The formation of the oxide ion O(g)2- requires first an exothermic and then an endothermic step as shown below: beginmatrixO(g)+e- arrow O(g)-;&Δ H=-142 kJ mol-1 O(g)+e arrow O(g)2-;&Δ H=844 KJ mol-1 endmatrix This is because:

Question

The formation of the oxide ion O(g)2- requires first an exothermic and then an endothermic step as shown below: beginmatrixO(g)+e- arrow O(g)-;&Δ H=-142 kJ mol-1 O(g)+e arrow O(g)2-;&Δ H=844 KJ mol-1 endmatrix This is because: (A) O- ion has comparatively larger size than oxygen atom (B) Oxygen has high electron affinity (C) O- ion will tend to resist the addition of another electron (D) Oxygen is more electronegative

Tardigrade Admin · Accepted Answer

O- acquires negative charge and thus shows repulsion for another electron

Q. The formation of the oxide ion $O_{(g)}^{2 -}$ requires first an exothermic and then an endothermic step as shown below:
$O_{(g) + e^{-} \to O_{(g)}^{-};} O (g) + e \to O_{(g)}^{2 -}; Δ H = - 142 k J m o l^{- 1} Δ H = 844 K J m o l^{- 1}$
This is because:

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

Oxygen has high electron affinity

$O^{-}$ ion will tend to resist the addition of another electron

Oxygen is more electronegative

$O^{-}$ acquires negative charge and thus shows repulsion for another electron

Q. 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)−​;​O(g)+e→O(g)2−​;​ΔH=−142kJmol−1ΔH=844KJmol−1​ This is because:

O− ion has comparatively larger size than oxygen atom