A white crystalline salt [A] reacts with dilute HCl to liberate a suffocatig gas [B] and also forms a yellow precipitate. The gas [B] turns potassium dichromatic acidified with dilute `H_(2)SO_(4)` to a green coloured solution [C]. The compound A, B and C are respectively

Options

A. `Na_(2)SO_(3), SO_(2), Cr_(2)(SO_(4))_(3)`
B. `Na_(2)S_(2)O_(3), SO_(2), Cr_(2)(SO_(4))_(3)`
C. `Na_(2)S, SO_(2), Cr_(2)(SO_(4))_(3)`
D. `Na_(2)SO_(4), SO_(2), Cr_(2)(SO_(4))_(3)`

Explanation

To solve the question step by step, we need to identify the compounds A, B, and C based on the given reactions and observations.

### Step 1: Identify Compound A
The question states that a white crystalline salt [A] reacts with dilute HCl to liberate a suffocating gas [B] and forms a yellow precipitate.

- A common white crystalline salt that fits this description is **sodium thiosulfate (Na2S2O3)**.

### Step 2: Reaction of Compound A with HCl
When sodium thiosulfate reacts with dilute hydrochloric acid (HCl), the following reaction occurs:

\[ \text{Na}_2\text{S}_2\text{O}_3 + 2 \text{HCl} \rightarrow 2 \text{NaCl} + \text{S} + \text{SO}_2 + \text{H}_2\text{O} \]

- This reaction produces sodium chloride (NaCl), sulfur (S), and sulfur dioxide (SO2), which is the suffocating gas [B].

### Step 3: Identify Compound B
From the reaction, we see that the suffocating gas [B] is **sulfur dioxide (SO2)**.

### Step 4: Formation of Yellow Precipitate
The yellow precipitate formed during the reaction is due to the formation of elemental sulfur (S). Thus, the yellow precipitate is sulfur.

### Step 5: Identify Compound C
The question states that the gas [B] (SO2) turns potassium dichromate (K2Cr2O7) acidified with dilute sulfuric acid (H2SO4) to a green colored solution [C].

- When sulfur dioxide reacts with potassium dichromate in an acidic medium, it reduces the dichromate ion (Cr2O7^2-) to chromium(III) ion (Cr^3+), which is responsible for the green color.

The reaction can be represented as follows:

\[ \text{K}_2\text{Cr}_2\text{O}_7 + 3 \text{SO}_2 + 8 \text{H}_2\text{SO}_4 \rightarrow 2 \text{Cr}_2\text{(SO}_4\text{)}_3 + 3 \text{K}_2\text{SO}_4 + 4 \text{H}_2\text{O} \]

- The green colored solution [C] is due to **chromium(III) sulfate (Cr2(SO4)3)**.

### Summary of Compounds
- Compound A: Sodium thiosulfate (Na2S2O3)
- Compound B: Sulfur dioxide (SO2)
- Compound C: Chromium(III) sulfate (Cr2(SO4)3)

### Final Answer
- A = Sodium thiosulfate (Na2S2O3)
- B = Sulfur dioxide (SO2)
- C = Chromium(III) sulfate (Cr2(SO4)3)

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