IBr Lewis structure

IBr (iodine monobromide) has one iodine atom and one bromine atom.

In the IBr Lewis structure, there is a single bond between the iodine and bromine atom, and on both iodine and bromine atoms, there are three lone pairs.

Alternative method: Lewis structure of IBr

Contents

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 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

Learn how to find: Bromine valence electrons

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

IBr Lewis Structure (Step 1) — Rough sketch of IBr Lewis structure | Image: Learnool

Lone pair

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:

IBr Lewis Structure (Step 2) — Lone pairs marked, and got the stable Lewis structure of IBr | Image: Learnool

Formal charge

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.

Final structure

The final structure of IBr consists of an iodine atom linked to a bromine atom through a single covalent bond. Within this diatomic configuration, both halogen atoms successfully complete their octets by sharing one electron pair and retaining three lone pairs each. This arrangement results in formal charges of zero for both atoms, representing the most stable and energetically favorable state for the molecule. Consequently, this simple bonding pattern serves as the definitive and most accurate Lewis representation of iodine monobromide.

Next: Cl Lewis structure

External video

IBr Lewis Dot Structure – YouTube • Wayne Breslyn

External links

Deep