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प्रश्न
A compound X, of boron reacts with NH3 on heating to give another compound Y which is called inorganic benzene. The compound X can be prepared by treating BF3 with Lithium aluminium hydride. The compounds X and Y are represented by the formulas.
विकल्प
B2H6, B3N3H6
B2O3, B3N3H6
BF3, B3N3H6
B3N3H6, B2H6
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उत्तर
B2H6, B3N3H6
Explanation:
A compound X, of boron, reacts with NH3 on heating to give another compound YY which is called inorganic benzene.
\[\ce{\underset{X(Diborance)}{3H2H6} + 6NH3 -> 3[BH2(NH3)2]+ [BH4]- ->[heat] \underset{Y(Borazole Inorganic Benzene)}{2B3N3H6} + 12H2}\]
\[\ce{4BF3 + 3LiAlH4 -> \underset{X}{2B2H6} + 3LiF + 3AlF3}\]
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संबंधित प्रश्न
What happens when BF3 is reacted with ammonia?
In the structure of diborane ______.
The most commonly used reducing agent is ______.
Dry ice is ______.
Which of the following statements are correct. Answer on the basis of Figure.
(i) The two birdged hydrogen atoms and the two boron atoms lie in one plane;
(ii) Out of six B – H bonds two bonds can be described in terms of 3 centre 2-electron bonds.
(iii) Out of six B – H bonds four B – H bonds can be described in terms of 3 centre 2 electron bonds;
(iv) The four-terminal B – H bonds are two centre-two electron regular bonds.
Explain why the following compounds behave as Lewis acids?
AlCl3
Aluminium dissolves in mineral acids and aqueous alkalies and thus shows amphoteric character. A piece of aluminium foil is treated with dilute hydrochloric acid or dilute sodium hydroxide solution in a test tube and on bringing a burning matchstick near the mouth of the test tube, a pop sound indicates the evolution of hydrogen gas. The same activity when performed with concentrated nitric acid, reaction doesn’t proceed. Explain the reason.
Match the species given in Column I with the properties mentioned in Column II.
| Column I | Column II |
| (i) \[\ce{BF^{-}4}\] | (a) Oxidation state of central atom is +4 |
| (ii) AICI3 | (b) Strong oxidising agent |
| (iii) SnO | (c) Lewis acid |
| (iv) PbO2 | (d) Can be further oxidised |
| (e) Tetrahedral shape |
Match the species given in Column I with the hybridisation given in Column II.
| Column I | Column II |
| (i) Boron in [B(OH)4]– | (a) sp2 |
| (ii) Aluminium in [Al(H2O)6]3+ | (b) sp3 |
| (iii) Boron in B2H6 | (c) sp3d2 |
| (iv) Carbon in Buckminsterfullerene | |
| (v) Silicon in \[\ce{SiO^{4-}4}\] | |
| (vi) Germanium in [GeCl6]2– |
Describe the general trends in the following properties of the elements in Groups 13 and 14.
Oxidation states
Account for the following observations:
Though fluorine is more electronegative than chlorine yet BF3 is a weaker Lewis acid than BCl3
Three pairs of compounds are given below. Identify that compound in each of the pairs which has group 13 element in more stable oxidation state. Give reason for your choice. State the nature of bonding also.
TlCl3, TlCl
Three pairs of compounds are given below. Identify that compound in each of the pairs which has group 13 element in more stable oxidation state. Give reason for your choice. State the nature of bonding also.
InCl3, InCl
BCl3 exists as monomer whereas AlCl3 is dimerised through halogen bridging. Give reason. Explain the structure of the dimer of AlCl3 also.
A nonmetallic element of group 13, used in making bullet proof vests is extremely hard solid of black colour. It can exist in many allotropic forms and has unusually high melting point. Its trifluoride acts as Lewis acid towards ammonia. The element exihibits maximum covalency of four. Identify the element and write the reaction of its trifluoride with ammonia. Explain why does the trifluoride act as a Lewis acid.
Boron compounds behave as Lewis acids because of their ______.
A group 13 element ‘X’ reacts with chlorine gas to produce a compound XCl3. XCl3 is electron deficient and easily reacts with NH3 to form \[\ce{Cl3X –> NH3}\] adduct; however, XCl3 does not dimerize X is ______.
Taking stability as the factor, which one of the following represents the correct relationship?
