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Since radicals already violate the octet rule, does it really matter where the extra lone electron is placed? for example in the Methyl radical, CH3, the extra electron is put on the carbon, but could it be placed on one of the hydrogens?
It does matter as different atoms are more electronegative than others and have different needs to become stable. In your example with CH3, H already has a bond thus giving it a full shell of electrons, making it improbable that it would have that radical. Carbon on the other hand is lacking a bond, as carbon normally has four binds but here has three, and thus it make some it logical to place the lone electron there.
An easy way to help figure out where the lone electron goes is to see which element will have a formal charge close to zero depending on the placement of the lone electron. For instance, in CH3, the lone electron would go on the C since the formal charge would be 0 as it is more favorable too. H can only have one bonding pair. Hope that helps!
One useful thing to keep in mind is that hydrogen will almost never have the radical because its valence is already full when it is involved in a single bond. This goes along with the formal charge point that the others above have mentioned.
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