Question 7.EP.5: Draw the Lewis structure of the sulfate ion, SO4^2-

Step 1: Count the total number of valence electrons. The sulfur atom has six valence electrons, and each oxygen atom also has six. The ion as a whole carries a charge of 2−, which means it must contain two additional electrons.

\begin{matrix} S & 1\times 6=6 & \\ O & 4\times 6=24 & \\ & +2 \ extra \ e^{-} \begin{matrix} \underline{=2} & \ \ \ \ \ \ \ \end{matrix} \\ &Total= 32 \end{matrix}

Step 2: Draw the skeletal structure of the molecule. Sulfur is the least electronegative atom, and appears first in the formula, so it should be the central atom. We’ll draw square brackets around the entire ion and indicate its charge.

\left[\begin{matrix} & \underset{}{O} & \\ O & \ S \ & O \\ & \overset{\underset{}{} }{O} & \end{matrix}\right] ^{2-}

Step 3: Place single bonds between all connected atoms.

\left[\begin{matrix}\underset{\mid}{O}\\ O—S—O \\\overset{\mid}{O} \end{matrix}\right]^{2-}

Step 4: Place the remaining valence electrons not accounted for in Step 3 on individual atoms until the octet rule is satisfied. By drawing bonds to the sulfur atoms, we have accounted for 8 of the 32 electrons, and 24 remain to be assigned. Placing three lone pairs around each oxygen atom will account for all of the available electrons.

\left[\begin{matrix}\underset{\mid}{:\overset{..}{O}: }\\ \overset{..}{:\underset{..}{O}}—S— \overset{..}{\underset{..}{O}:} \\\overset{\mid}{:\underset{..}{O}: } \end{matrix}\right]^{2-}

Step 5: Create multiple bonds for any atoms that do not have a full octet of valence electrons. We have fulfilled the octet rule and have accounted for all of the available electrons. Multiple bonds are not required in this ion.

Discussion
This is similar to the earlier examples, but the fact that we are working with an ion requires extra care in determining how many valence electrons to include.