Q.Explain structures of diborane and boric acid.

In diborane each boron is $sp^3$ hybridised: two terminal hydrogens per boron form ordinary 2-centre-2-electron bonds, while two hydrogens bridge both borons through unusual 3-centre-2-electron ("banana") bonds, since boron does not have enough electrons for normal bonding to all six hydrogens. Boric acid consists of planar, triangular BO$_3$ units (boron $sp^2$ hybridised) that link together via hydrogen bonding between –OH groups into a layered, shee …

Diborane, B$_2$H$_6$

1. Electron count. Each boron atom has only 3 valence electrons, so B$_2$H$_6$ has $2(3) + 6(1) = 12$ valence electrons — but 7 bonds (if all B–H were normal 2c-2e bonds) would need 14 electrons. Diborane is therefore "electron-deficient."
2. Structure. Each boron is $sp^3$ hybridised. Four hydrogens (two on each boron) are terminal, forming ordinary 2-centre-2-electron B–H bonds lying in a plane. The remaining two hydrogens are bridging, lying above and below this plane, each connecting both boron atoms.
3. Bridge bonding. Each bridge is a 3-centre-2-electron (3c-2e) bond: one B–H–B unit is held together by just 2 electrons shared across all three nuclei (a "banana bond"), rather than two separate 2-electron bonds. This uses the remaining orbitals/electrons economically and satisfies boron's electron deficiency.

Boric acid, H$_3$BO$_3$

4. Basic unit. Boron is $sp^2$ hybridised, bonded to three –OH groups in a planar, triangular BO$_3$ unit. …

For diborane: count valence electrons to show boron is electron-deficient, then describe $sp^3$ boron with 4 terminal 2c-2e B–H bonds and 2 bridging 3c-2e B–H–B bonds. For boric acid: describe the planar $sp^2$ BO$_3$ unit and how hydrogen bonding between –OH groups b …

• Drawing diborane with 6 identical, normal 2-centre bonds instead of recognising the 2 bridging bonds are 3-centre-2-electron bonds.
• Forgetting that boron in both diborane and boric acid does not have a complete octet / has only 6 electrons in its normal bonding shell. …