# S2O Lewis structure

S2O (disulfur monoxide) has two sulfur atoms and one oxygen atom.

In the S2O Lewis structure, there are two double bonds around the sulfur atom, with one other sulfur and oxygen atoms attached to it. The left sulfur atom has two lone pairs, the right sulfur atom has one lone pair, and the oxygen atom also has two lone pairs.

Contents

## Steps

To properly draw the S2O Lewis structure, follow these steps:

#1 Draw a rough sketch of the structure
#2 Next, indicate lone pairs on the atoms
#3 Indicate formal charges on the atoms, if necessary
#4 Minimize formal charges by converting lone pairs of the atoms
#5 Repeat step 4 if necessary, until all charges are minimized

Let’s break down each step in more detail.

### #1 Draw a rough sketch of the structure

• 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 S2O has two sulfur atoms and one oxygen atom, so…

Valence electrons of two sulfur atoms = 6 × 2 = 12
Valence electrons of one oxygen atom = 6 × 1 = 6

And the total valence electrons = 12 + 6 = 18

• Second, find the total electron pairs

We have a total of 18 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 = 18 ÷ 2 = 9

• 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.

Here, there are two sulfur atoms, so we can assume any one as the central atom.

Let’s assume that the central atom is right sulfur.

Therefore, place sulfurs in the center and oxygen outside.

• And finally, draw the rough sketch

### #2 Next, indicate lone pairs on the atoms

Here, we have a total of 9 electron pairs. And two bonds are already marked. So we have to only mark the remaining seven electron pairs as lone pairs on the sketch.

Also remember that sulfur is a period 3 element, so it can also 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 left sulfur and oxygen.

So for left sulfur and oxygen, there are three lone pairs, and for right sulfur, there is one lone pair.

Mark the lone pairs on the sketch as follows:

### #3 Indicate formal charges on the atoms, if necessary

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

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

For left sulfur atom, formal charge = 6 – 6 – ½ (2) = -1

For right sulfur atom, formal charge = 6 – 2 – ½ (4) = +2

For 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 formal charges by converting lone pairs of the atoms

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

### #5 Repeat step 4 (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 right sulfur atom as follows:

In the above structure, you can see that the central atom (right sulfur) forms an octet. And the outside atoms (left sulfur and oxygen) also form an octet. Hence, the octet rule is satisfied.

Also, the above structure is more stable than the previous structures. Therefore, this structure is the stable Lewis structure of S2O.