CHEMISTRY COMMUNITY

I've been trying to do this homework problem and seriously struggling. I understand what a radical is but I cannot seem to figure out why NO2- is a radical. I drew the lewis structure but I don't see any unpaired electrons. Can anyone explain why NO2- is a radical?

In the answer key, it says it’s not a radical.

Are radicals that important for the midterm?

Jorge Ramirez_4H wrote:Are radicals that important for the midterm?

I'd like to know this as well. Will we be expected to select what compounds are radicals or not? And does this knowledge have some other application?

NO2- is not a radical. It has 18 electrons, all of which are paired.

Alexis Webb 1B wrote:In the answer key, it says it’s not a radical.

Weird. On my sapling online textbook answer key it said it is a radical:

2C.1 Only (a) and (b) are radicals

But I understand that it isn't a radical.. maybe just a mistake?

For this question, it is evident that only B and C are radicals because they have one electron leftover that isn't paired. The answer key must be a mistake since NO2- does not have any radicals in it. It is important to note that when drawing lewis structures, you should be able to see an odd number of electrons for one to be leftover for a radical.

NO2- has 18 electrons. Since 18 is an even number, there aren't going to be any unpaired electrons, so it would not be a radical. NO2 on the other hand would have an unpaired electron, so it would be a radical.

Can someone explain how d isn't a radical. The structure should be of HOCO. I did it and i got H-O-C-O with 2 pairs of electrons on each O and with one lone pair on C, which should make it a radical right?