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Firstly we count the number of valence electrons for the XeF4 molecule using the modern periodic table. Once calculated, we can distribute them around the central atom and try to fill the outer shells of each atom present. XeF4 thus contains ( 8+ 4x 7 ) = 36 valence electrons. Formal charge is taken into account to get the best Lewis structure for the molecule. We must keep in mind that xenon (Xe) can have more than 8 valence electrons and fluorine (F) cannot form a double bond since it is highly electronegative.
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