Identify the products of reaction of $Cl_2$ with $NaOH$

Chlorine reacts differently with cold-dilute and hot-concentrated $NaOH$.

(i) With cold, dilute $NaOH$: $Cl_2 + 2NaOH \rightarrow NaCl + NaOCl + H_2O$ Here, $NaOCl$ is sodium hypochlorite, the sodium salt of hypochlorous acid $(HClO)$.

So, $P = \text{sodium salt of hypochlorous acid}$

(ii) With hot, concentrated $NaOH$: $3Cl_2 + 6NaOH \rightarrow 5NaCl + NaClO_3 + 3H_2O$ Here, $NaClO_3$ is sodium chlorate, the sodium salt of chloric acid $(HClO_3)$.

So, $Q = \text{sodium salt of chloric acid}$

Check the intermediate sequence given in the question

Chlorine reacts with sulfur dioxide in presence of charcoal to give thionyl chloride: $SO_2 + Cl_2 \xrightarrow{charcoal} SOCl_2$ Hence, $R = SOCl_2$

Reaction of $R$ with white phosphorus

Thionyl chloride reacts with white phosphorus to form phosphorus trichloride: $P_4 + 6SOCl_2 \rightarrow 4PCl_3 + 6SO_2$ Thus, $S = PCl_3$

Hydrolysis of $S$

$PCl_3 + 3H_2O \rightarrow H_3PO_3 + 3HCl$ So the oxoacid of phosphorus formed is $T = H_3PO_3$ This is consistent with the reaction chain, confirming our identification.

Determine the correct option

Therefore, $P$ and $Q$ are the sodium salts of:

• Hypochlorous acid
• Chloric acid

Hence, the correct option is: $\boxed{\text{A}}$