The formation of the oxide ion $\mathrm{O}^{2-}(\mathrm{g})$ requires first an exothermic and then an endothermic step as shown below $\mathrm{O}(\mathrm{g})+\mathrm{e}^{-} \rightarrow \mathrm{O}^{-}(\mathrm{g})$ $\Delta \mathrm{H}^{\circ}=-142 \mathrm{kJmol}^{-1}$ $\mathrm{O}^{-}(\mathrm{g})+\mathrm{e}^{-} \rightarrow \mathrm{O}^{2-}(\mathrm{g})$ $\Delta \mathrm{H}^{\circ}=844 \mathrm{kJmol}^{-1}$

(C) $\mathrm{O}^{-}$ ion will tend to resist the addition of another electron

(A) Oxygen is more electronegative

(B) $\mathrm{O}^{-}$ ion has comparatively larger size than oxygen atom

(D) Oxygen has high electron affinity

JEE Main · 2004 · 24 Apr 2004 · Offline · Classification of Elements and Periodicity

The formation of the oxide ion

O^{2 -} (g)

requires first an exothermic and then an endothermic step as shown below

O (g) + e^{-} \to O^{-} (g)

Δ H^{\circ} = - 142 kJmol^{- 1}

O^{-} (g) + e^{-} \to O^{2 -} (g)

Δ H^{\circ} = 844 kJmol^{- 1}

Answer: (C)

Adding the first electron to a neutral oxygen atom gives out energy because the nucleus draws the incoming electron in, hence the exothermic

- 142 kJ mol^{- 1}

. The second electron, though, has to come up against an already negative

O X^{-}

ion and is pushed away by the charge already there, so energy must now be put in, giving the endothermic

+ 844 kJ mol^{- 1}

. The governing point is that

O X^{-}

resists accepting a further electron because of this electron-electron repulsion.

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option