# What is the product of this combination reaction? Al(s) + I_2(s) ->?

$A l \left(s\right) + \frac{3}{2} {I}_{2} \left(s\right) \rightarrow A l {I}_{3} \left(s\right)$
While a subvalent aluminum iodide is known, the material you would put into a bottle is $\text{aluminium(III) iodide}$. Like the chloride, $A l {I}_{3}$ is dimeric in the solid state to give $A {l}_{2} {I}_{6}$, i.e. ${\left\{{I}_{2} A l\right\}}_{2} {\left(\mu - I\right)}_{2}$.