Question

Decreasing order of stability of $O_2, O^-_2, O_2^+ $ and ${ O_2^{2-}}$ is :

• A. ${O_2 > O_2^+ > O_2^{2-} > O_2^- }$
• B. $O^-_2 > O_2^{2-} > O_2^{+} > O_2 $
• C. ${O_2^+ > O_2 > O_2^- > O_2^{2-} }$
• D. ${O_2^{2-} > O_2^- > O_2 > O_2^+ }$

Answer 1

The Correct Option is C

To determine the stability of the given oxygen species $O_2, O_2^-, O_2^+, O_2^{2-}$, we need to look at their molecular orbital (MO) configurations and bond orders. Bond order is a significant factor affecting molecular stability; a higher bond order generally correlates with greater stability.

1. Molecular Orbital Configuration:
   • $O_2$: The electronic configuration is (\sigma_{1s})^2(\sigma^*_{1s})^2(\sigma_{2s})^2(\sigma^*_{2s})^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi^*_{2p_x})^1(\pi^*_{2p_y})^1.
   • $O_2^+$: The removal of one electron usually occurs from the antibonding \pi^* orbital, increasing its bond order.
   • $O_2^-$: Addition of one electron to the antibonding \pi^* orbital decreases its bond order.
   • $O_2^{2-}$: Adds two electrons to the antibonding \pi^* orbitals significantly reducing its bond order.
2. Bond Order Calculation:
   • $O_2$: Bond order = 2 (calculated as \frac{8-4}{2}).
   • $O_2^+$: Bond order = 2.5 (one less electron in the antibonding \pi^* orbital makes it more stable).
   • $O_2^-$: Bond order = 1.5 (additional electron in \pi^* orbital reduces stability).
   • $O_2^{2-}$: Bond order = 1 (two additional electrons in antibonding orbitals result in the least stable species).
3. Analysis of Stability:
   • Among the given species, $O_2^+$ has the highest bond order, thus is the most stable.
   • $O_2$ follows, with a bond order of 2.
   • $O_2^-$ has a lower bond order than $O_2$, making it less stable.
   • $O_2^{2-}$ is the least stable due to its lowest bond order.

Therefore, the decreasing order of stability is: $O_2^+ \gt O_2 \gt O_2^- \gt O_2^{2-}$. This matches the correct answer: ${O_2^+ \gt O_2 \gt O_2^- \gt O_2^{2-} }$.