{"id":1566,"date":"2023-11-21T11:46:24","date_gmt":"2023-11-21T06:16:24","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=1566"},"modified":"2023-11-22T12:26:16","modified_gmt":"2023-11-22T06:56:16","slug":"ts-inter-1st-year-chemistry-study-material-chapter-10","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-inter-1st-year-chemistry-study-material-chapter-10\/","title":{"rendered":"TS Inter 1st Year Chemistry Study Material Chapter 10 p-Block Elements: Group 13"},"content":{"rendered":"

Telangana TSBIE TS Inter 1st Year Chemistry Study Material<\/a> 10th Lesson p-Block Elements: Group 13 Textbook Questions and Answers.<\/p>\n

TS Inter 1st Year Chemistry Study Material 10th Lesson p-Block Elements: Group 13<\/h2>\n

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

Question 1.
\nDiscuss the pattern of variation in the oxidation states of Boron to Thallium.
\nAnswer:
\nThe elements of boron family exhibit two oxidation states +1 and +3. From boron to thallium the stability of +1 oxidation state increases while the stability of +3 oxidation state decreases. This is due to inert pair effect.<\/p>\n

Question 2.
\nHow do you explain higher stability of Tl Cl3<\/sub>?
\nAnswer:
\nChlorine is good oxidising agent and Cl–<\/sup> ion is weak reducing agent. So chlorine can oxidise thallium to higher oxidation state i.e., to +3 and form TlCl3<\/sub>. As Cl–<\/sup> ion cannot reduce Tl3+<\/sup>, TlCl3<\/sub> is stable.<\/p>\n

Question 3.
\nWhy does BF3<\/sub> behave as a Lewis acid?
\nAnswer:
\nBF3<\/sub> is an electron deficient compound as there is no octet around boron. So it can accept a pair of electrons to get octet. Thus it behaves as a Lewis acid.
\nH3<\/sub>N \u2192 BF3<\/sub><\/p>\n

Question 4.
\nIs boric acid a protic acid? Explain.
\nAnswer:
\nBoric acid is not protic acid but it is a Lewis acid. It accepts electrons from OH” ion of water and releases proton.
\nB(OH)3<\/sub> + H2<\/sub>O \u2192 [B(OH)4<\/sub>]–<\/sup> + H+<\/sup><\/p>\n

Question 5.
\nWhat happens when boric acid is heated?
\nAnswer:
\nOn heating orthoboric acid above 370K forms metaboric acid HB02 which on further heating yields boric oxide B2<\/sub>O3<\/sub>
\n\"TS<\/p>\n

\"TS<\/p>\n

Question 6.
\nDescribe the shapes of BF3<\/sub> and BH4<\/sub>. Assign the hybridization of boron in these species.
\nAnswer:<\/p>\n

    \n
  1. Shape of BF3<\/sub> molecule is Trigonal planar. Hybridization of ‘B’ in BF3<\/sub> is sp\u00b2.<\/li>\n
  2. Shape of BH–<\/sup>4<\/sub> molecule is Tetrahedral, Hybridizaion of ‘B’ in BH–<\/sup>4<\/sub> is sp\u00b3.<\/li>\n<\/ol>\n

    Question 7.
    \nExplain why atomic radius of Ga is less than that of Al.
    \nAnswer:
    \nGallium comes immediately after the first transition series in which 3d orbital is filled with electrons. Due to the poor shielding effect of these d electrons, effective nuclear charge of gallium increases. Consequently, the atomic radius of gallium is less than that of aluminium.<\/p>\n

    Question 8.
    \nExplain inert pair effect.
    \nAnswer:
    \nThe pair of electrons in s orbital of valence shell are reluctant to participate in bond formation. That pair of (ns\u00b2) electrons is called inert pair. Due to this inert pair the stability of +1 oxidation state among boron family elements increases. This is known as inert pair effect.<\/p>\n

    Question 9.
    \nWrite balanced equations for
    \n\"TS
    \nAnswer:
    \n\"TS<\/p>\n

    Question 10.
    \nWhy is boric acid polymeric?
    \nAnswer:
    \nBoric acid has layer structure, in which the BO3<\/sub> units are joined by hydrogen bonds. Hence, boric acid is polymeric.<\/p>\n

    Question 11.
    \nWhat is the hybridization of B in diborane and borazine?
    \nAnswer:
    \nIn diborane boron is involved in sp3 hybridisation while in borazine it is in sp2 hybridisation.<\/p>\n

    Question 12.
    \nWrite the electronic configuration of group -13 elements.
    \nAnswer:
    \nElectronic configurations:
    \n\"TS<\/p>\n

    Question 13.
    \nGive the formula of borazine. What is its common name?
    \nAnswer:
    \nThe formula of borazine is B3<\/sub>N3<\/sub>H6<\/sub>. It is commonly known as inorganic benzene.<\/p>\n

    Question 14.
    \nGive the formulae of
    \na) Borax
    \nb) Colemanite
    \nAnswer:
    \na) Formula of Borax = Na2<\/sub>B4<\/sub>O7<\/sub>.10H2<\/sub>O.
    \nb) Formula of Colemanite = Ca2<\/sub> B6<\/sub> O11<\/sub>. 5H2<\/sub>O<\/p>\n

    \"TS<\/p>\n

    Question 15.
    \nGive two uses of aluminium.
    \nAnswer:<\/p>\n

      \n
    1. Aluminium is used in making electrical cables.<\/li>\n
    2. It is used in making trays, picture frames etc.<\/li>\n
    3. Alloys of ‘Al’ are used in making parts of aircrafts, automobiles etc.<\/li>\n
    4. ‘Al’ powder is used in Aluminothermite<\/li>\n<\/ol>\n

      Question 16.
      \nWhat happens when
      \na) Li AlH4<\/sub> and BCl3<\/sub> mixture in dry ether is warmed and
      \nb) orax is heated with H2<\/sub>SO4<\/sub>?
      \nAnswer:
      \na) Diborane is formed.
      \n\"TS<\/p>\n

      b) Boric acid is formed.
      \nNa2<\/sub>B4<\/sub>O7<\/sub> + H2<\/sub>SO4<\/sub> + 5H2<\/sub>O \u2192 4H3<\/sub>BO3<\/sub> + Na2<\/sub>SO4<\/sub><\/p>\n

      Question 17.
      \nSketch the structure of Orthoboric acid.
      \nAnswer:
      \n\"TS<\/p>\n

      Structure of boric acid :
      \nThe dotted lines represent hydrogen bonds<\/p>\n

      Question 18.
      \nWrite the structure of AlCl3<\/sub> as a dimer.
      \nAnswer:
      \n\"TS<\/p>\n

      Question 19.
      \nMetal borides (having B) are used as protective shield – why?
      \nAnswer:
      \n10<\/sup>B isotope has high ability to absorb neutrons. So metal borides are used in nuclear industry as protective shields and control rods.<\/p>\n

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

      Question 1.
      \nWrite reactions to justify amphoteric nature of aluminium.
      \nAnswer:
      \nAluminium dissolves in both acids and alkalies. Aluminium dissolves in dilute HCl and liberates dihydrogen.
      \n2Al + 6HCl \u2192 2 AlCl3<\/sub> + 3H2.<\/p>\n

      Aluminium also dissolves in aqueous alkali and liberates dihydrogen.
      \n2 Al + 2NaOH + 6H2<\/sub>O \u2192 2 Na+<\/sup> [Al(OH)4<\/sub>]–<\/sup> + 3H2<\/sub>
      \nThese reactions indicate the amphoteric nature of aluminium.<\/p>\n

      Question 2.
      \nWhat are electron deficient compounds? Is BCl3<\/sub>an electron deficient species? Explain.
      \nAnswer:
      \nThe compounds in which the central atom is having less than octet electrons are called electron deficient compounds.<\/p>\n

      BCl3<\/sub> is an electron deficient compound since it is not having octet around boron.
      \n\"TS<\/p>\n

      BCl3<\/sub> molecule accepts an electron pair to achieve stable electronic configuration and thus behave as Lewis acid.<\/p>\n

      Eg : BCl3<\/sub> molecule easily accepts alone pair of electrons from ammonia to form BCl3<\/sub>. NH3<\/sub>.
      \n\"\"<\/p>\n

      Question 3.
      \nSuggest reasons why the B-F bond lengths in BF3<\/sub> (130 pm) and BF–<\/sup>4<\/sub> (143 pm) differ.
      \nAnswer:
      \nIn BF3<\/sub> there is back bonding. Boron accepts a pair of electrons from fluorine to complete octet. BF3<\/sub> exists in the following resonance structures due to back bonding. Due to the resonance hybridisation each B-F bond gets some double bond character. So the B-F length will be less than the sum of the covalent radii of boron and fluorine.
      \n\"TS<\/p>\n

      In BF–<\/sup>4<\/sub> there is no back bonding. So there is no double bond character. This is because in BF–<\/sup>4<\/sub> boron already had octet. Hence the B – F bond length in BF–<\/sup>4<\/sub> is greater than in BF3<\/sub>.<\/p>\n

      Question 4.
      \nB – Cl bond has a bond moment. Explain why BCl3<\/sub> molecule has zero dipole moment.
      \nAnswer:
      \nDipolemoment of a molecule is vector sum of the dipole moments of all the bonds in the molecule. Although B – Cl bonds have bond moments, since the three bonds are oriented at an angle 120\u00b0 to one another, the three bond moments give a net sum of zero as the resultant of any two is equal and opposite to the third.<\/p>\n

      So the dipole moment of BF3<\/sub> is zero.<\/p>\n

      \"TS<\/p>\n

      Question 5.
      \nExplain the structure of boric acid.
      \nAnswer:
      \nIn boric acid boron is in sp\u00b2 hybridisation forming planar borate units. Between the planar borate units the hydrogen atoms of the one OH group of boric acid form hydrogen bond with oxygen atom of another OH group of another boric acid unit. Thus several boric acid molecules associate through hydrogen bonds forming a layered lattice structure.
      \n\"TS<\/p>\n

      Structure of boric acid :
      \nThe dotted lines represent hydrogen bonds.<\/p>\n

      Question 6.
      \nWhat happens when
      \na) Borax is heated strongly?
      \nb) Boric acid is added to water?
      \nc) Aluminium is heated with dilute NaOH?
      \nd) BF3<\/sub> is treated with ammonia?
      \ne) Hydrated alumina is treated with aq.NaOH solution?
      \nAnswer:
      \na) Borax on heating strongly decomposes into NaBO2<\/sub> and B2<\/sub>O3<\/sub> forming a transparent glassy bead.
      \n\"TS<\/p>\n

      b) When boric acid is added to water it accepts a pair of electrons from OH–<\/sup> ion and releases proton, thus forming acidic solution.
      \nB(OH)3<\/sub> + H2<\/sub>O \u2192 [B (OH)4<\/sub>]–<\/sup> + H+<\/sup><\/p>\n

      c) When aluminium is heated with dilute NaOH, aluminium dissolves forming tetrahydroxo aluminate (III) and liberates hydrogen gas.
      \n2Al + 2 NaOH + 6H2<\/sub>O \u2192 2 Na+<\/sup> [Al (OH)4<\/sub>]–<\/sup> + 3H2<\/sub><\/p>\n

      d) BF3<\/sub> combines with ammonia by accepting a lone pair of electrons from ammonia.
      \nH3<\/sub>N: \u2192 BF3<\/sub><\/p>\n

      e) Hydrated alumina when heated with NaOH, dissolves in NaOH forming tetrahydroxo aluminate (III).
      \nAl (OH)3<\/sub> + NaOH \u2192 Na [Al(OH)4<\/sub>]<\/p>\n

      Question 7.
      \nGive reasons
      \na) Cone. HNO3<\/sub> can be transported in aluminium container.
      \nb) A mixture of dil. NaOH and aluminium pieces is used to open drain.
      \nc) Aluminium alloys are used to make aircraft body.
      \nd) Aluminium utensils should not be kept in water overnight.
      \ne) Aluminium wire is used to make transmission cables.
      \nAnswer:
      \na) Cone. HNO3<\/sub> can be transported in aluminium container:
      \nConcentrated nitric acid renders the aluminium passive by forming a protective oxide layer on the surface. So aluminium do not react with concentrated nitric acid. Thus the aluminium containers can be used to carry concentrated nitric acid.<\/p>\n

      b) A mixture of dil. NaOH and aluminium pieces is used to open drain:
      \nWhen aluminium pieces react with dilute NaOH, dihydrogen gas will be evolved, with high pressure, which can open the drain.<\/p>\n

      c) Aluminium alloys are used to make aircraft body:
      \nBecause of light weight and hardness aluminium alloys are used to make aircraft body.<\/p>\n

      d) Aluminium utensils should not be kept in water overnight :
      \nAluminium does not react with pure water but when salts are present in water they remove the oxide layer on the surface making aluminium reactive. Then aluminium reacts with water slowly. So aluminium vessels should not be kept in water overnight.<\/p>\n

      e) Aluminium wire is used to make transmission cables:
      \nAluminium is a good conductor of electricity. So it is used in transmission cables. ‘<\/p>\n

      Question 8.
      \nExplain why the electronegativity of Ga, In and Tl will not vary very much.
      \nAnswer:
      \nDue to the poor screening effect of d electrons in Ga and In and due to poor screening effect of d – and f-electrons in Thallium the atomic sizes of Gallium, Indium and Thallium do not vary very much. Further the effective nuclear charge increases. So the electronegativities of Ga, In and T1 will not vary very much.<\/p>\n

      Question 9.
      \nExplain borax bead test with a suitable example.
      \nAnswer:
      \nBorax Bead Test:
      \nThis test is useful for the identification of coloured metal radicals in qualitative analysis. On heating, borax swells into a white, opaque mass of anhydrous sodium tetraborate. When it is fused, borax glass is obtained. This contains sodium metaborate and B2<\/sub>O3<\/sub>. The boric anhydride combines with metal oxides to form metal metaborates as coloured beads. The reactions are as follows.
      \n\"TS<\/p>\n

      Question 10.
      \nExplain the structure of diborane. (AP Mar. 17; AP & TS 16, ’15; IPE ’13, ’11, ’09) AP Mar. ’19; Mar. 18 (TS)]
      \nAnswer:
      \nStructure of diborane:
      \nElectron diffraction studies have shown that diborane contains two coplanar BH2<\/sub> groups. Its structure can be represented as
      \n\"TS<\/p>\n

      The four H atoms present in the BH2<\/sub> groups are known as terminal hydrogen atoms. The remaining two H atoms are called Bridge hydrogens. The two bridge hydrogens lie in a plane perpendicular to the plane of the two BH2<\/sub> groups. Out of the two bridge hydrogens, one H atom lies above the plane and the other H atom lies below the plane.<\/p>\n

      Orbital structure of Diborane :
      \nIn Diborane each boron atom undergoes sp\u00b3 hybridisation resulting in four equivalent sp\u00b3 hybrid orbitals. Three of these orbitals have one electron each and the fourth hybrid orbital is vacant. In the formation of B – H – B bridge, sp\u00b3 hybrid orbital with one electron from one boron atom, Is orbital of one bridge hydrogen and the vacant sp\u00b3 hybrid orbital of the second boron atom overlap as shown below.
      \n\"TS<\/p>\n

      The two B – H – B bridges present in Diborane are abnormal bonds. They are considered as three centered two electron bonds. This type of bond is called banana bond or tau bond. In this type of bond a pair of electrons holds three atoms. In Diborane there are two such bridges. This Diborane structure can also be represented as shown here.
      \n\"TS<\/p>\n

      Question 11.
      \nExplain the reactions of aluminium with acids.
      \nAnswer:
      \nAluminium reacts with dilute or cone. HCl liberating H2<\/sub>.
      \n2 Al + 6HCl \u2192 2 AlCl3<\/sub> + 3H2<\/sub><\/p>\n

      With dil. H2<\/sub>SO4<\/sub> it reacts slowly in cold condition but reacts fast in hot condition.
      \n2 Al + 3H2<\/sub>SO4<\/sub> \u2192 Al2<\/sub> (SO4<\/sub>)3<\/sub> + 3H2<\/sub><\/p>\n

      With cone. H2<\/sub>SO4<\/sub> it liberates SO2<\/sub>.
      \n2 Al + 6H2<\/sub>SO4<\/sub> \u2192 Al2<\/sub>(SO4<\/sub>)3<\/sub> + 6H2<\/sub>O + 3SO2<\/sub><\/p>\n

      With very dil. HNO3<\/sub> it gives NH4<\/sub>NO3<\/sub>.
      \n8Al + 30 HNO3<\/sub> \u2192 8A\/(NO3<\/sub>)3<\/sub> + 3 NH4<\/sub> NO3<\/sub> + 9H2<\/sub>O<\/p>\n

      Concentrated nitric acid renders the aluminium passive.<\/p>\n

      \"TS<\/p>\n

      Question 12.
      \nWrite a short note on the anomalous behaviour of boron in the group – 13.
      \nAnswer:<\/p>\n

        \n
      1. Boron is a non-metal. ‘Al’ is amphoteric. Ga, In and Tl are metals.<\/li>\n
      2. Boron always forms covalent compounds, while the others form ionic compounds.<\/li>\n
      3. B2<\/sub>O3<\/sub> is an acidic oxide. The trioxides of others are either amphoteric or basic in nature.<\/li>\n
      4. B(OH)3<\/sub> is an acid while the hydroxides of other elements are either amphoteric or basic.<\/li>\n
      5. Simple borates and silicates can polymerise readily forming polyacids while others do not form such polymers.<\/li>\n<\/ol>\n

        Question 13.
        \nAluminium reacts with dil. HNO3<\/sub> but not with cone. HNO3<\/sub> – explain.
        \nAnswer:
        \nCone, nitric acid oxidises the aluminium to aluminium oxide. This forms as a layer on the surface of the metal which prevents the further reaction. This oxide layer acts as a protective layer. Thus aluminium becomes passive with cone. HNO3<\/sub> and does not react with cone. HNO3<\/sub>.<\/p>\n

        In dilute HNO3<\/sub> oxide layer on the surface will be dissolved, then the hydrogen liberated from nitric acid reduces it to ammonium nitrate.<\/p>\n

        Question 14.
        \nGive two methods of preparation of diborane.
        \nAnswer:
        \n1) Diborane can be prepared by treating boron trifluoride with LiAlH4<\/sub> in diethyl ether.
        \n4 BF3<\/sub> + 3 LiAlH4<\/sub> \u2192 2 B2<\/sub>H6<\/sub> + 3 LiF + 3 AlF3<\/sub><\/p>\n

        2) In the laboratory it is conveniently prepared by the oxidation of sodium boro-hydride with iodine.
        \n2 Na BH4<\/sub> + I2<\/sub> \u2192 B2<\/sub>H6 + 2NaI + H2<\/sub><\/p>\n

        3) On large scale it is prepared by reaction of BF3 with sodium hydride.
        \n\"TS<\/p>\n

        Question 15.
        \nHow does diborane react with
        \na) H2<\/sub>O b) CO c) N(CH3<\/sub>)3<\/sub>?
        \nAnswer:
        \na) Diborane hydrolyses in water liberating hydrogen gas with the formation of boric acid.
        \nB2<\/sub>H6<\/sub> (g) + 6H2<\/sub>O \u2192 2 B(OH)3<\/sub> (aq) + 6H2<\/sub>(g)<\/p>\n

        b) Diborane forms an addition product with CO. In this reaction diborane undergoes symmetric cleavage.
        \nB2<\/sub>H6<\/sub> + 2 CO \u2192 2 BH3<\/sub>. CO<\/p>\n

        c) While reacting with N(CH3<\/sub>)3<\/sub> also diborane undergoes symmetric cleavage forming the addition compound.
        \nB2<\/sub>H6<\/sub> + 2 N (CH3<\/sub>)3<\/sub> \u2192 2 BH3<\/sub> . N (CH3<\/sub>)3<\/sub><\/p>\n

        Question 16.
        \nAl2<\/sub>O3<\/sub> is amphoteric – explain with suitable reactions.
        \nAnswer:
        \nAl2<\/sub>O3<\/sub> reacts with both acids and bases. While reacting with acids it behaves like base. In the reaction with bases it behaves like acid. So it is amphoteric.<\/p>\n

        Reaction with acid:
        \nAl2<\/sub>O3<\/sub> + 6HCl \u2192 2 AlCl3<\/sub> + 3H2<\/sub>O<\/p>\n

        Reaction with base:
        \nAl2<\/sub>O3<\/sub> + 2NaOH \u2192 2Na AlO2<\/sub> + H2<\/sub>O<\/p>\n

        Question 36.
        \n\"TS
        \nIdentify A and B.
        \nHint: A = H3<\/sub>BO3<\/sub>; B = (C2<\/sub>H5<\/sub>)3<\/sub> BO3<\/sub>.
        \nAnswer:
        \nWhen borax is heated with cone. H2S04 boric acid is formed.
        \nNa2<\/sub>B4<\/sub>O7<\/sub> + H2<\/sub>SO4<\/sub> \u2192 Na2<\/sub>SO4<\/sub> + H2<\/sub>B4<\/sub>O7<\/sub>
        \nH2<\/sub> B4<\/sub>O7<\/sub> + 5H2<\/sub>O \u2192 4 H3<\/sub>BO3<\/sub>
        \nNa2<\/sub> B4<\/sub>O7<\/sub> + H2<\/sub>SO4<\/sub> + 5H2<\/sub>O \u2192 Na2<\/sub>SO4<\/sub> + 4H3<\/sub>BO3<\/sub><\/p>\n

        The boric acid forms triethyl borate when heated with cone. H2<\/sub>SO4<\/sub> and ethyl alcohol. This burns with green edged flame.
        \n\"TS<\/p>\n

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

        Question 1.
        \nHow are borax and boric acid prepared? Explain the action of heat on them.
        \nAnswer:
        \n1) Borax occurs naturally as tincol, associated with impurities in sediments of lakes. Borax is highly soluble in hot water but crystallises in cold water. So the tincol is dissolved in hot water and filtered to remove insoluble impurities. Then the filtrate is cooled to get the pure borax.<\/p>\n

        Borax on heating decomposes into sodium metaborate and boron trioxide which form a transparent glassy bead.
        \n\"TS<\/p>\n

        2) When borax is heated with cone. H2<\/sub>SO4<\/sub>, boric acid is formed.
        \nNa2<\/sub>B4<\/sub>O7<\/sub> + H2<\/sub>SO4<\/sub> + 5H2<\/sub>O \u2192 Na2<\/sub>SO4<\/sub> + 4 H3<\/sub> BO3<\/sub>.<\/p>\n

        Boric acid on heating first gives meta-boric acid at low temperature but at red heat condition forms boron trioxide.
        \n\"TS<\/p>\n

        Question 2.
        \nHow is diborane prepared? Explain its structure.
        \nAnswer:
        \nPreparation of Diborane :
        \n1) Boron trifluoride on reduction with LiH at 450K gives Diborane.
        \n2BF3<\/sub> + 6 LiH \u2192 B2<\/sub>H6<\/sub> + 6LiF<\/p>\n

        2)vOn passing silent electric discharge through a mixture of boron trichloride and hydrogen, diborane is formed.
        \n\"TS<\/p>\n

        Molecules in which the central atom gets less than octet configuration are called electron-deficient molecules.
        \nEx : (1) Diborane B2<\/sub>H6<\/sub>
        \n(TrBoron trifluoride BF3<\/sub>.<\/p>\n

        Structure of diborane:
        \nElectron diffraction studies have shown that diborane contains two coplanar BH2<\/sub> groups. Its structure can be represented as
        \n\"TS<\/p>\n

        The four H atoms present in the BH2<\/sub> groups are known as terminal hydrogen atoms. The remaining two H atoms are called Bridge hydrogens. The two bridge hydrogens lie in a plane perpendicular to the plane of the two BH2<\/sub> groups. Out of the two bridge hydrogens, one H atom lies above the plane and the other H atom lies below the plane.<\/p>\n

        Orbital structure of Diborane :
        \nIn Diborane each boron atom undergoes sp\u00b3 hybridisation resulting in four equivalent sp\u00b3 hybrid orbitals. Three of these orbitals have one electron each and the fourth hybrid orbital is vacant. In the formation of B – H – B bridge, sp\u00b3 hybrid orbital with one electron from one boron atom, Is orbital of one bridge hydrogen and the vacant sp\u00b3 hybrid orbital of the second boron atom overlap as shown below.<\/p>\n

        The two B – H – B bridges present in Diborane are abnormal bonds. They are considered as three centered two electron bonds. This type of bond is called banana bond or tau bond. In this type of bond a pair of electrons holds three atoms. In Diborane there are two such bridges. This Diborane structure can also be represented as shown here.<\/p>\n

        \"TS<\/p>\n

        Question 3.
        \nWrite any two methods of preparation of diborane. How does it react with
        \ni) Carbon monoxide and ii) Ammonia?
        \nAnswer:
        \ni) Preparation of diborane :
        \n1) Diborane can be prepared by treating BF3 with Li A\/ H4 in diethyl ether.
        \n4 BF3<\/sub> + 3 Li AlH4<\/sub> \u2192 2 B2<\/sub>H6<\/sub> + 3 LiF + 3 AlF3<\/sub><\/p>\n

        2) In the laboratory it is conveniently prepared by the oxidation of sodium borohydride with iodine.
        \n2 Na BH4<\/sub> + I2<\/sub> \u2192 B2<\/sub>H6<\/sub> + 2 Nal + H2<\/sub><\/p>\n

        3) On a large scale it is prepared by the reaction of BF3<\/sub> with sodium hydride.
        \n\"TS<\/p>\n

        ii) a) Diborane forms an additional product with CO with symmetric cleavage.
        \nB2<\/sub>H6<\/sub> + 2CO \u2192 2BH3<\/sub>.CO<\/p>\n

        b) With ammonia it forms an addition product with asymmetric cleavage which on heating forms borazine which is known as inorganic benzene.
        \n\"TS<\/p>\n","protected":false},"excerpt":{"rendered":"

        Telangana TSBIE TS Inter 1st Year Chemistry Study Material 10th Lesson p-Block Elements: Group 13 Textbook Questions and Answers. TS Inter 1st Year Chemistry Study Material 10th Lesson p-Block Elements: Group 13 Very Short Answer Type Questions Question 1. Discuss the pattern of variation in the oxidation states of Boron to Thallium. Answer: The elements … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/1566"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=1566"}],"version-history":[{"count":3,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/1566\/revisions"}],"predecessor-version":[{"id":2747,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/1566\/revisions\/2747"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=1566"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=1566"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=1566"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}