Answer:
[tex]\boxed{O_2 \: ^{2-} (\sigma2s)^2(\sigma∗2s)^2(\sigma2pz)^2(π2py,π2px)^4(π∗2py,π∗2px)^4}[/tex]
[tex]\boxed{O_2 \: ^{1-}(\sigma2s)^2(\sigma∗2s)^2(\sigma2pz)^2(π2py,π2px)^4{ (π∗2py)}^{2} {(π∗2px)}^{1}}[/tex]
Explanation:
[tex]\sf \: Electronic \: configuration \: of \: O_2 \: ^{2-} and \: O_2 \: ^{1-}[/tex]
Electronic configuration of any diatomic molecule can be determined using MOT(molecular orbital theory).
To know the electronic configuration of above two molecules, draw the MOT of O2 molecule.(refer the diagram)
There are 16 electrons in O2 molecule out of which Four electrons are in non bonded state. Remaining 12 electrons of 16 will be filled according to MOT, since O2 is not a S-P mixing case, in without S-P mixing case the sigma orbital(σ2pz) lies below the Pi orbital(π2px,π2py) in bonding state as you can see in the diagram, on other hand the other diatomic molecules like B2, C2 & N2 are S-P mixing case where Pi orbital lies below and sigma orbital lies above.
Now, the electronic configuration of O2 molecule is
[tex](\sigma2s)^2(\sigma∗2s)^2(\sigma2pz)^2(π2py,π2px)^4(π∗2py,π∗2px)^2 [/tex]
There are two electrons unpaired electrons in MOT hence the O2 molecule is paramagnetic which contradicts VBT, Thats why MOT dominant VBT.
in case of O2^2-(peroxide linkage) the two electrons enters in the Anti Bonding of pi orbital. hence the electronic configuration of O2^2- is
[tex](\sigma2s)^2(\sigma∗2s)^2(\sigma2pz)^2(π2py,π2px)^4(π∗2py,π∗2px)^4[/tex]
since the HOMO (highest occupied molecular orbital) is paired the O2^2- is diamagnetic in nature.
Coming to O2^1- (superoxide linkage) the one electron enters in the Anti Bonding of pi orbital hence the electronic configuration comes out,
[tex](\sigma2s)^2(\sigma∗2s)^2(\sigma2pz)^2(π2py,π2px)^4{ (π∗2py)}^{2} {(π∗2px)}^{1} [/tex]
The HOMO of O2^1- is unpaired hence it is Paramagnetic in nature.
Image source- Web
Thanks for joining brainly community!
