Lewis Structure of BrO- (With 6 Simple Steps to Draw!)

Ready to learn how to draw the lewis structure of BrO- ion?

Awesome!

Here, I have explained 6 simple steps to draw the lewis dot structure of BrO- ion (along with images).

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

Lewis structure of BrO- ion contains one single bond between the Bromine (Br) atom and Oxygen (O) atom. The Bromine atom and Oxygen atom, both have 3 lone pairs. And the Oxygen atom has -1 formal charge.

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 BrO-).

6 Steps to Draw the Lewis Structure of BrO-

Step #1: Calculate the total number of valence electrons

Here, the given ion is BrO-. In order to draw the lewis structure of BrO- ion, first of all you have to find the total number of valence electrons present in the BrO- ion.
(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 BrO-

• For Bromine:

Bromine is a group 17 element on the periodic table. ^[1]

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

• For Oxygen:

Oxygen is a group 16 element on the periodic table. ^[2]

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

Hence in a BrO- ion, 

Valence electrons given by Bromine (Br) atom = 7
Valence electrons given by Oxygen (O) atom = 6
Electron due to -1 charge, 1 more electron is added
So, total number of Valence electrons in BrO- ion = 7 + 6 + 1 = 14

Step #2: Select the center atom

While selecting the atom, you have to put the least electronegative atom at the center. 

But here in the BrO- ion, there are only two atoms. So you can consider any of the atoms as a center atom.

So, let’s assume that the bromine atom is a central atom. (You should assume the less electronegative atom as a central atom.)

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

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

These pairs of electrons present between the Bromine (Br) and Oxygen (O) atom form a chemical bond, which bonds the bromine and oxygen atom with each other in a BrO- ion.

Step #4: Complete the octet (or duplet) on outside atom. 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 BrO- ion, we have just assumed the bromine atom as a central atom and so the oxygen atom is an outer atom.

So now, we have to complete the octet on this oxygen atom.

Now, you can see in the above image that the oxygen atom forms an octet.

Also, only 8 valence electrons of BrO- are used in the above structure.

But there are total 14 valence electrons in BrO- ion (as calculated in step #1).

So the number of electrons left to be kept on the bromine atom = 14 – 8 = 6.

So let’s keep these six electrons (i.e three electron pairs) on the bromine atom.

Now, let’s move to the next step.

Step #5: Check whether the central atom has octet or not

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

In simple words, we have to check whether the central Bromine (Br) atom is having 8 electrons or not.

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

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 bromine atom (Br) as well as each oxygen atom (O).

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

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

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

Formal charge	=	Valence electrons	–	Nonbonding electrons	–	(Bonding electrons)/2
Br	=	7	–	6	–	2/2	=	0
O	=	6	–	6	–	2/2	=	-1

So you can see above that the formal charges on bromine is 0 and the formal charge on the oxygen atoms is -1.

Let’s keep these charges on the atoms in the above lewis structure of BrO.

As you can see in the above sketch, there is one -ve charge on the bromine atom, which indicates the -1 formal charge on the BrO molecule.

Hence, the above lewis structure of BrO- ion is the stable lewis structure.

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

Related lewis structures for your practice:
Lewis Structure of SeOF2
Lewis Structure of SBr6
Lewis Structure of IO3-
Lewis Structure of HOFO
Lewis Structure of BrF 

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