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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 π*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 π2p 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 π*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 π2p 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

Bond energy has been defined in the text as the amount of energy required to break a chemical bond, so we have come to think of the addition of energy as breaking bonds. However, in some cases the addition of energy can cause the formation of bonds. For example, in a sample of helium gas subjected to a high-energy source, some \(\mathrm{He}_{2}\) molecules exist momentarily and then dissociate. Use MO theory (and diagrams) to explain why \(\mathrm{He}_{2}\) molecules can come to exist and why they dissociate.

Use the localized electron model to describe the bonding in \(\mathrm{CCl}_{4}\).

The \(\mathrm{N}_{2} \mathrm{O}\) molecule is linear and polar. a. On the basis of this experimental evidence, which arrangement, NNO or NON, is correct? Explain your answer. b. On the basis of your answer to part a, write the Lewis structure of \(\mathrm{N}_{2} \mathrm{O}\) (including resonance forms). Give the formal charge on each atom and the hybridization of the central atom. c. How would the multiple bonding in \(: \mathrm{N} \equiv \mathrm{N}-\mathrm{O}:\) be described in terms of orbitals?

Carbon monoxide (CO) forms bonds to a variety of metals and metal ions. Its ability to bond to iron in hemoglobin is the reason that \(\mathrm{CO}\) is so toxic. The bond carbon monoxide forms to metals is through the carbon atom: \(\mathrm{M}-\mathrm{C} \equiv \mathrm{O}\) a. On the basis of electronegativities, would you expect the carbon atom or the oxygen atom to form bonds to metals? b. Assign formal charges to the atoms in CO. Which atom would you expect to bond to a metal on this basis? c. In the MO model, bonding MOs place more electron density near the more electronegative atom. (See the HF molecule in Figs. \(9.42\) and \(9.43 .\) Antibonding MOs place more electron density near the less electronegative atom in the diatomic molecule. Use the MO model to predict which atom of carbon monoxide should form bonds to metals.

Use the localized electron model to describe the bonding in \(\mathrm{C}_{2} \mathrm{H}_{2}\) (exists as \(\mathrm{HCCH}\) ).

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