Question

Which of the following compound has reducing character?

• A. \(SO_2\)
• B. \(TeO_2\)
• C. \(SO_3\)
• D. \(TeO_3\)

Hint:

A species shows reducing character if its central atom can be oxidized to a higher oxidation state. In \(SO_2\), sulphur is in \(+4\) oxidation state and can be oxidized to \(+6\), so \(SO_2\) acts as a reducing agent.

Answer 1

The Correct Option is A

Step 1: Understand reducing character.
A reducing agent is a substance that can donate electrons and itself gets oxidized.
A compound shows reducing character when the central atom can increase its oxidation state further.

Step 2: Find the oxidation state of sulphur in \(SO_2\).
Let the oxidation state of sulphur be \(x\).
In \(SO_2\): \[ x+2(-2)=0 \] \[ x-4=0 \] \[ x=+4 \] Sulphur in \(SO_2\) can be further oxidized from \(+4\) to \(+6\), for example in \(SO_3\) or sulphate.
Therefore, \[ SO_2 \] can act as a reducing agent.

Step 3: Compare with \(SO_3\) and \(TeO_3\).
In \(SO_3\), sulphur is in the \(+6\) oxidation state: \[ S=+6 \] This is the highest common oxidation state of sulphur, so it cannot be easily oxidized further.
Similarly, in \(TeO_3\), tellurium is also in the \(+6\) oxidation state.
Thus, these compounds do not show reducing character here.

Step 4: Final conclusion.
Among the given compounds, the compound with reducing character is \[ \boxed{SO_2} \] Hence, the correct option is \[ \boxed{(1)} \]