite has a two-dimensional sheet like structure and each carbon atom makes a use of sp2 hybridisation. The above layer structure of graphite is less compact than that of diamond. Further, since the bonding between the layers involving only weak van der Waals forces, these layers can slide over each other. This gives softness, greasiness and lubricating character of graphite. 64. (c) : NO2 and O3 both have unsymmetrical structures, so they have permanent dipole moment. 65. (c) : Ammonia is a Lewis base, accepting proton to form ammonium ion as it has tendency to donate an electron pair. O – O – Si – O – Si – O – O O 56. (d): CaO-basic, CO2 and SiO2-acidic, SnO2-amphoteric, as it reacts both with acids and bases. SnO2 + 4HCl  SnCl4 + 2H2O SnO2 + 2NaOH  Na2SnO3 + H2O 57. (b) : Grey tin is very brittle and easily crumbles down to a powder in very cold climates. Grey tin White tin (Cubic) (Tetragonal) The change of white tin to grey tin is accompanied by increase in volume. This is called tin disease or tin plague. 58. (b) : Al2Cl6 + 12H2O 2[Al(H2O)6]3+ + 6Cl– 59. (d) : Acidity of the oxides of non metals increases with the electronegativity and oxidation number of the element. Al2O3 < SiO2 < P2O3 < SO2 Al2O3 is amphoteric. SiO2 is slightly acidic whereas P2O3 and SO2 are the anhydrides of the acids H3PO3 and H2SO3. H .. . sp3 O . H .. .. O sp3 + H–NH H H 66. (a) : ZnO is an amphoteric oxide and dissolves readily in acids forming corresponding zinc salts and alkalies forming zincates. ZnO + H2SO4  ZnSO4 + H2O Zinc sulphate ZnO + 2NaOH  Na2ZnO2 + H2O Sodium zincate 67. (c) : HCl gas in presence of moisture in air forms droplets of liquid solution in the form of cloudy smoke. 68. (c) : Phosphine burns in the atmosphere of chlorine and forms phosphorus pentachloride. PH3 + 4Cl2  PCl5 + 3HCl 69. (b) : The solubility product of AgCl, AgBr and AgI at the room temperature are 2.8 × 10–10, 5.0 × 10–13 and 8.5 × 10–17 respectively. Thus, AgI is the least soluble silver halide. The lattice energies of AgBr and AgI are even higher because of greater number of electrons in their anions. Consequently, they are even less soluble than AgCl. Due to greater solubility of AgCl than AgI, ammonia solution dissolves only AgCl and forms a complex. AgCl + 2NH4OH  [Ag(NH3)2]Cl + 2H2O Diammine silver chloride 2 B sp 60. (b) : H H . H – N . + H+ .. H O .. sp3 61. (a) : Helium is twice as heavy as hydrogen, its lifting power is 92 percent of that of hydrogen. Helium has the lowest melting and boiling points of any element which makes liquid helium an ideal coolant for many extremely low-temperature applications such as superconducting magnets, and cryogenic research where temperatures close to absolute zero are needed. 62. (b) : Glass is a transparent or translucent amorphous supercooled solid solution (supercooled liquid) of silicates and borates, having a general formula R2O·MO·6SiO2 where R = Na or K and M = Ca, Ba, Zn or Pb. 70. (c) : The negatively charged colloidal particles of impurities get neutralised by the Al3+ ions and settle down and pure water can be decanted off. 71. (a) : 7N = 1s 2 2s 2 3p 3 2 2 6 2 3 15P = 1s 2s 2p 3s 3p In phosphorus the 3d-orbitals are available. 72. (d) : XeF2 XeF4 XeF6 sp 3d sp 3d 2 sp 3d 3 3 lone pairs 2 lone pairs 1 lone pair 73. (c) : Due to the higher electronegativity of F, HF is more polar than HBr pure. Water contains H+ and OH– ions. In covalency, sharing of electrons between two non-metal atoms takes place. 74. (c)