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Chapter 9: Problem 103

Use the MO model to determine which of the following has the smallest ionization energy: \(\mathrm{N}_{2}, \mathrm{O}_{2}, \mathrm{N}_{2}^{2-}, \mathrm{N}_{2}^{-}, \mathrm{O}_{2}^{+} .\) Explain your answer.

Short Answer

Expert verified
Among the molecules and ions, \(\mathrm{N}_2\) has the highest HOMO energy level (occupied by \(\pi_3\) orbitals) compared to \(\mathrm{O}_2\), \(\mathrm{N}_2^{2-}\), \(\mathrm{N}_2^{-}\), and \(\mathrm{O}_2^{+}\), all of which have \(\sigma_4\) as their HOMO. Therefore, \(\mathrm{N}_2\) has the smallest ionization energy.

Step by step solution

01

Determine the bond order and electron configuration of each molecule/ion

Using MO theory, we first determine the bond order and electron configuration of each molecule/ion to understand its electronic structure. Bond order can be determined with this formula: Bond Order = \(\frac{1}{2} \times (\mathrm{Number\,of\,electrons\,in\,bonding\,orbitals} - \mathrm{Number\,of\,electrons\,in\,antibonding\,orbitals})\) We will list the occupancy of bonding orbitals (BO) and antibonding orbitals (ABO) for \(\sigma\), \(\sigma^{*}\), \(\pi\) and \(\pi^{*}\). For \(\mathrm{N}_2\) (14 electrons): BO: \(\sigma_1\) (2), \(\sigma_2\) (2), \(\pi_3\) (4) ABO: \(\sigma^{*}_1\) (0), \(\sigma^{*}_2\) (0), \(\pi^{*}_3\) (0) For \(\mathrm{O}_2\) (16 electrons): BO: \(\sigma_1\) (2), \(\sigma_2\) (2), \(\pi_3\) (4), \(\sigma_4\) (2) ABO: \(\sigma^{*}_1\) (0), \(\sigma^{*}_2\) (0), \(\pi^{*}_3\) (2), \(\sigma^{*}_4\) (0) For \(\mathrm{N}_2^{2-}\) (14 + 2 = 16 electrons): BO: \(\sigma_1\) (2), \(\sigma_2\) (2), \(\pi_3\) (4), \(\sigma_4\) (2) ABO: \(\sigma^{*}_1\) (0), \(\sigma^{*}_2\) (0), \(\pi^{*}_3\) (2), \(\sigma^{*}_4\) (0) For \(\mathrm{N}_2^{-}\) (14 + 1 = 15 electrons): BO: \(\sigma_1\) (2), \(\sigma_2\) (2), \(\pi_3\) (4), \(\sigma_4\) (1) ABO: \(\sigma^{*}_1\) (0), \(\sigma^{*}_2\) (0), \(\pi^{*}_3\) (2), \(\sigma^{*}_4\) (0) For \(\mathrm{O}_2^{+}\) (16 - 1 = 15 electrons): BO: \(\sigma_1\) (2), \(\sigma_2\) (2), \(\pi_3\) (4), \(\sigma_4\) (1) ABO: \(\sigma^{*}_1\) (0), \(\sigma^{*}_2\) (0), \(\pi^{*}_3\) (2), \(\sigma^{*}_4\) (0)
02

Identify the highest occupied molecular orbitals

Now, we will identify the highest occupied molecular orbitals (HOMO) for each molecule/ion. HOMO for \(\mathrm{N}_2\): \(\pi_3\) HOMO for \(\mathrm{O}_2\): \(\sigma_4\) HOMO for \(\mathrm{N}_2^{2-}\): \(\sigma_4\) HOMO for \(\mathrm{N}_2^{-}\): \(\sigma_4\) HOMO for \(\mathrm{O}_2^{+}\): \(\sigma_4\)
03

Compare the HOMO energy levels and determine the smallest ionization energy

Now that we have identified the HOMO for each molecule/ion, we need to compare their energy levels. A molecule/ion with the highest HOMO energy level will have the smallest ionization energy. Based on the HOMO energy levels mentioned above, we can deduce the following ordering of HOMO energy levels: \(\pi_3 < \sigma_4\) We can now conclude that \(\mathrm{N}_2\) has the highest HOMO energy level. As a result, \(\mathrm{N}_2\) has the smallest ionization energy among the given molecules and ions.

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Most popular questions from this chapter

Using the molecular orbital model, write electron configurations for the following diatomic species and calculate the bond orders. Which ones are paramagnetic? Place the species in order of increasing bond length and bond energy. $$ \text {a} \mathrm{CO} \quad \text { b. } \mathrm{CO}^{+} \quad \text { c. } \mathrm{CO}^{2+} $$

Describe the bonding in \(\mathrm{NO}^{+}, \mathrm{NO}^{-},\) and \(\mathrm{NO}\) using both the localized electron and molecular orbital models. Account for any discrepancies between the two models.

Cyanamide \(\left(\mathrm{H}_{2} \mathrm{NCN}\right),\) an important industrial chemical, is produced by the following steps: $$ \begin{array}{c}{\mathrm{CaC}_{2}+\mathrm{N}_{2} \longrightarrow \mathrm{CaNCN}+\mathrm{C}} \\ {\mathrm{CaNCN} \stackrel{\mathrm{Acid}}{\longrightarrow} \mathrm{H}_{2} \mathrm{NCN}} \\\ {\mathrm{Cyanamide}}\end{array} $$ Calcium cyanamide (CaNCN) is used as a direct-application fertilizer, weed killer, and cotton defoliant. It is also used to make cyanamide, dicyandiamide, and melamine plastics: a. Write Lewis structures for \(\mathrm{NCN}^{2-}, \mathrm{H}_{2} \mathrm{NCN}\) , dicyandiamide, and melamine, including resonance structures where appropriate. b. Give the hybridization of the C and N atoms in each species. c. How many \(\sigma\) bonds and how many \(\pi\) bonds are in each species? d. Is the ring in melamine planar? e. There are three different \(C-N\) bond distances in dicyandiamide, \(\mathrm{NCNC}\left(\mathrm{NH}_{2}\right)_{2}\) , and the molecule is nonlinear. Of all the resonance structures you drew for this molecule, predict which should be the most important.

As the head engineer of your starship in charge of the warp drive, you notice that the supply of dilithium is critically low. While searching for a replacement fuel, you discover some diboron, B. a. What is the bond order in \(\mathrm{Li}_{2}\) and \(\mathrm{B}_{2} ?\) b. How many electrons must be removed from \(\mathrm{B}_{2}\) to make it isoelectronic with \(\mathrm{Li}_{2}\) so that it might be used in the warp drive? c. The reaction to make \(\mathrm{B}_{2}\) isoelectronic with \(\mathrm{Li}_{2}\) is generalized (where \(n=\) number of electrons determined in part \(\mathrm{b}\) ) as follows: $$ \mathrm{B}_{2} \longrightarrow \mathrm{B}_{2}^{n+}+n \mathrm{e}^{-} \quad \Delta H=6455 \mathrm{kJ} / \mathrm{mol} $$ How much energy is needed to ionize 1.5 \(\mathrm{kg} \mathrm{B}_{2}\) to the desired isoelectronic species?

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