The formation of the oxide ion `O_(g)^(2-)` requires first an exothermic and then an endothermic step as shown below:
`O_(g)+e^(-) rarr O_(g)^(-)`, `DeltaH=-142 kJ mol^(-1)`
`O(g)+e rarr O_(g)^(2-)`, `DeltaH=844kJ mol^(-1)`
This is because:

The formation of the oxide ion O_((g))^(2-) requires first an exothermic and then an endothermic step as shown below. O_((g)) +e^(-) = O_((g))^(-) DeltaH^(@) =- 142 kJ mol^(-1) O_((g))^(-) + e^(-) = O_((g))^(2-) DeltaH^(@) = 844 kJ mol^(-1) This is because of :

The formation of the oxide ion O^(2-) (g) requires first an exothermic and then an endothermic step as shown below, O(g)+ e^(-) = O^(-)(g), DeltaH^(@)= -142kJmol^(-1) O^(-)(g) +e^(-) =O^(2-)(g), DeltaH^(@)= 844 kJmol^(-1) This is because : oxygen is more electronegative, oxygen has high electron affinity, O^(-) ion will tend to resist the addition of another electron , O^(-) ion has comparatively larger size than oxygen atom

The formation of the oxide ion O^(2-) (g) requires first an exothermic and then an endothermic step as shown below : O (g) + e^(-)= O^(-) (g), DeltaH^(@) = - 142 kJ moI^(-1) O(g)+e^(-)toO^(2-)(g),DeltaH^(@)=844kJmol^(-1) This is because