# What salt should be formed by the interaction of aluminum metal and iodine?

Jan 12, 2017

This is a poor question. They ask you to predict a formula of ${\text{AlI}}_{3}$, $\text{aluminum triiodide}$.

#### Explanation:

$\text{Aluminum}$ is a Group 13 metal; it is commonly oxidized to the $A {l}^{3 +}$ cation:

$A l \rightarrow A {l}^{3 +} + 3 {e}^{-}$

On the other hand, iodine is a Group 17 non-metal. It is commonly reduced to give the iodide anion:

$\frac{1}{2} {I}_{2} + {e}^{-} \rightarrow {I}^{-}$

When aluminum and iodine make music together, i.e. combine in a redox reaction, 3 equiv of electrons of electrons are transfered:

$A l + \frac{3}{2} {I}_{2} \rightarrow \left\{A {l}^{3 +} {I}_{3}^{-}\right\}$. And $\left\{A {l}^{3 +} {I}_{3}^{-}\right\} \equiv A l {I}_{3}$

We know that most matter is electrically neutral, and all I have done here is to draft enough iodide ions, formally anionic, to neutralize the charge of the $A {l}^{3 +}$ cation.