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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
Using the MO energy-level diagram, F2 would have a lower first ionization energy than atomic fluorine. This is because F2's highest occupied molecular orbital (HOMO) is the 蟺*2p orbital, which has a higher energy than fluorine's 2p orbital. Consequently, it takes less energy to remove an electron from F2.

Step by step solution

01

Recall the Molecular Orbital (MO) theory

The Molecular Orbital theory helps us understand the bonding, anti-bonding, and non-bonding energies in a molecule. In an MO energy-level diagram, the x-axis represents the energy level of the orbitals, and the y-axis represents the positions of the atomic nuclei.
02

Draw the MO energy-level diagram for F2

Consider the electron configurations of fluorine atoms: Fluorine (F): 1s虏 2s虏 2p鈦� Since there are two fluorine atoms in F2, the total number of electrons is 14. The energy order for molecular orbitals in the second period of the periodic table is: 蟽1s < 蟽*1s < 蟽2s < 蟽*2s < 蟺2p < 蟺*2p < 蟽2p < 蟽*2p Start filling the molecular orbitals based on the energy levels, and follow the Pauli Exclusion Principle and Hund's rule: F2 MO electron configuration: (蟽1s)虏 (蟽*1s)虏 (蟽2s)虏 (蟽*2s)虏 (蟺2p)鈦� (蟺*2p)虏
03

Compare the ionization energies

To determine whether F2 has a lower or higher first ionization energy than atomic fluorine, consider the valence electrons in each species: - F2: The highest occupied molecular orbital is 蟺*2p and this is the orbital where electrons would be removed from F2. - Atomic fluorine: The electron configuration of atomic fluorine is 1s虏 2s虏 2p鈦�, so an electron will be removed from the 2p orbital. Now, let's compare the energies: - 蟺*2p in F2: In F2, 蟺*2p is an anti-bonding orbital with higher energy than the bonding 蟺2p orbital. - 2p in atomic Fluorine: In atomic fluorine, the 2p orbital is a valence shell electron with intermediate energy.
04

Conclusion

Since the HOMO (highest occupied molecular orbital) of F2 is the 蟺*2p orbital that has higher energy than fluorine's 2p orbital, it will require less energy to remove an electron from F2. Therefore, F2 has a lower first ionization energy than atomic fluorine.

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Most popular questions from this chapter

Cyanamide \(\left(\mathrm{H}_{2} \mathrm{NCN}\right)\), an important industrial chemical, is produced by the following steps: $$ \begin{aligned} \mathrm{CaC}_{2}+\mathrm{N}_{2} & \longrightarrow \mathrm{CaNCN}+\mathrm{C} \\\ \mathrm{CaNCN} & \stackrel{\text { Acid }}{\longrightarrow} \mathrm{H}_{2} \mathrm{NCN} \end{aligned} $$ Cyanamid 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 \(\mathrm{C}\) and \(\mathrm{N}\) 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 \(\mathrm{C}-\mathrm{N}\) 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.

Using the molecular orbital model to describe the bonding in \(\mathrm{F}_{2}{ }^{+}\), \(\mathrm{F}_{2}\), and \(\mathrm{F}_{2}^{-}\), predict the bond orders and the relative bond lengths for these three species. How many unpaired electrons are present in each species?

Describe the bonding in the \(\mathrm{CO}_{3}^{2-}\) ion using the localized electron model. How would the molecular orbital model describe the \(\pi\) bonding in this species?

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, \(\mathrm{B}_{2}\). 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 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?

In the molecular orbital model, compare and contrast \(\sigma\) bonds with \(\pi\) bonds. What orbitals form the \(\sigma\) bonds and what orbitals form the \(\pi\) bonds? Assume the \(z\) -axis is the internuclear axis.
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