Knowing the electron gain enthalpy values for $O \to O^{-}$ and $O \to O^{2 -}$ as $- 141$ and $702 k J m o l^{- 1}$ respectively, how can you account for the formation of a large number of oxides having $O^{2 -}$ species and not $O^{-}$ ?

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Solution

Oxygen has negative first electron gain enthalpy and positive second ionization enthalpy . Based on this, we should expect the formation of oxides with $O^{-}$ species and not with $O^{2 -}$ species. However, in reality, most oxides have $O^{2 -}$ ion due to following reasons:
(i) $O^{2 -}$ has stable noble gas electronic configuration.
(ii) The lattice energy of $O^{2 -}$ is much higher than the lattice energy of $O^{-}$ .
This more than compensates the higher energy required for removal of second electron from $O^{-}$ to form $O^{2 -}$ and makes $O^{2 -}$ ion stable.

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Similar questions

Knowing the electron gain enthalpy values for $O \to O^{-}$ and $O \to O^{- 2}$ as -141 and 702 kJ $m o l^{- 1}$ respectively, how can you account for the formation of a large number of oxides having $O^{2 -}$ species and not $O^{-}$ ?

Knowing the electron gain enthalpy values for O → O⁻ and O → O²⁻ as −141

and 702 kJ mol⁻¹ respectively, how can you account for the formation of a

large number of oxides having O²⁻ species and not O⁻?

(Hint: Consider lattice energy factor in the formation of compounds).

Q. Following are the values of the Electron gain enthalpy (in kJ mol

^{- 1}

) of the formation of

O^{-}

and

O^{2 -}

from

O

Q. Following are the values of the electron affinities of (in

k J m o l^{- 1}

) the formation of

O^{-}

and

O^{2 -}

from

O

Q. The formation of the oxide ion

O^{2 -}

involves first an exothermic step followed by an endothermic step as shown below:

O + e^{-} \to O^{-}; Δ H = - 142 k J m o l^{- 1}

O^{-} + e^{-} \to O^{2 -}; Δ H = 844 k J m o l^{- 1}

This is because :

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Knowing the electron gain enthalpy values for O→O− and O→O2− as −141 and 702 kJ mol−1 respectively, how can you account for the formation of a large number of oxides having O2− species and not O−?

Knowing the electron gain enthalpy values for $O \to O^{-}$ and $O \to O^{2 -}$ as $- 141$ and $702 k J m o l^{- 1}$ respectively, how can you account for the formation of a large number of oxides having $O^{2 -}$ species and not $O^{-}$ ?