/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 81 Using an MO energy-level diagram... [FREE SOLUTION] | 91影视

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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

Describe the bonding in the \(\mathrm{O}_{3}\) molecule and the \(\mathrm{NO}_{2}^{-}\) ion using the localized electron model. How would the molecular orbital model describe the \(\pi\) bonding in these two species?

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 \((1402 \mathrm{kJ} / \mathrm{mol})\)

Although nitrogen trifluoride \(\left(\mathrm{NF}_{3}\right)\) is a thermally stable compound, nitrogen triodide \(\left(\mathrm{N} \mathrm{I}_{3}\right)\) is known to be a highly explosive material. \(\mathrm{NI}_{3}\) can be synthesized according to the equation $$ \mathrm{BN}(s)+3 \mathrm{IF}(g) \longrightarrow \mathrm{BF}_{3}(g)+\mathrm{NI}_{3}(g) $$ a. What is the enthalpy of formation for \(\mathrm{NI}_{3}(s)\) given the enthalpy of reaction \((-307 \mathrm{kJ})\) and the enthalpies of formation for \(\mathrm{BN}(s)(-254 \mathrm{kJ} / \mathrm{mol}), \operatorname{IF}(g)(-96 \mathrm{kJ} / \mathrm{mol})\) and \(\mathrm{BF}_{3}(g)(-1136 \mathrm{kJ} / \mathrm{mol}) ?\) b. It is reported that when the synthesis of \(\mathrm{NI}_{3}\) is conducted using 4 moles of IF for every 1 \(\mathrm{mole}\) of \(\mathrm{BN}\) , one of the by-products isolated is \(\left[\mathrm{IF}_{2}\right]^{+}\left[\mathrm{BF}_{4}\right]^{-} .\) What are the molecular geometries of the species in this by-product? What are the hybridizations of the central atoms in each species in the by-product?

Indigo is the dye used in coloring blue jeans. The term navy blue is derived from the use of indigo to dye British naval uniforms in the eighteenth century. The structure of the indigo molecule is a. How many \(\sigma\) bonds and \(\pi\) bonds exist in the molecule? b. What hybrid orbitals are used by the carbon atoms in the indigo molecule?

Show how two 2\(p\) atomic orbitals can combine to form a \(\sigma\) or a \(\pi\) molecular orbital.
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