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Q.
$B_2H_6$ consists of
The p-Block Elements
Solution:
According to molecular orbital theory, each of the two boron atoms is in $sp^3$ hybrid state. Of the four hybrid orbitals, three have one electron each while the fourth is empty. Two of the four orbitals of each of the boron atom overlap with two terminal hydrogen atoms forming two normal $B - H \,σ$-bonds. One of the remaining hybrid orbtial (either filled or empty) of one of the boron atoms; 1s orbital of hydrogen atom (bridge atom) and one of hybrid orbitals of the other boron atom overlap to form a delocalised orbital covering the three nuclei with a pair of electrons. Such a bond is known as three centre two electron bonds.