Question

${\text{S}}{{\text{O}}_2}$ can act as both oxidant and reductant.
(A) True
(B) False

Answer 1

First determine the oxidation number of sulphur in ${\text{S}}{{\text{O}}_2}$ molecule. Then obtain general oxidation states of sulphur. Then determine if the sulphur atom of ${\text{S}}{{\text{O}}_2}$ molecule can increase/decrease its oxidation number.

Complete answer:
Let X be the oxidation number of Sulphur in ${\text{S}}{{\text{O}}_2}$ molecule. The oxidation number of oxygen is -2. In a neutral molecule, the sum of the oxidation number of all the elements is zero.
X + 2(- 2) = 0
X - 4 = 0
X = + 4
Hence the oxidation number of Sulphur in ${\text{S}}{{\text{O}}_2}$ molecules is +4. The maximum oxidation number of sulphur is +6 as in ${\text{S}}{{\text{O}}_3}$ molecule or in ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ molecule. Thus, the sulphur atom in ${\text{S}}{{\text{O}}_2}$ can be oxidized. So ${\text{S}}{{\text{O}}_2}$ acts as reductant (reducing agent ) and reduces other species.
Sulphur forms compounds with various oxidation states such as -2,0,+2,+4 and +6.
Thus, sulphur having +4 oxidation state in ${\text{S}}{{\text{O}}_2}$ molecules can be reduced to a lower oxidation state. Hence, ${\text{S}}{{\text{O}}_2}$acts as an oxidant and can oxidize other substances.
Thus, ${\text{S}}{{\text{O}}_2}$ can act as both oxidant and reductant.

The given statement is true and the option (A) is the correct option.

Note: Increase in the oxidation number is oxidation. Decrease in the oxidation number is reduction. An oxidant (or an oxidizing agent) oxidizes another substance and itself is reduced. A reductant (or reducing agent) reduces other substances and itself is oxidized. In a redox reaction, one species is oxidized whereas other species are reduced.