# What would the chemical formula of the compound look like if you combined Calcium and Iodine?

Mar 4, 2017

The formation reaction is:

$C a \left(s\right) + {I}_{2} \left(s\right) \rightarrow C a {I}_{2} \left(s\right)$

#### Explanation:

$C a \left(s\right) + {I}_{2} \left(s\right) \rightarrow C a {I}_{2} \left(s\right)$

This is in fact a redox reaction: calcium metal metal is oxidized up to $C {a}^{2 +}$, and ${I}_{2}$ is reduced to iodide anion.

How do you know this occurs save by memory? Another way is to consider their positions in the Periodic Table. Calcium is a Group 2 metal, that has 2 valence electrons; its reactivity is chatacterized by its tendency to LOSE its valence electrons to form the $C {a}^{2 +}$ ion. And iodine, as an oxidizing Group 17 non-metal, tends to GAIN electrons to form the iodide ion, ${I}^{-}$. Electrical neutrality demands that the salt of $C {a}^{2 +}$ and ${I}^{-}$ form an a $1 : 2$ ratio. Why?