# Why (SO4)^2- does not create 4 double bonds

I looked this question up and still couldn't understand. Why, in $$\ce{SO4^2-}$$ don't the 4 oxygens create double bonds.

In that case the all the oxygens will have 0 formal charge while the sulfur will have -2.

In what I've seen only 2 oxygens create double bonds making the sulfur have no formal charge, 2 oxygens have -1 formal charge and 2 others no formal charge.

When comparing formal charge of -2 on the sulfur, it is indeed less stable than no formal charge at all, and that is why it should be the most common resonance structure. But everywhere I looked that was not the case, $$\ce{SO4^2-}$$ only created 2 double bonds and I cant understand why. Cant the oxygens create coordination bonds with the sulfur?

• Having formal charges greater than 1 on a single atom will tend to be less stable than spreading it around. Also, oxygen is more electronegative than sulfur, so we would expect more of the negative formal charge to reside on the oxygens than the sulfur. @guesting – Tyberius Mar 21 at 4:06

When it undergoes chemical reactions, it typically donates both hydrogens as $$\ce{H+}$$ ions. This leaves behind the sulfate ion: $$\ce{H2SO4 -> 2H+ + SO4^2-}$$ When the $$\ce{H^+}$$ ion departs, it leaves its electron behind, so it has to go somewhere (it remains with the $$\ce{O}$$ atom).
Hypothetically if $$\ce{SO4}$$ existed, with all $$\ce{O}$$ atoms double-bonded to the $$\ce{S}$$, then the sulfur would have a total of 16 electrons in its valence shell, which would make it more unstable. But the main reason is that sulfur only has 6 valence electrons in the first place, so it can only form up to 6 covalent bonds. This gives it a total of 12 valence electrons.