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Chapter 9: Problem 43

Using molecular orbital theory, explain why the removal of one electron in \(\mathrm{O}_{2}\) strengthens bonding, while the removal of one electron in \(\mathrm{N}_{2}\) weakens bonding.

Short Answer

Expert verified
In summary, using molecular orbital theory, the removal of one electron in O2 strengthens bonding because the electron is removed from the anti-bonding 蟺*2p orbital, which increases the net bonding strength. On the other hand, the removal of one electron in N2 weakens bonding since the electron is removed from the bonding 蟺2p orbital, reducing the overall bonding strength between the nitrogen atoms.

Step by step solution

01

Draw the molecular orbital diagrams for O2 and N2

Begin by drawing the molecular orbital diagrams for both oxygen and nitrogen molecules. This diagram will show the energy levels and orbital configurations for the molecules. For O2, the electron configuration is \[ \sigma_{1s}^{2} \sigma_{1s*}^{2} \sigma_{2s}^{2} \sigma_{2s*}^{2} \sigma_{2p}^{2} \pi_{2p}^{4} \pi_{2p*}^{2} \] For N2, the electron configuration is \[ \sigma_{1s}^{2} \sigma_{1s*}^{2} \sigma_{2s}^{2} \sigma_{2s*}^{2} \sigma_{2p}^{2} \pi_{2p}^{4} \]
02

Understand the effect of removing one electron

Now let's see how removing one electron changes the electron configuration of these molecules: For O2, after removing one electron, the configuration becomes: \[ \sigma_{1s}^{2} \sigma_{1s*}^{2} \sigma_{2s}^{2} \sigma_{2s*}^{2} \sigma_{2p}^{2} \pi_{2p}^{4} \pi_{2p*}^{1} \] For N2, after removing one electron, the configuration becomes: \[ \sigma_{1s}^{2} \sigma_{1s*}^{2} \sigma_{2s}^{2} \sigma_{2s*}^{2} \sigma_{2p}^{2} \pi_{2p}^{3} \]
03

Analyze the effect on bonding

Now let's analyze how the new electron configurations affect the bonding strength: In O2, after removing one electron, there is a decrease in the number of electrons in the anti-bonding 蟺*2p orbital. This will strengthen the bonding because anti-bonding orbitals have a higher energy level and oppose the bonding between atoms. Therefore, by removing one electron from an anti-bonding orbital, we are increasing the net bonding strength in O2. In N2, after removing one electron, there is a decrease in the number of electrons in the bonding 蟺2p orbital. This will weaken the bonding because fewer electrons are present in the bonding orbital, which reduces the overall bonding strength between the nitrogen atoms.
04

Conclusion

According to molecular orbital theory, the removal of one electron in O2 strengthens bonding because the electron is removed from an anti-bonding orbital, whereas the removal of one electron in N2 weakens bonding because the electron is removed from a bonding orbital.

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Most popular questions from this chapter

In the molecular orbital model, compare and contrast \(\sigma\) bonds with \(\pi\) bonds. What orbitals form the \(\sigma\) bonds and what orbitals form the \(\pi\) bonds? Assume the \(z\) -axis is the internuclear axis.

What are the relationships among bond order, bond energy, and bond length? Which of these quantities can be measured?

In terms of the molecular orbital model, which species in each of the following two pairs will most likely be the one to gain an electron? Explain. a. CN or NO b. \(\mathrm{O}_{2}^{2+}\) or \(\mathrm{N}_{2}{ }^{2+}\)

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\(\mathrm{FClO}_{2}\) and \(\mathrm{F}_{3} \mathrm{ClO}\) can both gain a fluoride ion to form stable anions. \(\mathrm{F}_{3} \mathrm{ClO}\) and \(\mathrm{F}_{3} \mathrm{ClO}_{2}\) will both lose a fluoride ion to form stable cations. Draw the Lewis structures and describe the hybrid orbitals used by chlorine in these ions.
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