- A$KMnO_4,\, O_3,\, SO_3$
- B$HClO_4,\, HNO_2,\, H_2O_2$
- ✓$HNO_2,\, SO_2,\, H_2O_2$
- D$HNO_3,\, SO_2,\, H_2SO_4$
1.1n KMnO $_{4}, \mathrm{Mn}$ is in +7 oxidation state so it cannot acts as reducing agent.
$\mathrm{O}_{3}$ can only act as an oxidising agent due to its unstable nature.
In $\mathrm{SO}_{3}, \mathrm{S}$ is in +6 oxidation state so it cannot acts as reducing agent.
2.In $\mathrm{HClO}_{4}, \mathrm{Cl}$ is in +7 oxidation state so it cannot acts as reducing agent.
HN $\mathrm{O}_{2}$ can act both oxidizing as well as reducing agent because N is in +3 oxidation state.
In $\mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{O}$ is in -1 oxidation state so can act both oxidizing as well as reducing agent.
3.In HN O can act both oxidizing as well as reducing agent because N is in +3 oxidation state.
In $\mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{O}$ is in -1 oxidation state so can act both oxidizing as well as reducing agent.
ln $\mathrm{SO}_{2}, \mathrm{S}$ is in +4 oxidation state so it can act as both oxidizing agent as well as reducing agent.
4.HN O $_{2}$ can act both oxidizing as well as reducing agent because $\mathrm{N}$ is in +3 oxidation state.
In $\mathrm{SO}_{2}, \mathrm{S}$ is in +4 oxidation state so it can act as both oxidizing agent as well as reducing agent.
But $\mathrm{H}_{2} \mathrm{SO}_{4}$ cannot acts as reducing agent because S is present in its higher oxidation state.
So $(\mathrm{C})$ is a correct answer.
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