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Chapter 9: Problem 81

Using an MO energy-level diagram, would you expect \(\mathrm{F}_{2}\) to have a lower or higher first ionization energy than atomic fluorine? Why?

Short Answer

Expert verified
The first ionization energy of molecular fluorine (\(\mathrm{F}_{2}\)) is lower than that of atomic fluorine. This is because the highest occupied molecular orbital (HOMO) for \(\mathrm{F}_{2}\) has a lower energy level compared to the 2p atomic orbital of atomic fluorine. Consequently, less energy is required to remove an electron from \(\mathrm{F}_{2}\) compared to atomic fluorine.

Step by step solution

01

Understand ionization energy

Ionization energy is the energy required to remove an electron from an atom or molecule. The first ionization energy is the energy required to remove one electron from a neutral species. In this case, we are comparing the first ionization energy of fluorine atom and molecular fluorine.
02

MO energy-level diagram for atomic fluorine

In atomic fluorine, the electron configuration is given by \(1s^2 2s^2 2p^5\). Thus, the outermost electron is in the 2p orbital. An MO energy-level diagram for atomic fluorine would show the energy levels of different atomic orbitals, with the highest energy level being the 2p orbital.
03

MO energy-level diagram for molecular fluorine

For molecular fluorine (\(\mathrm{F}_{2}\)), we need to consider the interaction between the atomic orbitals of the two fluorine atoms. The resulting MO diagram will show the energy levels of the various molecular orbitals formed. As \(\mathrm{F}_{2}\) has a total of 18 electrons, these electrons will fill the molecular orbitals in order of increasing energy. The highest occupied molecular orbital (HOMO) is the one with the highest energy level that contains at least one electron.
04

Compare ionization energies using the MO diagrams

The first ionization energy of both species depends on the energy of their highest occupied orbitals, as this is the energy needed to remove the electron from these orbitals. Comparing the MO diagrams of atomic fluorine and \(\mathrm{F}_{2}\), we can look at the energy level of the 2p atomic orbital for atomic fluorine and the HOMO for \(\mathrm{F}_{2}\).
05

Conclusion

From the MO energy-level diagrams, we can see that the energy level of the HOMO for molecular fluorine (\(\mathrm{F}_{2}\)) is lower than the energy level of the 2p atomic orbital for atomic fluorine. Lower energy levels require less energy to remove an electron, meaning that the first ionization energy for \(\mathrm{F}_{2}\) is lower than that for atomic fluorine.

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Most popular questions from this chapter

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. \(\mathrm{He}_{2}^{2+}, \mathrm{He}_{2}^{+}, \mathrm{He}_{2}\)

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. $$ \text {a} C N \text { or } N O \qquad \text { b. } O_{2}^{2+} \text { or } N_{2}^{2+} $$

For each of the following molecules, write the Lewis structure(s), predict the molecular structure (including bond angles), give the expected hybrid orbitals on the central atom, and predict the overall polarity $$ \text {a} C F_{4} \quad \text { e. BeH }_{2} \quad \text { i. } \operatorname{KrF}_{4} $$ $$ \text {b} \mathrm{NF}_{3} \quad \text { f. } \mathrm{TeF}_{4} \quad \text { j. SeF }_{6} $$ $$ \text {c} \mathrm{OF}_{2} \quad \text { g. AsF_ } \quad \text { k. } \mathrm{IF}_{5} $$ $$ \text {d} \mathrm{BF}_{3} \quad \text { h. } \mathrm{KrF}_{2} \quad \text { L. } \mathrm{IF}_{3} $$

Explain the difference between the \(\sigma\) and \(\pi\) MOs for homo- nuclear diatomic molecules. How are bonding and antibonding orbitals different? Why are there two \(\pi\) MOs and one \(\sigma\) MO? Why are the \(\pi\) MOs degenerate?

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?
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