The **Lewis structure of ICl**_{2}** ^{–}** contains two single bonds, with iodine in the center, and two chlorines on either side. There are three lone pairs on each 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 ICl_{2}^{–}.

#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 ICl_{2}^{–} has one iodine atom and two chlorine atoms.

So the total number of valence electrons = valence electrons of iodine atom + (valence electrons of chlorine atom × 2)

And ICl_{2}^{–} has a negative (-1) charge, so we have to add one more electron.

Therefore, the **total number of valence electrons** = 7 + 14 + 1 = 22

- 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 ICl_{2}^{–} something like this:

**Also read:** How to draw Lewis structure of HCO_{3}^{–} (5 steps)

### #2 Show chemical bond

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

**Also read:** How to draw Lewis structure of HNO_{2} (5 steps)

### #3 Mark lone pairs

As calculated earlier, we have a total of 22 valence electrons. And in the above structure, we have already used four valence electrons. Hence, eighteen valence electrons are remaining.

Two valence electrons represent one lone pair. So eighteen valence electrons = **nine 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 also get three lone pairs.

So the Lewis structure of ICl_{2}^{–} 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 I_{2} (4 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 = 6

Bonding electrons = 4

Formal charge = 7 – 6 – ½ (4) = -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 ICl_{2}^{–} 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 ICl**_{2}** ^{–}**.

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

Now ICl_{2}^{–} 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 ICl_{2}^{–} looks something like this:

## Related

- Lewis structure of HCO
_{3}^{–} - Lewis structure of HNO
_{2} - Lewis structure of I
_{2} - Lewis structure of C
_{3}H_{8} - Lewis structure of IF
_{4}^{–}

## External links

- ICl2- Lewis Structure in 6 Steps (With Images) – Pediabay
- Chemical Bonding: ICl2- Lewis Structure – The Geoexchange
- ICl2- lewis structure, molecular geometry, bond angle, polarity, electrons – Topblogtenz
- ICL2- lewis structure: Drawings, Hybridization, Shape, Charges, Pairs – Lambda Geeks
- Why does the correct Lewis structure of ICl2 have 3 lone pairs on iodine and not 2 (i.e why does iodine have formal charge -1 and not chlorine)? – Quora
- ICl2- Lewis Structure (Iodine Dichloride) – Pinterest
- ICl2 – lewis structure – Laurence Lavelle
- In the Lewis structure for ICl2-, how many lone pairs of electrons are around the central iodine atom? – Homework.Study.com
- Drawing Lewis structure of ICl2- – Reddit
- draw the lewis structure for ICl2 – Chegg
- In the Lewis structure for ICl2–, how many lone pairs of electrons are around the central iodine atom – Brainly
- In the Lewis structure for ICl2-, how many lone pairs of electrons are around the central iodine atom? – Studocu
- In the Lewis structure for ICl2–, how many lone pairs of electrons are around the central iodine atom? – Bartleby
- Draw the Lewis structure of ICl2 – OneClass
- In the Lewis dot formula for ICl2-, the number of lone pairs of electrons around the central iodine atom is – Course Hero

Deep

Rootmemory.com was founded by Deep Rana, who is a mechanical engineer by profession and a blogger by passion. He has a good conceptual knowledge on different educational topics and he provides the same on this website. He loves to learn something new everyday and believes that the best utilization of free time is developing a new skill.