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Periodic Table and Periodicity question

2004 · Shift 0 · Q61
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Periodic Table and Periodicity question

2004 · Shift 0 · Q61

JEE MainChemistryPeriodic Table and PeriodicityMCQ+4 / −1
The formation of the oxide ion O2−O^{2-}O2−(g) requires first an exothermic and then an endothermic step as shown below OOO(g) + e−e^-e− = O(g)−ΔO_{(g)}^{-}\DeltaO(g)−​Δ Ho = -142 kJmol-1 O(g)−O_{(g)}^{-}O(g)−​+ e−e^-e− =O(g)2−ΔO_{(g)}^{2-}\DeltaO(g)2−​Δ Ho = 844 kJmol-1 This because
  1. A
    O−O^-O− ion will tend to resist the addition of another electron
  2. B
    Oxygen has high electron affinity
  3. C
    Oxygen is more electronegative
  4. D
    O−O^-O− ion has comparatively larger size than oxygen atom
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