Question

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?

Short answer

In conclusion, the bond orders for \(\mathrm{F}_{2}^{+}\), \(\mathrm{F}_{2}\), and \(\mathrm{F}_{2}^{-}\) are 1.5, 2, and 2, respectively. The relative bond lengths are longest for \(\mathrm{F}_{2}^{+}\), and the same shorter length for \(\mathrm{F}_{2}\) and \(\mathrm{F}_{2}^{-}\). There is one unpaired electron in \(\mathrm{F}_{2}^{+}\) and no unpaired electrons in \(\mathrm{F}_{2}\) and \(\mathrm{F}_{2}^{-}\).

Step-by-step solution

Write electron configurations for each species

We start by writing the electron configurations of the three fluorine species based on the molecular orbital diagrams for homonuclear diatomic molecules. The order of molecular orbitals is as follows: \(\sigma_{1s}, \sigma_{1s}^{*}, \sigma_{2s}, \sigma_{2s}^{*}, \sigma_{2p}, \pi_{2p_x}, \pi_{2p_y}, \pi_{2p_x}^{*}, \pi_{2p_y}^{*}, \sigma_{2p}^{*}\). The electron configuration for \(\mathrm{F}_2\) (with 14 valence electrons since each F atom has 7 valence electrons) is: \(\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^{*}{}^{\ }\pi_{2p_y}^{*} {}^{\ }\sigma_{2p}^{*}{}^{\ }\). For \(\mathrm{F}_2^+\) (with 13 valence electrons), the electron configuration is: \(\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^{*}{}^{\ }\pi_{2p_y}^{*} {}^{\ }\sigma_{2p}^{*}{}^{\ }\). And for \(\mathrm{F}_2^-\) (with 15 valence electrons), the electron configuration is: \(\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^{*}{}^{\ }\pi_{2p_y}^{*}{}^{\ }\sigma_{2p}^{*1}\).

Calculate bond order for each species

The bond order (B.O.) is calculated as follows: \[\mathrm{B.O.}=\frac{1}{2} (\text{number of bonding electrons}-\text{number of antibonding electrons})\] For \(\mathrm{F}_2\): B.O. = \( \frac{1}{2} (10 - 6) = 2\) For \(\mathrm{F}_2^+\): B.O. = \( \frac{1}{2} (9 - 6) = 1.5\) For \(\mathrm{F}_2^-\): B.O. = \( \frac{1}{2} (11 - 7) = 2\)

Compare bond orders to predict relative bond lengths

Bond length is inversely proportional to bond order. As the bond order increases, the bond length decreases. In this case, \(\mathrm{F}_2^+:\) B.O. = 1.5, and \(\mathrm{F}_2, \mathrm{F}_2^-:\) B.O. = 2. Thus, the relative bond lengths are as follows: \(\mathrm{F}_2^+\) has the longest bond length, and \(\mathrm{F}_2\) and \(\mathrm{F}_2^-\) have the relatively same, shorter bond lengths.

Identify unpaired electrons in each molecular orbital

We analyze the electron configurations for each species to find any unpaired electrons. For \(\mathrm{F}_2\), there are zero unpaired electrons. For \(\mathrm{F}_2^+\), there is one unpaired electron (in the \(\pi_{2p_y}\) orbital). For \(\mathrm{F}_2^-\), there are zero unpaired electrons. In conclusion, the bond orders are 1.5 for \(\mathrm{F}_2^+\), and 2 for both \(\mathrm{F}_2\) and \(\mathrm{F}_2^-\). The relative bond lengths are \(\mathrm{F}_2^+\) (longest), and \(\mathrm{F}_2\) and \(\mathrm{F}_2^-\) are the same (shorter). There is one unpaired electron in \(\mathrm{F}_2^+\) and no unpaired electrons in \(\mathrm{F}_2\) and \(\mathrm{F}_2^-\).