`SO_(3)^(2-)+S^(***) overset("boil")to S S^(***)O_(3)^(2-),S S^(***)O_(3)^(2-)+2H^(+)toH_(2)SO_(3)+S^(***)`
The above reaction sequence proves:

Options

A. Two sulphur atoms of thiosulphate are not equivalent
B. Both are equivalent
C. Both of the above are correct
D. None of these

Explanation

To analyze the given reaction sequence and determine what it proves, let's break it down step by step.

### Step 1: Identify the Reactants and Products
The reaction sequence provided is:
1. \( SO_3^{2-} + S^{***} \) (boiling) → \( S S^{***} O_3^{2-} \)
2. \( S S^{***} O_3^{2-} + 2H^+ \) → \( H_2SO_3 + S^{***} \)

Here, we have:
- \( SO_3^{2-} \): Sulfite ion
- \( S^{***} \): A sulfur atom with a different oxidation state or form
- \( H_2SO_3 \): Sulfurous acid

### Step 2: Analyze the First Reaction
In the first reaction, \( SO_3^{2-} \) reacts with \( S^{***} \) to form \( S S^{***} O_3^{2-} \). This indicates that two different sulfur species are combining. The notation \( S^{***} \) suggests that this sulfur has a different oxidation state or is in a different form compared to the sulfur in \( SO_3^{2-} \).

### Step 3: Analyze the Second Reaction
In the second reaction, \( S S^{***} O_3^{2-} \) reacts with \( 2H^+ \) to form \( H_2SO_3 + S^{***} \). This shows that the compound formed in the first reaction can further react with protons (H+) to yield sulfurous acid and regenerate the \( S^{***} \) sulfur.

### Step 4: Determine the Nature of Sulfur
The key point here is that the two sulfurs (one from \( SO_3^{2-} \) and the other \( S^{***} \)) are not of the same nature. The reaction demonstrates that:
- The sulfur in \( SO_3^{2-} \) is in a higher oxidation state compared to the sulfur represented by \( S^{***} \).
- The formation of \( S S^{***} O_3^{2-} \) implies a bond between two different sulfur atoms, which can be attributed to their differing oxidation states.

### Step 5: Conclusion
The reaction sequence proves that the two sulfur atoms involved in the reaction are of different oxidation states or forms. This is evident from the formation of \( S S^{***} O_3^{2-} \) and the subsequent reaction with protons leading to sulfurous acid.

### Summary of Findings
- The two sulfurs are not of the same nature.
- The oxidation states of the sulfurs differ, which affects their reactivity and the products formed.