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.

The term inert pair effect is often used in relation to the increasing stability of oxidation states that are 2 less than the group valency for the heavier elements of groups $13,14,15$ and 16. In group 13 all the elements show +3 oxidation state whereas Ga , In and TI show +1 oxidation state also. Boron, being small in size can lose its valence electrons to form $\mathrm{B}^{3+}$ ion and shows +3 oxidation state. +1 oxidation state of $\mathrm{TI}, \mathrm{Ga}$ is due to inert pair effect. The outer shell $s$-electrons ( $n s^2$ ) penetrate to $(n-1) d$-electrons and thus becomes closer to nucleus and are more effectively pulled towards the nucleus. This results in less availability of $n s^2$ electrons pair for bonding or $n s^2$ electron pair becomes inert. This reluctance in the participation of $n s^2$ eletrons in bonding is termed as inert pair effect. The inert pair is more effective after $n \geq 4$ and increases with increasing value of $n$. For groups 14 , in spite of four valence electrons, they do not form $M^{4+}$ or $M^{4-}$ ionic compounds. They form covalent compounds with four bonds. $\mathrm{Ge}, \mathrm{Sn}$ and Pb also exhibit +2 oxidation state due to inert pair effect. $\mathrm{Sn}^{2+}$ and $\mathrm{Pb}^{2+}$ show ionic nature. The tendency to form +2 ionic state increases on moving down the group due to inert pair effect.