IBr4- 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.

IBr4- lewis structure has an Iodine atom (I) at the center which is surrounded by four Bromine atoms (Br). There are 4 single bonds between the Iodine atom (I) and each Bromine atom (Br). There are 2 lone pairs on the Iodine atom (I) and 3 lone pairs on all the four Bromine atoms (Br).

If you haven’t understood anything from the above image of IBr4- lewis structure, then just stick with me and you will get the detailed step by step explanation on drawing a lewis structure of IBr4- ion.

So let’s move to the steps of drawing the lewis structure of IBr4- ion.

Steps of drawing IBr4- lewis structure

Step 1: Find the total valence electrons in IBr4- ion

In order to find the total valence electrons in IBr4- ion, first of all you should know the valence electrons present in iodine atom as well as bromine 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 iodine as well as bromine using a periodic table.

Total valence electrons in IBr4- ion

→ Valence electrons given by iodine atom:

Iodine is a group 17 element on the periodic table. ^[1] Hence the valence electrons present in iodine is 7.

You can see the 7 valence electrons present in the iodine atom as shown in the above image.

→ Valence electrons given by bromine atom:

Bromine is a group 17 element on the periodic table. ^[2] 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.

Hence, 

Total valence electrons in IBr4- ion = valence electrons given by 1 iodine atom + valence electrons given by 1 bromine atom + 1 more electron is added due to 1 negative charge = 7 + 7(4) + 1 = 36.

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 IBr4- ion and it contains iodine atom (I) and bromine atoms (Br).

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

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

So here the iodine atom (I) is the center atom and the bromine atoms (Br) are the outside atoms.

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

Now in the IBr4 molecule, you have to put the electron pairs between the iodine atom (I) and bromine atoms (Br).

This indicates that the iodine (I) and bromine (Br) are chemically bonded with each other in a IBr4 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 IBr4 molecule, you can see that the outer atoms are bromine atoms.

These outer bromine 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 IBr4- ion.

The IBr4- ion has a total 36 valence electrons and out of these, only 32 valence electrons are used in the above sketch.

So the number of electrons which are left = 36 – 32 = 4.

You have to put these 4 electrons on the central iodine atom in the above sketch of IBr4 molecule.

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 IBr4.

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 iodine (I) atom as well as bromine (Br) atoms present in the IBr4 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 IBr4 molecule in the image given below.

For Iodine (I) atom:
Valence electron = 7 (because iodine is in group 17)
Bonding electrons = 8
Nonbonding electrons = 4

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

Formal charge	=	Valence electrons	–	(Bonding electrons)/2	–	Nonbonding electrons
I	=	7	–	8/2	–	4	=	-1
Br	=	7	–	2/2	–	6	=	0

From the above calculations of formal charge, you can see that the iodine (I) atom has -1 charge, while the bromine atom has 0 charge.

So let’s keep these charges on the respective atoms in the IBr4 molecule.

This overall -1 charge on the IBr4 molecule is represented in the image given below.

In the above lewis dot structure of IBr4- ion, you can also represent each bonding electron pair (:) as a single bond (|). By doing so, you will get the following lewis structure of IBr4- 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:

HBrO2 Lewis Structure	HBrO3 Lewis Structure
HBrO4 Lewis Structure	PO2- Lewis Structure
TeF5- Lewis Structure	SeCl6 Lewis Structure