# SeS3 Lewis structure

SeS3 has one selenium atom and three sulfur atoms.

In SeS3 Lewis structure, there are three double bonds around the selenium atom, with three sulfur atoms attached to it, and on each sulfur atom, there are two lone pairs.

Contents

## Steps

To properly draw the SeS3 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 selenium and sulfur lie in group 16.

Hence, both selenium and sulfur have six valence electrons.

Since SeS3 has one selenium atom and three sulfur atoms, so…

Valence electrons of one selenium atom = 6 × 1 = 6
Valence electrons of three sulfur atoms = 6 × 3 = 18

And the total valence electrons = 6 + 18 = 24

• Second, find the total electron pairs

We have a total of 24 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 = 24 ÷ 2 = 12

• Third, determine the central atom

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

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

Therefore, place selenium in the center and sulfurs on either side.

• And finally, draw the rough sketch

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

Here, we have a total of 12 electron pairs. And three Se — S bonds are already marked. So we have to only mark the remaining nine electron pairs as lone pairs on the sketch.

Also remember that selenium is a period 4 element, so it can keep more than 8 electrons in its last shell. And sulfur is a period 3 element, so it can also keep more than 8 electrons in its last shell.

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

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

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 selenium atom, formal charge = 6 – 0 – ½ (6) = +3

For each sulfur atom, formal charge = 6 – 0 – ½ (6) = -1

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

The above structure is not a stable Lewis structure because both selenium and sulfur 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 sulfur atom to make a new Se — S bond with the selenium atom as follows:

### #5 Repeat step 4 (minimize charges again)

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

### #6 Minimize charges again

There are still charges on selenium and sulfur atoms, so again convert a lone pair of the sulfur atom to make a new Se — S bond with the selenium atom as follows:

In the above structure, you can see that the central atom (selenium) forms an octet. And the outside atoms (sulfurs) 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 SeS3.

Next: IBr5 Lewis structure