Question

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

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

Hint:

Using a simple frame or just bolding for the box Key Points: O$_2$ (Oxygen): Oxidation State = 0 O$_2^-$ (Superoxide): Oxidation State = -1/2 (average) O$_2^{2-$ (Peroxide): Oxidation State = -1 O$^{2-$ (Oxide): Oxidation State = -2 Four-electron reduction of O$_2$ (2 atoms) means each atom gains 2 electrons, forming 2 O$^{2-$ ions. Ozone (O$_3$) is an allotrope, not a simple reduction product.

Answer 1

The Correct Option is C

The goal is to identify the species resulting from molecular oxygen (O₂) gaining four electrons. Reduction is defined as electron gain. The stepwise reduction of O₂ is outlined as follows: (A) Initial state: O₂ (molecular oxygen), where each oxygen atom has an oxidation state of 0. (B) One-electron reduction: O₂ + 1e^- → O₂^- (Superoxide ion). The average oxygen oxidation state is -1/2. (C) Two-electron reduction: O₂ + 2e^- → O₂^2- (Peroxide ion). The oxidation state of each oxygen atom is -1. (D) Four-electron reduction: O₂ + 4e^- → 2O^2- (Oxide ions). Each oxygen atom has an oxidation state of -2. In the four-electron reduction, 4 electrons are added to O₂. Since O₂ has two oxygen atoms, each effectively gains two electrons (4 electrons / 2 atoms = 2 electrons/atom). This yields two oxide ions (O^2-), with oxygen at an oxidation state of -2. Thus, the four-electron reduction of O₂ produces oxide ions.