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Write & Insert Fields Preview Results Finish 5. Based on Molecular Orbital Theory MOT i) Write the ground state electron conf
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Answer #1

Bond Order(BO) = 1/2(Na - Nb) where Na and Nb are number of electrons in antibonding MO and in bonding MO respectively.

Be2 :  (205)?(20*s)

Here, Na = 2 and Nb = 2

BO = 1/2(2-2) = 0

B2: (205)?(20*s)?(112px)+(112py)

Here, Na = 2 and Nb = 4

BO = 1/2(4-2) = 1

F2 : (205)?(20*s)?(02px)?([12py)? (112pz)?(112px)?(12pm)?

Here, Na = 8 and Nb = 10

BO = 1/2(8-6) = 1

For C22- : Nb = 10 and Na = 4

BO = 1/2(10-4) = 3

It will exist

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