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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.
पर्याय
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
Suggest reasons why the B–F bond lengths in BF3 (130 pm) and `"BF"_4^(-)` (143 pm) differ.
How would you explain the lower atomic radius of Ga as compared to Al?
In some of the reactions thallium resembles aluminium, whereas in others it resembles with group I metals. Support this statement by giving some evidences.
Write a balanced equation for Al + NaOH → ?
The exhibition of highest co-ordination number depends on the availability of vacant orbitals in the central atom. Which of the following elements is not likely to act as central atom in \[\ce{MF^{3-}6}\]?
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
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 properties given in Column II.
| Column I | Column II |
| (i) Diborane | (a) Used as a flux for soldering metals |
| (ii) Galluim | (b) Crystalline form of silica |
| (iii) Borax | (c) Banana bonds |
| (iv) Aluminosilicate | (d) Low melting, high boiling, useful for measuring high temperatures |
| (v) Quartz | (e) Used as catalyst in petrochemical industries |
Describe the general trends in the following properties of the elements in Groups 13 and 14.
Ionisation enthalpy
Describe the general trends in the following properties of the elements in Groups 13 and 14.
Metallic character
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The +1 oxidation state of thallium is more stable than its +3 state.
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Boron fluoride exists as BF3 but boron hydride doesn’t exist as BH3. Give reason. In which form does it exist? Explain its structure.
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.
Taking stability as the factor, which one of the following represents the correct relationship?
