The formation of oxide ion O^{2-} from oxygen atom requires an exothermic reaction, followed by an endothermic step as shown below. The process of formation of O^{2-} in gas phase is unfavorable ( \\DeltaH^{\⊖}=+v e), even though it has stable configuration of the nearest noble gas, Neon. This is because ______ 0_{(g)}+e^{-} \\to0_{(g)}^{-} ; \\DeltaH^{\⊖}=-141 { kJ} . {mol}^{-1} 0_{(g)}^{-}+e^{-} \\to0_{(g)}^{2-} ; \\DeltaH^{\⊖}=+760 {kJ} . {mol}^{-1}

Correct Answer is : Electron repulsion in oxide ion is more which over comes the stability achieved by noble gas configuration

O^{2-} has larger size than that of Neon

Oxygen is more electronegative than Neon

O^{-}has larger size than oxygen atom

Electron repulsion in oxide ion is more which over comes the stability achieved by noble gas configuration

AP EAPCET 25-Aug-2021 Shift 2 Paper

Section: Chemistry

Question : 125 of 160

Marks: +1, -0

The formation of oxide ion O2− from oxygen atom requires an exothermic reaction, followed by an endothermic step as shown below. The process of formation of O2− in gas phase is unfavorable ( ΔH⊖=+ve), even though it has stable configuration of the nearest noble gas, Neon. This is because ______
0(g)+e−→0(g)−; ΔH⊖=−141 kJ. mol−1
0(g)−+e−→0(g)2−; ΔH⊖=+760 kJ. mol−1

O2− has larger size than that of Neon
Oxygen is more electronegative than Neon
O−has larger size than oxygen atom
Electron repulsion in oxide ion is more which over comes the stability achieved by noble gas configuration

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