{"id":35316,"date":"2022-09-09T09:46:09","date_gmt":"2022-09-09T04:16:09","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35316"},"modified":"2022-09-09T09:46:09","modified_gmt":"2022-09-09T04:16:09","slug":"ap-inter-2nd-year-chemistry-study-material-chapter-6b","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-inter-2nd-year-chemistry-study-material-chapter-6b\/","title":{"rendered":"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements"},"content":{"rendered":"<p>Andhra Pradesh BIEAP <a href=\"https:\/\/apboardsolutions.guru\/ap-inter-2nd-year-chemistry-study-material\/\">AP Inter 2nd Year Chemistry Study Material<\/a> Lesson 6(b) Group-16 Elements Textbook Questions and Answers.<\/p>\n<h2>AP Inter 2nd Year Chemistry Study Material Lesson 6(b) Group-16 Elements<\/h2>\n<p><span style=\"color: #0000ff;\">Very Short Answer Questions<\/span><\/p>\n<p>Question 1.<br \/>\nWhy is dioxygen a gas but sulphur a solid?<br \/>\nAnswer:<br \/>\nDioxygen is a gas but sulphur a solid.<br \/>\nExplanation:<\/p>\n<ul>\n<li>Due to small atomic size and high electronegativity oxygen forms P\u03c0 &#8211; P\u03c0 multiple bond and exists as O<sub>2<\/sub> molecules held together by weak vander waal\u2019s forces. Thus oxygen exists as a gas at room temperature.<\/li>\n<li>Due to large atomic size and less electronegativity sulphur forms strong S\u2014S single bonds and exists as S<sub>8<\/sub> molecules with puckered ring structure. Hence sulphur is a solid at room temperature.<\/li>\n<\/ul>\n<p>Question 2.<br \/>\nWhat happens when<br \/>\na) KClO<sub>3<\/sub> is heated with MnO<sub>2<\/sub><br \/>\nb) O<sub>3<\/sub> is passed through KI soultion<br \/>\nAnswer:<br \/>\na) When KClO<sub>3<\/sub> is heated with Mn02 and liberates oxygen gas.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23260\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6b-Group-16-Elements-1.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements 1\" width=\"219\" height=\"50\" \/><br \/>\nb) O<sub>3<\/sub> is passed through KI solution I<sub>2<\/sub> is liberated<br \/>\n2KI + O<sub>3<\/sub> + H<sub>2<\/sub>O \u2192 2KOH + I<sub>2<\/sub> + 2O<sub>2<\/sub><\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 3.<br \/>\nGive two examples each for amphoteric oxides and neutral oxides.<br \/>\nAnswer:<\/p>\n<ul>\n<li>Examples of amphoterc oxides &#8211; Al<sub>2<\/sub>O<sub>3<\/sub>, SiO<sub>2<\/sub>, PbO.<\/li>\n<li>Examples of neutral oxides &#8211; CO, NO and N<sub>2<\/sub>O.<\/li>\n<\/ul>\n<p>Question 4.<br \/>\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 &#8211; explain.<br \/>\nAnswer:<\/p>\n<ul>\n<li>Oxygen exhibits -2 oxidation state due to its high electronegativity. The tendency of exhibiting -2 oxidation state decrease down the group.<\/li>\n<li>Due to decrease of electronegativity down the group the other elements exhibit +2, +4 and +6 oxidation states also.<\/li>\n<\/ul>\n<p>Question 5.<br \/>\nWrite any two compounds, in which oxygen shows an oxidation state different from -2. Give the oxidation states of oxygen in them.<br \/>\nAnswer:<br \/>\nOF<sub>2<\/sub> and O<sub>2<\/sub> F<sub>2<\/sub> are two compounds in which oxygen shows an oxidation state different from-2.<\/p>\n<ul>\n<li>In OF<sub>2<\/sub> the oxidation state of oxygen is +2<\/li>\n<li>In O<sub>2<\/sub>F<sub>2<\/sub> the oxidation state of oxygen is +1.<\/li>\n<\/ul>\n<p>Question 6.<br \/>\nOxygen molecule has the formula O<sub>2<\/sub> while sulphur has S<sub>8<\/sub> &#8211; explain.<br \/>\nAnswer:<br \/>\nDue to small atomic size and high electronegativity oxygen forms P\u03c0 &#8211; P\u03c0 multiple bond and exists as O<sub>2<\/sub> molecule.<br \/>\nDue to large atomic size and less electronegativity sulphur forms strong S &#8211; S single bonds and exists S<sub>8<\/sub> molecule.<\/p>\n<p>Question 7.<br \/>\nWhy is H<sub>2<\/sub>O a liquid while H<sub>2<\/sub>S is a gas ?<br \/>\nAnswer:<br \/>\nH<sub>2<\/sub>O is liquid due to the presence of intermolecular hydrogen bonding. While H<sub>2<\/sub>S is gas because it is not having such type of bonding.<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 8.<br \/>\nH<sub>2<\/sub>O is neutral while H<sub>2<\/sub>S is acidic &#8211; explain.<br \/>\nAnswer:<br \/>\nH<sub>2<\/sub>O is neutral while H<sub>2<\/sub>S is acidic.<br \/>\nReason: The O-H bond dissociation Enthalpy is greater than the S &#8211; H bond dissociation Enthalpy.<\/p>\n<p>Question 9.<br \/>\nName the most abundant element present in earth&#8217;s crust.<br \/>\nAnswer:<br \/>\nThe most abundant element present in earth&#8217;s crust is oxygen (about 46.6%).<\/p>\n<p>Question 10.<br \/>\nWhich element of group-16 shows highest catenation ?<br \/>\nAnswer:<br \/>\nSulphus shows highest catenation among group &#8211; 16 elements and exists as S<sub>9<\/sub> molecule with puckered ring structure.<\/p>\n<p>Question 11.<br \/>\nAmong the hydrides of chalcogens, which is most acidic and which is most stable ?<br \/>\nAnswer:<\/p>\n<ul>\n<li>Among hydrides of chalcogens, H<sub>2<\/sub>Te is most acidic.<\/li>\n<li>Among hydrides of chalcogens, H<sub>2<\/sub>O is most stable.<\/li>\n<\/ul>\n<p>Question 12.<br \/>\nGive the hybridization of sulphur in the following.<br \/>\na) SO<sub>2<\/sub><br \/>\nb) SO<sub>3<\/sub><br \/>\nc) SF<sub>4<\/sub><br \/>\nd)SF<sub>6<\/sub><br \/>\nAnswer:<br \/>\na) Hybridisation of&#8217;S&#8217; in SO<sub>2<\/sub> is Sp<sup>2<\/sup><br \/>\nb) Hybridisation of &#8216;S&#8217; in SO<sub>3<\/sub> is Sp<sup>2<\/sup><br \/>\nc) Hybridisation of &#8216;S&#8217; in SF<sub>4<\/sub> is Sp<sup>3<\/sup>d<br \/>\nd) Hybridisation of &#8216;S&#8217; in SF<sub>6<\/sub> is Sp<sup>3<\/sup>d<sup>2<\/sup><\/p>\n<p>Question 13.<br \/>\nWrite the names and formulae of any two oxyacids of sulphur. Indicate the oxidation state of sulphur in them.<br \/>\nAnswer:<\/p>\n<ul>\n<li>Peroxy mono sulphuric acid &#8211; H<sub>2<\/sub>SO<sub>5<\/sub> &#8216;S&#8217; oxidation state +6 .<\/li>\n<li>Peroxy di sulphuric acid &#8211; H<sub>2<\/sub>S<sub>2<\/sub>O<sub>8<\/sub> &#8216;S&#8217; oxidation state +6<\/li>\n<\/ul>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 14.<br \/>\nExplain the structures of SF<sub>4<\/sub> and SF<sub>6<\/sub>.<br \/>\nAnswer:<br \/>\nStructure of SF<sub>4<\/sub>:<\/p>\n<ul>\n<li>It SF<sub>4<\/sub> \u2018S\u2019 undergoes sp<sup>3<\/sup>d hybndisation.<\/li>\n<li>It has trigonal bipyramidal structure in which one of the equitorial positions is occupied by a lone pair of electrons. This geometry is also known as see &#8211; saw geometry.<\/li>\n<\/ul>\n<p>Structure of SF<sub>6<\/sub>:<\/p>\n<ul>\n<li>In SF<sub>6<\/sub> \u2018S\u2019 undergoes sp<sup>3<\/sup>d<sup>2<\/sup> hybridisation.<\/li>\n<li>It has octahedral structure.<\/li>\n<\/ul>\n<p>Question 15.<br \/>\nGive one example each for<br \/>\na) a neutral oxide<br \/>\nb) a peroxide<br \/>\nc) a super oxide<br \/>\nAnswer:<br \/>\na) CO, N<sub>2<\/sub>O are neutral oxides.<br \/>\nb) Na<sub>2<\/sub>O<sub>2<\/sub>, BaO<sub>2<\/sub> are pernxides.<br \/>\nc) KO<sub>2<\/sub>, RbO<sub>2<\/sub> are super oxides.<\/p>\n<p>Question 16.<br \/>\nWhat is tailing of mercury? How is it removed? [A.P. &amp; T.S. (Mar. 15)]<br \/>\nAnswer:<br \/>\nMercury loses it\u2019s lustreness, meniscus and consequently sticks to the walls of glass vessel when it reacts with ozone. This phenomenon is called tailing of mercury.<br \/>\n2Hg + O<sub>3<\/sub> \u2192 Hg<sub>2<\/sub>O + O<sub>2<\/sub><br \/>\nIt is removed by shaking it with water which dissolves Hg<sub>2<\/sub>O.<\/p>\n<p>Question 17.<br \/>\nWrite the principle involved in the quantitative estimation of ozone gas.<br \/>\nAnswer:<br \/>\nWhen ozone reacts with an excess of Kl solution buffered with a borate buffer (pH 9.2) iodine is liberated which can be titrated against a standard solution of sodium thiosulphate. In this way O<sub>3<\/sub> is estimated quantitatively.<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 18.<br \/>\nWrite the structure of ozone.<br \/>\nAnswer:<br \/>\nStructure of Ozone:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23261\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6b-Group-16-Elements-2.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements 2\" width=\"356\" height=\"84\" \/><\/p>\n<ul>\n<li>O<sub>3<\/sub> is angular molecule with bond angle 117\u00b0<\/li>\n<li>O &#8211; O bond length is 128 pm.<\/li>\n<\/ul>\n<p>Question 19.<br \/>\nSO<sub>2<\/sub> can be used as an anti-chior. Explain.<br \/>\nAnswer:<br \/>\nSO<sub>2<\/sub> gas is used as anti-chlor. Anti-chior means the substance which removes the excess of on clothes. SO<sub>2<\/sub> reacts with chlorine in presence of charcoal to give suiphuryl chloride<br \/>\nSO<sub>2(g)<\/sub> + Cl<sub>2(g)<\/sub> \u2192 SO<sub>2<\/sub>Cl<sub>2l<\/sub><\/p>\n<p>Question 20.<br \/>\nHow is ozone detected ?<br \/>\nAnswer:<br \/>\nOzone is a pale blue gas, dark blue liquid and violet black solid and it has characterstic smell.<\/p>\n<ul>\n<li>It is detected by the reaction with &#8216;Hg&#8217; which is called as tailing of mercury.<br \/>\n2Hg + O<sub>3<\/sub> \u2192 Hg<sub>2<\/sub>O + O<sub>2<\/sub><\/li>\n<li>Ozone turns benzidene paper to brown colour.<\/li>\n<\/ul>\n<p>Question 21.<br \/>\nHow does ozone react with Ethylene ?<br \/>\nAnswer:<br \/>\nEthylene reacts with ozone to form Ethylene ozonoid followed by the hydrolysis to form formaldehyde.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23262\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6b-Group-16-Elements-3.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements 3\" width=\"495\" height=\"122\" \/><\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 22.<br \/>\nOut of O<sub>2<\/sub> and O<sub>3<\/sub>, which is paramagnetic ?<br \/>\nAnswer:<\/p>\n<ul>\n<li>O<sub>2<\/sub> is paramagnetic due to presence of unpaired electrons<\/li>\n<li>O<sub>3<\/sub> (gaseous) is diamagnetic due to absence of unpaired electrons.<\/li>\n<\/ul>\n<p>Question 23.<br \/>\nBetween O<sub>3<\/sub> and O<sub>2<\/sub>, ozone is a better oxidizing agent &#8211; why ?<br \/>\nAnswer:<br \/>\nOzone is next to fluorine in the oxidising capacity. It is best oxidising agent than O<sub>2<\/sub>. (Fluorine is the powerful oxidising agent). Ozone liberates nascent oxygen easily.<\/p>\n<p>Question 24.<br \/>\nWrite any two uses each for O<sub>3<\/sub> and H<sub>2<\/sub>SO<sub>4<\/sub>.<br \/>\nAnswer:<br \/>\nUses of O<sub>3<\/sub>:<\/p>\n<ul>\n<li>Ozone is used in sterilisation of water.<\/li>\n<li>Ozone is used in manufacture of artificial silk and camphor etc.<\/li>\n<li>Ozone is used to identify unsaturation in carbon compounds.<\/li>\n<\/ul>\n<p>Uses of H<sub>2<\/sub>SO<sub>4<\/sub>:<\/p>\n<ul>\n<li>H<sub>2<\/sub>SO<sub>4<\/sub> is used in the manufacture of fertilisers.<\/li>\n<li>H<sub>2<\/sub>SO<sub>4<\/sub> is used in petrol refining.<\/li>\n<li>H<sub>2<\/sub>SO<sub>4<\/sub> is used in detergent industry.<\/li>\n<\/ul>\n<p>Question 25.<br \/>\nWhich form of sulphur shows paramagnetism ?<br \/>\nAnswer:<br \/>\nIn vapour state sulphur partly exists as S<sub>2<\/sub> molecule which has two unpaired electrons in the antibonding \u03c0 (\u03c0*)orbitals like O<sub>2<\/sub>. Hence exhibits paramagnetism.<\/p>\n<p>Question 26.<br \/>\nHow is the presence of SO<sub>2<\/sub> detected ?<br \/>\nAnswer:<br \/>\nSO<sub>2<\/sub> has a pungent odour SO<sub>2<\/sub> presence can be detected by the following tests.<\/p>\n<ol>\n<li>SO<sub>2<\/sub> changes the colour of acidified potassium dichromate solution from orange to green.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23263\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6b-Group-16-Elements-4.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements 4\" width=\"371\" height=\"53\" \/><\/li>\n<li>SO<sub>2<\/sub> decolourises acidified KMnO<sub>4<\/sub> solution.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23264\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6b-Group-16-Elements-5.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements 5\" width=\"411\" height=\"51\" \/><\/li>\n<\/ol>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 27.<br \/>\nWhy are group &#8211; 16 elements called chalcogens ?<br \/>\nAnswer:<br \/>\nChalcogens means mineral forming (or) ore forming elements. Most of elements exist in earth crust as oxides, sulphides, selinides, telurids etc. So Group &#8211; 16 elements are called as chalcogens.<\/p>\n<p>Question 28.<br \/>\nAmong chalcogens, which has highest eletronegativity and which has highest electron gain enthalpy?<br \/>\nAnswer:<\/p>\n<ul>\n<li>Among chalcogens Oxygen has high electronegativity.<\/li>\n<li>Among chalcogens Sulphur has high electron gain Enthalpy.<\/li>\n<\/ul>\n<p>Question 29.<br \/>\nWhich hydride of group &#8211; 16 has highest boiling point and weakest acidic character ?<br \/>\nAnsw:<\/p>\n<ul>\n<li>Among group &#8211; 16 hydrides water (H<sub>2<\/sub>O) has high bioling point.<\/li>\n<li>Among group &#8211; 16 hydrides water (H<sub>2<\/sub>O) has weakest acidic, character.<\/li>\n<\/ul>\n<p><span style=\"color: #0000ff;\">Short Answer Questions<\/span><\/p>\n<p>Question 1.<br \/>\nJustify the placement of O, S, Se, Te and Po in the same group of the periodic table in terms of electronic configuration, oxidation states and hydride formation.<br \/>\nAnswer:<br \/>\n1) Electronic configurations :<br \/>\nOxygen (O) &#8211; [He] 2s<sup>2<\/sup> 2p<sup>2<\/sup><br \/>\nSulphur (S) &#8211; [Ne] 3s<sup>2<\/sup> 3p<sup>2<\/sup><br \/>\nSelenium (Se) &#8211; [Ar] 3d<sup>10<\/sup> 4s<sup>2<\/sup>4p<sup>4<\/sup><br \/>\nTellurium (Te) &#8211; [Kr] 4d<sup>10<\/sup> 5s<sup>2<\/sup> sp<sup>4<\/sup><br \/>\nPolonium (Po) &#8211; [Xe] 4f<sup>14<\/sup> 5d<sup>10<\/sup> 6s<sup>2<\/sup> 6p<sup>4<\/sup><br \/>\nAll the above elements has general outer electronic configuration ns<sup>2<\/sup>np<sup>4<\/sup>.<\/p>\n<p>2) Oxidation states :<br \/>\nAll the gives elements (chalcogens) exhibits common oxidation state of-2<br \/>\nO<sup>-2<\/sup>, S<sup>-2<\/sup>, Se<sup>-2<\/sup> etc.<\/p>\n<p>3) Hydride formation:<br \/>\nAll these elements (chalcogens) forms hydrides of type EH<sub>2<\/sub>(E = chalcogen)<br \/>\nEg : H<sub>2<\/sub>O, H<sub>2<\/sub>S, H<sub>2<\/sub>Se, H<sub>2<\/sub>Te, H<sub>2<\/sub>Po.<br \/>\nThe above mentioned concepts evident that the elements O, S, Se, Te and Po are present in the same group of periodic table.<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 2.<br \/>\nDescribe the manufacture of H<sub>2<\/sub>SO<sub>4<\/sub> by contact process.<br \/>\nAnswer:<br \/>\nManufacture of H<sub>2<\/sub>SO<sub>4<\/sub> by contact process:<br \/>\nManufacturing of H<sub>2<\/sub>SO<sub>4<\/sub> involves three main steps.<br \/>\nStep &#8211; 1.<br \/>\nSO<sub>2<\/sub> production : The required SO<sub>2<\/sub> for this process is obtained by burning S(or) Iron pyrites in oxygen.<br \/>\nS + O<sub>2<\/sub> \u2192 SO<sub>2<\/sub><br \/>\n4FeS<sub>2<\/sub> + 15O<sub>2<\/sub> \u2192 2Fe<sub>2<\/sub>O<sub>3<\/sub> + 8SO<sub>3<\/sub><br \/>\nStep &#8211; 2<br \/>\nSO<sub>3<\/sub> formation: SO<sub>2<\/sub> is oxidised in presence of catalyst with atmosphric air to form SO<sub>3<\/sub><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23265\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6b-Group-16-Elements-6.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements 6\" width=\"261\" height=\"49\" \/><br \/>\nStep &#8211; 3<br \/>\nFormation of H<sub>2<\/sub>SO<sub>4<\/sub>: SO<sub>3<\/sub> formed in the above step absorbed in 98% H<sub>2<\/sub>SO<sub>4<\/sub> to get oleum (H<sub>2<\/sub>S<sub>2<\/sub>O<sub>7<\/sub>). This oleum is diluted to get desired concentration of H<sub>2<\/sub>SO<sub>4<\/sub>.<br \/>\nSO<sub>3<\/sub> + H<sub>2<\/sub>SO<sub>4<\/sub> \u2192 H<sub>2<\/sub>S<sub>2<\/sub>O<sub>7<\/sub><br \/>\nH<sub>2<\/sub>S<sub>2<\/sub>O<sub>7<\/sub> + H<sub>2<\/sub>O \u2192 2H<sub>2<\/sub>SO<sub>4<\/sub><\/p>\n<p>Question 3.<br \/>\nHow is ozone prepared ? How does it react with the following? [Mar. 14]<br \/>\na) PbS<br \/>\nb) KI<br \/>\nc) Hg<br \/>\nd) Ag<br \/>\nAnswer:<br \/>\nPreparation of Ozone:<br \/>\nA slow dry stream of oxygen under silent electric discharge to form ozone (about 10%). The product obtained is known as ozonised oxygen.<br \/>\n3O<sub>2<\/sub> \u2192 2O<sub>3<\/sub> \u2206 H\u00b0 = 142kJ\/mole<\/p>\n<ul>\n<li>The formation of ozone is an endothermic reaction.<\/li>\n<li>It is necessary to use silent electric discharge in the preparation of O<sub>3<\/sub> to prevent its decomposition<\/li>\n<\/ul>\n<p>a) Reaction with PbS : Black lead suiphide oxidised to white lead sulphate in presence of ozone.<br \/>\nPbS + 4O<sub>3<\/sub> \u2192 PbSO<sub>4<\/sub> + 4O<sub>2<\/sub><\/p>\n<p>b) Reaction with KI: Moist Kl is oxidised to Iodine in presence of ozone.<br \/>\n2KI + H<sub>2<\/sub>O + O<sub>3<\/sub> -\u2192 2KOH + I<sub>2<\/sub> + O<sub>2<\/sub><\/p>\n<p>c) Reaction with Hg : Mercury loses it&#8217;s lustreness, meniscus and consequently sticks to the walls of glass vessel when it reacts with ozone. This phenomenon is called tailing of mercury.<br \/>\n2Hg + O<sub>3<\/sub> \u2192 Hg<sub>2<\/sub> O + O<sub>2<\/sub><br \/>\nIt is removed by shaking it with water which dissolves Hg<sub>2<\/sub>O.<\/p>\n<p>d) Reaction with Ag : Ag metal oxidised to Ag<sub>2<\/sub>O (Ag metal is tarnished):<br \/>\n2Ag + O<sub>3<\/sub> \u2192 Ag<sub>2<\/sub>O + O<sub>2<\/sub><\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 4.<br \/>\nWrite a short note on the allotropy of sulphur.<br \/>\nAnswer:<br \/>\nThe important allotropes of sulphur are<br \/>\na) yellow rhombic (\u03b1. sulphur)<br \/>\nb) Monoclinic (\u03b2 &#8211; sulphur)<\/p>\n<ul>\n<li>The stable from is \u03b1-sulphur (at room temperature)<\/li>\n<\/ul>\n<p>Rhombic sulphur (\u03b1 &#8211; Sulphur):<\/p>\n<ul>\n<li>Colour : Yellow. ,<\/li>\n<li>Melting point: 385.8K.<\/li>\n<li>Specific gravity : 2.06.<\/li>\n<li>It is insouble in water and partially soluble in alcohol, benzene etc. and readily soluble in CS<sub>2<\/sub>.<\/li>\n<\/ul>\n<p>Monoclinic sulphur (\u03b2 &#8211; Sulphur):<\/p>\n<ul>\n<li>Melting point: 392K<\/li>\n<li>Specific gravity : 1.98.<\/li>\n<li>It is soluble in CS<sub>2<\/sub>.<\/li>\n<li>Rhombic sulphur transforms to monoclinic sulphur by heating above 369K. This temperature is called transition temperature.<\/li>\n<\/ul>\n<p>Question 5.<br \/>\nHow does SO<sub>2<\/sub> react with the following ?<br \/>\na) Na<sub>2<\/sub>SO<sub>3<\/sub>(aq)<br \/>\nb) Cl<sub>2<\/sub><br \/>\nc) Fe<sup>+3 <\/sup>ions<br \/>\nd) KMnO<sub>4<\/sub><br \/>\nAnswer:<br \/>\na) Sodium sulphite (aq) reacts with So<sub>2<\/sub> to form sodium hydrogen sulphite.<br \/>\nNa<sub>2<\/sub>SO<sub>3<\/sub> + H<sub>2<\/sub>O + SO<sub>2<\/sub> \u2192 2NaHSO<sub>3<\/sub><\/p>\n<p>b) SO<sub>2<\/sub> as reacts with chlorine gas in the presence of charcoal to form sulphuryl chloride.<br \/>\nSO<sub>2(g)<\/sub> + Cl<sub>2(g)<\/sub> \u2192 SO<sub>2<\/sub>Cl<sub>2(l)<\/sub><\/p>\n<p>c) Fe<sup>+3<\/sup> ions are reduced to Fe<sup>+2<\/sup> ions by SO<sub>2<\/sub>.<br \/>\n2Fe<sup>+3<\/sup> + SO<sub>2<\/sub> + 2H<sub>2<\/sub>O \u2192 2Fe<sup>+2<\/sup> + SO<sup>-2<\/sup><sub>4<\/sub> + 4H<sup>+<\/sup><\/p>\n<p>d) SO<sub>2<\/sub> gas decolourises acidified potassium permanganate (VII) solution.<br \/>\n5SO<sub>2<\/sub> + 2MnO<sup>&#8211;<\/sup><sub>4<\/sub> + 2H<sub>2<\/sub>O \u2192 5SO<sup>-2<\/sup><sub>4<\/sub> + 4H<sup>+<\/sup>+ 2Mn<sup>+2<\/sup><\/p>\n<p>Question 6.<br \/>\nStarting from elemental sulphur, how is H<sub>2<\/sub>SO<sub>4<\/sub> prepared ?<br \/>\nAnswer:<br \/>\nManufacture of H<sub>2<\/sub>SO<sub>4<\/sub> by contact process :<br \/>\nManufacturing of H<sub>2<\/sub>SO<sub>4<\/sub> involves three main steps.<br \/>\nStep &#8211; 1.<br \/>\nSO<sub>2<\/sub> production : The required SO<sub>2<\/sub> for this process is obtained by burning S(or) Iron pyrites in oxygen.<br \/>\nS + O<sub>2<\/sub> \u2192 SO<sub>2<\/sub><br \/>\n4FeS<sub>2<\/sub> + 15O<sub>2<\/sub> \u2192 2Fe<sub>2<\/sub>O<sub>3<\/sub> + 8SO<sub>3<\/sub><br \/>\nStep &#8211; 2<br \/>\nSO<sub>3<\/sub> formation: SO<sub>2<\/sub> is oxidised in presence of catalyst with atmosphric air to form SO<sub>3<\/sub><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23265\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6b-Group-16-Elements-6.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements 6\" width=\"261\" height=\"49\" \/><br \/>\nStep &#8211; 3<br \/>\nFormation of H<sub>2<\/sub>SO<sub>4<\/sub>: SO<sub>3<\/sub> formed in the above step absorbed in 98% H<sub>2<\/sub>SO<sub>4<\/sub> to get oleum (H<sub>2<\/sub>S<sub>2<\/sub>O<sub>7<\/sub>). This oleum is diluted to get desired concentration of H<sub>2<\/sub>SO<sub>4<\/sub>.<br \/>\nSO<sub>3<\/sub> + H<sub>2<\/sub>SO<sub>4<\/sub> \u2192 H<sub>2<\/sub>S<sub>2<\/sub>O<sub>7<\/sub><br \/>\nH<sub>2<\/sub>S<sub>2<\/sub>O<sub>7<\/sub> + H<sub>2<\/sub>O \u2192 2H<sub>2<\/sub>SO<sub>4<\/sub><\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 7.<br \/>\nDescribe the structures (shapes) of SO<sup>-2<\/sup><sub>4<\/sub> and SO<sub>3<\/sub>.<br \/>\nAnswer:<br \/>\nStructure of SO<sub>3<\/sub>:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23266\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6b-Group-16-Elements-7.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements 7\" width=\"212\" height=\"137\" \/><\/p>\n<ul>\n<li>In SO<sub>3<\/sub> sulphur undergoes sp<sup>2<\/sup> hybridisation.<\/li>\n<li>Shape : Trigonal planait s<\/li>\n<li>Bondangle: 120\u00b0.<\/li>\n<li>S -0 bond length: 143 pm.<\/li>\n<\/ul>\n<p>Structure of SO<sup>-2<\/sup><sub>4<\/sub>:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23267\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6b-Group-16-Elements-8.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements 8\" width=\"142\" height=\"135\" \/><\/p>\n<ul>\n<li>In SO<sup>-2<\/sup><sub>4<\/sub> sulphur undergo sp<sup>3<\/sup> hybridisation.<\/li>\n<li>Shape : Tetrahedral.<\/li>\n<li>It has several resonance structures.<\/li>\n<li>In this two P\u03c0 &#8211; d\u03c0 bonds are present.<\/li>\n<\/ul>\n<p>Question 8.<br \/>\nWhich oxide of sulphur can act as both oxidizing and reducing agent? Give one example each.<br \/>\nAnswer:<br \/>\nSulphur dioxide (SO2) acts as both oxidising as well as reducing agent.<br \/>\nSO<sub>2<\/sub> as Oxidising agent:<br \/>\nSodium suiphide oxidises to hypo with SO<sub>2<\/sub>.<br \/>\n2Na<sub>2<\/sub>S + 3SO<sub>2<\/sub> \u2192 2Na<sub>2<\/sub>S<sub>2<\/sub>O<sub>2<\/sub> + S<br \/>\nSO<sub>2<\/sub> as Reducing agent:<br \/>\nSO<sub>2<\/sub> reduces Fe<sup>+3<\/sup> ions to Fe<sup>+2<\/sup> ions.<br \/>\n2Fe<sup>+3<\/sup> + SO<sub>2<\/sub> + 2H<sub>2<\/sub>O \u2192 2Fe<sup>+2<\/sup> + SO<sup>-2<\/sup><sub>4<\/sub> + 4H<sup>+<\/sup><\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 9.<br \/>\nExplain the conditions favourable for the formation of SO<sub>3<\/sub> from SO<sub>2<\/sub> in the contact process of H<sub>2<\/sub>SO<sub>4<\/sub>.<br \/>\nAnswer:<br \/>\nLe Chatlier\u2019s principle &#8211; Application to produce SO<sub>3<\/sub>:<br \/>\nThe oxidation of SO<sub>2<\/sub> to SO<sub>3<\/sub> in the presence of a catalyst is a reversible reaction. The thermochemical equation for the conversion is written as<br \/>\n2SO<sub>2(g)<\/sub> + O<sub>2(g)<\/sub> \u21cc 2SO<sub>3(g)<\/sub>; \u2206H = -196 kJ<br \/>\nThe equation reveals the following points :<\/p>\n<ol>\n<li>3 volumes of the reactants convert into 2 volumes of SO<sub>3<\/sub>. i.e., a decrease of volume accompanies the reaction.<\/li>\n<li>the reaction is an exothermic change.<\/li>\n<li>the catalyst may be present to increase the SO<sub>3<\/sub> yields.<\/li>\n<\/ol>\n<p>According to Le Chatlier&#8217;s principle,<br \/>\ni) a decrease in volume of the system is favoured at high pressures. But in practice only about 2 bar pressure is used. The reason for not using high pressures is acid resisting towers which can withstand high pressures cannot be built.<\/p>\n<p>ii) exothermic changes are favoured at low temperatures. It is not always convenient in the industry to work at low temperatures. In such situations an optimum temperature is maintained. At this temperature considerable amounts of the product are obtained. In the manufactrue of H<sub>2<\/sub>SO<sub>4<\/sub>, the optimum temperature suitable for the conversion of SO<sub>2<\/sub> into SO<sub>3<\/sub> is experimentally found to be 720K.<\/p>\n<p>iii) The rate of formation of SO<sub>3<\/sub> is enhanced by the use of a catalyst. (V<sub>2<\/sub> O<sub>5<\/sub> (or) Pt &#8211; asbestos).<br \/>\nFavourable Conditions :<br \/>\nTemperature : 720K<br \/>\nPressure : 2 bar<br \/>\nCatalyst: V<sub>2<\/sub>O<sub>5<\/sub> (or) platinized asbestos.<\/p>\n<p>Question 10.<br \/>\nComplete the following<br \/>\na) KCl + H<sub>2<\/sub>SO<sub>4<\/sub> (Conc) \u2192<br \/>\nb) Sucrose <img loading=\"lazy\" class=\"alignnone size-full wp-image-23268\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6b-Group-16-Elements-9.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements 9\" width=\"127\" height=\"36\" \/><br \/>\nc) Cu + H<sub>2<\/sub>SO<sub>4<\/sub> (Conc) \u2192<br \/>\nd) C + H<sub>2<\/sub>SO<sub>4<\/sub> (Cone) \u2192<br \/>\nAnswer:<br \/>\na) 2KCl + Cone. H<sub>2<\/sub>SO<sub>4<\/sub> \u2192 2HCl + K<sub>2<\/sub>SO<sub>4<\/sub><br \/>\nb) C<sub>12<\/sub>H<sub>22<\/sub>O<sub>11<\/sub> <img loading=\"lazy\" class=\"alignnone size-full wp-image-23268\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6b-Group-16-Elements-9.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements 9\" width=\"127\" height=\"36\" \/> 12C + nH<sub>2<\/sub>O<br \/>\nc) Cu + 2H<sub>2<\/sub>SO<sub>4<\/sub>(Conc) \u2192 CuS0<sub>4<\/sub> + SO<sub>2<\/sub> + 2H<sub>2<\/sub>O<br \/>\nd) C + 2H<sub>2<\/sub>SO<sub>4<\/sub>(Conc) \u2192 CO<sub>2<\/sub> + 2SO<sub>2<\/sub> + 2H<sub>2<\/sub>O<\/p>\n<p>Question 11.<br \/>\nWhich is used for drying ammonia ?<br \/>\nAnswer:<br \/>\nFor drying ammonia quick line (CaO) is used.<br \/>\nFor drying ammonia cone. H<sub>2<\/sub>SO<sub>4<\/sub>, P<sub>4<\/sub>O<sub>10<\/sub> and anhydrous CaCl<sub>2<\/sub> cannot be used because they react with ammonia and forms (NH<sub>4<\/sub>)<sub>2<\/sub>SO<sub>4<\/sub>, (NH<sub>4<\/sub>)<sub>3<\/sub>PO<sub>4<\/sub> and CaCl<sub>2<\/sub>. 8NH<sub>3<\/sub><br \/>\nH<sub>2<\/sub>SO<sub>4<\/sub> + 2NH<sub>3<\/sub> \u2192 (NH<sub>4<\/sub>)<sub>2<\/sub> SO<sub>4<\/sub><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23269\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6b-Group-16-Elements-10.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements 10\" width=\"208\" height=\"48\" \/><br \/>\nCaCl<sub>2<\/sub> + 8NH<sub>3<\/sub> \u2192 CaCl<sub>3<\/sub>8NH<sub>3<\/sub>.<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 12.<br \/>\nWhy cone H<sub>2<\/sub>SO<sub>4<\/sub>, P<sub>4<\/sub>O<sub>10<\/sub> and anhydrous CaCl<sub>2<\/sub> cannot be used to dry ammonia ?<br \/>\n(Hint: ammonia reacts with them forming (NH<sub>4<\/sub>)<sub>2<\/sub> SO<sub>4<\/sub>; (NH<sub>4<\/sub>)<sub>3<\/sub> PO<sub>4<\/sub> and CaCl<sub>2<\/sub>, 8NH<sub>2<\/sub>)<br \/>\nAnswer:<br \/>\nFor drying ammonia quick line (CaO) is used.<br \/>\nFor drying ammonia cone. H<sub>2<\/sub>SO<sub>4<\/sub>, P<sub>4<\/sub>O<sub>10<\/sub> and anhydrous CaCl<sub>2<\/sub> cannot be used because they react with ammonia and forms (NH<sub>4<\/sub>)<sub>2<\/sub>SO<sub>4<\/sub>, (NH<sub>4<\/sub>)<sub>3<\/sub>PO<sub>4<\/sub> and CaCl<sub>2<\/sub>. 8NH<sub>3<\/sub><br \/>\nH<sub>2<\/sub>SO<sub>4<\/sub> + 2NH<sub>3<\/sub> \u2192 (NH<sub>4<\/sub>)<sub>2<\/sub> SO<sub>4<\/sub><br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23269\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6b-Group-16-Elements-10.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements 10\" width=\"208\" height=\"48\" \/><br \/>\nCaCl<sub>2<\/sub> + 8NH<sub>3<\/sub> \u2192 CaCl<sub>3<\/sub>8NH<sub>3<\/sub>.<\/p>\n<p><span style=\"color: #0000ff;\">Long Answer Questions<\/span><\/p>\n<p>Question 1.<br \/>\nExplain in detail the manufacture of sulphuric acid by contact process. [T.S. Mar. 18, 16]<br \/>\nAnswer:<br \/>\nManufacture of H<sub>2<\/sub>SO<sub>4<\/sub> by contact process :<br \/>\nManufacturing of H<sub>2<\/sub>SO<sub>4<\/sub> involves three main steps.<br \/>\nStep &#8211; 1.<br \/>\nSO<sub>2<\/sub> production : The required SO<sub>2<\/sub> for this process is obtained by burning S(or) Iron pyrites in oxygen.<br \/>\nS + O<sub>2<\/sub> \u2192 SO<sub>2<\/sub><br \/>\n4FeS<sub>2<\/sub> + 15O<sub>2<\/sub> \u2192 2Fe<sub>2<\/sub>O<sub>3<\/sub> + 8SO<sub>3<\/sub><br \/>\nStep &#8211; 2<br \/>\nSO<sub>3<\/sub> formation: SO<sub>2<\/sub> is oxidised in presence of catalyst with atmosphric air to form SO<sub>3<\/sub><\/p>\n<p>Le Chatlier&#8217;s principle &#8211; Application to produce SO<sub>3<\/sub> :<br \/>\nThe oxidation of SO<sub>2<\/sub> to SO<sub>3<\/sub> in the presence of a catalyst is a reversible reaction. The thermochemical equation for the conversion is written as<br \/>\n2SO<sub>2(g)<\/sub> + O<sub>2(g)<\/sub> \u21cc 2SO<sub>3(g)<\/sub>; \u2206H = -196 kJ<br \/>\nThe equation reveals the following points : .<\/p>\n<ol>\n<li>3 volumes of the reactants convert into 2 volumes of SO<sub>3<\/sub>. i.e., a decrease of volume accompanies the reaction.<\/li>\n<li>the reaction is an exothermic change.<\/li>\n<li>the catalyst may be present to increase the SO<sub>3<\/sub> yields.<\/li>\n<\/ol>\n<p>According to Le Chatlier&#8217;s principle, .<br \/>\ni) a decrease in volume of the system is favoured at high pressures. But in practice only about 2 bar pressure is used. The reason for not using high pressures is acid resisting towers which can withstand high pressures cannot be built.<\/p>\n<p>ii) exothermic changes are favoured at low temperatures. It is not always convenient in the industry to work at low temperatures. In such situations an optimum temperature is maintained. At this temperature considerable amounts of the product are obtained. In the manufactrue of H<sub>2<\/sub>SO<sub>4<\/sub>, the optimum temperature suitable for the conversion of SO<sub>2<\/sub> into SO., is experimentally found to be 720K.<\/p>\n<p>iii) The rate of formation of SO<sub>3<\/sub> is enhanced by the use of a catalyst. (V<sub>2<\/sub> O<sub>5<\/sub> (or) Pt &#8211; asbestos).<br \/>\nFavourable Conditions :<br \/>\nTemperature : 720K .<br \/>\nPressure : 2 bar<br \/>\nCatalyst: V<sub>2<\/sub>O<sub>5<\/sub> (or) platinized asbestos.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23270\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6b-Group-16-Elements-11.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements 11\" width=\"665\" height=\"313\" \/><br \/>\nStep &#8211; 3 :<br \/>\nFormation of H<sub>2<\/sub>SO<sub>4<\/sub>: SO<sub>3<\/sub> formed in the above step absorbed in 98% H<sub>2<\/sub>SO<sub>4<\/sub> to get oleum (H<sub>2<\/sub>S<sub>2<\/sub>O<sub>7<\/sub>). This oleum is diluted to get desired concentration of H<sub>2<\/sub>SO<sub>4<\/sub>.<br \/>\nSO<sub>3<\/sub> + H<sub>2<\/sub>SO<sub>4<\/sub> \u2192 H<sub>2<\/sub>S<sub>2<\/sub>O<sub>7<\/sub><br \/>\nH<sub>2<\/sub>S<sub>2<\/sub>O<sub>7<\/sub> + H<sub>2<\/sub>O \u2192 2H<sub>2<\/sub>SO<sub>4<\/sub><\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 2.<br \/>\nHow is ozone prepared from oxygen ? Explain its reaction with [A.P. Mar. 19, 18, 17, 16] [Mar. 14]<br \/>\na) C<sub>2<\/sub>H<sub>4<\/sub><br \/>\nb) KI<br \/>\nc) Hg<br \/>\nd) PbS.<br \/>\nAnswer:<br \/>\nPreparation of Ozone :<br \/>\nA slow dry stream of oxygen under silent electric discharge to form ozone (about 10%). The product obtained is known as ozonised oxygen.<br \/>\n3O<sub>2<\/sub> \u2192 2O<sub>3<\/sub>; \u2206H\u00b0 = 142kJ\/mole<\/p>\n<ul>\n<li>The formation of ozone is an endothermic reaction.<\/li>\n<li>It is necessary to use silent electric discharge in the preparation of O<sub>3<\/sub> to prevent its decomposition<\/li>\n<\/ul>\n<p>a) Reaction with C<sub>2<\/sub>H<sub>3<\/sub> : Ethylene reacts with ozone to form Ethylene ozonoid followed by the hydrolysis to form formaldehyde.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23271\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6b-Group-16-Elements-12.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements 12\" width=\"529\" height=\"120\" \/><\/p>\n<p>) Reaction with KI: moist KI is oxidised to Iodine in presence of ozone.<br \/>\n2KI + H<sub>2<\/sub>O + O<sub>3<\/sub> \u2192 2KOH + I<sub>2<\/sub> + O<sub>2<\/sub><\/p>\n<p>c) Reaction with Hg : Mercury loses it&#8217;s lustreness, meniscus and consequently sticks to the walls of glass vessel when it reacts with ozone. This phenomenon is called tailing of mercury.<br \/>\n2Hg + O<sub>3<\/sub> \u2192 Hg<sub>2<\/sub>O + O<sub>2<\/sub><br \/>\nIt is removed by shaking it with water which dissolves Hg<sub>2<\/sub>O.<\/p>\n<p>d) Reaction with PbS : Black lead sulphide oxidised to white lead sulphate in presence of ozone.<br \/>\nPbS + 4O<sub>3<\/sub> \u2192 PbSO<sub>4<\/sub> + 4O<sub>2<\/sub>.<\/p>\n<p><span style=\"color: #0000ff;\">Intext Questions<\/span><\/p>\n<p>Question 1.<br \/>\nList the important sources of sulphur.<br \/>\nAnswer:<br \/>\nOccurrence or sources of sulphur in the earth&#8217;s crust, percentage of sulphur is only 0.03 to 0. 1%. In combined state, it occurs<\/p>\n<ol>\n<li>In the form of sulphates, eg., gypsum (CaSO<sub>4<\/sub> . 2H<sub>2<\/sub>O). epsom salt (MgSO<sub>4<\/sub> . 7H<sub>2<\/sub>O), baryte<br \/>\n(BaSO<sub>4<\/sub>).<\/li>\n<li>In the form of sulphides, eg., galena (PbS), zinc blende (ZnS), Copper pyrites (CuFeS<sub>2<\/sub>). In volcanoes, traces of sulphur occur as H2S. In organic materials such as eggs, proteins, garlic, onion,mustard, hair and wool, sulphur is present in trace amounts.<\/li>\n<\/ol>\n<p>Question 2.<br \/>\nWrite the order of thermal stability of the hydrides of group 16 elements.<br \/>\nAnswer:<br \/>\nThe thermal stability of hydrides of group 16 elements is directly proportional to the bond dissociation enthalpy of H &#8211; E bond. On moving down the group, bond dissociation energy decreases because bond length increases.<br \/>\nThus, the order of bond dissociation energy is<br \/>\nH<sub>2<\/sub>O &gt; H<sub>2<\/sub>S &gt; H<sub>2<\/sub>Se &gt; H<sub>2<\/sub>Te &gt; H<sub>2<\/sub>Po<br \/>\nThis is also the order of thermal stability.<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 3.<br \/>\nWhy is H<sub>2<\/sub>O a liquid and H<sub>2<\/sub>S a gas ?<br \/>\nAnswer:<br \/>\nThe difference in electronegativity values of 0(3.5) and H(2.1) is more than the difference between the electronegativity values of H(2.1) and S(2.5) i.e., O &#8211; H bond is more polar than S &#8211; H bond. That is why H-bonding is present among water molecules but absent in H<sub>2<\/sub>S. Thus,<br \/>\nstrong intermolecular interactions causes water to exist as a liquid but due to weak van der waals&#8217; force H<sub>2<\/sub>S exists as a gas.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23272\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6b-Group-16-Elements-13.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements 13\" width=\"207\" height=\"61\" \/><\/p>\n<p>Question 4.<br \/>\nWhich of the following does not react with oxygen directly ? Zn, Ti, Pt, Fe.<br \/>\nAnswer:<br \/>\nPlatinum (Pt).<\/p>\n<p>Question 5.<br \/>\nComplete the following reactions.<br \/>\ni) C<sub>2<\/sub>H<sub>4<\/sub> + O<sub>2<\/sub> \u2192<br \/>\nii) 4Al + 3O<sub>2<\/sub> \u2192<br \/>\nAnswer:<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23273\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6b-Group-16-Elements-14.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements 14\" width=\"442\" height=\"63\" \/><\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 6.<br \/>\nWhy does O<sub>3<\/sub> act as a powerful oxidising agent ?<br \/>\nAnswer:<br \/>\nBecause ozone liberates nascent oxygen very easily.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23274\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6b-Group-16-Elements-15.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements 15\" width=\"349\" height=\"62\" \/><\/p>\n<p>Question 7.<br \/>\nHow is Oa estimated quantitatively ?<br \/>\nAnswer:<br \/>\nWhen ozone reacts with an excess of potassium iodide solution buffered with a borate buffer (pH 9.2), iodine is liberated which can be titrated against a standard solution of sodium thiosulphate. In this way O<sub>3<\/sub> can be estimated quantitatively.<\/p>\n<p>Question 8.<br \/>\nWhat happens when sulphur dioxide is passed through an aqueous solution of Fe (III) salt ?<br \/>\nAnswer:<br \/>\nWhen SO<sub>2<\/sub> is passed through an aqueous solution of Fe (III), i.e., ferric salt, it is reduced to Fe (II) i,e., ferrous salt. Here, SO<sub>2<\/sub> acts as a reducing agent.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23275\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6b-Group-16-Elements-16.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements 16\" width=\"568\" height=\"87\" \/><\/p>\n<p>Question 9.<br \/>\nComment on the nature of two S &#8211; O bonds formed in SO<sub>2<\/sub> molecule. Are the two S &#8211; O bonds in this molecule equal ?<br \/>\nAnswer:<br \/>\nThe two S &#8211; O bonds in SO<sub>2<\/sub> molecule are covalent in nature. These are equal with bond length = 143 pm. The resonating structures of SO<sub>2<\/sub> are as follows.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23276\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6b-Group-16-Elements-17.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements 17\" width=\"497\" height=\"120\" \/><br \/>\nSO<sub>2<\/sub> is a resonance hybrid of these two canonical forms.<\/p>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 10.<br \/>\nHow is the presence of SO<sub>2<\/sub> detected ?<br \/>\nAnswer:<br \/>\nSO<sub>2<\/sub> has a pungent odour. Two tests to detect the presence of SO<sub>2<\/sub> are as follows :<\/p>\n<ol>\n<li>SO<sub>2<\/sub> decolourises acidified KMnO<sub>4<\/sub> solution.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23277\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6b-Group-16-Elements-18.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements 18\" width=\"432\" height=\"59\" \/><\/li>\n<li>SO<sub>2<\/sub> Changes the colour of acidified potassium dichromate solution from orange to green.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23278\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6b-Group-16-Elements-19.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements 19\" width=\"382\" height=\"57\" \/><\/li>\n<\/ol>\n<p>Question 11.<br \/>\nMention three areas in which H2S04 plays an important role.<br \/>\nAnswer:<br \/>\nUses of sulphuric acid.<\/p>\n<ol>\n<li>It is used in the manufacture of pigments, paints and dyestuff intermediate.<\/li>\n<li>It is used in petroleum refining.<\/li>\n<li>It is also used in fertilizer industry.<\/li>\n<\/ol>\n<p><img loading=\"lazy\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/07\/AP-Board-Solutions.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements\" width=\"196\" height=\"17\" \/><\/p>\n<p>Question 12.<br \/>\nWrite the conditions to maximize the yield of H<sub>2<\/sub>SO<sub>4<\/sub> by contact process.<br \/>\nAnswer:<br \/>\nThe key step in the manufacture of H<sub>2<\/sub>SO<sub>4<\/sub> is catalytic oxidation of SO<sub>2<\/sub> to produce SO<sub>3<\/sub> in presence of V<sub>2<\/sub>O<sub>5<\/sub>.<br \/>\n<img loading=\"lazy\" class=\"alignnone size-full wp-image-23280\" src=\"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2022\/08\/AP-Inter-2nd-Year-Chemistry-Study-Material-Chapter-6b-Group-16-Elements-20.png\" alt=\"AP Inter 2nd Year Chemistry Study Material Chapter 6(b) Group-16 Elements 20\" width=\"452\" height=\"83\" \/><br \/>\nThe reaction is exothermic, reversible and the forward reaction results in the decrease in volume. Thus according to Le-Chatelier&#8217;s principle, the forward reaction should be favoured by low temperature and high pressure. But the temperature should not be very low otherwise the. rate of reaction will become very slow.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material Lesson 6(b) Group-16 Elements Textbook Questions and Answers. AP Inter 2nd Year Chemistry Study Material Lesson 6(b) Group-16 Elements Very Short Answer Questions Question 1. Why is dioxygen a gas but sulphur a solid? Answer: Dioxygen is a gas but sulphur a solid. Explanation: Due &hellip;<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[14],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35316"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=35316"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35316\/revisions"}],"predecessor-version":[{"id":35317,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35316\/revisions\/35317"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35316"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35316"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35316"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}