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Chapter 9: Problem 75

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 \(\mathrm{F}_{2}\) is higher than that of atomic fluorine. This is because the highest occupied molecular orbital (HOMO) in \(\mathrm{F}_{2}\) is the nonbonding π*(2p) orbital, which has a higher energy level than the 2p orbital in atomic fluorine. It is easier to remove an electron from the lower-energy 2p orbital in atomic fluorine, resulting in a lower ionization energy compared to \(\mathrm{F}_{2}\).

Step by step solution

01

Understand Molecular Orbital Theory and Ionization Energy

To solve this problem, we need to understand the Molecular Orbital Theory, which explains the formation of bonds between atoms in a molecule. In this theory, atomic orbitals combine to form molecular orbitals that belong to the entire molecule. Electrons in these orbitals help determine the bond order and stability of the molecule. The ionization energy is the energy required to remove an electron from an atom or molecule—higher ionization energy means that it is harder to remove an electron.
02

Construct the MO energy-level diagram for F₂

To create the MO energy-level diagram for \(\mathrm{F}_{2}\), we need to know the atomic orbitals of F and how they combine to form molecular orbitals. For fluorine, the valence shell electron configuration is 2s²2p⁵. When two F atoms bond, the 2s and 2p orbitals combine to form molecular orbitals. In the MO diagram, the 2s orbitals combine to form σ and σ* molecular orbitals, while the 2p orbitals form following molecular orbitals: σ, π, π*, and σ*. The 2s orbitals are lower in energy than the 2p orbitals. Electrons will fill the molecular orbitals from lowest to highest energy according to the Aufbau principle.
03

Determine the electron configuration of F₂

Each F atom has 7 valence electrons, which means \(\mathrm{F}_{2}\) has a total of 14 valence electrons. Following the Aufbau principle, electrons fill the molecular orbitals from lowest to highest energy level: - 2 electrons in σ(2s) - 2 electrons in σ*(2s) - 2 electrons in σ(2p) - 4 electrons in π(2p) (2 in each degenerate orbital) - 4 electrons in π*(2p) (2inedegenerate orbital) Now we have filled all 14 valence electrons in the MO diagram.
04

Comparing the first ionization energy of F₂ and F

From the MO diagram, the highest occupied molecular orbital (HOMO) in \(\mathrm{F}_{2}\) is the nonbonding π*(2p) orbital. In atomic fluorine, the outermost electron is found in the 2p orbital. To remove an electron from \(\mathrm{F}_{2}\), we need to remove one of the electrons from the nonbonding π*(2p) orbital, which has a higher energy level than the bonding orbitals. In atomic fluorine, electrons are in the 2p orbital which is lower in energy than the nonbonding orbitals of \(\mathrm{F}_{2}\), making it easier to remove an electron from atomic fluorine. Therefore, the first ionization energy of atomic fluorine is lower than that of \(\mathrm{F}_{2}\).

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Most popular questions from this chapter

Consider the molecular orbital electron configurations for \(\mathrm{N}_{2}\), \(\mathrm{N}_{2}^{+}\), and \(\mathrm{N}_{2}^{-}\). For each compound or ion, fill in the table below with the correct number of electrons in each molecular orbital.

Complete the following resonance structures for \(\mathrm{POCl}_{3}\). (A) (B) a. Would you predict the same molecular structure from each resonance structure? b. What is the hybridization of \(\mathrm{P}\) in each structure? c. What orbitals can the \(\mathrm{P}\) atom use to form the \(\pi\) bond in structure \(\mathrm{B}\) ? d. Which resonance structure would be favored on the basis of formal charges?

Determine the molecular structure and hybridization of the central atom \(\mathrm{X}\) in the polyatomic ion \(\mathrm{XY}_{3}+\) given the following information: A neutral atom of \(\mathrm{X}\) contains 36 electrons, and the element \(Y\) makes an anion with a \(1-\) charge, which has the electron configuration \(1 s^{2} 2 s^{2} 2 p^{6}\).

Draw the Lewis structures for \(\mathrm{SO}_{2}, \mathrm{PCl}_{3}, \mathrm{NNO}, \mathrm{COS}\), and \(\mathrm{PF}_{3}\). Which of the compounds are polar? Which of the compounds exhibit at least one bond angle that is approximately 120 degrees? Which of the compounds exhibit \(s p^{3}\) hybridization by the central atom? Which of the compounds have a linear molecular structure?

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. a. CN or \(\mathrm{NO}\) b. \(\mathrm{O}_{2}^{2+}\) or \(\mathrm{N}_{2}^{2+}\)
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