BrO4- Lewis Structure in 5 Steps (With Images)

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

Let me explain the above image in short.

BrO4- lewis structure has a Bromine atom (Br) at the center which is surrounded by four Oxygen atoms (O). There are 3 double bonds and 1 single bond between the Bromine atom (Br) and each Oxygen atom (O). There are 2 lone pairs on double bonded Oxygen atoms (O) and 3 lone pairs on single bonded Oxygen atom (O).

If you haven’t understood anything from the above image of BrO4- lewis structure, then just stick with me and you will get the detailed step by step explanation on drawing a lewis structure of BrO4^– ion.

So let’s move to the steps of drawing the lewis structure of BrO4^– ion.

Steps of drawing BrO4- lewis structure

Step 1: Find the total valence electrons in BrO4- ion

In order to find the total valence electrons in BrO4- ion, first of all you should know the valence electrons present in 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 BrO4- 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 BrO4- ion = valence electrons given by 1 bromine atom + valence electrons given by 4 oxygen atoms + 1 more electron is added due to one negative charge = 7 + 6(4) + 1 = 32.

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 BrO4- ion and it contains bromine atom (Br) and oxygen atoms (O).

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

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

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

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

Now in the BrO4 molecule, you have to put the electron pairs between the bromine atom (Br) and oxygen atoms (O).

This indicates that the bromine (Br) and oxygen (O) are chemically bonded with each other in a BrO4 molecule.

Step 4: Make the outer atoms stable

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

Here in the sketch of BrO4 molecule, you can see that the outer atoms are oxygen atoms.

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

Also, in step 1 we have calculated the total number of valence electrons present in the BrO4- ion.

The BrO4- ion has a total 32 valence electrons and all these valence electrons are used in the above sketch.

Hence there are no remaining electron pairs to be kept on the central atom. 

So now let’s proceed to the next step.

Step 5: 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 BrO4- ion.

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 bromine (Br) atom as well as oxygen (O) atoms present in the BrO4 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 BrO4 molecule in the image given below.

For Bromine (Br) atom:
Valence electrons = 7 (because bromine is in group 17)
Bonding electrons = 8
Nonbonding electrons = 0

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)/2	–	Nonbonding electrons
Br	=	7	–	8/2	–	0	=	+1
O	=	6	–	2/2	–	6	=	-1

From the above calculations of formal charge, you can see that the bromine (Br) atom has +1 charge while the oxygen atoms has -1 charges.

So let’s keep these charges on the respective atoms of the BrO4 molecule.

The above image shows that the lewis structure of BrO4 is not stable.

So we have to minimize these charges by shifting the electron pairs from the oxygen atoms to the bromine atom.

After shifting the electron pairs from the oxygen atom to the bromine atom, the charges on central bromine atom as well as three oxygen atoms becomes zero. And this is a more stable lewis structure. (see below image).

There is one -ve charge left on the oxygen atom, which gives -1 formal charge on the BrO4 molecule.This overall -1 charge on the BrO4 molecule is represented in the image given below.

In the above lewis dot structure of BrO4- ion, you can also represent each bonding electron pair (:) as a single bond (|). By doing so, you will get the following lewis structure of BrO4- ion.

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:

NO2F Lewis Structure	XeOF4 Lewis Structure
ICl Lewis Structure	H2SO3 Lewis Structure
HSO4- Lewis Structure	CCl2F2 Lewis Structure