# What is the net ionic equation for rection of liquid bromide with sodium iodide crystals?

I have this problem,

Liquid bromine is added to a container of sodium iodide crystals

I interpreted it with this molecular formula $\ce{Br_{2(l)} + 2NaI_{(s)} \rightarrow 2NaBr_{(s)} + I_{2(g)}}$.

From my understanding, since none of the products nor any of the reactants is aqueous, there shouldn't be a net ionic equation. However, my friend has proceeded with the problem as follows:

$\ce{Br_2 + 2Na^+ + 2I^- \rightarrow 2Na^+ + 2Br^- + I_2}$

$\ce{Br_2 + 2I^- \rightarrow 2Br^- + I_2}$.

Could somebody explain which one of us is correct and why?

• First, iodine probably shouldn't be a gas, because it exists as a solid under standard conditions. As for how the net ionic equation should be written, I think it's a matter of convention. When I took general chemistry I, we were instructed to separate soluble ionic compounds and cancel spectator ions ($\ce{Na+}$) irrespective of whether the reaction was explicitly taking place in aqueous solution. This is a redox reaction, and sodium is not undergoing change in oxidation state, so that seems right to me. Hopefully somebody weighs in with a conclusive answer as to what the convention is. – Greg E. Aug 21 '13 at 22:51