Chemistry - Redox Reactions Question with Solution | TestHub

(A) S in ${\mathrm{H}}_2\mathrm{S}$ is -2, its minimum, so it can only be oxidized: reducing agent only (Q). (B) N in $\mathrm{HN}{\mathrm{O}}_3$ is +5, its maximum, so it can only be reduced: oxidizing agent only (P). (C) N in $\mathrm{HN}{\mathrm{O}}_2$ is +3 (intermediate), so it can be oxidized to +5 or reduced to lower states, and on standing $3HN{\mathrm{O}}_2$ $\to$ $\mathrm{HN}{\mathrm{O}}_3$ + 2NO + ${\mathrm{H}}_2\mathrm{O}$, a disproportionation (P, Q, R). (D) $\mathrm{N}{\mathrm{a}}^{+}$ is not reduced in ordinary chemistry and S in $\mathrm{S}{\mathrm{O}}_4^{2-}$ is already +6 and very stable, so $\mathrm{N}{\mathrm{a}}_2\mathrm{S}{\mathrm{O}}_4$ is neither (S).