Lewis Structure of BrO2- (With 6 Simple Steps to Draw!)

Ready to learn how to draw the lewis structure of BrO2- ion?

Awesome!

Here, I have explained 6 simple steps to draw the lewis dot structure of BrO2- 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 BrO2- contains one double bond and one single bond between the Bromine (Br) atom and Oxygen (O) atom. The Bromine atom (Br) is at the center and it is surrounded by 2 Oxygen atoms (O). The Bromine atom has 2 lone pairs, double bonded Oxygen atom has 2 lone pairs and the single bonded oxygen atom has 3 lone pairs.

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 BrO2-).

6 Steps to Draw the Lewis Structure of BrO2-

Step #1: Calculate the total number of valence electrons

Here, the given ion is BrO2- ion. In order to draw the lewis structure of BrO2- ion, first of all you have to find the total number of valence electrons present in the BrO2- 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 BrO2-

• 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 BrO2- ion, 

Valence electrons given by Bromine (Br) atom = 7
Valence electrons given by each Oxygen (O) atom = 6
Electron due to -1 charge, 1 more electron is added
So, total number of Valence electrons in BrO2- ion = 7 + 6(2) + 1 = 20

Step #2: Select the center atom

While selecting the 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). ^[3]

Here in the BrO2- ion, if we compare the Bromine atom (Br) and oxygen atom (O), then the bromine is less electronegative than oxygen.

So, bromine should be placed in the center and the remaining 2 oxygen atoms will surround it.

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

Now in the above sketch of BrO2- molecule, put the two electrons (i.e electron pair) between each bromine atom and oxygen atom to represent a chemical bond between them.

These pairs of electrons present between the Bromine (Br) and Oxygen (O) atoms form a chemical bond, which bonds the bromine and oxygen atoms with each other in a BrO2- ion.

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 BrO2-, the outer atoms are oxygen atoms.

So now, you have to complete the octet on these oxygen atoms (because oxygen requires 8 electrons to have a complete outer shell).

Now, you can see in the above image that all the oxygen atoms form an octet.

Also, only 16 valence electrons of BrO2- are used in the above structure.

But there are total 20 valence electrons in BrO2- ion (as calculated in step #1).

So the number of electrons left to be kept on the central atom = 20 – 16 = 4.

So let’s keep these four electrons (i.e two 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

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 = 4
  Bonding electrons = 4
• 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	–	4	–	4/2	=	+1
O	=	6	–	6	–	2/2	=	-1

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

This indicates that the above lewis structure of BrO2 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 bromine atom to form a bond.

Now, in the above structure, you can see that the charges are minimized and the above lewis structure of BrO2- is the final stable structure.

There is a -ve charge left on the oxygen atom, which indicates the -1 formal charge on the BrO2- ion.

Each electron pair (:) in the lewis dot structure of BrO2- ion represents the single bond ( | ). So the above lewis dot structure of BrO2- ion can also be represented as shown below.

Related lewis structures for your practice:
Lewis Structure of N2O4
Lewis Structure of COF2
Lewis Structure of SCl4
Lewis Structure of PBr5
Lewis Structure of SiS2 

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