Fluorine, F₂, has 18 electrons. Write its configuration and find its bond order…

Step 1. Configuration. $\sigma 1s^2$ $\sigma^* 1s^2$ $\sigma 2s^2$ $\sigma^* 2s^2$ $\sigma 2p_z^2$ $\pi 2p_x^2$ $\pi 2p_y^2$ $\pi^* 2p_x^2$ $\pi^* 2p_y^2$.

Step 2. $\text{F}_2$. Bonding $10$, antibonding $8$: $\dfrac{10 - 8}{2} = 1$.

Step 3. $\text{F}_2^+$. It loses a $\pi^*$ electron: $\dfrac{10 - 7}{2} = 1.5$.

Step 4. Why. The lost electron was antibonding, so the pull holding the nuclei grows.