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We first consider the molecular orbital diagram for \(\mathrm{F}_{2}\). The atomic number of fluorine is 9, so each fluorine atom has 9 electrons. Using Hydrogen as a base, counting up in atomic number we have that F has \(1s, 2s, 2p\) orbitals. In the molecular orbital diagram, the sigma bonds formed from the overlap of the atomic orbitals are lower in energy than the atomic orbitals, while the pi bonds formed from the overlap of the atomic orbitals are higher in energy than the atomic orbitals. Using the Aufbau principle and Hund's rule, we fill in the molecular orbitals from the lowest to the highest energy levels. For \(\mathrm{F}_{2}\), the molecular orbital electron configuration is: 1. \(\sigma_{1s}^2\sigma_{1s}^{*2}\sigma_{2s}^2\sigma_{2s}^{*2}\sigma_{2p}^2\pi_{2p}^4\pi_{2p}^{*2}\)

Now that we have the molecular orbital electron configuration for \(\mathrm{F}_{2}\), we can determine the electron configurations for \(\mathrm{F}_{2}^{+}\) and \(\mathrm{F}_{2}^{-}\) by adding or removing electrons: 1. \(\mathrm{F}_{2}^{+}\): Remove 1 electron from the highest occupied molecular orbital (HOMO): \(\sigma_{1s}^2\sigma_{1s}^{*2}\sigma_{2s}^2\sigma_{2s}^{*2}\sigma_{2p}^2\pi_{2p}^4\pi_{2p}^{*1}\) 2. \(\mathrm{F}_{2}^{-}\): Add 1 electron to the lowest unoccupied molecular orbital (LUMO): \(\sigma_{1s}^2\sigma_{1s}^{*2}\sigma_{2s}^2\sigma_{2s}^{*2}\sigma_{2p}^2\pi_{2p}^4\pi_{2p}^{*3}\)

The bond order is calculated using the following formula: Bond order = \(\frac{(\text{number of bonding electrons} - \text{number of antibonding electrons})}{2}\) Using the molecular orbital electron configurations determined, we can calculate the bond orders for the three species: 1. \(\mathrm{F}_{2}^{+}\): Bond order = \(\frac{(10 - 7)}{2} = 1.5\) 2. \(\mathrm{F}_{2}\): Bond order = \(\frac{(10 - 6)}{2} = 2\) 3. \(\mathrm{F}_{2}^{-}\): Bond order = \(\frac{(10 - 9)}{2} = 0.5\)

The bond length is inversely proportional to the bond order. Therefore, as bond order increases, the bond length decreases. Based on the bond orders calculated above: 1. \(\mathrm{F}_{2}^{+}\) has a bond order of 1.5, so its bond length would be shorter than \(\mathrm{F}_{2}^{-}\) and longer than \(\mathrm{F}_{2}\). 2. \(\mathrm{F}_{2}\) has a bond order of 2, so its bond length would be the shortest among the three species. 3. \(\mathrm{F}_{2}^{-}\) has a bond order of 0.5, so its bond length would be the longest among the three species.

Unpaired electrons are found in the molecular orbitals with an odd number of electrons. We can now determine the number of unpaired electrons for the three species based on their molecular orbital electron configurations: 1. \(\mathrm{F}_{2}^{+}\): There is 1 unpaired electron in the \(\pi_{2p}^{*}\) orbital. 2. \(\mathrm{F}_{2}\): There are no unpaired electrons. 3. \(\mathrm{F}_{2}^{-}\): There is 1 unpaired electron in the \(\pi_{2p}^{*}\) orbital.