What is the lewis dot structure of Na2SO4?

We will confine the answer to $\text{sulfate ion.......}$
For $S {O}_{4}^{2 -}$, a reasonable Lewis structure to accommodate the 32 electrons is.....${\left(O =\right)}_{2} S {\left(- {O}^{-}\right)}_{2}$, and clearly this is a derivative of the parent $\text{sulfuric acid}$, .....${\left(O =\right)}_{2} S {\left(- O H\right)}_{2}$. Given this Lewis structure, there are 6 valence electrons associated with the doubly bound oxygens (and hence these are neutral), 6 valence electrons with the central sulfur (and hence a neutral sulfur), and 7 valence electrons associated with each singly bound oxygen atoms, and hence these centres bear the 2 required formal NEGATIVE charges.
Note that all the oxygen atoms are equivalent, and we could even propose a Lewis structure of ""^(2+)S(-O^(-))_4.....${\left(O =\right)}_{2} S {\left(- {O}^{-}\right)}_{2}$ is a more conventional representation.