When it comes to the reaction with halogens, we're going to say the Bourne family metals react with halogens to produce ionic halid solids. Now if we took a look here, we have in Group 3A all our metals and then thalium is different. So it's separate. They are plus 3IN charge. And remember halogens on group 7/8, so they're -1 in charge. This helps us to predict the product that will form.

The three would come here and the one would come here. Our product will be M3. We'd have to balance this out. We'd put a 2 here to give us 6 halogens on the product side. So we'd have to put a three here and A2 here. Thalium. Remember thallium is different. It's preferred oxidation state is not +3 but plus one, and then the halogens are still -1. The numbers in the charges are the same, so they just combine together to give us TLX.

Now here we'd have to balance this out. We have two halogens on the reactant side, but only one here. So we put a 2 here, which means I'd have to put a 2 here with thallium and then we could put A1 here for the diatomic halogen. This would represent our different reactions when these boron family metals react with our diatomic halogens. In all the reactions, we're going to produce an Ionic Halid solid.