In ozone, O3, the two oxygen atoms on the ends of the molecule are equivalent to one another. (d) How many electrons are delocalized in the p system of ozone?

Video transcript

Hello everyone in this video. We're going to determine the number of the localized electrons and the price system of format. So let's first start off by drawing the LewiS structure for format. It's going to have a central atom of carbon Singapore into a hydrogen. Take up to an oxygen. Of course it has three lone pairs Wearing a formal church and double bonded to an oxygen with two phone pairs. Alright, so of course there's going to be resident structures. The structures in a different color to our first breed. Draw the lower structure for this for meat. So the carbon bond to an oxygen with two lone pairs, single bond to one oxygen With three lone pairs and a hydrogen. Alright, so join the arrow, pushing so we'll have one of them, the loan paris here come and make a double bond here and then we'll have one of these pie pons. Go ahead and break off to make a long pair. So be something like this because of that sense. We'll have our resident structure to look like this. So these are our resonance structures. And this present structure involves one C. O bond or pi bond and one long pair from the oxygen to do this long to do this resonant structure. So the C. O bond here and this long pair here of oxygen. So the pi bond is of course two electrons and then a lone pair has two electrons. So the pi bond that we're dealing with from the carbon and oxygen here that has two electrons. And then the lone pair. So they lone pair from this oxygen here. First that's going to have two electrons as well, So two plus 2, It goes to four. So that's a total of four total de localized electrons. So our answer here is four. Do you localized electrons? Because the two electrons from the pi bond and the two electrons from our lone pair are what are being used to create our resident structure. And so this is the final answer for this problem.