The formation of the oxide ion, $O_{(g)}^{2 -}$ from oxygen atom requires first an exothermic and then an endothermic step as shown below:
$O_{(g)} + e^{-} \to O_{(g)}^{-}; Δ_{f} H^{\circ} = - 141 k J m o l^{- 1} O_{(g)}^{-} + e^{-} \to O_{(g)}^{2 -}; Δ_{f} H^{\circ} = + 780 k J m o l^{- 1}$
Thus, process of formation of $O^{2 -}$ in gas phase is unfavourable even though $O^{2 -}$ is isoelectronic with neon. It is due to the fact that,

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

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oxygen is more electronegative

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addition of electron in oxygen results in larger size of the ion.

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electron repulsion outweighs the stability gained by achieving noble gas configuration.

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Solution

The correct option is D
electron repulsion outweighs the stability gained by achieving noble gas configuration.

When the oxygen atom gains one electron, it makes it more stable, which is reflected by the first equation. Adding another electron is really difficult due to the repulsion faced by it. This is shown by the second equation.

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Q. The formation of oxide ion,

O^{2 -} (g)

, from oxygen atom requires first an exothermic and then an endothermic step as shown below:

O (g) + e^{-} \to O_{(g)}^{-};_{Δ f} H^{⊖} = - 141 {kJ mol}^{- 1}

O^{-} (g) + e^{-} \to O_{(g)}^{2 -};_{Δ f} H^{⊖} = + 780 {kJ mol}^{- 1}

Thus the process of formation of

O^{2 -}

in gas phase is unfavourable even thought

O^{2 -}

is isoelectric with neon. It is due to the fact that,

The formation of oxide ion 0^{2-} (g) from oxygen atom requires first an exothermic and then an endothermic step as shown below
$O (g) + e^{-} \to O^{-} (g); △ H^{⊝} = - 141 k j m o l^{- 1} O^{-} (g) + e^{-} \Rightarrow O^{2} (g); △ H^{⊝} = + 780 k j m o l^{- 1}$
Thus, process of formation of $O^{2 -}$ in gas phase is unfavourable even through $O^{2 -}$ is isoelectronic with neon. It is due to the fact that
(a) Oxygen si more electronegative
(b) addition of electron in oxygen results in larger size of the ion
(c) electron repulsiion outweighs the stability gained by achieving noble gas configuration
(d) $O^{-}$ ion has comparativly smaller size than oxygen atom

The formation of the oxide ion, $O_{(g)}^{2 -}$ from oxygen atom requires first an exothermic and then an endothermic step as shown below:
$O_{(g)} + e^{-} \to O_{(g)}^{-}; Δ_{f} H^{\circ} = - 141 k J m o l^{- 1} O_{(g)}^{-} + e^{-} \to O_{(g)}^{2 -}; Δ_{f} H^{\circ} = + 780 k J m o l^{- 1}$
Thus, process of formation of $O^{2 -}$ in gas phase is unfavourable even though $O^{2 -}$ is isoelectronic with neon. It is due to the fact that,