The formation of the oxide ion ${{\rm{O}}^{2 - }}\left( g \right)$ requires first an exothermic and then an endothermic step as shown below: ${\rm{O}}\left( {\rm{g}} \right) + {e^ - } = {{\rm{O}}^ - }\left( {\rm{g}} \right);{\rm{\;}}\Delta H^\circ = - 142{\rm{\;kJmo}}{1^{ - 1}}$ , ${\rm{O}}{\left( {\rm{g}} \right)^ - } + {e^ - } = {{\rm{O}}^{2 - }}\left( {\rm{g}} \right);\Delta H^\circ = 844{\rm{\;kJmo}}{1^{ - 1}}$ This is because

Oxygen has high electron affinity

${\text{O}^ - }$ has comparatively larger size than oxygen atom

NEET UG CHEMISTRY MCQ: CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES

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The formation of the oxide ion

{\mathrm{O}^{2 - }}\left( g \right)

requires first an exothermic and then an endothermic step as shown below:

\mathrm{O}\left( \mathrm{g} \right) + {e^ - } = {\mathrm{O}^ - }\left( \mathrm{g} \right);\mathrm{\;}\Delta H^\circ = - 142\mathrm{\;kJmo}{1^{ - 1}}

\mathrm{O}{\left( \mathrm{g} \right)^ - } + {e^ - } = {\mathrm{O}^{2 - }}\left( \mathrm{g} \right);\Delta H^\circ = 844\mathrm{\;kJmo}{1^{ - 1}}

This is because

View answer and explanation

Correct option: C

{\text{O}^ - }

ion will tend to resist the addition of another electron

Option C (

{\text{O}^ - }

ion will tend to resist the addition of another electron) is correct because adding an electron to a negatively charged ion (

{\text{O}^ - }

) requires energy to overcome the electrostatic repulsion, making the process endothermic.