How do you draw a Lewis structure for the sulfate ion, #SO_4"^(2-)#?

1 Answer
Jan 25, 2016

Answer:

#(O=)_2S(-O^-)_2#

Explanation:

There are #5xx6+2# #=# #32# electrons to distribute over 5 centres (why? because all of the atoms are from Group VI, and thus each has 6 valence electrons, + 2 electrons for the negative charges).

Of course this is a resonance structure, in that ALL the sulfur oxygen bonds are equivalent. If we keep (arbitrarily) to the given resonance structure, around each doubly bound oxygen there are 6 electrons, around the central sulfur, there are 6 electrons (i.e. sulfur "owns" 6 electrons), and around each singly bound oxygen there are 7 electrons. As is usual, the electrons that form the bonds are SHARED by participating atoms, and lone pairs devolve solely to the relevant atom.

So (finally), the formal charge on the doubly bound oxygens, and the sulfur is ZERO. Around the singly bound oxygens, there are 7 electrons, and hence each oxygen BEARS A SINGLE NEGATIVE CHARGE.

Thus we assume we might represent the parent molecule, sulfuric acid as #(O=)_2S(-OH)_2#, #H_2SO_4#. Capisce?