Lewis structure of ICl4-

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The Lewis structure of ICl₄^– contains four single bonds, with iodine in the center, and four chlorines on either side. There are three lone pairs on each chlorine atom, and two lone pairs on the iodine atom.

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

Steps

By using the following steps, you can easily draw the Lewis structure of ICl₄^–:

#1 Draw skeleton
#2 Show chemical bond
#3 Mark lone pairs
#4 Calculate formal charge and check stability (if octet is already completed on central atom)

Let’s one by one discuss each step in detail.

#1 Draw skeleton

In this step, first calculate the total number of valence electrons. And then, decide the central atom.

• Let’s calculate the total number of valence electrons

We know that… both iodine and chlorine are the group 17 elements. Hence, both iodine and chlorine have seven valence electrons.

Now ICl₄^– has one iodine atom and four chlorine atoms.

So the total number of valence electrons = valence electrons of iodine atom + (valence electrons of chlorine atom × 4)

And ICl₄^– has a negative (-1) charge, so we have to add one more electron.

Therefore, the total number of valence electrons = 7 + 28 + 1 = 36

• Now decide the central atom

The atom with the least electronegative value is placed at the center. By looking at the periodic table, we get the electronegativity values for iodine and chlorine as follows:

Electronegativity value of iodine = 2.66
Electronegativity value of chlorine = 3.16

Obviously, iodine is less electronegative than chlorine. Hence, assume that iodine is the central atom.

So now, put iodine in the center and chlorines on either side. And draw the rough skeleton structure for the Lewis structure of ICl₄^– something like this:

Also read: How to draw Lewis structure of SiCl₄ (4 steps)

#2 Show chemical bond

Place two electrons between the atoms to show a chemical bond. Since iodine is surrounded by four chlorines, use eight electrons to show four chemical bonds as follows:

Also read: How to draw Lewis structure of SiF₄ (4 steps)

#3 Mark lone pairs

As calculated earlier, we have a total of 36 valence electrons. And in the above structure, we have already used eight valence electrons. Hence, twenty-eight valence electrons are remaining.

Two valence electrons represent one lone pair. So twenty-eight valence electrons = fourteen lone pairs.

Note 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.

Also, make sure that you start marking these lone pairs on outside atoms first. And then, on the central atom.

The outside atoms are chlorines, so each chlorine will get three lone pairs. And the central atom (iodine) will get two lone pairs.

So the Lewis structure of ICl₄^– looks something like this:

In the above structure, you can see that the octet is completed on the central atom (iodine), and also on the outside atoms. Therefore, the octet rule is satisfied.

After completing the octet, one last thing we need to do is, calculate the formal charge and check the stability of the above structure.

Also read: How to draw Lewis structure of OCN^– (5 steps)

#4 Calculate formal charge and check stability

The following formula is used to calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

Collect the data from the above structure and then, write it down below as follows:

• For iodine atom

Valence electrons = 7
Nonbonding electrons = 4
Bonding electrons = 8

Formal charge = 7 – 4 – ½ (8) = -1

• For each chlorine atom

Valence electrons = 7
Nonbonding electrons = 6
Bonding electrons = 2

Formal charge = 7 – 6 – ½ (2) = 0

Mention the formal charges of atoms on the structure. So the Lewis structure of ICl₄^– looks something like this:

In the above structure, you can see that the formal charges of atoms are closer to zero. Therefore, this is the most stable Lewis structure of ICl₄^–.

And each horizontal line drawn in the above structure represents a pair of bonding valence electrons.

Now ICl₄^– is an ion having a negative (-1) charge, so draw brackets around the above Lewis structure and mention that charge on the top right corner. And then, the Lewis structure of ICl₄^– looks something like this:

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Related

• Lewis structure of SiCl₄
• Lewis structure of SiF₄
• Lewis structure of OCN^–
• Lewis structure of CH₃F
• Lewis structure of BrO₃^–

External video

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

External links

• Drawing the Lewis Structure for ICl4 – The University of Maryland
• Chemical Bonding: ICl4- Lewis Structure – The Geoexchange
• ICl4- Lewis Structure in 5 Steps (With Images) – Pediabay
• ICl4- lewis structure, molecular geometry, bond angle, hybridization – Topblogtenz
• ICl4- Lewis Structure (Tetrachloroiodide Ion) – Pinterest
• Draw the Lewis structure for ICl4 – Homework.Study.com
• What is the molecular geometry of ICl4? – Quora
• What is the formal charge of i in icl4 – Brainly
• Lecture 11: Basic Shapes – Predict The Shape Of Icl4(-) – iLectureOnline
• Write the Lewis structure for ICl4- and identify its shape and hybridization scheme – Numerade
• Sketch a good Lewis structure for ICl4 – Bartleby