IBr4- Lewis structure

IBr₄^– has one iodine atom and four bromine atoms.

In IBr₄^– Lewis structure, there are four single bonds around the iodine atom, with four bromine atoms attached to it. Each bromine atom has three three lone pairs, and the iodine atom has two lone pairs.

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

Alternative method: Lewis structure of IBr₄^–

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 IBr₄^– has one iodine atom and four bromine atoms, so…

Valence electrons of one iodine atom = 7 × 1 = 7
Valence electrons of four bromine atoms = 7 × 4 = 28

Now the IBr₄^– has a negative (-1) charge, so we have to add one more electron.

So the total valence electrons = 7 + 28 + 1 = 36

Learn how to find: Bromine valence electrons

• Second, find the total electron pairs

We have a total of 36 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 = 36 ÷ 2 = 18

• 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 18 electron pairs. And four I — Br bonds are already marked. So we have to only mark the remaining fourteen 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 (8) = -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 IBr₄^– contains a central iodine atom connected to four bromine atoms through single covalent bonds. In this configuration, the iodine atom utilizes an expanded valence shell to accommodate twelve electrons, consisting of four bonding pairs and two lone pairs. Within this layout, each of the four bromine atoms fulfills the octet rule by maintaining three lone pairs alongside its single shared bond. This arrangement represents the most stable state for the ion because it results in a formal charge of -1 on the central iodine atom, while all four bromine atoms carry a formal charge of zero. Consequently, this specific electronic distribution serves as the definitive and most accurate Lewis representation of the IBr₄^– ion.

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: HCONH₂ Lewis structure

External video

• How to Draw the Lewis Dot Structure for IBr4 – – YouTube • Wayne Breslyn

External links

• https://lambdageeks.com/ibr4-lewis-structure/
• https://homework.study.com/explanation/what-is-the-molecular-geometry-of-ibr4.html
• https://oneclass.com/homework-help/chemistry/7046307-ibr4-lewis-structure.en.html