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 $s p^3$.

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.

(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.

(a) $\mathrm{CO}_2$ has a linear structure. Its dipole moment is zero. It is believed that $\mathrm{CO}_2$ molecule is a resonance hybrid has the following structures.

$$\mathrm{O}=\mathrm{C}=\mathrm{O} \leftrightarrow{ }^{-} \mathrm{O}-\mathrm{C} \equiv \mathrm{O}^{+} \leftrightarrow \mathrm{O}^{+} \equiv \mathrm{C}-\mathrm{O}^{-}$$

The $\mathrm{CO}_2$ molecules are held together by weak van der Waals' forces and thus, it exists as gas. In $\mathrm{SiO}_2$, due to large electronegative difference between Si and O , the $\mathrm{Si}-\mathrm{O}$ bonds have considerable ionic nature.

Therefore, silica has three dimensional network like structure in which Si-atom is tetrahedrally bonded to four oxygen atoms and each oxygen atom is bonded to two silicon atoms by covalent bonds.

There is no discrete $\mathrm{SiO}_2$ molecule. It is a network solid with octahedral coordination.

(b) In silicon, vacant 3d orbitals are available due to which it can accomodate electrons from 6 fluorine atoms, thereby forming $\mathrm{Si}_6^{2-}$ ion. However, in case of C only $2 p^2$ filled orbitals are available thus, it cannot expand their covalency more than 4. Thus, $\mathrm{CCl}_6^2$ is not known.