ICl4- Lewis structure

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ICl₄^– (tetrachloroiodide) has one iodine atom and four chlorine atoms.

In the ICl₄^– Lewis structure, there are four single bonds around the iodine atom, with four chlorine atoms attached to it. Each chlorine atom has three lone pairs, and the iodine atom has two lone pairs.

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

Steps

Here’s how you can easily draw the ICl₄^– Lewis structure step by step:

#1 Draw a rough skeleton structure
#2 Mention lone pairs on the atoms
#3 If needed, mention formal charges on the atoms

Now, let’s take a closer look at each step mentioned above.

#1 Draw a rough skeleton structure

• First, determine the total number of valence electrons

In the periodic table, both iodine and chlorine lie in group 17.

Hence, both iodine and chlorine have seven valence electrons.

Since ICl₄^– has one iodine atom and four chlorine atoms, so…

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

Now the ICl₄^– 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: Chlorine 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 chlorine, assume that the central atom is iodine.

Therefore, place iodine in the center and chlorines on either side.

• And finally, draw the rough sketch

#2 Mention lone pairs on the atoms

Here, we have a total of 18 electron pairs. And four I — Cl 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 chlorine is a period 3 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 chlorines.

So for each chlorine, there are three lone pairs, and for iodine, there are two lone pairs.

Mark the lone pairs on the sketch as follows:

#3 If needed, mention formal charges on the atoms

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 chlorine 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 (chlorines) 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 chlorine.

But if we convert a lone pair of the iodine atom to make a new I — Cl bond with the chlorine 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 ICl₄^–.

And since the ICl₄^– has a negative (-1) charge, mention that charge on the Lewis structure by drawing brackets as follows:

Next: SiF₄ Lewis structure

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External links

• https://www.thegeoexchange.org/chemistry/bonding/Lewis-Structures/ICl4-lewis-structure.html