IBr3 Lewis structure

IBr₃ (iodine tribromide) has one iodine atom and three bromine atoms.

In IBr₃ Lewis structure, there are three single bonds around the iodine atom, with three bromine atoms attached to it. Each bromine atom has three lone pairs, and the iodine atom has two lone pairs.

Alternative method: Lewis structure of IBr₃

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 IBr₃ has one iodine atom and three bromine atoms, so…

Valence electrons of one iodine atom = 7 × 1 = 7
Valence electrons of three bromine atoms = 7 × 3 = 21

And the total valence electrons = 7 + 21 = 28

Learn how to find: Bromine valence electrons

• Second, find the total electron pairs

We have a total of 28 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 = 28 ÷ 2 = 14

• 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

Lone pair

Here, we have a total of 14 electron pairs. And three I — Br bonds are already marked. So we have to only mark the remaining eleven 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 are two lone pairs.

Mark the lone pairs on the sketch as follows:

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 – 4 – ½ (6) = 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.

Final structure

The final structure of IBr₃ features a central iodine atom linked to three bromine atoms through single covalent bonds. In this configuration, the iodine atom utilizes an expanded valence shell to accommodate ten electrons, which consist of three bonding pairs and two lone pairs. Within this layout, each of the three bromine atoms successfully satisfies the octet rule by retaining three lone pairs alongside its single shared bond. This arrangement represents the most stable state for the molecule because it results in a formal charge of zero for every atom involved. Accordingly, this specific electronic distribution serves as the definitive and most accurate Lewis representation of iodine tribromide.

Next: B₂ Lewis structure

External video

• How to Draw the Lewis Dot Structure for IBr3: Iodine tribromide – YouTube • Wayne Breslyn

External links

• https://www.bartleby.com/questions-and-answers/draw-the-lewis-structure-for-iodine-tribromide-ibr3.-what-is-the-molecular-geometry-of-this-molecule/2133af36-150a-4690-ac13-bb395fa9a45f
• https://www.chegg.com/homework-help/questions-and-answers/lewis-structure-ibr3-central-atom–molecular-polar-nonpolar-identify-intermolecular-forces-q28406979
• https://www.numerade.com/ask/question/what-is-the-lewis-structure-for-ibr3-with-the-central-atom-of-i-is-the-molecular-polar-or-nonpolar-identify-the-intermolecular-forces-present-63972/