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

We are looking for the species formed when molecular oxygen (O₂) gains four electrons. Reduction is the gain of electrons. Let's consider the stepwise reduction of O₂: (A) Starting material: O₂ (molecular oxygen). The oxidation state of each oxygen atom is 0. (B) One-electron reduction:

O₂ + 1e^- → O₂^- (Superoxide ion). The average oxidation state of oxygen 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). The oxidation state of each oxygen atom is -2. The four-electron reduction process involves adding 4 electrons to the O₂ molecule. Since there are two oxygen atoms in O₂, each oxygen atom effectively gains two electrons (4 electrons / 2 atoms = 2 electrons/atom). This leads to the formation of two separate ions where oxygen has an oxidation state of -2, which is the oxide ion (O^2-). Therefore, the four-electron reduced form of O₂ consists of oxide ions.