Lewis Structure of BrO4- (With 5 Simple Steps to Draw!)

I’m super excited to teach you the lewis structure of BrO4- ion in just 5 simple steps.

Infact, I’ve also given the step-by-step images for drawing the lewis dot structure of BrO4- ion.

So, if you are ready to go with these 5 simple steps, then let’s dive right into it!

Lewis structure of BrO4- ion contains three double bonds and one single bond between the Bromine (Br) atom and Oxygen (O) atoms. The Bromine atom (Br) is at the center and it is surrounded by 4 Oxygen atoms (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 BrO4^– ion).

5 Steps to Draw the Lewis Structure of BrO4-

Step #1: Calculate the total number of valence electrons

Here, the given ion is BrO4^–. In order to draw the lewis structure of BrO4^– ion, first of all you have to find the total number of valence electrons present in the BrO4^– 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 BrO4- ion

• 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 BrO4- 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 BrO4^– ion = 7 + 6(4) + 1 = 32

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). ^[3]

Here in the BrO4 molecule, if we compare the bromine atom (Br) and oxygen atom (O), then bromine is less electronegative than oxygen.

So, bromine should be placed in the center and the remaining 4 oxygen atoms will surround it.

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

Now in the above sketch of BrO4, 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 BrO4^– ion.

Step #4: Complete the octet (or duplet) on outside atoms

In the Lewis structure of BrO4^– ion, 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, all the 32 valence electrons of BrO4^– ion (as calculated in step #1) are used in the above structure. So there are no remaining electron pairs.

Hence there is no change in the above sketch of BrO4^– ion.

Let’s move to the next step.

Step #5: Check the formal charge

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

But, in order to get the most stable lewis structure, we have to check the formal charge on BrO4^– ion.

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 = 0
  Bonding electrons = 8
• 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	–	0	–	8/2	=	+3
O	=	6	–	6	–	2/2	=	-1

Let’s keep these charges on the atoms in the above lewis structure of BrO4^– ion.

As we know that bromine can hold more than 8 electrons, we can further reduce the charges as mentioned below.

By doing this, the formal charges on the three oxygen atoms as well as the central bromine atom becomes “zero”, and this gives us the most stable lewis structure of BrO4-.

There is one -ve charge left on the oxygen atom, which indicates the -1 formal charge on the BrO4^– ion.

Each electron pair (:) in the lewis dot structure of BrO4^– ion represents the single bond ( | ). So the above lewis dot structure of BrO4^– ion can also be represented as shown below.

Related lewis structures for your practice:
Lewis Structure of NO2F
Lewis Structure of XeOF4
Lewis Structure of ICl
Lewis Structure of H2SO3
Lewis Structure of HSO4- 

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