InterviewSolution
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InterviewSolution
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Give the change in bond order in the following ionisation process? i. `O_(2) rarr O_(2)^(o+)+e^(-)` , ii.`N_(2) rarr N_(2)^(o+)+e^(-)` |
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Answer» According to molecular orbital theory, electronic configuration and bond order of `N_(2), N_(2)^(+), O_(2)` and `O_(2)^(+)` species are as follows. `N_(2)(14e^(-))=sigma1s^(2),sigma^(star)1s^(2), sigma2s^(2), sigma^(star)2s^(2), (pi2p_(x)^(2)=pi2p_(y)^(2)).sigma2p_(z)^(2))` Bond order =`1/2[N_(b)-N_(a)]=1/2[10-4]=3` `N_(2)^(+)(13e^(-))=sigma1s^(2), sigma^(star)1s^(2),sigma2s^(2),sigma^(star)2s^(2),(pi2p_(x)^(2)=pi2p_(y)^(2))sigma2p_(z)^(1)` Bond order =`1/2[N_(b)-N_(a)]=1/2(9-4)=2.5` `O_(2)16e^(-)=sigma1s^(2),sigma^(star)1s^(2),sigma2s^(2),sigma2p_(z)^(2), (pi2p_(x)^(2)=pi2p_(y)^(2)),(pi^(star)2p_(x)^(1)=pi^(star)2p_(y)^(1))` Bond order =`1/2[N_(b)-N_(a)]=1/2(10-6)=2` `O_(2)^(+)(15e^(-)= sigma1s^(2), sigma^(star)1s^(2), sigma2s^(2), sigma^(star)2s^(2), sigma2p_(z)^(2), (pi2p_(x)^(2)=pi2p_(y)^(2)), (pi^(star)2p_(x)^(1)=pi^(star)2p_(y))` Bond order =`1/2[N_(b)-N_(a)]=1/2[10-5]=2.5` a) `underset(B.O=3)N_(2) to underset(B.O=2.5)(N_(2)^(+)+e^(-)` Thus, bond order of decreases. b) `underset(B.O=2)O_(2) to underset(B.O=2.5)(O_(2)^(+) + e^(-)` Thus, bond order increases. |
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