Where to place radicals

[David Chibukhchian 2G]

Hey guys, I had a question about octet exceptions. Whenever a molecule has a radical, how do we decide which atom gets that one electron? I think we would have to use formal charge, but I'm still unsure because I don't know if we would still have to use that process. If someone could give advice for figuring out where to place the radical that would be really helpful!

[Zaid Bustami 1B]

Hey David,
Yea it's kind of weird because for the textbook problems, it seemed like formal charge wasn't what determined which atom has the unpaired electron. However, from what I've seen it seems like the most electronegative atom will want the full octet, so the less electronegative species will likely have the unpaired electron. Determining which atom gets less than an octet isn't about minimizing formal charge; the atoms' favoring of the unpaired e- takes precedent over minimizing formal charge. This is a good example from the textbook:

Even if formal charge would be minimized having O with the unpaired e-, I think O is so much more electronegative that it is more unfavorable for it to have an unpaired electron than to have a formal charge of -1. Considering the unpaired e- favorability for the Cl, formal charges are as stable as they can be because at least O, which is more electronegative, has the negative formal charge.
I think the OH radical is an exception here though, because I don't think it's favorable for H to get an extra e- past its valence shell and steal it away from the O. There simply aren't enough electrons to go around for even Oxygen to have a full valence shell, so the O will get as close as it can to having a full octet, becoming the atom that has the unpaired electron in the process.
I guess a general rule to follow is that the most electronegative atom will pull electrons to itself such that filling its valence shell takes priority.

[David Chibukhchian 2G]

Okay, that makes sense. I was also confused with that example, because I'd expect oxygen to have the radical and thus allow for a formal charge of 0 for both atoms. However, I guess that in this situation we just attach the radical to whichever atom is less electronegative like you said. I also see why OH would be an exception to this. I'll definitely keep that in mind, thank you for the visual!

[t_rasul2I]

Hey David,
Yea it's kind of weird because for the textbook problems, it seemed like formal charge wasn't what determined which atom has the unpaired electron. However, from what I've seen it seems like the most electronegative atom will want the full octet, so the less electronegative species will likely have the unpaired electron. Determining which atom gets less than an octet isn't about minimizing formal charge; the atoms' favoring of the unpaired e- takes precedent over minimizing formal charge. This is a good example from the textbook:
radical.jpg
Even if formal charge would be minimized having O with the unpaired e-, I think O is so much more electronegative that it is more unfavorable for it to have an unpaired electron than to have a formal charge of -1. Considering the unpaired e- favorability for the Cl, formal charges are as stable as they can be because at least O, which is more electronegative, has the negative formal charge.
I think the OH radical is an exception here though, because I don't think it's favorable for H to get an extra e- past its valence shell and steal it away from the O. There simply aren't enough electrons to go around for even Oxygen to have a full valence shell, so the O will get as close as it can to having a full octet, becoming the atom that has the unpaired electron in the process.
I guess a general rule to follow is that the most electronegative atom will pull electrons to itself such that filling its valence shell takes priority.

I liked how you explained this problem! I thought O would have the radical because it would make the formal charge of the molecule 0-0. How do we know when radicals will be affected by electronegativity?

[Zaid Bustami 1B]

I think the way to think about it is to prioritize giving the most electronegative atom an octet. If you use this approach to problems I think it will work (at least it works both with the ClO and OH radicals), but I'm not sure as to when electronegativity stops becoming a factor or what happens when you get to expanded octets. If you have a nonpolar covalent bond, I really don't know which atom would be left with the lone electron, but my guess would be that it would be the less electronegative one even if the difference might be small. Then again I don't know for sure if this rule works for every problem, so who knows maybe there's a better way to predict which atom gets the unpaired electron.

[Aaina 2D]

Hi! I believe when given a radical, the atom which is the most electronegative gets the full octet. So if you are deciding between giving one electron to which atom, go with the most electronegative one. This is because their pulling power of electrons is stronger, so therefore, the more electronegative an atom is, the more likely it would form an octet.