# Lewis Structure of SeOBr2 (With 6 Simple Steps to Draw!)

I’m super excited to teach you the lewis structure of SeOBr2 in just 6 simple steps.

Infact, I’ve also given the step-by-step images for drawing the lewis dot structure of SeOBr2 molecule.

So, if you are ready to go with these 6 simple steps, then let’s dive right into it!

Lewis structure of SeOBr2 contains a double bond between the Selenium (Se) & Oxygen (O) atom and two single bonds between Selenium (Se) & Bromine (Br) atoms. The Selenium atom (Se) is at the center and it is surrounded by two Bromine (Br) and one Oxygen atom (O).

Let’s draw and understand this lewis dot structure step by step.

(Note: Take a pen and paper with you and try to draw this lewis structure along with me. I am sure you will definitely learn how to draw lewis structure of SeOBr2).

## 6 Steps to Draw the Lewis Structure of SeOBr2

### Step #1: Calculate the total number of valence electrons

Here, the given molecule is SeOBr2. In order to draw the lewis structure of SeOBr2, first of all you have to find the total number of valence electrons present in the SeOBr2 molecule.
(Valence electrons are the number of electrons present in the outermost shell of an atom).

So, let’s calculate this first.

Calculation of valence electrons in SeOBr2

• For Selenium:

Selenium is a group 16 element on the periodic table.

Hence, the valence electrons present in selenium is 6 (see below image).

• For Oxygen:

Oxygen is a group 16 element on the periodic table.

Hence, the valence electron present in oxygen is 6 (see below image).

• For Bromine:

Bromine is a group 17 element on the periodic table.

Hence, the valence electrons present in bromine is 7 (see below image).

Hence in a SeOBr2 molecule,

Valence electrons given by Selenium (Se) atom = 6
Valence electrons given by Oxygen (O) atom = 6
Valence electrons given by each Bromine (Br) atom = 7
So, total number of Valence electrons in SeOBr2 molecule = 6 + 6 + 7(2) = 26

### Step #2: Select the center atom

While selecting the center atom, always put the least electronegative atom at the center.

(Remember: Fluorine is the most electronegative element on the periodic table and the electronegativity decreases as we move right to left in the periodic table as well as top to bottom in the periodic table).

Here in the SeOBr2 molecule, if we compare the selenium atom (Se), oxygen atom (O) and bromine atom (Br), then the selenium is less electronegative.

So, selenium should be placed in the center and the oxygen atom as well as bromine atom will surround it.

### Step #3: Put two electrons between the atoms to represent a chemical bond

Now in the above sketch of SeOBr2 molecule, put the two electrons (i.e electron pair) between the selenium-oxygen atoms and selenium-bromine atoms to represent a chemical bond between them.

These pairs of electrons present between the Selenium & Oxygen atoms as well as between the Selenium & Bromine atoms form a chemical bond, which bonds these atoms with each other in a SeOBr2 molecule.

### Step #4: Complete the octet (or duplet) on outside atoms. If the valence electrons are left, then put the valence electrons pair on the central atom

Don’t worry, I’ll explain!

In the Lewis structure of SeOBr2, the outer atoms are bromine atoms and oxygen atom.

You also have to see whether the oxygen atom as well as bromine atoms are forming an octet or not! (because oxygen and bromine both require 8 electrons to have a complete outer shell).

You can see in the above image that both the bromine atoms as well as one oxygen atom forms an octet.

Also, only 24 valence electrons of SeOBr2 molecule are used in the above structure.

But there are total 26 valence electrons in SeOBr2 molecule (as calculated in step #1).

So the number of electrons left to be kept on the central atom = 26 – 24 = 2.

So let’s keep these 2 electrons (i.e 1 electron pairs) on the central atom.

Now, let’s move to the next step.

### Step #5: Check whether the central atom has octet or not. If it does not have an octet, then move the electron pair from the outer atom to form a double bond or triple bond

In this step, we have to check whether the central atom (i.e selenium) has an octet or not.

In simple words, we have to check whether the central Selenium (Se) atom has 8 electrons or not.

As you can see from the above image, the central atom (i.e selenium), has 8 electrons. So it fulfills the octet rule and the selenium atom is stable.

### Step #6: Final step – Check the stability of lewis structure by calculating the formal charge on each atom

Now, you have come to the final step and here you have to check the formal charge on selenium atom (Se), oxygen (O) atom as well as bromine atoms (Br).

For that, you need to remember the formula of formal charge;

Formal charge = Valence electrons – Nonbonding electrons – (Bonding electrons)/2

• For Selenium:
Valence electrons = 6 (as it is in group 16)
Nonbonding electrons = 2
Bonding electrons = 6
• For Oxygen:
Valence electron = 6 (as it is in group 16)
Nonbonding electrons = 6
Bonding electrons = 2
• For Bromine:
Valence electron = 7 (as it is in group 17)
Nonbonding electrons = 6
Bonding electrons = 2

So you can see above that the formal charges on selenium is +1 and the formal charge on the oxygen atom is -1.

This indicates that the above lewis structure of SeOBr2 is not stable and so we have to minimize the charges to get a more stable lewis structure.

This can be done by shifting the lone pair from negatively charged oxygen atom to the positively charged selenium atom to form a bond.

Now, in the above structure, you can see that the charges are minimized and the above lewis structure of SeOBr2 is the final stable structure.

Each electron pair (:) in the lewis dot structure of SeOBr2 represents the single bond ( | ). So the above lewis dot structure of SeOBr2 can also be represented as shown below.

Related lewis structures for your practice:
Lewis Structure of HBrO2
Lewis Structure of HBrO3
Lewis Structure of HBrO4
Lewis Structure of PO2-
Lewis Structure of TeF5-