IBr2- Lewis structure

IBr₂^– has one iodine atom and two bromine atoms.

In IBr₂^– Lewis structure, there are two single bonds around the iodine atom, with two bromine atoms attached to it, and each atom has three lone pairs.

Also, there is a negative (-1) charge on the iodine atom.

Alternative method: Lewis structure of IBr₂^–

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 IBr₂^– has one iodine atom and two bromine atoms, so…

Valence electrons of one iodine atom = 7 × 1 = 7
Valence electrons of two bromine atoms = 7 × 2 = 14

Now the IBr₂^– has a negative (-1) charge, so we have to add one more electron.

So the total valence electrons = 7 + 14 + 1 = 22

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.

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 7 electron pairs. And two I — Br bonds are already marked. So we have to only mark the remaining five 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 atom, there are three 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 – 6 – ½ (4) = -1

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

Here, the iodine atom has a charge, so mark it on the sketch as follows:

Final structure

The final structure of IBr₂^– has a central iodine atom linked to two bromine atoms through single covalent bonds. In this arrangement, the iodine atom utilizes an expanded octet to accommodate ten valence electrons, which includes two bonding pairs and three lone pairs. Each bromine atom satisfies the octet rule by maintaining three lone pairs of its own. This configuration is the most stable because it results in a formal charge of -1 on the central iodine atom and zero on each of the bromine atoms, representing the most energetically favorable distribution for the ion. Consequently, this specific electronic pattern serves as the definitive and most accurate Lewis representation for this anion.

To complete the representation, draw square brackets around the entire Lewis structure and place a “-” or “-1” sign as a superscript outside the upper right bracket. This notation signifies that the negative charge is a property of the whole ion.

Next: C₃H₄ Lewis structure

External video

• IBr2- Lewis Structure: How to Draw the Lewis Structure for IBr2 – – YouTube • Wayne Breslyn

External links

• https://www.thegeoexchange.org/chemistry/bonding/Lewis-Structures/IBr2-lewis-structure.html
• https://lambdageeks.com/ibr2-lewis-structure/
• https://homework.study.com/explanation/draw-out-the-lewis-structure-for-the-given-ion-ibr2.html
• https://www.chegg.com/homework-help/questions-and-answers/lewis-structure-ibr2-q560282