Question

What is the four-electron reduced form of O$_2$?

• A. Superoxide
• B. Peroxide
• C. Oxide
• D. Ozone

Hint:

Oxygen reduction ladder: \[ \text{O}_2 \rightarrow \text{O}_2^- \rightarrow \text{O}_2^{2-} \rightarrow \text{O}^{2-} \] More electrons $\Rightarrow$ deeper reduction.

Answer 1

The Correct Option is C

Concept: Reduction of oxygen occurs stepwise: \begin{itemize} \item 1 electron reduction $\rightarrow$ Superoxide (O$_2^-$) \item 2 electron reduction $\rightarrow$ Peroxide (O$_2^{2-}$) \item 4 electron reduction $\rightarrow$ Oxide (O$^{2-}$) \end{itemize} Explanation: Complete 4-electron reduction breaks O–O bond and forms oxide ions: \[ \text{O}_2 + 4e^- \rightarrow 2\text{O}^{2-} \] Conclusion: Four-electron reduced form = oxide.