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

I’m super excited to teach you the lewis structure of IBr in just 6 simple steps.

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

So, if you are ready to go with these 6 simple steps, then let’s dive right into it!

Lewis structure of IBr (Iodine monobromide) contains one single bond between the Iodine (I) and Bromine (Br) atom. And both the Iodine and Bromine atoms have three lone pairs on it.

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 IBr).

6 Steps to Draw the Lewis Structure of IBr

Step #1: Calculate the total number of valence electrons

Here, the given molecule is IBr (Iodine monobromide). In order to draw the lewis structure of IBr, first of all you have to find the total number of valence electrons present in the IBr molecule.
(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 IBr

• For Iodine:

Iodine is a group 17 element on the periodic table. ^[1]

Hence, the valence electrons present in iodine is 7 (see below image).

• For Bromine:

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

Hence, the valence electrons present in bromine is 7 (see below image).

Hence in a IBr molecule, 

Valence electrons given by Iodine (I) atom = 7
Valence electrons given by Bromine (Br) atom = 7
So, total number of Valence electrons in IBr molecule = 7 + 7 = 14

Step #2: Select the center atom

While selecting the atom, you have to put the least electronegative atom at the center.

But here in the IBr molecule, there are only two atoms. So you can consider any of the atoms as a center atom.

So, let’s assume that the Iodine atom is a central atom. (You should assume the less electronegative atom as a central atom.)

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

Now in the above sketch of IBr molecule, put the two electrons (i.e electron pair) between the iodine and bromine atoms to represent a chemical bond between them.

These pair of electrons present between the Iodine (I) and Bromine (Br) atoms form a chemical bond, which bonds both these atoms with each other in a IBr molecule.

Step #4: Complete the octet (or duplet) on outside atom. 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 IBr, we have just assumed the iodine atom as a central atom and so the bromine atom is an outer atom.

So now, we have to complete the octet on the bromine atom.

Now, you can see in the above image that the bromine atom forms an octet.

Also, only 8 valence electrons of IBr molecule are used in the above structure.

But there are total 14 valence electrons in IBr molecule (as calculated in step #1).

So the number of electrons left to be kept on the central atom = 14 – 8 = 6.

So let’s keep these six electrons (i.e 3 electron pairs) on the central atom (i.e iodine 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 iodine atom) has an octet or not. 

In simple words, we have to check whether the Iodine (I) atom is having 8 electrons or not.

As you can see from the above image, the central atom (i.e iodine atom) has 8 electrons. So it fulfills the octet rule and this iodine atom is also stable.

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 the iodine atom (I) as well as bromine atom (Br).

For that, you need to remember the formula of formal charge;

Formal charge = Valence electrons – Nonbonding electrons – (Bonding electrons)/2

• For Iodine:
  Valence electron = 7 (as it is in group 17)
  Nonbonding electrons = 6
  Bonding electrons = 2
• For Bromine:
  Valence electron = 7 (as it is in group 17)
  Nonbonding electrons = 6
  Bonding electrons = 2

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

So you can see above that the formal charges on iodine as well as bromine are “zero”.

Hence, there will not be any change in the above structure and the above lewis structure of IBr is the final stable structure only.

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

Related lewis structures for your practice:
Lewis Structure of SeCl4
Lewis Structure of HOF
Lewis Structure of XeO2F2
Lewis Structure of XeH4
Lewis Structure of S2Cl2 

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