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Class 12 NEET Chemistry P Block Elements Questions
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Question : 29 of 101
Marks:
+1
,
-0
The formation of the oxide ion,
O
2
−
\mathrm{O}^{2-}
O
2
−
from oxygen atom requires first an exthermic and then an endothermic step as shown below:
O
g
)
+
e
−
→
O
−
(
g
)
;
Δ
f
H
∘
=
−
141
kJ
mol
−
1
\mathrm{O}_{\mathrm{g})}+e^{-} \rightarrow \mathrm{O}^{-}{ }_{( \mathrm{g})} ; \Delta_{f} H^{\circ}=-141 \text{kJ} \text{mol}^{-1}
O
g
)
+
e
−
→
O
−
(
g
)
;
Δ
f
H
∘
=
−
141
kJ
mol
−
1
O
−
(
g
)
+
e
−
→
O
2
−
(
g
)
;
Δ
f
H
∘
=
+
780
kJ
mol
−
1
\mathrm{O}^{-}{ }_{( \mathrm{g})}+e^{-} \rightarrow \mathrm{O}^{2-}{ }_{( \mathrm{g})} ; \Delta_{f} H^{\circ}=+780 \text{kJ} \text{mol}^{-1}
O
−
(
g
)
+
e
−
→
O
2
−
(
g
)
;
Δ
f
H
∘
=
+
780
kJ
mol
−
1
Thus, process of formation of
O
2
−
\mathrm{O}^{2-}
O
2
−
is isoelectronic with neon. It is due to the fact that,
[NEET 2015]
O
−
\mathrm{O}^{-}
O
−
ion has comparatively smaller size than oxygen atom
oxygen is more electronegative
addition of electron in oxygen results in larger size of the ion
electron repulsion outweighs the stability gained by achieving noble gas configuration.
Validate
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