Sodium can react with oxygen to form sodium peroxide $\left(\mathrm{Na}_{2} \mathrm{O}_{2}\right),\( which is composed of \)\mathrm{Na}^{+}$ and \(\mathrm{O}_{2} 2-\) ions. Potassium can react with oxygen to form potassium superoxide \(\left(\mathrm{KO}_{2}\right)\) which is composed of \(\mathrm{K}^{+}\) and \(\mathrm{O}_{2}-\) ions. Does the peroxide ion or the superoxide ion have the shorter bond length? Explain.

Short Answer

Expert verified
The superoxide ion \(\mathrm{O}_{2}-\) has a shorter bond length compared to the peroxide ion \(\mathrm{O}_{2}2-\), because it has a higher bond order (2.5) than the peroxide ion (2), and a higher bond order corresponds to a shorter and stronger bond.

Step by step solution

01

Determine the bond order of the peroxide ion

To calculate the bond order of the peroxide ion \(\mathrm{O}_{2}2-\), first consider the electron configuration of a neutral oxygen molecule \(\mathrm{O}_{2}\): $$\sigma_{1s}^{2}\sigma_{1s*}^{2}\sigma_{2s}^{2}\sigma_{2s*}^{2}\sigma_{2p}^{2}\pi_{2p}^{4}\pi_{2p*}^{2}$$ The peroxide ion has two additional electrons, so its electron configuration becomes: $$\sigma_{1s}^{2}\sigma_{1s*}^{2}\sigma_{2s}^{2}\sigma_{2s*}^{2}\sigma_{2p}^{2}\pi_{2p}^{4}\pi_{2p*}^{4}$$ The bond order can be calculated as follows: Bond Order = (Number of bonding electrons - Number of antibonding electrons) / 2 Bond Order = (10 - 6) / 2 = 2
02

Determine the bond order of the superoxide ion

Now, we need to calculate the bond order of the superoxide ion \(\mathrm{O}_{2}-\). The superoxide ion has one additional electron compared to a neutral oxygen molecule. Its electron configuration becomes: $$\sigma_{1s}^{2}\sigma_{1s*}^{2}\sigma_{2s}^{2}\sigma_{2s*}^{2}\sigma_{2p}^{2}\pi_{ GUIDATA2*pi}^{4}\pi_{2p*}^{3}$$ The bond order can be calculated as follows: Bond Order = (Number of bonding electrons - Number of antibonding electrons) / 2 Bond Order = (10 - 5) / 2 = 2.5
03

Compare bond lengths based on bond orders

We have determined that the bond order of the peroxide ion \(\mathrm{O}_{2}2-\) is 2, whereas the bond order of the superoxide ion \(\mathrm{O}_{2}-\) is 2.5. Recall that a higher bond order corresponds to a shorter and stronger bond.
04

Conclusion

Based on our determination of bond orders, we can conclude that the superoxide ion \(\mathrm{O}_{2}-\) has a shorter bond length compared to the peroxide ion \(\mathrm{O}_{2}2-\).

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