$$HO.S{O_2}OH + 2PC{l_5} \to $$      $$\mathop {ClS{O_2}Cl + 2POC{l_3}}\limits_{{\text{Sulphuryl chloride}}} + 2HCl$$

$$Cu + 2{H_2}S{O_4}\left( {conc} \right) \to $$      $$CuS{O_4} + S{O_2} + 2{H_2}O$$

No explanation is given for this question. Let's discuss the answer together.

$$A$$ is $$PC{l_5}$$  ( yellowish white powder ).
$${P_4} + 10C{l_2} \to 4PC{l_5}$$
$$B$$ is $$PC{l_3}$$  ( colourless oily liquid ).
$${P_4} + 6C{l_2} \to 4PC{l_3}$$
The products of hydrolysis of $$PC{l_3}$$  and $$PC{l_5}$$  are :
$$\eqalign{ & PC{l_3} + 3{H_2}O \to \mathop {{H_3}P{O_3}}\limits_{\left( B \right)} + 3HCl \cr & PC{l_5} + 4{H_2}O \to \mathop {{H_3}P{O_4}}\limits_{\left( A \right)} + 5HCl \cr} $$

Key Idea Geometry is determined by electron pair arrangement whereas shape is determined by arrangement of atoms around the centre atom.

Geometry — octahedral, Hybridisation — $$s{p^3}{d^2}$$
Thus, option (A) is correct.

Two dimensional sheet structures are formed when three oxygen atoms of each $${\left[ {Si{O_4}} \right]^{4 - }}$$  tetrahedral are shared.

$$C{O_2}$$  is an acidic oxide, $$CO$$  is neutral and $$BeO$$  is an amphoteric oxide.

$$AlC{l_3}$$  exists as a dimer $$\left( {A{l_2}C{l_6}} \right)$$  . It is a strong Lewis acid as it has an incomplete octet and has a tendency to gain electrons. $$AlC{l_3}$$  undergoes hydrolysis easily and forms an acidic solution.
$$AlC{l_3} + 3{H_2}O \to Al{\left( {OH} \right)_3} + 3HCl\,$$
Option (C) is true that $$AlC{l_3}$$  sublimes at $${100^ \circ }C$$  under vacuum.
$$AlC{l_3}$$  is a Lewis acid.

Boron carbide $${B_4}C$$  is a very hard substance like $$SiC.$$

Let oxidation states of phosphorus in  $${H_3}P{O_2},{H_3}P{O_4},{H_3}P{O_3}\,{\text{and}}\,{H_4}{P_2}{O_6}\,\,{\text{be}}\,\,w,x,y,\,\,{\text{and}}\,z\,\,{\text{respectively}}$$
Thus, in $${H_3}P{O_2}$$ :
$$3\, \times \left( { + 1} \right) + w + 2 \times \left( { - 2} \right) = 0\,\,\,\,\therefore w = + 1$$
In $${H_3}P{O_4}:$$
$$3\, \times \left( { + 1} \right) + x + 4 \times \left( { - 2} \right) = 0\,\,\,\,\therefore x = + 5$$
In $${H_3}P{O_3}:$$
$$3\, \times \left( { + 1} \right) + y + 3 \times \left( { - 2} \right) = 0\,\,\,\,\therefore y = + 3$$
In $${H_4}{P_2}{O_6}:\,$$
$$4\, \times \left( { + 1} \right) + 2z + 6 \times \left( { - 2} \right) = 0\,\,\,\,\therefore z = + 4$$
Thus, the order of oxidation state is:
$${H_3}P{O_4} > {H_4}{P_2}{O_6} > {H_3}P{O_3} > \,{H_3}P{O_2}$$