Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • The formation of the oxide ion, O^{2-}(g), from oxygen atom requires first an exothermic and then an endothermic step as shown below: Thus process of formation of O^{2-} in gas phase is unfavourable even though O^{2-} is isoelectronic with neon.

The formation of the oxide ion, O^{2-}(g), from oxygen atom requires first an exothermic and then an endothermic step as shown below: Thus 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,

O(g)+e^{-}\rightarrow O^{-}(g);\Delta H^{\ominus }= - 141 kJ\; mol^{-1}

O^{-}(g)+e^{-}\rightarrow O^{2-}(g);\Delta H^{\ominus }= +780\; kJ\; mol^{-1}

(i) oxygen is more electronegative.

(ii) addition of electron in oxygen results in larger size of the ion.

(iii) electron repulsion outweighs the stability gained by achieving noble gas configuration.

(iv) O– ion has comparatively smaller size than oxygen atom.

Answers (1)

The answer is the option (iii) electron repulsion outweighs the stability gained by achieving noble gas configuration.

Explanation:  Energy is released whenever an electron is added to the O atom to form O-  ion. Thus the first electron gain enthalpy of O is negative. Contrary to this whenever an electron is added to O- ion to form O2- ion, energy has to be given out in order to overcome the very strong electronic repulsions, and that is why the second gain enthalpy of O is positive.

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question