Question

Is the bond order in superoxide ion more or less than in peroxide ion.
Explain on the basis of MO theory .

Answer 1

Bond Order in superoxide ion = (1/2)[Nb - Na]

" " " " = (1/2)[8 - 5]=1.5

Similarly,

Bond Order in Peroxide ion = [8 - 6]/2 = 1.0

So, From here Bond order of superoxide ion has more bond order than peroxide ion.

I hope it will help you.

Answer 2

Peroxide ion is

$O_{2}^{2-}$

Super oxide ion is

$O_{2}^-$

We have the configuration in Oxygen.

1s² 2s² 2p^4

The Molecular orbital theory, Electronic configuration would be,

σ1s² σ*1s² σ2s² σ*2s² σ2pz² π2px² π2py² π*2px¹π*2py¹

Now For Peroxide ion, The Molecular orbital configuration would be,

σ1s² σ*1s² σ2s² σ*2s² σ2pz² π2px² π2py² π*2px²π*2py²

Number of electrons in Bonding orbitals = 10

Number of electrons in Antibonding orbitals = 8

Bond order = 1/2 ( 10 - 8) = 1.

Now for Superoxide ion, The Molecular orbital configuration would be, σ1s² σ*1s² σ2s² σ*2s² σ2pz² π2px² π2py² π*2px²π*2py¹

Number of electrons in Bonding orbitals = 10

Number of electrons in Antibonding orbitals = 7

Bond order = 1/2 ( 10 - 7) = 3/2 = 1.5

Therefore, Bond order of Super oxide is more than Peroxide.