Identification of Products from Chlorine Sulfur Dioxide and Phosphorus Reactions

To determine the chemical species $R$, $S$, and $T$, we trace each chemical reaction described in the passage step-by-step:

1. Formation of compound $R$: Chlorine gas ($\text{Cl}_2$) reacts with sulfur dioxide gas ($\text{SO}_2$) in the presence of a charcoal catalyst to yield sulfuryl chloride ($\text{SO}_2\text{Cl}_2$): $\text{SO}_2(g) + \text{Cl}_2(g) \xrightarrow{\text{charcoal}} \text{SO}_2\text{Cl}_2(l)$ Therefore, the product $R$ is $\text{SO}_2\text{Cl}_2$.

2. Formation of compound $S$: Sulfuryl chloride ($\text{SO}_2\text{Cl}_2$) acts as a strong chlorinating agent. When white phosphorus ($\text{P}_4$) reacts with excess sulfuryl chloride, it is chlorinated to form phosphorus pentachloride ($\text{PCl}_5$): $\text{P}_4 + 10\text{SO}_2\text{Cl}_2 \rightarrow 4\text{PCl}_5 + 10\text{SO}_2$ Therefore, the compound $S$ is $\text{PCl}_5$.

3. Formation of oxoacid $T$: Upon complete hydrolysis, phosphorus pentachloride ($\text{PCl}_5$) reacts with water to form orthophosphoric acid ($\text{H}_3\text{PO}_4$): $\text{PCl}_5 + 4\text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_4 + 5\text{HCl}$ Therefore, the oxoacid $T$ is $\text{H}_3\text{PO}_4$.

Thus, $R$, $S$, and $T$ are given respectively by $\text{SO}_2\text{Cl}_2$, $\text{PCl}_5$, and $\text{H}_3\text{PO}_4$.

Correct Option: A