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Question

O(g)+eO(g) ΔH1
O(g)+eO2(g) ΔH2
O2(g)O+2(g)+e ΔH3
O(g)O+(g)+e ΔH4
H2(g)H+2(g)+e ΔH5
H(g)H+(g)+e ΔH6
O(g)+2eO2 ΔH7
Which of the following option(s) is/are correct?

A
|H1|>|H2|
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B
|H4|>|H3|
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C
|H5|>|H6|
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D
|H7| is +ve
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Solution

The correct options are
B |H4|>|H3|
C |H5|>|H6|
D |H7| is +ve
A) O(g)+eO(g) H1=142 kJ
O(g)+eO2(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)+2eO2(g) H7=702 kJ

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