# SeI2 Lewis structure

SeI2 has one selenium atom and two iodine atoms.

In SeI2 Lewis structure, there are two single bonds around the selenium atom, with two iodine atoms attached to it. Each iodine atom has three lone pairs, and the selenium atom has two lone pairs.

Contents

## Steps

To properly draw the SeI2 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

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, selenium lies in group 16, and iodine lies in group 17.

Hence, selenium has six valence electrons and iodine has seven valence electrons.

Since SeI2 has one selenium atom and two iodine atoms, so…

Valence electrons of one selenium atom = 6 × 1 = 6
Valence electrons of two iodine atoms = 7 × 2 = 14

And the total valence electrons = 6 + 14 = 20

• Second, find the total electron pairs

We have a total of 20 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 = 20 ÷ 2 = 10

• Third, determine the central atom

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

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

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

• And finally, draw the rough sketch

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

Here, we have a total of 10 electron pairs. And two Se — I bonds are already marked. So we have to only mark the remaining eight 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 iodine is a period 5 element, so it can keep more than 8 electrons in its last shell.

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

So for each iodine, there are three lone pairs, and for selenium, there are two lone pairs.

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 – 4 – ½ (4) = 0

For each iodine atom, formal charge = 7 – 6 – ½ (2) = 0

Here, both selenium and iodine atoms do not have charges, so no need to mark the charges.

In the above structure, you can see that the central atom (selenium) forms an octet. And the outside atoms (iodines) also form an octet. Hence, the octet rule is satisfied.

Therefore, this structure is the stable Lewis structure of SeI2.

Next: H2Te Lewis structure