{"id":8940,"date":"2019-11-28T11:56:15","date_gmt":"2019-11-28T06:26:15","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=8940"},"modified":"2021-07-02T15:44:32","modified_gmt":"2021-07-02T10:14:32","slug":"1st-puc-chemistry-question-bank-chapter-11","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-chemistry-question-bank-chapter-11\/","title":{"rendered":"1st PUC Chemistry Question Bank Chapter 11 The P-Block Elements"},"content":{"rendered":"

You can Download Chapter 11 The P-Block Elements Questions and Answers, Notes, 1st PUC Chemistry Question Bank with Answers<\/a> Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 1st PUC Chemistry Question Bank Chapter 11 The P-Block Elements<\/h2>\n

1st PUC Chemistry The P-Block Elements One Mark Questions and Answers<\/h3>\n

Question 1.
\nWhy does carbon show maximum catenation among group 14 elements?
\nAnswer:<\/p>\n

    \n
  1. Small size of carbon atoms.<\/li>\n
  2. C-C bond energy is comparable with the bond energy between carbon and other element<\/li>\n
  3. 4 valency electrons are present in four orbitals and the tetravalency of carbon is fully saturated.<\/li>\n<\/ol>\n

    Question 2.
    \nName different types of charcoal.
    \nAnswer:
    \nWood charcoal, Animal charcoal, Sugar charcoal.<\/p>\n

    Question 3.
    \nGive one use of black diamond.
    \nAnswer:
    \nDue to their hardness, black diamonds are used in various cutting tools.<\/p>\n

    Question 4.
    \nWhat type of hybridization does carbon undergo in diamond and graphite?
    \nAnswer:
    \nIn diamond sp3<\/sup>, in graphite sp2<\/sup>.<\/p>\n

    Question 5.
    \nSilicon has become a vital element in the modern electronics industry. Why?
    \nAnswer:
    \nIt can function as semiconductors especially as n-type and p-type extrinsic semiconductors lubricants and insulators, Silicons are used for making water-proof papers by coating them with a thin layer of silicons.<\/p>\n

    \"KSEEB<\/p>\n

    Question 6.
    \nWhat is catenation?
    \nAnswer:
    \nCarbon atoms have a remarkable property of joining with one another in a large number to form a long chain and rings. This property is known as catenation or self linkage.<\/p>\n

    Question 7.
    \nWhat is the technique used in preparing very pure silicon?
    \nAnswer:
    \nZone refining.<\/p>\n

    Question 8.
    \nGive an example for intrinsic semiconductor other than silicon.
    \nAnswer:
    \nGermanium in the pure state.<\/p>\n

    Question 9.
    \nWho discovered fullerene 7
    \nAnswer:
    \nRichard E. Smalley, Robert F. Curl Jr. and Harold W. Kroto.<\/p>\n

    Question 10.
    \nWhat kind of semiconductor is obtained when a little of phosphorus is added to silicon?
    \nAnswer:
    \nn- type semiconductor.<\/p>\n

    Question 11.
    \nName the oxidation states exhibited by Ge, Sn, and Pb.
    \nAnswer:
    \n+2 and +4<\/p>\n

    Question 12.
    \nWhat type of semiconductor is obtained by doping germanium with indium?
    \nAnswer:
    \np-type semiconductor<\/p>\n

    Questuon 13.
    \nMention the catenation capacity of Si.
    \nAnswer:
    \n16.<\/p>\n

    Question 14.
    \nWhy is boron metalloid?
    \nAnswer:
    \nBoron resembles both with metals and non-metals therefore, it is metalloid<\/p>\n

    Question 15.
    \nGive two important ores of boron.
    \nAnswer:<\/p>\n

      \n
    1. Borax, Na2<\/sub>B4<\/sub>O7<\/sub>10H2<\/sub>O<\/li>\n
    2. Kemite, Na2<\/sub>B4<\/sub>O7<\/sub>2H2<\/sub>O<\/li>\n<\/ol>\n

      Question 16.
      \nName two isotopes of boron
      \nAnswer:
      \n\"1st are isotopes of boron.<\/p>\n

      Question 17.
      \nName two allotropes of boron.
      \nAnswer:
      \nAmorphous and crystalline.<\/p>\n

      Question 18.
      \nWhy does boron not form B3+<\/sup> ion?
      \nAnswer:
      \nBoron does not form B3+<\/sup> ion due to smallest atomic size and highest ionization energy<\/p>\n

      Question 19.
      \nWhy is crystalline boron hard solid?
      \nAnswer:
      \nIt is due to strong covalent bonds.<\/p>\n

      Question 20.
      \nWhy is boron used in nuclear reactors?
      \nAnswer:
      \nBonor can absorb neutrons.<\/p>\n

      \"KSEEB<\/p>\n

      Question 21.
      \nWhat is tincal ? Give its chemical formula.
      \nAnswer:
      \nTincal (or Borax) is an ore of boron. Its formula is NaB4<\/sub> Or 10H2<\/sub>O<\/p>\n

      Question 22.
      \nWhy does boron resemble Si?
      \nAnswer:
      \nBoth have similar charge over radius ratio, i.e., similar polarizing power.<\/p>\n

      Question 23.
      \nWhy does boron form stable electron deficient compounds?
      \nAnswer:
      \nBoron has three valence electrons, it will share three electrons with other elements to form electron deficient compounds which are stable.<\/p>\n

      Question 24.
      \nWhy is boric acid (H3<\/sub>BO3<\/sub>) monobasic acid?
      \nAnswer:
      \nIt accepts a pair of electrons from OH–<\/sup> ion of H2<\/sub>O therefore, it is monobasic acid.<\/p>\n

      Question 25.
      \nWhat type of glass is obtained when borax is added ?
      \nAnswer:
      \nPyrex glass is a glass which is heat resistant. It can withstand high temperature.<\/p>\n

      Question 26.
      \nWhat is use of diborane? Why BHs exists in form of diborane ?
      \nAnswer:
      \nBH3<\/sub> exists in form of diborane (B4<\/sub>H6<\/sub>) because it is electron deficient. Diborane is used as reducing agent.<\/p>\n

      Question 27.
      \nComplete the following
      \n(i) B + O2<\/sub> \u2192
      \n(ii) B + N2<\/sub> \u2192
      \n(iii) B + Cl2<\/sub> \u2192
      \n(iv) BF3 + NH2<\/sub> \u2192
      \n(v) Na2<\/sub>B4<\/sub>O7<\/sub>+H2<\/sub>O \u2192
      \n(vi) Na2<\/sub>B4<\/sub>O7<\/sub> \u2192
      \nAnswer:
      \n(i) 4B + 3O2<\/sub> \u2192 2B2<\/sub>O3<\/sub>
      \n\"1st<\/p>\n

      Question 28.
      \nWhy does \\(\\mathrm{BF}_{6}^{3-}\\) not exist?
      \nAnswer:
      \nIt is because \u2018B\u2019 does not have vacant d-orbitals.<\/p>\n

      Question 29.
      \nWhich type of bonds are formed by boron and why?
      \nAnswer:
      \nBoron forms covalent bonds because it cannot lose electrons or gain electron easily.<\/p>\n

      Question 30.
      \nWhat is basic structural unit of orthoboric acid? Name type of bond present in them.
      \nAnswer:
      \n\\(\\mathrm{BO}_{3}^{3-}\\) is basic structural unit of boric acid. Covalent bonds are present.<\/p>\n

      Question 31.
      \nGive structure of diborane.
      \nAnswer:
      \n\"1st<\/p>\n

      Question 32.
      \nWhy do boron halides form addition compounds with ammonia and amines?
      \nAnswer:
      \nIt is because they are electron deficient.<\/p>\n

      Question 33.
      \nIn which pure form does carbon exists in nature?
      \nAnswer:
      \nFullerene<\/p>\n

      Question 34.
      \nGive structure of CO2<\/sub>.
      \nAnswer:
      \nO = C = O, It has linear structure.<\/p>\n

      Question 35.
      \nWhat name is given to the compounds formed by more electropositive elements with carbon?
      \nAnswer:
      \nIonic compounds<\/p>\n

      \"KSEEB<\/p>\n

      Question 36.
      \nIs carbon dioxide poisonous or not?
      \nAnswer:
      \nNo, it is not poisonous.<\/p>\n

      Question 37.
      \nWhat is producer gas ?
      \nAnswer:
      \nProducer gas is mixture of CO and N2<\/sub> in the ratio of 2 : 1<\/p>\n

      Question 38.
      \nBuckminster fullerene is a crystalline allotrope of which element ?
      \nAnswer:
      \nCarbon<\/p>\n

      Question 39.
      \nWhat is state of hybridisation of C in \\(\\mathrm{co}_{3}^{3-}\\) ?
      \nAnswer:
      \n\u2018C\u2019 in \\(\\mathrm{co}_{2}^{3-}\\) has sp2<\/sup> hybridisation.<\/p>\n

      Question 40.
      \nHow does BF3<\/sub> act as catalyst in industrial process?
      \nAnswer:
      \nBF3<\/sub> is electron deficient, therefore acts as catalyst in industrial processes.<\/p>\n

      Question 41.
      \nWhy is BF3<\/sub> weaker Lewise acid than BCl3<\/sub>
      \nAnswer:
      \nBF3<\/sub> is weaker Lewis acid than BCl3<\/sub> because of more effective bonding in case of F due to smaller size, than Cl.<\/p>\n

      Question 42.
      \nName the elements present in boranes.
      \nAnswer:
      \nBoranes are made up of boron and hydrogen.<\/p>\n

      Question 43.
      \nWrite the state of hybridization in BF3<\/sub>.
      \nAnswer:
      \n\u2018B\u2019 in BF3<\/sub> has sp2<\/sup> hybridisation<\/p>\n

      Question 44.
      \nWhy does boric acid act as Lewis acid ?
      \nAnswer:
      \nIt is because in boric acid, boron does not have its octet complete. It accepts OH- from
      \nwater in aqueous solution B(OH)3<\/sub> + H2<\/sub>O \u2192 [B(OH)4<\/sub>]–<\/sup> + H+<\/sup><\/p>\n

      Question 45.
      \nWhy does boron form electron deficient compounds ?
      \nAnswer:
      \nIt is because boron has three valence electrons. It shares three electrons and gets six electrons, i.e., its octet is not complete. Therefore, it forms electron deficient compounds.<\/p>\n

      Question 46.
      \nMention the chief reason for the anomalous behaviour of boron in group 13 of the periodic table.
      \nAnswer:
      \nIt is due to small size and higher ionization energy and high charge\/size ratio of boron.<\/p>\n

      Question 47.
      \nMention the type of hybrid orbitals of silicon in \\(\\mathrm{SiF}_{6}^{2-}\\) ion.
      \nAnswer:
      \nsp3<\/sup>d2<\/sup><\/p>\n

      Question 48.
      \nMention the state of hybridization of B in \\(\\mathrm{BH}_{4}^{-}\\)
      \nAnswer:
      \nsp3<\/sup><\/p>\n

      Question 49.
      \nHow is that silicon atoms can have a co-ordination number more than four but carbon atoms cannot?
      \nAnswer:
      \nSilicon has vacant d-orbitals, therefore, it can have coordination number more than four but carbon cannot have because it does not have vacant d-orbitals.<\/p>\n

      Question 50.
      \nCO2<\/sub> is gas while SiO4<\/sub> is solid at room temperature. State a reason for this.
      \nAnswer:
      \nCO2<\/sub> exists as decrete molecules, therefore, it has weak Van der Walls\u2019 forces for attraction whereas SiO2<\/sub> is three dimensional covalent solid.<\/p>\n

      \"KSEEB<\/p>\n

      Question 51.
      \nBetween AIF3<\/sub> and AICl3<\/sub> which one will have a higher melting point.
      \nAnswer:
      \nAIF3<\/sub> is more ionic, therefore, has higher melting point.<\/p>\n

      Question 52.
      \nWhy is that aluminium metal cannot be obtained by electrolysis of an aqueous solution of a salt of aluminium?
      \nAnswer:
      \nIt is because aluminium metal reacts with water at high temperature.<\/p>\n

      Question 53.
      \nWhich one of the following elements +1 oxidation state as well: Al, B, Ca, Tl, Be?
      \nAnswer:
      \nTI shows +1 oxidation acid.<\/p>\n

      Question 54.
      \nCarbon and Silicon are mainly tetravalent but Ge, Sn and Pb show divalency. Give reason.
      \nAnswer:
      \nGe, Sn, Pb are divalent due to inert pair effect which is not there in carbon and silicon.<\/p>\n

      Question 55.
      \nWhat property of anhydrous AlCl3<\/sub> makes it a very good preparative reagent in organic chemistry?
      \nAnswer:
      \nIt acts as Lewis acid.<\/p>\n

      Question 56.
      \nAlthough pure silicon is an insulator, then how does it behave as semiconductor on heating?
      \nAnswer:
      \nSilicon becomes semiconductor on just heating because electrons becomes free to move.<\/p>\n

      Question 57.
      \nWhat is name given to elements which are neither metals nor non-metals?
      \nAnswer:
      \nMetalloids<\/p>\n

      Question 58.
      \nWhat is general formula of silicones ?
      \nAnswer:
      \nR2<\/sub>SiO is general formula of silicones.<\/p>\n

      Question 59.
      \nWhat is the oxidation state of Ni in [Ni(CO)4<\/sub>]?
      \nAnswer:
      \nNi has oxidation state of zero in [Ni(CO)4<\/sub>]<\/p>\n

      Question 60.
      \nWhat is inert pair effect ?
      \nAnswer:
      \nThe pair of electrons in valence s-orbital is reluctant to take part in bond formation due to poor shielding effect of -d and f-electron in heavier elements. It is called inert pair effect due to which lower oxidation state becomes more stable than higher oxidation state in case of p-block elements.<\/p>\n

      Question 61.
      \nHow are linear silicones obtained ?
      \nAnswer:
      \nLinear silicones are obtained by hydrolysis of R2<\/sub>SiCl2<\/sub> (chlorosilanes)
      \n\"1st<\/p>\n

      Question 62.
      \nIn terms of atomic size, ionization energy and charge, explain why are ionic compounds containing B3+<\/sup> not formed.
      \nAnswer:
      \nBoron has smaller atomic size, therefore, it cannot lose three electrons because its third ionization energy is very high.<\/p>\n

      Question 63.
      \nWhy does C differ from rest of elements ?
      \nAnswer:
      \nCarbon has smallest size, highest ionization energy and high electro negativity, therefore, it differs from rest of the elements.<\/p>\n

      Question 64.
      \nWhat are silicates?
      \nAnswer:
      \nSilicates are minerals which consist of \\(\\mathrm{SiO}_{4}^{4-}\\) units arranged in different ways.<\/p>\n

      Question 65.
      \nWrite the formula of inorganic benzene.
      \nAnswer:
      \nB3<\/sub>N3<\/sub>H6<\/sub><\/p>\n

      \"KSEEB<\/p>\n

      Question 66.
      \nExplain the following statement with reason.
      \nThe fullerene is considered as purest allotrope of carbon.
      \nAnswer:
      \nIt is because it does not have edges, therefore, impurities cannot be absorbed on it.<\/p>\n

      Question 67.
      \nMention the nature of an aqueous solution of borax.
      \nAnswer:
      \nBasic in nature.<\/p>\n

      Question 68.
      \nBoric acid is polymeric. Why?
      \nAnswer:
      \nThe presence of hydrogen bonds makes it polymeric.<\/p>\n

      Question 69.
      \nMention the type of hybridization of boron in diborane.
      \nAnswer:
      \nSp3<\/sup><\/p>\n

      Question 70.
      \nWhich is thermodynamically most stable form of carbon ?
      \nAnswer:
      \nGraphite<\/p>\n

      Question 71.
      \nState the trends observed in case of each of the following: Oxidation state of the elements of Group 14.
      \nAnswer:
      \nGroup 14 elements show +2 and +4 oxidation states. Tendency to show +2 oxidation state increases down the group due to inert pair effect.<\/p>\n

      Question 72.
      \nHow does the boric acid polymerise ?
      \nAnswer:
      \nThrough hydrogen bonding.<\/p>\n

      Question 73.
      \nDoes BH3<\/sub> exists?
      \nAnswer:
      \nNo.<\/p>\n

      Question 74.
      \nHow does boron interact with NaOH ?
      \nAnswer:
      \n2B + 6NaOH \u2192 2Na3<\/sub>B03<\/sub> +3H2<\/sub><\/p>\n

      Question 75.
      \nWhat is the oxidation state of C in
      \n(a) CO
      \n(b) HCN
      \n(c) H2<\/sub>CO3<\/sub>
      \n(d) CaC2<\/sub>.
      \nAnswer:
      \n(a) +2, (b) +2, (c) +4, (d) -1<\/p>\n

      Questuon 76.
      \nGive two examples of electron-deficient compounds.
      \nAnswer:
      \nBF3<\/sub> and B3<\/sub>H6<\/sub>.<\/p>\n

      Question 77.
      \nArrange the following halides of boron in the increasing order of acidic character. BF3<\/sub>, BCI3<\/sub>, BBr3<\/sub>, BI3<\/sub>.
      \nAnswer:
      \nBF3<\/sub> < BCI3<\/sub> < BBr3<\/sub> < BI3<\/sub>.<\/p>\n

      Question 78.
      \nWhich form of carbon is used to decolourise sugar ?
      \nAnswer:
      \nBone charcoal.<\/p>\n

      Question 79.
      \nWhat does we get when Cone. H2<\/sub>SO4<\/sub> is dropped on sugar ?
      \nAnswer:
      \nSugar Charcoal.<\/p>\n

      Question 80.
      \nWhat happens when a borax solution is acidified ? Write a balanced equation for the reaction.
      \nAnswer:
      \nBoric acid is formed (Na2<\/sub>B3<\/sub>O7<\/sub> + 2HC1 + 5H2<\/sub>)O \u21922Nacl + 4H3<\/sub>BO3<\/sub><\/p>\n

      \"KSEEB<\/p>\n

      1st PUC Chemistry The P-Block Elements Two Marks Questions and Answers<\/h3>\n

      Question 1.
      \nGive the chemical reaction as an evidence for each of the following observation?<\/p>\n

        \n
      • Tin (II) is a reducing agent whereas lead (II) is not<\/li>\n
      • Gallium (I) undergoes disproportionation reaction.<\/li>\n<\/ul>\n

        Answer:<\/p>\n

          \n
        • It is because Pb2+<\/sup> is more stable than Pb4+<\/sup> due to inert pair effect whereas Sn4+<\/sup> is more stable than Sn2+<\/sup>. Therefore Sn2+<\/sup> is good reducing agent 3Ga+<\/sup> \u2014> 2Ga + Ga3+<\/sup><\/li>\n
        • It is because Ga3+<\/sup> is more stable than Ga+<\/sup><\/li>\n<\/ul>\n

          Question 2.
          \nWrite the chemical equation for the reactions which occur when : R2<\/sub>SiCl2<\/sub> is subjected to hydrolysis.
          \nAnswer:
          \n\"1st<\/p>\n

          Question 3.
          \nComplete the following chemical reaction :
          \n1. \\(\\mathrm{RCI}+\\mathrm{Si} \\frac{\\text { Cupower }}{570 \\mathrm{K}}\\)
          \nAnswer:
          \n2RC1 + Si \u2192 R2<\/sub>SiCl2<\/sub><\/p>\n

          Questuon 4.
          \nDescribe two similarities and two dissimilarities between B and Al.
          \nAnswer:
          \nSimilarities :<\/p>\n

            \n
          1. Both have similar electronic configuration<\/li>\n
          2. Both have same number of valence electrons<\/li>\n<\/ol>\n

            Dissimilarities:<\/p>\n

              \n
            1. B is non-metal; Al is metal<\/li>\n
            2. B forms acidic oxide (B2<\/sub>O3<\/sub>) whereas Al form Al2<\/sub>O2<\/sub> which is amphoteric<\/li>\n<\/ol>\n

              Question 5.
              \nDraw a labelled diagram to show the dimeric structure of Al(III) chloride. State one use of anhydrous AICl3<\/sub>.
              \nAnswer:
              \nThe structure of AICl3<\/sub> is shown below.
              \n\"1st
              \nIt issued in preparation of organic compounds in Friedal Crafts reaction.<\/p>\n

              Question 6.
              \nArrange the following compounds in decreasing order of property indicated against each. Give reason for your answer : BCI3<\/sub>, AICl3<\/sub>, GaGL3<\/sub>, InCl3<\/sub>, TiCl3<\/sub> (Stability of +3 oxidation state)
              \nAnswer:
              \nBCl3<\/sub> > AICl3<\/sub>> GaCl3<\/sub> > InCl3<\/sub> > TiCl3<\/sub> because of inert pair effect stability of +3 oxidation state decreases.<\/p>\n

              Question 7.
              \nDiscuss the diagonal relationship of Be and Al with regard to<\/p>\n

                \n
              • the action of alkali and<\/li>\n
              • the structure of the chlorides.<\/li>\n<\/ul>\n

                Answer:
                \nBe and Al resemble with each other due to same charge over radius ratio.<\/p>\n

                  \n
                • Both B and A1 amphoteric oxides<\/li>\n
                • Be and A1 form covalent compounds<\/li>\n
                • BeCl2<\/sub> exists as polymer whereas AICl3<\/sub> exists as dimmer<\/li>\n
                • Both Be and A1 react with NaOH to form similar compounds.<\/li>\n<\/ul>\n

                  Question 8.
                  \nC and Si are always tetravalent but Ge, Sn, Pb show divalency.
                  \nAnswer:
                  \nGe, Sn, Pb show divalency due to inert pair effect, Pb2+<\/sup> is more stable than Pb4+<\/sup>.<\/p>\n

                  Question 9.
                  \nWhich of the following is acidic and why: SiCO2<\/sub>, AI2<\/sub>O3<\/sub>, PbCO2<\/sub>, SiCO2<\/sub>?
                  \nAnswer:
                  \nSiCO2<\/sub> is acidic oxide because Si is non-metal.<\/p>\n

                  Question 10.
                  \nAs we move down in group 13 elements increase in atomic size is comparatively very less. Explain.
                  \nAnswer:
                  \nIt is due to poor shielding effect of d and f-electrons, as result of which effective nuclear charge increase.<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 11.
                  \nWrite the chemistry involved in etching of glass.
                  \nSiO2<\/sub> + 4HF \u2192 SiF4<\/sub> + 2H2<\/sub>O
                  \nAnswer:
                  \nSiF4<\/sub> +2HF \u2192 H2<\/sub>SiF6<\/sub>(Fluorosilicicacid)<\/p>\n

                  Question 12.
                  \nElemental silicon does not form graphite like structure as carbon does. Give reason.
                  \nAnswer:
                  \nIt is because \u2018Si\u2019 cannot form p\u03c0-p\u03c0 bond like carbon in graphite due to larger atomic size.<\/p>\n

                  Question 13.
                  \nWhat are silicones ?
                  \nAnswer:
                  \nSilicones are rubber like polymers.<\/p>\n

                  Question 14.
                  \nGive the shapes of
                  \n(a) SiF4,
                  \n(b) \\(\\mathrm{SiF}_{6}^{2-}\\)
                  \nAnswer:
                  \n\"1st<\/p>\n

                  Questuon 15.
                  \nArrange the following in decreasing order of boiling point giving reason CH4<\/sub>, GeH4<\/sub>, SnH4<\/sub>, SiH4<\/sub>
                  \nAnswer:
                  \nSnH4<\/sub> > GeH4<\/sub> > SiH4<\/sub> > CH4<\/sub>
                  \nAs we go down the group from C to Sn surface area of hydrides increases therefore, Van der Waals\u2019 force increases hence boiling point increases.<\/p>\n

                  Question 16.
                  \nExplain the following: (a) CaO reacts with SiCO2<\/sub> to form a slag.
                  \nAnswer:
                  \n(a)
                  \n\"1st<\/p>\n

                  1st PUC Chemistry The P-Block Elements Threer \/ Four \/ Five Marks Questions and Answers<\/h3>\n

                  Question 1.
                  \nWhat are allotropy ? Mention allotropy of Pb, Sn, Ge, Si.
                  \nAnswer:
                  \nAllotropy: The existence of same element in two or more forms in the same physical state but having different properties called allotropy. The different forms are called allotrope.
                  \nPb : \u2014> No allotrope exists.
                  \nSn : \u2014> Grey ( ) = White ( ) = ; Rhombic { Tin Cry : the sound when crystals of tin is Bent or deformed}
                  \nGe : \u2014> Two (Both are crystalline form.)
                  \nSi: \u2014> Two ( Both are crystalline form.)<\/p>\n

                  Question 2.
                  \nExplain the oxidation property of group 14 elements.
                  \nAnswer:
                  \nGroup 14 elements show commonly +2 and +4 oxidation state. The carbon family also shows both positive and negative oxidation state. It shows +2 oxidation state when np2<\/sup> electrons involved in the bond formation Ex-SnCl2<\/sub>, PbSO4<\/sub>. This show +4 oxidation state when ns2<\/sup>np2<\/sup> electrons involved in the bond formation. Ex-SnCl4<\/sub><\/p>\n

                  Question 3.
                  \nExplain semiconductors with its types and uses.
                  \nAnswer:
                  \nSemiconductors are the materials their conductance is lies between those of conductors and insulators. They allow only a certain portion of the current to pass through them. Materials used for semiconductors should be extremely pure and is obtained by zone refining.<\/p>\n

                  Question 4.
                  \nExplain the types of semiconductors.
                  \nAnswer:
                  \n1. Intrinsic Semiconductors: If Si of Ge is sufficiently heated the energy supplied breaks some of the covalent bonds and electrons become free to carry current. When the electrons are get free are released, positive holes are left behind in the lattice. In intrinsic semiconductors, the number of holes and electrons will be equal.<\/p>\n

                  2. Extrinsic semiconductors: These are formed doping (addition of impurities of certain elements) to the insulators.<\/p>\n

                  Question 5.
                  \nExplain the types Of Extrinsic Semiconductors
                  \nAnswer:
                  \nI. n-type semiconductor: Doping of (Group 14 elements) Si, Ge, with pentavalent impurities (group 15 elements) like P. As, produces excess of electron and they are called n-type semiconductors. The symbol \u2018n\u2019 indicates the flow of negative charge in them.<\/p>\n

                  Question 6.
                  \nHow does metallic character vary among the group 14 elements? How is it related to ionisation energy?
                  \nAnswer:
                  \nCarbon and silicon are non metals. Germanium is a metalloid and tin and lead are metals. Metallic character increases from carbon to lead. The change from non metallic to metallic character is due to the increase in atomic size from carbon to lead. Consequently the ionization potential decreases from carbon to lead.<\/p>\n

                  Hence the electropositive character increases favouring ionic bonding especially with oxidation state +2. Whereas carbon and silicon show +4 oxidation state forming covalent bonding with electron sharing.<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 7.
                  \nDistinguish between n- type and p – type semiconductors.
                  \nAnswer:
                  \nn – type semiconductors are formed by doping pure Si or Ge with pentavalent elements like P, As, Sb, etc, Of the five valence electrons of the added elements, four of them form covalent bonds with Si (or Ge). The fifth electron, conducts the current.<\/p>\n

                  In p – type conductor Si or Ge is doped with trivalent elements like Boron, aluminum or indium. The three valence electron of the added element form three covalent bonds with Si (or Ge) results as deficiency of one electron in the added element causes a positive hole. This type of extrinsic semiconductors are known as p – type semiconductors ( p – stands for positive hole).<\/p>\n

                  Question 8.
                  \nHow do you account for the tetravalency of carbon even though there are only 2 unpaired electrons ?
                  \nAnswer:
                  \nElectronic configurations of carbon atom in the ground state is \\(1 \\mathrm{s}^{2} 2 \\mathrm{s}^{2} 2 \\mathrm{P}_{2}^{1} 2 \\mathrm{P}_{2}^{1} 2 \\mathrm{P}_{2}^{0}\\) During the formation of a compound the activation energy of the reactants excites the electrons to give the configuration as \\(1 \\mathrm{s}^{2} 2 \\mathrm{s}^{1} 2 \\mathrm{p}_{\\mathrm{x}}^{1} 2 \\mathrm{p}_{\\mathrm{y}}^{1} 2 \\mathrm{p}_{\\mathrm{z}}^{1}\\) Then four orbitals each containing an impaired electron, mix together and form four hybridized sp3 <\/sup>orbitals directed from the carbon atom tetrahedrally. Hence carbon exhibits tetravalency.<\/p>\n

                  Question 9.
                  \nHow do the conductivity or semic conduction can be increased?
                  \nAnswer:
                  \nAs the temperature increases, the number of covalent bonds ruptured in an intrinsic semiconductor increases. This results in the increase of the electron hole pairs. As a result, the conductivity of the intrinsic semiconductor increases with rise in temperature. When electron field is applied across these materials electron migrate in one direction and the positive holes migrate is the opposite direction and thus the material conducts. At a given temperature the conductivity of an intrinsic semiconductor is constant.<\/p>\n

                  Conductivity of a semiconductor can be increased by adding either pentavalent or trivalent elements. The elements added to intrinsic semiconductors to increase conductivity are called dopants, the process of addition of dopants (impurities) is called doping. The dopped semiconductors are called extrensic Semiconductors.<\/p>\n

                  Question 10.
                  \nWrite a note on structure of Fullerene. Mention any two of its application.
                  \n\"1st
                  \nAnswer:
                  \nRichard E. Smalley; Robert F. Curl Jr and Sir Harold W. Kroto the discoverers of this new form of pure carbon laked fullerene, they shared the 1996 Nobel Prize in chemistry for their discovery.<\/p>\n

                  This third crystalline form of carbon discovered was named as buckminster fullerene which is commonly called fullerenes. They were named so because their structure resembles geodestic domes designed by the American scientist, engineer and philosopher R. Buckminster Fuller, They are also philosopher R. Buckminster Fuller. They are also nicknamed \u201cBucky ball\u201d because their shape resembles a soccer ball.<\/p>\n

                  Fullerenes are large spheroidal Cn<\/sub> molecules with n = 60 or more. When an electric arc is struck between graphite electrodes in an inert atmosphere a large quantity of soot is formed together with a significant quantity of C60<\/sub> and much smaller quantities of inter related fullerenes such as C70<\/sub> ; C76<\/sub> ; C78<\/sub>.<\/p>\n

                  When graphite is vapourised using a high power laser, fullerene C60<\/sub> is obtained. The fullerene C60<\/sub> bucky ball, has the same form as in soccer ball. In this each carbon atom is sp2<\/sup> hybridised. Each carbon atom has three neighbouring carbon atoms and forms two single and a double bond. C60<\/sub> has 32 faces. 20 of which are hexagons and 12 of which are pentagons. Each pentagon is surrounded by five hexagons and each hexagon is linked to three pentagons and each three hexagons. It has 60 vertices with a carbon atom at each vertex.<\/p>\n

                  Question 11.
                  \nWrite the characteristic uses of fullerene.
                  \nAnswer:
                  \nC60<\/sub> dissolves in hydrocarbon solvents such as hexane, benzene, toluene etc giving magenta solutions. Fullerenes are the least stable of the carbon allotropes and graphite is the most stable. Fullerenes react with alkali metals to produce solids having composition such as K3<\/sub> C60<\/sub>. This compound acts as a superconductor below 18K. Fullerene C114<\/sub> has more surface area and can withstand a very high temperature. Hence it can be used as a lubricant in satellites. Fullerenes are used as industrial catalyst and as lubricant in the treatment of cancer.<\/p>\n

                  \"KSEEB<\/p>\n

                  Question 12.
                  \nExplain the structure of graphite giving the reason for its softness as well as its electrical conductivity.
                  \nAnswer:
                  \nGood conductor of electricity : In. graphite carbon atom is sp2<\/sup> hybridised. Out of four valiancy electrons in each carbon atom only three electrons, one from 2s and two from 2p are involved in hybridisation. One spare electron in its 2pz<\/sub> orbital is left free. These 2pz<\/sub> orbital overlap to form delocalised \u03c0 system which extends above and below each layer. In this delocalised \u03c0 electrons are free to move within the layer and hence graphite is a good conductor of electricity.<\/p>\n

                  The conductivity of graphite perpendicular to the plane of the layers of hexagons is low and increases with increase in temperature, signifying that graphite is a semiconductor in that direction. The electrical conductivity is much higher parallel to the plane but decreases as the temperature is raised.<\/p>\n

                  Soft, flaky and slippery substance : In graphite each layers of hexagon are held together by a weak Vander Waal\u2019s force of attraction at a large distance of 3.40 \u00c5. Even the slight pressure causes the layers to slide over one another. Hence graphite is soft, flaky and slippery. Graphite is used as a lubricant.<\/p>\n

                  Question 13.
                  \nExplain the structure of diamond giving the reason for its hardness as well as non – conductivity of electricity.
                  \nAnswer:
                  \nNon-conductor of electricity: All the four electrons present in the valence shell of carbon atoms are used up in sp3<\/sup> hybridisation and form sigma bonds. No mobile electrons are left in the diamond crystals to allow the conduction of electricity. Hence diamond is a non-conductor of electricity.<\/p>\n

                  Extremely hard: Due to high strength of covalent bonds holding all the carbon atoms in diamond together, diamond is very hard. Infact, diamond is the hardest substance known.<\/p>\n

                  Question 14.
                  \nGive the difference in structure of the following pair of compounds : CO2<\/sub> and SiO2<\/sub>.
                  \nAnswer:
                  \nCO2<\/sub> linear molecule and exists as monomer. It is gas while SiO2<\/sub> is solid at room temperature due to three dimensional network in which each Si atom is covalently bonded to four oxygen atoms tetrahedrally. In CO2<\/sub>, \u2018C\u2019 is sp-hybridised while in SiO2<\/sub>, \u2018Si\u2019 is sp3<\/sup> hybridised. CO2<\/sub> is discrete molecule where as SiO2<\/sub> is 3-dimensional solid.<\/p>\n

                  Question 15.
                  \nExplain the formation of<\/p>\n

                  (i) water gas
                  \n(ii) producer gas. Give their uses.
                  \nAnswer:
                  \n\"1st
                  \nWater gas and producer gas are used as fuel<\/p>\n

                  Question 16.
                  \nWhat happens when CO2<\/sub> is passed through lime water
                  \n(i) for short duration
                  \n(ii) for a long duration ?
                  \nAnswer:
                  \n\"1st<\/p>\n

                  Question 17.
                  \nThe first element in every group of representative elements shows properties different from the characteristic properties of the group. Name three such elements and mention two abnormal properties of each one of them.
                  \nAnswer:
                  \nBoron carbon Nitrogen
                  \nBoron<\/p>\n

                    \n
                  • W, forms acuhc. oxides -whereas owners item. am-phofork and.bask, oxide.<\/li>\n
                  • It cannot form [BFe]3′<\/sup> whereas others can form such type of complexes.<\/li>\n<\/ul>\n

                    Carbon:<\/p>\n

                      \n
                    • It shows property of catenation to maximum extent.<\/li>\n<\/ul>\n

                      Nitrogen<\/p>\n

                        \n
                      • N2<\/sub> is gas others are solid<\/li>\n
                      • NH2<\/sub> is liquid, other hydrides are gases.<\/li>\n<\/ul>\n

                        Question 18.
                        \nGive equations for the following:<\/p>\n

                          \n
                        • Preparation of CO2<\/sub>,<\/li>\n
                        • Basic properties of Na2<\/sub>CO3<\/sub>,<\/li>\n
                        • Formation of acetylene.<\/li>\n<\/ul>\n

                          Answer:<\/p>\n

                            \n
                          • CaCO3<\/sub> + 2HC1 \u2014> CaCl2<\/sub> + CO2<\/sub> + H2<\/sub>O<\/li>\n
                          • Na2<\/sub>CO3<\/sub> + 2H2<\/sub>0 \u2014> 2Na0H + H2<\/sub>CO3<\/sub><\/li>\n
                          • CaC2<\/sub>(calcium carbide) + 2H2<\/sub>0 \u2014>Ca(OH)2<\/sub> + C2<\/sub>H2<\/sub>(Acetylene)<\/li>\n<\/ul>\n

                            Question 19.
                            \nWhat is dry ice ? Why is it called so? How will you prepare pure sample of CO (Carbon monoxide)?
                            \nAnswer:
                            \nSolid CO2<\/sub> is called dry ice. It is called dry ice because it directly changes into gaseous state without becoming liquid.
                            \n\"1st<\/p>\n

                            Question 20.
                            \nWhat happens when ?<\/p>\n

                              \n
                            • Quick lime is heated with coke ?<\/li>\n
                            • Carbon monoxide reacts with Cl2<\/sub>?<\/li>\n
                            • Plants absorb CO2<\/sub>?<\/li>\n<\/ul>\n

                              Answer:<\/p>\n

                                \n
                              • CaO + 3C \u2014> CaC2<\/sub>(CalciumCarbide) + CO,<\/li>\n
                              • CO + Cl2<\/sub> \u2014> COCl2<\/sub> (Phosgene)<\/li>\n
                              • 6CO2<\/sub> + 6H2<\/sub>O \u2014> C6<\/sub>H12<\/sub>O6<\/sub>(glucose) + 6O2<\/sub><\/li>\n<\/ul>\n

                                \"KSEEB<\/p>\n

                                Question 21.
                                \nWrite balanced equations for the following reactions and name the main product formed in each case,<\/p>\n

                                  \n
                                • NaBH4<\/sub> +I2<\/sub> \u2014><\/li>\n
                                • B2<\/sub>h6<\/sub> +NaH \u2014><\/li>\n
                                • \\(\\mathrm{BF}_{3}+\\mathrm{LiH} \\stackrel{450 \\mathrm{K}}{\\longrightarrow}\\)<\/li>\n
                                • SiCl4<\/sub> + H2<\/sub>O \u2014><\/li>\n<\/ul>\n

                                  Answer:<\/p>\n

                                    \n
                                  • 2NaBH4<\/sub> +I2<\/sub> \u2014> 2Nal + +H2<\/sub> + B2<\/sub>H6<\/sub> (Diborane)<\/li>\n
                                  • B2<\/sub>H6<\/sub>+2NaH \u2014> 2NaBH4<\/sub> (sodium borohydride)<\/li>\n
                                  • 2BF3<\/sub> + 6LiH \u2014> 6LiF + B2<\/sub>H6<\/sub> (Diborane)<\/li>\n
                                  • SiCl4<\/sub> + 2H2<\/sub>O \u2014> 4HCl + Si(OH)4<\/sub> (Silicicacid)<\/li>\n<\/ul>\n

                                    Question 22.
                                    \nGive a comparative account of the chemistry of carbon and silicon with regard to their<\/p>\n

                                      \n
                                    • property of catenation and<\/li>\n
                                    • stability of hybrids and oxides.<\/li>\n<\/ul>\n

                                      Answer:<\/p>\n

                                        \n
                                      • Carbon shows property of catenation to more extent than silicon due to small size and tendency to form p\u03c0-p\u03c0 multiple bonds with itself<\/li>\n
                                      • Stability of hybrids : CH4<\/sub> is more stable than SiH4 due to small size of \u2018C\u2019.<\/li>\n
                                      • CO4<\/sub> is gas whereas SiO2<\/sub> is three dimensional covalent solid, therefore highly stable as compared to CO2<\/sub>.<\/li>\n<\/ul>\n

                                        Question 23.
                                        \nA certain salt X, gives the following results.
                                        \n(i) Its aqueous solution shows alkaline to litmus paper
                                        \n(ii) It swells up to a glassy material Y on strong heating
                                        \n(iii) When cone H2<\/sub>SO4<\/sub> is added to a hot solution of X, white crystal of an acid Z separates out-Write equations for all the above reactions and identify X, Y and Z.
                                        \nAnswer:
                                        \n\"1st<\/p>\n

                                        Question 24.
                                        \nWrite balanced equations for:<\/p>\n

                                          \n
                                        1. BF3<\/sub> + LiH \u2192<\/li>\n
                                        2. B2<\/sub>H6<\/sub> + H2<\/sub>O \u2192<\/li>\n
                                        3. NaH + B2<\/sub>H6<\/sub> \u2192<\/li>\n
                                        4. \\(\\mathrm{H}_{3} \\mathrm{BO}_{3} \\stackrel{\\Delta}{\\longrightarrow}\\)<\/li>\n
                                        5. A1 + NaOH \u2192<\/li>\n
                                        6. B2<\/sub>H6<\/sub> + NH3<\/sub> \u2192<\/li>\n<\/ol>\n

                                          Answer:<\/p>\n

                                          1. 2BF3<\/sub> +6LiH \u2192B2<\/sub>H6<\/sub> +6LiF<\/p>\n

                                          2. B2<\/sub>H6<\/sub> + 6H2<\/sub>O \u2192 2B(OH)3<\/sub> + 6H2<\/sub><\/p>\n

                                          3. 2NaH + B2<\/sub>H6<\/sub> \u2192 2NaBH4<\/sub><\/p>\n

                                          4. \\(\\begin{array}{l}{\\mathrm{H}_{3} \\mathrm{BO}_{3} \\stackrel{\\Delta}{\\longrightarrow} \\mathrm{HBO}_{2}+\\mathrm{H}_{2} \\mathrm{O}} \\\\ {4 \\mathrm{HBO}_{4} \\stackrel{\\Delta}{\\longrightarrow} 2 \\mathrm{B}_{2} \\mathrm{O}_{3}+2 \\mathrm{H}_{2} \\mathrm{O}}\\end{array}\\)<\/p>\n

                                          5. Al + 3NaOH \u2192 A1(0H)3<\/sub> +3Na<\/p>\n

                                          6. \\(3 \\mathrm{B}_{2} \\mathrm{H}_{6}+6 \\mathrm{NH}_{3} \\stackrel{\\text { heat }}{\\longrightarrow} 2 \\mathrm{B}_{3} \\mathrm{N}_{3} \\mathrm{H}_{6}+12 \\mathrm{H}_{2}\\)<\/p>\n

                                          Question 25.
                                          \nGive one method for industrial preparation and one for laboratory preparation of CO and CO2 each.
                                          \nAnswer:
                                          \n\"1st<\/p>\n

                                          Question 26.
                                          \n(a) Carbon dioxide is non-polar while water is polar. What conclusion do you draw about their structures from these.
                                          \n(b) Classify the following compounds into acidic, basic and amphoteric oxides. Al2<\/sub>O3<\/sub>, Cl2<\/sub>O7<\/sub>
                                          \nAnswer:
                                          \n\"1st is linear, bond moments are equal and opposite, net dipole moment is zero. Water is bent molecule, it has net dipole moment .
                                          \n\"1st
                                          \nAl2<\/sub>O3<\/sub> is amphoteric where Cl2<\/sub>O7<\/sub> is acidic oxide.<\/p>\n

                                          Question 27.
                                          \nA white crystalline solid \u2018A\u2019 dissolves in water to give an alkaline solution. On heating \u2018A\u2019 first loses water molecules and swells up. On further heating it turns into a transparent liquid which solidifies into a glassy bead.
                                          \nAnswer:
                                          \n\u2018A\u2019 is Borax i.e. Na2<\/sub>[B4<\/sub>O5<\/sub>(OH)4<\/sub>].8H2<\/sub>O which is also written as.
                                          \nThe reactions are:
                                          \n\"1st<\/p>\n

                                          Question 28.
                                          \nBoric acid is a weak acid. It acts as an acid only on first receiving OH- from the medium and then releasing H+ ions. From the following organic compounds; select the one in presence of which it acts as a strong acid. Glycerol, Ethylene, Ethylalcohol, Acetic Acid.
                                          \nAnswer:
                                          \nIt acts as a strong acid in presence of glyceol.
                                          \nReason: Acetic acid itself is an acid and cannot give OH–<\/sup> ion, ethyl alcohol being a primary alcohol is alos acidic. Ethylene also can\u2019t furnish OH–<\/sup> ion. But glycerol that has 3-OH groups. One of which is a secondary -OH group can assist boric acid to release H+ ions.<\/p>\n

                                          \"KSEEB<\/p>\n

                                          Question 29.
                                          \nA compound \u2018X\u2019 or reduction with LiAH4 gives a hydride \u2018Y that contains 21.72% hydrogen along with other products. The compound \u2018Y\u2019 reacts with air explosively resulting in boron trioxide. Identify \u2018X\u2019 and \u2018Y\u2019. Give balanced reactions involved in the formation of \u2018Y\u2019 and its reactions with air. Draw the structure of Y.
                                          \nAnswer:
                                          \n\u201cX\u2019 is BCl3<\/sub> Boron Trichloride Y is B2<\/sub>H6<\/sub> as percentage of hydrogen given is 21.72%
                                          \n\"1st<\/p>\n

                                          Question 30.
                                          \nMention three important uses of borax.
                                          \nAnswer: It is used :<\/p>\n

                                            \n
                                          • as a flux soldering and welding industry<\/li>\n
                                          • in the manufacture of borosilicate glass (or pyrex glass)<\/li>\n
                                          • in making enamels and glazes<\/li>\n
                                          • in stiffening of candle wicks<\/li>\n
                                          • in softening of water<\/li>\n
                                          • in qualitative analysis for borax bead test in laboratory.<\/li>\n<\/ul>\n

                                            Question 31.
                                            \nMention some important properties of carbon monoxide.
                                            \nAnswer:
                                            \n1. It is colourless, odourless gas, slightly soluble in water.<\/p>\n

                                            2. It is highly poisonous. It combines with haemoglobin in the red blood cells to form carboxy-haemoglobin which cannot absorb oxygen and thus supply of oxygen to the body is reduced.<\/p>\n

                                            3. It burns with a pale blue flame forming CO2<\/sub>; i.e., 2CO + O2<\/sub> \u2192 2CO2<\/sub><\/p>\n

                                            4. It is a reducing-agent. It reduced some metal oxides into metal,
                                            \nFe2<\/sub>O3<\/sub> + 3CO \u2192 2Fe + 3CO2<\/sub><\/p>\n

                                            5. It combines with transition metals like iron, cobalt, nickel to form their carbonyl compounds;
                                            \n\"1st<\/p>\n

                                            Question 32.
                                            \nWhat are halides of carbon ? Give few examples.
                                            \nAnswer:
                                            \nCarbon combines with halogens to form both simple and mixed tetrahalides. In case of simple halides all the four expected tetrahalides (e.g. CF4<\/sub>, Cl4<\/sub>, CBr4<\/sub> and Cl4<\/sub>) are known to exist. The stability of the simple tetrahalides (CF4<\/sub> > CCl4<\/sub> > CBr4<\/sub> > Cl4<\/sub>)
                                            \nAmongst the mixed halides the better known compounds are
                                            \n(CFCl3<\/sub>,CF2<\/sub>Cl2<\/sub> and CCl3<\/sub>Br)<\/p>\n

                                            Question 33.
                                            \nWhat is allotropy ? Give examples of allotropes.
                                            \nAnswer:
                                            \nTwo or more forms of the same elements in same physical state which differ in their physical properties but have same chemical properties are called allotropic forms or (allotropes) and the phenomenon is called allotropy.
                                            \nCarbon, phosphorus and sulphur are some elements which exhibit allotropy.<\/p>\n

                                              \n
                                            • Diamond and graphite are allotropic forms of carbon<\/li>\n
                                            • Red phosphorus and white phosphorus are allotropes of phosphorus.<\/li>\n
                                            • Rhombic sulphur, monoclinic sulphur and plastic sulphur are allotropic forms of sulphur.<\/li>\n<\/ul>\n

                                              Question 34.
                                              \nDraw the structure of
                                              \n(a) anion cyclic silicates
                                              \n(b) anion sheet silicates.
                                              \nAnswer:
                                              \n\"1st<\/p>\n

                                              Question 35.
                                              \nGive three uses of different allotropic forms of carbon.
                                              \nAnswer:<\/p>\n\n\n\n\n\n\n\n\n
                                              Forms of carbon<\/strong><\/td>\nUses<\/strong><\/td>\n<\/tr>\n
                                              Diamond<\/td>\nGemstone, cutting, drilling, grinding, polishing, industry.<\/td>\n<\/tr>\n
                                              Graphite<\/td>\nReducing agent, refractories, pencils, high temperature, crucibles, electrode making, moderator in nuclear reactors, high strength composite materials.<\/td>\n<\/tr>\n
                                              Activated carbon<\/td>\nRubber industry, pigments in ink, paints and plastics<\/td>\n<\/tr>\n
                                              Coke<\/td>\nFuel, strut manufacture<\/td>\n<\/tr>\n
                                              Charcoal<\/td>\nFuel, reducing agent, Adsorption.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                                              Question 36.
                                              \nWrite balanced equation for the following:
                                              \n(i) BF3<\/sub> + LiH \u2192
                                              \n(ii) B2<\/sub>H6<\/sub>+H2<\/sub>O \u2192
                                              \n(iii) NaH + B2<\/sub>H6<\/sub> \u2192
                                              \n(iv) H3<\/sub>BO3<\/sub> \u2192
                                              \n(v) A1 + NaOH + H2<\/sub>O \u2192
                                              \n(vi) B2<\/sub>H6 + NH3<\/sub> \u2192
                                              \nAnswer:
                                              \n\"1st<\/p>\n

                                              \"KSEEB<\/p>\n

                                              Question 37.
                                              \nDraw the structure of fullerene C60<\/sub>
                                              \nAnswer:
                                              \n\"1st<\/p>\n

                                              Question 38.
                                              \nDraw the structure of graphite.
                                              \nAnswer:
                                              \n\"1st<\/p>\n

                                              Question 39.
                                              \nHow does electron deficient compound BF3<\/sub> achieve electronic saturation i.e., the fully occupied outer electron shells ?
                                              \nAnswer:
                                              \n1. By multiple bonding, e.g., BF3 in which a lone pair of electrons of each fluorine atom may be used in a B \u2192 F \u03c0-bond (back bonding) involving the vacant orbital on the boron atom.<\/p>\n

                                              2. By formation of complexes in which electrons are received from a donor molecule, e.g., F3<\/sub>B \u2192 NH3<\/sub>. Boron compounds, thus, behave as Lewis acids.<\/p>\n

                                              Question 40.
                                              \nCompound X on reduction with LiAH4 gives a hydride Y containing 21.72% hydrogen along with other products. The compound Y reacts with air explosively resulting in boron trioxide. Identify X and Y. Give balanced equations involved in the formation of Y and its reaction with air. Draw the structure of Y.
                                              \nAnswer:
                                              \nStep 1: To determine the molecular formula and structure of compound Y.
                                              \n1. Since the hydride Y reacts with air forming boron trioxide, therefore, Y must be an hydride of boron.
                                              \n2. %H = 21.72% (Given)
                                              \n\"1st
                                              \n\u2234 Emperical formula of Y = BH3<\/sub>
                                              \nSince boron forms two types of hydrides, i.e., Bn<\/sub>Hn+4<\/sub> (nidoboranes) and Bn<\/sub>Hn+6<\/sub> (arachnoboranes), therefore, Y must be a nidoborane with n = 2. Thus, M.F of Y = B2<\/sub>H6<\/sub> . If Y is B2<\/sub>H6<\/sub> (diborane), then its structure must be as follows :
                                              \n\"1st
                                              \nBridges B…H = 134 pm Terminal B – H = 119 pm<\/p>\n

                                              Step 2: To determine the structure of the compound X.
                                              \nSince compound Y i.e., B2<\/sub>H6<\/sub> is formed by reduction of compound X with LiAH4<\/sub> ,
                                              \ntherefore, X must either BCl3<\/sub> or BF3<\/sub>.
                                              \n\"1st
                                              \nThe equation representing the reaction of Y with O2 may be written as follows :
                                              \nB2<\/sub>H6<\/sub> + 3O2<\/sub> \u2192 2B2<\/sub>O3<\/sub> + 3H2<\/sub>O
                                              \nDiborane,Y
                                              \nThus X = BF3<\/sub> and Y = B2<\/sub>H6<\/sub><\/p>\n

                                              Question 41.
                                              \nHow is boron obtained from boran? Give chemical equations with reaction conditions and its reaction with HCl1<\/sub>
                                              \nAnswer:
                                              \n\"1st<\/p>\n

                                              Question 42.
                                              \nWhich one is more soluble in diethyl ether, anhydrous AlCl3<\/sub> or hydrous AICl3<\/sub> ? Explain in terms of bonding ?
                                              \nAnswer: An hydrous AICI3<\/sub> is an electron-deficient compound while hydrated AICI3<\/sub> is not. Therefore, anhyd. AICI3<\/sub> is more soluble in diethyl ether because the oxygen atom of either donates a pair or electrons to the vacant p-orbital on the A1 atom in AICI3<\/sub> forming a coordinate bond.
                                              \n\"1st
                                              \nIn case of hydrated AlCl3<\/sub>. Al is not electron deficient since H2<\/sub>O has already donated a pair of electrons to it.<\/p>\n

                                              Question 43.
                                              \nMention the structures of (CH3<\/sub>)3<\/sub>N and (Me3<\/sub>Si)3<\/sub> N. Are they isostructural? Justify your answer.
                                              \nAnswer: (CH3<\/sub>)3<\/sub> N is pyramidal involving sp3<\/sup>. hybridisation of N atom. Nitrogen in (Me3<\/sub>Si)3<\/sub> is however, sp2<\/sup> -hybridized. The reason being that the p-orbit of N containing the lone pair of electrons overlaps with a vacant d-orbital of silicon. As a result of this p\u03c0 – d\u03c0 back bonding, (Me3<\/sub>Si)3<\/sub> N is planar.<\/p>\n

                                              \"KSEEB<\/p>\n

                                              Question 44.
                                              \n1. Describe with equation what happens when carbon is heated with cone. H2<\/sub>SO4<\/sub> ?
                                              \n2. State how will you separate C02<\/sub> and S02<\/sub> from the mixture?
                                              \nAnswer:
                                              \n1. H2<\/sub>S04<\/sub> oxidies C to CO2<\/sub> and itself is reduced to SO2<\/sub>
                                              \n\"1st<\/p>\n

                                              2. Since the O.N. of C is CO2<\/sub> is +4 which is the maximum for C, therefore, it cannot act as a reducing agent. In contrast the O.N. of S is SO2<\/sub> is also +4 but its maximum. O.N. +6, therefore, it can act as reducing agent. This reducing gas can be removed by passing the mixture of these gases through acidified K2<\/sub>Cr2<\/sub>O7<\/sub> solution when orange solution turns green due to reduction of \\(\\mathrm{Cr}_{2} \\mathrm{O}_{7}^{2-}\\) ion to Cr3+<\/sup> ions while CO2<\/sub> passes out unreacted.<\/p>\n

                                              Question 45.
                                              \nWhat is inorganic benzene ? Why is it so called ? How will you get it from diborane?
                                              \nAnswer:
                                              \nBorazine or borazole (B3<\/sub>N3<\/sub>H6<\/sub>) is known as inorganic benzene. It is also called Because the structure of borazine is similar to that of benzene.
                                              \nIt is also isoelectronic and isosteric with benzene.
                                              \n\"1st
                                              \nLike carbon is benzene, both N and B in borazine are sp2<\/sup> hybridized. Each N has a p-orbtal which is perpendicular to the a-bonding orbitals and contains a lone pair of electrons. In contrast, each B has an empty p-orbitals which is also perpendicular to the plane of the ring. Thus, the x-bonding in borazine is dative and it arises from the sideways overlap of fully filled orbitals of N and empty p-orbitals of B.<\/p>\n

                                              Preparation of borazine : At low temperature, diborane combines with ammonia to form an addition compound.
                                              \n\\(\\mathrm{B}_{2} \\mathrm{H}_{6}+2 \\mathrm{NH}_{3} \\stackrel{\\text { Low Temp. }}{\\longrightarrow} \\mathrm{B}_{2} \\mathrm{H}_{6} \\cdot 2 \\mathrm{NH}_{3}\\)
                                              \nWhen this addition compound is heated to 473 K, it decomposes to form a volatile compound called borazine.
                                              \n\"1st<\/p>\n

                                              Question 46.
                                              \nState with equations what happens when borax is heated on a platinum wire loop and to the resulting transparent mass a minute amount of CuO is added and the mixture is again heated first in the oxidizing flame and then in the reducing flame of a Bunsen Burner ?
                                              \nAnswer:
                                              \nWhen borax is heated on a platinum where loop, a transparent glassy bead is formed
                                              \n\"1st
                                              \nWhen this transparent glassy bead is heated with CuO in the oxidizing flame, it imparts blue colour to the bead due to the formation of cupric metaborate
                                              \n\\(\\mathrm{CuO}+\\mathrm{B}_{2} \\mathrm{O}_{3} \\frac{\\text { oxidising }}{\\text { flame }} \\mathrm{Cu}\\left(\\mathrm{BO}_{2}\\right)_{2}\\)(Cupric metaborate (blue)
                                              \nHowever, when cupric metaborate is heated in the reducing flame of the Bunsen burner, the blue cupric metaborate is reduced by carbon present in the flame first to colourless cuprous metaborate and then to metal which produces red colour.
                                              \n\"1st<\/p>\n","protected":false},"excerpt":{"rendered":"

                                              You can Download Chapter 11 The P-Block Elements Questions and Answers, Notes, 1st PUC Chemistry Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations. Karnataka 1st PUC Chemistry Question Bank Chapter 11 The P-Block Elements 1st PUC Chemistry The P-Block Elements One Mark …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[81],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/8940"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=8940"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/8940\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=8940"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=8940"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=8940"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}