The formation of the oxide ion,$O^{2-}_{(g)}$,from oxygen atom requires first an exothermic and then an endothermic step as shown below:
$O_{(g)} + e^- \to O^{-}_{(g)} ; \Delta_f H^{\Theta} = -141 \ kJ \ mol^{-1}$
$O^{-}_{(g)} + e^- \to O^{2-}_{(g)} ; \Delta_f H^{\Theta} = +780 \ kJ \ mol^{-1}$
Thus,the process of formation of $O^{2-}$ in gas phase is unfavourable even though $O^{2-}$ is isoelectronic with neon. It is due to the fact that,

  • A
    Oxygen is more electronegative
  • B
    Addition of electron in oxygen results in larger size of the ion
  • C
    Electron repulsion outweighs the stability gained by achieving noble gas configuration
  • D
    $O^{-}$ ion has comparatively smaller size than oxygen atom

Explore More

Similar Questions

In which of the following is the electron gain enthalpy of elements correctly arranged?

The second electron gain enthalpy of an element:

Identify the set of elements in which they are arranged in the increasing order of electron gain enthalpies.

Knowing the electron gain enthalpy values for $O \rightarrow O^{-}$ and $O \rightarrow O^{2-}$ as $-141 \ kJ \ mol^{-1}$ and $702 \ kJ \ mol^{-1}$ respectively,how can you account for the formation of a large number of oxides having $O^{2-}$ species and not $O^{-}$?
(Hint: Consider lattice energy factor in the formation of compounds).

Which of the following electronic configurations will have the lowest electron affinity?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo