# IBr Lewis Structure

IBr (iodine monobromide) has one iodine atom and one bromine atom. In the lewis structure of IBr, there is a single bond between the iodine and bromine atom, and on both iodine and bromine atoms, there are three lone pairs.

## Steps

Here’s how you can draw the IBr lewis structure step by step.

Step #1: draw sketch
Step #2: mark lone pairs
Step #3: mark charges (if there are)

Let’s break down each step in detail.

### #1 Draw 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 IBr has one iodine atom and one bromine atom, so…

Valence electrons of one iodine atom = 7 × 1 = 7
Valence electrons of one bromine atom = 7 × 1 = 7

And the total valence electrons = 7 + 7 = 14

• Second, find the total electron pairs

We have a total of 14 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 = 14 ÷ 2 = 7

• 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.

• And finally, draw the rough sketch

### #2 Mark Lone Pairs

Here, we have a total of 7 electron pairs. And one I — Br bond is already marked. So we have to only mark the remaining six 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 atom is bromine.

So for each atom, there are three lone pairs.

Mark the lone pairs on the sketch as follows:

### #3 Mark Charges

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 – 6 – ½ (2) = 0

For 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.

And in the above structure, you can see that the central atom (iodine) forms an octet. Hence, the octet rule is satisfied.

Therefore, this structure is the stable lewis structure of IBr.

Next: Cl Lewis Structure