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?
Suggest reasons why the B–F bond lengths in BF3 (130 pm) and `"BF"_4^(-)` (143 pm) differ.
In some of the reactions thallium resembles aluminium, whereas in others it resembles with group I metals. Support this statement by giving some evidences.
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 ______.
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 the following:
Boron does not exist as B3+ ion.
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.
Account for the following observations:
The +1 oxidation state of thallium is more stable than its +3 state.
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.
AlCl3 , AlCl
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 ______.
Which one of the following is the correct statement?
