Bond Order(BO) = 1/2(Na - Nb) where Na and Nb are number of electrons in antibonding MO and in bonding MO respectively.
Be2 :
Here, Na = 2 and Nb = 2
BO = 1/2(2-2) = 0
B2:
Here, Na = 2 and Nb = 4
BO = 1/2(4-2) = 1
F2 :
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
Write & Insert Fields Preview Results Finish 5. Based on Molecular Orbital Theory MOT i) Write...