# SO42- Lewis Structure

SO42- (sulfate) has one sulfur atom and four oxygen atoms. In the lewis structure of SO42-, there are two double bonds and two single bonds around the sulfur atom, with four oxygen atoms attached to it. The oxygen atom with double bonds has two lone pairs, and the oxygen atom with single bonds has three lone pairs.

Also, there is a negative (-1) charge on the oxygen atom with single bonds.

## Steps

Here’s how you can draw the SO42- lewis structure step by step.

Step #1: draw sketch
Step #2: mark lone pairs
Step #3: mark charges
Step #4: minimize charges
Step #5: minimize charges again (if there are)

Let’s break down each step in detail.

### #1 Draw Sketch

• First, determine the total number of valence electrons

In the periodic table, both sulfur and oxygen lie in group 16.

Hence, both sulfur and oxygen have six valence electrons.

Since SO42- has one sulfur atom and four oxygen atoms, so…

Valence electrons of one sulfur atom = 6 × 1 = 6
Valence electrons of four oxygen atoms = 6 × 4 = 24

Now the SO42- has a negative (-2) charge, so we have to add two more electrons.

So the total valence electrons = 6 + 24 + 2 = 32

• Second, find the total electron pairs

We have a total of 32 valence electrons. And when we divide this value by two, we get the value of total electron pairs.

Total electron pairs = total valence electrons ÷ 2

So the total electron pairs = 32 ÷ 2 = 16

• Third, determine the central atom

We have to place the least electronegative atom at the center.

Since sulfur is less electronegative than oxygen, assume that the central atom is sulfur.

Therefore, place sulfur in the center and oxygens on either side.

• And finally, draw the rough sketch

### #2 Mark Lone Pairs

Here, we have a total of 16 electron pairs. And four S — O bonds are already marked. So we have to only mark the remaining twelve electron pairs as lone pairs on the sketch.

Also remember that sulfur is a period 3 element, so it can keep more than 8 electrons in its last shell. And oxygen is a period 2 element, so it can not keep more than 8 electrons in its last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are oxygens.

So for each oxygen, there are three lone pairs, and for sulfur, there is zero lone pair because all twelve electron pairs are over.

Mark the lone pairs on the sketch as follows:

### #3 Mark Charges

Use the following formula to calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

For sulfur atom, formal charge = 6 – 0 – ½ (8) = +2

For each oxygen atom, formal charge = 6 – 6 – ½ (2) = -1

Here, both sulfur and oxygen atoms have charges, so mark them on the sketch as follows:

The above structure is not a stable lewis structure because both sulfur and oxygen atoms have charges. Therefore, reduce the charges (as below) by converting lone pairs to bonds.

### #4 Minimize Charges

Convert a lone pair of the oxygen atom to make a new S — O bond with the sulfur atom as follows:

### #5 Minimize Charges Again

Since there are charges on sulfur and oxygen atoms, again convert a lone pair of the oxygen atom to make a new S — O bond with the sulfur atom as follows:

In the above structure, you can see that the central atom (sulfur) forms an octet. Hence, the octet rule is satisfied.

Now there is still a negative (-1) charge on the two oxygen atoms.

This is okay, because the structure with a negative charge on the most electronegative atom is the best lewis structure. And in this case, the most electronegative element is oxygen.

Also, the above structure is more stable than the previous structures. Therefore, this structure is the most stable lewis structure of SO42-.

And since the SO42- has a negative (-2) charge, mention that charge on the lewis structure by drawing brackets as follows:

Next: N2O4 Lewis Structure