Back to Directory
NEET CHEMISTRYMedium

The formation of the oxide ion O²⁻(g) from oxygen atom requires first an exothermic and then an endothermic step as shown below: O(g) + e⁻ → O⁻(g); \Delta H° = -141 kJ mol⁻¹ O⁻(g) + e⁻ → O²⁻(g); \Delta H° = +780 kJ mol⁻¹ Thus, the process of formation of O²⁻ in gas phase is unfavourable even though O²⁻ is isoelectronic with neon. It is due to the fact that:

A

Electron repulsion outweighs the stability gained by achieving noble gas configuration

B

O⁻ ion has comparatively smaller size than oxygen atom

C

Oxygen is more electronegative

D

Addition of electron in oxygen result in large size of the ion

Step-by-Step Solution

The formation of the divalent oxide ion (O2−O^{2-}) involves two steps:

  1. First Electron Gain: O+e−→O−O + e^- \rightarrow O^-. This is exothermic (ΔH=−141\Delta H = -141 kJ/mol) because the neutral oxygen atom attracts the electron.
  2. Second Electron Gain: O−+e−→O2−O^- + e^- \rightarrow O^{2-}. This step is strongly endothermic (ΔH=+780\Delta H = +780 kJ/mol). Even though O2−O^{2-} achieves the stable noble gas configuration of Neon (2s22p62s^2 2p^6), energy is required to overcome the strong electrostatic repulsion between the negatively charged O−O^- ion and the incoming negatively charged electron.

This repulsive force dominates the energy landscape, making the overall process in the gas phase energetically unfavourable (endothermic), despite the stability of the octet.

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started