That’s excellent — formal charges: Se (0), two O (0), two O (–1). Total = –2. The double bonds can be placed between Se and any two of the four O atoms . All such arrangements are equivalent. Thus, the actual ion is a resonance hybrid with four equivalent Se–O bonds each having bond order 1.5. 3. Final Lewis Structure Representation :O: :O: || || :O: — Se — :O: | | :O: :O: (with lone pairs shown, and double bonds on two oxygens, but resonance delocalized) Better representation with formal charges: