Please write a dissociation-in-water equation for the compounds $\ce{BeI2}$ and $\ce{LiI}$. Make sure to add the states of matter after each compound.
Currently, for $\ce{BeI2}$ I have the equation $$\ce{BeI2 (s) -> Be^2+ (s) + I2^2- (g)}.$$ I have yet to attempt the second one.
Your ionic charges are not correct for iodine. Looking at your attempt.
As $\ce{Be}$ is in group 2, the ionic charge for beryllium ion is fine, but iodine is in group 17, so its ion is $\ce{I-}$. When the ions dissociate, they become aqueous or (aq) as the state of matter.
Then the ionic charges need to balance, thus:
$$\ce{BeI2 (s) -> Be^2+ (aq) + 2I- (aq)}$$
To balance the ionic charges in this example, you need 2 $\ce{I-}$ (iodine ions) to balance the $\ce{Be^{2+}}$ (beryllium ion).
A similar example (and further explanations) are provided on the UC Davis ChemWiki page Unique Features of Aqueous Solutions (including an example of the dissolution of $\ce{MgCl2}$ - another compound with group 2 and 17 elements).
So,
Now, use the process to determine the dissociation of $\ce{LiI}$
By definition, a dissociation-in-water reaction results in aqueous, independent ions, not negatively charged diatomic molecules like the $\ce{I_2^2-}$ in the original question (not to mention that diatomic molecules are generally not negatively charged).
The answer should result in solely single-atom ions, as so:
$$\ce{BeI2 (s) -> Be^2+ (aq) + 2I- (aq)}$$
And using this same logic for $\ce{LiI}$:
$$\ce{LiI (s) -> Li+ (aq) + I- (aq)}$$
