The formation of the oxide ion- O2-g- from oxygen atom requires first an exothermic and then an endothermic step as shown below- Og-e- O-g- -fH-141kJ mol-1 O-g-e- O2-g- -fH-780 kJ mol-1Thus- process of formation of O2- in gas phase is unfavourable even though O2- is isoelectronic with neon- It is due to the fact that-

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Electronegativity

The formation...

Question

The formation of the oxide ion, $O^{2 -} (g)$ , from oxygen atom requires first an exothermic and then an endothermic step as shown below:
$O (g) + e^{-} \to O^{-} (g); Δ_{f} H^{⊖} = - 141 k J m o l^{- 1}$
$O^{-} (g) + e^{-} \to O^{2 -} (g); Δ_{f} H^{⊖} = + 780 k J m o l^{- 1}$
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:

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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O^{-}

ion has comparatively smaller size than oxygen atom

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Solution

The correct option is B electron repulsion outweighs the stability gained by achieving noble gas configuration
The formation of the oxide ion, $O^{2 -} (g)$ , from oxygen atom requires first an exothermic and then an endothermic step as shown below:
$O (g) + e^{-} \to O^{-} (g); Δ_{f} H^{⊖} = - 141 k J m o l^{- 1}$
$O^{-} (g) + e^{-} \to O^{2 -} (g); Δ_{f} H^{⊖} = + 780 k J m o l^{- 1}$
Thus, the process of formation of $O^{2 -}$ in the gas phase is unfavorable even though $O^{2 -}$ is isoelectronic with neon. It is due to the fact that electron repulsion outweighs the stability gained by achieving noble gas configuration.

When an electron is added to $O^{-}$ anion, there is strong electrostatic repulsion between the two negative charges. Due to this, the second electron gain enthalpy of oxygen is positive.

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Similar questions

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,

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

Q. 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 H^{\circ} = - 141 k J m o l^{- 1}

O_{(g)}^{-} + e^{-} \to O_{(g)}^{2 -}; Δ 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:

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