BrO- Lewis Structure in 6 Steps (With Images)

BrO- Lewis Structure

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

Let me explain the above image in short.

BrO- (hypobromite ion) lewis structure has one Bromine atom (Br) and one Oxygen atom (O) which contain a single bond between them. There are 3 lone pairs on both the Bromine atom (Br) as well as Oxygen atom (O). There is a -1 formal charge on the Oxygen atom (O).

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

So let’s move to the steps of drawing the lewis structure of BrO- ion.

Steps of drawing BrO- lewis structure

Step 1: Find the total valence electrons in BrO- ion

In order to find the total valence electrons in a BrO- (hypobromite ion), first of all you should know the valence electrons present in a single bromine atom as well as oxygen 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 bromine as well as oxygen using a periodic table.

Total valence electrons in BrO- ion

→ 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 BrO- ion = valence electrons given by 1 bromine atom + valence electrons given by 1 oxygen atom + 1 more electron is added due to 1 negative charge = 7 + 6 + 1 = 14.

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 ion is BrO- ion. It has only two atoms, so you can select any of the atoms as a center atom.

BrO- step 1

Let’s assume the bromine atom as a central atom.
(You should assume the less electronegative atom as a center atom).

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

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

BrO- step 2

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

Here in the sketch of BrO molecule, we have assumed the bromine atom as a center atom. So the oxygen is the outer atom.

Hence you have to make the oxygen atom stable.

You can see in the below image that the oxygen atom is forming an octet and hence it is stable.

BrO- step 3

Also, in step 1 we have calculated the total number of valence electrons present in the BrO- ion.

The BrO- ion has a total 14 valence electrons and out of these, only 8 valence electrons are used in the above sketch.

So the number of electrons which are left = 14 – 8 = 6.

You have to put these 6 electrons on the bromine atom in the above sketch of BrO molecule.

BrO- 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 bromine atom (Br) is stable or not.

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

BrO- step 5

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

And hence the bromine atom is stable.

Now let’s proceed to the final step to check whether the lewis structure of BrO- ion 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 BrO.

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 the bromine atom (Br) as well as oxygen atom (O) present in the BrO 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 BrO molecule in the image given below.

BrO- step 6

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

For Oxygen (O) atom:
Valence electrons = 6 (because oxygen is in group 16)
Bonding electrons = 2
Nonbonding electrons = 6

Formal charge=Valence electrons(Bonding electrons)/2Nonbonding electrons
Br=72/26=0
O=62/26=-1

From the above calculations of formal charge, you can see that the bromine (Br) atom has 0 charge and the oxygen (O) atom has -1 charge.

So let’s keep these charges on the respective atoms in the BrO molecule.

BrO- step 7

This overall -1 charge on the BrO molecule is represented in the image given below.

BrO- step 8

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

lewis structure of BrO-

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:

SBr6 Lewis StructureIO3- Lewis Structure
HOFO Lewis StructureBrF Lewis Structure
AlH3 Lewis StructureMgF2 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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