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NCERT Class XII Chemistry
Chapter - The p-Block Elements
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Question : 19 of 74
Marks: +1, -0
Knowing the electron gain enthalpy values for O → O– and O →O2^{2–} as –141 and 702 kJ mol1^{–1} respectively, how can you account for the formation of large number of oxides having O2^{2–} species and not O^{–} ?
Hint : Consider lattice energy factor in the formation of compound.
Solution:  
This can be explained with the help of electronic configuration.
As O2^{2–} has most stable configuration amongst these. So, formation of O2^{2–} ismuch more easier. In solid state, large amount of energy (lattice enthalpy)is released to form divalent O2^{2–} ions.It is greater lattice enthalpy of O2^{2–}which compensates for the high energy required to remove the secondelectron.
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