{"id":38336,"date":"2022-12-03T16:44:25","date_gmt":"2022-12-03T11:14:25","guid":{"rendered":"https:\/\/tsboardsolutions.com\/?p=38336"},"modified":"2022-12-03T16:45:07","modified_gmt":"2022-12-03T11:15:07","slug":"ts-inter-2nd-year-chemistry-study-material-chapter-6b","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.com\/ts-inter-2nd-year-chemistry-study-material-chapter-6b\/","title":{"rendered":"TS Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements"},"content":{"rendered":"

Telangana TSBIE\u00a0TS Inter 2nd Year Chemistry Study Material<\/a> Lesson 6(b) Group-16 Elements Textbook Questions and Answers.<\/p>\n

TS Inter 2nd Year Chemistry Study Material Lesson 6(b) Group-16 Elements<\/h2>\n

Very Short Answer Questions (2 Marks)<\/span><\/p>\n

Question 1.
\nWhy is dioxygen a gas but sulphur a solid?
\nAnswer:
\nDue to small size and high electronegativity oxygen forms multiple bond (=) and exists as O2<\/sub> molecule. These O2<\/sub> molecules are held together by weak van der Waals forces. Hence oxygen exists as gas.<\/p>\n

Due to large size and less electronegativity Sulphur forms strong S – S bonds and exists as S8<\/sub> molecule. Hence Sulphur exists as solid.<\/p>\n

Question 2.
\nWhat happens when
\na) KClO3<\/sub> is heated with MnO2<\/sub>,
\nb) O2<\/sub> is passed through KI solution.
\nAnswer:
\na) Oxygen is evolved.
\n2KClO3<\/sub> \u2192 2KCl + 3O2<\/sub><\/p>\n

b) Iodide is oxidised to I2<\/sub>.
\n2KI + H2<\/sub>O + O3<\/sub> \u2192 2K0H + I2<\/sub> + O2<\/sub><\/p>\n

Question 3.
\nGive two examples each for amphoteric oxides and neutral oxides.
\nAnswer:
\nExamples of amphoteric oxides are Al2<\/sub>O3<\/sub> and ZnO. Examples of neutral oxides are CO, NO and N2<\/sub>O.<\/p>\n

\"TS<\/p>\n

Question 4.
\nOxygen generally exhibits an oxidation state of – 2 only while the other members of the group show oxidation states of +2, + 4 and + 6 also – Explain.
\nAnswer:
\nThe multiple oxidation states in the case of elements other than oxygen are due to the availability of d-orbitals in the valence shell of the atoms.<\/p>\n

Oxygen exhibits -2 oxidation state only generally except with F, where it shows + I in F2<\/sub>O2<\/sub> and + II in F2<\/sub>O.<\/p>\n

Question 5.
\nWrite any two compounds in which oxygen shows an oxidation state different from -2. Give the oxidation states of oxygen in them.
\nAnswer:<\/p>\n

    \n
  1. In peroxides oxidation state of oxygen is – 1. Ex: H2<\/sub>O2<\/sub><\/li>\n
  2. In super oxides oxidation state of oxygen is -1\/2. Ex : KO2<\/sub><\/li>\n
  3. In F2<\/sub>O, oxidation state of oxygen is + 2.<\/li>\n
  4. In F2<\/sub>O2<\/sub>, oxidation state of oxygen is + 1.<\/li>\n<\/ol>\n

    Question 6.
    \nOxygen molecule has the formula O2<\/sub> while sulphur has S8<\/sub> – explain.
    \nAnswer:
    \nDue to small atomic size and high electro-negativity in oxygen molecule, each oxygen atom is linked to other oxygen atom by a double bond. Hence its formula is O2<\/sub>.
    \n\\(: \\ddot{\\mathrm{O}}=\\ddot{\\mathrm{O}}:\\)
    \nDue to large atomic size and less electro-negativity in sulphur molecule, eight ‘S’ atoms are linked together by single covalent bonds forming.
    \nPuckered S8<\/sub> rings with crown configuration. Hence formula of sulphur is S8<\/sub>.
    \n\"TS<\/p>\n

    Question 7.
    \nWhy is H2<\/sub>O a liquid while H2<\/sub>S is a gas? [IPE ’14]
    \nAnswer:
    \nThe O-H bond in H2<\/sub>O is highly polar. There are hydrogen bonds among the molecules of H2<\/sub>O. Hence it is present as a liquid.
    \n\"TS
    \nThere are no hydrogen bonds among H2<\/sub>S molecules. So it exists as a gas at room temperature.<\/p>\n

    \"TS<\/p>\n

    Question 8.
    \nH2<\/sub>O is neutral while H2<\/sub>S is acidic-explain.
    \nAnswer:
    \nBond dissociation enthalpy of H-S bond is less than H – O bond. Hence H2<\/sub>S is acidic.<\/p>\n

    Question 9.
    \nName the most abundant element present in earth’s crust.
    \nAnswer:
    \nOxygen<\/p>\n

    Question 10.
    \nWhich element of group-16 shows highest catenation?
    \nAnswer:
    \nSulphur<\/p>\n

    Question 11.
    \nAmong the hydrides of chalcogens, which is most acidic and which is most stable?
    \nAnswer:<\/p>\n

      \n
    1. Most acidic hydride of chalcogens is H2<\/sub>Te.<\/li>\n
    2. Most stable hydride of chalcogens is H2<\/sub>O.<\/li>\n<\/ol>\n

      Question 12.
      \nGive the hybridisation of sulphur in the following.
      \na) SO2<\/sub>
      \nb) SO3<\/sub>
      \nc) SF4<\/sub>
      \nd) SF6<\/sub>
      \nAnswer:
      \nHybridisation of S in SO2<\/sub> is sp2<\/sup>.
      \nHybridisation of S in SO3<\/sub> is sp2<\/sup>.
      \nHybridisation of S in SF4<\/sub> is sp3<\/sup>d.
      \nHybridisation of S in SF2<\/sub> is sp3<\/sup>d2<\/sup>.<\/p>\n

      \"TS<\/p>\n

      Question 13.
      \nWrite the names and formulae of any two oxyacids of sulphur. Indicate the oxidation state of sulphur in them.
      \nAnswer:<\/p>\n

        \n
      1. Sulphurous acid = H2<\/sub>SO3<\/sub>
        \nOxidation state of sulphur is + 4.<\/li>\n
      2. Sulphuric acid = H2<\/sub>SO4<\/sub>
        \nOxidation state of sulphur is + 6.<\/li>\n
      3. Pyrosulphuric acid = H2<\/sub>S2<\/sub>O7<\/sub>
        \nOxidation state of sulphur is + 6.<\/li>\n<\/ol>\n

        Question 14.
        \nExplain the structures of SF4<\/sub> and SF6<\/sub>.
        \nAnswer:
        \nStructure of SF4<\/sub> :
        \nExcited state configuration of S is
        \n\"TS
        \nHybridisation of Sulphur is sp3<\/sup>d.
        \nSF4<\/sub> has distorted trigonal bipyramidal structure with one orbital being occupied by a lone pair of electrons.
        \n\"TS<\/p>\n

        Structure of SF6<\/sub>:
        \nExcited state configuration of Sulphur is
        \n\"TS
        \nSF6<\/sub> has octahedral symmetry.<\/p>\n

        Question 15.
        \nGive one example each for
        \na) a neutral oxide
        \nb) a peroxide
        \nc) a super oxide
        \nAnswer:
        \na) Nitric oxide NO is a neutral oxide.
        \nb) Na2<\/sub>O2<\/sub> is a peroxide. Peroxides contain O – O bond.
        \nc) KO2<\/sub> is potassium super oxide.<\/p>\n

        \"TS<\/p>\n

        Question 16.
        \nWhat is tailing of mercury? How is it removed ? [AP & TS ’15]
        \nAnswer:
        \nOzone reacts with mercury to form Hg2<\/sub>O. Due to dissolution of Hg2<\/sub>O in Hg mercury loses its meniscus and sticks to the sides of glass. This is called tailing of mercury.
        \nThe menisus can be regained by shaking with water which dissolves Hg2<\/sub>O.<\/p>\n

        Question 17.
        \nWrite the principle involved in the quantitative estimation of ozone gas.
        \nAnswer:
        \nOzone liberates I2<\/sub> from KI solution which can be titrated against a standard solution of Hypo using starch as an indicator.
        \n2KI + H2<\/sub>O + O3<\/sub> \u2192 2KOH + I2<\/sub> + O2<\/sub>
        \n2Na2<\/sub>S2<\/sub>O3<\/sub> + I2<\/sub>\u2192 Na2<\/sub>S4<\/sub>O6<\/sub> + 2NaI<\/p>\n

        Question 18.
        \nWrite the structure of Ozone.
        \nAnswer:
        \n\"TS
        \nO-O bond lengths in ozone are identical (128pm) and molecule is angular with a bond angle of 117\u00b0.<\/p>\n

        Question 19.
        \nSO2<\/sub> can be used as anti-chlor. Explain.
        \nAnswer:
        \nSulphur dioxide reacts with chlorine in presence of charcoal to give sulphuryl chloride.
        \nSO2<\/sub>(g) + Cl2<\/sub> (g) \u2192 SO2<\/sub>Cl2<\/sub> (l)
        \nSo it can remove chlorine and can be used as anti-chlor.<\/p>\n

        \"TS<\/p>\n

        Question 20.
        \nHow is ozone detected?
        \nAnswer:<\/p>\n

          \n
        1. Ozone turns startch iodide paper blue.<\/li>\n
        2. It tails mercury.<\/li>\n<\/ol>\n

          Question 21.
          \nHow does ozone react with ethylene? [Mar. 18 – A.P.]
          \nAnswer:
          \nWhen ozone is bubbled through the solution of ethylene in an inert solvent like CCl4<\/sub>, at 195K, ethylene ozonide is formed.
          \n\"TS
          \nNO + O3<\/sub> \u2192 NO2<\/sub> + O2<\/sub><\/p>\n

          Question 22.
          \nOut of O2<\/sub> and O3<\/sub>, which is paramagnetic?
          \nAnswer:
          \nO2<\/sub> is paramagnetic. It contains two unpaired electrons in its molecular form O2<\/sub>.<\/p>\n

          Question 23.
          \nBetween O3<\/sub> and O2<\/sub>, ozone is a better oxidising agent – why?
          \nAnswer:
          \nDue to the ease with which it liberates atoms of nascent oxygen O3<\/sub> \u2192 O2<\/sub> + (O). Ozone acts as a powerful oxidising agent.<\/p>\n

          \"TS<\/p>\n

          Question 24.
          \nWrite any two uses each for O3<\/sub> and H3<\/sub>SO4<\/sub>.
          \nAnswer:
          \nUses of Ozone :<\/p>\n

            \n
          1. It is used as a germicide, disinfectant, and for sterilising water.<\/li>\n
          2. It is used for bleaching oils, ivory, flour, starch, etc.<\/li>\n<\/ol>\n

            Uses of H2<\/sub>SO4<\/sub> :<\/p>\n

              \n
            1. H2<\/sub>SO4<\/sub> is used in the manufacture of fertilisers e.g. ammonium sulphate, super phosphate.<\/li>\n
            2. Petroleum refining<\/li>\n<\/ol>\n

              Question 25.
              \nWhich form of sulphur shows paramagnetism?
              \nAnswer:
              \nIn vapour state sulphur partly exists as S2<\/sub> molecule which has two unpaired electrons in the antibonding \u03c0* orbitals. Hence exhibits paramagnetism.<\/p>\n

              Question 26.
              \nHow is the presence of SO2<\/sub> detected?
              \nAnswer:
              \nSO2<\/sub> decolourises KMnO4<\/sub> solution in acid medium.
              \n5SO2<\/sub> + 2\\(\\mathrm{MnO}_4^{-}\\) + 2H2<\/sub>O \u2192 5\\(\\mathrm{SO}_4^{2-}\\) + 4H+<\/sup> + 2Mn++<\/sup><\/p>\n

              \"TS<\/p>\n

              Question 27.
              \nWhy are group-16 elements called chal- cogens ?
              \nAnswer:
              \nThe elements Oxygen, Sulphur, Selenium, Tellurium of group 16- are collectively known as chalcogens meaning ‘ore forming’. Since many metals occur as oxides or sulphides in nature. The name is derived from the Greek word for brass. It indicates the association of sulphur and its congeners with copper.<\/p>\n

              Question 28.
              \nAmong chalcogens, which has highest electronegativity and which has highest electron gain enthalpy? ‘
              \nAnswer:
              \nOxygen has highest electronegativity (3.5 Pauling scale) among chalcogens. Sulphur has highest electron gain enthalpy among chalcogens.<\/p>\n

              Question 29.
              \nWhich hydride of group-16 has highest boiling point and weakest acidic character ?
              \nAnswer:
              \nH2<\/sub>O (water)<\/p>\n

              Short Answer Questions (4 Marks)<\/span><\/p>\n

              Question 30.
              \nJustify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms of electronic configuration, oxida-tion states and hydride formation.
              \nAnswer:
              \nElectronic configuration: All the elements of group 16 have six electrons in the outermost shell and have ns2<\/sup>np4<\/sup> general electronic configuration.<\/p>\n

              Oxidation states:
              \nGeneral oxidation states of Group 16 elements are -2, +2, +4 and +6. Oxygen cannot show higher oxidation states +4, +6 because of non-availability of d-orbitals in the valence shell. Sulphur, Selenium, and Tellurium usually show +4 oxidation state in their compounds with oxygen and +6 with fluorine. The stability of +6 oxidation state decreases down the group and stability of +4 oxidation state increases. This is due to inert pair effect.<\/p>\n

              Hydride formation :
              \nAll the chalcogens form covalent hydrides of the formula H2<\/sub>E. (E = S, Se, Te, Po)<\/p>\n

              Thermal stability of the hydrides decreases from H2<\/sub>O to H2<\/sub>Po. This is due to an increase in E – H bond length.<\/p>\n

              Water is a liquid while others are gases. Water exists as associated liquid due to hydrogen bonding.<\/p>\n

              The acidic character in aqueous solution increases from H2<\/sub>O to H2<\/sub>Te. This is due to decrease in charge density on conjugate bases OH–<\/sup>, SH–<\/sup>, SeH–<\/sup>, TeH–<\/sup>.<\/p>\n

              All the hydrides except water possess reducing property. The reducing property increases from H2<\/sub>O to H2<\/sub>Po. This trend can be attributed to decrease in thermal stability of hydrides from H2<\/sub>O to H2<\/sub>Po.<\/p>\n

              Thus there is a regular gradation in properties of these elements. Hence the inclusion of these elements in the same group is justified.<\/p>\n

              \"TS<\/p>\n

              Question 31.
              \nDescribe the manufacture of H2<\/sub>SO4<\/sub> by contact process. [TS ’16]
              \nAnswer:
              \nManufacture of H2<\/sub>SO4<\/sub> by Contact process involves three steps.
              \ni) Burning of sulphur or sulphide ores in air to generate SO2<\/sub>.
              \nS + O2<\/sub> \u2192 SO2<\/sub>
              \n4FeS2<\/sub> + 11O2<\/sub> \u2192 2Fe2<\/sub>O3<\/sub> + 8SO2<\/sub><\/p>\n

              ii) Conversion of SO2<\/sub> to SO3<\/sub> by the reaction with oxygen in the presence of a catalyst (V2<\/sub>O5<\/sub>).
              \n\"TS<\/p>\n

              iii) Absorption of SO3<\/sub> in H2<\/sub>SO4<\/sub> to give oleum.
              \nSO3<\/sub> + H2<\/sub>SO4<\/sub> \u2192 H2<\/sub>S2<\/sub>O7<\/sub><\/p>\n

              iv) Dilution of oleum with water gives sul-phuric acid.
              \nH2<\/sub>S2<\/sub>O7<\/sub> + H2<\/sub>O \u2192 2H2<\/sub>SO4<\/sub><\/p>\n

              Question 32.
              \nHow is ozone prepared? How does it react with the following ? [Mar. 19, 18, 17, AP]
              \na) PbS
              \nb)KI
              \nc) Hg
              \nd) Ag
              \nAnswer:
              \nPreparation of Ozone :
              \nWhen a slow dry stream of oxygen is passed through a silent electrical discharge, oxygen is converted to ozone. This product is called ozonised oxygen.
              \n3O2<\/sub> \u2192 2O3<\/sub>, \u2206H\u00b0 (298K) = +142 kJ\/mol.
              \nIf high concentration of ozone (> 10%) is required, a battery of ozonisers can be used. Pure ozone can be condensed in a vessel surrounded by liquid oxygen.
              \nProperties:
              \na) Pbs is oxidised to PbSO4<\/sub>.
              \nPbS (s) + 4O3<\/sub> (g) \u2192 PbSO4<\/sub> + 4O2<\/sub>
              \nb) I2<\/sub> is liberated. I–<\/sup> is oxidised to I2<\/sub>.
              \n2KI(aq)<\/sub> + H2<\/sub>O (l) + O3<\/sub>(g) \u2192 2KOH(aq) + I2<\/sub>(s) + O3<\/sub>(g)<\/p>\n

              c) Hg2<\/sub>O formed sticks to the glass surface and mercury loses its meniscus due to dissolution of Hg2<\/sub>O in Hg. This is called tailing of mercury.
              \n2Hg + O3<\/sub> \u2192 Hg2<\/sub>O + O2<\/sub><\/p>\n

              d) Silver metal is blackened. This is due to oxidation of the metal to its oxide
              \n2Ag + O3<\/sub> \u2192 Ag2<\/sub>O + O2<\/sub> (oxidation)<\/p>\n

              \"TS<\/p>\n

              Question 33.
              \nWrite a short note on the allotropy of sulphur.
              \nAnswer:
              \nSulphur forms numerous allotropes. The two common crystalline forms are \u03b1 (alpha) or rhombic sulphur and \u03b2 (beta) or monoclinic sulphur. The stable form at room temperature is rhombic sulphur. It transforms to monoclinic sulphur when heated above 369K.<\/p>\n

              Rhombic sulphur:
              \nThis is yellow in colour. It is formed by evaporating the solution of roll sulphur in CS2<\/sub>. It is soluble in CS2<\/sub>. m.p. 385.8K.<\/p>\n

              Monoclinic sulphur:
              \nThis form of sulphur is prepared by melting rhombic sulphur in a dish and cooling the solution till crust is formed. Two holes are made in the crust and remaining liquid poured out.<\/p>\n

              \u03b1 – sulphur transforms to \u03b2 – sulphur above 369K. At 369K both are stable. This temperature is called transition temperature.
              \nBoth forms have S8<\/sub> puckered ring forms. It has a crown shape.
              \n\"TS
              \nIn cyclo – S6<\/sub>, the ring adopts chair form. At 1000K, S2<\/sub> is the dominant species and is paramagnetic like O2<\/sub>.
              \n\"TS<\/p>\n

              Question 34.
              \nHow does S02 react with the following?
              \na) Na2<\/sub>SO3<\/sub> (aq)
              \nb) Cl2<\/sub>
              \nc) Fe3+<\/sup> ions
              \nd) KMnO4<\/sub>
              \nAnswer:
              \na) When SO2<\/sub> is passed into sodium sulphite solution, sodium hydrogen sulphite is formed.
              \nNa2<\/sub>SO3<\/sub> + H2<\/sub>O + SO2<\/sub> \u2192 2NaHSO3<\/sub><\/p>\n

              b) SO2<\/sub> reacts with chlorine in the presence of charcoal to give sulphuryl chloride.
              \nSO2<\/sub>(g) + Cl2<\/sub>(g) \u2192 SO2<\/sub>Cl2<\/sub> (l)<\/p>\n

              c) SO2<\/sub> reduces Fe3+<\/sup> ions to Fe2+<\/sup> ions.
              \n2Fe3+<\/sup> + SO2<\/sub> + 2H2<\/sub>O \u2192 2Fe2+<\/sup> + SO4<\/sub>2-<\/sup> + 4H+<\/sup><\/p>\n

              d) It decolourises KMnO4<\/sub>.
              \n5SO2<\/sub> + 2MnO4<\/sub>–<\/sup> + 2H2<\/sub>O \u2192 5SO4<\/sub>2-<\/sup> + 4H+<\/sup> + 2Mn++<\/sup><\/p>\n

              \"TS<\/p>\n

              Question 35.
              \nStarting from elemental sulphur, how is H2S04 prepared?
              \nAnswer:
              \nManufacture of H2<\/sub>SO4<\/sub> by Contact process involves three steps.
              \ni) Burning of sulphur or sulphide ores in air to generate SO2<\/sub>.
              \nS + O2<\/sub> \u2192 SO2<\/sub>
              \n4FeS2<\/sub> + 11O2<\/sub> \u2192 2Fe2<\/sub>O3<\/sub> + 8SO2<\/sub><\/p>\n

              ii) Conversion of SO2<\/sub> to SO3<\/sub> by the reaction with oxygen in the presence of a catalyst (V2<\/sub>O5<\/sub>).
              \n\"TS<\/p>\n

              iii) Absorption of SO3<\/sub> in H2<\/sub>SO4<\/sub> to give oleum.
              \nSO3<\/sub> + H2<\/sub>SO4<\/sub> \u2192 H2<\/sub>S2<\/sub>O7<\/sub><\/p>\n

              iv) Dilution of oleum with water gives sul-phuric acid.
              \nH2<\/sub>S2<\/sub>O7<\/sub> + H2<\/sub>O \u2192 2H2<\/sub>SO4<\/sub><\/p>\n

              Question 36.
              \nDescribe the structures (shapes) of SO4<\/sub>-2<\/sup> and SO3<\/sub>.
              \nAnswer:
              \nStructure of SO3<\/sub>: SO3<\/sub> has planar triangular shape in the gas phase. S is in sp2<\/sup> hybridisation. Bond angle is 120\u00b0. S-O bond length 143pm,<\/p>\n

              In SO3<\/sub>, sulphur is in second excited state.
              \n\"TS
              \nIn the solid state it may be cyclic or chain.<\/p>\n

              Structure of Sulphite ion SO4<\/sub>– –<\/sup>
              \nSO4<\/sub>– –<\/sup> is tetrahedral in shape. Bond angle is 109\u00b028′: S – O bond length 149pm.
              \nSulphur undergoes sp3<\/sup> hybridisation.
              \n\"TS<\/p>\n

              \"TS<\/p>\n

              Question 37.
              \nWhich oxide of sulphur can act as both oxidising and reducing agent? Give one example each.
              \nAnswer:
              \nSO2<\/sub> acts both as oxidising and reducing agent.<\/p>\n

                \n
              1. H2<\/sub>S is oxidised to sulphur.
                \nSO2<\/sub> + 2H2<\/sub>S \u2192 2H2<\/sub>O + 3S<\/li>\n
              2. Fe3+<\/sup> is reduced to Fe2+<\/sup>
                \n2Fe3+<\/sup> + SO2<\/sub> + 2H2<\/sub>O \u2192 2Fe2+<\/sup> + SO4<\/sub>2-<\/sup> + 4H+<\/sup><\/li>\n<\/ol>\n

                Question 38.
                \nExplain the conditions favourable for the formation of SO3<\/sub> from SO2<\/sub> in the contact process of H2<\/sub>SO4<\/sub>.
                \nAnswer:
                \n2SO2<\/sub> (g) + O2<\/sub>(g) \u2192 2SO3<\/sub> (g); \u2206H = -196 kj\/mol
                \nThe reaction is exothermic and reversible. Forward reaction leads to a decrease in volume. Therefore, low temperature and high pressure are favourable for maximum yield.<\/p>\n

                In practice, the plant is operated at a pressure of 2 bar and a temperature of 720K.<\/p>\n

                Question 39.
                \nComplete the following.
                \na) KCl + H2<\/sub>SO4<\/sub>(conc.) \u2192
                \nb) Sucrose \"TS
                \nc) Cu + H2<\/sub>SO4<\/sub>(conc.) \u2192
                \nd) C + H2<\/sub>SO4<\/sub> (conc.) \u2192
                \nAnswer:
                \na) HCl gas evolves.
                \n2KCl + H2<\/sub>SO4<\/sub> \u2192 K2<\/sub>SO4<\/sub> + 2HCl<\/p>\n

                b) It removes water from carbohydrates.
                \n\"TS<\/p>\n

                c) SO2<\/sub> gas evolves. Cu is oxidised.
                \nCu . 2H2<\/sub>SO4<\/sub> \u2192 CuSO4<\/sub> + SO2<\/sub> + 2H2<\/sub>O<\/p>\n

                d) C is oxidised to CO2<\/sub>.
                \nC + 2H2<\/sub>SO4<\/sub> \u2192CO2<\/sub> + 2SO2<\/sub> + 2H2<\/sub>O<\/p>\n

                \"TS<\/p>\n

                Question 40.
                \nWhich is used for drying ammonia?
                \nAnswer:
                \nAmmonia is dried using quick lime CaO.<\/p>\n

                Question 41.
                \nWhy cone. H2<\/sub>SO4<\/sub>, P4<\/sub>O10<\/sub> and anhydrous CaCl2<\/sub> cannot be used for dry ammonia? (Hint: ammonia reacts with them forming (NH4<\/sub>)2<\/sub>SO4<\/sub> : (NH4<\/sub>)3<\/sub> PO4<\/sub> and CaCl2<\/sub>. 8 NH3<\/sub>
                \nAnswer:
                \nAmmonia forms (NH4<\/sub>)3<\/sub> PO4<\/sub> with P4<\/sub>O10<\/sub>
                \nAmmonia reacts with CaCl2<\/sub> forming CaCl2<\/sub> 8NH3<\/sub>. Hence they cannot be used for drying ammonia.<\/p>\n

                Long Answer Questions (8 Marks)<\/span><\/p>\n

                Question 42.
                \nExplain in detail the manufacture of sulphuric acid by contact process. [Mar. 2018 – TS]
                \nAnswer:
                \nContact process consists of three stages,
                \ni) Burning of suphur or sulphide ores in air to generate SO2<\/sub>.
                \nS + O2<\/sub> \u2192 SO2<\/sub>
                \n4FeS2<\/sub> + 11O2<\/sub> \u2192 2Fe2<\/sub>O3<\/sub> + 8SO2<\/sub><\/p>\n

                ii) SO2<\/sub> is oxidised to SO3<\/sub> in presence of V2<\/sub>O5<\/sub>.
                \n\"TS
                \n\u2206H = 196 kJ, mol-1<\/sup><\/p>\n

                iii) Absorption of SO3<\/sub> in H2<\/sub>SO4<\/sub> to give oleum.
                \nSO3<\/sub> + H2<\/sub>SO4<\/sub> \u2192 H2<\/sub>S2<\/sub>O7<\/sub>
                \nLow temperature and high pressure are favourable for high yield of SO3<\/sub>. In practice, a pressure of 2 bar and a temperature of 720K are employed.<\/p>\n

                The SO3<\/sub> gas from the catalytic converter is absorbed in concentrated H2<\/sub>SO4<\/sub> to produce oleum. Dilution of oleum with water gives H2<\/sub>SO4<\/sub> of the desired concentration.
                \nSO3<\/sub> + H2<\/sub>SO4<\/sub> \u2192 H2<\/sub>S2<\/sub>O7<\/sub>
                \nH2<\/sub>S2<\/sub>O7<\/sub> + H2<\/sub>O \u2192 2H2<\/sub>SO4<\/sub>
                \n\"TS<\/p>\n

                Question 43.
                \nHow is ozone prepared from oxygen? Explain its reaction with [AP Mar. ’18, 17, 16; 1PE ‘ 13,’ 14]
                \na) C2<\/sub>H4<\/sub>
                \nb) KI
                \nc) Hg
                \nd) PbS
                \nAnswer:
                \nPreparation of Ozone : When a slow dry stream of oxygen is passed through a silent electrical discharge, oxygen is converted to ozone. This product is called ozonised oxygen.
                \n3O2<\/sub> \u2192 2O3<\/sub>, \u2206H\u00b0 (298K) = +142 kJ\/mol.
                \nIf high concentration of ozone (> 10%) is required, a battery of ozonisers can be used. Pure ozone can be condensed in a vessel surrounded by liquid oxygen.<\/p>\n

                \"TS<\/p>\n

                a) Ethylene reacts with ozone to form ethylene ozonide.
                \n\"TS<\/p>\n

                b) Iodide is oxidised to Iodine.
                \n2KI + H2<\/sub>O + O3<\/sub> \u2192 2KOH + I2<\/sub> + O2<\/sub><\/p>\n

                c) Mercury is oxidised to Hg2<\/sub>O by ozone.
                \nAs Hg2<\/sub>O dissolves in Hg, Hg loses its meniscus and sticks to glass surface. It is called ‘tailing of mercury’.
                \n2Hg + O3<\/sub> \u2192 Hg2<\/sub>O + O2<\/sub><\/p>\n

                d) PbS is oxidised to PbSO4<\/sub> by O3<\/sub>.
                \nPbS + 4O3<\/sub> \u2192 PbSO4<\/sub> + 4O2<\/sub><\/p>\n

                Intext Questions – Answers<\/span><\/p>\n

                Question 1.
                \nList the important sources of sulphur.
                \nAnswer:<\/p>\n

                  \n
                1. Sulphur exists primarily such as gypsum CaSO4<\/sub> . 2H2<\/sub>O.<\/li>\n
                2. Epsom salt MgSO4<\/sub> . 7H2<\/sub>O<\/li>\n
                3. Sulphides such as
                  \ngalena PbS
                  \nzinc blende ZnS<\/li>\n
                4. eggs, proteins, garlic, onion ……………<\/li>\n<\/ol>\n

                  Question 2.
                  \nWrite the order of thermal stability of the hydrides of Group 16 elements
                  \nAnswer:
                  \nThe thermal stability of hydrides of Group -16 decreases from H2<\/sub>O to H2<\/sub>PO.<\/p>\n

                  \"TS<\/p>\n

                  Question 3.
                  \nWhy is H2<\/sub>O a liquid and H2<\/sub>S a gas at room temperature and pressure ?
                  \nAnswer:
                  \nThere is molecular association through hydrogen bonds in H2<\/sub>O. But, there is no such molecular association through hydrogen bond in H2<\/sub>S. H and H2<\/sub>O exists as liquid while H2<\/sub>S exists as gas.<\/p>\n

                  Question 4.
                  \nWhich of the following does not react with oxygen directly ? Zn, Ti, Pt, Fe
                  \nAnswer:
                  \nPlatinum.<\/p>\n

                  Question 5.
                  \nComplete the following reactions :
                  \ni) C2<\/sub>H4<\/sub> + O2<\/sub> \u2192
                  \nii) 4 Al + 3O2<\/sub> \u2192
                  \nAnswer:
                  \ni) C2<\/sub>H4<\/sub> + 3O2<\/sub> \u2192 2CO2<\/sub> + 2H2<\/sub>O
                  \nii) 4 Al + 3O2<\/sub> \u2192 2 Al2<\/sub>O3<\/sub><\/p>\n

                  Question 6.
                  \nWhy does O3<\/sub> act as a powerful oxidizing agent?
                  \nAnswer:
                  \nO3<\/sub> can easily release nascent oxygen. Hence, it acts as powerful oxidizing agent.<\/p>\n

                  \"TS<\/p>\n

                  Question 7.
                  \nHow is O3<\/sub> estimated quantitatively ?
                  \nAnswer:
                  \nO3<\/sub> reacts with excess of Kl solution and liberates Iodine. The liberated Iodine is titrated against a standard solution of sodium Thiosulphate.<\/p>\n

                  Question 8.
                  \nWhat happens when SO2<\/sub> is passed through an aqueous solution of Fe(III) salt ?
                  \nAnswer:
                  \nIt reduces Fe(lII) salts to Fe(II) salts.
                  \n2Fe+3<\/sup> + SO2<\/sub> + 2H2<\/sub>O \u2192 2Fe+2<\/sup> + SO4<\/sub>-2<\/sup> + 4H+<\/sup><\/p>\n

                  Question 9.
                  \nComment on the nature of two S-O bonds formed in S02 molecule. Are the two S-O bonds in this molecule equal ?
                  \nAnswer:
                  \nThe two S-O bonds in SO2<\/sub> molecule are covalent and they are not equal. One oxygen is linked by single bond and the other oxygen is linked by double bond. The S-O bond length is larger than S = O bond length.<\/p>\n

                  Question 10.
                  \nHow is the presence of SO2<\/sub> detected ?
                  \nAnswer:
                  \nDue to the strong pungent smell, the presence of SO2<\/sub> can be detected. It decolourises acidified KMNO4<\/sub> solution.<\/p>\n

                  \"TS<\/p>\n

                  Question 11.
                  \nMention three areas in which H2<\/sub>SO4<\/sub> plays an important role.
                  \nAnswer:<\/p>\n

                    \n
                  1. H2<\/sub>SO4<\/sub> is used for manufacture of fertilisers. e.g.: ammoium sulphate<\/li>\n
                  2. Petroleum refining<\/li>\n
                  3. Detergent industry<\/li>\n
                  4. Metallurgical applications e.g.: electro-plating and galvanising.<\/li>\n<\/ol>\n

                    Question 12.
                    \nWrite the conditions to maximise the yield of H2<\/sub>SO4<\/sub> by Contact process ?
                    \nAnswer:
                    \nThe yield of H2<\/sub>SO4<\/sub> can be maximised by maintaining the following conditions.
                    \na) low temperatures (720 k)
                    \nb) high pressures (2 bar).<\/p>\n

                    Question 13.
                    \nWhy is Ka2<\/sub><\/sub> < < Ka1<\/sub><\/sub> for in water?
                    \nAnswer:
                    \nH2<\/sub>SO4<\/sub> is a very strong acid in water largely because of its first ionisation to H3<\/sub>O+<\/sup> and \\(\\mathrm{HSO}_4^{-}\\). The ionisation of \\(\\mathrm{HSO}_4^{-}\\) to H3<\/sub>O+<\/sup> and SO4<\/sub>2-<\/sup> is very very small.
                    \n\"TS
                    \nHence , Ka2<\/sub><\/sub> < < Ka1<\/sub><\/sub><\/p>\n","protected":false},"excerpt":{"rendered":"

                    Telangana TSBIE\u00a0TS Inter 2nd Year Chemistry Study Material Lesson 6(b) Group-16 Elements Textbook Questions and Answers. TS Inter 2nd Year Chemistry Study Material Lesson 6(b) Group-16 Elements Very Short Answer Questions (2 Marks) Question 1. Why is dioxygen a gas but sulphur a solid? 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