# What is the valence molecular orbital of the ions of a fluorine molecule?

I have the following assignment:

Draw the molecular orbital diagram of $\ce{F2}$, $\ce{F2+}$ and $\ce{F2-}$. Calculate the bond order and say, if there is a bond, which compound has the strongest bond and which one has the longest one.

As my understanding goes, $\ce{F}$ has $7$ valence electrons in the $2s^2$ and $2p^5$ subshells. By combining the atomic orbitals, one will obtain:

$2$ $\sigma_{2s}$ orbitals (a bonding and an antibonding), 2 $\sigma_{2p}$ a bonding and an antibonding, and $4$ $\pi_{2p}$ (2 degenerate bonding and 2 degenerate antibonding).

By filling the MO, I obtained the following bond orders: 1 for $\ce{F2}$ and $\frac{1}{2}$ for $\ce{F2+}$ and $\ce{F2-}$.

First of all, I would say that there is definitely a bond in all the three compounds since the bond order is positive. Then I would say that $\ce{F2}$ has the strongest bond (since it has the highest bond order), then $\ce{F2+}$ and $\ce{F2-}$ have longer bonds than $\ce{F2}$ due to the lower bond order, however I do not know which one has the longest bond.

### As for the question:

I never came across $\ce{F2+}$ and $\ce{F2-}$. Is my reasoning correct? Do these compounds exist? Furthermore, I draw the MO using an atom of $\ce{F}$ and an ion of $\ce{F}$ (cation and anion depending on the case). Is the procedure correct?

• Your procedure sounds correct. If a compound has some nonzero bond order, it should be possible to form so you can say $\ce{F2+}$ and $\ce{F2-}$ could exist. – Tyberius Dec 9 '17 at 19:54