Hey, everyone. Here it says, draw a protonated form of imidazole, hint: draw both resonance structures. Alright. So remember, we need 6 pi electrons in order to make an aromatic structure. This nitrogen down here is single bonded, no double bonds.

Its lone pair will be used as part of the pi system. Therefore, they're not available to be used to accept a proton. The nitrogen over here, its lone pairs are on the outside. We don't need them because we already have our 6 pi electron system, so they are going to be free to accept an H⁺. So if I'm going to draw the protonated form, I'm going to show that nitrogen as being protonated.

This nitrogen here is not. Right now, this nitrogen is making four bonds, so therefore it's positively charged. This will be our first resonance structure. Now if we want to draw our second resonance structure, just remember that this lone pair here could resonate here to make a double bond, kicking this double bond to the nitrogen. This will result in our second resonance structure.

The lone pairs got kicked to the nitrogen as a lone pair. Here goes our hydrogen. This nitrogen is now making a double bond. It's making four bonds, so now it's positively charged. And this would represent our second resonance structure.

But again, it all starts out with drawing this first resonance structure. And remember, whether our lone pair acts as part of the pi system or not determines if it'll accept a proton. If it's not part of the pi system, it's free to accept a proton, which we saw the top right nitrogen do to give us our 1st resonance structure or 1st protonated form, and then we just draw its resonance structure to give us our second form. Alright. So these will be our two final answers.