The formation of the oxide ion $$O_{\left( g \right)}^{2 - }$$ requires first an exothermic and then an endothermic step as shown below
$$\eqalign{
& {O_{\left( g \right)}} + {e^ - } = O_{\left( g \right)}^ - \,\,\Delta {H^ \circ } = - 142\,kJmo{l^{ - 1}} \cr
& {O^ - }\left( g \right) + {e^ - } = O_{\left( g \right)}^{2 - }\,\,\Delta {H^ \circ } = 844\,kJmo{l^{ - 1}} \cr} $$
This is because
A.
$${O^ - }$$ on will tend to resist the addition of another
electron
B.
Oxygen has high electron affinity
C.
Oxygen is more elecronegative
D.
$${O^ - }$$ ion has comparatively larger size than oxygen atom
Answer :
$${O^ - }$$ on will tend to resist the addition of another
electron
Solution :
$${O^ - }$$ ion exerts a force of repulsion on the incoming electron. The energy is required to overcome it.
Releted MCQ Question on Inorganic Chemistry >> Classification of Elements and Periodicity in Properties
Releted Question 1
The correct order of second ionisation potential of carbon, nitrogen, oxygen and fluorine is