The formation of the oxide ion, O²⁻(g), from
oxygen atom requires first an exothermic and then
an endothermic step as shown below :
O(g) + e⁻ ⟶ O⁻ (g);Δ HΘ = - 141 kJ mol⁻¹
O⁻(g) + e⁻ ⟶ O²⁻ (g);Δ HΘ = + 780 kJ mol⁻¹
Thus process of formation of O²⁻ in gas phase is
unfavourable even though O²⁻ is isoelectronic
with neon. It is due to the fact that,
(a) oxygen is more electronegative.
(b) addition of electron in oxygen results in larger size of the ion.
(c) electron repulsion outweighs the stability
gained by achieving noble gas configuration.
(d) O⁻ion has comparatively smaller size than oxygen atom.
Answers
Explanation:
CHEMISTRY
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
−
→O
−
(g);Δ
f
H
⊖
=−141kJmol
−1
O
−
(g)+e
−
→O
2−
(g);Δ
f
H
⊖
=+780kJmol
−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:
October 15, 2019avatar
Omkar Shelke
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VIDEO EXPLANATION
ANSWER
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
−
→O
−
(g);Δ
f
H
⊖
=−141 kJmol
−1
O
−
(g)+e
−
→O
2−
(g);Δ
f
H
⊖
=+780 kJmol
−1
Thus, the process of formation of O
2−
in the gas phase is unfavorable even though O
2−
is isoelectronic with neon. It is due to the fact that electron repulsion outweighs the stability gained by achieving noble gas configuration.
When an electron is added to O
−
anion, there is strong electrostatic repulsion between the two negative charges. Due to this, the second electron gain enthalpy of oxygen is positive.
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Answer:
(c)
There is a lot of repulsion when similar charges approach each other as O–(g), and electron are both negatively charged. To add an electron under such situation, the force of repulsion is to be overcome by applying external energy.