Question

Which of the following can be used both as an oxidant and a reductant?
A. ${\text{HN}}{{\text{O}}_{\text{2}}}$
B. ${\text{S}}{{\text{O}}_{\text{2}}}$
C. ${{\text{O}}_{\text{2}}}$
D. ${\text{CO}}$

Answer 1

Hint: To find a species that can be used both as an oxidant and reductant it must possess stable higher as well as lower oxidation state which can be attained when it undergoes a redox reaction. The species acts as an oxidant as well as a reductant in its intermediate oxidation state.

Complete step by step solution:
We are given four species ${\text{HN}}{{\text{O}}_{\text{2}}}$, ${\text{S}}{{\text{O}}_{\text{2}}}$, ${{\text{O}}_{\text{2}}}$ and ${\text{CO}}$.
a. Consider ${\text{HN}}{{\text{O}}_{\text{2}}}$: In ${\text{HN}}{{\text{O}}_{\text{2}}}$, the oxidation state of nitrogen atom is $ + {\text{3}}$.
The oxidation states exhibited by nitrogen are $ - {\text{3}}$ to $ + {\text{5}}$. In ${\text{HN}}{{\text{O}}_{\text{2}}}$, the oxidation state of nitrogen is $ + {\text{3}}$ which is an intermediate oxidation state. Thus, ${\text{HN}}{{\text{O}}_{\text{2}}}$ can be used both as an oxidant and a reductant.
For example consider the reactions,
${\text{HN}}{{\text{O}}_{\text{2}}}$ acts as an oxidant in the reaction: ${\text{2HN}}{{\text{O}}_{\text{2}}} + {{\text{H}}_2}{\text{S}} \to {\text{2NO}} + 2{{\text{H}}_2}{\text{O}} + {\text{S}}$
${\text{HN}}{{\text{O}}_{\text{2}}}$ acts as a reductant in the reaction: ${\text{HN}}{{\text{O}}_{\text{2}}} + {\text{B}}{{\text{r}}_2} + {{\text{H}}_2}{\text{O}} \to {\text{HN}}{{\text{O}}_3} + 2{\text{HBr}}$
b. Consider ${\text{S}}{{\text{O}}_{\text{2}}}$: In ${\text{S}}{{\text{O}}_{\text{2}}}$, the oxidation state of sulphur atom is $ + {\text{4}}$.
The oxidation states exhibited by sulphur are $ - {\text{2,0,}} + {\text{2,}} + {\text{4 and }} + {\text{6}}$. In ${\text{S}}{{\text{O}}_{\text{2}}}$, the oxidation state of sulphur is $ + {\text{4}}$ which is an intermediate oxidation state. Thus, ${\text{S}}{{\text{O}}_{\text{2}}}$ can be used both as an oxidant and a reductant.
For example consider the reactions,
${\text{S}}{{\text{O}}_{\text{2}}}$ acts as an oxidant in the reaction: ${\text{S}}{{\text{O}}_{\text{2}}} + {\text{C}}{{\text{l}}_{\text{2}}} \to {\text{S}}{{\text{O}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}$
${\text{S}}{{\text{O}}_{\text{2}}}$ acts as a reductant in the reaction: $2{\text{NaOH}} + {\text{S}}{{\text{O}}_2} \to {\text{N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_3} + {{\text{H}}_2}{\text{O}}$
c. Consider ${{\text{O}}_{\text{2}}}$: In ${{\text{O}}_{\text{2}}}$, the oxidation state of oxygen atom is ${\text{0}}$.
The oxidation states exhibited by oxygen are $ - {\text{2,0 and }} + {\text{2}}$. In ${\text{S}}{{\text{O}}_{\text{2}}}$, the oxidation state of oxygen is ${\text{0}}$ which is an intermediate oxidation state. Thus, ${{\text{O}}_{\text{2}}}$ can be used both as an oxidant and a reductant.
For example consider the reactions,
${{\text{O}}_{\text{2}}}$ acts as an oxidant in the reaction: ${{\text{O}}_{\text{2}}} + 2{\text{ electrons}} \to {\text{O}}_2^{2 - }$
${{\text{O}}_{\text{2}}}$ acts as a reductant in the reaction: $4{\text{Fe}} + {\text{3}}{{\text{O}}_2} \to 2{\text{F}}{{\text{e}}_2}{{\text{O}}_3}$
d. Consider ${\text{CO}}$: In ${\text{CO}}$, the oxidation state of the carbon atom is $ + 2$.
The oxidation states exhibited by carbon are $ - 4$ to $ + 4$. In ${\text{CO}}$, the oxidation state of carbon is $ + 2$ which is an intermediate oxidation state. Thus, ${\text{CO}}$ can be used both as an oxidant and a reductant.
For example consider the reactions,
${\text{CO}}$ acts as an oxidant in the reaction: ${\text{CO}} + {{\text{H}}_{\text{2}}} \to {\text{C}} + {{\text{H}}_2}{\text{O}}$
${\text{CO}}$ acts as a reductant in the reaction: $2{\text{CO}} + {{\text{O}}_2} \to 2{\text{C}}{{\text{O}}_2}$
Thus, we can conclude that ${\text{HN}}{{\text{O}}_{\text{2}}}$, ${\text{S}}{{\text{O}}_{\text{2}}}$, ${{\text{O}}_{\text{2}}}$ and ${\text{CO}}$ can be used both as an oxidant and a reductant.

Thus, the correct options are (A), (B), (C) and (D).

Note:
Oxidation number is also known as oxidation state. It is the total number of electrons that an atom can gain or lose to form a chemical bond. It can be positive, negative or zero.