Advertisements
Advertisements
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.
Advertisements
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.
APPEARS IN
संबंधित प्रश्न
How can you explain higher stability of BCl3 as compared to TlCl3?
If B–Cl bond has a dipole moment, explain why BCl3 molecule has zero dipole moment.
Aluminium trifluoride is insoluble in anhydrous HF but dissolves on the addition of NaF. Aluminium trifluoride precipitates out of the resulting solution when gaseous BF3 is bubbled through. Give reasons.
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 \[\ce{MF^{3-}6}\]?
The most commonly used reducing agent is ______.
Explain the following:
Pb4+ acts as an oxidising agent but Sn2+ acts as a reducing agent.
Explain the following:
Electron gain enthalpy of chlorine is more negative as compared to fluorine.
Identify the compounds A, X and Z in the following reactions:
\[\ce{X ->[Δ][370 K] HBO2 ->[Δ][> 370 K] Z}\]
Complete the following chemical equations:
\[\ce{Z + 3 LiAlH4 -> X + 3LiF + 3AlF_3}\]
\[\ce{X + 6H2 -> Y + 6H2}\]
\[\ce{3X + 3O2 ->[Δ] B2O3 + 3H2O}\]
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.
Ionisation enthalpy
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
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.
InCl3, InCl
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.
