When it comes to the reaction with halogens, we're going to say the nitrogen family elements react with halogens to produce trihalids and pentahalids. Now recall nonmetals from periods 3 or lower, so periods 3 and lower can have expanded octets. So if we're looking at this periods 3 and lower, well, nitrogen is in. 2 So nitrogen can't form A pentahalid can't have an expanded octet.
Also, we said nonmetals bismuth is metallic, so bismuth would not be included. Arsenic and antimony are metalloid, so they have nonmetal characteristics with them, so they can fit under this idea of having expanded octets. Now here, if we take a look at reactions with halogens, we have E here, which represents basically anything within that group, and here it's reacting with a diatomic element. So here we're going to say that E reacts with X2. It'll create EX3.
We need to balance this out. So we put a three here and A2 here and we have to put a 2 here. In this way, we've just created a Tri halid. We have our group 5A element and we have 3 halogens with it. So this is our Tri halid. This Tri halid, if it wanted, could continue onward and react with another mole of diatomic halogen. And it's here that those two additional halogens could be combined with this structure to give us our pentahalid here. It's already balanced, just coefficients of one across.
So what do we take from this example? Well, what we take from this is that we start out with two moles of our group 5/8 element reacting with three moles of our diatomic halogen in order to create our Tri halid. And then that Tri halid, if it want it could react with a fourth mole of our diatomic halogen to create this pentahalid. So All in all, it takes about a total of 4 moles of diatomic halogens to take a group 5/8 element all the way to a penta halid. So just look at these patterns and remember them in the creation of trihalids and pentahalids.