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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.
Related Questions: - Which of the following lanthanide is commonly used
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- Among the following pairs of ions, the lower oxidation state in aqueous solution
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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 of the following lanthanide is commonly used
- The secondary precursors of photochemical smog are
- Among the following pairs of ions, the lower oxidation state in aqueous solution
- The number of gram molecules of oxygen in 6.02 x 10²⁴ CO molecules is
- An orbital in which n = 4 and l = 2 is expressed by
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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