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Basically a Lewis structure with one lone pair would be considered a radical. They tend to be really reactive because electrons like to form bonds instead of being a lone pair. You would say your compound is radical when your Lewis structure ended up having one lone electron.
If you are asked to do a Lewis structure of a molecule with a radical, one of the biggest things to consider is the formal charge. Even though you have an odd number of electrons, it is still possible for some molecules to do a Lewis structure without any formal charges and you should make sure to write your Lewis structure off of that.
As far as radical placement goes, I believe we can look at electronegativity. According to the textbook, an atom of the element with the higher electronegativity has a stronger pulling power on electrons and tends to pull them away from the atom of the element with lower electronegativity. Also, I remember at the end of the T.A-led Friday lecture they mentioned we can also use formal charge to decide where to place the radical. I'm not sure if it's better to use electronegativity or formal charge to determine radical placement though.
Just to add to this discussion, from my knowledge, if you were to count the number of valence electrons in a compound such as NO, nitrogen has 5 valence electrons and oxygen has 6 valence electrons. In total, the entire compound's total valence electron is 11. Therefore, there must be an electron that is unpaired since electrons usually come in pairs. So there will be 5 pairs of electrons and 1 electron that is unpaired, making 11 electrons. So, there is a radical. I am not sure if the NO compound even exists unless it's some sort of ion, but on a problem these are the steps you should follow to find if there is or is not a radical.
I believe you place the radical where it makes sense to due to formal charge.alexagreco1A wrote:I understand the concept of a radical, but when drawing Lewis structures, how do we decide which element has the unpaired electron?
The way to tell that something is a radical is that it will only have a lone electron (as stated above). How it affects formal charge is now you have an odd number of electrons. Say you have nitrogen (which wants 5 electrons) and it is bonded to some other element, has a lone pair, and a single electron (note: this is some hypothetical, I am not sure if there is such a molecule that exists to make this situation possible. It is only to demonstrate the effect radicals have on formal charge). The formal charge of the nitrogen in this situation would then be +1. You get this by subtracting the number of electrons nitrogen wants (5), by the number of electrons involved. You have two electrons from the lone pair, 2(1/2) from the bond, and one electron from the lone electron. This gives you 5-(2+1+1) = +1. If nitrogen in this situation was not a radical, and just had the one bond and lone pair, the formal charge would work out to be 5-(1+2) = +2.
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