Sure, if you have an odd number of electrons, there is no chance for it to be close-shell.

yes, it cannot be closed shell.

So it is to be considered as a radical or a charged species?@

If the C atom is replaced with a Li atom without any charge adding or decreasing. It should be considered as a radical. The command line of charge and spin multiplicity should be changed to "0 2".

it is a doping, so charges can be considered, right? Yijian Ma

If you calculate it as a charged species, then for a real material another counterion would be required. Otherwise you calculate a hypothetical [LiC59]+/- species which would not exist outside of an ion beam, and even there it would be tough to make since fullerenes prefer to fragment by C2n loss.

I would expect that in reality this species will oligomerize quite fast. Non-IPR fullerenes like C58 do not exist as isolated cages in condensed matter, and these aren't even necessarily radicals. See

Article Desorption of Fullerene Dimers upon Heating Non-IPR Fulleren...

and references therein for that comparison.

Therefore, if you want to calculate a hypothetical free-standing LiC59 cage, you will have to choose a doublet/quartet/sextet/... multiplicity, just like Yijian Ma wrote.

Edit: if you want to simulate alkali metal doped fullerenes, that's a whole different thing. In that case you do not replace a carbon atom by Li, you just add it, so you still have the C60 cage intact and it will just end up with a charge on it. For LiC60, that might pose the question whether in a bulk solid you would have a 50/50 mixture of C600 and C602- or really C60- which would be a radical again (close-shell Li+). Maybe someone already simulated that, checking for literature on that aspect would be a good idea before starting lengthy calculations.

I need to consider the Li doped fullerene C59Li, by replacing one C atom Jürgen Weippert

OK, then ignore everything after the "Edit" in my previous post.

Thank you Jürgen Weippert Yijian Ma