Is F2 two fluoride ions or just one?

I'm attempting to write the balanced equation of the combination of lithium metal and fluorine gas. I understand that this will produce an ionic solid because it is a reaction of a metal and a nonmetal. However, I'm confused with the fluoride ion having a charge of -1 and fluorine only occurring as a diatomic molecule ($\ce{F2}$). Is $\ce{F2}$ two fluoride ions or just one?

• Oxidation state of $F_2$ is 0 but in $F^-$ it has -1 oxidation state hence the $\ce{F2 + 2e- ->2F-}$ May 22 '15 at 8:26

$\ce{F2}$ will produce $2\space \ce{F-}$ ions.

You can write the balanced half cell reactions as follows:

$\ce{F2 + 2e- ->2F-}$

$\ce{Li -> Li+ + e-}$

Multiply the second equation by 2 to get $\ce{2Li -> 2Li+ + 2e-}$. This is to ensure that the final equation does not have any electrons on either side.

Now add this equation with the fluorine one.

You'll get $\ce{2Li + F2 -> 2Li+ + 2F-}$ which you can write as $\ce{2Li + F2 -> 2LiF}$

The fluorine atom is far to reactive to exist on its own, so naturally it bonds with other atoms in its vicinity to stabilize itself. It should be noted that an atom is electrically neutral, meaning that it has an equal number of protons as electrons. A fluorine atom can share electrons with another nonmetal to form a covalent bond. When sharing electrons in a covalent bond, the fluorine is still called an atom, because it has not gained complete control over the electron that is completing its $2p$ orbital. In the $\ce{F2}$ molecule, there are to fluorine atoms.

The fluoride ion is a result of a fluorine atom completely gaining control over an electron, which is usually donated by a metal. Electrons can be transferred from a metal to the fluorine atoms to form metal ions and fluoride ions. The reaction you are talking about would look like this:

$$\ce{2Li^{(0)} +F2^{(0)} -> 2Li^{(+1)}F^{(-1)}}$$

I included oxidation numbers to show how each species has gained/lost electrons, though this would normally not be included in the formula.

Fluorine gas in the diatomic $\ce{F2}$ is essentially the the covalent pair of fluorine atoms (as fluorine is a non-metal), as seen in the diagram below: Image source

In the reaction you describe:

$$\ce{2Li(s) + F2(g) ->2LiF(s)}$$

requires the fluoride ion be in an ionic bond with the lithium ion, twice (to balance the equation). Image source

• States of aggregation should not be subscripted, it is not wrong, but the recommendations (Sec. 2.1.) are different. May 22 '15 at 3:23
• @Martin-マーチン no problems, easily fixed :)
– user15489
May 22 '15 at 3:26
• Just pointing that out, I did not know it a month ago, so I want other people to profit from my knowledge :D And I think it's easier to read ;) May 22 '15 at 3:38
• It is a very good point too - I learnt something new and I agree, it is clearer
– user15489
May 22 '15 at 3:44

Fluorine in its elemental form is $\ce{F2}$, a neutral molecule. Every fluorine has seven valence electrons and after formation of the molecule, there is one covalent, i.e. electron sharing bond, and the remaining six valence electrons at each fluorine can be regarded as lone pairs.

When elemental fluorine reacts with a metal it will form an ionic compound just as you stated. For this to happen, both elements involved in the reaction have to be ionised, i.e. they have to gain or lose electrons. This is called a redox reaction.

In this particular case, lithium will transfer one of the electrons to fluorine, resulting in the following equations: \begin{align} \ce{Li &~<=> Li+ + e-}&&|\cdot 2\\ \ce{F2 + 2e- &~<=> 2F- }\\\hline \ce{2Li + F2 &~<=> 2LiF}\\ \end{align}