Br2O Lewis Structure in 6 Steps (With Images)

Br2O Lewis Structure

So you have seen the above image by now, right?

Let me explain the above image in short.

Br2O lewis structure has an Oxygen atom (O) at the center which is surrounded by two Bromine atoms (Br). There are 2 single bonds between the Oxygen atom (O) and each Bromine atom (Br). There are 2 lone pairs on the Oxygen atom (O) and 3 lone pairs on both the Bromine atoms (Br).

If you haven’t understood anything from the above image of Br2O lewis structure, then just stick with me and you will get the detailed step by step explanation on drawing a lewis structure of Br2O.

So let’s move to the steps of drawing the lewis structure of Br2O.

Steps of drawing Br2O lewis structure

Step 1: Find the total valence electrons in Br2O molecule

In order to find the total valence electrons in Br2O molecule, first of all you should know the valence electrons present in oxygen atom as well as bromine atom.
(Valence electrons are the electrons that are present in the outermost orbit of any atom.)

Here, I’ll tell you how you can easily find the valence electrons of oxygen as well as bromine using a periodic table.

Total valence electrons in Br2O molecule

→ Valence electrons given by bromine atom:

Bromine is a group 17 element on the periodic table. [1] Hence the valence electrons present in bromine is 7.

You can see the 7 valence electrons present in the bromine atom as shown in the above image.

→ Valence electrons given by oxygen atom:

Oxygen is group 16 element on the periodic table. [2] Hence the valence electrons present in oxygen is 6.

You can see the 6 valence electrons present in the oxygen atom as shown in the above image.

Hence, 

Total valence electrons in Br2O molecule = valence electrons given by 1 oxygen atom + valence electrons given by 2 bromine atoms = 6 + 7(2) = 20.

Step 2: Select the central atom

For selecting the center atom, you have to remember that the atom which is less electronegative remains at the center.

Now here the given molecule is Br2O and it contains oxygen atom (O) and bromine atoms (Br).

You can see the electronegativity values of oxygen atom (O) and bromine atom (Br) in the above periodic table.

If we compare the electronegativity values of oxygen (O) and bromine (Br) then the oxygen atom is less electronegative.

So here the oxygen atom (O) is the center atom and the bromine atoms (Br) are the outside atoms.

Br2O step 1

Step 3: Connect each atoms by putting an electron pair between them

Now in the Br2O molecule, you have to put the electron pairs between the oxygen atom (O) and bromine atoms (Br).

Br2O step 2

This indicates that the oxygen (O) and bromine (Br) are chemically bonded with each other in a Br2O molecule.

Step 4: Make the outer atoms stable. Place the remaining valence electrons pair on the central atom.

Now in this step, you have to check the stability of the outer atoms.

Here in the sketch of Br2O molecule, you can see that the outer atoms are bromine atoms.

These outer bromine atoms are forming an octet and hence they are stable.

Br2O step 3

Also, in step 1 we have calculated the total number of valence electrons present in the Br2O molecule.

The Br2O molecule has a total 20 valence electrons and out of these, only 16 valence electrons are used in the above sketch.

So the number of electrons which are left = 20 – 16 = 4.

You have to put these 4 electrons on the central oxygen atom in the above sketch of Br2O molecule.

Br2O step 4

Now let’s proceed to the next step.

Step 5: Check the octet on the central atom

In this step, you have to check whether the central oxygen atom (O) is stable or not.

In order to check the stability of the central oxygen (O) atom, we have to check whether it is forming an octet or not.

Br2O step 5

You can see from the above picture that the oxygen atom is forming an octet. That means it has 8 electrons.

And hence the central oxygen atom is stable.

Now let’s proceed to the final step to check whether the lewis structure of Br2O is stable or not.

Step 6: Check the stability of lewis structure

Now you have come to the final step in which you have to check the stability of lewis structure of Br2O.

The stability of lewis structure can be checked by using a concept of formal charge.

In short, now you have to find the formal charge on oxygen (O) atom as well as bromine (Br) atoms present in the Br2O molecule.

For calculating the formal charge, you have to use the following formula;

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

You can see the number of bonding electrons and nonbonding electrons for each atom of Br2O molecule in the image given below.

Br2O step 6

For Oxygen (O) atom:
Valence electrons = 6 (because oxygen is in group 16)
Bonding electrons = 4
Nonbonding electrons = 4

For Bromine (Br) atom:
Valence electron = 7 (because bromine is in group 17)
Bonding electrons = 2
Nonbonding electrons = 6

Formal charge=Valence electrons(Bonding electrons)/2Nonbonding electrons
O=64/24=0
Br=72/26=0

From the above calculations of formal charge, you can see that the oxygen (O) atom as well as bromine (Br) atom has a “zero” formal charge.

This indicates that the above lewis structure of Br2O is stable and there is no further change in the above structure of Br2O.

In the above lewis dot structure of Br2O, you can also represent each bonding electron pair (:) as a single bond (|). By doing so, you will get the following lewis structure of Br2O.

lewis structure of Br2O

I hope you have completely understood all the above steps.

For more practice and better understanding, you can try other lewis structures listed below.

Try (or at least See) these lewis structures for better understanding:

NF5 Lewis StructureSiO Lewis Structure
AlI3 Lewis StructurePF2- Lewis Structure
SI4 Lewis StructureGaCl3 Lewis Structure
About author

Jay is an educator and has helped more than 100,000 students in their studies by providing simple and easy explanations on different science-related topics. He is a founder of Pediabay and is passionate about helping students through his easily digestible explanations.

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