The structure of diborane (B$$_2$$H$$_6$$) contains

Each boron atom in diborane, $$B_2H_6$$, has three valence electrons. To attain an octet it needs to share five more electrons, but only four hydrogen atoms are available for ordinary $$B{-}H$$ bonding. Hence a special bonding scheme is required.

Step-1: Formation of the normal terminal bonds.
Each boron forms two ordinary sigma bonds with two terminal hydrogen atoms. These bonds are 2-centre-2-electron (2c-2e) bonds. Therefore, there are $$2 \times 2 = 4$$ terminal 2c-2e bonds in the molecule.

Step-2: Bridging hydrogens and multicentre bonding.
The remaining two hydrogens bridge the two boron atoms. Each bridge involves one hydrogen atom simultaneously bonded to both boron atoms. This is achieved through a 3-centre-2-electron (3c-2e) bond, often called a banana bond, because the electron pair is shared over three nuclei (B-H-B) but still contains only two electrons.

Step-3: Counting the 3c-2e bonds.
Since there are two bridging hydrogens, there are exactly two such 3c-2e bonds.

Step-4: Final bond count.
Total = four 2c-2e bonds (terminal) + two 3c-2e bonds (bridging).

Hence the structure of diborane contains four 2c-2e bonds and two 3c-2e bonds.

Option A which is: four 2c-2e bonds and two 3c-2e bonds