Here it says when setting up the steps of the Bornhaber cycle for potassium oxide, how many ionization energies IE and how many electron affinities EA do you need? Remember we have to get them into their ionic forms right? And we need two potassium ions as gas to combine with one oxide ion as a gas. Then they will combine to give me potassium oxide solid.

Realize here that we need oxygen to get a 2 minus charge. So we're not adding one electron, we're adding two electrons. That means that we're going to need to do 2 electron affinities O we need 2 electron affinities, which means this is out, this is out, and this is out. And then finally, what else do we need?

Well, we have one potassium solid initially, but at the end we're going to need 2 potassiums. So we're going to have to do ionization energy twice, one for each of the potassium solids, and then both of them would then undergo, oops, we'll have them as gases actually, and then each one will have its electron plucked off. So we need to do 2 ionization energies, 1 for each Potassium as a neutral gas.

So we need to do 2 electron affinities and two ionization energies, giving us E as our final answer.