SeBr4 Lewis structure

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SeBr4 Lewis Structure
SeBr4 Lewis structure

SeBr4 (selenium tetrabromide) has one selenium atom and four bromine atoms.

In SeBr4 Lewis structure, there are four single bonds around the selenium atom, with four bromine atoms attached to it. Each bromine atom has three lone pairs, and the selenium atom has one lone pair.


Here’s how you can easily draw the SeBr4 Lewis structure step by step:

#1 Draw a rough skeleton structure
#2 Mention lone pairs on the atoms
#3 If needed, mention formal charges on the atoms

Now, let’s take a closer look at each step mentioned above.

#1 Draw a rough skeleton structure

  • First, determine the total number of valence electrons
Periodic table

In the periodic table, selenium lies in group 16, and bromine lies in group 17.

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

Since SeBr4 has one selenium atom and four bromine atoms, so…

Valence electrons of one selenium atom = 6 × 1 = 6
Valence electrons of four bromine atoms = 7 × 4 = 28

And the total valence electrons = 6 + 28 = 34

  • Second, find the total electron pairs

We have a total of 34 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 = 34 ÷ 2 = 17

  • Third, determine the central atom

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

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

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

  • And finally, draw the rough sketch
SeBr4 Lewis Structure (Step 1)
Rough sketch of SeBr4 Lewis structure

#2 Mention lone pairs on the atoms

Here, we have a total of 17 electron pairs. And four Se — Br bonds are already marked. So we have to only mark the remaining thirteen electron pairs as lone pairs on the sketch.

Also remember that both (selenium and bromine) are the period 4 elements, so they can keep more than 8 electrons in their last shell.

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

So for each bromine, there are three lone pairs, and for selenium, there is one lone pair.

Mark the lone pairs on the sketch as follows:

SeBr4 Lewis Structure (Step 2)
Lone pairs marked, and got the stable Lewis structure of SeBr4

#3 If needed, mention formal charges on the atoms

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 – 2 – ½ (8) = 0

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

Here, both selenium and bromine 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 (bromines) also form an octet. Hence, the octet rule is satisfied.

Therefore, this structure is the stable Lewis structure of SeBr4.

Next: BrCl2 Lewis structure

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