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

Options

A. `Na_(2)SO_(3),SO_(2),Cr_(2)(SO_(4))_(3)`
B. `Na_(2)S_(3)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 substances A, B, and C based on the clues provided.

### Step 1: Identify the white crystalline salt (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.

Upon considering common white crystalline salts that react with dilute HCl, we can identify **sodium thiosulfate (Na2S2O3)** as the salt.

### Step 2: Reaction of A with dilute HCl
When sodium thiosulfate (Na2S2O3) reacts with dilute hydrochloric acid (HCl), the reaction can be represented as follows:

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

From this reaction, we can see that:
- The suffocating gas (B) produced is **sulfur dioxide (SO2)**.
- A yellow precipitate of sulfur (S) is also formed.

### Step 3: Identify the suffocating gas (B)
From the reaction, we have identified that the suffocating gas (B) is **sulfur dioxide (SO2)**.

### Step 4: Identify the green colored solution (C)
The problem states that the gas (B) turns potassium dichromate (K2Cr2O7) acidified with dilute sulfuric acid (H2SO4) into a green colored solution (C).

The reaction of sulfur dioxide with potassium dichromate in acidic medium can be represented as follows:

\[ 3 \text{SO}_2 + \text{K}_2\text{Cr}_2\text{O}_7 + 4 \text{H}_2\text{SO}_4 \rightarrow \text{Cr}_2(\text{SO}_4)_3 + K_2\text{SO}_4 + 4 \text{H}_2\text{O} \]

In this reaction, the potassium dichromate is reduced, and the chromium changes from +6 oxidation state (orange) to +3 oxidation state (green), resulting in a green colored solution (C).

### Conclusion
Based on the above analysis, we can summarize:
- A (white crystalline salt) = **Sodium thiosulfate (Na2S2O3)**
- B (suffocating gas) = **Sulfur dioxide (SO2)**
- C (green colored solution) = **Chromium(III) sulfate (Cr2(SO4)3)**

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