WBJEE 2025 Physics and Chemistry Papers

Let's break down how molecular oxygen $(\mathrm{O}_2)$ is reduced step-by-step:Adding one electron to $\mathrm{O}_2$ gives the superoxide ion:$\mathrm{O}_2 + e^{-} \longrightarrow \mathrm{O}_2^{\bullet-}$(This is a one-electron reduction.)Adding a second electron gives the peroxide ion:$\mathrm{O}_2 + 2e^{-} \longrightarrow \mathrm{O}_2^{2-}$ (This is a two-electron reduction.)When $\mathrm{O}_2$ undergoes a four-electron reduction, the process is:$\mathrm{O}_2 + 4e^{-} \longrightarrow 2 \mathrm{O}^{2-}$Here, each oxygen atom ends up as an oxide ion ( $\mathrm{O}^{2-}$ ).So, the four-electron reduced form of $\mathrm{O}_2$ is oxide, which corresponds to Option C .