Question

What is the oxidation state of \(S\) in the sulphur containing product of the following reaction?
\[ SO_3^{2-}(aq) + Br_2(l) + H_2O \longrightarrow \]

• A. \(+6\)
• B. \(+4\)
• C. \(+2.5\)
• D. \(+2\)

Hint:

In oxyanions, calculate oxidation state using the total charge of the ion. Sulphite \((SO_3^{2-})\) has sulphur in \(+4\) state, while sulphate \((SO_4^{2-})\) has sulphur in \(+6\) state.

Answer 1

The Correct Option is A

Step 1: Identify the reactants involved in the reaction.
The given sulphur containing species is sulphite ion:
\[ SO_3^{2-} \] and bromine \((Br_2)\) is present along with water.
Bromine acts as an oxidizing agent and oxidizes sulphite ion into sulphate ion.
Thus, the reaction is:
\[ SO_3^{2-} \longrightarrow SO_4^{2-} \]

Step 2: Calculate the oxidation state of sulphur in \(SO_3^{2-}\).
Let the oxidation state of sulphur be \(x\).
Since oxygen has oxidation state \(-2\), we write:
\[ x + 3(-2) = -2 \] \[ x - 6 = -2 \] \[ x = +4 \] Therefore, sulphur has oxidation state \(+4\) in sulphite ion.

Step 3: Determine the sulphur containing product.
Sulphite ion gets oxidized to sulphate ion:
\[ SO_4^{2-} \] Now calculate the oxidation state of sulphur in sulphate ion.
Let oxidation state of sulphur be \(x\):
\[ x + 4(-2) = -2 \] \[ x - 8 = -2 \] \[ x = +6 \] Thus, sulphur has oxidation state \(+6\) in the product.

Step 4: Understand the oxidation process.
The oxidation state of sulphur increases from:
\[ +4 \longrightarrow +6 \] An increase in oxidation state indicates oxidation.
Therefore, bromine oxidizes sulphite ion to sulphate ion.

Step 5: Match with the given options.
The oxidation state of sulphur in the sulphur containing product is:
\[ +6 \] which corresponds to option (1).

Step 6: Final conclusion.
Hence, the correct answer is:
\[ \boxed{+6} \]