ICl5 Lewis structure

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ICl5 Lewis Structure
ICl5 Lewis structure

ICl5 (iodine pentachloride) has one iodine atom and five chlorine atoms.

In the ICl5 Lewis structure, there are five single bonds around the iodine atom, with five chlorine atoms attached to it. Each chlorine atom has three lone pairs, and the iodine atom has one lone pair.


Use these steps to correctly draw the ICl5 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
Periodic table

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

Hence, both iodine and chlorine have seven valence electrons.

Since ICl5 has one iodine atom and five chlorine atoms, so…

Valence electrons of one iodine atom = 7 × 1 = 7
Valence electrons of five chlorine atoms = 7 × 5 = 35

And the total valence electrons = 7 + 35 = 42

  • Second, find the total electron pairs

We have a total of 42 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 = 42 ÷ 2 = 21

  • 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
ICl5 Lewis Structure (Step 1)
Rough sketch of ICl5 Lewis structure

#2 Mark lone pairs on the atoms

Here, we have a total of 21 electron pairs. And five I — Cl bonds are already marked. So we have to only mark the remaining sixteen 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 is one lone pair.

Mark the lone pairs on the sketch as follows:

ICl5 Lewis Structure (Step 2)
Lone pairs marked, and got the stable Lewis structure of ICl5

#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 – 2 – ½ (10) = 0

For each chlorine atom, formal charge = 7 – 6 – ½ (2) = 0

Here, both iodine and chlorine atoms do not have charges, so no need to mark the charges.

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.

Therefore, this structure is the stable Lewis structure of ICl5.

Next: SBr2 Lewis structure

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