Practice Problems 2C Number 1

[Jessica Phung 1E]

For part A of number 1, can someone please explain why NO2- is a radical? I thought radicals had an odd number of electrons in a compound meaning that there would be an unpaired electron however NO2- has an even number of electrons and the answer key says it's a radical. Also, in the same problem, part c is OH and the book doesn't recognize it as a radical however there is also an uneven number of electrons. I also checked the manual error...Thank you!

[Katelyn Alvarado 3K]

I'm not sure why the book considers NO2- a radical, but for 0H-, that isn't a radical because there isn't an unpaired electron anywhere in the molecule. There are a total of 8 electrons, 6 from O, 1 from H, and the extra one which contributes the negative charge of the molecule.

[Ruben Adamov 1E]

I agree with Katelyn about OH- not being a radical however the NO2- one kind of bewildered me. Maybe there was a typo and the problem was supposed to ask whether NO2 (without the minus) is a radical or not. In that case the answer is yes it is.

[Rose Arcallana 2B]

Jessica, I agree with you! I tried drawing it out manually and even looking it up. NO2- is a stable compound because the electron gives nitrogen a formal charge of 0 instead of a +1 if it didn't have it. I would say NO2 is a radical, but not NO2-. OH should also be radical because the Wednesday lecture gave it as an example to be a radical. OH- is not a radical, so I'm confused too!!

[Tess_Fleser_1D]

I'm having the exact same thought process. OH is definitely a radical and NO2- is definitely not. Maybe it's an error that not in the manual?