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

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Solution

Answer: (c) electron repulsion outweighs the stability gained by achieving the noble gas configuration

Although $O^{2 -}$ has noble gas configuration isoelectronic with neon its formation is unfavourable due to the strong electronic repulsion between the negatively charged $O^{-}$ ion and the second electron being added.
Hence, the electron repulsion outweighs the stability gained by achieving a noble gas configuration.

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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 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,