Illustrate the formation of oxide ion from oxygen atom.

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Solution

Formation of oxide ion from oxygen atom:
Oxygen atom $(O)$ has atomic number $8$ . Thus, electronic configuration of oxygen atom is $(2, 6)$ .
Oxygen atom loses two electrons from the outermost shell.
On losing two electrons, a negatively charged oxide ion $(O^{2 -})$ having two negative charges is formed.
Thus, oxygen atom $(O)$ loses two electrons and forms oxide ion $(O^{2 -})$ .

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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

Show the formation of oxide ion.(Atomic number of oxygen = 8)

Illustrate the formation of magnesium ion from magnesium atom.

Q. Which oxide has maximum negative heat of formation per oxygen atom?

Q. Assertion (A): In the formation of oxide ion,

O^{2 -} (g)

, first an exothermic reaction and then an endothermic reaction take place.

Reason (R):

(O^{-})

ion has comparatively large size than oxygen atom.

The correct answer is:

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