Explain structures of diborane and boric acid.

$B _{2} H _{6}$ is an electron-deficient compound. $B_2H_6$ has only $12$ electrons $-6 e ^{-}$ from $6 H$ atoms and $3 e ^{-}$ each from $2 B$ atoms. Thus, after combining with $3 H$ atoms, none of the boron atoms has any electrons left. $x$ -ray diffraction studies have shown the structure of diborane as: (figure)

$2$ boron and $4$ terminal hydrogen atoms $(Ht)$ lie in one plane, while the other two bridging hydrogen atoms $\left( H _{ b }\right)$ lie in a plane perpendicular to the plane of boron atoms. Again, of the two bridging hydrogen atoms, one $H$ atom lies above the plane and the other lies below the plane. The terminal bonds are regular two-centre two-electron $\left(2 c-2 e^{-}\right)$ bonds, while the two bridging $( B - H - B )$ bonds are three-centre two-electron $\left(3 c -2 e ^{-}\right)$ bonds. (figure)

$(b)$ Boric acid

Boric acid has a layered structure. Each planar $BO _{3}$ unit is linked to one another through $H$ atoms. The $H$ atoms form a covalent bond with a $BO _{3}$ unit, while a hydrogen bond is formed with another $BO _{3}$ unit. In the given figure, the dotted lines represent hydrogen bonds. (figure)