Based on $M O$ theory compare the relative stabilities of $O_{2}$ & $O_{2}^{2 -}$ and indicate their magnetic properties.

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Solution

Bond order is directly proportional to stability.
Magnetic nature depends paired and unpaired electrons.
If molecule has unpaired electrons it means molecule is paramagnetic nature.
And if the molecule has no unpaired electron{ e.g., all are paired electrons } then, a molecule is diamagnetic nature.
Electronic configuration of $O_{2}$ (16 electrons):

$σ 1 s ², σ * 1 s ², σ 2 s ², σ * 2 s ², (π 2 p x ² \approx π 2 P y ²), (π * 2 P x ¹ \approx π * 2 P y ¹)$

Na(Anti bonding molecular orbitals) = 6 , Nb(Bonding molecular orbitals) = 10

Bond order= $\frac{1}{2}$ [bonding molecular orbotals - anti bonding
molecular orbotals]

now, B.O = $\frac{1}{2} [10 - 6] = 2$
It has two unpaired electrons.so, $O_{2}$ molecule is paramagnetic.
Electronic configuration of $O_{2}^{2 -}$ (17 electrons):

$σ 1 s ², σ * 1 s ², σ 2 s ², σ * 2 s ², (π 2 p x ² \approx π 2 P y ²), (π * 2 P x ² \approx π * 2 P y ²)$

now, B.O = $\frac{1}{2} [10 - 8] = 1$

It has no unpaired electron.so, it is diamagnetic
Now, the bond order is directly proportional to stability so, higher the bond order will be a higher stable molecule or ion.
Hence, increasing order of stability is

$O_{2} > O_{2}^{2 -}$

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Similar questions

Compare the relative stability of the following species and indicate their magnetic properties;

O₂,, (superoxide), (peroxide)

\to O_{2}, O_{2}^{+}, O_{2}^{-}, O_{2}^{2 -}

O_{2}, O_{2}^{-}, O_{2}^{+}

O_{2}^{2 -}

O_{2}^{2 +}

O_{2}^{+}

O_{2}

O_{2}^{-}

O_{2}^{2 -}

O_{2}^{2 -}, N O

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