ICl4- Lewis structure

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.

Alternative method: Lewis structure of ICl₄^–

Rough sketch

• 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

Lone pair

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:

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 chlorine 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 ICl₄^– features a central iodine atom connected to four chlorine atoms through single covalent bonds. In this configuration, the iodine atom utilizes an expanded valence shell to accommodate twelve electrons, which include four bonding pairs and two lone pairs. Within this layout, each of the four chlorine atoms satisfies 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 each chlorine atom carries a formal charge of zero. Thus, this specific electronic distribution serves as the definitive and most accurate Lewis representation of this ion.

To properly represent this as a polyatomic ion, the entire Lewis structure is enclosed within square brackets. The overall charge of 1- is then written as a superscript outside the brackets at the top right, indicating that the structure possesses one additional electron beyond the valence count of the neutral atoms.

Next: SiF₄ Lewis structure

External video

• ICl4- Lewis Structure – How to Draw the Lewis Structure for ICl4- – YouTube • Wayne Breslyn

External links

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