Free radicals are important in many environmentally significant reactions (see the Chemistry in the Environment box on free radicals in this chapter). For example, photochemical smog— smog that results from the action of sunlight on air pollutants— forms in part by these two steps:

The product of this reaction, ozone, is a pollutant in the lower atmosphere. (Upper atmospheric ozone is a natural part of the atmosphere that protects life on Earth from ultraviolet light.) Ozone is an eye and lung irritant and also accelerates the weathering of rubber products. Rewrite the given reactions using the Lewis structure of each reactant and product. Identify the free radicals.

Video transcript

Hey everyone, we're given information regarding chlorofluorocarbons, which are abbreviated as CFCs. And these are derivatives of hydrocarbons. Where all of the hydrogen atoms have been replaced by chlorine and flooring were asked to identify which species in the above reactions are radicals. And then rewrite these reactions using the Lewis structures first. Let's go ahead and determine the total number of valence electrons for each of our compounds or adam for chlorine. We know that this is going to give us seven valence electrons since it's in our group seven A for ozone, we're going to multiply six times three since oxygen is in our group six A and we have three of oxygen and this will get us to a total of 18 valence electrons. Now looking at our chlorine monoxide, We know that chlorine is going to give seven valence electrons and oxygen is going to give six. So this will get us to a total of 13 valence electrons. And lastly looking at her oxygen, We're going to multiply six valence electrons times two since we have two of oxygen and this will get us to a total of 12. Now looking at these values, we can recognize that chlorine and chlorine monoxide both have odd numbers as their total number of valence electrons. So this means that these are going to be our radicals now that we've identified our radicals. Let's go ahead and draw out our lewis dot structures. Starting with our first reaction, we have our chlorine and we know that this is going to be one of our radicals. So we will draw out seven vans electrons drawing out our ozone. We have one oxygen that is double bonded to one oxygen and single bonded to another oxygen. And the oxygen that is double bonded will have two lone pairs while the other one that is double bonded and single bonded to an oxygen will have a plus one charge, While the last oxygen will have three lone pairs and a -1 charge. So our overall formal charge for this is going to be zero. Next we have our chlorine monoxide and we have chlorine attached to an oxygen chlorine will have three lone pairs and oxygen will have two lone pairs plus that one radical. And lastly we have our oxygen and we're going to have to oxygen's double bonded to one another and each with two lone pairs. Now let's go ahead and draw out our next reaction. This time we have chlorine monoxide as our reactant and we're going to draw out the same structures as we did in the previous reaction. So chlorine will have three lone pairs and it's single bonded to an oxygen with two lone pairs and one radical. We're going to add our ozone, which will be one oxygen attached to two oxygen's. The oxygen in the middle will be double bonded to one oxygen with two lone pairs and the oxygen in the middle will have a plus one charge. And the oxygen that is single bonded to that middle oxygen will have three lone pairs and a minus one formal charge. Moving on to our product side, we have our chlorine with seven valence electrons and this is going to have that radical and we add on our two oxygen's, so we'll write that coefficient of two since we have it listed in our reaction, and we have an oxygen double bonded to another oxygen, and these two oxygen's have two lone pairs. So this is going to be our final answer for this question. And again, our radicals were chlorine and chlorine monoxide. So I hope that made sense and let us know if you have any questions.