{"id":38590,"date":"2022-12-05T11:36:46","date_gmt":"2022-12-05T06:06:46","guid":{"rendered":"https:\/\/tsboardsolutions.com\/?p=38590"},"modified":"2022-12-05T11:37:32","modified_gmt":"2022-12-05T06:07:32","slug":"ts-inter-2nd-year-chemistry-study-material-chapter-6d","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.com\/ts-inter-2nd-year-chemistry-study-material-chapter-6d\/","title":{"rendered":"TS Inter 2nd Year Chemistry Study Material Chapter 6(d) Group-18 Elements"},"content":{"rendered":"

Telangana TSBIE\u00a0TS Inter 2nd Year Chemistry Study Material<\/a> Lesson 6(d) Group-18 Elements Textbook Questions and Answers.<\/p>\n

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

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

Question 1.
\nWhat inspired Bartlett for carrying out reaction between Xe and PtF6<\/sub>?
\nAnswer:
\nNeil Bartlett, first prepared a red compound which is formulated as Q2<\/sub>+<\/sup> PtF6<\/sub>–<\/sup>. He, then realised that the first ionisation enthalpy of molecular oxygen was almost identical with that of Xenon. He then made another red colour compound XePtF6<\/sub> by mixing PtF6<\/sub> and Xenon.<\/p>\n

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

Question 3.
\nWhy do noble gases have comparatively large atomic sizes?
\nAnswer:
\nThe atomic radii of the noble gas elements are all very large. Because these have van der Waals radii which Is larger than Ionic and covalent radii.<\/p>\n

\"TS<\/p>\n

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

    \n
  1. Neon is used in discharge tubes and fluorescent bulbs for advertisement display purposes.<\/li>\n
  2. Neon bulbs are used in botanical gardens and in green houses.<\/li>\n<\/ol>\n

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

      \n
    1. Argon is used to provide an inert atmosphere in high temperature metallurgical processes.<\/li>\n
    2. For filling electric bulbs.<\/li>\n
    3. Used in laboratory for handling substances that are air sensitive.<\/li>\n<\/ol>\n

      Question 6.
      \nIn modern diving apparatus, a mixture of He and O2<\/sub> is used. Why? [AP ’16]
      \nAnswer:
      \nHelium has very low solubility in blood. So it is used as a diluent for oxygen in modern diving apparatus.<\/p>\n

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

      \"TS<\/p>\n

      Question 8.
      \nHow is XeO3<\/sub> prepared ?
      \nAnswer:
      \nXeO3<\/sub> is prepared by the hydrolysis of XeF6<\/sub>.
      \nXeF6<\/sub> + 3H2<\/sub>O \u2192 XeO3<\/sub> + 6HF<\/p>\n

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

      Question 10.
      \nExplain the structure of XeO3<\/sub>. [TS ’16 ; IPE ’14]
      \nAnswer:
      \nXe undergoes sp3<\/sup> hybridisation.
      \n\"TS
      \nUnpaired electrons in d-orbital form d\u03c0 – p\u03c0 bonds with oxygen, sp3<\/sup> orbitals form 3\u03c3 bonds with oxygen atoms. Because of the presence of lone pair, the molecule assumes pyramidal shape with a bond angle of 103\u00b0.<\/p>\n

      Question 11.
      \nNoble gases are inert- Explain.
      \nAnswer:
      \nThe inertness to chemical reactivity is due to –<\/p>\n

        \n
      1. The noble gases except helium (1 s2<\/sup>) have completely filled ns2<\/sup>np6<\/sup> electronic configuration in their valence shell.<\/li>\n
      2. They have high ionisation enthalpy and high positive electron gain enthalpy.<\/li>\n<\/ol>\n

        \"TS<\/p>\n

        Question 12.
        \nWrite the name and formula of the first noble gas compound prepared by Bertlett.
        \nAnswer:
        \nXe+<\/sup> [PtF6<\/sub>]–<\/sup>
        \nXenon hexa fluoro platinate<\/p>\n

        Question 13.
        \nExplain the shape of XeF4<\/sub> on the basis of use VSEPR theory. [Mar. 2018 – AP]
        \nAnswer:
        \nXeF4<\/sub> [Xenon tetrafluoride]:
        \nshape is octahedral. But due to the presence of two lone pairs, the molecule assumes square planar structure.
        \n\"TS
        \nTotal number of electron pairs is 6. The Xe atom undergoes sp3<\/sup>d2<\/sup> Hybridisation. Hence shape is octahedral. But due to the presence of two lone pairs, the molecule assumes square planar structure.
        \n\"TS<\/p>\n

        Question 14.
        \nGive the outer electronic configuration of noble gases.
        \nAnswer:
        \nns2<\/sup>np6<\/sup><\/p>\n

        Question 15.
        \nWhy do noble gases form compounds with fluorine and oxygen only?
        \nAnswer:
        \nBecause fluorine and oxygen are highly electronegative, they can form compounds with noble gases.<\/p>\n

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

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

        \"TS<\/p>\n

        Question 18.
        \nWhich noble gas is radioactive? How is it formed?
        \nAnswer:
        \nRadon. Radon is obtained as a decay product of Ra226<\/sup>.
        \n\\({ }_{88}^{226} \\mathrm{Ra}\\) \u2192 \\({ }_{86}^{222} \\mathrm{Rn}\\) + \\({ }_2^4 \\mathrm{He}\\)<\/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
        \nb) Radon
        \nc) Helium
        \nd) Xenon
        \ne) Radon<\/p>\n

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

        Question 20.
        \nHow are xenon fluorides XeF2<\/sub>, XeF4<\/sub> and XeF6<\/sub> obtained ? [AP ’17 ; IPE ’14] [Mar. 2018 – TS]
        \nAnswer:
        \nXenon fluorides are obtained by the direct combination of elements under appropriate conditions.
        \n\"TS<\/p>\n

        \"TS<\/p>\n

        Question 21.
        \nHow are XeO3<\/sub> and XeOF4<\/sub> prepared?
        \nAnswer:
        \nHydrolysis of XeF4<\/sub> and XeF6<\/sub> gives 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 22.
        \nGive the formulae and describe the struc-tures of noble gas sp>ecies, isoelectronic with
        \na) ICl4<\/sub>
        \nb) IBr2<\/sub>–<\/sup>
        \nc) BrO3<\/sub>–<\/sup>
        \nAnswer:
        \na) XeF4<\/sub>
        \nb) XeF2<\/sub>
        \nc) XeF6<\/sub><\/p>\n

        a) ICl4<\/sub> is isoelectronic with XeF4<\/sub>. XeF4<\/sub> is square planar in which Xe undergoes sp3<\/sup>d2<\/sup> hybridisation. If has two lone pairs.<\/p>\n

        Structure of XeF4<\/sub>: Square planar.
        \n\"TS<\/p>\n

        \"TS<\/p>\n

        b) IBr2<\/sub>–<\/sup> is isoelectronic with XeF2<\/sub>. XeF2<\/sub> is linear. Xe undergoes sp3<\/sup>d hybridisation. Three lone pairs occupy three vertices of triangle.
        \n\"TS<\/p>\n

        c) BrO3<\/sub>–<\/sup> is isoelectronic with XeF6<\/sub>. In XeF6<\/sub>, Xe undergoes sp3<\/sup>d3<\/sup> hybridisation. XeF6<\/sub> has seven electron pairs (6 bonding pairs and one lone pair) XeF6<\/sub> thus has a distorted octahedral structure
        \n\"TS<\/p>\n

        Question 23.
        \nExplain the reaction of the following with water. [AP ’15 ; IPE ’14]
        \na) XeF2<\/sub>
        \nb) XeF4<\/sub>
        \nc) XeF6<\/sub>
        \nAnswer:
        \na) XeF2<\/sub> reacts with water and give Xe, HF and O2<\/sub>
        \n2XeF2<\/sub> + 2H2<\/sub>O \u2192 2Xe + 4HF + O2<\/sub><\/p>\n

        b) XeF4<\/sub> reacts with water and give Xe, XeO3<\/sub>, HF and O2<\/sub>
        \n6XeF4<\/sub> + 12H2<\/sub>O \u2192 4 Xe + 2XeO3<\/sub> + 24 HF + 3O2<\/sub><\/p>\n

        c) XeF6<\/sub> reacts with water and give XeO3<\/sub>, HF
        \nXeF6<\/sub> + 3H2<\/sub>O \u2192 XeO3<\/sub> + 6 HF<\/p>\n

        Question 24.
        \nExplain the structures of [(AP ’17) (Mar. 2018 – T.S)]
        \na) XeF2<\/sub> and
        \nb) XeF4<\/sub>
        \nAnswer:
        \na) In XeF2<\/sub>, Xe undergoes sp3d hybridisation.
        \n\"TS
        \nThere are three bond pairs and two lone pairs. The two bond pairs overlap end-end with 2pz<\/sub> orbitals of two F atoms. Thus XeF2<\/sub> molecule has linear structure.<\/p>\n

        b) XeF4<\/sub> is square planar. Xe undergoes sp3<\/sup>d2<\/sup> hybridisation. [Mar. 2018 – A.P.]
        \n\"TS
        \nThere are two lone pairs and four bond pairs. The four bond pairs overlap end-end with 2pz<\/sub> orbitals of four F atoms.
        \n\"TS<\/p>\n

        \"TS<\/p>\n

        Question 25.
        \nExplain the structure of [TS ’15]
        \na) XeF6<\/sub> and
        \nb) XeOF4<\/sub>.
        \nAnswer:
        \na) XeF6<\/sub> has distorted octohedral shapes. XeF6<\/sub> has seven electron pairs. (6 bonding and one lone pair). It has a distorted octahedral structure in the gas phase.
        \n\"TS<\/p>\n

        b) XeOF4<\/sub> has square pyramidal structure.
        \n\"TS
        \nXe in excited state undergoes sp3<\/sup>d2<\/sup> hybridisation. One hybrid orbital is occupied by a lone pair. Other five orbitals are occupied by unpaired electrons. Five orbitals form five \u03c3 bonds. The unpaired electrons in unhybridised’d’ orbital forms a \u03c0 bond with oxygen. As the lone pair occupies axial position, XeOF4<\/sub> assumes square pyramidal structure.
        \n\"TS<\/p>\n

        Question 26.
        \nComplete 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
        \n(Hint: NaF + XeF6<\/sub> \u2192 Na+<\/sup>[XeF7<\/sub>]–<\/sup>)
        \nAnswer:
        \na) XeF2<\/sub> is hydrolysed to give Xe, HF and O2<\/sub>.
        \n2XeF2<\/sub>(s) + 2H2<\/sub>O (l) \u2192 2Xe (g) + 4HF (aq) + O2<\/sub>(g)<\/p>\n

        b) Xenon fluorides react with fluoride ion acceptors to form cationic species.
        \nXeF2<\/sub> + PF5<\/sub> \u2192 [XeF]+<\/sup> [PF6<\/sub>]–<\/sup><\/p>\n

        c) Xenon tetrafluoride reacts with fluoride ion to form cationic species.
        \nXeF4<\/sub> + SbF5<\/sub> \u2192 [XeF3<\/sub>]+<\/sup> [SbF6<\/sub>]–<\/sup><\/p>\n

        d) XeF6<\/sub> + ASF5<\/sub> \u2192 [XeF5<\/sub>]+<\/sup> [AsF6<\/sub>]–<\/sup><\/p>\n

        e) XeF6<\/sub> is prepared by the interaction of XeF6<\/sub> and O2<\/sub>F2<\/sub> at 143 K.
        \nXeF4<\/sub> + O2<\/sub>F2<\/sub> \u2192 XeF6<\/sub> + O2<\/sub><\/p>\n

        f) XeF6<\/sub> reacts with fluoride ion donors to form fluoroanions.
        \nXeF4<\/sub> + NaF \u2192 Na+<\/sup> [XeF7<\/sub>]–<\/sup><\/p>\n

        \"TS<\/p>\n

        Question 27.
        \nHow are XeF2<\/sub> and XeF4<\/sub> prepared? Give their structures. [AP ’17]
        \nAnswer:
        \nXenon fluorides are obtained by the direct combination of elements under appropriate conditions.
        \n\"TS<\/p>\n

        Hydrolysis of XeF4<\/sub> and XeF6<\/sub> gives 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

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

        Question 28.
        \nHow are XeF2<\/sub>, XeF2<\/sub>. Explain their reaction with water. Discuss their structures.
        \nAnswer:
        \nXeF2<\/sub> is prepared by heating Xenon and Fluorine in a seated Nickel tube in the ratio 1 : 2 at 673K for 8hrs.
        \n\"TS
        \nReaction with water:
        \nXeF2<\/sub> is hydrolysed to give Xe, HF and O2<\/sub>
        \n2XeF2<\/sub> (s) + 2H2<\/sub>O (l) \u2192 2Xe(g) + 4HF (aq) + O2<\/sub>(g)<\/p>\n

        Structure:
        \nXe in the first excited state
        \n\"TS
        \nXe undergoes sp3<\/sup>d hybridisation. The sp3<\/sup>d orbitals assume trigonal bipyramidal geometry. Two fluorine atoms form 2\u03c0 bonds with Xe.<\/p>\n

        \"TS
        \nThe bond pairs occupy axial positions. XeF4<\/sub>: XeF4<\/sub> is formed when Xe and F2<\/sub> mixed in the ratio 1 : 5, heated to 873K at 7 bar.
        \n\"TS
        \nReaction with water: XeF4<\/sub> hydrolysed with water to XeO3<\/sub>.
        \n6XeF4<\/sub> + 12H2<\/sub>O \u2192 4Xe + 2XeO3<\/sub> + 24HF + 3O2<\/sub>
        \nStructure of XeF4<\/sub> : XeF4<\/sub> is square planar Xe in the second excited state
        \n\"TS
        \nXe undergoes sp3<\/sup> d2<\/sup> hybridisation. Xe forms four o bonds with 4 fluorine atoms. Two lone pairs occupy axial positions. The four fluorine atoms occupy corners of a square.
        \n\"TS<\/p>\n

        \"TS<\/p>\n

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

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

        Question 2.
        \nNoble gases have very low boiling points. Why?
        \nAnswer:
        \nNoble gases are monoatomic. There are no inter atomic forces except weak dispersion forces. Hence they are liquified at very low temperatures. Hence they have low boiling points.<\/p>\n

        Question 3.
        \nDoes the hydrolysis of XeF6<\/sub> lead to a redox reaction?
        \nAnswer:
        \nNo, the products of hydrolysis are XeF4<\/sub> and XeO2<\/sub>F2<\/sub> where the oxidation states of all the elements remain the same.<\/p>\n

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

        Question 1.
        \nWhy is Helium used in diving apparatus ?
        \nAnswer:
        \nHelium acts as a diluent for oxygen. Hence it is used in diving apparatus.<\/p>\n

        \"TS<\/p>\n

        Question 2.
        \nBalance the following equation.
        \nXeF6<\/sub> + H6<\/sub>O > XeO2<\/sub> F2<\/sub> + HF
        \nAnswer:
        \nXeF6<\/sub> + 2H2<\/sub>O > XeO2<\/sub>F2<\/sub> + 4HF<\/p>\n

        Question 3.
        \nWhy has it been difficult to study the chemistry of Radon?
        \nAnswer:
        \nRadon is radioactive element with very short Half-life period. Hence its study is difficult.<\/p>\n","protected":false},"excerpt":{"rendered":"

        Telangana TSBIE\u00a0TS Inter 2nd Year Chemistry Study Material Lesson 6(d) Group-18 Elements Textbook Questions and Answers. TS Inter 2nd Year Chemistry Study Material Lesson 6(d) Group-18 Elements Very Short Answer Questions (2 Marks) Question 1. What inspired Bartlett for carrying out reaction between Xe and PtF6? 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