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JEE Main & Advanced AIEEE Solved Paper-2004

  • question_answer
    The formation of the oxide ion\[{{O}^{2-}}(g)\]requires first an exothermic and then an exothermic step as shown below \[O(g)+{{e}^{-}}={{O}^{-}}(g);\]                 \[\Delta {{H}^{o}}=-142\,kJ\,mo{{l}^{-1}}\] \[{{O}^{-}}{{(g)}^{-}}+{{e}^{-}}={{O}^{2}}(g);\]   \[\Delta {{H}^{o}}=844\,kJ\,mo{{l}^{-1}}\] This is because

    A) oxygen is more electronegative

    B) oxygen has high electron affinity

    C) \[{{O}^{-}}\]ion will tend to resist the addition of another electron  

    D) \[{{O}^{-}}\]ion has comparatively larger size than oxygen atom

    Correct Answer: C

    Solution :

    \[{{O}^{-}}(g)+{{e}^{-}}\xrightarrow[{}]{{}}{{O}^{2-}}(g),\]\[\Delta {{H}^{o}}=844\,KJ\,mo{{l}^{-1}}\] This process is unfavourable in the gas phase because the resulting increase in electron-electron repulsion over weighs the stability gained by achieving the noble gas configuration.


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