In the $\mathrm{BCl}_3$, due to small size of $\mathrm{B}^{3+}, \mathrm{BCl}_3$ is covalent. Still, the octet of B remains incomplete. In $\mathrm{NH}_3, \mathrm{~N}$ has a lone pair of electrons. Hence, N shares lone pair of electron with B to complete the octet. Hence, $\mathrm{BCl}_3$ acts as Lewis acid and $\mathrm{NH}_3$ as Lewis bases.

$$\mathrm{H}_3 \mathrm{~N}:+\mathrm{BCl}_3 \longrightarrow \mathrm{H}_3 \mathrm{~N} \longrightarrow \mathrm{BCl}_3 \text { or } \mathrm{BCl}_3 \cdot \mathrm{NH}_3$$

$\mathrm{AlCl}_3$ forms dimer by completing octet of Al involving $p$-orbitals to accept electron pair from chlorine.

Orthoboric acid is less soluble in cold water but highly soluble in hot water. It is a monobasic acid. It does not liberate $\mathrm{H}^{+}$ion but accepts $\mathrm{OH}^{-}$from water, behaving as a Lewis acid.

$$\mathrm{H}_3 \mathrm{BO}_3+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{B}(\mathrm{OH})_4^{-}+\mathrm{H}^{+}$$

The structure of H$_3$BO$_3$ is

Octet of boron remains incomplete. Oxygen atom contains lone pair of electrons in water molecule. Hence, instead of donating proton $\left(\mathrm{H}^{+}\right)$, boric acid accepts $\mathrm{OH}^{-}$from water forming $\mathrm{B}(\mathrm{OH})_4^{-}$to complete octet.

Since, electron acceptor substance behaves as Lewis acid, therefore, boric acid acts as a Lewis acid in water.

Orthoboric acid $\mathrm{H}_3 \mathrm{BO}_3$, in solid state possesses a layer structure made up of $\mathrm{B}(\mathrm{OH})_3$ units forming hexagonal rings of H -bonding as given below

Each H -atom acts as a bridge between two oxygen atoms of different $\mathrm{BO}_3^{3-}$ units. Boric acid when dissolved in water, acts as Lewis acid forming $\mathrm{B}(\mathrm{OH})_4^{-}$

$$\mathrm{H}_3 \mathrm{BO}_3+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{B}(\mathrm{OH})_4^{-}+\mathrm{H}^{+}$$

The hybridisation of boron in $\mathrm{B}(\mathrm{OH})_4^{-}$is $s p^3$.