IBr 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.

IBr lewis structure has one Iodine atom (I) and one Bromine atom (Br) which contain a single bond between them. There are 3 lone pairs on both the Iodine atom (I) as well as Bromine atom (Br).

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

So let’s move to the steps of drawing the lewis structure of IBr.

Steps of drawing IBr lewis structure

Step 1: Find the total valence electrons in IBr molecule

In order to find the total valence electrons in an IBr molecule, first of all you should know the valence electrons present in the 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 IBr molecule

→ 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 IBr molecule = valence electrons given by 1 iodine atom + valence electrons given by 1 bromine atom = 7 + 7 = 14.

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 molecule is IBr. It has only two atoms, so you can select any of the atoms as a center atom.

Let’s assume the iodine atom as a central atom.
(You should assume the less electronegative atom as a center atom).

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

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

This indicates that the iodine (I) and bromine (Br) are chemically bonded with each other in a IBr molecule.

Step 4: Make the outer atoms stable. Place the remaining valence electrons pair on the central atom.

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

Here in the sketch of IBr molecule, we have assumed the iodine atom as a center atom. So the bromine is the outer atom.

Hence you have to make the bromine atom stable.

You can see in the below image that the bromine atom is forming an octet and hence it is stable.

Also, in step 1 we have calculated the total number of valence electrons present in the IBr molecule.

The IBr molecule has a total 14 valence electrons and out of these, only 8 valence electrons are used in the above sketch.

So the number of electrons which are left = 14 – 8 = 6.

You have to put these 6 electrons on the iodine atom in the above sketch of IBr 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 IBr.

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 IBr 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 IBr molecule in the image given below.

For Iodine (I) atom:
Valence electrons = 7 (because iodine is in group 17)
Bonding electrons = 6
Nonbonding electrons = 4

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

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

From the above calculations of formal charge, you can see that the iodine (I) atom as well as bromine (Br) atom has a “zero” formal charge.

This indicates that the above lewis structure of IBr is stable and there is no further change in the above structure of IBr.

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

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:

SeCl4 Lewis Structure	HOF Lewis Structure
XeO2F2 Lewis Structure	XeH4 Lewis Structure
S2Cl2 Lewis Structure	N2O5 Lewis Structure