Chemistry - CHEMICAL BONDING Question with Solution | TestHub
The simplest boron hydride known is diborane (B2H6). As implied by the mechanism of hydrolysis, diborane and many other light boron hydrides act as Lewis acids and they are cleaved by reaction with Lewis bases. Two different cleavage patterns have been observed, namely, symmetric cleavage and unsymmetric cleavage. In symmetric cleavage, B2H6 is broken symmetrically into two BH3 fragments with NMe3 . While with NH3, MeNH2, MeNH2 results in unsymmetrical cleavage, which is a cleavage leading to an ionic product.
Select the CORRECT statement regarding .
(i) It is not a electron deficient compound
(ii) It involves participation of vacant orbital in hybridization
(iii) type of overlapping is present
(iv) Bridge bond is longer than terminal bond
(v) Both the boron are hybridized
(vi) bond is not present in the molecule
Answer:
Solution:
Diborane (B₂H₆) is an electron-deficient compound because it has only 12 valence electrons for 8 bonding positions, so (i) is false.
Each boron atom uses sp³ hybridisation, and bonding involves vacant p-orbitals participating in multicenter bonding, so (ii) and (v) are true.
It does not have normal sp³–sp³–sp³ type 2c–2e overlap; instead it has 3-center–2-electron (banana) bonds, so (iii) is false.
Bridging B–H–B bonds are stronger and hence longer than terminal B–H bonds, so (iv) is true.
Diborane contains 3c–2e bonds, not 3c–4e bonds, so (vi) is true.
