Question

In diborane, the Tau-bonds can be formed by the overlapping of orbitals:
(A) $s{{p}^{3}}-s-s{{p}^{3}}$
(B) $s{{p}^{3}}-s{{p}^{3}}$
(C) $s{{p}^{2}}-s{{p}^{3}}$
(D) $s{{p}^{2}}-s-s{{p}^{2}}$

Answer 1

Hint: In Diborane, each boron atom exhibits tetrahedral structure, bonded to 4 hydrogen atoms. Two of them are bridged hydrogen atoms, and the rest two of them are terminal hydrogens. The bond length of the two types of hydrogen atoms are different from each other.

Complete step by step answer:
In diborane, the bonding between the boron atoms and the bridging hydrogen atoms is, however, different from that in molecules such as hydrocarbons. Each boron uses two electrons in bonding to the terminal hydrogen atoms, and has one valence electron remaining for additional bonding. The bridging hydrogen atoms provide one electron each. The ${{B}_{2}}{{H}_{2}}$ ring is held together by four electrons which form two 3-center 2-electron bonds. This type of bond is sometimes called a 'banana bond'.
Boron has an electronic configuration $2{{s}^{2}}2{{p}^{1}}$ of valence shell, so three covalent bonds gives it an incomplete octet.$B{{H}_{3}}$has an empty 2p orbital. This orbital overlaps the existing B−H $\sigma -$ bond cloud (in a nearby $B{{H}_{3}}$), and forms a 3c-2e bond.
In diborane, $s{{p}^{3}}$ hybrid orbital from each boron overlaps with 1s orbital of H atom to form 3 center-2 electron (3c-2e) bonds.
So, the correct answer is “Option B”.

Note: A bent bond, also known as a banana bond, is a type of covalent chemical bond with a geometry somewhat reminiscent of a banana. The term itself is a general representation of electron density or configuration resembling a similar "bent" structure within small ring molecules, such as cyclopropane $({{C}_{3}}{{H}_{6}})$ or as a representation of double or triple bonds within a compound that is an alternative to the sigma and pi bond model.