{"id":35320,"date":"2022-09-09T09:47:53","date_gmt":"2022-09-09T04:17:53","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=35320"},"modified":"2022-09-09T09:47:53","modified_gmt":"2022-09-09T04:17:53","slug":"ap-inter-2nd-year-chemistry-study-material-chapter-6d","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-inter-2nd-year-chemistry-study-material-chapter-6d\/","title":{"rendered":"AP Inter 2nd Year Chemistry Study Material Chapter 6(d) Group-18 Elements"},"content":{"rendered":"

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material<\/a> Lesson 6(d) Group-18 Elements Textbook Questions and Answers.<\/p>\n

AP Inter 2nd Year Chemistry Study Material Lesson 6(d) Group-18 Elements<\/h2>\n

Very Short Answer Questions<\/span><\/p>\n

Question 1.
\nWhat inspired Bartlett for carrying out reaction between Xe and PtF6<\/sub> ?
\nAnswer:<\/p>\n

    \n
  1. At first Bartlett prepared a red compound \\(\\mathrm{O}_2^{+}+\\mathrm{PtF}_6^{-}\\)<\/li>\n
  2. As the first Ionisation Enthalpy of molecular oxygen is identical with that the Xe.<\/li>\n
  3. He made efforts to prepare same type of compound with Xe and was successful in preparing another red colour compound Xe+<\/sup> \\(\\mathrm{PtF}_6^{-}\\) by mixing PtF6<\/sub> and Xe.<\/li>\n<\/ol>\n

    Question 2.
    \nWhich of the following does not exist ?
    \na) XeOF4<\/sub>
    \nb) NeF2<\/sub>
    \nc) XeF2<\/sub>
    \nd) XeF6<\/sub>
    \nAnswer:
    \nGiven compounds are XeOF4<\/sub>, NeF2<\/sub>, XeF2<\/sub>, XeF6<\/sub>.
    \nNeF2 <\/sub>does not exist among the above compounds.<\/p>\n

    Reason : Due to small size and high I.E. of ‘Ne’. It does not form chemical compounds.<\/p>\n

    \"AP<\/p>\n

    Question 3.
    \nWhy do noble gases have comparatively large atomic sizes ?
    \nAnswer:
    \nNoble gases have comparatively large atomic sizes as they have vander waals radii which is larger than both the ionic and covalent radii.<\/p>\n

    Question 4.
    \nList out the uses of Neon. [A.P. Mar. 18]
    \nAnswer:
    \nUses of Ne:<\/p>\n

      \n
    1. Ne is used in discharge tubes and fluor escent bulbs for advertisement display purpose.<\/li>\n
    2. ‘Ne’ – bulbs are used in botanical gardens and in green houses.<\/li>\n<\/ol>\n

      Question 5.
      \nWrite any two uses of argon.
      \nAnswer:
      \nUses of Ar :<\/p>\n

        \n
      1. ‘Ar’ is used to create inert atmosphere in high temperature net allurgical process.<\/li>\n
      2. ‘Ar’ is ued in filling electric bulbs.<\/li>\n<\/ol>\n

        Question 6.
        \nIn modern diving apparatus, a mixture of He and O2<\/sub> is used – Why ? [A.P. Mar. 16]
        \nAnswer:
        \nIn modem diving apparatus, a mixture of He and O2<\/sub> is used because He is very low soluble in blood.<\/p>\n

        Question 7.
        \nHelium is heavier than hydrogen. Yet helium is used (instead of H2<\/sub>) in filling baloons for meteorological observations – Why ?
        \nAnswer:
        \n‘He’ is a non-inflammable and light gas. Hence it is used in filling baloons for meterological observations.<\/p>\n

        \"AP<\/p>\n

        Question 8.
        \nHow is XeO3<\/sub> prepared ?
        \nAnswer:
        \nXeF6<\/sub> on hydrolysis produce XeO3<\/sub>
        \nXeF6<\/sub> + 3H2<\/sub>O \u2192 XeO3<\/sub> + 6HF<\/p>\n

        Question 9.
        \nGive the preparation of
        \na) XeOF4<\/sub> and
        \nb) XeOaF2<\/sub>
        \nAnswer:
        \nPartial hydrolysis of XeF6<\/sub> gives oxy fluorides XeOF4<\/sub> and XeO2<\/sub>F2<\/sub>
        \nXeF6<\/sub> + H2<\/sub>O \u2192 XeOF4<\/sub> + 2HF
        \nXeF6<\/sub> + 2H2<\/sub>O \u2192 XeO2<\/sub>F2<\/sub> + 4HF<\/p>\n

        Question 10.
        \nExplain the structure of XeO3<\/sub>. [T.S. Mar. 16]
        \nAnswer:
        \nStructure of XeO3<\/sub><\/p>\n

          \n
        1. Central atom is ‘Xe’<\/li>\n
        2. ‘Xe’ under goes sp3<\/sup> hybridisation in 3rd<\/sup> excited state.
          \n\"AP<\/li>\n
        3. ‘Xe’ forms 3\u03c3-bonds and 3\u03c0-bonds with three oxygens.<\/li>\n
        4. Shape of molecule is pyramidal with bond angle 103\u00b0.<\/li>\n<\/ol>\n

          Question 11.
          \nNoble gases are inert – explain.
          \nAnswer:
          \nNoble gases are chemically inert:<\/p>\n

            \n
          1. Noble gases have stable electronic configuration coctet configuration except He)<\/li>\n
          2. Noble gas have high Ionisation energy values and have large positive values of electron gain Enthalpy.<\/li>\n<\/ol>\n

            Question 12.
            \nWrite the name and formula of the first noble gas compound prepared by Bertlett.
            \nAnswer:
            \nThe first noble gas compound prepared by Bertlett is XePtF6. Name of the compound is xenon hexa fluoro platinate.<\/p>\n

            \"AP<\/p>\n

            Question 13.
            \nExplaIn the shape of XeF4<\/sub> on the basis of VSEPR theory
            \nAnswer:
            \nShape of XeF4<\/sub>:<\/p>\n

              \n
            1. Central atom in XeF4<\/sub> is \u2018Xe\u2019.<\/li>\n
            2. Xe undergoes sp3<\/sup>d2<\/sup> hybridisation in it\u2019s 2nd<\/sup> excited. state.
              \n\"AP<\/li>\n
            3. Shape of the molecule is square planar with bond angle 90\u00b0 and bond length 1 .95\u00c5.
              \n\"AP<\/li>\n
            4. Xe – forms four o-bonds by the overlap of sp3<\/sup>d2<\/sup> – 2pz<\/sub>(F) orbitals.<\/li>\n<\/ol>\n

              Question 14.
              \nGive the outer electronic configuration of noble gases.
              \nAnswer:
              \nThe outer electronic configuration of noble gases is ns2<\/sup>np6<\/sup> (except He (1s2<\/sup>)<\/p>\n

              Question 15.
              \nWhy do noble gases form compounds with fluorine and oxygen only ?
              \nAnswer:
              \nNoble gases form compounds with flourine and oxygen only.
              \nReason : Oxygen and Fluorine are most electronegative elements.<\/p>\n

              Question 16.
              \nHow is XeOF4<\/sub> prepared ? Describe its molecular shape.
              \nAnswer:
              \nPartial hydrolysis of XeF6<\/sub> gives XeOF4<\/sub>
              \n\"AP
              \nXeF6<\/sub> + H2<\/sub>O \u2192 XeOF4<\/sub> + 2HF
              \nXeOF4<\/sub> is a colourless volatile liquid it has a square pyramidal shape.<\/p>\n

              \"AP<\/p>\n

              Question 17.
              \nWhat is the major source of helium ?
              \nAnswer:
              \nThe major source of helium is natural gas.<\/p>\n

              Question 18.
              \nWhich noble gas is radioactive ? How is it formed ?
              \nAnswer:
              \nRadon (Rn) is radio active noble gas. Radon is obtained as decay product of 86<\/sub>Ra226<\/sup>
              \n86<\/sub>Ra226<\/sup> \u2192 86<\/sub>Rn222<\/sup> + 2<\/sub>He4<\/sup><\/p>\n

              Question 19.
              \nName the following :
              \na) most abundant noble gas in atmosphere
              \nb) radioactive noble gas
              \nc) noble gas with least boiling point
              \nd) noble gas forming large number of compounds
              \ne) noble gas not present in atmosphere
              \nAnswer:
              \na) Argon is the most abundant noble gas in atmosphere.
              \nb) Radon is the radio active noble gas.
              \nc) Helium has lowest boiling point (4.2K)
              \nd) Xenon forms large number of compounds.
              \ne) Radon is not present in atmosphere.<\/p>\n

              Short Answer Questions<\/span><\/p>\n

              Question 1.
              \nHow are Xenon fluorides XeF2<\/sub>, XeF4<\/sub> and XeF6<\/sub> obtained?
              \nAnswer:
              \nXenon forms the binary fluorides XeF2<\/sub>, XeF4<\/sub>, XeF6<\/sub> as follows. These are formed by direct combination of Xe and F2<\/sub>.
              \n\"AP<\/p>\n

              Question 2.
              \nHow are XeO3<\/sub> and XeOF4<\/sub> prepared? [Mar. 14]
              \nAnswer:
              \nXeF6<\/sub> on hydrolysis produce XeO3<\/sub>
              \nXeF6<\/sub> + 3H2<\/sub>O \u2192 XeO3<\/sub> + 6HF
              \nPartial hydrolysis of XeF6<\/sub> gives XeOF4<\/sub>
              \nXeF6<\/sub> + H2<\/sub>O \u2192 XeOF4<\/sub> + 2HF<\/p>\n

              Question 3.
              \nGive the formulae and describe the structures of a noble gas species, isoelectronic with
              \na) ICl4<\/sub>
              \nb) IBr2<\/sub>
              \nc) BrO–<\/sup>3<\/sub>
              \nAnswer:
              \na) ICl4<\/sub> is also electronic with XeF4<\/sub> and it has square planar shape.
              \nb) IBr–<\/sup>2<\/sub> is also electronic with XeF2<\/sub> and it has linear shape.
              \nc) BrO–<\/sup>3<\/sub> is also electronic with XeO4<\/sub> and it has tetrahedral shape.<\/p>\n

              \"AP<\/p>\n

              Question 4.
              \nExplain the reaction of the following with water.
              \na) XeF2<\/sub>
              \nb) XeF4<\/sub>
              \nc) XeF6<\/sub>
              \nAnswer:
              \na) XeF2<\/sub> is hydrolysed to form Xe, HF and O2<\/sub>
              \n2XeF2<\/sub> + 2H2<\/sub>O \u2192 2Xe + 4HF + O2<\/sub><\/p>\n

              b) XeF4<\/sub> is hydrolysed to give XeO3<\/sub>
              \n6XeF4<\/sub> + 12H2<\/sub>O \u2192 4Xe + 2XeO3<\/sub> + 24 HF + 3O2<\/sub><\/p>\n

              c) XeF6<\/sub> undergo hydrolysis to form XeO3<\/sub>
              \nXeF6<\/sub> + 3H2<\/sub>O \u2192 XeO3<\/sub> + 6HF
              \nXeF6<\/sub> undergo partial hydrolysis to form XeOF4<\/sub> + XeO2<\/sub>F2<\/sub>
              \nXeF6<\/sub> + H2<\/sub>O \u2192 XeOF4<\/sub> + 2HF
              \nXeF6<\/sub> + 2H2<\/sub>O \u2192 XeO2<\/sub>F2<\/sub> + 4HF<\/p>\n

              Question 5.
              \nExplain the structures of [A.P. Mar. 18] [Mar. 14]
              \na) XeF2<\/sub> and
              \nb) XeF4<\/sub>
              \nAnswer:
              \na) Structure of XeF2<\/sub> :<\/p>\n

                \n
              1. In XeF2<\/sub> central atom is ‘Xe’.<\/li>\n
              2. ‘Xe’ undergoes sp3<\/sup>d hybridisation in it’s 1st<\/sup> excited state
                \n\"AP<\/li>\n
              3. Shape of molecule is linear<\/li>\n
              4. Xe form two \u03c3 – bonds with two fluorines (sp3<\/sup> – 2pz<\/sub> overlap)
                \n\"AP<\/li>\n<\/ol>\n

                b) Structure of XeF4<\/sub> :<\/p>\n

                  \n
                1. Central atom in XeF4<\/sub> is ‘Xe’.<\/li>\n
                2. Xe undergoes sp3<\/sup>d2<\/sup> hybridisation in it’s 2nd<\/sup> excited, state.
                  \n\"AP<\/li>\n
                3. Shape of the molecule is square planar with bond angle 90\u00b0 and bond length 1.95\u00c5.
                  \n\"AP<\/li>\n
                4. Xe – forms four a – bonds by the overlap of sp3<\/sup>d2<\/sup> – 2pz<\/sub>(F) orbitals.<\/li>\n<\/ol>\n

                  Question 6.
                  \nExplain the structures of
                  \na) XeF6<\/sub> and
                  \nb) XeOF4<\/sub>
                  \nAnswer:
                  \na) Structure of XeF6<\/sub> :<\/p>\n

                    \n
                  1. Central atom in XeF6<\/sub> is ‘Xe’<\/li>\n
                  2. Xe under goes sp3<\/sup>d3<\/sup> hybridisation in it’s 3rd<\/sup> excited state.
                    \n\"AP<\/li>\n
                  3. Shape of molecule is distorted octahedral.
                    \n\"AP<\/li>\n<\/ol>\n

                    \"AP<\/p>\n

                    b) Structure of \u03b1 XeOF4<\/sub><\/p>\n

                      \n
                    1. In XeOF4<\/sub> molecule \u2018Xe\u2019 under goes sp3<\/sup>d2<\/sup> hybridisation.<\/li>\n
                    2. Shape of the molecule is square pyramid.<\/li>\n
                    3. There is one Xe-O double bond containing
                      \np\u03c0 = d\u03c0 overlaping.
                      \nPartial hydrolysis of XeF6<\/sub> gives XeOF4<\/sub>
                      \nXeF6<\/sub> + H2<\/sub>O \u2192 XeOF4<\/sub> + 2HF
                      \nXeOF4<\/sub> is a colourless volatile liquid it has a square pyramidal shape.
                      \n\"AP<\/li>\n<\/ol>\n

                      Question 7.
                      \nComplete the following.
                      \na) XeF2<\/sub> + H2<\/sub>O \u2192
                      \nb) XeF2<\/sub> + PF5<\/sub> \u2192
                      \nc) XeF4<\/sub> + SbF5<\/sub> \u2192
                      \nd) XeF6<\/sub> + ASF5<\/sub> \u2192
                      \ne) XeF4<\/sub> + O2<\/sub>F2<\/sub> \u2192
                      \nf) NaF + XeF6<\/sub> \u2192
                      \nAnswer:
                      \na) 2XeF2<\/sub> + 2H2<\/sub>O \u2192 2Xe + 4HF + O2<\/sub>
                      \nb) XeF2<\/sub> + PF5<\/sub> \u2192 [XeF]+<\/sup> + A[F6<\/sub>]–<\/sup>
                      \nc) XeF4<\/sub> + SbF5<\/sub> \u2192 [XeF3<\/sub>]+<\/sup> [SbF6<\/sub>]–<\/sup>
                      \nd) XeF6<\/sub> + ASF5<\/sub> \u2192 [Xe2<\/sub>F11<\/sub>]+<\/sup> [ASF6<\/sub>]–<\/sup>
                      \ne) XeF4<\/sub> + O2<\/sub>F2<\/sub> \u2192 XeF6<\/sub> + O2<\/sub>
                      \nf) NaF + XeF6<\/sub> \u2192 Na+<\/sup> [XeF7<\/sub>]–<\/sup><\/p>\n

                      Question 8.
                      \nHow are XeF2<\/sub> and XeF4<\/sub> prepared ? Give their structures. [T.S. Mar. 18]
                      \nAnswer:
                      \nXenon forms the binary fluorides XeF2<\/sub>, XeF4<\/sub>, XeF6<\/sub> as follows. These are formed by direct combination of Xe and F2<\/sub>.
                      \n\"AP
                      \nStructure of XeF2<\/sub><\/p>\n

                        \n
                      1. In XeF2<\/sub> central atom is Xe’.<\/li>\n
                      2. ‘Xe’ undergoes sp3<\/sup>d hybridisation in it’s 1st<\/sup> excited state
                        \n\"AP<\/li>\n
                      3. Shape of molecule is linear.<\/li>\n
                      4. Xe form two \u03c3 bonds with two fluorines.<\/li>\n<\/ol>\n

                        \"AP<\/p>\n

                        b) Structure of XeF4<\/sub> :<\/p>\n

                          \n
                        1. Central atom in XeF4<\/sub> is ‘Xe’.<\/li>\n
                        2. Xe undergoes sp3<\/sup>d2<\/sup> hybridisation in it’s 2nd<\/sup> excited state.
                          \n\"AP<\/li>\n
                        3. Shape of the molecule is square planar with bond angle 90\u00b0 and bond length 1.95\u00c5.
                          \n\"AP<\/li>\n
                        4. Xe – forms four \u03c3-bonds by the overlap of sp3<\/sup>d2<\/sup> – 2pz<\/sub> (F) orbitals.<\/li>\n<\/ol>\n

                          Long Answer Question<\/span><\/p>\n

                          Question 1.
                          \nHow are XeF2<\/sub>, XeF4<\/sub> and XeF6<\/sub> prepared ? Explain their reaction with water. Discuss their structures. [A.P. Mar. 15]
                          \nAnswer:
                          \nXenon forms the binary fluorides XeF2<\/sub>, XeF4<\/sub>, XeF6<\/sub> as follows. These are formed by direct combination of Xe and F2<\/sub>.
                          \n\"AP
                          \nReaction with water:<\/p>\n

                            \n
                          1. XeF2<\/sub> is hydrolysed to form Xe, HF and O2<\/sub>
                            \n2XeF2<\/sub> + 2H2<\/sub>O \u2192 2Xe + 4HF + O2<\/sub><\/li>\n
                          2. XeF4<\/sub> is hydrolysed to give XeO3<\/sub>
                            \n6XeF4<\/sub> + 12H2<\/sub>O \u2192 4Xe + 2XeO3<\/sub> + 24 HF + 3O2<\/sub><\/li>\n
                          3. XeFg undergo hydrolysis to form XeO3<\/sub>
                            \nXeF6<\/sub> + 3H2<\/sub>O \u2192 XeOF3<\/sub> + 6HF<\/li>\n
                          4. XeF6<\/sub> undergo partial hydrolysis to form XeOF4<\/sub> and XeO2<\/sub>F2<\/sub>
                            \nXeF6<\/sub> + H2<\/sub>O \u2192 XeOF4<\/sub> + 2HF
                            \nXeF6<\/sub> + 2H3<\/sub>O \u2192 XeO2<\/sub>F2<\/sub> + 4HF<\/li>\n<\/ol>\n

                            Structure of XeF2<\/sub>:<\/p>\n

                              \n
                            1. In XeF2<\/sub> central atom is ‘Xe’.<\/li>\n
                            2. ‘Xe’ undergoes sp3<\/sup>d hybridisation in it’s 1st<\/sup> excited state
                              \n\"AP<\/li>\n
                            3. Shape of molecule is linear.<\/li>\n
                            4. Xe form two \u03c3-bonds with two fluorines.
                              \n\"AP<\/li>\n<\/ol>\n

                              b) Structure of XeF4<\/sup> :<\/p>\n

                                \n
                              1. Central atom in XeF4<\/sup> is ‘Xe’.<\/li>\n
                              2. Xe undergoes sp3<\/sup>d2<\/sup> hybridisation in it’s 2nd<\/sup> excited state.
                                \n\"AP<\/li>\n
                              3. Shape of the molecule is square planar with bond angle 90\u00b0 and bond length 1.95\u00c5.
                                \n\"AP<\/li>\n
                              4. Xe – forms four \u03c3-bonds by the overlap of sp3<\/sup>d2<\/sup> – 2pz<\/sub> (F) orbitals.<\/li>\n<\/ol>\n

                                \"AP<\/p>\n

                                Structure of XeF6<\/sub> :<\/p>\n

                                  \n
                                1. Central atom in XeF6<\/sub> is ‘Xe’<\/li>\n
                                2. Xe under goes sp3<\/sup>d3<\/sup> hybridisation in it’s 3rd<\/sup> excited state.
                                  \n\"AP<\/li>\n
                                3. Shape of molecule is distorted octahedral.
                                  \n\"AP<\/li>\n<\/ol>\n

                                  Textual Examples<\/span><\/p>\n

                                  Question 1.
                                  \nWhy are the elements of group 18 known as noble gases ?
                                  \nSolution:
                                  \nThe elements present in Group 18 have their valence shell orbitals completely filled and, therefore, react with a few elements only under certain conditions. Therefore, they are now known as noble gases.<\/p>\n

                                  Question 2.
                                  \nNoble gases have very low boiling points. Why ?
                                  \nSolution:
                                  \nNoble gases being monoatomic have no interatomic forces except weak dispersion forces and therefore, they are liquefied at very low temperatures. Hence, they have low boiling points.<\/p>\n

                                  \"AP<\/p>\n

                                  Question 3.
                                  \nDoes the hydrolysis of XeF6<\/sub> lead to a redox reaction ?
                                  \nSolution:
                                  \nNo, the products of hydrolysis are XeOF4<\/sub> and XeO2<\/sub>F2<\/sub> where the oxidation states of all the elements remain the same as it was in the reacting state.<\/p>\n

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

                                  Question 1.
                                  \nWhy is helium used in diving apparatus ?
                                  \nSolution:
                                  \nHelium is used in diving apparatus due to its very low solubility in blood.<\/p>\n

                                  Question 2.
                                  \nBalance the following equation :
                                  \nXeF6<\/sub> + H2<\/sub>O \u2192 XeO2<\/sub>F2<\/sub> + HF
                                  \nSolution:
                                  \nXeF6<\/sub> + 2H2<\/sub>O \u2192 XeO2<\/sub>F2<\/sub> + 4HF<\/p>\n

                                  Question 3.
                                  \nWhy has it been difficult to study the chemistry of radon ?
                                  \nSolution:
                                  \nRadon is a radioactive element with very short half-life of 3.82 days. That is why, the study of chemistry of radon is a difficult task.<\/p>\n","protected":false},"excerpt":{"rendered":"

                                  Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material Lesson 6(d) Group-18 Elements Textbook Questions and Answers. AP Inter 2nd Year Chemistry Study Material Lesson 6(d) Group-18 Elements Very Short Answer Questions Question 1. What inspired Bartlett for carrying out reaction between Xe and PtF6 ? Answer: At first Bartlett prepared a red compound …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[14],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35320"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=35320"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35320\/revisions"}],"predecessor-version":[{"id":35321,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/35320\/revisions\/35321"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=35320"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=35320"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=35320"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}