The formation of the oxide ion- Og2- from oxygen atom requires first an exothermic and then an endothermic step as shown below-Og-e- O-g- -f H- -141 kJ mol-1O-g-e- O2-g- -f H- -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-

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XIII

Chemistry

Nature of Oxides

The formation...

Question

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.

No worries! We‘ve got your back. Try BYJU‘S free classes today!

oxygen is more electronegative

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

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.

Suggest Corrections

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,

Q. 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:

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:

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,

Join BYJU'S Learning Program