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it's because F it does not follow the rules to have an expanding octet meaning that it is not possible for F to take anymore electrons since it is in the second row (elements from third row and down can get expanding octets from d orbital)
Assuming you found the formal charges of both the normal structure and one with double bonds, the structure with 2 lone pairs (normal) will have a lower energy state than the one with double bonds, which is more appropriate. However, you can also explain this with the fact that Fluorine is in the 2nd row (2nd energy level). It can only fill itself with 8 electrons max because of the p orbital. Since F cannot expand into the d orbital because of its nature as a 2nd energy level element, it cannot acquire more than an octet of electrons.
The expanded octet rule can only apply to elements on the third row of the periodic table or higher due to the presence of the D-orbitals. Florine does not fit those criteria so there is only one option for the extra electrons, placing them on Xenon.
I understand that for this example, the additional electrons go around Xe because Xe has a d-orbital, enabling an expanded octet, whereas F does not, but in a case where more than one atom in a molecule had a d-orbital, how would you know which atom to draw those extra electrons around?
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