Draw the structures of $\mathrm{BC}_3 \cdot \mathrm{NH}_3$ and $\mathrm{Ald}_3$ (dimer).
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.
Explain the nature of boric acid as a Lewis acid in water.
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.
Draw the structure of boric acid showing hydrogen bonding. Which species is present in water ? What is the hybridisation of boron in this species ?
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 $s p^3$.
Explain why the following compounds behave as Lewis acids ?
(a) $\mathrm{Bd}_3$
(b) $\mathrm{Ald}_3$
In trivalent state, the number of electrons around the central atom in a molecule of compounds $\mathrm{BCl}_3$ and $\mathrm{AlCl}_3$ will be only six. Such electron deficient molecules have tendency to accept a pair of electron to achieve stable electronic configuration and thus, act as Lewis acids. The tendency to behave as Lewis acid decreases with the increase in the size down the group.
Give reasons for the following
(a) $\mathrm{CCl}_4$ is immiscible in water, whereas $\mathrm{Sid}_4$ is easily hydrolysed.
(b) Carbon has a strong tendency for catenation compared to silicon.
(a) Carbon tetrachloride $\left(\mathrm{CCl}_4\right)$ is a covalent compound while $\mathrm{H}_2 \mathrm{O}$ is a polar compound. $\mathrm{CCl}_4$ does not form H -bond with water molecule. Hence, it is immiscible in water. Further more, $\mathrm{CCl}_4$ is not hydrolysed by water because of the absence of $d$-orbitals in carbon while $\mathrm{SiCl}_4$ is readily hydrolysed by water.
$\mathrm{SiCl}_4+4 \mathrm{H}_2 \mathrm{O} \longrightarrow \underset{\text { Silicic acid }}{\mathrm{Si}(\mathrm{OH})_4}+4 \mathrm{HCl}$
The hydrolysis of $\mathrm{SiCl}_4$ occurs due to coordination of $\mathrm{OH}^{-}$with empty $3 d$ orbitals in silicon atom of $\mathrm{SiCl}_4$ molecule.
(b) Carbon atoms have the tendency to link with one another through covalent bonds to form chains and rings. This property is called catenation. This is because $\mathrm{C}-\mathrm{C}$ bonds are very strong.
Down the group, the size increases and electronegativity decreases and thereby, tendency to show catenation decreases. Thus, carbon has a strong tendency for catenation as compared to silicon.