{"id":34762,"date":"2022-01-25T17:53:49","date_gmt":"2022-01-25T12:23:49","guid":{"rendered":"https:\/\/infinitylearn.com\/surge\/?p=34762"},"modified":"2025-05-28T16:49:08","modified_gmt":"2025-05-28T11:19:08","slug":"important-questions-for-class-12-chemistry-chapter-7-the-p-block-elements-class-12-important-questions","status":"publish","type":"post","link":"https:\/\/infinitylearn.com\/surge\/study-material\/important-questions\/class-12\/chemistry\/chapter-7\/the-p-block-elements\/class-12\/important-questions\/","title":{"rendered":"Important Questions for Class 12 Chemistry Chapter 7 The p-Block Elements Class 12 Important Questions"},"content":{"rendered":"<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_37 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs 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type=\"checkbox\" id=\"item\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1' style='display:block'><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/infinitylearn.com\/surge\/study-material\/important-questions\/class-12\/chemistry\/chapter-7\/the-p-block-elements\/class-12\/important-questions\/#Important_Questions_for_Class_12_Chemistry_Chapter_7_The_p-Block_Elements_Class_12_Important_Questions\" title=\"Important Questions for Class 12 Chemistry Chapter 7 The p-Block Elements Class 12 Important Questions\">Important Questions for Class 12 Chemistry Chapter 7 The p-Block Elements Class 12 Important Questions<\/a><ul class='ez-toc-list-level-4'><li class='ez-toc-heading-level-4'><ul class='ez-toc-list-level-4'><li class='ez-toc-heading-level-4'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/infinitylearn.com\/surge\/study-material\/important-questions\/class-12\/chemistry\/chapter-7\/the-p-block-elements\/class-12\/important-questions\/#The_p-Block_Elements_Class_12_Important_Questions_Very_Short_Answer_Type\" title=\"The p-Block Elements Class 12 Important Questions Very Short Answer Type\">The p-Block Elements Class 12 Important Questions Very Short Answer Type<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-4'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/infinitylearn.com\/surge\/study-material\/important-questions\/class-12\/chemistry\/chapter-7\/the-p-block-elements\/class-12\/important-questions\/#The_p-Block_Elements_Class_12_Important_Questions_Short_Answer_Type_-I_SA_-_I\" title=\"The p-Block Elements Class 12 Important Questions Short Answer Type -I [SA \u2013 I]\">The p-Block Elements Class 12 Important Questions Short Answer Type -I [SA \u2013 I]<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-4'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/infinitylearn.com\/surge\/study-material\/important-questions\/class-12\/chemistry\/chapter-7\/the-p-block-elements\/class-12\/important-questions\/#The_p-Block_Elements_Class_12_Important_Questions_Short_Answer_Type_-II_SA-II\" title=\"The p-Block Elements Class 12 Important Questions Short Answer Type -II [SA-II]\">The p-Block Elements Class 12 Important Questions Short Answer Type -II [SA-II]<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-4'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/infinitylearn.com\/surge\/study-material\/important-questions\/class-12\/chemistry\/chapter-7\/the-p-block-elements\/class-12\/important-questions\/#The_p-Block_Elements_Class_12_Important_Questions_long_Answer_Type_LA\" title=\"The p-Block Elements Class 12 Important Questions long Answer Type [LA]\">The p-Block Elements Class 12 Important Questions long Answer Type [LA]<\/a><\/li><\/ul><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"Important_Questions_for_Class_12_Chemistry_Chapter_7_The_p-Block_Elements_Class_12_Important_Questions\"><\/span>Important Questions for Class 12 Chemistry Chapter 7 The p-Block Elements Class 12 Important Questions<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h4><span class=\"ez-toc-section\" id=\"The_p-Block_Elements_Class_12_Important_Questions_Very_Short_Answer_Type\"><\/span>The p-Block Elements Class 12 Important Questions Very Short Answer Type<span class=\"ez-toc-section-end\"><\/span><\/h4>\n<p>Question 1.<br \/>\nWhy is Bi(v) a stronger oxidant than Sb(v)? (Delhi 2009)<br \/>\nAnswer:<br \/>\nThe stability of +5 oxidation state decreases and that of +3 state increases due to inert pair effect down the group therefore Bi(v) accepts two electrons and gets reduced to Bi (v).<br \/>\nBi<sup>5+<\/sup> + 2e<sup>\u2013<\/sup> \u2192 Bi<sup>3+<\/sup><\/p>\n<p>Question 2.<br \/>\nWhich is a stronger oxidizing agent Bi(v) or Sb(v)? (Delhi 2009)<br \/>\nAnswer:<br \/>\nBi(v) is stronger oxidizing agent due to inert pair effect.<\/p>\n<p>Question 3.<br \/>\nWhy is red phosphorus less reactive than white phosphorus? (All India 2009)<br \/>\nAnswer:<br \/>\nBecause white phosphorus has angular strain in its P4 molecules where the angle is only 60\u00b0.<\/p>\n<p>Question 4.<br \/>\nWhy does NO<sub>2<\/sub> dimerise? (Delhi 2010)<br \/>\nAnswer:<br \/>\nNO<sub>2<\/sub> contains 7 + 2 \u00d7 8 i.e. 23 odd electrons. In the valence shell N has seven electrons and hence less stable. To acquire stability it dimerizes to form N<sub>2<\/sub>O<sub>4<\/sub><\/p>\n<p>Question 5.<br \/>\nWhat is the oxidation number of phosphorus in H<sub>3<\/sub>PO<sub>2<\/sub> molecule? (Delhi 2010)<br \/>\nAnswer:<br \/>\nH<sub>3<\/sub>PO<sub>2<\/sub><br \/>\n3 + x- 4 = 0 or x \u2013 1 = 0 \u2234 x = + 1<br \/>\nThus oxidation number of P in H<sub>3<\/sub>PO<sub>2<\/sub> = +1.<\/p>\n<p>Question 6.<br \/>\nDraw the structure of 03 molecule. (Delhi 2010)<br \/>\nAnswer:<br \/>\nStructure of Ozone (Os) : Angular structure<\/p>\n<p>Question 7.<br \/>\nFluorine does not exhibit any positive oxidation state. Why? (All India 2010)<br \/>\nAnswer:<br \/>\nSince fluorine is the most electronegative element and does not have d-orbitals in its valence shell, therefore, it cannot expand its octet and hence does not show positive oxidation state (O.S.) while other halogens have d-orbitals and therefore exhibit many oxidation states.<\/p>\n<p>Question 8.<br \/>\nGive the IUPAC name of the following compound : (All India 2010)<\/p>\n<p>Answer:<br \/>\nIUPAC name : 2-Bromo-3-methylpent-3-ene<\/p>\n<p>Question 9.<br \/>\nNitrogen is relatively inert as compared to phosphorus. Why? (All India 2010)<br \/>\nAnswer:<br \/>\nBecause P -P single bond is much weaker than N = N triple bond and the bond length of nitrogen is small and bond dissociation energy is very large which makes it inert and urtreactive and thus phosphorus becomes more reactive.<\/p>\n<p>Question 10.<br \/>\nDraw the structure of XeF<sub>2<\/sub> molecule. (Delhi 2011)<br \/>\nAnswer:<br \/>\nXeF<sub>2<\/sub>:<\/p>\n<p>Question 11.<br \/>\nDraw the structure of XeF<sub>4<\/sub> molecule. (Delhi 2011)<br \/>\nAnswer:<br \/>\nXeF<sub>4<\/sub> : sp<sup>3<\/sup>d<sup>2<\/sup> hybridization<br \/>\nShape \u2192 Square planar<\/p>\n<p>Question 12.<br \/>\nDraw the structure of BrF3 molecule. (Delhi 2011)<br \/>\nAnswer:<br \/>\nBrF<sub>3<\/sub><\/p>\n<p>Question 13.<br \/>\nWhich one of \\(\\mathbf{P C l}_{4}^{+}\\) and \\(\\mathbf{P C l}_{4}^{-}\\) is not likely to exist and why? (Delhi 2012)<br \/>\nAnswer:<br \/>\n\\(\\mathbf{P C l}_{4}^{-}\\), because P has 10 electrons which cannot be accommodated in sp<sup>3<\/sup> hybrid orbitals.<\/p>\n<p>Question 14.<br \/>\nOf PH<sub>3<\/sub> and H<sub>2<\/sub>S which is more acidic and why? (Delhi 2012)<br \/>\nAnswer:<br \/>\nH<sub>2<\/sub>S, because of higher electronegativity of sulphur.<\/p>\n<p>Question 15.<br \/>\nWhich is a stronger reducing agent, SbH<sub>3<\/sub> or BiH<sub>3<\/sub>, and why? (All India 2012)<br \/>\nAnswer:<br \/>\nBiH<sub>3<\/sub> : Because it is stronger reducing agent as its tendency to liberate H is maximum.<\/p>\n<p>Question 16.<br \/>\nWhat is the basicity of H<sub>3<\/sub>PO<sub>2<\/sub> acid and why? (All India 2012)<br \/>\nAnswer:<br \/>\nH<sub>3<\/sub>PO<sub>2<\/sub> has one replaceable H atom so it is monobasic.<\/p>\n<p>Question 17.<br \/>\nThough nitrogen exhibits +5 oxidation state, it does not form pentahalide. Why? (Comptt. Delhi 2012)<br \/>\nAnswer:<br \/>\nDue to non-availability of d-orbitals in its valence electronic configuration nitrogen does not form pentahalide.<\/p>\n<p>Question 18.<br \/>\nNoble gases have low boiling points. Why? (Comptt. Delhi 2012)<br \/>\nAnswer:<br \/>\nNoble gases being monoatomic have no interatomic forces except weak dispersion force and therefore, they are liquified at very low temperature. Hence, they have low boiling point.<\/p>\n<p>Question 19.<br \/>\nWrite a reaction to show the reducing behaviour of H<sub>3<\/sub>PO<sub>2<\/sub>. (Comptt. Delhi 2012)<br \/>\nAnswer:<br \/>\nH<sub>3<\/sub>PO<sub>2<\/sub> reduces AgNO<sub>3<\/sub> to metallic Ag :<br \/>\n4AgNO<sub>3<\/sub> + 2H<sub>2<\/sub>O + H<sub>3<\/sub>PO<sub>2<\/sub> \u2192 4Ag + 4HNO<sub>3<\/sub> + H<sub>3<\/sub>PO<sub>4<\/sub><\/p>\n<p>Question 20.<br \/>\nWhy does PCl<sub>3<\/sub> fume in moisture? (Comptt. Delhi 2012)<br \/>\nAnswer:<br \/>\nPCl<sub>3<\/sub> hydrolyses in moisture giving fumes of HCl.<br \/>\nPCl<sub>3<\/sub> + 3H<sub>2<\/sub>O \u2192 H<sub>3<\/sub>PO<sub>3<\/sub> + 3HCl<\/p>\n<p>Question 21.<br \/>\nWhy does NO<sub>2<\/sub> dimerise ? (Comptt. All India 2012)<br \/>\nAnswer:<br \/>\nNO<sub>3<\/sub> contains odd number of valence electrons. It behaves as a typical odd molecule. On dimerisation, it is converted to stable N<sub>2<\/sub>O<sub>4<\/sub> molecule with even number of electrons.<\/p>\n<p>Question 22.<br \/>\nWhy is ICl more reactive than I<sub>2<\/sub>? (Comptt. All India 2012)<br \/>\nAnswer:<br \/>\nICl is more reactive than I<sub>2<\/sub> because I-Cl bond is weaker than I-I bond of I<sub>2<\/sub>.<\/p>\n<p>Question 23.<br \/>\nWhy is BiH<sub>3<\/sub> the strongest reducing agent amongst all the hydrides of group 15 elements? (Comptt. All India 2012)<br \/>\nAnswer:<br \/>\nReducing nature depends upon the stability of M- H bond. As the stability of the bond decreases from N to Bi hydrides, BiH<sub>3<\/sub> is the strongest reducing agent.<\/p>\n<p>Question 24.<br \/>\nWhat is the covalency of nitrogen in N<sub>2<\/sub>O<sub>5<\/sub>? (Delhi 2013)<br \/>\nAnswer:<br \/>\nThe covalency of nitrogen in N<sub>2<\/sub>O<sub>5<\/sub> is 4 because each nitrogen atom has four shared pairs of electrons.<\/p>\n<p>Question 25.<br \/>\nWhat inspired N.Bartlett for carrying out reaction between Xe and PtF<sub>6<\/sub>? (Delhi 2013)<br \/>\nAnswer:<br \/>\nSince PtF6 oxidises Osub&gt;2 to O<sub>2<\/sub><sup>+<\/sup>, Bartlett thought that PtF<sub>6<\/sub> should also oxidise Xe to Xe<sup>+<\/sup> because the ionization enthalpies of O<sub>2<\/sub> (1175 kj mol<sup>-1<\/sup>) and Xe (1170 kj mol<sup>-1<\/sup>) are quite close.<\/p>\n<p>Question 26.<br \/>\nWhat happens when ethyl chloride is treated with aqueous KOH? (Delhi 2013)<br \/>\nAnswer:<\/p>\n<p>Question 27.<br \/>\nName two poisonous gases which can be prepared from chlorine gas. (All India 2013)<br \/>\nAnswer:<br \/>\n(i) Phosgene gas (COCl<sub>2<\/sub>) and (ii) Chloropicrin or tear gas (CCl<sub>3<\/sub>NO<sub>2<\/sub>).<\/p>\n<p>Question 28.<br \/>\nWhich aerosol depletes ozone layer? (All India 2013)<br \/>\nAnswer:<br \/>\nAerosols like foams, sprays etc. contain freons which are responsible for depletion of ozone layer.<\/p>\n<p>Question 29.<br \/>\nWhat is the basicity of H<sub>3<\/sub>PO<sub>5<\/sub> and why? (All India 2013)<br \/>\nAnswer:<br \/>\nThe basicity of H<sub>3<\/sub>PO<sub>5<\/sub> is 2 because it contains only two ionizable H-atoms which are present as OH groups.<\/p>\n<p>Questions 30.<br \/>\nWhy does PCl<sub>3<\/sub> fume in moisture? (Comptt. Delhi 2013)<br \/>\nAnswer:<br \/>\nPhosphorus trichloride reacts readily with water giving phosphorus acid and hydrochloric acid. Reaction is very quick and exothermic<\/p>\n<p>Question 31.<br \/>\nDraw the structure of H<sub>3<\/sub>PO<sub>2<\/sub> molecule. (Comptt. Delhi 2013)<br \/>\nAnswer:<br \/>\nCl<sub>2<\/sub> + H<sub>2<\/sub>O \u2192 [HCl + HOCl] \u2192 2HCl + [O]\n<p>Question 32.<br \/>\nFluorine exhibits only -1 oxidation state whereas other halogens exhibit +1, +3, +5 and +7 oxidation states also. Why is it so? (Comptt. Delhi 2013)<br \/>\nAnswer:<br \/>\nIt is because fluorine is the most electronegative element and it does not have d-orbitals.<\/p>\n<p>Question 33.<br \/>\nThough nitrogen exhibits +5 oxidation state, it does not form pentahalide. Why? (Comptt. All India 2013)<br \/>\nAnswer:<br \/>\nDue to absence of d-subshell in N atom.<\/p>\n<p>Question 34.<br \/>\nBond enthalpy of fluorine is lower than that of chlorine. Why? (Comptt. All India 2013)<br \/>\nAnswer:<br \/>\nBecause F<sub>2<\/sub> is very small and its interelectronic repulsions between the lone pairs of electrons are very large.<\/p>\n<p>Question 35.<br \/>\nWrite the structural formula of PCl<sub>5<\/sub>(s). (Comptt. All India 2013)<br \/>\nAnswer:<br \/>\nPCl<sub>5<\/sub>(s)<\/p>\n<p>Question 36.<br \/>\nHF is a weaker acid than HCI. Why? (Comptt. All India 2013)<br \/>\nAnswer:<br \/>\nBecause of higher bond dissociation energy and strong H-bonding in HF.<\/p>\n<p>Question 37.<br \/>\nDraw the structure of XeF2 molecule. (Comptt. All India 2013)<br \/>\nAnswer:<br \/>\nXeF<sub>2<\/sub> :<\/p>\n<p>Question 38.<br \/>\nWhat is the basicity of H<sub>3<\/sub>PO<sub>3<\/sub>? (All India 2014)<br \/>\nAnswer:<br \/>\nBasicity of H<sub>3<\/sub>PO<sub>3<\/sub> = 2<br \/>\nBecause basicity is the number of replaceable H<sup>+<\/sup> ions in an acid and H<sub>3<\/sub>PO<sub>3<\/sub> is a Dibasic acid.<\/p>\n<p>Question 39.<br \/>\nWhy does NOz dimerise? (All India 2014)<br \/>\nAnswer:<br \/>\nNO<sub>2<\/sub> contains 7 + 2 \u00d7 8 i.e. 23 odd electrons. In the valence shell N has seven electrons and hence less stable. To acquire stability it dimerizes to form<br \/>\nN<sub>2<\/sub>O<sub>4<\/sub>.<\/p>\n<p>Question 40.<br \/>\nWhy does NH<sub>3<\/sub> act as a Lewis base? (All India 2014)<br \/>\nAnswer:<br \/>\nDue to presence of lone pair on nitrogen NH<sub>3<\/sub> acts as .a Lewis base.<\/p>\n<p>Question 41.<br \/>\nWhy is F<sub>2<\/sub> a stronger oxidising agent than Cl<sub>2<\/sub>? (Comptt. All India 2014)<br \/>\nAnswer:<br \/>\nDue to low bond dissociation enthalpy and high electronegativity of Fluorine, it has strong tendency to accept electrons and thus get reduced.<br \/>\nF + e<sup>\u2013<\/sup> \u2192 F<sup>\u2013<\/sup><br \/>\nTherefore F<sub>2<\/sub> acts as strong oxidising agent, while Cl<sub>2<\/sub> is weak oxidising agent due to low electronegativity.<\/p>\n<p>Question 42.<br \/>\nWhat is the basicity of H<sub>3<\/sub>PO<sub>4<\/sub>? (Delhi 2015)<br \/>\nAnswer:<br \/>\nSince there are 3 OH groups present in II H<sub>3<\/sub>PO<sub>4<\/sub>, its basicity is 3.<\/p>\n<p>Question 43.<br \/>\nWrite the formulae of any two oxoacids of sulphur. (All India 2015)<br \/>\nAnswer:<br \/>\nH<sub>2<\/sub>SO<sub>3<\/sub> and H<sub>2<\/sub>SO<sub>4<\/sub><\/p>\n<p>Question 44.<br \/>\nOn adding NaOH to ammonium sulphate, a colourless gas with pungent odour is evolved which forms a blue coloured complex with Cu<sup>2+<\/sup> ion. Identify the gas. (Delhi 2016)<br \/>\nAnswer:<br \/>\nThe gas with a pungent odour is Ammonia (NH<sub>3<\/sub>) and the blue coloured complex is Tetra-ammine copper (II) sulphate monohydrate.<\/p>\n<p>Question 45.<br \/>\nPb(NO<sub>3<\/sub>)<sub>2<\/sub> on heating gives a brown gas which undergoes dimerization on cooling. Identify the gas. (All India 2016)<br \/>\nAnswer:<br \/>\nThe brown gas is nitrogen dioxide (NO<sub>2<\/sub>) which can dimerize to N<sub>2<\/sub>O<sub>4<\/sub><br \/>\n2Pb(NO<sub>3<\/sub>)<sub>2<\/sub> \\(\\stackrel{\\Delta}{\\longrightarrow}\\) 2PbO + 4NO<sub>2<\/sub> + O<sub>2<\/sub><\/p>\n<p>Question 46.<br \/>\nWrite the formula of the compound of phosphorus which is obtained when cone. HNO<sub>3<\/sub> oxidises P<sub>4<\/sub>. (All India 2017)<br \/>\nAnswer:<\/p>\n<p>Question 47.<br \/>\nWrite the formula of the compound of sulphur which is obtained when cone. HNO<sub>3<\/sub> oxidises S<sub>8<\/sub> (All India 2017)<br \/>\nAnswer:<br \/>\nS<sub>8<\/sub> + 48HNO<sub>3<\/sub> \u2192 8H<sub>3<\/sub>SO<sub>4<\/sub> + 48NO<sub>3<\/sub> + 16H<sub>3<\/sub>O<\/p>\n<p>Question 48.<br \/>\nWrite the formula of the compound of iodine which is obtained when cone. HNO<sub>3<\/sub> oxidises I<sub>2<\/sub> . (All India 2017)<br \/>\nAnswer:<br \/>\nI<sub>2<\/sub> \u2013 10HNO<sub>3<\/sub> \u2192 2HIO<sub>3<\/sub> + 10NO<sub>2<\/sub> + 4H<sub>3<\/sub> O<\/p>\n<h4><span class=\"ez-toc-section\" id=\"The_p-Block_Elements_Class_12_Important_Questions_Short_Answer_Type_-I_SA_-_I\"><\/span>The p-Block Elements Class 12 Important Questions Short Answer Type -I [SA \u2013 I]<span class=\"ez-toc-section-end\"><\/span><\/h4>\n<p>Question 49.<br \/>\nDraw the structures of the following molecules :<br \/>\n(i) XeF<sub>4<\/sub>(it) BrF<sub>3<\/sub> (Delhi 2009)<br \/>\nAnswer:<br \/>\n(i) XeF<sub>4<\/sub><\/p>\n<p>(ii) BrF<sub>3<\/sub><\/p>\n<p>Question 50.<br \/>\nComplete the following chemical reaction equations : (Delhi 2009)<br \/>\n(i) P<sub>4<\/sub> (s) + NaOH (aq) + H<sub>2<\/sub>O (1) \u2192<br \/>\n(ii) I<sup>\u2013<\/sup> (aq) + H<sub>2<\/sub>O (1) + O<sub>3<\/sub> (g) \u2192<br \/>\nAnswer:<\/p>\n<p>Question 51.<br \/>\nComplete the following chemical reaction equations : (All India 2009)<br \/>\n(i) XeF<sub>2<\/sub> (s) + H<sub>2<\/sub>O (l) \u2192<br \/>\n(ii) PH<sub>3<\/sub> + HgCl<sub>2<\/sub> \u2192<br \/>\nAnswer:<\/p>\n<p>Question 52.<br \/>\nDraw the structures of white phosphorus and red phosphorus, phosphorus is more reactive and why? Which one of these two types of (Delhi 2010)<br \/>\nAnswer:<\/p>\n<p>White phosphorus is more reactive due to its discrete tetrahedral structure and angular strain.<\/p>\n<p>Question 53.<br \/>\nDraw the structural formulae of molecules of following compounds :<br \/>\n(i) BrF<sub>3<\/sub> and (ii) XeF<sub>4<\/sub> (Delhi 2010)<br \/>\nAnswer:<br \/>\n(i) BrF<sub>3<\/sub><\/p>\n<p>(ii) XeF<sub>4<\/sub><\/p>\n<p>Question 54.<br \/>\nComplete the following chemical reaction equations : (All India 2010)<br \/>\n(i) I<sub>2<\/sub> + HNO<sub>3<\/sub>(Conc.) \u2192 (ii) HgCl<sub>2<\/sub> + PH<sub>3<\/sub> \u2192<br \/>\nAnswer:<\/p>\n<p>Question 55.<br \/>\nDraw the structural formulae of the following compounds :<br \/>\n(i) H<sub>4<\/sub>P<sub>2<\/sub>O<sub>5<\/sub><br \/>\n(ii) XeF<sub>4<\/sub> (All India 2010)<br \/>\nAnswer:<\/p>\n<p>Question 56.<br \/>\nComplete the following chemical reaction equations : (All India 2010)<\/p>\n<p>Answer:<\/p>\n<p>Question 57.<br \/>\nState reasons for each of the following :<br \/>\n(i) The N-O bond in \\(\\mathrm{NO}_{2}^{-}\\) is shorter than the N-O bond in \\(\\mathrm{NO}_{3}^{-}\\).<br \/>\n(ii) SF<sub>6<\/sub> is kinetically an inert substance. (Delhi 2011)<br \/>\nAnswer:<br \/>\n(i) The resonating structure of \\(\\mathrm{NO}_{2}^{-}\\) and \\(\\mathrm{NO}_{3}^{-}\\) show that in \\(\\mathrm{NO}_{2}^{-}\\) two bonds are sharing a double bond while in \\(\\mathrm{NO}_{3}^{-}\\), 3 bonds are sharing a double bond. That\u2019s why \\(\\mathrm{NO}_{2}^{-}\\) has shorter bond than that of \\(\\mathrm{NO}_{3}^{-}\\).<br \/>\nAnswer:<\/p>\n<p>(ii) Because SF<sub>6<\/sub> is showing steric hindrance due to 6 (six) fluorine atoms which make it unable to react further with any other atom.<\/p>\n<p>Question 58.<br \/>\nState reasons for each of the following :<br \/>\n(i) All the P-Cl bonds in PCl<sub>5<\/sub> molecule are not equivalent.<br \/>\n(ii) Sulphur has greater tendency for catenation than oxygen. (Delhi 2011)<br \/>\nAnswer:<br \/>\n(i) The PCl<sub>5<\/sub> molecule has sp<sup>3<\/sup>d hybridization and trigonal bipyramidal geometry. Therefore it has 3 equatorial P \u2013 Cl bonds and two axial P-Cl bonds. Since two axial P-Cl bonds are repelled by 3 bond pairs while 3 equatorial bonds are repelled by two bond pairs, so axial bonds are longer than equatorial bonds.<\/p>\n<p>(ii) The greater catenation tendency of sulphur is due to two reasons :<br \/>\n(a) The lone pair of electrons feels more repulsion in 0-0 bond than S-S bond due to its small size and thus S-S forms strong bond.<br \/>\n(b) As the size of atom increases down the group from O \u2013 PO, the strength of bond increases and therefore catenation tendency also increases.<\/p>\n<p>Question 59.<br \/>\nExplain the following giving an appropriate reason in each case. (Delhi 2012)<br \/>\n(i) O<sub>2<\/sub> and F<sub>2<\/sub> both stabilize higher oxidation states of metals but O<sub>2<\/sub> exceeds F<sub>2<\/sub> in doing so.<br \/>\n(ii) Structures of Xenon fluorides cannot be explained by Valence Bond approach.<br \/>\nAnswer:<br \/>\n(i) This is due to the ability of oxygen to form multiple bonds to metals.<br \/>\n(ii) This is because the energy required for the promotion of electrons in Xenon is very high. Energy factor does not favour VB approach.<\/p>\n<p>Question 60.<br \/>\nExplain the following facts giving appropriate reason in each case :<br \/>\n(i) NF<sub>3<\/sub> is an exothermic compound whereas NCl<sub>3<\/sub> is not.<br \/>\n(ii) All the bonds in SF<sub>4<\/sub> are not equivalent. (All India 2012)<br \/>\nAnswer:<br \/>\n(i) The bond energy of F \u2013 F bond is lower than that of N-F bond so NF<sub>3<\/sub> is an exothermic . compound. On the other hand bond energy<br \/>\nof Cl \u2013 Cl bond is higher than that of N \u2013 Cl bond so NCl<sub>3<\/sub> is an endothermic as well as unstable compound.<\/p>\n<p>(ii) SF<sub>4<\/sub> has trigonal bipyramidal structure in which one position of equitorial is occupied by a lone pair of electrons. Sulphur undergoes sp<sup>3<\/sup>d hybridisation and has sea-saw geometry.<\/p>\n<p>Question 61.<br \/>\nExplain the following :<br \/>\n(a) Xenon does not form such fluorides as XeF<sub>3<\/sub> and XeF<sub>5<\/sub>.<br \/>\n(b) Out of noble gases, only Xenon is known to form real chemical compounds. (Comptt. Delhi 2012)<br \/>\nAnswer:<br \/>\n(a) By impairing of one paired orbital, two singly occupied orbitals come into existence.Thus, either two or four or six singly occupied orbitals can be formed instead of one, three or five singly occupied orbitals. Hence XeF, XeF<sub>3<\/sub> or XeF<sub>5<\/sub> are not formed.<br \/>\n(b) Xe atom has a large size and lower ionisation potential and hence the force of nucleus over the electrons is weak and hence very small energy can excite the electrons and hence it is easier for Xenon to form compounds than other noble gases.<\/p>\n<p>Question 62.<br \/>\n(a) Which form of sulphur shows paramagnetic behaviour and why ?<br \/>\n(b) Fluorine exhibits only -1 oxidation state whereas other halogens exhibit +1, +3, +5 or +7 oxidation states also. Explain as to why. (Comptt All India 2012)<br \/>\nAnswer:<br \/>\n(a) In vapour state sulphur partly exists as S<sub>2<\/sub> molecule which has two unpaired electrons in the antibonding \u03c0* orbitals like O<sub>2<\/sub> and hence exhibits paramagnetism.<br \/>\n(b) It is because fluorine is the most electronegative element and it does not have d-orbitals.<\/p>\n<p>Question 63.<br \/>\nHow is XeO<sub>3<\/sub> obtained? Write the related chemical equations. Draw the structure of XeO<sub>3<\/sub> (Comptt. All India 2012)<br \/>\nAnswer:<br \/>\nHydrolysis of XeF<sub>4<\/sub> and XeF<sub>6<\/sub> with water gives XeO<sub>3<\/sub><br \/>\n6XeF<sub>4<\/sub> + 12H<sub>2<\/sub>O \u2192 4Xe + 2XeO<sub>3<\/sub> + 24HF + 3O<sub>2<\/sub><br \/>\nXeF<sub>6<\/sub> + 3H<sub>2<\/sub>O \u2192 XeO<sub>3<\/sub> + 6HF<\/p>\n<p>Question 64.<br \/>\nWhat happens when<br \/>\n(i) PCl<sub>5<\/sub> is heated? (ii) H<sub>3<\/sub>PO<sub>3<\/sub> is heated? Write the reactions involved. (Delhi 2013)<br \/>\nAnswer:<br \/>\n(i) On heating PCl<sub>5<\/sub> decomposes into PCl<sub>3<\/sub> + Cl<sub>2<\/sub><\/p>\n<p>(ii) Orthophosphorous acid on heating disproportionates to give orthophosphoric acid and phosphine.<\/p>\n<p>Question 65.<br \/>\nDraw the structures of the following molecules: (All India 2013)<br \/>\n(i) XeOF<sub>4<\/sub> (ii) H<sub>3<\/sub>PO<sub>3<\/sub><br \/>\nAnswer:<br \/>\n(i) XeOF<sub>4<\/sub><\/p>\n<p>Question 66.<br \/>\nHow are interhalogen compounds formed? What general compositions can be assigned to them? (All India 2013)<br \/>\nAnswer:<br \/>\nInterhalogen compounds: Halogens react with each other to form a number of compounds called interhalogen compounds, whose general formula is XX\u2019n.<br \/>\nWhere X = less electronegative atom (have larger size)<br \/>\nX\u2019 = more electronegative atom (have smaller size)<br \/>\nn = no. of more electronegative atoms\/high<br \/>\nThey are of four types :<br \/>\nXX\u2019 = ClF, BrF, IF, BrCl, ICl, IBr<br \/>\nXX\u2019<sub>3<\/sub> = ClF<sub>3<\/sub> BrF<sub>3<\/sub>, IF<sub>3<\/sub>, ICl<sub>3<\/sub><br \/>\nXX\u2019<sub>5<\/sub> = ClF<sub>5<\/sub>, BrF<sub>5<\/sub>, IF<sub>5<\/sub><br \/>\nXX\u2019<sub>7<\/sub> = IF<sub>7<\/sub><br \/>\nNaming: The halogen with positive oxidation state named first and with negative oxidation state named after first with suffix \u2018ide\u2019.<br \/>\nExample : BrCl<sub>3<\/sub> \u2192 Bromine trichloride<br \/>\nIF<sub>7<\/sub> \u2192 Iodine heptafluoride<br \/>\nPreparation of Interhalogen Compounds :<br \/>\nBy direct combination ;<br \/>\nExample:<\/p>\n<p>Question 67.<br \/>\nDraw the structures of the following molecules :<br \/>\n(i) XeF<sub>6<\/sub> (ii) H<sub>2<\/sub>S<sub>2<\/sub>O<sub>7<\/sub> (All India 2013)<br \/>\nAnswer:<br \/>\n(i) XeF<sub>6<\/sub> :<\/p>\n<p>(ii) H<sub>2<\/sub>S<sub>2<\/sub>O<sub>7<\/sub> :<\/p>\n<p>Question 68.<br \/>\nDraw the structures of the following molecules :<br \/>\n(i) N<sub>2<\/sub>O<sub>5<\/sub> (ii) XeF<sub>2<\/sub> (All India 2013)<br \/>\nAnswer:<br \/>\n(i) N<sub>2<\/sub>O<sub>5<\/sub>:<\/p>\n<p>(ii) XeF<sub>2<\/sub> :<\/p>\n<p>Shape : Linear<br \/>\nAngle : F-Xe \u2013 F &gt; 90\u00b0<\/p>\n<p>Question 69.<br \/>\nExplain the following :<br \/>\n(a) NO<sub>2<\/sub> readily forms a dimer.<br \/>\n(b) BiClj is more stable than BiCl<sub>5<\/sub>. (Comptt. Delhi 2013)<br \/>\nAnswer:<br \/>\n(a) NO<sub>2<\/sub> contains 7 + 2 \u00d7 8 i.e. 23 odd electrons. In the valence shell N has seven electrons and hence less stable. To acquire stability it dimerizes to form N<sub>2<\/sub>O<sub>4<\/sub><\/p>\n<p>(b) BiCl<sub>3<\/sub> is more stable than BiCl<sub>5<\/sub> due to inert pair effect because as we move down the group, the stability of +3 oxidation state increases and of +5 decreases.<\/p>\n<p>Question 70.<br \/>\nComplete the following chemical equations :<br \/>\n(i) Ca<sub>3<\/sub>P<sub>2<\/sub> + H<sub>2<\/sub>O \u2192<br \/>\n(ii) Cu + H<sub>2<\/sub>SO<sub>4<\/sub> (cone.) \u2192 (Delhi 2014)<br \/>\nAnswer:<br \/>\n(i) Ca<sub>3<\/sub>P<sub>2<\/sub> + 6H<sub>2<\/sub>O \u2192 2PH<sub>3<\/sub> + 3Ca(OH)<sub>2<\/sub><br \/>\n(ii) Cu + 2H<sub>2<\/sub>SO<sub>4<\/sub> (cone.) \u2192 CuS0<sub>4<\/sub> + 2H<sub>2<\/sub>O + SO<sub>2<\/sub><\/p>\n<p>Question 71.<br \/>\nArrange the following in the order of property indicated against each set :<br \/>\n(i) HF, HCl, HBr, HI \u2013 increasing bond dissociation enthalpy.<br \/>\n(ii) H<sub>2<\/sub>O, H<sub>2<\/sub>S, H<sub>2<\/sub>Se, H<sub>2<\/sub>Te \u2013 increasing acidic character. (Delhi 2014)<br \/>\nAnswer:<br \/>\n(i) HI &lt; HBr &lt; HCl &lt; HF<br \/>\n(ii) H<sub>2<\/sub>O &lt; H<sub>2<\/sub>S &lt; H<sub>2<\/sub>Se &lt; H<sub>2<\/sub>Te<\/p>\n<p>Question 72.<br \/>\nComplete the following equations :<br \/>\n(i) P<sub>4<\/sub> + H<sub>2<\/sub>O \u2192<br \/>\n(ii) XeF<sub>4<\/sub> + O<sub>2<\/sub>F<sub>2 <\/sub> \u2192 (All India 2014)<br \/>\nAnswer:<br \/>\n(i) P<sub>4<\/sub> + 6H<sub>2<\/sub>O \u2192 2PH<sub>3<\/sub>+ 2H<sub>3<\/sub>PO<sub>3<\/sub><br \/>\n(ii) XeF<sub>4<\/sub> + O<sub>2<\/sub>F<sub>2<\/sub> \\(\\underrightarrow { 143K } \\) XeF<sub>6<\/sub> + O<sub>2<\/sub><\/p>\n<p>Question 73.<br \/>\nDraw the structures of the following :<br \/>\n(i) XeF<sub>2<\/sub> (ii) BrF<sub>3<\/sub> (All India 2014)<br \/>\nAnswer:<br \/>\n(i) XeF<sub>2<\/sub> : Refer to Q. 84 (ii), Page 125<br \/>\n(ii) BrF<sub>3<\/sub> (SP<sup>3<\/sup>d hybridization)<\/p>\n<p>Question 74.<br \/>\nComplete the following equations :<br \/>\n(i) Ag + PCl<sub>5<\/sub> \u2192<br \/>\n(ii) CaF<sub>2<\/sub> + H<sub>2<\/sub>SO<sub>4<\/sub> \u2192 (All India 2014)<br \/>\nAnswer:<br \/>\n(i) Ag + PCl<sub>5<\/sub> \u2192 2AgCl + PCl<sub>3<\/sub><br \/>\n(ii) CaF<sub>2<\/sub> + H<sub>2<\/sub>SO<sub>4<\/sub> \u2192 CaSO<sub>4<\/sub>+ 2HF<\/p>\n<p>Question 75.<br \/>\nDraw the structures of the following :<br \/>\n(i) XeF<sub>4<\/sub><br \/>\n(ii) HClO<sub>4<\/sub> (All India 2014)<br \/>\nAnswer:<br \/>\n(i) XeF<sub>4<\/sub>:<\/p>\n<p>(ii) HClO<sub>4<\/sub>:<\/p>\n<p>Question 76.<br \/>\nComplete the following equations :<br \/>\n(i) C + conc. H<sub>2<\/sub>SO<sub>4<\/sub> \u2192<br \/>\n(ii) XeF<sub>2<\/sub> + H<sub>2<\/sub>O \u2192 (All India 2014)<br \/>\nAnswer:<\/p>\n<p>Question 77.<br \/>\nDraw the structures of the following :<br \/>\n(i) XeO<sub>3<\/sub> (ii) H<sub>2<\/sub>SO<sub>4<\/sub> (All India 2014)<br \/>\nAnswer:<br \/>\n(i) XeO<sub>3<\/sub> :<\/p>\n<p>(ii) H<sub>2<\/sub>SO<sub>4<\/sub>:<\/p>\n<p>Question 78.<br \/>\nDraw the structure of each of the following:<br \/>\n(i) H<sub>2<\/sub>SO<sub>4<\/sub><br \/>\n(ii) Solid PCl<sub>5<\/sub> (Comptt. Delhi 2014)<br \/>\nAnswer:<br \/>\n(i) H<sub>2<\/sub>SO<sub>4<\/sub>:<\/p>\n<p>(ii) PCl<sub>5<\/sub> (s):<\/p>\n<p>Question 79.<br \/>\nDraw the structures of the following compounds :<br \/>\n(i) H<sub>2<\/sub>SO<sub>3<\/sub><br \/>\n(ii) N<sub>2<\/sub>O<sub>5<\/sub> (Comptt. Delhi 2014)<br \/>\nAnswer:<\/p>\n<p>Question 80.<br \/>\nComplete the following chemical equations :<br \/>\n(i) PCl<sub>5<\/sub> \\(\\underrightarrow { Heat } \\)<br \/>\n(ii) NaHCO<sub>3<\/sub> + HCl \u2192 (Comptt. Delhi 2014)<br \/>\nAnswer:<\/p>\n<p>Question 81.<br \/>\nComplete the following chemical equations : (Comptt. All India 2014)<br \/>\n(i) SO<sub>2<\/sub> + MnO<sub>4<\/sub><sup>\u2013<\/sup> + H<sub>2<\/sub>O \u2192<br \/>\n(ii) F<sub>2<\/sub> (g) + H<sub>2<\/sub>O (l) \u2192<br \/>\nAnswer:<\/p>\n<p>Question 82.<br \/>\nComplete the following chemical reaction equations : (Comptt. All India 2014)<\/p>\n<p>Answer:<\/p>\n<p>Question 83.<br \/>\nComplete the following chemical equations : (Comptt. All India 2014)<\/p>\n<p>Answer:<\/p>\n<p>Question 84.<br \/>\nDraw the structures of the following :<br \/>\n(i) H<sub>2<\/sub>SO<sub>4<\/sub><br \/>\n(ii) XeF<sub>2<\/sub> (Comptt. Delhi 2015)<br \/>\nAnswer:<\/p>\n<p>Question 85.<br \/>\nExplain the following :<br \/>\n(i) Nitrogen is much less reactive than phosphorus.<br \/>\n(ii) NF<sub>3<\/sub> is an exothermic compound but NCl<sub>3<\/sub> is an endothermic compound. (Comptt All India 2015)<br \/>\nAnswer:<br \/>\n(i) Due to presence of weak single bond in P \u2013 P than N = N, phosphorous is more reactive than nitrogen and also because of high bond dissociation enthalpy of N = N.<\/p>\n<p>(ii) Due to smaller size of F as compared to Cl, the N \u2013 F bond is much stronger than N \u2013 Cl bond while bond dissociation energy of F2 is much lower than that of Cl<sub>2<\/sub>. Therefore, energy released during the formation of NF<sub>3<\/sub> molecule is more than the energy needed to break N<sub>2<\/sub> and F<sub>2<\/sub> molecules into individual atoms. In other words, formation of NF<sub>3<\/sub> is an exothermic reaction.<br \/>\nThe energy released during the formation of NCl<sub>3<\/sub> molecule is less than the energy needed to break N<sub>2<\/sub> and Cl<sub>2<\/sub> molecules into individual atoms. Thus formation of NCl<sub>3<\/sub> is an endothermic reaction.<\/p>\n<p>Question 86.<br \/>\nWhat happens when:<br \/>\n(i) SO<sub>2<\/sub> gas is passed through an aqueous solution of Fe<sup>3+<\/sup> salt?<br \/>\n(ii) XeF<sub>4<\/sub> reacts with SbF<sub>5<\/sub>? (All India 2016)<br \/>\nAnswer:<br \/>\n(i) In this sulphur dioxide acts as a reducing agent and reduces Fe<sup>3+<\/sup> to Fe<sup>2+<\/sup>.<br \/>\n2Fe<sup>3+<\/sup> + SO<sub>2<\/sub> + 2H<sub>2<\/sub>O \u2192 2Fe<sup>2+<\/sup> + SO<sub>4<\/sub><sup>2-<\/sup> + 4H<sup>+<\/sup><br \/>\n(ii) XeF<sub>4<\/sub> + SbF<sub>5<\/sub> \u2192 [XeF<sub>3<\/sub>]<sup>+<\/sup> [SbF<sub>6<\/sub>]<sup>\u2013<\/sup><\/p>\n<p>Question 87.<br \/>\nDraw the structures of the following compounds<br \/>\n(i) BrF<sub>3<\/sub> (ii) XeF<sub>4<\/sub> (All India, Comptt. Delhi 2016)<br \/>\nAnswer:<\/p>\n<p>Question 88.<br \/>\nWhat happens when :<br \/>\n(i) Concentrated H<sub>2<\/sub>SO<sub>4<\/sub> is added to calcium fluoride?<br \/>\n(ii) SO<sub>3<\/sub> is passed through water? (Comptt. All India 2016)<br \/>\nAnswer:<br \/>\n(i) Cone. H<sub>2<\/sub>SO<sub>4<\/sub> reacts with CaF<sub>2<\/sub> giving Hydrogen Fluoride<br \/>\nCaF<sub>2<\/sub> + H<sub>2<\/sub>SO<sub>4<\/sub> \u2192 CaSO<sub>4<\/sub> + 2HF<br \/>\n(ii) SO<sub>3<\/sub> passed in water giving Sulphuric Acid<br \/>\nSO<sub>3<\/sub> + H<sub>2<\/sub>O \u2192 H<sub>2<\/sub>SO<sub>4<\/sub><\/p>\n<p>Question 89.<br \/>\nComplete the following reactions:<br \/>\n(i) NH<sub>3<\/sub> + 3CI<sub>3<\/sub> (excess) \u2192<br \/>\n(ii) XeF<sub>6<\/sub> + 2H<sub>2<\/sub>O \u2192 (Delhi 2017)<br \/>\nAnswer:<\/p>\n<p>Question 90.<br \/>\nWhat happens when H<sub>3<\/sub>PO<sub>3<\/sub> is heated? (Delhi 2017)<br \/>\nAnswer:<\/p>\n<p>Question 91.<br \/>\nDraw the structures of the following:<br \/>\n(i) H<sub>2<\/sub>S<sub>2<\/sub>O<sub>7 <\/sub>(ii) XeF<sub>6<\/sub><br \/>\nAnswer:<br \/>\n(i) H<sub>2<\/sub>S<sub>2<\/sub>O<sub>7<\/sub> (Pyrosulphuric acid or Oleum)<\/p>\n<p>(ii)<\/p>\n<p>Question 92.<br \/>\nDraw the structures of the following:<br \/>\n(i) H<sub>3<\/sub>PO<sub>2<\/sub> (ii) XeF<sub>4<\/sub> (Delhi 2017)<br \/>\nAnswer:<\/p>\n<p>Question 93.<br \/>\nComplete the following reactions: (Delhi 2017)<br \/>\n(i) Cl<sub>2<\/sub> + H<sub>2<\/sub>O \u2192<br \/>\n(ii) XeF<sub>6<\/sub> + 3H<sub>2<\/sub>O \u2192<br \/>\nAnswer:<br \/>\n(i) Cl<sub>2<\/sub> + H<sub>2<\/sub>O \u2192 [HCl + HOCl] \u2192 2HCl + [O]\n(ii) XeF<sub>6<\/sub> + 3H<sub>2<\/sub>O \u2192 XeO<sub>3<\/sub> + 6HF<\/p>\n<p>Question 94.<br \/>\nWhat happens when<br \/>\n(i) cone. H<sub>2<\/sub>SO<sub>4<\/sub> is added to Cu?<br \/>\n(ii) SO<sub>3<\/sub> is passed through water? Write the equations. (Delhi 2017)<br \/>\nAnswer:<br \/>\n(i) Cu + 2H<sub>2<\/sub>SO<sub>4<\/sub> (cone) \u2192 CuSO<sub>4<\/sub> + 2H<sub>2<\/sub>O+SO<sub>2<\/sub><br \/>\n(ii) SO<sub>3<\/sub> + H<sub>2<\/sub>O \u2192 H<sub>2<\/sub>SO<sub>4<\/sub><\/p>\n<p>Question 95.<br \/>\nDraw the structures of the following:<br \/>\n(i) H<sub>4<\/sub>P<sub>2<\/sub>O<sub>7<\/sub><br \/>\n(ii) XeOF<sub>4<\/sub><br \/>\nAnswer:<br \/>\n(i) H<sub>4<\/sub>P<sub>2<\/sub>O<sub>7<\/sub><\/p>\n<p>(ii)<\/p>\n<p>Question 96.<br \/>\nComplete the following chemical equations:<br \/>\n(i) F<sub>2<\/sub> + 2Cl<sup>\u2013<\/sup> \u2192<br \/>\n(ii) 2XeF<sub>2<\/sub> + 2H<sub>4<\/sub>O \u2192 (Delhi 2017)<br \/>\nAnswer:<br \/>\n(i) F<sub>2<\/sub>+ 2Cl<sup>\u2013<\/sup> \u2192 Cl<sub>2<\/sub> + 2F<sup>\u2013<\/sup><br \/>\n(ii) 2XeF<sub>2<\/sub> + 2H<sub>2<\/sub>O \u2192 2Xe + 4HF + O<sub>2<\/sub><\/p>\n<p>Question 97.<br \/>\nWhat happens when<br \/>\n(i) HCl is added to MnO<sub>2<\/sub>?<br \/>\n(it) PCl<sub>5<\/sub>is heated? Write the equations involved. (Delhi 2017)<br \/>\nAnswer:<br \/>\n(i) When HCl is added to MnO<sub>2<\/sub>, chlorine gas is formed, along with other products.<br \/>\nMnO<sub>2<\/sub> + 4HCl \u2192 MnCl<sub>2<\/sub> + 2H<sub>2<\/sub>O + Cl<sub>2<\/sub><br \/>\n(ii) PCl<sub>5<\/sub> \\(\\stackrel{\\Delta}{\\longrightarrow}\\) PCl<sub>3<\/sub> + Cl<sub>2<\/sub><\/p>\n<p>Question 98.<br \/>\nDraw the structures of the following:<br \/>\n(i) H<sub>2<\/sub>SO<sub>3<\/sub> (ii) HClO<sub>3<\/sub> (All India 2017)<br \/>\nAnswer:<br \/>\n(i) Structure of H<sub>2<\/sub>SO<sub>3<\/sub> (Sulphurous acid):<\/p>\n<p>Question 99<br \/>\nDraw the structures of the following:<br \/>\n(a) H<sub>2<\/sub>S<sub>2<\/sub>O<sub>8<\/sub> (b) ClF<sub>3<\/sub> (All India 2017)<br \/>\nAnswer:<br \/>\n(a) Structure of H<sub>2<\/sub>S<sub>2<\/sub>O<sub>8<\/sub>:<\/p>\n<p>Question 100.<br \/>\nDraw the structures of the following:<br \/>\n(a) XeF<sub>4<\/sub> (b) BrF<sub>5<\/sub> (All India 2017)<br \/>\nAnswer:<br \/>\n(i) Structure of BrF<sub>5<\/sub><\/p>\n<p>Question 101.<br \/>\n\u201cOrthophosphoric acid (H<sub>3<\/sub>PO<sub>4<\/sub>) is non-reducing whereas hypophosphorus acid (H<sub>3<\/sub>PO<sub>2<\/sub>) is a strong reducing agent.\u201d Explain and justify the above statement with suitable example. (Comptt. Delhi 2017)<br \/>\nAnswer:<br \/>\nOrthophosphoric add (H<sub>3<\/sub>PO<sub>4<\/sub>) is not a reducing agent because it doesn\u2019t contain any P-H bond whereas hypophosphorus acid (H<sub>3<\/sub>PO<sub>2<\/sub>) is a strong reducing agent as it contains two P-H bonds. H<sub>3<\/sub>PO<sub>2<\/sub> can reduce silver nitrate (AgNO<sub>3<\/sub>) into metallic silver which H<sub>3<\/sub>PO<sub>4<\/sub> can not.<br \/>\n4AgNO<sub>3<\/sub> + H<sub>3<\/sub>PO<sub>2<\/sub> + 2H<sub>2<\/sub>O \u2192 4Ag \u2193 + H<sub>3<\/sub>PO<sub>4<\/sub> + 4HNO<sub>3<\/sub><\/p>\n<p>Question 102.<br \/>\n(a) What is the covalence of nitrogen in N<sub>2<\/sub>O<sub>5<\/sub>?<br \/>\n(b) BiH<sub>3<\/sub> is a stronger reducing agent than SbH<sub>3<\/sub>, why? (Comptt. Delhi 2017)<br \/>\nAnswer:<br \/>\n(a) The covalency of nitrogen in N<sub>2<\/sub>O<sub>5<\/sub> is 4 because each nitrogen atom has four shared pairs of electrons.<br \/>\n(b) BiH<sub>3<\/sub>: Because it is a stronger reducing agent as its tendency to liberate H is maximum.<\/p>\n<p>Question 103.<br \/>\nAccount for the following:<br \/>\n(i) The two oxygen-oxygen bond lengths in ozone molecule are identical.<br \/>\n(ii) Most of the reactions of fluorine are exothermic. (Comptt. Delhi 2017)<br \/>\nAnswer:<br \/>\n(i) Due to resonance the two oxygen atoms have partial double bond character and thus have same bond length i.e. 128 pm<\/p>\n<p>(ii) Due to much higher electrode potential, high electro-negativity and low bond dissociation enthalpy of F<sub>2<\/sub>.<\/p>\n<p>Question 104.<br \/>\nAccount for the following :<br \/>\n(i) Two S-O bond lengths in SO<sub>2<\/sub> are equal.<br \/>\n(ii) Fluorine shows only -1 oxidation state in its compounds. (Comptt. Delhi 2017)<br \/>\nAnswer:<br \/>\n(i) Due to resonance in SO<sub>2<\/sub> the double bond (\u03c0) electrons are distributed equally in both resonating structures as a result of which the bond length of two S-O becomes equal.<\/p>\n<p>Because it is most electronegative element and does not have d-orbitals for octet expansion, therefore it shows only a negative oxidation state of -1.<\/p>\n<p>Question 105.<br \/>\nAccount for the following:<br \/>\n(i) Bond angle is \\(\\mathrm{NH}_{4}^{+}\\) is higher than that in NH<sub>3<\/sub>.<br \/>\n(ii) ICl is more reactive than I<sub>2<\/sub>. (Comptt. Delhi 2017)<br \/>\nAnswer:<br \/>\n(i) Because in \\(\\mathrm{NH}_{4}^{+}\\) ion there is no lone pair of electrons which is present in NH<sub>3<\/sub> due to which lone pair-bond pair repulsion occurs and bond angle decreases from 109\u00b028\u2032 to 107.3\u00b0.<br \/>\n(ii) Because I-Cl bond is weaker than I-1 bond as a result of which ICl breaks easily to form halogen atoms which readily bring about the reaction, hence more reactive.<\/p>\n<p>Question 106.<br \/>\n\u201cOrthophosphoric acid (H<sub>3<\/sub>PO<sub>4<\/sub>) is not a reducing agent whereas hypophosphorus acid (H<sub>3<\/sub>PO<sub>2<\/sub>) is a strong reducing agent.\u201d Explain and justify the above statement with the help of a suitable example. (Comptt. All India 2017)<br \/>\nAnswer:<br \/>\nOrthophosphoric add (H<sub>3<\/sub>PO<sub>4<\/sub>) is not a reducing agent because it doesn\u2019t contain any P-H bond whereas hypophosphorus acid (H<sub>3<\/sub>PO<sub>2<\/sub>) is a strong reducing agent as it contains two P-H bonds. H<sub>3<\/sub>PO<sub>2<\/sub> can reduce silver nitrate (AgNO<sub>3<\/sub>) into metallic silver which H<sub>3<\/sub>PO<sub>4<\/sub> can not.<br \/>\n4AgNO<sub>3<\/sub> + H<sub>3<\/sub>PO<sub>2<\/sub> + 2H<sub>2<\/sub>O \u2192 4Ag \u2193 + H<sub>3<\/sub>PO<sub>4<\/sub> + 4HNO<sub>3<\/sub><\/p>\n<h4><span class=\"ez-toc-section\" id=\"The_p-Block_Elements_Class_12_Important_Questions_Short_Answer_Type_-II_SA-II\"><\/span>The p-Block Elements Class 12 Important Questions Short Answer Type -II [SA-II]<span class=\"ez-toc-section-end\"><\/span><\/h4>\n<p>Question 107.<br \/>\nAccount for the following :<br \/>\n(i) NH<sub>3<\/sub> is a stronger base than PH<sub>3<\/sub>.<br \/>\n(ii) Sulphur has a greater tendency for catenation than oxygen.<br \/>\n(iii) Bond dissociation energy of F<sub>2<\/sub> is less than that of Cl<sub>3<\/sub>. (Delhi 2009)<br \/>\nAnswer:<br \/>\n(i) Since both P and N contain lone pairs of electrons but due to small size and high electronegativity of Nitrogen in NH<sub>3<\/sub>, the electron density is much higher than PH<sub>3<\/sub>, therefore it can easily donate electrons and acts as strong Lewis base than PH<sub>3<\/sub>.<\/p>\n<p>(ii) The greater catenation tendency of sulphur is due to two reasons :<br \/>\n(a) The lone pair of electrons feels more repulsion in 0-0 bond than S-S bond due to its small size and thus S-S forms strong bond.<br \/>\n(b) As the size of atom increases down the group from O \u2013 PO, the strength of bond increases and therefore catenation tendency also increases.<\/p>\n<p>(iii) Due to smaller size of F than Cl as a result of which electron-electron repulsions between the lone pairs of electrons are very large than that of Cl, hence bond dissociation energy of F<sub>2<\/sub> is less than that of Cl<sub>2<\/sub>.<\/p>\n<p>Question 108.<br \/>\nExplain the following situations :<br \/>\n(i) In the structure of HNO<sub>3<\/sub> molecule, the N-O bond (121 pm) is shorter than N \u2013 OH bond (140 pm).<br \/>\n(ii) SF<sub>4<\/sub> is easily hydrolysed whereas SF<sub>6<\/sub> is not easily hydrolysed.<br \/>\n(iii) XeF<sub>2<\/sub> has a straight linear structure and not a bent angular structure. (Delhi 2009)<br \/>\nAnswer:<br \/>\n(i)<\/p>\n<p>In the structure the bond length of N-O is shorter due to formation of coordinate bond and double bond while in N-OH the bond is single covalent due to which its bond length is greater than other N-O bond.<br \/>\n(ii) In SF<sub>4<\/sub>, due to less steric hindrance by four F atoms, H<sub>2<\/sub>O molecules can attack easily while in SF<sub>6<\/sub> tire S atom is completely protected by six F atoms and does not allow H<sub>2<\/sub>O molecules to attack the S atom.<br \/>\n(iii) In XeF<sub>2<\/sub> there are 2 bond pairs and 3 lone pairs and thus show sp<sup>3<\/sup> d hybridization. It has linear geometry.<\/p>\n<p>Question 109.<br \/>\nExplain the following observations :<br \/>\n(i) Fluorine does not exhibit any positive oxidation state.<br \/>\n(ii) The majority of known noble gas compounds are those of Xenon.<br \/>\n(iii) Phosphorus is much more reactive than nitrogen. (Delhi 2009)<br \/>\nAnswer:<br \/>\n(i) Because it is most electronegative element and does not have d-orbitals for octet expansion, therefore it shows only a negative oxidation state of -1.<br \/>\n(ii) Because xenon has least ionization energy among noble gases and hence it readily forms chemical compounds particularly with oxygen and fluorine.<br \/>\n(iii) Because P-P single bond is much weaker than N = N triple bond and the bond length of nitrogen is small and bond dissociation energy is very large which makes it inert and unreactive and thus phosphorus becomes more reactive.<\/p>\n<p>Question 110.<br \/>\nHow would you account for the following :<br \/>\n(i) NCl<sub>3<\/sub> is an endothermic compound while NF<sub>3<\/sub> is an exothermic one.<br \/>\n(ii) XeF<sub>2<\/sub> is a linear molecule without a bend.<br \/>\n(iii) The electron gain enthalpy with negative sign for fluorine is less than that for chlorine, still fluorine is a stronger oxidising agent than chlorine. (All India 2010)<br \/>\nAnswer:<br \/>\n(i) F is more electronegative than Cl. The difference in the electronegativity between N and F is much more than the difference between electronegativity of N and Cl. So there is need of much more energy to break the N-F bond.<br \/>\n(ii) In XeF<sub>2<\/sub> there are 2 bond pairs and 3 lone pairs and thus show sp<sup>3<\/sup> d hybridization. It has linear geometry.<\/p>\n<p>(iii) Because of small size of flourine atom and strong electron-electron repulsions in its compact 2p orbitals.<\/p>\n<p>Question 111.<br \/>\nHow would you account for the following :<br \/>\n(i) The electron gain enthalpy with negative sign is less for oxygen than that for sulphur.<br \/>\n(ii) Phosphorus shows greater tendency for catenation than nitrogen.<br \/>\n(iii) Fluorine never acts as the central atom in polyatomic interhalogen compounds. (All India 2010)<br \/>\nAnswer:<br \/>\n(i) The least negative electron gain enthalpy of oxygen is due to small size and more interelectronic repulsion with coming electron.<br \/>\n(ii) The bond strength of P-P is more than N-N, therefore phosphorus shows more tendency for catenation than nitrogen.<br \/>\n(iii) Because F being smaller, it cannot accomodate larger sized other halogen atoms around it. Due to the absence of d-orbitals, F does not show positive oxidation state of +3, +5, +7 needed for the formation of polyatomic interhalogen compounds.<\/p>\n<p>Question 112.<br \/>\nHow would you account for the following :<br \/>\n(i) H<sub>2<\/sub>S is more acidic than H<sub>2<\/sub>O.<br \/>\n(ii) The N-O bond in \\(\\mathbf{N O}_{2}^{-}\\) is shorter than the N-O bond in \\(\\mathbf{N O}_{3}^{-}\\).<br \/>\n(iii) Both O<sub>2<\/sub> and F<sub>2<\/sub> stabilize high oxidation states but the ability of oxygen to stabilize the higher oxidation state exceeds that of fluorine. (All India 2011)<br \/>\nAnswer:<br \/>\n(i) Since the size of sulphur is more than oxygen, S-H bond length increases and hence bond dissociation energy of S-H is less than O-H. Therefore S-FI easily loses H+ and thus is more acidic than H<sub>2<\/sub>O.<br \/>\n(ii) The resonating structure of \\(\\mathrm{NO}_{2}^{-}\\) and \\(\\mathrm{NO}_{3}^{-}\\) show that in \\(\\mathrm{NO}_{2}^{-}\\) two bonds are sharing a double bond while in \\(\\mathrm{NO}_{3}^{-}\\), 3 bonds are sharing a double bond. That\u2019s why \\(\\mathrm{NO}_{2}^{-}\\) has shorter bond than that of \\(\\mathrm{NO}_{3}^{-}\\).<br \/>\nAnswer:<\/p>\n<p>(iii) Oxygen stabilizes the highest oxidation state even more than fluorine.<br \/>\nExample : Highest fluoride of Mn is MnF<sub>4<\/sub> whereas highest oxide is Mn<sub>2<\/sub>O<sub>7<\/sub>. It is due to ability of oxygen to form multiple bonds with the metal atoms.<\/p>\n<p>Question 113.<br \/>\nHow would you account for the following :<br \/>\n(i) NF<sub>3<\/sub> is an exothermic compound but NCl<sub>3<\/sub> is not.<br \/>\n(ii) The acidic strength of compounds increases in the order :<br \/>\nPH<sub>3<\/sub> &lt; H<sub>2<\/sub>S &lt; HCl.<br \/>\n(iii) SF<sub>6<\/sub> is kinetically inert. (All India 2011)<br \/>\nAnswer:<br \/>\n(i) F is more electronegative than Cl. The difference in the electronegativity between N and F is much more than the difference between electronegativity of N and Cl. So there is need of much more energy to break the N-F bond.<br \/>\n(ii) As the electronegativity increases in the same period from left to right so their electronegativity are in the increasing order, P &lt; S &lt; Cl.<br \/>\nIn the same way the acid strength is also in the increasing order i.e. PH<sub>3<\/sub> &lt; H<sub>2<\/sub>S &lt; HCl.<br \/>\n(iii) Because SF<sub>6<\/sub> is showing steric hindrance due to 6 (six) fluorine atoms which make it unable to react further with any other atom.<\/p>\n<p>Question 114.<br \/>\nGive reasons for the following:<br \/>\n(i) Where R is an alkyl group, R<sub>3<\/sub>P = O exists but R<sub>3<\/sub>N = O does not.<br \/>\n(ii) PbCl<sub>4<\/sub> is more covalent than PbCl<sub>2<\/sub>.<br \/>\n(iii) At room temperature, N<sub>2<\/sub> is much less reactive. (All India 2013)<br \/>\nAnswer:<br \/>\n(i) Due to presence of d-orbitals in P, it can form p\u03c0 -d\u03c0 bonds and can extend its covalency beyond 4 while N cannot do so due to absence of d-orbitals.<br \/>\n(ii) According to Fajan\u2019s rule, highly charged Pb<sup>4+<\/sup> can polarise the anion i.e., Cl <sup>\u2013<\/sup> more effectively than Pb<sup>2+<\/sup> and hence PbCl<sub>4<\/sub> becomes more covalent than PbCl<sub>2<\/sub>.<br \/>\n(iii) Due to presence of triple bonds between 2 N atoms, their bond length decreases and hence bond dissociation energy increases which makes N<sub>2<\/sub> lesser reactive. While in phosphorus due to presence of single bond, more bond length, bond dissociation energy is low, hence very reactive.<\/p>\n<p>Question 115.<br \/>\nGive reasons for the following :<br \/>\n(i) Though nitrogen exhibits +5 oxidation state, it does not form pentahalide.<br \/>\n(ii) Electron gain enthalpy with negative sign of fluorine is less than that of chlorine.<br \/>\n(iii) The two oxygen-oxygen bond lengths in ozone molecule are identical. (All India 2013)<br \/>\nAnswer:<br \/>\n(i) Due to absence of empty d-orbitals, N<sub>2<\/sub> does not form pentahalide.<br \/>\n(ii) Because of small size of flourine atom and strong electron-electron repulsions in its compact 2p orbitals.<br \/>\n(iii) Due to resonance the two oxygen atoms have partial double bond character and thus have same bond length i.e. 128 pm<\/p>\n<p>Question 116.<br \/>\n(a) Name the gas evolved on heating ammonium nitrate. Write the chemical reaction.<br \/>\n(b) Write two uses of ammonium nitrate. (Comptt. All India 2013)<br \/>\nAnswer:<br \/>\nThe gas evolved on heating is Nitrous oxide<\/p>\n<p>(b) Uses of NH<sub>4<\/sub>NO<sub>3<\/sub><\/p>\n<ul>\n<li>It is used in fertilizers.<\/li>\n<li>It is used in explosives.<\/li>\n<\/ul>\n<p>Question 117.<br \/>\nAccount for the following :<br \/>\n(i) NF<sub>3<\/sub> is an exothermic compound but NCl<sub>3<\/sub> is an endothermic compound.<br \/>\n(ii) HF is not stored in glass bottles but is kept in wax-coated bottles.<br \/>\n(iii) Bleaching of flowers by Cl<sub>2<\/sub> is permanent while that of SO<sub>2<\/sub> is temporary. (Comptt. All India 2013)<br \/>\nAnswer:<br \/>\n(i) F is more electronegative than Cl. The difference in the electronegativity between N and F is much more than the difference between electronegativity of N and Cl. So there is need of much more energy to break the N-F bond.<br \/>\n(ii) HF is highly corrosive and etches glass hence it is kept in wax-coated bottles.<br \/>\n(iii) Chlorine bleaches the material by oxidation hence it is permanent while SO<sub>2<\/sub> bleaches the material by reduction and as the material is exposed to air, it gets oxidised and the colour is restored, hence it is temporary.<\/p>\n<p>Question 118.<br \/>\n(a). With the help of chemical equations explain the principle of contact process in brief for the manufacture of sulphuric acid by contact process.<br \/>\n(b) Bismuth is a strong oxidizing agent in the pentavalent state. Explain. (Comptt. All India 2013)<br \/>\nAnswer:<br \/>\n(a) Contact Process : Burning sulphur in an excess of air<br \/>\nS + O<sub>2<\/sub> \u2192 SO<sub>2<\/sub> (g)<br \/>\nor, By heating sulphide ores like pyrites in an excess of air :<br \/>\n4FeS<sub>2<\/sub> + 11O<sub>2<\/sub> \u2192 2Fe<sub>2<\/sub>O<sub>3<\/sub> + 8SO<sub>2<\/sub><br \/>\nIn either case, an excess of air is used so that the SO<sub>2<\/sub> produced is already mixed with oxygen for the next stage. This is reversible reaction and the formation of SO<sub>3<\/sub> is exothermic in the presence of catalyst V<sub>2<\/sub>O<sub>5<\/sub> at 720 K<br \/>\n2SO<sub>2<\/sub> + O<sub>2<\/sub> \u21cc 2SO<sub>3<\/sub> \u0394H = -196 KJ\/mol<br \/>\nThis cannot be done by simply adding water to the S03. The reaction is so uncontrollable that it creates a fog of H2S04. Instead, the S03 is first dissolved in cone. H2S04.<br \/>\nH<sub>2<\/sub>SO<sub>4<\/sub> + SO<sub>3<\/sub> \u2192 H<sub>2<\/sub>S<sub>2<\/sub>O<sub>7<\/sub><br \/>\nThe product is known as fuming sulphuric acid or oleum to which water is added to get H<sub>2<\/sub>SO<sub>4<\/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>(b) The stability of +5 oxidation state decreases and that of +3 state increases due to inert pair effect down the group therefore Bi (V) accepts two electrons and gets reduced to Bi (III).<br \/>\nBi<sup>5+<\/sup> + 2e<sup>\u2013<\/sup> \u2192 Bi<sup>3+<\/sup> :<br \/>\nSo, Bi(V) is more stronger oxidising agent.<\/p>\n<p>Question 119.<br \/>\n(a) Draw the structures of the following molecules :<br \/>\n(i) XeOF<sub>4<\/sub> (ii) H<sub>2<\/sub>SO<sub>4<\/sub><br \/>\n(b) Write the structural difference between white phosphorus and red phosphorus. (Delhi 2014)<br \/>\nAnswer:<br \/>\n(a)<\/p>\n<p>(b) White phosphorus exists as discrete P4 units with SP3 hybridized phosphorus atom, arranged tetrahedrally but in red phosphorus all P<sub>4<\/sub> tetrahedral units are linked with each other to form polymeric structure.<\/p>\n<p>Question 120.<br \/>\nAccount for the following :<br \/>\n(i) PCl<sub>5<\/sub> is more covalent than PCl<sub>3<\/sub>.<br \/>\n(ii) Iron on reaction with HCl forms FeCl<sub>2<\/sub> and not FeCl<sub>3<\/sub>.<br \/>\n(iii) The two 0-0 bond lengths in the ozone molecule are equal. (Delhi 2014)<br \/>\nAnswer:<br \/>\n(i) In PCl<sub>5<\/sub>, phosphorus has +5 oxidation state and has less tendency to loose electrons than in +3 of PCl<sub>3<\/sub>. Therefore, PCl<sub>5<\/sub> has more tendency to share e<sup>-1<\/sup>s than PCl<sub>3<\/sub>.<br \/>\n(ii) Because HCl on reaction with iron liberates H<sub>2<\/sub> gas which prevents the formation of ferric chloride.<br \/>\n(iii) Due to resonance the two oxygen atoms have partial double bond character and thus have same bond length i.e. 128 pm<\/p>\n<p>Question 121.<br \/>\nAccount for the following :<br \/>\n(i) Bi(V) is a stronger oxidizing agent than Sb(V).<br \/>\n(ii) N \u2013 N single bond is weaker than P \u2013 P single bond.<br \/>\n(iii) Noble gases have very low boiling points. (Delhi 2014)<br \/>\nAnswer:<br \/>\n(i) Bi(V) is a stronger oxidizing agent than Sb(V) due to inert pair effect as the stability of lower oxidation state (+3) increases down the group.<br \/>\n(ii) Due to smaller size of Nitrogen, their lone pairs repel the bond pair of N \u2013 N bond while P \u2013 P due to bigger size does not show more repulsion.<br \/>\n(iii) Due to presence of weak Van der waal forces of attraction, noble gases have very low boiling point.<\/p>\n<p>Question 122.<br \/>\nAccount for the following :<br \/>\n(i) Sulphur in vapour form exhibits paramagnetic behaviour.<br \/>\n(ii) SnCl<sub>4<\/sub> is more covalent than SnCl<sub>2<\/sub>.<br \/>\n(iii) H<sub>3<\/sub>PO<sub>2<\/sub> is a stronger reducing agent than H<sub>3<\/sub>PO&lt;sub3. (Delhi 2014)<br \/>\nAnswer:<br \/>\n(i) In vapour state sulphur partly exists as S2 molecule which has two unpaired electrons in the antibonding II orbitals and hence exhibits paramagnetism.<br \/>\n(ii) Sn<sup>+4<\/sup> in SnCl<sub>4<\/sub> has more polarising power than SnCl<sub>2<\/sub><br \/>\n(iii) H<sub>3<\/sub>PO<sub>2<\/sub> contains two P-H bonds while H<sub>3<\/sub>PO<sub>3<\/sub> contains only one P-H bond therefore H<sub>3<\/sub>PO<sub>2<\/sub> is stronger reducing agent.<\/p>\n<p>Question 123.<br \/>\nGive reasons for the following :<br \/>\n(i) (CH<sub>3<\/sub>)<sub>3<\/sub> P = O exists but (CH<sub>3<\/sub>)<sub>3<\/sub> N = O does not.<br \/>\n(ii) Oxygen has less electron gain enthalpy with negative sign than sulphur.<br \/>\n(iii) H<sub>3<\/sub>PO<sub>2<\/sub> is a stronger reducing agent than H<sub>3<\/sub>PO<sub>3<\/sub>. (All India 2014)<br \/>\nAnswer:<br \/>\n(i) (CH<sub>3<\/sub>)<sub>3<\/sub>P = 0 exists due to presence of empty d-orbitals and thus can expand its covalency upto 6 but (CH<sub>3<\/sub>)<sub>3<\/sub> N = O cannot expand its covalency due to absence of d-orbitals.<br \/>\n(ii) The least negative electron gains enthalpy of oxygen is due to small size and more interelectronic repulsion with coming electron.<br \/>\n(iii) H<sub>3<\/sub>PO<sub>2<\/sub> contains two P-H bonds while H<sub>3<\/sub>PO<sub>3<\/sub> contains only one P-H bond therefore H<sub>3<\/sub>PO<sub>2<\/sub> is a stronger reducing agent.<\/p>\n<p>Question 124.<br \/>\nGive reasons:<br \/>\n(i) SO<sub>2<\/sub> is reducing while TeO<sub>2<\/sub> is an oxidizing agent.<br \/>\n(ii) Nitrogen does not form pentahalide.<br \/>\n(iii) ICl is more reactive than I<sub>2<\/sub>. (All India 2016)<br \/>\nAnswer:<br \/>\n(i) SO<sub>2<\/sub> is reducing while TeO<sub>2<\/sub> is an oxidising agent because sulphur can expand its covalency upto +6 from +4 due to presence of empty d-orbital but as we move down the group the stability of +6 oxidation state decreases and of +4 oxidation state increases due to inert pair effect. Hence SO<sub>2<\/sub> acts as reducing agent while TeO<sub>2<\/sub> acts as an oxidising agent.<br \/>\n(ii) Due to absence of empty d-orbitals, N<sub>2<\/sub> does not form pentahalides.<br \/>\n(iii) Because ICl bond is weaker than I -I bond as a result of which ICl breaks easily to form halogen atoms which readily bring about the reaction, hence more reactive.<\/p>\n<p>Question 125.<br \/>\nGive reasons:<br \/>\n(i) Thermal stability decreases from H<sub>2<\/sub>O to H<sub>2<\/sub>Te.<br \/>\n(ii) Fluoride ion has higher hydration enthalpy than chloride ion.<br \/>\n(iii) Nitrogen does not form pentahalide. (Delhi 2017)<br \/>\nAnswer:<br \/>\n(i) Thermal stability decreases from H<sub>2<\/sub>O to H<sub>2<\/sub>Te due to weakening of bond between hydrogen and the atom from O to T<sub>e<\/sub> as size is increasing down the group.<br \/>\n(ii) Fluoride ion has higher hydration enthalpy than chloride ion due to stronger attractions of smaller in size fluoride ion.<br \/>\n(iii) Nitrogen does not contain\u2019d\u2019 orbitals.<\/p>\n<p>Question 126.<br \/>\nGive reasons for the following:<br \/>\n(a) Red phosphorus is less reactive than white phosphorus.<br \/>\n(b) Electron gain enthalpies of halogens are largely negative.<br \/>\n(c) N<sub>2<\/sub>O<sub>5<\/sub> is more acidic than N<sub>2<\/sub>O<sub>3<\/sub>.(All India 2017)<br \/>\nAnswer:<br \/>\n(a) Red phosphorus is less reactive than white phosphorus because white phosphorus possess angle strain where long angles are only 60\u00b0 making it more reactive. Also, red phosphorus being polymeric is less reactive than white phosphorus which has discrete tetrahedral structure.<br \/>\n(b) Electron gain enthalpies of halogens are largely negative due to high effective nuclear charge and smaller size among period. They readily accept an electron to attain noble gas configuration.<br \/>\n(c) N<sub>2<\/sub>O<sub>5<\/sub> is more acidic than N<sub>2<\/sub>O<sub>3<\/sub> because higher the oxidation state, higher will be acidic character. N<sub>2<\/sub>O<sub>5<\/sub> has +5 oxidation state and N<sub>2<\/sub>O<sub>3<\/sub> has +3 oxidation state.<\/p>\n<p>Question 127.<br \/>\n(a) Arrange the hydrides of group 16 in increasing order of their acidic character. Justify your answer.<br \/>\n(b) Draw structure of XeOF4. (Comptt. Delhi 2017)<br \/>\nAnswer:<br \/>\n(a) H<sub>2<\/sub>O &lt; H<sub>2<\/sub>S &lt; H<sub>2<\/sub>Se &lt; H<sub>2<\/sub>Te<br \/>\nAs we move down the group the bond dissociation enthalpy decreases due to increase in bond length and size of the central atom.<br \/>\n(b)<\/p>\n<p>Question 128.<br \/>\n(a) Account for the the following :<br \/>\n(i) PCl<sub>5<\/sub> is more covalent than PCl<sub>3<\/sub><br \/>\n(ii) Iron on reaction with HC1 forms FeCl<sub>2<\/sub> and not FeCl<sub>3<\/sub><br \/>\n(b) Draw structure of XeO<sub>3<\/sub> (Comptt. Delhi 2017)<br \/>\nAnswer:<br \/>\n(a) (i) In PCl<sub>5<\/sub>, phosphorus has +5 oxidation state and has less tendency to loose electrons than in +3 of PCl<sub>3<\/sub>. Therefore, PCl<sub>5<\/sub> has more tendency to share e<sup>-1<\/sup>s than PCl<sub>3<\/sub>.<br \/>\n(ii) Because HCl on reaction with iron liberates H<sub>2<\/sub> gas which prevents the formation of ferric chloride.<br \/>\n(iii) Due to resonance the two oxygen atoms have partial double bond character and thus have same bond length i.e. 128 pm<\/p>\n<p>(b)<\/p>\n<h4><span class=\"ez-toc-section\" id=\"The_p-Block_Elements_Class_12_Important_Questions_long_Answer_Type_LA\"><\/span>The p-Block Elements Class 12 Important Questions long Answer Type [LA]<span class=\"ez-toc-section-end\"><\/span><\/h4>\n<p>Question 129.<br \/>\n(a) Draw the structures of the following :<br \/>\n(i) H<sub>2<\/sub>S<sub>2<\/sub>O<sub>8<\/sub> (ii) HClO<sub>4<\/sub><br \/>\n(b) How would you account for the following :<br \/>\n(i) NH<sub>3<\/sub> is a stronger base than PH<sub>3<\/sub><br \/>\n(ii) Sulphur has a greater tendency for catenation than oxygen.<br \/>\n(iii) F<sub>2<\/sub> is a stronger oxidising agent than Cl<sub>2<\/sub>. (All India 2009)<br \/>\nAnswer:<br \/>\n(a) (i) H<sub>2<\/sub>S<sub>2<\/sub>O<sub>8<\/sub> (Peroxodisulphuric acid) or Marshall\u2019s acid :<\/p>\n<p>(b)<br \/>\n(i) Since both P and N contain lone pairs of electrons but due to small size and high electronegativity of Nitrogen in NH<sub>3<\/sub>, the electron density is much higher than PH<sub>3<\/sub>, therefore it can easily donate electrons and acts as strong Lewis base than PH<sub>3<\/sub>.<\/p>\n<p>(ii) The greater catenation tendency of sulphur is due to two reasons :<br \/>\n(a) The lone pair of electrons feels more repulsion in 0-0 bond than S-S bond due to its small size and thus S-S forms strong bond.<br \/>\n(b) As the size of atom increases down the group from O \u2013 PO, the strength of bond increases and therefore catenation tendency also increases.<\/p>\n<p>(iii) Due to low bond dissociation enthalpy and high electronegativity of Fluorine, it has strong tendency to accept electrons and thus get reduced.<br \/>\nF + e<sup>\u2013<\/sup> \u2192 F<sup>\u2013<\/sup><br \/>\nTherefore F<sub>2<\/sub> acts as strong oxidising agent, while Cl<sub>2<\/sub> is weak oxidising agent due to low electronegativity.<\/p>\n<p>Question 130.<br \/>\n(a) Draw the structures of the following :<br \/>\n(i) H<sub>2<\/sub>S<sub>2<\/sub>O<sub>7<\/sub> (ii) HClO<sub>3<\/sub><br \/>\n(b) Explain the following observations :<br \/>\n(i) In the structure of HNO<sub>3<\/sub> the N-O bond (121 pm) is shorter than the N- OH bond (140 pm).<br \/>\n(ii) All the P-Cl bonds in PCl<sub>5<\/sub> are not equivalent.<br \/>\n(iii) ICl is more reactive than I<sub>2<\/sub>. (All India 2009)<br \/>\nAnswer:<br \/>\n(a) (i) H<sub>2<\/sub>S<sub>2<\/sub>O<sub>7<\/sub> (Pyrosulphuric acid) or oleum :<\/p>\n<p>(b) (i) The N-O bond has partial double bond character while the N-OH bond is a single bond in both resonance of HNO<sub>3<\/sub><br \/>\n(ii) All the P-Cl bonds in PCl<sub>5<\/sub> are not equivalent due to the fact that the axial bond pairs suffer more repulsion as compared to equatorial bond pairs.<br \/>\n(iii) Because ICl bond is weaker than I-I bond as a result of which ICl breaks easily to form halogen atoms which readily bring about the reaction, hence more reactive.<\/p>\n<p>Question 131.<br \/>\n(a) Draw the structures of the following :<br \/>\n(i) H<sub>3<\/sub>PO<sub>2<\/sub> (ii) BrF<sub>3<\/sub><br \/>\n(b) How would you account for the following observations :<br \/>\n(i) Phosphorus has a greater tendency for catenation than nitrogen.<br \/>\n(ii) Bond dissociation energy of fluorine is less than that of chlorine.<br \/>\n(iii) No chemical compound of helium is known. (All India 2009)<br \/>\nAnswer:<br \/>\n(a) (i) H<sub>3<\/sub>PO<sub>2<\/sub> (Hypophosphorous acid)<\/p>\n<p>(b) (i) The bond strength of P-P is more than N-N, therefore phosphorous shows more tendency for catenation than nitrogen.<br \/>\n(ii) Due to smaller size of F than Cl as a result of which electron-electron repulsions between the lone pairs of electrons are very large than that of Cl, hence bond dissociation energy of F<sub>2<\/sub> is less than that of Cl<sub>2<\/sub>.<br \/>\n(iii) Because the ionization energy of Helium is very high and the empty d-orbitals are also absent in it.<\/p>\n<p>Question 132.<br \/>\n(a) Draw the structures of the following :<br \/>\n(i) N<sub>2<\/sub>O<sub>5<\/sub> (ii) XeOF<sub>4<\/sub><br \/>\n(b) Explain the following observations :<br \/>\n(i) The electron gain enthalpy of sulphur atom has a greater negative value than that of oxygen atom.<br \/>\n(ii) Nitrogen does not form pentahalides.<br \/>\n(iii) In aqueous solutions HI is a stronger acid than HCl. (All India 2009)<br \/>\nAnswer:<br \/>\n(a) (i) N<sub>2<\/sub>O<sub>5<\/sub><\/p>\n<p>(ii) XeOF<sub>4<\/sub><\/p>\n<p>(b) (i) Because enthalpy of dissociation of S-S bond is higher than 0-0 bond and the hydration energy of S<sup>2-<\/sup> is less than that of O<sup>2-<\/sup> ion.<br \/>\n(ii) Due to absence of empty d-orbitals, N<sub>2<\/sub> does not form pentahalides.<br \/>\n(iii) Due to lower bond dissociation energy and higher degree of ionization, HI acts as stronger acid than HCl in aqueous solution.<\/p>\n<p>Question 133.<br \/>\n(a) Draw the structures of the following :<br \/>\n(i) XeF<sub>4<\/sub> (ii) H<sub>2<\/sub>S<sub>2<\/sub>O<sub>7<\/sub><br \/>\n(b) Explain the following observations :<br \/>\n(i) Phosphorus has a greater tendency for catenation than nitrogen.<br \/>\n(ii) The negative value of electron gain enthalpy is less for fluorine than that for chlorine.<br \/>\n(iii) Hydrogen fluoride has a much higher boiling point than hydrogen chloride. (All India 2009)<br \/>\nAnswer:<br \/>\n(a) (i) XeF<sub>4<\/sub> :<\/p>\n<p>(b) (i) The bond strength of P-P is more than N-N, therefore phosphorous shows more tendency for catenation than nitrogen.<br \/>\n(ii) Because of small size of flourine atom and strong electron-electron repulsions in its compact 2p orbitals.<br \/>\n(iii) Hydrogen fluoride (HF) has higher boiling point than HC1 due to extensive intermolecular hydrogen bonding while HCl doesn\u2019t show this H-bonding.<\/p>\n<p>Question 134.<br \/>\n(a) Draw the structures of the following :<br \/>\n(i) PCl<sub>5<\/sub>(s)<br \/>\n(ii) \\(\\mathrm{SO}_{3}^{2-}\\)<br \/>\n(b) Explain the following observations :<br \/>\n(i) Ammonia has a higher boiling point than phosphine.<br \/>\n(ii) Helium does not form any chemical compound.<br \/>\n(iii) Bi (V) is a stronger oxidising agent than Sb (V). (All India 2009)<br \/>\nAnswer:<br \/>\n(a) (i) PCl<sub>5<\/sub> (s)<\/p>\n<p>Angle: The angle O \u2013 S \u2013 O is greater than 90\u00b0<br \/>\n(b) (i) Due to intermolecular H-bonding in NH<sub>3<\/sub> it has higher boiling point than PH<sub>3<\/sub> which does not have any H-bonding.<br \/>\n(ii) Because the ionization energy of Helium is very high and very high positive electrons gain enthalpy.<br \/>\n(iii) The stability of +5 oxidation state decreases and that of +3 state increases due to inert pair effect down the group therefore Bi(v) accepts two electrons and gets reduced to Bi (v).<br \/>\nBi<sup>5+<\/sup> + 2e<sup>\u2013<\/sup> \u2192 Bi<sup>3+<\/sup><\/p>\n<p>Question 135.<br \/>\n(a) Complete the following chemical equations :<br \/>\n(i) NaOH (aq) + Cl<sub>2<\/sub> (g) \u2192<br \/>\n(Hot and cone.)<br \/>\n(ii) XeF<sub>6<\/sub> (s) + H<sub>2<\/sub>O(l) \u2192<br \/>\n(b) How would you account for the following?<br \/>\n(i) The value of electron gain enthalpy with negative sign for sulphur is higher than that for oxygen.<br \/>\n(ii) NF<sub>3<\/sub> is an exothermic compound but NClj is endothermic compound.<br \/>\n(iii) ClF<sub>3<\/sub> molecule has a T-shaped structure and not a trigonal planar one. (Delhi 2010)<br \/>\nAnswer:<br \/>\n(a) (i) 6NaOH + 3Cl<sub>2<\/sub> \u2192 5NaCl + 1NaClO<sub>3<\/sub> + 3H<sub>2<\/sub>O<br \/>\n(ii) XeF<sub>6<\/sub> (s) + 3H<sub>2<\/sub>O (1) \u2192 XeO<sub>3<\/sub> (s) + 6HF (aq)<\/p>\n<p>(b) (i) Because of enthalpy of dissociation of S- S bond is higher than O-O bond and the hydration energy of S<sup>2-<\/sup> is less than that of O<sup>2-<\/sup> ion.<\/p>\n<p>(ii) Due to smaller size of F as compared to Cl, the N \u2013 F bond is much stronger than N-Cl bond while bond dissociation energy of F<sub>2<\/sub> is much lower than that of Cl<sub>2<\/sub>. Therefore, energy released during the formation of NF<sub>3<\/sub> molecule is more than the energy needed to break N<sub>2<\/sub> and F<sub>2<\/sub> molecules into individual atoms. In other words, formation of NF<sub>3<\/sub> is an exothermic reaction.<br \/>\nThe energy released during the formation of NCl<sub>3<\/sub> molecule is less than the energy needed to break N<sub>2<\/sub> and Cl<sub>2<\/sub> molecule into individual atoms. Thus formation of NCl<sub>3<\/sub> is an endothermic reaction.<\/p>\n<p>(iii) The electronic configuration of Cl is 1s<sup>2<\/sup> 2s<sup>2<\/sup> 2p<sup>6<\/sup> 3s<sup>2<\/sup> \\(3 \\mathrm{P}_{x}^{2} 3 \\mathrm{P}_{y}^{2} 3 \\mathrm{P}_{z}^{1}\\). It has only one half filled orbital. But to form three Cl \u2013 F bonds, we need three half filled orbitals. To achieve this, one of the \\(3 P_{y}^{2}\\) electrons gets excited to 3d orbital. The resulting five orbitals of the third shell of Cl in the first excited state undergoes SP<sup>3<\/sup>d hybridization to give T-shaped structure to ClF<sup>3<\/sup> molecule. Since Cl does not undergo SP<sup>2<\/sup> hybridization, therefore ClF<sup>3<\/sup> does not have trigonal planar structure.<\/p>\n<p>Question 136.<br \/>\n(a) Complete the following chemical reaction equations :<br \/>\n(i) P<sub>4<\/sub> + SO<sub>2<\/sub>Cl<sub>2<\/sub> \u2192<br \/>\n(ii) XeF<sub>4<\/sub> + H<sub>2<\/sub>O \u2192<br \/>\n(b) Explain the following observations giving appropriate reasons :<br \/>\n(i) The stability of +5 oxidation state decreases down the group in group 15 of the periodic table.<br \/>\n(ii) Solid phosphorus pentachloride behaves as an ionic compound.<br \/>\n(iii) Halogens are strong oxidizing agents. (Delhi 2010)<br \/>\nAnswer:<br \/>\n(a)<\/p>\n<p>(b) (i) Group 15 first elements are pentavalent, therefore they can show positive oxidation state +3 (due to P-electron) and +5 (due to P and S electrons). In a group the +5 oxidation state stability decreases but +3 oxidation state increases due to inter pair effect which results the 5- orbitals electrons to participate in bonding hence shows +3 oxidation state as their stable oxidation state.<br \/>\n(ii) PCl<sub>5<\/sub> conducts electricity in the molten state. This means that in solid state it exists as [PCl<sub>4<\/sub>]<sup>+<\/sup> [PCl<sub>6<\/sub>]<sup>\u2013<\/sup> in which the cation is tetrahedral and the anion is octahedral.<br \/>\n2PCl<sub>5<\/sub> \u2192 [PCl<sub>4<\/sub>]<sup>+<\/sup> [PCl<sub>6<\/sub>]<sup>\u2013<\/sup><br \/>\nOn melting, these ions become free to move and hence PCl<sub>5<\/sub> conducts electricity in the molten state.<br \/>\n(iii) As halogens are strong elctron acceptors and change to negative ions and thus undergo reduction, so they are strong oxidising agent.<\/p>\n<p>Question 137.<br \/>\n(a) Explain the following :<br \/>\n(i) NF<sub>3<\/sub> is an exothermic compound \u2019 whereas NCl<sub>3<\/sub> is not.<br \/>\n(ii) F<sub>2<\/sub> is most reactive of all the four common halogens.<br \/>\n(b) Complete the following chemical equations :<br \/>\n(i) C + H<sub>2<\/sub>SO<sub>4<\/sub> (cone) \u2192<br \/>\n(ii) P<sub>4<\/sub> + NaOH + H<sub>2<\/sub>O \u2192<br \/>\n(iii) Cl<sub>2<\/sub> + F<sub>2<\/sub> (excess) \u2192 (Delhi 2010)<br \/>\nAnswer:<br \/>\n(a) (i) F is more electronegative than Cl. The difference in the electronegativity between N and F is much more than the difference between electronegativity of N and Cl. So there is need of much more energy to break the N-F bond.<br \/>\n(ii) Because of the low bond dissociation energy F<sub>2<\/sub> readily dissociates into atoms and reacts with other substances readily.<\/p>\n<p>(b) (i) C + 2H<sub>2<\/sub>SO<sub>4<\/sub> (cone.) \u2192 CO<sub>2<\/sub> + 2SO<sub>2<\/sub> + 2H<sub>2<\/sub>O<br \/>\n(ii) P<sub>4<\/sub> + 3NaOH + 3H<sub>2<\/sub>O \u2192 PH<sub>2<\/sub> + 3NaH<sub>2<\/sub>PO<sub>2<\/sub><br \/>\n(iii) Cl<sub>2<\/sub> + 3F<sub>2<\/sub> (excess) \u2192 2ClF<sub>3<\/sub><\/p>\n<p>Question 138.<br \/>\n(a) Account for the following :<br \/>\n(i) The acidic strength decreases in the order HCl &gt; H<sub>2<\/sub>S &gt; PH<sub>3<\/sub><br \/>\n(ii) Tendency to form pentahalides<br \/>\ndecreases down the group in group 15 of the<br \/>\n(b) Complete the following chemical equations :<br \/>\n(i) P<sub>4<\/sub> + SO<sub>2<\/sub>Cl<sub>2<\/sub> \u2192<br \/>\n(ii) XeF<sub>2<\/sub> + H<sub>2<\/sub>O \u2192<br \/>\n(iii) I<sub>2<\/sub> + HNO<sub>3<\/sub> (conc) \u2192 (Delhi 2010)<br \/>\nAnswer:<br \/>\n(a) (i) Because of decrease in electronegativity from chlorine to phosphorous, the bond dissociation enthalpy from HCl to H-P increases and their tendency to release H<sup>+<\/sup> decreases and thus acidic strength decreases.<br \/>\n(ii) Down the group, the tendency of next \u2018s\u2019 orbital\u2019s electron to jump to previous\u2019d\u2019 orbital decreases very much due to inert pair effect.<\/p>\n<p>Question 139.<br \/>\n(a) Draw the structures of the following molecules :<br \/>\n(i) (HPO<sub>3<\/sub>)<sub>3<\/sub><br \/>\n(ii) BrF<sub>3<\/sub><br \/>\n(b) Complete the following chemical equations :<br \/>\n(i) HgCl<sub>2<\/sub> + PH<sub>3<\/sub> \u2192<br \/>\n(ii) SO<sub>3<\/sub> + H<sub>2<\/sub>SO<sub>4<\/sub> \u2192<br \/>\n(iii) XeF<sub>4<\/sub> + H<sub>2<\/sub>O \u2192 (All India 2010)<br \/>\nAnswer:<br \/>\n(a) (i) (HPO<sub>3<\/sub>)<sub>3<\/sub> :<\/p>\n<p>Question 140.<br \/>\n(a) What happens when<br \/>\n(i) chlorine gas is passed through a hot concentrated solution of NaOH?<br \/>\n(ii) sulphur dioxide gas is passed through an aqueous solution of a Fe (III) salt?<br \/>\n(b) Answer the following :<br \/>\n(i) What is the basicity of H<sub>3<\/sub>PO<sub>3<\/sub> and why?<br \/>\n(ii) Why does fluorine not play the role of a central atom in interhalogen compounds?<br \/>\n(iii) Why do noble gases have very low boiling points? (All India 2010)<br \/>\nAnswer:<\/p>\n<p>(b) (i) Basicity of H<sub>3<\/sub>PO<sub>3<\/sub> = 2<br \/>\nBecause basicity is the number of replaceable H<sup>+<\/sup> ions in an acid and H<sub>3<\/sub>PO<sub>3<\/sub> is a Dibasic acid.<br \/>\n(ii) Because F being smaller, it cannot accomodate larger sized other halogen atoms around it. Due to the absence of d-orbitals, F does not show positive oxidation state of +3, +5, +7 needed for the formation of polyatomic interhalogen compounds.<br \/>\n(iii) Because the atoms of these elements are held together by weak van der Waal\u2019s forces of attraction.<\/p>\n<p>Question 141.<br \/>\n(a) Complete the following chemical reaction equations :<br \/>\n(i) P<sub>4<\/sub> + SO<sub>2<\/sub>Cl<sub>2<\/sub> \u2192<br \/>\n(ii) XeF<sub>6<\/sub> + H<sub>2<\/sub>O \u2192<br \/>\n(b) Predict the shape and the asked angle (90\u00b0 or more or less) in each of the following cases:<br \/>\n(i) \\(\\mathrm{SO}_{3}^{2-}\\) and the angle O \u2013 S \u2013 O<br \/>\n(ii) ClF<sub>3<\/sub> and the angle F \u2013 Cl \u2013 F<br \/>\n(iii) XeF<sub>2<\/sub> and the angle F \u2013 Xe \u2013 F (Delhi 2012)<br \/>\nAnswer:<br \/>\n(a) (i) P<sub>4<\/sub> + 10SO<sub>2<\/sub>Cl<sub>2<\/sub> \u2192 4PCl<sub>5<\/sub> + 10SO<sub>2<\/sub><br \/>\nor P<sub>4<\/sub> + 8SO<sub>2<\/sub>Cl<sub>2<\/sub> \u2192 4PCl<sub>3<\/sub> + 4SO<sub>2<\/sub> + 2S<sub>2<\/sub>Cl<sub>2<\/sub><br \/>\n(ii) XeF<sub>6<\/sub> + H<sub>2<\/sub>O \u2192 XeOF<sub>4<\/sub> + 2HF<br \/>\nor XeF<sub>6<\/sub> + 2H<sub>2<\/sub>O \u2192 XeOF<sub>4<\/sub> + 4HF<br \/>\nor XeF<sub>6<\/sub> + 3H<sub>2<\/sub>O \u2192 XeO<sub>3<\/sub> + 6HF<\/p>\n<p>(b) (i) \\(\\mathrm{SO}_{3}^{2-}\\)<\/p>\n<p>Angle : The angle F-Cl-F is lesser than 90\u00b0<br \/>\n(iii) XeF<sub>2<\/sub>; Structure :<\/p>\n<p>Shape : Linear<br \/>\nAngle ; F-Xe \u2013 F &gt; 90\u00b0<\/p>\n<p>Question 142.<br \/>\n(a) Complete the following chemical equations :<br \/>\n(i) NaOH (hot and cone.) + Cl<sub>2<\/sub> \u2192<br \/>\n(ii) XeF<sub>4<\/sub> + O<sub>2<\/sub>F<sub>2<\/sub> \u2192<br \/>\n(b) Draw the structures of the following molecules :<br \/>\n(i) H<sub>3<\/sub>PO<sub>2<\/sub> (ii) H<sub>2<\/sub>S<sub>2<\/sub>O<sub>7<\/sub> (iii) XeOF<sub>4<\/sub> (Delhi 2012)<br \/>\nAnswer:<br \/>\n(a) (i) 3Cl<sub>2<\/sub><sup>+<\/sup> 6NaOH \u2192 5 NaCl + NaClO<sub>3<\/sub> + 3H<sub>2<\/sub>O<br \/>\n(ii) XeF<sub>4<\/sub> + O<sub>2<\/sub>F<sub>2<\/sub> \u2192 XeF<sub>6<\/sub> + O<sub>2<\/sub><br \/>\n(b) (i) H<sub>3<\/sub>PO<sub>2<\/sub>:<\/p>\n<p>(ii) H<sub>2<\/sub>S<sub>2<\/sub>O<sub>7<\/sub>:<\/p>\n<p>(iii) XeOF<sub>2<\/sub>:<\/p>\n<p>Question 143.<br \/>\n(a) Draw the molecular structure of following compounds :<br \/>\n(i) XeF<sub>6<\/sub> (ii) H<sub>2<\/sub>S<sub>2<\/sub>O<sub>8<\/sub><br \/>\n(b) Explain the following observations :<br \/>\n(i) The molecules NH<sub>3<\/sub> and NF<sub>3<\/sub> have dipole moments which are of opposite direction.<br \/>\n(ii) All the bonds in PCl<sub>5<\/sub> molecule are not equivalent.<br \/>\n(iii) Sulphur in vapour state exhibits paramagnetism. (Delhi 2012)<br \/>\nAnswer:<br \/>\n(a) (i) XeF<sub>6<\/sub>:<\/p>\n<p>(b) (i) Because F is more electronegative than N in NF<sub>3<\/sub> whereas N is more electronegative than H in NH<sub>3<\/sub>.<br \/>\n(ii) Because PCl<sub>5<\/sub> has a trigonal bipyramidal structure in which the three equatorial P-Cl bonds are equivalent while the two axial bonds are longer than equatorial bonds.<br \/>\n(iii) Because sulphur in vapour state has two unpaired electrons in the antibonding \u03c0* orbitals like O<sub>2<\/sub>.<\/p>\n<p>Question 144.<br \/>\n(a) Complete the following chemical equations<br \/>\n(i) XeF<sub>4<\/sub> + SbF<sub>5<\/sub> \u2192<br \/>\n(ii) Cl<sub>2<\/sub> + F<sub>2<\/sub> (excess) \u2192<br \/>\n(b) Explain each of the following :<br \/>\n(i) Nitrogen is much less reactive than phosphorus.<br \/>\n(ii) The stability of +5 oxidation state decreases down group 15.<br \/>\n(iii) The bond angles (O \u2013 N \u2013 O) are not of the same value in \\(\\mathrm{NO}_{2}^{-}\\) and \\(\\mathrm{NO}_{2}^{+}\\). (Delhi 2012)<br \/>\nAnswer:<br \/>\n(a) (i) XeF<sub>4<\/sub> + SbF<sub>5<\/sub> \u2192 [XeF<sub>3<\/sub>]<sup>+<\/sup> [SbF<sub>6<\/sub>]<sup>\u2013<\/sup><br \/>\n(ii) Cl<sub>2<\/sub> + 3F<sub>2<\/sub> (excess) \u2192 2ClF<sub>3<\/sub><\/p>\n<p>(b) (i) Due to presence of triple bond between two nitrogen atoms\/high bond dissociation enthalpy of nitrogen, whereas in phosphorus molecule there is single bond between four atoms of phosphorus so have low bond dissociation enthalpy.<br \/>\n(ii) Because the participation of outer s- electron pair goes on decreasing down the group due to inert pair effect.<br \/>\n(iii) Because \\(\\mathrm{NO}_{2}^{-}\\) has sp<sup>2<\/sup> hybridisation whereas \\(\\mathrm{NO}_{2}^{+}\\) has sp hybridisation.<\/p>\n<p>Question 145.<br \/>\n(a) Draw the molecular structures of the following compounds :<br \/>\n(i) N<sub>2<\/sub>O<sub>5<\/sub> (ii) XeOF<sub>4<\/sub><br \/>\n(b) Explain the following observations :<br \/>\n(i) Sulphur has a greater tendency for catenation than oxygen.<br \/>\n(ii) IC1 is more reactive than I<sub>2<\/sub>.<br \/>\n(iii) Despite lower value of its electron gain<br \/>\nenthalpy with negative sign, fluorine (F<sub>2<\/sub>) is a stronger oxidising agent than Cl<sub>2<\/sub>. (All India 2012)<br \/>\nAnswer:<br \/>\n(a)<\/p>\n<p>(i) N<sub>2<\/sub>O<sub>5<\/sub><\/p>\n<p>(ii) XeOF<sub>4<\/sub><\/p>\n<p>(b) (i) It is because S \u2013 S bond is stronger than 0-0 bonds as there is more interelectronic repulsion in O \u2013 O due to small size than in S \u2013 S.<br \/>\n(ii) ICl is more reactive than I<sub>2<\/sub> because I-Cl bond is polar and weaker while I-I bond is stronger and non-polar.<br \/>\n(iii) It is due to (a) low enthalpy of dissociation of F \u2013 F bond, (b) high hydration enthalpy of F.<\/p>\n<p>Question 146.<br \/>\n(a) Complete the following chemical equations :<br \/>\n(i) Cu + HNO<sub>3<\/sub> (dilute) \u2192<br \/>\n(ii) XeF<sub>4<\/sub> + O<sub>2<\/sub>F<sub>2<\/sub> \u2192<br \/>\n(b) Explain the following observations :<br \/>\n(i) Phosphorus has greater tendency for catenation than nitrogen.<br \/>\n(ii) Oxygen is a gas but sulphur a solid.<br \/>\n(iii) The halogens are coloured. Why? (All India 2012)<br \/>\nAnswer:<br \/>\n(a) (i) 3Cu + 8HNO<sub>3<\/sub> (dil) \u2192 3CU(NO<sub>3<\/sub>)<sub>2<\/sub> + 2NO + 4H<sub>2<\/sub>O<br \/>\n(ii) XeF<sub>4<\/sub> + O<sub>2<\/sub>F<sub>2<\/sub> \u2192 XeF<sub>6<\/sub> + O<sub>2<\/sub><\/p>\n<p>(b) (i) The self linking property or catenation property of nitrogen is less than that of phosphorus because N \u2013 N bond is weaker than P \u2013 P bond.<\/p>\n<p>(ii) Oxygen forms a stable diatomic molecule. In 02 molecule two atoms of oxygen have joined together through double bond 0 = 0. The multiple bonding is possible due to small size of oxygen atoms. So oxygen is a gas. S has eight atoms arranged in the form of a puckered ring per molecule in which two sulphur atoms are joined by covalent bonds. So sulphur is a solid at room temperature.<\/p>\n<p>(iii) All halogens are coloured. This is due to absorption of radiation in visible region which results in the excitation of outer electrons to higher energy level. So they display different colours.<\/p>\n<p>Question 147.<br \/>\n(a) Draw the structures of the following molecules :<br \/>\n(i) H<sub>3<\/sub>PO<sub>2<\/sub> (ii) ClF<sub>3<\/sub><br \/>\n(b) Explain the following observations :<br \/>\n(i) Nitrogen is much less reactive than phosphorus.<br \/>\n(ii) Despite having greater polarity, hydrogen fluoride boils at a lower temperature than water.<br \/>\n(iii) Sulphur has greater tendency for catenation than oxygen in the same group. (All India 2012)<br \/>\nAnswer:<br \/>\n(a) (i) H<sub>3<\/sub>PO<sub>2<\/sub> :<\/p>\n<p>(ii) ClF<sub>3<\/sub> :<\/p>\n<p>(b) (i) Because P-P bond is weaker than N \u2261 N bond, N<sub>2<\/sub> is inert due to the presence of triple bond.<br \/>\n(ii) In water O is larger than F therefore Van- der-Waals forces of attraction increase with increase in size and hence the boiling point increases.<br \/>\n(iii) It is because S \u2013 S bond is stronger than 0-0 bond as there is more interelectronic repulsion in O \u2013 O than in S-S.<\/p>\n<p>Question 148.<br \/>\n(a) Draw the structures of the following molecules :<br \/>\n(i) N<sub>2<\/sub>O<sub>5<\/sub> (ii) HClO<sub>4<\/sub><br \/>\n(b) Explain the following observations :<br \/>\n(i) H<sub>2<\/sub>S is more acidic than H<sub>2<\/sub>O.<br \/>\n(ii) Fluorine does not exhibit any positive oxidation state.<br \/>\n(iii) Helium forms no real chemical compound. (All India 2012)<br \/>\nAnswer:<br \/>\n(a) (i) N<sub>2<\/sub>O<sub>5<\/sub> :<\/p>\n<p>(ii) HClO<sub>4<\/sub> :<\/p>\n<p>(b) (i) H<sub>2<\/sub>S is more acidic than H<sub>2<\/sub>O because bond dissociation enthalpy of H \u2013 S bond in H<sub>2<\/sub>S is less than that of H \u2013 O bond in<br \/>\nH<sub>2<\/sub>O.<br \/>\n(ii) F is the most electronegative element. It has no d-orbitals and therefore, there is no scope for any electron promotion. So it can only show oxidation state of -1 in its compounds.<br \/>\n(iii) Helium has very high ionisation enthalpy and no vacant d-orbitals, therefore no chemical compound of helium is known.<\/p>\n<p>Question 149.<br \/>\n(a) Describe the conditions and the steps involved in the manufacture of sulphuric acid by contact process. Write the necessary reactions. (No diagram is required)<br \/>\n(b) Give reasons :<br \/>\nii) Bond dissociation energy of F<sub>2<\/sub> is less than that of Cl<sub>2<\/sub>.<br \/>\n(ii) Nitric oxide becomes brown when released in air. (Comptt. Delhi 2012)<br \/>\nAnswer:<br \/>\n(a) Contact Process : Burning sulphur in an excess of air<br \/>\nS + O<sub>2<\/sub> \u2192 SO<sub>2<\/sub> (g)<br \/>\nor, By heating sulphide ores like pyrites in an excess of air :<br \/>\n4FeS<sub>2<\/sub> + 11O<sub>2<\/sub> \u2192 2Fe<sub>2<\/sub>O<sub>3<\/sub> + 8SO<sub>2<\/sub><br \/>\nIn either case, an excess of air is used so that the SO<sub>2<\/sub> produced is already mixed with oxygen for the next stage. This is reversible reaction and the formation of SO<sub>3<\/sub> is exothermic in the presence of catalyst V<sub>2<\/sub>O<sub>5<\/sub> at 720 K<br \/>\n2SO<sub>2<\/sub> + O<sub>2<\/sub> \u21cc 2SO<sub>3<\/sub> \u0394H = -196 KJ\/mol<br \/>\nThis cannot be done by simply adding water to the S03. The reaction is so uncontrollable that it creates a fog of H2S04. Instead, the S03 is first dissolved in cone. H2S04.<br \/>\nH<sub>2<\/sub>SO<sub>4<\/sub> + SO<sub>3<\/sub> \u2192 H<sub>2<\/sub>S<sub>2<\/sub>O<sub>7<\/sub><br \/>\nThe product is known as fuming sulphuric acid or oleum to which water is added to get H<sub>2<\/sub>SO<sub>4<\/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>(b) (i) Bond dissociation energy of F<sub>2<\/sub> is less than that of Cl<sub>2<\/sub> because of large electron- electron repulsion among the lone pair in F<sub>2<\/sub> molecule.<br \/>\n(ii) Nitric oxide becomes brown when released in air because of the formation of NO<sub>2<\/sub> gas.<\/p>\n<p>Question 150.<br \/>\nAccount for the following :<br \/>\n(a) Thermal stability of water is much higher than that of H<sub>2<\/sub>S.<br \/>\n(b) White phosphorus is more reactive than red phosphorus.<br \/>\n(c) Ammonia acts as a ligand.<br \/>\n(d) Bismuth is a strong oxidizing agent in pentavalent state.<br \/>\n(e) Concentrated sulphuric acid is a strong dehydrating agent. (Comptt. Delhi 2012)<br \/>\nAnswer:<br \/>\n(a) The thermal stablility of hydrides decrease from H<sub>2<\/sub>O to H<sub>2<\/sub>S because as the size of the atom increases, the bond becomes weaker and thus breaks on heating.<br \/>\n(b) Red phosphorus is regarded as a polymer consisting of chains of P<sub>4<\/sub> tetrahedral linked together. This makes phosphorus denser and less reactive<\/p>\n<p>But in white phosphorus (P<sub>4<\/sub>), the four phosphorus atoms lie at the corners of a regular tetrahedron. Each phosphorus atom is linked to each of the other three atoms by covalent bond. The bond angle is equal to 60\u00b0 which suggests that the molecule is under strain and hence active in nature.<\/p>\n<p>(c) Due to the presence of a lone pair of electrons on nitrogen atom, it has a tendency to donate an electron pair, hence acts as a ligand.<br \/>\n(d) In Bismuth, the inert pair effect is very prominent. Thus +5 oxidation state is less stable in comparison to +3 oxidation state i.e<br \/>\nit readily accepts two electrons in pentavalent state and gets reduced to trivalent state. Therefore, it acts as a strong oxidising agent.<br \/>\n(e) Cone. H<sub>2<\/sub>SO<sub>4<\/sub> has great affinity for water molecule i.e it acts as a dehydrating agent.<\/p>\n<p>Question 151.<br \/>\nAccount for the following :<br \/>\n(i) Chlorine water loses its yellow colour on standing.<br \/>\n(ii) BrCl<sub>3<\/sub> is more stable than BrCl<sub>5<\/sub>.<br \/>\n(iii) Fluorine does not form oxoacids.<br \/>\n(iv) PCl<sub>5<\/sub> acts as an oxidising agent.<br \/>\n(v) SO<sub>2<\/sub> is an air pollutant. (Comptt. All India 2012)<br \/>\nAnswer:<br \/>\n(i) Cl<sub>2<\/sub> water on standing loses its yellow colour due to the formation of HCl and HOCl. HOCl is unstable and decomposes to HCl. As a result, yellow colour disappears.<\/p>\n<p>(ii) In BrCl<sub>3<\/sub>, Br can accommodate three chlorine atoms around it and hence it is stable but in BrCl<sub>5<\/sub>, five Cl atoms cannot be accommodated around Br and hence it is unstable.<br \/>\n(iii) Due to high electonegativity and small size F does not form oxoacids.<br \/>\n(iv) In PCl<sub>5<\/sub>, the oxidation state of phosphorus is +5. Thus, it cannot increase its oxidation state beyond +5. So PCl<sub>5<\/sub> cannot act as a reducing agent. However, it can decrease its oxidation state from +5 to +3 or to some lower value. Thus, it acts as an oxidising agent.<br \/>\n(v) SO<sub>2<\/sub> is a pungent and irritating gas. It acts as an air pollutant due to the following reasons:<\/p>\n<ul>\n<li>It causes throat and eye irritation as it is absorbed readily by respiratory\u2019 tract.<\/li>\n<li>It combines with moisture forming sulphurous acid. It is then converted into H<sub>2<\/sub>SO<sub>4<\/sub>. Both these acids cause acid rain and destroy the marble, corrode metals, deteriorate fabrics, paper, leather etc.<\/li>\n<\/ul>\n<p>Question 152.<br \/>\n(a) With the help of chemical equations explain the principle of contact process in brief for the manufacture of sulphuric acid.<br \/>\n(No diagram)<br \/>\n(b) Account for the following :<br \/>\n(i) Bond dissociation energy of F<sub>2<\/sub> is less than that of Cl<sub>2<\/sub><br \/>\n(ii) Nitric oxide (NO) becomes brown when released in air. (Comptt. All India 2012)<br \/>\nAnswer:<br \/>\n(a) Principle of contact process : The process involves the oxidation of sulphur dioxide by air in the presence of a catalyst.<\/p>\n<p>Sulphur trioxide is dissolved in 98% H<sub>2<\/sub>SO<sub>4<\/sub> acid and oleum is formed<br \/>\nH<sub>2<\/sub>SO<sub>2<\/sub> + SO<sub>3<\/sub> \u2192 H<sub>2<\/sub>S<sub>2<\/sub>O<sub>7<\/sub> (oleum)<br \/>\nOleum is diluted with water to form sulphuric acid<br \/>\nH<sub>2<\/sub>S<sub>2<\/sub>07 + H<sub>2<\/sub>O \u2192 2H<sub>2<\/sub>S0<sub>4<\/sub><\/p>\n<p>(b) (i) The low bond dissociation energy of F is due to high inter electronic repulsions between non-bonding electrons in 2p- orbitals as the size of F atoms is small. As a result, F-F bond is weaker than Cl-Cl bond.<br \/>\n(ii) It readily combines with 02 to form brown fumes of NO<sub>2<\/sub>.<br \/>\n2NO + O<sub>2<\/sub>\u2192 2NO<sub>2<\/sub> (brown fumes)<\/p>\n<p>Question 153.<br \/>\n(a) Give reasons for the following:<br \/>\n(i) Bond enthalpy of F<sub>2<\/sub> is lower than that of Cl<sub>2<\/sub>.<br \/>\n(ii) PH<sub>3<\/sub> has lower boiling point than NH<sub>3<\/sub>,<br \/>\n(b) Draw the structures of the following molecules:<br \/>\n(i) ClF<sub>3<\/sub> (ii) (HPO<sub>3<\/sub>)<sub>3<\/sub> (iii) XeF<sub>4<\/sub> (Delhi 2013)<br \/>\nAnswer:<br \/>\n(a) (i) Due to smaller size of F than Cl as a result<br \/>\nof which electron-electron repulsions between the lone pairs of electrons are very large than that of Cl, hence bond dissociation enthalpy of F<sub>2<\/sub> is less than that of Cl<sub>2<\/sub><br \/>\n(ii) PH<sub>3<\/sub> has lower boiling point than NH<sub>3<\/sub> because PH<sub>3<\/sub> cannot form hydrogen bonds like NH<sub>3<\/sub>.<\/p>\n<p>(b) (i) BrF<sub>3<\/sub> :<\/p>\n<p>(ii) (HPO<sub>3<\/sub>)<sub>3<\/sub>: Cyclotrimetaphosphoric acid<\/p>\n<p>(iii) XeF4:<\/p>\n<p>Question 154.<br \/>\n(a) Account for the following:<br \/>\n(i) Helium is used in diving apparatus.<br \/>\n(ii) Fluorine does not exhibit positive oxidation state.<br \/>\n(iii) Oxygen shows catenation behaviour less than sulphur.<br \/>\n(b) Draw the structures of the following molecules:<br \/>\n(i) XeF<sub>2<\/sub> (ii) H<sub>2<\/sub>S<sub>2<\/sub>O<sub>8<\/sub> (Delhi 2013)<br \/>\nAnswer:<br \/>\n(i) Helium is used in diving apparatus because of its very low solubility in blood and therefore an oxygen-helium mixture is used for artificial respiration. (ii) Because it is most electronegative element and does not have d-orbitals for octet expansion, therefore it shows only a negative oxidation state of -1.<br \/>\n(ii) The greater catenation tendency of sulphur is due to two reasons :<br \/>\n(a) The lone pair of electrons feels more repulsion in 0-0 bond than S-S bond due to its small size and thus S-S forms strong bond.<br \/>\n(b) As the size of atom increases down the group from O \u2013 PO, the strength of bond increases and therefore catenation tendency also increases.<\/p>\n<p>(b) (i) XeF<sub>2<\/sub> :<\/p>\n<p>(ii) H<sub>2<\/sub>S<sub>2<\/sub>O<sub>8<\/sub> :<\/p>\n<p>Question 155.<br \/>\nGive reasons for the following :<br \/>\n(i) Oxygen is a gas but sulphur is a solid.<br \/>\n(ii) O<sub>3<\/sub> acts as a powerful oxidising agent.<br \/>\n(iii) BiH<sub>3<\/sub> is the strongest reducing agent<br \/>\namongst all the hydrides of Group 15 elements. (All India 2013)<br \/>\nAnswer:<br \/>\n(i) Due to small size and high electronegativity, oxygen forms p\u03c0 \u2013 p\u03c0 multiple bonds and thus forms diatomic. O<sub>2<\/sub> molecule are held together by weak van der Waals forces of attraction which can be easily overcome by collisions at room temperature. Therefore O<sub>2<\/sub> is gas at room temperature. On the other hand due to higher tendency for catenation and lower tendency for p\u03c0 \u2013 p\u03c0 multiple bonds, intra-atomic S<sub>8<\/sub> has strong forces of attraction which cannot be overcome by collisions. Therefore S is solid at room temperature.<\/p>\n<p>(ii) O<sub>3<\/sub> is a powerful oxidising agent due to its high energy content than oxygen and hence decomposes to give diatomic oxygen and atomic oxygen<\/p>\n<p>(iii) As we move down the group, the E\u2014H bond length increases and their strength decreases. Bi\u2014H bond is the weakest. It can break easily and evolves H<sub>2<\/sub> gas which acts as the reducing agent.<\/p>\n<p>Question 156.<br \/>\n(a) Write the balanced chemical equations for obtaining XeO<sub>3<\/sub> and XeOF<sub>4<\/sub> from XeF<sub>6<\/sub>.<br \/>\n(b) Account for the following :<br \/>\n(i) H<sub>2<\/sub>S is less acidic than H<sub>2<\/sub>Te.<br \/>\n(ii) H&lt;sub3PO<sub>2<\/sub> has reducing nature.<br \/>\n(iii) SO<sub>2<\/sub> is an air pollutant. (Comptt. Delhi 2013)<br \/>\nAnswer:<br \/>\n(a) XeF<sub>6<\/sub> + 3H<sub>2<\/sub>O \u2192 XeO<sub>3<\/sub> + 6HF<br \/>\nXeF<sub>6<\/sub> + H<sub>2<\/sub>O \u2192 XeOF<sub>4<\/sub> + 2HF<\/p>\n<p>(b) (i) In the group with increase in size of the element and increased bond distance, the bond dissociation energy decreases, therefore H-S bond dissociation energy is higher than H-Te and hence H-S bond breaks less easily than H-Te bond and H<sub>2<\/sub>S is a weaker acid than H<sub>2<\/sub>Te.<br \/>\n(ii) Because it contains two P-H bonds and thus reduces AgNO<sub>3<\/sub> to metallic silver<br \/>\n4AgNO<sub>3<\/sub> + H<sub>3<\/sub>PO<sub>2<\/sub> + 2H<sub>2<\/sub>O \u2192 4Agi \u2193 + H<sub>3<\/sub>PO<sub>4<\/sub> + 4HNO<sub>3<\/sub><br \/>\n(iii) SO<sub>2<\/sub> is a pungent and irritating gas. It acts as an air pollutant due to the following reasons:<\/p>\n<ul>\n<li>It causes throat and eye irritation as it is absorbed readily by respiratory tract.<\/li>\n<li>It combines with moisture forming sulphurous acid. It is then converted into H<sub>2<\/sub>SO<sub>4<\/sub>. Both these acids cause acid rain and destroy the marble, corrode metals, deteriorate fabrics, paper, leather etc.<\/li>\n<\/ul>\n<p>Question 157.<br \/>\nAccount for the following :<br \/>\n(a) Phosphorus shows high tendency for catenation.<br \/>\n(b) F<sub>2<\/sub> is more reactive than ClF<sub>3<\/sub> but ClF<sub>3<\/sub> is more reactive than Cl<sub>2<\/sub>.<br \/>\n(c) Nitrogen is found in gaseous state.<br \/>\n(d) Decomposition of ozone molecule is a spontaneous process.<br \/>\n(e) SF<sub>6<\/sub> is inert towards hydrolysis. (Comptt. Delhi 2013)<br \/>\nAnswer:<br \/>\n(a) The bond strength of P-P is more tan N-N, therefore, phosphorous shows more tendency for catenation than nitrogen.<br \/>\n(b) Because the bond between Cl-F is weaker than the bond between Cl-Cl due to less effective overlapping of orbitals of Cl-F than Cl-Cl.<br \/>\nThe F<sub>2<\/sub> is most reactive due to its high electrode potential.<br \/>\n(c) Due to small size and high electronegativity nitrogen forms p\u03c0 \u2013 p\u03c0 multiple bonds, which are held together by weak Van-der Waal\u2019s forces of attraction which can be easily broken. Hence N<sub>2<\/sub> exists as a gas at room temperature.<br \/>\n(d) Because O<sub>3<\/sub> is thermodynamically unstable and its decomposition results in liberation of heat (\u0394H is -ve) and increase in entropy (\u0394S is +ve) so (\u0394G becomes -ve and process becomes spontaneous.<br \/>\n(c) In SF<sub>6<\/sub>, S is protected by 6F atoms and does not allow H<sub>2<\/sub>O molecules to attach on it. So SF<sub>6<\/sub> inerts towards hydrolysis.<\/p>\n<p>Question 158.<br \/>\n(a) Complete the following chemical equations:<br \/>\n(i) P<sub>4<\/sub> + NaOH + H<sub>2<\/sub>O \u2192<br \/>\n(ii) XeF<sub>4<\/sub> + O<sub>2<\/sub>F<sub>2<\/sub> \u2192<br \/>\n(b) How would you account for the following situations?<br \/>\n(i) The acidic strength of these<br \/>\ncompounds increases in the following order :<br \/>\nPH<sub>3<\/sub> &gt; H<sub>2<\/sub>S &gt; HCl<br \/>\n(ii) The oxidising power of oxoacids of chlorine follows the order :<br \/>\nHClO<sub>4<\/sub> &gt; HClO<sub>3<\/sub> &gt; HClO<sub>2<\/sub> &gt; HClO<br \/>\n(iii) In vapour state sulphur exhibits paramagnetic behaviour. (Comptt. Delhi 2014)<br \/>\nAnswer:<br \/>\n(a)<\/p>\n<p>(b) (i) As the electronegativity increases in the same period from left to right so their electronegativity are in the increasing order, P &lt; S &lt; Cl.<br \/>\nIn the same way the acid strength is also in the increasing order i.e. PH<sub>3<\/sub> &lt; H<sub>2<\/sub>S &lt; HCl.<\/p>\n<p>Acidic strength of oxoacids of the same halogen increases with increase in oxidation number of the halogen because of the relative stability of the anions left after removal of proton. Thus as the number of oxygen atoms in the anion increases, the dispersal of the negative charge through pn-pn back bonding also increases and hence stability and acidic strength increases<\/p>\n<p>(iii) In vapour state sulphur partly exists as S<sub>2<\/sub> molecule which has two unpaired electrons in the antibonding \u03c0 orbitals and hence exhibits paramagnetism.<\/p>\n<p>Question 159.<br \/>\n(a) Using VSEPR theory predict the probable structures of the following :<br \/>\n(i) N<sub>2<\/sub>O<sub>3<\/sub> (ii) BrF<sub>3<\/sub><br \/>\n(b) Arrange the following groups of substances in the order of the property indicated against each group :<br \/>\n(i) NH<sub>3<\/sub>, PH<sub>3<\/sub>, AsH<sub>3<\/sub>, SbH<sub>3<\/sub> \u2013 increasing order of boiling points.<br \/>\n(ii) O, S, Se, Te \u2013 increasing order of electron gain enthalpy with negative sign.<br \/>\n(iii) F<sub>2<\/sub>, Cl<sub>2<\/sub>, Br<sub>2<\/sub>, I<sub>2<\/sub> \u2013 increasing order of bond dissociation enthalpy. (Comptt. Delhi 2014)<br \/>\nAnswer:<br \/>\n(a)<\/p>\n<p>(b) (i) PH<sub>3<\/sub> &lt; AsH<sub>3<\/sub> &lt; NH<sub>3<\/sub> &lt; SbH<sub>3<\/sub><br \/>\n(ii) O &lt; Te &lt; Se &lt; S<br \/>\n(iii) The increasing order of bond dissociation enthalpy is :<br \/>\nCl<sub>2<\/sub> &gt; Br<sub>2<\/sub> &gt; F<sub>2<\/sub> &gt; I<sub>2<\/sub><\/p>\n<p>Question 160.<br \/>\n(a) Write the formula and describe the structure of a noble gas species which is isostructural with<br \/>\n(i) IBr<sub>2<\/sub> (ii) BrO<sub>3<\/sub><sup>\u2013<\/sup><br \/>\n(b) Assign reasons for the following :<br \/>\n(i) SF<sub>6<\/sub> is kinetically inert.<br \/>\n(ii) NF<sub>3<\/sub> is an exothermic compound whereas NCl<sub>3<\/sub>is not.<br \/>\n(iii) HCl is a stronger acid than HF though flourine is more electronegative than chlorine. (Comptt. All India 2014)<br \/>\nAnswer:<br \/>\n(a) (i) IBr<sub>2<\/sub><sup>\u2013<\/sup> : It has 2 bond pairs and 3 lone pairs.<br \/>\nTherefore according to VSEPR theory it should have Linear structure.<br \/>\nIBr2- has (7 + 2 \u00d7 7 + 1) i.e. 22 valence electrons.<br \/>\nNo noble gas has 22 electrons<\/p>\n<p>\u2234 The isoelectronic species for IBr<sub>2<\/sub><sup>\u2013<\/sup> is XeF<sub>2<\/sub><br \/>\n(ii) BrO<sub>3<\/sub><sup>\u2013<\/sup>: It has 3 bond pairs and 1 lone pair of electrons. Therefore according to VSEPR theory it has pyramidal structure.<br \/>\nIt has 26 valence electrons i.e. (7 + 3 \u00d7 6 + Pyramidal shape 1 = 26).<\/p>\n<p>A noble gas having 26 valence electrons is XeO<sub>3<\/sub> (8 + 3 \u00d7 6 = 26).<br \/>\nThus XeO<sub>3<\/sub> also has pyramidal structure.<br \/>\n(b) (i) Because SF<sub>6<\/sub> is showing steric hindrance due to 6 (six) fluorine atoms which make it unable to react further with any other atom.<br \/>\n(ii)<\/p>\n<ul>\n<li>Because of enthalpy of dissociation of S- S bond is higher than 0-0 bond and the hydration energy of S<sup>2-<\/sup> is less than that of O<sup>2-<\/sup> ion.<\/li>\n<li>Due to smaller size of F as compared to Cl, the N \u2013 F bond is much stronger than N-Cl bond while bond dissociation energy of F<sub>2<\/sub> is much lower than that of Cl<sub>2<\/sub>. Therefore, energy released during the formation of NF<sub>3<\/sub> molecule is more than the energy needed to break N<sub>2<\/sub> and F<sub>2<\/sub> molecules into individual atoms. In other words, formation of NF<sub>3<\/sub> is an exothermic reaction.<br \/>\nThe energy released during the formation of NCl<sub>3<\/sub> molecule is less than the energy needed to break N<sub>2<\/sub> and Cl<sub>2<\/sub> molecule into individual atoms. Thus formation of NCl<sub>3<\/sub> is an endothermic reaction.<\/li>\n<\/ul>\n<p>(iii) Because HCl has large size and less bond strength than HF which makes easier liberation of H<sup>+<\/sup>.<\/p>\n<p>Question 161.<br \/>\n(a) How is ammonia prepared on a large scale?<br \/>\nName the process and mention the optimum conditions for the production of ammonia by this process.<br \/>\n(b) Assign reasons for the following :<br \/>\n(i) H<sub>2<\/sub>S is more acidic than H<sub>2<\/sub>O.<br \/>\n(ii) NH<sub>3<\/sub> is more basic than PH<sub>3<\/sub>.<br \/>\n(iii) Sulphur has a greater tendency for catenation than oxygen. (Comptt. All India 2014)<br \/>\nAnswer:<br \/>\n(a) According to Le Chatelier\u2019s principle the favourable conditions for the production of NH<sub>3<\/sub> by Haber\u2019s process are<\/p>\n<ul>\n<li>pressure of 200 atmosphere<\/li>\n<li>temperature of ~ 700 K<\/li>\n<li>use of catalyst like Fe<sub>2<\/sub>O<sub>3<\/sub> with small amount of K<sub>2<\/sub>O and Al<sub>3<\/sub>O<sub>3<\/sub><br \/>\nN<sub>2<\/sub> + 3H<sub>2<\/sub> \u21cc 2NH<sub>3<\/sub><\/li>\n<\/ul>\n<p>(b) (i) H<sub>2<\/sub>S is more acidic than H20 because bond dissociation enthalpy of H \u2013 S bond in H<sub>2<\/sub>S is less than that of H \u2013 O bond in H<sub>2<\/sub>O.<br \/>\n(ii) Since both P and N contain lone pairs of electrons but due to small size and high electronegativity of Nitrogen in NH<sub>3<\/sub>, the electron density is much higher than PH<sub>2<\/sub> therefore it can easily donate electrons and acts as strong Lewis base than PH<sub>2<\/sub>.<br \/>\n(iii) The greater catenation tendency of sulphur is due to two reasons :<\/p>\n<ul>\n<li>The lone pair of electrons feels more repulsion in O-O bond than S-S bond due to its small size and thus S \u2013 S forms strong bond.<\/li>\n<li>As the size of atom increases down the group from O-PO, the strength of bond increases and therefore catenation tendency also increases.<\/li>\n<\/ul>\n<p>Question 162.<br \/>\n(a) Account for the following :<br \/>\n(i) Acidic character increases from HF to HI.<br \/>\n(ii) There is large difference between the melting and boiling points of oxygen and sulphur.<br \/>\n(iii) Nitrogen does not form pentahalide.<br \/>\n(b) Draw the structures of the following :<br \/>\n(i) ClF<sub>3<\/sub> (ii) XeF<sub>4<\/sub> (Delhi, All India 2015)<br \/>\nAnswer:<br \/>\n(a) (i) Acidic character increases from HF to HI because bond dissociation enthalpy decreases from HF to HI.<br \/>\n(ii) Oxygen exists as diatomic O<sub>2<\/sub> molecule while sulphur exists as polyatomic S<sub>8<\/sub> molecule which has very high molecular mass therefore sulphur has much high melting and boiling points.<br \/>\n(iii) Nitrogen does not contain\u2019d\u2019 orbitals.<\/p>\n<p>(b) (i) ClF<sub>3<\/sub><\/p>\n<p>Question 163.<br \/>\n(i) Which allotrope of phosphorus is more reactive and why?<br \/>\n(ii) How the supersonic jet aeroplanes are responsible for the depletion of ozone layer?<br \/>\n(iii) F<sub>2<\/sub> has lower bond dissociation enthalpy than Cl<sub>2<\/sub>. Why?<br \/>\n(iv) Which noble gas is used in filling balloons for meteorological observations?.<br \/>\n(v) Complete the equation :<br \/>\nXeF<sub>2<\/sub> + PF<sub>5<\/sub> \u2192 (Delhi, All India 2015)<br \/>\nAnswer:<br \/>\n(i) White phosphorus is more reactive because it is less stable due to angular strain.<br \/>\n(ii) The oxides of nitrogen released by the exhausts of supersonic jetplanes are causing the depletion of ozone layer.<br \/>\n(iii) F<sub>2<\/sub> has lower dissociation enthalpy as due to small size of F there is strong repulsive forces between F-F electrons.Bond dissociation enthalpy of F<sub>2<\/sub> is less than that of Cl<sub>2<\/sub> because of large electron-electron repulsion among the lone pairs in small size F<sub>2<\/sub>molecule.<br \/>\n(iv) Helium is filled in the balloons for meteorological observations.<br \/>\n(v) XeF<sub>2<\/sub> + PF<sub>5<\/sub> \u2192 [XeF]<sup>+<\/sup> [PF]<sup>\u2013<\/sup><\/p>\n<p>Question 164.<br \/>\nAccount for the following :<br \/>\n(i) Acidic character increases from HF to HI.<br \/>\n(ii) There is large difference between the melting and boiling points of oxygen and sulphur.<br \/>\n(iii) Nitrogen does not form pentahalide.<br \/>\nAnswer:<br \/>\n(i) Acidic character increases from HF to HI bond because dissociation enthalpy decreases from HF to HI due to which H+ ion releases easily.<br \/>\n(ii) Oxygen exists as diatomic molecule O<sub>2<\/sub> while sulphur exists as S<sub>8<\/sub> molecule which has very high molecular mass therefore sulphur has high BP and MP.<br \/>\n(iii) Nitrogen does not contain\u2019d\u2019 orbitals.<\/p>\n<p>Question 165.<br \/>\n(a) Complete the following chemical equations :<br \/>\n(i) Cu + HNO<sub>3<\/sub> (dilute) \u2192<br \/>\n(ii) P<sub>4<\/sub> + NaOH + H<sub>2<\/sub>O \u2192<br \/>\n(b) (i) Why does R<sub>3<\/sub>P = O exist but R<sub>3<\/sub>N = O does not? (R = alkyl group)<br \/>\n(ii) Why is dioxygen a gas but sulphur a solid?<br \/>\n(iii) Why are halogens coloured? (Comptt. Delhi 2015)<br \/>\nAnswer:<br \/>\n(a)<\/p>\n<p>(b) (i) Nitrogen in R<sub>3<\/sub>N = O cannot form p\u03c0 \u2013 d\u03c0 multiple bonds because it cannot expand its covalency beyond 4 due to absence of d-orbital, so it will not exist. But R<sub>3<\/sub>P = O can do so due to presence of d-orbitals and formation of p\u03c0 \u2013 d\u03c0 multiple bonds which can expand its covalency up to 5.<\/p>\n<p>(ii) Because of bigger size and the strong forces of attraction holding 8 atoms, their bonds cannot be broken easily and hence sulphur exists as solid while oxygen due to high electro-negativity and tendency to form p\u03c0 \u2013 d\u03c0 multiple bonds through Vander-waals forces of attraction can be broken easily and hence exists as gas.<\/p>\n<p>(iii) All halogens are coloured due to absorption of light in the visible region as a result of which their electrons get excited to higher energy levels and while returning to lower level transmit energy of corresponding colour.<\/p>\n<p>Question 166.<br \/>\n(a) Write balanced equations for the following reactions :<br \/>\n(i) Chlorine reacts with dry slaked lime.<br \/>\n(ii) Carbon reacts with concentrated H<sub>2<\/sub>SO<sub>4<\/sub><br \/>\n(b) Describe the contact process for the manufacture of sulphuric acid with special reference to the reaction conditions, catalysts used and the yield in the process. (Comptt. Delhi 2015)<br \/>\nAnswer:<br \/>\n(a)<\/p>\n<p>(b) Contact process of sulphuric acid :<br \/>\nIt involves the following steps :<br \/>\n(i) Formation of sulphur dioxide by burning either sulphur or iron pyrites in excess of air.<br \/>\nS + O<sub>2<\/sub> \u2192 SO<sub>2<\/sub><br \/>\n4FeS<sub>2<\/sub> + 11O<sub>2<\/sub> \u2192 2Fe<sub>2<\/sub>O<sub>3<\/sub> + 8SO<sub>2<\/sub><br \/>\n(ii) Catalytic oxidation of SO<sub>2<\/sub> into SO<sub>3<\/sub> by using V<sub>2<\/sub>O<sub>5<\/sub> as catalyst<\/p>\n<p>According to Le-Chatelier\u2019s principle the reaction conditions are :<br \/>\n(a) High concentration of reactants<br \/>\n(b) Low temperature (623-723 K)<br \/>\n(c) High pressure (2 bar)<br \/>\nBy obeying above conditions the yield of H<sub>2<\/sub>SO<sub>4<\/sub> will be 96 \u2013 98%<br \/>\n(iii) Absorption of S03 in 98% H2S04 to give Oleum (H2S207)<\/p>\n<p>(iv) Dilution of Oleum to give sulphuric acid<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 167.<br \/>\n(a) Elements of Gr. 16 generally show lower value of first ionization enthalpy compared to the corresponding periods of Gr. 15. Why?<br \/>\n(b) What happens when<br \/>\n(i) Concentrated H<sub>2<\/sub>SO<sub>4<\/sub> is added to CaF<sub>2<\/sub>?<br \/>\n(ii) Sulphur dioxide reacts with chlorine in the presence of charcoal?<br \/>\n(iii) Ammonium chloride is treated with Ca(OH)<sub>2<\/sub>? (Comptt. All India 2015)<br \/>\nAnswer:<br \/>\n(a) Elements of group 16, i.e., oxygen family have general electronic configuration of ns<sup>2<\/sup>np<sup>4<\/sup> while elements of group 15, i.e., nitrogen family have general electronic configuration of ns<sup>2<\/sup>np<sup>3<\/sup> which is a relatively stable half-filled configuration with high exchange energy and therefore require more ionization energy to release electrons from this stable configuration.<br \/>\n(b)<\/p>\n<p>Question 168.<br \/>\n(a) Draw the structure of the following :<br \/>\n(i) BrF<sub>3<\/sub> (ii) XeO<sub>3<\/sub><br \/>\n(b) Answer the following :<br \/>\n(i) Why is NH<sub>2<\/sub> more basic than PH<sub>3<\/sub>?<br \/>\n(ii) Why are halogens strong oxidising agents?<br \/>\n(iii) Draw the structure of XeOF<sub>4<\/sub>. (Comptt. All India 2015)<br \/>\nAnswer:<br \/>\n(a)<\/p>\n<p>(b) (i) Both P and N contains lone pairs of electrons but due to small size and high electronegativity of nitrogen in NH<sub>3<\/sub>, the electron density is much higher than PH<sub>3<\/sub>, therefore it can easily donate electrons and acts as a strong Lewis base than PH<sub>3<\/sub>.<br \/>\n(ii) Halogens have strong tendency to accept an electron due to high negative electron gain enthalpies. Hence halogens act as strong oxidising agents.<br \/>\n(iii) XeOF<sub>4<\/sub>:<\/p>\n<p>Question 169.<br \/>\nAccount for the following:<br \/>\n(a) (i) Ozone is thermodynamically unstable.<br \/>\n(ii) Solid PCl<sub>5<\/sub> is ionic in nature.<br \/>\n(iii) Fluorine forms only one oxoacid HOF.<br \/>\n(b) Draw the structure of (i) BrF<sub>5<\/sub> (ii) XeF<sub>4<\/sub> (Delhi 2016)<br \/>\nAnswer:<br \/>\n(a) (i) Ozone is thermodynamically very unstable because:<br \/>\n\u2022 The decomposition of ozone into oxygen is exothermic in nature. (\u0394H = -ve)<br \/>\n\u2022 There is also increase in entropy which in turn makes \u0394G -ve and reaction spontaneous.<br \/>\n(ii) In solid state, PCl<sub>5<\/sub> exists as [PCl<sub>3<\/sub>]<sup>+<\/sup> [PCl<sub>6<\/sub>]<sup>\u2013<\/sup> in which the cation is tetrahedral and anion is octahedral. Because of the presence of strong attractive forces, it is a solid.<br \/>\n(iii) Due to absence of d-orbitals in fluorine, it can only form one oxoacid i.e., HOF.<\/p>\n<p>(b) (i) Structure of BrF<sub>5<\/sub><\/p>\n<p>Question 170.<br \/>\n(i) Compare the oxidizing action of F<sub>2<\/sub> and Cl<sub>2<\/sub> by considering parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy.<br \/>\n(ii) Write the conditions to maximize the yield of H<sup>2<\/sup>SO<sup>4<\/sup> by contact process.<br \/>\n(iii) Arrange the following in the increasing order of property mentioned:<br \/>\n(a) H<sub>3<\/sub>PO<sub>3<\/sub>, H<sub>3<\/sub>PO<sub>4<\/sub>, H<sub>3<\/sub>PO<sub>2<\/sub> (Reducing character)<br \/>\n(b) NH<sub>3<\/sub>, PH<sub>3<\/sub>, AsH<sub>3<\/sub>, SbH<sub>3<\/sub>, BiH<sub>3<\/sub> (Base strength) (Delhi 2016)<br \/>\nAnswer:<br \/>\n(i) Since the standard reduction potential value of fluorine is more than that of chlorine, so fluorine is a stronger oxidising agent.<\/p>\n<ul>\n<li>Bond dissociation enthalpy of F<sub>2<\/sub> is less as compared to that of chlorine.<\/li>\n<li>The negative electron gain enthalpy of fluorine is slightly less than that of chlorine.<\/li>\n<li>The hydration enthalpy of flouride ion is much higher than that of Cl<sup>\u2013<\/sup> ion due to smaller size.<\/li>\n<\/ul>\n<p>(ii) Favourable conditions for manufacturing sulphuric acid by contact process are:<\/p>\n<ul>\n<li>high pressure of about 2 bars<\/li>\n<li>low temperature 720 K<\/li>\n<li>presence of V<sub>2<\/sub>O<sub>5<\/sub> catalyst<\/li>\n<\/ul>\n<p>(iii) (a) H<sub>3<\/sub>PO<sub>4<\/sub> &lt; H<sub>3<\/sub>PO<sub>3<\/sub> &lt; H<sub>3<\/sub>PO<sub>2<\/sub> (Reducing character)<br \/>\n(b) BiH<sub>3<\/sub> &lt; SbH<sub>3<\/sub> &lt; AsH<sub>3<\/sub> &lt; PH<sub>3<\/sub> &lt; NH<sub>3<\/sub> (Basic strength)<\/p>\n<p>Question 171.<br \/>\n(a) What happens when :<br \/>\n(i) HCl reacts with finely powdered iron.<br \/>\n(ii) Cl<sub>3<\/sub> reacts with hot concentrated solution of NaOH.<br \/>\n(b) Give appropriate reason for each of the following :<br \/>\n(i) Sulphur vapour exhibits some paramagnetic character.<br \/>\n(ii) NH<sup>3<\/sup> is more basic than PH<sup>3<\/sup><br \/>\n(iii) Dioxygen is a gas but sulphur a solid. (Comptt. Delhi 2016)<br \/>\nAnswer:<br \/>\n(a) (i) It forms FeCl<sub>2<\/sub> and H<sub>2<\/sub> gas<\/p>\n<p>(b) (i) Sulphur above 1000 K exists partially as S<sub>2<\/sub> molecule which has two unpaired<br \/>\nelectrons in its Anti Bonding orbitals.<br \/>\n(ii) Due to small size of N, NH<sub>3<\/sub> molecule has small surface area and high electron density in comparison to PH<sup>3<\/sup>. Hence it can donate lone pair of electron more readily.<br \/>\n(iii) In dioxygen there is P<sub>\u03c0<\/sub> \u2013 P<sub>\u03c0<\/sub> multiple bonding and no cation hence it is a gas.<\/p>\n<p>Question 172.<br \/>\n(a) Complete the following equations :<\/p>\n<p>(b) Explain giving reason in each case :<br \/>\n(i) Halogens are strong oxidising agents<br \/>\n(ii) Fluorine form only one oxoacid, HOF<br \/>\n(iii) Helium is used in diving apparatus (Comptt. Delhi 2016)<br \/>\nAnswer:<br \/>\n(a)<\/p>\n<p>(b) (i) Due to their high electron affinity. As they have seven electrons, they accept one electron readily.<br \/>\n(ii) Due to high electronegativity and small size of fluorine.<br \/>\n(iii) Due to low solubility in blood.<\/p>\n<p>Question 173.<br \/>\nDraw the structure of<br \/>\n(i) BrF<sub>3<\/sub> (ii) XeOF<sub>4<\/sub><br \/>\nExplain giving reason in each case :<br \/>\n(i) Why H<sub>2<\/sub>Te is more acidic than H<sub>2<\/sub>S?<br \/>\n(ii) Why are halogens strong oxidising agents?<br \/>\n(iii) Why does nitrogen show catenation tendency less than phosphorus? (Comptt. All India 2016)<br \/>\nAnswer:<br \/>\n(a)<\/p>\n<p>(b) (i) Because of low bond dissociation enthalpy of H-Te bond as Te has larger size than S.<br \/>\n(ii) Because of their strong electron affinity they have seven valence electrons.<br \/>\n(iii0 Because N-N single bond is weaker than P-P single bond.<\/p>\n<p>Question 174.<br \/>\n(i) Why PCl<sub>5<\/sub> gives fumes in moisture?<br \/>\n(ii) Why interhalogens are more reactive than pure halogens?<br \/>\n(b) Draw the structures of the following :<br \/>\n(i) PCl<sub>5(s)<\/sub> (ii)H<sub>2<\/sub>S<sub>2<\/sub>O<sub>8<\/sub> (iii) XeF<sub>4<\/sub> (Comptt. All India 2016)<br \/>\nAnswer:<br \/>\n(a) (i) Because it reacts with moisture present in the air giving HCl.<br \/>\n(ii) Because of low bond dissociation enthalpy of interhalogens.<br \/>\n(b) (i) PCl<sub>5<\/sub>(s)<\/p>\n<p>Question 175.<br \/>\n(a) When concentrated sulphuric acid was added to an unknown salt present in a test tube a brown gas (A) was evolved. This gas intensified when copper turnings were added to this test tube. On cooling the gas (A) changed into a colourless solid (B).<br \/>\nIdentify (A) and (B). Write chemical reactions involved.<br \/>\n(b) Draw structure of XeOF<sub>4<\/sub>. (Comptt. All India 2017)<br \/>\nAnswer:<br \/>\n(a) The salt is sodium nitrate which on heating with cone. H<sub>2<\/sub>SO<sub>4<\/sub> evolves a brown gas i.e. NO<sub>2<\/sub> which gets intensified when Cu turnings are added.<\/p>\n<p>On cooling, NO<sub>2<\/sub> condenses as a brown liquid which turns paler on cooling and eventually becomes a colourless solid due to formation of dimerized NO<sub>2<\/sub> i.e. N<sub>2<\/sub>O<sub>4<\/sub> (Dinitrogen tetraoxide).<br \/>\nA: Nitrogen dioxide (NO<sub>2<\/sub>)<br \/>\nB : Dinitrogen tetraoxide (Na<sub>2<\/sub>O<sub>4<\/sub>)<br \/>\n(b) Structure of XeOF<sub>4<\/sub>:<\/p>\n<p>Question 176.<br \/>\n(a) Account for the following :<br \/>\n(i) Reducing character decreases from SO<sub>2<\/sub> to TeO<sub>2<\/sub>.<br \/>\n(ii) HClO<sub>3<\/sub> is a stronger acid than HCIO.<br \/>\n(iii) Xenon forms compounds with fluorine and oxygen only.<br \/>\n(b) Complete the following equations :<br \/>\n(i) 4NaCl + MnO<sub>2<\/sub> + 4H<sub>2<\/sub>SO<sub>4<\/sub> \u2192<br \/>\n(ii) 6XeF<sub>4<\/sub> + 12H<sub>2<\/sub>O \u2192 (Comptt. All India 2017)<br \/>\nAnswer:<br \/>\n(a) (i) Reducing character decreases from SO<sub>2<\/sub> to TeO<sub>2<\/sub> because the p\u03c0 \u2013 p\u03c0 bonds in them become weaker with increase in size and bond length along the group.<br \/>\n(ii) HClO<sub>3<\/sub> is a stronger acid than HClO because with increase in oxidation state and oxidation number the acidic character increases i.e. HClO<sub>3<\/sub> (+5) and HCIO (+1)<br \/>\n(iii) Xenon forms compounds with fluorine and oxygen only due to their high electronegativity and reactivity. The first ionisation energy of it is fairly close to that of O<sub>2<\/sub> and F<sub>2<\/sub>.<\/p>\n<p>(b) (i) 4NaCl + MnO<sub>2<\/sub> + 4H<sub>2<\/sub>SO<sub>4<\/sub> \u2192 4NaHSO<sub>4<\/sub> + MnCl<sub>2<\/sub> + 2H<sub>2<\/sub>O + Cl<sub>2<\/sub><br \/>\n(ii) 6XeF<sub>4<\/sub> + 12H<sub>2<\/sub>O \u2192 4Xe + 2XeO<sub>3<\/sub> + 24HF + 3O<sub>2<\/sub><\/p>\n<h4><\/h4>\n","protected":false},"excerpt":{"rendered":"<p>Important Questions for Class 12 Chemistry Chapter 7 The p-Block Elements Class 12 Important Questions The p-Block Elements Class 12 [&hellip;]<\/p>\n","protected":false},"author":7,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_yoast_wpseo_focuskw":"","_yoast_wpseo_title":"","_yoast_wpseo_metadesc":"Get Important Questions for Class 12 Chemistry Chapter 7 The p-Block Elements Class 12 Important Questions to infinity learn.","custom_permalink":"study-material\/important-questions\/class-12\/chemistry\/chapter-7\/the-p-block-elements\/class-12\/important-questions\/"},"categories":[93,21],"tags":[],"table_tags":[],"acf":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v17.9 - 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