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Chapter 10: Problem 57

Use molecular orbital theory to compare the relative stabilities of \(\mathrm{F}_{2}\) and \(\mathrm{F}_{2}^{+}\)

Short Answer

Expert verified
Using molecular orbital theory, we can determine that \(\mathrm{F}_{2}^{+}\) is more stable than \(\mathrm{F}_{2}\) due to it having a lesser number of electrons in an antibonding orbital and a higher bond order.

Step by step solution

01

Construct the MO diagrams

The first thing to do is to construct the molecular orbital diagram for \(\mathrm{F}_{2}\) and \(\mathrm{F}_{2}^{+}\). Both \(\mathrm{F}_{2}\) and \(\mathrm{F}_{2}^{+}\) have similar MO diagrams because they are made up from the same atomic orbitals of fluorine atoms. There are five molecular orbitals formed: - two bonding orbitals (蟽 and 蟽*)- two non-bonding orbitals (two degenerate p orbitals)- one antibonding orbital (蟺*)
02

Fill the electrons

The second step is to fill these molecular orbitals with the available electrons. \(\mathrm{F}_{2}\) has 18 electrons and the filling order is 蟽, two degenerate p orbitals then 蟽*. This results in all orbitals being filled. For \(\mathrm{F}_{2}^{+}\), it has 17 electrons. The filling order is the same but the 蟽* orbital will have one electron less compared to \(\mathrm{F}_{2}\).
03

Compare the final MO diagrams

With all the electrons filled in the MO diagrams, we then compare the electron configuration of both species. The \(\mathrm{F}_{2}^{+}\) ion, with one electron less in an antibonding orbital, will have a lower energy and hence be more stable than \(\mathrm{F}_{2}\).
04

Confirm stability with bond order

Lastly, to confirm the relative stabilities, the bond order for both species can be calculated. Bond order refers to the number of chemical bonds between two atoms. The bond order can be found by subtracting the number of electrons in antibonding orbitals from the number in bonding orbitals and divide by 2. The bond order for \(\mathrm{F}_{2}\) is 1, and for \(\mathrm{F}_{2}^{+}\), it's 1.5, which further confirms \(\mathrm{F}_{2}^{+}\) is more stable than \(\mathrm{F}_{2}\).

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