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.

Alternative method: Lewis structure of IBr₅

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

Learn how to find: Bromine valence electrons

• 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

Lone pair

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:

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 – 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.

Final structure

The final structure of IBr₅ contains a central iodine atom connected to five bromine atoms through single covalent bonds. In this layout, the iodine atom serves as an exception to the octet rule, utilizing an expanded valence shell to accommodate twelve electrons, which consist of five bonding pairs and one lone pair. Each bromine atom fulfills its octet by maintaining three lone pairs of its own alongside the single shared bond. This arrangement is the most stable because it results in formal charges of zero for all atoms involved, representing the most energetically favorable state for the molecule. Thus, this specific electronic distribution serves as the definitive and most accurate Lewis representation of IBr₅.

Next: BCl₂ Lewis structure

External links

• https://www.chegg.com/homework-help/questions-and-answers/draw-lewis-dot-structure-ibr5-scratch-paper-answer-following-questions-molecule-molecule-i-q68487041