IBr4- Lewis structure

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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.

Steps

Use these steps to correctly draw the IBr₄^– Lewis structure:

#1 First draw a rough sketch
#2 Mark lone pairs on the atoms
#3 Calculate and mark formal charges on the atoms, if required

Let’s discuss each step in more detail.

#1 First draw a 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

#2 Mark lone pairs on the atoms

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:

#3 Calculate and mark formal charges on the atoms, if required

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:

In the above structure, you can see that the central atom (iodine) forms an octet. And the outside atoms (bromines) also form an octet. Hence, the octet rule is satisfied.

Now there is still a negative (-1) charge on the iodine atom.

This is not okay, right? Because the structure with a negative charge on the most electronegative atom is the best Lewis structure. And in this case, the most electronegative element is bromine.

But if we convert a lone pair of the iodine atom to make a new I — Br bond with the bromine atom, and calculate the formal charge, then we do not get the formal charges on atoms closer to zero.

And the structure with the formal charges on atoms closer to zero is the best Lewis structure.

Therefore, this structure is the most stable Lewis structure of IBr₄^–.

And since the IBr₄^– has a negative (-1) charge, mention that charge on the Lewis structure by drawing brackets as follows:

Next: HCONH₂ Lewis structure

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