The reactions of `CI_(2)` gas with cold-dilute and hot-concentrated NaOH in water give sodium salts of two different oxioacids of chlorine, P and Q, respectively. The `CI_(2)` gas reacts with `SO_(2)` gas, in presence of charocal, to give a product R reacts with white phosphorus to give a compound S. On hydrolysis, S gives an oxoacid of phosphorus.
R, S and T, respectively, are

Options

A. `SO_(2)Cl_(2), PCl_(5) and H_(3)PO_(4)`
B. `SO_(2)Cl_(2), PCl_(3) and H_(3)PO_(3)`
C. `SOCl_(2), PCl_(3) and H_(3)PO_(2)`
D. `SOCl_(2), PCl_(5) and H_(3)PO_(4)`

Answer

To solve the problem step by step, we will analyze the reactions involving chlorine gas (Cl₂) and identify the products formed at each stage.

### Step 1: Reaction of Cl₂ with Cold Dilute NaOH
When chlorine gas (Cl₂) reacts with cold dilute sodium hydroxide (NaOH), it forms sodium hypochlorite (NaOCl) and sodium chloride (NaCl). The reaction can be represented as follows:

\[ \text{Cl}_2 + 2 \text{NaOH} \rightarrow \text{NaOCl} + \text{NaCl} + \text{H}_2\text{O} \]

- **Product P**: Sodium hypochlorite (NaOCl)

### Step 2: Reaction of Cl₂ with Hot Concentrated NaOH
When chlorine gas (Cl₂) reacts with hot concentrated sodium hydroxide (NaOH), it forms sodium chlorate (NaClO₃) and sodium chloride (NaCl). The reaction can be represented as follows:

\[ \text{Cl}_2 + 6 \text{NaOH} \rightarrow 2 \text{NaCl} + \text{NaClO}_3 + 3 \text{H}_2\text{O} \]

- **Product Q**: Sodium chlorate (NaClO₃)

### Step 3: Reaction of Cl₂ with SO₂ in the Presence of Charcoal
Chlorine gas (Cl₂) reacts with sulfur dioxide (SO₂) in the presence of charcoal to produce sulfuryl chloride (SO₂Cl₂). The reaction can be represented as follows:

\[ \text{Cl}_2 + \text{SO}_2 \rightarrow \text{SO}_2\text{Cl}_2 \]

- **Product R**: Sulfuryl chloride (SO₂Cl₂)

### Step 4: Reaction of R with White Phosphorus
Sulfuryl chloride (SO₂Cl₂) reacts with white phosphorus (P₄) to produce phosphorus pentachloride (PCl₅) and sulfur dioxide (SO₂). The reaction can be represented as follows:

\[ \text{SO}_2\text{Cl}_2 + \text{P}_4 \rightarrow 4 \text{PCl}_5 + 2 \text{SO}_2 \]

- **Product S**: Phosphorus pentachloride (PCl₅)

### Step 5: Hydrolysis of S
When phosphorus pentachloride (PCl₅) undergoes hydrolysis, it reacts with water to form phosphoric acid (H₃PO₄). The reaction can be represented as follows:

\[ \text{PCl}_5 + 4 \text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_4 + 5 \text{HCl} \]

- **Product T**: Phosphoric acid (H₃PO₄)

### Summary of Products
- **R**: Sulfuryl chloride (SO₂Cl₂)
- **S**: Phosphorus pentachloride (PCl₅)
- **T**: Phosphoric acid (H₃PO₄)

### Final Answer
R, S, and T are:
- R = SO₂Cl₂
- S = PCl₅
- T = H₃PO₄