# IBr5 Lewis structure

IBr5 has one iodine atom and five bromine atoms.

In IBr5 Lewis structure, there are five single bonds around the iodine atom, with five bromine atoms attached to it. Each bromine atom has three lone pairs, and the iodine atom has one lone pair.

Contents

## Steps

Use these steps to correctly draw the IBr5 Lewis structure:

#1 First draw a rough sketch
#2 Mark lone pairs on the atoms
#3 Calculate and mark formal charges on the atoms, if required

Let’s discuss each step in more detail.

### #1 First draw a rough sketch

• First, determine the total number of valence electrons

In the periodic table, both iodine and bromine lie in group 17.

Hence, both iodine and bromine have seven valence electrons.

Since IBr5 has one iodine atom and five bromine atoms, so…

Valence electrons of one iodine atom = 7 × 1 = 7
Valence electrons of five bromine atoms = 7 × 5 = 35

And the total valence electrons = 7 + 35 = 42

• Second, find the total electron pairs

We have a total of 42 valence electrons. And when we divide this value by two, we get the value of total electron pairs.

Total electron pairs = total valence electrons ÷ 2

So the total electron pairs = 42 ÷ 2 = 21

• Third, determine the central atom

We have to place the least electronegative atom at the center.

Since iodine is less electronegative than bromine, assume that the central atom is iodine.

Therefore, place iodine in the center and bromines on either side.

• And finally, draw the rough sketch

### #2 Mark lone pairs on the atoms

Here, we have a total of 21 electron pairs. And five I — Br bonds are already marked. So we have to only mark the remaining sixteen electron pairs as lone pairs on the sketch.

Also remember that iodine is a period 5 element, so it can keep more than 8 electrons in its last shell. And bromine is a period 4 element, so it can keep more than 8 electrons in its last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are bromines.

So for each bromine, there are three lone pairs, and for iodine, there is one lone pair.

Mark the lone pairs on the sketch as follows:

### #3 Calculate and mark formal charges on the atoms, if required

Use the following formula to calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

For iodine atom, formal charge = 7 – 2 – ½ (10) = 0

For each bromine atom, formal charge = 7 – 6 – ½ (2) = 0

Here, both iodine and bromine atoms do not have charges, so no need to mark the charges.

In the above structure, you can see that the central atom (iodine) forms an octet. And the outside atoms (bromines) also form an octet. Hence, the octet rule is satisfied.

Therefore, this structure is the stable Lewis structure of IBr5.

Next: BCl2 Lewis structure