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प्रश्न
Boron fluoride exists as BF3 but boron hydride doesn’t exist as BH3. Give reason. In which form does it exist? Explain its structure.
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उत्तर
In BF3, due to n – pπ back bonding between the vacant p-orbital of boron and filled p-orbital of fluorine. This π – pπ back bonding is absent in case of hydrogen as it is a single electron element.
Two BH3 molecules dimerise to form diborane.
In B2H6 There are two types of hydrogens present.
(i) Four hydrogens that are terminally bonded to each of two boron atoms.
(ii) Two hydrogens that are bonded to both boron atoms forming a bridge in between.
The four-terminal hydrogen atoms and two boron atoms lie in the same plane while bridging hydrogen lies in a plane perpendicular to them.
Two hydrogens forming a bridge in B2H6 are peculiar in bonding and can be termed as 3 -centered-2-electron bond or banana bond. 1sorbital of each hydrogen overlaps with the hybrid orbital of one of the boron then delocalising the 2e– over three atoms making 3-centred- 2 -electron bond.
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संबंधित प्रश्न
How would you explain the lower atomic radius of Ga as compared to Al?
Write a balanced equation for Al + NaOH → ?
Write a balanced equation for B2H6 + NH3 → ?
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 \[\ce{MF^{3-}6}\]?
Ionisation enthalpy (∆iH1kJ mol–1) for the elements of Group 13 follows the order.
Dry ice is ______.
Cement, the important building material is a mixture of oxides of several elements. Besides calcium, iron and sulphur, oxides of elements of which of the group (s) are present in the mixture?
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.
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.
Explain the following:
Electron gain enthalpy of chlorine is more negative as compared to fluorine.
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.
Metallic character
Describe the general trends in the following properties of the elements in Groups 13 and 14.
Nature of halides
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AlCl3 , AlCl
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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.
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?
