Question

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); ∆fHϴ = –141 kJ mol–1
O⁻ (g) + e⁻ → O²⁻ (g); ∆fHϴ = +780 kJ mol–1
Thus process of formation of O²⁻ in gas phase is
unfavourable even though O²⁻ is isoelectronic with neon. It
is due to the fact that [2015 RS]
(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 results in larger size of
the ion.

Answer 1

Explanation:

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); ∆fHϴ = –141 kJ mol–1

O⁻ (g) + e⁻ → O²⁻ (g); ∆fHϴ = +780 kJ mol–1

Thus process of formation of O²⁻ in gas phase is

unfavourable even though O²⁻ is isoelectronic with neon.

Answer 2

The process of formation of O²⁻ in the gas phase is unfavorable 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.

-When 1 electron is added to the O⁻ anion, there comes into play a strong electrostatic repulsion between the two electrons.

- this electrostatic repulsion is very much strong as compared to the stability achieved due to noble gas configuration.

- Due to this reason, the second electron gain enthalpy of oxygen is positive after the first being negative.

- Thus despite being isoelectronic with neon, the process of formation of O²⁻ in the gas phase is unfavorable