Question

write the electronic configuration of oxygen molecule and calculate bond order and periodic its magnetic property​

Answer 1

Answer:B.O=2 , Paramagnetic in nature

Explanation:As O2 has 16 electrons its MOT elecectronic configuration will be :

KK* MM* (σ2pz2) (π2px2=2py2)(π*2px1=π*2py1)

So, B.O = 1/2 ( 6-2) = 2

And As it as unpaired electrons it is paramagnetic in nature.