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Chapter 9: Problem 79

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

Short Answer

Expert verified
In O\(_2\), removing an electron from the highest occupied molecular orbital (HOMO), which is \(\pi_{2p*}\) (antibonding), results in a higher bonding order (\(\frac{3}{2}\) vs 2) and thus stronger bonding. In N\(_2\), removing an electron from the HOMO, which is \(\pi_{2p}\) (bonding), leads to a lower bonding order (\(\frac{5}{2}\) vs 3) and weaker bonding. This difference occurs due to the different electron configurations in the molecular orbitals.

Step by step solution

01

Write down the molecular orbital configurations for O\(_2\) and N\(_2\).

To analyze the effect of removing an electron from O\(_2\) and N\(_2\), we first need to determine the molecular orbital configurations of both molecules. For this, we can use the following diagram: ![Molecular Orbital Diagram](https://ibchem.com/IB/ibfiles/14/images/142/O2-N2-diag.gif) From the diagram, we have the following molecular orbital configurations: O\(_2\): \(\sigma_{1s}^{2}\sigma_{1s*}^{2}\sigma_{2s}^{2}\sigma_{2s*}^{2}\sigma_{2p}^{2}\pi_{2p}^{4}\pi_{2p*}^{2}\) N\(_2\): \(\sigma_{1s}^{2}\sigma_{1s*}^{2}\sigma_{2s}^{2}\sigma_{2s*}^{2}\sigma_{2p}^{2}\pi_{2p}^{4}\pi_{2p*}^{0}\)
02

Calculate the bonding orders for O\(_2\) and N\(_2\).

We calculate the bonding order using the following formula: Bonding Order = \(\frac{1}{2}(n_{b} - n_{a})\) where \(n_{b}\) is the number of bonding electrons, and \(n_{a}\) is the number of anti-bonding electrons. For O\(_2\), Bonding Order = \(\frac{1}{2}(10 - 6) = 2\) For N\(_2\), Bonding Order = \(\frac{1}{2}(10 - 4) = 3\)
03

Remove one electron from O\(_2\) and N\(_2\) and write down their new molecular orbital configurations.

Removing one electron from O\(_2\), we get O\(_2^+\) with the molecular orbital configuration: O\(_2^+\): \(\sigma_{1s}^{2}\sigma_{1s*}^{2}\sigma_{2s}^{2}\sigma_{2s*}^{2}\sigma_{2p}^{2}\pi_{2p}^{4}\pi_{2p*}^{1}\) Removing one electron from N\(_2\), we get N\(_2^+\) with the molecular orbital configuration: N\(_2^+\): \(\sigma_{1s}^{2}\sigma_{1s*}^{2}\sigma_{2s}^{2}\sigma_{2s*}^{2}\sigma_{2p}^{2}\pi_{2p}^{3}\pi_{2p*}^{0}\)
04

Calculate the bonding orders for O\(_2^+\) and N\(_2^+\)

For O\(_2^+\), Bonding Order = \(\frac{1}{2}(9 - 6) = \frac{3}{2}\) For N\(_2^+\), Bonding Order = \(\frac{1}{2}(9 - 4) = \frac{5}{2}\)
05

Compare the bonding orders to determine whether the bonding is strengthened or weakened.

Comparing the bonding orders, we can conclude: For O\(_2\) bonding is strengthened upon removal of an electron: O\(_2\): Bonding Order = 2 O\(_2^+\): Bonding Order = \(\frac{3}{2}\) For N\(_2\) bonding is weakened upon removal of an electron: N\(_2\): Bonding Order = 3 N\(_2^+\): Bonding Order = \(\frac{5}{2}\) Therefore, using molecular orbital theory, we have established that the removal of an electron from O\(_2\) strengthens bonding, whereas the removal of an electron from N\(_2\) weakens bonding.

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Most popular questions from this chapter

What modification to the molecular orbital model was made from the experimental evidence that \(\mathrm{B}_{2}\) is paramagnetic?

Biacetyl and acetoin are added to margarine to make it taste more like butter Complete the Lewis structures, predict values for all \(\mathrm{C}-\mathrm{C}-\mathrm{O}\) bond angles, and give the hybridization of the carbon atoms in these two compounds. Must the four carbon atoms and two oxygen atoms in biacetyl lie the same plane? How many \(\sigma\) bonds and how many \(\pi\) bonds are there in biacetyl and acetoin?

Use the localized electron model to describe the bonding in $\mathrm{H}_{2} \mathrm{O}$ .

The molecules \(\mathrm{N}_{2}\) and \(\mathrm{CO}\) are isoelectronic but their properties are quite different. Although as a first approximation we often use the same MO diagram for both, suggest how the \(\mathrm{MOs}\) in \(\mathrm{N}_{2}\) and \(\mathrm{CO}\) might be different.

A flask containing gaseous \(\mathrm{N}_{2}\) is irradiated with 25 -nm light. a. Using the following information, indicate what species can form in the flask during irradiation. $$ \begin{array}{ll}{\mathrm{N}_{2}(g) \longrightarrow 2 \mathrm{N}(g)} & {\Delta H=941 \mathrm{kJ} / \mathrm{mol}} \\ {\mathrm{N}_{2}(g) \longrightarrow \mathrm{N}_{2}^{+}(g)+\mathrm{e}^{-}} & {\Delta H=1501 \mathrm{kJ} / \mathrm{mol}} \\ {\mathrm{N}(g) \longrightarrow \mathrm{N}^{+}(g)+\mathrm{e}^{-}} & {\Delta H=1402 \mathrm{kJ} / \mathrm{mol}}\end{array} $$ b. What range of wavelengths will produce atomic nitrogen in the flask but will not produce any ions? c. Explain why the first ionization energy of $\mathrm{N}_{2}(1501 \mathrm{kJ} /\( mol) is greater than the first ionization energy of atomic nitrogen \)(1402 \mathrm{kJ} / \mathrm{mol})$
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