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Chapter 4: Problem 121

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
The molecule/ion with the smallest ionization energy is \(\mathrm{O}_{2}^{+}\). This is because the HOMO for all species is \(\pi_{2p}\), and the \(\pi_{2p}\) orbitals are higher in energy for \(\mathrm{O}_{2}\) than for \(\mathrm{N}_{2}\). Comparing \(\mathrm{N}_{2}^{-}\) and \(\mathrm{O}_{2}^{+}\), the HOMO in \(\mathrm{N}_{2}^{-}\) is slightly stabilized due to the extra electron, so \(\mathrm{O}_{2}^{+}\) has the smallest ionization energy.

Step by step solution

01

1. Determine the number of valence electrons

For each molecule/ion, we need to find the number of valence electrons involved. - For \(\mathrm{N}_{2}\), there are 10 valence electrons since nitrogen has 5 valence electrons each. - For \(\mathrm{O}_{2}\), there are 12 valence electrons since oxygen has 6 valence electrons each. - For \(\mathrm{N}_{2}^{2-}\), there are 12 valence electrons (10 from the nitrogen atoms and 2 from the negative charges). - For \(\mathrm{N}_{2}^{-}\), there are 11 valence electrons (10 from the nitrogen atoms and 1 from the negative charge). - For \(\mathrm{O}_{2}^{+}\), there are 11 valence electrons (12 from the oxygen atoms and 1 less due to the positive charge).
02

2. Examine the MO diagrams

Look at the MO diagrams for both Nitrogen and Oxygen diatomic molecules. Note that the MO energy levels for \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) are slightly different. For \(\mathrm{N}_{2}\), the \(\sigma_{2p}\) orbitals are lower in energy than the \(\pi_{2p}\) orbitals, while for \(\mathrm{O}_{2}\), the \(\pi_{2p}\) orbitals are lower in energy than the \(\sigma_{2p}\) orbitals.
03

3. Determine the HOMO

Based on the MO diagrams and the number of valence electrons calculated in step 1, we can determine the HOMO for each molecule/ion. - \(\mathrm{N}_{2}\): 10 electrons, the HOMO is \(\pi_{2p}\) (3 different MOs degenerate in energy). - \(\mathrm{O}_{2}\): 12 electrons, the HOMO is \(\pi_{2p}\) (3 different MOs degenerate in energy). - \(\mathrm{N}_{2}^{2-}\): 12 electrons, the HOMO is \(\pi_{2p}\) (3 different MOs degenerate in energy). - \(\mathrm{N}_{2}^{-}\): 11 electrons, the HOMO is \(\pi_{2p}\) (2 different MOs degenerate in energy). - \(\mathrm{O}_{2}^{+}\): 11 electrons, the HOMO is \(\pi_{2p}\) (2 different MOs degenerate in energy).
04

4. Compare HOMO energies and determine the smallest ionization energy

Comparing the HOMO energies, we can see that the \(\pi_{2p}\) orbitals are higher in energy for \(\mathrm{O}_{2}\) than for \(\mathrm{N}_{2}\), so \(\mathrm{O}_{2}\) and its ions will have smaller ionization energies than \(\mathrm{N}_{2}\) and its ions. Now compare the two ions with HOMOs in the \(\pi_{2p}\) orbitals: \(\mathrm{N}_{2}^{-}\) and \(\mathrm{O}_{2}^{+}\). Due to the extra electron in \(\mathrm{N}_{2}^{-}\) compared to \(\mathrm{O}_{2}^{+}\), the energy of the HOMO would be slightly stabilized in \(\mathrm{N}_{2}^{-}\). Therefore, \(\mathrm{O}_{2}^{+}\) has the smallest ionization energy out of these molecules and ions.

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Most popular questions from this chapter

What are molecular orbitals? How do they compare with atomic orbitals? Can you tell by the shape of the bonding and antibonding orbitals which is lower in energy? Explain.

Which of the following are predicted by the molecular orbital model to be stable diatomic species? a. \(\mathrm{H}_{2}^{+}, \mathrm{H}_{2}, \mathrm{H}_{2}^{-}, \mathrm{H}_{2}^{2-}\) b. \(\mathrm{He}_{2}^{2+}, \mathrm{He}_{2}^{+}, \mathrm{He}_{2}\)

Using the molecular orbital model, write electron configurations for the following diatomic species and calculate the bond orders. Which ones are paramagnetic? a. \(\mathrm{Li}_{2}\) b. \(C_{2}\) c. \(S_{2}\)

Why are \(d\) orbitals sometimes used to form hybrid orbitals? Which period of elements does not use \(d\) orbitals for hybridization? If necessary, which \(d\) orbitals \((3 d, 4 d, 5 d, \text { or } 6 d)\) would sulfur use to form hybrid orbitals requiring \(d\) atomic orbitals? Answer the same question for arsenic and for iodine.

The structure of \(\operatorname{Te} \mathrm{F}_{5}^{-}\) is Draw a complete Lewis structure for \(\operatorname{TeF}_{5}^{-},\) and explain the distortion from the ideal square pyramidal structure. (See Exercise 26.)
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