1. Identify the oxoacids formed with $NaOH$

Chlorine reacts with alkali differently under different conditions:

• With cold, dilute $NaOH$: $Cl_2 + 2NaOH \rightarrow NaCl + NaOCl + H_2O$ This gives sodium hypochlorite, corresponding to hypochlorous acid $HOCl$.

• With hot, concentrated $NaOH$: $3Cl_2 + 6NaOH \rightarrow 5NaCl + NaClO_3 + 3H_2O$ This gives sodium chlorate, corresponding to chloric acid $HClO_3$.

So the two different oxoacids are of chlorine, but we mainly need to identify $R, S, T$.

2. Find product $R$ from reaction of $Cl_2$ with $SO_2$ in presence of charcoal

In presence of charcoal, chlorine reacts with sulfur dioxide to form sulfuryl chloride: $SO_2 + Cl_2 \xrightarrow[\text{charcoal}]{} SO_2Cl_2$

Thus, $R = SO_2Cl_2$

3. Reaction of $R$ with white phosphorus

Sulfuryl chloride acts as a chlorinating agent. White phosphorus reacts with it to form phosphorus pentachloride: $P_4 + 10SO_2Cl_2 \rightarrow 4PCl_5 + 10SO_2$

Hence, $S = PCl_5$

4. Hydrolysis of $S$

Now hydrolyze phosphorus pentachloride: $PCl_5 + 4H_2O \rightarrow H_3PO_4 + 5HCl$

So the oxoacid of phosphorus formed is $T = H_3PO_4$

5. Match with options

We found:

• $R = SO_2Cl_2$
• $S = PCl_5$
• $T = H_3PO_4$

This matches Option A.

6. Check against stored correct answer

Stored correct answer: A

Our derived answer: A

So they agree.