# Lewis Structure of IBr5 (With 5 Simple Steps to Draw!)

I’m super excited to teach you the lewis structure of IBr5 in just 5 simple steps.

Infact, I’ve also given the step-by-step images for drawing the lewis dot structure of IBr5 molecule.

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

Lewis structure of IBr5 contains five single bonds between the Iodine (I) atom and each Bromine (Br) atom. The Iodine atom (I) is at the center and it is surrounded by 5 Bromine atoms (Br). The Iodine atom has 1 lone pair while all the five Bromine atoms have 3 lone pairs.

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 IBr5).

## 5 Steps to Draw the Lewis Structure of IBr5

### Step #1: Calculate the total number of valence electrons

Here, the given molecule is IBr5. In order to draw the lewis structure of IBr5, first of all you have to find the total number of valence electrons present in the IBr5 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 IBr5

• For Iodine:

Iodine is a group 17 element on the periodic table.

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

• For Bromine:

Bromine is a group 17 element on the periodic table.

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

Hence in a IBr5 molecule,

Valence electrons given by Iodine (I) atom = 7
Valence electrons given by each Bromine (Br) atom = 7
So, total number of Valence electrons in IBr5 molecule = 7 + 7(5) = 42

### Step #2: Select the center atom

While selecting the 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).

Here in the IBr5 molecule, if we compare the Iodine atom (I) and Bromine atom (Br), then the Iodine is less electronegative than bromine.

So, Iodine should be placed in the center and the remaining 5 bromine atoms will surround it.

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

Now in the above sketch of IBr5 molecule, put the two electrons (i.e electron pair) between each Iodine atom and bromine atom to represent a chemical bond between them.

These pairs of electrons present between the Iodine (I) and Bromine (Br) atoms form a chemical bond, which bonds the iodine and bromine atoms with each other in a IBr5 molecule.

### Step #4: Complete the octet (or duplet) on outside atoms. 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 IBr5, the outer atoms are bromine atoms.

So now, you have to complete the octet on these bromine atoms (because bromine requires 8 electrons to have a complete outer shell).

Now, you can see in the above image that all the bromine atoms form an octet.

Also, only 40 valence electrons of IBr5 molecule are used in the above structure.

But there are total 42 valence electrons in IBr5 molecule (as calculated in step #1).

So the number of electrons left to be kept on the central atom = 42 – 40 = 2.

So let’s keep these two electrons (i.e electron pair) on the central atom.

Now, let’s move to the next step.

### Step #5: 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 iodine atom (I) as well as each 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 electrons = 7 (as it is in group 17)
Nonbonding electrons = 2
Bonding electrons = 10
• For Bromine:
Valence electron = 7 (as it is in group 17)
Nonbonding electrons = 6
Bonding electrons = 2

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 IBr5 is the final stable structure only.

Each electron pair (:) in the lewis dot structure of IBr5 represents the single bond ( | ). So the above lewis dot structure of IBr5 can also be represented as shown below.

Related lewis structures for your practice:
Lewis Structure of AsI3
Lewis Structure of SbF6-
Lewis Structure of SbCl3
Lewis Structure of C2H4F2
Lewis Structure of Br2O