IBr5 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.

IBr5 lewis structure has an Iodine atom (I) at the center which is surrounded by five Bromine atoms (Br). There are 5 single bonds between the Iodine atom (I) and each Bromine atom (Br). There are 3 lone pairs on all the Bromine atoms (Br) and 1 lone pair on the Iodine atom (I).

If you haven’t understood anything from the above image of IBr5 lewis structure, then just stick with me and you will get the detailed step by step explanation on drawing a lewis structure of IBr5.

So let’s move to the steps of drawing the lewis structure of IBr5.

Steps of drawing IBr5 lewis structure

Step 1: Find the total valence electrons in IBr5 molecule

In order to find the total valence electrons in a IBr5 molecule, 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 IBr5 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 IBr5 molecule = valence electrons given by 1 iodine atom + valence electrons given by 5 bromine atoms = 7 + 7(5) = 42.

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 IBr5 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 IBr5 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 IBr5 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 atoms.

Here in the sketch of IBr5 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 IBr5 molecule.

The IBr5 molecule has a total 42 valence electrons and out of these, only 40 valence electrons are used in the above sketch.

So the number of electrons which are left = 42 – 40 = 2.

You have to put these 2 electrons on the central iodine atom in the above sketch of IBr5 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 IBr5.

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 IBr5 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 IBr5 molecule in the image given below.

For Iodine (I) atom:
Valence electrons = 7 (because iodine is in group 17)
Bonding electrons = 10
Nonbonding electrons = 2

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

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

This indicates that the above lewis structure of IBr5 is stable and there is no further change in the above structure of IBr5.

In the above lewis dot structure of IBr5, you can also represent each bonding electron pair (:) as a single bond (|). By doing so, you will get the following lewis structure of IBr5.

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:

About author

Jay Rana

Jay is an educator and has helped more than 100,000 students in their studies by providing simple and easy explanations on different science-related topics. He is a founder of Pediabay and is passionate about helping students through his easily digestible explanations.

Read more about our Editorial process.