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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
- A gas can be liquefied
- The number of hydroxide ions produced by one molecule of sodium carbonate
- The temperature above which a gas cannot be liquefied is called as
- How many grams of dibasic acid (mol.wt.200) should be present in 100 mL.
- Which of the following hydrogen bonds are strongest in vapour phase
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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