The correct options are
B |△H4|>|△H3|
C |△H5|>|△H6|
D |△H7| is +ve
A) O(g)+e−→O−(g) △H1=−142 kJ
O−(g)+e−→O−2(g) △H2=+844 kJ
since electron is added to negatively charged species, it will experience repulsion from incoming electron, hence the process is very endothermic.
B) O2(g)→O+2(g)+e−
(Electron removed from antibonding molecular orbitals and hence it requires less energy than an electron from atomic orbital π2Py or π2Px)
Electronic configuration of O2 is:
σ(1s)2σ∗(1s)2σ(2s)2σ∗(2s)2σ(2pz)2π(2px)2π(2py)2π∗(2px)1π∗(2py)1
O(g)→O+(g)+e− (Electron removed from 2P)
C) H2(g)→H+2(g)+e−
(Electron removed from bonding molecular orbital and it requires more energy than the electron from an atomic orbital1s
Electronic configuration of H2 is:
σ(1s)2
H(g)→H+(g)+e− (Electron removed from1s
D) O(g)+2e−→O−2(g) △H7=702 kJ