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The Formation of the oxide ion, O²⁻ from oxygen atom requires first an exothermic and then an endothermic step as shown below:
O(g) + e⁻ → O⁻(g) ; Δf H° = -141 kJ mol⁻¹
O⁻(g) + e⁻ → O²⁻(g) ; Δf H° = +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,
Options
(a) O⁻ ion has comparatively smaller size than oxygen atom
(b) oxygen is more electronegative
(c) addition of electron in oxygen results in larger size of the ion
(d) electron repulsion outweighs the stability gained by achieving noble gas configuration
Correct Answer:
electron repulsion outweighs the stability gained by achieving noble gas configuration
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
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Topics: P Block Elements in Group 15
(89)
Subject: Chemistry
(2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Which type of bond is there in H₂S molecules
- Volume occupied by one molecule of water (density= 1 g cm⁻³) is
- In acetylene molecule, the two carbon atoms are linked by
- In aqueous solution of H₂O₂ oxidises H₂S
- Molar volume of CO₂ is maximum at
Topics: P Block Elements in Group 15 (89)
Subject: Chemistry (2512)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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