Which one of the following is used to remove most of plutonium from spent nuclear fuel?

The removal of plutonium from spent nuclear fuel is achieved by converting insoluble plutonium dioxide, $$PuO_2, into a volatile hexafluoride, PuF_6. This allows gas-solid separation since PuF_6$$ sublimes at low temperatures.
Thus we require a reagent capable of strong fluorination at cryogenic temperature.

BrO$$_3^- (Option A) and I_2O_5 (Option B) are oxygen-containing oxidizers but contain no fluorine, so they cannot form PuF_6.
Chlorine trifluoride, ClF_3 (Option C), is a strong fluorinating agent but requires relatively high temperature and is less effective at -196\,^\circ\text{C}$$.

Dioxygen difluoride, $$O_2F_2$$ (Option D), is an extremely powerful oxidizing and fluorinating agent. It can convert $$PuO_2 to PuF_6 even at -196\,^\circ\text{C}. The balanced fluorination reaction is:
PuO_2 + 3\,O_2F_2 \rightarrow PuF_6 + 4\,O_2 Therefore, the reagent used to remove most of plutonium from spent nuclear fuel is Option D: O_2F_2.