Iodine oxidises `S_(2)O_(3)^(2-)` ion to 'X', change in oxidation state of sulphur 

Options

A. `+3`
B. `+2`
C. `+0.5`
D. `+4`

Answer

To solve the problem of determining the change in oxidation state of sulfur when iodine oxidizes the thiosulfate ion \((S_2O_3^{2-})\) to a product \(X\), we can follow these steps:

### Step 1: Identify the oxidation state of sulfur in \((S_2O_3^{2-})\)

1. The formula for thiosulfate is \((S_2O_3^{2-})\).
2. The total charge of the ion is \(-2\).
3. Oxygen typically has an oxidation state of \(-2\). In \((S_2O_3^{2-})\), there are 3 oxygen atoms:
\[ 3 \times (-2) = -6 \]
4. Let the oxidation state of sulfur be \(x\). Since there are 2 sulfur atoms, we have:
\[ 2x + (-6) = -2 \]
5. Rearranging gives:
\[ 2x - 6 = -2 \implies 2x = 4 \implies x = +2 \]
6. Therefore, the oxidation state of sulfur in \((S_2O_3^{2-})\) is \(+2\).

### Step 2: Identify the product formed after oxidation

1. The product formed when iodine oxidizes thiosulfate is tetrathionate, \((S_4O_6^{2-})\).
2. The formula for tetrathionate is \((S_4O_6^{2-})\).
3. Again, the total charge of the ion is \(-2\).
4. There are 6 oxygen atoms, so:
\[ 6 \times (-2) = -12 \]
5. Let the oxidation state of sulfur in \((S_4O_6^{2-})\) be \(y\). With 4 sulfur atoms, we have:
\[ 4y + (-12) = -2 \]
6. Rearranging gives:
\[ 4y - 12 = -2 \implies 4y = 10 \implies y = +2.5 \]
7. Therefore, the oxidation state of sulfur in \((S_4O_6^{2-})\) is \(+2.5\).

### Step 3: Calculate the change in oxidation state

1. The change in oxidation state of sulfur can be calculated as:
\[ \text{Change} = \text{Final oxidation state} - \text{Initial oxidation state} = +2.5 - +2 = +0.5 \]

### Conclusion

The change in oxidation state of sulfur when iodine oxidizes the thiosulfate ion \((S_2O_3^{2-})\) to tetrathionate \((S_4O_6^{2-})\) is \(+0.5\).

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